Chapt05

Chapt05 - Wind tunnel test of an F-18 fighter plane model....

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Unformatted text preview: Wind tunnel test of an F-18 fighter plane model. Testing of models is imperative in the design of complex, expensive fluids-engineering devices. Such tests use the principles of dimensional analysis and modeling from this chapter. (Courtesy of Mark E. Gibson/Visuals Unlimited) | v v 276 | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 5 Dimensional Analysis and Similarity Motivation. In this chapter we discuss the planning, presentation, and interpretation of experimental data. We shall try to convince you that such data are best presented in dimensionless form. Experiments which might result in tables of output, or even multiple volumes of tables, might be reduced to a single set of curves—or even a single curve—when suitably nondimensionalized. The technique for doing this is dimensional analysis. Chapter 3 presented gross control-volume balances of mass, momentum, and energy which led to estimates of global parameters: mass flow, force, torque, total heat transfer. Chapter 4 presented infinitesimal balances which led to the basic partial differential equations of fluid flow and some particular solutions. These two chapters covered analytical techniques, which are limited to fairly simple geometries and welldefined boundary conditions. Probably one-third of fluid-flow problems can be attacked in this analytical or theoretical manner. The other two-thirds of all fluid problems are too complex, both geometrically and physically, to be solved analytically. They must be tested by experiment. Their behavior is reported as experimental data. Such data are much more useful if they are expressed in compact, economic form. Graphs are especially useful, since tabulated data cannot be absorbed, nor can the trends and rates of change be observed, by most engineering eyes. These are the motivations for dimensional analysis. The technique is traditional in fluid mechanics and is useful in all engineering and physical sciences, with notable uses also seen in the biological and social sciences. Dimensional analysis can also be useful in theories, as a compact way to present an analytical solution or output from a computer model. Here we concentrate on the presentation of experimental fluid-mechanics data. 5.1 Introduction Basically, dimensional analysis is a method for reducing the number and complexity of experimental variables which affect a given physical phenomenon, by using a sort of compacting technique. If a phenomenon depends upon n dimensional variables, dimensional analysis will reduce the problem to only k dimensionless variables, where the reduction n k 1, 2, 3, or 4, depending upon the problem complexity. Generally n k equals the number of different dimensions (sometimes called basic or pri- | v v 277 | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 5 Dimensional Analysis and Similarity mary or fundamental dimensions) which govern the problem. In fluid mechanics, the four basic dimensions are usually taken to be mass M, length L, time T, and temperature , or an MLT system for short. Sometimes one uses an FLT system, with force F replacing mass. Although its purpose is to reduce variables and group them in dimensionless form, dimensional analysis has several side benefits. The first is enormous savings in time and money. Suppose one knew that the force F on a particular body immersed in a stream of fluid depended only on the body length L, stream velocity V, fluid density , and fluid viscosity , that is, F f(L, V, , ) (5.1) Suppose further that the geometry and flow conditions are so complicated that our integral theories (Chap. 3) and differential equations (Chap. 4) fail to yield the solution for the force. Then we must find the function f(L, V, , ) experimentally. Generally speaking, it takes about 10 experimental points to define a curve. To find the effect of body length in Eq. (5.1), we have to run the experiment for 10 lengths L. For each L we need 10 values of V, 10 values of , and 10 values of , making a grand total of 104, or 10,000, experiments. At $50 per experiment—well, you see what we are getting into. However, with dimensional analysis, we can immediately reduce Eq. (5.1) to the equivalent form F V2L2 or CF g VL (5.2) g(Re) i.e., the dimensionless force coefficient F/( V2L2) is a function only of the dimensionless Reynolds number VL / . We shall learn exactly how to make this reduction in Secs. 5.2 and 5.3. The function g is different mathematically from the original function f, but it contains all the same information. Nothing is lost in a dimensional analysis. And think of the savings: We can establish g by running the experiment for only 10 values of the single variable called the Reynolds number. We do not have to vary L, V, , or separately but only the grouping VL / . This we do merely by varying velocity V in, say, a wind tunnel or drop test or water channel, and there is no need to build 10 different bodies or find 100 different fluids with 10 densities and 10 viscosities. The cost is now about $500, maybe less. A second side benefit of dimensional analysis is that it helps our thinking and planning for an experiment or theory. It suggests dimensionless ways of writing equations before we waste money on computer time to find solutions. It suggests variables which can be discarded; sometimes dimensional analysis will immediately reject variables, and at other times it groups them off to the side, where a few simple tests will show them to be unimportant. Finally, dimensional analysis will often give a great deal of insight into the form of the physical relationship we are trying to study. A third benefit is that dimensional analysis provides scaling laws which can convert data from a cheap, small model to design information for an expensive, large prototype. We do not build a million-dollar airplane and see whether it has enough lift force. We measure the lift on a small model and use a scaling law to predict the lift on | v v 278 | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.1 Introduction 279 the full-scale prototype airplane. There are rules we shall explain for finding scaling laws. When the scaling law is valid, we say that a condition of similarity exists between the model and the prototype. In the simple case of Eq. (5.1), similarity is achieved if the Reynolds number is the same for the model and prototype because the function g then requires the force coefficient to be the same also: If Rem Rep then CFm CFp (5.3) where subscripts m and p mean model and prototype, respectively. From the definition of force coefficient, this means that Fp Fm p m Vp Vm 2 Lp Lm 2 (5.4) for data taken where pVp Lp / p mVm Lm / m. Equation (5.4) is a scaling law: If you measure the model force at the model Reynolds number, the prototype force at the same Reynolds number equals the model force times the density ratio times the velocity ratio squared times the length ratio squared. We shall give more examples later. Do you understand these introductory explanations? Be careful; learning dimensional analysis is like learning to play tennis: There are levels of the game. We can establish some ground rules and do some fairly good work in this brief chapter, but dimensional analysis in the broad view has many subtleties and nuances which only time and practice and maturity enable you to master. Although dimensional analysis has a firm physical and mathematical foundation, considerable art and skill are needed to use it effectively. EXAMPLE 5.1 A copepod is a water crustacean approximately 1 mm in diameter. We want to know the drag force on the copepod when it moves slowly in fresh water. A scale model 100 times larger is made and tested in glycerin at V 30 cm/s. The measured drag on the model is 1.3 N. For similar conditions, what are the velocity and drag of the actual copepod in water? Assume that Eq. (5.1) applies and the temperature is 20°C. Solution From Table A.3 the fluid properties are: Water (prototype): p 0.001 kg/(m s) p 998 kg/m3 Glycerin (model): m 1.5 kg/(m s) m 1263 kg/m3 The length scales are Lm 100 mm and Lp 1 mm. We are given enough model data to compute the Reynolds number and force coefficient Rem mVmLm m CFm Fm Vm2Lm2 m (1263 kg/m3)(0.3 m/s)(0.1 m) 1.5 kg/(m s) 1.3 N (1263 kg/m3)(0.3 m/s)2(0.1 m)2 25.3 1.14 | v v Both these numbers are dimensionless, as you can check. For conditions of similarity, the prototype Reynolds number must be the same, and Eq. (5.2) then requires the prototype force coefficient to be the same | e-Text Main Menu | Textbook Table of Contents | Study Guide 280 Chapter 5 Dimensional Analysis and Similarity Rep or Rem Vp CFp or 998Vp(0.001) 0.001 25.3 0.0253 m/s CFm 1.14 Fp 7.31 2.53 cm/s Ans. Fp 998(0.0253)2(0.001)2 10 7 N Ans. It would obviously be difficult to measure such a tiny drag force. Historically, the first person to write extensively about units and dimensional reasoning in physical relations was Euler in 1765. Euler’s ideas were far ahead of his time, as were those of Joseph Fourier, whose 1822 book Analytical Theory of Heat outlined what is now called the principle of dimensional homogeneity and even developed some similarity rules for heat flow. There were no further significant advances until Lord Rayleigh’s book in 1877, Theory of Sound, which proposed a “method of dimensions” and gave several examples of dimensional analysis. The final breakthrough which established the method as we know it today is generally credited to E. Buckingham in 1914 [29], whose paper outlined what is now called the Buckingham pi theorem for describing dimensionless parameters (see Sec. 5.3). However, it is now known that a Frenchman, A. Vaschy, in 1892 and a Russian, D. Riabouchinsky, in 1911 had independently published papers reporting results equivalent to the pi theorem. Following Buckingham’s paper, P. W. Bridgman published a classic book in 1922 [1], outlining the general theory of dimensional analysis. The subject continues to be controversial because there is so much art and subtlety in using dimensional analysis. Thus, since Bridgman there have been at least 24 books published on the subject [2 to 25]. There will probably be more, but seeing the whole list might make some fledgling authors think twice. Nor is dimensional analysis limited to fluid mechanics or even engineering. Specialized books have been written on the application of dimensional analysis to metrology [26], astrophysics [27], economics [28], building scale models [36], chemical processing pilot plants [37], social sciences [38], biomedical sciences [39], pharmacy [40], fractal geometry [41], and even the growth of plants [42]. 5.2 The Principle of Dimensional Homogeneity In making the remarkable jump from the five-variable Eq. (5.1) to the two-variable Eq. (5.2), we were exploiting a rule which is almost a self-evident axiom in physics. This rule, the principle of dimensional homogeneity (PDH), can be stated as follows: If an equation truly expresses a proper relationship between variables in a physical process, it will be dimensionally homogeneous; i.e., each of its additive terms will have the same dimensions. All the equations which are derived from the theory of mechanics are of this form. For example, consider the relation which expresses the displacement of a falling body S S0 V0t 1 2 gt2 (5.5) | v v Each term in this equation is a displacement, or length, and has dimensions {L}. The equation is dimensionally homogeneous. Note also that any consistent set of units can be used to calculate a result. | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.2 The Principle of Dimensional Homogeneity 281 Consider Bernoulli’s equation for incompressible flow p 12 V 2 gz const (5.6) Each term, including the constant, has dimensions of velocity squared, or {L2T 2}. The equation is dimensionally homogeneous and gives proper results for any consistent set of units. Students count on dimensional homogeneity and use it to check themselves when they cannot quite remember an equation during an exam. For example, which is it: S 1 2 gt2? or S 12 2 g t? (5.7) By checking the dimensions, we reject the second form and back up our faulty memory. We are exploiting the principle of dimensional homogeneity, and this chapter simply exploits it further. Equations (5.5) and (5.6) also illustrate some other factors that often enter into a dimensional analysis: Dimensional variables are the quantities which actually vary during a given case and would be plotted against each other to show the data. In Eq. (5.5), they are S and t; in Eq. (5.6) they are p, V, and z. All have dimensions, and all can be nondimensionalized as a dimensional-analysis technique. Dimensional constants may vary from case to case but are held constant during a given run. In Eq. (5.5) they are S0, V0, and g, and in Eq. (5.6) they are , g, and C. They all have dimensions and conceivably could be nondimensionalized, but they are normally used to help nondimensionalize the variables in the problem. Pure constants have no dimensions and never did. They arise from mathematical manipulations. In both Eqs. (5.5) and (5.6) they are 1 and the exponent 2, both 2 of which came from an integration: t dt 1 t2, V dV 1 V2. Other common 2 2 dimensionless constants are and e. Note that integration and differentiation of an equation may change the dimensions but not the homogeneity of the equation. For example, integrate or differentiate Eq. (5.5): S dt S0t dS dt 1 2 V0 V0t2 1 6 gt3 gt (5.8a) (5.8b) | v v In the integrated form (5.8a) every term has dimensions of {LT}, while in the derivative form (5.8b) every term is a velocity {LT 1}. Finally, there are some physical variables that are naturally dimensionless by virtue of their definition as ratios of dimensional quantities. Some examples are strain (change in length per unit length), Poisson’s ratio (ratio of transverse strain to longitudinal strain), and specific gravity (ratio of density to standard water density). All angles are dimensionless (ratio of arc length to radius) and should be taken in radians for this reason. The motive behind dimensional analysis is that any dimensionally homogeneous equation can be written in an entirely equivalent nondimensional form which is more | e-Text Main Menu | Textbook Table of Contents | Study Guide 282 Chapter 5 Dimensional Analysis and Similarity compact. Usually there is more than one method of presenting one’s dimensionless data or theory. Let us illustrate these concepts more thoroughly by using the falling-body relation (5.5) as an example. Ambiguity: The Choice of Variables and Scaling Parameters1 Equation (5.5) is familiar and simple, yet illustrates most of the concepts of dimensional analysis. It contains five terms (S, S0, V0, t, g) which we may divide, in our thinking, into variables and parameters. The variables are the things which we wish to plot, the basic output of the experiment or theory: in this case, S versus t. The parameters are those quantities whose effect upon the variables we wish to know: in this case S0, V0, and g. Almost any engineering study can be subdivided in this manner. To nondimensionalize our results, we need to know how many dimensions are contained among our variables and parameters: in this case, only two, length {L} and time {T}. Check each term to verify this: {S} {S0} {L} {t} {T} {V0} {LT 1 } {g} {LT 2 } Among our parameters, we therefore select two to be scaling parameters, used to define dimensionless variables. What remains will be the “basic” parameter(s) whose effect we wish to show in our plot. These choices will not affect the content of our data, only the form of their presentation. Clearly there is ambiguity in these choices, something that often vexes the beginning experimenter. But the ambiguity is deliberate. Its purpose is to show a particular effect, and the choice is yours to make. For the falling-body problem, we select any two of the three parameters to be scaling parameters. Thus we have three options. Let us discuss and display them in turn. Option 1: Scaling parameters S0 and V0: the effect of gravity g. First use the scaling parameters (S0, V0) to define dimensionless (*) displacement and time. There is only one suitable definition for each:2 S S0 S* t* V0t S0 (5.9) Substitute these variables into Eq. (5.5) and clean everything up until each term is dimensionless. The result is our first option: S* 1 t* 1 t*2 2 gS0 2 V0 (5.10) This result is shown plotted in Fig. 5.1a. There is a single dimensionless parameter , which shows here the effect of gravity. It cannot show the direct effects of S0 and V0, since these two are hidden in the ordinate and abscissa. We see that gravity increases the parabolic rate of fall for t* 0, but not the initial slope at t* 0. We would learn the same from falling-body data, and the plot, within experimental accuracy, would look like Fig. 5.1a. 1 | v v I am indebted to Prof. Jacques Lewalle of Syracuse University for suggesting, outlining, and clarifying this entire discussion. 2 Make them proportional to S and t. Do not define dimensionless terms upside down: S0/S or S0/(V0t). The plots will look funny, users of your data will be confused, and your supervisor will be angry. It is not a good idea. | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.2 The Principle of Dimensional Homogeneity 5 8 g S0 =2 V02 1 0.5 gS V02 0 S ** = S S0 g S0 =2 V02 1 6 0.5 0.2 S *= 4 3 283 0 4 2 2 1 0 0 3 2 0 1 Vt t* = 0 S0 2 gt t ** = V0 (a) (b) 1 3 10 8 V0 S *** = S S0 √gS0 Fig. 5.1 Three entirely equivalent dimensionless presentations of the falling-body problem, Eq. (5.5): the effect of (a) gravity, (b) initial displacement, and (c) initial velocity. All plots contain the same information. =2 1 6 0.5 0 4 2 0 0 1 2 3 t *** = t √g / S0 (c) Option 2: Scaling parameters V0 and g: the effect of initial displacement S0. Now use the new scaling parameters (V0, g) to define dimensionless (**) displacement and time. Again there is only one suitable definition: S** Sg 2 V0 t** t g V0 (5.11) | v v Substitute these variables into Eq. (5.5) and clean everything up again. The result is our second option: 12 gS0 S** t** t** (5.12) 2 2 V0 | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 5 Dimensional Analysis and Similarity This result is plotted in Fig. 5.1b. The same single parameter again appears and here shows the effect of initial displacement, which merely moves the curves upward without changing their shape. Option 3: Scaling parameters S0 and g: the effect of initial speed V0. Finally use the scaling parameters (S0, g) to define dimensionless (***) displacement and time. Again there is only one suitable definition: S S0 S*** t*** t g S0 1/2 (5.13) Substitute these variables into Eq. (5.5) and clean everything up as usual. The result is our third and final option: S*** 1 t*** 1 t***2 2 1 V0 gS0 (5.14) This final presentation is shown in Fig. 5.1c. Once again the parameter appears, but we have redefined it upside down, 1/ , so that our display parameter V0 is in the numerator and is linear. This is our free choice and simply improves the display. Figure 5.1c shows that initial velocity increases the falling displacement and that the increase is proportional to time. Note that, in all three options, the same parameter appears but has a different meaning: dimensionless gravity, initial displacement, and initial velocity. The graphs, which contain exactly the same information, change their appearance to reflect these differences. Whereas the original problem, Eq. (5.5), involved five quantities, the dimensionless presentations involve only three, having the form S gS0 2 V0 fcn(t , ) (5.15) The reduction 5 3 2 should equal the number of fundamental dimensions involved in the problem {L, T}. This idea led to the pi theorem (Sec. 5.3). The choice of scaling variables is left to the user, and the resulting dimensionless parameters have differing interpretations. For example, in the dimensionless drag-force formulation, Eq. (5.2), it is now clear that the scaling parameters were , V, and L, since they appear in both the drag coefficient and the Reynolds number. Equation (5.2) can thus be interpreted as the variation of dimensionless force with dimensionless viscosity, with the scaling-parameter effects mixed between CF and Re and therefore not immediately evident. Suppose that we wish to study drag force versus velocity. Then we would not use V as a scaling parameter. We would use ( , , L) instead, and the final dimensionless function would become CF F 2 fcn(Re) Re VL (5.16) In plotting these data, we would not be able to discern the effect of or , since they appear in both dimensionless groups. The grouping CF again would mean dimension- | v v 284 | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.2 The Principle of Dimensional Homogeneity 285 less force, and Re is now interpreted as either dimensionless velocity or size.3 The plot would be quite different compared to Eq. (5.2), although it contains exactly the same information. The development of parameters such as CF and Re from the initial variables is the subject of the pi theorem (Sec. 5.3). Some Peculiar Engineering Equations The foundation of the dimensional-analysis method rests on two assumptions: (1) The proposed physical relation is dimensionally homogeneous, and (2) all the relevant variables have been included in the proposed relation. If a relevant variable is missing, dimensional analysis will fail, giving either algebraic difficulties or, worse, yielding a dimensionless formulation which does not resolve the process. A typical case is Manning’s open-channel formula, discussed in Example 1.4: 1.49 2/3 1/2 V RS (1) n Since V is velocity, R is a radius, and n and S are dimensionless, the formula is not dimensionally homogeneous. This should be a warning that (1) the formula changes if the units of V and R change and (2) if valid, it represents a very special case. Equation (1) in Example 1.4 (see above) predates the dimensional-analysis technique and is valid only for water in rough channels at moderate velocities and large radii in BG units. Such dimensionally inhomogeneous formulas abound in the hydraulics literature. Another example is the Hazen-Williams formula [30] for volume flow of water through a straight smooth pipe dp 0.54 Q 61.9D2.63 (5.17) dx where D is diameter and dp/dx is the pressure gradient. Some of these formulas arise because numbers have been inserted for fluid properties and other physical data into perfectly legitimate homogeneous formulas. We shall not give the units of Eq. (5.17) to avoid encouraging its use. On the other hand, some formulas are “constructs” which cannot be made dimensionally homogeneous. The “variables” they relate cannot be analyzed by the dimensional-analysis technique. Most of these formulas are raw empiricisms convenient to a small group of specialists. Here are three examples: B S 0.0147DE 25,000 100 R (5.18) 140 130 API (5.19) 3.74 DE 0.26tR 172 tR (5.20) Equation (5.18) relates the Brinell hardness B of a metal to its Rockwell hardness R. Equation (5.19) relates the specific gravity S of an oil to its density in degrees API. | v v 3 We were lucky to achieve a size effect because in this case L, a scaling parameter, did not appear in the drag coefficient. | e-Text Main Menu | Textbook Table of Contents | Study Guide 286 Chapter 5 Dimensional Analysis and Similarity Equation (5.20) relates the viscosity of a liquid in DE, or degrees Engler, to its viscosity tR in Saybolt seconds. Such formulas have a certain usefulness when communicated between fellow specialists, but we cannot handle them here. Variables like Brinell hardness and Saybolt viscosity are not suited to an MLT dimensional system. 5.3 The Pi Theorem There are several methods of reducing a number of dimensional variables into a smaller number of dimensionless groups. The scheme given here was proposed in 1914 by Buckingham [29] and is now called the Buckingham pi theorem. The name pi comes from the mathematical notation , meaning a product of variables. The dimensionless groups found from the theorem are power products denoted by 1, 2, 3, etc. The method allows the pis to be found in sequential order without resorting to free exponents. The first part of the pi theorem explains what reduction in variables to expect: If a physical process satisfies the PDH and involves n dimensional variables, it can be reduced to a relation between only k dimensionless variables or ’s. The reduction j n k equals the maximum number of variables which do not form a pi among themselves and is always less than or equal to the number of dimensions describing the variables. Take the specific case of force on an immersed body: Eq. (5.1) contains five variables F, L, U, , and described by three dimensions {MLT}. Thus n 5 and j 3. Therefore it is a good guess that we can reduce the problem to k pis, with k n j 5 3 2. And this is exactly what we obtained: two dimensionless variables 1 CF and Re. On rare occasions it may take more pis than this minimum (see Example 2 5.5). The second part of the theorem shows how to find the pis one at a time: Find the reduction j, then select j scaling variables which do not form a pi among themselves.4 Each desired pi group will be a power product of these j variables plus one additional variable which is assigned any convenient nonzero exponent. Each pi group thus found is independent. To be specific, suppose that the process involves five variables 1 f ( 2, 3, 4, 5) Suppose that there are three dimensions {MLT} and we search around and find that indeed j 3. Then k 5 3 2 and we expect, from the theorem, two and only two pi groups. Pick out three convenient variables which do not form a pi, and suppose these turn out to be 2, 3, and 4. Then the two pi groups are formed by power products of these three plus one additional variable, either 1 or 5: 1 ( 2)a( 3)b( 4)c 1 M0L0T0 2 ( 2)a( 3)b( 4)c 5 M0L0T0 Here we have arbitrarily chosen 1 and 5, the added variables, to have unit exponents. Equating exponents of the various dimensions is guaranteed by the theorem to give unique values of a, b, and c for each pi. And they are independent because only 1 | v v 4 | Make a clever choice here because all pis will contain these j variables in various groupings. e-Text Main Menu | Textbook Table of Contents | Study Guide 5.3 The Pi Theorem Table 5.1 Dimensions of FluidMechanics Properties 287 Dimensions Quantity Symbol Length Area Volume Velocity Acceleration Speed of sound Volume flow Mass flow Pressure, stress Strain rate Angle Angular velocity Viscosity Kinematic viscosity Surface tension Force Moment, torque Power Work, energy Density Temperature Specific heat Specific weight Thermal conductivity Expansion coefficient MLT L A FLT L L2 L3 LT 1 LT 2 LT 1 L3T 1 MT 1 ML 1T 2 T1 None T1 ML 1T 1 L2T 1 MT 2 MLT 2 ML2T 2 ML2T –3 ML2T 2 ML–3 L2T 2 ML–2T MLT –3 V dV/dt a Q m ˙ p, ˙ F M P W, E T cp, c k L L2 L3 LT 1 LT 2 LT 1 L3T 1 FTL 1 FL 2 T1 None T1 FTL 2 L2T 1 FL 1 F FL FLT 1 FL FT2L–4 L2T FL FT 1 2 1 1 2 1 3 1 1 1 contains 1 and only 2 contains 5. It is a very neat system once you get used to the procedure. We shall illustrate it with several examples. Typically, six steps are involved: 1. 2. 3. 4. | v v 5. | List and count the n variables involved in the problem. If any important variables are missing, dimensional analysis will fail. List the dimensions of each variable according to {MLT } or {FLT }. A list is given in Table 5.1. Find j. Initially guess j equal to the number of different dimensions present, and look for j variables which do not form a pi product. If no luck, reduce j by 1 and look again. With practice, you will find j rapidly. Select j scaling parameters which do not form a pi product. Make sure they please you and have some generality if possible, because they will then appear in every one of your pi groups. Pick density or velocity or length. Do not pick surface tension, e.g., or you will form six different independent Weber-number parameters and thoroughly annoy your colleagues. Add one additional variable to your j repeating variables, and form a power product. Algebraically find the exponents which make the product dimensionless. Try to arrange for your output or dependent variables (force, pressure drop, torque, power) to appear in the numerator, and your plots will look better. Do e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 5 Dimensional Analysis and Similarity 6. this sequentially, adding one new variable each time, and you will find all n j k desired pi products. Write the final dimensionless function, and check your work to make sure all pi groups are dimensionless. EXAMPLE 5.2 Repeat the development of Eq. (5.2) from Eq. (5.1), using the pi theorem. Solution Write the function and count variables: Step 1 F f(L, U, , ) there are five variables (n 5) List dimensions of each variable. From Table 5.1 Step 2 F {MLT L 2 } U {L} {LT 1 {ML 3} } {ML 1T 1 } Step 3 Find j. No variable contains the dimension , and so j is less than or equal to 3 (MLT). We inspect the list and see that L, U, and cannot form a pi group because only contains mass and only U contains time. Therefore j does equal 3, and n j 5 3 2 k. The pi theorem guarantees for this problem that there will be exactly two independent dimensionless groups. Step 4 Select repeating j variables. The group L, U, Step 5 Combine L, U, with one additional variable, in sequence, to find the two pi products. First add force to find 1. You may select any exponent on this additional term as you please, to place it in the numerator or denominator to any power. Since F is the output, or dependent, variable, we select it to appear to the first power in the numerator: 1 LaUb cF (L)a(LT we found in step 3 will do fine. 1b ) (ML 3)c( MLT 2 ) M0L0T0 Equate exponents: Length: a b 3c Time: 1 0 c 1 0 2 Mass: 0 b We can solve explicitly for a 2 b 2 c 1 F U2L2 CF Therefore 1 L 2U 2 1 F Ans. This is exactly the right pi group as in Eq. (5.2). By varying the exponent on F, we could have found other equivalent groups such as UL 1/2/F1/2. | v v 288 | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.3 The Pi Theorem 289 Finally, add viscosity to L, U, and to find 2. Select any power you like for viscosity. By hindsight and custom, we select the power 1 to place it in the denominator: 2 LaUb c 1 La(LT 1b ) (ML 3)c(ML 1T 1 ) 1 M0L0T0 Equate exponents: Length: a b 3c 1 0 c 1 0 1 0 Mass: Time: b from which we find a Therefore 2 b L1U1 c 1 1 UL 1 Re Ans. We know we are finished; this is the second and last pi group. The theorem guarantees that the functional relationship must be of the equivalent form F U2L2 g UL Ans. which is exactly Eq. (5.2). EXAMPLE 5.3 Reduce the falling-body relationship, Eq. (5.5), to a function of dimensionless variables. Why are there three different formulations? Solution Write the function and count variables S f (t, S0, V0, g) five variables (n 5) List the dimensions of each variable, from Table 5.1: S {L} t S0 {T} {L} V0 LT g 1 } {LT 2 } | v v There are only two primary dimensions (L, T), so that j 2. By inspection we can easily find two variables which cannot be combined to form a pi, for example, V0 and g. Then j 2, and we expect 5 2 3 pi products. Select j variables among the parameters S0, V0, and g. Avoid S and t since they are the dependent variables, which should not be repeated in pi groups. There are three different options for repeating variables among the group (S0, V0, g). Therefore we can obtain three different dimensionless formulations, just as we did informally with the falling-body equation in Sec. 5.2. Take each option in turn: | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 5 Dimensional Analysis and Similarity 1. Choose S0 and V0 as repeating variables. Combine them in turn with (S, t, g): ab S1S 0V0 1 f cd t1S 0V 0 2 e g1S 0V 0 3 Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f ). Please allow us to give the results here, and you may check the algebra as an exercise: a 1 b 0 c S S0 S* 1 1 d e V0t S0 t* 2 1 1 f 2 Ans. gS0 2 V0 3 Thus, for option 1, we know that S* fcn(t*, ). We have found, by dimensional analysis, the same variables as in Eq. (5.10). But here there is no formula for the functional relation — we might have to experiment with falling bodies to establish Fig. 5.1a. 2. Choose V0 and g as repeating variables. Combine them in turn with (S, t, S0): a S1V 0 gb 1 c t1V 0gd 2 e S1V 0gf 0 3 Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f ). Once more allow us to give the results here, and you may check the algebra as an exercise. a 2 b 1 c Sg 2 V0 S** 1 1 d t** 2 1 e tg V0 1 f 2 Ans. gS0 2 V0 3 Thus, for option 2, we now know that S** fcn(t**, ). We have found, by dimensional analysis, the same groups as in Eq. (5.12). The data would plot as in Fig. 5.1b. 3. Finally choose S0 and g as repeating variables. Combine them in turn with (S, t, V0): a S1S 0gb 1 c t1S0 gd 2 1e V0S0g f 3 Set each power product equal to L0T0, and solve for the exponents (a, b, c, d, e, f ). One more time allow us to give the results here, and you may check the algebra as an exercise: a 1 1 b S*** 0 S S0 1 2 c 2 t*** d 1 2 g t S0 1 2 e f 1 2 V0 gS0 3 Ans. Thus, for option 3, we now know that S*** fcn(t***, 1/ ). We have found, by dimensional analysis, the same groups as in Eq. (5.14). The data would plot as in Fig. 5.1c. Dimensional analysis here has yielded the same pi groups as the use of scaling parameters with Eq. (5.5). Three different formulations appeared, because we could choose three different pairs of repeating variables to complete the pi theorem. EXAMPLE 5.4 At low velocities (laminar flow), the volume flow Q through a small-bore tube is a function only of the tube radius R, the fluid viscosity , and the pressure drop per unit tube length dp/dx. Using the pi theorem, find an appropriate dimensionless relationship. | v v 290 | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.3 The Pi Theorem 291 Solution Write the given relation and count variables: Q f R, , dp dx four variables (n 4) Make a list of the dimensions of these variables from Table 5.1: Q {L3T R 1 } dp/dx {ML 1T {L} 1 {ML 2T } 2 } There are three primary dimensions (M, L, T), hence j 3. By trial and error we determine that R, , and dp/dx cannot be combined into a pi group. Then j 3, and n j 4 3 1. There is only one pi group, which we find by combining Q in a power product with the other three: 1 Ra b dp c 1 Q dx (L)a(ML 1T 1b ) (ML 2T 2c ) (L3T 1 ) M0L0T0 Equate exponents: Mass: b c b 2c 3 0 Time: b 2c 1 0 Solving simultaneously, we obtain a 4, b Length: a 1 or 1 R 0 1, c 4 1 1. Then dp dx Q R 4(dp/dx) 1 Q const Ans. Since there is only one pi group, it must equal a dimensionless constant. This is as far as dimensional analysis can take us. The laminar-flow theory of Sec. 6.4 shows that the value of the constant is /8. EXAMPLE 5.5 Assume that the tip deflection of a cantilever beam is a function of the tip load P, beam length L, area moment of inertia I, and material modulus of elasticity E; that is, f (P, L, I, E). Rewrite this function in dimensionless form, and comment on its complexity and the peculiar value of j. Solution List the variables and their dimensions: P | v v {L} | e-Text Main Menu {MLT | L 2 } {L} I E 4 {ML 1T {L } Textbook Table of Contents | 2 } Study Guide 292 Chapter 5 Dimensional Analysis and Similarity There are five variables (n 5) and three primary dimensions (M, L, T), hence j 3. But try as we may, we cannot find any combination of three variables which does not form a pi group. This is because {M} and {T} occur only in P and E and only in the same form, {MT 2}. Thus we have encountered a special case of j 2, which is less than the number of dimensions (M, L, T). To gain more insight into this peculiarity, you should rework the problem, using the (F, L, T) system of dimensions. With j 2, we select L and E as two variables which cannot form a pi group and then add other variables to form the three desired pis: LaEbI1 1 (L)a(ML 1T 2b ) (L4) from which, after equating exponents, we find that a LaEbP1 2 from which we find a 2, b from which a or 1, b 1, or LaEb 3 0, or (L)a(ML 1T 1 2 M0L0T0 4, b 2b ) (MLT 0, or 2 1 I/L4. Then M0L0T0 ) P/(EL2), and (L)a(ML 1T 2b ) (L) M0L0T0 /L. The proper dimensionless function is 3 L f P I , EL2 L4 2, 1), Ans. 3 f( (1) This is a complex three-variable function, but dimensional analysis alone can take us no further. We can “improve” Eq. (1) by taking advantage of some physical reasoning, as Langhaar points out [8, p. 91]. For small elastic deflections, is proportional to load P and inversely proportional to moment of inertia I. Since P and I occur separately in Eq. (1), this means that 3 must be proportional to 2 and inversely proportional to 1. Thus, for these conditions, L or (const) P L4 EL2 I (const) PL3 EI (2) This could not be predicted by a pure dimensional analysis. Strength-of-materials theory predicts that the value of the constant is 1 . 3 5.4 Nondimensionalization of the Basic Equations We could use the pi-theorem method of the previous section to analyze problem after problem after problem, finding the dimensionless parameters which govern in each case. Textbooks on dimensional analysis [for example, 7] do this. An alternate and very powerful technique is to attack the basic equations of flow from Chap. 4. Even though these equations cannot be solved in general, they will reveal basic dimensionless parameters, e.g., Reynolds number, in their proper form and proper position, giving clues to when they are negligible. The boundary conditions must also be nondimensionalized. Let us briefly apply this technique to the incompressible-flow continuity and momentum equations with constant viscosity: | v v Continuity: | e-Text Main Menu V | 0 Textbook Table of Contents (5.21a) | Study Guide 5.4 Nondimensionalization of the Basic Equations dV 2 g p V dt Typical boundary conditions for these two equations are Momentum: Fixed solid surface: V Inlet or outlet: Free surface, z (5.21b) 0 Known V, p : w d dt p 293 pa (5.22) 1 (Rx Ry 1) We omit the energy equation (4.75) and assign its dimensionless form in the problems (Probs. 5.42 and 5.45). Equations (5.21) and (5.22) contain the three basic dimensions M, L, and T. All variables p, V, x, y, z, and t can be nondimensionalized by using density and two reference constants which might be characteristic of the particular fluid flow: Reference velocity U Reference length L For example, U may be the inlet or upstream velocity and L the diameter of a body immersed in the stream. Now define all relevant dimensionless variables, denoting them by an asterisk: V* x L x* t* tU L y* V U y z* L p p* z L (5.23) gz U2 All these are fairly obvious except for p*, where we have slyly introduced the gravity effect, assuming that z is up. This is a hindsight idea suggested by Bernoulli’s equation (3.77). Since , U, and L are all constants, the derivatives in Eqs. (5.21) can all be handled in dimensionless form with dimensional coefficients. For example, u x (Uu*) (Lx*) U u* L x* Substitute the variables from Eqs. (5.23) into Eqs. (5.21) and (5.22) and divide through by the leading dimensional coefficient, in the same way as we handled Eq. (5.12). The resulting dimensionless equations of motion are: Continuity: * V* dV* dt* Momentum: 0 *p* UL (5.24a) *2(V*) The dimensionless boundary conditions are: Fixed solid surface: | v v Inlet or outlet: | e-Text Main Menu | V* 0 Known V*, p* Textbook Table of Contents | Study Guide (5.24b) 294 Chapter 5 Dimensional Analysis and Similarity Free surface, z* *: p* w* pa U2 gL z* U2 d* dt* U2L (5.25) 1 (R* x R* 1) y These equations reveal a total of four dimensionless parameters, one in the momentum equation and three in the free-surface-pressure boundary condition. Dimensionless Parameters In the continuity equation there are no parameters. The momentum equation contains one, generally accepted as the most important parameter in fluid mechanics: Reynolds number Re UL It is named after Osborne Reynolds (1842 – 1912), a British engineer who first proposed it in 1883 (Ref. 4 of Chap. 6). The Reynolds number is always important, with or without a free surface, and can be neglected only in flow regions away from high-velocity gradients, e.g., away from solid surfaces, jets, or wakes. The no-slip and inlet-exit boundary conditions contain no parameters. The freesurface-pressure condition contains three: Euler number (pressure coefficient) Eu pa U2 This is named after Leonhard Euler (1707 – 1783) and is rarely important unless the pressure drops low enough to cause vapor formation (cavitation) in a liquid. The Euler number is often written in terms of pressure differences: Eu p/( U2). If p involves vapor pressure p , it is called the cavitation number Ca (pa p )/( U2). The second pressure parameter is much more important: Froude number Fr U2 gL It is named after William Froude (1810 – 1879), a British naval architect who, with his son Robert, developed the ship-model towing-tank concept and proposed similarity rules for free-surface flows (ship resistance, surface waves, open channels). The Froude number is the dominant effect in free-surface flows and is totally unimportant if there is no free surface. Chapter 10 investigates Froude number effects in detail. The final free-surface parameter is Weber number We U2L | v v It is named after Moritz Weber (1871 – 1951) of the Polytechnic Institute of Berlin, who developed the laws of similitude in their modern form. It was Weber who named Re and Fr after Reynolds and Froude. The Weber number is important only if it is of order unity or less, which typically occurs when the surface curvature is comparable in size to the liquid depth, e.g., in droplets, capillary flows, ripple waves, and very small hydraulic models. If We is large, its effect may be neglected. | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.4 Nondimensionalization of the Basic Equations 295 If there is no free surface, Fr, Eu, and We drop out entirely, except for the possibility of cavitation of a liquid at very small Eu. Thus, in low-speed viscous flows with no free surface, the Reynolds number is the only important dimensionless parameter. Compressibility Parameters In high-speed flow of a gas there are significant changes in pressure, density, and temperature which must be related by an equation of state such as the perfect-gas law, Eq. (1.10). These thermodynamic changes introduce two additional dimensionless parameters mentioned briefly in earlier chapters: Mach number Ma U a Specific-heat ratio k cp c The Mach number is named after Ernst Mach (1838 – 1916), an Austrian physicist. The effect of k is only slight to moderate, but Ma exerts a strong effect on compressibleflow properties if it is greater than about 0.3. These effects are studied in Chap. 9. Oscillating Flows If the flow pattern is oscillating, a seventh parameter enters through the inlet boundary condition. For example, suppose that the inlet stream is of the form u U cos t Nondimensionalization of this relation results in u U u* cos L t* U The argument of the cosine contains the new parameter Strouhal number St L U | v v The dimensionless forces and moments, friction, and heat transfer, etc., of such an oscillating flow would be a function of both Reynolds and Strouhal numbers. This parameter is named after V. Strouhal, a German physicist who experimented in 1878 with wires singing in the wind. Some flows which you might guess to be perfectly steady actually have an oscillatory pattern which is dependent on the Reynolds number. An example is the periodic vortex shedding behind a blunt body immersed in a steady stream of velocity U. Figure 5.2a shows an array of alternating vortices shed from a circular cylinder immersed in a steady crossflow. This regular, periodic shedding is called a Kármán vortex street, after T. von Kármán, who explained it theoretically in 1912. The shedding occurs in the range 102 Re 107, with an average Strouhal number d/(2 U) 0.21. Figure 5.2b shows measured shedding frequencies. Resonance can occur if a vortex shedding frequency is near a body’s structuralvibration frequency. Electric transmission wires sing in the wind, undersea mooring lines gallop at certain current speeds, and slender structures flutter at critical wind or vehicle speeds. A striking example is the disastrous failure of the Tacoma Narrows suspension bridge in 1940, when wind-excited vortex shedding caused resonance with the natural torsional oscillations of the bridge. | e-Text Main Menu | Textbook Table of Contents | Study Guide 296 Chapter 5 Dimensional Analysis and Similarity (a) 0.4 Data spread St = ω d 2π U 0.3 0.2 0.1 Fig. 5.2 Vortex shedding from a circular cylinder: (a) vortex street behind a circular cylinder (from Ref. 33, courtesy of U.S. Naval Research Laboratory); (b) experimental shedding frequencies (data from Refs. 31 and 32). 0 10 10 2 10 3 10 4 ρ Ud Re = µ 10 5 10 6 107 (b) Other Dimensionless Parameters We have discussed seven important parameters in fluid mechanics, and there are others. Four additional parameters arise from nondimensionalization of the energy equation (4.75) and its boundary conditions. These four (Prandtl number, Eckert number, Grashof number, and wall-temperature ratio) are listed in Table 5.2 just in case you fail to solve Prob. 5.42. Another important and rather sneaky parameter is the wallroughness ratio /L (in Table 5.2).5 Slight changes in surface roughness have a strik- | v v 5 Roughness is easy to overlook because it is a slight geometric effect which does not appear in the equations of motion. | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.4 Nondimensionalization of the Basic Equations Table 5.2 Dimensionless Groups in Fluid Mechanics Parameter Qualitative ratio of effects Definition 297 Importance UL Inertia Viscosity Always Ma U a Flow speed Sound speed Compressible flow Froude number Fr U2 gL Inertia Gravity Free-surface flow Weber number We Inertia Surface tension Free-surface flow Cavitation number (Euler number) Ca Pressure Inertia Cavitation Prandtl number Pr cp k Dissipation Conduction Heat convection Eckert number Ec U2 cpT0 Kinetic energy Enthalpy Dissipation Specific-heat ratio k Enthalpy Internal energy Compressible flow Strouhal number St Oscillation Mean speed Oscillating flow Wall roughness Body length Turbulent, rough walls Buoyancy Viscosity Natural convection Wall temperature Stream temperature Heat transfer Static pressure Dynamic pressure Aerodynamics, hydrodynamics Lift force Dynamic force Aerodynamics, hydrodynamics Drag force Dynamic force Aerodynamics, hydrodynamics Reynolds number Re Mach number Roughness ratio U2L p p U2 cp c L U L Grashof number Temperature ratio Tw T0 Pressure coefficient Cp Lift coefficient CL Drag coefficient CD TgL3 Gr 2 p 1 2 p U2 L 1 2 U2A D 1 2 U2A 2 | v v ing effect in the turbulent-flow or high-Reynolds-number range, as we shall see in Chap. 6 and in Fig. 5.3. This book is primarily concerned with Reynolds-, Mach-, and Froude-number effects, which dominate most flows. Note that we discovered all these parameters (except /L) simply by nondimensionalizing the basic equations without actually solving them. If the reader is not satiated with the 15 parameters given in Table 5.2, Ref. 34 contains a list of over 300 dimensionless parameters in use in engineering. See also Ref. 35. | e-Text Main Menu | Textbook Table of Contents | Study Guide 298 Chapter 5 Dimensional Analysis and Similarity Dimensional analysis is fun, but does it work? Yes; if all important variables are included in the proposed function, the dimensionless function found by dimensional analysis will collapse all the data onto a single curve or set of curves. An example of the success of dimensional analysis is given in Fig. 5.3 for the measured drag on smooth cylinders and spheres. The flow is normal to the axis of the cylinder, which is extremely long, L/d → . The data are from many sources, for both liquids and gases, and include bodies from several meters in diameter down to fine wires and balls less than 1 mm in size. Both curves in Fig. 5.3a are entirely experimental; the analysis of immersed body drag is one of the weakest areas of modern fluidmechanics theory. Except for some isolated digital-computer calculations, there is no theory for cylinder and sphere drag except creeping flow, Re 1. The Reynolds number of both bodies is based upon diameter, hence the notation Red. But the drag coefficients are defined differently: A Successful Application 5 4 Cylinder length effect Transition to turbulent boundary layer (10 4 < Re < 10 5) L/d CD 2 Cylinder (two – dimensional) 1 CD ∞ 40 20 10 5 3 2 1 3 1.20 0.98 0.91 0.82 0.74 0.72 0.68 0.64 Sphere 0 10 10 2 10 3 10 4 ρ Ud Red = µ (a) 10 5 10 6 10 7 1.5 1.0 CD 0.7 | v v Fig. 5.3 The proof of practical dimensional analysis: drag coefficients of a cylinder and sphere: (a) drag coefficient of a smooth cylinder and sphere (data from many sources); (b) increased roughness causes earlier transition to a turbulent boundary layer. 0.5 0.3 10 4 Cylinder: ε = 0.02 − d 0.009 0.007 0.004 0.002 0.0005 _ L=∞ d Smooth 10 5 Red 10 6 (b) | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.4 Nondimensionalization of the Basic Equations drag 1 2 CD 1 2 299 cylinder U2Ld drag U2 1 d2 4 (5.26) sphere They both have a factor 1 as a traditional tribute to Bernoulli and Euler, and both are 2 based on the projected area, i.e., the area one sees when looking toward the body from upstream. The usual definition of CD is thus CD 1 2 drag U2(projected area) (5.27) However, one should carefully check the definitions of CD, Re, etc., before using data in the literature. Airfoils, e.g., use the planform area. Figure 5.3a is for long, smooth cylinders. If wall roughness and cylinder length are included as variables, we obtain from dimensional analysis a complex three-parameter function CD f Red, L , dd (5.28) To describe this function completely would require 1000 or more experiments. Therefore it is customary to explore the length and roughness effects separately to establish trends. The table with Fig. 5.3a shows the length effect with zero wall roughness. As length decreases, the drag decreases by up to 50 percent. Physically, the pressure is “relieved” at the ends as the flow is allowed to skirt around the tips instead of deflecting over and under the body. Figure 5.3b shows the effect of wall roughness for an infinitely long cylinder. The sharp drop in drag occurs at lower Red as roughness causes an earlier transition to a turbulent boundary layer on the surface of the body. Roughness has the same effect on sphere drag, a fact which is exploited in sports by deliberate dimpling of golf balls to give them less drag at their flight Red 105. Figure 5.3 is a typical experimental study of a fluid-mechanics problem, aided by dimensional analysis. As time and money and demand allow, the complete threeparameter relation (5.28) could be filled out by further experiments. EXAMPLE 5.6 The capillary rise h of a liquid in a tube varies with tube diameter d, gravity g, fluid density , surface tension , and the contact angle . (a) Find a dimensionless statement of this relation. (b) If h 3 cm in a given experiment, what will h be in a similar case if the diameter and surface tension are half as much, the density is twice as much, and the contact angle is the same? Solution Part (a) Step 1 Write down the function and count variables: | v v h | e-Text Main Menu | f(d, g, , ,) Textbook Table of Contents n | 6 variables Study Guide Chapter 5 Dimensional Analysis and Similarity Step 2 List the dimensions {FLT} from Table 5.2: h Step 3 d {L} {L} g {LT 2 {FT2L 4} } {FL 1} none Find j. Several groups of three form no pi: , , and g or , g, and d. Therefore j 3, and we expect n j 6 3 3 dimensionless groups. One of these is obviously , which is already dimensionless: Ans. (a) 3 If we had carelessly chosen to search for it by using steps 4 and 5, we would still find Step 4 3 . Select j repeating variables which do not form a pi group: , g, d. Step 5 Add one additional variable in sequence to form the pis: abc Add h: (FT 2L 4)a(LT gdh 1 2b ) (L)c(L) F 0L0T 0 Solve for a b Finally add 0 c 1 h d 00 Therefore g d 1h 1 Ans. (a) , again selecting its exponent to be 1: abc gd 2 (FT 2L 4)a(LT 2b ) (L)c(FL 1) F0L0T0 Solve for a Step 6 b 1 1 Therefore 2 c g 1d 2 2 Ans. (a) gd2 The complete dimensionless relation for this problem is thus h d F gd2 , Ans. (a) (1) This is as far as dimensional analysis goes. Theory, however, establishes that h is proportional to . Since occurs only in the second parameter, we can slip it outside h d actual gd2 F1( ) Example 1.9 showed theoretically that F1( ) h gd or F1( ) 4 cos . 1 We are given h1 for certain conditions d1, 1, 1, and 1. If h1 3 cm, what is h2 for d2 2 d1, 2 1 2 1, and 2 2 1, 2 1? We know the functional relation, Eq. (1), must still hold at condition 2 Part (b) h2 d2 2 2 2gd 2 , F 2 But 2 2 2gd 2 | v v 300 | e-Text Main Menu | 1 2 1 2 1g( 1 d1)2 2 1 2 1gd1 Textbook Table of Contents | Study Guide 5.5 Modeling and Its Pitfalls 301 Therefore, functionally, h2 d2 F 1 2 1gd1 , 1 h1 d1 We are given a condition 2 which is exactly similar to condition 1, and therefore a scaling law holds h2 h1 d2 d1 (3 cm) 1 21 d d1 1.5 cm Ans. (b) If the pi groups had not been exactly the same for both conditions, we would have had to know more about the functional relation F to calculate h2. 5.5 Modeling and Its Pitfalls So far we have learned about dimensional homogeneity and the pi-theorem method, using power products, for converting a homogeneous physical relation to dimensionless form. This is straightforward mathematically, but there are certain engineering difficulties which need to be discussed. First, we have more or less taken for granted that the variables which affect the process can be listed and analyzed. Actually, selection of the important variables requires considerable judgment and experience. The engineer must decide, e.g., whether viscosity can be neglected. Are there significant temperature effects? Is surface tension important? What about wall roughness? Each pi group which is retained increases the expense and effort required. Judgment in selecting variables will come through practice and maturity; this book should provide some of the necessary experience. Once the variables are selected and the dimensional analysis is performed, the experimenter seeks to achieve similarity between the model tested and the prototype to be designed. With sufficient testing, the model data will reveal the desired dimensionless function between variables 1 f( 2, 3, ... k) (5.29) With Eq. (5.29) available in chart, graphical, or analytical form, we are in a position to ensure complete similarity between model and prototype. A formal statement would be as follows: Flow conditions for a model test are completely similar if all relevant dimensionless parameters have the same corresponding values for the model and the prototype. This follows mathematically from Eq. (5.29). If 2m 2p, 3m 3p, etc., Eq. (5.29) guarantees that the desired output 1m will equal 1p. But this is easier said than done, as we now discuss. Instead of complete similarity, the engineering literature speaks of particular types of similarity, the most common being geometric, kinematic, dynamic, and thermal. Let us consider each separately. | v v Geometric Similarity Geometric similarity concerns the length dimension {L} and must be ensured before any sensible model testing can proceed. A formal definition is as follows: | e-Text Main Menu | Textbook Table of Contents | Study Guide 302 Chapter 5 Dimensional Analysis and Similarity A model and prototype are geometrically similar if and only if all body dimensions in all three coordinates have the same linear-scale ratio. Note that all length scales must be the same. It is as if you took a photograph of the prototype and reduced it or enlarged it until it fitted the size of the model. If the model is to be made one-tenth the prototype size, its length, width, and height must each be onetenth as large. Not only that, but also its entire shape must be one-tenth as large, and technically we speak of homologous points, which are points that have the same relative location. For example, the nose of the prototype is homologous to the nose of the model. The left wingtip of the prototype is homologous to the left wingtip of the model. Then geometric similarity requires that all homologous points be related by the same linearscale ratio. This applies to the fluid geometry as well as the model geometry. All angles are preserved in geometric similarity. All flow directions are preserved. The orientations of model and prototype with respect to the surroundings must be identical. Figure 5.4 illustrates a prototype wing and a one-tenth-scale model. The model lengths are all one-tenth as large, but its angle of attack with respect to the free stream is the same: 10° not 1°. All physical details on the model must be scaled, and some are rather subtle and sometimes overlooked: 1. 2. 3. The model nose radius must be one-tenth as large. The model surface roughness must be one-tenth as large. If the prototype has a 5-mm boundary-layer trip wire 1.5 m from the leading edge, the model should have a 0.5-mm trip wire 0.15 m from its leading edge. If the prototype is constructed with protruding fasteners, the model should have homologous protruding fasteners one-tenth as large. 4. And so on. Any departure from these details is a violation of geometric similarity and must be justified by experimental comparison to show that the prototype behavior was not significantly affected by the discrepancy. Models which appear similar in shape but which clearly violate geometric similarity should not be compared except at your own risk. Figure 5.5 illustrates this point. Homologous points a * 40 m 1m a 10° Vp 4m 10° 8m Vm v v Fig. 5.4 Geometric similarity in model testing: (a) prototype; (b) one-tenth-scale model. | (a) | 0.1 m e-Text Main Menu | Textbook Table of Contents 0.8 m (b) | Study Guide * 5.5 Modeling and Its Pitfalls V2 V1 Huge sphere V3 303 V4 Large sphere Medium sphere Tiny sphere (a) V1 V2 V3 Large 4:1 ellipsoid Fig. 5.5 Geometric similarity and dissimilarity of flows: (a) similar; (b) dissimilar. Medium 3.5:1 ellipsoid Small 3:1 ellipsoid (b) The spheres in Fig. 5.5a are all geometrically similar and can be tested with a high expectation of success if the Reynolds number or Froude number, etc., is matched. But the ellipsoids in Fig. 5.5b merely look similar. They actually have different linear-scale ratios and therefore cannot be compared in a rational manner, even though they may have identical Reynolds and Froude numbers, etc. The data will not be the same for these ellipsoids, and any attempt to “compare” them is a matter of rough engineering judgment. Kinematic Similarity Kinematic similarity requires that the model and prototype have the same length-scale ratio and the same time-scale ratio. The result is that the velocity-scale ratio will be the same for both. As Langhaar [8] states it: The motions of two systems are kinematically similar if homologous particles lie at homologous points at homologous times. Length-scale equivalence simply implies geometric similarity, but time-scale equivalence may require additional dynamic considerations such as equivalence of the Reynolds and Mach numbers. One special case is incompressible frictionless flow with no free surface, as sketched in Fig. 5.6a. These perfect-fluid flows are kinematically similar with independent length and time scales, and no additional parameters are necessary (see Chap. 8 for further details). Frictionless flows with a free surface, as in Fig. 5.6b, are kinematically similar if their Froude numbers are equal Frm 2 Vm gLm 2 Vp gLp Frp (5.30) Note that the Froude number contains only length and time dimensions and hence is a purely kinematic parameter which fixes the relation between length and time. From Eq. (5.30), if the length scale is | v v Lm | e-Text Main Menu | Textbook Table of Contents Lp | (5.31) Study Guide 304 Chapter 5 Dimensional Analysis and Similarity V1p V1m = β V1p Dm = α Dp V∞ m = βV∞ p Dp V∞ p Model V2 m = βV2 p V2 p Prototype (a) λp Prototype waves: Cp Hp Period Tp Fig. 5.6 Frictionless low-speed flows are kinematically similar: (a) Flows with no free surface are kinematically similar with independent length- and time-scale ratios; (b) free-surface flows are kinematically similar with length and time scales related by the Froude number. Vp λm = α λ p Hm = α Hp Cm = C p √α Model waves: Vm = V p √α Period Tm = T p √α (b) where is a dimensionless ratio, the velocity scale is 1/ 2 Vm Vp Tm Tp and the time scale is Lm Lp Lm/Vm Lp/Vp (5.32) (5.33) These Froude-scaling kinematic relations are illustrated in Fig. 5.6b for wave-motion modeling. If the waves are related by the length scale , then the wave period, propagation speed, and particle velocities are related by . If viscosity, surface tension, or compressibility is important, kinematic similarity is dependent upon the achievement of dynamic similarity. Dynamic Similarity | v v Dynamic similarity exists when the model and the prototype have the same lengthscale ratio, time-scale ratio, and force-scale (or mass-scale) ratio. Again geometric sim- | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.5 Modeling and Its Pitfalls 305 ilarity is a first requirement; without it, proceed no further. Then dynamic similarity exists, simultaneous with kinematic similarity, if the model and prototype force and pressure coefficients are identical. This is ensured if: 1. For compressible flow, the model and prototype Reynolds number and Mach number and specific-heat ratio are correspondingly equal. 2. For incompressible flow a. With no free surface: model and prototype Reynolds numbers are equal. b. With a free surface: model and prototype Reynolds number, Froude number, and (if necessary) Weber number and cavitation number are correspondingly equal. Mathematically, Newton’s law for any fluid particle requires that the sum of the pressure force, gravity force, and friction force equal the acceleration term, or inertia force, Fp Fg Ff Fi The dynamic-similarity laws listed above ensure that each of these forces will be in the same ratio and have equivalent directions between model and prototype. Figure 5.7 shows an example for flow through a sluice gate. The force polygons at homologous points have exactly the same shape if the Reynolds and Froude numbers are equal (neglecting surface tension and cavitation, of course). Kinematic similarity is also ensured by these model laws. Discrepancies in Water and Air Testing The perfect dynamic similarity shown in Fig. 5.7 is more of a dream than a reality because true equivalence of Reynolds and Froude numbers can be achieved only by dramatic changes in fluid properties, whereas in fact most model testing is simply done with water or air, the cheapest fluids available. First consider hydraulic model testing with a free surface. Dynamic similarity requires equivalent Froude numbers, Eq. (5.30), and equivalent Reynolds numbers VmLm m VpLp (5.34) p Fpp Fgp Fip Ffp Fim v v Fig. 5.7 Dynamic similarity in sluice-gate flow. Model and prototype yield identical homologous force polygons if the Reynolds and Froude numbers are the same corresponding values: (a) prototype; (b) model. | | e-Text Main Menu Fpm a a' (a) | Textbook Table of Contents (b) | Study Guide Fgm Ffm 306 Chapter 5 Dimensional Analysis and Similarity But both velocity and length are constrained by the Froude number, Eqs. (5.31) and (5.32). Therefore, for a given length-scale ratio , Eq. (5.34) is true only if m p Lm Vm Lp Vp 3/2 (5.35) For example, for a one-tenth-scale model, 0.1 and 3/2 0.032. Since p is undoubtedly water, we need a fluid with only 0.032 times the kinematic viscosity of water to achieve dynamic similarity. Referring to Table 1.4, we see that this is impossible: Even mercury has only one-ninth the kinematic viscosity of water, and a mercury hydraulic model would be expensive and bad for your health. In practice, water is used for both the model and the prototype, and the Reynolds-number similarity (5.34) is unavoidably violated. The Froude number is held constant since it is the dominant parameter in free-surface flows. Typically the Reynolds number of the model flow is too small by a factor of 10 to 1000. As shown in Fig. 5.8, the low-Reynolds-number model data are used to estimate by extrapolation the desired high-Reynolds-number prototype data. As the figure indicates, there is obviously considerable uncertainty in using such an extrapolation, but there is no other practical alternative in hydraulic model testing. Second, consider aerodynamic model testing in air with no free surface. The important parameters are the Reynolds number and the Mach number. Equation (5.34) should be satisfied, plus the compressibility criterion Vm am Vp ap (5.36) Elimination of Vm /Vp between (5.34) and (5.36) gives Lm am Lp ap m p (5.37) Since the prototype is no doubt an air operation, we need a wind-tunnel fluid of low viscosity and high speed of sound. Hydrogen is the only practical example, but clearly it is too expensive and dangerous. Therefore wind tunnels normally operate with air as the working fluid. Cooling and pressurizing the air will bring Eq. (5.37) into better Range of Rem Range of Re p Power-law extrapolation log CD | v v Fig. 5.8 Reynolds-number extrapolation, or scaling, of hydraulic data with equal Froude numbers. Uncertainty in prototype data estimate Model data: 105 106 107 108 log Re | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.5 Modeling and Its Pitfalls 307 Fig. 5.9 Hydraulic model of a barrier-beach inlet at Little River, South Carolina. Such models of necessity violate geometric similarity and do not model the Reynolds number of the prototype inlet. (Courtesy of U.S. Army Engineer Waterways Experiment Station). agreement but not enough to satisfy a length-scale reduction of, say, one-tenth. Therefore Reynolds-number scaling is also commonly violated in aerodynamic testing, and an extrapolation like that in Fig. 5.8 is required here also. Finally, a serious discrepancy of another type occurs in hydraulic models of natural flow systems such as rivers, harbors, estuaries, and embayments. Such flows have large horizontal dimensions and small relative vertical dimensions. If we were to scale an estuary model by a uniform linear length ratio of, say, 1 1000, the resulting model would be only a few millimeters deep and dominated by entirely spurious surfacetension or Weber-number effects. Therefore such hydraulic models commonly violate geometric similarity by “distorting” the vertical scale by a factor of 10 or more. Figure 5.9 shows a hydraulic model of a barrier-beach inlet in South Carolina. The horizontal scale reduction is 1 300, but the vertical scale is only 1 60. Since a deeper channel flows more efficiently, the model channel bottom is deliberately roughened more than the natural channel to correct for the geometric discrepancy. Thus the friction effect of the discrepancy can be corrected, but its effect on, say, dispersion of heat and mass is less well known. EXAMPLE 5.7 | v v The pressure drop due to friction for flow in a long smooth pipe is a function of average flow velocity, density, viscosity, and pipe length and diameter: p fcn(V, , , L, D). We wish to know how p varies with V. (a) Use the pi theorem to rewrite this function in dimensionless form. (b) Then plot this function, using the following data for three pipes and three fluids: | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 5 Dimensional Analysis and Similarity D, cm L, m Q, m3/h 1.0 1.0 1.0 2.0 2.0 2.0 3.0 3.0 3.0 5.0 7.0 9.0 4.0 6.0 8.0 3.0 4.0 5.0 0.3 0.6 1.0 1.0 2.0 3.1 0.5 1.0 1.7 *V Q/A, A †Gasoline. ‡Water. §Mercury. p, Pa , kg/m3 4,680 22,300 70,800 2,080 10,500 30,400 540 2,480 9,600 680† 680† 680† 998‡ 998‡ 998‡ 13,550§ 13,550§ 13,550§ , kg/(m s) 2.92 E-4† 2.92 E-4† 2.92 E-4† 0.0010‡ 0.0010‡ 0.0010‡ 1.56 E-3§ 1.56 E-3§ 1.56 E-3§ V, m/s* 1.06 2.12 3.54 0.88 1.77 2.74 0.20 0.39 0.67 D2/4. (c) Suppose it is further known that p is proportional to L (which is quite true for long pipes with well-rounded entrances). Use this information to simplify and improve the pi-theorem formulation. Plot the dimensionless data in this improved manner and comment upon the results. Solution There are six variables with three primary dimensions involved {MLT}. Therefore we expect that j 6 3 3 pi groups. We are correct, for we can find three variables which do not form a pi product, for example, ( , V, L). Carefully select three ( j) repeating variables, but not including p or V, which we plan to plot versus each other. We select ( , , D), and the pi theorem guarantees that three independent power-product groups will occur: a b Dc 1 or D2 1 d p 2 p e D fV g 3 VD 2 2 h D iL L D 3 We have omitted the algebra of finding (a, b, c, d, e, f, g, h, i) by setting all exponents to zero M 0, L0, T 0. Therefore we wish to plot the dimensionless relation D2 p 2 fcn VD L , D Ans. (a) We plot 1 versus 2 with 3 as a parameter. There will be nine data points. For example, the first row in the data above yields D2 p 2 VD (680)(1.06)(0.01) 2.92 E-4 (680)(0.01)2(4680) (2.92 E-4)2 24,700 L D 3.73 E9 500 The nine data points are plotted as the open circles in Fig. 5.10. The values of L /D are listed for each point, and we see a significant length effect. In fact, if we connect the only two points which have the same L /D ( 200), we could see (and cross-plot to verify) that p increases linearly with L, as stated in the last part of the problem. Since L occurs only in 3 L /D, the function 1 fcn( 2, 3) must reduce to 1 (L /D) fcn( 2), or simply a function involving only two parameters: | v v 308 | e-Text Main Menu | Textbook Table of Contents | Study Guide 5.5 Modeling and Its Pitfalls 309 1011 900 700 L = 200 D Fig. 5.10 Two different correlations of the data in Example 5.7: Open circles when plotting D2 p/ 2 versus ReD, L/D is a parameter; once it is known that p is proportional to L, a replot (solid circles) of D3 p/(L 2) versus ReD collapses into a single power-law curve. 1010 Π1 133 300 500 200 400 100 109 108 Π1 7 Π3 10 0.155 ReD1.75 10 6 10 4 105 ReD D3 L p 2 VD fcn flow in a long pipe Ans. (c) We now modify each data point in Fig. 5.10 by dividing it by its L /D value. For example, for the first row of data, D3 p/(L 2) (3.73 E9)/500 7.46 E6. We replot these new data points as solid circles in Fig. 5.10. They correlate almost perfectly into a straight-line power-law function: D3 L p 0.155 2 VD 1.75 Ans. (c) All newtonian smooth pipe flows should correlate in this manner. This example is a variation of the first completely successful dimensional analysis, pipe-flow friction, performed by Prandtl’s student Paul Blasius, who published a related plot in 1911. For this range of (turbulent-flow) Reynolds numbers, the pressure drop increases approximately as V1.75. EXAMPLE 5.8 The smooth-sphere data plotted in Fig. 5.3a represent dimensionless drag versus dimensionless viscosity, since ( , V, d) were selected as scaling or repeating variables. (a) Replot these data to display the effect of dimensionless velocity on the drag. (b) Use your new figure to predict the terminal (zero-acceleration) velocity of a 1-cm-diameter steel ball (SG 7.86) falling through water at 20C. Solution To display the effect of velocity, we must not use V as a repeating variable. Instead we choose ( , , d) as our j variables to nondimensionalize Eq. (5.1), F fcn(d, V, , ). (See Example 5.2 for an alternate approach to this problem.) The pi groups form as follows: a | v v 1 | e-Text Main Menu | bc dF F 2 Textbook Table of Contents e 2 | fg dV Vd Study Guide Ans. (a) 310 Chapter 5 Dimensional Analysis and Similarity That is, a 1, b 2, c 0, e 1, f 1, and g 1, by using our power-product techniques of Examples 5.2 to 5.6. Therefore a plot of F/ 2 versus Re will display the direct effect of velocity on sphere drag. This replot is shown as Fig. 5.11. The drag increases rapidly with velocity up to transition, where there is a slight drop, after which it increases more quickly than ever. If the force is known, we may predict the velocity from the figure. For water at 20°C, take 998 kg/m3 and 0.001 kg/(m s). For steel, s 3 7.86 water 7840 kg/m . For terminal velocity, the drag equals the net weight of the sphere in water. Thus F ( Wnet w)g s 6 d3 (7840 998)(9.81) (0.01)3 6 0.0351 N Therefore the ordinate of Fig. 5.11 is known: (998 kg/m3)(0.0351 N) [0.001 kg/(m s)]2 F Falling steel sphere: 2 3.5 E7 From Fig. 5.11, at F/ 2 3.5 E7, a magnifying glass reveals that Red estimate of the terminal fall velocity is Vd 20,000 or V 20,000[0.001 kg/(m s)] (998 kg/m3)(0.01 m) 2 E4. Then a crude 2.0 m s 1011 1010 Transition: 109 108 µ2 ρF = π CD Re2 8 107 106 10 5 10 4 103 102 10 1 0.1 | v v Fig. 5.11 Cross-plot of sphere-drag data from Fig. 5.3a to isolate diameter and velocity. | 1 e-Text Main Menu 10 | 10 2 10 3 ρ Vd Re = µ 10 4 10 5 10 6 Textbook Table of Contents | Study Guide Ans. (b) Problems 311 Better accuracy could be obtained by expanding the scale of Fig. 5.11 in the region of the given force coefficient. However, there is considerable uncertainty in published drag data for spheres, so the predicted fall velocity is probably uncertain by at least 5 percent. Note that we found the answer directly from Fig. 5.11. We could use Fig. 5.3a also but would have to iterate between the ordinate and abscissa to obtain the final result, since V is contained in both plotted variables. Summary Chapters 3 and 4 presented integral and differential methods of mathematical analysis of fluid flow. This chapter introduces the third and final method: experimentation, as supplemented by the technique of dimensional analysis. Tests and experiments are used both to strengthen existing theories and to provide useful engineering results when theory is inadequate. The chapter begins with a discussion of some familiar physical relations and how they can be recast in dimensionless form because they satisfy the principle of dimensional homogeneity. A general technique, the pi theorem, is then presented for systematically finding a set of dimensionless parameters by grouping a list of variables which govern any particular physical process. Alternately, direct application of dimensional analysis to the basic equations of fluid mechanics yields the fundamental parameters governing flow patterns: Reynolds number, Froude number, Prandtl number, Mach number, and others. It is shown that model testing in air and water often leads to scaling difficulties for which compromises must be made. Many model tests do not achieve true dynamic similarity. The chapter ends by pointing out that classic dimensionless charts and data can be manipulated and recast to provide direct solutions to problems that would otherwise be quite cumbersome and laboriously iterative. Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems labeled with an EES icon, for example, Prob. 5.61, will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 5.1 to 5.91 (categorized in the problem list below) are followed by word problems W5.1 to W5.10, fundamentals of engineering exam problems FE5.1 to FE5.10, comprehensive applied problems C5.1 to C5.4, and design projects D5.1 and D5.2. Problem distribution 5.1 5.2 5.2 5.3 Topic Problems Introduction Choosing proper scaling parameters The principle of dimensional homogeneity The pi theorem 5.1 – 5.6 5.7 – 5.9 5.10 – 5.17 5.18 – 5.41 | v v Section | e-Text Main Menu | 5.4 5.4 5.5 5.5 5.5 Nondimensionalizing the basic equations Data for spheres and cylinders Scaling of model data Froude- and Mach-number scaling Inventive rescaling of the data 5.42 – 5.47 5.48 – 5.57 5.58 – 5.74 5.75 – 5.84 5.85 – 5.91 P5.1 For axial flow through a circular tube, the Reynolds number for transition to turbulence is approximately 2300 [see Eq. (6.2)], based upon the diameter and average velocity. If d 5 cm and the fluid is kerosine at 20°C, find the volume flow rate in m3/h which causes transition. P5.2 In flow past a thin flat body such as an airfoil, transition to turbulence occurs at about Re 1 E6, based on the distance x from the leading edge of the wing. If an airplane flies at 450 mi/h at 8-km standard altitude and undergoes transition at the 12 percent chord position, how long is its chord (wing length from leading to trailing edge)? Textbook Table of Contents | Study Guide 312 Chapter 5 Dimensional Analysis and Similarity P5.3 An airplane has a chord length L 1.2 m and flies at a Mach number of 0.7 in the standard atmosphere. If its Reynolds number, based on chord length, is 7 E6, how high is it flying? P5.4 When tested in water at 20°C flowing at 2 m/s, an 8-cm-diameter sphere has a measured drag of 5 N. What will be the velocity and drag force on a 1.5-m-diameter weather balloon moored in sea-level standard air under dynamically similar conditions? P5.5 An automobile has a characteristic length and area of 8 ft and 60 ft2, respectively. When tested in sea-level standard air, it has the following measured drag force versus speed: V, mi/h 20 40 31 115 P5.12 P5.13 60 Drag, lbf P5.11 249 The same car travels in Colorado at 65 mi/h at an altitude of 3500 m. Using dimensional analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag. *P5.6 SAE 10 oil at 20°C flows past an 8-cm-diameter sphere. At flow velocities of 1, 2, and 3 m/s, the measured sphere drag forces are 1.5, 5.3, and 11.2 N, respectively. Estimate the drag force if the same sphere is tested at a velocity of 15 m/s in glycerin at 20°C. P5.7 A body is dropped on the moon (g 1.62 m/s2) with an initial velocity of 12 m/s. By using option 2 variables, Eq. (5.11), the ground impact occurs at t** 0.34 and S** 0.84. Estimate (a) the initial displacement, (b) the final displacement, and (c) the time of impact. P5.8 The Bernoulli equation (5.6) can be written in the form p p0 1 2 V2 gz (1) where p0 is the “stagnation” pressure at zero velocity and elevation. (a) State how many scaling variables are needed to nondimensionalize this equation. (b) Suppose that we wish to nondimensionalize Eq. (1) in order to plot dimensionless pressure versus velocity, with elevation as a parameter. Select the proper scaling variables and carry out and plot the resulting dimensionless relation. P5.9 Modify Prob. 5.8 as follows. Suppose that we wish to nondimensionalize Eq. (1) in order to plot dimensionless pressure versus gravity, with velocity as a parameter. Select the proper scaling variables and carry out and plot the resulting dimensionless relation. P5.10 Determine the dimension {MLT } of the following quantities: (d) u x 2 2 (b) (p 1 p0) dA (c) cp T xy u dx dy dz t | v v (a) u | e-Text Main Menu | P5.14 P5.15 P5.16 All quantities have their standard meanings; for example, is density. For a particle moving in a circle, its centripetal acceleration takes the form a fcn(V, R), where V is its velocity and R the radius of its path. By pure dimensional reasoning, rewrite this function in algebraic form. The velocity of sound a of a gas varies with pressure p and density . Show by dimensional reasoning that the proper form must be a (const)(p/ )1/2. The speed of propagation C of a capillary wave in deep water is known to be a function only of density , wavelength , and surface tension . Find the proper functional relationship, completing it with a dimensionless constant. For a given density and wavelength, how does the propagation speed change if the surface tension is doubled? Consider laminar flow over a flat plate. The boundary layer thickness grows with distance x down the plate and is also a function of free-stream velocity U, fluid viscosity , and fluid density . Find the dimensionless parameters for this problem, being sure to rearrange if neessary to agree with the standard dimensionless groups in fluid mechanics, as given in Table 5.2. It is desired to measure the drag on an airplane whose velocity is 300 mi/h. Is it feasible to test a one-twentieth-scale model of the plane in a wind tunnel at the same pressure and temperature to determine the prototype drag coefficient? Convection heat-transfer data are often reported as a heattransfer coefficient h, defined by . Q hA T . where Q heat flow, J/s A surface area, m2 T temperature difference, K The dimensionless form of h, called the Stanton number, is a combination of h, fluid density , specific heat cp, and flow velocity V. Derive the Stanton number if it is proportional to h. P5.17 In some heat-transfer textbooks, e.g., J. P. Holman, Heat Transfer, 5th ed., McGraw-Hill, 1981, p. 285, simplified formulas are given for the heat-transfer coefficient from Prob. 5.16 for buoyant or natural convection over hot surfaces. An example formula is h 1.42 T L 1/4 where L is the length of the hot surface. Comment on the dimensional homogeneity of this formula. What might be the SI units of constants 1.42 and 1 ? What parameters might 4 be missing or hidden? P5.18 Under laminar conditions, the volume flow Q through a small triangular-section pore of side length b and length L Textbook Table of Contents | Study Guide Problems 313 is a function of viscosity , pressure drop per unit length p/L, and b. Using the pi theorem, rewrite this relation in dimensionless form. How does the volume flow change if the pore size b is doubled? P5.19 The period of oscillation T of a water surface wave is assumed to be a function of density , wavelength , depth h, gravity g, and surface tension . Rewrite this relationship in dimensionless form. What results if is negligible? Hint: Take , , and g as repeating variables. P5.20 The power input P to a centrifugal pump is assumed to be a function of the volume flow Q, impeller diameter D, rotational rate , and the density and viscosity of the fluid. Rewrite this as a dimensionless relationship. Hint: Take , , and D as repeating variables. P5.21 In Example 5.1 we used the pi theorem to develop Eq. (5.2) from Eq. (5.1). Instead of merely listing the primary dimensions of each variable, some workers list the powers of each primary dimension for each variable in an array: F 1 1 2 M L T L 0 1 0 U 0 1 1 1 3 0 v v | P5.27 P5.28 P5.29 H ,N L Viscosity effects are negligible. Find appropriate pi groups for this problem and rewrite the function above in dimensionless form. P5.23 The period T of vibration of a beam is a function of its length L, area moment of inertia I, modulus of elasticity E, density , and Poisson’s ratio . Rewrite this relation in dimensionless form. What further reduction can we make if E and I can occur only in the product form EI? Hint: Take L, , and E as repeating variables. P5.24 The lift force F on a missile is a function of its length L, velocity V, diameter D, angle of attack , density , viscosity , and speed of sound a of the air. Write out the dimensional matrix of this function and determine its rank. | P5.26 1 1 1 This array of exponents is called the dimensional matrix for the given function. Show that the rank of this matrix (the size of the largest nonzero determinant) is equal to j n k, the desired reduction between original variables and the pi groups. This is a general property of dimensional matrices, as noted by Buckingham [29]. P5.22 When freewheeling, the angular velocity of a windmill is found to be a function of the windmill diameter D, the wind velocity V, the air density , the windmill height H as compared to the atmospheric boundary layer height L, and the number of blades N: fcn D, V, , P5.25 e-Text Main Menu | P5.30 P5.31 (See Prob. 5.21 for an explanation of this concept.) Rewrite the function in terms of pi groups. When a viscous fluid is confined between two long concentric cylinders as in Fig. 4.17, the torque per unit length T required to turn the inner cylinder at angular velocity is a function of , cylinder radii a and b, and viscosity . Find the equivalent dimensionless function. What happens to the torque if both a and b are doubled? A pendulum has an oscillation period T which is assumed to depend upon its length L, bob mass m, angle of swing , and the acceleration of gravity. A pendulum 1 m long, with a bob mass of 200 g, is tested on earth and found to have a period of 2.04 s when swinging at 20°. (a) What is its period when it swings at 45°? A similarly constructed pendulum, with L 30 cm and m 100 g, is to swing on the moon (g 1.62 m/s2) at 20°. (b) What will be its period? In studying sand transport by ocean waves, A. Shields in 1936 postulated that the threshold wave-induced bottom shear stress required to move particles depends upon gravity g, particle size d and density p, and water density and viscosity . Find suitable dimensionless groups of this problem, which resulted in 1936 in the celebrated Shields sandtransport diagram. A simply supported beam of diameter D, length L, and modulus of elasticity E is subjected to a fluid crossflow of velocity V, density , and viscosity . Its center deflection is assumed to be a function of all these variables. (a) Rewrite this proposed function in dimensionless form. (b) Suppose it is known that is independent of , inversely proportional to E, and dependent only upon V 2, not and V separately. Simplify the dimensionless function accordingly. Hint: Take L, , and V as repeating variables. When fluid in a pipe is accelerated linearly from rest, it begins as laminar flow and then undergoes transition to turbulence at a time ttr which depends upon the pipe diameter D, fluid acceleration a, density , and viscosity . Arrange this into a dimensionless relation between ttr and D. In forced convection, the heat-transfer coefficient h, as defined in Prob. 5.16, is known to be a function of stream velocity U, body size L, and fluid properties , , cp, and k. Rewrite this function in dimensionless form, and note by name any parameters you recognize. Hint: Take L, , k, and as repeating variables. The heat-transfer rate per unit area q to a body from a fluid in natural or gravitational convection is a function of the temperature difference T, gravity g, body length L, and three fluid properties: kinematic viscosity , conductivity k, and thermal expansion coefficient . Rewrite in dimensionless form if it is known that g and appear only as the product g . Textbook Table of Contents | Study Guide 314 Chapter 5 Dimensional Analysis and Similarity P5.32 A weir is an obstruction in a channel flow which can be calibrated to measure the flow rate, as in Fig. P5.32. The volume flow Q varies with gravity g, weir width b into the paper, and upstream water height H above the weir crest. If it is known that Q is proportional to b, use the pi theorem to find a unique functional relationship Q(g, b, H). P5.37 H Q P5.38 Weir P5.39 P5.32 EES P5.33 A spar buoy (see Prob. 2.113) has a period T of vertical (heave) oscillation which depends upon the waterline cross-sectional area A, buoy mass m, and fluid specific weight . How does the period change due to doubling of (a) the mass and (b) the area? Instrument buoys should have long periods to avoid wave resonance. Sketch a possible long-period buoy design. P5.34 To good approximation, the thermal conductivity k of a gas (see Ref. 8 of Chap. 1) depends only upon the density , mean free path , gas constant R, and absolute temperature T. For air at 20°C and 1 atm, k 0.026 W/(m K) and 6.5 E-8 m. Use this information to determine k for hydrogen at 20°C and 1 atm if 1.2 E-7 m. P5.35 The torque M required to turn the cone-plate viscometer in Fig. P5.35 depends upon the radius R, rotation rate , fluid viscosity , and cone angle . Rewrite this relation in dimensionless form. How does the relation simplify it if it is known that M is proportional to ? P5.40 P5.41 P5.42 Ω R θ P5.43 S t θ Fluid P5.35 v v . P5.36 The rate of heat loss, Qloss through a window or wall is a function of the temperature difference between inside and outside T, the window surface area A, and the R value of the window which has units of (ft2 h °F)/Btu. (a) Using | | e-Text Main Menu Buckingham pi theorem, find an expression for rate of heat loss as a function of the other three parameters in the problem. (b) If the temperature difference T doubles, by what factor does the rate of heat loss increase? The pressure difference p across an explosion or blast wave is a function of the distance r from the blast center, time t, speed of sound a of the medium, and total energy E in the blast. Rewrite this relation in dimensionless form (see Ref. 18, chap. 4, for further details of blast-wave scaling). How does p change if E is doubled? The size d of droplets produced by a liquid spray nozzle is thought to depend upon the nozzle diameter D, jet velocity U, and the properties of the liquid , , and . Rewrite this relation in dimensionless form. Hint: Take D, , and U as repeating variables. In turbulent flow past a flat surface, the velocity u near the wall varies approximately logarithmically with distance y from the wall and also depends upon viscosity , density , and wall shear stress w. For a certain airflow at 20°C and 1 atm, w 0.8 Pa and u 15 m/s at y 3.6 mm. Use this information to estimate the velocity u at y 6 mm. Reconsider the slanted-plate surface tension problem (see Fig. C1.1) as an exercise in dimensional analysis. Let the capillary rise h be a function only of fluid properties, gravity, bottom width, and the two angles in Fig. C1.1. That is, h fcn( , , g, L, , ). (a) Use the pi theorem to rewrite this function in terms of dimensionless parameters. (b) Verify that the exact solution from Prob. C1.1 is consistent with your result in part (a). A certain axial-flow turbine has an output torque M which is proportional to the volume flow rate Q and also depends upon the density , rotor diameter D, and rotation rate . How does the torque change due to a doubling of (a) D and (b) ? Nondimensionalize the energy equation (4.75) and its boundary conditions (4.62), (4.63), and (4.70) by defining T* T/T0, where T0 is the inlet temperature, assumed constant. Use other dimensionless variables as needed from Eqs. (5.23). Isolate all dimensionless parameters you find, and relate them to the list given in Table 5.2. The differential equation of salt conservation for flowing seawater is | u S x S y w S z 2 S x2 2 S y2 2 S z2 where is a (constant) coefficient of diffusion, with typical units of square meters per second, and S is the salinity in parts per thousand. Nondimensionalize this equation and discuss any parameters which appear. P5.44 The differential energy equation for incompressible two-dimensional flow through a “Darcy-type” porous medium is approximately Textbook Table of Contents | Study Guide Problems 315 p x cp T x 2 pT yy cp k T y2 0 where is the permeability of the porous medium. All other symbols have their usual meanings. (a) What are the appropriate dimensions for ? (b) Nondimensionalize this equation, using (L, U, , T0) as scaling constants, and discuss any dimensionless parameters which arise. P5.45 In natural-convection problems, the variation of density due to the temperature difference T creates an important buoyancy term in the momentum equation (5.30). To first-order accuracy, the density variation would be T), 0(1 where is the thermal-expansion coefficient. The momentum equation thus becomes 0 dV dt (p 0gz) 2 T gk 0 V where we have assumed that z is up. Nondimensionalize this equation, using Eqs. (5.23), and relate the parameters you find to the list in Table 5.2. P5.46 The differential equation for compressible inviscid flow of a gas in the xy plane is 2 2 2 t t (u2 2 (u2 ) a2) x2 2 ( 2 a2) 2 2 y 2u xy 0 where is the velocity potential and a is the (variable) speed of sound of the gas. Nondimensionalize this relation, using a reference length L and the inlet speed of sound a0 as parameters for defining dimensionless variables. P5.47 The differential equation for small-amplitude vibrations y(x, t) of a simple beam is given by 2 A y 2 t 4 EI y x4 0 beam material density cross-sectional area area moment of inertia Young’s modulus where A I E | v v Use only the quantities , E, and A to nondimensionalize y, x, and t, and rewrite the differential equation in dimensionless form. Do any parameters remain? Could they be removed by further manipulation of the variables? P5.48 A smooth steel (SG 7.86) sphere is immersed in a stream of ethanol at 20°C moving at 1.5 m/s. Estimate its drag in N from Fig. 5.3a. What stream velocity would quadruple its drag? Take D 2.5 cm. P5.49 The sphere in Prob. 5.48 is dropped in gasoline at 20°C. Ignoring its acceleration phase, what will its terminal (constant) fall velocity be, from Fig. 5.3a? | e-Text Main Menu | P5.50 When a micro-organism moves in a viscous fluid, it turns out that fluid density has nearly negligible influence on the drag force felt by the micro-organism. Such flows are called creeping flows. The only important parameters in the problem are the velocity of motion U, the viscosity of the fluid , and the length scale of the body. Here assume the micro-organism’s body diameter d as the appropriate length scale. (a) Using the Buckingham pi theorem, generate an expression for the drag force D as a function of the other parameters in the problem. (b) The drag coefficient discussed in this chapter CD D/( 1 U2A) is not appropriate 2 for this kind of flow. Define instead a more appropriate drag coefficient, and call it Cc (for creeping flow). (c) For a spherically shaped micro-organism, the drag force can be calculated exactly from the equations of motion for creeping flow. The result is D 3 Ud. Write expressions for both forms of the drag coefficient, Cc and CD, for a sphere under conditions of creeping flow. P5.51 A ship is towing a sonar array which approximates a submerged cylinder 1 ft in diameter and 30 ft long with its axis normal to the direction of tow. If the tow speed is 12 kn (1 kn 1.69 ft/s), estimate the horsepower required to tow this cylinder. What will be the frequency of vortices shed from the cylinder? Use Figs. 5.2 and 5.3. P5.52 A 1-in-diameter telephone wire is mounted in air at 20°C and has a natural vibration frequency of 12 Hz. What wind velocity in ft/s will cause the wire to sing? At this condition what will the average drag force per unit wire length be? P5.53 Vortex shedding can be used to design a vortex flowmeter (Fig. 6.32). A blunt rod stretched across the pipe sheds vortices whose frequency is read by the sensor downstream. Suppose the pipe diameter is 5 cm and the rod is a cylinder of diameter 8 mm. If the sensor reads 5400 counts per minute, estimate the volume flow rate of water in m3/h. How might the meter react to other liquids? P5.54 A fishnet is made of 1-mm-diameter strings knotted into 2 2 cm squares. Estimate the horsepower required to tow 300 ft2 of this netting at 3 kn in seawater at 20°C. The net plane is normal to the flow direction. P5.55 The radio antenna on a car begins to vibrate wildly at 500 Hz when the car is driven at 55 mi/h. Estimate the diameter of the antenna. P5.56 A wooden flagpole, of diameter 5 in and height 30 ft, fractures at its base in hurricane winds at sea level. If the fracture stress is 3500 lbf/in2, estimate the wind velocity in mi/h. P5.57 The simply supported 1040 carbon-steel rod of Fig. P5.57 is subjected to a crossflow stream of air at 20°C and 1 atm. For what stream velocity U will the rod center deflection be approximately 1 cm? Textbook Table of Contents | Study Guide 316 Chapter 5 Dimensional Analysis and Similarity tion speed for which the power will not exceed 300 W? What will the pressure rise be for this condition? *P5.63 The pressure drop per unit length p/L in smooth pipe flow is known to be a function only of the average velocity V, diameter D, and fluid properties and . The following data were obtained for flow of water at 20°C in an 8-cm-diameter pipe 50 m long: D = 1 cm, L = 60 cm δ = 1 cm? U Q, m3/s P5.58 For the steel rod of Prob. 5.57, at what airstream velocity U will the rod begin to vibrate laterally in resonance in its EES first mode (a half sine wave)? Hint: Consult a vibration text under “lateral beam vibration.” P5.59 We wish to know the drag of a blimp which will move in 20°C air at 6 m/s. If a one-thirtieth-scale model is tested in water at 20°C, what should the water velocity be? At this velocity, if the measured water drag on the model is 2700 N, what is the drag on the prototype blimp and the power required to propel it? P5.60 A prototype water pump has an impeller diameter of 2 ft and is designed to pump 12 ft3/s at 750 r/min. A 1-ft-diameter model pump is tested in 20°C air at 1800 r/min, and Reynolds-number effects are found to be negligible. For similar conditions, what will the volume flow of the model be in ft3/s? If the model pump requires 0.082 hp to drive it, what horsepower is required for the prototype? P5.61 If viscosity is neglected, typical pump-flow results from Prob. 5.20 are shown in Fig. P5.61 for a model pump tested EES in water. The pressure rise decreases and the power required increases with the dimensionless flow coefficient. Curve-fit expressions are given for the data. Suppose a similar pump of 12-cm diameter is built to move gasoline at 20°C and a flow rate of 25 m3/h. If the pump rotation speed is 30 r/s, find (a) the pressure rise and (b) the power required. Pressur e ris e P ≈ 0.5 + 3Q ρ Ω 3D 5 ΩD3 r we Po ∆p Q ≈ 6.0 – 120 ρ Ω 2D 2 ΩD3 (( 2 Pump data (Ω in r/s) 0.005 0.01 0.015 0.020 p, Pa P5.57 5800 20,300 42,100 70,800 Verify that these data are slightly outside the range of Fig. 5.10. What is a suitable power-law curve fit for the present data? Use these data to estimate the pressure drop for flow of kerosine at 20°C in a smooth pipe of diameter 5 cm and length 200 m if the flow rate is 50 m3/h. P5.64 The natural frequency of vibration of a mass M attached to a rod, as in Fig. P5.64, depends only upon M and the stiffness EI and length L of the rod. Tests with a 2-kg mass attached to a 1040 carbon-steel rod of diameter 12 mm and length 40 cm reveal a natural frequency of 0.9 Hz. Use these data to predict the natural frequency of a 1-kg mass attached to a 2024 aluminum-alloy rod of the same size. ω M L Stiffness EI P5.64 P5.65 In turbulent flow near a flat wall, the local velocity u varies only with distance y from the wall, wall shear stress w, and fluid properties and . The following data were taken in the University of Rhode Island wind tunnel for airflow, 0.0023 slug/ft3, 3.81 E-7 slug/(ft s), and w 0.029 lbf/ft2: Q = flow coefficient ΩD3 P5.61 | v v *P5.62 Modify Prob. 5.61 so that the rotation speed is unknown but D 12 cm and Q 25 m3/h. What is the maximum rota- | e-Text Main Menu | 0.021 0.035 0.055 0.080 0.12 0.16 u, ft/s 0 y, in 50.6 54.2 57.6 59.7 63.5 65.9 (a) Plot these data in the form of dimensionless u versus dimensionless y, and suggest a suitable power-law curve fit. (b) Suppose that the tunnel speed is increased until u 90 ft/s at y 0.11 in. Estimate the new wall shear stress, in lbf/ft2. Textbook Table of Contents | Study Guide Problems 317 | v v Use these data to predict the drag force of a similar 15-in P5.66 A torpedo 8 m below the surface in 20°C seawater cavitates diamond placed at similar orientation in 20°C water flowat a speed of 21 m/s when atmospheric pressure is 101 kPa. ing at 2.2 m/s. If Reynolds-number and Froude-number effects are negligible, at what speed will it cavitate when running at a depth P5.71 The pressure drop in a venturi meter (Fig. P3.165) varies only with the fluid density, pipe approach velocity, and diof 20 m? At what depth should it be to avoid cavitation at ameter ratio of the meter. A model venturi meter tested in 30 m/s? water at 20°C shows a 5-kPa drop when the approach veP5.67 A student needs to measure the drag on a prototype of charlocity is 4 m/s. A geometrically similar prototype meter is acteristic dimension dp moving at velocity Up in air at stanused to measure gasoline at 20°C and a flow rate of dard atmospheric conditions. He constructs a model of char9 m3/min. If the prototype pressure gage is most accurate acteristic dimension dm, such that the ratio dp/dm is some at 15 kPa, what should the upstream pipe diameter be? factor f. He then measures the drag on the model at dynamically similar conditions (also with air at standard at- P5.72 A one-fifteenth-scale model of a parachute has a drag of 450 lbf when tested at 20 ft/s in a water tunnel. If Reynoldsmospheric conditions). The student claims that the drag number effects are negligible, estimate the terminal fall veforce on the prototype will be identical to that measured on locity at 5000-ft standard altitude of a parachutist using the the model. Is this claim correct? Explain. prototype if chute and chutist together weigh 160 lbf. NeP5.68 Consider flow over a very small object in a viscous fluid. glect the drag coefficient of the woman. Analysis of the equations of motion shows that the inertial terms are much smaller than the viscous and pressure P5.73 The yawing moment on a torpedo control surface is tested on a one-eighth-scale model in a water tunnel at 20 m/s, usterms. It turns out, then, that fluid density drops out of ing Reynolds scaling. If the model measured moment is the equations of motion. Such flows are called creeping 14 N m, what will the prototype moment be under similar flows. The only important parameters in the problem are conditions? the velocity of motion U, the viscosity of the fluid , and the length scale of the body. For three-dimensional bod- P5.74 A one-tenth-scale model of a supersonic wing tested at 700 m/s in air at 20°C and 1 atm shows a pitching moment of ies, like spheres, creeping flow analysis yields very good 0.25 kN m. If Reynolds-number effects are negligible, what results. It is uncertain, however, if such analysis can be will the pitching moment of the prototype wing be if it is applied to two-dimensional bodies such as a circular flying at the same Mach number at 8-km standard altitude? cylinder, since even though the diameter may be very small, the length of the cylinder is infinite for a two-di- P5.75 A one-twelfth-scale model of an airplane is to be tested at 20°C in a pressurized wind tunnel. The prototype is to fly mensional flow. Let us see if dimensional analysis can at 240 m/s at 10-km standard altitude. What should the tunhelp. (a) Using the Buckingham pi theorem, generate an nel pressure be in atm to scale both the Mach number and expression for the two-dimensional drag D2-D as a functhe Reynolds number accurately? tion of the other parameters in the problem. Use cylinder . diameter d as the appropriate length scale. Be careful — *P5.76 A 2-ft-long model of a ship is tested in a freshwater tow tank. The measured drag may be split into “friction” drag the two-dimensional drag has dimensions of force per unit (Reynolds scaling) and “wave” drag (Froude scaling). The length rather than simply force. (b) Is your result physimodel data are as follows: cally plausible? If not, explain why not. (c) It turns out that fluid density cannot be neglected in analysis of 0.8 1.6 2.4 3.2 4.0 4.8 creeping flow over two-dimensional bodies. Repeat the Tow speed, ft/s dimensional analysis, this time with included as a pa- Friction drag, lbf 0.016 0.057 0.122 0.208 0.315 0.441 rameter. Find the nondimensional relationship between Wave drag, lbf 0.002 0.021 0.083 0.253 0.509 0.697 the parameters in this problem. P5.69 A one-sixteenth-scale model of a weir (see Fig. P5.32) has The prototype ship is 150 ft long. Estimate its total drag a measured flow rate Q 2.1 ft3/s when the upstream wawhen cruising at 15 kn in seawater at 20°C. ter height is H 6.3 in. If Q is proportional to weir width P5.77 A dam spillway is to be tested by using Froude scaling with b, predict the prototype flow rate when Hproto 3.2 ft. a one-thirtieth-scale model. The model flow has an average P5.70 A diamond-shaped body, of characteristic length 9 in, has velocity of 0.6 m/s and a volume flow of 0.05 m3/s. What the following measured drag forces when placed in a wind will the velocity and flow of the prototype be? If the meaEES tunnel at sea-level standard conditions: sured force on a certain part of the model is 1.5 N, what will the corresponding force on the prototype be? V, ft/s 30 38 48 56 61 P5.78 A prototype spillway has a characteristic velocity of 3 m/s F, lbf 1.25 1.95 3.02 4.05 4.81 and a characteristic length of 10 m. A small model is con- | e-Text Main Menu | Textbook Table of Contents | Study Guide 318 P5.79 P5.80 P5.81 P5.82 P5.83 P5.84 Chapter 5 Dimensional Analysis and Similarity structed by using Froude scaling. What is the minimum scale ratio of the model which will ensure that its minimum Weber number is 100? Both flows use water at 20°C. An East Coast estuary has a tidal period of 12.42 h (the semidiurnal lunar tide) and tidal currents of approximately 80 cm/s. If a one-five-hundredth-scale model is constructed with tides driven by a pump and storage apparatus, what should the period of the model tides be and what model current speeds are expected? A prototype ship is 35 m long and designed to cruise at 11 m/s (about 21 kn). Its drag is to be simulated by a 1-mlong model pulled in a tow tank. For Froude scaling find (a) the tow speed, (b) the ratio of prototype to model drag, and (c) the ratio of prototype to model power. An airplane, of overall length 55 ft, is designed to fly at 680 m/s at 8000-m standard altitude. A one-thirtieth-scale model is to be tested in a pressurized helium wind tunnel at 20°C. What is the appropriate tunnel pressure in atm? Even at this (high) pressure, exact dynamic similarity is not achieved. Why? A prototype ship is 400 ft long and has a wetted area of 30,000 ft2. A one-eightieth-scale model is tested in a tow tank according to Froude scaling at speeds of 1.3, 2.0, and 2.7 kn (1 kn 1.689 ft/s). The measured friction drag of the model at these speeds is 0.11, 0.24, and 0.41 lbf, respectively. What are the three prototype speeds? What is the estimated prototype friction drag at these speeds if we correct for Reynolds-number discrepancy by extrapolation? A one-fortieth-scale model of a ship’s propeller is tested in a tow tank at 1200 r/min and exhibits a power output of 1.4 ft lbf/s. According to Froude scaling laws, what should the revolutions per minute and horsepower output of the prototype propeller be under dynamically similar conditions? A prototype ocean-platform piling is expected to encounter currents of 150 cm/s and waves of 12-s period and 3-m P5.85 P5.86 P5.87 P5.88 P5.89 P5.90 P5.91 height. If a one-fifteenth-scale model is tested in a wave channel, what current speed, wave period, and wave height should be encountered by the model? Solve Prob. 5.49, using the modified sphere-drag plot of Fig. 5.11. Solve Prob. 5.49 for glycerin at 20°C, using the modified sphere-drag plot of Fig. 5.11. In Prob. 5.62 it was difficult to solve for because it appeared in both power and flow coefficients. Rescale the problem, using the data of Fig. P5.61, to make a plot of dimensionless power versus dimensionless rotation speed. Enter this plot directly to solve Prob. 5.62 for . Modify Prob. 5.62 as follows: Let 32 r/s and Q 24 m3/h for a geometrically similar pump. What is the maximum diameter if the power is not to exceed 340 W? Solve this problem by rescaling the data of Fig. P5.61 to make a plot of dimensionless power versus dimensionless diameter. Enter this plot directly to find the desired diameter. Knowing that p is proportional to L, rescale the data of Example 5.7 to plot dimensionless p versus dimensionless diameter. Use this plot to find the diameter required in the first row of data in Example 5.7 if the pressure drop is increased to 10 kPa for the same flow rate, length, and fluid. Knowing that p is proportional to L, rescale the data of Example 5.7 to plot dimensionless p versus dimensionless viscosity. Use this plot to find the viscosity required in the first row of data in Example 5.7 if the pressure drop is increased to 10 kPa for the same flow rate, length, and density. Develop a plot of dimensionless p versus dimensionless viscosity, as described in Prob. 5.90. Suppose that L 200 m, Q 60 m3/h, and the fluid is kerosine at 20°C. Use your plot to determine the minimum pipe diameter for which the pressure drop is no more than 220 kPa. Word Problems W5.2 W5.3 W5.4 In 98 percent of data analysis cases, the “reducing factor” j, which lowers the number n of dimensional variables to n j dimensionless groups, exactly equals the number of relevant dimensions (M, L, T, ). In one case (Example 5.5) this was not so. Explain in words why this situation happens. Consider the following equation: 1 dollar bill 6 in. Is this relation dimensionally inconsistent? Does it satisfy the PDH? Why? In making a dimensional analysis, what rules do you follow for choosing your scaling variables? In an earlier edition, the writer asked the following question about Fig. 5.1: “Which of the three graphs is a more effective presentation?” Why was this a dumb question? | v v W5.1 | e-Text Main Menu | W5.5 W5.6 W5.7 This chapter discusses the difficulty of scaling Mach and Reynolds numbers together (an airplane) and Froude and Reynolds numbers together (a ship). Give an example of a flow which would combine Mach and Froude numbers. Would there be scaling problems for common fluids? What is different about a very small model of a weir or dam (Fig. P5.32) which would make the test results difficult to relate to the prototype? What else are you studying this term? Give an example of a popular equation or formula from another course (thermodynamics, strength of materials, etc.) which does not satisfy the principle of dimensional homogeneity. Explain what is wrong and whether it can be modified to be homogeneous. Textbook Table of Contents | Study Guide Comprehensive Problems W5.8 W5.9 Some colleges (e.g., Colorado State University) have environmental wind tunnels which can be used to study, e.g., wind flow over city buildings. What details of scaling might be important in such studies? If the model scale ratio is Lm / Lp, as in Eq. (5.31), and the Weber number is important, how must the model 319 and prototype surface tension be related to for dynamic similarity? W5.10 For a typical incompressible velocity potential analysis in Chap. 4 we solve 2 0, subject to known values of / n on the boundaries. What dimensionless parameters govern this type of motion? Fundamentals of Engineering Exam Problems FE5.1 FE5.2 FE5.3 FE5.4 FE5.5 Given the parameters (U, L, g, , ) which affect a certain liquid flow problem, the ratio V 2/(Lg) is usually known as the (a) velocity head, (b) Bernoulli head, (c) Froude number, (d ) kinetic energy, (e) impact energy A ship 150 m long, designed to cruise at 18 kn, is to be tested in a tow tank with a model 3 m long. The appropriate tow velocity is (a) 0.19 m/s, (b) 0.35 m/s, (c) 1.31 m/s, (d ) 2.55 m/s, (e) 8.35 m/s A ship 150 m long, designed to cruise at 18 kn, is to be tested in a tow tank with a model 3 m long. If the model wave drag is 2.2 N, the estimated full-size ship wave drag is (a) 5500 N, (b) 8700 N, (c) 38,900 N, (d ) 61,800 N, (e) 275,000 N A tidal estuary is dominated by the semidiurnal lunar tide, with a period of 12.42 h. If a 1 500 model of the estuary is tested, what should be the model tidal period? (a) 4.0 s, (b) 1.5 min, (c) 17 min, (d ) 33 min, (e) 64 min A football, meant to be thrown at 60 mi/h in sea-level air ( 1.22 kg /m3, 1.78 E-5 N s /m2), is to be tested using a one-quarter scale model in a water tunnel ( 998 kg/m3, 0.0010 N s/m2). For dynamic similarity, what is the proper model water velocity? (a) 7.5 mi/h, (b) 15.0 mi/h, (c) 15.6 mi/h, (d ) 16.5 mi/h, (e) 30 mi/h FE5.6 A football, meant to be thrown at 60 mi/h in sea-level air ( 1.22 kg/m3, 1.78 E-5 N m2), is to be tested using a one-quarter scale model in a water tunnel ( 998 kg/m3, 0.0010 N s/m2). For dynamic similarity, what is the ratio of prototype force to model force? (a) 3.86 1, (b) 16 1, (c) 32 1, (d ) 56 1, (e) 64 1 FE5.7 Consider liquid flow of density , viscosity , and velocity U over a very small model spillway of length scale L, such that the liquid surface tension coefficient is important. The quantity U2L / in this case is important and is called the (a) capillary rise, (b) Froude number, (c) Prandtl number, (d ) Weber number, (e) Bond number FE5.8 If a stream flowing at velocity U past a body of length L causes a force F on the body which depends only upon U, L, and fluid viscosity , then F must be proportional to (a) UL / , (b) U 2L2, (c) U/L, (d ) UL, (e) UL/ FE5.9 In supersonic wind tunnel testing, if different gases are used, dynamic similarity requires that the model and prototype have the same Mach number and the same (a) Euler number, (b) speed of sound, (c) stagnation enthalpy, (d) Froude number, (e) specific heat ratio FE5.10 The Reynolds number for a 1-ft-diameter sphere moving at 2.3 mi/h through seawater (specific gravity 1.027, viscosity 1.07 E-3 N s/m2) is approximately (a) 300, (b) 3000, (c) 30,000, (d) 300,000, (e) 3,000,000 Comprehensive Problems Estimating pipe wall friction is one of the most common tasks in fluids engineering. For long circular rough pipes in turbulent flow, wall shear w is a function of density , viscosity , average velocity V, pipe diameter d, and wall roughness height . Thus, functionally, we can write w fcn( , , V, d, ). (a) Using dimensional analysis, rewrite this function in dimensionless form. (b) A certain pipe has d 5 cm and 0.25 mm. For flow of water at 20°C, | v v C5.1 | e-Text Main Menu | measurements show the following values of wall shear stress: Q, gal/min w, Pa 1.5 3.0 6.0 9.0 12.0 14.0 0.05 0.18 0.37 0.64 0.86 1.25 Plot these data using the dimensionless form obtained in part (a) and suggest a curve-fit formula. Does your plot reveal the entire functional relation obtained in part (a)? Textbook Table of Contents | Study Guide 320 C5.2 Chapter 5 Dimensional Analysis and Similarity When the fluid exiting a nozzle, as in Fig. P3.49, is a gas, instead of water, compressibility may be important, especially if upstream pressure p1 is large and exit diameter d2 is small. In this case, the difference p1 p2 is no longer controlling, and the gas mass flow m reaches a maximum ˙ value which depends upon p1 and d2 and also upon the absolute upstream temperature T1 and the gas constant R. Thus, functionally, m fcn(p1, d2, T1, R). (a) Using dimensional ˙ analysis, rewrite this function in dimensionless form. (b) A certain pipe has d2 1 mm. For flow of air, measurements show the following values of mass flow through the nozzle: analysis. Let the vertical velocity be a function only of distance from the plate, fluid properties, gravity, and film thickness. That is, w fcn(x, , , g, ). (a) Use the pi theorem to rewrite this function in terms of dimensionless parameters. (b) Verify that the exact solution from Prob. 4.80 is consistent with your result in part (a). C5.4 The Taco Inc. model 4013 centrifugal pump has an impeller of diameter D 12.95 in. When pumping 20°C water at 1160 r/min, the measured flow rate Q and pressure rise p are given by the manufacturer as follows: Q, gal/min T1, K 300 300 500 800 p1, kPa 200 250 300 300 0.037 0.046 0.055 0.043 0.034 200 300 400 500 600 700 36 35 34 32 29 23 300 ˙ m, kg/s C5.3 300 2 p, lb/in Plot these data in the dimensionless form obtained in part (a). Does your plot reveal the entire functional relation obtained in part (a)? Reconsider the fully developed draining vertical oil-film problem (see Fig. P4.80) as an exercise in dimensional (a) Assuming that p fcn( , Q, D, ), use the pi theorem to rewrite this function in terms of dimensionless parameters and then plot the given data in dimensionless form. (b) It is desired to use the same pump, running at 900 r/min, to pump 20°C gasoline at 400 gal/min. According to your dimensionless correlation, what pressure rise p is expected, in lbf/in2? Design Projects We are given laboratory data, taken by Prof. Robert Kirchhoff and his students at the University of Massachusetts, for the spin rate of a 2-cup anemometer. The anemometer was made of ping-pong balls (d 1.5 in) split in half, facing in opposite directions, and glued to thin ( 1 -in) rods 4 pegged to a center axle. (See Fig. P7.91 for a sketch.) There were four rods, of lengths 0.212, 0.322, 0.458, and 0.574 ft. The experimental data, for wind tunnel velocity U and rotation rate , are as follows: 0.212 U, ft/s 18.95 22.20 25.90 29.94 38.45 0.322 , r/min U, ft/s 435.00 545.00 650.00 760.00 970.00 18.95 23.19 29.15 32.79 38.45 0.458 , r/min U, ft/s 225.00 290.00 370.00 425.00 495.00 20.10 26.77 31.37 36.05 39.03 0.574 , r/min U, ft/s 140.00 215.00 260.00 295.00 327.00 23.21 27.60 32.07 36.05 39.60 , r/min 115.00 145.00 175.00 195.00 215.00 Assume that the angular velocity of the device is a function of wind speed U, air density and viscosity , rod length , and cup diameter d. For all data, assume air is at 1 atm and 20°C. Define appropriate pi groups for this problem, and plot the data above in this dimensionless manner. | v v D5.1 | e-Text Main Menu | Comment on the possible uncertainty of the results. As a design application, suppose we are to use this anemometer geometry for a large-scale (d 30 cm) airport wind anemometer. If wind speeds vary up to 25 m/s and we desire an average rotation rate 120 r/min, what should be the proper rod length? What are possible limitations of your design? Predict the expected (in r/min) of your design as affected by wind speeds from 0 to 25 m/s. D5.2 By analogy with the cylinder-drag data in Fig. 5.3b, spheres also show a strong roughness effect on drag, at least in the Reynolds number range 4 E4 ReD 3 E5, which accounts for the dimpling of golf balls to increase their distance traveled. Some experimental data for roughened spheres [43] are given in Fig. D5.2. The figure also shows typical golf-ball data. We see that some roughened spheres are better than golf balls in some regions. For the present study, let us neglect the ball’s spin, which causes the very important side-force or Magnus effect (See Fig. 8.11) and assume that the ball is hit without spin and follows the equations of motion for plane motion (x, z): mx ˙˙ where F cos F CD 24 Textbook Table of Contents m˙˙ z D2(x 2 ˙ | F sin z 2) ˙ W tan Study Guide ˙ 1z x ˙ References 321 0.6 Golf ball 0.5 Drag coefficient, CD The ball has a particular CD(ReD) curve from Fig. D5.2 and is struck with an initial velocity V0 and angle 0. Take the ball’s average mass to be 46 g and its diameter to be 4.3 cm. Assuming sea-level air and a modest but finite range of initial conditions, integrate the equations of motion to compare the trajectory of “roughened spheres” to actual golfball calculations. Can the rough sphere outdrive a normal golf ball for any conditions? What roughness-effect differences occur between a low-impact duffer and, say, Tiger Woods? 0.4 0.3 900 0.2 10 5 1250 10 5 500 10 0.1 D 150 Rough spheres 5 10 Smooth sphere 5 0 2 104 105 106 4 106 Reynolds number, UD/ D5.2 References 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. P. W. Bridgman, Dimensional Analysis, Yale University Press, New Haven, CT, 1922, rev. ed., 1931. A. W. Porter, The Method of Dimensions, Methuen, London, 1933. F. M. Lanchester, The Theory of Dimensions and Its Applications for Engineers, Crosby-Lockwood, London, 1940. R. Esnault-Pelterie, L’Analyse dimensionelle, F. Rouge, Lausanne, Switzerland, 1946. G. W. Stubbings, Dimensions in Engineering Theory, CrosbyLockwood, London, 1948. G. Murphy, Similitude in Engineering, Ronald, New York, 1950. H. E. Huntley, Dimensional Analysis, Rinehart, New York, 1951. H. L. Langhaar, Dimensional Analysis and the Theory of Models, Wiley, New York, 1951. W. J. Duncan, Physical Similarity and Dimensional Analysis, Arnold, London, 1953. C. M. Focken, Dimensional Methods and Their Applications, Arnold, London, 1953. L. I. Sedov, Similarity and Dimensional Methods in Mechanics, Academic, New York, 1959. E. C. Ipsen, Units, Dimensions, and Dimensionless Numbers, McGraw-Hill, New York, 1960. E. E. Jupp, An Introduction to Dimensional Methods, CleaverHume, London, 1962. R. Pankhurst, Dimensional Analysis and Scale Factors, Reinhold, New York, 1964. S. J. Kline, Similitude and Approximation Theory, McGrawHill, New York, 1965. B. S. Massey, Units, Dimensional Analysis, and Physical Similarity, Van Nostrand Reinhold, New York, 1971. J. Zierep, Similarity Laws and Modeling, Dekker, New York, 1971. | v v 1. | e-Text Main Menu | 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. W. E. Baker et al., Similarity Methods in Engineering Dynamics, Spartan, Rochelle Park, NJ, 1973. E. S. Taylor, Dimensional Analysis for Engineers, Clarendon Press, Oxford, England, 1974. E. de St. Q. Isaacson and M. de St. Q. Isaacson, Dimensional Methods in Engineering and Physics, Arnold, London, 1975. P. LeCorbeiller, Dimensional Analysis, Irvington, New York, 1966. V. J. Skoglund, Similitude — Theory and Applications, International, Scranton, PA, 1967. M. S. Yalin, Theory of Hydraulic Models, Macmillan, London, 1971. J. J. Sharp, Hydraulic Modeling, Butterworth, London, 1981. G. I. Barenblatt, Dimensional Analysis, Gordon and Breach, New York, 1987. R. Esnault-Pelterie, Dimensional Analysis and Metrology, F. Rouge, Lausanne, Switzerland, 1950. R. Kurth, Dimensional Analysis and Group Theory in Astrophysics, Pergamon, New York, 1972. F. J. de-Jong, Dimensional Analysis for Economists, North Holland, Amsterdam, 1967. E. Buckingham, “On Physically Similar Systems: Illustrations of the Use of Dimensional Equations,” Phys. Rev., vol. 4, no. 4, 1914, pp. 345 – 376. “Flow of Fluids through Valves, Fittings, and Pipe,” Crane Co. Tech. Pap. 410, Chicago, 1957. A. Roshko, “On the Development of Turbulent Wakes from Vortex Streets,” NACA Rep. 1191, 1954. G. W. Jones, Jr., “Unsteady Lift Forces Generated by Vortex Shedding about a Large, Stationary, Oscillating Cylinder at High Reynolds Numbers,” ASME Symp. Unsteady Flow, 1968. O. M. Griffin and S. E. Ramberg, “The Vortex Street Wakes of Vibrating Cylinders,” J. Fluid Mech., vol. 66, pt. 3, 1974, pp. 553 – 576. Textbook Table of Contents | Study Guide 322 Chapter 5 Dimensional Analysis and Similarity Encyclopedia of Science and Technology, 8th ed., McGrawHill, New York, 1997. 35. H. A. Becker, Dimensionless Parameters, Halstead Press (Wiley), New York, 1976. 36. D. J. Schuring, Scale Models in Engineering, Pergamon Press, New York, 1977. 37. M. Zlokarnik, Dimensional Analysis and Scale-Up in Chemical Engineering, Springer-Verlag, New York, 1991. 38. W. G. Jacoby, Data Theory and Dimensional Analysis, Sage, Newbury Park, CA, 1991. | v v 34. | e-Text Main Menu | 39. B. Schepartz, Dimensional Analysis in the Biomedical Sciences, Thomas, Springfield, IL, 1980. 40. A. J. Smith, Dosage and Solution Calculations: The Dimensional Analysis Way, Mosby, St. Louis, MO, 1989. 41. J. B. Bassingthwaighte et al., Fractal Physiology, Oxford Univ. Press, New York, 1994. 42. K. J. Niklas, Plant Allometry: The Scaling of Form and Process, Univ. of Chicago Press, Chicago, 1994. 43. R. D. Blevins, Applied Fluid Dynamics Handbook, van Nostrand Reinhold, New York, 1984. Textbook Table of Contents | Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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