2010_08_14_NotesOnFluidMechanicsAndGasDynamics_Wassgren

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Unformatted text preview: Notes on Fluid Mechanics and Gas Dynamics Carl Wassgren, Ph.D. School of Mechanical Engineering Purdue University wassgren@purdue.edu 16 Aug 2010 Chapter 01: Chapter 02: Chapter 03: Chapter 04: Chapter 05: Chapter 06: Chapter 07: Chapter 08: Chapter 09: Chapter 10: Chapter 11: Chapter 12: The Basics Fluid Statics The Integral Approach The Differential Approach Potential Flows Dimensional Analysis Navier-Stokes Solutions Turbulence Boundary Layers Pipe Flows Fluid Machinery Gas Dynamics Chapter 01: The Basics 1. 2. 3. 4. 5. 6. 7. 8. Symbolic vs. Numeric Approach Dimensions and Units Taylor Series Expansion Approximation Experimental Uncertainty Statistical vs. Continuum Approach Fluid Properties Flow Visualization Some Basic Definitions and Concepts C. Wassgren Chapter 01: The Basics 1 Last Updated: 05 Sep 2008 1. Symbolic vs. Numeric Calculations Consider the following example. You need to determine the trajectory of a projectile fired from a cannon. The projectile has a mass of 10 kg and the cannon is tilted at an angle of 30° from the horizontal. The initial velocity of the projectile from the cannon is 100 m/s. Determine: 1. the distance the projectile will travel and 2. how long the projectile is in flight. We can approach this problem a couple of different ways. The first is to start with the given numbers and immediately begin the calculations. The second approach is to solve the problem symbolically and then substitute the numbers at the end. Numerical Solution: Draw a free body diagram (FBD) of the projectile. y 100 m/s 30° x 10 kg*9.81 m/s2= 98.1 N Assume this height is negligible. Use Newton’s 2nd law to determine the acceleration of the projectile: ⇒ x = 0 m/s 2 ∑ Fx = mx ⇒ 0 N = (10 kg ) x ∑F y = my ⇒ (10 kg ) ( −9.81 m/s 2 ) = (10 kg ) y ⇒ y = −9.81 m/s 2 Integrate with respect to time to determine the projectile’s velocity and position given the projectile’s initial x and y velocities and positions: x = x0 = (100 m/s ) ( cos 30 ) = 86.6 m/s y = ( −9.81 m/s 2 ) t + y0 = ( −9.81 m/s 2 ) t + (100 m/s ) ( sin 30 ) = ( −9.81 m/s 2 ) t + 50 m/s x = ( 86.6 m/s ) t (1.1) y = ( −4.91 m/s ) t + ( 50 m/s ) t 2 2 (1.2) The projectile will hit the ground when y = 0 m so that by rearranging Eqn. (1.2) we find that the time aloft is: 50.0 m/s t= = 10.2 s (1.3) 4.91 m/s 2 Substituting into Eqn. (1.1) gives the distance traveled as x = ( 86.6 m/s ) (10.2 s ) = 883 m (1.4) As you can see, we’ve made a number of calculations along the way to finding the answers. Now let’s address some additional questions based on these answers. How does the maximum time aloft depend on the mass of the projectile? If the initial speed from the cannon doubles, how is the range affected? What angle maximizes the distance the projectile travels? The answers to these questions are not obvious from Eqns. (1.1) – (1.4); we would need to perform additional calculations. Also, consider how many calculations would need to be made if we had to determine the range and time aloft for a variety of cannon angles, initial velocities, and cannon ball masses. C. Wassgren Chapter 01: The Basics 2 Last Updated: 05 Sep 2008 Now let’s try working the same problem using symbols rather than numbers. We’ll plug in the numbers at the very end of the problem. Symbolic Solution: Draw the FBD as before: y V θ mg x Assume this height is negligible. Follow the same approach as before but now with symbols rather than with numbers. ⇒ x=0 ∑ Fx = mx ⇒ 0 = mx ∑F y = my ⇒ − mg = my ⇒ y = −g x = x0 = V cos θ y = − gt + y0 = − gt + V sin θ x = (V cos θ ) t y=− 1 (x0 = 0) (1.5) gt + (V sin θ ) t (y0 = 0) (1.6) 2 2 The time aloft is found by setting y = 0: 2V sin θ t= g and the distance traveled is: 2 2V 2 cos θ sin θ V sin ( 2θ ) = x= g g (1.7) (1.8) We can now plug in the given numbers to get our numerical answers: 2 (100 m/s ) sin ( 30 ) t= = 10.2 s 9.81 m/s 2 Same answers as before! 2 (100 m/s ) sin ( 60 ) x= = 883 m 9.81 m/s 2 Using these results for t and x we can easily calculate the time aloft and distance traveled for a variety of values of θ, V, and m. Note that nowhere in Eqns. (1.5) – (1.8) does the mass appear so we conclude that the mass of the cannon ball is unimportant to our calculations. We also observe that if we double the initial velocity, the time aloft will double and the distance traveled will quadruple. This information is easily lost in our calculations where numbers were used right away (refer to Eqns. (1.3) and (1.4)). C. Wassgren Chapter 01: The Basics 3 Last Updated: 05 Sep 2008 Lastly, if we wanted to determine the angle that will maximum the distance traveled for a given velocity, we observe from Eqn. (1.8) that we want sin(2θ) to be as large as possible. Thus, we should tilt our cannon at an angle of θ = 45°. Substituting this back into Eqns. (1.7) and (1.8) gives: 2V tmax = g xmax = V2 g We can also easily double-check the dimensions of the equations and verify that they are dimensionally homogeneous. L [t ] = L T = T T2 OK! 2 L [ x] = L T = L T2 where L and T represent length and time, respectively. () We can conclude from this exercise the following: 1. More information is contained in our solutions when using the symbolic approach than when using the numeric approach. 2. If several calculations must be made using different values of the parameters, solving the problem first symbolically rather than starting the problem immediately with the numbers can save considerably on the number of computations required. Furthermore, it’s much easier to correct numerical mistakes at the end of the problem rather than at the beginning or in the middle of the problem. ⇒ You’re almost always better off working out a problem using symbols than with numbers! Be Sure To: 1. Work out problems symbolically and wait to substitute numerical values until the final relation has been derived. 2. Try to physically interpret your equations. 3. Make sure any relations you derive and the numbers you calculate are physically reasonable. 4. Double check that the dimensions (or units) of your answers are correct. C. Wassgren Chapter 01: The Basics 4 Last Updated: 05 Sep 2008 2. Dimensions and Units A dimension is a qualitative description of the physical nature of some quantity. Notes: 1. 2. 3. 4. A basic or primary dimension is one that is not formed from a combination of other dimensions. It is an independent quantity. A secondary dimension is one that is formed by combining primary dimensions. Common dimensions include: [M] = mass [L] = length [T] = time [θ] = temperature [F] = force If [M], [L], and [T] are primary dimensions, then [F = ML/T2] is a secondary dimension. If [F], [L], and [T] are primary dimensions, then [M = FT2/L] is a secondary dimension. A unit is a quantitative description of a dimension. A unit gives “size” to a dimension. Common systems of units include: dimension [L], length [T], time [θ], temperature [M], mass [F = ML/T2], force SI (Systéme International d’ Unités) m s K kg N BG (British Gravitational) ft s °R slug lbf EE (English Engineering) ft s °R lbm lbf Notes: 1. 1 N = 1 kg⋅m/s2 ; 1 lbf = 1 slug⋅ft/s2 ; 1 lbf = 32.2 lbm⋅ft/s2 2. The kilogram-force (kgf) is (unfortunately) a commonly used quantity. The conversion between kgf and Newtons is: 1 kgf = 9.81 N. 3. A helpful reference for unit conversions and constant values is: Avallone, E.A. and Baumeister, T., Mark’s Standard Handbook for Mechanical Engineers, McGraw-Hill. Dimensional homogeneity is the concept whereby only quantities with similar dimensions can be added (or subtracted). It is essentially the concept of “You can’t add apples and oranges.” For example, consider the following equation: 10 kg + 16 °C = 26 m/s This equation doesn’t make sense since it is not dimensionally homogeneous. How can one add mass to temperature and get velocity?!? Note that dimensional homogeneity is a necessary, but not sufficient, condition for an equation to be correct. In other words, an equation must be dimensionally homogeneous to be correct, but a dimensionally homogeneous equation isn’t always correct. For example, 10 kg + 10 kg = 25 kg The equation has the right dimensions but the wrong answer! Be Sure To: 1. Verify that equations are dimensionally homogeneous. 2. Carefully evaluate unit conversions. (A unit conversion error caused the loss of the $125M Mars Climate Observer spacecraft in 1999!) C. Wassgren Chapter 01: The Basics 5 Last Updated: 05 Sep 2008 3. Taylor Series Expansion Approximation If we know the value of some quantity, y, at some location, x, then how can we determine the value of y at a nearby location x+dx? y y(x+dx) y( x) x x+dx x SOLUTION: We can use a Taylor series expansion for y about location x: y ( x + δ x) = y ( x) + dy dx (δ x ) + x d2y dx 2 (δ x ) x 2! 2 + (1.9) As δx becomes very small, (δx)→(dx), and the higher order terms become negligibly small: (δx) >> (δx)2 > > ( δ x ) 3: dy y ( x + dx ) = y ( x ) + (1.10) ( dx ) dx x Note that Eqn. (1.10) is simply the equation of a line. We can see what’s happening more clearly if we consider the following plot: y y(x+dx) ≈ y(x) + dy/dx|x(dx) ≈ y(x+dx) dy/dx|x y( x) x x+dx x We’ll use this approximation often, especially when examining how quantities vary over small distances. Be Sure To: 1. Make sure you understand how this procedure works. It will be used frequently in the remainder of the notes. C. Wassgren Chapter 01: The Basics 6 Last Updated: 05 Sep 2008 4. Experimental Uncertainty In any experimental (or even computational) study, attention must be paid to the uncertainties involved in making measurements. Including the uncertainty allows one to judge the validity or accuracy of the measurements. Uncertainty analysis can also be useful when designing an experiment so that the propagation of uncertainties can be minimized (this will be covered in an example). Consider a measurement of a flow rate through a pipe. Let’s say that one measures a flow rate of 1.6 kg/s. Now consider a theoretical calculation that predicts a flow rate of 1.82 kg/s. Are the theory and measurement inconsistent? The answer depends upon the uncertainty in the measurement. If the experimental uncertainty is ±0.3 kg/s, then the true measured value could very well be equal to the theoretical value. However, if the experimental uncertainty is ±0.1 kg/s, then the two results are very likely to be inconsistent. There are two parts to uncertainty analysis. These include: 1. estimating the uncertainty associated with a measurement and 2. analyzing the propagation of uncertainty in subsequent analyses. Both of these parts will be reviewed in the following sections. There are many texts (such as Holman, J.P., Experimental Methods for Engineers, McGraw-Hill) that can be referred to for additional information concerning experimental uncertainty. Estimation of Uncertainty There are three common types of error. These include “blunders,” systematic (or fixed) errors, and random errors. 1. “Blunders” are errors caused by mistakes occurring due to inattention or an incorrectly configured experimental apparatus. Examples: Blatant blunder: An experimenter looks at the wrong gauge or misreads a scale and, as a result, records the wrong quantity. Less blatant blunder: A measurement device has the wrong resolution (spatial or temporal) to measure the parameter of interest. For example, an experimenter who uses a manometer to measure the pressure fluctuations occurring in an automobile piston cylinder will not be able to capture the rapid changes in pressure due to the manometer’s slow response time. Subtle blunder: A measurement might affect the phenomenon that is being measured. For example, an experimenter using an ordinary thermometer to make a very precise measurement of a hot cavity’s temperature might inadvertently affect the measurement by conducting heat out of the cavity through the thermometer’s stem. 2. Systematic (or fixed) errors occur when repeated measurements are in error by the same amount. These errors can be removed via calibration or correction. Example: The error in length caused by a blunt ruler. This error could be corrected by calibrating the ruler against a known length. C. Wassgren Chapter 01: The Basics 7 Last Updated: 05 Sep 2008 3. Random errors occur due to unknown factors. These errors are not correctable in general. Blunders and systematic errors can be avoided or corrected. It is the random errors that we must account for in uncertainty analyses. How we quantify random errors depends on whether we conduct a single experiment or multiple experiments. Each case is examined in the following sections. 1. Single Sample Experiments A single sample experiment is one in which a measurement is made only once. This approach is common when the cost or duration of an experiment makes it prohibitive to perform multiple experiments. The measure of uncertainty in a single sample experiment is ±1/2 the smallest scale division (or least count) of the measurement device. For example, given a thermometer where the smallest discernable scale division is 1 °C, the uncertainty in a temperature measurement will be ±0.5 °C. If your eyesight is poor and you can only see 5 °C divisions, then the uncertainty will be ±2.5 °C. One should use an uncertainty within which they are 95% certain that the result lies. Example: The least count for the ruler to the left is 1 mm. Hence, the uncertainty in the length measurement will be ±0.5 mm. Example: You use a manual electronic stop watch to measure the speed of a person running the 100 m dash. The stop watch gives the elapsed time to 1/1000th of a second. What is the least count for the measurement? SOLUTION: Although the stop watch has a precision of 1/1000th of a second, you cannot respond quickly enough to make this the limiting uncertainty. Most people have a reaction time of 1/10th of a second. (Test yourself by having a friend drop a ruler between your fingers. You can determine your reaction time by where you catch the ruler.) Hence, to be 95% certain of your time measurement, you should use an uncertainty of ±1/2(0.1 sec) = ±0.05 sec. Be Sure To: 1. Always indicate the uncertainty of any experimental measurement. 2. Carefully design your experiments to minimize sources of error. 3. Carefully evaluate your least count. The least count is not always ±1/2 of the smallest scale division. C. Wassgren Chapter 01: The Basics 8 Last Updated: 05 Sep 2008 2. Multiple Sample Experiment A multiple sample experiment is one where many different trials are conducted in which the same measurement is made. Example: - making temperature measurements in many “identical” hot cavities (shown below) or making temperature measurements in the same cavity many different times many identical cavities and thermometers We can use statistics to estimate the random error associated with a multiple sample experiment. For truly random errors, the distribution of errors will follow a Gaussian (or normal) distribution which has the following qualitative histogram: # of measurements with a particular measurement value measurement value (e.g. temperature) To quantify the set of measurement data, we commonly use the mean of the data set and its standard deviation or variance. For example, consider N measurements of some parameter x: x1, x2, …, xN. sample mean, x (a type of average) 1N x = ∑ xn N n =1 (1.11) sample standard deviation, σ (a measure of how precise the measurements are: as σ ↓, the precision ↑) 1 N σ = ∑ ( xn − x ) N − 1 n =1 C. Wassgren Chapter 01: The Basics 2 1 2 (Note: The variance is σ2.) 9 (1.12) Last Updated: 05 Sep 2008 Notes: 1. It is not possible to comprehensively discuss statistical analyses of data within the scope of these notes. The reader is encouraged to look through an introductory text on statistics for additional information (see, for example, Vardeman, S.B., Statistics for Engineering Problem Solving, PWS Publishing, Boston). 2. The square of the standard deviation (σ2) is known as the variance. 3. The coefficient of variation, CoV or CV (also rsd = relative standard deviation), is defined as the ratio of the standard deviation to the mean, i.e. CoV ≡ σ x . A small CoV means that the scatter in your measurements is small compared to the mean. 4. For random data (a Gaussian or normal distribution) and a very large number of measurements: 68% x ± 1σ 95% of measurements fall between x ± 2σ x ± 3σ 99% used in most engineering situations as a measure of the uncertainty # of measurements with a particular measurement value measurement value (e.g. temperature) x ±1σ ±2σ ±3σ 4. If the number of measurements is not very large (N < 30 for example), it is better to use the Student’s t-distribution for estimating the uncertainty (refer to an introductory text on statistics such as Vardeman, S.B., Statistics for Engineering Problem Solving, PWS Publishing, Boston): x ± tσ (1.13) where t is a factor related to the degree of confidence desired (again, a 95% uncertainty is typically desired in engineering applications), σ is the standard deviation given in Eqn. (1.12), and n is the number of measurements made. The following table gives the value of t for various values of N and a 95% confidence level. Note that as N → ∞ the t factor approaches the large sample size value (1.96). N t95% 2 12.71 3 4.30 4 3.18 5 2.78 6 2.57 7 2.45 8 2.36 9 2.31 10 2.26 15 2.14 20 2.09 30 2.04 ∞ 1.96 5. Often one presents data in terms of its true (rather than sample) mean, µ, and a confidence interval. For random data, the true mean lies within the interval tσ µ=x± (1.14) N where x and σ are the sample mean (Eqn. (1.11)) and standard deviation (Eqn. (1.12)), respectively, t is the confidence interval factor (found from the Student t-distribution as in the table above), and N is the number of data. Be Sure To: 1. Use a t-distribution for evaluating the uncertainty when you do not have sufficient data for using the standard deviation as a meaningful measure of the uncertainty. C. Wassgren Chapter 01: The Basics 10 Last Updated: 05 Sep 2008 Example: Eight measurements are made of a fluid’s density and are presented in the following table. Determine the interval in which the mean density lies with a confidence level of 95%. Measurement # 1 2 3 4 5 6 7 8 density [g/cc] 1.20 1.24 1.25 1.22 1.17 1.19 1.21 1.20 SOLUTION: The sample mean and standard deviation are: 1 x = (1.20 + 1.24 + 1.25 + 1.22 + 1.17 + 1.19 + 1.21 + 1.20 ) g/cc = 1.21 g/cc 8 2 2 2 2 1 (1.20 − 1.21) + (1.24 − 1.21) + (1.25 − 1.21) + (1.22 − 1.21) 2 ( g/cc ) = 0.03 g/cc σ= 8 − 1 + (1.17 − 1.21)2 + (1.20 − 1.19 )2 + (1.20 − 1.21)2 + (1.20 − 1.20 )2 The number of samples is N = 8 and the confidence interval is 95%. Thus, the t factor should be 2.36 as given on the table from the previous page. The true mean then lies within the interval (to within 95% confidence): ( 2.36 ) ( 0.03) tσ = 1.21 ± g/cc (95% CI) µ=x± N 8 ∴ µ = 1.21 ± 0.02 g/cc (95% CI) C. Wassgren Chapter 01: The Basics 11 Last Updated: 05 Sep 2008 Propagation of Uncertainty Let R be a result that depends on several measurements: x1, …, xN, or, in mathematical terms: R = R ( x1 ,… , xN ) For example, the volume of a cylinder is: V = π r 2h ⇒ V = V ( r, h ) How do we determine the uncertainty in the result R due to the uncertainties in the measurements x1, …, xN? In the example above, what is the uncertainty in the volume V given the uncertainties in the radius, r, and height, h? Consider how a small variation in parameter, xn, call it δxn, causes a variation in R, call this variation δRxn: δ Rxn = R ( x1, … , xn + δ xn ,… xN ) − R ( x1, … , xn ,… xN ) δ Rx = R ( x1, … , xn + δ xn ,… xN ) − R ( x1, … , xn ,… xN ) δ xn n δ Rx n ≈ ∂R ∂xn δ xn δ xn (Note that an “=” is only strictly true as δxn → dxn.) uncertainty partial derivative in measurement xn of R w/r/t xn uncertainty in R due to uncertainty in xn The total uncertainty in R, δR, due to uncertainties in all measurements x1,…,xN, assuming that the xn are independent so that the variations in one parameter do not affect the variations in the others, is estimated as: 1 1 2 2 N ∂R (1.15) = ∑ ∂x δ xn n =1 n The relative uncertainty in R, uR, is given by: δR uR = (1.16) R 2 For example, the uncertainty in the cylinder volume, V=πr h, due to uncertainties in the radius, r, and height, h, is: N δ R = ∑ δ Rxn n =1 ( ) 2 2 2 2 ∂V ∂V δ V = δr + δh ∂h ∂r 1 2 1 2 2 2 = ( 2π rhδ r ) + (π r 2δ h ) and the relative uncertainty is: 1 2 2 δV 1 2 uV = = 2 ( 2π rhδ r ) + (π r 2δ h ) V πr h δ r 2 δ h 2 = 2 + r h 2 2 = ( 2ur ) + ( uh ) 1 1 2 2 Note: 1. Use absolute quantities when calculating the uncertainty. For example, use °R or K as opposed to °F or °C for temperature, and use absolute pressures rather than gage pressures. 2. In an uncertainty analysis the uncertainty of some quantities may be so small compared to the uncertainties in the remaining quantities that they can be considered “exactly” known. This is generally the case for well characterized constants and material parameters, e.g. the acceleration due to gravity. C. Wassgren Chapter 01: The Basics 12 Last Updated: 05 Sep 2008 Be Sure To: 1. Use absolute quantities when evaluating uncertainties, e.g. absolute temperature and pressure. 2. Review your uncertainty analyses to determine which measurements result in the greatest error in a derived quantity. Design your experiments to reduce these uncertainties. C. Wassgren Chapter 01: The Basics 13 Last Updated: 05 Sep 2008 Example: A resistor has a nominal stated value of 10±0.1 Ω. A voltage difference occurs across the resister and the power dissipation is to be calculated in two different ways: a. from P=E2/R b. from P=EI In (a) only a voltage measurement will be made while both current and voltage will be measured in (b). Calculate the uncertainty in the power for each case when the measured values of E and I are: E = 100±1 V (for both cases) I = 10±0.1 A R E I SOLUTION: Perform an uncertainty analysis using the first formula for power. P=E 2 (1.17) R The relative uncertainty in P is: 1 2 2 2 u P = u P , E + u P , R where 1 ∂P R 2E δE δE = 2 uP , E = = 2u E δ E = 2 P ∂E R E E (1.18) (1.19) 1 ∂P R −E 2 δR δ R = 2 2 δ R = − = −u R R P ∂R R E Substitute into Eqn. (1.18). uP , R = (1.20) 1 2 22 uP = 4uE + uR The relative uncertainties in the voltage and resistance are: δE 1V uE = = = 1% E 100 V δ R 0.1 Ω uR = = = 1% R 10 Ω ⇒ uP = 2.24% C. Wassgren Chapter 01: The Basics 14 (1.21) (1.22) (1.23) Last Updated: 05 Sep 2008 Now perform an uncertainty analysis using the second relation for power. P = EI The relative uncertainty in P is: (1.24) 1 2 2 2 u P = u P , E + u P , I where 1 ∂P 1 δE δE = uP , E = ( I ) δ E = = uE P ∂E EI E 1 ∂P 1 δI δR = uP , I = ( E ) δ I = = uI P ∂R EI I Substitute into Eqn. (1.18). (1.25) (1.26) 1 2 2 uP = uE + uI2 The relative uncertainties in the voltage and resistance are: δE 1V uE = = = 1% E 100 V δ I 0.1 A uI = = = 1% I 10 A ⇒ uP = 1.41% (1.27) (1.28) (1.29) We observe that using the second relation (P = EI) gives a smaller uncertainty for the given values. C. Wassgren Chapter 01: The Basics 15 Last Updated: 05 Sep 2008 Example: A certain obstruction-type flowmeter is used to measure the flow of air at low velocities. The relation describing the flow rate is: 1/ 2 2p m = CA 1 ( p1 − p 2 ) RT1 where C is an empirical discharge coefficient, A is the flow area, p1 and p2 are the upstream and downstream pressures, T1 is the upstream temperature, and R is the gas constant for air. Calculate the relative uncertainty in the mass flow rate for the following conditions: C = 0.92±0.005 (from calibration data) p1 = 25±0.5 psia T1 = 530±2 °R ∆p = p1-p2 = 1.4±0.005 psia A = 1.0±0.001 in2 What factors contribute the most to the uncertainty in the mass flow rate? SOLUTION: The relative uncertainty in the mass flow rate is given by: 2 2 2 2 2 um = um,C + um, A + um, p1 + um,T1 + um,∆p where 1 ∂m δC δC = u m ,C = = uC m ∂C C 1 ∂m δA δA= um , A = = uA m ∂A A 1 ∂m 1 δ p1 1 δ p1 = um, p1 = = 2 u p1 m ∂p1 2 p1 um ,T1 = 1 2 (1.30) (1.31) (1.32) (1.33) 1 ∂m 1 δ T1 δ T1 = − = − 1 uT1 2 m ∂T1 2 T1 (1.34) 1 ∂m 1 δ ( ∆p ) 1 = 2 u∆p (1.35) δ ( ∆p ) = m ∂∆p 2 ∆p Note that the there is negligible uncertainty in the gas constant R since it is presumed to be known to a high degree of accuracy. um,∆p = C. Wassgren Chapter 01: The Basics 16 Last Updated: 05 Sep 2008 Substitute into Eqn. (1.30). 2 2 2 2 um = uC + u A + 1 u 21 + 1 uT1 + 1 u∆p 4p 4 4 where the relative uncertainties are: δ C 0.005 uC = = = 0.54% C 0.92 1 2 (1.36) (1.37) δ A 0.001 in 2 = = 0.10% A 1.0 in 2 δ p 0.5 psia u p1 = 1 = = 2.0% p1 25 psia uA = uT1 = δ T1 T1 = 2R 530 R (1.38) (1.39) = 0.38% (1.40) δ ( ∆p ) 0.005 psia = = 0.36% 1.4 psia ∆p ⇒ um = 1.2% u∆p = (1.41) Examine the contributions of each term on the right hand side of Eqn. (1.36) to determine which uncertainty has the greatest influence on the uncertainty in m . ( ) = 2.9 *10 u = (1.0 *10 ) = 1.0 *10 u = ( 2.0 *10 ) = 1.0 *10 u = ( 3.8*10 ) = 3.6 *10 u = ( 3.6 *10 ) = 3.2 *10 2 −5 −3 2 −6 2 uC = 5.4 *10−3 2 A 2 p1 1 4 12 4 T1 1 4 1 4 1 4 2 ∆p 1 4 −2 2 −4 −3 2 −6 −3 2 −6 The uncertainty in the p1 measurement contributes the most to the uncertainty in m . C. Wassgren Chapter 01: The Basics 17 Last Updated: 05 Sep 2008 5. Statistical vs. Continuum Approach Most substances consist of a collection of molecules or atoms. In studying how substances behave, we can either explicitly account for the molecular behavior, referred to as the statistical approach, or instead treat the substance as being continuous, referred to as the continuum approach. Statistical Approach In the statistical approach we treat the substance of interest as a collection of individual objects (e.g. molecules). The interactions between molecules are modeled (e.g. using Newton’s laws) and the macroscopic behavior of the substance is determined by utilizing probability and statistics. This approach is useful in understanding the behavior of material properties (e.g. fluid viscosity) but it is not very practical for modeling typical engineering applications. Statistical Mechanics and Kinetic Theory utilize the statistical approach. Continuum Approach The continuum approach ignores individual molecules and instead treats the substance of interest as being continuously distributed in space. The macroscopic behavior of the substance is modeled using basic conservation laws (e.g. mass, momentum, energy). The continuum method requires that the smallest length scale of interest, referred to as the macroscopic length scale (e.g. the length scale over which significant changes in properties occur), be much larger than the microscopic length scale, typically the mean free path of a molecule (for gases). The mean free path, λ, of a molecule is the average distance a molecule travels before colliding with another molecule. By requiring that the (macroscopic length scale) >> (microscopic length scale), there will be enough molecules at a “point” (the smallest macroscopic length scale of interest) so that meaningful averages of the molecular behavior can be made at that point. Consider the following experiment. Let’s measure the density, ρ, of a fluid in a small cube of length, L. The local density of the fluid is defined as the total mass of molecules in the cube, Σmi, divided by the cube’s volume, L3, as the volume approaches the smallest macroscopic length scale of interest, ε: The following WWW page demonstrates this experiment: http://widget.ecn.purdue.edu/~meapplet mi ρ ≡ lim 3 L →ε L ∑ Σmi/L3 L statistical fluctuations local value ε of the density spatial fluctuations L At large length scales, L, the density will not have very good spatial resolution and, hence, may not be a very useful quantity. At very small length scales the number of molecules within the box will vary significantly with time since molecules continuously enter and exit the box. Since the small box can contain only a few molecules to begin with, the density fluctuations will be very large. There is an intermediate region between the previously discussed extremes that will have few fluctuations but good spatial resolution. C. Wassgren Chapter 01: The Basics 18 Last Updated: 05 Sep 2008 Notes: 1. In most engineering applications the length scale condition is easily satisfied. The mean free path of a gas can be estimated using the following simple analysis. Consider gas molecules with an effective cross-sectional area of A and a mean (molecular) speed of c . The volume swept out by a molecule per A unit time is: V = cA (1.42) c The expected number of collisions with other molecules per unit time is: n = νV (1.43) where ν is the number of molecules per unit volume. The average time between collisions, t’, is the inverse of the collision rate: t′ = 1 = 1 =1 (1.44) n (ν V ) (ν cA) Thus, the typical distance between collisions, aka the mean free path λ, is: λ = ct ′ = 1 (ν A) (1.45) For air at standard conditions: ν ≈ 2.7 *1019 cm-3 and A ≈ 1*10−15 cm 2 ⇒ λ ≈ 3.7 *10−5 cm This mean free path is certainly much smaller than any length scale in which we are typically interested. Examples of where the continuum assumption may not be valid include: a. flow in the upper atmosphere where the mean free path is large (e.g. at an altitude of 100 miles, the mean free path is approximately 80 m!), b. flow within a shock wave where the macroscopic length scale is very small (the width of a shock wave is on the order of 1 µm), and c. granular flows (e.g. flowing sand) where the microscopic length scale ≈ the macroscopic length scale. 2. The Knudsen number, Kn, is a dimensionless parameter that indicates when the continuum assumption is valid: (microscopic length scale) Kn ≡ 1 ⇒ the continuum assumption valid (1.46) (macroscopic length scale) Four flow regimes are typically defined based on the Knudsen number. These are (according to Zucrow, M.J. and Hoffman, J.D., Gas Dynamics Vol. 1, Wiley): Kn < 0.01 continuum flow This is the flow regime that occurs in most engineering applications. 0.01 < Kn < 0.1 slip flow In this regime, the fluid may still be treated as a continuous substance but the no-slip condition at boundaries does not hold. Instead, fluid may slip at boundaries. 0.1 < Kn < 3.0 transitional flow The flow in this regime is very difficult to analyze since the fluid cannot be considered a continuum and molecules still interact to a considerable degree. 3.0 < Kn free molecular flow In this regime molecules are spaced so far apart that they rarely interact. The flow may be modeled as a collection of non-interacting molecular impacts. 3. The continuum assumption is valid in the vast majority of engineering applications. For example, the Knudsen number for a typical engineering flow in which the mean free path is ~1*10-5 cm (refer to the calculation in Note #1) and the macroscopic length of interest is 1 mm, is Kn ~ 0.0001. C. Wassgren Chapter 01: The Basics 19 Last Updated: 05 Sep 2008 4. The Knudsen number can be related to the Reynolds and Mach numbers using some results from kinetic theory. From kinetic theory, the kinematic viscosity of a gas, ν, is related to the mean free path, λ, and mean molecular speed, c , by: ν = 1 λc (1.47) 2 The mean molecular speed is related (via kinetic theory) to the speed of sound, c, by: πγ c=c (1.48) 8 where γ is the specific heat ratio (= cp/cv). Substituting Eqn. (1.47) into Eqn. (1.48) and simplifying gives: c= λ= 2ν πγ λ 8 πγ ν (1.49) 2c From the definition of the Knudsen, Reynolds, and Mach numbers: Kn L = λ L = πγ ν 2 cL = πγ ν V 2 VL c (where L is the macroscopic length scale of interest) = 1 Re = Ma L ∴ Kn L = πγ Ma 2 Re L Hence, flows occurring at large Mach numbers and small Reynolds numbers tend toward noncontinuum flows. (1.50) Note that for large Reynolds numbers over an object, the significant length scale that should be used in the defining the Reynolds number is typically the boundary layer thickness, δ. A later set of notes will show that for (laminar) boundary layers, the boundary layer thickness is related to the macroscopic length scale, L, by: δ 1 ∝ ⇒ Reδ ∝ Re L (where ReL = VL/ν and Reδ = Vδ/ν) (1.51) L Re L Hence, Eqn. (1.50) for boundary flows (occurring at large Reynolds numbers) is: Kn δ = 5. πγ Ma 2 Reδ ∝ Ma (1.52) Re L To learn more about non-continuum flows, refer to the following texts: a. Schaaf, S.A. and Chambré, P.L., Flow of Rarefied Gases, Princeton University Press. b. Nguyen, N-T. and Wereley, S.T., Fundamentals and Applications of Microfluidics, Artech House. C. Wassgren Chapter 01: The Basics 20 Last Updated: 05 Sep 2008 6. Fluid Properties Some good fluid property references include: a. Avallone, E.A. and Baumeister III, T., Marks’ Standard Handbook for Mechanical Engineers, McGraw-Hill. b. Kestin, J., and Wakeham, W.A., Transport Properties of Fluids, CINDAS Data Series on Material Properties, C.Y. Ho, ed., Hemisphere Publishing. density, ρ {M/L3}, [kg/m3, slugs/ft3, lbm/ft3] - The density of a substance is a measure of how much mass there is of the substance per unit volume. ρ H 0@4 C = 1000 kg/m3 = 1.94 slugs/ft 3 = 62.4 lb m /ft 3 2 ρair@15 C ,1atm = 1.23 kg/m3 = 2.38*10−3 slugs/ft 3 = 7.68*10−2 lb m /ft 3 - Density does not vary greatly with temperature for liquids in general. Density does change considerably with temperature and pressure for gases. A substance where the density remains constant for all conditions is considered incompressible. - A good engineering rule of thumb is that if there are no significant temperature changes and for fluid velocities less than approximately 1/3 the speed of sound in the fluid, the fluid can be approximated as incompressible (the proof of this is examined when discussing compressible fluid flow). - In air, the speed of sound at standard conditions (p = 1 atm, T = 59 °F = 15 °C), is approximately 1100 ft/s (340 m/s). - The speed of sound in water is approximately 4800 ft/s (1500 m/s). - The speed of sound in steel is approximately 16400 ft/s (5000 m/s). - In most instances, liquids and gases flowing at low speeds can be approximated as incompressible. - specific gravity, SG {no units – dimensionless} - The specific gravity of a liquid is the ratio of the liquid’s density to the density of water at some specified condition (typically at 4°C). SGliquid ≡ ρ liquid ρH O @ 4 C 2 For example, the density of mercury (Hg) at 20°C is 13.6*103 kg/m3. Hence, SGHg = 13.6. - specific weight, γ {F/L3} [N/m3, lbf/ft3] - The specific weight of a substance is the weight of the substance per unit volume. W mg ρVg γ≡ = = V V V ∴γ = ρ g For example, the specific weight of water at 4°C is 9.81*103 N/m3 = 62.4 lbf/ft3. - specific volume, v {L3/M} [m3/kg, ft3/slug, ft3/lbm] - The specific volume of a substance is how much volume the substance occupies per unit mass. - The specific volume is simply the inverse of the density: 1 v= ρ - In thermodynamics, the specific volume is commonly used in place of density. Be Sure To: 1. Be careful when using the incompressible flow assumption. Make sure that the assumption is reasonable for your flow situation. For liquids, the assumption is usually reasonable. For gases, you need to check the flow velocities and temperatures. C. Wassgren Chapter 01: The Basics 21 Last Updated: 05 Sep 2008 pressure, p {F/L2} [Pa=N/m2, psf=lbf/ft2, psi=lbf/in2] - Pressure is the inward acting force per unit area acting normal to a surface. ˆ dFn = − p * dA * n negative sign pressure acting since pressure on surface acts inward on surface small normal force acting on surface area of surface outward pointing unit normal vector for the surface pressure force is normal (at right angles) to the surface ˆ dFn = − pdAn ˆ n surface with small area, dA, and outward ˆ pointing unit normal vector, n - Pressure is a scalar quantity. It has no direction (it’s a number – just like temperature). The surface orientation is what gives the pressure force its direction. Proof: Consider the pressure forces acting on a small triangular wedge of fluid at rest. Assume the wedge extends a depth of 1 (unit depth) into the page. Also assume that the pressure depends on direction. ∑F ( x p dy = mx sin θ ) (1) sin θ − p ( dy )(1) = ρ ( x p(dy/sinθ)(1) 1 2 dxdy )(1) x ( p − px ) dy = ρ ( 1 2 dxdy ) x θ y dy/sinθ As dx, dy → 0, p = px x A similar approach can be taken in the y-direction to find: p = py θ dy dx py(dx)(1) Thus, we observe that the pressure is the same in every direction. - Pressure is the result of molecules colliding against (a real or imaginary) surface. Every time a particle bounces off a surface, its momentum, mu, changes (u changes direction). From Newton’s 2nd Law, F=d/dt(mu), there must be a force exerted on the molecule by the wall (and conversely, on the wall by the molecule) in order to cause the change in the molecule’s momentum. These collisions result in what we call pressure. - Absolute pressure [psia] is referenced to zero pressure (a perfect vacuum has pabs=0). - There are no molecules in a perfect vacuum thus there is no pressure (due to molecular collisions). - Atmospheric pressure at standard conditions: patm, abs = 14.696 psia = 101.33*103 Pa (abs) = 1 atm = 1.0133 bar - Gage pressure [psig] is referenced to atmospheric pressure: pgage = pabs - patm - Atmospheric pressure at standard conditions: patm = 0 psi (or psig). - A perfect vacuum has p = -14.7 psig (referencing to standard atmospheric conditions). - An important equation of state for an ideal gas is the ideal gas law: p = ρ RT Note that p and T are absolute quantities, e.g. [p] = psia or Pa (abs), [T] = °R or K C. Wassgren Chapter 01: The Basics 22 Last Updated: 05 Sep 2008 px(dy)(1) - The vapor pressure, pv of a liquid is the pressure at which the liquid is in equilibrium with its own vapor. The pressure in the vapor is the vapor pressure, pv. vapor liquid A closed container with the liquid and vapor in equilibrium. - The vapor pressure for a liquid will increase with increasing temperature. - If the pressure in the liquid falls below the vapor pressure, then the liquid will turn to vapor. This can occur via boiling or cavitation. - Boiling occurs when the temperature of the liquid increases so that the vapor pressure equals the surrounding atmospheric pressure. For example, the vapor pressure of water at 20 °C is 0.023 atm while the vapor pressure at 100 °C is 1 atm. Hence, one method of turning liquid water to vapor is to bring the water temperature to 100 °C while holding the surrounding pressure at 1 atm. This is known as boiling and is shown schematically in the phase plot shown below on the left-hand side. - Cavitation occurs when the surrounding pressure drops below the vapor pressure. Using the previous example, we can also turn liquid water to water vapor by dropping the surrounding pressure to 0.023 atm at 20 °C. This is shown schematically in the phase plot shown below on the right-hand side. pressure pressure liquid liquid solid solid cavitation (T =constant, p ↓ ) boiling (p=constant, T ↑ ) gas gas temperature - temperature Cavitation can cause considerable damage to surfaces. When cavitation occurs in a liquid, pockets of vapor form (either as bubbles or large “voids”). As the vapor pockets travel into a region where the surrounding pressure is greater than the vapor pressure, the vapor region rapidly collapses. This collapse can be so rapid that shock waves propagate from the collapsing region and impact on nearby surfaces causing small bits of the surface to erode away. Hence, cavitation is typically avoided when designing pumps, pipe bends, and underwater propellers. Be Sure To: 1. Use an absolute pressure, and not a gage pressure, in the ideal gas law. 2. Be careful not to mix gage and absolute pressures when evaluating pressure forces. 3. Use the correct area when calculating a pressure force. 4. Integrate to find a pressure force when the pressure is not uniform over the area over which the pressure acts. 5. Check for cavitation in low pressure flows of liquids. C. Wassgren Chapter 01: The Basics 23 Last Updated: 05 Sep 2008 temperature, T {θ} [°R, K, °F, °C] An object’s temperature is a quantitative way of describing how “hot” the object is (temperature is a measure of the agitation or random kinetic energy of the molecules). We typically measure an object’s temperature using a device called a “thermometer.” Experience tells us that when two objects are placed in contact with each other and have different temperatures, the hotter object (i.e., the one with a larger temperature) will become cooler while the cooler object becomes hotter. When the two objects have the same temperature, they no longer change temperature. The objects are then said to be in thermal equilibrium. A simple but fundamental concept concerning thermal equilibrium is the: Zeroth Law of Thermodynamics: If two bodies are in thermal equilibrium with a third body, then the two bodies will also be in thermal equilibrium. This concept is key when comparing the temperatures of two objects not in contact using a thermometer since the thermometer acts as the third body. To use the concept of temperature, we must first define some scale on which we’ll measure temperatures. Perhaps the easiest scale to define would be one where we reference all temperatures to some reproducible and unique physical phenomena such as the freezing or boiling points of water. Two point scales use two phenomena to define the temperature scale. For example, let’s define the ice point of water (when ice and water are in equilibrium at a pressure of 1 atm) as our 0° temperature and the steam point of water (when water and water vapor are in equilibrium at a pressure of 1 atm) as our 100° temperature. We can now measure all temperatures relative to this scale. Examples of two-point scales include the Celsius and Fahrenheit scales: ice point of H2O steam point of H2O Celsius scale 0 °C 100 °C Fahrenheit scale 32 °F 212 °F Long ago, researchers noticed something curious when measuring the temperature of various gases at different pressures (and constant volume). They found that the temperature of a gas at low pressures is proportional to its pressure at a constant volume: T = a + bp (1.53) where a and b are constants. p gas #1 gas #2 gas #3 gas #4 T -273.15 °C = -459.67 °F By extrapolating the temperature of the gases at zero pressure, we find that the lowest possible temperature, or the absolute zero temperature, is –273.15 °C on the Celsius scale which corresponds to –459.67 °F on the Fahrenheit scale (i.e., a= -273.15 °C = -459.67 °F). This temperature scale, based on the behavior of ideal gases, is referred to as the ideal gas temperature scale. C. Wassgren Chapter 01: The Basics 24 Last Updated: 05 Sep 2008 To make things a bit easier, let’s redefine our scale so that absolute zero is the zero point of our temperature scale, i.e.: T = bp (a=0) p gas #1 gas #2 gas #3 gas #4 T 0K This is called an absolute temperature scale since the lowest temperature is zero. Note that so far our temperature scales have been based on the behavior of a particular substance (e.g. water) or a particular class of substances (e.g. gases). A better way to define a temperature scale is to make the scale independent of substances. Such a scale is called a thermodynamic temperature scale. In order to define such a temperature scale, we would need to first learn about the 2nd Law of Thermodynamics; a topic not covered in these notes (refer to [3] for this topic). Suffice it to say here that the scale using this method gives the same result as that using the ideal gas temperature scale. To summarize, the lowest possible temperature is: 0 °R = 0 K and ∆(1 K) = ∆(1 °C) ∆(1 °R) = ∆(1 °F) Some additional helpful conversions are given below (the “θ” refers to temperature): θ(K) = 1.8 θ(°R) (1.8 = 9/5) θ(°C) = [θ(°F)-32]/1.8 θ(°C) = θ(K) – 273.15 θ(°F) = θ(°R) – 459.67 Another convenient conversion formula: 10 °C = 50 °F (for every 5 °C increase, add 9 °F ) Be Sure To: 1. Use an absolute temperature when using the ideal gas law. References 3. Moran, M.J. and Shapiro, H.N., Fundamentals of Engineering Thermodynamics, Wiley. C. Wassgren Chapter 01: The Basics 25 Last Updated: 05 Sep 2008 (dynamic) viscosity, µ {F⋅T/L2} [N⋅s/m2, lbf⋅s/ft2] 10 Poise = 1 N⋅s/m2 = 1 kg/(m⋅s) 1 centipoise = 0.01 Poise = 0.001 N⋅s/m2 (Note: “Poise” is pronounced “’pwäz”.) - Another common unit: - Viscosity is the “internal friction” within the fluid. It’s a measure of how easily a fluid flows. - The viscous stresses in a fluid will be related to the deformation rate of a small element of fluid. Recall that for solids, the force is related to the amount of deformation, e.g. Hooke’s Law for springs. For fluids, however, the forces are related to the deformation rates. Consider the deformation of a small piece of fluid with area (dxdy) as shown below. The top of the fluid element is subject to a shear stress, τ, over a short time, dt. During this time, the top of the fluid element moves with a small velocity, du, with respect to the bottom of the element. The total distance the top moves relative to the bottom will be (du*dt). The angular deformation of the fluid element can be measured by the angle the vertical sides of the element have deformed. τ (du*dt) dθ dy dx Here, the small angle, dθ, is found from simple trigonometry to be: dudt tan dθ = dy Since the angle is very small, tan(dθ) ≈ dθ. The rate at which the element deforms is found by dividing both sides by dt: dθ du = shear or angular deformation rate dt dy Hence, the rate of angular deformation of the element, dθ/dt, is equal to the velocity gradient, du/dy, in the fluid. Furthermore, since the shear stress, τ, will be a function of deformation rate (as stated previously), we have: du dθ τ = fcn = fcn dt dy where the term “fcn” refers to the fact that we don’t yet know how the shear stress depends on the deformation rate, we just know that it does. - A Newtonian fluid is one in which the shear stress varies proportionally with the deformation rate. The constant of proportionality is called the dynamic viscosity, µ. du τ =µ shear stress relation for a Newtonian fluid (1.54) dy - Air and water are two examples of Newtonian fluids. - A more precise definition of the shear stress in Eqn. (1.54) is: du τ yx = µ x dy C. Wassgren Chapter 01: The Basics 26 Last Updated: 05 Sep 2008 where the subscript on the stress indicates that the shear stress acts on a y-plane in the xdirection. The x-subscript on the velocity indicates that it is the x-component. We will review this sign convention in greater detail later in the notes when discussing stresses (Chapter 03). - In a non-Newtonian fluid the shear stress is not proportional to the deformation rate but instead varies in some other way. - Non-Newtonian fluids are further classified by how the shear stress varies with deformation rate. The apparent viscosity, µapp, is the viscosity at the local conditions as shown in the plot below (for a Newtonian fluid the apparent viscosity remains constant). non-Newtonian shear thinning Newtonian shear stress, τ non-Newtonian shear thickening apparent viscosity, µapp rate of shearing strain (deformation rate), du/dy - - - In shear thinning (aka psuedoplastic) fluids, the apparent viscosity decreases as the shear stress increases. Examples of shear thinning fluids include blood, latex paint, and cookie dough. In shear thickening (aka dilatant) fluids, the apparent viscosity increases as the shear stress increases. An example of a shear thickening fluid is quicksand or a thick cornstarch-water mixture. Viscosity is weakly dependent on pressure but is sensitive to temperature. For liquids, the viscosity generally decreases as the temperature increases and increases as pressure increases. Changes in temperature and pressure can be very significant in lubrication problems. For gases, the viscosity increases as the temperature increases (in fact, from kinetic theory one can show that µ ∝ T1/2). The following plot shows the variation in dynamic viscosity with temperature for several fluids. µH20@20 °C = 1.00*10-3 N⋅s/m2 = 1 cP µair@20 °C = 1.81*10-5 N⋅s/m2 = 0.018 cP (Plot from Fox, R.W. and McDonald, A.T., Introduction to Fluid Mechanics, 5th ed., Wiley.) C. Wassgren Chapter 01: The Basics 27 Last Updated: 05 Sep 2008 - The kinematic viscosity, ν, is a quantity that appears often in fluid mechanics. It is defined as: νH20@20 °C = 1.00*10-6 m2/s = 1 cSt µ ν≡ νair@20 °C = 1.50*10-5 m2/s = 15 cSt ρ The dimensions of kinematic viscosity are {L2/T} with common units of [m2/s, ft2/s]. Another common unit for kinematic viscosity is the Stoke: 1 Stoke = cm2/s. Note that kinematic viscosity has the dimensions of a diffusion coefficient. The kinematic viscosity is a measure of how rapidly momentum diffuses into a flow. This will be discussed again in a later set of notes when reviewing solutions to the Navier-Stokes equations (Chapter 06). - There are several techniques for measuring the viscosity of a fluid. Several devices include the: - Couette viscometer - capillary tube viscometer - sedimentation rate viscometer - falling ball or falling cylinder viscometer - rotating disk or oscillating disk viscometer A good reference on experimental viscometry is: Dinsdale, A. and Moore, F., Viscosity and its Measurement, Chapman and Hall. - Let’s examine a common flow situation shown in the figure below. A fluid is contained between two, infinitely long parallel plates separated by a distance H. The bottom plate is fixed while the top plate moves at a constant velocity V. There are no pressure gradients in the fluid. This type of flow situation is called a planar Couette flow. top plate moves at constant velocity, V H fluid fluid velocity, u(y) = (V/H)y y x fixed bottom plate - One of the first things we would notice when conducting this experiment is that the fluid sticks to the solid boundaries, i.e. there is zero relative velocity between the fluid and the boundary. This is referred to as the no-slip condition. The no-slip condition occurs for all real, viscous fluids under normal conditions. (The no-slip condition may be violated in rarefied flows where the motion of individual molecules becomes significant, i.e. when the Knudsen number is Kn > 0.1.) - The second thing we would notice is that the fluid velocity profile is linear with the velocity given by: y u =V H Note that the velocity at the bottom plate is zero and at the top plate the velocity is V; thus, the no-slip condition is satisfied. We will derive this velocity profile later in these notes when discussing solutions to the Navier-Stokes equations (Chapter 06). C. Wassgren Chapter 01: The Basics 28 Last Updated: 05 Sep 2008 - If the fluid is Newtonian, then the shear stress acting on the fluid is: du V τ yx = µ x = µ dy H Note that the shear stress is a constant everywhere in the fluid, i.e. there is no y dependence. Additionally, the shear stress will act to resist the motion of the top plate and try to carry the bottom plate along with the fluid. shear stress acting on wall due to fluid positive shear stress acting on fluid element next to wall (refer to the stress sign convention in Chapter 03) - An inviscid fluid is one that has zero viscosity (µ = 0). Although there are no typical fluids that are completely inviscid (a counter-example: superfluid helium is completely inviscid!), there are many cases where assuming that the fluid viscosity is negligible is a reasonable approximation. In addition, many analyses of fluid flow rely on an inviscid assumption in order to make the mathematics of the analyses tractable without the use of computers. Note that the no-slip condition does not hold when the fluid is inviscid. - An ideal fluid is one that is incompressible and inviscid. The ideal fluid approximation is often reasonable and is commonly used in fluid mechanics analyses. We’ll consider ideal fluid flow in Chapter 04. Be Sure To: 1. Get your shear stress directions correct. Eqn. (1.54) is the shear stress acting on the fluid element. 2. Use the correct area when evaluating shear forces. 3. Integrate to determine a shear force on an area where the shear stress is not uniform over the area. 4. Evaluate the velocity gradient in Eqn. (1.54) at the location where you’re interested in determining the shear stress. C. Wassgren Chapter 01: The Basics 29 Last Updated: 05 Sep 2008 Example: Determine the magnitude and direction of the shear stress that the water applies: a. to the base b. to the free surface free surface U u y y = 2 − U h h water h 2 y u x SOLUTION: The shear stress, τyx, acting on a Newtonian fluid is given by: du τ yx = µ (1.55) dy where du 1 2y =U 2 − 2 (1.56) dy h h Evaluating the shear stress at the base and free surface gives: 2µU τ yx y = 0 = base (y = 0): (1.57) h This is the stress the wall exerts on the fluid. The fluid will exert an equal but opposite stress on the wall. y x free surface (y = h): τ yx y =h fluid element adjacent to the wall with positive τyx shown =0 (1.58) The air at the free surface does not exert a stress on the water. Although in reality the air will exert a small shear stress on the water, assuming that the shear stress is negligible is reasonable in most engineering applications. C. Wassgren Chapter 01: The Basics 30 Last Updated: 05 Sep 2008 Example: Magnet wire is to be coated with varnish for insulation by drawing it through a circular die of 0.9 mm diameter. The wire diameter is 0.8 mm and it is centered in the die. The varnish, with a dynamic viscosity of 20 cP, completely fills the space between the wire and the die for a length of 20 mm. The wire is drawn through the die at a speed of 50 m/s. Determine the force required to pull the wire. SOLUTION: Apply Newton’s 2nd Law in the x-direction to the wire shown in the diagram below. R0 R i rx Since the gap width is small compared to the wire radius, curvature effects may be neglected and a linear velocity profile may be reasonably assumed (i.e. (Ro – Ri)/Ri << 1). L ∑F x F, V = F + τ rx ( 2π Ri L ) = 0 (Note that the wire is not accelerating.) (1.59) where τrx is the shear stress the fluid exerts on the wire and (2πRiL) is the area over which the shear stress acts. The shear stress the wire exerts on the fluid (assumed to be Newtonian) is: du τ rx = µ (1.60) dr where du 0 −V −V = = (Note that u(r = Ro) = 0 and u(r = Ri) = V.) (1.61) dr Ro − Ri Ro − Ri Recall that the shear stress the fluid exerts on the wire will be equal to, but opposite, the value given by Eqn. (1.60). Substitute Eqns. (1.60) and (1.61) into Eqn. (1.59) and simplify. V F −µ ( 2π Ri L ) = 0 Ro − Ri 2π Ri L µV ∴F = (1.62) Ro − Ri Use the numerical values given in the problem statement. Ri = 4.0*10-4 m Ro = 4.5*10-4 m L = 2.0*10-2 m µ = 20 cP = 0.02 kg/(m⋅s) (Note: 100 cP = 0.1 kg/(m⋅s).) V = 50 m/s ⇒ F = 1.0 N C. Wassgren Chapter 01: The Basics 31 Last Updated: 05 Sep 2008 bulk modulus, Eν {F/L2} [Pa=N/m2, psi=lbf/in2] - The bulk modulus is a measure of the compressibility of a substance. The larger the bulk modulus, the less compressible the substance. An incompressible substance has an infinite bulk modulus. - The bulk modulus is defined as the ratio of the differential pressure, dp, over the relative change in the differential volume, dV/V, caused by the differential pressure change: dp Eν ≡ − bulk modulus dV dp V original volume, V - If dp>0 (increase in pressure) and dV<0 (decrease in volume), then Eν>0. change in volume, dV - The bulk modulus can also be written in terms of density rather than volume: dp Eν = dρ ( ) ( ρ) - - since m=ρV and dm=0=dρV+ρdV ⇒ dρ/ρ = -dV/V. Water at 20 °C, Eν = 2.21 GPa. For an ideal gas under isothermal conditions: dp p = ρ RT ⇒ = RT ⇒ Eν T = ρ RT = p dρ T For an ideal gas under isentropic conditions: dp p = ( const ) ρ γ ⇒ = ( const ) γρ γ −1 dρ S ⇒ Eν S = γ p = γρ RT where γ=cp/cv. C. Wassgren Chapter 01: The Basics 32 Last Updated: 05 Sep 2008 Example: What is the fractional change in density of water in the ocean over a depth of 1000 m (3281 ft)? The change in pressure over this depth, ∆p, is 9.81*106 Pa. SOLUTION: Assume that the bulk modulus for water remains constant at Eν = 2.2*109 Pa (found from a fluid properties table). Recall that the bulk modulus is defined as: dp Eν = (1.63) dρ ρ ρf ∫ dρ ρ0 1 = Eν ρ pf ∫ dp p0 ( ) p f − p0 ρf ∆p ln = = Eν Eν ρ0 ρf ∆p = exp ρ0 Eν Note that ρf = ρ0 + ∆ρ so that: ρ f ρ 0 + ∆ρ ∆ρ = = 1+ ρ0 ρ0 (1.64) (1.65) ρ0 Substitute Eqn. (1.65) into Eqn. (1.64): ∆p ∆ρ 1+ = exp ρ0 Eν ∆ρ ρ0 ∆p = exp −1 Eν In the given problem: ∆p = 9.81*106 Pa Eν = 2.2*109 Pa ∆ρ −3 ⇒ ρ = 4.5*10 = 0.45% 0 (1.66) Water remains essentially incompressible over this depth! Note that in most engineering problems (where pressure variations are not enormous), it is reasonable to assume that water is incompressible. C. Wassgren Chapter 01: The Basics 33 Last Updated: 05 Sep 2008 - The speed of sound in a substance is related to how compressible the substance is. It can be shown that the speed of sound, c, is given by: c= ∂p ∂ρ (note that the “S” indicates that the process occurs isentropically) S (This derivation is covered in the notes regarding compressible flows in Chapter 11.) Since the bulk modulus can be written as: dp dp Eν = = ρ dρ dρ ( ρ) we have: c= Eν S ρ Hence, we observe that the speed of sound in a substance is related to the ratio of the substance’s compressibility to its density. - Recall that for an isentropic process involving an ideal gas: Eν = γp = γρRT, so that: c = γ RT speed of sound for an ideal gas C. Wassgren Chapter 01: The Basics 34 Last Updated: 05 Sep 2008 surface tension, σ {F/L} [N/m, lbf/ft] - Surface tension is caused by unbalanced molecular forces occurring at the interface between a liquid and a solid surface, two dissimilar liquids, or between a liquid and a gas. - For example, water “beads” up on a waxed surface because the intermolecular forces between the water molecules is greater than the intermolecular forces between the water and the wax. - Examples where we observe surface tension include: bubbles bugs walking on the surface of water trees soda in a straw Surface tension effects become much more significant as length scales (L) decrease since: Fsurface tension ~ L (proportional to interface length) Finertia, body forces ~ L3 (proportional to volume) C. Wassgren Chapter 01: The Basics 35 Last Updated: 05 Sep 2008 Example: Determine the relationship between the surface tension in a soap bubble and the pressure difference between the inside and outside of the bubble. SOLUTION: Analyze the problem by cutting the spherical bubble with radius, R, in half. The free body diagram of the half-bubble is shown in the figure below. The forces acting on the bubble include the surface tension force holding the two bubble halves together, and the pressure force acting to push the two bubble halves apart. surface tension force, FST = (2πR)σ exterior pressure force, FPE = pE (πR2) R ∑F = interior pressure force, FPI = pI (πR2) FPE + FST − FPI = 0 = pE π R 2 ∴ ( pI − pE ) = = 2π Rσ = pI π R 2 2σ R C. Wassgren Chapter 01: The Basics 36 Last Updated: 05 Sep 2008 - The contact angle between a liquid and a surface is the angle the liquid surface makes at the contact point with the surface as shown in the figure below. liquid θ - contact angle (always measured through the liquid) If θ > 90°, then the surface is considered hydrophobic (it doesn’t like the fluid). If θ < 90°, then the surface is considered hydrophilic (it does like the fluid). - For example: water on metal: hydrophilic mercury on metal - hydrophobic Precise measurements of contact angle are difficult to make due to the sensitivity of the contact angle to surface chemical variations. Be Sure To: 1. Consider the effects of surface tension at fluid interfaces, especially at small length scales. C. Wassgren Chapter 01: The Basics 37 Last Updated: 05 Sep 2008 Example: Determine the height a liquid will rise in a cylindrical tube due to surface tension. The contact angle of the liquid on the tube surface is θ. SOLUTION: Draw a free body diagram of the liquid contained in the tube above the free surface. The two forces acting on the liquid will be its weight and the surface tension force. These forces are indicated in the figure on the bottom right. 2R θ θ surface tension force in y-direction: = σ(2πR)cosθ θ H patm H weight ≈ ρ(πR2H)g y liquid 2R From Newton’s 2nd Law in the y-direction: ∑ Fy = 0 = σ ( 2π R ) cos θ − ρ (π R 2 H ) g ∴H = 2σ cos θ Note that as R ↓, H ↑. ρ gR C. Wassgren Chapter 01: The Basics 38 Last Updated: 05 Sep 2008 7. Flow Kinematics A good reference on the experimental aspects of this topic is: Merzkirch, W., Flow Visualization, Academic Press. In 1986, the Chernobyl Nuclear Power Plant in the Soviet Union released radioactive fallout into the atmosphere as a result of an explosion in one of the reactors (see the figure to the right). The radioactive plume covered regions of the western Soviet Union, Europe, and even parts of eastern North America. If such an accident were to happen again, how would you know which communities would be affected by the drifting radioactive cloud? If one had predictions of wind velocity measurements as a function of location and time, i.e. u = u(x, t), could you figure out what areas will be covered by the cloud? In this section, we’ll present three forms of flow kinematics: streamlines, streaklines, and pathlines. Each of these lines provides different information on the movement of fluid. For the toxic cloud release, one would be most interested in determining the streakline passing through the location of the damaged reactor. Damaged reactor at Chernobyl (photo from http://en.wikipedia.org/wiki/Chernobyl_disaster). Streamlines A streamline is a line that is everywhere tangent to the velocity field vectors. Note that since streamlines are parallel to the velocity vectors, there will be no flow across a streamline. streamline Experimentally, one can visualize streamlines using the Particle Image Velocimetry (PIV) technique. In PIV, fluid particles are “tagged,” usually by mixing in very small, neutrally buoyant bits of “paint,” and taking two photographs in rapid succession. Velocity vectors can then be produced by “connecting the dots.” photo at time t photo at time t+δt approximate velocity vectors at time t Note that the values of approximate velocity vectors can be found from the definition of a derivative: x (t + δ t ) − x (t ) x (t ) ≈ δt C. Wassgren Chapter 01: The Basics 39 Last Updated: 05 Sep 2008 We can determine the equation of a streamline given a velocity field by simply using the definition of a streamline. Let’s zoom in on a small part of a streamline (we’ll consider only a 2D flow for simplicity): Since the streamline is tangent to the velocity vector, the slope of the streamline will be equal to the slope of the velocity vector: uy dy (1.67) = dx ux u uy dx dy ux slope of streamline slope of velocity vector Similarly, in the x-z and y-z planes we have: dz u z dz u z = = dx u x dy u y (1.68) We can combine Eqns. (1.67) and (1.68) into a more compact form: dx dy dz = = Equation for a streamline. ux u y uz (1.69) Notes: 1. There is no flow across a streamline. 2. A stream tube is a tube made by all the streamlines passing through a closed curve. There is no flow through a stream tube wall. streamlines 3. A stream filiment is a stream tube with infinitesimally small cross-sectional area. Streaklines A streakline consists of a line that connects all fluid particles that have passed through the same point in space at a previous (or later) time. Experimentally a streamline can be visualized by injecting dye into a fluid flow at a particular point. fluid particle that passed through (x0, y0, z0) at time t0=t1 dye (x0, y0, z0) fluid particle that passed through (x0, y0, z0) at time t0=t2 streakline at time t passing through (x0, y0, z0) To determine the equation of a streakline at time, t, passing through the point (x0, y0, z0), we solve the differential equation describing a particle’s position: dx u= (1.70) dt where u is the velocity field subject to the initial condition that a particle pass through the point x0 = (x0, y0, z0) at some previous (or later) time, t0: x ( t = t0 ) = x 0 (1.71) C. Wassgren Chapter 01: The Basics 40 Last Updated: 05 Sep 2008 Note that t0 will be different for each particle, i.e. it varies (refer to the fluid particles in the figure above). The solution of Eqn. (1.70) subject to initial condition in Eqn. (1.71) will consist of a set of parametric equations in t0 (note that t is a known value since we want to know the streakline at a particular time, t). Pathlines A pathline is the line traced out by a particular particle as it moves from one point to another. It is the actual path a particle takes. Experimentally a pathline can be visualized by “tagging” a particular fluid particle and taking a longexposure photograph of the fluid motion. pathline of a particle passing through (x0, y0, z0) at time t0 particle passes through (x0, y0, z0) at time t0 location of tagged fluid particle at time t To determine the equation of a pathline at time, t, for a particle passing through the point (x0, y0, z0) at some previous time t0, we solve the differential equation describing the particle’s position: dx u= (1.72) dt where u is the velocity field subject to the initial condition that the particle pass through the point x0 = (x0, y0, z0) at some time, t0: x ( t = t0 ) = x 0 (1.73) Note that t0 will be a particular value, i.e. it’s fixed. The solution of Eqn. (1.72) subject to initial condition in Eqn. (1.73) will consist of a set of parametric equations in t (note that by varying t we can trace out the location of the particle for various times). Notes: 1. The streamline, streakline, and pathline passing through a particular location can be different in an unsteady flow, but will be identical in a steady flow. 2. The quantity t0 is the time when a fluid particle passes through the point x0. Hence, for a pathline t0 is fixed since there is only one fluid particle. However, for a streakline t0 varies since there are many fluid particles passing through the point x0, each at a different t0. 3. Why don’t we use a Lagrangian derivative (Chapter 03 material) when solving Eqn. (1.72) for a particle’s pathline (since the pathline is a Lagrangian concept)? It turns out that the Lagrangian derivative of a particle’s position is equal to its Eulerian derivative. Consider, for example, the change in the x position of the particle as we follow it. Note that the position, x, is an Eulerian quantity. Dx ∂x ∂x ∂x ∂x ∂x (1.74) = + ( u ⋅∇ ) x = + ux + uy + uz = ux Dt ∂t ∂t ∂x ∂y ∂z =0 4. =1 =0 =0 A demonstration of the differences between streamlines, streaklines, and pathlines can be found at: https://engineering.purdue.edu/~wassgren/applets Be Sure To: 1. Understand the definitions for streamlines, streaklines, and pathlines. 2. Understand what initial conditions to use when evaluating streaklines and pathlines. 3. Draw the direction of flow on the streamlines, streaklines, and pathlines. 4. It’s perfectly correct to represent the position of a fluid particle parametrically, i.e. x = x(t) and y = y(t). C. Wassgren Chapter 01: The Basics 41 Last Updated: 05 Sep 2008 Example: A velocity field is given by: −V y Vx u = 2 0 2 1 ˆ+ 2 0 2 1 ˆ i j 2 (x + y ) (x + y ) 2 where V0 is a positive constant, i.e. V0 > 0. Determine: a. where in the flow the speed is V0 b. the equation and sketch of the streamlines c. the equations for the streaklines and pathlines SOLUTION: The speed is given by: 2 2 u = ux + u y (1.75) where ux = uy = −V0 y 2 (x + y2 ) V0 x 1 (1.76) 2 (1.77) 1 ( x2 + y 2 ) 2 Substituting into Eqn. (1.75) gives: u= V02 y 2 ( x2 + y2 ) + V02 x 2 ( x2 + y2 ) ∴ u = V0 (1.78) The flow speed is everywhere equal to V0. The slope of the streamline is tangent to the slope of the velocity vector: dy u y = dx u x Substitute Eqns. (1.76) and (1.77) and solving the resulting differential equation. V0 x (1.79) 1 dy ( x 2 + y 2 ) 2 x = = −V0 y dx −y 1 ( x2 + y2 ) 2 y x ∫ ∫ xdx y0 x0 − ydy = ( ) (x 2 − 1 y 2 − y0 = 2 2 (where (x0, y0) is a point located on the streamline) 2 2 x0 1 2 2 2 − x0 ) 2 y0 x +y = + = constant (1.80) The streamlines are circles! Note that when x > 0 and y > 0, Eqns. (1.76) and (1.77) indicate that ux < 0 and uy > 0 (note that V0 > 0) so that the flow is moving in a counter-clockwise direction. y x Since the flow is steady, the streaklines and pathlines will be identical to the streamlines. C. Wassgren Chapter 01: The Basics 42 Last Updated: 05 Sep 2008 Example: Consider a 2D flow with a velocity field given by: u = x (1 + 2t )ˆ + yˆ i j Determine the equations for the streamline, streakline, and pathline passing through the point (x,y)=(1,1) at time t=0. SOLUTION: The slope of a streamline is tangent to the velocity vector. dy u y = dx u x where u x = x(1 + 2t ) (1.81) (1.82) uy = y (1.83) Substitute Eqns. (1.82) and (1.83) into Eqn. (1.81) and solve the resulting differential equation. dy y = dx x(1 + 2t ) y dy (1 + 2t ) = y x ∫ ∫ y0 x0 dx (where (x0, y0) is a point passing through the streamline) x y x (1 + 2t ) ln = ln y0 x0 (1+ 2t ) y x = y0 x0 For the streamline passing through the point (x0, y0) = (1, 1) at time t = 0: y=x (1.84) (1.85) A streakline is a line that connects all of the fluid particles that pass through the same point in space. The equation for the streakline can be found parametrically using Eqns. (1.82) and (1.83). dx ux = (1.86) = x(1 + 2t ) dt dy uy = =y (1.87) dt Solve the previous two differential equations. x ∫ x0 y ∫ y0 t dx = (1 + 2t )dt x ∫ ⇒ x 2 ln = t + t 2 − t0 − t0 x0 (1.88) ⇒ y ln = t − t0 y0 (1.89) t0 t dy = dt y ∫ t0 where t0 is the time at which a fluid particle passes through the point (x0, y0) on the streakline. Hence, the streakline passing through the point (x0, y0) = (1, 1) at time t = 0 is given parametrically (in t0) as: ( 2 ln ( x ) = −t0 − t0 ⇒ 2 x = exp −t0 − t0 ln ( y ) = −t0 ⇒ ) y = exp ( −t0 ) (1.90) (1.91) Recall that t0 is the time when a fluid particle passes through the point (x0, y0). C. Wassgren Chapter 01: The Basics 43 Last Updated: 05 Sep 2008 A pathline is a line traced out by a particular fluid particle as it moves through space. The equation for the pathline can be found parametrically using Eqns. (1.82) and (1.83). dx ux = = x(1 + 2t ) (1.92) dt dy uy = =y (1.93) dt Solve the previous two differential equations. x ∫ x0 y ∫ y0 t dx = (1 + 2t )dt x ∫ ⇒ x 2 ln = t + t 2 − t0 − t0 x0 (1.94) ⇒ y ln = t − t0 y0 (1.95) t0 t dy = dt y ∫ t0 where t0 is the time at which a fluid particle passes through the point (x0, y0) on the pathline. Hence, the pathline for a particle passing through the point (x0, y0) = (1, 1) at time t0 = 0 is given parametrically (in t) as: ( ln ( x ) = t + t 2 ⇒ x = exp t + t 2 ln ( y ) = t ⇒ ) y = exp ( t ) (1.96) (1.97) Note that the streamline, streakline, and pathline are all different. A plot of these lines through (1, 1) at t = 0 is shown below. 16 streamline streakline pathline 14 12 y 10 8 6 4 2 0 0 2 4 6 8 10 x C. Wassgren Chapter 01: The Basics 44 Last Updated: 05 Sep 2008 8. Some Basic Definitions and Concepts Before we begin to study the behavior of fluids, we need to first define a few commonly used terms. fluid A fluid is a substance that deforms continuously when subject to non-isotropic stresses. Notes: 1. An isotropic stress state is one in which the stresses are the same in all directions. 2. Liquids and gases are considered fluids. Their properties are different (e.g. the density of air at standard conditions is 1.23 kg/m2 while the density of water at standard conditions is 1000 kg/m3), but their bulk motion is similar. 3. Solids will not deform continuously when subject to unbalanced forces. They will deform up until a point and then resist further deformations (e.g. behavior similar to that of a spring). 4. Some substances are difficult to classify, e.g. slurries and granular materials. The study of material behavior (deformation and flow) is known as rheology. For additional information, refer to Steffe, J.F., Rheological Methods in Food Process Engineering, Freeman Press and Bird, R.B., Armstrong, R.C., and Hassager, O., Dynamics of Polymeric Liquids, Wiley. system and surroundings A system is a particular quantity of matter chosen for study. The surroundings include everything that is not the system. control volume (CV), control surface (CS), and outward-pointing unit normal vector A control volume (CV) is a particular volume or region in space. A control surface (CS) is the surface enclosing the control volume. The orientation of the CS at a particular location is given by the ˆ direction of its outward-pointing unit normal vector, n , at that location. The outward-pointing unit normal vector has a magnitude of one, is perpendicular to the control surface, and always points out of the CV. ˆ n ˆ ˆ n ˆ n ˆ CV ˆ n CS ˆ ˆ n ˆ n ˆ ˆ C. Wassgren Chapter 01: The Basics 45 Last Updated: 05 Sep 2008 property (extensive, intensive, and specific) Properties are macroscopic characteristics of a system. An extensive property is one that depends on the amount of mass in the system. An intensive property is one that is independent of the mass in the system. A specific property is an extensive property per unit mass (a specific property is also an intensive property). Some examples: mass, m, is an extensive property, pressure, p, is an intensive property, and specific volume, v≡V/m, is a specific property. An easy way to determine whether a property is extensive or intensive is to divide the system into two parts and see how the property is affected. state The state of a system is the system’s condition or configuration as described by its properties in sufficient detail so that it is distinguishable from other states. Often a state can be described by a subset of the system’s properties since the properties themselves may be related. scalar and vector fields A field representation of a quantity gives the value of that quantity at all locations and times, i.e. the field representation of the density, ρ, is: ρ = ρ ( x, y , z , t ) A scalar is a quantity that has only a magnitude, e.g. temperature, T. A vector is a quantity that has both magnitude and direction, e.g. velocity, u. steady and unsteady A steady flow is one in which flow properties at each location do not change with time, i.e.: ∂ ( )=0 ∂t An unsteady flow is a flow where properties do change with time. flow dimensionality The dimension of a flow is equal to the number of spatial coordinates required to describe the flow. For example: 0D flow: u=u(t) or u=constant 1D flow: u=u(t; x) or u=u(t; x) 2D flow: u=u(t; r, θ) or u=u(t; r, θ) 3D flow: u=u(t; x, y, z) or u=u(x, y, z) A flow is uniform if it does not vary in a spatial direction. A flow is non-uniform if it does vary in that spatial direction. For example, the velocity profile in the figure below at left is non-uniform in the ydirection. The figure at the right is uniform in the y-direction. y y u u Be Sure To: 1. Understand the previous definitions since they will be used frequently in the discussion of fluid mechanics. C. Wassgren Chapter 01: The Basics 46 Last Updated: 05 Sep 2008 Review Questions 1. What is the conversion between lbm and slugs? What is the conversion between lbf and lbm? Estimate, without a calculator, what 20 ° is in °F. C 2. What are the dimensions (in M, L, T) of power? 3. Do you understand the physical concept behind the Taylor series approximation given below? dy y ( x + dx ) = y ( x ) + ( dx ) dx x Can you draw a sketch corresponding to your explanation? 4. What is meant by “least count”? Give an example of a situation where the least count for a measurement may be different for two experimenters using the same equipment. 5. Set up and work out an example concerning propagation of experimental uncertainty. 6. What is the difference between relative and absolute uncertainty? 7. Why should uncertainty be included in experimental measurements? 8. Under what conditions can a fluid be considered a continuum? 9. Give an example of where the continuum approximation is not appropriate. 10. What is the no-slip assumption? 11. Discuss the definition and meaning of the Knudsen number. 12. What is meant (in words) by a “Newtonian fluid”? Write the shear stress-shear strain rate relationship for a Newtonian fluid. Is this the shear stress that acts on the fluid or that the fluid exerts on its neighbors? 13. Give an example of a shear thickening fluid. 14. How does viscosity vary with temperature for liquids? with pressure? 15. What is the difference between “dynamic” and “kinematic” viscosities? 16. What is meant by an “inviscid” fluid? 17. What is meant by an “incompressible” fluid? Under what conditions can a flow be considered incompressible? 18. What is meant by an “ideal” fluid? 19. What are the pressure and temperature at standard conditions? 20. What is the density of air at standard conditions (in kg/m3)? What is the approximate density of water under ordinary conditions (in kg/m3)? What is the specific gravity of mercury? 21. What is the difference between absolute and gage pressures? 22. An engineer wants to minimize the resistance on a block sliding over a thin film of liquid. Would it be better to increase the thickness of the film, decrease the thickness, or does it matter? Explain your reasoning in words and provide analytical support for your argument (i.e. supporting equations). 23. Estimate the absolute and kinematic viscosities of water under ordinary conditions. 24. What is meant by a Couette flow? Under what geometric conditions would it be reasonable to apply the Couette flow assumption to flow between two rotating concentric cylinders? 25. What is cavitation? Can cavitation be a problem in a pipeline containing steam? Give some examples of where cavitation occurs. Will a submarine propeller be more likely to cavitate near the ocean surface or deep in the ocean? Why? 26. How do you expect the speed of sound in water containing small air bubbles to compare with the speed of sound in water and the speed of sound in air? (Hint: Consider the mixture’s bulk modulus and density to the pure substance values.) 27. What is the approximate speed of sound in air at standard conditions? 28. Which bubble has greater internal pressure: one with a smaller radius or one with a larger radius? 29. What are the dimensions of surface tension? 30. Draw a picture defining the contact angle measurement. 31. A small cubicle object is place in water. Determine the maximum specific gravity of the object, in terms of the problems other pertinent parameters, that will keep the object afloat. 32. Describe, in words and equations, what is meant by a streamline, streakline, and pathline. 33. Under what conditions will a streamline, streakline, and pathline be identical, in general? 34. Experimentally, how would one produce a streamline, streakline, and pathline? 35. Describe analytically the difference between a streakline and pathline. 36. What is the definition of a fluid? What is the fundamental difference between fluids and solids? Give an example of a substance that has similarities with both fluids and solids. C. Wassgren Chapter 01: The Basics 47 Last Updated: 05 Sep 2008 37. What is meant by a “system”? What is meant by a “control volume”? 38. What is meant by “uniform”? What is meant by “steady”? Give examples of velocity fields that are uniform/non-uniform and steady/unsteady. 39. What is meant by an “extensive” property? Give an example of an extensive property. What is meant by an “intensive” property? Give an example of an intensive property. 40. How is the “dimension” of a flow determined? Give an example of a 2D, unsteady velocity field. Is it possible to have a 2D velocity field but a 3D temperature field? 41. What is meant by a “uniform” flow? Give an example of a velocity profile that is uniform in the xdirection but is unsteady. C. Wassgren Chapter 01: The Basics 48 Last Updated: 05 Sep 2008 Chapter 02: Fluid Statics 1. 2. 3. 4. Pressure Distribution in a Static Fluid a. Incompressible Fluids b. Compressible Fluids Manometers Forces on Submerged Surfaces a. Planar Surfaces b. Curved Surfaces Pressure Distribution Due to Rigid Body Motion C. Wassgren Chapter 02: Fluid Statics 49 Last Updated: 10 Sep 2008 Chapter 03: Integral Analysis 1. 2. 3. 4. 5. 6. 7. 8. 9. Lagrangian and Eulerian Perspectives The Reynolds’ Transport Theorem Conservation of Mass The Linear Momentum Equation The Angular Momentum Equation Basic Thermodynamic Definitions Discussion of Energy, Work, and Heat Conservation of Energy The 2nd Law of Thermodynamics C. Wassgren Chapter 03: Integral Analysis 50 Last Updated: 25 Aug 2008 1. Lagrangian and Eulerian Perspectives There are two common ways to study a moving fluid: 1. Look at a particular location and observe how all the fluid passing that location behaves. This is called the Eulerian point of view. 2. Look at a particular piece of fluid and observe how it behaves as it moves from location to location. This is called the Lagrangian point of view. Example: Let’s say we want to study migrating birds. We could either: 1. stand in a fixed spot and make measurements as birds fly by (Eulerian point of view), or 2. bird tag some of the birds and make measurements as they fly along (Lagrangian point of view). radio transmitter Lagrangian (aka Material, Substantial) Derivative (Go to http://widget.ecn.purdue.edu/~meapplet for an interactive Java applet on this topic.) If we follow a piece of fluid (Lagrangian viewpoint), how will some property of that particular piece of fluid change with respect to time? same piece of fluid piece of fluid time: t0+δt location: x(t0+δt) time: t0 location: x(t0) Let’s say we’re interested in looking at the time rate of change of temperature, T, that the particle observes as it moves from location to location. The particle may experience a temperature change because the temperature of the entire field of fluid may be changing with respect to time (i.e. the temperature field may be unsteady). In addition, the temperature field may have spatial gradients (different temperatures at different locations, i.e. non-uniform) so that as the particle moves from point to point it will experience a change in temperature. Thus, there are two effects that can cause a time rate of change of temperature that the particle experiences: unsteady effects, also known as local or Eulerian effects, and spatial gradient effects, also known as convective effects. We can describe this in mathematical terms by writing the temperature of the entire field as a function of time, t, and location, x: T = T (t, x ) (3.1) Note that the location of the fluid particle is a function of time: x=x(t) so that T = T (t, x (t )) (3.2) Taking the time derivative of the temperature, expanding the location vector into its x, y, and z components, and using the chain rule gives: dT ∂T ∂T dx ∂T dy ∂T dz = + + + (3.3) dt following a ∂t ∂x dt ∂y dt ∂z dt fluid particle =ux =u y =u z Note that dx/dt, dy/dt, and dz/dt are the particle velocities ux, uy, and uz respectively. Writing this in a more compact form: C. Wassgren Chapter 03: Integral Analysis 51 Last Updated: 25 Aug 2008 DT ∂T ∂T ∂T ∂T = + ux + uy + uz Dt ∂t ∂x ∂y ∂z (3.4) ∂T = + (u ⋅ ∇)T ∂t The notation, D/Dt, indicating a Lagrangian (also sometimes referred to as the material or substantial) derivative, has been used in Eqn. (3.4) to indicate that we’re following a particular piece of fluid. More generally, we have: D ∂ + () = () ( u ⋅∇ )( ) Dt ∂t Lagrangian rate of change (changes as we follow a fluid particle) local or Eulerian rate of change (changes due to unsteady effects) convective rate of change (changes due to a change in particle position) ∂ ∂ ∂ ( ) + ux ( ) + u y ( ∂t ∂x ∂y where (⋅⋅⋅) represents any field quantity of interest. = ) + uz ∂ ( ∂z (3.5) ) Notes: 1. 2. The Lagrangian derivatives in cylindrical and spherical coordinates are: ∂ ∂ uθ ∂ ∂ D = + ur + + uz cylindrical: Dt ∂t ∂r r ∂θ ∂z uφ ∂ ∂ uθ ∂ ∂ D spherical: = + ur + + Dt ∂t ∂r r ∂θ r sin θ ∂φ (3.6) (3.7) The acceleration experienced by a fluid particle is given by: Du ∂u ∂u ∂u ∂u ∂u Cartesian: = + (u ⋅ ∇) u = + ux + uy + uz Dt ∂t ∂t ∂x ∂y ∂z ∂ur ∂t ∂u aθ = θ ∂t ∂u z az = ∂t ∂u ar = r ∂t ar = cylindrical: spherical: aθ = aφ = C. Wassgren Chapter 03: Integral Analysis ∂uθ ∂t ∂uφ ∂t (3.8) ∂ur uθ ∂ur ∂u u2 + + uz r − θ ∂r r ∂θ ∂z r ∂u u ∂u ∂u uu + ur θ + θ θ + u z θ + r θ ∂r r ∂θ ∂z r ∂u z uθ ∂u z ∂u z + ur + + uz ∂r r ∂θ ∂z uφ ∂ur 1 2 u ∂ur ∂u 2 + ur r + θ + − uθ + uφ ∂r r ∂θ r sin θ ∂φ r uφ ∂uθ 1 ∂u u ∂u 2 + ur θ + θ θ + + ur uθ − uφ cot θ ∂r r ∂θ r sin θ ∂φ r ∂uφ uθ ∂uφ uφ ∂uφ 1 + ur + + + ur uφ + uθ uφ cot θ ∂r r ∂θ r sin θ ∂φ r + ur ( ) ( ( 52 (3.9) ) (3.10) ) Last Updated: 25 Aug 2008 Example: A fluid velocity field is given by: u = (cy 2 )ˆ + (cx 2 ) ˆ i j where c is a constant. Determine a. the components of the acceleration and b. the points in the flow field where the acceleration is zero. SOLUTION: The acceleration of a fluid element is given by: Du ∂u ∂u ∂u a= = + ux + uy Dt ∂t ∂x ∂y where ∂u = 0 (steady flow) ∂t ∂u ux = cy 2 2cxˆ = 2c 2 xy 2 ˆ j j ∂x ∂u uy = cx 2 2cyˆ = 2c 2 x 2 yˆ i i ∂y ∴ a = 2c 2 x 2 yˆ + 2c 2 xy 2 ˆ i j ( )( ) ( )( (3.11) ) (3.12) Set the acceleration equal to zero. a = 0 = 2c 2 x 2 yˆ + 2c 2 xy 2 ˆ i j ∴ either x = 0 or y = 0 (This is the locus of points where the total acceleration is zero.) C. Wassgren Chapter 03: Integral Analysis 53 (3.13) Last Updated: 25 Aug 2008 Example: A fluid velocity field is given by: ˆ u = 2te x Will a fluid particle accelerate in this flow? Why? SOLUTION: The acceleration is given by: Du ∂u ∂u ∂u ∂u a= = + ux + uy + uz Dt ∂t ∂x ∂y ∂z ˆ = 2e x 2t =0 =0 =0 Hence, for the given flow: ˆ a = 2e x Yes, fluid particles will accelerate due to the local (or Eulerian) derivative. Example: Now consider the following flow: ˆ u = xe x Will a fluid particle accelerate in this flow? Why? SOLUTION: The acceleration is given by: Du ∂u ∂u ∂u ∂u a= = + ux + uy + uz Dt ∂t ∂x ∂y ∂z =0 x ˆ =e x =0 =0 Hence, for the given flow: ˆ a = xe x Yes, fluid particles will accelerate due to the convective derivative. C. Wassgren Chapter 03: Integral Analysis 54 Last Updated: 25 Aug 2008 Example: The market price, P (in dollars), of used cars of a certain model is found to be: P = $1000 + ( $0.02 / mile ) x − ( $2 / day ) t where x is the distance in miles west of Detroit, MI and t is the time in days. If a car of this model is driven from Detroit at t = 0 towards the west at a rate of 400 miles per day, determine: a. whether the value of the car is increasing or decreasing, and b. how much of this change is due to depreciation and how much is due to moving into a better market. SOLUTION: To determine if the value of the car is decreasing, take the Lagrangian derivative of the market price. DP ∂P ∂P = +u = ( −$2/day ) + ( 400 miles/day )( $0.02/mile ) (where u is the speed of the car) (3.14) Dt ∂t ∂x = $8 / day ∴ DP = $6/day Hence, the value of the car is increasing. Dt The car depreciates at a rate of -$2/day (this is ∂P/∂t). The change in the car’s value increases at a rate of $8/day due to moving into a different market (this is u∂P/∂x). C. Wassgren Chapter 03: Integral Analysis 55 Last Updated: 25 Aug 2008 2. Reynolds’ Transport Theorem (RTT) Recall that we can look at the behavior of small pieces of fluid in two ways: the Eulerian perspective or the Lagrangian perspective. Often we’re interested in the behavior of an entire system of fluid (many pieces of fluid) rather than just an individual piece. How do we analyze this situation? We can use Eulerian and Lagrangian approaches for analyzing a macroscopic amount of fluid but we need to first develop an important tool called the Reynolds Transport Theorem. Why do we want to do this? It turns out that the behavior of fluids (most substances in fact) can be described in terms of a few fundamental laws. These laws include: − conservation of mass (COM), − Newton’s 2nd Law, − the angular momentum principle, − conservation of energy (COE, aka the First Law of Thermodynamics), and − the second law of thermodynamics. These laws are typically easiest to apply to a particular system of fluid particles (Lagrangian perspective). However, the Lagrangian forms of the laws are typically difficult to use in practical applications since we can’t easily keep track of many individual bits of fluid. It’s much easier to apply the laws to a particular volume in space instead (referred to as a control volume, an Eulerian perspective). For example, tracking the behavior of individual bits of gas flowing through a rocket nozzle would be difficult. It’s much easier to just look at the behavior of the gas flowing into, out of, and within the volume enclosed by the rocket nozzle. The Reynolds Transport Theorem is a tool that will allow us to convert from a system point of view (Lagrangian) to a control volume point of view (Eulerian). Let’s consider a system of fluid particles that is coincident (occupying the same region in space) as our control volume (CV) at some time, t: at time t system CV the fluid is moving At some later time, t+δt, the system may have moved relative to the CV. at time t+δt system CV C. Wassgren Chapter 03: Integral Analysis 56 Last Updated: 25 Aug 2008 Let B be some transportable property (i.e., some property that can be transported from one location to another, e.g. mass, momentum, energy) and β be the corresponding amount of B per unit mass, i.e.: small volume, dV, containing: dBsys=βρdV Bsys = ∫ βρ dV Vsys BCV = ∫ (3.15) βρ dV CV system with total volume, Vsys, containing Bsys where Bsys and BCV refer to the total amount of B in the system and control volume, respectively. Note that at time, t, the total amounts of B in the system and control volume are equal since the system and CV are coincident: Bsys ( t ) = BCV ( t ) (3.16) However, at time, t+δt, the system and CV no longer occupy the same region in space so that, in general, Bsys(t+δt) ≠ BCV(t+δt). at time t+δt Note that B may be changing with time so that, in general, Bsys(t+δt) ≠ Bsys(t). Bout(t+δt) In the figure at the left, Bout is the amount of B that has left the CV and Bin is the amount of B that has entered the CV. system Bin(t+δt) CV Utilizing the figure shown above, we see that: BCV ( t + δ t ) = Bsys ( t + δ t ) − Bout ( t + δ t ) + Bin ( t + δ t ) (3.17) Subtracting Bsys(t) from both sides and dividing through by δt gives: BCV ( t + δ t ) − Bsys ( t ) Bsys ( t + δ t ) − Bsys ( t ) Bout ( t + δ t ) Bin ( t + δ t ) = − + (3.18) δt δt δt δt Now let’s substitute BCV(t) = Bsys(t) on the left hand side, subtract Bout(t)/δt and Bin(t)/δt on the right-hand side (note that Bout(t)=Bin(t)=0), and then take the limit of the entire equation as δt→0: BCV ( t + δ t ) − BCV ( t ) lim = δ t →0 δt Bsys ( t + δ t ) − Bsys ( t ) Bout ( t + δ t ) − Bout ( t ) Bin ( t + δ t ) − Bin ( t ) lim − lim + lim (3.19) δ t →0 δ t →0 δ t →0 δt δt δt DBsys dBout dBin dB ⇒ CV = − + dt Dt dt dt C. Wassgren Chapter 03: Integral Analysis 57 Last Updated: 25 Aug 2008 Note that the D/Dt notation has been used to signify that the first term on the right hand side of Eqn. (3.19) represents the time rate of change as we follow a particular system of fluid (Lagrangian perspective). Re-arranging the equation and substituting in for BCV and Bsys using Eqn. (3.15): d (B − B ) D d out in βρ dV = βρ dV + (3.20) dt Dt dt CV Vsys ∫ ∫ The last term on the right hand side of Eqn. (3.20) represents the net rate at which B is leaving the control volume through the control surface (CS). Let’s examine this term more closely by zooming in on a small piece of the control surface and observing how much B leaves through this surface in time δt. The component of the fluid velocity out of the control volume through surface, dA, is given by: ˆ u rel ⋅ n where urel=usys-uCS is the velocity of the fluid small piece of control relative to the control surface. The volume of fluid surface, dA, with outward leaving through surface dA in time δt is then: ˆ point normal vector, n ˆ dV = ( u rel ⋅ n ) δ tdA = ( u rel ⋅ dA ) δ t ˆ (urel⋅ n )δt urel dA ˆ n Thus, the volumetric flowrate, dQ, (volume per unit time) through surface dA is given by: dQ = u rel ⋅ dA (3.21) system CV Now use Eqn. (3.21) to write the net rate at which B leaves the control volume: d ( Bout − Bin ) = βρ dQ = β ( ρ u rel ⋅ dA ) dt ∫ ∫ CS CS Combining Eqn. (3.22) with Eqn. (3.20) gives: D d βρ dV = βρ dV + β ( ρ u rel ⋅ dA ) dt Dt CS CV Vsys ∫ rate of increase of B within the system ∫ rate of increase of B within the CV ∫ (3.22) (3.23) net rate at which B leaves the CV through the CS The Reynolds Transport Theorem! C. Wassgren Chapter 03: Integral Analysis 58 Last Updated: 25 Aug 2008 3. Conservation of Mass (COM) In words and in mathematical terms, COM for a system is: D The mass of a system remains constant. ⇒ ρ dV = 0 (3.24) Dt Vsystem mass of the system where D/Dt is the Lagrangian derivative (implying that we’re using the rate of change as we follow the system), V is the volume, and ρ is the density. Using the Reynolds Transport Theorem, Eqn. (3.24) can be converted into an expression for a control volume: D d ρ dV = ρ dV + ( ρ u rel ⋅ dA ) = 0 (3.25) dt Dt Vsystem CV CS ∫ ∫ d dt ∫ ρ dV CV rate of increase of mass inside the CV ∫ + ∫ ( ρu ∫ rel ⋅ dA ) =0 COM for a CV (3.26) CS net rate at which mass leaves the CV through the CS Helpful Hints: 1. Carefully draw your control volume. Don’t neglect to draw a control volume or draw a control volume and then use a different one. 2. Make sure you understand what each term in conservation of mass represents. 3. Carefully evaluate the dot product in the mass flux term. 4. You must integrate the terms in conservation of mass when the density or velocity are not uniform. Let’s consider a few examples to see how COM for a CV is applied. C. Wassgren Chapter 03: Integral Analysis 59 Last Updated: 25 Aug 2008 Example: Calculate the mass flux through the control surface shown below. Assume a unit depth into the page. D V control surface SOLUTION: d A = Rd θ The mass flux through the surface is given by: m= ∫ θ =π 2 ρ u rel ⋅ dA = ∫ ) ρ Vˆ ⋅ − cos θ ˆ + sin θ ˆ ( Rdθ ) i i j = u rel θ =−π 2 CS ( ∫ θ cos θ dθ = − ρVR sin θ =−π 2 π 2 −π 2 R θ y = dA ˆ =n θ =π 2 = − ρVR ˆ n = − cos θ ˆ + sin θ ˆ i j x = −2 ρVR ∴ m = − ρVD We could have also determined the mass flux by noticing that any mass passing through the curved control surface must also pass through a vertical control surface as shown below. y x D ˆ n = −ˆ i V y=R m= ∫ CS ρ u rel ⋅ dA = ∫ () ρ Vˆ ⋅ −ˆ ( dy ) = −2 ρVR = − ρVD (The same answer as before!) i i y =− R C. Wassgren Chapter 03: Integral Analysis =u rel ˆ =n = dA 60 Last Updated: 25 Aug 2008 Example: Consider the flow of an incompressible fluid between two parallel plates separated by a distance 2H. If the velocity profile is given by: y2 u = u c 1 − 2 H where uc is the centerline velocity, determine the average velocity of the flow, u . Assume the depth into the page is w. y u u H H SOLUTION: The volumetric flow rate using the average velocity profile must give the same volumetric flow rate using the real velocity profile. y =+ H ∫ ∫ A A Qreal = u ⋅ dA = dQ = = dA 2 uc 1 − y dyw = 4 uc wH 3 H2 y =− H ∫ = dQ dy u ( y) (3.27) The velocity, u(y), is nearly constant over the small distance dy so we can write the volumetric flowrate over this small area as dQ = u(y)dA = u(y)(dyw). ∫ Qaverage = u ⋅ dA = u ( 2 Hw ) (There is no need to integrate since the velocity is uniform over y.) (3.28) A Qreal = Qaverage ⇒ 4 u wH 3c = u ( 2 Hw ) (3.29) ∴ u = 2 uc 3 C. Wassgren Chapter 03: Integral Analysis (3.30) 61 Last Updated: 25 Aug 2008 Example An incompressible flow in a pipe has a velocity profile given by: r2 u (r ) = u c 1 − 2 R where uc is the centerline velocity and R is the pipe radius. Determine the average velocity in the pipe. r u R SOLUTION: The volumetric flow rate using the average velocity profile must give the same volumetric flow rate using the real velocity profile. dr dA r=R 2 2 dA = 2πrdr 1 Qreal = u ⋅ dA = dQ = uc 1 − r 2 ( 2π rdr ) = 2 π uc R (3.31) R A A r =0 r = dQ ∫ ∫ ( ∫ ) The velocity, u(r), is nearly constant over the small annulus with radius dr so we can write the volumetric flow rate over this small area as dQ = u(r)dA = u(r)(2πrdr). ( ∫ Qaverage = u ⋅ dA = u π R 2 ) (There is no need to integrate since the velocity is uniform over r.) (3.32) A Qreal = Qaverage ⇒ 1 π uc R 2 2 ( = u π R2 ) (3.33) ∴ u = 1 uc 2 C. Wassgren Chapter 03: Integral Analysis (3.34) 62 Last Updated: 25 Aug 2008 Example: Water enters a cylindrical tank through two pipes at volumetric flow rates of Q1 and Q2. If the level in the tank remains constant, calculate the average velocity of the flow leaving the tank through a pipe with an area, A3. Q1 Q2 h=constant V3 = ? A3 SOLUTION: Apply conservation of mass to the fixed control volume shown below. Q1 Q2 h=constant V3 = ? A3 d dt where d dt ∫ ρ dV + ∫ ρ u CV rel ⋅ dA = 0 (3.35) CS ∫ ρ dV = 0 (steady flow, the mass in the control volume isn’t changing with time) CV ∫ ρu rel ⋅ dA = − ρ Q2 − ρ Q1 + ρV3 A3 CS Substitute and re-arrange. − ρ Q2 − ρ Q1 + ρV3 A3 = 0 ∴V3 = Q1 + Q2 A3 C. Wassgren Chapter 03: Integral Analysis (3.36) 63 Last Updated: 25 Aug 2008 Example: Water enters a cylindrical tank with diameter, D, through two pipes at volumetric flow rates of Q1 and Q2 and leaves through a pipe with area, A3, with an average velocity, V 3 . The level in the tank, h, does not remain constant. Determine the time rate of change of the level in the tank. D Q1 Q2 h≠constant V3 A3 SOLUTION: Apply conservation of mass to a control volume that deforms to follow the free surface of the liquid. D Q1 Q2 h≠constant V3 A3 d dt ∫ ρ dV + ∫ ρ u CV rel ⋅ dA = 0 (3.37) CS where d dt ∫ ρ dV = CV d π D2 dh π D 2 ρh =ρ dt 4 dt 4 = M CV ∫ ρu rel ⋅ dA = − ρ Q2 − ρ Q1 + ρV3 A3 CS Substitute and re-arrange. dh π D 2 − ρ Q2 − ρ Q1 + ρV3 A3 = 0 dt 4 dh Q2 + Q1 − V3 A3 = dt π D2 4 ρ C. Wassgren Chapter 03: Integral Analysis (3.38) 64 Last Updated: 25 Aug 2008 We could have also chosen a fixed control volume through which the free surface moves. Using this type of control volume, conservation of mass is given by: d ρ dV + ρ u rel ⋅ dA = 0 (3.39) dt ∫ where d dt ∫ CV CS ∫ ρ dV = 0 (the mass of fluid in the fixed control volume remains constant) CV =Vtop ∫ ρ u rel ⋅ dA = − ρ Q2 − ρ Q1 + ρV3 A3 + ρ CS dh π D 2 dt 4 mtop Substitute and re-arrange. dh ρ Q2 + ρ Q1 − ρV3 A3 = 2 dt ρπD 4 dh Q2 + Q1 − V3 A3 = (This is the same answer as before!) dt π D2 4 C. Wassgren Chapter 03: Integral Analysis 65 (3.40) Last Updated: 25 Aug 2008 Example: A spherical balloon is filled through an area, A1, with air flowing at velocity, V1, and constant density, ρ1. The radius of the balloon, R(t), can change with time, t. The average density within the balloon at any given time is ρb(t). Determine the relationship between the rate of change of the density within the balloon and the rest of the variables. R ( t) ρ1, V1, A1 ρb SOLUTION: Apply conservation of mass to a control volume that deforms to follow the interior surface of the balloon. R ( t) ρ1, V1, A1 ρb d dt where d dt ∫ ρ dV + ∫ ρ u CV rel d ∫ ρ dV = dt ρ CV ∫ ρu ⋅ dA = 0 (3.41) CS b dρ 4 dR 4 π R3 = π R3 b + 4πρb R 2 3 dt dt 3 = M CV rel ⋅ dA = − ρ1V1 A1 CS Substitute and simplify. dρ 4 dR π R3 b + 4πρb R 2 − ρ1V1 A1 = 0 3 dt dt dR ρ1V1 A1 − 4πρb R 2 d ρb dt = 4 π R3 dt 3 C. Wassgren Chapter 03: Integral Analysis (3.42) 66 Last Updated: 25 Aug 2008 Example: A box with a hole of area, A, moves to the right with velocity, ubox, through an incompressible fluid as shown in the figure. If the fluid has a velocity of ufluid which is at an angle, θ, to the vertical, determine how long it will take to fill the box with fluid. Assume the box volume is Vbox and that it is initially empty. ufluid θ area, A ubox volume, Vbox SOLUTION: Apply conservation of mass to a control volume fixed to the interior of the box. Change our frame of reference so the box appears stationary. ufluid area, A θ ubox y x volume, Vbox d dt ∫ ρ dV + ∫ ρ u CV rel ⋅ dA = 0 (3.43) CS where d dt ∫ ρ dV = CV d ( ρVCV ) dt =ρ dVCV dt ˆ−u ˆ−u ˆ ⋅ Aˆ = − ρ ( u + u ρ u rel ⋅ dA = ρ −ubox i fluid sin θ i fluid cos θ j i box fluid sin θ ) A = dA CS =u rel Substitute and simplify. dV ρ CV − ρ ( ubox + ufluid sin θ ) A = 0 dt dVCV = ( ubox + ufluid sin θ ) A dt ( ∫ VCV =VCV ∫ ) t =T dVCV = ( ubox + ufluid sin θ ) A VCV = 0 ∫ dt (Note that ubox, ufluid, θ, and A don’t change with time.) t =0 VCV = ( ubox + ufluid sin θ ) AT ∴T = ( ubox VCV + ufluid sin θ ) A C. Wassgren Chapter 03: Integral Analysis (3.44) 67 Last Updated: 25 Aug 2008 Example: Determine the rate at which fluid mass collects inside the room shown below in terms of ρ, V1, A1, V2, A2, Vc, R, and θ. Assume the fluid moving through the system is incompressible. 2R V1 room r2 V = Vc 1 − 2 R A1 A2 θ V2 SOLUTION: Apply conservation of mass to a control volume fixed to the interior of the room. 2R V1 room r2 V = Vc 1 − 2 R A1 y A2 x θ d dt where d dt ∫ ∫ ρ dV + ∫ ρ u CV ∫ ρ dV = CV rel V2 ⋅ dA = 0 (3.45) CS dM CV dt ( ) ( r=R ) ρ u rel ⋅ dA = ρ V1ˆ ⋅ − A1ˆ + ρ V2 sin θ ˆ − cos θ ˆ ⋅ − A2 ˆ + ρ i i i j j ∫ Vˆi ⋅ dAˆi r =0 CS r =R = − ρV1 A1 + ρV2 A2 cos θ + ρ ∫ V (1 − r c r =0 2 R2 ) 2π rdr = − ρV1 A1 + ρV2 A2 cos θ + π ρVc R 2 2 Substitute and simplify. dM CV − ρV1 A1 + ρV2 A2 cos θ + π ρVc R 2 = 0 2 dt dM CV ∴ = ρV1 A1 − ρV2 A2 cos θ − π ρVc R 2 2 dt C. Wassgren Chapter 03: Integral Analysis (3.46) 68 Last Updated: 25 Aug 2008 Example: Construct from first principles an equation for the conservation of mass governing the planar flow (in the xy plane) of a compressible liquid lying on a flat horizontal plane. The depth, h(x,t), is a function of position, x, and time, t. Assume that the velocity of the fluid in the positive x-direction, u(x,t), is independent of y. Also assume that the wavelength of the wave is much greater than the wave amplitude so that the horizontal velocities are much greater than the vertical velocities. free surface liquid h(x,t) y u(x,t) x SOLUTION: Apply conservation of mass to the fixed control volume shown below. Assume a unit depth into the page. h(x,t) Note that the mass flow rate at the center of the control volume is ρuh. u(x,t) y x dx d dt where d dt ∫ ρ dV + ∫ ρ u CV ⋅ dA = 0 (3.47) CS ∂ ∂ ∫ ρ dV = ∂t ( ρ hdx ) = ∂t ( ρ h ) dx = M CV CV ∫ ρu CS rel rel ∂ ∂ ⋅ dA = − ( ρ uh ) + ( ρ uh ) − 1 dx + ( ρ uh ) + ( ρ uh ) 2 ∂x ∂x ( ) = mleft (Note that: mleft/ = mcenter + right ∂mcenter ∂x = mright ( 1 2 dx ) where mcenter = ρ uh ) Substitute and simplify. ∂ ∂ ( ρ h ) dx + ( ρ uh ) dx = 0 ∂t ∂x ∂ ∂ ( ρ h ) + ( ρ uh ) = 0 ∂t ∂x (3.48) If the fluid is incompressible, then Eqn. (3.48) simplifies to: ∂h ∂ + ( uh ) = 0 ∂t ∂x C. Wassgren Chapter 03: Integral Analysis ( 1 dx ) = ∂∂x ( ρ uh ) dx 2 69 (3.49) Last Updated: 25 Aug 2008 4. Linear Momentum Equation (LME) In this section we’ll consider Newton’s 2nd law applied to a control volume of fluid. Recall that linear momentum is a vector quantity (it has both magnitude and direction) and is given by the mass*velocity. In words and in mathematical terms, Newton’s 2nd Law for a system is: The rate of change of a system’s linear momentum is equal to the net force acting on the system. system D u XYZ ρ dV = Fon system Dt Vsystem uXYZρdV Y LM of the system LM of a small Z X piece of fluid ∫ (3.50) where D/Dt is the Lagrangian derivative (implying that we’re using the rate of change as we follow the system), V is the volume, and ρ is the density. The quantity, uXYZ, represents the velocity of a small piece of fluid in the system with respect to an inertial (aka non-accelerating) frame of reference XYZ (recall that Newton’s 2nd law holds strictly for inertial frames of reference). Note that a frame of reference moving at a constant velocity in a straight line is non-accelerating and thus is inertial. The term, Fon system, represents the net forces acting on the system. These forces can be of two different types. The first are body forces, Fbody, and the second are surface forces, Fsurface. Body forces are those forces that act on each piece of fluid in the system. Examples include gravitational and electro-magnetic forces. Surface forces are those forces acting only at the surface of the system. Examples of surface forces include pressure and shear forces. Thus, Fon system = Fbody on system + Fsurface on system (3.51) Using the Reynolds Transport Theorem to convert the left hand side of Eqn. (3.24) from a system point of view to an expression for a control volume gives: D d u XYZ ρ dV = u XYZ ρ dV + u XYZ ( ρ u rel ⋅ dA ) (3.52) dt Dt Vsystem CV CS Since the Reynolds Transport Theorem is applied to a coincident system and control volume, the forces acting on the system will also act on the control volume. Thus, the LME for a CV becomes: ∫ ∫ Fbody on CV + Fsurface on CV = net body force acting on the CV net surface force acting on the CV C. Wassgren Chapter 03: Integral Analysis ∫ d dt ∫u XYZ ρ dV CV rate of increase of LM inside the CV + ∫u XYZ ( ρ u rel ⋅ dA ) LME for a CV (3.53) CS net rate at which LM leaves the CV through the CS 70 Last Updated: 25 Aug 2008 Notes: 1. Recall that the LME is a vector expression. There are actually three equations built into Eqn. (3.53). For example, in a rectangular coordinate system (Cartesian coordinates) we have: d FX ,body on CV + FX ,surface on CV = u X ρ dV + u X ( ρ u rel ⋅ dA ) dt ∫ FY ,body on CV + FY ,surface on CV = FZ ,body on CV + FZ ,surface on CV d dt d = dt ∫ CV CS ∫u Y ρ dV + CV ∫u ∫u Y ( ρ u rel ⋅ dA ) (3.54) CS Z ρ dV + CV ∫u Z ( ρ u rel ⋅ dA ) CS 2. It is important to distinguish between the two velocities uXYZ and urel. The velocity, uXYZ, represents the fluid velocity with respect to an inertial frame of reference XYZ (e.g. a frame of reference fixed with respect to the stars or moving at a constant velocity in a straight line). The velocity, urel, is the velocity of the fluid with respect to the control surface, i.e. urel = ufluid, XYZ – uCS, XYZ. 3. So far we’ve only discussed the LME for inertial (aka non-accelerating) frames of reference. We can also apply the LME to non-inertial (aka accelerating) frames of reference but we need to add additional acceleration terms. We’ll consider accelerating frames of reference a little later. 4. In order to avoid mistakes, be sure to do the following when applying the LME: a. draw the CV that the LME is being applied to, b. indicate the frame of reference that is being used, c. state your significant assumptions, d. draw a free body diagram (FBD) of the pertinent forces, and e. write down the LME and then indicate the value of each term in the LME While these things may seem trivial and unnecessary, writing them down in a clear and concise manner can greatly reduce the likelihood of mistakes. Helpful Hints: 1. Carefully draw your control volume. Don’t neglect to draw a control volume or draw a control volume and then use a different one. 2. Clearly indicate your frame of reference. Don’t neglect to indicate a frame of reference or draw a frame of reference and then use a different one. 3. Carefully draw free body diagrams for your control volume. This will facilitate your evaluation of forces in the linear momentum equation. 4. Make sure you understand what each term in the linear momentum equation represents. 5. Carefully evaluate the dot product in the mass flux term. 6. Be sure to use the correct velocity components in the CV and CS integral terms. 7. You must integrate the terms in the linear momentum equation when the density or velocity are not uniform. 8. Don’t forget to include pressure and shear forces in the surface force term. 9. Don’t forget to include the weight of everything inside the control volume when gravitational body forces are significant. Now let’s consider some examples to see how we apply the LME to a CV. C. Wassgren Chapter 03: Integral Analysis 71 Last Updated: 25 Aug 2008 Example: A jet of water is deflected by a vane mounted on a cart. The water jet has an area, A, everywhere and is turned by an angle θ with respect to the horizontal. The pressure everywhere within the jet is atmospheric. The incoming jet velocity with respect to the ground (axes XY) is Vjet. The cart has mass M. Determine: a. the force components, Fx and Fy, required to hold the cart stationary, and b. the horizontal force component, Fx, if the cart moves to the right at the constant velocity, Vcart (Vcart<Vjet) θ A Vjet Vcart Y Fx X Fy SOLUTION: Part (a): Apply conservation of mass and the linear momentum equation to a control volume surrounding the cart. Use an inertial frame of reference fixed to the ground (XY). Vout (this velocity is currently an unknown quantity) θ A Vjet Y Fx X Fy First apply conservation of mass to the control volume to determine Vout. d ρ dV + ρ u rel ⋅ dA = 0 dt ∫ ∫ CV CS C. Wassgren Chapter 03: Integral Analysis 72 (3.55) Last Updated: 25 Aug 2008 where d dt ∫ ρ dV = 0 (the mass within the control volume doesn’t change) CV = u rel =A =u rel = A ρ V ˆ ⋅ − Aˆ + ρ V ρ u rel ⋅ dA = jet i i out cos θ ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ i j i j CS ( ∫ )( left side ) right side ( = − ρV jet A + ρVout A cos θ + sin θ 2 2 ) =1 = − ρV jet A + ρVout A (Note that the jet area remains constant.) Substitute and re-arrange. − ρV jet A + ρVout A = 0 Vout = V jet (3.56) Now apply the linear momentum equation in the X-direction: d u X ρ dV + u X ( ρ u rel ⋅ dA ) = FB , X + FS , X dt ∫ where d dt ∫ CV CS ∫u X (3.57) ρ dV = 0 (the momentum within the control volume doesn’t change with time) CV =u X ∫u X ( ρ u rel ⋅ dA ) = (Vjet ) CS =u X =u rel = A ρ V ˆ ⋅ − Aˆ + V cos θ i jet jet i ( ) = u rel =A ˆ ˆ ˆ ˆ ρ V jet cos θ i + sin θ j ⋅ A cos θ i + sin θ j ( left side )( ) right side ( 2 = − ρVjet A + ρVj2 A cos θ cos 2 θ + sin 2 θ et ) =1 = 2 ρVjet A ( cos θ − 1) FB , X = 0 (no body forces in the x-direction) FS , X = − Fx (all of the pressure forces cancel out) Substitute and re-arrange. 2 ρVjet A ( cos θ − 1) = − Fx 2 Fx = ρVjet A (1 − cos θ ) C. Wassgren Chapter 03: Integral Analysis (3.58) 73 Last Updated: 25 Aug 2008 Now look at the Y-direction: d uY ρ dV + uY ( ρ u rel ⋅ dA ) = FB ,Y + FS ,Y dt ∫ where d dt ∫ CV CS ∫u Y ρ dV (3.59) = 0 (the momentum within the control volume doesn’t change with time) CV =u rel =A uY ( ρ u rel ⋅ dA ) = Vjet sin θ ρ Vjet cos θ ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ i j i j CS =uY ( ∫ ( ) )( ) right side = 2 ρV jet A sin θ ( cos 2 θ + sin 2 θ ) =1 2 = ρV jet A sin θ FB ,Y = − Mg (assume that the fluid weight in the CV is negligible compared to the cart weight) FS ,Y = Fy (all of the pressure forces cancel out) Substitute and re-arrange. 2 ρVjet A sin θ = − Mg + Fy 2 Fy = ρVjet A sin θ + Mg C. Wassgren Chapter 03: Integral Analysis (3.60) 74 Last Updated: 25 Aug 2008 Part (b): Apply the linear momentum equation to a control volume surrounding the cart. Use a frame of reference fixed to the cart (xy). Note that this is an inertial frame of reference since the cart moves in a straight line at a constant speed. In addition, in this frame of reference, the cart appears stationary and the jet velocity at the left is equal to Vjet-Vcart. Vout (this velocity is currently an unknown quantity) θ A Vjet - Vcart y x Fx Fy First apply conservation of mass to the control volume to determine Vout d ρ dV + ρ u rel ⋅ dA = 0 dt ∫ where d dt ∫ CV CS ∫ ρ dV = 0 (3.61) (the mass within the control volume doesn’t change) CV = u rel = u rel =A =A ˆ ⋅ − Aˆ + ρ V ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ ρ u rel ⋅ dA = ρ Vjet − Vcart i i out cos θ i j i j CS ( ∫ ( ) )( left side ( ) right side ) ( = − ρ V jet − Vcart A + ρVout A cos 2 θ + sin 2 θ ) =1 ( ) = − ρ V jet − Vcart A + ρVout A (Note that the jet area remains constant.) Substitute and re-arrange. ( ) − ρ Vjet − Vcart A + ρVout A = 0 Vout = Vjet − Vcart (3.62) Now apply the linear momentum equation in the x-direction: d u x ρ dV + u x ( ρ u rel ⋅ dA ) = FB , x + FS , x dt ∫ where d dt ∫ CV CS ∫ u ρ dV = 0 x (3.63) (the momentum within the control volume doesn’t change with time) CV C. Wassgren Chapter 03: Integral Analysis 75 Last Updated: 25 Aug 2008 =u rel =u rel =u X =A =A ˆ ⋅ − Aˆ + V − V ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ u x ( ρ u rel ⋅ dA ) = V jet − Vcart ρ V jet − Vcart i i j i j jet cart cos θ ρ V jet − Vcart cos θ i CS =u x ∫ ( )( ) ( ) )( ( left side ( = − ρ V jet − Vcart ) ) right side ( 2 )( A + ρ V jet − Vcart ) ( 2 A cos θ cos 2 θ + sin 2 θ ) =1 ( = ρ V jet − Vcart ) 2 A ( cos θ − 1) FB, x = 0 (no body forces in the x-direction) FS , x = − Fx (all of the pressure forces cancel out) Substitute and re-arrange. ρ (Vjet − Vcart ) A ( cos θ − 1) = − Fx 2 ( Fx = ρ Vjet − Vcart 2 ) A (1 − cos θ ) (3.64) Now solve the problem using an inertial frame of reference fixed to the ground (frame XY). From Eqn. (3.62) we know that using a frame of reference fixed to the cart, the jet velocity on the right hand side is: V = V −V cos θ ˆ + sin θ ˆ i j (3.65) out, relative to cart ( jet cart )( ) Hence, relative to the ground the jet velocity on the right hand side is: V =V + V = V −V cos θ ˆ + sin θ ˆ + V ˆ i j i out, relative to ground out, relative to cart ( cart jet cart )( ) (3.66) cart Now consider the linear momentum equation in the X direction. d u X ρ dV + u X ( ρ u rel ⋅ dA ) = FB , X + FS , X dt ∫ where d dt ∫ CV CS ∫u X (3.67) ρ dV = 0 (the momentum within the control volume doesn’t change with time) CV =u X = u rel = u rel =A =A ˆ ⋅ − Aˆ + V − V ρ V jet − Vcart cos θ ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ u X ( ρ u rel ⋅ d A ) = V jet ρ V jet − Vcart i i i j i j jet cart cos θ + Vcart CS =u X ∫ ( ) ( ) )( ( left side )( right side ( 2 = − ρV jet V jet − Vcart A + ρ V jet − Vcart cos θ + Vcart V jet − Vcart A cos 2 θ + sin 2 θ ( ) ( ) ( ) ) =1 2 2 2 = ρ −V jet + V jetVcart + V jet − Vcart cos θ + VcartV jet − Vcart A ( = ρ V jet − Vcart ( ( = ρ V jet − Vcart ) ) ) 2 ( 2 ) 2 ( cos θ − 1) A cos θ − V jet − Vcart A FB, X = 0 (no body forces in the x-direction) FS , X = − Fx (all of the pressure forces cancel out) Substitute and re-arrange. ρ (Vjet − Vcart ) A ( cos θ − 1) = − Fx 2 ( Fx = ρ Vjet − Vcart 2 ) A (1 − cos θ ) (Same answer as before!) (3.68) Note that using a frame of reference that is fixed to the control volume is easier than using one fixed to the ground. This is often the case. C. Wassgren Chapter 03: Integral Analysis 76 ) Last Updated: 25 Aug 2008 Example: A fluid enters a horizontal, circular cross-sectioned, sudden contraction nozzle. At section 1, which has diameter D1, the velocity is uniformly distributed and equal to V1. The gage pressure at 1 is p1. The fluid exits into the atmosphere at section 2, with diameter D2. Determine the force in the bolts required to hold the contraction in place. Neglect gravitational effects and assume that the fluid is inviscid. bolts D2 V1 D1 p1 atmosphere SOLUTION: Apply the linear momentum equation in the X-direction to the fixed control volume shown below. bolts D2 V1 D1 p1 atmosphere Fbolts The CS cuts through the bolts. So that Fbolts is the force one side of the bolts applies to the other side. X d dt where d dt ∫u X ∫ ρ dV + u X ( ρ u rel ⋅ dA ) = FB , X + FS , X CV ∫u (3.69) Cs X ρ dV = 0 (steady flow) CV =A =u rel π D2 u X ( ρ u rel ⋅ dA ) = ρ V1 V1ˆ ⋅ − 1 i 4 Cs (Note that V2 is unknown for now.) FB, X = 0 =u X ∫ FS , X = p1,gage =A =u X = urel 2 ˆ + ρ V V ˆ ⋅ π D2 i i 2 2 4 2 2 ˆ = − ρV 2 π D1 + ρV 2 π D2 i 1 2 4 4 π D12 + Fbolts 4 (Note that p2,gage = 0 since p2,abs = patm. We could have also worked the problem using absolute pressures everywhere. The pressure force on the left hand side would be p1,absπD12/4 and the pressure force on the right hand side would be patmπD12/4 (note that the diameter is D1 and not D2).) C. Wassgren Chapter 03: Integral Analysis 77 Last Updated: 25 Aug 2008 Substitute and re-arrange. 2 π D12 π D2 π D12 − ρV12 + ρV22 = p1,gage + Fbolts 4 4 4 2 π D12 π D2 π D12 Fbolts = − ρV12 + ρV22 − p1,gage 4 4 4 (3.70) To determine V2, apply conservation of mass to the same control volume. d ρ dV + ρ u rel ⋅ dA = 0 dt ∫ where d dt ∫ ∫ CV Cs (3.71) ∫ ρ dV = 0 CV ρ u rel ⋅ dA = − ρV1 π D12 4 Cs + ρV2 2 π D2 4 Substitute and simplify. − ρV1 π D12 4 + ρV2 D V2 = V1 1 D2 2 π D2 4 =0 2 (3.72) Substitute Eqn. (3.72) into Eqn. (3.70) and simplify. Fbolts = − ρV12 π D12 4 4 + ρV12 2 D1 π D2 π D12 − p1,gage 4 D2 4 π D12 (3.73) − 1 − p1,gage 4 D2 4 Note that Fbolts was assumed to be positive when acting in the +X direction (causing compression in the bolts). If Fbolts < 0 then the bolts will be in tension. Fbolts = ρV12 2 π D12 D1 C. Wassgren Chapter 03: Integral Analysis 78 Last Updated: 25 Aug 2008 Example: Water is sprayed radially outward through 180° as shown in the figure. The jet sheet is in the horizontal plane and has thickness, H. If the jet volumetric flow rate is Q, determine the resultant horizontal anchoring force required to hold the nozzle stationary. Q R H SOLUTION: Apply the linear momentum equation in the X direction to the fixed control volume shown below. Y Fy X Fx θ Fx V Q H R side view V top view d dt where d dt ∫u X ρ dV + CV ∫u ∫u X ( ρ u rel ⋅ dA ) = FB , X + FS , X (3.74) CS X ρ dV = 0 (steady flow) CV =u X = dA θ =π π ρV Rdθ H = ρV 2 RH u X ( ρ u rel ⋅ dA ) = sin θ dθ = − ρV 2 RH cos θ 0 (V sin θ ) CS θ =0 θ =0 ∫ θ =π ∫ ∫ = − ρV 2 RH ( −1 − 1) = 2 ρV 2 RH (Note that there is no X-momentum at the control volume inlet. Also, V is an unknown quantity at the moment.) FB, X = 0 FS , X = Fx (All of the pressure forces cancel and only the anchoring force remains.) C. Wassgren Chapter 03: Integral Analysis 79 Last Updated: 25 Aug 2008 Substitute. Fx = 2 ρV 2 RH (3.75) To determine V, apply conservation of mass to the same control volume. d ρ dV + ρ u rel ⋅ dA = 0 dt ∫ where d dt ∫ ∫ CV CS ∫ ρ dV = 0 (3.76) (steady flow) CV = dA θ =π ρ u rel ⋅ dA = − ρ Q + inlet CS ∫ ρV Rdθ H = − ρ Q + ρV π RH θ =0 outlet Substitute and simplify. − ρ Q + ρV π RH = 0 V= Q π RH (3.77) Substitute Eqn. (3.77) into Eqn. (3.75). 2 Q Fx = 2 ρ RH π RH (3.78) Note that Fy = 0 due to symmetry. C. Wassgren Chapter 03: Integral Analysis 80 Last Updated: 25 Aug 2008 Example: A variable mesh screen produces a linear and axi-symmetric velocity profile as shown in the figure. The static pressure upstream and downstream of the screen are p1 and p2 respectively (and are uniformly distributed). If the flow upstream of the mesh is uniformly distributed and equal to V1, determine the force the mesh screen exerts on the fluid. Assume that the pipe wall does not exert any force on the fluid. variable mesh screen Section 1 Section 2 p1 2R p2 V1 SOLUTION: First, note that the linear velocity profile at the outlet may be written as: r V = Vmax (3.79) R where Vmax is the flow velocity at r = R. Now apply conservation of mass to the fixed control volume shown below to find Vmax in terms of the upstream properties. d ρ dV + ρ u rel ⋅ dA = 0 (3.80) dt ∫ where d dt ∫ CV CV ∫ ρ dV = 0 (steady flow) dr CV r=R =V = dA r ρ u rel ⋅ dA = − ρV1π R + ρ Vmax ( 2π rdr ) R CV r =0 left side ∫ 2 ∫ r dA = 2πrdr right side = − ρV1π R 2 + ρ 2 π Vmax R 2 3 Substitute and simplify. − ρV1π R 2 + ρ 2 π Vmax R 2 = 0 3 Vmax = 3 V1 2 Section 1 variable mesh screen (3.81) Section 2 F p1 2R X p2 V1 C. Wassgren Chapter 03: Integral Analysis 81 The control volume weaves in and out of the mesh so that the mesh is not part of the control volume, and instead exerts a force, F, on the control volume. Last Updated: 25 Aug 2008 Now apply the linear momentum equation in the X-direction to the same control volume. d u X ρ dV + u X ( ρ u rel ⋅ dA ) = FB, X + FS , X dt ∫ where d dt ∫ ∫ CV CV ∫u X (3.82) ρ dV = 0 CV r=R u X ( ρ u rel ⋅ dA ) = − ρV12π R 2 + CV left side = dA 2 3 r 2 V1 ρ ( 2π rdr ) R r =0 ∫ right side = − ρV12π R 2 = − 1 ρV12π R 2 8 + ρπ 922 VR 81 FB, X = 0 FS , X = − F + p1π R 2 − p2π R 2 Substitute and simplify. − 1 ρV12π R 2 = − F + p1π R 2 − p2π R 2 8 F = − π ρV12 R 2 + ( p1 − p2 ) π R 2 8 (3.83) This is the force the mesh applies to the control volume (i.e. the fluid). The fluid applies an equal and opposite force to the mesh. C. Wassgren Chapter 03: Integral Analysis 82 Last Updated: 25 Aug 2008 Example: Incompressible fluid of negligible viscosity is pumped, at total volume flow rate Q, through a porous surface into the small gap between closely spaced parallel plates as shown. The fluid has only horizontal motion in the gap. Assume uniform flow across any vertical section. Obtain an expression for the pressure variation as a function of x. L x h V ( x) Assume a depth w into the page. Q SOLUTION: Apply conservation of mass to the following differential control volume. L Q X d ∫ ρ dV + CS ρ u rel ⋅ dA = 0 ∫ dt CV where d ρ dV = 0 (steady flow) dt C∫ V d d Q ⋅ dA = − ρVh ( w ) + ( ρVhw ) ( − 1 dx ) + ρVhw + ( ρVhw ) ( 1 dx ) − ρ wdx 2 2 dx dx Lw CS d Q dV Q wdx = ρ hw dx − ρ wdx = ( ρVhw ) dx − ρ dx Lw dx Lw Substituting and simplifying gives: dV Q ρ hw dx = ρ wdx dx Lw dV Q = dx Lhw V =V x=x Q dV = ∫ dx (V(x = 0) = 0 due to symmetry.) ∫0 Lhw V= x=0 ∫ ρu rel Vhw x Q x = or V = Q L hw L C. Wassgren Chapter 03: Integral Analysis (3.84) (3.85) 83 Last Updated: 25 Aug 2008 Now apply the linear momentum equation in the X-direction to the same control volume. d ∫ u X ρ dV + CS u X ( ρ urel ⋅ dA ) = FBX + FSX ∫ dt CV where d u X ρ dV = 0 (steady flow) dt C∫ V d d ⋅ dA ) = − ρV 2 hw + ( ρV 2 hw ) ( − 1 dx ) + ρV 2 hw + ( ρV 2 hw ) ( 1 dx ) 2 2 dx dx CS d dV = ( ρV 2 hw ) dx = 2 ρV hwdx dx dx (Note that the flux of mass from the porous surface has no X-momentum.) FBX = 0 ∫ u ( ρu X rel d d FSX = phw + ( phw ) ( − 1 dx ) − phw + ( phw ) ( 1 dx ) 2 2 dx dx d dp = − ( phw ) dx = − hwdx dx dx Substituting and simplifying gives: dV dp 2 ρV hwdx = − hwdx dx dx dV dp 2 ρV =− dx dx Substituting Eqns. (3.84) and (3.85) gives: 2 dp Q 2ρ x=− dx Lhw p = patm ∫ p= p Q dp = −2 ρ Lhw 2 x= 1 L 2 ∫ xdx x= x 2 Q 12 2 patm − p = − ρ (4 L −x ) Lhw 2 2 Q 1 x p − patm = ρ 4 − hw L p − patm ( hw) ρQ 2 = 1 x − 4 L 2 C. Wassgren Chapter 03: Integral Analysis (3.86) 84 Last Updated: 25 Aug 2008 The LME for Non-Inertial Frames of Reference Recall that Newton’s 2nd law holds strictly for inertial (non-accelerating) frames of reference. Now let’s consider frames of reference that are non-inertial (accelerating). First let’s examine how we can describe the motion of a particle in an accelerating frame of reference, call it frame xyz, in terms of a nonaccelerating frame of reference, call it frame XYZ. particle y rxyz θxyz/XYZ rXYZ Y x z rxyz/XYZ X Z The position of a particle in frame XYZ is given by rXYZ and in frame xyz the particle’s position is given by rxyz. The two position vectors are related by the position vector of the origin of frame xyz in frame XYZ, rxyz/XYZ: rXYZ = rxyz / XYZ + rxyz (3.87) The velocity of the particle in frame XYZ can be found by taking the time derivative of the position vector, rXYZ, with respect to frame XYZ (as indicated by the subscript XYZ in the equation below): drxyz / XYZ drxyz drXYZ = + (3.88) dt XYZ dt dt XYZ XYZ The time derivative of rxyz/XYZ is simply the velocity of the origin of frame xyz with respect to frame XYZ, uxyz/XYZ, i.e.: drxyz / XYZ = u xyz / XYZ (3.89) dt XYZ We must be careful, however, when evaluating the time derivative of rxyz in frame XYZ since both the magnitude of rxyz and the basis vectors of frame xyz can change with time (the basis vectors of xyz can change due to rotation of the frame xyz with respect to frame XYZ). To calculate the time derivative of rxyz in frame XYZ, let’s first write rxyz as a magnitude, rxyz, multiplied by the basis vectors of frame xyz, e xyz , ˆ then use the product rule to evaluate the time derivative: drxyz dt = XYZ Note that drxyz dt ( ˆ d rxyz e xyz ) = dt XYZ drxyz dt ˆ e xyz + rxyz XYZ ˆ e xyz = u xyz ˆ de xyz dt (3.90) XYZ (3.91) XYZ is the velocity of the particle in frame xyz. C. Wassgren Chapter 03: Integral Analysis 85 Last Updated: 25 Aug 2008 The time derivative of the xyz basis vectors is found from geometric considerations. Consider the drawing shown below of the change in the x-basis vector as a function of time. For simplicity, we’ll assume that the rotation only occurs in the xy plane (i.e. θx = θy = 0) : The time derivative of the basis vector is given by: ˆ ˆ ˆ e x ( t + ∆t ) − e x ( t ) de x = lim ∆t → 0 dt ∆t Note from the diagram that: ˆ ˆ ˆ ˆ ˆ e x ( t + ∆t ) − e x ( t ) = e x ( t ) cos ∆θ z + e y ( t ) sin ∆θ z − e x ( t ) ˆ e x ( t + ∆t ) ˆ ˆ e x ( t + ∆t ) − e x ( t ) ∆θz ˆ ex (t ) ˆ ˆ = e x ( t )( cos ∆θ z − 1) + e y ( t ) sin ∆θ z In addition, as ∆t → 0, ∆θz → 0 and: ( cos ∆θ z − 1) ≈ 1 − ( ∆θ z )2 / 2 − 1 = − 1 ( ∆θ z )2 2 and sin ∆θ z ≈ ∆θ z so that: ˆ ˆ ˆ ˆ − 1 ( ∆θ z ) e x ( t ) + ∆θ z e y e x ( t + ∆t ) − e x ( t ) ˆ de x = lim = lim 2 ∆t →0 ∆t → 0 dt ∆t ∆t dθ ˆ = z ey dt ˆ de ˆ ∴ x = ω z e y (where ωz = dθz/dt) dt In general, it can be shown that: ˆ de xyz ˆ = ω xyz / XYZ × e xyz dt XYZ 2 (3.92) (3.93) so that ˆ de xyz rxyz dt ( ) ˆ = rxyz ω xyz / XYZ × e xyz = ω xyz / XYZ × rxyz (3.94) XYZ Combining Eqns. (3.88) – (3.91), and (3.94) we find that the velocity of a fluid particle in the inertial frame XYZ is: u XYZ = u xyz / XYZ + u xyz + ω xyz / XYZ × rxyz (3.95) velocity of particle in frame XYZ velocity of frame xyz w/r/t frame XYZ velocity of particle in frame xyz velocity of particle in XYZ due to rotation of frame xyz w/r/t frame XYZ where uxyz is the particle velocity in non-inertial frame xyz, ω xyz / XYZ is the angular velocity of frame xyz with respect to frame XYZ, and rxyz is the position vector of the particle from the origin of frame xyz. The acceleration of a particle in frame XYZ in terms of xyz quantities can be found in a similar manner: du xyz / XYZ du xyz du XYZ d = + + ω xyz / XYZ × rxyz (3.96) dt XYZ dt dt XYZ dt XYZ XYZ ( =a XYZ = a xyz / XYZ = ( d ˆ u xyz e xyz dt ) XYZ =ω xyz / XYZ ×rxyz + ω xyz / XYZ × ) ( ˆ d rxyz e xyz ) dt XYZ where the results from Eqns. (3.90), (3.91), (3.93), and (3.94) are used to simplify the last two expressions in Eqn. (3.96): ˆ du xyz de xyz d ˆ ˆ u xyz e xyz = e xyz + u xyz dt dt dt (3.97) XYZ ( ) = a xyz + ω xyz / XYZ × u xyz and C. Wassgren Chapter 03: Integral Analysis 86 Last Updated: 25 Aug 2008 θ xyz / XYZ × ( ˆ d rxyz e xyz ) ( = ω xyz / XYZ × u xyz + ω xyz / XYZ × rxyz dt ) (3.98) XYZ ( = ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ) Substituting Eqns. (3.97) and (3.98) into Eqn. (3.96) and simplifying gives: a XYZ rectilinear acceleration of particle in frame XYZ = + a xyz / XYZ rectilinear acceleration of frame xyz w/r/t frame XYZ ( 2ω xyz / XYZ × u xyz ) + Coriolis acceleration of particle in frame XYZ due to rectilinear motion of particle in frame xyz ( ω xyz / XYZ × rxyz ) + a xyz rectilinear acceleration of particle in frame xyz tangential acceleration of particle in frame XYZ due to rotational acceleration of frame xyz ( ) + ω xyz / XYZ × ω xyz / XYZ × rxyz (3.99) centripital acceleration of particle in frame XYZ due to rotation of frame xyz Now let’s use these relations to determine an expression for the LME using a non-inertial frame of reference. Recall that the Lagrangian statement for the LME is (refer to Eqn. (3.50)): D u XYZ ρ dV = Fon system (3.100) Dt Vsystem Substitute Eqn. (3.95) into Eqn. (3.100) and re-arrange: D Fon system = u xyz / XYZ + u xyz + ω xyz / XYZ × rxyz ρ dV Dt Vsystem (3.101) D D = u xyz ρ dV + u xyz / XYZ + ω xyz / XYZ × rxyz ρ dV Dt Dt Vsystem Vsystem Now use the Reynolds Transport Theorem to convert the first term on the right hand side to a control volume perspective and re-arrange: D Fbody on CV + Fsurface on CV − u xyz / XYZ + ω xyz / XYZ × rxyz ρ dV Dt Vsystem (3.102) ∫ ∫( ) ∫( ∫ ) ∫( = d dt ∫u xyz ρ dV CV C. Wassgren Chapter 03: Integral Analysis + ∫u xyz ) ( ρ u rel ⋅ dA ) CS 87 Last Updated: 25 Aug 2008 The remaining Lagrangian term can be simplified by changing the volume integral to a mass integral and noting that the mass of the system doesn’t change with time: D u xyz / XYZ + ω xyz / XYZ × rxyz ρ dV Dt Vsystem ∫( ) D = u xyz / XYZ + ω xyz / XYZ × rxyz dm Dt M system D = u xyz / XYZ + ω xyz / XYZ × rxyz dm Dt ∫( ) ( ∫ (3.103) ) M system = ( ) D u xyz / XYZ + ω xyz / XYZ × rxyz ρ dV Dt ∫ Vsystem Since uxyz/XYZ and ωxyz/XYZ are functions only of time (these variables describe the motion of the reference frame xyz and not the fluid field), and because Drxyz/Dt=uxyz1, we can replace the Lagrangian time derivative with an Eulerian time derivative and substitute in our result from Eqn. (3.99): D u xyz / XYZ + ω xyz / XYZ × rxyz ρ dV Dt ( ∫ ) Vsystem ( ∫ = Vsystem (3.104) ( ∫ = ) d u xyz / XYZ + ω xyz / XYZ × rxyz ρ dV dt ) a xyz / XYZ + ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ρ dV Vsystem Substituting Eqns. (3.103) and (3.104) back into Eqn. (3.102) and noting that when we apply the Reynolds Transport Theorem the control volume and system volume are coincident (so that the system volume integral in Eqn. (3.104) can be replaced by a control volume integral), we find that the LME can be applied using a non-inertial frame of reference, xyz, if the following form is used: Fbody on CV + Fsurface on CV − ∫ {a xyz / XYZ ( )( ) ( )} + ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ρ dV (3.105) CV = d dt ∫u xyz ρ dV CV + ∫u xyz ( ρ u rel ⋅ dA ) CS Let’s consider a few examples to see how this form of the LME is applied. 1 Drxyz Dt = ∂rxyz ∂t =0 + ux ∂rxyz ∂x + uy ˆ =e x ∂rxyz ∂y ˆ =e y + uz ∂rxyz ∂z = u xyz ˆ =e z ˆ ˆ ˆ where rxyz = xe x + ye y + ze z C. Wassgren Chapter 03: Integral Analysis 88 Last Updated: 25 Aug 2008 Example: A jet of water is deflected by a vane mounted on a cart. The water jet has an area, A, everywhere and is turned an angle θ with respect to the horizontal. The pressure everywhere within the jet is atmospheric. The incoming jet velocity with respect to the ground (axes XY) is Vjet. The cart has mass M. Determine the horizontal acceleration of the cart at the instant when the cart moves with velocity Vcart (Vcart<Vjet) if no horizontal forces are applied θ A Vjet Vcart Y Fx X Fy SOLUTION: Apply the linear momentum equation to a control volume surrounding the cart. Use a frame of reference fixed to the cart (xy). Note that this is not an inertial frame of reference since the cart is accelerating. As in part (b) of this example problem worked out previously, in this frame of reference the cart appears stationary and the jet velocity at the left is equal to Vjet-Vcart. Following the analysis given in part (b) of this example problem worked out previously, conservation of mass indicates that the velocity on the right of the control volume is Vjet – Vcart. Vjet - Vcart θ A Vjet - Vcart y x Fy Apply the linear momentum equation in the x-direction: d u x ρ dV + u x ( ρ u rel ⋅ dA ) = FB, x + FS , x − ax / X ρ dV dt ∫ ∫ ∫ CV CS CV C. Wassgren Chapter 03: Integral Analysis 89 (3.106) Last Updated: 25 Aug 2008 where d dt ∫ u ρ dV ≈ 0 x CV (The cart has zero velocity in this frame of reference. The fluid in the control volume does accelerate in this frame of reference; however, its mass is assumed to be much smaller than the cart mass. Hence, the rate of change of the control volume momentum in this frame of reference is assumed to be zero.) =u rel =u rel =ux =A =A u x ( ρ u rel ⋅ dA ) = V jet − Vcart ρ V jet − Vcart ˆ ⋅ − Aˆ + V jet − Vcart cos θ ρ V jet − Vcart cos θ ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ i i i j i j CS =u x ∫ ( )( ) ( ) )( ( left side ( = − ρ V jet − Vcart ) 2 )( ) right side ( A + ρ V jet − Vcart ) 2 A cos θ ( cos 2 θ + sin 2 θ ) =1 ( = ρ V jet − Vcart ) 2 A ( cos θ − 1) FB, x = 0 (no body forces in the x-direction) FS , x = 0 (all of the pressure forces cancel out) ∫a x/ X ρ dV ≈ Ma (the mass within the CV is approximately equal to the cart mass) CV Substitute and re-arrange. ρ (Vjet − Vcart ) A ( cos θ − 1) = − Ma 2 ρ (Vjet − Vcart ) A (1 − cos θ ) 2 a= (3.107) M Now solve the problem using an inertial frame of reference fixed to the ground (frame XY). The velocity out of the right side of the cart is given by Eqn. (3.66). The linear momentum equation in the X direction gives: d u X ρ dV + u X ( ρ u rel ⋅ dA ) = FB , X + FS , X (3.108) dt ∫ where d dt ∫ CV CS ∫u X ρ dV ≈ Ma CV (The mass within the control volume is approximately equal to the cart mass since the fluid mass is assumed to be negligible.) =u X = u rel = u rel =A =A ˆ ⋅ − Aˆ + V − V u X ( ρ u rel ⋅ dA ) = V jet ρ V jet − Vcart i i cos θ + Vcart ρ V jet − Vcart cos θ ˆ + sin θ ˆ ⋅ A cos θ ˆ + sin θ ˆ i j i j cart jet CS ∫ =u X ( ) ( ) )( ( left side )( right side ( 2 = − ρV jet V jet − Vcart A + ρ V jet − Vcart cos θ + Vcart V jet − Vcart A cos 2 θ + sin 2 θ ( ) ( ) ( ) ) =1 2 2 2 = ρ −V jet + V jetVcart + V jet − Vcart cos θ + VcartV jet − Vcart A ( ) 2 2 = ρ V jet − Vcart cos θ − V jet − Vcart A ( ( = ρ V jet − Vcart ) ) 2 ( ) ( cos θ − 1) A FB, X = 0 (no body forces in the x-direction) FS , X = 0 (all of the pressure forces cancel out) C. Wassgren Chapter 03: Integral Analysis 90 ) Last Updated: 25 Aug 2008 Substitute and re-arrange. ( Ma + ρ Vjet − Vcart 2 ) A ( cosθ − 1) = 0 ρ (Vjet − Vcart ) A (1 − cos θ ) 2 a= (Same answer as before!) (3.109) M As in part (b), using a frame of reference that is fixed to the control volume is easier than using one fixed to the ground. C. Wassgren Chapter 03: Integral Analysis 91 Last Updated: 25 Aug 2008 Example: The tank shown rolls along a level track. Water received from a jet is retained in the tank. The tank is to accelerate from rest toward the right with constant acceleration, a. Neglect wind and rolling resistance. Find an algebraic expression for the force (as a function of time) required to maintain the tank acceleration at constant a. initial mass of cart and water, M0 A V F U SOLUTION: First apply conservation of mass to a control volume surrounding the cart (shown below) in order to determine how the cart mass changes with time. A The frame of reference xy is fixed to the cart. jet velocity relative to the y cart = (V – U) x F d dt where d dt ∫ ρ dV + ∫ ρ u CV ∫ ρ dV = CV ∫ ρu rel ⋅ dA = 0 (3.110) CS rel dM CV dt ⋅ dA = − ρ (V − U ) A CS Substitute and re-arrange. dM CV − ρ (V − U ) A = 0 dt dM CV = ρ (V − U ) A dt C. Wassgren Chapter 03: Integral Analysis (3.111) 92 Last Updated: 25 Aug 2008 Since the cart acceleration is constant (= a), we may write: U = at (Note that U(t = 0) = 0 since the cart starts from rest.) (3.112) Note that Eqn. (3.112) is only true when a = constant. Otherwise, if a = a(t) one must write the velocity as: t ∫ U = U 0 + adt (3.113) 0 Substitute Eqn. (3.112) into Eqn. (3.111) and solve the resulting differential equation. dM CV = ρ (V − at ) A dt M CV = M CV ∫ t =t dM CV = M CV = M 0 ∫ ρ (V − at ) Adt t =0 ( + ρ (Vt − ) )A M CV − M 0 = ρ Vt − 1 at 2 A 2 M CV = M 0 1 2 at 2 (3.114) Now apply the linear momentum equation in the x direction to the same control volume. Note that the frame of reference xy is not inertial since the cart is accelerating. d u x ρ dV + u x ( ρ u rel ⋅ dA ) = FB, x + FS , x − ax / X ρ dV (3.115) dt ∫ where d dt ∫ ∫ CV CS CV ∫ u ρ dV ≈ 0 x (most of the mass inside the CV has zero velocity in the given frame of reference) CV ∫u x ( ρ u rel ⋅ dA ) = − ρ (V − U )2 A CS FB, x = 0 FS , x = − F ∫a x/ X ρ dV ≈ aM CV CV Substitute and re-arrange. − ρ (V − U ) A = − F − aM CV 2 F = ρ (V − U ) A − aM CV 2 (3.116) Now substitute Eqns. (3.112) and (3.114) into Eqn. (3.116). 2 F = ρ (V − at ) A − a M 0 + ρ Vt − 1 at 2 A 2 ( ) (3.117) Note that we could have also used a frame of reference fixed to the ground (inertial) to solve this problem. C. Wassgren Chapter 03: Integral Analysis 93 Last Updated: 25 Aug 2008 Example: A model solid propellant rocket has a mass of 69.6 gm, of which 12.5 gm is fuel. The rocket produces 1.3 lbf of thrust for a duration of 1.7 sec. For these conditions, calculate the maximum speed and height attainable in the absence of air resistance. Plot the rocket speed and the distance traveled as functions of time. SOLUTION: Assume that the mass flow rate from the rocket is constant. Also assume that the thrust remains constant over the burn duration. Apply the linear momentum equation in the y-direction to the CV shown using a frame of reference attached to the rocket. y x g MCVg (pe – patm)Ae Ve d ∫ u y ρ dV + CS u y ( ρ u rel ⋅ dA ) = FB, y + FS , y − CV a y / Y ρ dV ∫ ∫ dt CV where d u y ρ dV ≈ 0 (Most of the fluid has zero velocity in this frame of reference.) dt C∫ V ∫ u ( ρu y rel ⋅ dA ) = −Ve ( ρ eVe Ae ) = − ρeVe2 Ae CS FB , y = − M CV g (weight) FS , y = ( pe − patm ) Ae (The exit pressure may be different from atmospheric pressure.) ∫a y /Y ρ dV ≈ aM CV (We’re using an accelerating frame of reference.) CV Substituting and simplifying: − ρeVe2 Ae = − M CV g + ( pe − patm ) Ae − M CV a a = −g + ρeVe2 Ae + ( pe − patm ) Ae (3.118) M CV C. Wassgren Chapter 03: Integral Analysis 94 Last Updated: 25 Aug 2008 Note that the thrust, T, is the force required to hold the rocket stationary (neglecting gravity). T Ve (pe – patm)Ae x d ∫ ux ρ dV + CS ux ( ρ u rel ⋅ dA ) = FB, x + FS , x ∫ dt CV where d u x ρ dV ≈ 0 (Most of the fluid has zero x-velocity.) dt C∫ V ∫ u ( ρu x rel ⋅ dA ) = Ve ( ρeVe Ae ) = ρeVe2 Ae CS FB , x = 0 FS , x = − ( pe − patm ) Ae + T Substituting and simplifying: ρeVe2 Ae = − ( pe − patm ) Ae + T T = ρ eVe2 Ae + ( pe − patm ) Ae (3.119) Substitute Eqn. (3.119) into Eqn. (3.118): T a = −g + M CV (3.120) Apply COM to the same CV: d ∫ ρ dV + CS ( ρ u rel ⋅ dA ) = 0 ∫ dt CV where dM CV d ∫V ρ dV = dt dt C ∫ ( ρu rel ⋅ dA ) = ρeVe Ae = m CS Substituting and simplifying: dM CV +m=0 dt Assuming the mass flow rate is a constant, solve Eqn. (3.121) subject to initial conditions: M CV ∫ M0 (3.121) t dM CV = − m ∫ dt 0 M CV = M 0 − mt where M0 is the initial mass of the CV. C. Wassgren Chapter 03: Integral Analysis (3.122) 95 Last Updated: 25 Aug 2008 Substitute Eqn. (3.122) into Eqn. (3.120) and solve the differential equation for the velocity: dU T a= = −g + dt M 0 − mt U t t 0 0 0 ∫ dU = ∫ − gdt + ∫ M Tdt 0 − mt U = − gt − T M 0 − mt ln m M0 U = − gt − T mt ln 1 − m M0 (3.123) Solve the differential equation given in Eqn. (3.123) for the height of the rocket. dh T mt U= = − gt − ln 1 − dt m M0 h 0 T mt ln 1 − dt M0 0m t t 0 ∫ dh = ∫ − gtdt − ∫ h = − 1 gt 2 + 2 T M0 mt ln 1 − m m M0 mt − t ln 1 − M0 + t (3.124) Note that Eqns. (3.120), (3.122) – (3.124) are written specifically for when the fuel is burning. When the fuel has been expended, the rocket equations of motion are: a = −g (3.125) U = U t =t ′ − g ( t − t ′ ) (3.126) h = − 1 g ( t − t ′ ) + U t = t ′ ( t − t ′ ) + ht = t ′ 2 (3.127) 2 where t’ is the time at which the fuel has been expended. For the given problem we’re told: M0 = 69.6 g Mfuel = 12.5 g T = 1.3 lbf = 5.79 N t’ = 1.7 sec giving a mass flow rate of: M m = fuel = 7.35 g/sec = 7.35*10-3 kg/sec t′ The maximum velocity will occur at the moment the fuel has been expended (neglecting the velocities as the rocket falls back to the ground). The maximum height will occur when the velocity is zero. Umax = U(t = t’ = 1.7 sec) = 139.2 m/s hmax = h(t = tm = 15.9 sec) = 1100 m (h(t = t’) = 114 m) The maximum height occurs when: U = U t = t ′ − g ( tm − t ′ ) = 0 tm = t ′ + U t =t ′ g C. Wassgren Chapter 03: Integral Analysis 96 Last Updated: 25 Aug 2008 The rocket speed and height are plotted below: 200 1200 150 1000 800 50 0 600 0 5 10 15 20 -50 25 30 35 height, h [m] velocity, U [m/s] 100 400 -100 200 -150 -200 0 time, t [sec] C. Wassgren Chapter 03: Integral Analysis 97 Last Updated: 25 Aug 2008 Example: Explain why hurricanes rotate in a counter-clockwise direction in the northern hemisphere and in a clockwise direction in the southern hemisphere. Do you think the flow down a bathtub drain or toilet will follow the same trend? Why or why not? SOLUTION: The acceleration of a fluid particle in a rotating frame of reference is: ( a XYZ = a xyz / XYZ + a xyz + ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz For the given problem: Ω ) a xyz / XYZ = 0 ˆ ez Y (3.128) R θ (the hurricane eye isn’t accelerating) a xyz = 0 ˆ er (fluid particles aren’t accelerating) ω xyz / XYZ = 0 X (constant Earth rotation speed) ˆ ω xyz / XYZ = ΩeY (constant Earth rotation speed) ˆ ˆ u xyz = Vr e r + Vφ eφ ˆ eY φ ˆ re r ˆ eφ u = V e + V e xyz r ˆr φ ˆφ fluid particle rxyz (velocity of a fluid particle assuming 2D motion, since fluid is pulled toward the low pressure hurricane eye, Vr < 0) ˆ = re r (radial distance out to a fluid particle) Note: ˆ ˆ ˆ ˆ eY = − cos φ cos θ e r + sin φ cos θ eφ + sin θ e z looking at the hurricane from above C. Wassgren Chapter 03: Integral Analysis 98 Last Updated: 25 Aug 2008 Substitute into Eqn. (3.128): ˆ ˆ ˆ ˆ ˆ ˆ a XYZ = 2ΩeY × Vr er + Vφ eφ + ΩeY × ( ΩeY × re r ) ( ) (3.129) centripital acceleration Coriolis acceleration Note that: ˆˆ ˆ ˆ ˆ ˆ eY × er = − cos φ cos θ e r + sin φ cos θ eφ + sin θ e z × e r ( ) ˆ ˆ = − sin φ cos θ e z + sin θ eφ ( ) ˆˆ ˆ ˆ ˆ ˆ eY × eφ = − cos φ cos θ er + sin φ cos θ eφ + sin θ e z × eφ ˆ ˆ = − cos φ cos θ e z − sin θ e r ( ) ˆˆ ˆ ˆ ˆ ˆ eY × e z = − cos φ cos θ er + sin φ cos θ eφ + sin θ e z × e z ˆ ˆ = cos φ cos θ eφ + sin θ cos θ er Hence, Eqn. (3.129) becomes (substituting sθ = sinθ , cθ = cosθ, etc.): ( ) ( ) ( ) 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = 2ΩVr ( − s φ cθ e z + s θ eφ ) + 2ΩVφ ( − c φ c θ e z − s θ er ) + Ω r − s φ cθ ( cφ cθ eφ + sθ cθ e r ) + s θ ( −cφ cθ e z − sθ e r ) ˆ ˆ ˆ ˆ ˆ ˆ a XYZ = 2ΩVr ( eY × e r ) + 2ΩVφ eY × eφ + Ω2 r eY × ( eY × er ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = 2ΩVr −sφ cθ e z + s θ eφ + 2ΩVφ ( − c φ c θ e z − s θ er ) + Ω2 reY × − s φ cθ e z + s θ eφ ( ) ( ˆ ( −2ΩVφ c φ cθ − 2ΩV s φ cθ − Ω rcφ sθ cθ ) e ) ˆ ˆ a XYZ = −2ΩVφ s θ − Ω 2 r s φ sθ c 2θ − Ω 2 rs 2θ e r + 2ΩVr s θ − Ω 2 r s φ cφ c 2θ eφ + 2 r (3.130) z Note that since the rotation rate of the Earth is very small, the Coriolis acceleration effects will be much larger than the centripetal acceleration effects, i.e.: 2ΩVr Ω 2 r (3.131) ˆ Looking at first eφ term in Eqn. (3.130) (this is the one that will dominate based on Eqn. (3.131)), we observe that when Vr < 0 (fluid moves toward the hurricane eye), θ > 0 (northern hemisphere), and Ω > 0 ˆ (the Earth rotates in a counter-clockwise direction when looking from above), the eφ term will be negative. ˆ When θ < 0 (southern hemisphere), the eφ term will be positive. Recall from the linear momentum equation: d u xyz ρ dV + dt ∫ CV ∫u xyz ( ρ u rel ⋅ dA ) = FB + FS − CS ∫a XYZ ρ dV CV ˆ dFCoriolis ≈ −2ΩVr s θ eφ ρ dV (3.132) Hence, the effective Coriolis effect produces an effective “force” acting on fluid particles pushing them in a ˆ ˆ counter-clockwise direction ( +eφ ) in the northern hemisphere (θ > 0) and in a clockwise direction ( −eφ ) in the southern hemisphere (θ < 0). C. Wassgren Chapter 03: Integral Analysis 99 Last Updated: 25 Aug 2008 Will the flow down a bathroom drain follow the same trend? Let’s investigate the ratio of a typical (convective) inertial force magnitude to a typical Coriolis force magnitude (this ratio is known as the Rossby #, Ro, and shows up often in geophysical flows). F ρ L2V 2 V Ro ≡ inertial ~ 3 ~ (3.133) FCoriolis ρ L ΩV LΩ where V is a typical flow velocity and L is a flow length scale. When the Rossby number is small, then the Coriolis effect becomes significant. However, when the Rossby number is large then we may neglect the Coriolis effect. For a typical hurricane: L ≈ 550 km (340 mi), V ≈ 10 m/s (23 mi/hr), Ω ≈ 2π rad/24 hr = 7.3*10-5 rad/s ⇒ Ro ≈ 0.25 ⇒ the Coriolis effect must be considered! For flow down a bathtub drain: L ≈ 1 m, V ≈ 1 m/s (2.2 mi/hr), Ω ≈ 2π rad/24 hr = 7.3*10-5 rad/s ⇒ Ro ≈ 13,800 ⇒ the Coriolis effect can be neglected ⇒ The Earth’s rotation has essentially no effect on the flow going down a bathtub drain. C. Wassgren Chapter 03: Integral Analysis 100 Last Updated: 25 Aug 2008 5. Angular Momentum Equation (AME) In words and in mathematical terms, the angular momentum principle for a system is: The rate of change of a system’s angular momentum (AM) is equal to the net moment (or torque) acting on the system. hXYZρdV AM of a small piece of fluid system rXYZ uXYZρdV LM of a small piece of fluid Y Z X D ( h XYZ + rXYZ × u XYZ ) ρ dV = M on system Dt Vsystem AM of the system ∫ (3.134) where D/Dt is the Lagrangian derivative (implying that we’re using the rate of change as we follow the system), V is the volume, and ρ is the density. The quantity hXYZ is the intrinsic specific angular momentum of a small piece of fluid resulting from the spin of fluid molecules contained within that small piece of fluid. In typical fluids (e.g. non-polar, non-magnetic fluids), the angular momentum vectors of the individual molecules are randomly oriented so that the sum of the intrinsic angular momentum vectors in a region containing many molecules is zero. Hence, we will neglect this contribution to the angular momentum of the fluid in the remainder of these notes. The quantity, uXYZ, represents the velocity of a small piece of fluid in the system with respect to an inertial (aka non-accelerating) frame of reference XYZ (recall that Newton’s 2nd law holds strictly for inertial frames of reference) and rXYZ is the distance from the inertial coordinate system origin to the fluid element. Note that a frame of reference moving at a constant velocity in a straight line is non-accelerating and thus is inertial. The term, Mon system, represents the net moments (or torques) acting on the system. These moments can be due to both body and surface forces, i.e. M body = rXYZ × Fbody M surface = rXYZ × Fsurface Note that if the fluid is magnetic, it is also possible to have an additional body moment that would induce the fluid molecules to change their intrinsic angular momentum (h). As stated before, we won’t consider such fluids in these notes. The study of magnetic fluids is known as magnetohydrodynamics. Using the Reynolds Transport Theorem to convert the left hand side of Eqn. (3.134) from a system point of view to an expression for a control volume gives: D = d rXYZ × u XYZ ρ dV rXYZ × u XYZ ρ dV + rXYZ × u XYZ ( ρ u rel ⋅ dA ) (3.135) dt Dt Vsystem CV CS Since the Reynolds Transport Theorem is applied to a coincident system and control volume, the moments acting on the system will also act on the control volume. Thus, the AME for a CV becomes: ∫ ∫ M body on CV + M surface on CV = net moment due to body forces acting on the CV net moment due to surface forces acting on the CV d dt ∫ ∫r XYZ × u XYZ ρ dV + CV ∫r XYZ × u XYZ ( ρ u rel ⋅ dA ) (3.136) CS rate of increase of AM inside the CV net rate at which the CV AM changes due to fluid leaving through the CS AME for a CV C. Wassgren Chapter 03: Integral Analysis 101 Last Updated: 25 Aug 2008 Notes: 1. Recall that the AME is a vector expression. There are actually three equations built into Eqn. (3.136). AME for Rotating (and not accelerating in translation) Frames of Reference Recall that Newton’s 2nd law holds strictly for inertial (non-accelerating) frames of reference. Often it is more convenient to use a rotating (non-inertial) reference frame when applying the AME. Y y uXYZ, uxyz rXYZ, rxyz x θxyz/XYZ X The Lagrangian statement for the AME is (refer to Eqn. (3.134)): D M on system = rXYZ × u XYZ ρ dV (3.137) Dt Vsystem The mass of the system remains constant so the Lagrangian derivative can be brought inside the integral: D D = rXYZ × u XYZ ρ dV (3.138) ( rXYZ × u XYZ ) ρ dV Dt Dt Vsystem Vsystem Since we are considering a reference frame that is only rotating and not accelerating in translation, the position and velocity vectors may be written as2: rXYZ = rxyz (3.139) u XYZ = u xyz + ω xyz / XYZ × rxyz ∫ ∫ ∫ Substituting into equation (3.138) gives: D rxyz × u xyz + ω xyz / XYZ × rxyz ρ dV Dt ( ∫ ) Vsystem = ∫ Vsystem = ∫ Vsystem D D rxyz × u xyz + r × ω xyz / XYZ × rxyz ρ dV xyz Dt Dt ( ) ( ( ) (3.140) ) D rxyz × u xyz ρ dV Dt + Drxyz D × ω xyz / XYZ × rxyz + rxyz × ω xyz / XYZ × rxyz ρ dV Dt Dt Vsystem ∫ ( ) ( ) Again, since the mass of the system is constant, the Lagrangian derivative can be brought outside of the first integral. In addition, we know from previous work1 that Drxyz/Dt=uxyz and: D ω xyz / XYZ × rxyz = ω xyz / XYZ × rxyz + ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz Dt ( ) ( ) Substituting and simplifying: 2 Refer to an earlier set of notes regarding the LME for non-inertial reference frames for the derivation of these quantities. C. Wassgren 102 Last Updated: 25 Aug 2008 Chapter 03: Integral Analysis ( Vsystem + = ) D rxyz × u xyz ρ dV Dt ∫ D Dt Drxyz D × ω xyz / XYZ × rxyz + rxyz × ω xyz / XYZ × rxyz ρ dV Dt Dt Vsystem ( ∫ ) ( ) (3.141) ∫ (rxyz × u xyz ) ρ dV Vsystem ( ) u xyz × ω xyz / XYZ × rxyz + ρ dV Vsystem rxyz × ω xyz / XYZ × rxyz + ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz To simplify things further, we can re-arrange the first term in the second integral and incorporate it into the second term: ∫ + ( ( ) ) u xyz × ω xyz / XYZ × rxyz + ( ) = rxyz × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ( ω xyz / XYZ × rxyz ) rxyz × ω xyz / XYZ × rxyz + ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz (3.142) Substituting Eqns. (3.139) – (3.142) into Eqn. (3.137) gives: D M on system = rxyz × u xyz ρ dV Dt Vsystem ∫ + ∫ {r xyz )} ( (3.143) × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ρ dV Vsystem Now use the Reynolds Transport Theorem to convert the first term on the right hand side to a control volume perspective and re-arrange. Also note that since the CV and system are coincident, the moments acting on the CV will be the same as the moments acting on the system and the CV mass will be the same as the system mass. M body on CV + M surface on CV − ∫ {r xyz ( )} × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ρ dV (3.144) CV = d dt ∫r xyz CV × u xyz ρ dV + ∫r xyz × u xyz ( ρ u rel ⋅ dA ) CS AME for a rotating frame of reference Let’s consider some examples to see how this form of the AME is applied. C. Wassgren Chapter 03: Integral Analysis 103 Last Updated: 25 Aug 2008 Example: A lawn sprinkler is constructed from pipe with an inner diameter of d with each arm having a length of R. Water flows through the sprinkler at a volumetric flow rate of Q. A force, F, is applied a distance, l, from the sprinkler hub on one of the sprinkler arms. If the water stream leaving the sprinkler arm is at an angle θ with respect to the tangent of the circle traced out by the sprinkler arms, determine: a. the force, F, required to hold the sprinkler stationary, b. the force, F, required to have the sprinkler rotate at a constant angular velocity, Ω, c. the angular acceleration of the sprinkler if the sprinkler’s moment of inertia (including the fluid inside the sprinkler) is I and it is rotating with angular velocity Ω, and d. the maximum angular velocity, Ωmax, of the sprinkler if no force is applied V F l Ω θ pipe diameter, d θ V TOP VIEW R SIDE VIEW Q C. Wassgren Chapter 03: Integral Analysis 104 Last Updated: 25 Aug 2008 SOLUTION: First consider the case where the sprinkler does not rotate (Ω = 0). For the fixed frame of reference and control volume shown, the angular momentum equation is: d (3.145) ( r × u ) ρ dV + ( r × u )( ρ u rel ⋅ dA ) = M S + M B ˆ eθ dt ∫ where d dt ∫ CV CS ∫ ( r × u ) ρ dV = 0 top view (steady flow) ˆ er CV ∫ ( r × u ) ( ρurel ⋅ dA ) = CS πd2 ˆ ˆ 2 Rer × V ( sin θ e r + cos θ eθ ) ρV ˆ 4 two arms = 2 ρ RV 2 cos θ πd2 ˆ ez 4 (Neglect the small bend in the pipe end when estimating the radius, r.) ˆ ˆ ˆ M S = le r × Feθ = Fle z MB = 0 Substitute and simplify: 2 ρ RV 2 cos θ ∴ F = 2 ρV 2 πd2 4 ˆ ˆ e z = Fle z πd2 R cos θ 4 l (3.146) Note that from conservation of mass on the same control volume: Q = 2V πd2 4 ⇒ V= 2Q (3.147) πd2 C. Wassgren Chapter 03: Integral Analysis 105 Last Updated: 25 Aug 2008 If the sprinkler is rotating, then use the same control volume (attached to and surrounding the sprinkler arms) but use a frame of reference that rotates with the control volume. The angular momentum equation for a rotating frame of reference is: d dt ∫ (r × u ) xyz ρ dV + CV ∫ (r × u) xyz ( ρ u rel ⋅ dA ) = M S + M B − CS ∫r ( ) xyz × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ρ dV CV where d dt ∫ (r × u ) xyz ρ dV = 0 (the flow is steady in the rotating frame of reference) CV ˆ ˆ ˆ ∫ (r × u )xyz ( ρurel ⋅ dA ) = 2 Rer ×V ( sin θ er + cosθ eθ ) ρV CS πd2 = 2 ρV 4 2 πd2 4 ˆ R cos θ e z (Neglect the small bend in the pipe end when estimating the radius, r.) ˆ ˆ ˆ M S = ler × Feθ = Fle z MB = 0 ∫r xyz ( ) × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ω xyz / XYZ × rxyz ρ dV = CV πd2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 re r × Ωe z × rer + 2Ωe z × Ver + Ωe z × ( Ωe z × re r ) ρ dr 4 =Ωre ˆ r =0 = 2 ΩVeθ ˆ ˆθ =Ωreθ ˆ =−Ω2 re r r=R ∫ R =2 ˆ ∫ ( Ωr e ) ˆ + 2ΩVre z ρ 2 z πd2 0 4 R ∫ ˆ dr = Ωe z 2 ρ 0 πd2 4 r 2 dr + ρ πd2 4 R ∫ ˆ 4ΩVe z rdr 0 =I ˆ = I Ωe z + 2 ρV πd2 4 ˆ ΩR 2 e z where I is the sprinklers moment of inertia. Note that the small, bent portion of the pipe has been neglected in the control volume integral. Substitute and simplify. 2 ρV 2 πd2 ˆ ˆ ˆ ˆ R cos θ e z = − I Ωe z + 2ΩR 2 ρV e z + Fle z 4 4 πd2 I Ω = Fl − 2 ρV πd2 4 R (V cos θ + ΩR ) (3.148) From Eqn. (3.148) the force required to maintain the sprinkler at a constant angular velocity is: F = 2 ρV πd2 R 4 l (V cos θ + ΩR ) (Note that if Ω = 0 this simplifies to Eqn. (3.146).) (3.149) If the force, F, is removed, the angular acceleration is: Ω = −2 ρV πd2 R 4 I (V cos θ + ΩR ) (3.150) The maximum angular velocity of the sprinkler if no force is applied is: V cos θ Ωmax = − R C. Wassgren Chapter 03: Integral Analysis 106 (3.151) Last Updated: 25 Aug 2008 Now let’s solve the same problem but using a fixed frame of reference as shown in the figure below. Consider the sprinkler in an arbitrary orientation. φ F ˆ eY l V θ φ φ ˆ eX top view d dt ∫ ( r × u ) ρ dV + ∫ ( r × u )( ρu CV rel ⋅ dA ) = M S + M B CS where d dt ∫ r=R d ˆ ˆ ˆ ˆ ˆ 2 r ( cos φ e X + sin φ eY ) × V ( cos φ e X + sin φ eY ) + Ωr ( − sin φ e X dt r =0 r=R d rV ( cos φ sin φ e − sin φ cos φ e ) + Ωr 2 cos 2 φ e + sin 2 φ e ˆZ ˆZ ˆZ ˆZ = 2 dt r =0 r=R d π d 2 r 2 dr ˆZ 2 ρ = Ωe 4 dt r =0 =I d ˆ = {I Ωe Z } dt ˆ = I Ωe Z ∫{ ( r × u ) ρ dV = CV ( ∫ ˆ + cos φ eY ) ρ π d 2 dr 4 ρ π d 2 dr 4 } ) ∫ ∫ ( r × u ) ( ρu ( (3.152) } ρV π d 2 ) 4 { ˆ ˆ ˆ ˆ ˆ ˆ rel ⋅ dA ) = 2 R ( cos φ e X + sin φ eY ) × V sin (θ − φ ) e X + cos (θ − φ ) eY + ΩR [ − sin φ e X + cos φ eY ] CS { } ˆ ˆ ˆ ˆ = 2 R V cos φ cos (θ − φ ) e Z − sin φ sin (θ − φ ) e Z + ΩR cos 2 φ e Z + sin 2 φ e Z ρV π d 2 4 (3.153) = 2 ρV π d 2 R V cos (φ + θ − φ ) + ΩR e Z ˆ 4 ˆ = 2 ρV π d 2 R (V cos θ + ΩR ) e Z 4 ˆ ˆ ˆ ˆ M S = l ( cos φ e X + sin φ eY ) × F ( − sin φ e X + cos φ eY ) ( ˆ ˆ = Fl cos 2 φ e Z + sin 2 φ e Z ) (3.154) ˆ = Fle Z MB = 0 Substitute and simplify (considering only the Z-direction). I Ω + 2 ρV πd2 4 R (V cos θ + ΩR ) = Fl C. Wassgren Chapter 03: Integral Analysis (3.155) This is the same result as Eqn. (3.148)! 107 Last Updated: 25 Aug 2008 Example: A pipe branches symmetrically into two legs of length, L, and the whole system rotates with angular speed, ω, around its axis. Each branch is inclined at angle, α, to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at the volume flow rate Q. The pipe diameter, D, is much smaller than L. a. Obtain an expression for the external torque required to turn the pipe. b. What additional torque would be required to impart angular acceleration ω ? V r L D Q α z frame of reference rotates with the arms ω V SOLUTION: Apply the angular momentum equation to the CV shown above using a frame of reference rotating with the arms (this is an accelerating frame of reference). Let V be the velocity of the fluid in the pipe arms (V πD2/4= 1/2Q). d ∫ ( rxyz × u xyz ) ρ dV + CS ( rxyz × u xyz ) ( ρ urel ⋅ dA ) ∫ dt CV = MB + MS − ∫r xyz × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ( ω xyz / XYZ × rxyz ) ρ dV CV where d ∫ ( rxyz × u xyz ) ρ dV = 0 (steady problem in the given frame of reference) dt CV ∫ (r xyz × u xyz ) ( ρ u rel ⋅ dA ) = CS 0 ˆ ˆ ˆ ˆ 2 + L ( cos α e z + sin α e r ) × V ( cos α e z + sin α e r ) ( 1 ρ Q ) incoming flow top arm ˆ ˆ ˆ ˆ 2 + L ( cos α e z − sin α e r ) × V ( cos α e z − sin α e r ) ( 1 ρ Q ) bottom arm =0 M B = 0 (neglect gravity) M S = T (this is the torque we must apply to the rotating arm) C. Wassgren Chapter 03: Integral Analysis 108 Last Updated: 25 Aug 2008 ∫r xyz × ω xyz / XYZ × rxyz + 2ω xyz / XYZ × u xyz + ω xyz / XYZ × ( ω xyz / XYZ × rxyz ) ρ dV = CV ω e z × s ( cos α e z + sin α e r ) + ˆ ˆ ˆ π D2 ˆ ˆ ˆ ˆ ˆ ∫0 s ( cos α e z + sin α er ) × 2ωe z × V ( cos α e z + sin α er ) + ρ 4 ds + s= ˆ ˆ ω e z × ω e z × s ( cos α e z + sin α e r ) ˆ ˆ s=L top arm ωe z × s ( cos α e z − sin α er ) + ˆ ˆ ˆ s=L π D2 ˆ ˆ ˆ ˆ ˆ s ( cos α e z − sin α e r ) × 2ωe z × V ( cos α e z − sin α e r ) + ds ρ ∫0 4 s= ˆ ˆ ωe z × ωe z × s ( cos α e z − sin α er ) ˆ ˆ bottom arm s= L =ρ =ρ =ρ =ρ πD 2 4 π D2 4 π D2 4 πD 2 ˆ ˆ ˆ ˆ ˆ ∫ s ( cos α e z + sin α e r ) × sω sin α eθ + 2ωV sin α eθ − sω sin α er ds + s=0 s= L s cos α e − sin α e × − sω sin α e − 2ωV sin α e + sω 2 sin α e ds ˆz ˆr ) ˆθ ˆθ ˆr ∫ ( s=0 2 2 2 2 L ˆ ˆ ˆ ˆ ˆ − s ω sin α cos α e r + s ω sin α e z − 2sωV sin α cos α e r + 2sωV sin α e z − s 2ω 2 sin α cos α eθ ∫ + s 2ω sin α cos α e + s 2ω sin 2 α e + 2sωV sin α cos α e + 2sωV sin 2 α e + s 2ω 2 sin α cos α e ds ˆr ˆz ˆr ˆz ˆθ 0 L ∫ 2s ω sin 2 2 ˆ ˆ α e z + 4sωV sin 2 α e z ds 0 2 2 ω L3 sin 2 α + 2ωVL2 sin 2 α e z ˆ 4 3 Combining together all of the terms in the angular momentum equation: 0 = T−ρ π D2 2 ω L3 sin 2 α + 2ωVL2 sin 2 α e z ˆ 4 3 π D2 2 ω L3 sin 2 α + 2ωVL2 sin 2 α e z ˆ 4 3 or, since V πD2/4= 1/2Q: T=ρ π D2 ˆ ω L3 sin 2 α + ρ Qω L2 sin 2 α e z T = ρ 6 The torque required to turn the pipe at a constant angular velocity, ω, is: ˆ T = ρ Qω L2 sin 2 α e z The additional torque required to impart an angular acceleration ω is: Tadditional = ρ π D2 6 ˆ ω L3 sin 2 α e z C. Wassgren Chapter 03: Integral Analysis 109 Last Updated: 25 Aug 2008 6. Basic Thermodynamic Definitions Before beginning our discussion of conservation of energy, we need to first review some basic thermodynamic definitions. system and surroundings A system is a particular quantity of matter chosen for study. The surroundings include everything that is not the system. control volume (CV), control surface (CS), and outward-pointing unit normal vector A control volume (CV) is a particular volume or region in space. A control surface (CS) is the surface enclosing the control volume. The orientation of the CS at a particular location is given by ˆ the direction of its outward-pointing unit normal vector, n , at that location. The outward-pointing unit normal vector has a magnitude of one, is perpendicular to the control surface, and always points out of the CV. ˆ n ˆ ˆ n n ˆ n ˆ n CV ˆ n CS ˆ n ˆ n ˆ n ˆ n ˆ n property (extensive, intensive, and specific) Properties are macroscopic characteristics of a system. An extensive property is one that depends on the amount of mass in the system. An intensive property is one that is independent of the mass in the system. A specific property is an extensive property per unit mass (a specific property is also an intensive property). Some examples: mass, m, is an extensive property, pressure, p, is an intensive property, and specific volume, v≡V/m, is a specific property. An easy way to determine whether a property is extensive or intensive is to divide the system into two parts and see how the property is affected. state The state of a system is the system’s condition or configuration as described by its properties in sufficient detail so that it is distinguishable from other states. Often a state can be described by a subset of the system’s properties since the properties themselves may be related. process A process is a transformation from one state to another. A few common processes include: isothermal process: A process that occurs at constant temperature. isobaric process: A process that occurs at constant pressure. isochoric or isometric process: A process that occurs at constant volume. adiabatic process: A process in which there is no heat transfer between the system and surroundings. isentropic process: A process that occurs at constant entropy. path A process path is the series of states that a system passes through during some process. C. Wassgren Chapter 03: Integral Analysis 110 Last Updated: 25 Aug 2008 cycle A cycle is a sequence of processes that begins and ends at the same state. At the conclusion of a cycle, all properties have the same values they had at the beginning of the cycle. Thus, there is no change in the system’s state at the end of a cycle. equilibrium An equilibrium state is one in which there are no unbalanced potentials (or driving forces) within the system driving the system to another state. Note that there are many types of equilibrium, e.g. thermal, mechanical, chemical, and phase equilibriums. quasi-equilibrium process A quasi-equilibrium process is one where the process proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all times. One can interpret a quasiequilibrium process as occurring slowly enough so that the system has time to adjust internally such that properties in part of the system do not change any faster than those properties in other parts of the system. reversible and irreversible processes A reversible process is one in which the system is in a state of equilibrium at all points in its path. In a reversible process, the system and the surroundings can be restored exactly to their initial states. An irreversible process is one where the system is not in a state of equilibrium at all points in its path. The system and surroundings cannot be returned to their exact initial states in an irreversible process. Note that all natural processes are irreversible. Several effects causing irreversibility include viscosity, heat conduction, and mass diffusion. equation of state An equation of state is a relationship between properties of a particular substance or class of substances. Equations of state cannot be obtained from thermodynamics but are obtained either from experimental measurements or from some molecular model. Note that there can be various types of equations of state, e.g. two equations of state for an ideal gas include a thermal equation of state which is the ideal gas law, p=ρRT , and a caloric equation of state which describes the relationship between the internal energy and temperature, du=cvdT. C. Wassgren Chapter 03: Integral Analysis 111 Last Updated: 25 Aug 2008 7. Discussion of Energy, Work, and Heat Now let’s move our discussion to the three basic thermodynamic concepts of energy, work, and heat. Energy The energy associated with some phenomenon is not a physical quantity but is, in fact, just a number resulting from a formula containing physically measurable quantities related to that phenomenon. For example, the energy associated with the macroscopic motion of a system of mass, m, moving with a speed, V, is equal to 1/2mV 2. By itself, the energy associated with a phenomenon is not a very useful quantity. However, experiments examining the total energy of a system, i.e. the sum of all the various energies, have resulted in a very remarkable observation. When the system does not interact with its surroundings, the total energy of the system remains constant. The energy associated with a particular phenomenon may change; however, it can only change at the expense of the energy associated with some other phenomenon. We’ll examine this observation in greater detail a little later but for now we will define the various types of energy that are most common in fluid flows. Kinetic Energy, KE The energy associated with the macroscopic motion of a system relative to a reference frame xyz is known as the kinetic energy, KE: 2 KE = 1 2 mVxyz (3.156) where m is the mass of the system and Vxyz is the velocity of the system in the reference frame, xyz. Potential Energy, PE The energy associated with a system’s ability to do work in an external force field such as a gravity field is known as the potential energy, PE. For example, the gravitational potential energy for a mass, m, located in a gravitational field with gravitational acceleration, g, pointing in the –z-direction is: (3.157) PE = mgz where z is the height of the mass above some reference plane. Internal Energy, U The internal energy of a system is comprised of a number of sub-classes of energy which include: a. sensible energy This is the energy associated with the internal molecular translational, rotational, and vibrational motion. Temperature is a measure of this type of internal energy. The larger the temperature of a system, the greater its sensible energy. b. latent energy This is the energy associated with the attraction between molecules. We concern ourselves with latent energy most often when examining processes that involve a change of phase such as going from a solid to a liquid or from a liquid to a gas (or vice versa). c. chemical energy This is the energy associated with the attraction between atoms. d. nuclear energy This is the energy associated with the attraction between particles within an atom such as the attraction between protons and neutrons. There are other forms of internal energy (e.g., the energy associated with electric and magnetic dipole moments) but we rarely encounter these in typical engineering applications. In these notes we’ll only concern ourselves with sensible energy. C. Wassgren Chapter 03: Integral Analysis 112 Last Updated: 25 Aug 2008 The total energy of a system is the sum of these various types of energy: E = U + KE + PE (3.158) Note that the total, internal, kinetic, and potential energies are extensive properties, i.e., the magnitude of these energies depends on the system mass. In terms of specific quantities (extensive properties per unit mass) we have: e = u + 12V 2 + G (3.159) where e is the specific total energy, u is the specific internal energy, V is the velocity magnitude, and G is a (conservative) potential energy function. Note that the force per unit mass resulting from a conservative potential energy function is found by taking the negative of its gradient, i.e. if G = gz where g is the gravitational acceleration and z is the height of the system above some reference plane, then fgravity = –∇G = –g. The values of the specific internal energy, u, at different states for various substances are tabulated in thermodynamic property tables. Most introductory thermodynamics books have such tables for steam, refrigerants, and a variety of gases. Of particular interest in these notes is the specific internal energy for ideal gases and incompressible substances. We’ll examine how to evaluate u for these substances a little later. C. Wassgren Chapter 03: Integral Analysis 113 Last Updated: 25 Aug 2008 Work Work is an energy interaction (a way to transfer energy) occurring at the boundary between a system and its surroundings. Thus, work is not a property of a system but rather is associated with a process that the system is undergoing. The work done on the system by its surroundings depends on the path of the process. A quantity that is also commonly encountered when discussing work is the power, defined as the work done per unit time. The work done on a system is equal to the dot product of the force acting on the system, Fon system, and the distance over which the force acts, ds: dWon system = Fon system ⋅ ds (3.160) small amount of work on the system force acting on the system small distance over which the force acts s2 ⇒ ∫ W1→2 = F ⋅ ds (3.161) s1 where W1→2 is the total work in going from state 1 (indicated by s1) to state 2 (s2). Note that the work depends on the path taken from s1 to s2. For example, consider the situation shown below: Example: A block with weight, w, is pushed on a frictional surface. The friction coefficient between the block and the surface is µ. Determine the amount of work done by the friction force on the block when moving the block from state 1 to state 2 using the paths shown. 1 1 2 2 L L (a) (b) SOLUTION: The work done by the friction force, µw, in case (a) is: W1→2 = − µ w ( L ) (note that the friction force is in the direction opposite to the displacement) The work done by friction in case (b) is: W1→2 = − µ w ( 3L ) C. Wassgren Chapter 03: Integral Analysis 114 Last Updated: 25 Aug 2008 Types of Work Now let’s consider a few different types of work that can be done by or on a system. The types of work we’ll present here include work due to gravity, acceleration, pressure, electricity, springs, and rotating shafts. In the drawings below, the system is enclosed by a dashed line. Gravitational Work (aka Potential Energy) Consider the work required to move an object with mass, m, to a higher elevation in a gravity field (assume a quasi-static process so that accelerations can be neglected): ∆h 2 F g object with weight, mg ∆h z Won system,1→2 = ∫F on system 0 ∆h = 1 ˆ ∫ ( mge z ⋅ ds ˆ ) ⋅ ( dze z ) 0 F ∴Won system,1→ 2 = mg ∆h (3.162) This is just the change in the potential energy of the system! Note that the work on the surroundings is equal to, but has the opposite sign, of the work done on the system. Acceleration Work (aka kinetic energy) Consider the work required to accelerate an object with mass, m, from velocity, V1, to velocity, V2: 1 F F 2 ∫ V2 V1 x 2 Won system,1→2 = Fon system ⋅ ds 1 V2 dV ˆ ˆ m = e x ⋅ Vdt e x dt = dx V1 Newton's 2nd law object with mass, m ∫ V2 ∫ = m VdV V1 ∴Won system,1→2 = 1 2m (V 2 2 − V12 ) (3.163) This is just the change in the kinetic energy of the system! C. Wassgren Chapter 03: Integral Analysis 115 Last Updated: 25 Aug 2008 Pressure Work Consider the work done by the expansion of a fluid (a gas or liquid) in a piston: 2 ∫ Won surr,1→2 = Fon surr ⋅ ds 12 fluid 1 x2 F = ˆ ∫ ( pAe x ˆ ) ⋅ ( dxe x ) x1 x2 dx x = ∫ pAdx x1 pressure force the system exerts on the face of the piston: F = pA where p is the pressure at the piston face and A is the area of the piston face V2 ∴Won surr,1→2 = ∫ pdV (3.164) V1 Note that dV=Adx. Note also that in this example, the work on the surroundings has been calculated instead of the work acting on the system. To get the work done on the system, we simply have: Won system = -Won surr If we plot how the pressure changes with volume we get a p-V diagram: p The area under the curve is equal to the work done by path 2 p2 the fluid on the surroundings in going from state 1 to 1 area = pdV state 2. p 1 V2 Won surr,1→2 = V1 V V2 ∫ pdV V1 dV Note that different paths from state 1 to state 2 will give different amounts of work: p p 2 2 1 1 V V (a) (b) W1→2(a) > W1→2(b) One example of a pressure-volume relationship is known as a polytropic process where the pressure and volume are related by: pV n = constant (3.165) C. Wassgren Chapter 03: Integral Analysis 116 Last Updated: 25 Aug 2008 Example: A gas in a piston assembly undergoes a polytropic expansion from an initial volume, Vi=0.1 m3, and initial pressure, pi=2 bar (1 bar = 1*105 Pa), to a final volume of Vf=0.5 m3. Determine the work the gas does on the piston for n=1.5 and n=1 (where pV n=constant). SOLUTION: The work the gas performs on the piston is given by: V = 0.5 m3 ∫ Wi → f = (3.166) pdV V = 0.1 m 3 where, for a polytropic expansion, pV n = constant=c where n is a constant. Substitute Eqn. (3.167) into Eqn. (3.166). c 1− n 0.5 m3 V = 0.5 m3 V n ≠1 0.1 m3 Wi → f = cV − n dV = 1 − n c ln V 0.5 m3 V = 0.1 m3 n =1 0.1 m3 ∫ (3.167) (3.168) When n = 1.5, the constant is ( )( c = 1*105 Pa * 0.1 m3 = pi ) 1.5 = 3.2 *103 N ⋅ m-0.5 (3.169) =Vi and the work performed by the gas, using Eqn. (3.168), is: Wi → f = 3.2*103 N ⋅ m-0.5 3 0.5 m −0.5 ( ) −0.5 ( − 0.1 m3 ) −0.5 ∴Wi → f = 1.1*104 N ⋅ m When n = 1, the constant is: ( )( (3.170) ) c = 1*105 Pa * 0.1 m3 = 1*104 N ⋅ m = pi (3.171) =Vi and the work performed by the gas, using Eqn. (3.168), is: 0.5 m3 Wi → f = 1*104 N ⋅ m ln 0.1 m3 ( ) ∴Wi → f = 1.6 *104 N ⋅ m C. Wassgren Chapter 03: Integral Analysis (3.172) 117 Last Updated: 25 Aug 2008 Electric Work Electrons moving across a system boundary can do work on a system since in an electric field, a force acts on an electron. When N Coulombs of electrons pass through a potential difference, V (the voltage), the electric work done on the system is: Won system = NV (3.173) The corresponding power is: Won system = VN = VI = I 2 R = V (3.174) 2 R where I is the current and R is the resistance of the system. Note that Ohm’s Law (V=IR) has been used in deriving the last two expressions on the right hand side of Eqn. (3.174). Example: A 12 V automotive battery is charged with a constant current of 1.5 A for 3 hrs. Determine the work done on the battery. SOLUTION: The work done on the battery is: t =T Won battery = ∫ VIdt = VIT = (12 V )(1.5 A ) ( 3 hr ⋅ 3600 s hr ) t =0 ∴Won battery = 0.2 kJ Spring Work Now let’s examine the work required to compress a spring with stiffness, k: 2 k ∫ Won system,1→ 2 = Fon system ⋅ ds F k 1 x2 F = ˆ ∫ ( −kxe x ˆ ) ⋅ ( dxe x ) x1 x x2 x1 ∴Won system,1→ 2 = 1 2k (x 2 1 2 − x2 ) (3.175) Note k is assumed constant in the Eqn. (3.175). Shaft Work Another method of transferring energy between a system and the surroundings is through shaft work. Shaft work is most often associated with rotating fluid machines such as compressors, pumps, turbines, fans, propellors, and windmills. The power acting on a system is given by: Won system = Ton system ⋅ ω (3.176) where Ton system is the torque acting on the system (assumed constant here) and ω is the angular velocity of the shaft. C. Wassgren Chapter 03: Integral Analysis 118 Last Updated: 25 Aug 2008 Heat Heat is another form of boundary energy interaction occuring between a system and its surroundings. The difference between heat and work is that heat transfer occurs due to differences in temperature. Heat moves from regions of high temperature to regions of low temperature. Like work, heat is not a property of a system but rather is associated with a process. Furthermore, the amount of heat transferred during a process depends on the path taken during the process. Modes of Heat Transfer Heat can be transferred between the system and surroundings via three methods: conduction, convection, and radiation. Conduction Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. The rate of heat transfer, Q , (this is a vector quantity since the heat travels in a particular direction) due to conduction through an area, A, of a substance is given by Fourier’s Law of Heat Conduction: Q = −kA∇T (3.177) where k is a material property of the substance known as the thermal conductivity, and ∇T is the thermal gradient in the substance. Note that the negative sign in the equation is required so that heat moves from regions of higher to lower temperature. Convection Convection is the mode of energy transfer between and solid surface and an adjacent fluid that is in motion; it involves the combined effects of conduction and relative fluid motion (also known as advection). The rate of heat transfer, Q , leaving a surface with area, Asurface, and entering the fluid due to convection is given by Newton’s Law of Cooling: Q = hAsurface (Tsurface − Tfluid ) (3.178) where h is the heat transfer coefficient for the situation, Tsurface is the temperature of the surface which is in contact with the fluid with temperature, Tfluid. The heat transfer coefficient will depend on the surface and fluid properties as well as the flow characteristics. It is generally an experimentally determined property for all but the simplest flow situations. Radiation Radiation is the energy emitted by matter in the form of electromagnetic waves as a result of changes in the electronic configurations of the atoms or molecules. Unlike conduction and convection, radiation does not require an intervening medium. The rate at which heat is emitted from a surface with area, Asurface, depends on the absolute temperature of the surface, Tsurface, as indicated by the modified Stefan-Boltzmann Law: 4 Qemitted = εσ AsurfaceTsurface (3.179) where Qemitted is the rate at which heat is emitted from the surface, ε is the emissivity of the surface (0≤ε≤1), and σ is the Stefan-Boltzmann constant (σ=5.67*10-8 W/(m2⋅K4)=0.1714*10-8 Btu/(hr⋅ft2⋅°R4)). A blackbody is an object with an emissivity of one: εblackbody=1, i.e. a blackbody is a perfect emitter of radiation. C. Wassgren Chapter 03: Integral Analysis 119 Last Updated: 25 Aug 2008 Surfaces can also absorb radiation. The heat flux absorbed by a surface via radiation is given by: Qabsorbed = α Qincident (3.180) where α is the absorptivity of the surface (0≤α≤1). Note that a blackbody has α=1; it is both a perfect emitter and absorber of radiation. Actual determination of the rate at which radiation is emitted and absorbed by a surface can be very complicated since the rate depends on factors such as surface orientation, the effects of the intervening medium, and the surface spectral characteristics. For the special case where a small surface interacts with a much larger surface, the intervening fluid has no affect on the radiation transfer, and α=ε (termed a grey body), the rate of heat transfer from the surface to the surroundings via radiation is: ( 4 4 Qemitted = εσ Asurface Tsurface − T∞ C. Wassgren Chapter 03: Integral Analysis ) (3.181) 120 Last Updated: 25 Aug 2008 Example: An insulated frame wall of a house has an average thermal conductivity of 0.0318 Btu/(hr⋅ft⋅°R). The thickness of the wall is 6 in. At steady state, the rate of energy transfer by conduction through an area of 160 ft2 is 400 Btu/hr, and the temperature decreases linearly from the inner surface to the outer surface. If the outside surface temperature of the wall is 30 °F, what is the inner surface temperature in °F? SOLUTION: A TC TH Qx k ∆x A Qx TC = = = = = 0.0318 Btu/(hr⋅ft⋅°R) 6 in = 0.5 ft 160 ft2 400 Btu/hr 30 °F = 490 °R ∆x x From Fourier’s Law, the heat transfer through the wall is: Qx = − kA dT T − TH = − kA C ∆x dx (assuming a linear temperature change) (3.182) Re-arrange to solve for TH. TH = TC + Q ∆x kA (3.183) Using the given parameters: k = 0.0318 Btu/(hr⋅ft⋅°R) ∆x = 6 in = 0.5 ft A = 160 ft2 Q = 400 Btu/hr TC = 30 °F = 490 °R ⇒ TH = 529 °R = 69 °F C. Wassgren Chapter 03: Integral Analysis 121 Last Updated: 25 Aug 2008 Example: A cartridge electrical heater is shaped as a cylinder of length 200 mm and outer diameter of 20 mm. Under normal operating conditions the heater dissipates 2 kW while submerged in a water flow which is at 20 °C and provides a convection heat transfer coefficient of 5000 W/(m2⋅K). Neglecting heat transfer from the ends of the heater, determine the heater’s surface temperature. If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air which is also at 20 °C but for which the heat transfer coefficient is 50 W/(m2⋅K). What is the corresponding surface temperature? What are the consequences of such an event? SOLUTION: Q D T∞ TS L Determine the surface temperature using Newton’s Law of Cooling. Q = hA (TS − T∞ ) (3.184) where A = πDL. Re-arranging gives: TS = T∞ + Q hA (3.185) Using the given parameters: T∞ = 20 °C = 293 K Q = 2000 W h = 5000 W/(m2⋅K) D = 20*10-3 m L = 200*10-3 m ⇒ A = 1.3*10-2 m2 TS = 325 K = 52 °C If instead, h = 50 W/(m2⋅K), then: = 3500 K = 3200 °C TS This temperature is probably large enough to melt the cartridge heater! C. Wassgren Chapter 03: Integral Analysis 122 Last Updated: 25 Aug 2008 Example: An uninsulated steam pipe passes through a room in which the air and walls are at 25 °C. The outside diameter of the pipe is 70 mm, and its surface temperature and emissivity are 200 °C and 0.8, respectively. If the coefficient associated with free convection heat transfer from the surface to the air is 15 W/(m2⋅K), what is the rate of heat loss from the surface per unit length of pipe? SOLUTION: The convective heat transfer rate, QC , is given by Newton’s Law of Cooling: QC = hAS ( TS − T∞ ) (3.186) where h is the convection heat transfer coefficient, AS is the surface area of the pipe, TS is the surface temperature of the pipe and T∞ is the ambient temperature. The radiative heat transfer rate, QR , is given by: QR = εσ AS (TS4 − T∞4 ) (3.187) where ε is the surface emissivity and σ is the Stefan-Boltzmann constant. The total heat transfer rate from the pipe is: QT = QC + QR (3.188) Using the given parameters: h = 15 W/(m2⋅K) D = 70*10-3 m ⇒ AS = πDL ⇒ AS/L = 0.22 m TS = 200 °C = 473 K = 15 °C = 288 K T∞ ε = 0.8 σ = 5.67*10-8 W/(m2⋅K4) ⇒ QC / L = 580 W/m QR / L = 420 W/m QT / L = 1 kW/m C. Wassgren Chapter 03: Integral Analysis 123 Last Updated: 25 Aug 2008 8. Conservation of Energy (COE) (aka 1st Law of Thermodynamics) Conservation of Energy for a System In words and in mathematical form, the first law of thermodynamics for a system is: The increase in total energy of a system is equal to the heat added to the system plus the work done on the system. δWon system system dEof system = δ Qinto system + δ Won system (3.189) δQinto system where dE is a small increase in the total energy of the system, δQinto system is a small amount of heat transferred into the system, and δWon system is a small amount of work done on the system by the surroundings. Notes: 1. Since energy is a property of a system, an exact differential (the “d” operator in dE) is used to specify the small change in the energy. In other words, the difference in energy between two states depends only upon the endpoint states and is independent of the path between the two states. The small change in heat and work are indicated using an inexact differential (the “δ” operator in δQ and δW) to signify that both heat and work are path dependent processes. 2. Note that different disciplines have different notations for COE. In particular, in thermodynamics work is usually discussed in terms of the work done by the system on the surroundings so that the 1st Law becomes: dEof system = δ Qinto system − δ Wby system (3.190) In order to avoid confusion regarding the proper sign for work, these notes will try to clearly specify whether work is being done on or by the system. Understanding that if one does work on a system, its energy will increase is generally sufficient to avoid most sign convention problems. 3. We can also write COE in terms of time rates of changes by taking the limit of the changes in the properties over a short amount of time as the time approaches zero: Eof system = Qinto system + Won system (3.191) Now let’s consider a few simple examples. C. Wassgren Chapter 03: Integral Analysis 124 Last Updated: 25 Aug 2008 Example: A rigid tank contains a hot fluid that is cooled while being stirred. Initially the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat and the stirring propeller does 100 kJ of work on the fluid. What is the final internal energy of the fluid? fluid stirring propeller rigid tank SOLUTION: Apply conservation of energy to the system of fluid contained within the tank. ∆Esys = E f − Ei = Qinto system + Won system where Ei = Ui = 800 kJ Qinto system = -500 kJ Won system = 100 kJ ⇒ Ef = Uf = 400 kJ C. Wassgren Chapter 03: Integral Analysis 125 Last Updated: 25 Aug 2008 Example: Four kilograms of a certain gas is contained within a piston-cylinder assembly. The gas undergoes a polytropic process where: pV1.5=constant. The initial pressure is 3 bars, the initial volume is 0.1 m3, and the final volume is 0.2 m3. The change in the specific internal energy of the gas in the process is ∆u = -4.5 kJ/kg. There are no significant changes in the kinetic or potential energies of the gas. What is the net heat transfer for the process? gas SOLUTION: Apply conservation of energy the system of gas as shown in the figure below. pA gas ∆Esys = Qadded + Won sys (3.192) to sys where V =V2 Won sys = ∫ V =V1 V =V2 − pdV = − p1V11.5 =c dV = 1.5 V =V1 V ∫ 1 0.5 p1V11.5 (V2−0.5 − V1−0.5 ) (c is the constant in pV1.5 = c) (3.193) and ∆Esys = msys ∆esys = msys ∆usys (The kinetic and potential energy changes are negligible.) (3.194) Re-arranging Eqn. (3.192) and substituting Eqns. (3.193) and (3.194) gives: 1 Qadded = msys ∆usys − 0.5 p1V11.5 (V2−0.5 − V1−0.5 ) (3.195) to sys Using the given values: msys = 4 kg ∆usys = -4500 J/kg p1 = 3*105 Pa V1 = 0.1 m3 V2 = 0.2 m3 ⇒ Qadded = -0.4 kJ (heat is leaving the system) C. Wassgren Chapter 03: Integral Analysis 126 Last Updated: 25 Aug 2008 Example: A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kJ/hr is maintained at 22 °C at all times during a winter night for 10 hr. The house is to be heated by 50 glass containers each containing 20 L of water that is heated to 80 °C during the day by absorbing solar energy. A thermostatcontrolled, 15 kW back-up electric resistance heater turns on whenever necessary to keep the house at 22 °C. a. How long will the electric heating system need to run during the night? b. How long would the electric heater run during the night if the house did not incorporate solar heating? heat loss of 50,000 kJ/hr house maintained at 22 °C 50 glass containers filled with 20 L of water each at an initial temperature of 80 °C SOLUTION: Apply conservation of energy to the house. Qinto house house maintained at 22 °C Won house ∆Esystem = Qinto + Won system (3.196) system The change in total energy of the house will consist of the change in the internal energy (potential and kinetic energy changes will be negligible). Furthermore, the total internal energy change will include the total energy change in the house structure, house air, and water tanks. ∆Esystem = ∆U system = ∆U house + ∆U air + ∆U water (3.197) structure Since the house structure and air are maintained at a constant temperature, ∆Uhouse = ∆Uair = 0. Hence, Eqn. (3.196) can be re-written as: ∆U water = Qinto + Won (3.198) system system C. Wassgren Chapter 03: Integral Analysis 127 Last Updated: 25 Aug 2008 The total change in the internal energy of the water (assuming an incompressible fluid) is given by: ∆U water = mwater cwater (T f ,water − Ti ,water ) (3.199) The total heat added to the house is: Qinto = ( −50, 000 kJ/hr ) (10 hr ) = −500,000 kJ (3.200) system and the total work done on the house by the electric heater is: Won = (15 kW ) ∆t (3.201) system where ∆t is the time over which the heater operates. Substitute Eqns. (3.199) – (3.201) into Eqn. (3.198). mwater cwater (T f ,water − Ti ,water ) = −500, 000 kJ+ (15 kW ) ∆t (3.202) Using the given parameters in Eqn. (3.202). mwater = 50(20 L)(0.001 m3/L)(1000 kg/m3) = 1000 kg cwater = 4.179 kJ/(kg⋅K) (from a thermodynamics table) Tf,water = 22 °C Ti,water = 80 °C ⇒ ∆t = 4.8 hrs Hence, the heater must be on for 4.8 hrs at night with the water tanks. If the water containers were not present, then the left-hand side of Eqn (3.202) would be zero (∆Uwater = 0) and: ⇒ ∆t = 9.3 hrs Hence, the heater must be on for 9.3 hrs at night without the water tanks. C. Wassgren Chapter 03: Integral Analysis 128 Last Updated: 25 Aug 2008 Conservation of Energy for a Control Volume To write COE for a control volume, we utilize the Reynolds Transport Theorem (RTT) to convert our system expression to a control volume expression. Let’s first rewrite Eqn. (3.191) using the Lagrangian derivative notation (we’re interested in how things change with respect to time as we follow a particular system of fluid) and write the total energy of a system in terms of an integral: D (3.203) e ρ dV = Qinto system + Won system Dt ∫ Vsys Esystem Applying the Reynolds Transport Theorem and noting that the system and control volume are coincident at the time we apply the RTT gives: d dt ∫ eρ dV + ∫ e ( ρu CV rel ⋅ dA ) = Qinto CV + Won CV (3.204) COE for a CV CS Note: e=u+1/2V2+G where G is a conservative potential energy function with the specific gravitational force given by fgravity = -∇G. For the remainder of these notes, G will be assumed to be G = gz (⇒ fgravity = -g) where g is the acceleration due to gravity which points in the negative z-direction. Now let’s expand the rate of work (power) term into rate of pressure work (pdV power), shaft power, and the power due to other effects such as viscous forces, electromagnetic forces, etc. Won CV = Wpressure,on CV + Wshaft,on CV + Wother,on CV (3.205) In particular, we can write the rate of pressure work term in the following way: urel dWpressure,on CV = dFon CV ⋅ u rel ˆ n ˆ = − pdAn ⋅ u rel pressure force, pdA = − p ( u rel ⋅ dA ) surface area, dA The rate of pressure work over the entire CS is: Wpressure, on CV = ∫ − p (u rel ⋅ dA ) (3.206) CS Equation (3.206) is the rate at which pressure work is performed on the fluid flux through the control surface. The rate at which pressure work is done on a moving solid boundary is included in the Wother term. Substituting Eqns. (3.206) and (3.205) into Eqn. (3.204), expanding the specific total energy term in the surface integral, and bringing the rate of pressure work term to the LHS gives: d e ρ dV + u + p ρ + 1 2 V 2 + gz ( ρ u rel ⋅ dA ) dt (3.207) ∫ CV ∫( ) CS = Qinto CV + Wshaft, on CV + Wother, on CV The quantity (u+p/ρ) shows up often in thermal-fluid systems and is given the special name of specific enthalpy, h: h ≡ u + p = u + pv specific enthalpy (3.208) ρ where v=V/m is the specific volume Note that just as with internal energy, tables of thermodynamic properties typically list the value of the specific enthalpy for various substances under various conditions. Substituting Eqn. (3.208) into Eqn. (3.207) gives: d e ρ dV + h + 1 2 V 2 + gz ( ρ u rel ⋅ dA ) = Qinto CV + Wshaft, on CV + Wother, on CV dt ∫ CV ∫( ) (3.209) CS C. Wassgren Chapter 03: Integral Analysis 129 Last Updated: 25 Aug 2008 Notes: 1. For a flow where the total energy within the CV does not change with time (a steady flow), we have: ∫ (h + 1 2V 2 ) + gz ( ρ u rel ⋅ dA ) = Qinto CV + Wshaft, on CV + Wother, on CV (3.210) CS Note that flows may be unsteady at the local level (e.g. detailed flow within a pump), but may be steady on a larger scale (e.g. the average conditions within the pump housing). 2. For a steady flow with a single inlet (call it state 1) and outlet (call it state 2) we can write COE as: (h + α 1 2V 2 + gz ) − (h + α 1 2 2V 2 + gz ) 1 = qinto CV + wshaft, on CV + wother, on CV (3.211) where q = Q m and w = W m (note that from COM the mass flow rate is a constant.) i. The quantity, α, is known as the kinetic energy correction factor. It is a correction factor accounting for the fact that the average velocity, V , does not contain the same kinetic energy as a non-uniform distribution. For example, consider the kinetic energy contained in the two flow profiles shown below: The average specific kinetic energy flux is: ke = ∫ 1 2 2V ( ρ V ⋅ dA ) ≠ 1 2 mV 2 A We define the kinetic energy correction factor, α: pipe with cross-sectional area, A ∫ V V α≡ The average velocity is found by: 1 V= VdA A 1 2V 2 ρ ( V ⋅ dA ) A 1 2 mV (3.212) 2 so that ∫ ke = α 1 2 mV 2 A For a laminar flow in a circular pipe, the velocity profile is parabolic resulting in α=2. For a turbulent flow, α→1 as the Reynolds number, defined as Re D ≡ ρVD µ , increases (suggesting that the velocity profile becomes more uniform as the Reynolds number increases). ii. The quantity: hT = h0 ≡ h + α 1 2 V 2 + gz (3.213) is referred to as the total specific enthalpy or the stagnation specific enthalpy. Note that for gases, the gz term is often much smaller than the other terms and thus is often neglected. iii. If the flow is adiabatic ( q = 0 ) and the rate of work by forces other than pressure can be neglected, then: hT = constant adiabatic flow with no shaft or other work (3.214) 3. Now let’s re-write Eqn. (3.211) but expand the specific enthalpy terms: u + p + α 1 V 2 + gz − u + p + α 1 V 2 + gz = q +w ( ρ 2 )( 2 ρ ) 2 1 into CV shaft, on CV + wother, on CV (3.215) Re-arranging terms and dividing through by the gravitational acceleration gives: C. Wassgren Chapter 03: Integral Analysis 130 Last Updated: 25 Aug 2008 p p V2 V2 +α + z = +α + z ρg 2g 2g 2 ρ g 1 − ( u2 − u1 − qinto CV ) (3.216) Wshaft, on CV Wother, on CV + + g mg mg Note that each term in this equation has the dimensions of length. The terms are also referred to as head quantities as defined below: V2 ≡ velocity head 2g p ≡ pressure head ρg z ≡ elevation head ( u2 − u1 − qinto CV ) head loss, H (head lost due to heating of the fluid) ≡ L g Wshaft on CV ≡ shaft head, HS (head added due to shaft work; recall that W = T ω ) mg Wother on CV ≡ other head, HO (head added due to other work on the fluid) mg The equation in this form is also known as the Extended Bernoulli Equation: p p V2 V2 +α + z = +α + z − H L + HS + H O ρg ρg 2g 2g 2 1 4. (3.217) Now let’s examine the “other” work term more closely. Specifically, let’s concern ourselves with the work done by viscous effects. Consider the rate of viscous work done on the CV shown below: urel ˆ n surface area, dA dWviscous,on CV = dFviscous on CV ⋅ u rel viscous force, dFviscous so that the total rate of viscous work acting on the CS is: Wviscous,on CV = ∫ dF viscous on CV ⋅ u rel CS i. Note that at a solid boundary, urel = 0 (due to the no-slip condition) so that the rate of viscous work is zero at solid surfaces. If the flow is inviscid then urel ≠ 0 but dFviscous = 0 and so the rate of viscous work is also zero. ii. If the control volume is oriented so that the velocity vectors are perpendicular to the normal vectors of the CS, then the rate of viscous work done on the CV will be zero. dFviscous on CV ⋅ u rel = 0 since the viscous force will be perpendicular to the velocity vector. iii. The rate of viscous work may not be negligible if the control volume is chosen as shown below: Viscous forces along streamline surfaces may be significant if the shear stress, τ, is large: Fv2 ∂u τ =µ ∂n Fv1 where n is the direction normal to the streamlines. streamlines C. Wassgren Chapter 03: Integral Analysis 131 Last Updated: 25 Aug 2008 5. The rate of work due to body and surface forces could have been written in a different form: Wbody, on CV = ∫ ( u ⋅ f B ) ρ dV CV Wsurface, on CV = ∫ ( u ⋅ f ) dA S CS where fB is the body force per unit mass acting on the CV and fS is the surface force per unit area. COE for a control volume using these relations is: d e ρ dV + e ( ρ u rel ⋅ dA ) = Qinto CV + ( u ⋅ f B ) ρ dV + ( u ⋅ f S ) dA + Wother, on CV dt ∫ ∫ ∫ ∫ CV CS CV CS where e=u+1/2V2. Note that the total energy does not include the potential energy since the potential energy is contained in the rate of work due to body forces term. The rate of gravitational work can be included in the LHS integral (i.e. we can write e=u+1/2V2+gz) assuming that the gravitational acceleration remains steady and is conservative (i.e. derivable from a potential function). Let’s examine a few examples. C. Wassgren Chapter 03: Integral Analysis 132 Last Updated: 25 Aug 2008 Example: Consider a large classroom on a hot summer day with 150 students, each dissipating 60 W of heat. All the lights, with 4.0 kW of rated power, are kept on. The room has no external walls, and thus heat transfer through the walls and the roof is negligible. Chilled air is available at 15 °C and the temperature of the return air is not to exceed 25 °C. Determine the required flow rate of air, in kg/s, that needs to be supplied to the room to keep the average temperature of the room constant. SOLUTION: Apply conservation of energy to a control volume that encloses the classroom. d e ρ dV + h + 1 2 V 2 + gz ( ρ u rel ⋅ dA ) minlet@15 C dt ∫( ∫ CV ) CS = Qinto CV + Wshaft, on CV + Wother, on CV where d dt ∫ eρ dV = 0 (steady state) Qlights Qstudents mreturn@25 CV ∫ (h + 1 2V 2 ) + gz ( ρ u rel ⋅ dA ) = − ( mh )inlet + ( mh )return = m ( hreturn − hinlet ) CS (Note that in steady operation: minlet = mreturn . In addition the changes in the airflow velocity and elevation will be negligible since the velocities and elevation differences are expected to be small.) Qinto CV = Qstudents + Qlights Wshaft, on CV = Wother, on CV = 0 Substitute and re-arrange. m ( hreturn − hinlet ) = Qstudents + Qlights m= Qstudents + Qlights (3.218) hreturn − hinlet Assuming air is a perfect gas: hreturn − hinlet = c p (Treturn − Tinlet ) (3.219) where cp is the specific heat of air at constant pressure. Note that if we don’t consider air as a perfect gas then we can look up the specific enthalpy of air (at 1 atm) in a thermodynamics table. Using the given data: Qstudents = 150 * 60 W = 9 kW Qlights = 4 kW Treturn = 25 C = 298 K Tinlet = 15 C = 288 K c p = 1000 J ( kg ⋅ K ) ⇒ m = 1.3 kg/s C. Wassgren Chapter 03: Integral Analysis 133 Last Updated: 25 Aug 2008 C Example: Determine the maximum pressure increase across the 10 hp pump shown in the figure. d2 = 1.5 in water V1=30 ft/s d 1 = 1 in pump 10 hp SOLUTION: Apply conservation of energy, in the form of Eqn. (3.217), across the pump. p p V2 V2 +α + z = +α + z − H L + HS + H O ρg ρg 2g 2g 2 1 where ∆p = p2 − p1 (This is the pressure rise we’re trying to determine.) 2 d V2 = V1 1 (from conservation of mass) d2 α 2 ≈ α1 ≈ 1 (assuming turbulent flow) z2 − z1 ≈ 0 (negligible elevation difference between the inlet and outlet) H L = 0 (no losses – this will give the maximum pressure rise across the pump) HS = WS (from the definition of shaft head) mg H O = 0 (there is no other work being performed other than shaft work) Substitute and re-arrange. 4 V12 WS ∆p V12 d1 + + = ρ g 2 g d2 2 g mg 4 ∆p V12 d1 WS 1 − + = ρ g 2 g d 2 mg C. Wassgren Chapter 03: Integral Analysis (3.220) 134 Last Updated: 25 Aug 2008 Using the given data: V1 = 30 ft/s d1 = 1.0 in = 8.3*10-2 ft d 2 = 1.5 in = 1.3*10-1 ft V2 = 13 ft/s ν H 2 0 = 1.1*10-5 ft 2 /s Re1 = V1d1 Re 2 = V2 d 2 m = ρV1 HS = ⇒ ν H 2 0 =230,000 (turbulent flow assumption ok!) ν H 2 0 =150,000 (turbulent flow assumption ok!) π d12 4 WS = mg ∆p ρg ( = 1.94 slug/ft 3 ) ( 30 ft/s ) (10 hp ⋅ 550 ft ⋅ lbf / (s ⋅ hp )) ( 3.8 slug / s ) ( 32.2 ft/s 2 ) ( π 8.3*10−2 ft 2 4 ) = 3.8 slug/s = 45 ft (Note: 1 lbf = 1 slug⋅ft/s2) = 56 ft ⇒ ∆p = 3500 psf = 24 psi C. Wassgren Chapter 03: Integral Analysis 135 Last Updated: 25 Aug 2008 Example: The velocity profile for a particular pipe flow is linear from zero at the wall to a maximum of uc at the centerline. Determine the average velocity and the kinetic energy correction factor. r uc R SOLUTION: The average velocity is found by setting the volumetric flow rate using the average velocity profile equal to the volumetric flow rate using the real profile. Qavg = Qreal (3.221) profile profile r =R ∫u uπ R2 = C r =0 r 1 − ( 2π rdr ) R r=R = 2π uC r2 r− ∫ R r =0 dr r =R 1 1 r3 = 2π uC r 2 − 3 R r =0 2 ∴ u = 1 uC 3 (3.222) The kinetic energy correction factor, α, is found by equating the kinetic energy flux using the average velocity with the kinetic energy flux using the actual velocity profile. α 1 2 ( ρuπ R2 ) u 2 = r =R ∫ 1 2 r =0 =m r r ρ uC 1 − R ( 2π rdr ) uC 1 − R 2 = dm r =R = ∫ 1 2 3 ρ uC 1 − r =0 r=R 3 = πρ uC 3 r ( 2π rdr ) R (3.223) 3 r ∫0 1 − R ( rdr ) r= where u = 1 uC . Solving the previous equation for α gives: 3 α= 27 = 2.7 10 C. Wassgren Chapter 03: Integral Analysis (3.224) 136 Last Updated: 25 Aug 2008 Example: Air, treated as an ideal gas, flows through the turbine and heat exchanger arrangement shown in the figure with the data for the flow streams also indicated. Heat transfer to the surroundings can be neglected, as can all kinetic and potential energy effects. Determine the temperature T3, in K, and the power output of the second turbine, in kW, at steady state. power out = 10,000 kW T1 T3 = ? p3 = p2 T2 = 1100 K p2 = 4 bars air T1 = 1400 K p1 = 12 bars T2 power out = ? T4 = 980 K p4 = 1 bar heat exchanger air T5 = 1480 K p5 = 1 bar mass flow rate = 1200 kg/min T6 = 1200 K p6 = p5 SOLUTION: Apply conservation of energy to each component. Assume steady, 1D flow with negligible heat transfer to the surroundings and negligible kinetic and potential energy changes. power out = 10,000 kW T1 T2 = 1100 K p2 = 4 bars air T1 = 1400 K p1 = 12 bars From conservation of energy: m12 ( h2 − h1 ) = Winto (3.225) CV where h1 = h2 = Winto = 1515.4 kJ/kg (from thermodynamics tables for air at T1 = 1400 K) 1161.1 kJ/kg (from thermodynamics tables for air at T2 = 1100 K) -10,000 kW CV and m12 is the mass flow rate at stations 1 and 2. Substitute and solve for the mass flow rate. m12 = 28.2 kg/s C. Wassgren Chapter 03: Integral Analysis 137 Last Updated: 25 Aug 2008 T3 = ? p3 = p2 T2 = 1100 K p2 = 4 bars heat exchanger air T5 = 1480 K p5 = 1 bar mass flow rate = 1200 kg/min T6 = 1200 K p6 = p5 From conservation of energy: m12 ( h3 − h2 ) = m56 ( h5 − h6 ) where h3 h2 m12 h6 h5 m12 = = = = = = (3.226) ? 1161.1 kJ/kg (from thermodynamics tables for air at T2 = 1100 K) 28.2 kg/s 1277.8 kJ/kg (from thermodynamics tables for air at T6 = 1200 K) 1611.8 kJ/kg (from thermodynamics tables for air at T5 = 1480 K) 1200 kg/min = 20 kg/s Substitute and solve for the enthalpy at station 3. h3 = 1398.0 kJ/kg ⇒ T3 = 1300 K (from interpolation of thermodynamics tables for air) T3 = ? p3 = p2 T2 power out = ? T4 = 980 K p4 = 1 bar From conservation of energy: m34 ( h4 − h3 ) = Won (3.227) CV where h4 h3 m34 = = = 1023.3 kJ/kg (from thermodynamics tables for air at T4 = 980 K) 1398.0 kJ/kg (from previous work) 28.2 kg/s (= m12 ) Substitute and solve for the mass flow rate. Won = −10,570 kW (work is extracted from the turbine) CV C. Wassgren Chapter 03: Integral Analysis 138 Last Updated: 25 Aug 2008 Example: In a proposed jet propulsion system for an automobile, air is drawn in vertically through a large intake in the roof at a rate of 3 kg/s, the velocity through this intake being small. Ambient pressure and temperature are 100 kPa (abs) and 30 °C. This air is compressed, heated, and then discharged horizontally out a nozzle at the rear of the automobile at a velocity of 500 m/s and a pressure of 140 kPa (abs). If the rate of heat addition to the air stream is 600 kW, find the nozzle discharge area and the thrust developed by the system. SOLUTION: The nozzle discharge area can be determined from the mass flow rate. Note that since the flow is steady, the outlet and inlet mass flow rates are the same. m0 = ρoVo Ao = mi (3.228) mi (3.229) ρoVo The outlet air density can be expressed in terms of the outlet pressure and temperature using the ideal gas law. m RTo Ao = i (3.230) Vo po Ao = The outlet temperature can be determined by applying COE to the following CV. mi , pi , Ti Qinto CV Vo , po F x d other ∫ eρ dV + CS hT ( ρ u rel ⋅ dA ) = Qinto + Won CV ∫ dt CV CV where d e ρ dV = 0 (steady flow) dt C∫ V ∫ h ( ρu T rel (3.231) ⋅ dA ) = mo ( ho + 1 Vo2 ) − mi ( hi ) 2 CS Note that: (1) since the flow is steady, mo = mi = m , (2) the potential energy differences between the inlet and outlet are negligible, and (3) the inlet kinetic energy is negligible. Qinto = Qinto (this quantity is given in the problem statement) CV CV Wother = 0 on CV Substitute and simplify. m ( ho + 1 Vo2 − hi ) = Qinto 2 (3.232) CV Assuming perfect gas behavior (∆h = cp∆T): Qinto V2 CV To = Ti + −o c p m 2c p C. Wassgren Chapter 03: Integral Analysis (3.233) 139 Last Updated: 25 Aug 2008 Using the given data: Ti = 30 °C = 303 K mi = 3 kg/s Qinto = 600 kW CV cp V0 ⇒ = 1005 J/(kg⋅K) = 500 m/s T0 = 378 K R To po V0 ⇒ ⇒ = = = = ρ0 Α0 287 J/(kg⋅K) 378 K 140 kPa 500 m/s = 1.29 kg/m3 = 4.6*10-3 m2 To determine the thrust, apply the linear momentum equation in the x-direction to the same control volume. d (3.234) ∫ ux ρ dV + CS ux ( ρ u rel ⋅ dA ) = FB, x + FS , x ∫ dt CV where d u x ρ dV = 0 dt C∫ V ∫ u ( ρu x rel ⋅ dA ) = Vo mo CS FB , x = 0 FS , x = F − ( po − patm ) Ao (the pressures are given in terms of absolute pressures) Substitute and simplify. Vo mo = F − ( po − patm ) Ao ∴ F = Vo mo + ( po − patm ) Ao (3.235) Using the given data: V0 = 500 m/s mo = 3 kg/s po = 140 kPa pi = 100 kPa Α0 = 4.6*10-3 m2 ⇒ F = 1670 N C. Wassgren Chapter 03: Integral Analysis 140 Last Updated: 25 Aug 2008 9. The Second Law of Thermodynamics While the first law of thermodynamics tells use that energy must be conserved during a process, it doesn’t tell us whether or not the process will actually occur. The Second Law of Thermodynamics states that processes occur in a certain direction. A process cannot take place unless it satisfies both the First and Second Laws of Thermodynamics. A detailed analysis of the Second Law is beyond the scope of these notes. A brief review of some of the more significant points will be given however. The Second Law of Thermodynamics can be stated in many ways. For example, two common and equivalent ways to state the 2nd Law are: Kelvin-Planck Statement of the 2nd Law: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. Clausius Statement of the 2nd Law: It is impossible to construct a device that operates on a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. An important corollary to the Clausius Statement is the Clausius Inequality: δ Qinto system ≤0 Clausius Inequality (3.236) T where Qinto system is the heat added to the system and T is the absolute temperature of the system. The equality sign holds strictly for reversible processes. This inequality is used in the derivation of the property known as entropy. Rather than show the derivation for this property, we will simply define it below. ∫ entropy, S Entropy, S, is an extensive system property defined in the following manner: δ Qinto system dSsystem ≡ T rev (3.237) where Q is the heat added to the system, T is the absolute temperature of the system, and the process by which the heat is added is reversible. The change in a system’s entropy as it undergoes a change of state is found by integrating Eqn. (3.237) : 2 S2 − S1 = ∫ 1 δ Qinto system T C. Wassgren Chapter 03: Integral Analysis (3.238) rev 141 Last Updated: 25 Aug 2008 Notes: 1. reversible and irreversible processes A reversible process is one in which the system is in a state of equilibrium at all points in its path. In a reversible process, the system and the surroundings can be restored exactly to their initial states. An irreversible process is one where the system is not in a state of equilibrium at all points in its path. The system and surroundings cannot be returned to their exact initial states in an irreversible process. Note that all natural processes are irreversible. Several effects causing irreversibility include viscosity, heat conduction, and mass diffusion. 2. Entropy is a system property so that the change in the entropy depends only on the endpoints of the process and not on the path taken during the process. Thus, the change in entropy between two states is independent of whether or not the process is reversible or irreversible. Equation (3.238) simply gives a method for calculating the entropy change using a reversible process. Note that if a process is irreversible, we have from the Clausius inequality (Eqn. (3.236)): ∫ δ Qinto system T 2 = ∫ δ Qinto system 1 T 1 + ∫ 2 δ Qinto system T ≤0 rev = S1 − S 2 2 S2 − S1 ≥ ∫ δ Qinto system (3.239) T 1 We can combine Eqns. (3.238) and (3.239) to give the general equation: 2 S2 − S1 ≥ ∫ δ Qinto system 1 (3.240) T or in differential form: δ Qinto system dSsystem ≥ T where the equality holds strictly for reversible process paths. (3.241) 3. A reversible, adiabatic process is also an isentropic process since dS=δQ/T (reversible) and δQ=0 (adiabatic) so that dS=0. 4. The value of entropy for various substances at various states can typically be found in thermodynamic property tables. 5. Statistical thermodynamics shows us that entropy can also be interpreted as a measure of the molecular disorder of a system. C. Wassgren Chapter 03: Integral Analysis 142 Last Updated: 25 Aug 2008 Second Law of Thermodynamics for a Control Volume As with COM, the LME, the AME, and COE, we will convert our system form of the 2nd Law to a control volume form using the Reynolds Transport Theorem. To do so, we should first write the 2nd Law in terms of time rates of change (refer to Eqn. (3.241)): DSsystem Qinto system ≥ (3.242) Dt T where the Lagrangian derivative notation has been used to remind us that we’re following a system. Now let’s write the equation in terms of an integral so that we can have variations in the properties of the system: δ qinto system D s ρ dV ≥ dV (3.243) Dt T ∫ ∫ Vsys Vsys where s is the specific entropy and δ qinto system is the rate of heat addition into the system per unit volume. The RHS of Eqn. (3.243) represents the sum of the heat transfer input to absolute temperature ratio for every particle in the system. After applying the Reynolds Transport Theorem to convert to a control volume perspective, we have: δ qinto CV d s ρ dV + s ( ρ u rel ⋅ dA ) ≥ dV (3.244) dt T ∫ ∫ ∫ CV CS CV where the equality holds strictly for reversible heat addition processes. C. Wassgren Chapter 03: Integral Analysis 143 Last Updated: 25 Aug 2008 Example: An inventor claims to have developed a device requiring no work input or heat transfer, yet able to produce steady state hot and cold air streams as shown in the figure. Evaluate this claim assuming the ideal gas model for air and ignoring kinetic and potential energy effects. air at 60 °C, 2.7 bars air at 20 °C, 3 bars air at 0 °C, 2.7 bars no heat or work addition SOLUTION: Apply conservation of mass, conservation of energy, and the 2nd Law to the control volume shown below. air at 60 °C, 2.7 bars 2 1 air at 20 °C, 3 bars 3 air at 0 °C, 2.7 bars no heat or work addition Conservation of Mass: d ∫ ρ dV + CS ρ u rel ⋅ dA = 0 ∫ dt CV where d ρ dV = 0 (steady flow) dt C∫ V ∫ ρu rel (3.245) ⋅ dA = m3 + m2 − m1 CS Substitute and re-arrange. m3 + m2 − m1 = 0 m3 m = 1− 2 m1 m1 C. Wassgren Chapter 03: Integral Analysis (3.246) 144 Last Updated: 25 Aug 2008 Conservation of Energy: d 2 on ∫ eρ dV + CS ( h + 12 V + gz ) ( ρ u rel ⋅ dA ) = Qinto + WCV ∫ dt CV CV where d e ρ dV = 0 (steady flow) dt C∫ V ∫ (h + 1 2 (3.247) V 2 + gz ) ( ρ u rel ⋅ dA ) = m3 h3 + m2 h2 − m1h1 CS Qinto = Won = 0 (no heat or work addition) CV CV Substitute and re-arrange. m3 h3 + m2 h2 − m1h1 = 0 m3 m h3 = h1 − 2 h2 m1 m1 Substitute Eqn. (3.246) into Eqn. (3.248) and simplify. m2 m2 h2 1 − h3 = h1 − m1 m1 (3.248) m2 h1 − h3 = m1 h2 − h3 (3.249) From thermodynamics tables for air at the given inlet and outlet temperatures: h1 = 293.2 kJ/kg (T1 = 20 °C = 293 K) h2 = 333.3 kJ/kg (T2 = 60 °C = 333 K) h3 = 273.1 kJ/kg (T3 = 0 °C = 273 K) Hence, m3 m2 = 0.334 and = 0.666 m1 m1 Note that the outgoing mass flow rates are both positive, consistent with the problem description. C. Wassgren Chapter 03: Integral Analysis 145 (3.250) Last Updated: 25 Aug 2008 2nd Law of Thermodynamics: δ qinto CV d ∫ s ρ dV + CS s ( ρ u rel ⋅ dA ) ≥ CV T dV ∫ ∫ dt CV where d s ρ dV = 0 (steady flow) dt C∫ V ∫ s ( ρu rel (3.251) ⋅ dA ) = m3 s3 + m2 s2 − m1 s1 CS δ qinto CV dV = 0 (no heat added to the control volume) T CV ∫ Substitute and re-arrange. ? m3 s3 + m2 s2 − m1 s1 ≥ 0 ? m3 m s3 + 2 s2 − s1 ≥ 0 m1 m1 m2 1 − m1 ? m2 s2 − s1 ≥ 0 s3 + m1 ? m ( s3 − s1 ) + 2 ( s2 − s3 ) ≥ 0 m1 (3.252) Note that for an ideal gas, the specific entropy at a given temperature and pressure can be determined (derived in Chapter 11) by: p s (TB , pB ) − s (TA , p A ) = s 0 (TB ) − s 0 (TA ) − R ln B (3.253) pA Substituting Eqn. (3.253) into Eqn. (3.252) gives: 0 0 p3 m2 0 0 p2 ? (3.254) s2 − s3 − R ln ≥ 0 s3 − s1 − R ln + p1 m1 p3 From thermodynamics tables for air at the given conditions: s01 = 1.6783 kJ/(kg⋅K) (T1 = 20 °C = 293 K) s02 = 1.8069 kJ/(kg⋅K) (T2 = 60 °C = 333 K) s03 = 1.6073 kJ/(kg⋅K) (T3 = 0 °C = 273 K) and from the given conditions: p1 = 3.0 bar p2 = 2.7 bar p3 = 2.7 bar Thus, Eqn. (3.254) simplifies to: 0.0259 kJ/ ( kg ⋅ K ) > 0 (3.255) nd and the 2 Law of Thermodynamics is satisfied. Since conservation of mass, conservation of energy, and the 2nd Law are all satisfied, the claim of the inventor is not unreasonable. C. Wassgren Chapter 03: Integral Analysis 146 Last Updated: 25 Aug 2008 Review Questions 1. What is meant by the Eulerian and Lagrangian perspectives? 2. Describe the Reynolds Transport Theorem in words? Why is it used? 3. State, both in words and in mathematics, the Lagrangian forms of conservation of mass, Newton’s 2nd Law, and conservation of energy. 4. Why is it important to draw a well-defined control volume when applying conservation of mass, the linear momentum equations, conservation of energy, or the 2nd Law of Thermodynamics? 5. What do each of the terms represent in the Lagrangian and Eulerian statements of conservation of mass? 6. Does conservation of mass depend upon the frame of reference? 7. Why is it important to draw a well-defined frame of reference when applying the linear momentum equations? 8. What does each of the terms represent in the Eulerian form of the linear momentum equation? 9. What restrictions are placed on the frame of reference when applying the LMEs? 10. In order to change the momentum of a flow, what must act on the flow? 11. Give examples of body and surface forces. 12. Explain what urel is. For what circumstances will urel and u be the same? 13. Why is the dot product used (urel⋅dA) in determining the flow rate out of a control volume? 14. Can one apply the non-inertial form of the LME to an inertial frame of reference? How about applying the inertial form of the LME to a non-inertial frame of reference? 15. What types of frames of reference can be considered inertial? Give examples of frames of reference that are not inertial. 16. Given the following where MCV is the mass in a control volume, m is a mass flow rate, and t is time: dM CV = −m dt will the following always be true (where M0 is the control volume mass at t = 0)? Explain your answer. M CV = M 0 − mt 17. Describe what each term represents in the Eulerian form of the angular momentum equation. 18. Why isn’t the intrinsic angular momentum of the fluid included in the angular momentum equation? 19. Consider a precessing, spinning top. Is its angular momentum conserved? 20. Describe what each term represents in the Eulerian form of conservation of energy. 21. What is meant by the term “adiabatic”? 22. Why are shear work terms often (but not always!) neglected in conservation of energy? 23. What is the definition of enthalpy? 24. In the following form of conservation of energy, where are the terms involving the work due to movement in a gravity field and the work due to pressure forces? d 2 on ∫ eρ dV + CS ( h + 12 V + gz ) ( ρ u rel ⋅ dA ) = Qinto + WCV ∫ dt CV CV 25. Describe what each term represents in the Eulerian form of the 2nd Law of Thermodynamics. 26. What is the definition of entropy in terms of heat and temperature? 27. What is meant by the term “reversible”? Give some examples of physical processes that result in irreversibility. 28. How are adiabatic, reversible processes related to isentropic ones? C. Wassgren Chapter 03: Integral Analysis 147 Last Updated: 25 Aug 2008 Chapter 04: Differential Analysis 1. Index Notation 2. Continuity Equation 3. Review of Stress 4. Momentum Equations 5. Fluid Element Deformations 6. Stress-Strain Rate Relations for a Newtonian Fluid 7. Acceleration in Streamline Coordinates 8. Euler’s Equations in Streamline Coordinates 9. Energy Equation 10. Entropy Equation 11. Vorticity Dynamics 12. Vorticity Transport Equation 13. Bernoulli’s Equation 14. Kelvin’s Theorem 15. Crocco’s Equation C. Wassgren Chapter 04: Differential Analysis 148 Last Updated: 05 Sep 2008 1. Index Notation (aka Tensor Notation) Index notation is a compact way of writing numbers and equations. Index notation is extremely useful for representing the equations of fluid mechanics, derivations, etc. Examples: 1. a1, a2, a3 can be written as: ai where i=1, 2, 3 2. A matrix of numbers can be written as: a11 a12 a13 a where i = 1, 2, 3 and j = 1, 2, 3 21 a22 a23 = aij a31 a32 a33 3. ai+bi represents three numbers: a 1+ b 1, a 2+ b 2, a 3+ b 3 4. Aij = Bij represents nine equations: A11=B11, A12=B12, A13=B13, A21=B21, …, A33=B33 5. Aijk=Bijk represents 27 equations: A111=B111, A112=B112, A113=B113, A211=B211, …, A333=B333 Free Indices A free index is an index that appears exactly once in a term. Each term in an equation must have the same free indices. A repeated index is one that appears twice in a term. No index may appear more than twice in a term. Examples: 1. aijk bj= cik CORRECT (two free indices: i and k, one repeated index: j) 2. aijbjk = clm INCORRECT (two free indices on each side of the equation (i, j and l, m) but they’re not the same!) 3. aijbj = cij INCORRECT (i is the only free index on the LHS while i and j are free indices on the RHS) C. Wassgren Chapter 04: Differential Analysis 149 Last Updated: 05 Sep 2008 Summation Convention If a subscript appears exactly twice in a term (i.e. it’s a repeated index), then summation over that subscript from 1 to 3 is implied. Examples: 3 1. aii = ∑a ii = a11 + a22 + a33 i =1 3 2. aij b j = ∑a b ij j = ai1b1 + ai 2 b2 + ai 3b3 j =1 3 3. Aij Bij = 3 ∑∑ A B ij ij = ( A11 B11 + A12 B12 + A13 B13 ) + + ( A31 B31 + A32 B32 + A33 B33 ) i =1 j =1 Notes: 1. Repeated indices are dummy indices: 2. No index may appear more than twice in a term: INCORRECT ai bi ci Aiii 3. aii = ajj = akk INCORRECT The summation convention is suspended by writing “no sum” or by underlining one of the repeated subscripts. σ ii ( no sum ) = σ 11 , σ 22 , σ 33 C. Wassgren Chapter 04: Differential Analysis 150 Last Updated: 05 Sep 2008 Kronecker’s Delta Kronecker’s Delta, δij, is the 2 index symbol defined by: 0 i ≠ j δ ij = 1 i = j Notes: 1. 2. δ12 = δ13 = δ 21 = δ 23 = δ 31 = δ 32 = 0 δ11 = δ 22 = δ 33 = 1 The Kronecker Delta is the identity matrix: 1 0 0 δ ij = 0 1 0 0 0 1 Examples: 1. Show that δ ij a j = ai . i = 1: δ1 j a j = δ11 a1 + δ12 a2 + δ13 a3 = a1 =1 =0 =0 i = 2 : δ 2 j a j = δ 21 a1 + δ 22 a2 + δ 23 a3 = a2 =0 =1 =0 i = 3 : δ 3 j a j = δ 31 a1 + δ 32 a2 + δ 33 a3 = a3 =0 2. =0 =1 Show that δ ii = 3 . δ ii = δ11 + δ 22 + δ 33 = 3 =1 =1 =1 C. Wassgren Chapter 04: Differential Analysis 151 Last Updated: 05 Sep 2008 Permutation (aka Alternating) Unit Tensor The permutation tensor, εijk, is the 3-index symbol defined as: +1 for ε123,ε 231,ε 312 ε ijk = −1 for ε 321,ε 213,ε132 0 for all other permutations Notes: 1. It’s convenient to remember the following pictures for determining the proper sign of the permutation tensor: 1 3 + 1 3 2 2 2. Switching any two indices changes the sign of the permutation tensor, e.g.: εijk = - εikj 3. A convenient identity: ε ijk ε ilm = δ jl δ km − δ jmδ kl Example: 1. Show that ε ijk ε ijk = 6 . ε ijk ε ijk = δ jj δ kk − δ jk δ kj = 9 − δ1k δ k1 + δ 2 k δ k 2 + δ 3k δ k 3 = 3 =3 =1 =1 =1 = 9−3 =6 C. Wassgren Chapter 04: Differential Analysis 152 Last Updated: 05 Sep 2008 Tensors A tensor of rank r is a quantity having nr components in n-dimensional space (e.g. a tensor of rank 2 in 3D has 32 = 9 components). The components of a tensor expressed in two different coordinate systems are related by: ′ Tijk…m = λis λ jt λku … λmvTstu…v where λis are the direction cosines between the e’i and es axes. Notes: 1. A tensor of rank 2 is often called a dyad, e.g. Aij (two free subscripts). 2. A tensor of rank 1 is called a vector, e.g. ai (one free subscript). 3. A tensor of rank 0 is called a scalar, e.g. c (zero free subscripts). 4. The vector notation for a dyad is often given as: A C. Wassgren Chapter 04: Differential Analysis 153 Last Updated: 05 Sep 2008 Basic Mathematical Operations Addition: Two tensors of equal rank can be added to yield a tensor of the same rank. Cij…k = Aij…k + Bij…k Multiplication: If a tensor, A, having rank, a, is multiplied by tensor, B, having rank, b, then a tensor, C, of rank, a+b, results. Cij…krs…t = Aij…k Brs…t Example: Aij Brs 2nd order 2nd order Transpose: = Cijrs 4th order The transpose of a tensor is given by: TijT…k = Tk… ji Example: T Aij = A ji If you represent the components of a 2nd order tensor in matrix form, then the transpose is equivalent to swapping the off-diagonal components. Symmetric: A symmetric tensor is one that has the property: Tij…k = Tk… ji A symmetric tensor is equal to its transpose. Anti-Symmetric: An anti-symmetric tensor is one that has the property: Tij…k = −Tk… ji Notes: 1. A tensor, symmetric in ij…k, is often indicated using the notation: T(ij…k). 2. A tensor, anti-symmetric in ij…k, is often indicated using the notation: T[ij…k]. 3. T( ij…k ) = 1 2 (Tij…k + Tk… ji ) 4. T[ij…k ] = 1 2 (Tij…k − Tk… ji ) 5. Tij…k = T( ij…k ) + T[ij…k ] 6. Tii = T11 + T22 + T33 = trace (Tij ) C. Wassgren Chapter 04: Differential Analysis 154 Last Updated: 05 Sep 2008 Basic Mathematical Operations (cont.) Dot Products (aka Inner Products): a ⋅ b = ai bi a ⋅ B = ai Bij A ⋅ b = Aij b j A ⋅ B = Aij B jk A : B = Aij B ji A : BT = Aij Bij T = ab ⇒ Tij = ai b j (Note: ab = ( ba ) .) T ci = ε ijk a j bk ⇒ Cross Product: c = a×b Gradient: ( ∇λ )i = Divergence: ∇ ⋅a = ∂ai = ai ,i ∂xi ∇⋅A = ∂λ = λ ,i ∂xi ∂Aij ∂xi = Aij ,i Curl: Laplacian (scalar): ∇2λ = ∂ak = ε ijk ak , j ∂x j ( ∇ × a )i = ε ijk ∂ 2λ = λ ,ii ∂xi ∂xi Gauss’ Theorem (aka Divergence Thm): ∫ ˆ n ∫ ˆ a ⋅ ndS = ∇ ⋅ adV S dS V V ∫ a n dS = ∫ a i ,i dV ii S V ˆ where S is the surface enclosing the volume V and n is the outward pointing unit normal vector for the surface area element dS. Stokes’ Theorem: ˆ n ˆ ∫ a ⋅ dl = ∫ ( ∇ × a ) ⋅ ndS C dl S ∫ a dl = ∫ ε i i C dS ijk ak , j ni dS C S where the curve C defines the surface S and dl is a vector tangent to the curve C at a particular point. (Note that the shape of the surface on which Stokes’ Theorem is to be applied must be known so that the relation between the surface area and contour is well defined.) References: 1. 2. 3. 4. Borg, S.F., Matrix-Tensor Methods in Continuum Mechanics, Van Nostrand, 1963. Myklestad, N.O., Cartesian Tensors, Van Nostrand, 1967. Synge, J.L. and Schild, A., Tensor Calculus, University of Toronto Press, 1949. Aris, R., Vectors, Tensors, and the Basic Equations of Fluid Mechanics, Prentice-Hall, 1962. C. Wassgren Chapter 04: Differential Analysis 155 Last Updated: 05 Sep 2008 Example: Prove that the following are true using index notation: (a × b) ⋅ c = a ⋅ (b × c) = (c × a) ⋅ b t × (u × v ) = u (t ⋅ v ) − v ( t ⋅ u ) u × v = −v × u SOLUTION: (a × b) ⋅ c = ε ijk a j bk ci = ε jki bk ci a j = (b × c) ⋅ a = ε kij ci a j bk = (c × a) ⋅ b [ t × (u × v)]i = ε ijk t j ε klmul vm = ε ijk ε klm t j ul vm = ε kij ε klm t j ul vm = (δ il δ jm − δ imδ jl ) t j ul vm = t j ui v j − t j u j vi = u ( t ⋅ v ) − v ( t ⋅ u ) i [u × v ]i = ε ijk u j vk = −ε ikj vk u j = [ − v × u ]i C. Wassgren Chapter 04: Differential Analysis 156 Last Updated: 05 Sep 2008 Example: Show, using index notation, that: ∇ × ∇θ = 0 SOLUTION: ( ∇ × ∇θ )i = ε ijk (θ , k ), j = ε ijk θ, kj (4.1) = −ε ikjθ , kj = −ε ijkθ , jk = −ε ijk θ , kj (the order of the differentiation doesn’t matter) (4.2) The only way for Eqns. (4.1) and (4.2) to be equal is if they both equal zero, i.e.: ε ijkθ , kj = 0 = −ε ijkθ , kj (4.3) Therefore: ( ∇ × ∇θ )i = 0 ⇒ ∇ × ∇θ = 0 C. Wassgren Chapter 04: Differential Analysis 157 Last Updated: 05 Sep 2008 Example: Solve: ai = ε ijk b jk for b[jk]. SOLUTION: ε imn ai = ε imn ε ijk b jk = (δ mj δ nk − δ mk δ nj ) b jk = bmn − bnm = 2 ⋅ 1 ( bmn − bnm ) 2 = 2b[mn] ∴ b[ jk ] = 1 ε ijk ai 2 C. Wassgren Chapter 04: Differential Analysis 158 Last Updated: 05 Sep 2008 2. Continuity Equation (aka COM for a differential CV ) The continuity equation, which is simply conservation of mass for a differential fluid element or control volume, can be derived several different ways. Two of these methods are given below. Method 1: Apply the integral approach to a differential control volume: y dy z x dz dx Assume that the density and velocity are ρ and u, respectively, at the control volume’s center. The mass fluxes through each of the side of the control volume are given by: ∂mx,center min through left = mx,center + − 1 dx 2 ∂x ∂ = ( ρ u x dydz ) + ( ρ u x dydz ) − 1 dx 2 ∂x ∂ = ρ u x + ( ρ u x ) − 1 dx ( dydz ) 2 ∂x (where mx ,center is the mass flux in the x-direction at the center of the control volume (Recall ( ) ( ( ) ) the Taylor Series approximation discussed in Chapter 01.)) ∂ mout through right = ρ u x + ( ρ u x ) 1 dx ( dydz ) 2 ∂x ∂ min through bottom = ρ u y + ρ u y − 1 dy ( dxdz ) 2 ∂y ( ( ) )( ) ∂ mout through top = ρ u y + ρ u y ( 1 dy ) ( dxdz ) 2 ∂y ∂ min through back = ρ u z + ( ρ u z ) ( − 1 dz ) ( dxdy ) 2 ∂z ∂ mout through front = ρ u z + ( ρ u z ) ( 1 dz ) ( dxdy ) 2 ∂z Thus, the net mass flow rate into the control volume is given by: ∂ ∂ ∂ mnet,into CV = − ( ρ u x ) + ρ u y + ( ρ u z ) ( dxdydz ) ∂y ∂z ∂x ( ) ( ) The rate at which mass is increasing within the control volume is given by: ∂m ∂ = ( ρ dxdydz ) ∂t within CV ∂t (4.4) where ρ is the density at the center of the control volume. Note that since the density varies linearly within the CV (from the Taylor Series approximation), the average density in the CV is ρ. C. Wassgren Chapter 04: Differential Analysis 159 Last Updated: 05 Sep 2008 Conservation of mass states that the rate of increase of mass within the control volume must equal the net rate at which mass enters the control volume: ∂m = mnet,into CV ∂t within CV ∂ ∂ρ ∂ ∂ ( dxdydz ) = − ( ρ u x ) + ρ u y + ( ρ u z ) ( dxdydz ) ∂t ∂x ∂y ∂z ( ) ∂ρ ∂ ∂ ∂ + ( ρux ) + ρu y + ( ρ uz ) = 0 ∂t ∂x ∂y ∂z Written in a more compact form: ∂ρ + ∇ ⋅ ( ρu) = 0 ∂t Or in index notation form: ∂ρ ∂ + ( ρ ui ) = 0 ∂t ∂xi ( ) (4.5) (4.6) (4.7) Method 2: Recall that the integral form of COM is given by: d ρ dV + ρ u rel ⋅ dA = 0 dt ∫ ∫ CV CS Consider a fixed control volume so that: d ∂ρ ρ dV = dV and u rel = u dt ∂t ∫ ∫ CV CV By utilizing Gauss’ Theorem (aka the Divergence Theorem), we can convert the area integral into a volume integral: ∫ ρu ⋅ dA = ∫ ∇ ⋅ ( ρu ) dV CS CV Substitute these expressions back into COM to get: ∂ρ ∂t + ∇ ⋅ ( ρ u ) dV = 0 CV ∫ Since the choice of control volume is arbitrary, the kernel of the integral must be zero, i.e.: ∂ρ + ∇ ⋅ ( ρu) = 0 same result as before! ∂t C. Wassgren Chapter 04: Differential Analysis 160 Last Updated: 05 Sep 2008 Notes: 1. For a fluid in which the density remains uniform and constant (ρ = constant), the continuity equation simplifies to: ∇ ⋅ u = 0 continuity equation for a constant density fluid 2. An incompressible fluid is one in which the density of a particular piece of fluid remains constant, i.e.: Dρ = 0 definition of an incompressible fluid Dt Note that an incompressible fluid does not necessarily imply that the density is the same everywhere in the flow (i.e. it’s not uniform). An example of such a flow would be a stratified flow in the ocean where the density of various layers of ocean water varies due to salinity and temperature variations. The density of fluid particles varies from layer to layer but remains constant within a layer. A flow with a constant and uniform density fluid, however, is an incompressible flow. The continuity equation for an incompressible fluid can be found by using the definition of an incompressible flow: Dρ ∂ρ ∂ρ =0= + ( u ⋅∇ ) ρ ⇒ = − ( u ⋅∇ ) ρ ∂t ∂t Dt Substituting into the continuity equation: ∂ρ + ∇ ⋅ ( ρu ) = 0 ∂t − ( u ⋅∇ ) ρ + ∇ ⋅ ( ρ u ) = 0 − ( u ⋅∇ ) ρ + ρ ( ∇ ⋅ u ) + ( u ⋅ ∇ ) ρ = 0 ∇ ⋅u = 0 3. continuity equation for an incompressible flow (4.8) Another useful form of the continuity equation is as follows: ∂u ∂ρ ∂ ∂ρ ∂ρ + ( ρ ui ) = 0 = + ui + ρ i ∂t ∂xi ∂t ∂xi ∂xi = Dρ Dt ∂u Dρ = −ρ i Dt ∂xi (4.9) 4. The continuity equation (Eqn. (4.7)) is valid for any continuous substance (e.g., a solid as well as a fluid). 5. Equation (4.7) is referred to as the conservative form of the continuity equation while Eqn. (4.9) is the non-conservative form. The conservative form implies that the equation represents an Eulerian viewpoint of the continuity equation. The non-conservative form represents the Lagrangian viewpoint. C. Wassgren Chapter 04: Differential Analysis 161 Last Updated: 05 Sep 2008 Example: The y-velocity component of a steady, 2D, incompressible flow is given by: u y = 3 xy − x 2 y Determine the most general velocity component in the x-direction for this flow. SOLUTION: Consider the continuity equation: ∂u x ∂u y + =0 ∂x ∂y (4.10) ∂u y ∂u x ∂ =− =− 3xy − x 2 y = −3 x + x 2 ∂x ∂y ∂y Integrate ux with respect to x. u x = − 3 x 2 + 1 x3 + f ( y ) 2 3 ( ) (4.11) where f(y) is an unknown function of y. C. Wassgren Chapter 04: Differential Analysis 162 Last Updated: 05 Sep 2008 Example: A piston compresses gas in a cylinder by moving at a constant speed, V. The gas density and the piston length are initially ρ0 and L0, respectively. Assume that the gas velocity varies linearly from velocity, V, at the piston face to zero velocity at the cylinder wall (at L). If the gas density varies only with time, determine ρ(t). V gas with density, ρ(t) L(t) SOLUTION: As given in the problem statement, assume the gas velocity, u, varies linearly with distance x from the piston face with the boundary conditions: u(x = 0) = V and u(x = L(t)) = 0. x ⇒ u ( x, t ) = V 1 − (4.12) L (t ) However, the piston moves at a constant speed so that: L ( t ) = L0 − Vt (4.13) Substituting Eqn. (4.13) into Eqn. (4.12) gives: x u ( x, t ) = V 1 − L0 − Vt (4.14) Apply the continuity equation assuming 1D flow. ∂ρ ∂ + ( ρu ) = 0 ∂t ∂x dρ ∂u = −ρ (Note that ρ = ρ(t).) dt ∂x (4.15) 1 =V dt ρ L0 − Vt dρ ρ =ρ ∫ ρρ = dρ ρ 0 t =t =V dt 0 − Vt ∫L t =0 ρ L − Vt ln = − ln 0 ρ0 L0 ∴ ρ Vt = 1 − ρ0 L0 −1 C. Wassgren Chapter 04: Differential Analysis (4.16) 163 Last Updated: 05 Sep 2008 3. Review of Stress Traction Vector (aka Stress Intensity, Stress Vector) Consider a small area on or within a deformable body subject to both surface and body forces as shown below. ∆M ∆Aν ∆F The net force and moment acting on the area ∆Aν, where ∆A is the magnitude of the area and ν is its corresponding unit normal vector, are denoted by ∆F and ∆M, respectively. The traction vector (aka stress intensity or stress vector) on the surface is defined as: ∆F T ν ≡ lim (a vector with dimensions of force per unit area) (4.17) ∆A→0 ∆A and the couple stress vector on the surface is defined as: ∆M C ν ≡ lim (a vector with dimensions of torque per unit area) (4.18) ∆A→0 ∆A Notes: 1. Usually Cν = 0 since body moments are rare. An example of a case in which body moments (and thus the couple stress vector) is not zero is in a material comprised of polar or magnetic elements (i.e. molecules or domains) subject to an external electric or magnetic field. In such a case, the material elements will try to orient themselves in a preferred direction. The couple stress vector may also be non-zero in powders where the component particles have an aspect ratio greater than one and, hence, tend to re-orient when under load. 2. To completely describe the traction at a “point”, we need to know Tν for all orientations, ν, of the differential surface area at that point. 3. Let σij be the components of the traction vector Tν. Consider, for example, the tractions on the faces of a differential cube as shown below. The traction on each face in terms of its components is: x2 ˆ ˆ ˆ T1 = σ 11e1 + σ 12 e2 + σ13e3 T2 ˆ ˆ ˆ T2 = σ 21e1 + σ 22 e2 + σ 23e3 ˆ ˆ ˆ T3 = σ 31e1 + σ 32e 2 + σ 33e3 ˆ where ei are the unit direction vectors of the axes. The quantity σij is known as the stress tensor. σ 22 σ 21 σ 23 σ 12 σ 32 σ 11 x1 σ 13 σ 31 σ 33 T1 x3 T3 C. Wassgren Chapter 04: Differential Analysis 164 Last Updated: 05 Sep 2008 Stress Sign Convention The sign conventions for stresses are as follows: 1. A positive face is a face that has a normal vector pointing in a positive direction. 2. Positive stresses on positive faces point in the positive direction. 3. Positive stresses on negative faces point in the negative direction. 4. The first subscript on the stress refers to the face on which the stress acts. The second subscript refers to the direction in which the stress acts. The figures below show positive stresses on all of the cube’s faces. x2 face σ 22 σ 21 σ 23 σ 32 σ 31 x2 direction σ 12 σ 13 σ 11 x1 σ 33 σ 13 σ 11 σ 32 σ 12 x3 σ 33 σ 31 x1 σ 23 σ 21 Positive Stresses on Positive Faces x3 σ 22 Positive Stresses on Negative Faces Notes: 1. σii (no sum) are referred to as normal stresses. 2. σij (i≠j) are referred to as shear stresses. 3. negative normal stresses ⇒ compression 4. positive normal stresses ⇒ tension C. Wassgren Chapter 04: Differential Analysis 165 Last Updated: 05 Sep 2008 Cauchy’s Formula Cauchy’s formula is used to determine the traction vector on an arbitrarily oriented surface with an orientation vector, ν, given the stress tensor. Consider the small tetrahedral element shown below. x2 Tν The areas of each face are: ˆ ˆ ˆ dA ν = dAν ν = dAν (ν 1e1 + ν 2 e2 + ν 3e3 ) C dA ν = dAν ν σ 31 σ 13 σ 11 σ 12 ˆ ˆ dAe1 = dA ν ⋅ e1 = dAνν 1 σ 33 σ 32 ˆ ˆ dAe2 = dA ν ⋅ e 2 = dAνν 2 A σ 23 ˆ ˆ dAe3 = dA ν ⋅ e3 = dAνν 3 x1 σ 21 σ 22 B x3 Apply Newton’s 2nd Law to the element in the x1 direction: ˆ ˆ ˆ ˆ ρ dVx1 = F1 = T ν ⋅ e1 dAν − σ 11 dA e1 − σ 21 dA e2 − σ 31 dA e3 + f1 ρ dV ( ∑ ) = σ 11dA ν 1 + σ 21dAνν 2 + σ 31dAνν 3 + ρ ( x1 − f1 ) dV ⇒ T1ν dAν ⇒ T1ν = σ 11ν 1 + σ 21ν 2 + σ 31ν 3 + ( x1 − f1 ) ρ dV but lim ν dAν →0 dA ν =0 ⇒ ⇒ dV dAν T1ν = σ11ν 1 + σ 21ν 2 + σ 31ν 3 T1ν = σ j1ν j A similar approach may be followed to derive expressions for the x2 and x3 directions. In general, we have: Ti ν = σ jiν j Cauchy’s Formula Cauchy’s formula may be used to determine the traction, Tν, on a surface with an orientation, ν, given the stress tensor, σji, at the point of interest. σyy T ˆ ey y σyx σyz x z C. Wassgren Chapter 04: Differential Analysis 166 Last Updated: 05 Sep 2008 Symmetry of the Stress Tensor Consider the Angular Momentum Principal for the small element of material shown below (only the stresses acting on the positive faces and causing rotations in the x3 axis are shown for clarity). Also note that no body couples are assumed to act on the element. x2 σ 21 + ∂σ 21 ∂x2 ( 1 dx2 ) 2 dx3 σ 12 + ∂σ 12 ∂x1 ( 1 dx1 ) 2 x1 dx2 x3 dx1 Using Newton’s 2nd Law for rotational moments in the x3-direction: ∂σ ∂σ θ3 = σ 12 + 12 1 dx1 dx2 dx3 1 dx1 + σ 12 + 12 − 1 dx1 dx2 dx3 I3 2 2 2 ∂x1 ∂x1 2 2 ( ( ) ( ) ( ) ) ( 1 dx1 ) 2 ∼ O dx1dx2 dx3 dx1 + dx2 ∂σ ∂σ − σ 21 + 21 1 dx2 dx1dx3 1 dx2 − σ 21 + 21 − 1 dx2 dx1dx3 2 2 2 ∂x2 ∂x2 Dividing through by the volume (dx1dx2dx3) and taking the limit as dx1,dx2,dx3→0 gives: σ 21 = σ 12 A similar approach can be taken in the x1 and x2 directions to arrive at the general result: σ ij = σ ji The stress tensor is symmetric (when no couple stresses are present). ( C. Wassgren Chapter 04: Differential Analysis ) ( 167 ) ( ) ( 1 dx2 ) 2 Last Updated: 05 Sep 2008 Another approach to proving symmetry of the stress tensor is to write the AME explicitly for the small element (again assuming no body couples): D ε ijk r j f k ρ dV + ε ijk r j σ lk nl dA = ε ijk r j uk ρ dV (4.19) Dt ∫ ∫ Vsys Ssys torque due to body forces ∫ =Tk Vsys torque due to surface forces time rate of change of the angular momentum of the element Using the divergence theorem the surface integral may be written as a volume integral: ∂ ε ijk r jσ lk nl dS = ε ijk r jσ lk dV = ε ijk δ jl σ lk + r jσ lk ,l dV (4.20) ∂xl σ Ssys Vsys Vsys jk Substituting into equation (4.19) and simplifying (note that the mass of the element remains constant): D ε ijk rj f k ρ dV + ε ijk σ jk + rjσ lk ,l dV = ε ijk rj uk ρ dV Dt ∫ ( ∫ ∫ ∫ Vsys ) ∫ ( ) Vsys ∫ Vsys D r j uk dV = 0 ε ijk rj ρ f k + σ jk + rjσ lk ,l − ρ Dt Vsys Duk Dr j rj uk + Dt Dt Note that Drj/Dt=uj and εijkujuk=0 (i.e. u×u=0). Re-arranging the previous equation gives: Du ε ijk r j ρ f k + σ lk ,l − ρ k + σ jk dV = 0 Dt ( ∫ ) ∫ Vsys As the volume of the cube becomes very small (dV→0), the r vector also becomes very small. Hence: Duk lim ε ijk r j ρ f k + σ lk ,l − + σ jk dV = ε ijk σ jk dV = 0 dV → 0 Dt V Vsys sys Since the volume is arbitrary: ε ijk σ jk = 0 (4.21) ∫ ∫ Now apply the permutation tensor to both side of Eqn. (4.21) and utilize an identity: ε ilmε ijk σ jk = 0 (δljδ mk − δ lk δ mj ) σ jk = 0 σ lm − σ ml = 0 ∴ σ ml = σ lm Thus, the stress tensor is symmetric (again, assuming no body couples). C. Wassgren Chapter 04: Differential Analysis 168 Last Updated: 05 Sep 2008 Example: The material element shown below has the following stress tensor components: y (0, 2, 0) m T 0 −4 3 0 kPa −4 0 5 1 ν [σ ] = 0 x (1, 0, 0) m z a. b. c. d. (0, 0, 1) m Find the components of the traction vector, T, on the plane described by the unit normal vector, ν. Determine the component of T parallel to ν. Determine the component of T perpendicular to ν. Determine the angle between T and ν. SOLUTION: First determine the components of the unit normal vector, ν, by taking the cross product of the vector pointing from (1, 0, 0) to (0, 2, 0) with the vector pointing from (1, 0, 0) to (0, 0, 1) then normalizing. ( 0, 2, 0 ) − (1, 0, 0 ) × ( 0, 0,1) − (1, 0, 0 ) = ( −1, 2, 0 ) × ( −1, 0,1) = ( 2,1, 2 ) ν= ( −1, 2, 0 ) × ( −1, 0,1) ( 2,1, 2 ) ( 0, 2, 0 ) − (1, 0, 0 ) × ( 0, 0,1) − (1, 0, 0 ) 2 ∴ν = (3, 1, 2) 33 Next, use Cauchy’s formula to determine the traction vector, T. Ti = σ jiν j 2 T1 = σ 11ν 1 + σ 21ν 2 + σ 31ν 3 = (1 kPa ) ( 3 ) + ( 0 kPa ) ( 1 ) + ( −4 kPa ) ( 2 ) 3 3 ⇒ T2 = σ 12ν 1 + σ 22ν 2 + σ 32ν 3 = ( 0 kPa ) ( 2 ) + ( 3 kPa ) ( 1 ) + ( 0 kPa ) ( 2 ) 3 3 3 T3 = σ 13ν 1 + σ 23ν 2 + σ 33ν 3 = ( −4 kPa ) ( 2 ) + ( 0 kPa ) ( 1 ) + ( 5 kPa ) ( 2 ) 3 3 3 2 ∴ T = ( −2,1, 3 ) kPa The component of T parallel to ν is found by taking the dot product T ⋅ ν. 2 2 T|| = T ⋅ ν = Tiν i = ( −2 kPa ) ( 2 ) + (1 kPa ) ( 1 ) + ( 3 kPa ) ( 3 ) 3 3 ∴ T|| = − 5 kPa 9 The component of T perpendicular to ν is found by taking the difference between T and T||ν. 28 2 2 44 T⊥ = T − T|| ν = ( −2,1, 2 ) − ( − 5 ) ( 3 , 1 , 3 ) kPa = ( − 27 , 32 , 27 ) kPa 3 9 3 27 T⊥ = 2.27 kPa The angle between T and ν can be determined from the dot product between the two vectors. T⋅ ν T ⋅ ν = T ν cos θ ⇒ cos θ = Tν cos θ = 5 − 9 kPa − 5 kPa T⋅ ν 5 = = 79 = − 21 2 Tν ( −2,1, 3 ) kPa (1) 3 kPa ∴θ = 103.8 C. Wassgren Chapter 04: Differential Analysis 169 Last Updated: 05 Sep 2008 4. Momentum Equations (aka the LME for a differential CV) The momentum equations, which are the simply the linear momentum equations for a differential fluid element or control volume, can be derived several different ways. Three of these methods are given below. Method 1: Apply the integral approach to a differential control volume: y dy z x dz dx Assume that the density and velocity are ρ and u, respectively, at the control volume’s center. Consider only the x-momentum equation first. The x-momentum fluxes through each of the side of the control volume are given by: ∂ ( mu x )center − 1 dx ( mu x )in through left = ( mu x )center + 2 ∂x ∂ = ( ρ u x dydzu x ) + ( ρ u x dydzu x ) ( − 1 2 dx ) ∂x ∂ = ρ u x u x + ( ρ u x u x ) ( − 1 2 dx ) ( dydz ) ∂x (where mx ,center is the mass flux in the x-direction at the center of the control volume (Recall ( ) the Taylor Series approximation discussed in Chapter 01.)) ∂ ( mux )out through right = ρ ux ux + ( ρ u xu x ) ( 1 2 dx ) ( dydz ) ∂x ( mux )in through bottom = ρ u y u x + ∂ ρ u y ux ∂y ( ) ( − 1 2 dy ) ( dxdz ) ) ( 1 2 dy ) ( dxdz ) ( mux )out through top = ρ u y ux + ∂ ρu yux ∂y ( mux )in through back = ρ uz u x + ∂ ( ρ uz u x ) ( − 1 2 dz ) ( dxdy ) ∂z ( ∂ ( ρ u z ux ) ( 1 2 dz ) ( dxdy ) ∂z Thus, the net x-momentum flux out of the control volume is given by: ∂ ∂ ∂ ( mux )net, out of CV = ( ρ ux ux ) + ρ u y ux + ( ρ u z ux ) ( dxdydz ) ∂y ∂z ∂x ( mux )out through front = ρ u z ux + ( ) (4.22) The rate at which the x-momentum is increasing within the control volume is given by: ∂ ∂ ∂ = ( u x ρ dxdydz ) = ( u x ρ ) ( dxdydz ) ( mux ) ∂t ∂t ∂t within CV (4.23) where ρ and ux are the density and x-velocity, respectively, at the center of the control volume. Note that since these quantities vary linearly within the CV (from the Taylor Series approximation), the averages within the CV are ρ and ux. C. Wassgren Chapter 04: Differential Analysis 170 Last Updated: 05 Sep 2008 The forces acting on the control volume include both body and surface forces. The body force acting on the CV in the x-direction, FB,x, can be written as: FB, x = f B, x ρ ( dxdydz ) (4.24) where fB,x is the body force per unit mass acting in the x-direction (e.g., for weight, the body force per unit mass acting in the x-direction is simply gx). The surface forces acting on the control volume include both normal and tangential forces. Writing the surface force acting in the x-direction, FS,x, in terms of stresses gives: ∂σ xx ∂σ FS , x = − σ xx + ( − 1 2 dx ) ( dydz ) + σ xx + ∂xxx ( 1 2 dx ) ( dydz ) ∂x normal force on left face normal force on right face ∂σ yx ∂σ − σ yx + ( − 1 2 dy ) ( dxdz ) + σ yx + yx ∂y ∂y shear force on bottom face ( 1 2 dy ) ( dxdz ) shear force on top face ∂σ zx ∂σ − σ zx + ( − 1 2 dz ) ( dxdy ) + σ zx + ∂zzx ∂z shear force on back face ( 1 2 dz ) ( dxdy ) shear force on front face ∂σ yx ∂σ zx ∂σ ∴ FS , x = xx + + ( dxdydz ) ∂y ∂z ∂x (4.25) The LME states that the rate of increase of linear momentum within the control volume plus the net rate at which momentum leaves the control volume must equal the net force acting on the control volume: ∂ + ( mu x ) = FB , x + FS , x (4.26) ( mux ) net, out of CV ∂t within CV Substituting Eqns. (4.83)-(4.25) into Eqn. (4.26) gives: ∂ ∂ ∂ ∂ ( u x ) ( dxdydz ) + ( ρ u x ux ) + ρ u y ux + ( ρ uz ux ) ( dxdydz ) = ∂t ∂y ∂z ∂x ( ) ∂σ yx ∂σ zx ∂σ f B, x ρ ( dxdydz ) + xx + + ( dxdydz ) ∂y ∂z ∂x ∂σ ∂σ ∂σ ∂ ∂ ∂ ∂ (4.27) ( u x ρ ) + ( ρ ux u x ) + ρ u y u x + ( ρ uz u x ) = ρ f B , x + xx + yx + zx ∂t ∂x ∂y ∂z ∂x ∂y ∂z A similar approach can be taken to determine the y- and z-components of the momentum equations. All three components of the momentum equations can be written in the following compact (index notation) form: ∂σ ji ∂ ∂ ρ u j ui = ρ f B,i + (4.28) ( ui ρ ) + ∂t ∂x j ∂x j ( ( ) ) In vector notation, Eqn. (4.28) is written as: ∂ ( uρ ) + ( u ⋅∇ )( ρ u ) = ρ f B +∇ ⋅ σ T ∂t (4.29) Note that σT = σ since the stress tensor is symmetric. C. Wassgren Chapter 04: Differential Analysis 171 Last Updated: 05 Sep 2008 Expand the left hand side of Eqn. (4.28) and utilize the continuity equation: ∂u ∂u ∂ ∂ ∂ρ ∂ + ρ i + ui ρ u j ui = ui ρu j + ρu j i ( ui ρ ) + ∂t ∂x j ∂t ∂t ∂x j ∂x j ( ) ( ∂ρ ∂u ∂u ∂ ρu j + ρ i + u j i = ui + ∂t ∂t ∂x j ∂x j ( ) = 0 (continuity eqn) = ) Dui Dt Substituting into back into Eqn. (4.28) gives: ∂σ ji Du ρ i = ρ f B ,i + momentum equations Dt ∂x j (4.30) Method 2: Apply Newton’s 2nd Law directly to a small piece of fluid: ∂σ D ( ui ρ dxdydz ) = f B,i ρ ( dxdydz ) + ji ( dxdydz ) Dt ∂x j (4.31) where the determination of the body and surface forces are described in the previous method. Expanding the Lagrangian derivative gives: Du D D ( ui ρ dxdydz ) = i ( ρ dxdydz ) + ui ( ρ dxdydz ) Dt Dt Dt but the second term on the RHS of this equation will be zero since the mass of the fluid element remains constant. Thus, Eqn. (4.31) can be simplified to: ∂σ ji Du ρ i = ρ f B ,i + same result as before! Dt ∂x j Method 3: Recall the integral form of the LME: d ui ρ dV + ui ( ρ u rel ⋅ dA ) = FB ,i + FS ,i dt ∫ ∫ CV CS Consider a fixed control volume so that: ∂ ( ui ρ ) d ui ρ dV = dV and dt ∂t ∫ ∫ CV CV u rel = u Note that the body force can be written as: FB,i = ∫ f B,i ρ dV CV and the surface force can be written as: FS ,i = ∫σ ji n j dA CS C. Wassgren Chapter 04: Differential Analysis 172 Last Updated: 05 Sep 2008 By utilizing Gauss’ Theorem (aka the Divergence Theorem), we can convert the area integrals into volume integrals: ∂ ui ( ρ u ⋅ dA ) = ∇ ⋅ ( ui ρ u ) dV = ρ u j ui dV ∂x j ∫ ∫ ∫ CS CV ( ) CV ∫σ CS ji n j dA = ∫ CV ∂σ ji ∂x j dV Substitute these expressions back into the LME to get: ∂ ∂σ ji ∂ ρ u j ui − ρ f B ,i − ( ui ρ ) + dV = 0 ∂x j ∂x j ∂t CV Since the choice of control volume is arbitrary, the kernel of the integral must be zero, i.e.: ∂σ ji ∂ ∂ ρ u j ui − ρ f B,i − =0 ( ui ρ ) + ∂t ∂x j ∂x j ( ∫ ( ) ) But this is the same expression as Eqn. (4.28) so we see that the final result will be the same! ∂σ ji Du ρ i = ρ f B ,i + Dt ∂x j Notes: 1. In order to be more useful to us, we need to have some way of relating the stresses acting on the fluid element (or CV) to other properties of the flow (namely the velocities). This is accomplished using a constitutive law which relates the stresses to the strain rates for a particular fluid or class of fluids. 2. Equation (4.30) is valid for any continuous substance. 3. Equation (4.28) is the conservative form of the LME. Equation (4.30) is the non-conservative form of the LME. C. Wassgren Chapter 04: Differential Analysis 173 Last Updated: 05 Sep 2008 Example Consider the flow of a mixture of liquid water and small water vapor bubbles. The bubble diameters are very small in comparison to the length scales of interest in the flow so that the properties of the mixture can be considered point functions. For example, the density of the mixture at a “point” can be written as: ρ M = αρ V + (1 − α )ρ L where ρM is the mixture density, ρL is the liquid density, ρV is the vapor density, and α is the “void fraction” or the fraction of volume that is vapor in a unit volume of the mixture. Assume that evaporation occurs at the bubble surface so that the liquid water turns to water vapor at a mass flow rate per unit volume denoted by s. a. b. c. What is the continuity equation for the mixture? What is the continuity equation for the liquid water phase? What are the momentum equations for the liquid water phase? SOLUTION: The continuity equation for the mixture will be the “normal” continuity equation: ∂ρ M ∂ + ( ρ M ui ) = 0 ∂t ∂xi To show that this relation is true, consider the control volume shown below. (4.32) dx y dy x The rate of change of mass within the control volume is: ∂ρ ∂ ( ρ M dxdydz ) = M dxdydz ∂t ∂t The net mass flux into the CV in the x-direction is: ∂ ∂ mx , net = ρ M ux dydz − ρ M u x + ( ρ M u x ) dx dydz = − ( ρ M ux ) dxdydz ∂x ∂x into CV Following a similar approach in the y and z directions gives: ∂ m y ,net = − ( ρ M u y ) dxdydz ∂y into CV mz,net into CV =− ∂ ( ρ M uz ) dxdydz ∂z (4.34) (4.35) (4.36) Thus, from conservation of mass: ∂ρ M ∂ ∂ ∂ dxdydz = − ( ρ M u x ) dxdydz − ( ρ M u y ) dxdydz − ( ρ M u z ) dxdydz ∂t ∂x ∂y ∂z ∂ρ M ∂ + ( ρ M ui ) = 0 ∂t ∂xi C. Wassgren Chapter 04: Differential Analysis (4.33) (4.37) (4.38) 174 Last Updated: 05 Sep 2008 To determine the continuity equation for the liquid water phase, consider the control volume drawn below where the CV surrounds each vapor bubble. dx y x dy The rate of change of liquid mass within the control volume is: ∂ ∂ ρ L (1 − α ) dxdydz = (1 − α ) ρ L dxdydz ∂t ∂t (4.39) The net liquid mass flux into the CV in the x-direction is: ∂ ∂ mx , net = (1 − α ) ρ L u x dydz − (1 − α ) ρ L u x + (1 − α ) ρ L u x dx dydz = − (1 − α ) ρ L u x dxdydz ∂x ∂x into CV (4.40) Following a similar approach in the y and z directions gives: ∂ m y ,net = − (1 − α ) ρ L u y dxdydz (4.41) ∂y into CV ∂ (1 − α ) ρ L u z dxdydz ∂z The rate at which liquid mass is being converted to vapor mass is: mout of CV = s (1 − α ) dxdydz =− mz,net (4.42) into CV (4.43) due to evap. Thus, from conservation of mass: ∂ (1 − α ) ρ L dxdydz = ∂t (4.44) ∂ ∂ ∂ − (1 − α ) ρ L ux dxdydz − (1 − α ) ρ L u y dxdydz − (1 − α ) ρ L u z dxdydz − s (1 − α ) dxdydz ∂x ∂y ∂z ∂ ∂ (1 − α ) ρ L + ∂x (1 − α ) ρ L ui = − (1 − α ) s (continuity eqn. for liquid phase) ∂t i (4.45) To determine the momentum equations for the liquid phase, apply the momentum equation to the same control volume used to derive the liquid phase continuity equation. The change in momentum of liquid within the CV is: d ∂ ∂ (4.46) ∫ uρ dV = ∂t ui ρ L (1 − α ) dxdydz = ∂t ui ρ L (1 − α ) dxdydz dt CV The net flux of linear momentum out of the CV through the sides of the CV is: ∫ uρ ( u rel ⋅ dA ) = CS = ∂ ∂ ∂ ui (1 − α ) ρ Lu x dxdydz + ui (1 − α ) ρ L u y dxdydz + ui (1 − α ) ρ L u z dxdydz + ui s (1 − α ) dxdydz ∂x ∂y ∂z ∂ ui (1 − α ) ρ Lu j dxdydz + ui s (1 − α ) dxdydz ∂x j (4.47) (Note that the term involving s is the rate at which momentum leaves the liquid phase due to the fact that the liquid is evaporating.) C. Wassgren Chapter 04: Differential Analysis 175 Last Updated: 05 Sep 2008 The surface forces acting on the control surface are: ∂σ ji FS + FB = − dxdydz + fVonL ,i ρ L (1 − α ) dxdydz + g i ρ L (1 − α ) dxdydz ∂x j (4.48) Note that the stress terms are the surfaces forces acting on the sides of the CV. The term fVonL,i is the force per unit mass that the vapor phase exerts on the liquid phase, and the last term in Eqn. (4.48) is the body force acting on the liquid phase where gi is the body force per unit mass. Substituting into the linear momentum equation and simplifying results in: ∂σ ji ∂ ∂ ui ρ L (1 − α ) + ∂x ui (1 − α ) ρ L u j = − ∂x + fVonL ,i ρ L (1 − α ) − ui s (1 − α ) + g i ρ L (1 − α ) ∂t j j (4.49) The continuity equation derived previously for the liquid phase (Eqn. (4.45)) could be used to further simplify the momentum equation, if desired. C. Wassgren Chapter 04: Differential Analysis 176 Last Updated: 05 Sep 2008 5. Fluid Element Deformations In order to relate the stresses acting on a fluid element to the other variables in the momentum equations, we need to determine a constitutive law. The constitutive law should relate the stresses to the rates of strain (or deformation rates) of a fluid element. For a solid, the necessary constitutive law relates the stresses to the strains (or deformations). In order to derive an appropriate constitutive law, we must first discuss the general types of deformations that can occur for a fluid element and then describe, in mathematical terms, the rates at which these deformations occur. Any general deformation can be decomposed into a combination of translation, dilation (aka dilatation), rigid body rotation, and angular deformation as shown in the figure below. + = general deformation translation + dilation + angular deformation rigid body rotation Now let’s describe the rate at which each of these deformations occurs. Translation dx The rate of translation is described by the time rate of change of the position of the element, i.e. the velocity. dx =u (4.50) rate of translation = dt C. Wassgren Chapter 04: Differential Analysis 177 Last Updated: 05 Sep 2008 Dilation (aka Dilatation) B’ ∂u2 dx2 dt ∂x2 C’ B C O A The rate of dilation can be described by the rate at which the relative volume of the element increases with time. 1 dV volumetric dilation rate = θ ≡ V dt d x2 d x1 A’ ∂u1 dx1dt ∂x1 The velocity of point A relative to point O in the x1 direction is: ∂u u1 = 1 dx1 ∂x1 Thus, point A “stretches” the element in the x1 direction over time dt a distance of: ∂u1 dx1dt ∂x1 The increase in volume of the element due to the relative movement of point A is: ∂u dVA = 1 dx1dt dx2 dx3 ∂x1 A similar approach can be followed for stretching in the x2 and x3 directions. The total increase in volume of the element is: This volume is comprised of ∂u ∂u ∂u C’ dV = 1 + 2 + 3 dx1dx2 dx3 dt B’ H.O.T.s and, hence, is ∂x1 ∂x2 ∂x3 neglected. ∂u2 dx2 dt dx1dx3 ∂x2 O ∂u1 dx dt dx dx A’ ∂x 1 2 3 1 Note that higher order volume terms have been neglected in deriving the previous result. The volumetric dilation rate is thus: ∂u1 ∂u2 ∂u3 + + dx1dx2 dx3 dt 1 dV ∂x1 ∂x2 ∂x3 θ≡ = V dt dx1dx2 dx3 dt θ= ∂u1 ∂u2 ∂u3 ∂ui + + = = ∇ ⋅u ∂x1 ∂x2 ∂x3 ∂xi (4.51) Notes: 1. For an incompressible fluid, the volumetric dilation rate is zero since if the volume of the element changes, the density must also change (COM). C. Wassgren Chapter 04: Differential Analysis 178 Last Updated: 05 Sep 2008 Angular Deformation ∂u1 dx2 dt ∂x2 B’ The rate of angular deformation in the 1-2 plane can be described as the average rate at which the sides of the element approach one another, i.e. the average rate at which the angles AOA’ and BOB’ increase. C’ B C dβ A’ d x2 O dα d x1 A ∂u 2 d x1 d t ∂ x1 The angle AOA’ (dα) is: ∂u2 dx1dt ∂x1 tan ( dα ) = dx1 But since the angle dα is very small, tan(dα) = dα: ∂u dα = 2 dt ∂x1 Similarly, the angle BOB’ (dβ) is: ∂u d β = 1 dt ∂x2 Define the rate of angular deformation, S12, (aka rate of shearing strain) in the 1-2 plane as the average time rate of change of these two angles: dα d β 1 ∂u2 ∂u1 + + S12 ≡ 1 = 2 dt 2 ∂x1 ∂x2 dt Similarly, we can determine the angular deformation rate in the 1-3 and 2-3 planes: ∂u ∂u ∂u ∂u S13 = 1 3 + 1 and S23 = 1 3 + 2 2 2 ∂x1 ∂x3 ∂x2 ∂x3 Combine the angular deformation rate and the dilation rate into one tensor quantity called the shearing strain tensor, Sij: ∂u ∂u j Sij = 1 i + (4.52) 2 ∂x j ∂xi Notes: 1. Dilation rate is given as the trace of Sij: θ = trace ( Sij ) = Sii = S11 + S22 + S33 = 2. ∂u1 ∂u2 ∂u3 + + ∂x1 ∂x2 ∂x3 The shearing strain tensor is symmetric! i.e. Sij = Sji C. Wassgren Chapter 04: Differential Analysis 179 Last Updated: 05 Sep 2008 Rigid Body Rotation − ∂u1 dx2 dt ∂x2 B’ The rate at which the fluid element rotates about the 3axis in rigid body motion can be described as the average rate at which the sides of the element rotate in the same direction. C’ B dβ d x2 O C d α A’ d x1 A ∂u2 dx1dt ∂x1 The rotation rate about the 3-axis, Ω3, is given by: dα d β 1 ∂u2 ∂u1 Ω3 ≡ 1 + − = 2 dt 2 ∂x1 ∂x2 dt Rotations about the 1 and 2 axes can be found in a similar manner: ∂u ∂u ∂u ∂u Ω1 = 1 3 − 2 and Ω 2 = 1 1 − 3 2 2 ∂x2 ∂x3 ∂x3 ∂x1 The rate of rotation of the element can be summarized using the rotation rate vector, Ω: ∂u ∂u ∂u ∂u ∂u ∂u ˆ2 ˆ ˆ = 1 3 − 2 e1 + 1 1 − 3 e 2 + 1 2 − 1 e3 2 2 ∂x2 ∂x3 ∂x3 ∂x1 ∂x1 ∂x2 (4.53) = 1 ∇×u 2 Notes: 1. The rotation rate vector, Ω, is written in index notation form as: ∂u Ωi = 1 ε ijk k 2 ∂x j 2. The rotation rate vector can also be written as an anti-symmetric rotation rate tensor, Rij: ∂u ∂u j (4.54) Rij ≡ 1 i − = −ε ijk Ω k (Note: Rij = − R ji ) 2 ∂x j ∂xi where Rij is the rotation rate in the i-j plane. Note that the diagonal elements of the tensor Rij are zero. 3. The vorticity, ω, of a fluid element is defined to be twice the rotation rate of the element: ω ≡ 2 = ∇×u or in index notation: ∂u ωi = ε ijk k ∂x j (4.55) An irrotational flow is one in which ω = 0. A rotational flow is one in which ω ≠ 0. C. Wassgren Chapter 04: Differential Analysis 180 Last Updated: 05 Sep 2008 Now that we have described the deformation rate components (e.g. dilation, angular deformation, and rigid body rotation; translations are treated separately) of a fluid element, let’s combine these into a single tensor quantity known as the deformation rate tensor, eij: ∂u ∂u j ∂u ∂u j ∂u eij ≡ i = Sij + Rij = 1 i + (4.56) + 1 i − 2 2 ∂x j ∂xi ∂x j ∂x j ∂xi We will use the deformation rate tensor when deriving the constitutive relations between stress and strain (deformation) rates in a fluid. C. Wassgren Chapter 04: Differential Analysis 181 Last Updated: 05 Sep 2008 Example A fluid has a velocity field given by u = 2 xˆ − 3 yˆ + zk i jˆ At the location (x, y, z) = (-2, -1, 2), calculate: a. the normal and shearing strain rates at the location, and b. the rotational velocity of the fluid. SOLUTION: The strain rate tensor is given by: ∂u ∂u j Sij = 1 i + 2 ∂x j ∂xi so that the normal strain rates are: ∂u ∂u y ∂u z S xx , S yy , S zz = x , , ∂x ∂y ∂z ( ) ( ) ∴ S xx , S yy , S zz = ( 2, −3,1) and the shearing strain rates are: ( S xy = S yx , S xz = S zx , S yz = S zy ) = 1 ∂∂uyx + 2 ( ∂u y 1 ∂u x ∂u z 1 ∂u y ∂u z + , + , ∂x 2 ∂z ∂x 2 ∂z ∂y ) ∴ S xy = S yx , S xz = S zx , S yz = S zy = ( 0, 0, 0 ) The rotational velocity of a fluid element is given by: ∂u Ωi = 1 ε ijk k (Note that the vorticity is twice the rotation rate, i.e. ω = 2Ω.) 2 ∂x j and, thus, for the given case: 1 ∂u ∂u y ∂u x ∂u z ∂u y ∂u x Ω x , Ω y , Ω z = z − − − , , 2 ∂y ∂z ∂z ∂x ∂x ∂y ( ) ( ) ∴ Ω x , Ω y , Ω z = ( 0, 0, 0 ) (Fluid elements are not rotating anywhere! The flow is irrotational.) C. Wassgren Chapter 04: Differential Analysis 182 Last Updated: 05 Sep 2008 Stress-Strain Rate Relations for a Newtonian Fluid The following assumptions are based on observation and intuition. The key assumptions in deriving the stress-strain rate constitutive relations for a Newtonian fluid are: 1. When the fluid is at rest, the pressure exerted by the fluid is the thermodynamic pressure, p. 2. For a Newtonian fluid, the stress tensor, σij, is linearly related to the deformation rate tensor, ekl, and depends only on that tensor. 3. There are no preferred directions in the fluid so that the fluid properties are point functions. This is the condition of isotropy. Now let’s examine how these assumptions aid us in deriving the appropriate constitutive law. Assumption 1: When the fluid is at rest, the pressure exerted by the fluid is the thermodynamic pressure, p. This assumption implies the following: σ ij = − pδ ij + τ ij (4.57) where τij is referred to as the viscous stress tensor (aka deviatoric stress tensor) and it is only a function of the fluid motion (i.e. τij=0 for a static fluid). Note that the pressure term is negative since compression of the fluid element is indicated by a negative normal stress. Assumption 2: For a Newtonian fluid, the stress tensor, σij, is linearly related to the deformation rate tensor, ekl, and depends only on that tensor. The 9 elements of τij can be written as a linear combination of the 9 elements of ekl: τ ij = Aijkl ekl (4.58) where Aijkl is a tensor of rank 4 (81 elements) that depends only on the local state of the fluid. Notes: 1. Recall that the deformation rate tensor is given as: ∂u ∂u ∂u ∂u ∂u ekl ≡ k = Skl + Rkl = 1 k + l + 1 k − l 2 2 ∂xl ∂xl ∂xk ∂xl ∂xk symmetric, S kl = Slk 2. anti-symmetric, Rkl =− Rlk Air and water are common examples of Newtonian fluids. Since the stress tensor, σij, is symmetric (refer to the notes reviewing stress), τij must also be symmetric. And since τij is symmetric, the components of the Aijkl tensor multiplied by the antisymmetric part of the deformation rate tensor, ekl, must be zero. Thus, ∂u ∂u τ ij = 1 Bijkl k + l (4.59) 2 ∂xl ∂xk where Bijkl is the Aijkl tensor with the Aijkl components multiplied by the components of the antisymmetric part of ekl set equal to zero. C. Wassgren Chapter 04: Differential Analysis 183 Last Updated: 05 Sep 2008 Assumption 3: There are no preferred directions in the fluid so that the fluid properties are point functions. This is the condition of isotropy. The condition of isotropy means that the fluid properties are the same in all directions. Examples of non-isotropic materials include fluids comprised of long chain molecules or oriented fibrous solids such as wood. It can be shown (out of the scope of these notes) that the most general 4th order isotropic tensor can be written as: ( )( Bijkl = λδ ij δ kl + µ δ ik δ jl + δ il δ jk + γ δ ik δ jl − δ il δ jk ) (4.60) where λ, µ, and γ are scalar quantities. Substitute Eqn. (4.60) into Eqn. (4.58) and simplify: ∂u ∂u τ ij = 1 λδ ij δ kl + µ (δ ik δ jl + δ il δ jk ) + γ (δ ik δ jl − δ il δ jk ) k + l 2 ∂x ∂x l k (4.61) where 1 2 1 2 ∂u ( λδijδ kl ) ∂uk + ∂x l = 1 λδ ij 2 ∂uk = λδ ij ∂uk 2 ∂x ∂x ∂x l k µ (δ ik δ jl + δ il δ jk ) 1 γ 2 k k ∂uk ∂ul 1 ∂ui ∂u j + + = 2 µ ∂x j ∂xi ∂xl ∂xk ∂u j ∂ui + + ∂xi ∂x j ∂u ∂u j = µ i + ∂x j ∂xi ∂u ∂u ∂u ∂u ∂u (δik δ jl − δ ilδ jk ) ∂uk + ∂x l = 1 γ ∂x i + ∂x j − ∂x j + ∂x i = 0 2 ∂x l k j i i j so that: τ ij = λ ∂u ∂u j ∂uk δ ij + µ i + ∂x j ∂xi ∂xk (4.62) Substituting Eqn. (4.62) into Eqn. (4.57) gives: σ ij = − pδ ij + τ ij = − pδ ij + λ ∂u ∂u j ∂uk δ ij + µ i + ∂x j ∂xi ∂xk ∂u ∂u j ∂u ∴ σ ij = − p + λ k δ ij + µ i + ∂x j ∂xi ∂xk stress-strain rate constitutive relation for a Newtonian fluid (4.63) Notes: 1. The quantity, µ, is referred to as the dynamic viscosity. 2. The quantity, λ, is referred to as the 2nd coefficient of viscosity. C. Wassgren Chapter 04: Differential Analysis 184 Last Updated: 05 Sep 2008 3. How is the thermodynamic pressure related to the normal stresses? Define the mechanical pressure, p , as the average of the normal stresses: () p ≡ − 1 trace σ ij = − 1 σ ii = − 1 (σ 11 + σ 22 + σ 33 ) 3 3 3 For a Newtonian fluid: p = − 1 (σ 11 + σ 22 + σ 33 ) 3 ∂u ∂u ∂u ∂u = − 1 − p + λ k + 2µ 1 + − p + λ k + 2 µ 2 3 ∂xk ∂x1 ∂xk ∂x2 = p−λ ∂uk ∂u + 2µ 3 +−p + λ ∂xk ∂x3 ∂uk 2 ∂u1 ∂u2 ∂u3 − µ + + ∂xk 3 ∂x1 ∂x2 ∂x3 ( ∴p = p− λ+ 2µ 3 =K ) ∂uk ∂x ( or ∴ p = p + λ + 2 µ 3 k =K ) ∂uk ∂x k 2 where K ≡ bulk viscosity = λ+ /3µ. In general, the thermodynamic pressure is not the same as the mechanical pressure. What then is the physical significance of the bulk viscosity, K, term? The mechanical pressure is a measure of the translational energy only. The thermodynamic pressure, however, is a measure of the total energy (translational, rotational, vibrational, etc.) The bulk viscosity, K, is a measure of the transfer of energy from the translational mode to the other modes. For example, when fluid flows through a shock wave, there is a considerable transfer of energy between the translational mode and the other modes; hence, the bulk viscosity cannot be neglected for such a flow process. For typical flows, however, the bulk viscosity is often neglected. For example: 1. For monatomic gases the only energy mode is the translational mode so that: K =0 ⇒ p= p For polyatomic gases and liquids, the bulk viscosity is often small so that we usually assume: K ≈0 ⇒ p= p The assumption that the bulk viscosity is zero (or equivalently, λ = –2/3µ) is known as Stokes’ Relation. 2. For an incompressible fluid the velocity divergence term is zero (from the continuity equation) so that the bulk viscosity is irrelevant: ∂uk =0 ⇒ p= p ∂xk 3. The bulk viscosity term is generally not negligible when there is a rapid expansion or contraction of the fluid such as when fluid passes through a shock wave or when considering acoustic absorption. C. Wassgren Chapter 04: Differential Analysis 185 Last Updated: 05 Sep 2008 4. The stress tensor given in Eqn. (4.63) can be substituted into the momentum equations to give the Navier-Stokes Equations: ∂σ ji Du ρ i= + ρ fi Dt ∂x j ∂u ∂u j µ i + ∂x j ∂xi Navier-Stokes equations for a Newtonian fluid ∴ρ Dui ∂p ∂ ∂uk ∂ =− + λ + Dt ∂xi ∂xi ∂xk ∂x j + ρ fi (4.64) For an incompressible fluid with constant dynamic viscosity: ∂ 2u ∂u j Dui ∂p ∂ i + ρ fi ρ =− +µ + Dt ∂xi ∂x j ∂x j ∂xi ∂x j =0 ∴ρ Dui ∂ 2 ui ∂p Du =− +µ + ρ f i or ρ = −∇p + µ∇ 2 u + ρ f ∂xi ∂x j ∂x j Dt Dt (4.65) Navier-Stokes equations for an incompressible, Newtonian fluid with constant dynamic viscosity For an inviscid (µ=0) fluid that follows Stokes’ Relation or is incompressible: Dui ∂p Du ∴ρ =− + ρ f i or ρ = −∇p + ρ f ∂xi Dt Dt These relations are known as Euler’s Equations. (4.66) As a reminder, the Navier-Stokes equations are the momentum equations for a Newtonian fluid. Euler’s equations are the momentum equations for an inviscid fluid. C. Wassgren Chapter 04: Differential Analysis 186 Last Updated: 05 Sep 2008 Example Consider a 3D steady flow of an incompressible, Newtonian liquid with a velocity field given by: ˆ u = axˆ + ayˆ − 2azk i j There are no body forces acting on the flow and the pressure at the origin is p0. a. Show that the continuity equation is satisfied, b. Determine the pressure field. c. Determine the vorticity field. SOLUTION: The continuity equation is: ∂u x ∂u y ∂u z + + = a + a − 2a = 0 ∂x ∂y ∂z Continuity is satisfied! The pressure field may be found using the Navier-Stokes equations. Note that the body forces are zero. Du ∂ 2ui ∂p ρ i =− +µ Dt ∂xi ∂x j ∂x j ∂ 2u ∂u ∂u ∂ 2u ∂ 2u ∂u ∂p = − ρ u x x + u y x + u z x + µ 2x + 2x + 2x ∂x ∂x ∂x ∂y ∂z ∂y ∂z ∂p = −ρ a2 x ⇒ ∂x ⇒ p ( x, y , z ) = − 1 ρ a 2 x 2 + f ( y , z ) 2 ∂ 2u y ∂ 2 u y ∂ 2u y ∂u y ∂u y ∂u y ∂p = − ρ ux + uy + uz +µ 2 + +2 ∂x ∂y ∂x ∂y ∂z ∂y 2 ∂z ∂p ⇒ = −ρ a2 y ∂y ⇒ p ( x, y , z ) = − 1 ρ a 2 y + g ( x, z ) 2 ∂ 2u ∂u ∂u ∂u ∂ 2u ∂ 2u ∂p = − ρ u x z + u y z + u z z + µ 2z + 2z + 2z ∂x ∂z ∂x ∂y ∂z ∂y ∂z ∂p = −4 ρ a 2 z ⇒ ∂z ⇒ p ( x, y, z ) = −2 ρ a 2 z 2 + h ( x, y ) Combining the previous expressions and noting that p(0, 0, 0) = p0: p ( x, y, z ) = p0 − 1 ρ a 2 x 2 − 1 ρ a 2 y − 2 ρ a 2 z 2 2 2 The vorticity field is: ∂u ωi = ε ijk k ∂x j ωx = ∂u z ∂u y − =0 ∂y ∂z ∂u x ∂u z − =0 ∂z ∂x ∂u y ∂u x ωz = − =0 ∂x ∂y ωy = ω ( x, y, z ) = 0 The flow is irrotational! Note that the viscous force terms in the Navier-Stokes equation are zero (µui,jj = 0 ). C. Wassgren Chapter 04: Differential Analysis 187 Last Updated: 05 Sep 2008 6. Acceleration of a Fluid Particle in Streamline Coordinates Often it’s helpful to use streamline coordinates (s,n) instead of Cartesian coordinates (x, y) when describing the motion of a fluid particle. Let’s determine a fluid particle’s acceleration parallel (s-direction) and normal (n-direction) to a streamline for a steady, 2D flow. Consider the figure shown below. fluid particle s = s3 s = s2 streamlines s = s1 ˆ s Notes: 1. The coordinates (s, n) are just like (x, y) coordinates. They specify the location of the fluid particle. 2. Lines of constant s and n are ⊥. ˆ 3. n points toward the center of curvature. ˆ n n = n1 n = n2 n = n3 The acceleration of the fluid particle is given by: Du a= Dt ˆ where u = us . Substituting and expanding gives: ˆ D ( us ) ˆ Du Ds ˆ a= =s +u Dt Dt Dt Now let’s expand the Lagrangian derivative terms keeping in mind that u=u(s, n): Du ∂u ∂u ∂u ∂u = + un + us =u Dt ∂t ∂n ∂s ∂s = 0 (flow tangent to streamline) = 0 (steady) (4.67) (4.68) (4.69) = u (flow tangent to streamline) and ˆ Ds = Dt ˆ ∂s ∂t = 0 (steady) + un = 0 (flow tangent to streamline) ˆ ∂s + ∂n C. Wassgren Chapter 04: Differential Analysis us =u (flow tangent to streamline) ˆ ˆ ∂s ∂s =u ∂s ∂s (4.70) 188 Last Updated: 05 Sep 2008 ˆ To determine how s varies with the s-coordinate, consider the following figures: O dθ radius of curvature, R streamline ˆ n ( s) B ds A ˆ s ( s + ds ) ˆ s ( s + ds ) O’ dθ B’ ˆ ds ˆ s ( s ) A’ ˆ s(s) Note that the triangles OAB and O’A’B’ are similar. Hence, ˆ ˆ ds 1 ds ds ˆ = = ds ⇒ = ˆ R ds R s (4.71) =1 ˆ ˆ Also as ds→0, ds points in the n direction so: ˆ ds 1 ˆ =n ds R Substituting Eqn. (4.72) into Eqn. (4.70) gives: ˆ Ds u ˆ =n Dt R (4.72) (4.73) Substituting Eqns. (4.73) and (4.69) into equation (4.68) gives the fluid particle acceleration in streamline coordinates: u2 ∂u ˆ ˆ a = u s + n (4.74) R ∂s tangential acceleration normal acceleration C. Wassgren Chapter 04: Differential Analysis 189 Last Updated: 05 Sep 2008 Example: Water flows through the curved hose shown below with an increasing speed of u = 10t ft/s, where t is in seconds. For t = 2 s determine: a. the component of acceleration along the streamline, b. the component of acceleration normal to the streamline, and c. the net acceleration (magnitude and direction). u R = 20 ft SOLUTION: The acceleration component in the streamline direction is: ∂u ∂u as = +u ∂t ∂s where ∂u = 10 ft 2 (The flow is unsteady.) s ∂t ∂u = 0 (The flow velocity doesn’t change with respect to position along the streamline.) ∂s ∴ as = 10 ft 2 s (4.75) The acceleration component normal to the streamline is: an = u2 R (4.76) where u 2 (10 * 2 ft/s ) = = 20 ft 2 (The velocity is evaluated at t = 2 s.) s R 20 ft ∴ an = 20 ft 2 (The acceleration is toward the center of curvature.) s 2 The net acceleration is: ˆ ˆ a = an n + as s ˆ ˆ a = ( 20n + 10s ) ft s2 as (4.77) an a a = 22.4 ft s2 C. Wassgren Chapter 04: Differential Analysis 190 Last Updated: 05 Sep 2008 7. Euler’s Equations in Streamline Coordinates Recall from previous analyses that the differential equations of motion for a fluid particle in an inviscid flow in a gravitational field are: Du ρ = −∇p + ρ g (Euler’s Equations) (4.78) Dt For simplicity, further assume that we’re dealing with a 2D, steady flow. Now write Eqn. (4.78) in streamline coordinates (s, n): fluid particle streamlines ˆ s ˆ n ∂p + ρ gs ∂s ∂p n-direction: ρ an = − + ρ g n ∂n s-direction: ρ as = − (4.79) (4.80) Recall that in streamline coordinates: u2 ∂u and an = R ∂s so that Eqns. (4.79) and (4.80) become: ∂u 1 ∂p u =− + gs ∂s ρ ∂s 2D, steady Euler’s equations in streamline coordinates 1 ∂p u2 =− + gn ρ ∂n R as = u (4.81) We can draw an important and very useful conclusion from the normal component of Eqn. (4.81). For a flow moving in a straight line (R → ∞ ) and neglecting gravity (gn = 0) we have: ∂p =0 i.e., the pressure does not change normal to the direction of the flow! ∂n This is very helpful when considering the pressure in a free jet (shown in the figure below). Since free jets typically have negligible curvature and gravitational effects, the pressure everywhere normal to the free jet will be the same! p = patm ˆ n g free jet p = patm R→∞ ˆ In n direction: 0 = − 1 ∂p ∂p +g ⇒ = ρg ρ ∂n ∂n u2 1 ∂p ∂p u2 =− ⇒ = −ρ ⇒ p ↑ as n ↓ R ρ ∂n ∂n R The largest pressure is on the outside bend while the smallest pressure is on the inside bend. If the fluid is a liquid and the inside bend pressure reaches the vapor pressure of the liquid, cavitation will occur. ˆ In n direction: R ˆ n C. Wassgren Chapter 04: Differential Analysis 191 Last Updated: 05 Sep 2008 8. Energy Equation (aka COE for a differential CV) The energy equation, which is simply conservation of energy for a differential fluid element or control volume, can be derived several different ways. Two of these methods are given below. Method 1: Apply the integral approach to a differential control volume: y dy z x dz dx Assume that the density, specific total energy, and velocity are ρ, e, and u, respectively, at the control volume’s center. The total energy fluxes through each of the side of the control volume are given by: ∂ ( mx e )center 1 − 2 dx ( me )in through left = ( mx e )center + ∂x ∂ = ( ρ u x dydze ) + ( ρ u x dydze ) ( − 1 2 dx ) ∂x ∂ = ρ u x e + ( ρ u x e ) ( − 1 2 dx ) ( dydz ) ∂x (where mx ,center is the mass flux in the x-direction at the center of the control volume (Recall ( ) the Taylor Series approximation discussed in Chapter 01.)) ∂ ( me )out through right = ρ u x e + ( ρ u x e ) ( 1 2 dx ) ( dydz ) ∂x ( me )in through bottom = ρ u y e + ∂ ρ u y e ( − 1 2 dy ) ( dxdz ) ∂y ( ) ) ( 1 2 dy ) ( dxdz ) ( me )out through top = ρ u y e + ∂ ρu y e ∂y ( me )in through back = ρ u z e + ∂ ( ρ u z e ) ( − 1 2 dz ) ( dxdy ) ∂z ( ∂ ( ρ uz e ) ( 1 2 dz ) ( dxdy ) ∂z where the specific total energy (not including the potential energy), e, is given by: e = u + 12 u ⋅ u and u is the specific internal energy. ( me )out through front = ρ uz e + Thus, the flux of total energy out of the control volume is: ∂ ∂ ∂ ( me )net, out of CV = ( ρ u x e ) + ρ u y e + ( ρ u z e ) ( dxdydz ) ∂y ∂z ∂x ( C. Wassgren Chapter 04: Differential Analysis ) 192 (4.82) (4.83) Last Updated: 05 Sep 2008 The rate at which the total energy is increasing within the control volume is given by: ∂ ∂ ∂ = ( eρ dxdydz ) = ( e ρ )( dxdydz ) ( me ) ∂t ∂t ∂t within CV (4.84) where ρ and e are the density and specific internal energy, respectively, at the center of the control volume. Note that since these quantities vary linearly within the CV (from the Taylor Series approximation), the averages within the CV are ρ and e. The rate at which heat is added to the control volume is given by: Qinto CV = δ qinto CV ( dxdydz ) (4.85) where δ qinto CV is the rate of heat transfer into the control volume per unit volume. Note that the mode of heat transfer is not indicated at this point in the derivation. The rate at which work is done on the control volume due to body forces is given by: WB,on CV = ( f B ⋅ u ) ( ρ dxdydz ) = f B ,i ui ( ρ dxdydz ) ( ) (4.86) Note that the potential energy is not included in Eqn. (4.82) since that term is included in the rate of body force work term in Eqn. (4.86). The rate at which work is done on the control volume due to surface forces is given by: WS ,on CV = dF ⋅ u σxydA ∂ (σ xx u x ) ∂ (σ u ) WS ,on CV = σ xx u x + ( 1 2 dx ) − σ xx ux + xx x ( − 1 2 dx ) ( dydz ) ∂x ∂x u σxxdA ( ∂ σ xy u y + σ xy u y + ∂x ) ( 1 2 dx ) − σ xy ux + ( ∂ σ xy u y ∂x ) ( − 1 2 dx ) ( dydz ) ∂ (σ xz u z ) ∂ (σ u ) + σ xz u z + ( 1 2 dx ) − σ xz uz + ∂xz z ( − 1 2 dx ) ( dydz ) ∂x x + ∂ = σ ji ui ( dxdydz ) ∂x j ( (4.87) ) Conservation of energy states that the rate of increase of total energy within the control volume plus the net rate at which total energy leaves the control volume must equal the rate at which heat is added to the control volume plus the rate at which work is done on the control volume. ∂ + ( me ) = Qinto CV + WB,on CV + WS ,on CV (4.88) ( me ) net, out of CV ∂t within CV Substituting Eqns. (4.83) – (4.87) into Eqn. (4.26) gives: ∂ ∂ ∂ ∂ ( eρ )( dxdydz ) + ( ρ u x e ) + ρ u y e + ( ρ uz e ) ( dxdydz ) = ∂t ∂y ∂z ∂x ( ) + δ qinto ( dxdydz ) + f B,i ui ( ρ dxdydz ) + ∂ ∂ ∂ e ρ u j = δ qinto CV + ρ ui f B ,i + σ ji ui ( eρ ) + ∂t ∂x j ∂x j ( ) C. Wassgren Chapter 04: Differential Analysis ( 193 ) ∂ σ ji ui ∂x j ( ) ( dxdydz ) (4.89) Last Updated: 05 Sep 2008 Expand the left hand side of Eqn. (4.89) and utilize the continuity equation: ∂ ∂ ∂ρ ∂e ∂ ∂e ρu j + ρu j eρ u j = e +ρ +e ( eρ ) + ∂t ∂x j ∂t ∂t ∂x j ∂x j ( ) ( ) ∂ρ ∂e ∂ ∂e ρu j + ρ + u j = e + ∂t ∂x j ∂t ∂x j ( ) = 0 (continuity eqn) Substituting back into Eqn. (4.89) gives: De ∂ ρ = δ qinto CV + ρ ui f B,i + σ ji ui Dt ∂x j ( C. Wassgren Chapter 04: Differential Analysis = ) De Dt energy equation 194 (4.90) Last Updated: 05 Sep 2008 Method 2: Recall the integral form of COE: d e ρ dV + e ( ρ u rel ⋅ dA ) = Qinto CV + WB, on CV + WS , on CV dt ∫ ∫ CV CS Consider a fixed control volume so that: ∂ ( eρ ) d e ρ dV = dV and dt ∂t ∫ ∫ CV CV u rel = u Note that the heat transfer into the CV can be written as: ∫ δq Qinto CV = into CV dV CV The work on the CV due to body forces is written as: WB,on CV = ∫ (f ⋅ u ) ρ dV = B CV ∫(f B ,i ui ) ρ dV CV and the work on the CV due to surface forces is written as: WS ,on CV = ∫ (f S ⋅ u ) dA = CS ∫(f S ,i ui ) dA = ∫ σ ji n j ui dA CS CS where the surface forces have been written in terms of the stresses. By utilizing Gauss’ Theorem (aka the Divergence Theorem), we can convert the area integrals into volume integrals: ∂ e ρ ( u ⋅ dA ) = ∇ ⋅ ( e ρ u ) dV = e ρ u j dV ∂x j ∫ ∫ CV ( ∫ CS ) CV ∫ σ ji n j ui dA = CS ∫ ( ∂ uiσ ji ∂x j CV ) dV Substitute these expressions back into COE to get: ∂ ∂ ∂ e ρ u j − δ qinto CV − ρ f B,i ui − σ ji ui dV = 0 ( eρ ) + ∂x j ∂x j ∂t CV Since the choice of control volume is arbitrary, the kernel of the integral must be zero, i.e.: ∂ ∂ ∂ e ρ u j − δ qinto CV − ρ f B ,i ui − σ ji ui = 0 ( eρ ) + ∂t ∂x j ∂x j ( ∫ ( ) ( ) ( ) ) But this is the same expression as Eqn. (4.89) so we see that the final result will be the same! De ∂ ρ = δ qinto CV + ρ ui f B,i + σ ji ui Dt ∂x j ( C. Wassgren Chapter 04: Differential Analysis ) 195 Last Updated: 05 Sep 2008 Notes: 1. The rate of heat transfer may be re-written in terms of the rate of heat transfer per unit area out of the control volume through the control surface, q. In terms of the Method 1 approach, the rate of total heat transfer into the control may be written as: ∂q ∂q δ Qinto CV = − − qx + x ( − 1 2 dx ) + qx + x ( 1 2 dx ) ( dydz ) ∂qx 1 x ∂x ∂x qx + ∂x 2 dx ∂q y ∂q y − − q y + ( − 1 2 dy ) + q y + ( 1 2 dy ) ( dxdz ) ∂y ∂y ∂q − qx + x − 1 dx ∂qz ∂qz 1 2 1 dz ) + q + ∂x − − qz + ( − 2 z ∂z ( 2 dz ) ( dxdy ) ∂z Simplifying the previous relation gives: ∂q y ∂qz ∂q j ∂q δ Qinto CV = − x + + ( dxdydz ) ( dxdydz ) = − ( ∇ ⋅ q )( dxdydz ) = − ∂y ∂z ∂x j ∂x The rate of total heat transfer in terms of the rate of heat transfer per unit area may also be derived using the Method 2 approach and the Divergence Theorem: ∂q j δ Qinto CV = − ( q ⋅ dA ) = − q j n j dA = − dV ∂x j ( ( ∫ ∫ ∫ CS CS ) CV Substituting the previous expressions into the energy equation (Eqn. (4.90)) gives: ∂q De ∂ ρ σ ji ui = − i + ρ ui f B ,i + Dt ∂xi ∂x j ( ) (4.91) energy equation in terms of the heat transfer per unit area 2. The energy equation (either Eqn. (4.90) or (4.91)) may be simplified further by noting that: De D ∂u ∂ ∂u ∂ = ( u + 1 2 u ⋅ u ) = ∂t + ∂t 1 ui ui + u j ∂x + u j ∂x 1 ui ui 2 2 Dt Dt j j ( ) ( u ∂u ∂u ∂u = +uj + ui i + u j ui i ∂t ∂x j ∂t ∂x j = Du Dt = ui ) (4.92) Dui Dt and ∂σ ji ∂u ∂ σ ji ui = ui + σ ji i ∂x j ∂x j ∂x j ( ) (4.93) Substituting these expressions into Eqn. (4.90) (or (4.91)) gives: ∂σ ji Dui ∂ui Du + ui ρ = δ qinto CV + ρ ui f B ,i +ui ∂x + σ ji ∂x Dt Dt j j ∂σ ji ∂u Dui Du = δ qinto CV + σ ji i + ui ρ f B,i + −ρ Dt ∂x j ∂x j Dt but the terms in brackets are just the momentum equations! ρ C. Wassgren Chapter 04: Differential Analysis 196 Last Updated: 05 Sep 2008 ) The dot product of the momentum equations with the velocity is known as the mechanical energy equation. ∂σ ji Dui ui ρ = ρ ui f B ,i + ui mechanical energy equation (4.94) Dt ∂x j rate of increase of kinetic energy of fluid element rate at which work is done on the fluid element due to body forces rate at which work is done on the fluid element due to stress gradients Note that: Du D1 ( 2 ui ui ) = ui Dti Dt The energy equation without the mechanical energy equation terms is known as the thermal energy equation and is given by: ∂u Du ρ = δ qinto CV + σ ji i Dt ∂x j or, in terms of the heat transfer per unit area: ∂q Du ρ = −i + Dt ∂xi rate of increase of internal energy within the fluid element 3. rate at which heat is added to the fluid element through the surface area σ ji ∂ui ∂x j thermal energy equation (4.95) rate at which mechanical energy is converted to thermal energy due to deformations of the fluid element The heat flux term, q, may be written in terms of a temperature gradient using Fourier’s Law of Conduction (assuming that conduction is the dominant mode of heat transfer): ∂T q = −k ∇T or qi = −k (4.96) ∂xi where k is the thermal conductivity (in its most general form, the thermal conductivity is a tensor quantity) of the substance and T is the temperature. Note that the negative sign is included in Eqn. (4.96) to account for the fact that heat flows from regions of high temperature to regions of low temperature. Thus, the thermal energy equation can be written as: ∂ui Du ∂ ∂T ρ = (4.97) k + σ ji Dt ∂xi ∂xi ∂x j Thermal Energy Eqn using Fourier’s Law of Conduction C. Wassgren Chapter 04: Differential Analysis 197 Last Updated: 05 Sep 2008 4. The rate of work term in the thermal energy equation (Eqn. (4.95)) includes both reversible and irreversible work terms. Consider the rate of work term using the stress tensor for a Newtonian fluid: ∂u j ∂ui ∂ui ∂u ∂u σ ji i = − p + λ k δ ij + µ + ∂xi ∂x j ∂x j ∂x j ∂xk ∂u j = −p ∂x j 2 ∂u j ∂ui ∂u + +λ k +µ ∂xi ∂x j ∂xk reversible pressure work ∂u i ∂x j irreversible viscous work The irreversible rate of work term (the rate at which mechanical energy is being converted into thermal energy) is often referred to as the energy dissipation function, Φ: 2 ∂u j ∂ui ∂u Φ = λ k +µ + ∂xi ∂x j ∂xk ∂u i ∂x j energy dissipation function Thus, the thermal energy equation can be written as: ∂u j Du ρ = δ qinto CV − p +Φ ∂x j Dt or, if conduction is the significant mode of heat transfer: ∂u j Du ∂ ∂T ρ = k − p +Φ ∂x j Dt ∂x j ∂x j 5. (4.98) (4.99) (4.100) Note that for an incompressible fluid, the continuity equation: ∂ui =0 ∂xi can be used to simplify the thermal energy equation and the energy dissipation function to the following forms: Du ρ = δ qinto CV + Φ (4.101) Dt thermal energy equation for an incompressible fluid where the energy dissipation function is given by: ∂u j ∂ui ∂ui Φ = µ + (4.102) ∂xi ∂x j ∂x j energy dissipation function for an incompressible, Newtonian fluid C. Wassgren Chapter 04: Differential Analysis 198 Last Updated: 05 Sep 2008 6. The energy dissipation function for a Newtonian fluid is a positive definite quantity which means that viscosity always acts to convert mechanical energy into thermal energy. 2 ∂u i ∂x j 2 1 ∂u ∂u j i + 2 ∂x j ∂xi ∂u j ∂ui ∂u Φ = λ k +µ + ∂xk ∂xi ∂x j ∂u j ∂ui ∂u = λ k +µ + ∂xk ∂xi ∂x j 1 ∂u ∂u j + i − 2 ∂x j ∂xi 2 2 1 ∂u j ∂ui ∂u j ∂u j ∂ui ∂ui ∂ui ∂u j − + − + µ 2 ∂xi ∂x j ∂xi ∂xi ∂x j ∂x j ∂x j ∂xi 2 1 + µ 2 ∂u 1 ∂u j ∂ui = λ k + µ + 2 ∂xi ∂x j ∂xk ∂u 1 ∂u j ∂ui = λ k + µ + 2 ∂xi ∂x j ∂xk 2 ∂u j ∂u j ∂ui ∂ui + − ∂xi ∂xi ∂x j ∂x j = 0( since i and j are dummy indices ) 2 ∂u 1 ∂u j ∂ui ∴Φ = λ k + µ + 2 ∂xi ∂x j ∂xk 2 (4.103) Note that for an incompressible fluid: ∂uk =0 ∂xk so that: Φ= 1 ∂u j ∂ui µ + 2 ∂xi ∂x j 2 >0 since µ>0. For a compressible, Newtonian fluid, Stokes’ hypothesis states that: λ = − 23 µ After simplifying Eqn. (4.103) it can also be shown that Φ>0. C. Wassgren Chapter 04: Differential Analysis 199 Last Updated: 05 Sep 2008 6. The total energy equation may also be written in a form using the specific enthalpy, h, which is defined as: p h≡u+ (4.104) ρ Consider the thermal energy equation (Eqn. (4.99)) using the definition of the energy dissipation function (equation (4.98)): ∂u j Du ρ = δ qinto CV − p (4.105) +Φ ∂x j Dt Re-write the pressure term in the previous equation utilizing the continuity equation: ∂u j 1 Dρ D p Dp (4.106) p = p − =ρ − ∂x j Dt ρ Dt ρ Dt Substituting into Eqn. (4.99) gives: Du D p Dp ρ = δ qinto CV − ρ +Φ + Dt Dt ρ Dt D p Dp +Φ u + = δ qinto CV + Dt Dt ρ Dh Dp ∴ρ = δ qinto CV + +Φ Dt Dt ρ (4.107) For an incompressible fluid, Eqn. (4.105) is: Du ρ = δ qinto CV + Φ Dt For an incompressible fluid the internal energy is a function only of temperature, i.e., u = u(T) = cT. Thus, the energy equation for an incompressible fluid becomes: DT ρc = δ qinto CV + Φ energy eqn for an incompressible fluid (4.108) Dt Note: For an incompressible substance, cv = cp = c. Thus, the energy equation is uncoupled from the continuity and momentum equations for an incompressible flow. In other words, the unknowns of velocity, ui , and pressure, p, (4 unknowns) can be solved using the continuity and momentum equations (4 equations). Once these quantities are determined, the internal energy (i.e. temperature) can be calculated using the thermal energy equation (Eqn. (4.108) ). C. Wassgren Chapter 04: Differential Analysis 200 Last Updated: 05 Sep 2008 7. For an adiabatic, inviscid flow, Eqn. (4.107) reduces to: Dh Dp ρ = (4.109) Dt Dt The mechanical energy equation (Eqn. (4.94)) can be used to re-write the Lagrangian derivative of the pressure: Dui ∂p ρ ui = −ui + ρ ui f B ,i Dt ∂xi ρ ui ∂p Dui ∂p ∂p − = − + ui + ρ ui f B ,i Dt ∂t ∂t ∂xi = Dp Dt Dui Dp ∂p = − ρ ui + ρ ui f B ,i Dt ∂t Dt Substituting into Eqn. (4.109): Dui ∂p Dh ∂p D1 = − ρ ui + ρ ui f B ,i = −ρ ρ ( 2 ui ui ) + ρ ui f B,i Dt ∂t Dt ∂t Dt (4.110) D ∂p ρ ( h + 1 2 ui ui ) = + ρ ui f B,i ∂t Dt If the body force is conservative, i.e. fB = -∇G, then: ∂G DG ∂G ρ ui f B,i = − ρ ui = −ρ +ρ (4.111) ∂xi Dt ∂t Furthermore, if the body force is also independent of time (obviously a good assumption if gravity is the only body force considered), then Eqn. (4.111) may be substituted into Eqn. (4.110) and then simplified to give: D ∂p DG ρ ( h + 1 2 ui ui ) = ∂t − ρ Dt Dt D 1 ∂p h + 1 2 ui ui + G = (4.112) ρ ∂t Dt = h0 , stagnation enthalpy energy equation for the adiabatic flow of an inviscid fluid with conservative, time-independent body forces With the additional assumption that the flow is steady, we observe that the total specific enthalpy of a fluid particle will remain constant, i.e. Dh0/Dt = 0. In particular, the total specific enthalpy will remain constant along a streamline. ∴ C. Wassgren Chapter 04: Differential Analysis 201 Last Updated: 05 Sep 2008 9. Entropy Equation (aka the 2nd Law for a differential CV) The entropy equation, which is simply the 2nd Law of Thermodynamics for a differential fluid element or control volume, can be derived several different ways. Three of these methods are given below. Method 1: Apply the integral approach to a differential control volume: y dy z x dz dx Assume that the density, specific entropy, and velocity are ρ, s, and u, respectively, at the control volume’s center. The entropy fluxes through each of the side of the control volume are given by: ∂ ( mx s )center 1 − 2 dx ( ms )in through left = ( mx s )center + ∂x ∂ = ( ρ u x dydzs ) + ( ρ u x dydzs ) ( − 1 2 dx ) ∂x ∂ = ρ u x s + ( ρ u x s ) ( − 1 2 dx ) ( dydz ) ∂x (where mx ,center is the mass flux in the x-direction at the center of the control volume (Recall ( ) the Taylor Series approximation discussed in Chapter 01.)) ∂ ( ms )out through right = ρ u x s + ( ρ u x s ) ( 1 2 dx ) ( dydz ) ∂x ∂ ( ms )in through bottom = ρ u y s + ρ u y s ( − 1 2 dy ) ( dxdz ) ∂y ( ( ms )out through top = ρ u y s + ) ∂ ρu y s ∂y ( ) ( 1 2 dy ) ( dxdz ) ∂ ( ms )in through back = ρ u z s + ( ρ u z s ) ( − 1 2 dz ) ( dxdy ) ∂z ∂ ( ms )out through front = ρ u z s + ( ρ u z s ) ( 1 2 dz ) ( dxdy ) ∂z Thus, the net entropy flux out of the control volume is given by: ∂ ∂ ∂ ( ms )net, out of CV = ( ρ u x s ) + ρ u y s + ( ρ u z s ) ( dxdydz ) ∂y ∂z ∂x The rate at which the entropy is increasing within the control volume is given by: ∂ ∂ ∂ = ( s ρ dxdydz ) = ( s ρ )( dxdydz ) ( ms ) ∂t ∂t ∂t within CV ( ) (4.113) (4.114) where ρ and s are the density and specific entropy, respectively, at the center of the control volume. Note that since these quantities vary linearly within the CV (from the Taylor Series approximation), the averages within the CV are ρ and s. C. Wassgren Chapter 04: Differential Analysis 202 Last Updated: 05 Sep 2008 The 2nd Law states that the rate of increase of entropy within the control volume plus the net rate at which entropy leaves the control volume must be greater than or equal to the rate of heat transfer into the control volume divided by the temperature of the fluid element to which the heat is transferred: δq ∂ + ( ms ) net, out of CV ≥ into CV ( dxdydz ) (4.115) ( ms ) ∂t T within CV where δ qinto CV is the rate of heat transfer into the CV per unit volume. Substituting Eqns. (4.113) and (4.114) into Eqn. (4.115) gives: δq ∂ ∂ ∂ ∂ ( s ρ )( dxdydz ) + ( ρ u x s ) + ρ u y s + ( ρ u z s ) ( dxdydz ) ≥ into CV ( dxdydz ) ∂t ∂y ∂z T ∂x ( ) ∂ δq ∂ ∂ ∂ ( s ρ ) + ( ρ u x s ) + ρ u y s + ( ρ u z s ) ≥ into CV T ∂t ∂x ∂y ∂z This equation can be written in the following compact forms: δq ( s ρ ) + ∇ ⋅ ( ρ us ) ≥ into CV ∂t T or δq ∂ ∂ ρ u j s ≥ into CV ( sρ ) + T ∂t ∂x j ( ( ) (4.116) ) (4.117) Expand the left hand side of Eqn. (4.117) and utilize the continuity equation: ∂ ∂ ∂ρ ∂s ∂ ∂s ρu j s = s + ρ + s ρu j + ρu j ( sρ ) + ∂t ∂x j ∂t ∂t ∂x j ∂x j ( ) ( ) ∂ρ ∂s ∂ ∂s = s + ρu j + ρ + u j ∂t ∂x j ∂t ∂x j ( ) = 0 (continuity eqn) = Ds Dt Substituting back into Eqn. (4.117) gives: Ds δ qinto CV ρ ≥ entropy equation Dt T (4.118) Method 2: Apply the 2nd Law of Thermodynamics directly to a small piece of fluid: δq D (4.119) ( s ρ dxdydz ) ≥ into CV ( dxdydz ) Dt T Expanding the Lagrangian derivative gives: D Ds D ( s ρ dxdydz ) = ( ρ dxdydz ) + s ( ρ dxdydz ) Dt Dt Dt but the second term on the RHS of this equation will be zero since the mass of the fluid element remains constant. Thus, Eqn. (4.119) can be simplified to: Ds δ qinto CV ρ ≥ same result as before! Dt T C. Wassgren Chapter 04: Differential Analysis 203 Last Updated: 05 Sep 2008 Method 3: Recall the integral form of the 2nd Law: δ qinto CV d s ρ dV + s ( ρ u rel ⋅ dA ) ≥ dV dt T ∫ ∫ ∫ CV CS CV Consider a fixed control volume so that: ∂ ( sρ ) d s ρ dV = dV and dt ∂t ∫ ∫ CV CV u rel = u By utilizing Gauss’ Theorem (aka the Divergence Theorem), we can convert the area integral into a volume integral: ∂ s ( ρ u ⋅ dA ) = ∇ ⋅ ( s ρ u ) dV = ρ u j s dV ∂x j ∫ ∫ ∫ CS CV ( ) CV Substitute these expressions back into the 2nd Law to get: ∂ δq ∂ ρ u j s − into CV dV = 0 ( sρ ) + ∂x j T ∂t CV ( ∫ ) Since the choice of control volume is arbitrary, the kernel of the integral must be zero, i.e.: δq ∂ ∂ ρ u j s − into CV = 0 ( sρ ) + ∂t ∂x j T ( ) But this is the same expression as Eqn. (4.118) so we see that the final result will be the same! Ds δ qinto CV ρ ≥ Dt T Notes: 1. For a reversible, adiabatic flow (aka an isentropic flow): Ds =0 Dt 2. Recall that for a simple, compressible system where the only surface forces are reversible pressure forces (so that the “=” may be used in Eqn. (4.118)), the 1st and 2nd Laws of Thermodynamics may be combined to give: De 1 Dv = δ qinto sys − p Dt ρ Dt (i.e. de = vδq – pdv where δq is the heat per unit volume, not per unit mass) where v is the specific volume and for a simple, compressible system the total specific energy is equal to the specific internal energy, i.e. e = u. Re-writing the specific volume in terms of the density, utilizing Eqn. (4.118), and simplifying gives: Du Ds D1 =T −p Dt Dt Dt ρ Ds Du p D ρ = − (4.120) Dt Dt ρ 2 Dt This expression may also be written in terms of the specific enthalpy, h, by re-writing the 2nd term on the RHS of the equation: D 1 1 Dp D p Ds Du D p 1 Dp −p − = + = ⇒T − Dt ρ ρ Dt Dt ρ Dt Dt Dt ρ ρ Dt T = Dh Dt T Ds Dh 1 Dp = − Dt Dt ρ Dt C. Wassgren Chapter 04: Differential Analysis (4.121) 204 Last Updated: 05 Sep 2008 10. Vorticity Dynamics Recall that the vorticity, ω, of a fluid element is equal to twice the rotation rate of the element: ∂u ω = ∇ × u or ωi = ε ijk k ∂x j Notes: 1. A rotational flow is defined as one in which the vorticity is not zero. 2. An irrotational flow is defined as one in which there is no vorticity, i.e. irrotational ⇒ ω = ∇ × u = 0 A useful concept when discussing vorticity is the vortex line. A vortex line is a line that is everywhere tangent to the flow’s vorticity vectors. Notes: 1. A vortex line is analogous to a streamline. 2. A vortex tube is a tube made by all the vortex lines passing through a closed curve. vortex lines 3. A vortex filament is a vortex tube with infinitesimally small cross-section. 4. There are no vortex lines in an irrotational flow. 5. There can be no sources or sinks of vorticity in a flow. This follows from the following vector identity: ∇ ⋅ (∇ × u ) = 0 ∴∇ ⋅ ω = 0 (4.122) Zero divergence of vorticity means that there are no sources or sinks of vorticity which in turn means that vorticity is neither created nor destroyed in a flow. So how then is vorticity generated in a flow? It must be introduced at a boundary (e.g. a fluid or solid boundary). According to Eqn. (4.122), vortex lines must either form closed curves or start and end at boundaries. C. Wassgren Chapter 04: Differential Analysis 205 Last Updated: 05 Sep 2008 6. Another useful quantity for the discussion of vorticity dynamics is the circulation, Γ: Γ≡ ∫ u ⋅ ds ds (4.123) u flow C A C The relationship between the vorticity and the circulation about a curve, C, enclosing an area, A, with ˆ unit normal, n , is found using Stokes’ Theorem: Γ≡ ˆ ∫ u ⋅ ds = ∫ (∇ × u ) ⋅ ndA C A dΓ ˆ = ω ⋅n dA ∫ ˆ ∴Γ = ω ⋅ ndA or A (4.124) Notes: a. The circulation around any cross-section of the same vortex tube remains constant. Recall that: ∇ ⋅ω = 0 so that ω2 ∫ ( ∇ ⋅ ω ) dV = 0 V ⇒ ˆ ∫ ( ω ⋅ n ) dA = 0 divergence theorem S where V is the volume enclosed within the vortex tube and S is the surface area of this volume. A2 ω1 A1 ˆ n Aside Breaking the total area into the area of the top, bottom, and sides: ˆ ˆ ˆ ˆ ∫ ( ω ⋅ n ) dA = 0 = ∫ (ω ⋅ n ) dA + ∫ ( ω ⋅ n ) dA + ∫ ( ω ⋅ n ) dA S A1 A2 Aside On the sides of the vortex tube, the normal vector for the area is perpendicular to the vorticity vectors (from the definition of a vortex tube) so that: ˆ ( ω ⋅ n )sides = 0 Thus, 0= ˆ ˆ ∫ ( ω ⋅ n ) dA + ∫ ( ω ⋅ n ) dA A1 − A2 ˆ ˆ ∫ ( ω ⋅ n ) dA = ∫ (ω ⋅ n ) dA A1 A2 Using Eqn. (4.124) and noting that the outward pointing normal vector on area A1 points in the opposite direction as the vorticity vector there, we have: Γ1 = Γ 2 Hence, the circulation around any cross-section of the same vortex tube remains constant. This observation is also known as Helmholtz’s Third Law. C. Wassgren Chapter 04: Differential Analysis 206 Last Updated: 05 Sep 2008 11. Vorticity Transport Equations (aka Helmholtz Eqn) The vorticity transport equation is an alternate expression of the Navier-Stokes equations. Consider the Navier-Stokes equations for an incompressible fluid with constant dynamic viscosity: Du ρ = −∇p + µ∇ 2 u + ρ f Dt Divide through by the density, ρ, (note that it is a constant here since we’re considering an incompressible fluid) and also write the body force, f, as the gradient of a potential function, G (allowable if f is a conservative body force): g p µ Du ˆ = −∇ + ∇ 2u − ∇G ( f B = −∇G , e.g. Let G = gz so that f B = − ge z .) z (4.125) Dt ρ ρ =ν Note that the kinematic viscosity, ν, is the ratio of the dynamic viscosity to the density. Now expand the acceleration term on the LHS: Du ∂u = + ( u ⋅∇ ) u Dt ∂t The second term in the previous equation can be expanded using the following vector identity: (u ⋅ ∇ ) u = 1 ∇ (u ⋅ u ) − u × (∇ × u ) 2 =ω Substituting these relations into equation (4.125) gives: p ∂u 1 + ∇ ( u ⋅ u ) − u × ω = −∇ +ν∇ 2 u − ∇G ∂t 2 ρ Now take the curl of Eqn. (4.126) and simplify: ∂u p ∇ × + 1 ∇ ( u ⋅ u ) − u × ω = −∇ + ν∇ 2 u + ∇G 2 ρ ∂t where (4.126) ∂u ∂ ∂ω = (∇ × u ) = ∂t ∂t ∂t ∇ × 1 ∇ ( u ⋅ u ) = 0 (From the vector identity: ∇×∇φ=0.) 2 ∇× ∇ × (u × ω ) = u (∇ ⋅ ω ) −ω = 0, vorticity is divergence free (∇ ⋅ u ) − ( u ⋅∇ ) ω + ( ω ⋅∇ ) u (using a vector identity) = 0, continuity equation p ∇×∇ = 0 ρ ∇ ×ν∇ 2 u = ν∇ 2 ( ∇ × u ) = ν∇ 2 ω ∇ × ∇G = 0 Substituting and simplifying: ∂ω + ( u ⋅∇ ) ω − ( ω ⋅∇ ) u = ν∇ 2 ∂t = Dω Dt Dω = ( ω ⋅∇ ) u + ν∇ 2 ω (4.127) Dt Vorticity Transport Equation for an incompressible, Newtonian fluid (aka Helmholtz Eqn) ∴ C. Wassgren Chapter 04: Differential Analysis 207 Last Updated: 05 Sep 2008 Notes: 1. Let’s interpret what each of the terms in the vorticity transport equation means: Dω = ( ω ⋅∇ ) u + ν∇ 2 ω Dt diffusion of rate of change of fluid element vorticity stretching and turning of a vortex line vorticity 2. The vorticity transport equations do not contain pressure or body force terms explicitly. Assuming uniform density, the pressure and body forces act through the center of mass of the element and thus cannot produce rotation. Only the shear stresses may produce vorticity. Note that in a stratified flow where the density gradient results in a non-coincident geometric center and center of mass, the pressure forces can produce rotation of the fluid element. Hence, Eqn. (4.127) should not be used for stratified flows. 3. The vorticity transport equations are sometimes used in numerical calculations in place of the NavierStokes equations. 4. For a 2D flow, the Vorticity Transport Equation simplifies to: Dω = ν∇ 2ω (since the vorticity points in a direction perpendicular to the stream lines) Dt Hence, vorticity can only diffuse (and not stretch) in a 2D flow. If the flow is inviscid, then: Dω = 0 (2D inviscid flow) Dt and the vorticity remains constant for each fluid element. Hence, in a 2D inviscid flow, if the flow starts off irrotational, then it must remain irrotational! This is a very important result that will be explored more fully when discussing Kelvin’s Theorem. C. Wassgren Chapter 04: Differential Analysis 208 Last Updated: 05 Sep 2008 12. Bernoulli’s Equation Euler’s equations (the momentum equations for an inviscid fluid) can be simplified to an expression known as Bernoulli’s Equation for the conditions given below: Steady flow of an inviscid fluid in a conservative force field along either a streamline or a vortex line: dp 1 + 2 ( u ⋅ u ) + G = constant (4.128) ∫ρ Irrotational flow of an inviscid fluid in a conservative force field: ∂φ dp 1 + + 2 ( ∇φ ⋅∇φ ) + G = F ( t ) ρ ∂t where u=∇φ and F(t) is a function only of time. ∫ (4.129) Derivation of Bernoulli’s Equation: To begin, first consider Euler’s equations (recall that Euler’s equations are the momentum equations for an inviscid fluid): Du 1 = − ∇p − ∇G Dt ρ where a conservative body force (fB = -∇G) has been assumed. Re-write the convective acceleration term using the following vector identity: (u ⋅ ∇) u = ∇ ( 12 u ⋅ u ) − u × (∇ × u ) ∂u 1 + ∇ ( 1 2 u ⋅ u ) − u × ( ∇ × u ) = − ∇p − ∇G ∂t ρ Collect gradient terms on the left hand side of the equation: 1 ∂u ∇p + ∇ ( 1 2 u ⋅ u ) + ∇G = − + u × (∇ × u ) (4.130) ρ ∂t Note that the pressure gradient term can be re-written in a slightly different form as shown below. 1 dp dp dp Note: =d ∇p ⋅ dx = = ∇ ⋅ dx ρ ρ ρ ρ ∂a ∂a ∂a ˆ ˆ ˆ ˆ ˆ ˆ ∇a ⋅ dx = e x + e y + e z ⋅ ( dxe x + dye y + dze z ) dp 1 ∂y ∂z ∂x ∴ ∇p = ∇ ρ ρ ∂a ∂a ∂a dx + dy + dz = ∂x ∂y ∂z = da ⇒ ∫ ∫ ∫ Substituting this expression into Eqn. (4.130), simplifying, and noting that the vorticity, ω, is given by ω=∇×u: dp 1 ∂u ∇ + 2 u ⋅u + G = − + u×ω (4.131) ρ ∂t Now consider two particular cases. ∫ C. Wassgren Chapter 04: Differential Analysis 209 Last Updated: 05 Sep 2008 Steady flow along a streamline or a vortex line A steady flow results in: ∂u =0 ∂t Taking the dot product of Eqn. (4.131) with a little length of line, dx, that is along either a streamline or a vortex line gives: dp 1 ∇ + 2 u ⋅ u + G ⋅ dx = ( u × ω ) ⋅ dx (4.132) ρ Since the vector (u×ω) is perpendicular to both the streamline and the vortex line, the dot product with dx ω will be zero: ( u × ω ) ⋅ dx = 0 ∫ Furthermore, the dot product of the gradient on the left hand side of Eqn. (4.132) with dx results in an ordinary differential: dp 1 dp 1 ∇ + 2 u ⋅ u + G ⋅ dx = d + 2 u ⋅u + G ρ ρ Thus, dp 1 d + 2 u ⋅u + G = 0 ρ dp 1 + 2 u ⋅ u + G = constant (4.133) ∫ ∫ ∫ ∫ρ Irrotational flow In an irrotational flow, the velocity can be written as the gradient of a velocity potential function, φ, i.e., u=∇φ, since in an irrotational flow, ω=∇×u=0, and from the vector identity, ∇×∇φ=0, thus u can be written as u=∇φ. Substituting into Eqn. (4.131) and noting that the vorticity is zero in an irrotational flow: ∂ ( ∇φ ) dp 1 ∂φ ∇ + 2 ∇φ ⋅∇φ + G = − = −∇ (4.134) ∂t ρ ∂t ∫ Combining gradient terms and simplifying: ∂φ dp 1 ∇ + + 2 ∇φ ⋅ ∇φ + G = 0 (4.135) ρ ∂t Now take the dot product of the previous equation with a short distance in any direction, dx, and integrate the resulting expression along that path: ∂φ dp 1 ∇ + + 2 ∇φ ⋅ ∇φ + G ⋅ dx = 0 ρ ∂t ∫ ∫ ∂φ dp 1 + 2 ∇φ ⋅∇φ + G = 0 d + ρ ∂t ∂φ dp 1 + + 2 ∇φ ⋅∇φ + G = F ( t ) (4.136) ρ ∂t where F(t) is a function only of time (this is introduced in the integration since the terms in the equation may vary with both position and time). ∫ ⇒ ∫ C. Wassgren Chapter 04: Differential Analysis 210 Last Updated: 05 Sep 2008 Notes: 1. For a fluid with constant density, i.e. ρ=constant: dp 1 = dp ∫ρ ∴ 2. ρ dp ∫ρ = ∫ p fluid with constant density ρ For an ideal gas (p=ρRT): isothermal case: T = T0 dp ∫ ρ =∫ d ( ρ RT0 ) ρ = RT0 ∫ dρ ρ ρ = RT0 ln ideal gas with isothermal conditions ρ0 where T0 and ρ0 are a reference temperature and density, respectively ∴ dp ∫ρ isentropic case (constant specific heats, i.e. a perfect gas): dp ∫ρ ∫ = ( − d p0 ρ0 γ ρ γ ρ ) = p ρ γγ − 00 ∫ ρ γ −2 d ρ = γ p0 γ γ − 1 ρ0 γ p p0 = ( ρ ρ0 ) γ −1 ρ γ −1 = γ p0 ρ γ − 1 ρ0 ρ0 where p0 and ρ0 are a reference pressure and density, respectively This expression can be simplified further by noting that for a perfect gas: p0 γR = RT0 cp = ρ0 γ −1 and for a perfect gas undergoing an isentropic process: γ −1 T ρ = T0 ρ0 so that dp ∴ = c pT ∫ρ C. Wassgren Chapter 04: Differential Analysis perfect gas with isentropic conditions 211 Last Updated: 05 Sep 2008 Example: A water tank has an orifice in the bottom of the tank: g h area of orifice, A(0) y cross-sectional area, A(y) The height, h, of water in the tank is kept constant by a supply of water which is not shown. A jet of water emerges from the orifice; the cross-sectional area of the jet, A(y), is a function of the vertical distance, y. Neglecting viscous effects and surface tension, find an expression for A(y) in terms of A(0), h, and y. SOLUTION: Apply conservation of mass to the following CV: 1 0 y 2 d ∫ ρ dV + CS ρ u rel ⋅ dA = 0 ∫ dt CV where d ρ dV = 0 (The flow is steady.) dt C∫ V ∫ ρu rel ⋅ dA = − ρV0 A0 + ρV2 A2 CS Substitute and simplify: A V2 = V0 0 A2 (4.137) Now apply Bernoulli’s equation from point 1 to point 0 and from point 1 to point 2: ( p + 12 ρV 2 − ρ gy ) = ( p + 12 ρV 2 − ρ gy ) = ( p + 12 ρV 2 − ρ gy ) 1 0 2 where p1 = p0 = p2 = patm (These points are all at free surfaces.) V1 = 0 and V0 and V2 are related through Eqn. (4.137). y1 = −h, y0 = 0, y2 = y C. Wassgren Chapter 04: Differential Analysis 212 Last Updated: 05 Sep 2008 Substitute and simplify: ρ gh = 1 ρV02 = 1 ρV22 − ρ gy 2 2 2 A ρ gh = ρV = ρV 0 − ρ gy A2 The first two equations in the previous expression state that: V0 = 2 gh Equation (4.138) combined with the second two equations gives: 1 2 2 0 1 2 2 0 (4.138) 2 A0 ρ gy = 1+ 1 2 A2 2 ρV0 2 A0 ρ gy y = 1+ = 1+ A2 h ρ gh A2 1 = A0 y 1+ h C. Wassgren Chapter 04: Differential Analysis 213 Last Updated: 05 Sep 2008 Example: a. b. c. Using an integral approach, write the differential equation governing the motion of an inviscid, incompressible fluid (with density ρ) oscillating within the U-tube manometer shown. The manometer cross-sectional area is A. What is the natural frequency of the fluid motion? What are the implications of this result for making time-varying pressure measurements using a manometer? tube ends are open to the atmosphere g h1 h2 incompressible, inviscid fluid with density ρ Assume this distance is negligible compared to h1+h2. SOLUTION: Apply the LME in the y-direction to the following two CVs. h1 CV 1 CV 2 C. Wassgren Chapter 04: Differential Analysis h2 Y 214 Last Updated: 05 Sep 2008 CV 1: d ∫ uY ρ dV + CS uY ( ρ u rel ⋅ dA ) = FBY + FSY ∫ dt CV where d 2 h dh 2 d d dh uY ρ dV = 1 ρ Ah1 = ρ A h1 21 + 1 ∫ dt CV dt dt dt dt ∫ u ( ρu Y rel ⋅ dA ) = 0 CS FBY = − ρ Ah1 g FSY = pbottom A Substitute and simplify: d 2 h dh 2 ρ A h1 21 + 1 = − ρ Ah1 g + pbottom A dt dt 2 d 2h p dh h1 21 = −h1 g − 1 + bottom ρ dt dt (4.139) CV 2: d ∫ uY ρ dV + CS uY ( ρ u rel ⋅ dA ) = FBY + FSY ∫ dt CV where d 2 h dh 2 d d dh uY ρ dV = 2 ρ Ah2 = − ρ A h2 22 + 2 ∫ dt CV dt dt dt dt ∫ u ( ρu Y rel ⋅ dA ) = 0 CS FBY = − ρ Ah2 g FSY = pbottom A Substitute and simplify: d 2 h dh 2 ρ A h2 22 + 2 = − ρ Ah2 g + pbottom A dt dt 2 h2 d 2 h2 p dh = −h2 g − 2 + bottom 2 ρ dt dt C. Wassgren Chapter 04: Differential Analysis (4.140) 215 Last Updated: 05 Sep 2008 From conservation of mass considering both CVs combined together: dh2 dh =− 1 dt dt (One side moves down at the same rate that the other side moves up.) (4.141) Subtract Eqn. (4.140) from (4.139) and make use of Eqn. (4.141). 2 2 h1 p p d 2 h1 d 2h dh dh − h2 22 = −h1 g − 1 + bottom + h2 g + 2 − bottom 2 dt dt ρ ρ dt dt h1 d 2 h1 d dh dh dh − h2 − 1 = ( h2 − h1 ) g − 1 + 1 2 dt dt dt dt dt 2 ( h1 + h2 ) d 2 h1 + g ( h1 − h2 ) = 0 dt 2 2 (4.142) Let: L = h1 + h2 z = h1 − h2 dh dz dh1 dh2 dh1 dh1 = − = + =2 1 dt dt dt dt dt dt 2 2 dh dz = 2 21 dt 2 dt so that Eqn. (4.142) becomes: d 2 z 2g + z=0 dt 2 L (4.143) The general solution to this differential equation is: 2g 2g z = A sin t + B cos t L L Specifying the following initial conditions: z ( t = 0 ) = z0 (4.144) (4.145) dz (t = 0) = V dt gives the solution: z =V 2g 2g L sin t + z0 cos t 2g L L (4.146) The natural (radian) frequency of the manometer, ω, is: ω= 2g L (4.147) The practical implication of this result is that one must make sure that the manometer fluid length, L, is sufficiently small so that the manometer’s natural frequency is large enough to accurately capture the temporal variations in the pressure measurements. In other words, a manometer with a large L will not be able to capture rapid pressure fluctuations. C. Wassgren Chapter 04: Differential Analysis 216 Last Updated: 05 Sep 2008 This problem may also be solved using a accelerating frames of reference. y1 y2 These FORs are fixed to the free surfaces. h1 CV 1 CV 2 h2 Y LMEs in the y-direction using the indicated accelerating FORs: d ∫ u y ρ dV + CS u y ( ρ u rel ⋅ dA ) = FBy + FSy − CV ay / Y ρ dV ∫ ∫ dt CV where d d u y ρ dV = 0 u y ρ dV = 0 dt C∫ dt C∫ V1 V2 ∫ u ( ρu y rel ⋅ dA ) = 0 ∫ u ( ρu y rel ⋅ dA ) = 0 CS1 CS2 FBy1 = − ρ h1 Ag FBy 2 = − ρ h2 Ag FSy1 = pbottom A FSy 2 = pbottom A ∫ CV1 a y / Y ρ dV = d 2 h1 ρ h1 A dt 2 d 2h ∴ 0 = − ρ h1 Ag + pbottom A − 21 ρ h1 A dt Subtract the two previous equations and simplify: d 2h d 2h 0 = ( h2 − h1 ) g − 21 h1 + 22 h2 dt dt Make use of the Eqn. (4.141) and re-arrange to get: d 2h ( h1 + h2 ) 21 + g ( h1 − h2 ) = 0 dt This is the same equation as Eqn. (4.142)! C. Wassgren Chapter 04: Differential Analysis ∫ CV2 a y / Y ρ dV = d 2 h2 ρ h2 A dt 2 ∴ 0 = − ρ h2 Ag + pbottom A − 217 d 2 h2 ρ h2 A dt 2 Last Updated: 05 Sep 2008 This problem may also be solved using the unsteady Bernoulli equation. Assuming the flow is irrotational, inviscid, and incompressible, Bernoulli’s equation may be written at an instant in time as: y ∂φ p 1 2 ∂φ p 1 2 g (4.148) + + 2 V + gy = + + 2 V + gy ∂t ρ 1 ∂t ρ 2 where the point 1 is on the free surface of the left leg of the manometer and point 2 is on the free surface of the right leg of the manometer. In addition: V= ∂φ ∂ = ∂t 1 ∂t ∂φ ⇒ ∂y ∂φ ∂t = 2 ∂ ∂t y = h1 dV dh d 2h dh d (V1h1 ) = h1 1 + V1 1 = h1 21 + 1 dt dt dt dt dt Vdy = dV dh d 2h dh d (V2 h2 ) = h2 1 + V1 2 = h2 22 + 2 dt dt dt dt dt ∫ y=0 y = h2 ∫ 2 Vdy = y =0 2 p1 = p2 = patm V12 = V22 y1 = h1 y2 = h2 Substituting and simplifying gives: 2 2 d 2 h dh d 2 h dh h1 21 + 1 + gh1 = h2 22 + 2 + gh2 dt dt dt dt Making use of Eqn. (4.141) gives: d 2h (4.149) ( h1 + h2 ) 1 + g ( h1 − h2 ) = 0 dt Eqn. (4.149) is identical to Eqn. (4.142) so the solution will be the same as that derived previously. C. Wassgren Chapter 04: Differential Analysis 218 Last Updated: 05 Sep 2008 Another Approach to Deriving Bernoulli’s Equation We can also derive Bernoulli’s equation by considering LME and COM applied to a differential control volume as shown below. In the following analysis, we’ll make the following simplifying assumptions: 1. steady flow 2. inviscid fluid 3. incompressible fluid Note that the control volume shown follows the streamlines. streamlines A+1/2dA (pA) (p+1/2dp)(A+1/2dA) V+1/2dV θ A-1/2dA ρdsAg (p-1/2dp) (A-1/2dA) V-1/2dV θ-1/2dθ g z θ+1/2dθ y z θ (pA) ds x dz = ds sinθ First apply COM to the CV shown above: d ρ dV + ( ρ u rel ⋅ dA ) = 0 dt ∫ where d dt ∫ CV CS ∫ ρ dV = 0 (steady flow) CV ∫ ( ρu rel ⋅ dA ) = ρ (V + 1 2 dV ) ( A + 1 2 dA ) CS − ρ (V − 1 2 dV ) ( A − 1 2 dA ) = ρVdA + ρ AdV + H .O.T . (Note that there is no flow across a streamline.) ∴VdA = − AdV C. Wassgren Chapter 04: Differential Analysis 219 Last Updated: 05 Sep 2008 Now apply the LME in the s-direction to the same CV: d us ρ dV + us ( ρ u rel ⋅ dA ) = FB , s + FS , s dt ∫ where d dt ∫ CV CS ∫ u ρ dV = 0 (steady flow) s CV ∫ u ( ρu s rel ⋅ dA ) = − ρ (V − 1 2 dV ) 2 ( A − 1 2 dA) CS + ρ (V + 1 2 dV ) 2 ( A + 1 2 dA) = 2 ρVAdV + ρV 2 dA + H .O.T . FB, s = ρ dsA ( − g sin θ ) = − ρ Ag ds sin θ = − ρ Agdz = dz FS , s = ( p − 1 2 dp )( A − 1 2 dA ) − ( p + 1 2 dp ) ( A + 1 2 dA) + pdA = − Adp + H .O.T . ∴ 2 ρVAdV + ρV 2 dA = − ρ Agdz − Adp Substitute the result from COM into the result from the LME and simplify: 2 ρVAdV + ρV 2 dA = − ρ Agdz − Adp =− ρVAdV dp ρ + VdV + gdz = 0 We can integrate this equation to get: p12 + 2 V + gz = constant (4.150) ρ Again, it’s important to review the assumptions built into the derivation for Eqn. (4.150): 1. steady flow 2. inviscid fluid 3. incompressible fluid 4. flow along a streamline C. Wassgren Chapter 04: Differential Analysis 220 Last Updated: 05 Sep 2008 13. Kelvin’s Theorem In an inviscid flow of a fluid with constant density, or a fluid where the pressure is a function only of the density, where the only body forces are conservative, the vorticity of each fluid element is preserved. Notes: 1. A conservative body force is one that can be written as the gradient of a potential function, i.e.: f = −∇G The force due to gravity is an example of a conservative body force: ˆ f = −∇ ( gz ) = − ge z 2. A fluid in which the pressure is a function only of the density, i.e., p=p(ρ), is called a barotropic fluid. An example of such a fluid would be an ideal gas undergoing an isentropic flow process: 3. p ρ = p0 ρ0 An important result of Kelvin’s Theorem is that if a flow starts off irrotational, viscous forces are negligible, and the fluid has either constant density or is a barotropic fluid, then the flow will always remain irrotational. γ Proof of Kelvin’s Theorem: Consider the flow of an inviscid fluid where only conservative body forces are considered. The time rate of change of the circulation about a specific collection of fluid particles is given by: DΓ D D = u ⋅ dl = ui dxi (4.151) Dt Dt Dt ∫ ∫ C C where Γ is the circulation, C is the contour about the fluid particles, u is the fluid velocity, and dl=dxiei is a small displacement along the contour. The Lagrangian time derivative can be brought inside the contour integral (refer to Pneuli, D. and Gutfinger, C., Fluid Mechanics, pp. 310-312 for the proof to this) to give: Du D ( dxi ) DΓ = i dxi + ui (4.152) Dt Dt Dt ∫ C Note that: D ( dxi ) ∂x ∂x Dx = d i = d i + u j i = dui ∂t Dt ∂x j Dt =0 =δ ij We can also substitute in for the fluid particle acceleration using Euler’s equations: Dui 1 ∂p ∂G =− − Dt ρ ∂xi ∂xi Note that in the previous equation conservative body forces have been assumed where: ∂G fi = − ∂xi Substituting Eqns. (4.153) and (4.154) into Eqn. (4.152) gives: 1 ∂p DΓ ∂G = − dxi − dxi + ui dui Dt ∂xi ρ ∂xi ∫ (4.153) (4.154) (4.155) C C. Wassgren Chapter 04: Differential Analysis 221 Last Updated: 05 Sep 2008 Several of the terms in this equation may be simplified since: ∂p dxi = dp ∂xi ∂G dxi = dG ∂xi (4.156) ui dui = d ( 1 2 ui ui ) Substituting Eqn. (4.156) into Eqn. (4.155) gives: dp DΓ dp = − − dG + d ( 1 2 ui ui ) = − − Dt ρ ρ ∫ ∫ ∫ dG + ∫ d ( C C C 1 2 ui ui ) (4.157) C The second and third terms in this equation are zero since these functions are single-valued (i.e. at each location the quantities have a unique value) and the contour C is a closed curve: ∫ dG = 0 C ∫ d( 1 2 ui ui )=0 C Thus, DΓ =− Dt ∫ C dp (4.158) ρ If the density is a constant then: DΓ 1 =− dp Dt ρ ∫ C and since the pressure is a single-valued function: DΓ =0 Dt Since the circulation about the fluid particles remains unchanged, the vorticity of the fluid particles will also remain unchanged. Also, if the pressure is a function only of the density then: dp p = p ( ρ ) ⇒ dp = d ρ ⇒ dp = f ( ρ ) d ρ dρ DΓ dp dρ =− = − f (ρ) ρ ρ Dt ∫ ∫ C C DΓ =0 Dt since the density is a single-valued function. ∴ Therefore we see that for the flow of an inviscid fluid in a conservative force field where either the density of the fluid is constant or where the pressure is a function only of the density, the vorticity of a collection of fluid particles will remain unchanged. C. Wassgren Chapter 04: Differential Analysis 222 Last Updated: 05 Sep 2008 Notes: 1. When the pressure is a function of variables other than the density, the contour integral will not, in general, be zero. 2. As might be anticipated, the vorticity of a fluid element may be changed through the action of viscosity, non-conservative forces, or density variations that are not a function solely of the pressure variations. 3. Kelvin’s Theorem applies strictly to a simply-connected region, i.e. a contour that does not intersect itself and contains only fluid. A contour that surrounds some object, e.g. an airfoil, is not a simplyconnected region and therefore Kelvin’s Theorem does not hold for such a contour. This fact is significant when examining the lift on an airfoil since it is possible to have circulation around an airfoil (i.e., around a non-simply connected region) in an otherwise irrotational flow. C. Wassgren Chapter 04: Differential Analysis 223 Last Updated: 05 Sep 2008 14. Crocco’s Equation Crocco’s equation relates the vorticity of a flow field to the gradients in the entropy and stagnation enthalpy of the fluid in a flow where viscosity and body forces are negligible. Crocco’s equation is given as: ∂u u × ω + T ∇s = ∇ ( h + 1 2 u ⋅ u ) + (4.159) ∂t Derivation of Crocco’s Equation: Consider the momentum equations for an inviscid fluid (Euler’s equations) for a flow in which body forces are negligible: Du ∂u 1 = + ( u ⋅∇ ) u = − ∇p (4.160) Dt ∂t ρ Re-write the convective derivative term using the following vector identity: (u ⋅ ∇) u = ∇ ( 12 u ⋅ u ) − u × (∇ × u ) Utilize the definition of vorticity, ω: ω = ∇×u and substitute into equation (4.160) to get: ∂u 1 + ∇ ( 1 2 u ⋅ u ) − u × ω = − ∇p ∂t ρ (4.161) Now consider the 1st Law of Thermodynamics for a fluid element (assumed to be a pure substance) where only reversible pressure work is considered: 1 1 du = δ q − pd (4.162) ρ ρ Note that δq is the amount of heat added to the fluid element per unit volume. Substituting the definition of entropy for a reversible (zero viscosity has been assumed) process: 1 Tds = δ q ρ and enthalpy: p 1 dp dh = du + d = du + pd + ρ ρ ρ into Eqn. (4.162) and simplifying gives: dp dh = Tds + (4.163) ρ Note that we can write Eqn. (4.163) in a slightly different form by utilizing the following: dh = ∇ h ⋅ dx ds = ∇s ⋅ dx dp = ∇p ⋅ dx where dx is a small length in any direction. Substituting these relations into Eqn. (4.163) and simplifying gives: ∇p − = T ∇s − ∇h (4.164) ρ C. Wassgren Chapter 04: Differential Analysis 224 Last Updated: 05 Sep 2008 Substituting Eqn. (4.164) into Eqn. (4.161) gives: ∂u + ∇ ( 1 2 u ⋅ u ) − u × ω = T ∇s − ∇h ∂t Re-arranging this equation results in Crocco’s Equation: ∂u u × ω + T ∇s = ∇ ( h + 1 2 u ⋅ u ) + ∂t Recall that this equation holds for a flow in which viscous and body forces are negligible. (4.165) Notes: 1. Consider a steady, inviscid flow in which body forces are negligible so that Eqn. (4.159) becomes: u × ω + T ∇s = ∇h0 where h0 = h + 1 2 u ⋅ u (4.166) Let’s restrict our investigation to flow along a streamline by taking the dot product of the previous equation with u. u ⋅ ( u × ω ) + T ( u ⋅∇ ) s = ( u ⋅∇ ) h0 (4.167) = 0 (vector identity) Note that since we’re concerned here with steady flows, Dh0 Ds = ( u ⋅∇ ) s and = ( u ⋅∇ ) h0 (4.168) Dt Dt and Eqn. (4.167) becomes: Ds Dh0 = (4.169) T Dt Dt Hence, if there is no dissipation along a streamline (i.e., Ds/Dt = 0), then the stagnation enthalpy, h0, must remain constant along the streamline (since Dh0/Dt = 0). 2. Now consider the case where the stagnation enthalpy in a steady flow is uniform along the streamlines so that ∇h0 = 0. For this case, Eqn. (4.159) becomes: u × ω = −T ∇s (4.170) For such a flow we can conclude the following important statement. For the steady flow of an inviscid fluid in which body forces are negligible and where the stagnation enthalpy is constant, an irrotational flow will be isentropic and an isentropic flow will be irrotational. 3. Consider the uniform, supersonic flow in front of a blunt-nosed object. The incoming flow will have a constant stagnation enthalpy and will also be irrotational (and thus isentropic). A curved shock wave will stand in front of the object. Across the shock wave the stagnation enthalpy will remain constant but the entropy will change (flow across a shock wave is a non-isentropic process). Since the shock wave is curved, there will be a gradient in the entropy normal to the downstream streamlines. Thus, from Crocco’s equation we observe that vorticity will also be generated downstream of the curved shock wave and the flow will, by definition, be rotational. Although there is a change in entropy across a normal shock wave and an oblique shock wave, there is no entropy gradient in the normal direction (the entropy change across the shock is uniform along the length of the shock) and, thus, the flow will remain irrotational downstream of normal and oblique shocks if the upstream flow is irrotational. Note that Crocco’s equation does not strictly apply across a shock wave since the large velocity gradient within the shock means that viscous effects are also significant there. However, Crocco’s equation can be applied just upstream and downstream of the shock wave. C. Wassgren Chapter 04: Differential Analysis 225 Last Updated: 05 Sep 2008 Review Questions 1. Describe what each term represents in the Lagrangian derivative. 2. What is the formal definition of an incompressible fluid? Give an example of an incompressible flow where the fluid density is not uniform? 3. Write the continuity equation for an incompressible fluid? 4. Write the continuity equation for a fluid with constant and uniform density? 5. Describe the naming and sign convention for stresses, σij. 6. Will the stress tensor always be symmetric? 7. Describe the various ways in which a fluid element can deform. 8. What is the vorticity of a fluid element (in words and in mathematical form)? 9. What is the deformation rate tensor (in words and in mathematical form)? 10. What is meant by an “irrotational” flow? 11. What three key assumptions are made in deriving the stress-strain rate constitutive relation for a Newtonian fluid? 12. How is the mechanical pressure related to the thermodynamic pressure? 13. Why doesn’t the bulk viscosity enter into incompressible fluid flow problems? 14. Describe what each term represents in the Navier-Stokes equations. 15. What are Euler’s equations? 16. Describe the various ways in which a fluid element can deform. 17. What is the vorticity of a fluid element (in words and in mathematical form)? 18. What is the deformation rate tensor (in words and in mathematical form)? 19. What is meant by an “irrotational” flow? 20. What three key assumptions are made in deriving the stress-strain rate constitutive relation for a Newtonian fluid? 21. How is the mechanical pressure related to the thermodynamic pressure? 22. Why doesn’t the bulk viscosity enter into incompressible fluid flow problems? 23. What does each term represent in the energy equation? 24. What does each term represent in the mechanical energy equation? 25. What does each term represent in the thermal energy equation? 26. What is Fourier’s Law of Conduction (in mathematical terms)? 27. What does the energy dissipation function represent? 28. Is the energy dissipation function always positive? 29. Is the thermal energy equation required to solve for the flow velocity and pressure in incompressible flows? How about for compressible flows? 30. Under what conditions can vorticity be generated within a flow? 31. How are vorticity and circulation related? 32. What are the assumptions that go into the following form of Bernoulli’s equation? p V2 + + z = constant ρ g 2g 33. What are the assumptions that go into the following form of Bernoulli’s equation? ∂φ p V 2 ++ + gz = F ( t ) ∂t ρ 2 34. What does Bernoulli’s equation look like for an ideal gas flowing isentropically? 35. What is Kelvin’s Theorem? What is its significance? C. Wassgren Chapter 04: Differential Analysis 226 Last Updated: 05 Sep 2008 Chapter 05: Potential Flows 1. Stream Functions 2. Potential Functions 3. Complex Variable Methods for Potential Flows 4. Blasius Integral Laws 5. Kutta-Joukowski Theorem 6. Conformal Mappings 7. Joukowski Transformation 8. Method of Images 9. Added Mass 10. Numerical Methods for Solving Potential Flows 11. Doublet Distributions C. Wassgren Chapter 05: Potential Flows 227 Last Updated: 14 Aug 2010 1. Stream Functions A stream function is a special scalar function that is useful when analyzing 2D flows. As will be shown, a stream function has the following properties: 1. A stream function satisfies the continuity equation. 2. A stream function is a constant along a streamline. 3. The flow rate between two streamlines is equal to the difference in the streamlines’ stream functions. First, let’s define the stream function. Define a scalar function, , called a stream function, such that the continuity equation is automatically satisfied for 2D (planar and axi-symmetric) flows. For a 2D incompressible flow in rectangular coordinates, define =(x, y) such that: ux and u y (5.1) y x If the stream function is defined in this manner, then the continuity equation will automatically be satisfied: u u y 2 2 u x 0 x y xy yx For a 2D incompressible flow in polar coordinates, =(r, ): 1 and u ur r r The continuity equation for a 2D, incompressible flow in polar coordinates is: 1 1 u 1 2 1 2 0 rur r r r r r r r (5.2) Now let’s consider some of the additional properties of the stream function. C. Wassgren Chapter 05: Potential Flows 228 Last Updated: 14 Aug 2010 The Stream Function is Constant Along a Streamline Let’s determine the curve along which the stream function remains constant. We’ll only consider an incompressible flow in rectangular coordinates here for simplicity (the same result holds for polar coordinates and compressible flows). The total change in the stream function, d, where =(x, y), over some displacement (dx, dy) is given by: x, y d dx dy u y dx u x dy x y where the definition of the stream function has been used to write d in terms of the velocities. To find the slope of the curve along which =constant, we let d=0. d 0 u y dx u x dy uy dy dx constant u x Notice that the slope of the curve along which the stream function is constant is exactly the same as the slope of a streamline. Thus, we conclude that the stream function is constant along a streamline! =1 =2 =3 where 1, 2, and 3 are constants. Note that these constants may vary from streamline to streamline. streamlines C. Wassgren Chapter 05: Potential Flows 229 Last Updated: 14 Aug 2010 Example: A particular planar, incompressible flow can be described with the following stream function: Axy where A is a constant. a. Sketch the streamlines for the flow. b. Determine the velocity components for the flow. SOLUTION: The stream function is a constant along a streamline so the equation of the streamlines will be: 1 y (hyperbolas!) Ax A plot of the streamlines is shown below. Note that A has been assumed to be a positive constant (i.e. A > 0) in determining the direction of the flow. y x A>0 The velocities are determined from the definition of the stream function. Ax and u y Ay ux x y C. Wassgren Chapter 05: Potential Flows 230 Last Updated: 14 Aug 2010 The Flow Rate Between Two Streamlines is Equal to the Difference in their Stream Functions Now let’s examine how the flow rate between two streamlines is related to the stream function. Consider the sketch below. A Q B dy dA dx dx=dA sin dy=dA cos ˆ ˆ ˆ n cos e x sin e y The volumetric flow rate passing between the two streamlines, and thus crossing through a line drawn between the two streamlines can be found by calculating the volumetric flow rate through a small piece of the line and then integrating from one streamline to the other. dy dx ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ dQ u dA u x e x u y e y cos e x sin e y dA u x e x u y e y e x e y dA dA dA dy dx u x dy u y dx y x dQ d Integrating from streamline A to streamline B gives: QAB B A The volumetric flow rate between two streamlines is equal to the difference in the streamline stream functions! Note that if B>A then the flow is from left to right. If B<A then the flow is from right to left. C. Wassgren Chapter 05: Potential Flows 231 Last Updated: 14 Aug 2010 Example: The velocity field for a planar, incompressible flow is given by: ˆ ˆ u 2( x 2 y 2 )e x 4 xye y a. b. Determine the stream function for this flow field if (0,0)=0. Determine the volumetric flow rate across the line AB shown in the figure. y B: (0,1) x A: (1,0) SOLUTION: Recall that the velocities are determined from the stream function in the following manner. ux 2 x2 y 2 y uy 4 xy x 2 x2 y 1 y3 f x 3 (5.3) 2 x 2 y g y (5.4) where f and g are, at this point, unknown functions of x and y, respectively. Comparing Eqns. (5.3) and (5.4) indicates that: f x c and g y 2 y 3 c 3 where c is a constant. Hence, the stream function is: 2 x2 y 1 y3 c 3 (5.5) Knowing that (0, 0) = 0 we can conclude that c = 0 and: 2 x2 y 1 y3 3 (5.6) Recall that the volumetric flow rate between two streamlines is equal to the difference in the streamline stream functions. QAB B A (5.7) where B 0,1 2 3 A 1, 0 0 so that QAB 2 3 C. Wassgren Chapter 05: Potential Flows (5.8) 232 Last Updated: 14 Aug 2010 “Building Block” Stream Functions The properties of stream functions described previously are enough to justify their use. There are additional reasons to use stream functions, however. Models of real flows can be produced by combining “building block” stream functions. The significance of this topic will be discussed in greater detail when examining the potential function (especially the complex potential); a topic covered later in the notes. For now, however, it is sufficient to present some “building block” stream functions and discuss how they can be combined to produce models of actual flows. First, let’s examine a few basic “building block” stream functions. uniform stream shown for U0,V0> 0 V0 y x V0 x U 0 y u x U 0 ; u y V0 U0 line source (m>0) or sink (m<0) y r free line vortex y shown for m> 0 m is referred to as the source (sink) strength. x u x forced line vortex y r u is referred to as the circulation. ur 0; u 1 2 r shown for > 0 Kr 2 2 ur 0; u Kr r ln r 2 shown for > 0 r m 0 2 2 m1 ; u 0 ur 2 r x r shear flow shown for > 0 y u x 2 Ay; u y 0 x extensional flow y shown for > 0 x C. Wassgren Chapter 05: Potential Flows Ay 2 Can also be used to model flow in a corner and stagnation point flow. 233 Axy u x Ax; u y Ay Last Updated: 14 Aug 2010 Superposition of Stream Functions Because the continuity equation is a linear PDE, the “building block” stream functions just presented can be combined together to produce new stream function flows. Proof: LetT=1+2 where 1 and 2 are stream functions that satisfy the continuity equation. The velocities determined from the new stream function are given by: T 1 2 1 2 ux y y y y 1 2 T 2 1 x x x x Substitute these velocities into the continuity equation: u u y 2 1 2 u x 1 x y x y y y x x 1 2 1 2 xy xy yx yx 0 Thus, the new stream function, T, formed by the superposition of the original stream functions also satisfies the continuity equation. uy C. Wassgren Chapter 05: Potential Flows 234 Last Updated: 14 Aug 2010 Example: The doublet is formed by superposing a source and sink of equal strength separated by an infinitesimal distance. y r2 2 a r r1 1 x a The stream function for the source/sink pair is given by (m>0): m m m 2 1 2 1 2 2 2 Re-arrange and take the tangent of both sides: tan 2 tan 1 2 tan (5.9) tan 2 1 m 1 tan 21 Note that a trig identity has been used in deriving the previous expression. From the figure we observe that: r sin r sin tan 1 and tan 2 r cos a r cos a Substitute these expressions into Eqn. (5.9) and simplify: r sin r sin 2 r cos a r cos a tan m 1 r sin r sin r cos a r cos a r 2 sin cos ar sin r 2 sin cos ar sin r 2 cos 2 a 2 r 2 sin 2 1 2 r cos 2 a 2 2ar sin r 2 cos 2 a 2 2 r cos 2 a 2 r 2 sin 2 r 2 cos 2 a 2 2ar sin 2 r a2 m 2ar sin tan 1 2 (5.10) 2 2 r a Stream function for a source/sink pair of equal strength each located a distance a from the origin along the x-axis. C. Wassgren Chapter 05: Potential Flows 235 Last Updated: 14 Aug 2010 The streamlines for the stream function given in Eqn. (5.10) are shown in the following figure. y a x a Note that as a0, Eqn. (5.10) becomes: m 2ar sin lim a 0 2 r 2 a 2 since the tangent of a very small angle approaches the value of the angle. If we let the source and sink approach each other (a0) while we let the source/sink strength approach infinity (m) such that the ratio ma/=K=constant, then the stream function becomes: K sin doublet oriented Note: 0<2 r along x -axis The streamlines for the doublet are circles passing through the origin as shown in the figure below. y shown for K>0 x Note: 1. Doublets have an orientation. The stream function for a doublet oriented in the y-direction is given by: K cos y doublet oriented r along y-axis shown for K>0 x C. Wassgren Chapter 05: Potential Flows 236 Last Updated: 14 Aug 2010 Example: The flow of a frictionless fluid (there can be slip at solid surfaces) around a non-rotating cylinder can be modeled as a uniform stream superimposed with a doublet: flow around uniform doublet cylinder stream If the cylinder radius is R, determine the velocity of the fluid on the surface of the cylinder as a function of angular position, . y uniform stream with velocity, U x doublet with strength k R U Note that inside the cylinder the streamlines look like: C. Wassgren Chapter 05: Potential Flows 237 Last Updated: 14 Aug 2010 SOLUTION: The stream function for flow around a non-rotating cylinder is given by combining the stream function for a uniform stream with the stream function for a (horizontally-oriented) doublet. flow around uniform doublet cylinder stream Uy K sin r K sin r cylinder At the moment, K is an unknown constant. It can be determined by noting that there is no flow through the cylinder surface. Hence, the radial velocity, ur, at r = R should be zero. 1 K cos ur U cos r r2 K sin u U sin r r2 No flow through the cylinder surface at r = R: K cos ur r R 0 U cos R2 2 K UR Hence, the stream function and flow velocities for flow around a non-rotating cylinder are: R 2 flow around Ur sin 1 cylinder r flow around Ur sin ur R 2 1 Ur cos 1 r r u R 2 U sin 1 r r On the cylinder surface (r = R): ur r R 0 and u r R 2U sin Note that there are stagnation points at = 0, A maximum speed of 2U occurs at = /2, -/2. C. Wassgren Chapter 05: Potential Flows 238 Last Updated: 14 Aug 2010 Notes: 1. Stream functions can also be defined for steady, compressible, 2D flows. For example, in rectangular coordinates: ux 0 and u y 0 where 0 is a reference density y x The continuity equation for these conditions is: u x u y 0 0 u 0 x x y x y y 2. Stream functions also exist for axi-symmetric, incompressible flows (referred to as Stokes’ stream functions): =(r, z) 1 1 ur and u z r z r r The continuity equation for these conditions is: 1 rur u z 0 r r z 3. Stream functions cannot be defined for arbitrary 3D flows. C. Wassgren Chapter 05: Potential Flows 239 Last Updated: 14 Aug 2010 2. Potential Functions The velocity field for an irrotational flow can be written as the gradient of a potential function, : u since, from a vector identity: 0 and because an irrotational flow is defined as one with zero vorticity, i.e.: ω 0 (5.11) Now let’s ensure that the continuity equation is satisfied for an incompressible fluid (compressible potential flows will be considered in a separate set of notes dedicated specifically to compressible flows): u 0 0 2 0 (5.12) This is Laplace’s Equation!, a well studied, linear, elliptic partial differential equation that appears in many other disciplines such as electromagnetics and conduction heat transfer. The momentum equations for a potential flow simplify to Bernoulli’s equation since the flow is everywhere irrotational (refer to an earlier set of notes concerning Bernoulli’s equation): p 1 2 G F t (5.13) t ˆ where G is a conservative body potential (e.g. for gravity, G = gz where g and e z point in opposite directions) and F(t) is a function of time. For a steady potential flow, Eqn. (5.13) simplifies to: p1 2 G constant (5.14) Note that the momentum equation (Eqn. (5.13) or (5.14)) need not be solved to determine the fluid kinematics. Solving Eqn. (5.12) subject to appropriate boundary conditions is sufficient to determine the flow velocity field. This occurs because we placed two restrictions on the flow field: the continuity equation and the irrotationality assumption. The momentum equation can be solved to determine the fluid pressure field once the velocity field is known. The appropriate boundary conditions for Laplace’s equation are either Dirichlet (the functions value is specified), Neumann (the functions gradient is specified), or mixed. At solid surfaces the appropriate boundary condition for the flow is that the flow velocity normal to the surface is equal to the surface velocity, i.e.: ˆ ˆ u n U n (5.15) ˆ where u is the fluid velocity, U is the boundary velocity, and n is the normal vector to the boundary. This is a Dirichlet or kinematic boundary condition. Note that the no-slip condition is not satisfied for potential flows. This occurs because potential flows have no viscous force contributions since the viscous terms in the Navier-Stokes equations (i.e. momentum equations) drop out due to the irrotationality assumption. As a result, the Navier-Stokes equations, which are normally 2nd order PDEs, simplify to the 1st order Euler’s equations (and can be simplified further to Bernoulli’s equation). Hence, only a single boundary condition must be specified. Neumann boundary conditions are specified at free surfaces, i.e. surfaces where the pressure is defined. These are sometimes called dynamic boundary conditions. Bernoulli’s Equation (Eqns. (5.13) or (5.14)) is used to relate free pressure boundary conditions to the velocity field. C. Wassgren Chapter 05: Potential Flows 240 Last Updated: 14 Aug 2010 Notes: 1. Incompressible potential flows are often referred to as ideal fluid flows since the fluid is incompressible and viscous forces are negligible. 2. Potential functions can be defined for 3D flows (as long as they’re irrotational). Recall that stream functions could only be specified for 2D flows. 3. The governing equation for potential flows (Laplace’s equation) is a linear PDE so that the principle of superposition can be used to combine potential flow solutions. The approach is similar to that discussed previously for stream functions. 4. Potential functions and stream functions are intimately related. This will become clear in the following section of notes concerning the complex potential function. 5. Streamlines ( = constant) and equi-potential lines ( = constant) are perpendicular everywhere in the flow. Consider the curves along which = constant (a streamline) and = constant: d 0 dx d 0 dx where dx is a small distance along the curves. Re-write these relations in terms of the velocities: dy u y dx 0 u y dx u x dy dx u x dx 0 u x dx u y dy u dy x dx uy From analytical geometry, two curves are perpendicular to each other if the slopes of the curves multiplied together equals -1. Hence, we see that the streamlines and equi-potential lines will always be perpendicular to each other. The resulting mesh of streamlines and equi-potential lines is known as the flow net. = 1 = 2 = 3 = 2 = 3 = 1 C. Wassgren Chapter 05: Potential Flows 241 Last Updated: 14 Aug 2010 3. Complex Variable Methods for Investigating Planar, Ideal, Irrotational Flows A good mathematics reference for this topic is: Churchill, R.V. and Brown, J.W., Complex Variables and Applications, McGraw-Hill. Let’s define the complex potential, f(z): where f ( z ) i z = x+iy = rexp(i) where 0 < 2and exp(i)=cos()+isin() is the velocity potential is the stream function y r x Why do this? Because it allows us to present information in a compact manner and because we can use tools from complex variable mathematics to analyze fluid flows. Notes: 1. A few complex variables preliminaries: a. z = x+iy = rexp(i) where 0 < 2and exp(i)=cos()+isin() b. |z| = (x2+y2)1/2 = r c. arg(z) = tan-1(y/x) = 2 d. zz z x 2 y 2 e. f. log(z) = ln(r) + i A function f of the complex variable z is analytic on an open set if it has a derivative at each point in that set. (counter-example: f(z) = |z|2 is not analytic anywhere since its derivative exists only at z = 0.) A function, h, is harmonic if it has continuous partial derivatives of the first and second order and satisfies Laplace’s equation: 2 h 0 If a function, f(z)=a(x,y)+ib(x,y), is analytic in D, then the first order partial derivatives of its component functions, a and b, must satisfy the Cauchy-Riemann equations throughout D. a b a b and x y y x If two functions, a and b, are harmonic in a domain D and their first-order partial derivatives satisfy the Cauchy-Riemann equations throughout D, b is said to be a harmonic conjugate of a. g. h. i. 2. If a function, f(z)=a(x,y)+ib(x,y), is analytic in a domain D then its component functions, a and b, are harmonic conjugates in D. Proof: Since f(z) is analytic in a domain D, then the first order partial derivatives of its component functions, a(x,y) and b(x,y), satisfy the Cauchy-Riemann equations throughout D (Note #1h). Differentiating the Cauchy-Riemann equations gives: 2 a 2b 2a 2b 2 x 2 xy xy x and 2 2 2 a b a 2b 2 2 yx y yx y But from advanced calculus: 2a 2a 2b 2b and xy yx xy yx Substituting and simplifying: C. Wassgren Chapter 05: Potential Flows 242 Last Updated: 14 Aug 2010 2a 2a 2b 2b 2 and 2 x 2 y y 2 x 2 a 0 and 2 b 0 Thus, a and b are harmonic (Note 1g). Since a and b are harmonic and satisfy the Cauchy-Riemann equations in D, they are harmonic conjugates of each other in D (Note 1h). Therefore, if f(z)=a(x,y)+ib(x,y) is analytic in a domain D then its component functions, a and b, are harmonic conjugates in D. 3. Any analytic function, f(z)=(x,y)+i(x,y), is a valid 2D, incompressible, irrotational flow field. Proof: Recall that for an irrotational flow, the velocity may be written as the gradient of a potential function, : ω u 0 from a vector identity: 0 for any u For the flow to satisfy the continuity equation for an incompressible fluid: u 0 2 0 is a harmonic function! Also recall that the stream function, , is defined for 2D flows such that continuity for an incompressible fluid is automatically satisfied: ux and u y y x 2 2 0 xy xy continuity is automatically satisfied! If the flow is also irrotational, the stream function must also satisfy: u y u x 0 ω u 0 x y so that u 2 2 2 0 x 2 y 2 is a harmonic function! In addition, ux x y uy y x Cauchy-Riemann equations and are harmonic conjugates! From Note 2, the components of any analytic function are harmonic conjugates. Thus, since the governing equations for the fluid are Laplace’s equation and since and are harmonic conjugates, then any analytic f(z)=+iwill be a valid flow field. Thus, by choosing various forms of f(z) that are analytic, we can produce various valid (incompressible and irrotational) flow fields. Whether or not the flow fields are interesting from an engineering perspective is another matter. C. Wassgren Chapter 05: Potential Flows 243 Last Updated: 14 Aug 2010 Notes…: 4. Some “building block” flows and their complex potentials: uniform stream f ( z ) U 0 iV0 z V0 y U 0 x V0 y V0 x U 0 y U0 x shown for U0,V0> 0 line source (m>0) or sink (m<0) y z0 r x f ( z) m log z z0 2 m m ln r 2 m 2 i log z z0 2 2 ln r 2 shown for m> 0 free line vortex y z0 f ( z) r x shown for > 0 line doublet (x-orientation) y f ( z) z0 r c z z0 c c cos r c sin r shown for c > 0 line doublet (y-orientation) y z0 r f ( z) ic z z0 c c sin r c cos r shown for c > 0 Note that in the table above: z=x+iy and z0=x0+iy0 0 < 2 1/ 2 2 2 r x x0 y y0 C. Wassgren Chapter 05: Potential Flows y y0 and tan 1 x x0 244 Last Updated: 14 Aug 2010 Notes…: 5. Fluid velocities are found via differentiation of the complex potential: df f ( z ) u x iu y dz f df z where f z x, y i x, y x dz x f i u x iu y and x x x z 1 and since z x iy x df f df z u x iu y (an identical result occurs if we consider ) dz y dz y A rectangular coordinates example: c f z (line doublet oriented in the x-direction and centered at the origin) z c x 2 2 xyi y 2 c x 2 y 2 df c cz 2 2cxy 2 i 2 2 2 22 2 22 dz z zz x y x y x2 y 2 ux c x 2 y 2 x 2 y 22 and u y 2cxy x 2 y2 2 A polar coordinate example: m log( z ) (source/sink at origin) f ( z) 2 1 df m1 m m exp(i ) m cos i sin 2 r dz 2 z 2 r exp(i ) 2 r m m cos and u y sin ux 2 r 2 r or in polar coordinates using some geometry and trig.: usin u tan-1(u/ur) 2 2 x 2 y C. Wassgren Chapter 05: Potential Flows -1 245 ur u tan-1(uy/ux) uy u tan uy -1 u u y ur u u u u and tan tan ur ux u x 1 u tan u r Substituting in for our values of ux and uy and simplifying: u tan 2 u m u r 2 2 1 tan 2 0 ur u u and tan u r 2 r 1 u tan r m u 0 and ur 2 r 2 r ucos ux urcos Last Updated: 14 Aug 2010 ursin In general the relation between the velocity components expressed in rectangular and polar coordinates is given by (refer to the figure shown above): u x ur cos u sin u y ur sin u cos df u x iu y ur cos u sin i ur sin u cos dz ur cos i sin iu cos i sin ur iu cos i sin 6. df u x iu y ur iu exp i dz (5.16) We can use the principle of superposition to combine complex potentials and form new complex potentials since if two functions are analytic in a domain D, then their sum is also analytic. C. Wassgren Chapter 05: Potential Flows 246 Last Updated: 14 Aug 2010 Notes…: 7. A few example flows created by superposition: Flow over a Rankine half-body: Combine the complex potentials for a uniform stream and a source (m>0): m f ( z ) Uz log z 2 y x source Flow over a Rankine oval: Combine the complex potentials for a uniform stream, a source, and a sink: m m f ( z ) Uz log( z a ) log( z a) 2 2 where m>0 and a > 0. y x sink source C. Wassgren Chapter 05: Potential Flows 247 Last Updated: 14 Aug 2010 Flow around a non-rotating cylinder of radius R: Combine the complex potentials for a uniform stream and a doublet: c f ( z ) Uz z where the constant c is found by not allowing any flow through the cylinder walls, i.e. ur r R 0 c cos c sin f ( z ) Ur cos i Ur sin r r c cos 1 U cos r2 r r c cos ur r R 0 U cos c UR 2 R2 so that the complex potential becomes: R2 f ( z) U z z ur (5.17) y x doublet Notes: 1. Real (viscous) flow over a sphere (a golf ball in the figure below) is shown below. The streamlines for flow over a cylinder look much the same. The streamlines over the front half of a cylinder are similar to those predicted by the potential flow analysis. In fact, the velocity and pressures field on the front half of the cylinder are also accurate (the pressures will be discussed in a moment.) The flow field downstream of the cylinder is not accurately predicted. The discrepancy between the potential flow analysis and real life occurs due to the formation of a viscous boundary layer on the cylinder surface. The boundary layer separates near the top/bottom points of the cylinder and forms a wake. Assuming irrotational flow in the boundary layer and wake are poor assumptions. However, outside the boundary layer and wake, the potential flow assumption is reasonable. We’ll discuss boundary layers in a later section of notes. C. Wassgren Chapter 05: Potential Flows 248 Last Updated: 14 Aug 2010 2. The pressure distribution on the cylinder surface can be predicted using Eqn. (5.17) and Bernoulli’s equation: R2 f ( z) U z z From Eqn. (5.16), the flow velocity field is: ur iu exp i R2 df U 1 2 dz z R2 U 1 2 r exp 2i R2 R2 U 1 2 exp 2i U exp i 2 exp i exp i r r R2 U cos i sin 2 cos i sin exp i r 2 2 R R U 1 2 cos i 1 2 sin exp i r r (Note that for this case it would be easier to determine the velocity using or directly rather than the complex potential). On the cylinder surface (r = R): ur r R 0 u r R 2U sin The pressure distribution on the cylinder surface is found via Bernoulli’s equation, and expressed in terms of a dimensionless pressure coefficient, cp: p p c p 1s 2 1 4sin 2 (5.18) 2 U Notes: a. The total drag (FD) and lift (FL) on the cylinder may be found by integrating the pressure distribution over the entire cylinder surface: FL p 2 FD p cos Rd 0 2 FL U R FD (5.19) p sin Rd 0 Either by actually evaluating Eqn. (5.19) or noting that the velocity field is symmetric over the front and back and upper and lower surfaces, the drag and lift forces on the cylinder are both zero. Of course in real flows we know that the drag is not zero. The fact that the potential flow model predicts zero drag while real flows have non-zero drag is known as d’Alembert’s Paradox. We, of course, now know that the discrepancies are explained by the formation of a boundary layer and boundary layer separation. d’Alembert’s paradox will be discussed again when reviewing Blasius’ integral law. C. Wassgren Chapter 05: Potential Flows 249 Last Updated: 14 Aug 2010 b. Equation (5.18) is compared to experimental data in the plot below (from Fox, R.W. and McDonald, A.T., Introduction to Fluid Mechanics, 5th ed., Wiley.) Again, the potential flow analysis predicts the pressure distribution reasonably well over the upstream part of the cylinder but does a poor job over the back half due to boundary layer separation. Flow around a rotating cylinder of radius R: Combine the complex potentials for a uniform stream, a doublet, and a free vortex: c i f ( z ) Uz log z z 2 where the constant c is found by not allowing any flow through the cylinder walls, just as in the previous example. Note that the addition of a vortex will not change the value of c since a vortex only produces tangential flow and not radial flow. As a result, the complex potential becomes: R 2 i f ( z) U z log z z 2 FL y x FD doublet and free vortex C. Wassgren Chapter 05: Potential Flows 250 Last Updated: 14 Aug 2010 Notes: 1. The corresponding velocity field is: R 2 i df ur iu exp i U 1 2 dz z 2 z 2 R i exp i U 1 2 exp 2i r 2 r Using the previous results for a non-rotating cylinder: R2 ur U 1 2 cos r (5.20) R2 1 u U 1 2 sin 2 r r On the cylinder surface (r = R): ur r R 0 1 2 R Using Bernoulli’s equation, the pressure coefficient over the surface is: 2 1 1 c p 1 4sin 2 4 sin 2 UR 2 UR The corresponding drag, FD, and lift, FL, are: FD 0 u r R 2U sin FL U (5.21) (5.22) (5.23) Notes: a. The drag again is zero and is not unexpected due to the fore/aft symmetry of the velocity field. b. The lift is non-zero and is related to the flow circulation. This type of lift is referred to as Magnus lift. Both drag and lift for potential flows will be discussed in detail when reviewing Blasius’ integral law and the Kutta-Joukowski theorem. c. The photo below shows the flow past a rotating golf ball. The flow is from left to right and the golf ball rotates in a clockwise manner ( < 0). From Eqn. (5.23), the lift on the golf ball will be in the positive vertical direction. In real (i.e. viscous) flows the lift on a rotating object comes primarily from deflection of the downstream wake (the fluid momentum is directed downward resulting in an upward force on the ball) rather than from the unbalanced pressure distribution on the object. The Magnus effect is often mistakenly referred to as the primary source of the lift force. C. Wassgren Chapter 05: Potential Flows 251 Last Updated: 14 Aug 2010 Notes…: 7. Flow in and around corners of varying angles can be modeled using the following complex potential: f ( z ) Az n where A and n are constants a. This produces flows between boundaries intersecting at an angle /n (only flows with n1/2 are of interest): n=2 n=3 n=1 y y y x x x or or y y x x y y x x n=2/3 n=1/2 b. The potential and stream functions are given by: f ( z ) Az n A r exp i Ar n exp in Ar n cos n iAr n sin n n Ar cos n and Ar n sin n n ur c. Anr n 1 cos n and r u 1 Anr n 1 sin n r The fluid speed at the origin is: lim u lim f ( z ) lim nAr n 1 exp i (n 1) lim nAr n 1 r 0 r 0 0 lim u A r 0 C. Wassgren Chapter 05: Potential Flows r 0 r 0 n 1 n 1 n 1 252 Last Updated: 14 Aug 2010 d. In a real fluid (one with viscosity), the flow along the surface streamline would: for n>1: separate before reaching the corner and produce a standing eddy for n<1: separate after reaching the corner unless the corner angle is small C. Wassgren Chapter 05: Potential Flows 253 Last Updated: 14 Aug 2010 4. Blasius Integral Law Consider the 2D, incompressible, inviscid, steady, irrotational flow around an arbitrary closed body: L Coutside y Cinside x D U uy dy dx ux p Using the LMEs, determine the lift, L, and drag, D, acting on the body: D pdy Coutside L Coutside pdx Coutside Coutside u x u x dy u y dx u dA u y u x dy u y dx u dA From Bernoulli’s equation (neglecting gravity): 2 2 2 2 p 1 2 ux u y c p c 1 2 ux u y where c is a constant. Substituting and re-arranging: 2 2 2 D cdy 1 2 u x u y dy u x dy u x u y dx Coutside L Coutside 2 2 2 cdx 1 2 u x u y dx u x u y dy u y dx Noting that: cdy Coutside cdx 0 Coutside and simplifying: 2 2 D 1 2 u x u y dy u x u y dx Coutside L Coutside 2 2 1 2 u x u y dx u x u y dy As shown below, the previous drag and lift relations may be written in terms of the complex potential. C. Wassgren Chapter 05: Potential Flows 254 Last Updated: 14 Aug 2010 i 2 2 df dz i 2 C ux iu y dx idy 2 Coutside dz outside i i 2 Coutside u 2 x 2 2iu x u y u y dx idy 2 2 2 2 ux dx u y dx 2ux u y dy i ux dy u y dy 2ux u y dx 2 Coutside 1 2 u 2 u 2 dy u u dx i 1 2 u 2 u 2 dx u u dy xy x y xy x y Coutside D L Substituting the expressions for lift, L, and drag, D, found previously: i 2 df dz D iL 2 Coutside dz BLASIUS’ INTEGRAL LAW How is this result used? Typically, it is applied using a theorem from complex variables referred to as the Residue Theorem (Churchill, R.V. and Brown, J.W., Complex Variables and Applications, McGraw-Hill, 5th ed., pg. 169) which states: n w( z )dz 2 i Res w( z ) C k 1 z zk A residue is the coefficient in front of the 1/(z-z0) term (the b1 term in the series below) in the Laurent series expansion of an analytic complex function about the point z0 (Churchill and Brown, pg. 144): bn n w( z ) an z z0 n n 0 n 1 z z0 where the coefficients an and bn are given by 1 w( z )dz 1 w( z )dz an z z n 1 and bn 2 i z z n 1 2 i C C 0 0 The details of the expressions above won’t concern us here and are only presented for completeness. Blasius’ Integral Law is used in deriving the Kutta-Joukowski theorem given in the following section which relates the lift (and drag) around any arbitrary, closed object to the circulation, , caused by the object. C. Wassgren Chapter 05: Potential Flows 255 Last Updated: 14 Aug 2010 Example: Determine, using the Blasius Integral Law, the lift acting on a rotating cylinder. SOLUTION: The complex potential function for flow around a rotating cylinder is: R 2 i 1 df R 2 i f z U z log z U 1 2 dz z 2 z z 2 (5.24) 2 2 R 2 i 1 R 2 2 1 df U 2 1 2 U 1 2 2 2 z z z 4 z dz Let the contour in the integral law, C, be the circle defined as: z R exp i dz iR exp i d (5.25) (5.26) where 0 < 2and R’ is an arbitrary radius greater than R. Thus, the Blasius Integral Law for this problem is: i 2 2 i 2 R2 1 R2 1 2 U 1 2 0 R exp 2i R exp i U 1 R2 exp 2i 4 2 R2 exp 2i iR exp i d 2 2 4 2 i R 2 exp i R R 2 U 1 2 exp 2i 2 R U exp i 1 2 2 exp 2i 4 exp 4i d R 2 2 0 R R R R 4 df dz i 2 2 dz C 2 2 i R U 2 R 2 df dz dz i U D 0 and L U 2C These are the same results that we found previously (Eqn. (5.23))! D iL i C. Wassgren Chapter 05: Potential Flows 256 (5.27) Last Updated: 14 Aug 2010 5. Kutta-Joukowski Theorem Now consider the flow around an arbitrary closed body (centered at the origin) in a uniform stream of horizontal velocity, U. Far from the body (z) the complex potential will be of the following form (a Laurent series expansion): bn m i f ( z ) Uz (5.28) log z n 2 n 1 z Note that the coefficients, an, for the terms involving zn (n 2) in the Laurent series (refer to the previous set of notes on the Blasius integral law) are all zero since we are considering external flows (recall that the velocity field is given by df/dz so that terms involving zn where n 2 will approach as z). Furthermore, since we are concerned only with closed bodies, the net source term, m, should also be zero. We, however, will continue to include the source term until the end of this analysis. Given the complex potential above, let’s apply Blasius’ Integral Law to determine the lift and drag about an arbitrary object: df m i 1 n 1 U nbn z dz 2 z n 1 2 2 df m i U m i 1 1 2 U 2 2 O 3 dz 2 z 2 z z Using the Residue Theorem to evaluate Blasius’ Integral Law: D iL i 2 2 df df m i dz i 2 2 i Res dz i 2 2 i 2 2 U z 0 2 Coutside dz mU i U Thus, we see that for a closed object (m=0): D 0 and L U KUTTA-JOUKOWSKI THEOREM For an object that is not closed (e.g. a Rankine half-body), we have: D mU and L U U (5.29) L Notes: 1. The result given above indicates that there is no drag around an arbitrary closed object in an a steady, incompressible, irrotational, inviscid flow. In real life of course there is always some drag on an object. The conflict between the derived value of zero drag and the real-life value of non-zero drag is referred to as d’Alembert’s Paradox. There is no paradox, in fact, if one realizes that it is viscous effects (skin friction drag and form, aka pressure, drag resulting from the formation of a wake (which in turn is a result of boundary layer separation)) which produces drag on an object. 2. Bodies of semi-infinite extent (e.g. a Rankine half-body) do have drag on them due to the fact that the net source term, m, is not zero. The drag is a result of a non-zero flux of horizontal momentum out through the control surface. 3. The Kutta-Joukowski theorem states that the lift on an object is directly proportional to the net circulation, , caused by the object. This is an important observation that is especially useful in aerodynamics when calculating the lift on an airfoil. As will be shown later, the circulation around an airfoil is dependent on the free stream velocity so that the lift turns out to be proportional to the circulation squared. C. Wassgren Chapter 05: Potential Flows 257 Last Updated: 14 Aug 2010 6. Conformal Mappings Conformal maps are analytic functions that transform curves (e.g. equi-potential lines and streamlines) in one complex plane, call this the z-plane, to similar curves, but expanded or contracted and rotated at each point, in a different complex plane, call this the -plane. Conformal maps are useful because they allow us to use the complex potential for a straightforward flow (e.g. flow around a rotating cylinder), after the proper mapping, as the complex potential for a more complex flow (e.g. flow around an airfoil). The complexity comes into play when trying to find the proper mapping that will give us the desired transformation. y x f(z) where z=x+iy C. Wassgren Chapter 05: Potential Flows mapping function: =F(z) 258 f() where =+i Last Updated: 14 Aug 2010 First, let’s examine some properties of a conformal map. Let =+i be an analytic function of z=x+iy given by: =F(z) Because the transforming function, F, is analytic, there is a connection between curves in the z-plane and corresponding curves in the -plane. Proof: Determine the derivative of the function, =F(z), at a point, z, by approaching the point from two different directions: z+z’ y z +’ z+z” x +” where: F z z F z z The length ratios in the z and planes are: z and z and the angles separating the lines are: arg( ) arg( ) arg( z ) arg( z ) and z arg arg z Also, because an analytic function has a unique derivative: dζ 2 z O z dz dζ 2 z O z dz So that the length ratios and angles between the lines are the same in each plane: z z and arg arg z z as z , z 0 Thus, lengths in the neighborhood of z are stretched by a scale factor, d dz , and are rotated by an angle, arg d dz , into the -plane. C. Wassgren Chapter 05: Potential Flows 259 Last Updated: 14 Aug 2010 Notes: 1. Curves of small linear dimension in the z-plane are mapped into curves of similar shape, but expanded or contracted and rotated, in the -plane. 2. 3. 4. 5. Note that a large region may be transformed into a region that bears no resemblance to the original one since the scale factor and angle of rotation vary, in general, from point to point. d At singular points of the map, i.e. points where 0 or , the mapping is not conformal. dz Since lines of constant and are in the z-plane, they will also be in the -plane, except at singular points. (Refer to Note 1.) Since an analytic function of another analytic function is also analytic (Churchill and Brown, p. 56), we are assured that the function resulting from the conformal map of a complex potential will also be a valid complex potential (it will be a 2D, incompressible, irrotational flow). 6. Velocities in the -plane are proportional to the velocities in the z-plane by the inverse of the scale factor: df df dz dz u x iu y u iu d dz d d 7. Singularities such as vortices and sources/sinks in the z-plane map to identical singularities in the -plane. This can be seen by considering the flow in a neighborhood of the singularity as the neighborhood shrinks to an infinitesimally small radius. m u dA u x dy u y dx C C u ds u x dx u y dy C C Consider the integral of the complex velocity around the contour C: df dz dz ux iu y dx idy ux dx u y dy i ux dy u y dx C C C C m df dz dz im C Thus, z imz Cz df df d df dz d dz dz C d d im dz Cz Source/sink and free vortex singularities in the z-plane map to similar singularities in the -plane. 8. Doublet singularities in the z-plane map to doublet singularities in the -plane but with the strength changed in d d magnitude by and an orientation rotated by arg (recall that a doublet is formed by bringing a dz dz source and sink of equal strength infinitesimally close to each other while keeping the product m/a constant where a is the separation distance between the source and sink). 9. Even though the occurrence of boundary layer separation in the real flow may limit the usefulness of a potential flow model, transformation of the flow field to a different flow field may produce a realistic flow. For example, although in a real flow boundary layer separation occurs for flow over a rotating cylinder, the mapping to an airfoil shape appears realistic. 10. Conformal maps are another tool we can use to produce realistic-looking flows using potential functions. C. Wassgren Chapter 05: Potential Flows 260 Last Updated: 14 Aug 2010 7. Joukowski Transformation An example of a particular conformal mapping is the Joukowski transformation: c2 z z where c. This mapping will transform flow around a rotating cylinder in the z-plane to flow around an airfoil in the -plane. This airfoil is referred to as a Joukowski airfoil. (5.30) y R Consider a circle of radius R centered at the point z0 such that the circumference of the circle passes through the point z=c: z0 c Rexp i z0 c The points defining the circle are given by: z z0 Rexp i c Rexp i Rexp i If we map the points of the circle in the z-plane to the -plane using the transformation given in Eqn. (5.30), the resulting figure looks like an airfoil: c =1.0, R /c =1.1, =5.0 deg 3.0 2.0 y or 1.0 0.0 -3 -2 -1 0 1 2 3 -1.0 -2.0 -3.0 x or C. Wassgren Chapter 05: Potential Flows 261 Last Updated: 14 Aug 2010 x Notes: 1. The geometry of the Joukowski airfoil is determined by the quantities R/c and. The camber of the airfoil is proportional to (camber as ). The thickness of the airfoil increases as (R/c) increases. The chord length of the airfoil is approximately equal to 4c (the exact chord length will also depend on the airfoil thickness). camber thickness mean camber line leading edge (LE) trailing edge (TE) chord An airfoil with no camber (=0): c =1.0, R /c =1.1, =0 deg 1.0 0.5 0.0 -3 -2 -1 0 1 2 3 -0.5 -1.0 An airfoil with no camber (=0) and larger thickness (R/c=2.0): c =1.0, R /c =2.0, =0 deg 2.0 1.0 0.0 -4 -2 0 2 4 -1.0 -2.0 C. Wassgren Chapter 05: Potential Flows 262 Last Updated: 14 Aug 2010 2. The trailing edge of the Joukowski airfoil will be cusp-shaped. Real airfoils typically end in a finite angle. angle > 0 2 3. Joukowski airfoils are not commonly used in practice; however, they provide a good model for predicting the general behavior of airfoils at small angles of attack (so that boundary layer separation won’t occur in the real-world airfoils). 4. Note that the trailing edge of the airfoil corresponds to the location where the cylinder intersects the location z=c. The transformation is not conformal at z=c and thus the angle between intersecting lines in the -plane is not necessarily the same angle between intersecting lines in the z-plane at the point z=c. C. Wassgren Chapter 05: Potential Flows 263 Last Updated: 14 Aug 2010 Now consider, in the z-plane, a uniform flow with velocity U around a rotating cylinder with circulation (an unknown value at this point) and radius R. The complex potential for this flow is given by: R 2 i z f z U z log (5.31) z 2 R Let’s rotate the flow so that the incoming stream is at an angle of attack, , with respect to the horizontal: z z exp i z z exp i Let’s also translate the origin of the cylinder so that it is centered at the position z0=c-Rexp(-i): z z z0 z z z0 The new complex potential is given by i z z0 R2 f z U z z0 exp i exp i log exp i z z0 2 R (5.32) Note that we haven’t yet determined the value of the circulation, . This will be found in the section below. First, however, let’s plot some streamlines for the case with zero circulation (=0): Of particular interest in the plot is the condition at the trailing edge of the airfoil. The streamlines at the trailing edge make a very sharp turn (the streamlines are not smooth at the very tip of the airfoil) resulting in infinite fluid accelerations and velocities at the trailing edge. This is not a very realistic flow and does not match what we observe in flows around real airfoils. We can avoid this infinite velocity problem by adjusting the circulation around the airfoil so that the flow leaves smoothly from the trailing edge. This is equivalent to moving the rear stagnation point to the tip of the trailing edge. This adjustment is referred to as the Kutta Condition. C. Wassgren Chapter 05: Potential Flows 264 Last Updated: 14 Aug 2010 To quantitatively determine what the value of the circulation must be to satisfy the Kutta Condition, let’s examine the complex velocity of fluid along the airfoil surface: df df dz dz dz u iu d dz dz dz d where df R2 i U 1 2 dz 2 z z dz exp i dz dz 1 dz 2 dz c 1 d z 1 z z z0 exp i z0 c Rexp i so that the complex velocity in the -plane is given by: 1 c 2 df R2 i U 1 u iu exp i 1 d z 2 2 z z (5.33) Note that at z=c the magnitude of the complex velocity approaches infinity due to the (dz/d term. To prevent infinite velocities from occurring, the term within the curly brackets {} must equal zero at z=c: R2 i 0 U 1 2 z 2 z z c R2 R2 2 iUz 1 2 iU Rexp i 2 Rexp i z z c where z z c c z0 exp i c c Rexp i exp i Rexp i 2 iUR cos i sin cos i sin 4 UR sin Thus, to prevent infinite velocities from occurring at the trailing edge of the airfoil, the circulation must be given by: 4 UR sin (5.34) Now that we know the circulation about the airfoil, we can use the Kutta-Joukowski Theorem to determine the lift of the airfoil: L U (5.35) L 4U 2 R sin C. Wassgren Chapter 05: Potential Flows 265 Last Updated: 14 Aug 2010 The lift is often presented in dimensionless form as the lift coefficient, cL: L R cL 2 sin 1 U 2 4c c 2 where 4c is the approximate chord length of the airfoil. (5.36) We can also determine the pressure distribution on the airfoil surface by using Bernoulli’s equation (recall that we’re dealing with an incompressible, irrotational flow so the same Bernoulli constant is used everywhere): ps 1 2 us2 p 1 2 U 2 where ps and us are the pressure and speed on the airfoil surface, and p and U are the pressure and speed far from the airfoil. The magnitude of the velocity on the surface of the airfoil can be found using the complex velocity given in Eqn. (5.33) with: z z z0 exp i Rexp i exp i Rexp i where defines the location on the airfoil surface. After some algebra, we arrive at: 2U sin sin us 2 c 1 z (5.37) with z=c-Rexp(-i)+Rexp(i). The pressure is often expressed non-dimensionally as the pressure coefficient, cP: cP p p u 1 s 1 U 2 U 2 2 (5.38) Notes: 1. The lift coefficient predicted by our potential flow analysis of a Joukowski airfoil is reasonably close to values found experimentally at small angles of attack and small camber (to avoid boundary layer separation). 2. Flow over a flat plate can be found by letting R/c=1 and =0. The resulting lift coefficient is: cL 2 sin The Joukowski transformation can also produce flow around curved plates (let R/c=0 and >0). 3. Note that increasing the angle of attack, , the camber, , and the thickness, R/c, all act to increase the lift of an airfoil. C. Wassgren Chapter 05: Potential Flows 266 Last Updated: 14 Aug 2010 4. The Joukowski transformation can also produce flow around ellipses. To produce this type of flow, we center the cylinder at the origin and choose R>c. The points on the cylinder surface are given by z Rexp i c =1.0, R =2.0 2.0 1.0 0.0 -3 -2 -1 0 1 2 3 -1.0 -2.0 where 0<2 C. Wassgren Chapter 05: Potential Flows 267 Last Updated: 14 Aug 2010 8. Method of Images In much of our previous work concerning potential flows, we investigated external flows in an infinite expanse of fluid. Since there are a number of phenomena that are of interest when there is flow near a boundary, we should find a method for modeling flows near walls. The Method of Images is such a method. Consider how we can model the flow from a source near a wall. To produce a horizontal streamline representing the wall, we can add to our original source, an “image source” an equal distance away from where we want our wall to be. m ln r1 ln r2 y 2 y 2 where r1 x 2 y a r1 r2 x 2 y a 2 a a x r r2 a a Notes: 1. There is a net upward velocity at the location of the original source of: m1 V 2 2a where m is the source strength due to the flow induced by the image source. 2. The vertical force acting on the source can be determined by first calculating the pressure force acting on the wall using Bernoulli’s equation and then noting that the force acting on the wall is equal, but in the opposite direction, to the force acting on the source. The pressure at the wall is given by: pw 1 2 Vw2 p where pw and Vw are the pressure and velocity magnitude at the wall and p is the pressure far from the wall (U approaches zero as we move far from the wall). The velocity along the wall is found from the potential function: m 2 2 ln x 2 y a ln x 2 y a 2 ux uy y 0 y 0 x y 0 y y 0 m x x2 a2 0 2 x m pw p 1 2 2 2 x a Note that without the sources, the pressure acting on the wall would be p∞. Hence, the increase in the force acting on the wall is: C. Wassgren Chapter 05: Potential Flows 268 (x, y) Last Updated: 14 Aug 2010 x x Fon wall due to source x pw p dx 2 x m dx 12 2 2 x a x x 2 1 m x 1 x 2 tan 1 2 a 4 x a a Fon wall due to source m2 4 a Consequently, the force on the source will be: m2 Fon source 4 a A particularly interesting application of the method of images is investigating the effect that the ground has on the lift of an airfoil. Let’s use a crude model consisting of a free vortex combined with a uniform stream to investigate this effect. Recall that in order to satisfy the Kutta condition, an airfoil must have some circulation which in turn develops lift (from the Kutta-Joukowski Theorem). The potential flow model is given below (drawn for >0). y Ux 1 2 2 y r1 ya where 1 tan 1 U x r a r2 ya 2 tan 1 a x a x a As before with the source example, the force acting on the source will have the same magnitude, but with opposite sign, as the force acting on the wall. The force acting on the wall is found by integrating the pressure force over the entire wall: a 2 x U a U ux y 0 2 a x y 0 2 x2 a2 1 x uy y 0 y 0 y 0 2 2 aU a pw p 1 2 U 2 1 2 U 2 x2 a2 x2 a2 The resulting force acting on the wall due to the vortices (again, subtracting out the pressure when no vortices are present) is then: C. Wassgren Chapter 05: Potential Flows 269 Last Updated: 14 Aug 2010 (x, y) x x Fon wall due to vortex x pw p dx 2 1 2 aU a 2 dx x2 a2 x2 a2 x x 2 2U 1 x 1 x x 12 tan 1 tan 1 2 2 a 2 x a a a 2 2U 2a 2 1 U 1 4 aU Thus, the force acting on the vortex is: Fon vortex U 1 4 aU Fon wall due to vortex Recall that the lift force acting on an object in an infinite expanse of fluid with circulation is given by the Kutta-Joukowski theorem as: Linfinite U expanse Keep in mind that from the previous discussion regarding Joukowski airfoils, the circulation around an airfoil is negative ( < 0) in order to satisfy the Kutta condition at the trailing edge of the airfoil. The difference between the lift generated when the wall is nearby versus the lift in an infinite expanse of fluid is: L Lwall Linfinite U 1 U 4 aU present expanse 2 4 a Hence, the wall acts to decrease the lift. Experience shows, however, that an airfoil near the ground, aka in ground effect, actually has increased lift (and decreased drag) rather than decreased lift. Why do we have this discrepancy? It’s because our analysis considers an infinitely long airfoil, i.e. one with no wing tips. At the end of a finite wing, “trailing” vortices (as opposed to the vortex “bound” to the airfoil resulting from the Kutta condition) are generated by the wing tips as shown below. These wingtip vortices occur because air in the high pressure region underneath the airfoil is pushed around the wing tips to the low pressure above the airfoil. U bound vortex trailing vortices starting vortex (This will be discussed later.) When viewed from behind, the trailing vortices appear as shown below. wingtip vortex wingtip vortex low pressure side of wing high pressure side of wing C. Wassgren Chapter 05: Potential Flows 270 Last Updated: 14 Aug 2010 The wingtip vortices induce a “downwash” along the wing surface and thus reduce the effective angle of attack that the airfoil sees. Dind L U∞ ind wind eff ind eff wing The lift acting on the wing per unit span (i.e. distance into the page), L, will be the lift calculated for the local effective angle of attack, eff, where: w eff ind with tan ind ind U Here, is the nominal angle of attack and ind is the induced angle of attack resulting from the wingtip vortices which induce a local downwash velocity of wind. Since the effective angle of attack is reduced, the lift on the airfoil will also be reduced. There is also an induced drag on the wing Dind since the local flow is at an angle of ind from the free stream which tilts the actual lift vector slightly downstream Dind L tan ind Again, Dind is the induced drag per unit span of the wing. Note that the induced drag on the wing is not due to viscous effects, but is due solely to the induced angle of attack resulting from the induced downwash. Hence, there is a drag on a finite wing, even in an ideal flow, due to the trailing vortices. The trailing vortices also drift downwards over time due to the flow induced at the center of each vortex by the other vortex. If the airfoil was turned upside down, the vortex orientation would be reversed and the vortices would drift upwards over time! b Vdownward Vdownward If this vortex has a circulation of , the induced velocity at the center of the other vortex is: 1 Vdownward 2 b where b is the distance between the wingtips. Near the ground, two image vortices must also be included in the analysis in order to make the horizontal ground streamline. As a result of the flow induced by the image vortices, the wingtip vortices drift outward when reaching the ground. downwash induced by original vortices original vortices upwash induced by image vortices image vortices C. Wassgren Chapter 05: Potential Flows 271 Last Updated: 14 Aug 2010 The image vortices also contribute an “upwash” along the wing which helps to counteract the downwash caused by the original vortices. This reduction in the downwash helps increase the lift and reduce the drag on the finite wing when it is located near the ground. This is the source of the observed “ground effect.” Notes: 1. Recall that from previous discussions regarding vorticity, vortex lines must either form closed loops or terminate on a boundary ( = 0). So how then do the vortex lines corresponding to the trailing vortices terminate? The bound vortex/trailing vortex lines actually form a closed loop through a “starting” vortex line (shown in a previous figure). The starting vortex occurs during the transient when the lift on the airfoil changes (e.g. at start up). Since the starting vortex is typically located far behind the bound vortex, its effects are typically neglected in steady airfoil analyses and the bound vortex/trailing vortex combination is treated as a horseshoe vortex. 2. The trailing vortices are not actually concentrated solely at the wingtips. Instead, there is a distribution of trailing vortices along the wing due to variations in the circulation which result from changes in airfoil geometry and local flow conditions. These variations must be included a finite wing analyses (see, for example, Kuethe, A.M. and Chow, C-Y., Foundations of Aerodynamics, Wiley.) bound vortex trailing vortices 3. Trailing vortices have been the source of several airline disasters (see, for example, http://www.asy.faa.gov/safety_products/wake.htm and http://aviation-safety.net/events/EFV.shtml). If an aircraft flies behind a preceding aircraft too closely, it can be caught in the trailing vortices and cause the pilot to lose control of the aircraft (this phenomenon is sometimes mistakenly referred to as “wake turbulence”). The strength of the vortices is proportional to the lift generated by the airfoil which in turn is related to the weight of the aircraft. Hence, the spacing between aircraft (near an airport for example) is a function of their relative size. 4. Ground effect has been used as a significant component in the design of several aircraft. The following web site lists a number of WIG (wing-in-ground-effect) aircraft: http://www.setechnology.com/wig/index.php). Even pelicans take advantage of ground effect! C. Wassgren Chapter 05: Potential Flows 272 Last Updated: 14 Aug 2010 Notes: 1. We may sometimes need an infinite number of reflections to properly model a flow. Consider for example, the flow resulting from a source located midway between two walls. An infinite number of images are required for perfect symmetry. C. Wassgren Chapter 05: Potential Flows 273 Last Updated: 14 Aug 2010 Example: a. b. Write the potential function that simulates the flow of a line source placed asymmetrically between two parallel walls as shown in the figure. Compute the dimensionless velocity, u’ = au/(4m), on the lower wall at (x/a, y/a) = (1,0) accurate to three significant digits. y 2a m a x SOLUTION: Use the Method of Images to create the given flow. The sequence of images is shown below. repeat reflecting images to ∞ a a y 2a 2a m a a x 2a 2a repeat reflecting images to ∞ The potential function for the given flow field is: 2 2 2 2 2 2 2 2 m ln x y a ln x y a ln x y 5a ln x y 5a (5.39) 2 2 2 2 2 2 2 2 2 ln x y 7a ln x y 7a ln x y 11a ln x y 11a C. Wassgren Chapter 05: Potential Flows 274 Last Updated: 14 Aug 2010 ln x 2 y 6k 1 a 2 ln x 2 y 6k 1 a 2 2 2 2 2 k 0 ln x y 6k 5 a ln x y 6k 5 a or, in dimensionless terms: 2 2 2 2 k ln x y 6k 1 ln x y 6k 1 2 2 2 2 2 k 0 ln x y 6k 5 ln x y 6k 5 4 ln a m 4 k (5.40) (5.41) where the dimensionless potential function is ’ = /(4m) and the dimensionless positions are x’ = x/a and y’ = y/a. The dimensionless velocities resulting from this potential function are: 1 1 2 2 2 2 k x y 6k 1 x y 6k 1 2 x ux 1 1 x k 0 x 2 y 6 k 5 2 x 2 y 6 k 5 2 y 6k 1 y 6k 1 2 2 2 2 k x y 6k 1 x y 6k 1 u y y k 0 y 6k 5 y 6k 5 2 2 2 x y 6k 5 x 2 y 6 k 5 where u’x = aux/(4m) and u’y = auy/(4m). The velocity at (x’, y’) = (1, 0) is: k 1 1 u x 1, 0 4 2 2 1 6k 5 k 0 1 6k 1 (5.43) (5.44) u 1, 0 0 (as expected since the point is on a wall) y C. Wassgren Chapter 05: Potential Flows (5.42) 275 (5.45) Last Updated: 14 Aug 2010 The value of the horizontal velocity component at (x’, y’) = (1,0) as a function of k is given in the table below. Note that “%diff from prev” is the percentage change in the value of u’x from the previous value of u’x, i.e. % diff = (u’x,k+1 – u’x,k)/u’x,k * 100%. k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 u' x % diff from prev 1.962 2.042 4.08% 2.066 1.15% 2.077 0.53% 2.083 0.31% 2.087 0.20% 2.090 0.14% 2.092 0.10% 2.094 0.08% 2.095 0.06% 2.097 0.05% 2.097 0.04% 2.098 0.04% 2.099 0.03% 2.099 0.03% 2.100 0.02% 2.100 0.02% 2.101 0.02% 2.101 0.02% 2.101 0.01% Hence, the velocity components at (x’,y’) = (1,0) are (u’x, u’y) = (2.10, 0). C. Wassgren Chapter 05: Potential Flows 276 Last Updated: 14 Aug 2010 9. Added Mass Added mass (aka apparent or virtual mass) is the concept whereby we add an extra “mass” to an object when we accelerate the object through a fluid. This added mass term accounts for the force required to accelerate the surrounding fluid to a higher velocity. To study this concept, let’s consider the potential function for a cylinder in a fluid that is stagnant far from the cylinder. To form this potential function, we first form the potential function for a uniform stream with velocity U flowing around a stationary cylinder of radius R. R2 stationary cylinder in uniform stream Ur cos 1 2 r To change our frame of reference so that the fluid far from the cylinder is stationary, we add in a uniform stream of velocity U in the opposite direction: cylinder moving through stagnant fluid stationary cylinder in uniform stream of velocity U Ur cos UR 2 cos r The resulting potential function describes the flow produced by a cylinder of radius R moving at velocity U through an otherwise quiescent fluid. cylinder moving through stagnant fluid Let’s now determine the total kinetic energy in the fluid (outside of the cylinder): r KEtotal 1 2 2 rdr ur2 u2 R r R where the fluid velocity components are given by: dA = 2rdr UR 2 cos 1 UR 2 sin ur and u r r2 r r2 U 2 R4 ur2 u2 4 r Note that the in the kinetic energy formula is the fluid density. Substituting the speed into the expression for the total kinetic energy: r KEtotal r R 1 U 2 R 4 U 2 R 4 2 2 rdr r4 2r 2 r R2 rR 2 U 2 If we apply a force such that it increases the velocity of the cylinder by a small amount U, the total kinetic energy of the fluid will increase by an amount (neglecting higher order terms): R2 R2 2 KEtotal U U U 2 R 2 U U 2 2 The average force we must apply to the cylinder over time t (the time over which the velocity goes from velocity U to velocity U+U) to increase the total kinetic of the fluid is: FU t KEtotal total work done in time t F required to increase KE of fluid R 2 U U U R2 U t t Thus, the total force required to accelerate a cylinder of mass, M, through a quiescent fluid is given by: 2 dU F M R cylinder added dt mass mass C. Wassgren Chapter 05: Potential Flows 277 Last Updated: 14 Aug 2010 The term added to the cylinder mass in the previous equation is referred to as the added mass (aka apparent or virtual mass). Thus, the added mass for a cylinder is Madded=R2. Again, the is the fluid density. Notes: 1. Note that added mass is only a factor for unsteady flows. There is no added mass term for a steady flow. 2. Added mass terms are typically only significant in flows of liquids since the added mass for gases is often small compared to the object’s mass (gas is typically very small). Added mass terms in a gas can be significant however if the object is large and has small mass (e.g. a parachute). 3. We could have also found the force on the cylinder by integrating the pressure force around the cylinder surface which is found using the unsteady Bernoulli equation (neglecting gravitational effects): p 1 2 V 2 p t on surface (r R ) dU dt Integrating the pressure force along the surface of the cylinder to find the total force on the cylinder: 2 2 2 2 dU F ps cos Rd R p cos d 1 2 U 2 cos d R cos 2 d dt 0 0 0 0 dU F R 2 (the same answer as before) dt ps p 1 2 U 2 R cos 4. The added mass is dependent on the shape of the object. It’s possible to have different values for the added mass depending on the orientation of the object (e.g. an ellipse will have different added masses depending on its orientation.) We can also have added mass effects due to rotational acceleration of an object. 5. An additional reference concerning added mass is: Yih, C.S., 1969, Fluid Dynamics, McGraw-Hill (Now published by West River Press, Ann Arbor, MI). 6. The added mass presented here was calculated for an inviscid fluid. For unsteady viscous flows an additional term referred to as the Bassett force also appears which takes into account unsteady viscous force terms. C. Wassgren Chapter 05: Potential Flows 278 Last Updated: 14 Aug 2010 10. Finite Difference Methods Recall that the governing equation for an incompressible, irrotational flow is: 2 0 (5.46) where is the velocity potential. Our goal here is to re-write Eqn. (5.46) using a finite difference approximation so that we can solve the equation numerically. We’ll assume a 2D flow in Cartesian coordinates to make the following analyses more straightforward; however, the same ideas can be applied to 3D and non-Cartesian (but still orthogonal!) coordinate systems. In particular, we’ll solve Eqn. (5.46) at the grid points shown in the figure below. Note that for simplicity, neighboring grid points are assumed to be separated by a distance h in both the x- and y-directions. The derivations given below may also be extended to non-uniform grid spacings. streamlines y i,j+1 i,j i+1,j h i-1,j finite difference grid x x h i,j-1 The values of the second order partial derivatives in Eqn. (5.46), e.g. 2i,j/x2, may be written in terms of the neighboring values of by using Taylor series expansions about the point i,j. For example, for determining 2i,j/x2, express i+1,j and i-1,j in terms of Taylor series expansions about i,j and then add the two Taylor series together. i 1, j i , j i 1, j i , j h 1! x h 2 2 2! x 2 i, j h 1! x h 3 3 2! x 2 h 2 2 2! x 2 2 i, j 3! x3 i, j h 2 2 i, j i 1, j i 1, j 2i , j 2 i, j h 4 4 4! x 4 h 3 3 i, j h 4 4 4! x 4 3! x3 (5.47) i, j h 4 4 i, j 4! x 4 (5.48) i, j (5.49) i, j The previous expression may be re-arranged to solve for the 2nd order derivative. i 1, j i 1, j 2i , j h 2 4 2 2 4! x 4 i , j h2 x 2 i , j A similar approach may be used in the y-direction to determine 2i,j/y2. i , j 1 i , j 1 2i , j 2 h 2 4 2 4! y 4 i , j y 2 i , j h2 (5.50) (5.51) Hence, at point (i,j), the solution to Eqn. (5.46) may be written as: 0 2 i, j 2 x 2 i, j 2 y 2 i 1, j i 1, j 2i , j h 2 i, j 2 h 2 4 4! x 4 i , j 1 i , j 1 2i , j i, j 0 i 1, j i 1, j i , j 1 i , j 1 4i , j 2 C. Wassgren Chapter 05: Potential Flows h h 4 4 4! x 4 2 2 i, j 279 2 h 4 4 4! y 4 h 2 4 4! y 4 i, j (5.52) i, j Last Updated: 14 Aug 2010 The previous equation is an exact solution to Eqn. (5.46) at the point (i,j) as long as all of the higher order terms are included in Eqn. (5.52). If the value of h is sufficiently small, then we may approximate Eqn. (5.52) by neglecting terms of order h4 and higher since they will be small in comparison to the remaining terms (as long as the higher order derivatives don’t simultaneously become very large). The resulting truncated equation is now only an approximate solution to Eqn. (5.46). (5.53) 0 i 1, j i 1, j i , j 1 i , j 1 4i , j where the truncation error in the previous equation is of order h4. Notes: 1. Note that at a vertical solid boundary, the horizontal velocity (ux) is zero. In terms of the velocity potential: i,j+1 y u x i, j 0 h x i , j i-1,j i,j x h i,j-1 Determining the potential at (i-1,j) in terms of the Taylor series expansion about point (i,j) gives: i 1, j i , j h 1! x i , j h 2 2 2! x 2 h 3 3 i, j 3! x3 i, j h 4 4 4! x 4 (5.54) i, j 0 Re-arranging this equation gives: 3 2 i 1, j i , j 2 Oh 3 x x 2 i , j h2 i, j Combine Eqn. (5.55) with Eqn. (5.51) to solve for 2i,j = 0. 0 2i , j 2 x 2 2 i 1, j i , j h2 i, j 2 y 2 i, j i, j 1 i, j 1 2i, j O h 3 4 O h2 x3 y 4 i, j h2 3 4 0 i , j 1 i , j 1 2i 1, j 4i , j O h3 3 O h 4 4 x y i, j If the previous equation is truncated, then it becomes: 0 i , j 1 i , j 1 2i 1, j 4i , j A similar approach can be used at a horizontal boundary to give: 3 4 0 i 1, j i 1, j 2i , j 1 4i , j O h3 3 O h 4 4 x y i, j 0 i 1, j i 1, j 2i , j 1 4i , j C. Wassgren Chapter 05: Potential Flows (5.55) 280 i, j i, j (5.56) (5.57) i, j y (5.58) h i-1,j i,j i,j-1 h i+1,j x (5.59) Last Updated: 14 Aug 2010 or at a corner: 0 2i 1, j 2i , j 1 4i , j 3 O h3 3 x 4 O h4 y 4 i, j 0 2i 1, j 2i , j 1 4i , j i, j h y i,j+1 0 i+1,j h (5.60) x (5.61) 2. Equation (5.46), i.e. Laplace’s equation, is an elliptic partial differential equation. In order to have a well-posed problem, i.e. the equation has a unique solution that depends continuously on the boundary and/or initial data, the gridded flow domain must be finite and continuous boundary conditions must be specified along the entire boundary. The boundary conditions may be either Dirichlet boundary conditions (where the value of is specified), Neumann boundary conditions (where the gradient of is specified), or a combination of both types of boundary conditions (known as mixed boundary conditions). 3. There are two common methods to solving the resulting finite difference approximations to Eqn. (5.46) at every point in the flow domain. Non-iterative, or direct, methods solve the equations directly (in “one step”) while iterative methods solve the equations after repeated calculations that (hopefully) converge on the answer. Examples of each of these methods are given in the following discussions using the simple flow field and grid shown below. j 4 3 2 1 = 0 uniform grid spacing = 1 1234 i Before numerically solving for the values of at each of the grid points, we can easily observe that for the uniform grid spacing shown, we anticipate that values for in row 2 (i,j = 2) should be 2/3 and the values for in row 3 (i, j = 3) should be 1/3. C. Wassgren Chapter 05: Potential Flows 281 Last Updated: 14 Aug 2010 For the given example, write the finite difference equations for each point on the grid using the expressions derived previously. At (i,j) = (1,1): 1,1 1 (a given boundary condition) At (i,j) = (2,1): 2,1 1 (a given boundary condition) At (i,j) = (3,1): 3,1 1 (a given boundary condition) At (i,j) = (4,1): 4,1 1 (a given boundary condition) At (i,j) = (1,2): 1,1 1,3 22,2 41,2 0 At (i,j) = (2,2): 1,2 3,2 2,1 2,3 42,2 0 At (i,j) = (3,2): 2,2 4,2 3,1 3,3 43,2 0 At (i,j) = (4,2): 23,2 4,1 4,3 44,2 0 At (i,j) = (1,3): 22,3 1,2 1,4 41,3 0 At (i,j) = (2,3): 1,3 3,3 2,2 2,4 42,3 0 At (i,j) = (3,3): 2,3 4,3 3,2 3,4 43,3 0 At (i,j) = (3,3): 23,3 4,2 4,4 44,3 0 At (i,j) = (1,4): 1,4 0 (a given boundary condition) At (i,j) = (2,4): 2,4 0 (a given boundary condition) At (i,j) = (3,4): 3,4 0 (a given boundary condition) At (i,j) = (4,4): 4,4 0 (a given boundary condition) Re-write the previous equations in matrix form. 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1,1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2,1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 3,1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 4,1 1 1 0 0 0 4 2 0 0 1 0 0 0 0 0 0 0 1,2 0 0 1 0 0 1 4 1 0 0 1 0 0 0 0 0 0 2,2 0 0 0 1 0 0 1 4 1 0 0 1 0 0 0 0 0 0 3,2 0 0 0 1 0 0 2 4 0 0 0 1 0 0 0 0 4,2 0 0 0 0 4 2 0 0 1 0 0 0 1,3 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 4 1 0 0 1 0 0 2,3 0 0 0 0 0 0 0 1 0 0 1 4 1 0 0 1 0 3,3 0 0 0 0 0 0 0 0 1 0 0 2 4 0 0 0 1 4,3 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1,4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 2,4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 3,4 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4,4 0 A C. Wassgren Chapter 05: Potential Flows 282 b (5.62) Last Updated: 14 Aug 2010 A direct method for solving Eqn. (5.62) is Gauss Elimination. The algorithm for Gauss Elimination is not presented here and instead the reader is encouraged to review the method in a numerical methods text (see, for example, Hoffman, J.D., Numerical Methods for Engineers and Scientists, 2nd ed., Marcel-Dekker). Solving Eqn. (5.62) using Gaussian Elimination gives: 1,1 1 2,1 1 3,1 1 4,1 1 1,2 2 3 2,2 2 3 2 3,2 3 4,2 2 3 These are the results we expected! 1 1,3 3 2,3 1 3 3,3 1 3 1 4,3 3 1,4 0 0 2,4 3,4 0 4,4 0 (5.63) Notes: a. Gaussian elimination is the preferred method for solving systems of linear equations. Modifications to the Gaussian elimination algorithm have been proposed that are optimized for banded matrices (where non-zero entries in the matrix occur in diagonal bands) such as the one in Eqn. (5.62). Thomas’ algorithm is one such algorithm that is particularly efficient for tridiagonal matrices. b. Direct methods, as opposed to iterative methods, will always converge to a solution (assuming that the given [A] matrix is non-singular, i.e. it has a non-zero determinant). c. For very large systems, direct methods are generally less efficient than iterative methods. C. Wassgren Chapter 05: Potential Flows 283 Last Updated: 14 Aug 2010 Another approach for determining the values of i,j in Eqn. (5.62) is to use an iterative scheme. With an iterative method, an initial guess for the solution to i,j is assumed. These initial values for i,j are then used to generate new values for i,j using a scheme that reduces the value between the current values of i,j and the actual solution values. The scheme is repeated using the new values for i,j until the values converge to the solution. Note that convergence of the iterative algorithm is not always guaranteed, which is the major drawback to iterative methods. One commonly used iterative algorithm is Gauss-Seidel Iteration with Successive Over-Relaxation. In this algorithm, the new value for i,jn+1, where the superscript “n+1” indicates the new value, is determined using the previous values, found at iteration step “n”, of at the surrounding points (this is actually known as the Jacobi Iteration Method – Gauss-Seidel Iteration will be discussed in a moment). For example, Eqn. (5.53) may be written in iterative form as: in 1 ,j 1 4 n i 1, j in 1, j in j 1 in j 1 , , (5.64) Iteration on all points (i,j) continues until the error between the current iteration value for and the previous iteration’s value for is less than some tolerance, i.e.: Repeat iterations until in 1 in j tolerance for all (i, j). ,j , (5.65) The difference between Gauss-Seidel iteration and Jacobi iteration is that Jacobi iteration determines the value for i,jn+1 based on all of the previous iteration values whereas Gauss-Seidel iteration makes use of the new values for as they become available. For example, if we iterate in the previous example moving in the direction of increasing i and increasing j, then the value for 3,3 will be: n 3,31 1 4 n 1 2,3 n n n 4,3 3,21 3,4 (5.66) By using the already updated values at the neighboring grid points, convergence is accelerated. Often, the rate of convergence of the iterations can be improved by implementing a relaxation scheme. With relaxation the value of i,jn+1 is found using a linear combination of Eqn. (5.64) (or rather an equation similar to Eqn. (5.66) depending on the direction of iteration) and the previous value for i,jn, i.e.: in 1 in j in 2 in j ,j , ,j , 1 (5.67) where the superscript n+1/2 refers to the intermediate value of i,j calculated using Eqn. (5.64) (or Eqn. (5.66)) and is referred to as the relaxation parameter. The effect of relaxation can be most easily understood when presented graphically. In many instances the iterative values of i,j approach the actual value of i,j from one direction as shown in the plot below where a particular i,j is shown as a function of the number of iterations, n. We observe from the plot that by using over-relaxation we are extrapolating the value of i,jn+1 using i,jn and i,jn+1/2 to help reach the converged value more quickly. actual value of i,j i1, j 1 i0 2 ,j i0 j , C. Wassgren Chapter 05: Potential Flows i,j over-relaxation no relaxation i0 2 i0 j ,j , 1 number of iterations, n 284 Last Updated: 14 Aug 2010 Notes: a. If > 1 then the process is known as over-relaxation, < 1 is under-relaxation, and when = 1 there is no relaxation. Under-relaxation is typically used when the iterations produce oscillatory values for i,j. b. c. Relaxation can reduce the convergence rate considerably, often by one to two orders of magnitude! d. 4. For over-relaxation, the iterative scheme can be shown to diverge if ≥ 2. The optimal choice for the relaxation parameter is not known a priori, in general, and multiple computations using differing values of need to be performed to determine opt. Despite the additional computations, determining opt is still a worthwhile effort, especially if the system of equations must be solved multiple times (if the boundary conditions change for example). As a rule of thumb, larger systems usually have a larger value for opt. Additional issues such as the effects of round-off and truncation errors should be considered in more depth for a better understanding of numerical solutions. The reader is encouraged to study numerical methods texts for more information on these topics. C. Wassgren Chapter 05: Potential Flows 285 Last Updated: 14 Aug 2010 11. Doublet Distributions So far we’ve approached potential flow problems by choosing potential functions and observing what types of flows result. Let’s now look at a method of specifying an object shape and determining what the potential function should be. We’ll just examine a simple method here but it should be noted that more sophisticated methods (although based on the same concepts) are addressed in most books on aerodynamics (see, for example, Kuethe, A.M. and Chow, C.-Y., Foundations of Aerodynamics, Wiley). Recall that when we combined a uniform flow with a doublet, flow around a cylinder resulted. Now let’s imagine combining on the x-axis a large number of doublets with varying strength. y U x The stream function evaluated at a point (x, y) for such a flow is given by L K yd y Uy 2 2 0 x y (x,y) r where K()d is the total strength of the doublets over a very small distance d, is the distance from the origin, and L is the total length of the line of doublets. x L Since we generally solve these types of problems numerically, re-write the integral as a summation: jN K j y Uy 2 2 j 1 x y j where =L/N . We’re usually interested in determining what the potential function should be for a specific object. To solve this inverse problem, we note that the object surface is a streamline so the stream function remains constant on the surface. Since we can arbitrarily adjust the value of the stream function (by adding in a constant – remember that only differences or derivatives of the stream function are of interest to us), we can adjust the stream function so that its value is zero on the object surface: jN i 0 Uyi cij K j j 1 where i is a point on the object surface and cij is referred to as the “influence coefficient” (the contribution of a doublet of unit density at the location j to the point i): yi cij (5.68) 2 xi j yi2 C. Wassgren Chapter 05: Potential Flows 286 Last Updated: 14 Aug 2010 The result is a system of equations we can solve numerically to determine the appropriate values of Kj: c11 K1 c12 K 2 c1N K N Uy1 cN 1 K1 cN 2 K 2 cNN K N Uy N Here the yi are known since the geometry is known, U is known, and the cij are known as discussed previously. Notes: 1. We could also have used a potential function in the previous analysis but instead of specifying the value of the potential function on the surface of the object, we would instead require that there is no flow through the surface: 0 un n This approach is generally more involved than if we use stream functions. 2. We can extend these ideas to asymmetric objects by distributing doublets along curved paths. 3. Instead of using a line of doublets (aka doublet panel), we could also use lines of sources (aka source panels), or lines of vortices (aka vortex panels). 4. There will be no lift on objects generated with doublets or sources since they produce no net circulation. Only vortex panels will produce circulation and lift. 5. These ideas can be extended to 3D using 3D source/doublet/vortex potentials. C. Wassgren Chapter 05: Potential Flows 287 Last Updated: 14 Aug 2010 Review Questions 1. Describe three properties of stream functions. 2. Can stream functions be used for rotational flows? How about irrotational flows? 3. What restrictions are there when using stream functions? 4. Can stream functions be superposed? 5. What is the governing equation for an incompressible potential flow? 6. What are the requirements for modeling a flow as a potential flow? 7. What are the appropriate boundary conditions for a potential flow? 8. Can potential functions be written for 3D flows? How about stream functions? 9. Can stream functions be written for rotational flows? 10. How are streamlines related to equipotential lines? 11. Under what conditions can one write a complex potential function to describe a flow? 12. How is a fluid velocity field determined from a complex potential function? 13. Describe the potential flow model for ideal fluid flow around a non-rotating cylinder. 14. Describe the potential flow model for ideal fluid flow around a rotating cylinder. 15. What is d’Alembert’s paradox? 16. What causes Magnus lift? Is this what causes baseballs or golf balls to curve? 17. Why does potential flow modeling fail to capture the behavior of real flows downstream of a cylinder? 18. What is the Blasius integral law? 19. What is the Kutta-Joukowski theorem? 20. What is d’Alembert’s paradox? 21. Describe the “method of images”. 22. What is meant by “ground effect”? 23. What is meant by “added mass”? Under what conditions will the added mass on an object be significant? C. Wassgren Chapter 05: Potential Flows 288 Last Updated: 14 Aug 2010 Chapter 06: Dimensional Analysis 1. 2. 3. 4. 5. 6. Dimensional Analysis Buckingham-Pi Theorem Method of Repeating Variables Dimensionless Governing Equations Modeling and Similarity Stokes Number for Small Particles in a Flow C. Wassgren Chapter 06: Dimensional Analysis 289 Last Updated: 16 Aug 2009 1. Dimensional Analysis Dimensional analysis is a method for reducing the number and complexity of variables used to describe a physical system. It’s a technique that can be applied to all fields, not just fluid mechanics. The mechanics of dimensional analysis are simple to learn and apply, and the benefits from using it are significant. Dimensional analysis can be used to present data in an efficient manner, reduce the number of experiments or simulations one needs to perform to investigate the relationship between variables, and scale results. However, dimensional analysis cannot tell us what the functional relationship is between variables. Additional experiments or theoretical analyses are required to determine this information. Motivating Example #1 To motivate the use of dimensional analysis, let’s consider a simple example involving a ball falling under the action of gravity in a vacuum. From basic physics, we know that the vertical position of the ball, y, is given by: y 1 gt 2 y0 t y0 (6.1) 2 g where g is the acceleration due to gravity, t is the time from when the ball was released, y y0 is the initial speed of the ball, and y0 is the initial position of the ball. Note that Eqn. (6.1) is dimensional. In other words, each term in the equation has dimensions of length [L]. For example, the dimension of the first term on the right hand side is length, [1/2gt2] = L, where the square brackets indicate “dimensions of”. If we were to plot the position, y, as a function of time, t, for varying g, y0 , and y0, we would have plots that look like the following. 35 35 y0 = 10 m, ydot0 = 0 y0 = 20 m, ydot0 = 0 y0 = 30 m, ydot0 = 0 y0 = 10 m, ydot0 = -1 m/s y0 = 20 m, ydot0 = -1 m/s y0 = 30 m, ydot0 = -1 m/s y0 = 10 m, ydot0 = -2 m/s y0 = 20 m, ydot0 = -2 m/s y0 = 30 m, ydot0 = -2 m/s y0 = 10 m, ydot0 = 0 g = 9.81 m/s 2 g = 4.91 m/s 2 y0 = 20 m, ydot0 = 0 30 30 y0 = 30 m, ydot0 = 0 y0 = 10 m, ydot0 = -1 m/s y0 = 20 m, ydot0 = -1 m/s 25 y0 = 30 m, ydot0 = -1 m/s position, y 0 [m] position, y 0 [m] 25 y0 = 10 m, ydot0 = -2 m/s 20 y0 = 20 m, ydot0 = -2 m/s y0 = 30 m, ydot0 = -2 m/s 15 20 15 10 10 5 5 Clearly there is 0 0.0 0.5 0 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 time, t [s] 1.5 2.0 2.5 time, t [s] 35 y0 = 10 m, ydot0 = 0 y0 = 20 m, ydot0 = 0 y0 = 30 m, ydot0 = 0 y0 = 10 m, ydot0 = -1 m/s y0 = 20 m, ydot0 = -1 m/s y0 = 30 m, ydot0 = -1 m/s y0 = 10 m, ydot0 = -2 m/s y0 = 20 m, ydot0 = -2 m/s y0 = 30 m, ydot0 = -2 m/s g = 3.27 m/s 2 30 position, y 0 [m] 25 20 15 10 5 0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time, t [s] C. Wassgren Chapter 06: Dimensional Analysis 290 Last Updated: 16 Aug 2009 3.0 Now let’s present the same information, but in dimensionless form. Starting with Eqn. (6.1), divide all terms by y0 (a length), to make each term dimensionless: y y g 1g t2 0t 1 (6.2) y0 y0 2 y0 gy0 or, in a slightly more compact form, y 1 t 2 y0 t 1 2 (6.3) where y y , y0 y0 y0 gy0 , and t t g y0 (6.4) are the dimensionless position, initial speed, and time, respectively. Note that Eqns. (6.3) and (6.1) are identical; they’re just written in dimensionless or dimensional form. Now if we were to plot Eqn. (6.3) for all of the various combinations of variables, we would have the following plot. 1.0 dimensionless position, y' = y /y 0 0.9 ydot0' = 0 ydot0' = -1 0.8 ydot0' = -2 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.5 1.0 1.5 0.5 dimensionless time, t' = t (g /y 0) This dimensionless plot contains all of the information that was contained in the previous dimensional plots. As you can see, presenting data in dimensionless form is very efficient! Now let’s assume we didn’t know that Eqn. (6.1) existed and we had to perform a series of experiments to try to find the functional relationship between the variables (6.5) y fcn1 g , t , y0 , y0 where fcn1 is the unknown function we’re trying to determine. Let’s say that we perform a series of experiments where we vary each of the variables independently five times. Since we have four independent variables (g, t, y0 , y0), this means we have a total of 54 = 625 experiments to perform! Not only is this a lot of experiments, but some of these experiments are likely to be difficult and expensive to carry out (e.g. varying g is not trivial). Now if we instead performed a dimensional analysis on Eqn. (6.5) (which you will learn how to do later in this set of notes), we could show that Eqn. (6.5) can be written in dimensionless form as: g y y , 0 fcn2 t (6.6) y0 gy y0 0 (Compare Eqn. (6.6) to Eqn. (6.2) to verify.) Equation (6.6) contains only two independent variables, hence, varying each parameter five times gives a total of 52 = 25 total experiments. Clearly, performing a dimensional analysis can reduce the number of experiments one needs to perform! Not only are the number of experiments reduced, but the experiments can be much easier to perform. For example, varying the two independent parameters in Eqn. (6.6) can be achieved by simply letting time vary, and varying the C. Wassgren Chapter 06: Dimensional Analysis 286 Last Updated: 16 Aug 2009 initial drop speed. We needn’t worry about varying gravity, g, independently since g is contained within the term t(g/y0)1/2. Finally, now let’s say that we are interested in launching an object on the Moon (indicated by the subscript “M”) from a specified height, y0,M = 1 m, with a specified speed, y0, M = 1 m/s, and want to know how long it will take for the object to impact the ground, yM(tM = ?) = 0. We know that the acceleration due to gravity on the Moon is gM = 1.62 m/s2. Again, assuming we don’t know that Eqn. (6.1) exists, we can still determine this time by performing a similar experiment on Earth, and then scaling the result. If we’re to perform this similar experiment on Earth, where gE = 9.81 m/s2, we need to first determine the initial drop height, yE,0, and speed, y0, E , for the Earth experiment. Since the same physical process holds for both the Moon and Earth, the dimensionless terms describing the process will be identical, i.e. Eqn. (6.6) will be the same for the Earth and Moon. Thus, we can equate dimensionless terms to determine the values that should be used on the Moon: lim y y yE 0 m y E , yM 0 y0,E = 1 m y0, E y0, M 1 m 0 m yM y0 E y0 M (6.7) y0 y y0, E g E 1 m 9.81 m s 2 y0, E = 2.46 m/s(6.8) 1 m s 0 y0, E y0, M gy gy 1 m 1.62 m s 2 y0, M g M 0 E 0 M When performing the drop test on Earth with the given initial conditions, the time required for the ball to hit the ground is tE = 0.77 s (which can be verified using Eqn. (6.1)). To determine the corresponding time for the Moon, we equate the last dimensionless term in Eqn. (6.6). g g g E y0, M 9.81 m s 2 1 m tM = 1.89 s (6.9) 0.77 s t t tM t E y y g M y0, E 1.62 m s 2 1 m 0 E 0 M This time is exactly what one would calculate from Eqn. (6.1) using y0,M = 1 m, y0, M = 1 m/s, and gM = 1.62 m/s2. Thus, we see that dimensional analysis can be used for scaling! Hopefully, you’re convinced that dimensional analysis is a worthwhile topic to study and apply. The remainder of this chapter presents the mechanics of performing a dimensional analysis along with examples. In addition, similarity and scaling issues are discussed. Motivating Example #2 Most fluids engineering problems are too complex to be amenable to analytic, closed-form solutions. As a result, experiments are used to determine relationships between the variables of interest (e.g. pressure and velocity). Let’s consider the following example. Say we want to measure the pressure difference, p = p2-p1, between two points separated by a distance, L, in a pipe. p2 p1 L C. Wassgren Chapter 06: Dimensional Analysis 287 Last Updated: 16 Aug 2009 On what variables do we expect the average pressure gradient, p/L, to depend? From experience and intuition we might expect the following parameters to be important: V average flow velocity D pipe diameter fluid density fluid dynamic viscosity We can write this relationship in the following, more mathematical form: p/L = fcn(V, D, , ) In order to determine the form of this function, it would be logical to design experiments where we vary just one of the parameters while holding the others constant and observe how p/L varies. For example: p/L D, , constant p/L V, , constant V p/L D V, D, constant p/L V, D, constant This procedure, although logical, can be very time consuming, expensive, and difficult (if not impossible) to perform (e.g. Can you find fluids that have the same viscosity but varying density?) As with the first motivating example, using dimensional analysis will greatly simplify our experimental procedure. This example will be used while presenting the various steps of performing a dimensional analysis in the following sections. C. Wassgren Chapter 06: Dimensional Analysis 288 Last Updated: 16 Aug 2009 2. Buckingham Pi Theorem The key component to dimensional analysis is the: If an equation involving k variables is dimensionally homogeneous, it Buckingham Pi Theorem: can be reduced to a relationship among k-r independent dimensionless products (referred to as terms), where r is the minimum number of reference dimensions required to describe the variables, i.e. of variables # of terms # # of reference dimensions k r (The proof to this theorem will not be presented here.) Notes: 1. Dimensionally homogeneous means that each term in the equation has the same units. For example, the following form of Bernoulli’s equation: p V2 z constant g 2g is dimensionally homogeneous since each term has units of length (L). 2. A dimensionless product, also commonly referred to as a Pi () term, is a term that has no dimensions. For example, p V 2 is a dimensionless product since both the numerator and denominator have the same dimensions. 3. Reference dimensions are usually basic dimensions such as mass (M), length (L), and time (T) or force (F), length (L), and time (T). We’ll discuss the “usually” modifier a little later when discussing the method of repeating variables. C. Wassgren Chapter 06: Dimensional Analysis 289 Last Updated: 16 Aug 2009 3. Method of Repeating Variables The Buckingham Pi Theorem merely states that a relationship among dimensional variables may be written, perhaps in a more compact form, in terms of dimensionless variables (terms). The Pi Theorem does not, however, tell us what these dimensionless variables are. The method of repeating variables is an algorithm that can be used to determine these dimensionless variables. The method of repeating variables algorithm is as follows: 1. List all variables involved in the problem. a. This is the most difficult step since it requires experience and insight. b. Variables are things like pressure, velocity, gravitational acceleration, viscosity, etc. c. List only independent variables. For example, you can list: (density) and g (gravitational acceleration), or and (specific weight), or g and but you should not list, g, and since one of the variables is dependent on the others. d. If you include variables that are unimportant to the system, then you’ll form terms that won’t have an impact in practice. This situation is the same one you’d have if dimensional variables were used. e. If you leave out an important variable, then you’ll find that in practice your relationship between dimensionless terms can’t fully describe the system behavior. Again, this situation is the same one you’d have if you used dimensional terms. 2. Express each variable in terms of basic dimensions. a. For fluid mechanics problems we typically will use mass (M), length (L), and time (T) or force (F), length (L), and time (T) as basic dimensions. We may occasionally need other basic dimensions such as temperature (). b. For example, the dimensions of density can be written as: M FT 2 (Note: The square brackets are used to indicate “dimensions of.”) 3 4 L L 3. Determine the number of terms using the Buckingham-Pi Theorem. a. (# of terms) = (# of variables) – (# of reference dimensions) b. Usually the # of reference dimensions will be the same as the # of basic dimensions found in step 2. There are (rare) cases where some of the basic dimensions always appear in particular combinations so that the # of reference dimensions is less than the # of basic dimensions. For example, say that the variables in the problem are A, B, and C, and their corresponding basic dimensions, are: M M MT A 3 B 3 2 C 3 L LT L The basic dimensions are M, L, and T (there are 3 basic dimensions). Notice, however, that the dimensions M and L always appear in the combination M/L3. Thus, we really only need two reference dimensions, M/L3 and T, to describe all of the variables’ dimensions. 4. Select repeating variables where the number of repeating variables is equal to the number of reference dimensions. a. The repeating variables should come from the list of variables. In our previous example, the list of variables is p/L, V, D, , and . b. Each repeating variable must have units independent of the other repeating variables. c. Don’t make the dependent variable one of the repeating variables. In our previous example, p/L is the dependent variable. d. All of the reference dimensions must be included in the group of repeating variables. C. Wassgren Chapter 06: Dimensional Analysis 290 Last Updated: 16 Aug 2009 5. Form a term by multiplying one of the non-repeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. a. This step is most clearly illustrated in an example and will not be discussed here. b. Repeat this step for all non-repeating variables. 6. Check that all terms are dimensionless. a. This is an important, and often overlooked, step to verify that your terms are, in fact, dimensionless. 7. Express the final form of the dimensional analysis as a relationship among the terms. a. For example, 1 = fcn(2, 3, …, k-r). C. Wassgren Chapter 06: Dimensional Analysis 291 Last Updated: 16 Aug 2009 Example: Let’s use our pipe flow experiment as an example. Step 1: The variables that are important in this problem are: p/L average pressure gradient over length L V average flow velocity D pipe diameter fluid density fluid dynamic viscosity p/L = fcn1(V, D, , ) Step 2: The basic dimensions of each variable are: p F M L L3 L2T 2 L V T D L M L3 FT M 2 LT L Step 3: (# of variables) = 5 (p/L, V, D, , ) (# of reference dimensions) = 3 (F, L, T or M, L, T) (# of terms) = (# of variables) – (# of reference dimensions) = 2 Thus, instead of having a relation involving 5 terms, we actually have a relationship involving just 2 terms! Step 4: Select the following 3 repeating variables (We require three since the # of reference dimensions is three.): , V, D Notes: 1. These three repeating variables have independent dimensions. 2. The dependent variable (p/L) is not one of the repeating variables. 3. We could have also selected (,, V) or (,, D) or (D,, V) as repeating variables. The choice of repeating variables is somewhat arbitrary (as long as they have independent reference dimensions and do not include the dependent variable). C. Wassgren Chapter 06: Dimensional Analysis 292 Last Updated: 16 Aug 2009 Step 5: Form terms from the remaining, non-repeating variables: 1 p aV b D c L a b 0 M M L L MLT 2 2 3 L T L T 1 0 1 M M Ma 0 1 a 2 3 a c a 1 L L L L L 0 2 3a b c b 2 T 0 T 2 T b 0 2 b c 1 0 bc p L D 1 V 2 2 aV b D c a b c 0 M M L L MLT 3 LT L T 1 M 0 M 1M a 0 1 a a 1 0 1 3 a b c L L L L L 0 1 3a b c b 1 T 0 T 1T b 2 0 1 b c 1 VD Step 6: Verify that each term is, in fact, dimensionless. p D L 1 2 V M 2 VD LT M L L3 T 2 M 0 L0T 0 22 2 LT 1 M L OK! L3 T 1 M 0 L0T 0 M LL OK! Step 7: Re-write the original relationship in dimensionless terms. p D L fcn2 2 V VD C. Wassgren Chapter 06: Dimensional Analysis 293 Last Updated: 16 Aug 2009 Notes: 1. Instead of having to run four different sets of experiments as was discussed at the beginning of this chapter, we only really need to run one set of experiments where we vary: VD p L D and measure: p D L V 2 V 2 Note: We can use 1/ in place of since we haven’t actually specified what the function, fcn2, looks like. This is a Reynolds number! VD All of the information contained in the previous four plots is contained within this single plot! This reduces the complexity, cost, and time required to determine the relationship between the average pressure gradient and the other variables. 2. Dimensional analysis is a very powerful tool because it tells us what terms really are important in an equation. For example, we started with the relation: p fcn V , D, , 1 L leading us to believe that V, D, , and are all important terms by themselves. However, dimensional analysis shows us that instead of the terms by themselves, it is the following grouping of terms: p D L fcn2 V 2 VD that is important in the relationship. This is a subtle but very important point. 3. Dimensional analysis tells us how many dimensionless terms are important in a relation. It does not tell us what the functional relationship is. 4. The dimensionless terms found via dimensionless analysis are not necessarily unique. Had we chosen different repeating variables in the previous example, we would have ended up with different terms. One can multiply, divide, or raise their set of terms to form the terms found by another. The number of terms, however, is unique. 5. After a bit of practice, one can quickly form terms by inspection rather than having to go through the method of repeating variables. C. Wassgren Chapter 06: Dimensional Analysis 294 Last Updated: 16 Aug 2009 Example: An open cylindrical tank having a diameter D is supported around its bottom circumference and is filled to a depth h with a liquid having a specific weight . The vertical deflection, , of the center of the bottom is a function of D, h, d, , and E where d is the thickness of the bottom and E is the modulus of elasticity of the bottom material. Form the dimensionless groups describing this relationship. SOLUTION: 1. Write the dimensional functional relationship. f1 D, h, d , , E 2. Determine the basic dimensions of each parameter. L h L D L d L M F 3 22 LT L M F E 2 2 LT L 3. Determine the number of terms required to describe the functional relationship. # of variables = 6 (, D, h, d, , E) # of reference dimensions = 2 (L, F/L2 or L, M/T2) (Note that the number of reference dimensions and the number of basic dimensions are not the same for this problem!) (# terms) = (# of variables) – (# of reference dimensions) = 6 – 2 = 4 4. Choose two repeating variables by which all other variables will be normalized (same # as the # of reference dimensions). D, (Note that the dimensions for D and are independent.) 5. Make the remaining non-repeating variables dimensionless using the repeating variables. 1 D a b 1 L1 F L F 0 L0 L a b 3 F: 0 b L: 0 1 a 3b a 1 1 D 2 hD a b 1 L1 F L F 0 L0 L a b 3 F: 0 b L: 0 1 a 3b a 1 2 h D C. Wassgren Chapter 06: Dimensional Analysis 295 Last Updated: 16 Aug 2009 3 dD a b 1 L1 F L a F 0 L0 L b 3 F: 0 b L: 0 1 a 3b a 1 3 d D 4 ED a b L L1 F L F 0 L0 F a 2 b 3 F: 0 1 b b 1 L: 0 2 a 3b a 1 4 E D 6. Verify that each term is, in fact, dimensionless. 1 D L L 1 OK! 2 h D L L 1 OK! 3 d D L L 1 OK! 4 E D F L2 1 L L 3 7. F 1 OK! Re-write the original relationship in dimensionless terms. h d E f2 , , D D D D C. Wassgren Chapter 06: Dimensional Analysis 296 Last Updated: 16 Aug 2009 Example: A viscous fluid is poured onto a horizontal plate as shown in the figure. Assume that the time, t, required for the fluid to flow a certain distance, d, along the plate is a function of the volume of fluid poured, V, acceleration due to gravity, g, fluid density, , and fluid dynamic viscosity, . Determine an appropriate set of dimensionless terms to describe this process. volume, V d SOLUTION: 1. Write the dimensional functional relationship. t f1 d ,V , g , , 2. Determine the basic dimensions of each parameter. t T d L V L3 g LT 2 M L3 M LT 3. Determine the number of terms required to describe the functional relationship. # of variables = 6 (t, d, V, g, , ) # of reference dimensions = 3 (T, L, M) (Note that the number of reference dimensions and the number of basic dimensions are equal for this problem.) (# terms) = (# of variables) – (# of reference dimensions) = 6 – 3 = 3 4. Choose three repeating variables by which all other variables will be normalized (same # as the # of reference dimensions). d, g, (Note that the dimensions for these variables are independent.) C. Wassgren Chapter 06: Dimensional Analysis 297 Last Updated: 16 Aug 2009 5. Make the remaining non-repeating variables dimensionless using the repeating variables. 1 td a g b c T L M 0 L0T 0 1 1 M: 0c L : 0 a b 3c 0 1 2b T: 1 t a b L M 2 3 T L a1 2 b c 1 2 c0 g d 2 Vd a g b c L3 L L M M 0 L0T 0 2 3 1 1 T L M: 0c a 3 L : 0 3 a b 3c b 0 0 2b T: c0 a c a 2 b b c V d3 3 d a g b c M L L M M 0 L0T 0 2 3 LT 1 T L a3 M: 0 1 c 2 L : 0 1 a b 3c b 1 2 0 1 2b T: c 1 3 6. d gd Verify that each term is, in fact, dimensionless. g T L1 2 1 1 t 1 OK! 1 2 d 1 T L 2 V L3 1 1 OK! 3 3 d 1 L M L3 1 T 1 1 OK! 1 1 d gd LT M L L 2 L 2 3 7. Re-write the original relationship in dimensionless terms. V g t f2 3 , d d gd d C. Wassgren Chapter 06: Dimensional Analysis 298 Last Updated: 16 Aug 2009 4. Dimensionless Form of the Governing Equations Consider the dimensional form of the governing equations for an incompressible fluid with constant viscosity in a gravity field: u j Continuity (COM) x j Navier-Stokes Eqns (LME): Thermal Energy Eqn: 0 ui u 2 ui p uj i gi t x j xi x j x j T u j ui T 2T c uj k t xi x j x j x j x j ui x j Note that in the thermal energy equation the internal energy has been written as the specific heat (assumed constant, note that for an incompressible flow cv = cp = c) multiplied by the temperature and the heat transfer has been assumed to be due solely to conduction (Fourier’s Law with a constant conduction coefficient). Let’s re-write these equations in dimensionless form using some characteristic flow quantities (to be discussed in a moment). The variables in the equations will be normalized using the following quantities: x xi* i xi Lxi* L u * ui i ui Uui* U t t* t t* p* p p0 p p0 p* T T T0T * T0 where the superscript “*” refers to a dimensionless quantity. The quantity L represents a characteristic length for the flow of interest (e.g. a pipe diameter or the diameter of a sphere), U is a characteristic velocity (e.g. the free stream velocity or the average velocity in a pipe), is a characteristic time scale (e.g. the period of an oscillating boundary), p0 is a characteristic pressure (e.g. the free stream pressure or the vapor pressure), and T0 is a characteristic temperature (e.g. the free stream temperature). These characteristic quantities give us an estimate of the typical magnitude of the various terms in the equations. They won’t be exact values but they will give us a feel for how a parameter might scale in a flow, e.g. we might expect the fluid velocities in a flow to scale with the incoming free stream velocity. T* Now let’s rewrite the governing equations using these dimensionless parameters. Continuity: u * j x* j 0 Navier-Stokes Eqns: p p* U 2 ui* U ui* U 2 * ui* uj * 0 * 2 gi t * L L xi x j L x* x* j j Thermal Energy Eqn: cT0 T * U cT0 * T * kT0 2T * U 2 2 uj * 2 t * L x j L x* x* L j j u* u* u* j i i xi* x* x* j j The continuity equation simplifies to an identical equation but in dimensionless form. The Navier-Stokes and thermal energy equations are still in dimensional form due to the dimensional terms in front of each dimensionless term. Note that each dimensional term represents a characteristic force/energy magnitude. C. Wassgren Chapter 06: Dimensional Analysis 299 Last Updated: 16 Aug 2009 For example, the term in front of the convective acceleration term in the momentum equations, i.e. the (U2/L) in front of u j ui x j , represents the magnitude of a typical (convective) inertial “force” (recall that each term in the N-S equations has dimensions of force per unit volume). Similarly, the (U/L2) term represents a characteristic viscous force. In order to make the N-S and thermal energy equations dimensionless, it is customary to divide through by the characteristic (convective) inertial force term in the N-S equations and by the convective term in the thermal energy equation: u* j continuity: x* j 0 * p p* ui* ui* g L L ui u* * 0 2 * i j U t * UL x* x* U 2 x j U xi j j Navier-Stokes: * * T * k 2T * U 2 u j ui* ui* L T u* * j U * x j c UL x* x* cT0 UL xi* x* x* t j j j j Now each term in the continuity, N-S, and thermal energy equations is dimensionless. The dimensionless quantities in front of each term have special meaning: Thermal Energy: Strouhal #, St L represents the ratio of (local or Eulerian) inertial forces to (convective) inertial forces. U The Strouhal number is often significant in unsteady, periodic flows. Euler #, Eu p0 represents the ratio of pressure forces to (convective) inertial forces. U 2 The Euler number is typically significant in flows where large changes in pressure occur. The Euler number is also often written as a pressure coefficient, cP: cP p p0 1 U 2 2 or in flows where cavitation occurs, as the cavitation number, Ca: Ca p pv 1 2 U 2 where pv is the vapor pressure of the fluid. Reynolds #, Re UL represents the ratio of (convective) inertial forces to viscous forces. The Reynolds number is significant in virtually all fluid flows. Froude #, Fr U represents the ratio of (convective) inertial forces to gravitational forces gL The Froude (pronounced “’früd”) number is typically significant in flows involving a free surface. Prandtl #, Pr c k represents the ratio of the momentum diffusivity, , to the thermal diffusivity, k/(c). The Prandtl number gives a measure of how rapidly momentum diffuses through a fluid compared to the diffusion of heat. Most gases have a Prandtl number near one (heat and momentum diffuse at nearly the same rate) while water has a Prandtl number near ten (momentum diffuses faster than heat). C. Wassgren Chapter 06: Dimensional Analysis 300 Last Updated: 16 Aug 2009 Thus, the dimensionless forms of the governing equations are: continuity: Navier-Stokes Eqns: Thermal Energy Eqns: u* j x* j ui* St St 0 t * T * t u* j ui* u* j * x* j Eu p* xi* 1 ui* 1g 2i ** Re x j x j Fr gi * 2T * U 2 1 u j ui* ui* 1 x* Pr Re x* x* c pT0 Re xi* x* x* j j j j j T * Additional dimensionless quantities occur when dealing with other equations of significance (e.g. the equations for a compressible fluid) and with the boundary conditions (e.g. surface tension effects or surface roughness). C. Wassgren Chapter 06: Dimensional Analysis 301 Last Updated: 16 Aug 2009 5. Modeling and Similarity Models are often used in fluid mechanics to predict the kinematics and dynamics of full-scale (often referred to as prototype) flows. From previous discussions of dimensional analysis, we observe that we can write the governing equations and boundary conditions of our flow in dimensionless terms ( terms). Thus, if we have two different flows (e.g., a large-scale, prototype flow and a small scale, model flow) that have identical dimensionless parameters, then the same solution, also in terms of dimensionless parameters, will hold for both. This is extremely helpful when modeling fluid systems. When a model and the prototype have the same dimensionless parameters, we say that they are similar. We typically discuss similarity in three categories: geometric, dynamic, and kinematic. Geometric similarity occurs when the model is an exact geometric replica of the prototype. In other words, all of the lengths in the model are scaled by exactly the same amount as in the prototype. For example: prototype WP model WM LM LP LP/LM = WP/WM Note that surface roughness may even need to be scaled if it is a significant factor in the flow. Dynamic similarity occurs when the ratio of forces in the model is the same as the ratio of forces in the prototype, i.e. [ReP = ReM] (ratio of inertial to viscous forces)P = (ratio of inertial to viscous forces)M [EuP = EuM] (ratio of pressure to inertial forces)P = (ratio of pressure to inertial forces)M [FrP = FrM] (ratio of inertial to grav. forces)P = (ratio of inertial to grav. forces)M etc. Kinematic similarity occurs when the prototype and model fluid velocity fields have identical streamlines (but scaled velocities). Since the forces affect the fluid motion, geometric similarity and dynamic similarity will automatically ensure kinematic similarity. Note: 1. When modeling, we need to maintain similarity between all of the dimensionless parameters that are important to the physics of the flow. This means that we do not necessarily need to have similarity between all terms, just the ones that significantly affect the flow physics. Knowing a priori what dimensionless terms are important can be difficult but with experience the task is often easier. 2. It is not uncommon to have the important physics of a system change at different scales. For example, surface tension forces become more pronounced at smaller geometric scales. If one was scaling up a small system in which surface tension was an important effect, but didn’t consider the dynamic similarity of the surface tension force at the larger scale, then the scaling experiments would result in incorrect results. C. Wassgren Chapter 06: Dimensional Analysis 302 Last Updated: 16 Aug 2009 Example: The drag characteristics of a blimp 5 m in diameter and 60 m long are to be studied in a wind tunnel. If the speed of the blimp through still air is 10 m/s, and if a 1/10 scale model is to be tested, what airspeed in the wind tunnel is needed for dynamic similarity? Assume the same air temperature and pressure for both the prototype and model. SOLUTION: For dynamic similarity, equate the model and prototype Reynolds numbers. Re P Re M VD VD P M Since both the model and prototype use air at the same temperature and pressure as the working fluid, P = M. D 10 VM VP P 10 m s 1 DM VM 100 m s Note that the model speed is still low enough that Mach number effects (i.e. compressibility effects) do not come into play. C. Wassgren Chapter 06: Dimensional Analysis 303 Last Updated: 16 Aug 2009 Example: The height of the free surface, h, in a tank of diameter, D, that is draining fluid through a small hole at the bottom with diameter, d, decreases with time, t. This change in free surface height is studied experimentally with a half-scale model. For the prototype tank: H = 16 in. (the initial height of the free surface) D = 4.0 in. d = 0.25 in. Experimental data is obtained from the prototype and half-scale model and is given below: Model Data h [in.] 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 1. 2. 3. Prototype Data h [in.] 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 t [s] 0.0 3.1 6.2 9.9 13.5 18.1 24.0 32.5 43.0 t [s] 0.0 4.5 8.9 14.0 20.2 25.9 32.8 45.7 59.8 Plot, on the same graph, the height data as a function of time for both the model and the prototype. Develop a set of dimensionless parameters for this problem assuming that: h = f(H, D, d, g, t) Replot, on the same graph, the height data as a function of time in non-dimensional form for both the model and prototype. D g H h hole with diameter, d C. Wassgren Chapter 06: Dimensional Analysis 304 Last Updated: 16 Aug 2009 SOLUTION: h [in.] First plot the model and prototype dimensional data. 18 16 14 12 10 8 6 4 2 0 model prototype 0 20 40 60 80 t [s] Now perform a dimensional analysis to determine the dimensionless terms describing the relationship. 1. Write the dimensional functional relationship. h f1 H , D, d , g , t 2. Determine the basic dimensions of each parameter. h L H L D L d L g LT 2 t T 3. Determine the number of terms required to describe the functional relationship. # of variables = 6 (h, H, D, d, g, t) # of reference dimensions = 2 (L, T) (Note that the number of reference dimensions and the number of basic dimensions are equal for this problem.) (# terms) = (# of variables) – (# of reference dimensions) = 6 – 2 = 4 4. Choose two repeating variables by which all other variables will be normalized (same # as the # of reference dimensions). H, g (Note that the dimensions for these variables are independent.) C. Wassgren Chapter 06: Dimensional Analysis 305 Last Updated: 16 Aug 2009 5. Make the remaining non-repeating variables dimensionless using the repeating variables. h 1 (Found via inspection.) H D 2 (Found via inspection.) H d 3 (Found via inspection.) H 4 tH a g b a T L L L0T 0 2 1 1 T a1 L: 0 ab 2 T : 0 1 2b b1 2 4 t 6. b g H Verify that each term is, in fact, dimensionless. h L1 1 H 1 L 1 OK! D L1 2 H 1 L 1 OK! d L 1 1 OK! 3 H 1 L g T L12 1 1 OK! 4 t 1 2 H 1 T L 7. Re-write the original relationship in dimensionless terms. D d h g f2 , , t H H H H Now plot the model and prototype data in dimensionless form. Note that since there is geometric similarity (the model is one-half the size of the prototype): d d D D and H M H P H M H P C. Wassgren Chapter 06: Dimensional Analysis 306 Last Updated: 16 Aug 2009 1.2 model prototype 1.0 h/H 0.8 0.6 0.4 0.2 0.0 0 100 200 300 400 t*sqrt(g/H) Notice that the data collapse to a single curve when plotted in dimensionless terms. C. Wassgren Chapter 06: Dimensional Analysis 307 Last Updated: 16 Aug 2009 Partial Similarity True similarity may be difficult to achieve in practice. In such cases, one must either: (a) acknowledge that model testing may not be possible, or (b) relax one or more similarity requirements and use a combination of experimentation and analysis to scale the measurements. For example, in modeling the flow around ships both Reynolds number and Froude number similarity are important; however, both are difficult to achieve simultaneously. In such cases, one of the similarity requirements is relaxed (in boat modeling it’s the Reynolds number similarity requirement) and a combination of experiments and analysis is utilized to scale the measurements. The two primary components of drag on a ship’s hull are viscous drag (i.e., the friction of the water against the hull’s surface) and wave drag (i.e., the force required to create the waves generated by the hull). The two significant dimensionless parameters corresponding to these phenomena are: VL Re (ratio of inertial to viscous forces) Reynolds number: Froude number: Fr V (ratio of inertial to gravitational forces) gL Maintaining both Reynolds number and Froude number similarity is difficult to achieve in practice. V V L g M gM g P L FrM FrP (6.10) VM VP M VM VP M gL gL LP g P LP M P using previous 3 result L 2 VL VL VL Re M Re P M P M M M P M (6.11) VP LP M P LP As an example, consider a scale model which has LP = 100LM, P = H2O = 1 cSt M = 0.001 cSt. There is no such common model fluid available! Thus, we cannot easily maintain both Froude number and Reynolds number similarity. How do we resolve this difficulty? In practice, Froude number similarity is maintained with water as both the prototype and model fluid (i.e., Eqn. (6.10) holds). Reynolds number similarity is neglected in the experiment and instead analysis or computation is used to estimate the viscous drag contribution. The procedure is as follows: 1. The total drag acting on the model is measured in the experiment. This drag force is usually expressed in terms of a dimensionless resistance coefficient. 2. The viscous drag contribution to the total drag is calculated using analysis (e.g., boundary layer analysis) or computation (e.g., computational fluid dynamics). 3. The difference between the total drag and the viscous drag is the wave drag. 4. The wave drag is a function of the Froude number, which is held constant during scaling. So, the experimental wave drag data can be scaled between the model and prototype. 5. Estimate the viscous drag contribution for the prototype using analysis or computation. 6. Sum the predicted viscous drag force (step 5) with the scaled wave drag force (step 4) to get the total prototype drag force. C. Wassgren Chapter 06: Dimensional Analysis 308 Last Updated: 16 Aug 2009 As a demonstration of how well the procedure works, consider the resistance coefficient data from a 1:80 scale model test of the U.S. Navy guided missile frigate Oliver Hazard Perry (FFG-7) as shown in the plots below. (Note that these plots are from Figs. 7.2 and 7.3 in Fox, R.W., Pritchard, P.J., and McDonald, A.T., 2008, Introduction to Fluid Mechanics, 7th ed., Wiley.) The error between the scaled and actual total drag force measurements is approximately 5%. from scale model testing predicted from boundary layer theory predicted from boundary layer theory Scale Model Prototype Experimental observations have shown that in many (but not all!) cases, Reynolds number similarity may be neglected for sufficiently large Reynolds numbers. (Figure from Fox, R.W., Pritchard, P.J., and McDonald, A.T., 2008, Introduction to Fluid Mechanics, 7th ed., Wiley.) Moody diagram – Friction factor as a function of Reynolds number and pipe relative roughness. Note that the friction factor becomes independent of Reynolds number for sufficiently large Reynolds numbers. C. Wassgren Chapter 06: Dimensional Analysis 309 Last Updated: 16 Aug 2009 (Figure from Fox, R.W., Pritchard, P.J., and McDonald, A.T., 2008, Introduction to Fluid Mechanics, 7th ed., Wiley.) Drag coefficients for a sphere and circular disk as a function of Reynolds number. Note that the drag coefficient is insensitive to Reynolds number over a wide range of Reynolds numbers. C. Wassgren Chapter 06: Dimensional Analysis 310 Last Updated: 16 Aug 2009 The Stokes Number (St) for Small Particles in a Flow The Stokes number, St, is defined as the ratio of the particle response time, p, to the fluid response time, f: p St (6.12) f A response time is a measure of how rapidly a quantity responds to rapid changes. The Stokes number for a particle is essentially a measure of how well the particle follows fluid streamlines. If St<<1 then the particle will be able to follow the fluid streamlines whereas if St>>1 then the particle will not be able to follow sudden changes in the fluid velocity. For example, consider driving down a country road late at night during the summer when a lot of bugs are out. If the Stokes number for a bug is small, then it will follow the fluid streamlines as you drive past it and it won’t impact your car. However, if the Stokes number for the bug is large, it will end up hitting your windshield since it won’t be able to follow the fluid streamlines that contour around your car. Splat! St >> 1 St << 1 The particle response time can be found by considering the particle equation of motion (assuming spherical particles) for a particle with a speed slower than the surrounding fluid (so the particle accelerates): 2 2 dp 3 du p (6.13) p dp CD 1 2 f u f u p 4 6 dt where p an f are the particle and fluid densities, dp is the particle diameter, up and uf are the particle and fluid velocities, t is time, and CD is the particle drag coefficient. Define the Reynolds number for the particle using the local relative velocity: f u f u p d p Re d (Note that uf > up is assumed.) (6.14) f Substitute Eqn. (6.14) into Eqn. (6.13) and simplify: du p f 2 3 f u f u p CD Re d 4 d dt f u f u p d p pp 3 f CD Re d u up 2 f 4 p d p For small Reynolds numbers the drag coefficient approaches the Stokes drag: 24 CD (We’re now assuming that we’re dealing with small particles.) Re d so that the particle equation of motion becomes: du p 18 f u up 2 f dt pdp The solution to this equation, assuming a constant fluid velocity and a particle released from rest (up(t = 0) = 0), is: u p u f 1 exp t / p where p is the particle response time given by: d2 p p p 18 f C. Wassgren Chapter 06: Dimensional Analysis 311 Last Updated: 16 Aug 2009 Let the fluid response time, f, for the flow geometry be: L f uf where L is a typical flow dimension (e.g. the effective frontal diameter of the car in the car/bug example discussed previously). Hence, the Stokes number for the particle is: 2 p pd p u f St f 18 f L Re-writing in terms of the Reynolds number, ReL, based on the typical flow dimension, L: f uf L Re L f gives: St 2 p pd p u f f Re L f 18 f L f uf L Re St L 18 p f dp L 2 For very small particles compared to the flow dimension, i.e. (dp/L << 1), and moderate flow Reynolds numbers and density ratios, we observe that St<<1 and the particle should follow the fluid streamlines. C. Wassgren Chapter 06: Dimensional Analysis 312 Last Updated: 16 Aug 2009 Review Questions 1. Describe some of the benefits to performing a dimensional analysis of a problem. 2. What does the Buckingham-Pi theorem state? Are the dimensionless terms resulting from the theorem unique? 3. Describe the method of repeating variables. Must this method always be followed to determine dimensionless terms? 4. What is the difference between “basic dimensions” and “reference dimensions”? 5. Describe the three types of similarity. 6. Must there be exact similarity between a model and prototype in order to perform engineering modeling? 7. In words, define the Reynolds, Froude, Strouhal, and Euler numbers. C. Wassgren Chapter 06: Dimensional Analysis 313 Last Updated: 16 Aug 2009 Chapter 07: Navier-Stokes Solutions 1. A Few Comments Regarding Exact Solutions to the Navier-Stokes Equations 2. Planar Couette-Poiseuille Flow 3. Poiseuille Flow 4. Starting Flow in a Circular Pipe 5. Impulsively Started Flat Plate (aka Stokes 1st Problem) 6. Oscillating Flat Plate (aka Stokes 2nd Problem) 7. Planar Stagnation Point Flow (aka Heimenz Flow) 8. Low Reynolds Number Flows 9. Stokes Flow Around a Sphere 10. Lubrication Flow C. Wassgren Chapter 07: Navier-Stokes Solutions 314 Last Updated: 16 Aug 2009 1. A Few Comments Regarding Exact Solutions to the Navier-Stokes Equations Because there is no general method for solving a system of non-linear, partial differential equations, there are only a few exact solutions to the governing equations of fluid mechanics. For an incompressible fluid with constant viscosity, the equations governing the fluid motion are the continuity and Navier-Stokes equations: u 0 u u u p 2 u f t In general, we must make a number of assumptions to simplify the governing equations so that they become manageable analytically. In particular, we often simplify the equations so that the non-linear convective term in the Navier-Stokes equations, (u)u, is zero. Although we need to make many assumptions in determining exact solutions, the resulting solutions are still of great engineering value. They are often good models for real-world flows and they are commonly used to validate numerical codes and experimental methods. One assumption we’ll make in all of the solutions is that the flow is laminar as opposed to being turbulent or transitional. A laminar flow means that the fluid moves in smooth layers (or lamina). A turbulent flow is one in which the fluid flows in an almost chaotic manner with a number of vortices of different size and nearly random spatial and temporal variations in the fluid velocity. A transitional flow is one between the laminar and turbulent states where the flow is mostly laminar but with occasional turbulent fluctuations. Boundary Conditions (BCs) When solving the governing equations of fluid dynamics, we’ll need to apply boundary conditions (BCs) for specific flow geometries. Two common types of BCs include kinematic and dynamic boundary conditions. Kinematic boundary conditions specify the fluid velocity. One example is the no-slip boundary condition which states that at either a solid boundary or fluid interface, the fluid velocity must be continuous: u fluid U boundary Another common kinematic boundary condition is that fluid velocities must remain finite. Dynamic boundary conditions specify that stresses must be continuous across solid or fluid interfaces: nn ,fluid nn ,boundary ns ,fluid ns ,boundary where the subscripts “nn” and “ns” refer to the normal and shear stresses at the boundary with the normal ˆ vector n . Now let’s investigate some exact solutions. Note that there are many graduate level texts that review more exact solutions than will be presented in these notes. Several good references include: White, F.M., Viscous Fluid Flow, McGraw-Hill. Panton, R.L., Incompressible Flow, Wiley. Currie, I.G., Fundamental Mechanics of Fluids, McGraw-Hill. C. Wassgren Chapter 07: Navier-Stokes Solutions 315 Last Updated: 16 Aug 2009 2. Planar Couette-Poiseuille Flow Consider the steady flow of an incompressible, constant viscosity Newtonian fluid between two infinitely long, parallel plates separated by a distance, h. g h y fluid x We’ll make the following assumptions: u z constant and 1. The flow is planar. 0 z 0 t u y u x 0 x x f x 0 and f y g 2. The flow is steady. 3. The flow is fully-developed in the x-direction. 4. The only body force is that due to gravity in the –y-direction. Let’s first examine the continuity equation: u x u y 0 x y From assumption #3 we see that: u y u x 0 0 x y Based on this result and assumptions #1 and #3 we see that the y-velocity is a constant: u y constant Since there is no flow through the walls, the y-velocity must be zero. (call this condition #5) uy 0 Now let’s examine the Navier-Stokes equation in the y-direction: 2u u u 2u u p y u x y u y y 2y 2y f y x y y y t x We can simplify this equation using our assumptions: 2u y u y u y u y 2u y p fy ux uy t x y y x 2 y 2 g (#4) 0 (#2,#5) 0 (#3,#5) 0 (#5) 0 (#3,#5) 0 (#5) p g y p x, y f x gy (7.1) where f(x) is an unknown function of x. C. Wassgren Chapter 07: Navier-Stokes Solutions 316 Last Updated: 16 Aug 2009 Now let’s examine the Navier-Stokes equation in the x-direction: 2u 2u u u u p x u x x u y x 2x 2x f x x y x y t x After simplifying: 2 u x u x p u x ux 2ux ux uy fx x x 2 y 2 t x 0 y 0 (#4) (#5) 0 (#3) 0 (#2) 0 (#3) 2 u x 1 p x y 2 Based on assumptions #1, #2, and #3 we can write: 2ux d 2ux y 2 dy 2 so that the simplified Navier-Stokes equation in the x-direction becomes: d 2 u x 1 p dy 2 x Integrating twice with respect to y (note that from Eqn. (7.1) we observe that p/x is not a function of y): 1 p 2 ux y c1 y c2 2 x where c1 and c2 are unknown constants that we find using our boundary conditions. Note that this is the equation of a parabola. Let’s examine the following case: fixed bottom boundary: ux y 0 0 top boundary moving with velocity, U: u x y h U After applying these boundary conditions to determine the constants c1 and c2 we find that the fluid velocity in the x-direction is given by: y h 2 p y y ux U 1 h 2 x h h This type of flow is often referred to as a planar Couette-Poiseuille flow (pronounced “’pwäz I”) Notes: 1. The stress acting on the fluid at any point can be found from the stress-strain rate constitutive relations for a Newtonian fluid. u u ij p ij i j x j xi 2. If we remove the pressure gradient and move the fluid using just the moving upper boundary, the velocity profile becomes linear: y ux U h This type of flow is referred to as a planar Couette flow. C. Wassgren Chapter 07: Navier-Stokes Solutions 317 Last Updated: 16 Aug 2009 3. If we hold both boundaries stationary and move the fluid using only a pressure gradient (note that flow in the positive x-direction occurs for dp/dx < 0), the velocity profile becomes: h 2 p y y ux 1 2 x h h This type of flow is referred to as a planar Poiseuille flow. 4. The average flow velocity may be found by setting the volumetric flow rate using the average velocity equal to the volumetric flow rate using the real velocity profile. For example, for planar Poiseuille flow the average velocity is: y h h 2 p y y h3 p Q uh 1 dy 2 x h h 12 x y 0 u 5. h 2 p 2 umax 12 x 3 (7.2) Recall that we assumed that these solutions only hold for laminar flows (the uy component is zero). Experimentally we observe that planar Couette-Poiseuille flow remains laminar for: uh Re 1500 where Re is the Reynolds number of the flow and u is the average flow velocity. It should be noted that the value of 1500 is only approximate and can vary considerably depending on how carefully the experiment is conducted. Its value is given only as an engineering rule-of-thumb. 5. Velocity profiles for the various conditions are sketched below: U dp/dx > 0 y x C. Wassgren Chapter 07: Navier-Stokes Solutions Couette flow (dp/dx = 0) dp/dx < 0 318 Last Updated: 16 Aug 2009 3. Poiseuille Flow Consider the steady flow of an incompressible, constant viscosity, Newtonian fluid within an infinitely long, circular pipe of radius, R. r R z We’ll make the following assumptions: 1. The flow is axi-symmetric and there is no “swirl” velocity. 2. The flow is steady. 3. The flow is fully-developed in the z-direction. 4. There are no body forces. 0 and u 0 0 t ur u z 0 z z f r f f z 0 Let’s first examine the continuity equation: 1 rur 1 u u z 0 r r r z From assumptions #1 and #3 we see that: rur constant Since there is no flow through the walls, the constant must be equal to zero and thus: ur 0 (call this condition #5) Now let’s examine the Navier-Stokes equation in the z-direction: 1 u z 1 2 u z 2 u z u u u u p u z ur z z u z z 2 fz r 2 2 r r z z z t r r r r We can simplify this equation using our assumptions: 2 2 u z u z u u z u z p 1 u z 1 u z u z r ur uz 2 fz r r r r r 2 t z z 2 z 0 (#5) r 0 (#4) 0 (#3) 0 (#1) 0 (#3) 0 (#1) 0 (#2) d du z r dp r dr dr dz du r 2 dp r z c1 dr 2 dz r 2 dp c1 ln r c2 4 dz Note that in the previous derivation the fact that uz is a function only of r has been used to change the partial derivatives to ordinary derivatives. Furthermore, examining the Navier-Stokes equations in the r and directions demonstrates that the pressure, p, is a function only of z and thus ordinary derivatives can be used when differentiating the pressure with respect to z. uz C. Wassgren Chapter 07: Navier-Stokes Solutions 319 Last Updated: 16 Aug 2009 Now let’s apply boundary conditions to determine the unknown constants c1 and c2. First, note that the fluid velocity in a pipe must remain finite as r0 so that the constant c1 must be zero (this is a type of kinematic boundary condition). Also, the pipe wall is fixed so that we have uz(r=R)=0 (no-slip condition). After applying boundary conditions we have: R 2 dp r2 Poiseuille Flow in a Circular Pipe uz 1 2 4 dz R Notes: 1. The velocity profile is a paraboloid with the maximum velocity occurring along the centerline. The average velocity in the pipe is found from: u 1 R2 rR u z 2 rdr r 0 R 2 dp 8 dz 1 2 umax where umax is the maximum fluid velocity. 2. As with planar Couette-Poiseuille flow, we can determine stresses using the constitutive relations for a Newtonian fluid. The shear stress that the pipe walls apply to the fluid, w, is: R dp 4 u w 2 dz R where u is the average velocity in the pipe. Note that an alternate method for determining the average wall shear stress, which in this case is equal to the exact wall shear stress, is to balance shear forces and pressure forces on a small slice of the flow as shown below. dp dz R 2 p dz p R 2 dz w 2 Rdz dp 2 dz R w 2 Rdz 0 p R 2 p dz R dp w (The same answer as before!) 2 dz F z z In engineering applications it is common to express the average shear stress in terms of a (Darcy) friction factor, fD, which is defined as: 4 64 f D 1 w 2 64 u uD Re 2 where D=2R is the pipe diameter and Re is the Reynolds number. The Darcy friction factor commonly appears in the Moody chart for incompressible, viscous pipe flow. Note again that this solution is only valid only for a laminar flow. The condition for the flow to remain laminar is found experimentally to be: uD Re 2300 3. We can also use the general solution (before applying boundary conditions) to determine the flow between two concentric cylinders by applying different boundary conditions. For example, two fixed cylinders will have the boundary conditions: uz(r=RI)=0 and uz(r=RO)=0 where RI and RO are the inner and outer cylinder radii. C. Wassgren Chapter 07: Navier-Stokes Solutions 320 Last Updated: 16 Aug 2009 4. Laminar flow in an elliptical cross-section pipe can be determined by considering the simplified Navier-Stokes equation in the z-direction but using Cartesian coordinates (assuming ux=uy=0): 2 u z 2 u z 1 dp (Poisson’s equation!) 2 x 2 y dz where z is the coordinate along the centerline of the pipe. Note that the pipe wall boundary is the y ellipse given by: 2 2 x y 1 a b x 2b 2a where a and b are the lengths of the major and minor axes. Since we must satisfy the no-slip boundary condition at the pipe walls, let’s guess that the solution has the form: x 2 y 2 u z 1 a b since this profile automatically satisfies the boundary condition. The quantity is an unknown constant. To determine if this is indeed a valid solution to the fluid equations, we first note that it satisfies the continuity equation (ux=uy=0 and uz is not a function of z). If we substitute into the zcomponent of the Navier-Stokes equations (Poisson’s equation above) we find that our guess for the velocity distribution is valid if the constant is given by: 2 1 dp 2 2 2 b dz a a 2b2 dp 2 a 2 b 2 dz which means that the velocity profile for an elliptical pipe is given by: 2 2 a 2b2 dp x y uz 1 2 2 2 a b dz a b velocity profile in a pipe of elliptical cross-section For very complex cross-sections, we can determine the velocity profile by solving Poisson’s equation numerically. C. Wassgren Chapter 07: Navier-Stokes Solutions 321 Last Updated: 16 Aug 2009 4. Starting Flow Between Two Parallel Plates Consider a flow starting from rest between two parallel flat plates. The bottom plate is fixed while the top plate moves impulsively at t > 0 with constant velocity, U. There are no pressure gradients in the flow. top plate moves with velocity, U h y fluid x fixed bottom plate We’ll make the following assumptions: u z constant and 1. The flow is planar. 0 z u x u y 0 x x f x f y =0 2. The flow is fully-developed in the x-direction. 3. There are no body forces. Let’s first examine the continuity equation: u x u y u z 0 x y z From assumptions #1 and #2 we see that: u y 0 y Based on this result and assumptions #1 and #2 we see that the y-velocity can be at most a function of time: uy f t Since there is no flow through the walls at any time, the y-velocity must be zero. (call this condition #4) uy 0 Now let’s examine the Navier-Stokes equation in the x-direction: 2ux 2ux 2ux u x u x u x u x p fx ux uy uz 2 2 2 x z x x z 0 (#4) y t y 0 (#3) 0 (#2) 0 0 #1 0 #1 0 (#2) 2 u x ux t y 2 The initial and boundary conditions for the flow are: no flow initially u x y, t 0 0 (7.3) (7.4) no slip at y = 0 u x y 0, t 0 (7.5) no slip at y = h u x y h, t 0 U (7.6) C. Wassgren Chapter 07: Navier-Stokes Solutions 322 Last Updated: 16 Aug 2009 Note that as t → ∞, the flow profile should approach the Couette flow profile derived previously, i.e.: y u x y, t U (7.7) h Hence, let’s investigate a solution of the form: y ux ux U (7.8) h Substituting back into Eqn. (7.3) and the boundary and initial conditions gives: u x 2ux (7.9) 2 t y u x y, t 0 U u x y 0, t 0 y h (7.10) (7.11) u x y h, t 0 0 (7.12) To solve Eqn. (7.3), let’s try a separation of variables approach where: ux y, t Y y T t (7.13) so that, upon substitution into Eqn. (7.3), we have: YT Y T T Y 2 (7.14) T T where is a constant since the only way the T and Y sides of the equation can be equal for any t and y is if both sides are equal to a constant. Solving for each part of the equation gives: T 2 T t c1 exp 2 t (7.15) T Y 2 Y y c2 sin y c3 cos y (7.16) Y Hence the solution has the form: u x y, t c2 sin y c3 cos y c1 exp 2 t u x y, t exp 2 t c4 sin y c5 cos y (7.17) In order to satisfy the boundary condition at y = 0 (Eqn. (7.11)), the constant c5 must equal zero and Eqn. (7.17) becomes: (7.18) u x y, t c exp 2 t sin y In order to satisfy the boundary condition at y = h (Eqn. (7.12)) without having c4 = 0, we must have: n where n is an integer (7.19) h Hence: t y u x , n y, t cn exp n 2 2 2 sin n (7.20) h h C. Wassgren Chapter 07: Navier-Stokes Solutions 323 Last Updated: 16 Aug 2009 Since Eqn. (7.9) and the boundary and initial conditions (7.10) - (7.12) are linear, we can add the together the solutions in Eqn. (7.20) so that they satisfy the given initial condition (Eqn. (7.10)). Note that we can add together the constants for the negative values of n with the positive values of n since the magnitude of the exponential and sine terms are identical, i.e.: 2 2 t y t y c n exp n 2 2 sin n c n exp n 2 2 sin n h h h h 2 t y c n c n exp n 2 2 sin n h h 2 t y d n exp n 2 2 sin n h h Furthermore, we needn’t include n = 0 since it will give u’x,n = 0 = 0 which doesn’t contribute to the summation. Hence, the solution to Eqn. (7.9) subject to the given boundary conditions (Eqns. (7.11) and (7.12)) is: t y u x y, t d n exp n 2 2 2 sin n (7.21) h h n 1 where the constants dn are found by forcing Eqn. (7.21) to satisfy the given initial condition (Eqn. (7.10)). Fourier sine series analysis at t = 0 gives the constants as: yh 2 y y 2U 2U n d n U sin n dy cos n (7.22) 1 h y 0 h h n n Combining Eqns. (7.8), (7.21), and (7.22) gives: u x y, t U y 2 1 t exp n 2 2 2 h n 1 n h n y sin n h (7.23) A plot of the dimensionless velocity profile for various dimensionless times is shown below. dimensionless position, y /h 1.0 0.9 0.8 0.7 0.6 t ' = t /h 0.5 0.4 2 t'=0.025 t'=0.100 t'=0.250 t'=0.500 0.3 0.2 0.1 0.0 0.0 0.2 0.4 0.6 0.8 1.0 dimensionless velocity, u x /U C. Wassgren Chapter 07: Navier-Stokes Solutions 324 Last Updated: 16 Aug 2009 6. Starting Flow in a Circular Pipe Consider the unsteady flow of an incompressible, constant viscosity, Newtonian fluid within an infinitely long, circular pipe of radius, R. r R z We’ll make the following assumptions: 1. The flow is axi-symmetric and there is no “swirl” velocity. 2. The flow is fully-developed in the z-direction. 3. There are no body forces. 0 and u 0 u u r z 0 z z f r f f z 0 Let’s first examine the continuity equation: 1 rur 1 u u z 0 r r r z From assumptions #1 and #2 we see that: rur constant Since there is no flow through the walls, the constant must be equal to zero and thus: ur 0 (call this condition #4) Now let’s examine the Navier-Stokes equation in the z-direction: 1 u z 1 2 u z 2 u z u u u u p u 2 fz z ur z z u z z r 2 2 r z z z r t r r r r We can simplify this equation using our assumptions: 2 2 u z u z u u z u z p 1 u z 1 u z u z ur uz 2 fz r r r r r r 2 z z 2 z t 0 (#4) r 0 (#3) 0 (#3) 0 (#1) 0 (#2) 0 (#1) u 1 d du z dp (7.24) z r t dz r dr dr Note that in the previous derivation the fact that uz is a function only of r has been used to change the partial derivatives to ordinary derivatives. Furthermore, examining the Navier-Stokes equations in the r and directions demonstrates that the pressure, p, is a function only of z and thus ordinary derivatives can be used when differentiating the pressure with respect to z. The initial and boundary conditions for the flow are: no flow initially uz r, t 0 0 no slip u z r R, t 0 C. Wassgren Chapter 07: Navier-Stokes Solutions (7.25) (7.26) 325 Last Updated: 16 Aug 2009 We know that as t → ∞ the flow should approach the Poiseuille flow solution found in the previous section, i.e.: uz r , t R 2 dp r2 1 2 4 dz R (7.27) Hence, let’s investigate a solution of the following form: R 2 dp r2 uz uz 1 2 4 dz R (7.28) THE REMAINDER OF THIS DERIVATION IS INCOMPLETE. C. Wassgren Chapter 07: Navier-Stokes Solutions 326 Last Updated: 16 Aug 2009 5. Impulsively Started Flat Plate (aka Stokes’ First Problem, aka the Rayleigh Problem) Now let’s consider the incompressible, constant viscosity, Newtonian fluid flow resulting from the sudden movement of an infinitely long flat plate. The geometry of the problem is shown below: y fluid x 0 U t U t 0 t 0 We’ll make the following assumptions for this unsteady flow: 1. The flow is planar. 2. The flow is fully-developed in the x-direction. 3. There are no body forces. 0 and u z constant z u y u x 0 x x fx f y fz 0 4. There is no pressure gradient in the x-direction. p 0 x Noting that the continuity equation gives uy=0 and simplifying the Navier-Stokes equations using the given assumptions we find: u x 2ux t y 2 where =/ is the kinematic viscosity (dimensions of L2/T). The (kinematic) boundary and initial conditions for this flow are: u x y 0, t U (for t 0) (no-slip condition at the plate) u x y , t remains finite and (kinematic condition far from the plate) u x y, t 0 0 (fluid is initially at rest) Note that there is no geometric length scale in the problem which suggests that we can use a similarity variable, , to reduce the number of independent variables from two (t and y) to one (=(y,t)) (i.e. convert the PDE into an ODE). We may anticipate this reduction in the number of variables by considering where and when the fluid velocity reaches some value, e.g. ux(y,t)=0.4U. It is reasonable to expect that the location, y, where the velocity reaches 0.4U will vary depending on t (e.g. the location y gets farther from the plate as t gets larger). Thus, ux will not depend on the parameters y and t independently but will instead depend on some combination of y and t. To determine this combination, let’s first re-write the velocity in dimensionless form using the plate velocity, U: u u* x U so that the original PDE becomes: u 2u u * 2u* 2 t y t y 2 with the boundary conditions: C. Wassgren Chapter 07: Navier-Stokes Solutions 327 Last Updated: 16 Aug 2009 0 for t 0 u * y 0, t 1 for t 0 * u y , t remains finite Note that since the velocity is dimensionless, it must depend only on dimensionless quantities. The only dimensional quantities in the PDE are t, y, and . We can form only one dimensionless variable, call it , the similarity variable, from these quantities: y 4 t Note that a “4” is added to the similarity variable, , for convenience in solving the resulting differential equation. Thus the dimensionless velocity will be a function only of the similarity variable: u u * x f U Re-writing the original PDE in terms of this similarity variable gives: f f t y 2 where f df d and f d2 f d 2 1 and t y 2t 4 t so that the final equation becomes: 1 f f 4t 2t f 2 f 0 subject to the boundary conditions: 0 for (Note that the initial condition is subsumed into the → ∞ condition.) f 1 for 0 Thus, we see that by using a similarity variable (justified based on dimensional arguments), the PDE with two independent variables is reduced into a (linear) ODE. Now we must solve the ODE. Fortunately, we can solve the resulting ODE without much difficulty: d f 2 f 0 ln f 2 d df c1 exp 2 d f c1 exp 2 d c2 0 where is a dummy variable of integration. Applying the boundary conditions to determine the constants c1 and c2: f 0 1 c2 f 0 c1 exp 2 d 1 c1 0 2 where the indefinite integral has been evaluated. Thus, the velocity distribution for this flow is given by: C. Wassgren Chapter 07: Navier-Stokes Solutions 328 Last Updated: 16 Aug 2009 y 4 t ux 2 y 2 1 exp d 1 erf 4 t impulsively started flat plate flow U 0 where the integral is also known as the error function (erf). Notes: 1. A plot of the flow profile in terms of dimensional quantities: 2.0 t=t1 t=t2 t=t3 y 1.5 t=t4 1.0 0.5 0.0 0.0 0.5 1.0 u/U and in dimensionless form: 2.0 =y /(4 t ) 1/2 1.5 1.0 0.5 0.0 0.0 0.5 1.0 u /U C. Wassgren Chapter 07: Navier-Stokes Solutions 329 Last Updated: 16 Aug 2009 2. As we can see from the plots shown in Note #1, the effect of the plate diffuses into the remainder of the fluid. An estimate of the depth of fluid that is affected by the movement of the plate may be found by determining the distance from the plate, , where the velocity is 1% that of the plate velocity, i.e. u/U=0.01. This can be found either by estimating from the plot or calculating: ux 0.01 when 1.8 3.6 t U Thus, we see that the thickness of the affected layer is proportional to the square root of the kinematic viscosity and to the square root of the time (note that it is not a function of the plate velocity, U, or the absolute viscosity, ). The distance, , which is also referred to as the shear layer thickness, is an important parameter that gives us an estimate of how far into the flow the effects of the boundary are felt. We will come across this parameter again, in terms of a boundary layer thickness, in a later set of notes. The shear layer thickness after 1 minute in: air is: =10.8 cm (air=0.150 cm2/sec) water is: =2.8 cm (water =0.010 cm2/sec) Thus we see that the effects of a boundary are felt further into a flow of air than into a flow of water in a given amount of time! 3. We can also use this solution to examine the flow resulting from a fluid with a uniform velocity U over a plate that has come to a sudden stop from an initial velocity U. To produce the resulting velocity profile, we note that this flow can be produced via a Galilean transformation of the problem we just investigated, (u/U)stopped plate = 1-(u/U)moving plate. The resulting flow profile is: ux y erf U 4 t 4. The vorticity in the flow is found via: u y z x U 1 erf y y 4 t y2 U exp t 4 t distance from plate, y 3.0 2.5 t1 2.0 t2 1.5 t3 1.0 t4 0.5 0.0 0.0 0.5 1.0 vorticity, z Vorticity is created at the wall through the no-slip condition and diffuses through the rest of the fluid through the action of viscosity. C. Wassgren Chapter 07: Navier-Stokes Solutions 330 Last Updated: 16 Aug 2009 6. Oscillating Flat Plate (aka Stokes’ Second Problem, aka the Rayleigh Problem) Consider the incompressible, constant viscosity, Newtonian fluid flow resulting from the sinusoidal oscillation of an infinitely long flat plate. The geometry of the problem is shown below: y fluid x U(t) = Ucos(t) We’ll make the following assumptions for this unsteady flow: 0 and u z constant z u y u x 0 x x fx f y fz 0 1. The flow is planar. 2. The flow is fully-developed in the x-direction. 3. There are no body forces. Noting that the continuity equation gives uy=0 and simplifying the Navier-Stokes equations using the given assumptions we find: u x 2ux t y 2 where =/ is the kinematic viscosity (dimensions of L2/T). The (kinematic) boundary and initial conditions for this flow are: u x y 0, t U cos t (no-slip condition at the plate) u x y , t remains finite (kinematic condition far from the plate) u x y, t 0 0 (fluid is initially at rest) Since the boundary condition is time dependent, we might expect that the fluid velocity will have the following (separation of variables) form: u x y, t f y exp i t where only the real part of the velocity is relevant to the solution. Substituting into the PDE and simplifying gives: i f exp it f exp it f exp it i f exp it 0 f i f 0 Solving for f we find: i f y A exp y i i u x y, t A exp y it exp it A exp y which can be simplified to: C. Wassgren Chapter 07: Navier-Stokes Solutions 331 Last Updated: 16 Aug 2009 1 i u x y, t A exp y it A exp y t exp i y 2 2 2 u x y, t A exp y t i sin y t cos y 2 2 2 u x y, t A exp y cos t y 2 2 where in the last step of the previous analysis, only the real part of the velocity component is relevant to the solution. Applying the boundary conditions we find: flow due to an oscillating plate u x y, t U exp y cos t y 2 2 Notes: 1. A plot of the flow looks like: y *sqrt( /(2 )) 5.0 wt = 0 4.0 wt = (1/5)*2pi wt = (2/5)*2pi wt = (3/5)*2pi 3.0 2.0 1.0 wt = (4/5)*2pi 0.0 -1.0 0.0 1.0 u /U 2. The velocity decreases exponentially with the distance from the plate. Also note that there is a phase lag in the velocity profile compared to the plate which is a function of distance from the plate. 3. The region of fluid affected by the plate can be estimated by determining the y location (y = ) at which u/U = 0.01. We’ll also assume the maximum value for the cosine function. ux 0.01 exp cos t U 2 2 1 2 ln 0.01 6.51 Again notice that . 4. We can also use this solution to investigate flow oscillating far from the plate and having a fixed plate by performing a Galilean transformation on the velocity profile. C. Wassgren Chapter 07: Navier-Stokes Solutions 332 Last Updated: 16 Aug 2009 7. (Planar) Stagnation Point Flow (aka Hiemenz Flow) Consider the flow in the vicinity of a stagnation point: y x We’ll make the following assumptions in the analysis of this flow: 0 and uz constant z u y u x 0 t t fx f y fz 0 1. The flow is planar. 2. The flow is steady. 3. There are no body forces. Recall from our earlier discussion of potential flows that the complex potential model for this type of flow is: f z Az 2 A x 2 y 2 i 2 Axy where A is a constant that is proportional to the velocity far from the body, U , divided by a characteristic length of the body, L: U A L The constant of proportionality depends on the exact shape of the body. The velocity components for the flow are given by: f z u x iu y 2 Az 2 Ax i 2 Ay u x 2 Ax and u y 2 Ay The pressure distribution at any point (x,y) in the flow is given by: 2 2 p( x, y ) p0 1 2 u x u y p0 2 A2 x 2 y 2 where p0 is the pressure at the stagnation point. Note that this potential flow solution satisfies our governing equations of fluid dynamics (continuity and Navier-Stokes) and it satisfies part of the no-slip condition (no flow through the surface). It does not, however, satisfy the tangential component of the no-slip condition. Thus, the potential flow solution is of limited use since it won’t be a good model close to the plate surface. To determine a valid solution close to the plate surface, let’s try modifying the potential flow model (we’ll work with the stream function since close to the surface the flow will be rotational) so that it does satisfy the no-slip boundary conditions. Let’s try the following stream function: 2 Axf u x 2 Axf and u y 2 Af where f=f(y) and f ’=df/dy. The function, f, is unknown at this point. We’ll place several constraints on the function, f, so that it satisfies the governing fluid equations (continuity and Navier-Stokes) and we’ll also make sure that far from the plate the flow has the same form as the original potential flow solution. C. Wassgren Chapter 07: Navier-Stokes Solutions 333 Last Updated: 16 Aug 2009 We know that since we’re using a stream function the continuity equation is automatically satisfied. To make sure we satisfy the momentum equations we substitute the velocity components into the NavierStokes equations (simplified using our assumptions): ux ux 2u 2u u x u 1 p uy x 2x 2x x y y x x u y x uy u y y 2u y 2u y 1 p 2 2 x y y 4 A2 x f 4 A2 xff 1 p 2 4 A2 ff 1 p y x 2 A xf 2 A f We need to say something about the pressure distribution before proceeding further. Let’s integrate the ycomponent of the Navier-Stokes equations with respect to y: 1 p 4 A2 ff 2 A f y p 4 A2 ff 2 A f y p x, y 2 A2 f 2 A f g ( x) where g(x) is an unknown function of x (since p is a function of both x and y). To determine the form of g(x) we recall that far from the plate the current solution should approach the potential flow solution where the pressure distribution is given by: p( x, y ) p0 2 A2 x 2 y 2 2 and the function f(y)y (which gives the original potential flow function). Thus the unknown function of x should be given by (as y becomes very large): p x, y 2 A2 y 2 A 1 g ( x) p0 2 A2 x 2 y 2 2 g ( x) p0 2 A2 x 2 2 A and the pressure distribution becomes: p x, y p0 2 A2 f 2 A 1 f 2 A2 x 2 2 Substituting this pressure distribution into the x-component of the Navier-Stokes equations gives: p -4 ρA2 x x 2 4 A2 x f 4 A2 xff 4 A2 x 2 A xf f ff f 1 0 2A Thus, the function f must satisfy the non-linear, 3rd order ODE given above in order to satisfy the xmomentum equation (note that we’ve already established that the original stream function will satisfy the continuity equation and y-momentum equation). The boundary conditions for the ODE are: no-slip (x -component): u x x, y 0 0 f y 0 0 2 no-slip (y -component): u y x, y 0 0 potential flow far from plate: f y y C. Wassgren Chapter 07: Navier-Stokes Solutions 334 f y 0 0 f y 1 Last Updated: 16 Aug 2009 Currently the ODE and boundary conditions are in dimensional form. To make the solution to the ODE general, let’s re-write it in terms of dimensionless parameters: f y 2A F where y f f d 2 f d d 2 dy 2 2A 2A F F 2A df df d dy d dy 2A 2A F 3 2A F 3/ 2 d 3 f d 2A 2A F F f 3 d dy 2A so that the original dimensional ODE is given in dimensionless form by: F FF F 1 0 2 subject to the boundary conditions: F 0 0 F 0 0 F 1 An exact analytical solution to this ODE has not been found so we solve it numerically (using, for example, a Runge-Kutta numerical scheme). Even though we solve the equation numerically, we still consider the result an “exact” solution since we can find the solution numerically to any precision. The velocity components and pressure distribution are found using the original assumed stream function: 2 A xF u x 2 AxF and u y 2 A F p x, y p0 A F 2 2 A 1 F 2 A2 x 2 Notes: 1. Plots of the functions F and F ’ as a function of look like (plot from Panton, R.L., Incompressible Flow, Wiley.) = (2A/)0.5y C. Wassgren Chapter 07: Navier-Stokes Solutions F, F’, F’’ 335 Last Updated: 16 Aug 2009 2. Note that for this flow the non-linear convective terms in the Navier-Stokes equations, (u)u, did not drop out as they have in the previous exact solutions. 3. In flows around objects with surface curvature (e.g. a torpedo-shaped object), this solution will still be valid in some vicinity of the stagnation point. As we zoom in very close to the stagnation point, the local object surface will be approximately flat. 4. Recall that far from the plate, the viscous flow solution should approach the potential flow solution. We can estimate this distance by determining the location, y=, at which the x-velocity is 99% that of the velocity far from the plate at the same x-location, U (for y> the vorticity will be very small since the velocity gradients are small): u x 2 AxF and u x U 2 Ax ux 2A F 0.99 when 2.4 U 2.4 2A The distance, , is referred to as the (99%) boundary layer thickness. Note that here the boundary layer thickness is a constant value and proportional to the square root of the kinematic viscosity. Since the boundary layer thickness is constant, we can interpret that the shear layer displaces the outer potential flow a constant distance from the surface. From the plot we note that as , F(-0.65) (recall that in the potential flow region “far” from the boundary, F is linear) so that this displacement thickness, D, is given by: 2A D 0.65 D 0.65 2A We’ll address the concept of a displacement thickness again when discussing boundary layer flow. 5. The pressure gradients for the flow are given by: dU p (using the U∞ defined in the previous note) 4 A2 x U dx x p 2 A FF F y The pressure gradient in the x-direction is the same as that given by Bernoulli’s equation using the outer potential flow velocity while the pressure gradient in the y-direction will be small if the kinematic viscosity is small (F, F ’, and F ” are all of order 1 near the surface). Thus, the pressure in the shear layer is nearly constant in the y-direction and it has the same magnitude as the pressure in the outer potential flow. This is an important result that will be discussed again when investigating boundary layer flows. 6. An exact solution for axi-symmetric stagnation point flow can also be found. Its solution was first presented by Homann (1936). The approach for the axi-symmetric problem is very similar to what was presented here for planar flow except a different stream function is used. Refer to White, F.M., Viscous Fluid Flow, McGraw-Hill for more details. The resulting velocity profiles, pressure, and shear stress distributions for the axi-symmetric case are similar to those found for the planar case but with slightly different magnitudes. C. Wassgren Chapter 07: Navier-Stokes Solutions 336 Last Updated: 16 Aug 2009 8. Very Low-Reynolds Number (aka Creeping, aka Stokes) Flows (Re << 1) Consider the governing equations for an incompressible fluid, neglecting body forces, in dimensional form: u 0 (7.29) u u u p 2 u t Recall that when the Reynolds number is very small, viscous forces dominate the inertial forces. Let’s rewrite these equations in dimensionless form keeping in mind that we’ll be investigating flows where viscous forces dominate (or where the fluid inertia is negligible). The variables in the equations are normalized in the following manner: x u x* x Lx * u* u Uu * L U (7.30) p tU Lt * t p U p * t* p* L L U U L where the superscript “*” refers to a dimensionless quantity and L and U represent, respectively, a characteristic length and velocity for the flow of interest. Note that the pressure has been made dimensionless using a characteristic viscous stress, U/L, rather than the usual dynamic pressure term, i.e. U2. This is because here we’re investigating flows where viscous forces dominate (or fluid inertia is negligible). Also note that we’ve assumed that the time scale is set by the flow velocity and length scale. This assumption is fine unless there is some other well-defined time scale in the problem such as an oscillation period (for acoustic applications, for example). Now let’s rewrite the governing equations using these dimensionless parameters. * u* 0 U 2 u* U * * U *2 * * * * * u u 2 p 2 u L t L L Dividing through by the characteristic viscous force term gives: UL u* *2 * * * * ** * u u p u t u* u* * u* * p* *2 u* t * Re (7.31) (7.32) where Re is the Reynolds number which is a ratio of typical fluid inertial forces to viscous forces in a flow. If the viscous forces dominate, then the Reynolds number should be small. For creeping flows we investigate the limit when Re0, i.e., the fluid has no inertial terms. Thus, for creeping flows the governing fluid equations simplify to: * u* 0 (7.33) * p* *2 u* or in dimensional form: u 0 Stokes’ Equations (7.34) p 2 u Note that does not appear in Eqn. (7.34) indicating that Stokes flows behave the same regardless of the surrounding fluid density! C. Wassgren Chapter 07: Navier-Stokes Solutions 337 Last Updated: 16 Aug 2009 Two additional useful relations can be found if we take the curl of both sides of the momentum equation: p 2 u 2ω 0 and if we take the divergence of both sides of the momentum equation (and using continuity): p 2 u (7.35) 2 p 0 Both the vorticity and pressure fields satisfy Laplace’s equation for a creeping flow. (7.36) Notes: 1. Examples where creeping flow might occur (Re << 1): a. small length dimensions (flow in small pipes or channels, around small particles, flow through small pores) b. very viscous fluids c. small velocities A good reference for this topic is Happel and Brenner (1965). 2. Since Laplace’s equation is a linear PDE, we can add together solutions to form new solutions (the principle of superposition). This is very similar to what we observed in potential flows where we could add together valid velocity fields (in the form of a potential function) to form new velocity fields. The difference however is that here we can also add together pressure or vorticity fields. Note that in potential flows we couldn’t add together pressure fields since the pressure was found using the non-linear Bernoulli’s equation. 3. If we consider a 2D flow and use a stream function to describe the velocity field, we find that the vorticity can be written in terms of the stream function as: u u 2 2 z y x 2 2 2 (7.37) x y x y Substituting Eqn. (7.37) into Eqn. (7.35) gives: 2 2 4 0 governing equation for a 2D Stokes flow (7.38) (In 2D Cartesian coordinates: 4 4 x 4 2 4 x 2 y 2 4 y 4 .) This is referred to as the biharmonic equation and is a common equation found other fields of study (e.g. solid mechanics where the Airy stress function is used to solve plane problems in elasticity). 4. The pressure increases proportionally with the dynamic viscosity of the fluid assuming that the viscosity is independent of pressure (refer to Eqn. (7.34)). Note that when the pressures become very large, such as in lubrication flows, the viscosity becomes a function of pressure (recall that the viscosity is also a function of temperature). 5. There are several approaches to finding solutions to creeping flow problems. These include: a. forming “building block” solutions that we can add together to form new solutions (we used a similar approach with potential flow problems) (Note that we can add together pressure fields since Eqn. (7.36) is linear! We can’t do this for potential flows since Bernoulli’s equation is nonlinear in terms of the velocities.) b. solving the boundary value problem for the given geometry and boundary conditions, c. borrowing solutions from other disciplines which have the same governing equations (e.g. the Airy stress function for solid mechanics), and d. using computational methods to solve the governing equations. C. Wassgren Chapter 07: Navier-Stokes Solutions 338 Last Updated: 16 Aug 2009 6. Note that if u is a Stokes flow solution, then u’ = -u is also a solution since: u 0 and p 2u p p In addition, 2ω 0 and 2 p 0 where ω u u . Hence, Stokes flows are kinematically reversible and flow around symmetric objects will produce symmetric streamlines. C. Wassgren Chapter 07: Navier-Stokes Solutions 339 Last Updated: 16 Aug 2009 9. Stokes Flow Around a Sphere Now let’s examine the creeping flow around a sphere of radius, R, in a uniform stream of velocity, U. For axi-symmetric creeping flows it is convenient to use a stream function in spherical polar coordinates, (r, , ), to describe the fluid velocity. The angle is zero when aligned with the incoming free stream. Since the flow is axi-symmetric, the stream function will be a function only of r and . After substituting the stream function into the biharmonic equation (in spherical coordinates and noting that for an axi-symmetric problem there is no variation in the -direction): r 4 0 2 2 1 2 cot U 2 2 2 0 2 r r r The velocity components are related to the spherical stream function by: R 1 1 and u ur 2 r sin r r sin These forms of the velocity in terms of the stream function can be verified by substituting them into the (incompressible) continuity equation in spherical polar coordinates (recall that the stream function is defined such that it automatically satisfies the continuity equation): 1 2 1 u sin 0 r ur 2 r sin r r 1 2 1 1 r 2 r 2 r r sin r sin ur 1 1 1 r 2 r sin r sin 1 r 2 sin 2 r 1 2 r sin 1 sin r r sin u 1 r r 2 0 r The no-slip condition at the surface means that: u r r R u r R 0 r R 0 and r R 0 r and far from the sphere, as r, the stream function approaches the stream function for a uniform stream: r2 r Usin 2 constant 2 ur U cos and u U sin Solve the differential equation with the given boundary conditions using separation of variables. Based on the form of the stream function far from the origin, let’s assume that the solution has the form: =f(r)sin2 2 2 1 2 cot 2 2 2 2 f r sin 0 r 2 r r After simplifying we get: 2 d2 2 2 2 f r 0 r dr In trying to solve this ODE, let’s try a solution of the form: f=rn: C. Wassgren Chapter 07: Navier-Stokes Solutions 340 Last Updated: 16 Aug 2009 2 d2 2 n n4 2 2 r n 2 n 3 2 n n 1 2 r 0 r dr n 1,1, 2,3 A f (r ) Br Cr 2 Dr 3 r The corresponding stream function and velocities are: A (r , ) Br Cr 2 Dr 3 sin 2 r A B ur 2 3 C Dr cos r r AB u 3 2C 3Dr sin r r Applying the boundary conditions we find that: UR 3 3UR U A B C D0 4 4 2 Consequently: R 2U R 3r 2r 2 2 (r , ) sin 4 r R R2 R 3 3R ur U cos 1 3 2r 2r R 3 3R u U sin 1 3 4r 4r The pressure, found using the momentum equation (p = 2u), is: 2 3 U R p p cos 2 R r The viscous stresses are found using the constitutive relations for a Newtonian fluid: u rr 2 r r 1 u ur 2 r r 1 u ur u cot r r r sin 2 u r r r r 1 ur r u 1 ur r r r r sin r sin r u 1 u sin r sin Evaluating at the sphere’s surface (r=R) gives: rr r R r R r R r r R r R 0 r r R 3 U sin 2R C. Wassgren Chapter 07: Navier-Stokes Solutions 341 Last Updated: 16 Aug 2009 The drag force acting on the sphere surface (r=R) is found by integrating the pressure and viscous forces in the horizontal direction over the entire sphere’s surface: r’ r R sin R dr R cos d F r r R sin dA p r R cos dA where dA 2 R 2 sin d dr dA 2 r 2 R 2 sin d 0 0 cos F 4UR 2UR Thus, the total force acting on the sphere consists of 2/3 viscous force and 1/3 pressure force giving a total force of: F 6UR Stokes Drag Notes: 1. Stokes drag is strictly valid only when Re0 but it is found experimentally to be a reasonable estimate up to Re=1. 2. Note that the drag is independent of the fluid density and is proportional to the velocity (and not velocity squared). 3. Stokes drag is usually presented in dimensionless form as a drag coefficient, cD. The usual form of this is: F 6UR 12 24 cD 1 U 2 R 2 1 U 2 R 2 2 UR UD 2 where D=2R is the sphere diameter. Note that the expression can be simplified further by using the Reynolds number based on the sphere diameter: 24 cD Stokes Drag Coefficient Re D 4. Oseen (1910) included first-order inertial effects in the drag analysis and found a drag coefficient of: 24 3 cD 1 Re D Re D 16 This drag coefficient is found to give good results up to ReD 5 5. Although the streamlines for flow around a sphere look similar between a potential flow and a Stokes flow (in the sphere’s frame of reference (FOR)), there are some important differences. The streamlines for potential flows are grouped closer together near the sphere than they are for a Stokes flow. More strikingly, if we plot the streamlines using a frame of reference where the fluid is at rest far from the sphere (and the sphere moves with a velocity -U), we find that in a potential flow the fluid is “pushed” out of the way while in a Stokes flow the fluid is “dragged” along with the sphere. FOR fixed to cylinder FOR fixed to ground creeping flow potential flow C. Wassgren Chapter 07: Navier-Stokes Solutions 342 Last Updated: 16 Aug 2009 6. We can also use the solution approach presented here to determine the drag on a spherical droplet of a fluid (with dynamic viscosity i) in a different fluid (of dynamic viscosity o). The general differential equation is the same but the boundary conditions are different. For the spherical droplet problem, the boundary conditions at the sphere radius consist of continuous velocity components (no-slip but the tangential velocity is not zero) and stresses between the droplet fluid and the outer fluid. The resulting drag force acting on the droplet becomes: 1 2 o 3 i F 6 R oU 1 o i For i >> o (e.g. solid droplet in a gas or liquid) we get the original Stokes drag equation: F 6oUR For i << o (e.g. gas bubble in a liquid) we get a smaller drag force (since the outer fluid can “slip” at the boundary surface): F 4 R oU 7. It can be shown that the drag on an irregular object in a Stoke’s flow is bounded by the drag on a sphere that inscribes the object and the drag on a sphere that circumscribes the object (refer to Hill and Power, 1956). This is a useful result for practical applications. Dinscribed sphere < Dirregular object < Dcircumscribed sphere 8. An interesting observation, referred to as the Stokes Paradox, can be made using dimensional analysis. Assuming that inertia is negligible for a Stokes flow, the force, F, on an object is: F fcn , U , L (7.39) where , U, and L are the dynamic viscosity, velocity, and characteristic length for the flow. For a 2D flow, the dimensions of the force will be F/L (force per unit depth) while for a 3D flow the dimension of the force will simply be F. Thus, from dimensional analysis: F (7.40) F2D 2 D constant U F (7.41) F3D 3 D constant UL where F’ is the dimensionless force. Equation (7.40) indicates that in a 2D Stokes flow the force on an object is independent of the object size. This contradicts what we observe in reality. Hence, our initial assumption that the fluid inertia is negligible must be incorrect. For 2D flows, the density must be a factor in determining the force on an object: F2 D fcn , , U , L (7.42) F2D F2 D UL fcn U C. Wassgren Chapter 07: Navier-Stokes Solutions (7.43) 343 Last Updated: 16 Aug 2009 10. Lubrication Flow One very important application of creeping flows is in the study of lubrication problems. Let’s consider the example of a simple, stationary, planar slipper pad bearing as shown in the figure below: L stationary slipper pad bearing p flow direction h0 p y h1 x bottom plate moving at constant horizontal velocity U To analyze this flow, let’s examine the typical magnitudes of various terms in the Navier-Stokes equations. First let’s consider the x-component of the Navier-Stokes equations for the steady flow of an incompressible fluid with negligible body forces (the gravitational body force term in lubrication problems is typically very small in comparison to the other term and so is neglected): 2u 2u u u p u v 2 2 (7.44) y x y x x Characteristic magnitudes: u ~ U x ~ L y ~ h0 The characteristic y-velocity can be determined from the continuity equation: Uh u v v u U v~ 0 0 ~ x y y x L L (7.45) (7.46) If we examine the magnitudes of the convective inertial forces we find: u U 2 u U 2 u ~ and v ~ (7.47) L L x y The magnitudes of the viscous forces are: 2 u U 2 u U 2~ 2 2~ 2 and (7.48) L h0 x y Since we’re investigating flows where h0/L<<1, the second term will dominate the magnitude of the viscous forces. Let’s examine the case where convective inertial terms can be neglected in comparison to the viscous terms ( a creeping flow): ρU 2 U 2 L h0 2 ρUL h0 1 L (7.49) 2 h Re L 0 1 L for a creeping flow (7.50) To check the value of this ratio for a typical lubrication problem, consider the following parameters: U=10 m/s, L=4 cm, h0=0.1 mm, =5*10-4 m2/s (SAE 30 oil) gives: ReL=800 but ReL(h0/L)2=0.005 C. Wassgren Chapter 07: Navier-Stokes Solutions 344 Last Updated: 16 Aug 2009 Thus, this flow could be considered a creeping flow. Using the simplifications just discussed, the Navier-Stokes equation in the x-direction reduces to: p 2u 2 x y Note that we expect the magnitude of the pressure gradient in the x-direction to be of the order: p U ~2 x h0 based on the previous dimensional arguments. (7.51) (7.52) Considering the Navier-Stokes equation in the y-direction we find that the pressure gradient in the ydirection should be of the order: p U p U ~ ~ 2 (7.53) y h0 L h0 x where h0/L<<1 has been assumed. Thus, it is reasonable to assume that the pressure remains essentially constant in the y-direction (in comparison to how the pressure changes in the x-direction). Solving the differential equation given in equation (7.51) gives: 1 p 2 u y c1 y c2 2 x Applying no-slip boundary conditions at the top and bottom walls: u y 0 U and u y h 0 (7.54) (7.55) results in the following velocity profile: 1 p 2 y u y yh U 1 h 2 x (7.56) Couette Flow Poiseuille Flow We see that the velocity profile is a combination of a Poiseuille flow and a Couette flow. We now have a relation relating two unknowns, the velocity profile and the pressure distribution. We must use another relation solve for the pressure distribution (or velocity distribution) in terms of known quantities. So far we’ve used the momentum equations to derive equation (7.56), now let’s consider the continuity equation; specifically, the mass flow rate at any cross-section must remain the same: y h x m (7.57) udy 0 x x y0 (Note that the fluid was assumed incompressible when writing Eqn. (7.57).) Substituting in for the velocity using equation (7.56) gives: x y h x y 0 udy x y h x y 0 1 p 2 y 2 x y yh U 1 h dy 0 h 1 p h h3 U h 0 2 x 2 x 3 2 3 p 3 h h 6U x x x Reynolds’ Equation for lubrication in a planar channel C. Wassgren Chapter 07: Navier-Stokes Solutions 345 (7.58) Last Updated: 16 Aug 2009 Notes: 1. We can use Reynolds Equation to solve for the pressure distribution, p(x), assuming we know the bearing geometry, h(x). Let’s consider the simple example using the slipper pad bearing design shown in the figure below: L stationary slipper pad bearing p h0 p y bottom plate moving at constant horizontal velocity h1 x U The bearing geometry, which is a straight line, is described by: x (7.59) h x h0 h1 h0 L Substituting into equation (7.58) and solving for the pressure gradient (using the boundary conditions: p(x=0) = p(x=L) = p) gives: 6 x L 1 x L 1 h1 h0 p p (7.60) 2 UL 2 1 h1 h0 1 1 h1 h0 x L h0 A plot of this dimensionless pressure distribution is shown below. 4.0 (p -p inf)/(UL /h 02) 3.5 As h1/h0 → 1, the flow approaches a Couette flow. 3.0 2.5 h1/h0=0.90 h1/h0=0.75 2.0 h1/h0=0.50 h1/h0=0.25 1.5 1.0 0.5 0.0 0.0 0.2 0.4 0.6 0.8 1.0 x /L The magnitude of the maximum pressure can be quite large. Consider for example the following parameters (the same parameters as used previously): U=10 m/s, L=4 cm, h0=0.1 mm, =4*10-1 N-sec/m2 (SAE 30 oil) C. Wassgren Chapter 07: Navier-Stokes Solutions 346 Last Updated: 16 Aug 2009 UL 1.6 MPa 160 atm! h02 A more accurate analysis of the flow would also include variations in the fluid viscosity due to the large pressure variations. 2. Note that a truly symmetric bearing and flow would result in zero lift on the bearing since the pressure increase on the upstream side of the bearing would be offset by an equivalent pressure reduction on the downstream side (the solid line in the drawing below). pgage x The reason real bearings can support a load is because the pressure distribution is, in fact, not symmetric. When a liquid flows past the centerline the pressure begins to decrease below zero gage pressure as shown in the figure above. However, the minimum pressure is limited by the cavitation (i.e. vapor) pressure of the liquid (dashed line in the figure above). Hence, there is a net positive gage pressure force acting on the bearing surface. C. Wassgren Chapter 07: Navier-Stokes Solutions 347 Last Updated: 16 Aug 2009 Review Questions 1. Describe several common assumptions used to simplify the Navier-Stokes equations. 2. Describe several common boundary conditions used when solving the Navier-Stokes equations. 3. At what (rule of thumb) Reynolds number does transition from laminar to turbulent flow occur for planar Couette flow? 4. At what (rule of thumb) Reynolds number does transition from laminar to turbulent flow occur for Poiseuille flow? 5. Sketch the velocity profiles for a planar Couette-Poiseuille flow with different pressure gradients. 6. What is meant by the “shear layer thickness”? 7. How does the shear layer thickness typically depend on the kinematic viscosity for laminar flows? C. Wassgren Chapter 07: Navier-Stokes Solutions 348 Last Updated: 16 Aug 2009 Chapter 08: Boundary Layers 1. 2. 3. 4. 5. 6. 7. 8. 9. Boundary Layer Introduction Boundary Layer Equations Solution for Laminar Boundary Layer Flow Over a Flat Plate (Blasius’ Solution) Falkner-Skan Solutions Approximate Methods: The Kármán Momentum Integral Equation Approximate Methods: Thwaites’ Correlation Turbulent Boundary Layer over a Flat Plate Boundary Layer Separation Forces on Objects Immersed in a Fluid Flow C. Wassgren Chapter 08: Boundary Layers 349 Last Updated: 14 Aug 2010 1. Boundary Layer Introduction Boundary layers are the regions near a boundary in which rotational effects are significant. The various flow field regions are indicated in the figure below. outer, potential flow (irrotational flow, negligible viscous effects) typical streamline wake region (viscous forces negligible since velocity gradients are small, vorticity is convected in from the boundary layer) boundary layer thickness inner, boundary layer flow (strong viscous effects, no-slip condition velocity gradients0 vorticity0) C. Wassgren Chapter 08: Boundary Layers 350 boundary layer separation point Last Updated: 14 Aug 2010 Boundary Layer Thickness Definitions Before continuing further, we should define what we mean by the “thickness” of a boundary layer. There are three commonly used definitions: 1. (99%) boundary layer thickness, : This thickness definition is the most commonly used definition. The boundary layer thickness, , is defined as the distance from the boundary at which the fluid velocity, u, is 99% that of the outer velocity, U: u y 0.99U (8.1) y U u=0.99U u 2. displacement thickness, D or *: The displacement thickness, D, is the distance at which the undisturbed outer flow is displaced from the boundary by a stagnant layer of fluid that removes the same mass flow as the actual boundary layer profile. y U y mass flow deficit U effective displaced boundary y U u dy y 0 D u mass flow deficit U D u U u dy U D 0 0 D 1 u u dy 1 dy U U 0 C. Wassgren Chapter 08: Boundary Layers (8.2) 351 Last Updated: 14 Aug 2010 Example: Consider the mass flow rate between two parallel plates in which a boundary layer has formed: free stream velocity, U H/2 H/2 boundary layer velocity, u(y) core velocity, U’ Let’s just consider the lower half of the flow since the upper and lower halves are symmetric. The mass flow rate in the lower half is: H udy m1 2 2 U dy mass flow rate in BL mass flow rate in outer flow 0 Note that the uniform, outer flow velocity at this cross-section is U’ which is larger than U (to conserve mass). The second integral can be re-written as: H H 2 U dy 2 U dy U dy 0 0 so that the mass flow rate is: H m1 2 udy 0 H 0 2 0 2 U dy U dy u U dy U dy 0 0 u U 1 dy U H 2 U 0 D m1 2 H 2 D U Thus, if we replace the real velocity profile by a uniform velocity profile at that cross-section (which has velocity U‘), we must decrease the effective cross-sectional area (per unit depth) of each half of the pipe at that particular location by the displacement thickness, D, to maintain the same mass flow rate. Note that we would need to have another relation for D in order to solve for U’ and visa-versa. C. Wassgren Chapter 08: Boundary Layers 352 Last Updated: 14 Aug 2010 3. momentum thickness, M or : The momentum thickness, M, is the thickness of a stagnant layer that has the same momentum deficit, relative to the outer flow, as the actual boundary layer profile. The concept is similar to that for the displacement thickness except instead of a mass deficit, the momentum thickness considers the momentum deficit. y U y momentum deficit y U effective displaced boundary u U u dy y 0 M u u 2 u U u dy U M 0 momentum deficit U 2 M u u u u 1 dy 1 dy U U 0U 0U M (8.3) Example: Consider the boundary layer flow over a flat plate. We’ll assume a steady flow with the pressure everywhere equal. top of CV follows a streamline so that there is y no flow out from the top of the CV U U h u x D x Let’s apply the LMEs in the x-direction to the CV shown to determine the drag acting on the fluid (or the plate) over the distance x. d D u x dV u x u rel dA D u 2 dy U 2 h dt CV CS 0 where the height, h, is found via COM on the same CV to be: u Uh udy h dy U 0 0 so that the drag is: u u u D u 2 dy U 2 dy D U 2 1 dy U U U 0 0 0 M D U M 2 (8.4) Thus we see that the drag is related to the momentum thickness. Note that this particular example is for a situation with no pressure gradients (the pressure is a constant). The drag expression will be different if the pressure varies for the flow but the resulting drag expression will still be in terms of the momentum thickness. Notes: 1. Usually: D M C. Wassgren Chapter 08: Boundary Layers 353 Last Updated: 14 Aug 2010 2. Boundary Layer Equations To determine the boundary layer velocity profile for a given flow, we need to go back to the governing equations of fluid motion. Let’s consider steady, incompressible, laminar flow over a sharp-edged flat plate as shown in the figure below. y U u x x In the following analysis, we’ll assume that the boundary layer thickness, , is much smaller than the distance from the leading edge of the plate, x: x 1 To help simplify the governing equations, let’s make some estimates of the magnitudes of some of the other parameters: u ~U x~x y~ From continuity: u v v u U U ~ ~ 0 v~ x x y y x x We can use these estimates in the momentum equations (Navier-Stokes equations) to determine if any terms are small in comparison to the other terms in the equations (note that body forces are neglected in the following equations since they are typically very small in comparison to the other terms). 2u 2u 1 p u u x-dir: u v 2 2 x y y x x y -dir: u 2v 2v 1 p v v v 2 2 x y y y x x-dir: U2 U2 1 p U U x x x x 2 2 y -dir: U2 U2 1 p U U x x x x y x 2 x 2 x Since we’re assuming that <<x, we see that the second viscous term in the x-momentum equation is much greater than the first viscous term: U U 2 2 x so that the first viscous term can be neglected. Also, since the fluid in the boundary layer is being accelerated in the x-direction despite strong viscous forces, we’ll assume that the inertial terms are of the same order as the viscous term: C. Wassgren Chapter 08: Boundary Layers 354 Last Updated: 14 Aug 2010 x U 2 U ~ 2 ~ U x Thus, we expect the boundary layer thickness to scale with Furthermore, since we assumed that <<x, we have: x for a laminar flow over a flat plate. 1 Re x 1 (8.5) x Ux The assumption that <<x is the same as saying that the Reynolds number based on the x-position must be much greater than one. 1 Now let’s examine the magnitude of the pressure gradient in the x-direction to the magnitude of the pressure gradient in the y-direction. From the x-momentum equation we see that: p U 2 U ~ ~2 x x From the y-momentum equation we see that: p U 2 U ~ ~ y x x 2 x Since <<x we see that: p p x y so that the pressure remains essentially constant in the y-direction in comparison to how the pressure changes in the x-direction, i.e. p=p(x). Thus, we can use the pressure in the outer, potential flow, determined using Bernoulli’s equation, to determine how p varies with x (since the pressure is essentially constant in the y-direction): p U p 1 2 U 2 constant U x x Substituting this relation into the x-momentum equation and simplifying gives the following: u u U 2u v U 2 x y x y u v continuity: 0 x y boundary conditions: momentum: u u x, y 0 0 v x, y 0 0 (no flow through boundary) u x, y U assuming: (no-slip at boundary) (inner flow matches outer flow far from wall) Re x Ux 1 These are known as the laminar boundary layer equations. Note that U is the outer flow velocity, i.e. the velocity just outside the boundary layer. It is not necessarily the upstream velocity, U∞. C. Wassgren Chapter 08: Boundary Layers 355 Last Updated: 14 Aug 2010 3. Exact Solution to Laminar Boundary Layer Flow over a Flat Plate with no Pressure Gradient (aka the Blasius Solution) Let’s consider steady, incompressible, laminar flow over a sharp-edged flat plate as shown in the figure below. y U u x x From the previous notes, the boundary layer equations are: u u U 2u v U 2 x y x y u v continuity: 0 x y boundary conditions: momentum: u u x, y 0 0 v x, y 0 0 (no flow through boundary) u x, y U assuming: (no-slip at boundary) (inner flow matches outer flow far from wall) Re x Ux 1 For flow over a flat plate, the outer potential flow velocity, U, will remain constant if the displacement thickness remains small (Rex>>1 so that the outer flow is not perturbed much): U p U constant 0 0 x x so that the boundary layer equations simplify to: u u 2u momentum: u v 2 x y y u v continuity: 0 x y boundary conditions: u x, y 0 0 v x, y 0 0 (no flow through boundary) u x, y U assuming: (no-slip at boundary) (inner flow matches outer flow far from wall) Re x Ux 1 Notice that there is no characteristic length scale in the x-direction so that we might expect that the flow profiles will have similar shape (but scaled in magnitude) as we move downstream. This being the case, let’s look for a solution to the boundary layer equations of the form: u y f U C. Wassgren 356 Last Updated: 14 Aug 2010 Chapter 08: Boundary Layers where the boundary layer thickness, , for this laminar flow will scale with (refer to the previous notes): x ~ U Thus, let’s try looking for a similarity solution to the original PDEs using the similarity variable, , defined as: u U f where y U 2 x Note that a factor of “2” has been included in the similarity variable merely for convenience. It will make the resulting differential equation a little easier to work with. One other simplification we can make since we are investigating a planar flow is to write the velocity components u and v in terms of a stream function: d u Uf y d y d y U f d 2 x f d constant U Since the constant is arbitrary (we really only care about the velocities so it doesn’t matter what the constant is), set it equal to zero. The resulting stream function becomes: 2 UxF = U (8.6) where F f . The velocities are found from the stream function: U 2νUxF 2νUxF 2 x y y u UF u v 1 x 2 v 2νU 1 F 2νUxF x x 2 (8.7) 2νU 1 U F 2νUxF y x 2 2 x3 νU F F 2x (8.8) Note that f = F ’. Also, UF y U U u UF F 3 2 2x x x 2 x U u UF UF 2 x y y 2u y 2 2 UF U UF 2 x 2 y 2 U v U F F F y 2 x y 2 x C. Wassgren Chapter 08: Boundary Layers 357 Last Updated: 14 Aug 2010 The continuity equation is automatically satisfied after substitution (as expected since a stream function has been used) and the original momentum PDE becomes: U U U U UF ' F F F UF UF 2x 2 x 2 x 2x F FF 0 and the boundary conditions become: u x, y 0 0 F 0 0 v x, y 0 0 F 0 0 u x, y U (8.9) F 1 (8.10) In summarizing these results, we see that the original boundary layer PDEs for laminar flow over a flat plate with no pressure gradient can be simplified to a non-linear ODE by using a similarity variable. The resulting equations and boundary conditions become: F FF 0 F 0 0 F 0 0 F 1 Blasius Equation for Boundary Layer Flow over a Flat Plate u UF ' v U y 2x F F U 2 x C. Wassgren Chapter 08: Boundary Layers 358 Last Updated: 14 Aug 2010 Notes: 1. There is no known closed-form solution to this ODE so we resort to solving it numerically (using a Runge-Kutta method for example). A plot of the solution looks like (plot from Panton, R.L., Incompressible Flow, 2nd ed., Wiley): u U y U x The analytical results found here match experimental data very well as shown in the figure. Note that the similarity variable in the plot above does not include the square root of two term, i.e.: plot 2 what's used in these notes 2. The boundary layer thickness, , is found from the numerical solution to occur at =3.5: u F ' 3.5 3.5 0.99 U The boundary layer thickness is then: U 5.0 3.5 1 x Re x2 2 x 3. The displacement and momentum thicknesses can also be found numerically using their definitions: D 1.72 M 0.664 and 1 1 2 x x Re x Re x2 4. The shear stress at the plate surface in dimensionless form is known as the friction coefficient, cf: d u U U d y 2 F yx y 0 0.664 y 0 0 cf 1 1 U 2 1 U 2 Re x Re x2 2 2 Note that the friction coefficient is a ratio of the shear stress to the dynamic pressure in the flow. C. Wassgren Chapter 08: Boundary Layers 359 Last Updated: 14 Aug 2010 5. The drag coefficient, cD, defined as the dimensionless drag acting on the plate between x=0 and x=L, is found by integrating the shear force over the plate area: xL cD x 0 yx y 0 dx 1.328 (where cD is the drag coefficient per unit depth) 1 Re L2 Note that although the boundary layer assumptions break down near the leading edge of the plate (Rex is not >> 1), the distance over which this is the case is small in comparison to the typical lengths of interest. This discrepancy is generally neglected in engineering applications. 6. 1 2 U 2 L This solution is only valid for laminar boundary layers. As a rule of thumb, the transition from laminar flow to turbulent flow occurs at: Rex 500,000. 7. Recall that he boundary layer equations on which the Blasius solution is based are valid only when Rex >> 1. In practice, it has been found that the Blasius solution is accurate when ReL > 1000. For 1 ≤ ReL ≤ 1000, the following relation developed by Imai (1957) is more appropriate: 1.328 2.3 cD 1 Re L2 Re L 8. In summary, for flow over a flat plate with no pressure gradient, the “exact” Blasius solution gives: D 1.72 M 0.664 5.0 1 1 1 x Re x2 x x Re x2 Re x2 cf 0.664 1 Re x 2 cD 1.328 1 Re L2 Re x 106 C. Wassgren Chapter 08: Boundary Layers 360 Last Updated: 14 Aug 2010 Example: A thin flat plate 55 by 110 cm is immersed in a 6 m/s stream of SAE 10 oil at 20 C. Compute the total skin friction drag if the stream is parallel to (a) the long side and (b) the short side. SOLUTION: U = 6 m/s W SAE 10 oil = 1.20*10-4 m2/s SAE 10 oil = 870 kg/m3 L Determine the Reynolds number at the trailing edge of the plate to see if it’s laminar. UL Re L (The flow is considered laminar if Re < 500,000.) (8.11) When L = 1.10 m then ReL = 55,200 laminar flow. When L = 0.55 m then ReL = 27,500 laminar flow. Now determine the drag on the plate using the drag coefficient, cD, for laminar flat plate flow (the Blasius solution). D 1 U 2 2 LW cD 2 top and bottom faces 1.328 D 1 U 2 2 LW 2 Re 1 2 L (8.12) When L = 1.10 m, W = 0.55 m, ReL = 55,200, and D = 107 N. When L = 0.55 m, W = 1.10 m, ReL = 27,500, and D = 152 N. Note that the drag is greater when the short side is aligned with the flow. Why? Because from Eqn. (8.12) we observe that the drag varies with L but is proportional to W. Hence the drag will increase more rapidly with increasing width than with increasing length. C. Wassgren Chapter 08: Boundary Layers 361 Last Updated: 14 Aug 2010 4. Falkner-Skan Boundary Layer Solutions Recall that a similarity approach was used to find solutions to stagnation point flow (Hiemenz Flow) and boundary layer flow over a flat plate (Blasius solution). Falkner and Skan (1931) investigated what other flows can be solved using similarity solutions. In their analysis they assumed that: (8.13) u x, y U x f where y (8.14) x Here, u is the velocity within the boundary layer, U is the outer flow (potential flow) velocity, is the similarity variable, and is a distance scaling function. Note that at this point, U and are unknown functions of x. It’s convenient to work in terms of a stream function rather than velocity components so let’s re-write the velocity in terms of a stream function, : y U x f U x f u d constant y U x x f (8.15) Note that the constant in the previous equation has been set to zero since only differences (or derivatives) in the stream function have any significance. Note also that the stream function automatically satisfies the continuity equation. The stream function must also satisfy the boundary layer momentum equation: u u U 2u u v U 2 x y x y (8.16) 2 2 U 3 U 3 y xy x y 2 x y where d y d dU d d dU f U f U f 2 f U f U f dx dx dx dx x dx dx 1 U f Uf y dU y d dU U d 2 f f U 2 f f dx dx dx xy dx 2 y 2 3 y 3 (8.17) Uf Uf 2 C. Wassgren Chapter 08: Boundary Layers 362 Last Updated: 14 Aug 2010 Substituting and simplifying… dU dU U d d d dU Uf Uf 2 f U f f U f Uf f U dx dx dx dx dx dx U dU U 2 d dU U 2 d U 2 d dU Uf f f U ff ff f f U 2 f 2 dx dx dx dx dx dx U Ud U ff dx dU Ud dU Uf f 2 U ff U 2 dx dx dx (8.18) Multiply through by 2/(U) and re-arrange to get: d 2 dU 2 f 1 f 0 U ff dx dx (8.19) If a similarity solution exists, Eqn. (8.19) should be an ODE for the function f in terms of . Thus, the coefficients in front of the second and third terms should be, at most, constants, i.e. for a similarity solution to exist, we must have: d 2 dU U and dx dx where and are constants. (8.20) Note that these two equations can be combined to form an expression that is more convenient to work with later in our analysis: d d dU U 2 2U 2 dx dx dx dU 2 dU d 2 U dx dx dx d dU 2 U 2 dx dx 2 d U 2 2 dx (8.21) Substituting Eqn. (8.20) into Eqn. (8.19) gives: 2 f ff 1 f 0 (8.22) The boundary conditions for this ODE are: u x, y 0 0 f 0 0 v x, y 0 0 f 0 0 u x, y U f 1 C. Wassgren Chapter 08: Boundary Layers (8.23) 363 Last Updated: 14 Aug 2010 The procedure for determining the exact solutions to the boundary layer equations can be found via the following procedure: a. Select values for the constants and . Note that at this point we don’t yet know what geometry we are investigating. The geometry will be determined in the next step. b. Determine the corresponding form for U(x) and (x) using Eqn. (8.20), or, more conveniently, part of Eqn. (8.20) and Eqn. (8.21): d dU U 2 2 and 2 (8.24) dx dx c. Determine the function f() from the ODE given in equation (8.22) subject to the boundary conditions specified in equation (8.23). This part is usually solved numerically. 2 (8.25) f ff 1 f 0 f 0 0 f 0 0 (8.26) f 1 d. Determine the stream function from Eqn. (8.15) and velocity components from the stream function. The wall shear stress may also be determined. U x x f (8.27) u Uf y (8.28) dU d d f U f f U x dx dx dx U f 0 u 2 2 w y y 0 y y 0 v (8.29) (Refer to Eqn. (8.17).) (8.30) Notes: 1. Recall that U(x) is the flow profile for the outer, potential flow and (x) is a distance scaling parameter. 2. Consider the case when =1 and is left arbitrary. From Eqn. (8.24) we have: d U 2 2 dx U 2 x 2 constant The scaling factor will be zero at x=0 so the constant will also be zero. Now divide the equation by the other equation given in Eqn. (8.24): 1 1 U 2 x 2 dU 2 dx 1 dU U dx x 2 ln U 2 U x cx ln x constant 2 C. Wassgren Chapter 08: Boundary Layers (8.31) 364 Last Updated: 14 Aug 2010 This outer flow velocity distribution has the same form as the potential flow over a wedge (refer back to an earlier set of notes). Recall that the complex potential for flow over a wedge is given by: f z Az n (8.32) df n 1 U iV Anz n 1 An x iy dz where A is a constant and the angle between the walls of the wedge is equal to /n as shown below. Note that this f (the complex potential) is not the same as the f in Eqn. (8.22). y /n x line of symmetry Making use of symmetry we can produce flow over a wedge shape with a wedge angle denoted by, : y /n wedge shape x line of symmetry wedge angle, Along the surface of the wedge, y=0, the potential flow horizontal velocity is: U Anx n 1 Comparing Eqn. (8.33) to Eqn. (8.31) we find that and n are related by: 2 n 1 n 2 The full angle of the wedge, , will be: 1 2 - 2 1 n n (8.33) (8.34) (8.35) The scaling function,(x), is found using Eqns. (8.24) and (8.31): dU dx 2 x w 2 U f 0 c 2 1 x 2 2 c (8.36) 1 x 2 cx 2 2 c C. Wassgren Chapter 08: Boundary Layers x 1 2 f 0 365 c3 22 1 x f 0 2 (8.37) Last Updated: 14 Aug 2010 Now let’s consider two special cases for =1. Flow over a Flat Plate (Blasius flow): n=1 (=0, =1): The outer flow velocity is given by equation (8.31): U x c U0 (8.38) where U0 is a constant velocity. The scaling function is found from Eqn. (8.36) to be: 2 x x U0 This is precisely the same scaling function found in our earlier investigation of the Blasius solution. The governing ODE is found from Eqn. (8.25): f ff 0 Again, this is identical to what was found during our investigation of the Blasius solution. (8.39) (8.40) Stagnation Point Flow (Heimenz flow): n=2 (=1, =1): The stagnation point flow solution can be recovered using =1 and =1. Using these parameters we find: U x cx x (8.41) c f ff 1 f 0 which are the same results we found in our previous analysis of stagnation point flow. 2 C. Wassgren Chapter 08: Boundary Layers 366 Last Updated: 14 Aug 2010 Notes: a. The figure below shows the velocity profile for various flows over wedge shapes (produced by letting =1 and varying .) (Figure 20.4 from Panton, R.L., Incompressible Flow, 2nd ed, Wiley. Note that in the figure m = n-1= /(2 - ). n>1 (accelerating flow) decelerating flows Blasius soln n<1 (decelerating flow) accelerating flows One item of significance observed from the figure is the fact that for accelerated flows (n>1 dU/dx > 0) there is no inflection point in the boundary layer profile. When the flow decelerates (n<1 dU/dx < 0) there is an inflection point. Of particular interest is the case when (n-1) = -0.0904. For this case we observe that the inflection point occurs at the wall and, as a result, the corresponding wall shear stress is zero. It is at this point when boundary layer separation occurs. Thus, we see that a laminar boundary layer is able to support only a very small deceleration without separation occurring. We will address the topic of boundary layer separation later in the notes. /n n = 1 (= 0) C. Wassgren Chapter 08: Boundary Layers n = 2 (= 1) 367 n = 2/3 (= -1) Last Updated: 14 Aug 2010 Example: A uniform stream of incompressible fluid flows over a planar wedge of half-angle, (/2), side length, L, and base length, H, as shown in the figure below. The upstream flow velocity and pressure are U and p. L U, p a. b. c. (/2) H Use a Faulkner-Skan boundary layer solution to determine the skin friction drag acting on the wedge assuming laminar flow. Check your solution to part (a) by calculating the drag for a flat plate (= 0) and comparing with the Blasius solution. Determine the form drag on the wedge assuming that the pressure at the back of the wedge is the same as the free stream pressure. SOLUTION: Recall that the Faulkner-Skan boundary layer solution for flow over a wedge with total angle, (/2), is: 2 f ff 1 f 0 (8.42) x y L (8.43) f 0 0 f 0 0 U, p (/2) f 1 U x cx 2 c 1 x 2 (8.46) y (8.47) x w (8.44) (8.45) 2 x H c3 22 1 x f 0 2 (8.48) The skin friction drag over both surfaces of length, L, (assuming unit depth into the page), and also taking into account the angle of the surfaces, is: xL Dskin 2 cos friction Dskin 2 cos friction 2 w dx 2 x 0 xL (8.49) c3 22 1 c3 x f 0 dx f 0 2 2 x 0 1 2 1 1 L c3 f 0 2 1 x 2 2 1 0 2 Dskin friction 2 cos 2 x 0 2 1 x 2 dx (8.50) 3 2 1 c 2 f 0 L 2 1 C. Wassgren Chapter 08: Boundary Layers xL 368 (8.51) Last Updated: 14 Aug 2010 Note that when = 0, we get Blasius flow over a flat plate. Dskin c3 2 2 friction f 0 2 L 2 1 (8.52) U x c (from Eqn. (8.45)) (8.53) Dskin friction 8U 3 L f 0 32 1 2 2 U L UL Blasius f 0 (8.54) To determine f 0 , we need to solve the ODE: f ff 0 (8.55) numerically. Performing this calculation (or using Table 4.1 from White, Viscous Fluid Flow, for example) gives: f 0 0.46960 (8.56) and thus: 2.656 Dskin friction 1 U 2 L 2 Re 1 2 Blasius L (8.57) The Blasius drag coefficient (considering two sides of the plate) is: Dskin fraction, 1.328 2.656 two-sides 2 Dskin fraction, 1 U 2 L CD ,Blasius, 2 1 U 2 L Re 1 2 Re 1 2 two-sides two-sides L L 2 This is the same as Eqn. (8.57)! (8.58) The form drag may be found by integrating the pressure force acting on the surface and taking the component in the direction of the incoming flow. Use gage pressures to simplify the calculation. The pressure may be found using Bernoulli’s equation in the outer, potential flow. Dform 2sin 2 xL 2 pgage dx where pgage 1 U U 2 2 (8.59) x 0 (Note that since a gage pressure is being used, the pressure force on the back of the wedge needn’t be considered.) xL U2 2 1 U 1 2 dx Dform 2sin 2 2 U x 0 xL U2 1 2 U x 0 dx sin 2 1 2 2 sin 2 sin c 2 2 U L U 2 Dform sin 2 U 2 1 2 1 L2 2 C. Wassgren Chapter 08: Boundary Layers x L 2 c 2 x 2 1 dx 2 U x 0 (8.60) c 2 2 2 U L 1 U 2 2 L (8.61) 369 Last Updated: 14 Aug 2010 2 2 2 U 5. Approximate Methods: The Kármán Momentum Integral Equation So far we’ve only examined boundary layer flows that lend themselves to similarity solutions. This, of course, is very restrictive. There are many non-similar boundary layer flows that we would also like to investigate. Since the majority of fluid mechanics problems are extremely complex, we often have to resort to empirical or semi-empirical methods for investigating the flows in greater detail. Here we’ll discuss one such semi-empirical method used for investigating boundary layers called the Kármán Momentum Integral Equation (KMIE). The idea is straightforward and relies on the Linear Momentum Equation. Consider a differential control volume as shown in the figure below. The top of the control volume is defined by the line separating the boundary layer region from the outer flow region (this is not a streamline). dp 1 p dx 2 dx d U U U p p d p dx dx +d w dx dx Let’s apply the linear momentum equation in the x-direction to the control volume: Fx,on CV d dt u dV u u x x CV rel dA (8.62) CS where d dp 1 Fx,on CV p p dx p p 2 dx d dx (assuming unit depth) w dx dx left bottom right d dt CS u dV 0 x top (steady flow) CV y y d d u 2 dy dx U udy dx dx dx y 0 y 0 y 0 y 0 u x u rel dA y y u 2 dy left C. Wassgren Chapter 08: Boundary Layers u 2 dy right 370 top Last Updated: 14 Aug 2010 Note that the mass flow rate through the top is found via conservation of mass on the same control volume: d mtop udy udy udy dx 0 dx 0 0 in through top 0 in at left (8.63) out through right mtop d udy dx dx 0 Substituting and simplifying (and neglecting higher order terms): dp d d u 2 dy dx U udy dx dx w dx dx dx dx 0 0 (8.64) dp d d u 2 dy U udy w dx dx dx 0 0 Recall that the pressure at a given x location remains constant with y position so that we can find dp/dx in terms of the outer (potential) flow velocity using Bernoulli’s equation outside of the boundary layer: dp dU p 1 2 U 2 constant U 0 dx dx (8.65) dp dU U dx dx In addition, we’ll re-write the boundary layer thickness in terms of an integral so that: dp dU dU U dy dx dx dx 0 Udy (8.66) 0 Additional re-arranging gives: d dU d udy uUdy udy U dx dx 0 dx 0 0 Substituting Eqns. (8.66) and (8.67) into Eqn. (8.64) gives: d dU dU d 2 udy Udy w u dy uUdy dx dx dx 0 dx 0 0 0 Additional re-arranging and simplifying gives: dU d w u U u dy U u dy dx dx 0 0 d 2 u u dU u U U 1 dy 1 dy dx U U dx U 0 0 M D (8.67) (8.68) Thus, if the fluid has constant density: w d 2 dU U M DU dx dx This is known as the Kármán Momentum Integral Equation (KMIE). C. Wassgren Chapter 08: Boundary Layers (8.69) 371 (8.70) Last Updated: 14 Aug 2010 Notes: 1. 2. If the pressure remains constant, then dU/dx=0 and d w U 2 M dx (8.71) The typical methodology for using the KMIE is as follows: a. Obtain an approximate expression for U=U(x) from inviscid flow theory (e.g. potential flow theory). Recall that Bernoulli’s equation can be used to relate the pressure and U. b. Assume a velocity profile in the boundary layer subject to the appropriate boundary conditions, i.e. assume a form for: u y (8.72) f U subject to the boundary conditions: uy uy (8.73) 0 0 and 1 1 U U The form of the approximate velocity profile is typically found based on curve fits to experimental measurements of the boundary layer velocity profile. Note that higher order profiles will have additional boundary conditions. For example, a cubic curve fit will also have a boundary condition that matches the slope of the velocity profile at the freestream boundary. c. The shear stress at the wall for a laminar flow can also be determined from the Newtonian stressstrain rate constitutive relations to be: u d U U (8.74) w d y y 0 (For a turbulent flow experimental data for the wall shear stress are used. This will be discussed in greater detail later in these notes.) This laminar wall shear stress must be the same shear stress as that found using the KMIE (Eqn. (8.70)). Thus, we can equate the two shear stress expressions. The resulting differential equation can then be solved for the boundary layer thickness, , as a function of x. C. Wassgren Chapter 08: Boundary Layers 372 Last Updated: 14 Aug 2010 Example: Consider laminar flow over a flat plate (U=constant). Approximate the boundary layer velocity distribution using a parabolic profile: y y 2 y for 0 1 2 u (8.75) U y 1 1 Evaluate the momentum thickness, M: 1 1 u u 2 2 1 U d 2 1 2 d U 0 0 M (8.76) 2 M 15 where in the above equation the substitution =y/ has been used. Now substitute this momentum thickness into Eqn. (8.70). Note that dU/dx=0 so that the second term on the right hand side in Eqn. (8.70) is zero. w d 2 dU U M DU dx dx 0 d d 2 w U 2 M U 2 dx dx 15 This should be the same shear stress found using Eqn. (8.74): Ud U 2 w 2 2 d 0 (8.77) (8.78) so that by equating the two we have: U 2 d 2 U 2 15 dx x d 0 15 U dx 0 15 12 x U 2 x x (8.79) 30 5.5 Ux Ux 5.5 1 Re x2 Recall that the exact (Blasius) solution for flow over a flat plate gave: 5.0 1 x Re x2 (8.80) so that the approximate solution is off by only 10%! C. Wassgren Chapter 08: Boundary Layers 373 Last Updated: 14 Aug 2010 Example: A measured dimensionless laminar boundary layer profile for flow past a flat plate is given in the table below. Use the momentum integral equation to determine the 99% boundary layer thickness. Compare your result with the exact (Blasius) result. y/ 0.00 0.08 0.16 0.24 0.32 0.40 0.48 0.56 0.64 0.72 0.80 0.88 0.96 1.00 u/U 0.00 0.133 0.265 0.394 0.517 0.630 0.729 0.811 0.876 0.923 0.956 0.976 0.988 1.000 SOLUTION: Apply the Kármán Momentum Integral Equation: w d dU M U 2 DU dx dx Assuming a flat plate flow with no pressure gradient: dU U constant 0 dx Simplifying Eqn. (8.81) gives: d w U 2 M dx (8.81) (8.82) (8.83) The momentum thickness is given by: y M u 0 U y u 1 U y 1 u dy U y 0 u 1 U y d Integrating the data numerically using the trapezoidal rule gives: M 0.131 Substitute into Eqn. (8.83). d w 0.131U 2 dx C. Wassgren Chapter 08: Boundary Layers (8.84) (8.85) 374 Last Updated: 14 Aug 2010 For a laminar flow, the shear stress can also be expressed as: w du dy y 0 u Ud U dy (8.86) y 0 Differentiating the data numerically using a 1st order finite difference scheme: U w 1.66 (8.87) Equating Eqns. (8.85) and (8.87) gives: d U 0.131U 2 1.66 dx 0 x x x U x 12 12.67 2 U dx d 12.67 5.034 x0 5.034 Ux Re x 1 (8.88) 2 Equation (8.88) is within 1% of the exact Blasius solution of x 5.0 / Re x2 . 1 Another approach to this problem is to fit a polynomial curve to the given data rather than numerically differentiating and integrating the data. 3. This approximate technique can be used for either laminar or turbulent flows. In fact, this method is especially useful for analyzing turbulent boundary layer profiles (to be discussed in later notes). C. Wassgren Chapter 08: Boundary Layers 375 Last Updated: 14 Aug 2010 6. Approximate Methods: Thwaites’ Correlation Thwaites’ method (1949) is generally considered the best available one parameter method for describing laminar boundary layers. The correlation uses the Kármán Momentum Integral Equation along with dimensionless experimental laminar boundary layer data. Recall from the previous notes that the momentum thickness, M, the displacement thickness, D, and the shear stress at the wall, w, can be related in an average sense via the Kármán Momentum Integral Equation (KMIE): W d dU (8.89) U 2 M DU dx dx (Note again that U is the outer velocity, i.e. the flow velocity just outside the boundary layer.) We can rearrange this equation to the following form: W d dU dU 2 M U U 2 M DU dx dx dx (8.90) d dU 2 M D U U 2 M dx dx Let’s write the KMIE using two dimensionless shape factors, H and T, defined in the following manner (following the approach of Holstein and Bohlen, 1940): H D M and T W W M U (8.91) U M where H is the shape correlation and T is the shear correlation. These shape factors only depend on the shape of the velocity profile. Re-writing the KMIE using these shape factors gives: d UT dU U 2 M 2 H M U (8.92) M dx dx Multiplying through by M/(U) and re-writing the dM/dx term gives: d UT dU U 2 M 2 H M U M dx dx T U M d M 2 dU 2 H M dx dx 2 2 U d M M dU T 2 H 2 dx dx 2 2 d M dU M U (8.93) 2 T 2 H dx dx After plotting several known velocity profiles, researchers found that H and T depend almost entirely on another dimensionless quantity, , where: 2 dU M (8.94) dx so that the shape of the boundary layer velocity profile will be determined entirely by . Substituting Eqn. (8.94) into Eqn. (8.93) gives: 2 2 d M dU M U 2 T 2 H dx dx d U (8.95) F 2 T 2 H dx U 0 C. Wassgren Chapter 08: Boundary Layers 376 Last Updated: 14 Aug 2010 Let’s plot F() as a function of for a number of known profiles. (Figure scanned from Figure 4-22, White, F.M., Viscous Fluid Flow, 2nd ed., McGraw-Hill. Note that in the figure below is the momentum thickness, M.) = 2[T-(2+H)] Thwaites used a simple linear curve fit to the data: 2 d M U F 0.45 6.0 dx 2 dU M 0.45 6.0 dx 0.45 U 6.0 2 d M dx 1d 2 5 MU 6 U dx 2 M x dU dx 2 M U 6 0.45 U 5 dx constant M x 0 0 constant 0 (8.96) 0 so that the momentum thickness is related to the outer flow velocity by: 2 M 0.45 U6 x U dx 5 (8.97) 0 C. Wassgren Chapter 08: Boundary Layers 377 Last Updated: 14 Aug 2010 Notes: 1. The functions H() and T() are found from plots of several known velocity profiles. The figure below shows Thwaites’ correlation data for the shape factors. The following two curve fits to the data are given by White, F.M., Viscous Fluid Flow, 2nd ed., McGraw-Hill. T 0.09 0.62 H 2.0 4.14 z 83.5 z 2 854 z 3 3337 z 4 4576 z 5 (8.98) where z 0.25 4.0 0.6 3.5 0.5 3.0 ( ) HH() 2.0 0.3 1.5 T(() ) T 0.4 2.5 0.2 1.0 H(lambda) T(lambda) 0.5 0.0 -0.15 -0.10 -0.05 0.1 0.0 0.00 = 0.05 0.10 0.15 0.20 0.25 0.30 (2M )(dU//dx ) M/ / )(dU dx) 2. The typical procedure for using Thwaites’ method is as follows: a. Determine U=U(x) for the outer potential flow. b. Determine M using Eqn. (8.97). c. Determine using Eqn. (8.94). d. Determine W and D using Eqn. (8.91) and the curve fits or figure shown in the first note. 3. This method is considered one of the best methods for predicting the behavior of laminar boundary layers. It is accurate to about 5% for favorable or mild adverse pressure gradients and is accurate to about 15% near the separation point. When more accurate calculations are necessary, one typically turns to numerical methods for solving the boundary layer equations. 4. Recall that the boundary layer separation point occurs when w=0 T()=0 so that at separation point 0.090 C. Wassgren Chapter 08: Boundary Layers 378 (8.99) Last Updated: 14 Aug 2010 5. Thwaites’ method, as presented here, is restricted to laminar, planar flows. A similar type of method can also be derived for laminar, axi-symmetric flows. 6. Note that the outer flow velocity, U, may in fact be significantly different than the expected potential flow velocity profile. For example, for flow around a cylinder (or any bluff body), boundary layer separation results in a large wake region. Thus, the potential flow prediction (using a uniform stream and a doublet to model flow around a cylinder) for the outer flow velocity may be greatly in error. Often we must resort to experimental data to obtain the outer flow velocity profile. C. Wassgren Chapter 08: Boundary Layers 379 Last Updated: 14 Aug 2010 Example: Consider the simple decelerating non-similar outer flow described by x U U 0 1 L Using Thwaites’ method, determine the point at which flow separation occurs. (8.100) SOLUTION: Using Thwaites’ correlation we have: 2 M x 0.45 5 U0 1 x L 0.075 1 x L U 6 U0 1 x 6 0 L L 5 dx (8.101) 1 0 so that the dimensionless parameter, , is: 6 2 dU M 0.075 1 x 1 L dx Separation will occur when =-0.090: 6 0.090 0.075 1 x 1 L xsep 0.123 L This value is within 3% of the “exact” result xsep/L=0.120 which is found numerically. 6 C. Wassgren Chapter 08: Boundary Layers 380 (8.102) (8.103) Last Updated: 14 Aug 2010 Example: Consider the boundary layer flow resulting from a sink located at the trailing edge of a thin, flat plate as shown in the figure. y sink strength, m x a a. b. Using Thwaites’ method, determine and plot the dimensionless momentum thickness, M/(a2/m)1/2 , over the top surface of the plate as a function of dimensionless distance along the plate, x/a. Will boundary layer separation occur on the plate? If so, determine the location of the separation point. If not, explain why. [Hint: No calculations are necessary for this part.] SOLUTION: Determine the outer flow velocity profile by modeling the outer, potential flow as a single line sink located at x = a. m f z log z a (where m > 0) (8.104) 2 The fluid velocity is given by: z a z a x a iy df m 1 m m m u x iu y 2 2 2 x y 2 2 xa a 2 dz 2 z a 2 z a z a 2 z za za a 2 u x iu y x a m m y i 2 2 2 x a y 2 x a 2 y 2 (8.105) On the plate surface, y = 0 so that: m 1 and u y 0 (Note that 0 x a on the plate surface.) ux 2 x a (8.106) Use Thwaites’ method to determine the momentum thickness, M: 2 M 0.45 U6 x x x x 0.45 U dx m 5 x 0 1 2 x a 2 0.45 x a 6 x x x a m 6 5 x 0 5 m 1 dx 2 x a dx x 0 2 0.45 x a 6 4 x a a 4 4m 2 6 2 0.45 a x x 1 1 2m a a M 0.45 a 2 2m x 2 x 6 1 1 a a C. Wassgren Chapter 08: Boundary Layers 381 Last Updated: 14 Aug 2010 M a2 m x 2 x 6 0.225 1 1 a a (8.107) Boundary layer separation will not occur on the plate since there is a favorable pressure gradient along the surface. This can be shown by examining Bernoulli’s equation in the outer flow. p U p 1 U 2 constant U 0 2 x x p U U (8.108) x x The velocity on the plate surface is given by Eqn. (8.106) so that: p m x 2 2 1 (8.109) x a 3 Since 0 x a, p/dx < 0 (a favorable pressure gradient). Another approach to showing that there will be no boundary layer separation is to use Thwaites’ method (applicable to laminar boundary layers). Boundary layer separation occurs when the wall shear stress is zero, i.e. w = 0. In Thwaites’ method this occurs when: 2 dU 0.090 (8.110) M dx Substituting Eqn. (8.107) for U gives: 2 m 1 0.090 M (8.111) 2 x a 2 Noting that the right hand side will always be positive (note that m > 0 as defined in Eqn. (8.104)), we must conclude that boundary layer separation will not occur. C. Wassgren Chapter 08: Boundary Layers 382 Last Updated: 14 Aug 2010 7. Turbulent Boundary Layer over a Flat Plate (no pressure gradient) To analyze a turbulent boundary layer we must use the momentum integral approach coupled with experimental data since no exact solutions are known. To approximate the velocity profile in a turbulent boundary layer, recall the Law of the Wall (from a previous set of notes on turbulence): * u * yu for yu*/ 5 u 1 yu * ln for yu*/ > 5 c u K where u* = (w/)1/2 is the “friction velocity.” We could substitute this velocity profile into the momentum thickness integral: y u u M 1 dy U U y 0 u * (8.114) and solve. Using the stated velocity profile, however, can get cumbersome. Instead, Prandtl suggested approximating the logarithmic turbulent velocity profile using a 1/7th power-law curve fit: 1 u y 7 for y U u 1 for y U Using this velocity profile, the momentum thickness becomes: 1 1 y y 7 7 u u y y M 1 dy 1 dy U U y 0 y 0 7 72 To determine the shear stress, recall that from the Karman momentum integral equation: 7 2 d 2 d w U M U dx dx 72 so that the friction coefficient becomes: w 7 d Cf 1 U 2 36 dx 2 M (8.117) (8.118) Experimental wall friction data for turbulent boundary layers can be fit using: 1 C f 0.020 Re 6 (8.119) where Re = (U/). Equating the two friction coefficients gives: 7 d U 0.020 36 dx 16 laminar U d 0.103 0 1 dx x0 x x0 Note: The shear stress relationship holds strictly for the part of the BL that is turbulent. If the laminar BL thickness and distance downstream are small in comparison to the current thickness and location, then we may assume that the turbulent BL starts approximately at the leading edge, i.e. 0 = 0 at x0 = 0. 1 U 6 x x0 Assuming 0 = 0 at x0 = 0 gives: 7 6 7 6 6 0.120 x U x 1 turbulent 16 x x 6 0 0.120 7 0 6 C. Wassgren Chapter 08: Boundary Layers 383 Last Updated: 14 Aug 2010 x 0.163 1 Re x7 (8.120) From this relation we can also determine the displacement thickness, momentum thickness, friction factor, and drag coefficient. These relations are summarized below. x 0.16 1 Re x7 Cf (8.121) 0.027 1 Re x7 (8.124) D x CD 0.02 1 Re x7 (8.122) 0.031 1 Re L7 (8.125) M x Re x 0.016 1 Re x7 (8.123) U x Notes: 1. The boundary layer thickness grows as ~ x6/7 for a turbulent boundary layer whereas it grows as ~ x1/2 for a laminar boundary layer. Hence, a boundary layer grows more rapidly with distance for turbulent flow than for a laminar flow. The momentum and displacement thicknesses also increase more rapidly for turbulent boundary layers. 2. The shear stress decreases more rapidly for laminar flow than for a turbulent flow. The drag does not increase as rapidly in a laminar flow as compared to a turbulent flow. 3. Another experimental friction curve fit that is commonly used is: 0.0466 Cf (8.126) 1 Re 4 which gives: D 0.0478 M 0.0371 0.382 (8.127) (8.128) (8.129) 1 1 1 5 5 x Re x x x Re x Re x5 Cf 0.0594 1 Re x5 (8.130) CD 0.0742 1 Re L5 (8.131) Re x U x nd White (in White, F.M., Viscous Fluid Flow, 2 ed., McGraw-Hill) states that the experimental curve fit given by Eqn. (8.126) is based on limited data and is not as accurate as the curve fit given by Eqn. (8.119). This argument is supported by the plot shown below (plot from White, F.M., Viscous Fluid Flow, 2nd ed., McGraw-Hill). a more accurate analytical relation not presented here due to its complexity Eqn. (8.130) Eqn. (8.124) C. Wassgren Chapter 08: Boundary Layers 384 Last Updated: 14 Aug 2010 Example: A thin smooth sign is attached to the side of a truck as shown. Estimate the skin friction drag on the sign when the truck speed is 55 mph. 5 ft 20 ft 3 ft 4 ft GO BOILERS !! SOLUTION: Assume that the boundary layer forms at the front of the trailer. U b L1 region 1 L2 region 2 L1 = 5 ft L2 = 25 ft b = 4 ft U = 55 mph = 80.7 ft/s air = 1.57*10-4 ft2/s air = 2.38*10-3 slugs/ft3 To find the drag on the sign, determine the drag on region 2 and subtract the drag from region 1. Dsign D2 D1 (8.132) where Di cDi 1 U 2 Li b (i = 1 or 2) 2 (8.133) Substitute and simplify. Dsign 1 U 2 b cD 2 L2 cD1 L1 2 (8.134) The drag coefficients are determined from the Reynolds numbers at each region’s trailing edge. 80.7 ft/s 5 ft UL 2.6 *106 (turbulent!) Re1 1 1.57*10-4 ft 2 /s Re 2 UL2 80.7 ft/s 25 ft 1.57*10-4 ft 2 /s 1.3*107 (turbulent!) (8.135) (8.136) Assume that the flow is fully turbulent throughout regions 1 and 2 (neglect any laminar flow contribution) so that: 0.0742 0.0742 cD1 3.87 *103 (8.137) 1 1 5 65 Re1 2.6 *10 C. Wassgren Chapter 08: Boundary Layers 385 Last Updated: 14 Aug 2010 cD 2 0.0742 1 Re 2 5 0.0742 1.3*10 7 1 2.80 *103 (8.138) 5 Substitute into Eqn. (8.134) and evaluate. Dsign 1 2 2.38*10 3 slugs/ft 3 80.7 ft/s 2 4 ft 2.80*10-3 25 ft 3.87*10-3 5 ft Dsign 1.57 lbf C. Wassgren Chapter 08: Boundary Layers (8.139) 386 Last Updated: 14 Aug 2010 Example: The flat plate formulas for turbulent flow over a flat plate assume that turbulent flow begins at the leading edge (x = 0). In reality, there is an initial region of laminar flow as shown in the figure. U y x turbulent flow laminar flow 1. Derive an expression for the 99% boundary layer thickness in the turbulent region by accounting for the laminar part of the flow. 2. Plot the dimensionless boundary layer thickness, /x, as a function of Reynolds number (104 Rex 108, use a log scale for the Rex axis) for your derived relation and for the turbulent relation that does not consider the laminar part. Assume a 1/7th power law velocity profile for the turbulent boundary layer and an experimental friction 1 coefficient correlation of C f 0.020 Re 6 . SOLUTION: First determine the boundary layer thickness in the laminar flow region using the Blasius solution: 5.0 (Rex < 500,000) (8.140) 1 x Re x2 Assume that the transition to turbulence occurs at a Reynolds number of 500,000 so that condition at the transition point is: trans 3.536 *103 (8.141) U where 500, 000 xtrans (8.142) U Now use the Karman Momentum Integral Equation to determine the boundary layer characteristics for the turbulent region. Assume that the velocity profile follows the following form: 1 y 7 y 1 u (8.143) U y 1 1 Using this velocity profile, the momentum thickness is: 1 1 y y 7 7 u u 7 y y M 1 dy 1 dy 72 U U y 0 y 0 (8.144) To determine the shear stress, recall that from the Karman Momentum Integral Equation, with a constant outer velocity: d d 7 w U 2 M U 2 (8.145) 72 dx dx so that the friction coefficient is: C. Wassgren Chapter 08: Boundary Layers 387 Last Updated: 14 Aug 2010 Cf 1 w 7 d 2 36 dx 2 U (8.146) Using the given experimental wall friction correlation: 1 C f 0.020 Re 6 (8.147) where Re = (U/), equate the two friction coefficients to give: 16 7 d U 0.020 36 dx U d 0.103 trans 1 16 (8.148) x x 6 dx (8.149) x xtrans 16 U (8.150) x xtrans where Eqns. (8.141) and (8.142) are used for trans and xtrans, respectively. Substituting and simplifying results in: trans 0.120 7 7 6 6 7 6 U 0.120 U 1.380 *104 7 6 1 7 6 16 x 500, 000 U 7 0.120 x 6.000 *104 1.380 *104 U U U 7 6 1 6 4 0.120 4.620 *10 x Ux Ux 6 6 0.120 4.620 *104 1 7 x Re x6 Re x6 6 7 (8.151) 7 6 (8.152) 6 (8.153) 7 Re x 500, 000 (8.154) Compare this result to one that assumes that the turbulent boundary layer starts from the leading edge: 0.16 (Rex > 500,000) (8.155) 1 x Re x7 dimensionless boundary layer thickness, /x 0.06 no laminar part includes laminar part 0.05 0.04 0.03 0.02 0.01 0.00 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 Reynolds number, Rex C. Wassgren Chapter 08: Boundary Layers 388 Last Updated: 14 Aug 2010 8. Boundary Layer Separation Consider flow around a cylinder as shown below: s Let s be the distance from the stagnation point along the cylinder surface. wake outer potential flow thin BL BL separates On the front side of the cylinder the boundary layer grows with increasing distance as we might expect. (Note: The pressure near the cylinder surface decreases with distance from the stagnation point on the front half dp/ds < 0.) At a particular location on the cylinder surface, the boundary layer no longer remains “attached” to the surface. This point is termed the boundary layer separation point. The occurrence of boundary layer separation is a result of energy dissipation in the boundary layer due to viscous effects and adverse pressure gradients. Define: an adverse pressure gradient as one in which dp/ds > 0, and a favorable pressure gradient as one in which dp/ds < 0 Consider the pressure near the surface of the cylinder determined from potential flow theory: adverse favorable p C p+ /2U2 A 1 B U A C potential flow theory B p-3/2U2 0 Imagine a fluid particle in the BL. It experiences approximately the same pressure as that in the outer potential flow (refer to the notes discussing the BL equations). As the fluid particle moves in the BL, it loses energy due to viscous effects. As a result, it does not have enough energy to “coast up the pressure hill” from point B to C. The fluid moves up the pressure hill as far as possible and then BL separation occurs. C. Wassgren Chapter 08: Boundary Layers 389 Last Updated: 14 Aug 2010 (From White, F.M., Fluid Mechanics, 3rd ed., McGraw-Hill.) Inviscid analysis works reasonably well on the upstream side but fails miserably on the downstream side due to boundary layer separation. Notice that the pressure in the wake is relatively uniform and equal to the pressure at the boundary layer separation point. Hence, delaying boundary layer separation helps to recover the pressure in the wake as a well as decreasing the area of the wake. As a result, the drag due to pressure forces (aka form drag) decreases as boundary layer separation is delayed. C. Wassgren Chapter 08: Boundary Layers 390 Last Updated: 14 Aug 2010 Let’s look at the velocity profile at different points along a plate for a flow with an adverse pressure gradient (dp/dx > 0): y y y y recirculation region velocity profile inflection point, i.e. 2u 0 y 2 w = 0 (zero slope at the wall) This is defined as the separation point. backflow at the wall (recirculation). The boundary layer thickens considerably the BL assumption that v << u is no longer valid In an adverse pressure gradient flow the boundary layer velocity profile will always have an inflection point. This can be shown by considering the boundary layer momentum equation: u u 1 dp 2u 1 dp dU .) 2 (Bernoulli’s equation has been used to write: u v U dx dx x y dx y Note that at the wall boundary (y = 0), u = v = 0 so that: dp 2u 2 dx y y 0 Thus, at the wall boundary in an adverse pressure gradient (dp/dx > 0), we must have: 2u 0 y 2 y 0 However, at the free stream interface (y = ) we have: 2u 0 y 2 y in order for the boundary layer profile to merge smoothly with the outer flow velocity profile. The change in sign of the boundary layer curvature indicates that somewhere within the boundary layer there must be a point of inflection. The inflection point moves toward the outer flow boundary as the flow moves downstream in an adverse pressure gradient flow. (From White, F.M., Fluid Mechanics, 3rd ed., McGraw-Hill.) C. Wassgren 391 Last Updated: 14 Aug 2010 Chapter 08: Boundary Layers C. Wassgren Chapter 08: Boundary Layers 392 Last Updated: 14 Aug 2010 Notes: 1. Turbulent BLs have more momentum than laminar BLs (consider the velocity profiles) fluid in a turbulent BL will travel further into an adverse pressure gradient than fluid in a laminar BL BL separation occurs later for a turbulent BL than for a laminar BL y/ L Note that w, laminar < w, turbulent since the velocity gradient for laminar flow is less than the velocity gradient for turbulent flow. T u/U (From Van Dyke, M., An Album of Fluid Motion, Parabolic Press.) C. Wassgren Chapter 08: Boundary Layers 393 Last Updated: 14 Aug 2010 (From White, F.M., Fluid Mechanics, 3rd ed., McGraw-Hill.) C. Wassgren Chapter 08: Boundary Layers 394 Last Updated: 14 Aug 2010 9. Forces on Objects Immersed in a Fluid Flow The force acting on an object immersed in a fluid flow is comprised of the force due to pressure variations over the surface and the force due to viscous shear stresses. If we know the pressure (p) and shear stress () distribution over the object, then: FL ˆ FP pndA ˆ n A FS ,i ji ni dA U dA FD A where FP is the force due to the pressure component, FS is the force due to the shear stress component, and A is the surface area of the object. The component of the force acting in the direction parallel to the incoming flow is known as the drag force, FD, and the component perpendicular to the incoming flow is known as the lift force, FL. Notes: 1. The pressure force component of the drag is known as form drag while the shear stress drag component is known as the skin friction drag. 2. A streamlined body is one in which the (skin friction drag) >> (form drag). small wake A bluff body is one in which the (form drag) >> (skin friction drag). large wake 3. The lift and drag are often expressed in dimensionless form as a lift and drag coefficient, CD and CL: FL FD CL CD 1 U 2 A 1 U 2 A 2 2 where A is usually the frontal projected area (area seen from the front) for a bluff body or the planform area (the area seen from above) for a streamlined body. bluff body (use the frontal projected area) C. Wassgren Chapter 08: Boundary Layers streamlined body (use the planform area) 395 Last Updated: 14 Aug 2010 Flow Around a Cylinder (or Sphere) As a Function of Reynolds Number Note: Re VD Re << 1 (creeping or Stoke’s flow) 5 < Re < 50 (fixed eddies) 60 < Re < 5000 (Karman Vortex Street, periodic shedding of vortices) 5000 < Re < 200,000 Re 200,000 (drag crisis) Re > 200,000 C. Wassgren Chapter 08: Boundary Layers 396 Last Updated: 14 Aug 2010 (From Batchelor, G.K., An Introduction to Fluid Dynamics, Cambridge University Press.) C. Wassgren Chapter 08: Boundary Layers 397 Last Updated: 14 Aug 2010 (From Van Dyke, M., An Album of Fluid Motion, Parabolic Press.) C. Wassgren Chapter 08: Boundary Layers 398 Last Updated: 14 Aug 2010 (From Van Dyke, M., An Album of Fluid Motion, Parabolic Press.) C. Wassgren 399 Chapter 08: Boundary Layers Last Updated: 14 Aug 2010 Notes: 1. The periodic shedding of vortices off the object results in periodic forces exerted on the object in the cross-stream direction. The Tacoma Narrows bridge disaster (figure shown below) occurred because a structural natural frequency of the bridge matched the frequency of the shedding vortices. The Tacoma Narrows bridge collapsed in 1940. 2. Experimental measurements have shown that the dimensionless frequency of the shedding vortices, f, expressed as a Strouhal number, i.e. St = fD/V, remains relatively constant at 0.2 for 100 < ReD < 1*106. St = fV/D Re = VD/ (Figure from White, F.M., Fluid Mechanics, McGraw-Hill.) The fact that the Strouhal number is insensitive to the Reynolds number over a wide range of Reynolds numbers has been used to design a flow velocity meter known as the vortex flow meter (shown below). By measuring the frequency of the forces acting on the obstruction (of known size) and knowing that the Strouhal number is approximately equal to 0.2, the flow velocity can be estimated. C. Wassgren Chapter 08: Boundary Layers 400 Last Updated: 14 Aug 2010 drag crisis Commonly used curve fits to the curve shown above are: ReD < 1: CD = 24/ReD (Stokes’ drag law) CD = 24/ReD (1+3/16ReD) (Oseen’s approximation) ReD < 5: 0 ReD 2*105: CD = 24/ReD + 6/(1+ReD0.5) + 0.4 5 ReD < 2*10 : CD = 0.44 (Newton’s Law) drag crisis C. Wassgren Chapter 08: Boundary Layers 401 Last Updated: 14 Aug 2010 From the data we observe that increasing the roughness of the surface causes the drag crisis to occur at a smaller Reynolds number. Note: The Reynolds number for a 95 mph baseball and a 170 mph golf ball are approximately 206,000 and 213,000, respectively. Hence, both are near the drag crisis! C. Wassgren Chapter 08: Boundary Layers 402 Last Updated: 14 Aug 2010 The following is from White, F.M, Fluid Mechanics, 3rd ed, McGraw-Hill. C. Wassgren Chapter 08: Boundary Layers 403 Last Updated: 14 Aug 2010 The following is from White, F.M, Fluid Mechanics, 3rd ed, McGraw-Hill. C. Wassgren Chapter 08: Boundary Layers 404 Last Updated: 14 Aug 2010 Review Questions 1. What scaling arguments are used in deriving the boundary layer equations? 2. What are the appropriate boundary conditions for the boundary layer equations? 3. What restrictions are there on the Reynolds number for using the boundary layer equations? 4. Describe how the pressure within a boundary layer is determined. 5. Describe, in words, the approach used in the Blasius solution to the boundary layer equations. 6. What assumptions are made in the Blasius boundary layer solution? (e.g. Reynolds number limitations, pressure gradients, free stream conditions, surface curvature, etc.) 7. At what Reynolds number (an engineering rule of thumb estimate) does a laminar boundary layer transition to a turbulent boundary layer? 8. How does the boundary layer thickness vary with the distance from the leading edge of the boundary layer for a flat plate, no pressure gradient boundary layer flow? 9. What is the expression for the 99% boundary layer thickness resulting from the Blasius solution? 10. What do the Falkner-Skan boundary layer solutions represent? 11. What are the boundary conditions used in the Falkner-Skan boundary layer solution? 12. Give two examples of practical boundary layer solutions that are embedded within the Falkner-Skan general solution. 13. Can the Karman momentum integral equation (KMIE) be used for flows with non-uniform pressure gradients? Turbulent flows? Compressible flows? Unsteady flows? 14. How might one find the outer flow velocity, U, when using the KMIE? 15. Describe the typical methodology used when applying the KMIE. 16. What is the 1/7th power law profile for a turbulent boundary layer? 17. In which type of boundary layer flow does the shear stress decrease most rapidly? Laminar or turbulent? In which type of flow does the drag increase most rapidly? 18. Give a physical description of why boundary layer separation occurs. 19. What defines the point at which boundary layer separation occurs? 20. Why can’t the boundary layer equations be used downstream of boundary layer separation point? 21. Why do turbulent boundary layers separate further downstream than laminar boundary layers? 22. What is meant by “favorable” and “adverse” pressure gradients? 23. Can a boundary layer separate in a favorable pressure gradient flow? 24. Must boundary layers always separate in an adverse pressure gradient flow? 25. What are the restrictions in using Thwaites’ correlation? 26. Describe the flow behavior as a function of Reynolds number for flow over a cylinder. 27. Sketch a plot of drag coefficient as a function of Reynolds number for flow over a sphere. Indicate points of particular interest on the plot. Identify whether the axes are linear or logarithmic. C. Wassgren Chapter 08: Boundary Layers 405 Last Updated: 14 Aug 2010 Chapter 09: Turbulence 1. Introduction to Turbulence 2. Time-Averaged Continuity and Navier-Stokes Equations and Reynolds Stresses 3. Law of the Wall C. Wassgren Chapter 09: Turbulence 406 Last Updated: 09 Sep 2008 1. Introduction to Turbulence Let’s consider the following simple experiment (this thought experiment is similar to the famous dye injection experiment performed by Osbourne Reynolds). At a particular point in a pipe flow, let’s measure a velocity component of the fluid as a function of time for varying Reynolds numbers. Re D D u uD measure velocity at this point At Reynolds numbers less than approximately 2300 we would find the following: velocity, u u laminar flow: smooth flow with no fluctuations ReD < 2300 time For 2300 < ReD < 4000 we would find a slightly different behavior: velocity, u u transitional flow: smooth laminar flow with random bursts of turbulent behavior 2300 < ReD < 4000 time For ReD>4000 we find a very different behavior: velocity, u u turbulent flow: complex flow with random fluctuations about some mean velocity ReD > 4000 time C. Wassgren Chapter 09: Turbulence 407 Last Updated: 09 Sep 2008 Notes: 1. Turbulence is a very difficult phenomenon to analyze. It is typically studied using semi-empirical analyses, i.e. those analyses that use both theory and experimental data. The transition region is even more difficult to analyze. 2. The uncertainties associated with the transitional regime are also reflected in the value for the friction factor, f, shown below in the Moody chart. Note that for 2300 < ReD < 4000 the value for the friction factor is not well defined since the friction factor varies considerably as the flow transitions between laminar behavior and turbulent behavior. (The plot below came from Fox, R.W. and McDonald, A.T., Introduction to Fluid Mechanics, 5th ed., Wiley). C. Wassgren Chapter 09: Turbulence 408 Last Updated: 09 Sep 2008 2. Time-Averaged Continuity and Navier-Stokes Equations Since the time-varying velocity data shown in Section 1’s example appears to consist of a fluctuating part superimposed on a mean value, let’s make the following definitions. First, express the instantaneous velocity component, ui, as the sum of a mean velocity, ui , and a fluctuating velocity, ui , i.e. ui ui ui (9.1) where the mean velocity over the time interval from t0 to t0+T is given by: 1 ui T t0 T (9.2) ui dt t0 Note that the time average of the fluctuating velocity will be zero: 1 ui T t0 T t0 t0 T t0 T 1 1 ui dt ui dt 0 ui ui dt T T t0 t0 ui (9.3) ui However, the mean of the square of the fluctuating velocities will, in general, be greater than zero: ui 2 1 T t0 T ui ui 2 dt 0 (9.4) t0 Similarly, 1 uiu j T t0 T ui ui u j u j dt 0 (in general) (i j) (9.5) t0 Now let’s look at the governing equations for an incompressible, Newtonian fluid (continuity and NavierStokes) where we’ll use the mean and fluctuating parts for our unknown variables: ui ui ui (9.6) p p p Substituting into the continuity equation gives: ui ui ui ui ui (9.7) 0 xi xi xi xi Now let’s take the time-average of the continuity equation: 1 T t0 T t0 ui ui dt 0 xi xi (9.8) where 1 T 1 T t0 T t0 t0 T t0 t0 T ui u 1 dt ui dt i xi xi xi T t0 (9.9) t0 T ui 1 dt uidt 0 xi xi T t0 Thus, the time-averaged continuity equation becomes: ui 0 xi (9.10) Note that the time-averaged continuity equation looks essentially the same as the instantaneous continuity equation except it is in terms of time-averaged velocities. C. Wassgren Chapter 09: Turbulence 409 Last Updated: 09 Sep 2008 Now let’s take the same approach with the Navier-Stokes equations: u u 2u p 2i fi i uk i xk xi xk t (9.11) To help with the upcoming analysis, let’s re-write the left-hand-side using the continuity equation: u u u 2u p 2i fi i uk i ui k t xk xk xi xk 0 (9.12) ui uk xk Write the velocities and pressure in terms of mean and fluctuating parts: ui ui uk uk ui ui p p 2 ui ui fi 2 t xk xi xk u u 2u 2u p p i i 2i 2i fi ui uk ui uk uk ui uiuk t t xk xi xi xk xk Time-average the entire previous equation noting that for any term F: F F x x and (9.14) ui p 2ui ui uk uk ui 2 0 t t xk so that we have: u 2u p i ui uk uiuk 2i fi t x x x k i (9.13) (9.15) (9.16) k We can further simplify this equation by using the time-averaged continuity equation: u u ui uk ui k uk i xk xk xk (9.17) 0 Thus, the time-averaged Navier-Stokes equations become: u u p ui uiuk fi i uk i xk xi xk xk t C. Wassgren Chapter 09: Turbulence 410 (9.18) Last Updated: 09 Sep 2008 If we compare Eqn. (9.18) to the instantaneous Navier-Stokes equation, we see that an extra term appears on the right hand side with the same dimensions as the laminar shear stress term: uiuk (9.19) xk These terms are referred to as Reynolds “stresses” (in fact, they are momentum flux terms). Notes: 1. There are both Reynolds normal and shear stresses. For example, the x-component of the timeaveraged Navier-Stokes equations is given by: Du p fx Dt x (9.20) u 2 u u u u v u w y y z z x x Reynolds "normal stress" Reynolds "shear stress" Reynolds "shear stress" 2. What, physically speaking, is a Reynolds “stress”? First let’s examine what causes viscosity in a laminar flow. Recall that a fluid is comprised of a collection of molecules. In a flow with a velocity gradient, the molecules in a particular “layer” will have an average velocity in addition to some random (thermal) motion. Since there is a velocity gradient in the fluid (u constant), molecules in adjacent layers will not have the same average velocity. Due to their random motion, molecules starting in a particular layer (with a particular average velocity) will move into layers with a different average velocity. When a molecule moves into a region of higher velocity, the molecules in that layer must accelerate the incoming molecule up to the new speed. In order to accelerate the new molecule, the molecules already in the layer must exert a force on the new molecule (via molecular collisions). From Newton’s 3rd law, the new molecule exerts an equal but opposite force on the layer. An identical process happens when a particle moves into a region of slower average velocity. Thus, we see that the laminar shear stress is a result of the random flux of molecules into neighboring layers with different average velocities. In a turbulent flow, we have this microscopic effect plus a macroscopic flux of fluid due to the random motion of the fluid. This macroscopic random fluid motion into regions of differing average motion gives rise to Reynolds stresses. y u y u u v y random macroscopic fluid motion total turbulent laminar C. Wassgren Chapter 09: Turbulence 411 Last Updated: 09 Sep 2008 3. Near a wall, the Reynolds shear stresses are small due to the wall restricting the random motion of the fluid. This region is termed the viscous sub-layer and (9.21) viscous sub-layer laminar Far from the wall, turbulent motion dominates. This region is termed the turbulent core and: turbulent core turbulent (9.22) The region between the laminar sub-layer and the turbulent core is referred to as the transitional region where the laminar and turbulent shear stresses are of the same order of magnitude. y y u turbulent core total transitional region laminar sub-layer 4. T The relative turbulence intensity is defined as the magnitude of the Reynolds stresses relative to some characteristic flow speed, U, e.g. an outer flow velocity: TI 5. L uiu j (9.23) U If the turbulence intensity is the same in all directions, then the turbulence is considered isotropic. Otherwise, the turbulence is anisotropic. As shown in the following figure for turbulent flow over a flat plate with no pressure gradient and Rex 107, the turbulence near the wall is found to be anisotropic with the relative turbulence intensity normal (v’) to the wall being smaller than the streamwise (u’) and lateral (w’) turbulence intensities. This effect is the result of the geometric constraint posed by the wall. Farther from the wall at approximately y/ = 0.8, the turbulence becomes nearly isotropic. u U (Figure 6.4, White, F.M., Viscous Fluid Flow, McGraw-Hill, 3rd ed.) C. Wassgren Chapter 09: Turbulence 412 Last Updated: 09 Sep 2008 6. The turbulent kinetic energy, K, is defined as the kinetic energy of the normal turbulent fluctuations: K 1 uiui (9.24) 2 The turbulent kinetic energy is used in models that examine the balance of energy associated with turbulent motion (beyond the scope of these notes). 3. Law of the Wall Let’s write the quantities that affect the mean velocity profile, u , near the wall: u f y, , , w (9.25) where y is the distance from the wall, and are the fluid density and kinematic viscosity, and w is the shear stress that the wall exerts on the fluid. By performing a dimensional analysis on these variables we find a relation termed the Law of the Wall: yu* u g (9.26) u* where g is an unknown function and w u* (9.27) where u* is referred to as the “friction velocity”. Nearest the wall, in the region known as the laminar sub-layer ( 0 yu*/ 5), the wall stress is given by: du w (9.28) dy Dividing through by the density and integrating, the mean velocity profile in the vicinity of the wall is: y 0 u w dy du u* u* u u * 2 y u (9.29) 0 2 u* y yu * mean velocity profile in the laminar sub-layer 0 5 C. Wassgren Chapter 09: Turbulence 413 (9.30) Last Updated: 09 Sep 2008 Now let’s examine the turbulent core region. One simple model, known as Prandtl’s Mixing Length Hypothesis, assumes that the fluctuating velocities, e.g. u’ and v’, are approximately equal to some typical eddy length, l, multiplied by the velocity gradient, du / dy : y u , v l u du dy l Thus, the Reynolds shear stress in the turbulent core is: du dy 2 u v l 2 (9.31) Note that l cannot be constant since u’, v’ 0 as y 0. This stipulation implies that l 0 as y 0. Let’s assume a simple relation that satisfies this condition: l Ky (9.32) where the constant K is known as Karman’s Universal Mixing Length Constant. Substituting equation (9.32) into equation (9.31) and simplifying gives: 2 du 2 2 du K y dy dy 2 l 2 u *2 du K y dy 2 u * Ky 2 since u * 2 w du dy dy K K u 1 * du ln y * u c * ln y c y u u u K u 1 yu * yu* ln 70 i.e. in the turbulent core region) c (for u* K (9.33) In summary, the Law of the Wall states: yu* u g u* 0 yu * 5 yu * yu* 5: 70 70 : C. Wassgren Chapter 09: Turbulence u u* y (laminar sub-layer) (9.34) (often use Eqn. (9.36) here) (transition region) (9.35) u 1 yu * ln c u* K (turbulent core) (9.36) u* 414 Last Updated: 09 Sep 2008 Notes: 1. Experimental curve fits indicate that: K’ 0.41 and c = 5.0. 2. Comparing laminar and turbulent pipe flows we observe that the turbulent velocity profile (a logarithmic curve) is blunter than the laminar profile (a parabolic curve). laminar 3. turbulent (The profile becomes blunter as Re.) A plot of the Law of the Wall velocity relations and corresponding experimental data is shown below. Fit 1 refers to: u u* y u* Fit 2 refers to the transition from laminar to turbulent behavior. Fit 3 refers to: yu * u 2.5ln 5.5 * u Fit 4 refers to: 1 u* y 7 8.74 u* Fit 5 refers to: u 1 u* y 10 11.5 u* u laminar sub-layer transition region turbulent region (Figure 20.4 from Schlichting, H., Boundary Layer Theory, McGraw-Hill.) C. Wassgren Chapter 09: Turbulence 415 Last Updated: 09 Sep 2008 Chapter 10: Pipe Flows NOT YET COMPLETE 1. 2. 3. 4. 5. 6. 7. Entrance Region Fully Developed Laminar Circular Pipe Flow Fully Developed Turbulent Circular Pipe Flow Moody Chart Other Losses Extended Bernoulli Equation Pipe Systems C. Wassgren Chapter 10: Pipe Flows 416 Last Updated: 10 Sep 2008 1. Entrance Region inviscid core pipe diameter, D entrance length, L fully developed flow The shaded regions are where viscous stresses are important (the boundary layer). The flow in the entrance region is complex and will not be investigated here. Experiments have shown that the dimensionless length of the entrance region depends on whether the entering flow is laminar or turbulent, with: laminar flow: L/D 0.06 ReD (10.1) turbulent flow: L/D 4.4 ReD1/6 (10.2) For many engineering flows: (10.3) 104 < ReD < 105 20 < L/D < 30 2. Fully Developed Laminar Circular Pipe Flow (Poiseuille Flow) Consider the steady flow of an incompressible, constant viscosity, Newtonian fluid within an infinitely long, circular pipe of radius, R. r R z We’ll make the following assumptions: 1. The flow is axi-symmetric and there is no “swirl” velocity. 2. The flow is steady. 3. The flow is fully-developed in the z-direction. 4. 0 and u 0 There are no body forces. f r f f z 0 0 t ur u z 0 z z Let’s first examine the continuity equation: 1 rur 1 u u z 0 r r r z From assumptions #1 and #3 we see that: rur constant Since there is no flow through the walls, the constant must be equal to zero and thus: ur 0 (call this condition #5) C. Wassgren Chapter 10: Pipe Flows 417 (10.4) (10.5) (10.6) Last Updated: 10 Sep 2008 Now let’s examine the Navier-Stokes equation in the z-direction: 1 u z 1 2 u z 2 u z u u u u p u 2 fz z ur z z u z z r 2 2 r z z z r t r r r r We can simplify this equation using our assumptions: u z u z u u z u z p 1 u z r ur uz r r r r t 0 r z z (#5) 0 (#3) 0 (#1) 0 (#2) 2 2 1 uz uz 2 2 fz r 2 z 0 (#4) 0 (#1) 0 (#3) r dp dz du z r 2 dp r c1 dr 2 dz (10.7) d du z r dr dr (10.8) r 2 dp c1 ln r c2 (10.9) 4 dz Note that in the previous derivation the fact that uz is a function only of r has been used to change the partial derivatives to ordinary derivatives. Furthermore, examining the Navier-Stokes equations in the r and directions demonstrates that the pressure, p, is a function only of z and thus ordinary derivatives can be used when differentiating the pressure with respect to z. uz Now let’s apply boundary conditions to determine the unknown constants c1 and c2. First, note that the fluid velocity in a pipe must remain finite as r0 so that the constant c1 must be zero (this is a type of kinematic boundary condition). Also, the pipe wall is fixed so that we have uz(r=R)=0 (no-slip condition). After applying boundary conditions we have: R 2 dp r2 (10.10) uz 1 2 Poiseuille Flow in a Circular Pipe 4 dz R Notes: 1. The velocity profile is a paraboloid with the maximum velocity occurring along the centerline. The average velocity in the pipe is found from: u 1 R2 r R u z 2 rdr r 0 R 2 dp D 2 dp 8 dz 32 dz 1 2 umax (10.11) where umax is the maximum fluid velocity and D is the pipe diameter. 2. The volumetric flow rate through the pipe is: D 4 dp Q u D2 4 128 dz C. Wassgren Chapter 10: Pipe Flows (10.12) 418 Last Updated: 10 Sep 2008 3. We can determine stresses using the constitutive relations for a Newtonian fluid. The shear stress that the pipe walls apply to the fluid, w, is: R dp 4 u (10.13) w 2 dz R where u is the average velocity in the pipe. Note that an alternate method for determining the average wall shear stress, which in this case is equal to the exact wall shear stress, is to balance shear forces and pressure forces on a small slice of the flow as shown below. dp dz R 2 p dz p R 2 dz w 2 Rdz z dp 2 dz R w 2 Rdz 0 p R 2 p dz R dp w (The same answer as before!) 2 dz F z (10.14) (10.15) In engineering applications, it is common to express the average shear stress in terms of a dimensionless (Darcy) friction factor, fD, which is defined as: 4 64 (10.16) f D 1 w 2 64 u uD Re 2 where D=2R is the pipe diameter and Re is the Reynolds number. The Darcy friction factor commonly appears in the Moody chart for incompressible, viscous pipe flow. Note again that this solution is only valid only for a laminar flow. The condition for the flow to remain laminar is found experimentally to be: uD (10.17) Re 2300 4. Let’s re-write Eqn. (10.11): p p+p D 2 dp D 2 p (10.18) u u L 32 dz 32 L where, in the fully developed region, the pressure gradient remains constant and so we may write dp/dz as p/L where p is the pressure drop over a length L of the pipe. Re-arranging Eqn. (10.18) and dropping the absolute value symbol for convenience: 32 uL (10.19) p D2 Make the previous equation dimensionless by dividing through by the dynamic pressure based on the average flow speed. 64 L 64 L p (10.20) 1 u 2 uD D Re D D 2 The dimensionless pressure drop is also referred to as the loss coefficient, k. Hence, for a laminar flow, the loss coefficient corresponding to the viscous stresses due to the pipe walls is: klaminar, wall stresses 64 Re D C. Wassgren Chapter 10: Pipe Flows L L fD D D (10.21) 419 Last Updated: 10 Sep 2008 3. Fully Developed Turbulent Circular Pipe Flow Turbulent Flow in a Smooth (but Frictional) Pipe The volumetric flow rate in a smooth pipe for turbulent flow may be estimated by integrating the time averaged velocity profile, modeled using the Law of the Wall, over the cross-sectional area of the pipe. As an engineering approximation, we’ll neglect the influence of the pipe curvature and the presence of the opposite side of the pipe in this estimation. r R R y 1 yu * R r (10.22) Q u 2 rdr 2 u * ln c R y dy r 0 0 where 0.41 and c = 5.0. Note that we’ve switched coordinates so we can integrate out from the wall to the centerline. Also, we’re neglecting the laminar sub-layer velocity profile since it is typically very thin. The average velocity in the pipe is thus: Ru * V Q * 2.44 ln V 1.34 u R2 (10.23) Recall that 8 u* 8 f D w2 V V2 Thus, V 8 u* fD 2 since u * w (10.24) (10.25) Also note that: Ru * Thus, Ru * R w 1 Re D 2 R f DV 2 1 2 RV 8 2 fD 8 (10.26) fD 8 (10.27) Substituting Eqns. (10.25) and (10.27) into Eqn. (10.23) gives: 1 fD 8 2.44 ln Re D 1.34 2 fD 8 1 1.99 log10 Re D fD (10.28) f D 1.02 (10.29) where a “log10” is used in place of the “ln” in the previous equation. Prandtl derived Eqn. (10.29) but altered the constants slightly to better fit the data: 1 2.0 log10 Re D fD f D 0.8 Friction factor for turbulent flow in a smooth pipe. (10.30) Equation (10.30) is implicit in fD, which means that an iterative approach must be taken in order to solve for fD as a function of ReD. A number of approximations to this relation have been proposed that are easier to solve. For example, Blasius, a student of Prandtl’s, suggested the following approximation: 0.316 fD Valid for 4000 < ReD < 105. (10.31) 1 Re D4 C. Wassgren Chapter 10: Pipe Flows 420 Last Updated: 10 Sep 2008 Turbulent Flow in a Very Rough Pipe The roughness of the pipe walls can significantly affect the friction factor for turbulent flows (roughness has a negligible effect on the friction factor for laminar flows). Recall from the Law of the Wall that the time averaged velocity in the laminar sub-layer is: u yu * yu * for 0 5 (10.32) u* Thus, the thickness of the laminar sub-layer, LSL, is: LSL u * 5 5 LSL * (10.33) u Since, f u * w V D (Refer to Eqn. (10.25).) (10.34) 8 we have 5 8 5 8 5 8 LSL LSL (10.35) V fD D VD f D Re D f D LSL D 14.1 Re D (10.36) fD Thus, if the wall roughness, , is much smaller than the laminar sub-layer thickness, then we’ll still have a laminar sub-layer and the flow won’t be significantly affected by the wall roughness, i.e. we may treat the wall as being smooth (but still frictional). However, if >> LSL, then the laminar sub-layer will be destroyed and the wall roughness becomes the new length scale for use in the Law of the Wall, i.e. u y (10.37) fcn * u Following the same analysis as that for turbulent flow in a smooth pipe, but using y/ in place of yu*/, we obtain: 1 2.0 log10 D Friction factor for turbulent flow in a very rough pipe. (10.38) 3.7 fD Note that Eqn. (10.38) is independent of ReD. Turbulent Flow in a Rough Pipe For the transitional regime where /D is between “smooth” and “very rough,” empirical formulas in which the friction factor is a function of both /D and ReD have been developed: 1 2.51 Friction factor for turbulent flow in a rough pipe. 2.0 log10 D (10.39) 3.7 Re D f D fD or, the explicit empirical formula, 1.11 1 6.9 D 1.8log10 Friction factor for turbulent flow in a rough pipe. (10.40) Re 3.7 fD D C. Wassgren Chapter 10: Pipe Flows 421 Last Updated: 10 Sep 2008 4. Moody Chart The previous friction factor relations have been summarized into a single plot known as the Moody chart, which is shown below. (Figure from Fox, et al., Introduction to Fluid Mechanics, Wiley.) C. Wassgren Chapter 10: Pipe Flows 422 Last Updated: 10 Sep 2008 Notes: 1. For Reynolds numbers less than 2300, one may use either the analytical expression for the friction factor: 64 fD Re D or the Moody chart. 2. Reynolds numbers between approximately 2300 and 4000 correspond to the transitional regime between laminar and turbulent flow. The gray region in the Moody chart reflects the fact that the friction factor can vary significantly in this region. 3. The fully rough zone in the Moody chart is a region where the friction factor is a weak function of the Reynolds number, but a strong function of the relative roughness (refer to the second half of Section 3 of these notes). If the Reynolds number of a flow is unknown, but is expected to be large, it is often helpful to assume that the flow is in the fully rough zone as an initial first guess. 4. The roughnesses of various types of pipe materials have been compiled into tables such as the following one. Material (new) [ft] [mm] riveted steel 0.003 – 0.03 0.9 – 9.0 concrete 0.001 – 0.01 0.3 – 3.0 wood stave 0.0006 – 0.003 0.18 – 0.9 cast iron 0.0085 0.26 galvanized iron 0.0005 0.15 asphalted cast iron 0.0004 0.12 commercial steel or wrought iron 0.00015 0.046 drawn tubing 0.000005 0.0015 glass smooth smooth C. Wassgren Chapter 10: Pipe Flows 423 Last Updated: 10 Sep 2008 Example: 1. 2. 3. Using the Moody chart, determine the friction factor for a Reynolds number of 105 and a relative roughness of 0.001. What is the friction factor for a Reynolds number of 1000? What is the friction factor for a Reynolds number of 106 in a smooth pipe? SOLUTION: 1. 2. 3. The friction factor is fD 0.0225 (Follow the red lines in the following figure.) Since the Reynolds number is less than 2300, we can use the exact laminar flow relation: 64 fD fD = 0.064. (10.41) Re D Alternately, we could use the Moody chart by following the blue lines in the following figure. The friction factor is fD 0.012. (Follow the green lines in the following figure.) C. Wassgren Chapter 10: Pipe Flows 424 Last Updated: 10 Sep 2008 5. Other Losses The loss due to the viscous resistance caused by the pipe walls is referred to as a major loss. Pressure losses may occur due to viscous dissipation resulting from fluid interactions with other parts of a pipe system such as valves, bends, contractions/expansions, inlets, and connectors. These losses are known as minor losses. The names can be misleading since it’s not uncommon in pipe systems to have most of the pressure loss resulting from the minor losses (e.g., a pipe system with a large number of bends and valves, but short sections of straight pipe). What causes these minor losses? The pressure loss results primarily from viscous dissipation in regions with large velocity gradients, such as in a recirculation zone as shown in the figure below. recirculation zone A closely related phenomenon known as the vena contracta acts to effectively reduce the diameter at entrances and bends. The recirculation zone also results in a pressure loss. recirculation zone Reffective Although minor loss coefficients can be determined analytically for certain situations, most frequently the loss coefficient for a particular device is found experimentally. Essentially, one measures the pressure drop across the device, p, and forms the loss coefficient, k, using: p k1 (10.42) 2 2 V where is the fluid density and V is the average speed through the device. Many tables with experimentally determined loss coefficients have been generated. Notes: 1. When using a loss coefficient, it is important to know what velocity has been used to form the coefficient. For example, the loss coefficient for a contraction is typically based on the speed downstream of the contraction, while the loss coefficient for an expansion is based on the speed upstream of the expansion. 2. Minor losses are sometimes given in terms of equivalent lengths of pipe. An equivalent minor loss of 10 pipe diameters worth of a particular type of pipe means that the major loss caused by a pipe of that type, 10 diameters in length will give the same pressure loss as the minor loss. Thus, a loss coefficient and equivalent pipe length, Le, can be related by: L (10.43) k fD e D C. Wassgren Chapter 10: Pipe Flows 425 Last Updated: 10 Sep 2008 6. Extended Bernoulli Equation Recall from previous notes that conservation of energy may be written as: z p p V2 V2 z z H L ,1 2 H S ,1 2 2g 2g g 2 g 1 g (10.44) where each of the terms in the equation has dimensions of length. The “1” and “2” subscripts refer to the inlet and outlet conditions, respectively. The individual terms are referred to as: p pressure head g V2 2g velocity or dynamic head z elevation head H L ,1 2 head loss H S ,1 2 shaft head Recall that the in the velocity head term is the kinetic energy correction factor which accounts for the fact that an average speed is used in the Extended Bernoulli Equation rather than the real velocity profile (refer back to earlier notes concerning conservation of energy). A value of = 2 is used for laminar flows while = 1 is typically assumed for turbulent flows (actually, 1 as ReD .) The head loss term, HL, accounts for both major and minor pressure losses, and may be written as: V2 (10.45) H L ,1 2 ki i 2g i where the subscript “i” accounts for every loss in the pipe system. Recall that the major loss coefficient may be written as: L (10.46) kmajor f D D The shaft head term, HS, accounts for the pressure addition (reduction) resulting from the inclusion of devices such as pumps, compressors, fans, turbines, and windmills. Those devices that add head to the flow are positive (e.g. pumps), while those that extract head are negative (e.g. turbines). The shaft head term may be written in terms of shaft power, WS , as: W (10.47) H S ,1 2 S ,1 2 mg where m is the mass flow rate through the device. C. Wassgren Chapter 10: Pipe Flows 426 Last Updated: 10 Sep 2008 7. Pipe Systems Most pipe flow problems can be classified as being of one of three types: Type I: The desired flow rate is specified and the required pressure drop must be determined. Type II: The desired pressure drop is specified and the required flow rate must be determined. Type III: The desired flow rate and pressure drop are specified and the required pipe diameter must be determined. Serial Systems Parallel Systems C. Wassgren Chapter 10: Pipe Flows 427 Last Updated: 10 Sep 2008 Chapter 11: Fluid Machinery 1. 2. 3. 4. 5. 6. Types of Fluid Machinery Elementary Pump Theory Net Positive Suction Head Pump Similarity Specific Speed System Characteristic Curves and Pump Selection C. Wassgren Chapter 11: Fluid Machinery 428 Last Updated: 16 Aug 2009 1. Types of Fluid Machinery Two basic categories of fluid machines: 1. those that do work on the fluid: a. pumps (used for liquids) b. fans (used for gases/vapor; p < a few inches of H20) c. blowers (used for gases/vapor; a few inches of H2O < p < 1 atm) d. compressors (used for gases/vapor; p > 1 atm) 2. those that extract work from the fluid: a. turbines Types of Pumps 1. Positive Displacement Pumps (PDPs) a. force fluid movement using changes in volume b. e.g. reciprocating piston engines, heart, gear pumps, rotating screw pumps, bellows c. typically produce a periodic flow rate d. large p (pressure rise) possible but usually have a small Q (flow rate) 2. Dynamic Pumps a. no closed volumes as in PDPs b. p due to changes in fluid momentum c. e.g. axial flow and radial flow pumps (aka turbomachines), jet pumps, electromagnetic pumps d. (pdynamic pumps) typically < (pPDP) e. (Qdynamic pumps) typically > (QPDP) The operating principles for a wide variety of pumps may be viewed at: http://www.animatedsoftware.com/elearning/All%20About%20Pumps/aapumps.html C. Wassgren Chapter 11: Fluid Machinery 429 Last Updated: 16 Aug 2009 Some Examples of Positive Displacement Pumps This section remains incomplete. Gear Pumps External Gear Pump (Image from: http://www.pumpschool.com/principles/external.htm ) Internal Gear Pump (Image from: http://media-2.web.britannica.com/eb-media/56/3656-004-5CA4041D.gif ) Gear pumps are often used in automatic transmissions. Lobe Pump (Image from: http://www.megator.co.uk/lobe_pump.htm ) Two and three lobes Commonly used in diesel superchargers. C. Wassgren Chapter 11: Fluid Machinery 430 Last Updated: 16 Aug 2009 Vane Pump (Image from: http://www.pumpschool.com/principles/vane.htm ) Centrifugal force or springs are used to push out the vanes. Often used in power steering pump, automatic transmissions Screw Pump (Image from: http://en.wikipedia.org/wiki/File:Archimedes_screw.JPG ) Archemiedes Screw Pump – 2000 yrs old, often used for irrigation C. Wassgren Chapter 11: Fluid Machinery 431 Last Updated: 16 Aug 2009 (Progressive) Cavity Pump (Top image from: http://www.animatedsoftware.com/pumpglos/progrssv.htm Bottom image from: http://www.roymech.co.uk/Related/Pumps/Rotary%20Positive%20Displacement.html ) Piston Pump Wobble Plate Piston Pump (Image from: http://www.roymech.co.uk/Related/Pumps/Rotary%20Positive%20Displacement.html ) C. Wassgren Chapter 11: Fluid Machinery 432 Last Updated: 16 Aug 2009 Wolfhart Principle Pump (Images from: http://www.allstar.fiu.edu/aero/wolfhart_pump_principle.htm ) Ball Piston Pump (Image from: http://www.animatedsoftware.com/pumpglos/ballpist.htm ) Bent-Axis Piston Pump (Image from: http://www.roymech.co.uk/Related/Pumps/Rotary%20Positive%20Displacement.html ) C. Wassgren Chapter 11: Fluid Machinery 433 Last Updated: 16 Aug 2009 Radial piston pump (Image from: http://www.roymech.co.uk/Related/Pumps/Rotary%20Positive%20Displacement.html ) Rotary cam pump (Image from: http://www.labpump.co.kr/data/aboutpump.htm ) Swash plate piston pump (Image from: http://www.roymech.co.uk/Related/Pumps/Rotary%20Positive%20Displacement.html ) C. Wassgren Chapter 11: Fluid Machinery 434 Last Updated: 16 Aug 2009 Diaphragm Pump (Image from: http://en.wikipedia.org/wiki/File:Bomba_diafragma.jpg ) Finger pump (Image from: http://www.animatedsoftware.com/pumpglos/fingerpu.htm ) Peristaltic pump (Left image from: http://en.wikipedia.org/wiki/File:Eccentric_pump.gif Right image from: http://www.roymech.co.uk/Related/Pumps/Rotary%20Positive%20Displacement.html ) Often used as fuel pumps. C. Wassgren Chapter 11: Fluid Machinery 435 Last Updated: 16 Aug 2009 Some Examples of Dynamic Pumps This section remains incomplete. Axial Pump Propeller pump motor discharge elbow propeller inlet plenum (Image from: http://www.sulzerpumps.com/Portaldata/9/Resources/brochures/power/vertical/JP_Vertical_E00635.pdf ) Radial Pump (Left image from: http://commons.wikimedia.org/wiki/File:CetriFugal_Pump.jpg Right image from: http://www.motorera.com/dictionary/pics/r/Radial-flow_pump.gif ) C. Wassgren Chapter 11: Fluid Machinery 436 Last Updated: 16 Aug 2009 Mixed Pump guide vanes volute (Image from: http://www.fao.org/docrep/010/ah810e/AH810E07.htm ) C. Wassgren Chapter 11: Fluid Machinery 437 Last Updated: 16 Aug 2009 Jet Pumps (Image from: http://www.fao.org/docrep/010/ah810e/AH810E07.htm ) Ram Pump (Image from: http://www.lifewater.ca/ram_pump.htm ) C. Wassgren Chapter 11: Fluid Machinery 438 Last Updated: 16 Aug 2009 Air Lift Pumps (Images from: http://www.airliftpump.com/index.htm ) C. Wassgren Chapter 11: Fluid Machinery 439 Last Updated: 16 Aug 2009 Elementary Pump Theory To determine the work that the pump does on the fluid passing through it, we’ll use the Moment of Momentum Equation which relates torque to momentum fluxes. Consider flow through the following rotating pump: Vrb2 front view of pump impeller 2 V2 V2 , outgoing fluid velocity w/r/t a fixed FOR U2 = r2 Vrb1 V1 V1, incoming fluid velocity w/r/t a fixed FOR 1 U1 = r1 r1 r2 Here: 1, 2 V1, V2 U1, U2 Vrb1, Vrb2 r2 r1 entrance/exit blade angles w/r/t the hub fluid velocities w/r/t a FOR fixed to the ground blade velocities w/r/t a FOR fixed to the ground fluid velocities w/r/t to the blade From geometry, we have: V1 U1 Vrb1 (11.1) V2 U 2 Vrb2 To determine the torque, T, that must be applied to the shaft in order to rotate the impeller at angular velocity, , we use the moment of momentum relation: V2 d (11.2) r F on CV r u dV r u u rel dA dt CV CS V 1 Assume the following: 1. steady flow 2. uniform flow between blades 3. incompressible flow 4. the only moment applied to the CV is that due to shaft torque, Ton CV (which points in the same direction as the shaft rotation, ) 5. the fluid velocities are measured w/r/t an inertial FOR Using these assumptions and simplifying the moment-of-momentum equation: Ton CV m r2Vt 2 rVt1 Euler’s Turbomachinery Equation 1 (11.3) where Vt1 and Vt2 are the absolute fluid velocities tangent to the impeller hub. C. Wassgren Chapter 11: Fluid Machinery 440 Last Updated: 16 Aug 2009 Notes: 1. The power required to drive the impeller is: Won CV Ton CV ω W m r V rV on CV 2 t2 1 t1 but U1 = r1 and U2 = r2 so that: Won CV m U 2Vt 2 U1Vt1 In terms of the head added to the fluid: W U V U1Vt1 H added on CV 2 t 2 mg g to CV (shaft head) (11.4) 2. Only the absolute fluid velocity tangential to the impeller contributes to the increase in fluid head. 3. For an idealized centrifugal pump, the incoming flow has no tangential component Vt1 = 0 UV H added 2 t 2 (shaft head for an idealized centrifugal pump) g to CV 4. (11.5) Consider the fluid velocity vector geometry at the exit of the blade: Vrb2 V2 Vt2 Vn2 Vrb2 U2 = r2 U2-Vt2 Vn 2 tan 2 U 2 Vt 2 Vt 2 U 2 Vn 2 cot 2 (11.6) Substituting into the shaft head relation for an idealized centrifugal pump (equation (11.5)) gives: 2 U U Vn 2 cot 2 U 2 U 2Vn 2 cot 2 (11.7) H added 2 2 g g g to CV Note that the volumetric flow rate through the pump is related to the radial fluid velocity from COM: Vn2 Vn2 Q Vn 2 2 r2 b2 (11.8) impeller impeller circumference thickness b2 C. Wassgren Chapter 11: Fluid Machinery 441 Last Updated: 16 Aug 2009 Substituting and noting that U2 = r2gives: 2 r2 r2Q cot 2 H added g g 2 r2 b2 to CV r2 2 cot 2 Q g to CV g 2 b2 theoretical head rise across an idealized centrifugal pump H added (11.9) Notes: a. Equation (11.9) is an equation of a line. H shut-off r2 g head 2 > 90 (forward curved) Forward curved blades are not used very often due to flow instabilities. 2 = 90 (radial) 2 2 < 90 (backward curved) Q b. In an actual flow, losses occur within the pump due to friction with the blades (which varies with Q2), flow separation, impeller blade-shroud clearance flows, and other 3D flow effects. H ideal case friction losses (~ Q2) actual other losses Q A quadratic curve is often used to fit experiment pump head curves: H = H0 – A Q2. 5. Pump efficiency is defined as: mgH water or hydraulic horsepower (power you get out) p T brake horsepower (power you put in) a. b. typical pump efficiencies: p = 85% (well-designed) – 60% (poorly-designed) As pump size ↓, the ratio of surface area to volume ↑ frictional losses ↑ p↓. C. Wassgren Chapter 11: Fluid Machinery 442 Last Updated: 16 Aug 2009 best efficiency point (BEP) power put in (From Munson, B.R., Young, D.F., and Okiishi, T.H., Fundamentals of Fluid Mechanics, 3rd ed., Wiley.) (From Munson, B.R., Young, D.F., and Okiishi, T.H., Fundamentals of Fluid Mechanics, 3rd ed., Wiley.) C. Wassgren Chapter 11: Fluid Machinery 443 Last Updated: 16 Aug 2009 Example: An idealized centrifugal water pump is shown below. The volumetric flow rate through the pump is 0.25 ft3/s and the angular speed of the impeller is 960 rpm. Calculate the power required to drive the pump. 55 960 rpm 3 in. 11 in. 0.75 in. 0.25 ft3/s SOLUTION: Apply conservation of angular momentum to the fixed control volume shown below. 2 r2 T r1 z b2 V2 d dt r u dV r u u CV CS C. Wassgren Chapter 11: Fluid Machinery rel dA r F CV,CS 444 Last Updated: 16 Aug 2009 where d dt r u dV 0 (steady flow) CV exiting fluid velocity w/r/t ground ˆ ˆ ˆ ˆ 2 r b2 V2 sin r u urel dA r2er V2 sin 2er cos 2e r2e 2 2 CS velocity of fluid exit area exiting the CV ˆ r22 r2V2 cos 2 V2 sin 2 2 r2b2 e z (Note that there is no flux of angular momentum at the inlet since r1 and V1 are parallel.) ˆ r F Te z CV,CS Substitute and simplify. T r22 r2V2 cos 2 V2 sin 2 2 r2b2 (11.10) The velocity V2 can be related to the volumetric flow rate, Q, by considering the flow at the impeller exit. Q V2 sin 2 2 r2b2 V2 Q 2 r2 b2 sin 2 (11.11) Substitute Eqn. (11.11) into Eqn. (11.10) and simplify. gQ Q T r22 2 b2 tan 2 g (11.12) Using the given data: (g)H20 = 62.4 lbf/ft3 g = 32.2 ft/s2 Q = 0.25 ft3/s b2 = 0.75 in = 6.25*10-2 ft = 960 rpm = 100.5 rad/s 2 = 55 r2 = 5.5 in = 4.58*10-1 ft T = 10.0 ftlbf The power required to drive the impeller is: W T Using the given data: W 1005 ftlbf/s = 1.83 hp C. Wassgren Chapter 11: Fluid Machinery 445 Last Updated: 16 Aug 2009 The problem could have also been worked out using velocity polygons. Since the absolute inlet velocity has no tangential component: UV H 2 t 2 where U 2 r2 g and W mgH mg r2Vt 2 W g Use a velocity polygon at the exit to determine Vt2. Vrb2 V2 Vn2 Vrb2 Vt2 U2 = r2 U2-Vt2 From the geometry: Vt 2 U 2 Vn 2 cot 2 where Vn2 is found from conservation of mass: Q Vn 2 2 r2 b2 Using the given data: U2 = 46.3 ft/s = 1.39 ft/s Vn2 H = 65.1 ft W 1005 ftlbf/s = 1.83 hp (Same answer as before!) C. Wassgren Chapter 11: Fluid Machinery 446 Last Updated: 16 Aug 2009 2. Net Positive Suction Head (NPSH) Along the suction side of the impeller blade near the pump inlet are regions of low pressure. rotation direction blade suction side, region of low pressure fluid velocity at inlet relative to blade If the local pressure is less than the vapor pressure of the liquid, then cavitation will occur: p pvapor cavitation Recall that cavitation is “boiling” (liquid turning to vapor) occurring when the pressure is less than the liquid’s vapor pressure. Cavitation can not only significantly decrease the performance of a pump, but it can also cause pump damage, vibration, and noise. Vapor bubbles caused by cavitation move into regions of higher pressure, collapse violently, and produce localized regions of very high pressure that can chip away at surfaces. The result is that the pump material erodes away. One can often hear when cavitation occurs because of the noise generated by the collapsing vapor bubbles. Let’s define a quantity that will aid us in determine when cavitation in a pump might occur: p V2 pv net positive suction head, NPSH g 2 g s g (11.13) Notes: 1. The first term in NPSH is the head at the suction side of the pump near the impeller inlet. This is the region where we expect to have the lowest head. (Note that we define our reference plane for elevation along the centerline of the pump z = 0.) 2. The second term in NPSH is the vapor pressure head. This is the head when the liquid turns to vapor. The vapor pressure is typically given in terms of an absolute pressure so the suction pressure should also be an absolute pressure. Note that vapor pressure increases as temperature increases. 3. (NPSH)R ≡ Net Positive Suction Head Required to avoid cavitation. This quantity is a pump property and is determined experimentally. 4. (NPSH)A ≡ Net Positive Suction Head Available to the pump. This quantity is a system property and can be determined via analysis or experiments. NPSHA is related to the total head available to the pump at the pump inlet (minus the vapor head). 5. We must have (NPSH)A > (NPSH)R to avoid cavitation. Regulations typically recommend at least a 10% margin for safety. For critical applications such as for power generation or flood control, a 100% margin is often used. 6. NPSHR increases with increasing flow rate since the pressure at the suction side of the pump blade near the pump inlet will decrease (consider Bernoulli’s equation). Similarly, the pressure will decrease with increasing blade rotation rate resulting in an increased NPSHR. C. Wassgren Chapter 11: Fluid Machinery 447 Last Updated: 16 Aug 2009 Example: Determine the NPSHA for the following system P H SOLUTION: Choose point 1 to be on the surface of the tank and point 2 to be just upstream of the pump. Apply the EBE from 1 to 2: p p V2 V2 z z H L12 H S 12 2g 2g g 2 g 1 where p2 ps p1 patm V2 Vs (assume turbulent flow = 1) V1 0 z2 z1 H H S 12 0 H L12 H L12 Substitute and simplify to get: p V2 patm z1 z2 H L12 g g 2g S H From the definition of NPSH we have: p p NPSHA atm H H L12 v g g Notes: 1. Increasing H, HL12, or pv decreases NPSHA and increases the likelihood of cavitation since the difference between NPSHA and NPSHR is reduced. 2. Increasing patm increases NPSHA and decreases the likelihood of cavitation. 3. pv varies with temperature 4. pv is usually given in terms of absolute pressure C. Wassgren Chapter 11: Fluid Machinery 448 Last Updated: 16 Aug 2009 3. Pump Similarity Most pump performance data (H-Q curves) are given only for one value of the pump rotational speed and one pump impeller diameter. Is there some way to determine the pump performance data for other speeds and diameters without requiring additional testing? There is – using dimensional analysis! Recall that we typically are interested in knowing the head rise across a pump, H (=h/g where h is the specific energy rise across the pump), power required to operate the pump, W (bhp), and pump efficiency, , as a function of the volumetric flowrate through the pump, Q: (11.14) h, W , fcns Q, , , D, where and are the fluid density and dynamic viscosity, D is the pump impeller diameter, and is the pump rotational speed. Performing a dimensional analysis we find the following: , , fcns , Re (11.15) where gH ≡ dimensionless head coefficient = ≡ dimensionless power coefficient = ≡ efficiency = ≡ dimensionless flow coefficient = Re ≡ 2 D2 W 3 D 5 QgH (11.17) (11.18) W Reynolds number = (11.16) Q D3 (11.19) D 2 (11.20) Notes: 1. In most pump flows, Re is very large the variations in viscous effects from one flow to another are small Re similarity can be neglected a. b. 2. If Re is considerably different from one flow to another, e.g. pump water vs. pumping molasses, then Re effects cannot be ignored. The flow physics for large Re, where viscous forces « inertial forces, are different than for small Re where viscous forces » inertial forces. Thus, for large Re: , , fcns (11.21) QgH W C. Wassgren Chapter 11: Fluid Machinery (11.22) 449 Last Updated: 16 Aug 2009 3. For similarity between geometrically similar flows (assuming large Re), we have the following pump scaling laws: Q Q 1 2 (11.23) 3 3 D 1 D 2 gH gH 1 2 2 2 2 2 (11.24) D 1 D 2 1 2 1 2 a. WW 3 5 3 5 D 1 D 2 (since 1 = 2, 1 = 2, 1 = 2 and = /) (11.25) For a given pump (D = constant) using the same fluid (, = constant) and the same gravity (g1 = g2): Q1 1 Q Q Q2 2 1 2 H1 1 H H 2 2 H 2 2 1 2 (11.26) 2 (11.27) 3 W W W 3 1 1 3 W2 2 1 2 b. (11.28) For a given pump speed ( = constant) but varying diameters (assuming a geometrically similar family of pumps), and using the same fluid (, = constant) and the same gravity (g1 = g2): Q1 D1 Q2 D2 3 H1 D1 H 2 D2 2 5 W1 D1 W2 D2 (11.29) Note that we are assuming that all length scales within the pump are scaled in the same way to maintain geometric similarity. This is not always true in practice since pump impellers with different diameters are often put in the same pump casing. Also, surface roughness isn’t scaled proportionally. The result is that the pump scaling laws are only approximations. C. Wassgren Chapter 11: Fluid Machinery 450 Last Updated: 16 Aug 2009 Example: A centrifugal pump with a 12 in. diameter impeller requires a power input of 60 hp when the flowrate is 3200 gpm against a 60 ft head. The impeller is changed to one with a 10 in. diameter. Determine the expected flowrate, head, and input power if the pump speed remains the same. SOLUTION: Since the pump speed remains the same and assuming geometrically similar pumps, the correct pump scaling laws are: Q1 D1 Q2 D2 3 H1 D1 H 2 D2 2 5 W1 D1 W2 D2 Using the given parameters: Q1 = 3200 gpm D1 = 12 in D2 = 10 in H1 = 60 ft W1 = 60 hp Q2 = H2 = W2 = 1850 gpm 41.7 ft 24.1 hp C. Wassgren Chapter 11: Fluid Machinery 451 Last Updated: 16 Aug 2009 4. Specific Speed, Ns A useful dimensionless term results from the following combination of previously defined terms: 1 1 2 Q 2 Ns specific speed = 3 3 4 gH 4 (11.30) Combining the terms in this manner eliminates the impeller diameter, D. Notes: 1. It is customary to characterize a machine by its specific speed at the design point, i.e. Ns is usually only given for the BEP operating conditions. a. low Q, high H low Ns centrifugal pumps b. high Q, low H high Ns axial pumps 2. In practice (especially in the US), a combination of units are used to describe , Q, and H such that Ns is dimensional (signified by Nsd): N sd rpm Q gpm H ft Ns and Nsd have the same physical meaning but are different in magnitude by a constant factor: Nsd = 2733 [rpm(gpm)1/2/(ft)3/4] Ns 3. 3 4 (11.31) Given , Q, and H, we can calculate Ns (or Nsd) and, using the following chart, determine which type of pump would be most efficient for the given conditions. (From Munson, B.R., Young, D.F., and Okiishi, T.H., Fundamentals of Fluid Mechanics, 3rd ed., Wiley.) Following are some rules of thumb: 1. Positive displacement pumps are used for small flow rates, Q, and large head rises, H. 2. Centrifugal pumps are for moderate H and large Q. 3. For very large head rises, pumps are often combined in series (aka multi-stage). 4. Axial flow pumps are for large Q and low H. C. Wassgren Chapter 11: Fluid Machinery 452 Last Updated: 16 Aug 2009 Example: A small centrifugal pump, when tested at 2875 rpm with water, delivered a flowrate of 252 gpm and a head of 138 ft at its best efficiency point (efficiency is 76%). Determine the specific speed of the pump at this test condition. Sketch the impeller shape you expect. Compute the required power input to the pump. SOLUTION: The dimensional specific speed is given by: N sd rpm Q gpm 3 H ft 4 Using the given data: Nsd = 1130 rpmgpm1/2/ft3/4 The dimensionless specific speed is: N sd Ns 1 rpm gpm 2 2733 3 ft 4 Ns = 0.414 The expected impeller shape is radial as shown in the figure below. (Figure from Munson, B.R., Young, D.F., and Okiishi, T.H., Fundamentals of Fluid Mechanics, 3rd ed., Wiley.) The power input to the pump is given by: W Wshaft fluid P where Wfluid mgH QgH Wfluid = 8.80 hp W = 11.6 hp (Note: 1 ft3 = 7.48 gal, 1 hp = 550 lbfft/s, and 1 lbf = 1 slugft/s2.) shaft C. Wassgren Chapter 11: Fluid Machinery 453 Last Updated: 16 Aug 2009 5. System Characteristic Curves and Pump Selection How do we select a pump for a given system? Analyze the system to determine the shaft head required to give a specified volumetric flow rate. Compare this result to a given pump performance curve (H-Q curve) to determine if the pump operates efficiently at this Q. If so, then the choice of pump is appropriate. Consider the following example: 2 1 z2 z1 P Apply the EBE from 1 to 2: p p V2 V2 z z H L12 H S 12 2g 2g g 2 g 1 where p1 = p2 = patm and V1 = V2 = 0. Thus, the head required from the pump is: H S 12 z2 z1 H L12 (11.32) (11.33) Recall that: H L12 K i i Vi 2 Q2 Ki i 2 2g 2 gAi i (11.34) so that: H S 12 z2 z1 CQ 2 (11.35) where C is a constant that incorporates the loss coefficients and area ratios, and an “” is used since the loss coefficients may depend on the flow velocities. C. Wassgren Chapter 11: Fluid Machinery 454 Last Updated: 16 Aug 2009 The conditions at which the system will operate will depend on the intersection of the system head curve with the pump performance curve as shown in the figure below. H system curve pump curve operating point Q Notes: 1. Ideally we would want the operating point to occur near the BEP for the pump. 2. 3. For laminar flow, 64 L 64 L c where c is a constant K major Re D VD D Q HL ~ Q system curve is: HS = c1 + c2 Q (a line instead of a parabola!) (11.36) The system curve may change over time due to fouling of the pipes and other factors increased losses system curve becomes steeper H system curve after fouling original system curve Q The pump curve may also change due to wear on the bearings, impeller, etc. H original pump curve pump curve after wear Q C. Wassgren Chapter 11: Fluid Machinery 455 Last Updated: 16 Aug 2009 4. Stability issues become significant when the pump has a flat or falling (defined as a performance curve where H as Q ) performance curve. H rising (H as Q ) falling (H as Q ) Q Consider perturbations to the system from the operating point. Hpump > Hsystem Q H Hpump < Hsystem Q stable system curve pump curve unstable Hpump > Hsystem Hpump < Hsystem Q Q Q H system curve falling pump curve pump curve Q For the system shown at the left there are three operation points! The system may oscillate between these points! H system curve flat pump/system curves pump curve For the system shown at the left, the system may drift over a wide range of Q! Q C. Wassgren Chapter 11: Fluid Machinery 456 Last Updated: 16 Aug 2009 Example: Water is to be pumped from one large open tank to a second large open tank. The pipe diameter throughout is 6 in. and the total length of the pipe between the pipe entrance and exit is 200 ft. Minor loss coefficients for the entrance, exit, and the elbow are shown on the figure and the friction factor can be assumed constant and equal to 0.02. A certain centrifugal pump having the performance characteristics shown is suggested as a good pump for this flow system. a. With this pump, what would be the flow rate between the tanks? b. Do you think this pump would be a good choice? Kelbow = 1.5 pump 10 ft pipe diameter = 6 in total pipe length = 200 ft Kexit = 1.0 Kentrance = 0.5 SOLUTION: Apply the extended Bernoulli’s equation from point 1 to point 2. 2 pump H z 1 C. Wassgren Chapter 11: Fluid Machinery 457 Last Updated: 16 Aug 2009 p p V2 V2 g 2g z g 2g z H L H S 2 1 where p2 p1 patm (free surface) (11.37) V2 V1 0 (large tanks) z2 z1 H V2 L f K entrance K exit K elbow (where V is the mean velocity in the pipe) 2g D Note that the mean pipe velocity can be expressed in terms of the volumetric flow rate. Q V D2 4 Substitute and simplify. 8Q 2 L H S H 2 4 f K entrance K exit K elbow gD D HL (11.38) (11.39) For the given problem: H = 10 ft g = 32.2 ft/s2 f = 0.02 D = 6 in = 0.5 ft L = 200 ft (Note: Kmajor = f(L/D) = 8.0) Kentrance = 0.5 Kexit = 1.0 Kelbow = 1.5 HS = (10 + 4.43Q2) ft Note that [Q] = ft3/s. (11.40) This is the head that must be added to the fluid by the pump in order to move the fluid at the volumetric flow rate Q. With [Q] = gpm, Eqn. (11.40) becomes: HS = (10 + 2.25*10-5Q2) ft Note that [Q] = gpm. C. Wassgren Chapter 11: Fluid Machinery 458 (11.41) Last Updated: 16 Aug 2009 Plot Eqn. (11.41) on the pump performance curve to determine the operating point. system curve (Eqn. (11.41)) From the figure we observe that the operating point occurs at: Q 1600 gpm corresponding to a head rise and efficiency of H 67 ft 84% The operating efficiency is close to the optimal efficiency of 86% so this is a good pump to use. C. Wassgren Chapter 11: Fluid Machinery 459 Last Updated: 16 Aug 2009 Chapter 12: Gas Dynamics 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. Introduction to Gas Dynamics Equations of State One-Dimensional Flow Speed of Sound and the Mach Cone Adiabatic, 1D Compressible Flow of a Perfect Gas Effects of Area Change on Steady, 1D, Isentropic Flow Adiabatic Flow with Friction (Fanno Flow) Flow with Heat Transfer (Rayleigh Flow) Normal Shock Waves Flow in Converging-Diverging Nozzles Flows with Mass Addition Generalized Steady, One-Dimensional Flow Oblique Shock Waves Expansion Waves Reflection and Interaction of Oblique Shock Waves Reflection and Interaction of Expansion Waves Equations of Motion in Terms of the Velocity Potential Small Perturbation Theory Method of Characteristics Flow Past a Wavy Wall Using Small Perturbation Theory Thin Airfoils in Supersonic Flow Unsteady, 1D Compressible Flow Description of a Shock Tube C. Wassgren Chapter 12: Gas Dynamics 460 Last Updated: 14 Aug 2010 1. Introduction to Gas Dynamics What is Gas Dynamics? Gas dynamics is a branch of fluid mechanics that examines the dynamics of compressible fluid flows and of gases in particular. What is the motivation for studying compressible fluids? Although topics regarding compressible fluid mechanics have been studied since the 1800s, few scientists and engineers were interested in the topic apart from those studying ballistics and steam turbine design. It wasn’t until WW II with the development of high speed planes, rockets, and energetic explosives that the study of compressible flows became widespread. Ever since, the understanding of compressible fluid mechanics has been important in the development of not only the previously mentioned topics, but also of jet engines, rocketry, re-entry spacecraft, gas pipelines, combustion, and gas turbines. What is special about compressible fluids? Compressibility of a fluid results in several important phenomena that are not observed in incompressible fluids. Two of the most significant of these phenomena are shock waves and “choked” flow conditions. Both of these phenomena are the result of the fact that in compressible fluids, pressure disturbances propagate at a finite speed. For example, if one claps their hands, the pressure disturbance caused by the colliding hands propagates into the surrounding atmosphere with a finite speed (equal to the speed of sound). Thus, a finite amount of time passes before the surrounding air recognizes the effects of the clapping hands. In contrast, in a truly incompressible fluid, pressure disturbances propagate at an infinite speed. Thus, pressure disturbances are felt instantly everywhere in the fluid domain. The fact that disturbances travel at finite speed raises the question of what happens if the cause of the pressure disturbance travels faster than the pressure disturbance itself? As an example, let’s consider an aircraft flying in the atmosphere. As the aircraft moves slower than the disturbances propagate, pressure disturbances travel ahead of the aircraft and “inform” the air in front that the aircraft is about to arrive. Thus, the air can move smoothly out of the way as the aircraft approaches. However, if the aircraft travels faster than the speed of propagation, then the air in front of the aircraft can’t move out of the way and begins to “pile up” in front of the aircraft. The result is the formation of a shock wave across which there is a rapid change in the air pressure, temperature, density and velocity. Now let’s consider a different situation. Imagine a large, pressurized tank with a converging nozzle which empties into another large tank (refer to the figure below). While holding the pressure in the left tank constant, let’s begin to reduce the pressure in the right hand tank. pressure in this tank remains constant pressure in this tank can vary When we lower the pressure in the right hand tank, a pressure disturbance propagates upstream to the constant pressure tank and “informs” the fluid upstream that the pressure in the right hand tank has dropped. As a result, the flow rate between tanks increases. As we continue to lower the pressure in the right hand tank, the flow rate continues to increase until we reach a velocity in the converging nozzle where the fluid velocity is equal to the speed at which pressure disturbances propagate. Now if we continue to lower the pressure in the right hand tank, that pressure information can no longer propagate upstream since C. Wassgren Chapter 12: Gas Dynamics 461 Last Updated: 14 Aug 2010 the fluid is flowing in the opposite direction at the same speed. Thus, we have a “choked” flow condition where we can no longer increase the flow rate between the tanks. In addition to these two phenomena, compressible flows have other counter-intuitive behaviors regarding how the fluid velocity varies with the area through which the fluid flows and how the velocity is affected by frictional effects. We’ll investigate all of these phenomena later in this course. What tools are required to study compressible fluid mechanics? Several basic concepts are used in studying compressible fluid mechanics. These include: conservation of mass linear momentum equation conservation of energy (1st law of thermodynamics) 2nd law of thermodynamics equations of state (e.g., the ideal gas law) various concepts from thermodynamics In addition, we’ll require knowledge of calculus, vector calculus, and differential equations (ODEs and PDEs). C. Wassgren Chapter 12: Gas Dynamics 462 Last Updated: 14 Aug 2010 2. Equations of State Data for the various properties of a particular substance or class of substances can typically be found in thermodynamic property tables. Often, it is more convenient and informative to establish analytical relations between these various data. Equations of state describe the relations between the various properties of a substance or class of substances. Note that there can be more than one equation of state. For example, when discussing an ideal gas, we’ll consider both a thermal and caloric equation of state. Before discussing equations of state for various substances, we should first define a few terms. First, we’ll be concerned only with pure substances in these notes: A pure substance is one that has a fixed chemical composition throughout. A pure substance can exist in more than one phase, but its chemical composition must be the same in each phase. Examples of pure substances include distilled water, nitrogen, helium, and carbon dioxide. A uniform mixture of non-reacting gases can be regarded as a pure substance. Note that if one of the component gases changes into a liquid phase (for example if the mixture is cooled), the liquid would have a different composition than the gas phase and thus the system could no longer be considered a pure substance since the system would not have a uniform composition. Another important point to address is the question of how many properties we need to specify before we uniquely define the system state. We know from experience that not every system property must be specified before we can identify the state of a system. In fact, we observe that many properties are related. Thus, only a subset of properties needs to be given to uniquely determine a system’s state. In our studies, we’ll consider simple, compressible systems, defined as those systems where electrical, magnetic, surface tension, gravitational, and motion effects are negligible. The state postulate for a simple, compressible system states that only two independent properties are required to uniquely define the system’s state. If additional effects are significant, e.g. gravitational forces and accelerations, then additional properties are required, e.g. elevation and velocity. C. Wassgren Chapter 12: Gas Dynamics 463 Last Updated: 14 Aug 2010 Relations for an Ideal Gas One particularly important class of substances that is very important in gas dynamics is the ideal gas. An ideal gas is a model describing the behavior of real gases in the limit of zero pressure and infinite temperature (i.e., zero density). It does not account for the interaction between molecules of the gas (e.g. inter-molecular forces). Nevertheless, the ideal gas model is a reasonably accurate model for gases where the system pressure is less than 0.05 times the critical pressure or the system temperature is more than twice the critical temperature (to be discussed). thermal equation of state The thermal equation of state for an ideal gas is given by what is commonly referred to as the ideal gas law: p RT (12.1) where p is the absolute pressure of the gas, is the gas density, R is the gas constant for the gas of interest, and T is the absolute temperature. Notes: 1. Absolute pressures and temperatures must be used when using the ideal gas law. 2. The gas constant, R, will be different for different gases. The gas constant can be determined in terms of the universal gas constant, Ru. R R u (12.2) M where Ru=8314 (Nm)/(kgmoleK)=1545.4 (ftlbf)/(lbmmoleR) and M is the molecular mass of the gas. The gas constant for air is R=287 (Nm)/(kgK)=53.3 (ftlbf)/(lbmR). Note that Mair = 28.98 kg/kgmol. 3. The compressibility factor, Z, is defined as: pv p (12.3) Z RT RT If Z 1 for a gas, then it can be modeled well with the ideal gas model. The compressibility factor, Z, is plotted below for a variety of substances as a function of the reduced pressure, p/pcr, and reduced temperature, T/Tcr, where pcr and Tcr are the critical pressure and temperature. Note that the critical temperature is the temperature above which a gas cannot be liquefied no matter how large a pressure is applied. The critical pressure is the minimum pressure for liquefying a gas at the critical temperature. Notes: 1. For values of p/pcr < 0.05 or T/Tcr > 2, Z 1 so that in this range the ideal gas model works well. 2. The table below gives the critical temperature and pressure values for various substances: TR = T/Tcr PR = p/pcr C. Wassgren Chapter 12: Gas Dynamics 464 Gas air He H2 N2 O2 CO2 CO Tcr [K] 132.41 5.19 33.24 126.2 154.78 304.20 132.91 pcr [atm] 37.25 2.26 12.80 33.54 50.14 72.90 34.26 Last Updated: 14 Aug 2010 T/Tcr > 2 Tair > 265 K p/pcr < 0.05 pair < 1.86 atm caloric (aka energy) equation of state Before discussing the caloric equation of state, we must first define the property known as specific heat. specific heat Recall that earlier it was mentioned that internal energy, sensible energy in particular, is related to temperature. Furthermore, we know from experience that some materials heat up at different rates than others. For example, 4.5 kJ of energy added to a 1 kg mass of iron will raise the iron’s temperature from 20 C to 30 C. To raise 1 kg of water from 20 C to 30 C, however, requires 41.8 kJ; about 9 times the amount of energy than is required to raise the iron’s temperature an equivalent amount. The property that quantifies the energy storage capability of matter is called the specific heat. The specific heat of a substance is the energy required to raise the temperature of a unit mass of a substance by one degree. In general, the energy required to raise the temperature of a substance will depend on the process path. Two particular processes of interest are where the system’s volume is held constant while energy is added and where the pressure in the system is held constant while energy is added. The specific heat at constant volume, cv, is the energy required to raise the temperature of a unit mass by one degree during a constant volume process. The specific heat at constant volume is defined as: u cv (12.4) T v The specific heat at constant pressure, cp, is the energy required to raise the temperature of a unit mass by one degree during a constant pressure process. The specific heat at constant pressure is defined as: h cp (12.5) T p Note that cp is always greater than cv since at constant pressure, the system is allowed to expand and the energy for this expansion work must also be supplied to the system. Now let’s return to our discussion of the caloric equation of state for an ideal gas. Since an ideal gas is considered a simple, compressible system, the internal energy, u, is uniquely determined by two properties. Here we’ll use the properties of temperature, T, and specific volume, v(=1/): u u T , v (12.6) so that any change in the internal energy is given by: u u (12.7) du dT dv T v v T Utilizing our definition for specific heat given in equation (12.4) we have: u (12.8) du cv dT dv v T The second term on the right hand side of equation (12.8) is zero for an ideal gas. (This can be shown using Maxwell’s relations; a topic not addressed in these notes. See for example, Moran and Shapiro, Fundamentals of Engineering Thermodynamics, 3rd ed., Wiley, Section 11.4.2.) Thus, the internal energy of an ideal gas, u, is a function only of the temperature: du u u T du dT (12.9) dT Using the definition given in equation (12.4) we have: C. Wassgren Chapter 12: Gas Dynamics 465 Last Updated: 14 Aug 2010 du cv dT (12.10) Integrating both sides and noting that the specific heat can be a function of temperature in general: T u uref c T dT (12.11) v Tref The enthalpy is also only a function of temperature for an ideal gas as shown below: p h u u RT h h(T ) (12.12) Taking the derivative of enthalpy with respect to the temperature and using the definition given in equation (12.5) we see that: dh h dT c p dT (12.13) dT T h href c p T dT (12.14) Tref where cp is the specific heat at constant pressure which can, in general, be a function of temperature. In addition, dh du RdT cv dT RdT cv R dT (12.15) Comparing equations (12.13) and (12.15) we see that the specific heats for an ideal gas are related in the following manner: c p cv R (12.16) The specific heat ratio, , defined as cp cv appears frequently in gas dynamics. Other helpful relations include: R R cp cv 1 1 C. Wassgren Chapter 12: Gas Dynamics 466 (12.17) (12.18) Last Updated: 14 Aug 2010 1st and 2nd Law Considerations for an Ideal Gas (Gibb’s Equation) Now let’s consider the 1st and 2nd laws for an ideal gas. Recall that the 1st Law (COE) for a system is: (12.19) de qinto system won system From the 2nd law we have: qinto system ds (12.20) T where the equality holds only for a reversible process. If we consider a simple, compressible system (so that de = du) where only reversible pdv work is considered, we can substitute equation (12.20) into equation (12.19) (note that we use the equality in equation (12.20) since we are considering a reversible process): du Tds pdv Tds du pdv du pdv dh vdp (Note: dh d u pv du pdv vdp ) T T T T This equation is also known as the “Tds” equation or Gibb’s equation. ds (12.21) Notes: 1. Even though Eqn. (12.21) was derived for a reversible process, it holds true even for irreversible processes. This is because entropy is a property and is thus independent of the process taken between the end states. For an ideal gas we can substitute in Eqns. (12.10) and (12.13) for du and dh : dT dv dT d dT dp R cv R cp R ds cv T v T T p (12.22) The Perfect Gas A perfect gas is an ideal gas with constant specific heats. The previous relations given for an ideal gas simplify to the following for a perfect gas: equation (12.11) u2 u1 cv T2 T1 (12.23) equation (12.14) h2 h1 c p T2 T1 (12.24) s2 s1 cv ln equation (12.22) c p ln T2 v T R ln 2 cv ln 2 R ln 2 1 T1 v1 T1 T2 p R ln 2 T1 p1 (12.25) isentropic process for a perfect gas Recall that for an isentropic process (a reversible, adiabatic process), ds = 0, so that Eqn. (12.25) becomes (note that density will be used instead of specific volume): T p T c p ln 2 R ln 2 cv ln 2 R ln 2 T1 p1 1 T1 R T2 2 cv T1 1 Using the relations in Eqn. (12.18) we have: C. Wassgren Chapter 12: Gas Dynamics cp R 1 T2 2 T1 1 p2 T2 p1 T1 p2 T2 1 p1 T1 (12.26) 467 (12.27) Last Updated: 14 Aug 2010 Substituting Eqn. (12.26) into Eqn. (12.27): p2 2 p1 1 (12.28) Equations (12.26) through (12.28) will be very useful in our analysis of gas dynamics. C. Wassgren Chapter 12: Gas Dynamics 468 Last Updated: 14 Aug 2010 The Imperfect Gas For most real gases, the compressibility factor is not equal to one, i.e. Z = pv/(RT) 1. However, in most engineering applications the pressure ratio, p/pcr, is small enough and the temperature ratio, T/Tcr, large enough so that the ideal gas model can be used with reasonable accuracy. The specific heats, though, may not be constant so that a perfect gas model cannot be assumed. The specific heats are a strong function of temperature and only weakly dependent on the pressure (over a typical range found in most engineering applications) as shown in the plots below for air. for air for air The figure on the left plots cp/R as a function of temperature for various substances. Note that for air the specific heat is a weak function of temperature for temperatures less than ~1000K. Hence, a perfect gas assumption for air flows with temperatures less than 1000 K is reasonable. The specific heat temperature dependence can be predicted from statistical mechanics models. The variations in specific heat are due to the activation of different energy storage modes (e.g. vibration and rotation) within the gas molecules at different temperatures. More detail on this topic can be found in Callen, H.B., Thermodynamics and an Introduction to Thermostatistics, Wiley. There are several relations used for describing the variation in the specific heats with temperature. It is often easier, however, to use tabulated values for determining the various thermodynamic properties such as the specific internal energy, enthalpy, and entropy. C. Wassgren Chapter 12: Gas Dynamics 469 Last Updated: 14 Aug 2010 Notes: 1. Recall that that for an ideal gas: T dh c p dT h T c p dT where h(T = 0) = 0 (12.29) 0 u h pv h RT (12.30) Values for h and u can typically be found in tables given at the back of most thermodynamics texts. 2. For an ideal gas we also have: T dT dp dT p ds c p R s sref c p R ln T p T pref Tref (12.31) Choosing absolute zero as the reference temperature and defining: T dT s0 cp T 0 gives: T T T2 dT Tref dT p 1 dT ref dT p s2 sref s1 sref c p c p R ln 2 c p c p R ln 1 T T pref 0 T T pref 0 0 0 p p dT 1 dT cp R ln 2 ln 1 T0 T pref pref 0 p 0 s2 s1 s2 s10 R ln 2 p1 0 Values for s can be found in thermodynamics tables. T2 s2 s1 c p C. Wassgren Chapter 12: Gas Dynamics (12.32) T 470 (12.33) Last Updated: 14 Aug 2010 Relations for an Incompressible Substance An incompressible system is one in which the density (or specific volume) of the system remains constant: d dv 0 (12.34) Since the state of a simple system can be determined using two properties, we can write: u=u(T,v) so that: u u (12.35) du dT dv cv dT T v v T 0 cv From the definition of enthalpy we have: dh du p dv vdp du vdp (12.36) 0 The enthalpy can also be written as: h=h(T,p) so that: h h dh dT dp T p p T (12.37) Equating Eqn. (12.37) with Eqn. (12.36) we have: h h dh dT dp du vdp T p p T cv dT c p c p cv for an incompressible substance Equation (12.21) for an incompressible substance reduces to: du dh vdp ds T T T C. Wassgren Chapter 12: Gas Dynamics 471 (12.38) (12.39) Last Updated: 14 Aug 2010 3. One-Dimensional Flow flow dimensionality The dimension of a flow is equal to the number of spatial coordinates required to describe the flow. For example: 0D flow: u=u(t) or u=constant u=u(t; x) or u=u(x) 1D flow: 2D flow: u=u(t; r,) or u=u(r,) 3D flow: u=u(t; x, y, z) or u=u(x, y, z) A flow is uniform if it does not vary in a spatial direction. A flow is non-uniform if it does vary in that spatial direction. For example, the velocity profile in the figure below at left is non-uniform in the y-direction. The figure at the right is uniform in the y-direction. y y u u Notes: 1. The flow of a real fluid through a pipe is not one-dimensional due to the no-slip condition at the pipe walls. If the Reynolds number of the flow is large enough, the flow may be approximated to be 1D within reasonable accuracy. As a flow’s Reynolds number increases, the velocity profile becomes more blunt-shaped and more closely approaches that of a uniform profile. Re small Reynolds number large Reynolds number (laminar) (turbulent) 2. VD where Re is the Reynolds number, V is the average velocity, D is the pipe diameter, and is the kinematic viscosity of the fluid. The one-dimensional flow assumption is exact for a stream filament. a. Recall that there is no flow across a streamline. b. A stream tube is a tube made by all the streamlines passing through a closed curve. There is no flow through a stream tube wall. streamlines c. A stream filiment is a stream tube with infinitesimally small cross-sectional area. C. Wassgren Chapter 12: Gas Dynamics 472 Last Updated: 14 Aug 2010 Governing Equations for One-Dimensional, Steady Flow Let’s write our governing equations (COM, LME, COE, 2nd Law) for a one-dimensional, steady flow. Conservation of Mass (COM) VA VA+d(VA) dx d dt where d dt dV u CV rel dA 0 CS dV 0 (steady flow) CV u rel dA VA VA d VA CS d VA 0 d VA C. Wassgren Chapter 12: Gas Dynamics m constant (12.40) 473 Last Updated: 14 Aug 2010 Linear Momentum Equation (LME) (p+1/2dp)dA V2A+d(V2A) 2 V A pA pA+d(pA) Let P be the pipe perimeter Pdx dx x Since the flow is 1D, consider only the x-component of the LME: d u x dV u x u rel dA Fx, body Fx , surface dt on CV on CV CV where d dt CS u dV 0 (steady flow) x CV u x CS u rel dA V 2 A V 2 A d V 2 A d V 2 A but, from COM we have: m VA =constant u x u rel dA d V 2 A d mV mdV VAdV CS Fx, body 0 on CV Fx, surface pA pA d pA p 1 2 dp dA Pdx on CV d ( pA) pdA Pdx Adp Pdx Note that higher-order terms have been neglected in the previous expression and that the friction force acts only in the x-direction since the boundaries vary smoothly (the slope is small, no discontinuities). We’ll re-write the friction force term using a hydraulic diameter, DH: 4A DH P and a friction factor, f: f F 1 2 V 2 fF, Fanning friction factor or 1 4 fD 1 2 V 2 fD, D’Arcy friction factor so that the momentum equation becomes, after substitution and re-arranging: 4f dp VdV 1 2 V 2 F dx 0 DH C. Wassgren Chapter 12: Gas Dynamics 474 (12.41) (12.42) (12.43) (12.44) Last Updated: 14 Aug 2010 Notes: 1. Gravitational effects have been neglected in the previous analysis since when dealing with gases, gravitational effects are typically very small compared to other terms in the momentum equation. 2. The D’Arcy friction factor, fD, is the friction factor found in the Moody diagram for pipe flows. 3. For a frictionless flow, equation (12.44) simplifies to: dp VdV 0 dp 1 2V 2 constant (12.45) Bernoulli’s Equation! The integral occurs because the pressure can, in general, be a function of the density. a. For an incompressible fluid,=constant, so that, after integrating along a streamline: p12 2 V constant BE for incompressible fluid b. (12.46) For an ideal gas: p=RT, so that, for an isothermal process: RT ln 12 2 V constant 0 BE for ideal gas (isothermal) (12.47) where 0 is a reference density. c. For a perfect gas undergoing an isentropic process, p-=constant: p c dp c 1 p c 1 1 p 1 1 1 1 p p0 dp c 1 but the constant, c, is given by: 1 1 1 c p p0 0 c p 1 p0 0 1 Substituting and simplifying we find that the momentum equation can be simplified to p 1 2 2 V constant 1 Note that from the ideal gas law, p/ = RT and from Eqn. (12.18): c pT 1 2 V 2 constant BE for perfect gas (isentropic) C. Wassgren Chapter 12: Gas Dynamics 475 (12.48) Last Updated: 14 Aug 2010 Conservation of Energy (COE) Won CV Qinto CV dx d dt where d dt e dV h CV 2V 2 gz u rel dA Qinto CV Wshaft, Wother, on CV CS e dV 0 CV h 1 1 2V 2 gz on CV (steady flow) u rel dA m h CS md h 1 1 2V 2V 2 2 h 1 2V 2 d h 1 2V 2 Note that the gravitational potential term has been neglected since this term is typically much smaller than the other terms in the equation for gases. Qinto CV Qinto CV W W on CV on CV Substituting and simplifying we have: d h 1 2 V 2 qinto won CV (12.49) CV Notes: 1. For an adiabatic flow ( qinto CV 0 ) with no work ( won CV 0 ): h 1 2 V 2 constant COE for adiabatic flow with no work (i.e. isentropic flow) This is the same as Eqn. (12.48) since for a perfect gas, h = cpT. C. Wassgren Chapter 12: Gas Dynamics 476 (12.50) Last Updated: 14 Aug 2010 2nd Law of Thermodynamics (aka Entropy Equation) Qinto CV ms d ms ms dx d dt where d dt s dV s u CV rel CS dA qinto CV T CV s dV 0 CV s u rel dA m s s ds mds CS CV qinto CV T qinto CV T Substitute and simplify. q ds into CV (where dqinto CV is the heat added to the CV per unit mass) T (12.51) Notes: 1. For an adiabatic flow ( qinto CV 0 ): ds 0 where the equality holds only for a reversible flow (adiabatic and reversible isentropic). C. Wassgren Chapter 12: Gas Dynamics 477 Last Updated: 14 Aug 2010 4. Speed of Sound The speed of sound, c, in a substance is the speed at which infinitesimal pressure disturbances propagate through the surrounding substance. To understand how the speed of sound depends on the substance properties, let’s examine the following model. Consider a wave moving at velocity, c, through a stagnant fluid. Across the wave, the fluid properties such as pressure, p, density, , temperature, T, and the velocity, V, can all change as shown in the figure below. downstream upstream V V=0 p T p+p + T+T c Now let’s change our frame of reference such that it moves with the wave: downstream upstream c-V V=c p T p+p + T+T stationary wave Now apply COM and the LME on a very thin CV surrounding the wave: downstream upstream c-V V=c p T C. Wassgren Chapter 12: Gas Dynamics p+p + T+T A 478 Last Updated: 14 Aug 2010 COM: d dt where d dt dV u CV rel dA 0 CS dV 0 CV u rel dA cA c V A CS Substitute and simplify. cA c V A c c V c V c V LME: d dt where d dt u dV u x CV x (12.52) u rel dA Fx, body Fx, surface CS on CV on CV u dV 0 x CV u x u rel dA mc m c V mV CS cAV Fx, body 0 on CV Fx , surface pA p p A pA on CV Substituting and simplifying: cAV pA p c V Substituting the result from equation (12.52): p c c c2 p 1 C. Wassgren Chapter 12: Gas Dynamics (12.53) 479 Last Updated: 14 Aug 2010 For a sound wave, the changes across the wave are infinitesimal so that: p p c 2 lim 1 p dp (12.54) d We also need to specify the process by which these changes occur. Since the changes across the wave are infinitesimal, we can regard the wave as a reversible process. Additionally, the temperature gradient on either side of the wave is small so there is negligible heat transfer into the control volume so the process is adiabatic. Thus, the changes across a sound wave occur isentropically (an adiabatic, reversible process is also isentropic): p c2 speed of sound in a continuous substance (12.55) s Notes: 1. Note that equation (12.55) is the speed of sound in any continuous substance. It’s not limited to fluids. 2. If the wave is not “weak” (i.e. the changes in the flow properties across the wave are not infinitesimal), then viscous effects and temperature gradients within the wave will be significant and the process can no longer be considered isentropic. We will discuss this situation later when examining shock waves. 3. Note that according to equation (12.54) the stronger the wave, i.e. the greater , the faster the wave will propagate. This will also be examined when discussing shock waves. 4. For an ideal gas undergoing an isentropic process (ds = 0): R d R dp p p = RT cv c p p s Substituting into equation (12.55), the sound speed for an ideal gas is: c RT speed of sound in an ideal gas (12.56) (12.57) Notes: a. The absolute temperature must be used when calculating the speed of sound. b. The speed of sound in air (=1.4, R=287 J/(kgK)) at standard conditions (T=288 K) is 340 m/s (=1115 ft/s 1/5 mile/s). c. It is not unexpected that the speed of sound is proportional to the square root of the temperature. Since disturbances travel through the gas as a result of molecular impacts, we should expect the speed of the disturbance to be proportional to the speed of the molecules. The temperature is equal to the random kinetic energy of the molecules and so the molecular speed is proportional to the square root of the temperature. Thus, the speed of sound is proportional to the square root of the temperature. C. Wassgren Chapter 12: Gas Dynamics 480 Last Updated: 14 Aug 2010 5. Equation (12.55) can also be written in terms of the bulk modulus. The bulk modulus, E of a substance is a measure of the compressibility of the substance. It is defined as the ratio of a differential applied pressure to the resulting differential change in volume of a substance at a given volume: p p E (12.58) V V Notes: 1. dp>0 dV<0 E>0 2. From COM: dV/V = -d 3. E compressibility dp volume before: volume after: V V-dV The isentropic bulk modulus, E|s is defined as: p p E s V V s (12.59) Thus, the speed of sound can also be written as: E s c2 alternate sound speed expression (12.60) Notes: a. The isentropic bulk modulus for air is E|s = RT. b. The isentropic bulk modulus for water is 2.19 GPa so that the speed of sound in water (=1000 kg/m3) is 1480 m/s (=4900 ft/s 1 mile/s 5X faster than the speed of sound in air at sea level). c. For solids, the bulk modulus, E, is related to the modulus of elasticity, E, and Poisson’s ratio, , by: E 3 1 2 E For many metals (e.g. steel and aluminum), Poisson’s ratio is approximately 1/3 so that EE 1. The speed of sound in stainless steel (E = 163 GPa; = 7800 kg/m3) is 4570 m/s (= 15000 ft/s 3 mile/s 3X faster than the speed of sound in water). C. Wassgren Chapter 12: Gas Dynamics 481 Last Updated: 14 Aug 2010 6. The Mach number, Ma, is a dimensionless parameter that is commonly used when discussing compressible flows. The Mach number is defined as: V Ma (12.61) c where V is the flow velocity and c is the speed of sound in the flow. Notes: a. Compressible flows are often classified by their Mach number: Ma < 1 subsonic Ma = 1 sonic Ma > 1 supersonic Additional sub-classifications include: Ma < 0.3 incompressible transonic Ma 1 Ma > 5 hypersonic b. The square of the Mach number, Ma2, is a measure of a flow’s macroscopic kinetic energy to its microscopic kinetic energy. 7. The change in the properties across the sound wave can be found from the following: COM (note that A = constant) cA d c dV A 0 dV cd 0 dV d c Speed of Sound (isentropic process): dp c2 d (12.62) (12.63) Ideal Gas: p RT dp d dT p T (12.64) Combine Eqns. (12.62) and (12.63) and simplify. 2 dp RT dp c c p p d RT p dV dV d 1 dp c p (12.65) Now combine Eqns. (12.65) and (12.64). dp 1 dp dT p p T dT 1 dp T p (12.66) Thus, across a compression wave (dp > 0): dV > 0, d > 0, and dT > 0. Across a rarefaction, or expansion wave (dp < 0): dV < 0, d < 0, and dT < 0. C. Wassgren Chapter 12: Gas Dynamics 482 Last Updated: 14 Aug 2010 The Mach Cone Consider the propagation of infinitesimal pressure waves, i.e., sound waves, emanating from an object at rest. The waves will travel at the speed of sound, c. object (V=0) location of sound pulse after time t c(t) c(3t) c(2t) Now consider an object moving at subsonic velocity, V<c: V(2t) The pressure pulses are more closely spaced in the direction of the object’s motion and more widely spaced behind the object. Thus, the frequency of the sound in front of the object increases, while the frequency behind the object decreases. This is known as the Doppler Shift. V(3t) V(t) c(3t) c(2t) c(t) direction of object motion Now consider an object traveling at the sonic speed, V=c: Since no wave fronts propagate ahead of the object, an observer standing in front of the object will not hear it approaching until the object reaches the observer. V(3t) V(2t) V(t) c(2t) c(3t) Note that the infinitesimal pressure changes in front of the object “pile up” on one another producing a sudden, finite pressure change (a shock wave!) c(t) locus of all wave fronts direction of object motion C. Wassgren Chapter 12: Gas Dynamics 483 Last Updated: 14 Aug 2010 Lastly, consider an object travelling at supersonic speeds, V>c: The object out runs the pressure pulses it generates. V(3t) V(2t) V(t) c(3t) c(2t) The locus of wave fronts forms a cone which is known as the Mach Cone. The object cannot be heard outside the Mach Cone. This region is termed the zone of silence. Inside the cone, the region known as the zone of action, the object can be heard. c(t) Mach Cone (locus of all wave fronts) zone of action zone of silence sin 1 1 Ma direction of object motion C. Wassgren Chapter 12: Gas Dynamics The angle of the cone, known as the Mach angle, , is given by: ct c 1 sin V t V Ma 484 (12.67) Last Updated: 14 Aug 2010 5. Adiabatic, 1D Compressible Flow of a Perfect Gas Now let’s consider the 1D, adiabatic flow of a compressible fluid. Recall that from the energy equation we have: h 1 2 V 2 constant (12.68) For a perfect gas we can re-write the specific enthalpy in terms of the specific heat at constant pressure and the absolute temperature: h=cpT: c pT 1 2 V 2 constant T V2 constant 2c p (12.69) We can re-write this equation in terms of the Mach number: V V Ma c RT V 2 RT Ma 2 T RTMa 2 2c p constant Substituting the following relation: R 1 cp results in: 1 T 1 Ma 2 constant 2 Adiabatic, 1D Flow of a Perfect Gas (12.70) If the flow can also be considered reversible, thus making the flow isentropic (adiabatic + reversible = isentropic), we can use the isentropic relations for a perfect gas: p constant T 1 1 and constant T 1 to give: 1 1 p 1 Ma 2 constant 2 (12.71) 1 1 1 1 Ma 2 constant 2 Isentropic, 1D Flow of a Perfect Gas C. Wassgren Chapter 12: Gas Dynamics (12.72) 485 Last Updated: 14 Aug 2010 Stagnation and Sonic Conditions It is convenient to choose some significant reference point in the flow where we can evaluate the constants in equations (12.70)-(12.72). Two such reference points are commonly used in compressible fluid dynamics. These are the stagnation conditions and sonic conditions. Stagnation Conditions Stagnation conditions are those conditions that would occur if the fluid is brought to rest (zero velocity Ma=0). We use the isentropic stagnation conditions where the flow is brought to rest isentropically. These conditions are typically indicated by the subscript “0”. Thus, equations (12.70)-(12.72) can be written in terms of stagnation conditions: T 1 1 Ma 2 T0 2 1 adiabatic, 1D flow of a perfect gas (12.73) isentropic, 1D flow of a perfect gas (12.74) isentropic, 1D flow of a perfect gas (12.75) p 1 1 Ma 2 1 p0 2 1 1 2 1 1 Ma 0 2 We can also determine the speed of sound at the stagnation conditions using the fact that c=(RT)1/2: c 1 1 Ma 2 c0 2 12 (12.76) adiabatic, 1D flow of a perfect gas Notes: 1. Stagnation conditions are also commonly referred to as total conditions (given by the subscript “T”). 2. Stagnation conditions can be determined even for a moving fluid. The fluid doesn’t necessarily have to be at rest to state its stagnation conditions. To determine stagnation conditions we only need to consider the conditions if the flow were brought to rest. 3. Equations (12.74) and (12.75) are for a flow brought to rest isentropically. 4. Tables listing the values of equations (12.73)-(12.76) for various Mach numbers are typically given in the back of most textbooks concerning compressible fluid flows. 5. Note that the stagnation temperature is greater than the flow temperature. This is because when the flow is decelerated to zero velocity, the macroscopic kinetic energy is converted to internal energy (microscopic kinetic energy) and thus the temperature increases. 1.2 1 ratio 0.8 T/T0 p/p0 0.6 r/r0 0.4 c/c0 0.2 0 0 1 2 3 Ma C. Wassgren Chapter 12: Gas Dynamics 486 4 5 6 =1.4 Last Updated: 14 Aug 2010 6. The stagnation pressure is a significant property for a flow because it is directly related to the amount of work we can extract from the flow. For example, imagine we bring a flow to rest so that we have stagnation conditions within the tank shown below. tank with stagnation pressure, p0 piston used for doing work flow The larger the stagnation pressure in the tank, the greater the force we can exert on a piston that can be used to perform useful work. C. Wassgren Chapter 12: Gas Dynamics 487 Last Updated: 14 Aug 2010 Sonic Conditions Another convenient reference point is where the flow has a Mach number of one (Ma=1). Conditions where the Mach number is one are known as sonic conditions and are typically specified using the superscript “*”. Equations (12.73)-(12.76) evaluated at sonic conditions are: T * 1 1 T0 2 1 adiabatic, 1D flow of a perfect gas (12.77) isentropic, 1D flow of a perfect gas (12.78) isentropic, 1D flow of a perfect gas (12.79) adiabatic, 1D flow of a perfect gas (12.80) T* 0.8333 T0 adiabatic, 1D flow of air (12.81) p* 0.5283 p0 isentropic, 1D flow of air (12.82) * 0.6339 0 isentropic, 1D flow of air (12.83) c* 0.9129 c0 adiabatic, 1D flow of air (12.84) p* 1 1 1 p0 2 1 * 1 1 1 0 2 c* 1 1 c0 2 12 Notes: 1. For air, =1.4 so that: On the Isentropic Flow Assumption In many engineering gas dynamics flows, the assumption that the entropy of the fluid remains constant (an isentropic process) is a good one. If viscous and heat transfer effects can be neglected, then we can reasonably assume that the flow is isentropic (isentropic = reversible + adiabatic). This is often the case for flows through short, insulated ducts or through stream tubes not passing through a boundary layer or a shock wave (strong viscous effects occur in both cases). Experiments have verified that the isentropic assumption under these conditions is reasonable. C. Wassgren Chapter 12: Gas Dynamics 488 Last Updated: 14 Aug 2010 Mollier (aka h-s) Diagrams Mollier diagrams are diagrams that plot the enthalpy (h) as a function of entropy (s) for a process. They are often useful in visualizing trends. Notes: 1. Sketches of constant pressure and constant volume (or density) curves are shown in the figure below. h p curves of constant p curves of constant v s 2. For a perfect gas, curves of constant volume (or density) and constant pressure have slopes given by: Tds du pdv Tds dh vdp cv c p dT pdv dh Tds T cp ds p dh 3. dh T ds v The difference between the flow enthalpy and the stagnation enthalpy for an isentropic process is equal to the specific kinetic energy: h p0 stagnation state h0 1 /2V2 p h0 h 1 2 V 2 h s0 isentropic process C. Wassgren Chapter 12: Gas Dynamics s 489 Last Updated: 14 Aug 2010 6. Effects of Area Change on Steady, 1D, Isentropic Flow Mass conservation states that for a steady, 1D, incompressible flow, a decrease in the area will result in an increase in velocity (and visa-versa). This is not necessarily true, however, for a compressible flow as will be shown below. Consider mass conservation for a steady, 1D flow: m AV constant d AV 0 AVd VdA AdV 0 d dA dV 0 AV (12.85) Notes: 1. If the flow is incompressible, then d=0 and we see that: dV/V = -dA/A. Thus, if the area decreases (dA<0), then the velocity must increase (dV>0). 2. For a compressible fluid, the density may change so we need an additional relationship between density and either area or velocity to draw any conclusions about how changes in area affect changes in velocity. Recall that the speed of sound is given by: p c2 s Let’s concern ourselves with an isentropic flow, s=constant, so we can re-write this expression as: dp d 2 c We’ll also make use of Bernoulli’s equation (COLM): dp VdV 0 (12.86) (12.87) Substituting equations (12.86) and (12.87) into equation (12.85) gives: 1 dp dA dV 0 AV c2 V 2 dV dA dV 0 AV c2 V V2 dV dA 2 1 c V A Ma 1 dV dA V A 2 (12.88) Note that the trends for pressure and density are opposite the trends for velocity: dp dV d dV From equations (12.88) and (12.85): From equation (12.87): Ma 2 2 V V V For an ideal gas: ds 0 cv dT d dT R d dT dV R 1 Ma 2 T T cv T V dV 0 dT 0 d Ma 0 dV 0 dT 0 d Ma 0 C. Wassgren Chapter 12: Gas Dynamics 490 Last Updated: 14 Aug 2010 Let’s interpret equation (12.88). Consider the following the cases: Ma < 1 (subsonic flow) dA < 0 dV > 0 dA > 0 dV < 0 (decreases in area result in increases in velocity) (increases in area result in decreases in velocity) A subsonic nozzle should have a decreasing area. A subsonic diffuser should have an increasing area. The area-velocity relationships for subsonic flow are similar to that for incompressible flow. Ma > 1 (supersonic flow) dA < 0 dV < 0 (decreases in area result in decreases in velocity) dA > 0 dV > 0 (increases in area result in increases in velocity) A supersonic nozzle should have an increasing area. A supersonic diffuser should have a decreasing area. The area-velocity relationships for supersonic flow are opposite to that for subsonic flow. Ma = 1 (sonic flow) dA = 0 (sonic conditions must occur at an inflection point in the area) The area at which Ma=1 must be a minimum based on the previous relationships for subsonic and supersonic flow. For example, if we assume that Ma=1 occurs at a maximum, then if the flow starts off subsonic and we increase the area, the Mach number will decrease and diverge from Ma=1. If the flow starts off supersonic and the area increases, then the Mach number will increase and diverge from Ma=1. Thus, we see that the Mach number will not occur at the maximum area. Notes: 1. Nothing can be said about how the velocity is changing when the Ma=1. The velocity can either be decreasing, remaining constant, or increasing. 2. A minimum area does not necessarily imply that the Mach number will be one, i.e.: Ma 1 minimum area (12.89) minimum area Ma 1 This is because if dA=0 in equation (12.88), then either Ma=1 or dV=0. Ma=1 Ma=1 Ma<1 Ma>1 Ma<1 or Ma>1 Ma<1 or Ma>1 area must be a minimum Now let’s examine some other consequences resulting from mass conservation. C. Wassgren Chapter 12: Gas Dynamics 491 Last Updated: 14 Aug 2010 Since the mass flow rate must be a constant in 1D, steady flow, we can write: m AV * A*V * where the “*” quantities are the conditions where Ma=1. Let’s re-arrange this equation and substitute the isentropic flow relations we derived previously. A * V * A* V * 0 c* cMa 0 * 0 c* c0 1 c c Ma 0 0 where * 1 1 0 2 1 1 1 2 1 Ma 0 2 c* 1 1 2 c0 1 1 12 c 1 Ma 2 1 2 c0 12 8 Substituting and simplifying we have: 1 2 1 6 5 A/A* 1 2 1 Ma A 1 2 A* Ma 1 1 2 7 (12.90) 4 3 2 1 0 0 1 2 3 Ma C. Wassgren Chapter 12: Gas Dynamics 492 Last Updated: 14 Aug 2010 1.4 4 Notes: 1. 2. 3. 4. Equation (12.90) tells us what area we would need to contract to to get sonic conditions (Ma=1, A=A*) given the current Mach #, Ma, and area, A. We could also interpret equation (12.90) as saying, given the area for sonic conditions, A*, the Mach number, Ma, and area, A, are directly related for an isentropic flow. Recall that this relationship results from conservation of mass and the assumption of an isentropic flow. Values for A/A* as a function of Mach # are typically included in compressible flow tables found in the appendices of most fluid mechanics textbooks. What happens if we constrict the area to a value less than A*? For a subsonic flow, the new area information can propagate upstream and downstream and, as a result, the conditions everywhere change (i.e., the Mach numbers change according to equation (12.90) where the new area would be A*.) If the upstream flow is supersonic, then some non-isentropic process must occur upstream (a shock wave) so that the constricted area is no longer less than A*. C. Wassgren Chapter 12: Gas Dynamics 493 Last Updated: 14 Aug 2010 Choked Flow Consider the flow of a compressible fluid from a large reservoir into the surroundings. Let the pressure of the surroundings, called the back pressure, pB, be controllable: stagnation conditions p0, T0, 0 throat conditions Ath, pth, Vth back pressure, pB When pB=p0 there will be no flow from the reservoir since there is no driving pressure gradient. When the back pressure, pB, is decreased, a pressure wave propagates through the fluid in the nozzle and into the tank. Thus, the fluid in the tank “is informed” that the pressure outside has been lowered and a pressure gradient is established resulting in fluid being pushed out of the tank. pressure wave propagating upstream Thus, when pB < p0, the fluid will begin to flow out of the reservoir. Furthermore, as pB/p0 , Vth , and the mass flow rate . Note that the flow through the nozzle will be subsonic (Ma<1) since the fluid starts off from stagnation conditions and does not pass through a minimum until reaching the throat. Additionally, since the flow is subsonic, the pressure at the throat will be the same as the back pressure, i.e., pth=pB. That this is so can be seen by noting that if pth>pB, then the flow would expand upon leaving the nozzle and as a result, the jet velocity would decrease, and the pressure would increase. Thus, the jet pressure would diverge from the surrounding pressure. But the jet must eventually reach the surrounding pressure so the assumption that pth>pB must be incorrect. A similar argument can be made for pth<pB. As we continue to decrease pB/p0, we’ll eventually reach a state where the velocity at the throat will reach Ma=1 (Vth=V*=c*). The pressure ratio at the throat will then be: pth p p* 1 B 1 2 p0 p0 p0 and the fluid velocity will be: 1 (12.91) Vth V * c* RT * C. Wassgren Chapter 12: Gas Dynamics (12.92) 494 Last Updated: 14 Aug 2010 Any further decrease in pB has no effect on the velocity at the throat since the pressure information can no longer propagate upstream into the reservoir. The fluid velocity out of the tank is the same as the speed of the pressure wave into the tank so that the pressure information can’t propagate upstream of the throat. Thus, all flow conditions upstream of the throat will remain unchanged. As a result, we can no longer increase the mass flow rate from the tank by changing the back pressure. This condition is referred to as choked flow conditions. The maximum, or choked, mass flow rate will be the same as the mass flow rate at the throat where sonic conditions occur: p* * * mchoked *V * A* VA RT * p* A* RT * where equation (12.92) has been utilized. Substituting the following isentropic relations: p* 1 p p0 1 2 p0 * 1 p0 1 T* 1 T0 1 T0 2 T0 and simplifying results in: T* 1 mchoked * 1 21 1 p0 A 2 RT0 (12.93) Notes: 1. The choked mass flow rate (equation (12.93)) is the maximum mass flow rate that can be achieved from the reservoir. 2. A quick check to see if the flow will be choked or not is to check if the back pressure is less than or equal to the sonic pressure, i.e.: pth pB p* 1 1 p* 1 (12.94) , then the flow will be choked and p p . p0 p0 2 0 0 What happens outside of the nozzle if the back pressure is less than the sonic pressure? The flow must eventually adjust to the surrounding pressure. It does so by expanding in a two-dimensional process known as an expansion fan; a topic we’ll address later in the course. If 3. Ma=1 pth/p0=p*/p0 expansion fans pB/p0<p*/p0 C. Wassgren Chapter 12: Gas Dynamics 495 Last Updated: 14 Aug 2010 4. The previously described processes are sketched in the following plots: pth/p0 m 1 mchoked p*/p0 p*/p0 1 pB/p0 p*/p0 1 pB/p0 p/p0 1 pB/p0 p*/p0 pressure variations in expansion fan throat C. Wassgren Chapter 12: Gas Dynamics 496 x Last Updated: 14 Aug 2010 Steady, 1D, Isentropic Flow of an Imperfect Gas Imperfect gas effects become significant when dealing with high temperature and high Mach number flows. Examples of imperfect gas effects include: 1. variations in specific heats due to the activation of intra-molecular energy modes (i.e. rotational, vibration, electronic modes), 2. dissociation of molecules, e.g. for T > 2000 K, O2 2O and for T > 4000 K, N 2 2 N 3. ionization of atoms, e.g. for T > 9000 K, O O e and N N e 4. chemical reactions. Non-equilibrium effects can also be significant at high Mach numbers. In these notes, we’ll focus only on the variations in specific heats due to temperature. Recall that the governing equations are: COM: d VA 0 LME: COE: 2nd Law: (12.95) dp VdV 0 (12.96) (12.97) (12.98) dh VdV 0 ds 0 Notes: 1. These equations are independent of the type of fluid being considered. 2. The COE and LME statements are equivalent for an isentropic flow. Recall that combining the first and second laws for a simple, compressible system with reversible pdv work gives: du Tds pdv du d pv Tds pdv d pv (12.99) dh Tds vdp but since ds = 0 and v = 1/: dp dh (12.100) Substituting into COE gives: dp VdV 0 (12.101) This is the same relation that we had from the LME! Hence, the COE and LME expressions are equivalent. 3. Equations (12.95)-(12.98) (3 equations, recall from note 2 that equations (12.96) and (12.97) are equivalent) have a total of 5 unknowns (, p, V, h, s). Note that area, A, is typically a known quantity. We need two additional extra relations to close the system of equations. These relations are the equations of state for the substance of interest. p, s h h p, s The equations of state may be given in equation, tabular, or graphical form. To solve the system of equations, one must generally employ an appropriate numerical scheme. a. Note that only two independent properties are required to fix the state of a simple compressible system at equilibrium. C. Wassgren Chapter 12: Gas Dynamics 497 Last Updated: 14 Aug 2010 4. This section remains incomplete. Empirical specific heat data is often expressed in terms of curve fits. Exponential and polynomial curve fits are common. For example, Zucrow and Hoffman use a 4th-order polynomial curve fit: cp (12.102) a bT cT 2 dT 3 eT 4 R where [T] = K and the constants a – e are given in the table below for various substances and for various temperature ranges. gas O2 N2 CO CO2 A H2 H2O CH4 C2H4 a. b*103 -1.87822 0.736183 -1.20815 1.51549 -1.61910 1.48914 8.73510 3.09817 0 0 2.67652 0.511195 -1.10845 2.94514 -3.97946 10.4168 11.3831 11.4918 a 3.62560 3.62195 3.67483 2.89632 3.71009 2.98407 2.40078 4.46080 2.50000 2.50000 3.05745 3.10019 4.07013 2.71676 3.82619 1.50271 1.42568 3.45522 c*106 7.05545 -0.196522 2.32401 -0.572353 3.69236 -0.578997 -6.60709 -1.23926 0 0 -5.80992 0.0526442 4.15212 -0.802243 24.5583 -3.91815 7.98900 -4.36518 d*109 -6.76351 0.0362016 -0.632176 0.0998074 -2.03197 0.103646 2.00219 0.227413 0 0 5.52104 -0.0349100 -2.96374 0.102267 -22.7329 0.677779 -16.2537 0.761551 e*1012 2.15560 -0.00289456 -0.225773 -0.00652236 0.239533 -0.00693536 0.000632740 -0.0155260 0 0 -1.81227 0.00369453 0.807021 -0.00484721 6.96270 -0.0442837 6.74913 -0.0501232 h0*10-3 -1.04752 -1.120198 -1.06116 -0.905862 -14.3563 -14.2452 -48.3775 -48.9614 -0.745375 -0.745375 -0.988905 -0.877380 -30.2797 -29.9058 -10.1450 -9.97871 5.33708 4.47731 s00 4.30528 3.61510 2.35804 6.16151 2.95554 6.34792 9.69515 -0.986360 4.36600 4.36600 -2.29971 -1.96294 -0.322700 6.63057 0.866901 10.7071 14.6218 2.69879 T range [K] 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 300-1000 1000-5000 Note that table also includes data for calculating the specific enthalpy, h, as a function of temperature: T h c dT p T0 h h0 aT 1 bT 2 1 cT 3 1 dT 4 1 eT 5 2 3 4 5 R where h = h0R when T = T0. b. (12.103) The table also includes data for calculating s0, a quantity that will be discussed in the following note (note #5). T 0 s c T0 p dT T 0 s 0 s0 a ln T bT 1 cT 2 1 dT 3 1 eT 4 2 3 4 R 0 0 where s = s 0 R when T = T0. c. (12.104) The specific internal energy can be determined using the definition for enthalpy and the ideal gas law: u h pv h RT (12.105) C. Wassgren Chapter 12: Gas Dynamics 498 Last Updated: 14 Aug 2010 d. The specific heat at constant volume, cv, and the specific heat ratio, , are determined from: cv c p R (12.106) e. cp (12.107) cv Include calculations for dry air – incomplete. The composition of clean dry atmospheric air near sea level. Component % by Volume N2 78.084 O2 20.9476 A 0.934 CO2 0.0314 H2 0.00005 Ne 0.001818 Kr 0.000114 Xe 0.0000087 He 0.000524 CH4 0.0002 NO 0.00005 f. 5. Alternate cp expressions for monatomic and diatomic molecules – incomplete. In order to simplify matters when dealing with isentropic flow of an ideal, but imperfect gas (one where the specific heats vary with temperature), we’ll define some additional useful properties. Note that from equation (12.99): dh dp ds T T For an ideal gas (dh = cpdT and p = RT): dT dp ds c p R T p Integrating and using some reference state as the lower limit of integration gives: T2 s2 s1 c p T1 p dT R ln 2 T p1 (12.108) (12.109) (12.110) Define s0: T s0 c Tref p dT T (12.111) where Tref is some reference state so that equation can be written as: p 0 0 s2 s1 s2 s1 R ln 2 p1 (12.112) Note that equation (12.111) is a function only of temperature. Tabulated values of s0 are commonly given in the tables of most thermodynamics or gas dynamics texts. For an isentropic process, equation (12.112) can be written as: s0 s0 p2 exp 2 1 R p1 C. Wassgren Chapter 12: Gas Dynamics 499 (12.113) Last Updated: 14 Aug 2010 If we define state 1 as the reference state, then equation (12.113) simplifies to: s0 p pr exp (12.114) R pref where pr is known as the relative pressure. Tabulated values of pr are commonly given in the tables of most thermodynamics or gas dynamics texts. Note that for an isentropic process: p2 pr 2 (12.115) p1 pr1 C. Wassgren Chapter 12: Gas Dynamics 500 Last Updated: 14 Aug 2010 Example: Consider the steady, 1D, isentropic flow of air from a tank. Within the tank the pressure and temperature are 10 MPa and 3000 K, respectively. Determine the temperature and velocity at a location downstream of the tank where the pressure is 100 kPa if: a. the air is assumed to behave as a perfect gas, and b. the air is ideal but has variable specific heats. C. Wassgren Chapter 12: Gas Dynamics 501 Last Updated: 14 Aug 2010 Example: A wind tunnel using air operates with a stagnation pressure and temperature of 10 atm and 1000 K, respectively. The test section of the tunnel is designed to operate at a Mach number of 5. Determine the pressure and temperature in the wind tunnel at this design Mach number: a. assuming the perfect gas behavior, and b. assuming imperfect gas behavior. C. Wassgren Chapter 12: Gas Dynamics 502 Last Updated: 14 Aug 2010 7. Adiabatic Flow with Friction (Fanno Flow) So far we’ve examined adiabatic, reversible flows (isentropic flows). For flows in long ducts, the frictional, or viscous, effects can be significant and the flow can no longer be modeled as isentropic. long duct: need to include friction non-isentropic (irreversible) short duct: friction negligible isentropic (reversible) To examine the effects of friction, let’s consider the case of a steady, 1D, adiabatic flow in a constant area duct that has irreversible effects due to friction (viscous) effects. First let’s derive the governing equations for the following differential control volume: V+dV V pA P is the perimeter of the duct (p+dp)A wPdx dx x From COM we have: d dV u rel dA 0 dt CV where d dt CS dV 0 (steady flow) CV u rel dA VA VA d VA CS d VA Thus, d VA AdV Vd A 0 (recall that A=constant) d dV 0 V C. Wassgren Chapter 12: Gas Dynamics (12.116) 503 Last Updated: 14 Aug 2010 From the LME in the x-direction we have: d u x dV u x u rel dA Fx,body Fx,surface dt on CV on CV CV where d dt CS u dV 0 (steady flow) x CV u x u rel dA mV m V dV CS mdV AVdV Fx ,body 0 on CV Fx ,surface pA p dp A w Pdx on CV dpA VAdV dpA where DH4A/P is a hydraulic diameter 4 w A dx DH 4 w A dx DH Thus, dp VdV 4 w dx 0 DH (12.117) From COE we have: d e dV h 12V 2 dt CV where d dt rel dA Qinto CV Won CV CS e dV 0 CV h u 1 2V 2 u CS (steady flow) rel dA m h 1 2 V 2 m h 1 2 V 2 d h 1 2 V 2 md h 1 2 V 2 m dh VdV Qinto CV 0 W 0 on CV (adiabatic conditions) (no work is done on the control volume, recall that although the walls exert a force on the control volume, they do no work since there is no displacement at the walls due to the no-slip condition) Thus, dh VdV 0 C. Wassgren Chapter 12: Gas Dynamics (12.118) 504 Last Updated: 14 Aug 2010 The Second Law gives: d s dV s u rel dA dt CV CS qinto CV CV T where d dt s dV 0 (steady flow) CV s u rel dA ms m s ds mds CS qinto CV T CV 0 (adiabatic conditions) Thus, ds 0 Note that only the “>” has been retained since the friction results in irreversible conditions. (12.119) Now let’s also include the equations of state. We’ll concern ourselves only with an ideal gas. The Thermal Equation of State (the ideal gas law) gives: p RT dp RTd RdT so that dp d dT p T The Caloric Equation of State for an ideal gas can be written as: dh c p dT We can also combine the 1st and 2nd Laws with the ideal gas law to write: dT dp ds c p R T p (12.120) (12.121) (12.122) Finally, from the Mach number relation for an ideal gas we have: V2 V2 Ma 2 2 RT c so that 2VdV V 2 dT 2Mad Ma RT RT 2 d Ma Ma d Ma Ma VdV RTMa 2 V 2 dT 2 RT 2 Ma 2 dV dT 2T V C. Wassgren Chapter 12: Gas Dynamics (12.123) 505 Last Updated: 14 Aug 2010 Now let’s combine equations (12.116) through (12.118) and (12.120) through (12.123) (we won’t go through all of the algebra here) to get the following relations: 1 Ma 2 1 Ma 2 d Ma 2 4 w dx (12.124) 2 V 2 D Ma 1 Ma H Ma 2 dV V 2 1 Ma 2 4 w dx V 2 D H 4 dT 1 Ma T 1 Ma 2 (12.125) 4 w dx V 2 D H (12.126) 2 2 dp Ma 1 1 Ma 4 w dx 2 2 V D p 1 Ma H d (12.127) Ma 2 4 w dx 2 2 1 Ma V DH (12.128) 4 w ds 1 Ma 2 dx V 2 D cp H (12.129) It is useful to consider how the isentropic stagnation pressure, i.e. the pressure we would have if we brought the flow to rest isentropically, varies in a Fanno flow. Recall that the isentropic stagnation pressure is given by: p 1 1 1 Ma 2 p0 2 so that, after differentiating and doing some algebra, dp0 Ma 2 4 w dx V 2 D p0 2 H (12.130) Note that we could also derive equation (12.130) by noting that equation (12.122) may be written in terms of stagnation quantities: dT dp ds c p 0 R 0 T0 p0 and incorporating COE (equation (12.118)) written for an ideal gas: dh VdV dh0 dT0 0 dT0 0 gives: dp0 ds p0 R C. Wassgren Chapter 12: Gas Dynamics (12.131) (12.132) 506 Last Updated: 14 Aug 2010 Now let’s examine the trends indicated by these equations. Note that: 4 w dx 0 V 2 DH Ma < 1 (subsonic flow) d(Ma) > 0 dV > 0 dT < 0 dp < 0 d< 0 ds > 0 dT0 = 0 dp0 < 0 Ma > 1 (supersonic flow) d(Ma) < 0 dV < 0 dT > 0 dp > 0 d > 0 ds > 0 dT0 = 0 dp0 < 0 eqn (12.124) eqn (12.125) eqn (12.126) eqn (12.127) eqn (12.128) eqn (12.129) eqn (12.131) eqn (12.130) Notes: 1. Usually we write the shear stress in terms of a friction factor, f: w fF w 1 1 4 fD fF 1 4 2 V 1 2 2 V Fanning friction factor 2 D’Arcy friction factor fD 2. The D’Arcy friction factor is used most often in the analysis of incompressible pipe flows while the Fanning friction factor is often used for compressible flows. 3. In general, the friction factor, f, is a function of the flow Reynolds number, Re, the relative roughness of the pipe walls, /DH, and the Mach number, Ma (recall the Moody chart from undergraduate incompressible pipe flow problems). Typically, Mach number effects are small in comparison to the effects of Re and /DH so they are often neglected. Furthermore, the Reynolds number for compressible flows is often quite large. As a result, the friction factor remains essentially constant for a compressible flow. 4. The stagnation temperature remains constant and the stagnation pressure always decreases with friction. C. Wassgren Chapter 12: Gas Dynamics 507 Last Updated: 14 Aug 2010 Substituting the (Fanning) friction factor into equation (12.124) and integrating over a particular length of pipe: Ma 2 Ma1 2 1 Ma 2 d Ma Ma 3 1 1 Ma 2 2 x2 4 fF D x1 dx H 1 2 1 2 Ma 2 4 fF L12 (12.133) DH 1 2 1 2 Ma1 where L12 = x2-x1. Note that the overbar (implying a mean value) on the friction factor will be dropped in subsequent equations for convenience. 1 1 1 2 2 Ma1 Ma 2 2 1 Ma1 ln 2 2 Ma 2 Equation (12.133) relates the conditions between two arbitrary points in a frictional duct flow. It’s often more convenient to make one of the points a well-defined reference point. The most logical reference would be where sonic conditions occur, i.e. Ma2=1, since in Fanno flow the Mach number approaches unity. Using this reference point equation (12.133) becomes: 1 Ma 2 4 f F * 1 1 Ma 2 1 L ln (12.134) Ma 2 2 DH 1 2 1 Ma 2 2 where L* is the length of duct required, real or imaginary, for the flow to reach Ma = 1. Equations (12.125) through (12.129) can be integrated in a similar manner: 1 1 2 1 2 Ma 1 2 Ma * V 2 V 12 (12.135) 2 c 1 1 2 1 2 Ma T * c* 2 T 1 1 1 2 1 2 2 1 2 Ma * Ma p p 1 1 * Ma 2 12 1 (12.136) 12 1 2 1 2 Ma (12.137) 1 2 (12.138) 1 2 1 s s* 1 2 1 2 Ma ln 1 R 2 Ma 1 1 1 * p0 Ma 2 p0 1 2 1 C. Wassgren Chapter 12: Gas Dynamics 1 2 1 2 Ma 1 2 1 508 (12.139) (12.140) Last Updated: 14 Aug 2010 Notes: 1. The length of duct, L12, required for the flow to go from a given initial Mach number, Ma1, to a given final Mach number, Ma2, is found by: 4 fF 4f 4f L12 F L* F L* 1 2 DH DH DH where 4fFL*1/DH and 4fFL*2/DH are found from Eqn. (12.134) for the given Mach numbers. Ma=1 Ma2 Ma1 L2* L12 L1* 2. In order to find the change in some flow property, e.g. the pressure, between two locations where the Mach numbers are known, simply take the ratio: p2 p* p2 p1 p1 p* 3. 4. 5. 6. Equations (12.134) through (12.140) are tabulated as a function of Mach number for air in the appendices of most compressible flow texts. What happens if the duct is longer than L*. Assuming the back pressure is low enough, for a subsonic flow the flow adjusts upstream so that the length of duct becomes L*. If the flow is supersonic, then a shock wave forms upstream at a location where the length from the shock to the end of the duct is L*. We’ll discuss this in more detail in the next set of notes. L* is typically on the order of 10-100 duct diameters for a supersonic flow, depending on the friction factor. For example, for a friction factor of fF = 0.0025, the length of duct required to reduce a flow with Ma = to Ma = 1 is 82 pipe diameters. Thus, supersonic flows reach sonic conditions in a short distance. Furthermore, large losses in stagnation pressure also occur. In a Fanno flow with no shock waves, there can be no transition between supersonic and subsonic flows since both types of flow tend toward sonic conditions. C. Wassgren Chapter 12: Gas Dynamics 509 Last Updated: 14 Aug 2010 The Fanno Line The Fanno Line is the locus of all possible Fanno flow states shown in a T-s plot. To determine how T and s are related in a Fanno flow, we can combine equations (12.118), (12.120), (12.121), and (12.122) then integrate to give: 1 1 2 s s1 T T0 T (12.141) ln T1 T0 T1 cp Equation (12.141) allows us to relate the entropy, s, to an arbitrarily chosen temperature, T, for a given stagnation temperature, T0. A plot of this relation is shown below: p01 T p02 T0 = constant 1 Ma<1 Ma>1 1’ 2 Ma=1 (maximum entropy) 2’ s The Fanno line shows all possible combinations of entropy and temperature that can exist in an adiabatic flow with friction in a constant area duct at a given stagnation temperature (recall that the stagnation temperature remains constant in an adiabatic flow). Notes: 1. The processes always move in a direction that increases the entropy. 2. Again for a Fanno flow: Ma > 1 (supersonic flow) Ma < 1 (subsonic flow) d(Ma) > 0 eqn (12.124) d(Ma) < 0 dV > 0 eqn (12.125) dV < 0 dT < 0 eqn (12.126) dT > 0 dp < 0 eqn (12.127) dp > 0 d< 0 eqn (12.128) d > 0 ds > 0 eqn (12.129) ds > 0 eqn (12.131) dT0 = 0 dT0 = 0 eqn (12.130) dp0 < 0 dp0 < 0 C. Wassgren Chapter 12: Gas Dynamics 510 Last Updated: 14 Aug 2010 Example: Air flows in a 0.100 m ID duct under adiabatic conditions. Calculate the length of duct required to raise the Mach number of the air from Ma1=0.5 to Ma2=0.9 if the average value of the Fanning friction coefficient is 0.005. C. Wassgren Chapter 12: Gas Dynamics 511 Last Updated: 14 Aug 2010 Example: An experiment is designed to measure friction coefficients for the supersonic flow of air. Measurements from an experimental apparatus, consisting of a well-insulated, converging-diverging nozzle attached to a smooth, round tube, give the following data: pressure upstream of the nozzle = 516 cm Hg abs temperature upstream of the nozzle = 107.3 F throat diameter = 0.2416 in diameter of nozzle exit and of tube = 0.5009 in pressure at a point 29.60 diameters from the tube inlet = 37.1 cm Hg abs What is the average Fanning friction factor for the tube with these conditions? C. Wassgren Chapter 12: Gas Dynamics 512 Last Updated: 14 Aug 2010 Choking in Fanno Flow Consider what happens if the duct length has its maximum value for a given duct inlet Mach number (so that the exit Mach number is unity) and then we increase the duct length beyond this value. We’ll assume that the back pressure remains low enough to produce sonic exit conditions, i.e. pB p*. Subsonic Flow: The duct inlet Mach number will decrease until a steady-state solution again becomes possible with sonic exit conditions. This decrease results in a reduction in the flow rate, i.e. the flow is “choked” by friction. The mass flow rate through the duct can be found by the conditions at the duct inlet which is assumed to have isentropic flow. Mai < 1 isentropic duct inlet Mae = 1 L* 1 c 1 21 Ma 2 m VA 0 Ma c0 A 0 c0 AMa 1 2 0 c0 (As the Mach number decreases, the mass flow rate also decreases, i.e. m Ma 0 .) Further increases in the duct length result in decreases in the duct inlet Mach number and mass flow rate. C. Wassgren Chapter 12: Gas Dynamics 513 Last Updated: 14 Aug 2010 Supersonic Flow: - First consider the case where the duct length has its maximum value for the inlet Mach number. The flow within the duct will be supersonic and the exit conditions will be sonic (refer to curve A shown in the plots below). - Now increase the duct length to position B as shown in the figure. A shock wave forms in the duct at a location such that the exit conditions are sonic. The mass flow rate through the device remains unaffected since the flow at the nozzle throat is sonic. - As the duct length is increased further, the shock wave moves upstream until finally at the duct length indicated by position C, the shock wave is located at the duct inlet. Now the flow throughout the duct is subsonic. The mass flow rate, however, remains unaffected since the throat conditions are sonic. - Further increases in the duct length cause the shock wave to move into the diverging section of the nozzle toward the nozzle throat. The duct inlet Mach number will continue to decrease but the mass flow rate is unaffected. - If the duct length is increased further, the shock vanishes at the nozzle throat and the flow throughout the entire device becomes subsonic and further increases in the duct length will result in decreasing duct inlet Mach number and decreasing mass flow rate through the device (refer to the previous discussion concerning subsonic flow). A Ma A,B B C A B 1 x C B C C p/p* C A,B C. Wassgren Chapter 12: Gas Dynamics B B 1 x A 514 Last Updated: 14 Aug 2010 Performance of Long Ducts at Various Pressure Ratios Now let’s consider the flow in long ducts with differing inlet nozzle conditions and varying back pressures. Ducts Fed by Converging Nozzles Consider the flow through a long duct fed by a converging nozzle as shown in the figure below. The stagnation conditions are assumed fixed while the back pressure can be varied. For a high enough back pressure (cases A and B shown in the plots below), the flow through the entire device, including the exit, will remain subsonic and the exit pressure will equal the back pressure. If the back pressure is decreased to precisely the sonic pressure (case C), the flow throughout the device will be subsonic except at the exit where sonic conditions exist. For this particular case the exit pressure is equal to the back pressure. If the back pressure is decreased further, the flow conditions within the device will remain the same as in case C since the flow is choked at the exit; however, upon leaving the exit the flow will expand through expansion fans to reach the back pressure. p0, T0 fixed pB (variable) pE frictionless nozzle constant area, frictional duct p/p0I 1 A B C subsonic (MaE < 1) subsonic (MaE < 1) sonic (MaE = 1) D sonic (MaE = 1) with expansion fans x C. Wassgren Chapter 12: Gas Dynamics 515 Last Updated: 14 Aug 2010 Ducts Fed by Converging-Diverging Nozzles Now consider flow through a duct with a converging-diverging inlet nozzle. The flow through this device is considerably more complicated than through the duct fed by a converging nozzle. In the following discussion we’ll assume that the stagnation conditions feeding the nozzle are constant and that the ratio of the nozzle exit/duct inlet area to the throat area is fixed, i.e. Ai/At = constant. Subsonic flow at the nozzle exit / duct inlet: We’ll first consider the cases where the Mach number at the nozzle exit/duct inlet is subsonic, i.e. MaI < 1. Refer to the following figure for the various cases described below. p0, T0 fixed pE frictionless nozzle pB (variable) constant area, frictional duct p/p0I 1 1a, 1bi 1bii p*/p0I 2a, 2bi 2bii 3a, 3bi 3bii throat exit inlet x As the back pressure decreases: 1. 2. 3. The flow in the nozzle and duct are subsonic. Consider cases (a) and (b) described below. The flow in the nozzle is subsonic except at the nozzle throat where sonic conditions occur: MaT = 1, pT = p*. The flow is now choked at the nozzle throat. Further decreases in back pressure will not affect conditions upstream of the throat. Note that the Mach number at the duct inlet, MaI, for this case is determined by the area ratio, AI/A* = AI/AT. Consider cases (a) and (b) described below. The flow downstream of the throat is supersonic for some distance then passes through a normal shock wave and becomes subsonic for the remainder of the nozzle. Hence, the flow into the duct is subsonic. Consider cases (a) and (b) described below. a. If the back pressure is greater than the sonic pressure for the inlet Mach number, then the exit Mach number will be subsonic and the back pressure will equal the exit pressure: MaE < 1 p E = p B > p *. b. If the back pressure is less than or equal to the sonic pressure, then the exit will be at sonic conditions: MaE = 1 pE = p* pB. Note that the flow is now choked at the duct exit. Further decreases in the back pressure will not affect the flow within the duct or nozzle. i. If pB = p* then the flow equals the back pressure when exiting the duct. ii. If pB < p* then expansion fans will form outside of the duct through which the flow pressure will equilibrate with the back pressure. C. Wassgren Chapter 12: Gas Dynamics 516 Last Updated: 14 Aug 2010 Supersonic flow at the nozzle exit / duct inlet: Now consider the cases when the Mach number at the nozzle exit/duct inlet is supersonic, i.e. MaI > 1. Refer to the following figure for the various cases described below. Note that the Mach number at the nozzle entrance is determined solely by the area ratio, AI/A*. Furthermore, since the Mach number is fixed, the sonic length is also fixed. p0, T0 fixed pE frictionless nozzle pB (variable) constant area, frictional duct p/p0I 1 p* is in this range p*/p0I pE,supersonic throat exit inlet x 1 2 3 4 5 First consider the cases where the duct length is less than the sonic length, L < L*I, i.e. the exit pressure, assuming supersonic flow throughout the duct, is less than the sonic pressure, pE,supersonic < p*. (Recall that the pressure in a supersonic Fanno flow increases with increasing distance down the duct). 1. 2. 3. 4. 5. The back pressure is sufficiently large that the flow cannot leave the duct under supersonic conditions and come into equilibrium with the back pressure. Instead, a normal shock wave forms within the duct causing a transition to subsonic flow. The shock wave forms at a location such that the exit pressure equals the back pressure: MaE < 1 pE = pB. Decreasing the back pressure causes the shock wave to move closer to the duct exit. The entire duct flow is supersonic and the flow pressure equilibrates with the back pressure through a normal shock wave at the duct exit. The flow downstream of the shock is subsonic so the pressure just downstream of the shock is equal to the back pressure. The flow within the duct is supersonic but pE < pB (also note that pE < p* as stated in the paragraph above). Oblique shock waves form outside the duct through which the flow pressure equilibrates with the back pressure. The flow within the duct is supersonic and the pressure at the duct exit is equal to the back pressure, pE = pB. Note that pE < p*. The flow within the duct is supersonic but pE > pB (also note that pE < p*). Expansion fans form outside the duct through which the flow pressure equilibrates with the back pressure. C. Wassgren Chapter 12: Gas Dynamics 517 Last Updated: 14 Aug 2010 p0, T0 fixed pE frictionless nozzle pB (variable) constant area, frictional duct p/p0I 1 As pB, the shock moves toward the exit. p*/p0I 1 pE,supersonic 2 p* throat exit inlet x 3 Now consider the cases where the duct length is greater than or equal to the sonic length, L ≥ L*I, i.e. the exit pressure, assuming supersonic flow throughout the duct, is greater than or equal to the sonic pressure, pE,supersonic p*. (Recall that the pressure in a supersonic Fanno flow increases with increasing distance down the duct). 1. 2. 3. For the case where the back pressure is greater than the sonic pressure, i.e. pB > p*, a normal shock wave will form at a location within the duct such that the exit flow (which is subsonic) will equal the back pressure: MaE < 1 pE = pB > p*. For the case where the back pressure is equal to the sonic pressure, i.e. pB = p*, a normal shock wave will form at a location within the duct such that the exit flow is sonic: MaE = 1 pE = pB = p*. For the case where the back pressure is less than the sonic pressure, i.e. pB < p*, a normal shock wave will form at a location within the duct such that the exit flow is sonic: MaE = 1 pE = p* > pB. Expansion fans will form outside the duct through which the flow pressure equilibrates with the back pressure. Notes: 1. Once the flow chokes at either the nozzle throat or duct exit, the mass flow rate through the system will no longer be a function of the downstream conditions and will depend only on the upstream stagnation conditions. 2. Shock waves within the duct will not be sharp discontinuities as idealized here. Instead, boundary layer effects will tend to smear the transition from supersonic to subsonic flow over a considerable distance. Similar shock smearing is observed when shock waves form in the diverging section of a converging-diverging nozzle. C. Wassgren Chapter 12: Gas Dynamics 518 Last Updated: 14 Aug 2010 Example: This example is not yet complete. Air flows through a converging-diverging nozzle, with an exit-to-throat area ratio of 2.0, into a long, well insulated duct that has a diameter of 0.10 m, (Fanning) friction factor of 0.001, and a length of 5 m. Determine: C. Wassgren Chapter 12: Gas Dynamics 519 Last Updated: 14 Aug 2010 Isothermal Flow with Friction In long pipelines there is too much surface area for the flow to be considered adiabatic. Instead a more reasonable assumption is that the flow is isothermal. The Mach numbers in such flows tend to be small but the pressure change, p, can be large we can’t treat the flow as being incompressible. Consider isothermal (dT=0), 1D, steady flow of an ideal gas in a constant area (dA=0) duct. Using the same approach as the differential analysis for adiabatic frictional flow, we determine the following relations: d dV 0 (12.142) COM: V Note: dA=0 4 w dx 0 DH LME: dp VdV COE: c p dT0 VdV qinto (12.143) (12.144) Notes: 1. dh=0 since dT=0. 2. dh0 = cpdT + VdV = qinto = Qinto m 3. We are not assuming adiabatic conditions so heat transfer must be included. 2nd Law: Ideal Gas Law: ds qinto T Since the flow is frictional, it is also irreversible. dp d p Note: dT=0 d Ma (12.145) (12.146) Mach # relation: dV Ma V Note: dT=0 (12.147) Entropy relation: dp p Note: dT=0 (12.148) ds R The local adiabatic stagnation temperature is given by: 1 T0 T 1 Ma 2 2 so that dT0 1 Mad Ma T where dT=0 T0 1 2 T 1 Ma 2 1 Ma 2 d Ma dT0 Ma T0 1 2 1 2 Ma C. Wassgren Chapter 12: Gas Dynamics (12.149) 520 Last Updated: 14 Aug 2010 The local stagnation pressure is given by: 1 p0 p 1 Ma 2 2 so that 1 1 dp 1 Ma 2 dp0 2 p0 1 1 1 Mad Ma p 1 Ma 2 2 1 p 1 Ma 2 2 dp0 dp 1 Ma 2 1 Mad Ma 1 p0 p 2 2 1 1 1 1 Combine equations (12.142) through (12.149) leaving (4fFdx/D) as the independent variable: 2 dp d dV ds Ma 2 4 f F 1 d Ma dx p V R 2 1 Ma 2 DH 2 Ma 2 dp0 p0 dT0 T0 (12.150) 1 Ma 2 Ma 2 1 2 4 f F dx 1 D 2 1 Ma 2 1 Ma 2 H 2 (12.151) 1 Ma 4 1 2 2 1 Ma 4 fF dx D 1 Ma 2 H 2 Ma2 < 1/(subsonic) d(Ma) > 0 dV > 0 dp < 0 dT = 0 d< 0 ds > 0 dT0 > 0 (heat added) dp0 < 0 C. Wassgren Chapter 12: Gas Dynamics (12.152) Ma2 > 1/ (subsonic/supersonic) d(Ma) < 0 dV < 0 dp > 0 dT = 0 d > 0 ds < 0 (Note: ds > qinto/T) dT0 < 0 (heat removed dp0 > 0 for Ma2 < 2/(+1) dp0 < 0 for Ma2 > 2/(+1) eqn (12.150) eqn (12.150) eqn (12.150) isothermal eqn (12.150) eqn (12.150) eqn (12.152) eqn (12.151) 521 Last Updated: 14 Aug 2010 Working equations can be found by integrating equations (12.150) through (12.152) using the location where Ma2=1/ as a reference point (indicated by the superscript “*t” to signify limiting conditions for isothermal, frictional flow). 4 f F L*t 1 Ma 2 ln Ma 2 DH Ma 2 V V *t p0 * p0t T0 T0*t (12.153) *t p*t Ma p 1 2 3 1 1 1+ Ma 2 1 2 Ma (12.154) 1 (12.155) 2 1 1+ Ma 2 3 1 2 (12.156) Notes: 1. Let’s determine the amount of heat that must be added (in the case of subsonic flow) or removed (in the case of supersonic flow) for the flow to go from a given Mach number, Ma, to the choked flow Mach number, Ma*t = 1/1/2. From equation (12.144) we have: c p dT0 qinto and from equation (12.152) we have 1 Ma 4 4 fF dT0 dx 1 2 DH T0 2 1 Ma 2 1 Ma 2 Combining these two relations gives: c pT0 1 Ma 4 4 fF dx qi nto (12.157) 1 D 2 1 Ma 2 1 Ma 2 H 2 From equation (12.157) we see that as the Mach number approaches its limiting value of 1/1/2, the local heat transfer required to maintain isothermal conditions becomes very large. Hence, the assumption of isothermal flow for Mach numbers in the neighborhood of the limiting Mach number may not be a good one since there may not be sufficient heat transfer to maintain isothermal conditions. C. Wassgren Chapter 12: Gas Dynamics 522 Last Updated: 14 Aug 2010 2. Consider the following flow situation. p0, T0 e i pback L Assume that for the given back pressure, pback, the duct length is the sonic length, i.e. L=L*t and Mae=1/1/2. Note that since the exit Mach number is subsonic, the exit pressure will equal the back pressure, i.e. pe = p*t = pback. What happens if we now increase the duct length or drop the back pressure? It can be shown that there is no solution using the isothermal, frictional flow equations for the new conditions. The isothermal flow assumption breaks down since we would need to supply an infinite amount of heat transfer to get to the new conditions (refer to equation (12.157)). C. Wassgren Chapter 12: Gas Dynamics 523 Last Updated: 14 Aug 2010 Example: Natural gas (=1.3, R=75 (ftlbf)/(lbmR)) flows through a pipeline of diameter 3 ft and (Fanning) friction factor 0.001. At a particular point in the pipeline the pressure is 200 psia, the temperature is 500 R, and the velocity is 50 ft/s. The pipeline is kept at constant temperature. a. Determine the maximum length of the pipe that one could have from this point given these conditions. b. What would be the pressure at the maximum length? C. Wassgren Chapter 12: Gas Dynamics 524 Last Updated: 14 Aug 2010 8. Flow with Heat Transfer (Rayleigh Flow) Now let’s consider compressible flows where heat transfer is significant. Examples of such flows include those in which combustion, evaporation, condensation, or wall heat exchange occurs. Note that we won’t worry about how the heat gets into (or out of) the flow. We’ll just assume that we know its value, otherwise we would need to include heat transfer analyses into our discussions. To analyze the effects of heat transfer on a compressible flow, consider frictionless, 1D, steady flow of an ideal gas in a constant area (dA=0) duct. qinto V+dV V (p+dp)A pA dx x Using the same approach as the differential analysis for frictional flow gives: d dV 0 COM: V Note: dA=0 (12.158) LME: dp VdV 0 (12.159) COE: dh VdV qinto Note: q Q m (heat addition per unit mass flow rate) (12.160) 2nd Law: ds qinto (12.161) T Note: The flow may be reversible if the temperature gradients are very small. Ideal Gas Law: dp d dT p T (12.162) caloric eqn of state: dh c p dT (12.163) c p dT0 c p dT VdV qinto d Ma Mach # relation: Entropy relation: Ma ds c p C. Wassgren Chapter 12: Gas Dynamics dV dT 2T V (12.164) dT dp R T p (12.165) 525 Last Updated: 14 Aug 2010 Local stagnation pressure and temperature: T 1 Ma 2 1 2 T0 1 (12.166) p 1 1 Ma 2 p0 2 1 (12.167) Combining these equations gives and using dT0 (= qinto/cP) as the independent variable: 1 1 Ma 2 1 Ma 2 d Ma 2 dT0 2 Ma T0 2 1 Ma dp p 1 Ma 2 2 dT0 2 1 Ma T0 Ma 2 1 1 Ma 1 2 1 Ma 1 Ma 2 dT T (12.168) 2 2 (12.169) dT 0 T0 (12.170) 1 1 Ma 2 dV d 2 dT0 2 V 1 Ma T0 (12.171) dp0 Ma 2 dT0 p0 2 T0 (12.172) ds 1 dT 1 Ma 2 0 2 cp T0 (12.173) C. Wassgren Chapter 12: Gas Dynamics 526 Last Updated: 14 Aug 2010 Note: for qinto > 0 dT0 > 0 for qinto < 0 dT0 < 0 Ma > 1 Ma < 1 heat heat addition removal dT0 < 0 dT0 > 0 d(Ma) > 0 d(Ma) < 0 dV > 0 dV < 0 dp < 0 dp > 0 dT < 0 for Ma2<1/ dT > 0 for Ma2<1/ dT < 0 for Ma2>1/ dT > 0 for Ma2>1/ d< 0 d > 0 dp0 > 0 dp0 < 0 ds > 0 ds < 0 heat addition dT0 > 0 d(Ma) < 0 dV < 0 dp > 0 dT > 0 heat removal dT0 < 0 d(Ma) > 0 dV > 0 dp < 0 dT < 0 d> 0 dp0 < 0 ds > 0 d < 0 dp0 > 0 ds < 0 Notes: 1. dT0 = qinto/cp 2. Heating always decreases the stagnation pressure a loss in pressure recovery and thus efficiency. Removing heat results in an increasing stagnation pressure; however, in practice other effects act to decrease the stagnation pressure. 3. For 1/0.5<Ma<1, heat addition decreases the flow temperature while heat removal increases the flow temperature. The energy input from the heat goes into the kinetic energy of the flow rather than the thermal energy for this range of Mach numbers. 4. Mathematically, it appears that we could transition from subsonic to supersonic flow by controlling the heat transfer rate. In practice, transitioning from subsonic to supersonic flow using heat transfer has not been observed. C. Wassgren Chapter 12: Gas Dynamics 527 Last Updated: 14 Aug 2010 Integrating equations (12.168) through (12.173) using sonic conditions as a reference point gives the following working equations: p p * 1 1 Ma 2 (12.174) V * 1 Ma 2 * V +1 Ma 2 T T* (12.175) 12 Ma 2 1 Ma 2 (12.176) 2 1 2 1 2 Ma 1 * 2 p0 1 Ma 2 1 p0 T0 T0* 1 (12.177) 2 1 Ma 2 1 1 Ma 2 2 2 1 Ma 2 (12.178) 1 s s* 1 2 ln Ma 2 cp 1 Ma (12.179) T q12 c pT01 02 1 T01 (12.180) Note that the maximum amount of heat that can be added to a flow for a given initial state is: T* q12,max c pT01 0 1 T01 where, using Eqn. (12.178), T0* T01 2 2 1 Ma 2 2 1 Ma 1 1 Ma 2 2 1 so that q12,max 2 2 1 Ma c pT01 2 1 Ma 2 C. Wassgren Chapter 12: Gas Dynamics 1 1 2 1 Ma 1 2 528 (12.181) Last Updated: 14 Aug 2010 Plotting the locus of all possible states for Rayleigh flow on a T-s diagram gives the Rayleigh Line: T * T s* s1 1 Ma 1 ln 1 cp * p T p 1 Ma=1 (maximum entropy) Ma<1 Ma>1 s The arrows are drawn for heat addition. Reverse the arrows for heat removal. C. Wassgren Chapter 12: Gas Dynamics 529 Last Updated: 14 Aug 2010 Example: Air flows with negligible friction through a duct of area 0.25 ft2. At section 1, flow properties are T1=600 R, p1=20 psia, and V1=360 ft/s. At section 2, p2=10 psia. The flow is heated between sections 1 and 2. Determine the properties at section 2, the energy added, and the entropy change. Finally, plot the process on a T-s diagram. C. Wassgren Chapter 12: Gas Dynamics 530 Last Updated: 14 Aug 2010 Example: Air flows in a constant-area duct. At the inlet the Mach number is 0.2, the static pressure is 90 kPa, and the static temperature is 27 C. Heat is added at a rate of 120 kJ/(kg of air). Assuming a perfect gas with constant specific heats, determine the properties of the air at the end of the duct. Assume also that the flow is frictionless and that cp=1000 J/(kgK). C. Wassgren Chapter 12: Gas Dynamics 531 Last Updated: 14 Aug 2010 Example: A gaseous mixture of air and fuel enters a ram-jet combustion chamber with a velocity of 200 ft/s, at a temperature of 120 F, and at a pressure of 5 psia. The heat of reaction H of the mixture for the particular fuel-air ratio employed is 500 Btu/(lbm of the mixture). It is desired to find the stream properties at the exit of the combustion chamber. It will be assumed that friction is negligible, that the cross-sectional area is constant, and that the properties of both the reactants and the products are equivalent to air in respect to molecular weight and specific heat. C. Wassgren Chapter 12: Gas Dynamics 532 Last Updated: 14 Aug 2010 9. Normal Shock Waves Consider the movement of a piston in a cylinder: stagnant gas x compression wave When we first move the piston, an infinitesimal (compression) pressure wave travels down the cylinder at the sonic speed. Behind the wave, the pressure, temperature, density, and increase slightly and the fluid has a small velocity following the wave. If we continue to increase the piston velocity, additional pressure waves will propagate down the cylinder. However, these waves travel at a slightly increased speed relative to a fixed observer due to the increased fluid temperature and fluid movement. The result is that the waves formed later catch up to the previous waves. When the waves catch up to the first wave, their effects add together so that the small changes across the individual waves now become a sudden and finite change called a shock wave. 1st wave 2nd wave u1 c2 T1 c1 u0 = 0 T0 c2 > c1 since T1 > T0 and u1 moves to the right t, time piston shock wave pressure wave (sound wave) x, position C. Wassgren Chapter 12: Gas Dynamics 533 Last Updated: 14 Aug 2010 Notes: 1. 2. 3. 4. 5. The velocity of a shock wave is greater than the speed of sound since: p c2 1 For a sound wave, d 0. For a shock wave, so that cshock wave > csound wave. A shock wave is a pressure wave across which there is a finite change in the flow properties. Shock waves only occur in supersonic flows. Shock waves are typically very thin with thicknesses on the order of 1 m. Thus, we consider the changes in the flow properties across the wave to be discontinuous. The sudden change in flow properties across the shock wave occurs non-isentropically since the thermal and velocity gradients are large within the shock wave itself. To analyze a shock wave, we’ll use an approach similar to that used to examine a sound wave. Let’s consider a fixed shock wave across which flow properties change: very thin CV so there is no flow out of the top and bottom p1, T1, 1, V1 upstream (isentropic flow) p2, T2, 2, V2 downstream (isentropic flow) fixed shock wave (non-isentropic process) COM: 1V1 A 2V2 A 1V1 2V2 (12.182) LME (in x-direction): mV2 mV1 p1 A p2 A 1V1 V2 V1 2V2 V2 V1 p1 p2 (12.183) COE: h1 1 2 V12 h2 1 2 V22 (12.184) Also, h01 h02 . Note that no heat is added to the CV, i.e. the process is adiabatic. 2nd Law: s2 s1 (since the process is adiabatic but irreversible) Thermal Equation of State (ideal gas law): p1 p 2 R 1T1 2T2 (12.186) Caloric Equation of State (for a perfect gas): h c pT C. Wassgren Chapter 12: Gas Dynamics (12.185) (12.187) 534 Last Updated: 14 Aug 2010 Combining equations (12.182) and (12.183): p1 p 2 V2 V1 1V1 2V2 Substituting equation (12.186): RT1 RT2 V2 V1 V1 V2 Substituting equations (12.184) and (12.187) then re-arranging: V2 R V2 R T0 1 T0 2 V2 V1 2c p V2 2c p V1 RV2T0 (12.188) RV2V12 RV V 2 RV1T0 1 2 V1V2 V2 V1 2c p 2c p VV V1V2 V2 V1 R V2 V1 T0 V2 V1 1 2 2c p VV V1V2 V2 V1 R V2 V1 T0 1 2 2c p VV V1V2 R T0 1 2 2c p R V1V2 1 RT0 2c p RT0 V1V2 R 1 2c p Finally, substituting the relation: R c p cv 1 cp cp and re-arranging we have: 2 RT0 V1V2 1 C. Wassgren Chapter 12: Gas Dynamics Prandtl’s Equation (12.189) 535 Last Updated: 14 Aug 2010 Dividing both sides of the equation by the sound speed on either side of the shock wave and utilizing the definition for the Mach # for a perfect gas: RT0 RT0 V1 V2 2 RT1 RT2 1 RT1 RT2 T0 T0 2 1 T1 T2 Recall that for the adiabatic flow of a perfect gas: Ma1Ma 2 T 1 1 Ma 2 T0 2 So that: T0 2 Ma1Ma 2 1 T1 1 T0 T2 1 2 1 2 2 1 2 1 2 Ma1 1 2 Ma 2 1 1 2 After additional algebra we can reduce this equation to the following: Ma 2 2 2 1 Ma1 2 2 2 Ma1 1 (12.190) Relation between upstream Mach # (Ma1) and downstream Mach # (Ma2) across a normal shock wave. Notes: 1. When Ma1>1, then Ma2<1 (supersonic to subsonic flow) and when Ma1<1, then Ma2>1 (subsonic to supersonic flow). 2. From experiments, we observe that shock waves never form in subsonic flows (Ma1<1) even though equation (12.190) does not give any indication of this. We’ll use the 2nd law in a moment to show that shock waves can only form in supersonic flows (Ma1>1). The temperature ratio across the shock wave can be determined using the adiabatic stagnation temperature relation for a perfect gas and noting that the stagnation temperature remains constant across a shock: 1 T2 1 1 Ma 2 2 T 2 0 1 T1 1 2 T 1 Ma1 0 2 1 2 1 Ma1 T2 2 T1 1 2 1 2 Ma 2 C. Wassgren Chapter 12: Gas Dynamics (12.191) 536 Last Updated: 14 Aug 2010 The pressure ratio across the shock can be determined by combining equations (12.191), (12.186), and (12.182) along with the definition of the Mach number for a perfect gas: p2 2T2 V1T2 1T1 V2T1 p1 Ma1 Ma 2 RT Ma T RT1 Ma1 T2 2 2 1 T2 T1 1 2 1 Ma1 p2 Ma1 2 p1 Ma 2 1 1 Ma 2 2 2 1 2 (12.192) 1 2 2 V1 p2T1 Ma1 1 2 Ma1 1 V2 p1T2 Ma 2 1 1 Ma 2 2 2 1 2 2 V1 Ma1 1 2 Ma1 1 V2 Ma 2 1 1 Ma 2 2 2 3 1 2 1 2 1 2 Ma1 1 1 Ma 2 2 2 2 (12.193) We can also determine the ratio of the isentropic stagnation pressures and densities across the shock wave: p 1 1 Ma 2 p0 2 1 1 2 p1 Ma1 p 1 01 2 1 2 p2 Ma 2 p 1 2 02 p02 p01 1 2 p2 1 2 Ma1 p1 1 1 Ma 2 2 2 p02 p01 1 2 1 Ma1 Ma1 2 Ma 2 1 1 Ma 2 2 2 21 C. Wassgren Chapter 12: Gas Dynamics 1 1 1 2 1 Ma1 Ma1 2 Ma 2 1 1 Ma 2 2 2 1 2 1 2 1 2 Ma1 1 2 Ma 2 1 2 1 1 (12.194) 537 Last Updated: 14 Aug 2010 02 p02 T01 01 p01 T02 but since T01 = T02 02 p02 01 p01 1 2 1 Ma1 Ma1 2 Ma 2 1 1 Ma 2 2 2 1 21 (12.195) The sonic area ratio across the shock can be determined from the fact that the mass flow rate across the shock must remain constant (COM): m1 m2 * * * * 1 V1* A1 2V2* A2 * A2 * 1 V1* * * A1 2 V2* The sonic ratios can be determined from the following: 1 * 1 * 1 1 2 1 2 01 02 * 1 01 * 2 02 1 2 1 Ma1 Ma1 2 Ma 2 1 1 Ma 2 2 2 1 2 1 and T1* T * 0 V1* c1 T* * 1* 1 * T2* V2 c2 T2 T 0 Note that T01 = T02 has been used in the previous equation. Substituting these two sonic ratios and simplifying: 1 2 1 Ma1 * A2 Ma 2 2 * Ma1 1 1 Ma 2 A1 2 2 C. Wassgren Chapter 12: Gas Dynamics 1 2 1 (12.196) 538 Last Updated: 14 Aug 2010 Note that we could have also used the isentropic area ratios on either side of the shock wave: 1 2 1 Ma1 A1 1 2 * A1 Ma1 1 1 2 so that: 1 2 1 1 2 1 Ma 2 A2 1 2 * A2 Ma 2 1 1 2 and A1 1 2 * 1 Ma1 * A2 A1 Ma 2 2 * A1 A2 Ma1 1 1 Ma 2 2 * 2 A2 1 2 1 1 2 1 where A1=A2 Notes: 1. Equations (12.191)-(12.196) may be written only in terms of Ma1 by substituting equation (12.190). The resulting equations are: Ma 2 2 2 1 Ma1 2 2 2 Ma1 1 (12.197) 2 Ma 2 1 T2 1 2 2 1 Ma1 2 2 T1 1 Ma1 (12.198) 1 Ma1 2 V1 2 1 V2 1 Ma1 2 (12.199) p2 2 1 2 Ma1 p1 1 1 (12.200) T02 1 T01 (12.201) 2 p02 02 p01 01 1 2 Ma1 * A1 2 * A2 1 1 Ma 2 1 2 C. Wassgren Chapter 12: Gas Dynamics 1 2 1 Ma 2 1 1 1 539 1 1 (12.202) Last Updated: 14 Aug 2010 2. Now let’s examine the change in entropy across the shock using: T p s2 s1 c p ln 2 R ln 2 T1 p1 If we substitute equations (12.198) and (12.200) into equation (12.203) and plot: (12.203) 0.10 0.05 (s2 -s 1)/ c p 0.00 -0.05 0.5 1.0 1.5 2.0 -0.10 -0.15 -0.20 -0.25 -0.30 Ma1 We observe that for Ma1 < 1 the entropy decreases across the shock. The 2nd law, however, states that the entropy must increase across the shock (refer to equation (12.185)). Thus, shock waves can only form when Ma1>1. Also note that as the upstream Mach number approaches one (Ma11), the flow through the shock approaches an isentropic process. An infinitesimally weak shock wave, one occurring when Ma1=1, results in an isentropic process. This type of shock is, in fact, just a sound wave. C. Wassgren Chapter 12: Gas Dynamics 540 Last Updated: 14 Aug 2010 3. Plots of equations (12.198)-(12.202) as a function of Ma1 are shown below: 5 values 4 Ma2 T2/T1 p2/p1 r2/r1 p02/p01 A2*/A1* 3 2 V2/V1 1 0 1.0 1.5 2.0 2.5 3.0 Ma1 The plot shows the following relations: T2 > T1 and T2/T1 as Ma1 p2 > p1 and p2/p1 as Ma1 2 > 1 and 2/1 as Ma1 V2 < V1 and V2/V1 as Ma1 T02 = T01 p02 < p01 and p02/p01 as Ma1 02 < 01 and 02/01 as Ma1 A2* > A1* and A2*/A1* as Ma1 1 Ma 2 Ma1 2 Ma2 as Ma1 Furthermore, lim 4. 1 2 The shock strength is defined as the change in pressure across the shock wave relative to the upstream pressure: p/p1=p2/p1-1. Viewing the trends shown in the previous plot, the larger the incoming Mach number the stronger the shock wave. C. Wassgren Chapter 12: Gas Dynamics 541 Last Updated: 14 Aug 2010 5. On a T-s diagram, the states across a shock wave correspond to the intersection of the Fanno and Rayleigh lines for the flow: T, temperature downstream state (state 2) Fanno line Rayleigh line upstream state (state 1) s, entropy The reason for this is because the flow across the shock satisfies the Fanno relations for COM (equation (12.182)), COE (equation (12.184)), and the ideal gas relations (equations (12.186) and (12.187)). The shock also satisfies the Rayleigh relations for COM, COLM (equation (12.183)), and the ideal gas relations. The shock states must therefore occur at the intersection of the Fanno and Rayleigh lines in order for the shock to satisfy all of the basic relations simultaneously. Furthermore, state 2 lies to the right of state 1 in the T-s diagram since entropy must increase across the shock (2nd law). C. Wassgren Chapter 12: Gas Dynamics 542 Last Updated: 14 Aug 2010 Example: According to a newspaper article, at the center of a 12,600 lbm “Daisy-Cutter” bomb explosion the overpressure in the air is approximately 1000 psi. Estimate: a. the speed of the resulting shock wave into the surrounding air, b. the wind speed following the shock wave, c. the temperature after the shock wave has passed, and d. the air density after the shock wave has passed. C. Wassgren Chapter 12: Gas Dynamics 543 Last Updated: 14 Aug 2010 10. Flow in Converging-Diverging Nozzles Consider flow through a converging-diverging nozzle (aka a deLaval nozzle) as shown below. p0, T0, 0 V0 pB pE x Athroat, pthroat Let’s hold the stagnation pressure, p0, fixed and vary the back pressure, pB. The plot below shows how the static pressure ratio, p/p0, varies with location, x, for various values of the back pressure ratio, pB/p0: p/p0 1 2 3 5 6 1 p*/p0 7 4 8 x throat exit Cases: 1. There is no flow through the device since pB = p0. 2. There is subsonic flow throughout the device and pE = pB. 3. There is subsonic flow throughout the device except at the throat where pthroat = p* (Ma=1). The flow is now choked; further decreases in pB will not affect the flow upstream of the throat. The exit pressure will equal the back pressure, pE = pB, since the flow is subsonic at the exit. 4. Subsonic flow will occur in the converging section, sonic flow will occur at the throat (pthroat = p*), and supersonic flow will occur in the diverging section. This type of flow is called correctly expanded flow or flow at design conditions since no shock waves form anywhere in the device and pE = pB. 5. Subsonic flow will occur in the converging section and sonic flow will occur at the throat (pthroat = p*). A portion of the diverging section will be supersonic with a normal shock wave occurring at a location such that the subsonic flow downstream of the flow will have an exit pressure equal to the back pressure: pE = pB. As the back pressure decreases, the shock wave moves downstream of the throat and toward the exit. The pressure rise across the shock wave also increases as the back pressure decreases. 6. This case is similar to case 5 except that the shock wave is precisely at the nozzle exit. The pressure just downstream of the shock wave equals the back pressure since the flow is subsonic there. The flow everywhere within the C-D nozzle is isentropic except right at the exit. 7. The flow within the C-D nozzle (and the exit) is isentropic. The normal shock that was located at the exit for case 6 has moved outside the device to form a complicated sequence of oblique shock waves alternating with expansion fans (these are 2D phenomena to be discussed in a following section of notes). This case is called the overexpanded case since the diverging section of the device has an area that overexpands the flow to a pressure that is lower than the back pressure. External shock waves are required to compress the flow to match the back pressure. 8. This case is similar to case 7 except that the flow outside of the device forms a sequence of expansion fans alternating with oblique shock waves (a sequence out of phase with the sequence mentioned in case 7). This case is called the underexpanded case since the diverging section of the device has an area that is not large enough to drop the exit pressure to the back pressure. External expansion waves are required expand the flow to match the back pressure. C. Wassgren Chapter 12: Gas Dynamics 544 Last Updated: 14 Aug 2010 Notes: 1. The critical back pressure ratio corresponding to case 3 can be found from the isentropic relations (the flow throughout the entire device is isentropic). Assume that the geometry, and hence the exit-tothroat area ratio, Ae/At, is given. Since for case 5 the flow is choked we know that At = A*. Furthermore, since the exit flow is subsonic we also know that pe = pb. From the area ratio we can determine the exit Mach number, Mae: 1 2 1 1 2 1 Ma e Ae Ae 1 2 (where the subsonic Mae is chosen) A* At Ma e 1 1 2 The back pressure ratio, pb/p0, for case 5 can be determined given the exit Mach number. 1 pb pe 1 2 1 Ma e 2 p0 p0 2. The critical back pressure ratio corresponding to case 4 can be determined in a manner similar to that described above in Note 1. For case 4 however, the supersonic value for Mae should be used when determining the exit Mach number from the area ratio. 3. The critical back pressure ratio corresponding to case 6 can be found by combining the isentropic relations with the normal shock wave relations. When the shock wave occurs right at the exit of the device, the flow just upstream of the exit can be found from the isentropic relations: 1 2 1 1 2 1 Ma e1 Ae Ae 1 2 A1* At Ma e1 1 1 2 (where the supersonic Mae1 is chosen) 1 pe1 1 2 1 Ma e1 2 p01 Note that the subscript ‘1’ denotes the conditions just upstream of the shock wave. To determine the conditions just downstream of the shock we use the normal shock wave relations: pe 2 2 1 2 Ma e1 1 pe1 1 where pe2 is the pressure just downstream of the shock. Since the downstream flow is subsonic and because we’re at the exit of the device, the downstream pressure, pe2, must also equal the back pressure, pb. Thus, pb p pp 2 1 1 2 1 2 Ma e1 1 Ma e1 e 2 e 2 e1 1 2 p01 p01 pe1 p01 1 C. Wassgren Chapter 12: Gas Dynamics 545 Last Updated: 14 Aug 2010 4. The location of a shock wave for a back pressure in the range corresponding to case 3 and case 5 can be determined through iteration. a. Assume a location for the shock wave (e.g. pick a value for A/At since the geometry is known). b. Determine the Mach number and pressure just upstream of the shock, Ma1 and p1, using the isentropic relations as discussed in Note 2. 1 2 1 Ma1 1 A A 2 A1* At Ma1 1 1 2 1 2 1 (where the supersonic Ma1 is chosen) c. 1 p1 1 2 1 Ma1 p01 2 Calculate the stagnation pressure ratio and sonic area ratio across the shock using the normal shock relations: d. 1 1 2 1 * 2 Ma1 2 p02 A1 1 1 2 Ma1 * p01 A2 1 1 Ma 2 1 1 1 2 Determine the exit Mach number and exit pressure ratio using the isentropic relations and the downstream sonic area and stagnation pressure: 1 2 1 1 2 1 Ma e Ae Ae A1* 1 2 (where the subsonic Mae is chosen) * * A2 At A2 Ma e 1 1 2 Note that since the flow is choked, the throat area is equal to the upstream sonic area, i.e. At = A1*. 1 pe 1 2 1 Ma e 2 p02 e. f. Note that since the exit Mach number is subsonic, the exit pressure will equal the back pressure, i.e. pe = pb. Calculate the ratio of the back pressure to the upstream stagnation pressure: pb pp e 02 p01 p02 p01 Check to see if the back pressure ratio calculated in step (e) matches with the given back pressure ratio. If so, then the assumed location of the shock is correct. If not, then the go back to step (a) and repeat. If the back pressure ratio calculated in part (e) is less than the given back pressure ratio, then the assumed shock location is too far downstream. If the back pressure ratio calculated in part (e) is greater than the given back pressure ratio, then the assumed shock location is too far upstream. p/p01 actual pb/p01 actual location exit C. Wassgren Chapter 12: Gas Dynamics 546 x Last Updated: 14 Aug 2010 (Figure from: Liepmann, H.W. and Roshko, A., Elements of Gasdynamics, Wiley.) C. Wassgren Chapter 12: Gas Dynamics 547 Last Updated: 14 Aug 2010 5. In real nozzles flows, the flow will typically separate from the nozzle walls as a result of the large adverse pressure gradient occurring across a shock wave. Interaction of the shock with the separated boundary layer results in a more gradual pressure rise than what is expected for the ideal, normal shock analysis. It is also possible that downstream pressure information can propagate upstream in the diverging section even when the core flow is supersonic. In a real flow, a boundary layer will form along the wall with the flow in part of this boundary layer being subsonic. Thus, pressure information can propagate upstream within the subsonic part of the boundary layer and affect the flow in the diverging section. When the back pressure is in the range corresponding to Case 7 (back pressure less than the exit pressure when a shock stands at the exit, and greater than the isentropic case corresponding to supersonic diverging section flow), oblique shocks will typically form within the diverging section and flow separation occurs as shown in the figures below. The exact pressure and location of the separation point are dependent on the boundary layer flow. C. Wassgren Chapter 12: Gas Dynamics 548 Last Updated: 14 Aug 2010 A converging-diverging nozzle with pressure taps along the length of the device. The flow is from left to right. The pressure ratio as a function of the axial distance in the CD nozzle for various back pressures. Note the gradual pressure rise due to the interaction between the shock and the separated boundary layer. C. Wassgren Chapter 12: Gas Dynamics 549 Last Updated: 14 Aug 2010 Example: A converging-diverging nozzle, with Ae/At = 1.633, is designed to operate with atmospheric pressure at the exit plane. Determine the range(s) of stagnation pressures for which the nozzle will be free from normal shocks. C. Wassgren Chapter 12: Gas Dynamics 550 Last Updated: 14 Aug 2010 Example: A converging-diverging nozzle, with an exit to throat area ratio, Ae/At, of 1.633, is designed to operate with atmospheric pressure at the exit plane, pe = patm. a. Determine the range(s) of stagnation pressures for which the nozzle will be free from normal shocks. b. If the stagnation pressure is 1.5patm, at what position, x, will the normal shock occur? The converging-diverging nozzle area, A, varies with position, x, as: 2 A x Ae 2x 1 1 1 At At L throat area, At exit area, Ae stagnation conditions x 1 /2L L C. Wassgren Chapter 12: Gas Dynamics 551 Last Updated: 14 Aug 2010 Supersonic Wind Tunnel Design There are three common designs for supersonic wind tunnels: 1. high-pressure gas storage tanks (and/or vacuum tanks) for blowdown wind tunnels, 2. a compressor and diffuser for continuous-duty wind tunnels, and 3. shock tubes for high-enthalpy wind tunnels. Here we’ll study only the first two categories: blowdown and continuous-duty wind tunnels. Blowdown Wind Tunnels A schematic for a typical blowdown wind tunnel is shown below. Another possible design would be to use atmospheric conditions at the inlet and use a vacuum tank at the exit (such a design is effectively the same as the one shown below). C-D nozzle with throat area, At model to be tested tank with pressure , p01, temperature, T01, and volume, V back pressure, pb test section with design Mach number, MaTS > 1, and area, ATS The wind tunnel will have supersonic flow in the test section as long as the back-to-tank pressure ratio, pB/p01, is less than the back pressure ratio corresponding to case 6 shown in the figure below (no shock waves anywhere within the device). If the back pressure ratio becomes too high, then a shock wave will form in the diverging section of the nozzle and there will be subsonic flow in the test section. p/p0 1 2 3 5 6 1 p*/p0 throat C. Wassgren Chapter 12: Gas Dynamics exit 552 7 4 8 x Last Updated: 14 Aug 2010 Image from: http://history.nasa.gov/SP-440/ch5-6.htm Notes: 1. There is a fixed amount of time for which the device will operate at the design test section Mach number, MaTS, since the tank mass will decrease with time. To extend the duration of the test, a diverging section can be added to the exit of the device as shown below. C-D nozzle with throat area, At tank with pressure , p01, temperature, T01, and volume, V diverging section with exit area, AD back pressure, pb test section with design Mach number, MaTS > 1, and area, ATS The presence of the diverging section allows the tank to drop to a lower pressure before a shock wave appears ahead of the test section. C. Wassgren Chapter 12: Gas Dynamics 553 Last Updated: 14 Aug 2010 Example: A blowdown wind tunnel exhausting to atmospheric pressure (14.7 psia) is to be designed. The test section cross-sectional area is specified to be 1 ft2, and the desired test section Mach number is 2.0. The supply tank can be pressurized to 150 psia and heated to 150 F. Determine the throat area and supply tank volume required for a testing time of 30.0 s. If a diverging section with an area ratio equal to 3.375 times that of the throat is added downstream of the test section, what is the new testing time? C. Wassgren Chapter 12: Gas Dynamics 554 Last Updated: 14 Aug 2010 Continuous-Duty Wind Tunnels Continuous duty wind tunnels utilize a compressor to produce the driving pressure gradient for the flow. In order to minimize the required compressor power, the wind tunnel should operate as efficiently as possible, i.e. as close to isentropic conditions as possible. Continuous-duty wind tunnels can be either open-circuit where air is drawn in and exhausted to the surroundings, or closed-circuit where the working gas is recycled through the system. The following schematics show examples of both types of systems. Image from: http://history.nasa.gov/SP-440/ch5-4.htm Image from: http://history.nasa.gov/SP-440/ch5-3.htm C. Wassgren Chapter 12: Gas Dynamics 555 Last Updated: 14 Aug 2010 Again, in order to minimize the compressor power requirements, the losses in the system should be minimized. The ideal case (shown below) is to have an isentropic deceleration from supersonic to subsonic speeds. Mat2 = 1 Mat1 = 1 MaTS > 1 p01 Ma < 1 Ma < 1 pback p/p01 Of course in reality there would be viscous boundary layer losses in the duct which would result in a loss of stagnation pressure in the duct. We’ll ignore those losses here and assume isentropic flow. 1 p*/p01 1st throat x 2nd throat TS Consider what happens if we design the wind tunnel such that the downstream throat has a smaller area than the upstream throat, i.e. At2 < At1. As we decrease the back pressure ratio, pback/p01, then we will have subsonic flow throughout the device until at a critical back pressure the flow through the 2nd throat will become choked. As we decrease the back pressure further, a shock wave will form in the diverging section of the downstream diffuser (refer to the figure shown below). The test section is considered blocked since further reductions in the back pressure will not cause any changes upstream of the second throat. Since the test section was subsonic before blocking occurred, it will remain subsonic after blocking. Mat2 = 1 Mat1 < 1 p01 MaTS < 1 Ma < 1 Ma < 1 pback At2 = A1* < At1 p/p01 1 p*/p01 1st throat C. Wassgren Chapter 12: Gas Dynamics 2nd throat TS 556 x Last Updated: 14 Aug 2010 Now consider what happens if we make the 2nd throat just a little bit larger than the 1st throat. As we decrease the back pressure we will reach a case where the flow in the 1st throat becomes choked and a shock wave forms in the diverging section of the first throat. The flow in the test section will be subsonic. As the back pressure decreases, the shock wave in the first throat moves further downstream and becomes stronger. Recall that as the shock becomes stronger, A2*/A1* increases. If the second throat area is smaller than the A2* for the strongest shock wave, i.e., one that stands at the test section entrance, then a second shock will appear downstream of the second throat and the flow is once again blocked. The figure shown below illustrates this condition. Mat2 = 1 Mat1 = 1 MaTS < 1 Ma < 1 p01 Ma < 1 pback At2 = A2* At1 > At2 At1 = A1* p/p01 1 p*/p02 p*/p01 1st throat x 2nd throat TS If the downstream throat has an area greater than the sonic area downstream of the shock when the shock wave stands at the entrance of the test section (see the figure below) Mat2 < 1 Mat1 = 1 p01 MaTS < 1 Ma < 1 Ma < 1 pback At1 = A1* * A2 At1 where A2*/A1* is found from MaTS A1* For example, if MaTS = 2.0, At2 > 1.39 At1. * At 2 A2 p/p01 1 p*/p01 p*/p02 1st throat C. Wassgren Chapter 12: Gas Dynamics TS 557 2nd throat x Last Updated: 14 Aug 2010 and we decrease the back pressure further, then the shock will be swallowed by the 2nd throat and the flow within the test section will, at last, be supersonic (shown below). Mat2 > 1 Mat1 = 1 p01 MaTS > 1 Ma < 1 Ma < 1 pback At1 = A1* p/p01 1 p*/p01 1st throat TS 2nd throat x Now let’s get back to the original discussion. Once the shock has been swallowed by the second throat, the shock will stand in the diverging section of the downstream diffuser. The wind tunnel is now considered running or started. In order to isentropically decelerate the flow we should now decrease the area of the 2nd throat so that it is approximately the same as the upstream throat area. Since there are no longer any shock waves upstream of the 2nd throat, we theoretically could approach the 1st throat area; however, due to boundary layer effects we will always have to make the 2nd throat slightly larger than the first. Decreasing the 2nd throat area too much results in shock waves in the diverging sections of both the 1st and 2nd throats (discussed previously) and the wind tunnel is once again blocked. Photo from: http://history.nasa.gov/SP-440/ch5-5.htm C. Wassgren Chapter 12: Gas Dynamics 558 Last Updated: 14 Aug 2010 Once the wind tunnel is running and we’ve decreased the 2nd throat area, we should try to minimize the stagnation pressure loss through the shock wave in the 2nd diverging section (and, hence, increase the tunnel efficiency). To do this we increase the back pressure, pback, so that the shock wave will move further toward the second throat thus decreasing the shock’s strength. Mat2 > 1 Mat1 = 1 p01 MaTS > 1 Ma < 1 Ma < 1 pback p/p01 1 p*/p01 1st throat TS 2nd throat x The ideal case is to have the shock positioned exactly at the second throat. In practice, however, a shock standing exactly at the second throat is unstable and could disgorge, i.e. move back into the diverging section of the first throat, and block the test section once again. Note: An excellent reference on the history of wind tunnel development at NACA/NASA can be found at: http://www.hq.nasa.gov/office/pao/History/SP-440/contents.htm C. Wassgren Chapter 12: Gas Dynamics 559 Last Updated: 14 Aug 2010 Example: Consider a supersonic wind tunnel starting as shown in the figure below. The upstream nozzle throat area is 1.25 ft2, and the test section design Mach number is 2.50. As the tunnel starts, a normal shock stands in the divergence of the nozzle where the area is 3.05 ft2. Upstream stagnation conditions are T01 = 1080 R and p01 = 115 psia. Find the minimum possible diffuser throat area at this instant. Calculate the entropy increase across the shock. What would be the minimum possible diffuser throat area to start this wind tunnel? p01, T01 C. Wassgren Chapter 12: Gas Dynamics 560 Last Updated: 14 Aug 2010 Supersonic Diffuser Design Another application where the efficient deceleration of a supersonic flow is of interest is a supersonic diffuser at the inlet of aircraft jet engines. The flow entering a jet engine typically needs to be subsonic in order to avoid shocks in the compressor section and give efficient combustion in the combustor. Obviously the most efficient deceleration of the incoming flow is desired since the thrust out of the device will decrease if the upstream stagnation pressure decreases. Many of the same ideas discussed previously for the design of supersonic wind tunnels are pertinent here as well. Consider a diffuser with a fixed inlet and throat area, Ai and At, respectively, in a supersonic flow with an upstream Mach number of Ma > 1. At design conditions, i.e. Ma = MaD, the flow through the diffuser will be shockless (isentropic) as shown in the figure below. (Ma > 1) = MaD Mat = 1 Ma < 1 At Ai In the ideal case the inlet-to-throat area ratio will be related to the design Mach number, MaD, by the isentropic relation for the sonic area ratio, i.e.: 1 1 Ma 2 D Ai Ai 1 2 * At Ma D 1 1 A 2 1 2 1 (12.204) However, we need to consider what happens as the aircraft comes up to the design Mach number from rest (Ma = 0). Consider the plot shown below: A/A* 1 1 MaD Ma For Ma < MaD, (Ai/A*) < (Ai/A*)D so that At = A*D < A*Ma < MaD. Thus, the diffuser cannot “swallow” all of the air flowing toward the inlet. A shock wave forms in front of the inlet to produce subsonic flow so that some of the air can spill over the inlet as shown in the figure below. (Ma > 1) < MaD C. Wassgren Chapter 12: Gas Dynamics 561 Last Updated: 14 Aug 2010 As the upstream Mach number increases, the sonic area approaches the throat area, i.e. A* At, and the shock moves closer to the inlet (the shock gets weaker and less flow needs to be diverted around the diffuser). Eventually we’ll reach design conditions but a normal shock will still appear ahead of the inlet since the sonic area after the shock, A2*, will be greater than the throat area, At, at design conditions. Ai Ai A* A A* * 1* i 1* * A2 A1 A2 At A2 (Ma > 1) = MaD 12 * A2 At 1 If we continue to increase Ma the shock wave will move closer to the inlet until at a critical upstream Mach number, Ma, crit, the shock will be positioned exactly at the inlet such that the diffuser can accommodate all of the mass flow heading toward it (no spill-over). This occurs when A2* = At: Ai Ai Ai A* * * 1* fcn Ma , crit At A2 A1 A2 where Ai/A1* is found from the isentropic relations and A1*/A2* is found from the normal shock relations. Ma,crit > 1 12 A further increase in the upstream Mach number will cause the shock wave to be swallowed by the diffuser where it will come to a steady state position within the diverging section. Ma > Ma,crit Now the upstream Mach number can be decreased back down to the design Mach number so that the shock wave will travel back upstream toward the throat and, hence, become weaker (less stagnation pressure drop across the shock). The ideal case is to bring the Mach number down exactly to the design Mach number so that the shock occurs exactly at the throat and has zero strength. Notes: 1. In practice we don’t want to operate the diffuser exactly at design conditions since any small decrease in the upstream Mach number will cause the shock wave to disgorge from the diffuser and the entire process for swallowing the shock must be repeated again. 2. One measure of the diffuser performance is the stagnation pressure recovery coefficient defined as: p 0,exit of diffuser p0 At design conditions under ideal conditions this coefficient should be unity. C. Wassgren Chapter 12: Gas Dynamics 562 Last Updated: 14 Aug 2010 3. Over-speeding the diffuser is often impractical. For example, consider a diffuser designed to operate at a Mach number of 1.7 (Ai/A* = Ai/At = 1.338). The critical Mach number for swallowing the shock will be: Ma,crit > 1 Note: A2* = At 12 1 1 Ma 2 2 Ai Ai 1 2 1.338 * Ma 2 1 1 A2 sub At 2 Ma ,crit 2.65 1 2 1 Ma 2 0.5 Ma1 2.65 Thus, to achieve isentropic flow through the diffuser, we would need to operate our diffuser at a Mach number just greater than 2.65 then decrease the Mach number down to just greater than 1.7. Designing an aircraft to achieve this over-speed Mach number is often impractical. 4. Since fixed geometry diffusers are often impractical, other diffuser types have been designed. These include designs that have variable areas so that the throat area can be increased to swallow the shock, then decreased again to the design conditions (very similar to what was discussed for supersonic wind tunnels). variable area actuators movable plug Oblique shock wave diffusers are also often used. The stagnation pressure loss across an oblique shock is less than that across a normal shock wave. The weaker the oblique shock wave, the smaller the stagnation pressure loss. Normal shock waves may still appear in the device but they’ll be weaker than if there wasn’t an oblique shock since the oblique shock helps to decelerate the flow. C. Wassgren Chapter 12: Gas Dynamics 563 Last Updated: 14 Aug 2010 C. Wassgren Chapter 12: Gas Dynamics 564 Last Updated: 14 Aug 2010 Note that engine inlets typically “bleed off” or remove boundary layers as shown in the F-16 engine inlet design. This is done in order to avoid exposing engine components to the unsteady conditions resulting from wakes formed by separated boundary layers. C. Wassgren Chapter 12: Gas Dynamics 565 Last Updated: 14 Aug 2010 11. Flows with Mass Addition Now let’s consider compressible flows where mass addition (or removal) occurs. Examples of such flows include those in which solid rocket propellant is burned, a gas coolant is added to the flow such as in film cooling of turbine blades, or the boundary layer flow is removed such as in some wind tunnels. We’ll make the following assumptions for our analysis of flow with mass addition: - no heat transfer - no other work - no friction - no area change - no significant elevation changes - steady flow - 1D flow (This isn’t a very good assumption in general, but we’ll make the assumption here for simplicity.) dmi , pi , i , Ti , Vix , Viy , si m, p , , T , V , s m dm, p dp, d , T dT , V dV , s ds fluid is assumed to be completely mixed in this region x Using the same control volume approach as in previous analyses gives: COM: m dm m dmi 0 dm d VA dmi d dV dm V m Note: dA=0 C. Wassgren Chapter 12: Gas Dynamics (12.205) 566 Last Updated: 14 Aug 2010 LME in x-direction: mV d mV mV dmiVix pA pA d pA d mV dmiVix dpA Vdm mdV dmiVix dpA dp V V Vix dm VmdV 0 VA VA dm 0 m Note: y Vix/V, and dm dmi dp VdV V 2 1 y (12.206) COE: mh0 d mh0 mh0 dmi h0i 0 dm dh0 h0 h0i 0 m dh0 0 (12.207) Note: For simple mass addition we assume that the inlet fluid has the same composition and stagnation enthalpy as the main stream, i.e. h0i = h0. In general, however, the addition stream may have a different stagnation enthalpy than the main flow. The general case will be considered when we investigate general 1D flows. 2nd Law: ms d ms ms dmi si 0 dm ds s si 0 (12.208) m Note: The mass addition stream may have, in general, a different entropy than the main stream. Moreover, mixing processes are generally irreversible. Ideal Gas Law: dp d dT p T (12.209) caloric eqn dh c p dT (12.210) of state When combined with COE (Eqn. (12.207)): c p dT0 0 dT0 0 (12.211) Mach # relation: Gibbs Eqn: d Ma Ma dV dT V 2T (12.212) dp dT dp R R 0 T p p0 Note: dT0 = 0 from eqn (12.211). ds c p C. Wassgren Chapter 12: Gas Dynamics (12.213) 567 Last Updated: 14 Aug 2010 Local stagnation pressure and temperature: 1 T0 T 1 Ma 2 2 1 p0 p 1 Ma 2 2 (12.214) 1 (12.215) Combining these equations so that dm is the driving potential, and using the substitution, y Vix/V, gives: d Ma Ma 1 2 1 2 Ma 2 2 dm 1 Ma y Ma 2 m 1 Ma dp Ma 2 p 1 Ma 2 d 1 dm Ma 2 1 y y 2 1 2 m 1 1 Ma 2 y Ma 2 dm m 1 Ma 2 dT 1 Ma dm 1 Ma 2 y Ma 2 2 m T 1 Ma (12.217) (12.218) 2 (12.216) (12.219) dV 1 1 Ma 2 y Ma 2 dm 2 m V 1 Ma (12.220) dp0 dm Ma 2 1 y p0 m (12.221) dT0 0 (12.222) ds dm 1 Ma 2 1 y cp m (12.223) C. Wassgren Chapter 12: Gas Dynamics 568 Last Updated: 14 Aug 2010 Notes: 1. The trends in equations (12.216)-(12.223) will depend on dm , y, and Ma. Equation (12.221) shows that for mass addition, the stagnation pressure will decrease if y < 1 and will increase if y > 1. Equation (12.223) shows that entropy has the opposite trend. 2. For y < 1, all factors that involve y are positive so that for mass addition ( dm 0 ) we have: Ma < 1 Ma > 1 d(Ma) > 0 d(Ma) < 0 dp < 0 dp > 0 d < 0 d > 0 dT < 0 dT > 0 dV > 0 dV < 0 dp0 < 0 dp0 < 0 ds > 0 ds > 0 The opposite trends occur for mass removal ( dm 0 ). Note that choking is possible with mass addition since the Mach number approaches unity for both subsonic and supersonic flows. 3. We cannot simply add mass to transition from a subsonic to a supersonic flow. Equation (12.216) indicates that as the flow approaches a sonic Mach number, the addition stream must have a large mass flow rate to continue approaching the sonic Mach number. In the limit as Ma → 1, dm . C. Wassgren Chapter 12: Gas Dynamics 569 Last Updated: 14 Aug 2010 4. For y = 0 (i.e. the flow comes in normal to the stream), equations (12.216)-(12.223) become exact differentials. Integrating using sonic conditions (denoted by the superscript “*”) as a reference gives: m Ma * m 1 Ma 2 p p * 1 Ma 2 2 +1 1 2 1 2 (12.224) 1 1 Ma 2 2 * 1 Ma 2 (12.225) 1 2 1 2 Ma 2 1 2 1 2 Ma * T 1 T 1 2 1 2 Ma 1 2 Ma * V +1 V 1 * p0 1 Ma 2 p0 (12.226) (12.227) 12 (12.228) 2 1 Ma 2 1 2 1 1 (12.229) T0 1 T0* p s s* ln 0 * R p0 5. (12.230) (12.231) The relationship between T and s may be found by considering equations (12.227), (12.229), and (12.231). Below is a T-s diagram for a flow with mass addition. arrows are drawn for mass addition (y = 0) T at Ma = 0, T/T* = (+1)/2 T* Ma < 1 Ma = 1 Ma > 1 as Ma → ∞, T/T* → 0 C. Wassgren Chapter 12: Gas Dynamics s* 570 s Last Updated: 14 Aug 2010 Example: Air with initial stagnation conditions of 600 K and 1 MPa (abs) flows at a Mach number of 0.3 at the entrance to a constant-area, porous-walled duct. During passage through the duct, the mass flow rate is increased by 50%. Determine the exit Mach number, pressure, temperature, and stagnation pressure. C. Wassgren Chapter 12: Gas Dynamics 571 Last Updated: 14 Aug 2010 12. Generalized Steady, One-Dimensional Flow In our previous analyses we have only considered simple flows where there is only one potential affecting the flow (e.g. area change, friction, heat transfer, or mass addition). Now we’ll discuss how to analyze flows where multiple effects are considered. Our approach is very similar to the approach we have used many times before: we’ll draw a control volume, apply our conservation laws, utilize definitions, and simplify our resulting equations. Qinto dm m D z x m dm Ff gx gz In our analysis, we’ll include the effects of: - area change, dA - friction, Ff - heat transfer, Q - mass addition, dm - “other” work, W - other forces, D - gravity body force, gAdx We’ll further assume that the flow is steady, one-dimensional, and that the fluid is well-mixed within the control volume. Now let’s apply our conservation laws: COM: m dm m dmi 0 dm d VA dmi dm d dV dA m V A C. Wassgren Chapter 12: Gas Dynamics (12.232) 572 Last Updated: 14 Aug 2010 LME (in x-direction): mV d mV mV dmiVix pA pA d pA p 1 2 dp dA D Ff g x A 1 2 dA dx dmV mdV dmiVix dpA D Ff g x Adx dpA mdV g x Adx Ff D dm V Vix 0 Substituting the following: m VA , y Vix/V, and Ff w Pdx f F 1 2 V 2 Pdx 1 2 V 2 4 Af F DH : 4 f dx D dm V 2 1 y 0 dp VdV g x dx 1 2 V 2 F m DH A (12.233) COE: mh0 d mh0 mh0 dmi h0i Qinto CV Won CV dm qinto CV won CV (12.234) m The term in parentheses represents the difference in the stagnation enthalpy of the main stream and the incoming flow. Note that: h0 h 1 V 2 g z z . 2 dh0 h0 h0i 2nd Law: ms d ms ms dmi si CV ds s si dm m Qinto CV T qinto CV CV (12.235) T Now let’s specify that we’re dealing with an ideal gas: p RT dh c p dT ds c p Ma (12.236) (12.237) dT dp R T p (12.238) V (12.239) RT Since we’re concerned with a gas, we’ll also assume that gravitational effects are negligible compared to the other terms in the equations (i.e. g 0). C. Wassgren Chapter 12: Gas Dynamics 573 Last Updated: 14 Aug 2010 We’ll also utilize the definitions of the isentropic stagnation pressure and the adiabatic stagnation temperature: p 1 1 Ma 2 p0 2 T 1 1 Ma 2 T0 2 1 (12.240) 1 (12.241) Equations (12.232)-(12.241) are a system of equations that can be combined and solved1 for the dependent variables: d Ma dp d dT dV dp0 ds ,, , , , , Ma p T V p0 c p in terms of the independent variables, or driving potentials: dA 4 f F dx 2 D dT0 dm , , , A DH Ma 2 pA T0 m The following table summarizes the resulting equations. 1 For details, refer to Zucrow and Hoffman, Gas Dynamics: Volume I, Wiley. C. Wassgren 574 Last Updated: 14 Aug 2010 Chapter 12: Gas Dynamics Table 1. Change in flow properties in terms of driving potentials. Driving Potentials Change in Flow Property d Ma Ma 4 f F dx 2 D 2 DH Ma pA dA A Ma 2 1 Ma 2 dp Ma 2 p 1 Ma 2 d Ma 2 1 Ma 2 dT 1 Ma 2 T 2 1 Ma 1 Ma 2 dV V dp0 1 Ma ds 2 Ma 2 Ma 2 1 Ma Ma 2 2 Ma 2 2 1 Ma 2 1 Ma 2 Ma 2 2 Ma 2 2 1 y y 1 Ma 2 1 Ma 2 y Ma 2 1 Ma 2 1 Ma 2 2 1 Ma 2 1 Ma 2 2 1 Ma 2 1 Ma 2 1 Ma 2 y Ma 2 1 Ma 2 1 Ma 2 y Ma 2 1 Ma 2 Ma 2 1 y 2 0 y Ma 2 1 Ma 2 1 Ma 2 0 cp where 1 Ma 1 2 1 Ma 4 2 1 Ma 2 p0 F Ma 2 2 1 Ma 2 2 0 dF 1 Ma 2 y Ma 2 2 2 1 Ma 2 T0 1 Ma 2 1 Ma Ma 2 1 1 Ma 2 1 1 Ma 2 dm m dT0 1 Ma 2 1 y 2 1 y 1 2 Ma 2 Vix V F impulse function pA mV C. Wassgren Chapter 12: Gas Dynamics 575 Last Updated: 14 Aug 2010 Notes: 1. The terms in the table are often referred to as influence coefficients. 2. The impulse function, F pA mV , is a convenient definition that is helpful in determining the reaction force for one-dimensional steady flows. For example, the thrust on the jet engine shown below can be determined from the difference between the outgoing and incoming impulse functions: m1V1 m2V2 p1A1 p2A2 T m2V2 m1V1 T p1 A1 p2 A2 T m2V2 p2 A2 m1V1 p1 A1 T F2 F1 3. As can be seen from equations (12.234) and (12.237), the effects of heat transfer, other work, and the difference in the main and incoming stream stagnation enthalpies are all included in the change in the stagnation temperature, dT0. 4. How one uses the table is best shown by example. Let’s say we’re interested in determining how the Mach number varies as a function of the driving potentials. From the table we see that the Mach number variation is given by: d Ma dA Ma 2 4 f F dx 2 D 2 2 Ma 1 Ma A 2 1 Ma DH Ma 2 pA (12.242) 1 Ma 2 dT0 1 Ma 2 y Ma 2 dm m 1 Ma 2 2 1 Ma 2 T0 We would now need to know how to model the driving potentials as we move downstream in flow (e.g. How does the area vary as we move downstream in the duct?) Finally, we would integrate the resulting equation (numerically if necessary) to solve for the Mach number variation. C. Wassgren Chapter 12: Gas Dynamics 576 Last Updated: 14 Aug 2010 5. We can also extract trends from the influence coefficients. For example, equation (12.242) may be written as: dA Ma 2 4 f F dx 2 D 2 2 DH Ma pA d Ma A 2 Ma 1 Ma 1 Ma 2 dT0 dm 1 Ma 2 y Ma 2 m T0 2 Note that is always positive. a. If < 0, then the Mach number will decrease for subsonic flow and increase for supersonic flow, i.e. the Mach number diverges from one. b. If > 0, then the Mach number will increase toward one for subsonic Mach numbers while the Mach number will decrease toward one for supersonic Mach numbers. Hence, choking is possible if > 0. c. If = 0, then the Mach number does not change since d(Ma) = 0 (the flow is at an inflection point). It is possible that a flow may have changes in the sign of as the flow moves downstream. For example, simple isentropic flow in a converging-diverging nozzle has > 0 in the converging section, = 0 at the throat, and < 0 in the diverging section. Note that sonic conditions (Ma = 1) can only occur where = 0 otherwise d(Ma) would be infinite. For example, consider the special case where only friction and area changes are present (D = dT0 = dm = 0) so that: dA Ma 2 4 f F dx A 2 DH Since the friction term will always be positive, and since sonic conditions must occur when = 0, the sonic point must occur in a diverging section (dA > 0). The exact location of the sonic point can be determined since we know how the area varies with x: 2 1 4 f F dx 1 dA dx 0 A 2 DH dx dA A 1 dA 4 f F (Since A(x), fF, and DH are given, one can solve for x.) A dx 2 DH C. Wassgren Chapter 12: Gas Dynamics 577 Last Updated: 14 Aug 2010 6. Note that many of the relations in Table 1 have a (1 – Ma2) in the denominator. Hence, one must proceed with care when integrating the relations near the sonic point since the gradients become very large there. At the sonic point in particular, terms with (1 – Ma2) in the denominator are undefined so they can not be integrated directly. If the independent variables can be expressed explicitly as a function of x, gradients at the sonic point can be resolved using l’Hopital’s rule. For example, let’s again consider the case for a frictional flow with area change (D = dT0 = dm = 0) so that the Mach number relation is given by: d Ma Ma 1 dA Ma 2 4 f F Ma x 2 DH dx 1 Ma 2 A dx 1 Ma 2 d Ma d Ma d Ma dx lim Ma 1 dx dx 2Ma d Ma x x* dx * Ma 1 d Ma dx * 2 1 d Ma 2 dx x x* Ma 1 1 d 4 d Ma 3 1 x Max*1 * dx x x 4 dx Ma 1 * 2 * d Ma * 3 1 d Ma 1 d * xx dx 4 Ma 1 dx 4 dx where 4f 1 dA 0 (12.243) (12.244) F xx Ma 1 A dx x x 2 DH * x x* Ma 1 * d dx 4 f d Ma d 1 dA F dx A dx x x* DH dx * x x* Ma 1 (12.245) Equations (12.243), (12.244), and (12.245) can be combined to give a single quadratic equation in terms of the unknown sonic Mach number gradient. Two solutions can be found with the appropriate one being determined by the downstream boundary conditions (Recall our previous discussions concerning isentropic flow through a C-D nozzle. Whether the flow remains subsonic or supersonic after the throat depends on the back pressure.) 7. The previous generalized flow table (Table 1) can be simplified to give our previous results for simple flows as shown in the following table. C. Wassgren Chapter 12: Gas Dynamics 578 Last Updated: 14 Aug 2010 Simple Frictional Flow A 1 2 1 T0 * 1 Ma 0 Ma DH m 1 m* 2 1 * 1 1 2 2 ln Ma 1 2 0 0 1 1 2 1 T * V p0 1 Ma 1 1 1 2 Ma 2 1 1 2 1 1 2 1 2 1 Ma 1 Ma 1 1 2 Ma 1 1 1 1 Ma 2 1 Ma cp 1 2 ln Ma 1 2 2 2 2 1 Ma 1 2 2 1 ln Ma 2 1 Ma 1 2 Ma 1 1 1 1 2 2 1 2 1 2 1 Ma 2 1 0 2 Ma 2 2 1 Ma 2 * 2 2 1 Ma 2 1 Ma Ma 2 1 * 1 Ma 2 1 Ma 2 2 2 1 1 Ma 1 Ma 2 F 1 2 Ma 1 1 * p0 1 1 2 Ma 1 * 1 2 1 2 1 * ss 2 2 1 2 1 Ma p F 1 Ma 2 4 fF L V 2 1 Ma 2 1 * T0 T 1 Ma 1 A 1 2 1 * p Simple Mass Addition, y 0 1 Simple Diabatic 1 Simple Area Change 1 2 2 2 1 Ma 1 1 1 1 2 1 1 Ma 2 ln 1 1 Table 2. Change in flow properties for simple flows. C. Wassgren Chapter 10: Gas Dynamics 579 1 Last Updated: 15 Aug 2005 1 Example: Consider steady air flow through a constant area circular duct which has a diameter of 10 cm and a length of 1 m. The Mach number, pressure, and temperature at the inlet to the duct are 0.3, 200 kPa, and 80 C. Heat is added at a uniform rate to the air as it flows through the duct causing the stagnation temperature to increase by 300 K. If the (Fanning) friction factor can be assumed to be 0.003, find the Mach number, pressure, and temperature at the outlet of the duct. C. Wassgren Chapter 11: Gas Dynamics 580 Last Updated: 16 Aug 2009 13. Oblique Shock Waves An oblique shock wave is a shock wave that forms at an angle with respect to the incoming flow (note that normal shock waves are a special case of an oblique shock wave). As with normal shock waves, the flow properties abruptly change when passing through an oblique shock. In addition, the flow is turned through the shock as shown in the figure below. p1, 1, T1 V2 V1 p2, 2, T2 We’ll refer to as the shock angle (the angle of the shock with respect to the upstream velocity) and as the flow turning (or deflection) angle. The working equations for oblique shock waves can be determined by analyzing a very thin control volume that straddles the shock wave (an approach identical to what we used to analyze normal shock waves). We’ll also break the upstream and downstream velocities into components that are normal and tangential to the shock wave. p1, 1, T1 V1 VN1 VT2 VT1 VN2 V2 p2, 2, T2 COM: 1VN 1 2VN 2 (12.246) LME: 2 2 N-dir: 2VN 2 1VN 1 p1 p2 T-dir: mVT 2 mVT 1 0 VT 2 VT 1 VT (The tangential velocity components on either side of the wave are equal!) COE: h1 1 2 V12 h2 1 2 V22 C. Wassgren Chapter 11: Gas Dynamics 2 2 h1 1 2 VN 1 h2 1 2 VN 2 (since VT1 = VT2) 581 (12.247) (12.248) (12.249) Last Updated: 16 Aug 2009 2nd Law: s2 s1 (The large gradients within a shock wave result in an irreversible process.) (12.250) Equations of state for a perfect gas: p RT h c pT (12.251) (12.252) For an ideal gas: V Ma RT (12.253) p 1 Ma 2 1 p0 2 T 1 Ma 2 1 T0 2 1 (12.254) 1 (12.255) From geometry: 2 V12 VN 1 VT2 (12.256) 2 V22 VN 2 VT2 VN 1 V1 sin VN 2 V2 sin (12.257) VT V1 cos V2 cos C. Wassgren Chapter 11: Gas Dynamics 582 Last Updated: 16 Aug 2009 Combining the previous equations and, after much simplifying, we have: Ma N 1 Ma1 sin Ma N 2 Ma 2 sin Ma 2 2 N 1 Ma 2 1 N 2 Ma 2 1 N (12.258) (12.259) 2 (12.260) 1 p2 2 1 Ma 2 1 N p1 1 1 (12.261) 1 Ma 2 1 2 VN 1 tan N 1 VN 2 tan 2 1 Ma 2 1 N (12.262) 2 Ma 2 1 1 T2 N 2 1 Ma 2 1 N 2 T1 1 Ma N1 (12.263) V2 sin 2 1 V1 sin 1 Ma 2 1 1 N (12.264) 2 p02 1 Ma N 1 p01 2 1 Ma 2 1 N 1 1 2 2 Ma N 1 1 1 1 (12.265) T02 1 T01 (12.266) tan 21 1 2 tan 1 Ma N 2 2 (12.267) 2 1 Ma1 1 1 tan 2 tan 2 Ma N 1 1 (12.268) C. Wassgren Chapter 11: Gas Dynamics 583 Last Updated: 16 Aug 2009 Notes: 1. Equations (12.260)-(12.266) are the same relations as for a normal shock wave except that the normal component of the velocity is used. Thus, we can use normal shock relations to determine flow properties across an oblique shock as long as we use the normal component of the velocity. 2. There are two free parameters in the equations (12.260) - (12.268). These typically are the incoming Mach number, Ma1, and the flow turning angle, . Note that the flow is always turned toward the shock wave. 3. Analysis using the 2nd Law (refer to the notes on normal shock waves) states that an oblique shock will only form if the incoming normal Mach number is greater than or equal to one, i.e. MaN1 1. The downstream normal Mach number will be less than one, i.e. MaN2 1. Note that the total downstream Mach number, Ma2, may not be less than one due to the tangential velocity component. 4. Based on the previous note, the minimum value for (the angle of the shock to the incoming flow) is sin-1(1/Ma1) which corresponds to a Mach wave (i.e. a sound wave). The maximum value for is 90 corresponding to a normal shock wave. For both of these shock angle limits we find (from equation (12.268)) that the turning angle of the flow is = 0. Note that will have a maximum value since it is zero for min = sin-1(1/Ma1), is positive for larger values of , then returns to zero for max = 90. We will discuss this maximum value for in a later note. C. Wassgren Chapter 11: Gas Dynamics 584 Last Updated: 16 Aug 2009 5. Since the equations relating the incoming Mach number, Ma1, the wave angle, , and the turning angle, , are complicated, it’s often more instructive and useful to present the data in graphical form rather than in equation form. The following figure plots the wave angle, , as a function of the incoming Mach number, Ma1, for different values of the turning angle, , for =1.4 using Eqn. (12.268) [plot from Zucrow and Hoffman, Gas Dynamics: Vol. I, Wiley]. The lines corresponding to the maximum turning angle, max, and where the downstream flow is sonic, Ma2 = 1, are also shown in the figure. Note that for a given Ma1, there are two values for , the larger corresponding to a “strong” shock with Ma2 < 1 and the smaller corresponding to a “weak” shock with Ma2 > 1. We’ll discuss these observations in greater detail in a subsequent note. strong shocks (Ma2 < 1) maximum turning angle, max weak shocks (Ma2 < 1) weak shocks (Ma2 > 1) C. Wassgren Chapter 11: Gas Dynamics 585 Last Updated: 16 Aug 2009 The following figure plots the downstream Mach number, Ma2, as a function of the incoming Mach number, Ma1, for different values of the turning angle, , for =1.4 using equations (12.267) and (12.268) [plot from Zucrow and Hoffman, Gas Dynamics: Vol. I, Wiley]. The line corresponding to the maximum turning angle, max, is also shown in the figure. weak shocks (Ma2 > 1) strong shocks (Ma2 < 1) C. Wassgren Chapter 11: Gas Dynamics 586 Last Updated: 16 Aug 2009 6. The plots shown in Note 5 indicate that there are two possible wave angles, , corresponding to a particular incoming Mach number, Ma1, and turning angle, . The smaller value of corresponds to a “weak” oblique shock and the larger value corresponds to a “strong” oblique shock. Flow downstream of the strong shock is always subsonic while the flow downstream of the weak shock is usually supersonic, except for a region where is close to max. Ma1 Ma2,strong < 1 Ma2,weak usually > 1 strong weak Both types of shocks occur in practice with the weak shock being more prevalent. If there is a blockage or a high-pressure condition downstream of the shock, the strong shock solution will typically occur (e.g. at the inlet of a supersonic jet engine diffuser where the internal flow is at a high pressure). For flows occurring in the atmosphere where the pressure far downstream of the deflection can only be infinitesimally different from the pressure far upstream of the deflection, the weak shock will occur (e.g. supersonic flow over the surface of an aircraft). To further complicate matters, since the governing equations of the fluid motion are nonlinear, it is possible to have multiple, stable flow solutions. Which solution is observed will depend on the path taken to get to the solution. The hysteresis associated with the starting/overspeeding of a supersonic jet engine diffuser is a good example of a flow situation having multiple, stable solutions (e.g. operating at the design speed with a shock in front of the inlet, or after overspeeding where the shock has been swallowed). The solution that occurs in this flow situation depends on the path taken to get to that solution. If no other information is available, it is generally reasonable to assume that the weak shock occurs. 7. The figures in Note 5 also indicate that there is a maximum turning angle, max, that can be achieved through an oblique shock (this maximum turning angle separates the weak and strong shock solutions). It can be shown that the maximum wave angle corresponding to the maximum turning angle for a given upstream Mach number is (refer to Ferri, A., Elements of Aerodynamics of Supersonic Flow, Macmillan, NY): 1 1 1 4 1 2 2 sin 2 max Ma1 1 1 1 Ma1 Ma1 (12.269) 2 2 16 Ma1 4 The corresponding maximum turning angle may be found using Eqns. (12.258) and (12.268). C. Wassgren Chapter 11: Gas Dynamics 587 Last Updated: 16 Aug 2009 The following figure plots the maximum turning angle, max, and the wave angle corresponding to this maximum turning angle, max, as a function of the incoming Mach number, Ma1 [plot from Zucrow and Hoffman, Gas Dynamics: Vol. I, Wiley]. C. Wassgren Chapter 11: Gas Dynamics 588 Last Updated: 16 Aug 2009 What happens if the flow is deflected by an angle larger than max? A curved, detached shock wave occurs with a strong shock below the sonic line and a weak shock above sonic line. Analysis of curved shock waves is very difficult due to the existence of subsonic, transonic, and supersonic flow, each of which has very different governing differential equations. In addition, the flow downstream of the shock system will have curved streamlines and be irrotational (we’ll discuss this in a different set of notes when investigating Crocco’s Theorem). weak Ma2 > 1 Ma1 > 1 Ma2 = 1 strong Ma2 < 1 >max Note that max increases with increasing upstream Mach number. Hence, it’s possible that a flow with a detached, curved shock at a low Mach number may produce an attached, oblique shock at larger Mach numbers. Ma1 > 1 C. Wassgren Chapter 11: Gas Dynamics > max Ma’1 > Ma1 589 < max Last Updated: 16 Aug 2009 Example: A uniform supersonic air flow traveling at Mach 2.0 passes over a wedge. An oblique shock, making an angle of 40 with the flow direction, is attached to the wedge for these flow conditions. If the static pressure and temperature in the uniform flow are 5 psia and 0 F, determine the static pressure and temperature behind the wave, the Mach number of the flow passing over the wedge, and the wedge half angle. C. Wassgren Chapter 11: Gas Dynamics 590 Last Updated: 16 Aug 2009 Example: Air having an initial Mach number of 3.0, a free stream static pressure of 101.3 kPa, and a free stream static temperature of 300 K is deflected by a wedge by an angle of 15. Assuming that a weak oblique shock occurs, calculate: a. the downstream static pressure, b. the downstream density, c. the downstream temperature, d. the change in the stagnation pressure through the shock, e. the downstream velocity, and f. the downstream Mach number. C. Wassgren Chapter 11: Gas Dynamics 591 Last Updated: 16 Aug 2009 Example: Consider a compression corner with a deflection angle of 28. How does the pressure ratio across the oblique shock change if the incoming Mach number is doubled from 3 to 6? Ma1, p1 Ma2, p2 28 C. Wassgren Chapter 11: Gas Dynamics 592 Last Updated: 16 Aug 2009 Example: An aircraft is to cruise at a Mach number of 3. The stagnation pressure in the flow ahead of the aircraft is 400 kPa. Compare the stagnation pressure recovery ratio for three possible intake scenarios. a. An intake that involves a normal shock in the free stream ahead of the intake followed by an isentropic deceleration of the subsonic flow behind the shock wave to an essentially zero velocity. b. An oblique shock wave diffuser in which the air flows through an oblique shock wave and then an isentropic deceleration of the subsonic flow behind a normal shock wave to an essentially zero velocity. c. An ideal shockless convergent-divergent diffuser in which the air is isentropically brought to an essentially zero velocity. Ma=3 p=400 kPa isentropic deceleration 15 Ma=3 p=400 kPa Ma=3 p=400 kPa C. Wassgren Chapter 11: Gas Dynamics isentropic deceleration isentropic deceleration 593 Last Updated: 16 Aug 2009 14. Expansion Waves Recall the piston experiment in our previous discussion regarding the formation of shock waves. Now let’s consider what happens when we move the piston toward the left (as shown in the figure below) so that an expansion (or rarefaction) wave propagates down the length of the tube. dV dV x stagnant gas rarefaction or expansion wave When we first move the piston, an infinitesimal strength pressure wave travels down the cylinder at the sonic speed. Behind the wave the pressure, temperature, and density decrease (refer to a previous set of notes concerning property changes across a sound wave). In addition, the flow velocity behind the wave will move in the direction of the piston (away from the wave). If we continue to increase the piston velocity, additional pressure waves will propagate down the cylinder. However, these waves travel at a slightly lower speed relative to a fixed observer due to the decreased fluid temperature and leftward fluid velocity behind each wave. Hence, the waves start to spread out. This is the opposite of what occurred for compression waves. 1st wave 2nd wave 2dV 2dV c2 dV T1 c1 V=0 T0 c2 < c1 since T1 < T0 and dV moving to left t, time piston pressure wave (sound wave) x, position Notes: 1. Since the waves do not coalesce, the change in the properties across each wave are infinitesimal. Hence, the flow through each wave is considered isentropic. 2. It is impossible to form an expansion shock wave since each subsequent wave travels slower than the previous wave. The waves will never coalesce. This can also be proven mathematically by showing that entropy would decrease across an “expansion shock.” C. Wassgren Chapter 11: Gas Dynamics 594 Last Updated: 16 Aug 2009 Prandtl-Meyer Expansion Fans Now let’s consider expanding a steady supersonic flow around a gradual, outward-turning corner as shown in the figure below. Ma1 1 Mach waves 2< 1 Ma2 > Ma1 The gradual curve can be approximated by a series of very small, discrete turns, each of which results in a small expansion Mach wave as shown in the previous figure. Recall that Mach waves exist only for supersonic flows, are infinitesimally weak pressure waves, and are inclined at an angle (known as the Mach angle), = sin-1(1/Ma), with respect to the flow. Note that across an expansion wave the Mach number will increase (refer to a previous set of notes concerning sound waves) so that successive expansion waves will have smaller Mach angles (Ma2 > Ma1 2 < 1 in the figure above). As a result, the waves diverge, remain infinitesimally weak, and thus the flow across the waves is isentropic. This type of expansion is sometimes referred to as a non-centered expansion fan. For a sharp corner, the waves comprising the expansion fan start at the corner point then diverge outward as shown in the figure below. This type of fan is known as a centered expansion fan. oblique shock centered expansion fan Notes: 1. The flow into and out of each Mach wave is uniform (i.e. 1D). 2. It is also possible to have isentropic compression waves resulting from gradual (infinitesimal), noncentered turns as shown in the figure below. Non-centered compression fans, however, will eventually converge to form oblique shock waves which are non-isentropic. A centered compression fan resulting from a finite angle corner is an oblique shock wave and thus is non-isentropic. non-isentropic oblique shock isentropic compression region gradual compression turn 3. A compression wave turns the flow toward the wave while an expansion wave turns the flow away from the wave. C. Wassgren Chapter 11: Gas Dynamics 595 Last Updated: 16 Aug 2009 Now let’s analyze the steady, 2D flow through a single Mach wave as shown in the figure below. Our analysis will be for a compression Mach wave (d > 0 in the figure below) but the analysis will also hold for an expansion Mach wave (d < 0). Note that the upstream Mach number must be supersonic in order for Mach waves to exist. p , T, VT + dVT VN VN + dVN d V d VT V+dV d p+dp, T+dT, +d Note: d is defined as being positive for counter-clockwise (i.e. compression) turns. For an expansion, d < 0. From geometry: VT V cos (12.270) VT dVT V dV cos d (12.271) From the LME in the tangential direction: m VT dVT mVT 0 dVT 0 (12.272) Combining equations (12.270)-(12.272): VT V cos V dV cos d V cos V dV cos cos d sin sin d Since the turning angle, d, is very small we can write: V cos V dV cos d sin V cos Vd sin dV cos dVd sin Neglecting H.O.T.s and simplifying: dV d tan V (12.273) Since the expansion wave is a Mach wave (with wave angle, sin = 1/Ma), we have: 1 tan Ma Ma 2 1 1 (1-Ma2)1/2 Substituting and simplifying: d dV (Note: Since Ma > 1, d > 0 dV < 0 and d < 0 dV > 0.) V Ma 2 1 C. Wassgren Chapter 11: Gas Dynamics 596 (12.274) Last Updated: 16 Aug 2009 For a perfect gas: V 2 RT Ma 2 1 RT0 Ma 1 Ma 2 2 1 2 1 dV 1 d Ma Ma 2 1 V 2 Ma (12.275) Substituting (12.275) into equation (12.274) gives: Ma 2 1 d Ma d 1 2 Ma 1 Ma 2 Note: Since Ma > 1, d > 0 d(Ma) < 0 and d < 0 d(Ma) > 0. Integrating equation (12.276) gives: 1 1 1 tan Ma 2 1 tan 1 Ma 2 1 constant 1 1 (12.276) (12.277) The constant of integration can be determined if the initial Mach number, Ma1, and flow deflection angle, 1, are known. For convenience, we define a reference state where, Ma = 1, and = 0. For these conditions, equation (12.277) becomes: 1 1 1 tan Prandtl-Meyer Angle (12.278) Ma 2 1 tan 1 Ma 2 1 1 1 Note: The symbol “” has been changed to a “” in order to signify that this is the Prandtl-Meyer angle. The Prandtl-Meyer angle is the angle, , that the flow needs to be turned to go from sonic conditions to get to the new Mach number, Ma. C. Wassgren Chapter 11: Gas Dynamics 597 Last Updated: 16 Aug 2009 Notes: 1. The angle, , is positive for counter-clockwise (compressive) rotations and negative for clockwise (expansive) rotations. The convention, however, is to drop the negative sign when reporting PrandtlMeyer angles. The Prandtl-Meyer angle is plotted as a function of Mach number for = 1.4 in the figure below. Prandtl-Meyer angle, | [deg] 2. 120 = 1.4 100 80 60 40 20 0 0 1 2 3 4 5 6 7 8 Mach number, Ma 3. For an arbitrary incoming Mach number (Ma1 > 1, 1 = fcn(Ma1)) we can imagine that there is some imaginary upstream corner that expands the flow from Ma0 = 1 (0 = 0) to the current Mach number, Ma1 (1 = fcn(Ma1)), as shown in the figure below. Ma1 > 1 1 0 imaginary expansion Ma0 = 1 0 = 0 To expand a flow from Ma1 to Ma2 (Ma2 > Ma1), we need to turn the flow by an angle of 2-1. For example, to get from sonic conditions (Ma0 = 1) to Ma1 = 2.0, we need to turn the flow by an angle of 1-0 = -26.38 - 0 = -26.38 and to get the flow from sonic conditions to Ma2 = 3.0 we need to turn the flow by an angle of 2-0 = -49.76 - 0 = -49.76. To go from Ma1 = 2.0 to Ma2 = 3.0, we need to turn the flow by an angle of 2-1 = -49.76 - (-26.38) = -23.38. Note that the negative signs indicate that we need to turn the flow away from the Mach wave, i.e. expand the flow. 4. Compressive Mach waves turn the flow toward the Mach wave while expansive Mach waves turn the flow away from the Mach wave. d d expansive Mach wave compressive Mach wave C. Wassgren Chapter 11: Gas Dynamics 598 Last Updated: 16 Aug 2009 5. Since the expansion process is isentropic, the isentropic flow relations can be used throughout the expansion fan. The isentropic relations may also be used in a limited region within a non-centered compression fan where the Mach waves do not intersect. An oblique shock forms where the compression Mach waves merge. 6. To expand the flow of air (= 1.4) from sonic conditions (Ma = 1) to an infinite Mach number, i.e. Ma , we must turn the flow by max = -130.4. 1 1 (12.279) max lim Ma 2 1 If the corner has an angle greater than this maximum angle, then a vacuum region forms. Note that the continuum assumption would break down in the region adjacent to the vacuum. Ma = 1 130.4 Ma = 1 > max vacuum Ma C. Wassgren Chapter 11: Gas Dynamics Ma 599 Last Updated: 16 Aug 2009 Example: A uniform supersonic flow at Mach 2.0, static pressure 10 psia, and temperature 400 R expands around a 10 corner. Determine the downstream Mach number, pressure, temperature, and the fan angle. fan angle 10 C. Wassgren Chapter 11: Gas Dynamics 600 Last Updated: 16 Aug 2009 Example: A wind tunnel nozzle is designed to yield a parallel uniform flow of air with a Mach number of 3.0. The stagnation pressure of the air supply reservoir is 7000 kPa, and the nozzle exhausts into the atmosphere (100 kPa). Calculate the flow angle at the exit lip of the nozzle. Ma=3.0 7000 kPa 100 kPa angle = ? C. Wassgren Chapter 11: Gas Dynamics 601 Last Updated: 16 Aug 2009 Example: Calculate the lift and drag coefficients (per unit depth into the page) for a flat-plate airfoil with a chord length of 1 m. The plate is at an angle of attack of 6 degree with respect to the incoming flow which has a Mach number of 2.5. Clearly sketch the wave patterns at both the leading and trailing edges of the airfoils. Note that the lift and drag coefficients are based on the planform area of the airfoil. 1m Ma=2.5 6 C. Wassgren Chapter 11: Gas Dynamics 602 Last Updated: 16 Aug 2009 15. Reflection and Interaction of Oblique Shock Waves In this set of notes we’ll consider the interaction between oblique shocks and solid boundaries, free surfaces, and with other oblique shocks. Reflection with a Solid Boundary When an oblique shock intersects a solid, straight boundary, it will reflect as another oblique shock in order for the flow to remain parallel to the boundary as shown in the figure below. Ma2, p2 1 Ma3, p3 Ma1, p1 1 2 wall Image from: Shapiro, A.H., The Dynamics and Thermodynamics of Compressible Fluid Flow Vol. I, Wiley. The flow properties through the incident and reflected shocks can be determined using the following procedure. 1. For the given Ma1 and , determine Ma2 and p2/p1. 2. For this value of Ma2 and since the turning angle of the second wave is also (in order to keep the flow parallel to the flat wall), one can determine Ma3,2, and p3/p2. 3. The pressure ratio across both waves, p3/p1, is found using: p3 p3 p2 p1 p2 p1 4. The angle that the reflected wave makes with the wall is 2-. Notes: 1. The wave angle, , and the turning angle, , are measured with respect to the incoming flow direction. 2. Without viscous effects the pressure change is discontinuous across the shock at the wall. Due to the boundary layer, the flow drops to zero velocity at the wall and so the flow adjacent to the wall is subsonic and thus cannot sustain pressure discontinuities. Thus the boundary layer causes the pressure distribution to “spread out” as shown in the figure below. 1 2 3 p3 p1 without BL p3 p1 with BL The details of the shock interaction with the wall depend on whether the boundary layer is laminar or turbulent, the thickness of the boundary layer, and the shock strength. A flow separation bubble may also occur due to an adverse pressure gradient. C. Wassgren Chapter 11: Gas Dynamics 603 Last Updated: 16 Aug 2009 3. Recall that there is a maximum possible angle through which the flow can be turned using an oblique shock. This maximum angle (max) decreases as the incoming Mach number decreases. Thus, when the flow passes through the initial oblique shock, there exists the possibility that the reflected flow must be turned by an angle that is greater than the maximum turning angle in order to remain parallel to the wall. In this case, a Mach reflection appears where a curved strong shock occurs adjacent to the wall behind which there is subsonic flow. Since this subsonic flow need not be parallel to the wall, the flow above the wall shock layer also does not have to be parallel to the wall. The flow behind the curved wall shock is divided from the flow behind the “reflected” oblique shock by a slipline (aka contact surface, slipstream, vortex sheet). Across the slipline there are changes in velocity, temperature, and entropy. Since the flow is subsonic, the details of the flow may be affected by downstream conditions and, as a result, the analytical modeling of the Mach reflection is difficult. incident shock reflected incident shock shock Ma > 1 flow is not parallel to the wall Ma < 1 slipline reflected shock flow direction Image from: Shapiro, A.H., The Dynamics and Thermodynamics of Compressible Fluid Flow Vol. I, Wiley. 4. Instead of a flat wall, consider the reflection from a wall that is turned at some angle. The flow must still remain parallel to the downstream wall. The oblique shock reflected from a wall turned away from the flow will be weaker than one reflected from a flat wall. It is even possible to “cancel” the reflected oblique shock by turning the downstream shock at the angle through which the flow is turned by the initial shock. Turning the wall by an angle greater than this results in an expansion fan. C. Wassgren Chapter 11: Gas Dynamics 604 Last Updated: 16 Aug 2009 Example: Air flowing with a Mach number of 2.5 with a pressure of 60 kPa (abs) and a temperature of 253 K passes over a wedge which turns the flow through an angle of 4 leading to the generation of an oblique shock wave. a. If this oblique shock wave impinges on a flat wall, which is parallel to the flow upstream of the wedge, determine the pressure and Mach number behind the reflected shock wave. b. If the wedge is 1 m long, what is the minimum height above the wall, hmin, that the wedge must be in order to not intersect the reflected shock. c. If the reflected shock did intersect with the wedge, what type of shock pattern would appear? d. How could we avoid producing the reflected shock? 1m 4 Ma1=2.5 p1=60 kPa (abs) hmin T1=253 K C. Wassgren Chapter 11: Gas Dynamics 605 Last Updated: 16 Aug 2009 Oblique Shock Reflection from a Free Surface When an oblique shock intersects a free surface, the reflection must be an expansion fan so that the flow pressure remains equal to the free surface pressure as shown in the figure below. free surface p0 free surface p0 p3 = p0 p1 = p0 Ma3 Ma1 oblique shock C. Wassgren Chapter 11: Gas Dynamics Ma2 p2 > p0 606 expansion fan Last Updated: 16 Aug 2009 Example: A symmetric converging-diverging nozzle is designed for an exit Mach number of 2.0. With the nozzle exhausting to a back pressure of 15 psia, however, and a reservoir pressure of 78.2 psia, the nozzle is overexpanded as is shown in the figure below. pb= 15 psia R1 p0=78.2 psia a. b. R2 R3 Determine the Mach number in regions R1, R2, and R3. Determine the angle of the flow with respect to the horizontal (in degrees) in regions R1, R2, and R3. C. Wassgren Chapter 11: Gas Dynamics 607 Last Updated: 16 Aug 2009 Interaction of Oblique Shock Waves When two oblique shock waves intersect, each wave will be transmitted through, but be affected by, the other wave. For example, consider the interaction of two oblique shocks as shown in the following diagram. 2 4 1 5 3 slipline The flow in regions 2 and 3 are found given the conditions in region 1 and the wall turning angles. The flow in regions 4 and 5 must be parallel to one another and, hence, from the linear momentum equation the pressures in regions 4 and 5 must be the same. An iterative procedure can be used to solve for the flows from regions 2 and 3 into regions 4 and 5. A slipline occurs between regions 4 and 5 since these regions, in general, have different entropy, temperature, and velocity. Notes: 1. There may be oblique shock intersection situations where no solution exists using oblique shocks. In this case, more complicated flow patterns may occur that include normal shock waves (Mach reflections). slipline Ma < 1 slipline Image from: Shapiro, A.H., The Dynamics and Thermodynamics of Compressible Fluid Flow Vol. I, Wiley. C. Wassgren Chapter 11: Gas Dynamics 608 Last Updated: 16 Aug 2009 Example: Air is flowing down a wide channel with a Mach number of 3, pressure of 30 kPa, and a temperature of 263 K. The upper wall of this channel turns through an angle of 4 towards the flow while the lower wall turns through an angle of 3 towards the flow leading to the generation of two oblique shock waves which intersect each other. Find the pressure and flow direction downstream of the shock intersection. 4 Ma1=3 p1=30 kPa T1=263 K C. Wassgren Chapter 11: Gas Dynamics 3 609 Last Updated: 16 Aug 2009 Now consider flow in a corner with several small, discrete changes in the wall angle as shown in the figure below. sliplines weak reflected waves required in order to equalize the pressure on both sides of the slipstream (can be compression or expansion depending on the flow situation) stronger oblique shock weak oblique shocks The oblique shocks will intersect and coalesce into a single oblique shock which is stronger than any of the initial oblique shock waves. Sliplines and weak reflected waves appear at the intersection of the waves. Notes: 1. An oblique shock is generally considered to form when two or more compression waves coalesce. 2. For a continuously inward curving wall, the slipstreams are spaced an infinitesimal distance apart so the downstream flow has continuously changing entropy, velocity, and temperature. The oblique shock that forms from the interacting compression waves will have a strength such that it turns the flow by the wall angle, , for the given incoming Mach number. Region of continuously varying velocity, entropy and temperature in the direction perpendicular to the wall. 3. When waves travel in the same direction as viewed by an observer oriented so that they’re looking downstream, the waves are said to be of the same family. The waves are further classified as being either left-running or right-running depending on what direction the waves are oriented. Examples of this notation are given below. flow direction observer looking downstream left-running compression wave right-running expansion wave These waves are of different families since one is left-running and the other is right running. C. Wassgren Chapter 11: Gas Dynamics 610 All of the compression waves shown in the figure above are of the same family since they are all left-running waves. Last Updated: 16 Aug 2009 16. Reflection and Interaction of Expansion Waves Reflection with a Solid Boundary When an expansion wave intersects a solid boundary, it will reflect as another expansion fan (of opposite family) in order for the flow to remain parallel to the boundary as shown in the figure below. Ma3, p3 Ma2, p2 Ma1, p1 non-simple region The flow through the incident and reflected expansion fans can be determined using the following procedure. 1. For the given (Ma1, 1) and 12, determine (Ma2, 2) using 2 = 1 + . 2. For this value of (Ma2, 2) and since the turning angle of the second wave is also (in order to keep the flow parallel to the flat wall), one can determine (Ma3,: 3 = 2+ . 3. The pressure ratio across both waves, p3/p1, is found using the isentropic relations: p2 1 1 Ma 2 2 p01 2 1 p3 1 2 1 Ma 3 p01 2 1 Notes: 1. The flow through the expansion fans is isentropic. 2. The region where the expansion fans interact is known as a non-simple region. This region is not amenable to our Prandtl-Meyer expansion fan analysis. Instead, we use can the Method of Characteristics (a topic that will be discussed in a later set of notes) to analyze the flow in this region. 3. An expansion wave may be canceled by turning the flow inward by the same amount as the flow turning angle. C. Wassgren Chapter 11: Gas Dynamics 611 Last Updated: 16 Aug 2009 Expansion Fan Reflection from a Free Surface When an expansion fan intersects a free surface, the reflection must be an oblique shock (of opposite family) so that the flow pressure remains equal to the free surface pressure as shown in the figure below. non-simple region p0 free surface p1 = p0 p0 Ma1 expansion fan Ma3 p2 < p0 free surface p3 = p0 Ma2 oblique shock Interaction of Expansion Waves When expansion fans intersect, each expansion wave will be transmitted through, but be affected by, the other waves. There can be many regions of non-simple flow as shown in the figure below. Notes: 1. No slipstreams occur with expansion fans since the flow properties change continuously through the fan. C. Wassgren Chapter 11: Gas Dynamics 612 Last Updated: 16 Aug 2009 Mach Diamonds A phenomenon known as Mach Diamonds (aka shock diamonds) can form when a supersonic stream exits a device and interacts with the surrounding atmosphere. Consider supersonic flow from an exit, with pressure pe, into the surrounding atmosphere with pressure, pb. Over-Expanded Case pe < pb Under-Expanded Case pe > pb Notes: 1. At design conditions, pe = pb, so additional expansion or compression of the flow is not required. Hence, Mach diamonds do not appear at design conditions. 2. If the flow turning angle is sufficiently large, Mach reflections (aka Mach disks in 3D) will appear as shown in the figure below. pe < pb 3. 4. 5. The bright regions in the Mach diamond shown in the photograph above are caused by heating of the gas as it passes through the oblique shocks. Viscous interaction with the external fluid results in dissipation of the Mach diamond pattern. Even without viscous effects, the sequence of oblique shocks and expansion fans will eventually dissipate since the stagnation pressure decreases after each shock. This can be shown by considering the Mach number in the regions downstream of an oblique shock (referred to by the subscript i) which are in contact with the free surface such that: pb 1 1 1 Ma i2 2 p0i Since p0i decreases after passing through an oblique shock and pb is a constant, the ratio pb/p0i increases and Mai decreases. Thus, the sequence of oblique shocks gets weaker. C. Wassgren Chapter 11: Gas Dynamics 613 Last Updated: 16 Aug 2009 Interaction Between An Oblique Shock and an Expansion Fan When an oblique shock interacts with an expansion fan of the same family, the shock will be weakened and become curved. Behind the shock wave the flow becomes rotational. In addition, the incident expansion waves will be reflected. straight shock curved, weakened shock region of rotational flow straight shock slip line Notes: 1. Waves will also be reflected from where waves intersect the slip line. These waves were left off the previous schematic for the sake of clarity. 2. The region of rotational flow is not isentropic (the entropy varies continuously behind this curved section of shock wave). This will be discussed in a later set of notes concerning Crocco’s Theorem. C. Wassgren Chapter 11: Gas Dynamics 614 Last Updated: 16 Aug 2009 17. Equations of Motion in Terms of the Velocity Potential Now let’s consider the irrotational, isentropic flow of a compressible fluid where body forces are negligible. Recall that the momentum equations for a fluid in which viscous and body forces are negligible are given by Euler’s Equations: u u u p t We can re-write the pressure gradient term in terms of the speed of sound, c, as is shown below: dp dp d p dx dp dx d d dp p d but since the flow is isentropic, we have: dp dp p c 2 d d s Thus, the momentum equations can be written as: u u u c 2 t Now take the dot product of this equation with the velocity and re-arrange: u c2 u u u u u t 1 c2 u u u u u u 2 t The continuity equation can be used to re-write the RHS of the previous equation: u u u 0 t t u u t Thus, the momentum equations become: 1 c 2 2 (12.280) u u u u u t c u 2 t The density in the previous equation may be eliminated using Bernoulli’s equation. Since the flow is irrotational, we can write Bernoulli’s equation as: dp 1 2 F t t where the velocity has been written in terms of a velocity potential, . C. Wassgren Chapter 11: Gas Dynamics 615 Last Updated: 16 Aug 2009 Taking the time derivative of Bernoulli’s equation gives: 2 dp 1 u u F ' t t 2 t 2 t But, dp dp d d c2 t t d t so that: 2 t 2 c2 d t c 2 t c 2 1 u u F ' t t 2 t 2 1 c 2 F ' t 2 u u t 2 t t Substituting into Eqn. (12.280) gives: 1 2 1 u u u u u F ' t 2 u u c 2 u 2 t 2 t t u 1 2 u u u u u F ' t 2 2 t c t 1 t u u F t u u u 2 c t Re-writing the velocities in terms of the potential function gives: 1 2 2 F t c t t (12.281) Governing equation for the isentropic, irrotational flow of a compressible fluid where body forces are negligible C. Wassgren Chapter 11: Gas Dynamics 616 Last Updated: 16 Aug 2009 Notes: 1. Consider the flow of an incompressible fluid. If the flow is incompressible, then the speed of sound in the fluid will be infinite. Thus, the governing equation for the irrotational flow of an incompressible fluid becomes: 2 0 (12.282) Governing equation for the irrotational flow of an incompressible fluid in which body forces are negligible Note that this is just Laplace’s equation! A significant point regarding Eqn. (12.282) is that it is a linear PDE which means that the principle of superposition may be used to add together “building block” solutions to form new and more complex solutions. 2. For a steady, compressible flow, Eqn. (12.281) simplifies to: 1 2 2 (12.283) c Governing equation for the steady, isentropic, irrotational flow of a compressible fluid in which body forces are negligible 3. Equations (12.281) and (12.283) are non-linear PDEs which are complex to solve by hand. There is currently no known method for analytically solving these equations in a general way (computational techniques can be used, however). Instead, we must resort to special cases for solving these equations. One of these methods, known as small-perturbation theory, is described in the following section. C. Wassgren Chapter 11: Gas Dynamics 617 Last Updated: 16 Aug 2009 18. Small Perturbation Theory Recall that the equation of motion for an irrotational flow where body and viscous forces are negligible is: 1 2 2 F t (12.284) c t t where the velocity is given in terms of a potential function, u=, and c is the speed of sound. For a steady flow, this equation simplifies to: 1 2 2 (12.285) c Suppose that a uniform flow approaches an object that is sufficiently slender so that it produces only a small perturbation to the incoming stream as shown in the figure below. The potential function for such a flow can be written as: U x (12.286) U where U is the velocity of the incoming flow and is the y potential function for the velocity perturbations, u’, i.e.: u ' x ˆ (Note: U x U e x ) u ' Substituting Eqn. (12.286) into Eqn. (12.285) gives: 1 ˆ ˆ ˆ 2 2 U e x U e x U e x (12.287) c The speed of sound, c, should also be written in terms of a velocity perturbation. This is accomplished by recalling that for a steady, isentropic flow, the stagnation enthalpy will remain constant: 2 h0 h0, h 1 2 U 2 h 1 2 U where h and U are the local enthalpy and velocity. If we consider the fluid to be a perfect gas, then: R T c 1 1 2 h c p T so that: c2 c2 2 12U 2 12U 1 1 1 U 2 2 2 c 2 c 1 Ma 2 2 c where ˆ ˆ U 2 U e x U e x C. Wassgren Chapter 11: Gas Dynamics (12.288) 618 Last Updated: 16 Aug 2009 Substituting Eqn. (12.288) into Eqn. (12.287) and expanding (refer to the Appendix): 2 2 2 1 Ma x y z 2 2 ' ux U + 1 1 2 2 u 'y 2 2 2 2 2 Ma 1 2 1 2 2 2 Ma y U x z 1 2 Ma ' u x U 2 u 'y x 2 U 2 1 2 Ma ' u x U 2 u' 2 2Ma x U u 'y U 2 2 2 2 u' z y 2 U 2 2 u 'y 2 2 z U y 2 u 'y 2 2Ma yx U u' z U 2 2 u' 2 2 Ma z U xy 2 xz 2 2 z 2 ' 2 2 uz 2 2 z U x u' 2 2 2Ma x zy U 2 2 2 2 2 y x ' uz U 2 zx Note that the small perturbation assumption has not been used in deriving the previous equation. Now, if we assume that the velocity perturbations are indeed small, i.e.: u 'y u' u' 2 2 2 , Ma z 1 Ma x , Ma U U U 2 Ma ' ux U 2 u 'y 2 , Ma U 2 u' 2 , Ma z U u ' u 'y u' 2 2 , Ma x Ma x U U U then the PDE simplifies to: 2 2 ' u z U 2 2 2 2 (12.289) u 'y 2 , Ma U 1 Ma x y z 1 Ma 2 2 1 2 u' x U u' z U 1 2 2 x (12.290) Eqn of Motion for Small Perturbations when Ma1 Note that the 2/x2 term on the RHS has been retained in the previous equation. This is because when Ma is near unity, the 2/x2 term on the LHS may be of the same order of magnitude as the RHS and thus the RHS can not be neglected. Thus, Eqn. (12.290) is the appropriate form for the equation of motion when the free stream Mach number is near unity. Note that Eqn. (12.290) is non-linear. If we can assume further that: ' 2 Ma u x 1 2 1 Ma U then Eqn. (12.290) simplifies to: 2 2 2 1 Ma x y z 0 2 2 2 (12.291) (12.292) 2 Eqn of Motion for Small Perturbations, Ma not near unity C. Wassgren Chapter 11: Gas Dynamics 619 Last Updated: 16 Aug 2009 Notes: 1. Note that the assumptions (Eqn. (12.289)) used in deriving Eqn. (12.292) also indicate that the Mach number cannot be too large (e.g. we can’t use Eqn. (12.292) to model hypersonic flows). 2. Equation (12.292) is a linear PDE. This means that we can use the principle of superposition to add together solutions of the equation to form new solutions. 3. It is instructive to examine the mathematics of Eqn. (12.292) in greater detail. For simplicity, let’s consider only 2D flows. The general form for a 2nd order, linear PDE (with x and y as the independent variables) is: 2 2 2 C 2 D E F G A 2 B xy x y x y The behavior of the PDE will vary significantly depending on the value of its principle part which is given by: 0 elliptic PDE 2 B 4 AC 0 parabolic PDE 0 hyperbolic PDE The details of the differences in behavior between the three types of PDEs will not be examined here except for dependence on boundary conditions. For an elliptic PDE, the solution at a particular point in the domain will depend on all of the boundary conditions. In the language of mathematics, this is the same as saying that there are no real characteristic directions (we will discuss characteristic curves later in the course). The prototype elliptic PDE is Laplace’s equation: 2 0 For a hyperbolic PDE, the solution at a particular point in the domain will depend only on a particular region of the boundary conditions (termed the zone of dependence). Furthermore, the solution at that point will have an effect only on a particular region termed the zone of influence. Mathematically speaking, there are two real characteristic directions for hyperbolic PDEs. The prototype equation for hyperbolic PDEs is the wave equation: 2u 2u c2 2 t 2 x We will not discuss parabolic PDEs here since we will only be concerned with elliptic and hyperbolic PDEs in the following analyses. In addition, we already stated that Eqn. (12.292) is for the case when Ma is not near a value of one, and -4AC only equals zero when Ma = 1. How does all of this relate to the small perturbation equation of motion? Consider Eqn. (12.292) simplified for 2D flow: 2 2 1 Ma x y 0 2 2 (12.293) 2 The behavior of the flow will vary significantly depending on the Mach number: 1 elliptic PDE Ma A = 1-Ma∞2, B = 0, C = 1 B2-4AC = 4(Ma∞2 - 1) 1 hyperbolic PDE Recall that when Ma1 (the transonic region), we must use equation (12.290) rather than equation (12.292). The analysis in the transonic region is complex due to the non-linearity of the governing equation. Based on our previous analyses of 1D, compressible flow, the behavior of subsonic and supersonic flows can be quite different. The same holds true here even though the governing equation (Eqn. (12.293)) looks fairly simple. A deeper investigation of the differences between subsonic and supersonic flows will be given in a following section. C. Wassgren Chapter 11: Gas Dynamics 620 Last Updated: 16 Aug 2009 4. Since we linearized the governing PDE using the small perturbation assumption, we should also linearize the boundary conditions. The appropriate boundary condition at a solid boundary for an inviscid flow is not the no-slip condition since the fluid can slip tangent to the surface. Instead, we specify that the flow will not penetrate the solid object, i.e.: 0 n where n is the direction normal to the object’s surface. This is the same as stating that the surface of the object is a streamline for the flow (recall that there is no flow across a streamline). Thus, we can write: u 'y y u 'y U dy ' U dx surface U u x u' x 1 x U Since we’re assuming that the perturbation velocities are small in comparison with the free stream velocity, we have: u 'y ' ux 1 U U surface Since the object is considered to be slender, we can use a Taylor series to show that the y-perturbation velocity at y=0 can be used rather using the velocity on the surface: dy dx surface u 'y xs , ys u 'y xs , 0 u 'y u 'y y ys xs ,0 xs , 0 where (xs, ys) are the coordinates of the object surface and ys is assumed to be very small. Thus, the appropriate boundary condition at the object surface becomes: dy dx surface u 'y xs ,0 U C. Wassgren Chapter 11: Gas Dynamics (12.294) 621 Last Updated: 16 Aug 2009 5. An additional quantity that is often helpful when analyzing external flows is the pressure coefficient, Cp, which is defined as: p p Cp 1 U2 2 This can be written for a perfect gas as: p 1 p (12.295) Cp 1 Ma 2 2 The pressure ratio can be written in terms of the free stream and perturbation velocities by using the energy equation for an isentropic (2D) flow: 2 2 1 1 ' ' 2 T U u x u y T 2c U 2c p p where cp R 1 so that T 1 2 ' ' U 2U u x u x 2 R u T 2 ' y ' u' u T 2 1 1 2 1 Ma 2 x x U U T For an isentropic flow of a perfect gas: 2 2 u 'y U 1 2 U 2 R 2 2 1 2 ' ' u x u 'y p T 1 2 ux 1 2 1 Ma 2 (12.296) U U U p T The previous relation may be simplified further by use of the binomial theorem which states that for x<1: n n 1 x 2 1 x n 1 nx 2! Since the perturbations are small: 1 2 ' u ' u 'y u 1 2 1 2 1 Ma 2 x x U U U 1 C. Wassgren Chapter 11: Gas Dynamics 1 2 2 Ma 2 1 2 ' ' ' 2 ux ux u y U U U 622 2 Last Updated: 16 Aug 2009 Using the assumption that the perturbation velocities are small in comparison to the free stream velocity, the remaining terms in the series can be neglected. Substituting this result into Eqn. (12.296) and then substituting this result back into the definition of the pressure coefficient (Eqn. (12.295)) gives: 2 2 ' ' u x u 'y 2 ux 1 2 Ma 2 2 U U U 2 ' ' ' 2 ux ux u y Cp U U U 1 Ma 2 2 Again, using the small perturbation assumption we observe that the squared terms in the previous equation will be very small compared to the remaining term so that the pressure coefficient is given by: u' C p 2 x (12.297) U C. Wassgren Chapter 11: Gas Dynamics 623 Last Updated: 16 Aug 2009 6. Equation (12.293) for a subsonic flow may be reduced to Laplace’s equation using an appropriate transformation of variables. This is useful since there has been a large body of work in determining the solution to Laplace’s equation for various boundary conditions. In particular, the equation of motion for an incompressible flow is Laplace’s equation. To transform Eqn. (12.293), define a new coordinate system using the variables (, ) and a transformed perturbation potential, , in the following manner: x 2 y where 1-Ma (12.298) Note that: 1 1 x x x x x 2 x x x x 1 0 1 0 2 (12.299) 1 2 2 1 1 y y y y y 2 y y y y 0 0 2 (12.300) 2 2 Substituting Eqns. (12.299) and (12.300) into Eqn. (12.293) gives: 1 2 2 2 2 2 0 2 2 0 (12.301) 2 2 This is Laplace’s Equation in the (, ) plane! This is the same equation of motion as for an incompressible fluid (2= 0). Since we transformed the governing equation, we also need to transform boundary conditions. Let the shape of a boundary surface in the (x, y) plane be described by: y s f xs (12.302) and the corresponding surface in the (, ) plane be: s f s C. Wassgren Chapter 11: Gas Dynamics 624 (12.303) Last Updated: 16 Aug 2009 Recall from Eqn. (12.294) that the boundary condition at a surface in the actual flow (in the (x, y) plane) is: dy (12.304) u U y x ,0 s y x ,0 dx surface s The same boundary condition expressed in the transformed flow, i.e. in the (, ) plane, is: df u U s ,0 d surface ,0 (12.305) s Note however that: 1 y x ,0 y y ,0 s s 0 s ,0 Hence, since the left hand sides of Eqns. (12.304) and (12.305) are equal, we also have: dy df df dx surface dx surface d surface (12.306) (12.307) Since the slope of the actual boundary surface is (df/dx)surface, Eqn. (12.307) tells us that the shape of the boundary surface in the transformed plane ((, )) is the same as that in the real ((x, y)) plane. Furthermore, since the boundary surface is the same in the transformed plane, and since Eqn. (12.301) is the equation of motion for an incompressible fluid, then must be the perturbation velocity potential for an incompressible flow past the same boundary surface. Hence, if we can determine assuming incompressible flow over the surface, the corresponding compressible flow solution can be determined using the scaling relationships given in Eqn. (12.298). Recall that the pressure coefficient (Eqn. (12.297)) is given by: u' 2 2 1 1 2 C p 2 x U x U x U U (12.308) C p Note that the last term in parentheses in the previous equation is the pressure coefficient determined from the incompressible (i.e., transformed plane) solution, C p 0 . Hence, the pressure coefficient for the subsonic, compressible flow can be determined by scaling the incompressible pressure coefficient by: Cp Cp (12.309) 2 1 Ma This scaling relationship is known as the Prandtl-Glauert Rule. C. Wassgren Chapter 11: Gas Dynamics 625 Last Updated: 16 Aug 2009 Notes: a. It can also be shown that the lift and moment coefficients are scaled in a similar manner for linearized, subsonic compressible flow (note that the drag coefficient is always zero for subsonic potential flow): CL CL (12.310) 2 1 Ma CM b. CM (12.311) 2 1 Ma The effect of compressibility on the flow perturbation velocities can be determined from Eqn. (12.298): 1 u u (12.312) 2 u 1 Ma Compressibility acts to increase the magnitude of the perturbations (since the denominator is always less than one). Hence, a disturbance to a compressible flow reaches further into the flow than for an incompressible flow. c. The Prandtl-Glauert rule tends to underpredict the pressure coefficient for real flows. It is only reasonably accurate up to a Mach number of about 0.7. This is because the rule is based on linearization of the governing equations. Other rules have been proposed (e.g. the Karman-Tsien rule and the Laitone rule) that incorporate non-linear flow effects to give improved predictions to the pressure coefficient. d. Recall that for an isentropic flow that 1 c p cT 1 dp dT 1 1 so that a decrease in pressure corresponds to a decrease in temperature. So, for subsonic flow over the suction side of an airfoil near sonic conditions, not only will the pressure drop, but the temperature will drop as well. This drop in temperature in humid conditions can result in water vapor turning to liquid water, i.e. condensation. This effect is often visible on the surface of high speed aircraft in humid environments. (Images from: http://en.wikipedia.org/wiki/Prandtl%E2%80%93Glauert_singularity ) C. Wassgren Chapter 11: Gas Dynamics 626 Last Updated: 16 Aug 2009 Appendix 1 ˆ ˆ ˆ U e x U e x U e x c2 1 ˆ ˆ ˆ 2 U ex ey ez x y z c 2 ˆ ˆ ˆ U x y y z z U x e x y e y z e z x 1 ˆ ˆ ˆ ex ey ez 2 U y z x c 2 = 1 c2 1 c2 1 c2 2 2 2 ˆ ˆ ˆ U 2 e x U x xy e y U x xz e z x x 2 2 2 ˆ ˆ ˆ ex ey ez y y 2 y yz y yx 2 2 2 ˆ ˆ ˆ ez ex ey z zx z zy z z 2 2 2 2 U U x 2 y yx z zx x x 2 2 2 U xy y 2 x z zy y y 2 2 2 U z x xz y yz z z 2 2 U U U U 2 x 2 2 2 2 U yx z zx x z zx 22 2 2 xy y y 2 y z zy 22 2 2 xz z y yz z z 2 2 2U 2 2 x x 2 x x 2 2 y yx x y 2 y xy y x 2 z xz z x 2 2 2 2 2 U 2 2U 2 y xy z xz x x x 22 22 22 2 2 2 z z y y x x 2 2 2 2 2 2 x y yx y z zy x z zx C. Wassgren Chapter 11: Gas Dynamics 627 Last Updated: 16 Aug 2009 2 2 2 2 2 U 2 2U 2 y xy z xz x x x 22 22 22 1 2 2 x x 2 y y 2 z z 2 c 2 2 2 2 2 2 x y yx y z zy x z zx Recall that from the speed of sound equation: 1 U 2 2 2 c 2 c 1 Ma 2 2 c 2 ˆ 1 2 U e x 2 Ma c 1 2 2 c (12.313) 2 2 2 2 U 2U x x y z 1 2 2 c 1 Ma 2 2 c 2 2 2 1 2U 1 2 c 1 2 2 2 c x c x y z 2 c 2 c Ma 2 1 2 1 2 2 2 2 1 U x U x y z 2 C. Wassgren Chapter 11: Gas Dynamics 628 (12.314) Last Updated: 16 Aug 2009 Substituting (12.314) into Eqn. (12.313) gives: Ma 2 1 2 1 2 2 2 2 2 2 c 1 U x U x y z 2 2 2 2 2 2 U 2 2U 2 y xy z xz x x x 22 22 22 x x 2 y y 2 z z 2 2 2 2 2 2 2 x y yx y z zy x z zx Ma 2 1 2 1 2 2 2 2 1 2 U x U x y z 2 2 2 2 2 2 2 2 y xy z xz U x x x 22 22 22 21 Ma 2 x x 2 y y 2 z z 2 U 2 2 2 2 2 U x y yx y z zy x z zx 2 2 2 2 Ma 1 2 Ma 1 2 Ma 1 2 U U U x x 2 x y 2 x z 2 2 2 2 Ma 1 2 2 Ma 1 2 2 Ma 1 2 2 x x x 2 2 2 2U 2U 2U x 2 y 2 z 2 2 2 2 Ma 1 2 Ma 1 2 Ma 1 2 2 2 2 2 2 2 2U 2U 2U y z y x y y 2 2 2 Ma 1 2 2 Ma 1 2 2 Ma 1 2 2 z z z 2 2 2 2U 2U 2U x 2 y 2 z 2 2 2 +Ma 2 x 2 2 2 2 2 2 2 Ma 2 2 Ma 2 2 Ma 2 U x x 2 U y xy U z xz 2 2 2 2 2 Ma 2 Ma 2 2 U x x 2 U 2 2 2 2Ma 2 2Ma 2 2Ma 2 2 2 2 U x y yx U y z zy U x z zx C. Wassgren Chapter 11: Gas Dynamics 2 2 Ma 2 2 2 U z z 2 y y 629 Last Updated: 16 Aug 2009 2 2 Ma 2 2 2 2 2 2 Ma 1 2 Ma 1 2 2 2 2 2 2 2 2 U x 2 2U z z x x x x y y 2 2 2 2 2 2 2 2 Ma 1 Ma 1 2 Ma 2 Ma U U U y xy U z xz x y 2 x z 2 2 2 Ma 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2U z y x z z x y x y 2 2 2 2 2Ma 2 2Ma 2 2Ma 2 2 2 2 U x y yx U y z z y U x z z x 2 2 1 Ma x y z 2 2 2 Ma 2 2 u' 1 x U ' 2 2 u 2 1 2 1 Ma x x U u' 2 1 Ma x U u 'y 2 2 2 2 2 2 Ma y U z ' 2 u 1 2 1 Ma x U u' 2 2Ma x U 2 2 2 2 u 'y x 2 U u 'y U 2 2 2 u 'y 2 2 y z U 2 u 'y 2 2Ma yx U 2 2 u' z y 2 U 2 u' 2 2 Ma z xy U 2 2 2 u' 2 z x 2 z U u' z U u' 2 2 2Ma x U zy 2 2 z 2 2 xz 2 2 2 2 2 x y ' uz U 2 zx 2 1 Ma x y z 2 2 ' ux U 2 2 u 'y 2 2 2 2 2 Ma 1 2 1 2 2 2 Ma U x z y 1 2 Ma ' u x U 2 u 'y x 2 U 1 2 1 2 Ma ' u x U + 1 2 u' 2 2Ma x U u 'y U C. Wassgren Chapter 11: Gas Dynamics 2 2 2 2 u' z y 2 U 2 2 u 'y 2 2 z U y 2 u 'y 2 2Ma yx U u' z U 630 2 2 u' 2 2 Ma z xy U 2 xz 2 2 2 z ' 2 2 uz 2 2 U z x u' 2 2 2Ma x zy U 2 2 2 2 2 y x ' uz U 2 z x Last Updated: 16 Aug 2009 19. Method of Characteristics The method of characteristics is a procedure by which non-linear, hyperbolic PDEs may be solved in an algorithmic fashion. Before beginning with the algorithm, we must first discuss a few details regarding hyperbolic PDEs. To begin, let’s first define a characteristic curve. A characteristic curve is a curve across which the value of some parameter is continuous, but the derivatives of that parameter are indeterminate. Consider for example the flow through a Mach wave. Recall from our previous analysis that velocity changes across a Mach wave are very small, i.e. the velocity across a Mach wave goes from V to V+dV. This infinitesimal velocity change (dV) occurs over zero distance since the Mach wave has no thickness. Thus, although the velocity changes continuously across the Mach wave, the velocity gradient is indeterminate (dV/0=?). Hence, a Mach wave is a characteristic curve according to our definition. Characteristic curves have additional useful properties but we will not discuss them at this time. Our first task will instead be to determine the slope of a characteristic curve in a general flow situation. Recall from our previous analysis that the equation of motion for a steady, irrotational, isentropic compressible flow with negligible body and viscous forces is given by: 1 2 2 c Expanding using Cartesian coordinates: 2 2 2 1 ˆ ˆ ˆ ex ey e z 2 2 2 2 y z x y z c x x y y z z x x 1 ˆ ˆ ˆ 2 ex ey ez y z x c x x 2 x 2 2 y 2 2 z 2 1 c2 1 c2 C. Wassgren Chapter 11: Gas Dynamics 2 2 ˆ ex y yx z zx x 2 2 2 ˆ e 2 y xy y y z zy 2 2 2 ˆ z e xz y yz z z 2 2 2 2 2 2 2 2 x y yx x z zx x x 22 2 2 2 y z zy y x xy y y 22 2 2 z x xz z y yz z z 2 22 22 22 2 2 2 x x z z y y 2 2 2 2 2 2 y z zy x z zx x y yx 631 (12.315) Last Updated: 16 Aug 2009 For 2D flow, the previous equation can be written as: 2 2 2 2 2 2 c c 2 2 2 2 0 x y xy x x y y This non-linear equation has the form: 2 2 2 A 2 2B C 2 0 xy x y (12.316) (12.317) where A, B, and C depend on /x and /y. For an isentropic flow, the velocities, i.e. /x and /y, are continuous. Since we’re interested in finding the velocity characteristic curves, we should consider differential changes in the velocities: 2 2 du x d 2 dx dy (12.318) xy x x 2 2 du y d dx 2 dy y y xy Now let’s rewrite equations (12.317) through (12.319) as a system of linear equations where the independent variables are the velocity gradients (2/x2, 2/xy, and 2/y2): 2 2 2 C 2 0 A 2 2 B xy x y dx 2 x 2 dy 2 xy 0 d x (12.319) (12.320) 2 2 dy 2 d xy y y Recall from linear algebra that Kramer’s rule states that a system of equations has a unique solution if and only if the determinant of the system is not equal to zero, i.e.: A 2B C (12.321) dx dy 0 Ady 2 2 Bdxdy Cdx 2 0 0 dx dy 0 dx Thus, if this determinant is equal to zero, then the velocity derivatives (or gradient) (2/x2, 2/y2, and 2/xy) will be indeterminate. The slope of such a curve is given by: Ady 2 2 Bdxdy Cdx 2 0 dy 2 B 4 B 2 4 AC B B 2 AC dx 2A A where 2 2 2 A c 2 c ux x B u x u y x y 2 2 2 C c 2 c uy y C. Wassgren Chapter 11: Gas Dynamics 632 Last Updated: 16 Aug 2009 Thus, 22 2 2 2 2 dy u x u y u x u y c u x c u y 2 dx c2 ux dy dx 2 2 2 u x u y u x u 2 c 4 c 2 u x c 2u 2 u x u 2 y y y c 2 2 ux 2 u x u y c 2 u x u 2 c 4 y c uxu y c 2 2 2 ux Ma 2 1 1 slope of characteristic curves 2 ux (12.322) c2 Recall from our previous discussion that a curve across which the velocity derivatives (or gradient) are indeterminate are referred to as characteristic curves. Notes: 1. For Ma<1 there are no real characteristic curves (elliptic PDE). For Ma=1 there is one characteristic curve (parabolic PDE), and for Ma>1 there are two characteristic curves (hyperbolic PDE). 2. Recall from our previous, linearized (small perturbation) analysis that the governing PDE was linear: 2 2 2 0 A 2 2B C x xy y 2 where 2 A 1 Ma B0 C 1 Thus, the slope of the characteristic curves are given by: dy B B 2 AC dx A 1 2 Ma 1 Mach lines! Thus, the characteristic curves for linearized flow are straight Mach lines. C. Wassgren Chapter 11: Gas Dynamics 633 Last Updated: 16 Aug 2009 3. It can be shown that the characteristic curves are inclined at the Mach angle to the local flow direction. To show this, first re-write the velocity components in terms of the velocity magnitude, V, and the local flow inclination, : y V u x V cos (12.323) u y V sin x Substituting into Eqn. (12.322) and noting that, sin= 1/Ma (where is the Mach angle), gives: dy Ma 2 sin cos Ma 2 1 dx 1 Ma 2 cos 2 1 Ma sin cos Ma 2 1 2 1 Ma 1 cos 2 Ma 2 (Ma2-1)1/2 sin 2 tan = (Ma2-1)-1/2 sin cos tan sin = 1/Ma 2 2 sin cos After considerable rearrangement and simplification using trig identities we find that: dy tan (12.324) dx The “-” corresponds to the right-running characteristic while the “+” corresponds to the left-running characteristic. Thus, the slope of the characteristic lines will be inclined at the Mach angle to the local flow direction. left-running characteristic curve left-running Mach line V y right-running characteristic curve right-running Mach line x Notes: a. The direction of a characteristic curve or Mach line is determined by facing downstream. The leftrunning waves are on the LHS while the right-running waves are on the RHS. b. The characteristic curves are the “Mach curves” for the flow. C. Wassgren Chapter 11: Gas Dynamics 634 Last Updated: 16 Aug 2009 c. Although we now have an expression for the slope of the characteristic curves (Eqn. (12.324)), we still need to have some relation for determining how the velocities change along a characteristic curve. Recall from the system of equations from Eqn. (12.320): 2 2 2 C 2 0 A 2 2 B xy x y dx 2 0 2 xy dx x 2 dy 2 xy d du x x 0 dy 2 y d du y y 2 where 2 2 A c 2 c2 ux x B u x u y x y 2 2 2 C c 2 c uy y The solution for the gradient of the x-velocity component in the x-direction (using Kramer’s rule) is: u x 2 2 x x 0 du x du y 2B C dy 0 dx dy A 2B C dx dy 0 0 dx dy Cdxdu x Cdydu y 2 Bdydu x Ady 2 2 Bdxdy Cdx 2 (Note: We could also consider the other velocity gradients such as uy/x and would get the same result given below.) Recall that along a characteristic curve the denominator in the previous expression will be zero. Thus, in order for ux/x to remain finite (we’re assuming that the velocities in the flow will be finite so that velocity gradients will also be finite), the numerator must also be zero: Cdxdu x Cdydu y 2 Bdydu x 0 dx 2 B du x dy C Substituting in for the characteristic curve slope (Eqn. (12.324)) and simplifying gives: uxu y Ma 2 1 du y c2 du x u2 y 1 2 c du y C. Wassgren Chapter 11: Gas Dynamics 635 (12.325) Last Updated: 16 Aug 2009 Re-write the velocity components in terms of the velocity magnitude, V, and direction, : u x V cos du x dV cos V sin d u y V sin du y dV sin V cos d Substituting into Eqn. (12.325) and simplifying: 1 dV 1 tan V d Ma 2 1 (12.326) We now need to integrate the previous expression as we move along a characteristic line. We’ve performed this same integration before when discussing Prandtl-Meyer expansion fans (refer to Eqn. (12.275) and (12.276)). Instead of re-deriving the result again, we’ll just copy the result here: b a b d Ma 2 1 a dV b a V b a b a (12.327) where a andb are the Prandtl-Meyer angles for the flow at locations a and b, both of which lie on the same characteristic curve. Again, a and b are the flow velocity directions at points a and b. If we let the point a be some reference point on the characteristic curve, then: constant along a right-running characteristic curve (12.328) constant along a left-running characteristic curve (12.329) Equations (12.328), (12.329), and (12.324) can be combined together in an algorithm for solving steady, supersonic, irrotational compressible flows. C. Wassgren Chapter 11: Gas Dynamics 636 Last Updated: 16 Aug 2009 Method of Characteristics Algorithm Suppose that the flow conditions are known along some curve CD: P C A B D Since the flow conditions are known on this curve, the flow angle, , and Prandtl-Meyer angle, , are known at the two points A and B. We now move along the right-running characteristic curve from point A until it intersects with the left-running characteristic curve from point B. Call this intersection point P. At point P we have: P P A A CR P P B B CL where CR and CL are constants. Re-arrange the previous expressions to get: P 1 2 CR CL P 1 2 CR C L Since we now know the Prandtl-Meyer angle and flow direction at point P, we can determine the remainder of the flow properties for this isentropic flow (e.g., Mach number, pressure, temperature, velocity components, etc.) However, we still don’t know the location of point P. Since the Mach number changes, in general, from point to point in the flow field, the characteristics curves will not necessarily be straight lines. However, if we choose the points A and B to be very close together, then we can approximate the position of point P by the intersection of the left- and right-running Mach lines passing through A and B (call this point P’). P’ Mach lines P C A B D Obviously the location of point P’ approaches P as the distance between A and B becomes smaller. We can determine the conditions at other points in the flow field by repeating this procedure for other points where the flow conditions are known. C. Wassgren Chapter 11: Gas Dynamics 637 Last Updated: 16 Aug 2009 Note that along a right-running characteristic, CR = constant so: 1 CR CL 1 CR CL 1 CL 2 2 2 0 1 CR CL 1 CR CL 1 CL 2 2 2 0 Similarly, along a left-running characteristic, CL = constant so: 1 CR CL 1 CR 2 2 0 1 1 CR CL 1 CR CL 2 CR 2 2 0 1 2 (12.330) CR CL C. Wassgren Chapter 11: Gas Dynamics 638 (12.331) Last Updated: 16 Aug 2009 Example: Uniform radial flow at Mach 2.0 enters a 2D diverging channel with straight walls. Compute the variation of Mach number in this radial flow field, assuming isentropic, steady flow. The walls are inclined at a total angle of 12. Compare your results at the centerline with those using 1D isentropic flow analysis. Ma=2.0 12 x C. Wassgren Chapter 11: Gas Dynamics 639 Last Updated: 16 Aug 2009 Method of Characteristics This section remains incomplete. Types of Points: Interior Points: CL and CR are known, determine and Solid Boundary Points: either CL or CR is known and is known Free Boundary Points: either CL or CR is known and is known (Since p is known can determine Ma can determine ) Interaction of Waves reflections off solid boundaries cancellation of waves reflections off free boundaries intersection of waves Region-to-Region Method Can cross characteristics too. Recall from our previous work that across a Mach wave (a characteristic curve), we have: where is the Prandtl-Meyer angle and is the flow orientation. These expressions can be written as: 2 2 1 1 across a right-running characteristic and 2 2 1 1 across a left-running characteristic (same as moving along the opposite characteristic) Angle of lines separating regions ij 1 i i j j (across R-running) 2 ik 1 i i k k (across L-running) 2 L k i u R j C. Wassgren Chapter 11: Gas Dynamics 640 Last Updated: 16 Aug 2009 Example: Flow at the exit of a Mach 1.5 supersonic nozzle is expanded from an exit-plane pressure of 200 kPa (abs) to a back pressure of 100 kPa (abs). Determine the flow just downstream of the nozzle exit in the regions indicated. pb=100 kPa (abs) Mae=1.50 pe=200 kPa (abs) 4 7 9 1 2 C. Wassgren Chapter 11: Gas Dynamics 3 641 5 6 10 8 Last Updated: 16 Aug 2009 Design of a Supersonic Wind Tunnel Nozzle One simple application of the Method of Characteristics is in the design of a supersonic wind tunnel nozzle. Recall that the function of a supersonic nozzle is to accelerate a flow from Ma=1 to some final supersonic Mach number. In order to simulate free flight, the flow through the test section of the wind tunnel should be parallel and uniform. Consider a symmetric nozzle so that only the upper half of the nozzle must be considered. For simplicity, assume that the incoming flow at the throat is uniform and at Mach 1.0 (of course in a real nozzle the incoming flow would not be uniform). The flow will first expand in the region from points a through b as shown in the figure below. Four of the expansion waves (right-running characteristics) are shown in the sketch. f e d c b a test section g i h j The expansion waves reflect off the centerline at points g through j as expansion waves which act to turn the flow back toward the horizontal (these are left-running characteristics). These waves impinge on the nozzle wall at points c through f. In order to avoid further reflections, the wall contour at points c through f should be sloped at the wave turning angle so that the waves are cancelled. Notes: 1. As more waves are included in the analysis, the nozzle contour will become smoother. 2. The initial expansion from a to b is arbitrary in the design. The critical points in the analysis are points c through f which must designed to provide wave cancellation. 3. To design a nozzle with the shortest length, the expansion from points a to b should take place as a centered Prandtl-Meyer expansion fan: d c e f a,b g 4. h i j The design of a real nozzle should also factor in boundary layer effects in order to give the correct flow area and avoid boundary layer separation. C. Wassgren Chapter 11: Gas Dynamics 642 Last Updated: 16 Aug 2009 Example: Design the diverging section of a supersonic nozzle to produce uniform Mach 1.8 flow. Assume that the length of the nozzle should be kept to a minimum. C. Wassgren Chapter 11: Gas Dynamics 643 Last Updated: 16 Aug 2009 20. Flow Past a Wavy Wall Using Small Perturbation Theory Recall that for a steady, irrotational, 2D flow with negligible body and viscous forces, the equation of motion assuming small velocity perturbations, is: 2 2 2 (12.332) 1 Ma x 2 y 2 0 The linearized boundary condition at the object surface is: ' dy dx surface uy x ,0 (12.333) U Far from the object (y), the velocity perturbations must remain finite. Recall that the pressure coefficient at the object surface is given by: ' 2u x Cp U (12.334) Now consider flow past a wavy wall. Let the profile of the wall be given by: 2 x ys A sin y Ma A x We’ll assume that the wall causes small perturbations in the flow, i.e.: A Thus, the boundary condition at the wall is: 2 AU dy 2 x ' cos U uy x ,0 dx surface C. Wassgren Chapter 11: Gas Dynamics (12.335) 644 Last Updated: 16 Aug 2009 Subsonic Flow First let’s examine the case when Ma2 < 1 so that (1-Ma2) > 0. Equation (12.332) will be an elliptic PDE for this case. One method for solving elliptic, linear PDEs is to use separation of variables where we assume that the solution can be written as some function of x multiplied by some function of y: x, y X x Y y Substituting into Eqn. (12.332) and simplifying gives: 1 Ma X Y XY 0 2 X X Y 1 1 Ma Y 2 k 2 Since the LHS of the previous equation is a function only of x, and the RHS is a function only of y, then in order for the two sides to be equal, they must equal a constant, which we’ll call -k2. Thus, we can write the following two equations: X k 2 X 0 2 Y k 2 1 Ma Y 0 The solution to the first differential equation involving X is: X c1 cos kx c2 sin kx and the solution to the differential equation involving Y is: Y c3 exp ky 1 Ma c4 exp ky 1 Ma 2 2 Note that the square root term results in a real quantity since the incoming flow is subsonic. Substituting these functions into the perturbation potential and determining the corresponding velocity perturbations gives: 2 2 c1 cos kx c2 sin kx c3 exp ky 1 Ma c4 exp ky 1 Ma ' ux ' uy k c sin kx c cos kx c exp ky 1 Ma c exp ky 1 Ma x k 1 Ma c cos kx c sin kx c exp ky 1 Ma c exp ky 1 Ma y 1 2 2 1 2 3 2 2 4 2 3 4 2 The constants c1, c2, c3, and c4 are determined from the boundary conditions at the wall and at y. In order for the perturbation velocities to remain finite as y we must have c3=0. At the wall (y = 0) we have (Eqn. (12.335)): 2 AU 2 x 2 u x ,0 cos c4 k 1 Ma c1 cos kx c2 sin kx y Thus we see that c2=0, k=(2/), and U AU 2 A c1c4 2 2 2 1 Ma 1 Ma C. Wassgren Chapter 11: Gas Dynamics 645 Last Updated: 16 Aug 2009 Thus, the perturbation potential and perturbation velocities become: AU 2 x 2 y 2 cos 1 Ma exp 2 1 Ma 2 AU ux ' 1 Ma uy ' 2 AU 2 2 x 2 y 2 1 Ma exp (12.336) sin 2 x 2 y 2 1 Ma exp cos The pressure coefficient at the wall (refer to Eqn. (12.334)) is: 4 A 2 x Cp sin 2 1 Ma (12.337) Notes: 1. The disturbance of the wall dies out as y. y streamlines x 2. The pressure peaks occur in the troughs of the wall and vice versa, i.e., the pressure is in-phase with the wall. As a result, there will be no drag force on the wall. y, Cp x pressure profile is symmetric net horizontal force is zero C. Wassgren Chapter 11: Gas Dynamics 646 Last Updated: 16 Aug 2009 3. We can also examine this compressible subsonic flow using the coordinate transformation discussed in a previous set of notes concerning the Prandtl-Glauert rule. Recall that if we examine flow in the (, ) plane where: x 2 y where 1-Ma (12.338) then the governing equation becomes Laplace’s equation, i.e.: 2 2 0 (This is the governing equation for an incompressible flow!) (12.339) 2 2 Furthermore, the shape of the boundary surface in the (, ) plane is the same as the boundary surface shape in the (x, y) plane. Solving (12.339) using separation of variables gives: (12.340) c1 cos k c2 sin k c3 exp k c4 exp k Since the flow must have finite velocities as → (this is true for both the compressible and incompressible flow cases), we can conclude that c3 must be zero. The boundary condition at the wavy wall surface: 2 AU 2 u ,0 cos (12.341) c4 k c1 cos k c2 sin k indicates that c2 must be zero, k must be (2/), and c1c4 must be -AU∞. Thus: 2 2 AU cos (12.342) exp Transforming back into the (x, y) plane using equation (12.338) gives: AU 1 2 x 2 y 2 (12.343) cos 1 Ma exp 2 1 Ma This is precisely the same relation derived in equation (12.336). The pressure coefficient corresponding to the incompressible flow (i.e. the (, ) plane) is: 2 4 A 2 Cp sin 0 U (12.344) u Using the Prandtl-Glauert rule, the pressure coefficient for the compressible flow (i.e. the (x, y) plane) is: Cp 4 A 2 x (12.345) Cp sin 2 2 1 Ma 1 Ma Equation (12.345) is the same as equation (12.337), as expected. C. Wassgren Chapter 11: Gas Dynamics 647 Last Updated: 16 Aug 2009 Supersonic Flow Now consider the case where Ma2 > 1 so that (1-Ma2) < 0 and Eqn. (12.332) becomes a hyperbolic PDE with a form identical to that of the wave equation. Thus, we’ll utilize d’Alembert’s solution to the wave equation which has the form: Recall that the wave equation has the form: f L x y Ma 1 f R x y Ma 1 or 2 2 f L f R where 2u 2 c2 2u t x 2 and has the (d’Alembert) solution: u x, t f x ct g x ct (12.346) where f and g are functions determined by the initial and boundary conditions. For our case, the governing equation is: x y Ma 1 2 2 2 x y Ma 1 y 2 2 Ma 1 2 2 2 x Note that the functions fL and fR are unknown at this point but will eventually be determined using the boundary conditions. To verify that Eqn. (12.346) is indeed a general solution, substitute it back into Eqn. (12.332) and simplify: 2 1 Ma x 2 2 2 y 2 0 where x 2 d f L 2 2 2 2 2 d f L d fL d fL 2 2 2 2 d x d x d d 2 2 2 2 d 2 fL d 2 fL d 2 f L d 2 f L 2 Ma 1 2 2 d 2 y d 2 y d 2 y d 2 Of particular interest for this solution are the curves corresponding to fL=constant and fR=constant (along these curves and, hence ux’ and uy’, will remain constant). Note that the form of these functions will be dictated by the boundary conditions. Thus, the value of the perturbation potential, , (as well as the perturbation velocities) will be propagated from the boundary conditions into the rest of the flow along the curves where fL and fR are constant. The shape of these curves can be determined from: dy 1 2 x y Ma 1 constant fL=constant 2 dx f L constant Ma 1 fR=constant 2 x y Ma 1 constant dy dx f R constant 1 2 Ma 1 Thus we see that the two curves are, in fact, lines with opposite slopes. Moreover, the slope of these lines is equal to the slope of a Mach line: Ma 1 1 sin tan 1 2 Ma Ma 1 (Ma-1)1/2 It is worthwhile to re-iterate these last two important points: 1. Information propagates along the curves where fL and fR are constant. 2. Curves along which fL and fR are constant correspond to Mach lines of opposite slope. C. Wassgren Chapter 11: Gas Dynamics 648 Last Updated: 16 Aug 2009 For the problem at hand (supersonic flow over a wavy wall), the perturbation potential should only include the function fL since if we included fR in the solution, then information could propagate upstream along the Mach line with negative slope – an impossibility in supersonic flows. fR=constant fL=constant Ma x Thus, the perturbation potential is: f L 2 x y Ma 1 where (12.347) The form of fL will be dependent on the boundary conditions. Recall that the boundary condition at the wall is given by Eqn. (12.335): 2 AU dy 2 x ' U uy cos x ,0 dx surface Substituting Eqn. (12.347) into the boundary condition gives: 2 x y Ma 1 df L df L 2 AU 2 x ' 2 cos Ma 1 uy 2 y 0 Ma 1 y y 0 d y y 0 dx y df L 2 AU 2 x cos 2 dx Ma 1 Note that in the previous equations the fact that at y=0, dfL/d = dfL/dx, has been used. Integrating gives: AU 2 x sin fL x constant 2 Ma 1 Thus, AU f L Ma 1 2 2 constant sin The resulting perturbation potential and corresponding perturbation velocities are: AU 2 2 sin x y Ma 1 constant 2 Ma 1 ' ux ' uy 2 AU Ma 1 2 2 AU 2 2 x y Ma 1 cos (12.348) 2 2 x y Ma 1 cos C. Wassgren Chapter 11: Gas Dynamics 649 Last Updated: 16 Aug 2009 The pressure coefficient at the wall (Eqn. (12.334)) for this flow is: Cp ' 2u x U 4 A Ma 1 2 2 x (12.349) cos Notes: 1. Along lines of x-y(Ma2-1)1/2=constant (recall that these are the Mach lines), the perturbation potential and velocities are constant. Thus, the disturbances produced by the wall are felt equally along the same Mach line. Recall that in the subsonic flow case, the disturbances die out as y. Mach lines y streamlines x 2. In the subsonic case the drag on the wall is zero. For supersonic flow, however, the drag is not zero. This is because the pressure coefficient is out of phase with the wall shape (shown below). This type of drag is commonly referred to as supersonic wave drag. pressure profile is asymmetric net horizontal force is not zero y, C p x 3. In an actual flow, the drag coefficient on the wall as a function of Mach number look like: CD perturbation model experimental data Ma 1 The drag occurring in real subsonic flows is due solely to viscous effects. For real supersonic flows around slender bodies, the wave drag is typically much larger than the viscous drag. Note that the assumptions for perturbation analysis do not allow for the investigation near Ma1. C. Wassgren Chapter 11: Gas Dynamics 650 Last Updated: 16 Aug 2009 21. Thin Airfoils in Supersonic Flow Recall that for steady, irrotational, supersonic, compressible flow with negligible body and surface forces and small perturbations, the perturbation potential is given by: 2 2 f R x y Ma 1 f L x y Ma 1 where fL and fR are arbitrary functions that are dependent on the boundary conditions. y fL=constant Ma x fR=constant Since information cannot be propagated upstream in a supersonic flow, the perturbation potential for flow over the top of the airfoil (y>0) should only include left-running Mach waves: 2 f L x y Ma 1 while under the bottom of the airfoil (y<0) the solution should contain only right-running Mach waves: 2 f R x y Ma 1 The boundary condition on the upper surface of the airfoil is: dy dx u 'y upper U upper 1 U y upper so dy dx upper ' fL 1 U y U 2 Ma upper dy 1 dx Ma 2 1 ' fL U (Note: df L .) y d y (12.350) upper The pressure coefficient on the upper surface is given by: ' 2 u x 2 2 ' df L (Note: .) fL x d x U U x y 0 U Substituting equation (12.350) into equation (12.351) gives: 2 U dy C p ,upper 2 U Ma 1 dx upper C p ,upper C p ,upper y 0 2 2 Ma dy 1 dx C. Wassgren Chapter 11: Gas Dynamics (12.351) (12.352) upper 651 Last Updated: 16 Aug 2009 A similar approach can be taken to determine that: 2 dy C p ,lower Ma 2 1 dx lower (12.353) Notes: 1. The slope of the upper and lower surfaces relative to the incoming flow depends not only on the airfoil shape, but also on the airfoil’s angle of attack, . slope at point on surface relative to flow is different y, y’ Ma Ma y’ y x x, x’ c c (a) x’ (b) For example, the local slope of the surface relative to the incoming flow, (dy/dx), for case (a) is different than the slope of the surface at the same point for case (b). In order to isolate the effects of the airfoil shape and angle of attack, let’s define the quantity, , as: dy (12.354) dx so that represents the airfoil’s surface slope relative to the airfoil’s chord line, i.e. with respect to the (x’, y’) axes in the figures above. The slope of the surface relative to the incoming flow, , will be: airfoil surface tan Note that and are slopes, not angles. dx dx’ dy’ dy dy dx dy dx (12.355) Note that tan since we’re concerned only with small perturbations to the flow. The pressure coefficients (equations (12.352) and (12.353)) may be written as: 2 U C p ,U (12.356) 2 Ma 1 C p, L 2 L (12.357) 2 Ma 1 where the subscripts “U” and “L” refer to the upper and lower airfoil surfaces, respectively. C. Wassgren Chapter 11: Gas Dynamics 652 Last Updated: 16 Aug 2009 2. The airfoil’s lift and drag can be determined by integrating the net pressure force acting over the entire airfoil surface. The pressures acting on the upper and lower airfoil surfaces are: pU p C p ,U 2 U C p ,U 1 2 2 Ma p 2 Ma 1 1 2 2 p Ma (12.358) U 2 Ma pL p p 2 Ma 1 L (12.359) Resolving the pressure force acting on the upper surface on a small area element, ds (Note that the airfoil distance into the page, also known as the span, is assumed to have unit depth.), into lift (L) and drag (D) components gives: dLU pU ds cos U pU dx pU dx (since the slopes are small, dx’ dx) dx dy dDU pU ds sin U pU dx pU U dx dx dy pds ds Similarly, for the lower surface: dLL pL ds cos L pL dx pL dx dy dx dy dDL pL ds sin L pL dx pL L dx dx dx’ dx dy’ dy Note that the small angle approximation has been used in the expressions above. The net lift and drag are determined by integrating over the entire airfoil surface. x c L x 0 c dLU dLL pU pL dx 0 p L p x c D x 0 c 2 Ma 2 Ma 1 0 c 2 Ma 2 Ma 1 0 L U L dx U 2 dx (12.360) c dDU dDL pU U pL L dx p D p 0 2 Ma 2 Ma 2 Ma 2 Ma c 1 U 0 c 1 C. Wassgren Chapter 11: Gas Dynamics 2 U 2 2 L dx 2 L 2 U L 2 2 dx (12.361) 0 653 Last Updated: 16 Aug 2009 y’ Note, however, that: c c dx 0 0 c dy dx dy 0 dx x’ c 0 so that Eqns. (12.360) and (12.361) become: 2 Ma L p 2 c 2 Ma 1 Recall that this is the chord length so that: y x 0 y x c 0 c 2 2 U L dx 2 2 c 2 Ma 1 0 Written in terms of the lift and drag coefficients (based on the chord length): L 4 CL 1 p Ma 2 c 2 Ma 1 2 2 Ma D p CD 3. D 1 2 2 p Ma c c 2 2 c Ma 1 0 2 2 U L dx 4 2 2 Ma 1 (12.362) (12.363) (12.364) (12.365) The lift coefficient is directly proportional to the angle of attack for thin supersonic airfoils at small angles of attack. Note that there is also a component to the drag coefficient that depends only on the angle of attack. This component is often referred to as the wave drag due to lift: 4 2 CD ,wave drag (12.366) 2 due to lift Ma 1 The other part of the drag coefficient depends only the shape of the airfoil and is known as the wave drag due to thickness since the thicker the airfoil, the larger the integral term (the slopes will be larger): CD , wave drag due to thickness 2 c 2 Ma c 1 2 U 2 L dx (12.367) 0 These previous results suggest that in order to minimize the drag acting on a supersonic airfoil, the airfoil should be as thin as possible. Furthermore, the lift acting on the airfoil will be unaffected by the shape of the airfoil. C. Wassgren Chapter 11: Gas Dynamics 654 Last Updated: 16 Aug 2009 4. Let’s now consider subsonic flow around the airfoil shown below. Ma∞ < 1 For sufficiently small free stream Mach numbers, Ma∞, the flow over the entire airfoil surface will remain subsonic and the corresponding drag coefficient for the airfoil will remain at a relatively small value (Point A on the plot shown below. Note that the total drag will be due to skin friction and form drag.). CD C A B Macr Madd 1 Ma∞ As the free stream Mach number is increased, a critical free stream Mach number, Macr, will be reached where a sonic Mach number will occur at the minimum pressure point on the airfoil surface. Ma∞ = Macr < 1 Ma = 1 At free stream Mach numbers slightly greater than Macr, a small region of supersonic flow will occur near the minimum pressure point. The drag coefficient for these flow conditions (point B in the diagram above) remains close to, but only slightly greater than, the drag coefficient for purely subsonic flow. (Ma∞ > Macr) < 1 Ma < 1 Ma > 1 At another critical Mach number, known as the drag divergence Mach number, Madd (Macr < Madd < 1), the drag on the airfoil increases suddenly (point C in the CD vs. Ma∞ plot shown above) due to the formation of a terminating shock wave. The shock wave on the airfoil surface causes the boundary layer to separate (due to the large adverse pressure gradient across the shock wave) resulting in a significant increase in the form drag. Ma < 1 (Ma∞ > Madd) < 1 separated flow Ma > 1 shock wave This sudden increase in the drag at Madd is the origin of the concept of the sound barrier. C. Wassgren Chapter 11: Gas Dynamics 655 Last Updated: 16 Aug 2009 In order to minimize the drag acting on the airfoil when operating near sonic conditions, the drag divergence Mach number should be pushed as close to sonic conditions as much as possible. a. One approach to increasing Madd is to decrease the thickness of the airfoil. Recall that the largest Mach numbers will occur in the vicinity of the minimum pressure region. If the minimum pressure can be brought closer to the free stream pressure, then the corresponding local Mach number will deviate less from the free stream Mach number (which is subsonic). Making the airfoil thinner will result in less expansion of the flow and hence the Mach numbers over the airfoil will be smaller. As a result, the drag divergence Mach number for the thin airfoil will occur at a higher free stream Mach number than a thicker airfoil. Thinner airfoils also produce less drag for supersonic conditions as discussed previously in Note 3. b. Another approach to reducing the drag is to use a supercritical airfoil (shown below) which is designed to give Madd 1. The airfoil shape is designed to give mostly supersonic flow and discourage the formation of shock waves. c. A third approach to delaying Madd is to use a swept-wing design. Consider flow over straight and swept wings (inclined by an angle from the straight wing) that have identical airfoil crosssections. Note that only the component of the flow normal to the airfoil will be important in determining the drag divergence Mach number since the flow tangential to the surface does not “see” variations in the airfoil geometry. Hence, the effective free stream Mach number for the swept-back wing is smaller than that for the straight wing (Ma∞eff = Ma∞cos). As a result, the negative effects associated with shock formation on the airfoil can be delayed until a higher free stream Mach number is reached. The downsides of swept wings are that the wing area must be increased to generate the same lift as a straight wing (since the lift decreases as a result of the lower effective free stream Mach number), and the structural design of the wing is more complex. Ma∞ Ma∞cos Ma∞ Swept-wings are also advantageous in supersonic flight since the wing may be subject to a subsonic effective Mach number and, as a result, the penalty of supersonic wave drag can be avoided. C. Wassgren Chapter 11: Gas Dynamics 656 Last Updated: 16 Aug 2009 Example: A symmetric strut of chord length, L, is placed at a small angle of attack, , in a supersonic flow of Mach number, Ma. L x’ Ma t(x’) The geometry of the strut is such that the centerline is straight and the foil thickness, t(x’), is given by: t x x x 4 1 tM L L where tM is the maximum thickness of the foil which occurs at x’/L=1/2. Assuming that the foil is slender and that the angle of attack is small, find expressions for the lift and drag coefficients for this strut as functions of Ma, , and tM/L. C. Wassgren Chapter 11: Gas Dynamics 657 Last Updated: 16 Aug 2009 22. Unsteady, 1D Compressible Flow Applications that might be approximated as being, 1D and unsteady: a. accelerating piston in a cylinder b. projectile moving through a cylinder c. shock tube d. start-up and shut-down transients in a wind tunnel Governing equations for 1D, unsteady flow (ignoring viscous forces, body forces, and area changes): u u 0 (12.368) u 0 continuity: t x x t x u u 1 p u t x x momentum: (12.369) Assume that the flow is isentropic: p dp p p c2 x x d x (12.370) c2 Note that in the previous expression, the sound speed, c, has been used because the flow is isentropic. To make our analysis here look similar in form to the analysis we used while investigating steady, 2D flows, let’s re-write the velocity in the following manner: u x x where the subscript “x” signifies a partial differentiation with respect to x. Using this notation and substituting Eqn. (12.370) into Eqn. (12.369) and simplifying, Eqns. (12.368) and (12.369) become: (12.371) x xx 0 t x c 2 (12.372) xt xxx 0 x Multiply Eqn. (12.372) with dx and integrate with respect to x (note that Eqn. (12.372) is a function of both x and t): 2 c t 1 2 x2 d f t (12.373) 0 where f(t) is an unknown function of time. Taking the partial derivative of Eqn. (12.373) with respect to t gives: c 2 f t (12.374) tt xxt t We can substitute Eqns. (12.372) and (12.374) into Eqn. (12.371) to give an expression that does not include the density: f t tt xxt xxt x2xx c 2xx 0 c 2 x2 xx 2xxt tt f t C. Wassgren Chapter 11: Gas Dynamics (12.375) 658 Last Updated: 16 Aug 2009 The unknown function of time may be eliminated by defining: f t dt x x u t t f t (12.376) xx xx tt tt f t xt xt so that Eqn. (12.375) becomes: c 2 u 2 xx 2uxt tt 0 (12.377) Note also that Eqns. (12.373) and (12.374) become: 2 c t 1 2 x2 d 0 (12.378) 0 tt xxt c 2 0 t (12.379) Now let’s examine the nature of Eqn. (12.377) more closely. This 2nd order PDE in two-independent variables (x and t) will always be hyperbolic since: B 2 4 AC 4u 2 4 c 2 u 2 1 4c 2 0 where A c 2 u 2 ; B 2u; C 1 . The slope of the characteristic curves for Eqn. (12.377) can be found using an approach similar to that used when investigating 2D, steady supersonic flows. We can write the following system of equations: c 2 u 2 2u 1 xx 0 dx du dt 0 xt dx where (12.380) dx dt d f d 0 dx dt tt t Across a characteristic curve, the derivatives of x and t are indeterminate. In order for this to occur, the determinant of the matrix on the LHS of the previous equation must be zero: c 2 u 2 2u 1 dx dt 0 c 2 u 2 dt 2 dx 2 2udxdt 0 dx dt 0 2 dx dx 2 2 2u c u 0 dt dt dx dt 1 u c and (12.381) dt dx u c Slope of the characteristic curves in the x-t plane. The “+” sign represent a right-running characteristic and the “-” represents a left-running characteristic. Since disturbances propagate along characteristic curves, we see that disturbances will propagate at the speed of sound relative to the local flow velocity. C. Wassgren Chapter 11: Gas Dynamics 659 Last Updated: 16 Aug 2009 In order to ensure that the velocity of the flow will always remain finite, we must also have finite velocity gradients. The velocity gradient can be found from the system of equations given in Eqn. (12.380) using Kramer’s rule: 0 2u 1 0 dx dt dt dx dt d dx dt dt 2udx dt u xx (12.382) 2 2 x2 2 2 2 x c u 2u 1 c u dt dx 2udxdt 0 dx dt 0 dx dt Since along a characteristic curve the denominator of the previous expression is zero, the only way for the velocity gradient to remain finite is for the numerator to also be zero. Hence, we have the following relation: dx dx dt dt 2udx dt 0 dx du 2u dt dt The RHS of the previous equation is found with the aid of Eqn. (12.378): d d udu c 2 dt x dx c 2 (12.383) Substituting back into Eqn. (12.383) and utilizing Eqn. (12.381): d du u c 2u udu c 2 du d 1 dp d 1 dp p (Note: since c 2 .) c c2 s c2 (12.384) Conditions along a characteristic curve. The “-” corresponds to a right-running characteristic while the “+” corresponds to a left-running characteristic. Notes: 1. We could have also solved Eqn. (12.380) for finite xt or tt to arrive at Eqn. (12.384). 2. The numerical algorithms given during our previous notes on the method of characteristics for 2D, steady flow may also be applied here. 3. For an ideal gas, Eqn. (12.384) can be written as: du d 1 dT 2 dc c 1 T 1 c 2 dc (12.385) 1 Integrating along the characteristic curve gives: 2 along a R-running characteristic 1 c 2 u2 u1 (12.386) c2 c1 or u 1 2 c along a L-running characteristic 1 Conditions along a characteristic curve in a perfect gas. The “+” corresponds to a rightrunning characteristic while the “-” corresponds to a left-running characteristic. du C. Wassgren Chapter 11: Gas Dynamics 660 Last Updated: 16 Aug 2009 4. We can also determine how properties change across a characteristic curve by considering that by moving along a right-running characteristic we cross left-running characteristics and vice-versa. du d 1 dp (Note: compression dp > 0, expansion dp < 0.) (12.387) c c2 du 2 dc 1 u2 u1 (12.388) 2 (12.389) c2 c1 1 Conditions across a characteristic curve in a perfect gas. The “+” corresponds to a rightrunning characteristic while the “-” corresponds to a left-running characteristic. 5. A sketch of characteristic curves on the t-x plane looks as follows: t left-running characteristics uninfluenced by initial data right-running characteristics influenced by initial data influenced by initial data uninfluenced by initial data 6. R-running uninfluenced by initial data initial data line Note: The orientation of left and right running characteristics are relative to the pathline direction of fluid particles. L-running pathline x For an isentropic flow of a perfect gas we have: 1 1 p c2 T2 2 c1 T1 p1 so that Eqn. (12.389) can be written as: u u 2 c2 2 1 1 c1 c1 1 c1 2 2 (12.390) 2 u u 2 p2 2 1 1 (12.391) c1 c1 1 p1 Conditions across a characteristic curve in a perfect gas. The “+” corresponds to a rightrunning characteristic while the “-” corresponds to a left-running characteristic. 1 7. The previous relations do not apply to a shock wave. Shock waves are non-isentropic processes and, hence, the isentropic assumption used in deriving the equations is not valid. In addition, shock waves travel at speeds greater than the sonic speed. C. Wassgren Chapter 11: Gas Dynamics 661 Last Updated: 16 Aug 2009 Simple Waves Flows involving simple waves contain either left or right-traveling waves but not both. A simple wave flow can be produced using an accelerating piston as shown in the figure below: waves undisturbed gas Let’s consider two cases of simple waves in a perfect gas using this piston geometry. First we’ll consider a piston moving to the left and then we’ll consider a piston moving to the right. Piston Moving to the Left The movement of the piston can be shown on the t-x diagram. Note that we’ll assume that the piston velocity increases with time. t right-running characteristics particle path piston path 1 u1 c1 region uninfluenced by the piston (dead zone) 1 u0 c0 0 x At t=0, a sound wave leaves the piston and travels into the undisturbed fluid. This wave will travel at the sound speed in the undisturbed fluid, c0, and thus is a straight line on the t-x plot. Across this right-running characteristic, we have from Eqn. (12.388): 2 2 (12.392) du dc (or u c along a left running characteristic) 1 1 Since du < 0 (the piston moves to the left so the fluid velocity should also move to the left), then dc < 0. Thus, the next pressure pulse will travel at a slightly slower speed which corresponds to a larger slope on the t-x plot. Each characteristic curve will, in fact, be a straight line since u and c are constant in the regions between the pressure waves (refer to Eqn. (12.381)). Thus, the characteristic lines diverge and the influence of the piston motion is “stretched” out as the waves propagate downstream. Notes: 1. The motion of an individual fluid particle, known as a pathline, can be determined using: dx u particle dt particle C. Wassgren Chapter 11: Gas Dynamics 662 (12.393) Last Updated: 16 Aug 2009 2. Now let’s consider what happens if we accelerate the piston to the left from rest (u0 = 0) to a very large speed. From Eqn. (12.389) (we’re crossing right-running characteristics) we have: umax 2 2 2 c 1 u u u0 1 c c0 1 2 c0 1 1 c0 c0 0 Since the piston is moving to the left, u < 0. Thus, the largest speed we can have that results in a nonnegative speed of sound (c 0) is: umax 2 (12.394) c0 1 This is known as the escape speed. If the piston continues to accelerate, a vacuum (called the cavitation zone) will form on the face of the piston as shown in the diagram below. t 1 1 umax cmin 2c0 0 vacuum (cavitation zone) 1/c0 piston path x 4. An impulsive withdrawal to the left results in an expansion fan as shown below: t 1 u P cP 1 1 1 1 up uP c0 u p c0 2 2 particle path expansion fan uniform flow region piston path 1/c0 1/uP C. Wassgren Chapter 11: Gas Dynamics 663 undisturbed region x Crossing R-running waves: uP – u0 = 2/(– 1) (cp – c0) where u0 = 0 so that cP = c0 + ½(– 1) up Last Updated: 16 Aug 2009 Piston Moving to the Right Now let’s consider the case where the piston moves to the right to produce compression waves (refer to the figure shown below). t piston path shock wave 1/c0 x Across each compression wave (right-running characteristics as shown in the diagram above) we have from Eqn. (12.388): 2 du dc 1 where du > 0. Hence, dc > 0 and the characteristic curves become steeper (refer to Eqn. (12.381)). Eventually these compression waves will intersect and the isentropic assumption breaks down since the velocity gradients across the wave are no longer infinitesimal and viscous (irreversible) losses become significant. The point of the first intersection is defined as the start of a shock wave (refer to the previous figure). Notes: 1. Consider the diagram shown below where the piston is accelerated in very small, discrete increments of du each at every time step dt. shock wave t piston path complex system of sliplines and reflected waves 1/c0 dt x This section remains incomplete. 2. An impulsive acceleration to the right will immediately form a shock wave as shown in the following figure. t piston path shock wave particle path x The strength of the shock wave will be such that the fluid velocity behind the shock will equal the piston velocity. C. Wassgren Chapter 11: Gas Dynamics 664 Last Updated: 16 Aug 2009 Interactions with Boundaries Stationary Boundaries Near a stationary wall the fluid velocity must equal zero. This implies that a wave must reflect in a similar sense as shown in the diagram below. For example, consider a right-running compression wave (dp > 0 and, according to Eqn. (12.387), du > 0) impinging against a stationary wall. In order for the velocity to remain zero at the wall we must have for the reflected wave (a left-running characteristic), du < 0. Hence, according to Eqn. (12.387) we see that dp > 0 and thus we have another compression wave. Note that a fluid particle follows a compression wave and moves away from an expansion wave. t particle path for expansion wave reflected wave particle path for compression wave stationary u=0 boundary Across: R-running: du/c = 1/c2 (dp/) L-running: -du/c = 1/c2 (dp/) incident wave u=0 x Free Surface Boundaries This section remains incomplete. Reflections from an open end are more complicated since we must consider four different cases: the flow may be either inflow or outflow and it may be either subsonic or supersonic. Subsonic Outflow/Inflow For low speed subsonic outflow or inflow, it is reasonable to assume that the pressure at the end of the duct is equal to the ambient pressure. As a result, waves will reflect in an unlike sense. For example, consider a right-running compression wave (dp > 0, du > 0) impinging on an open end. In order for the pressure to remain constant at the open end (and equal to the ambient pressure), we must have dp < 0 which implies that the reflected wave is an expansion wave. Across the left-running reflected wave we observe from Eqn. (12.387) that du > 0. A similar approach may be taken to determine the conditions for an incident expansion wave. t t u>0 reflected wave (exp.) free surface boundary (p = constant) reflected wave (comp.) incident wave (comp.) u<0 free surface boundary (p = constant) incident wave (exp.) u=0 particle path C. Wassgren Chapter 11: Gas Dynamics u=0 x particle path 665 x Last Updated: 16 Aug 2009 For high speed subsonic outflow or inflow, the unsteady effects at the open end must be included making the analysis much more difficult. Supersonic Outflow/Inflow Since for supersonic flow the flow velocity is larger than the propagation velocity of the reflected waves, the reflected waves are unable to propagate from the open end of the duct. Hence there are no reflected waves. C. Wassgren Chapter 11: Gas Dynamics 666 Last Updated: 16 Aug 2009 Example: A diaphragm at the end of a 4 m long pipe containing air at a pressure of 200 kPa and a temperature of 30 C suddenly ruptures causing an expansion wave to propagate down the pipe. Find the velocity at which the air is discharged from the pipe if the ambient air pressure is 103 kPa. Also find the velocity of the front and the back of the wave and hence find the time taken for the front of the wave to reach the end of the pipe. C. Wassgren Chapter 11: Gas Dynamics 667 Last Updated: 16 Aug 2009 23. Description of a Shock Tube A shock tube is a device consisting, in its simplest form, of a long tube of constant area divided into two sections by a diaphragm. One of the sections contains a high pressure gas (aka driver gas) while the other contains a low pressure gas (aka driven gas). A sketch of the device is shown below. high pressure low pressure diaphragm When the diaphragm between the two sections is broken, either by mechanical means or by increasing the pressure on the high pressure side and using a “scored” diaphragm designed to burst at a specified presssure, a shock wave propagates into the low pressure section and an expansion wave propagates into the high pressure section. Between the shock wave and the expansion wave is a region of uniform velocity. expansion wave shock wave Shock tubes are used: as an inexpensive, but short duration (usually on the order of milliseconds), wind tunnel, to study of transient aerodynamic effects, to study dynamic and thermal response, to study relaxation effects and reaction rates, and to generate high enthalpies for studying dissociation and ionization. C. Wassgren Chapter 11: Gas Dynamics 668 Last Updated: 16 Aug 2009 Analysis of the Flow in a Shock Tube The velocity and pressure behind the shock wave must be equal to the velocity and pressure behind the expansion wave as shown in the figure below. The temperatures (and densities and entropies) are not necessarily equal however. If the temperature in the tube is initially uniform, then behind the expansion fan the temperature will be lower than the initial temperature while behind the shock wave the temperature will be higher than the initial temperature. The interface separating the two regions is called the contact surface. The contact surface is the interface dividing the gases that were originally separated by the diaphragm. Over time, diffusion will cause this interface to spread out. contact shock surface wave expansion wave 1 2 3 4 diaphragm location velocity pressure temperature position time current time for drawing shown above position C. Wassgren Chapter 11: Gas Dynamics 669 Last Updated: 16 Aug 2009 We know the following about the flow: 1. The gas properties are assumed identical on either side of the diaphragm, i.e. 1 = 2 = 3 = 4 and R1 = R2 = R3 = R4. 2. The velocities in regions 1 and 4 are zero, i.e. u1 = u4 = 0. 3. The pressures and temperatures in regions 1 and 4 are known, i.e. p1, p4, T1, and T4 are known. 4. The pressure and velocity across the contact surface are equal, i.e. p2 = p3 and u2 = u3. The remainder of the flow field is most easily analyzed using an iterative procedure as described below. 1. Assume a value for p2 = p3. 2. The value of u2 can be determined from the pressure ratio, p2/p1, and a relationship developed previously in our notes concerning 1D, unsteady, and isentropic compressible flow: 1 2 u2 2 p2 1 (crossing L-running waves) (12.395) c1 1 p1 where c1 = (RT1)1/2. 3. Use the normal shock relations to determine MaS and u3: p3 2 1 Ma 2 (determine MaS) S p4 1 1 u3 – uS uS 1 Ma 2 u S S 2 u3 uS 1 Ma S 2 velocities are given w/r/t the ground where uS = MaS(RT4)1/2. 4. 5. (12.396) (12.397) Is u3 = u2? If not, then go back to step 1. If so, then we’ve correctly determine the pressure and velocity in regions 2 and 3. Now that the pressure and velocity in regions 1-4 are known, we can also determine the temperature in region 2 since we know the speed of sound there: u 2 c2 2 1 c1 1 c1 T2 2 c2 R or 1 T2 p2 T1 p1 We can also determine the temperature in region 3 using the normal shock relations: 2 Ma 2 1 T3 S 2 1 Ma 2 S T4 12 Ma 2 S 6. The speed of the contact surface can be determined by noting that it has the same velocity as the gas in regions 2 and 3, i.e. uCS = u2 = u3. C. Wassgren Chapter 11: Gas Dynamics 670 Last Updated: 16 Aug 2009 Example: A shock tube containing air has a high pressure section at 300 kPa and a low pressure section at 30 kPa. The temperature of the air is uniform at 15 C. The diaphragm separating the two sections is suddenly ruptured. Find: a. the velocity of the air between the shock wave and the expansion wave relative to the ground, b. the speed of the shock wave relative to the ground, and c. the speed of the front and back of the expansion fan relative to the ground. d. Sketch the process on a t-x diagram. C. Wassgren Chapter 11: Gas Dynamics 671 Last Updated: 16 Aug 2009 ...
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This note was uploaded on 10/01/2011 for the course ME 509 taught by Professor Wereley during the Spring '11 term at Purdue University-West Lafayette.

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