OF FLOW FLUIDS IN PIPES
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Trans-Alaskan pipeline amazing facts
http://en.wikipedia.org/wiki/TransAlaska_Pipeline_System 800 miles from Prudhoe Bay to Valdez! Single 48 inch diameter pipe (1.22m) 11 pump stations with 4 pumps each Oil emerges at 180F and travels at 120F 15 billion barrels of oil has been pumped
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Heat exchangers on vertical support members
Heat exchangers on the support members to dissipate the heat of the oil.
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Pipes on sliders where pipeline crosses Denali Fault.
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Zig-zag configuration of pipeline allows movement. Pipes connected with shoes that allow pipe movement.
Pipeline inspection gages (pigs) sent through the pipeline for checking (radar scans and pressure measurement using electronics) and cleaning the pipe!
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Definitions Volume flow rate (Q) = volume of fluid flowing past a section per unit time (m3/s) Weight flow rate (W) = weight of fluid flowing past a section per unit time Mass of flow rate (M) = mass of fluid flowing past a section per unit time Q = A*v ; where A is the area and v is the velocity. (m3/s) W = Q*; where Q is the flow rate and is the specific weight. (m3/s * N/ m3 = N/s) M = Q*; where Q is the flow rate, and is the density. ( = kg/ m3 * m3/s = kg/s)
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***** Principle of Continuity
Applicable for steady flow - when the quantity of fluid passing by a section does not change with time (for the given period of time) If no fluid is added or removed between two sections Then Principle Mass of fluid at section 1 = Mass of fluid at section 2 Principle of conservation of mass
Or
M1 = M2 8
Or
1*A1*v1 = 2*A2*v2
If the fluid is incompressible, then Or Or A1*v1 = A2*v2 Q 1 = Q2
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Example Problem 6.4: Two pipe sections 1 and 2 D1 = 50 mm = 0.050 m D2 = 100 mm = 0.10 m V1 = 8 m/s Water at 70C : = 9.59kN/m3 ; = 978 kg/m3 Find: V2 Volume flow rate - Q Weight flow rate - W Mass flow rate - M
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A = D2/4 A1 = 0.001963 m2 A2 = 0.007854 m2 To find V2, A1*V1 = A2*V2 Or V2 = (A1*V1)/A2 = (0.001963*8)/ 0.007854 = 2.0 m/s Volume flow rate = Q1 = A1*V1 = Q = 0.0157 m3/s Weight flow rate = Q* = 0.0157 m3/s * 9.59kN/m3 = W = 0.151 kN/s Mass Flow rate = = 0.0157 m3/s * 978 kg/m3 M = 15.36 kg/s 11
Conservation of Energy Bernoullis Equation Relates to the energy in of flow for fluids Energy can neither be created or destroyed it can be transformed from one form to another Referred to as the Conservation of Energy!
What are the various types of Energy?????
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Types of Energy Three forms of energy: 1. Potential energy relates to its elevation PE = w*z Where w is the weight and z is the elevation 2. Kinetic energy Energy due to motion of the fluid, due to the velocity KE = w*v2/2g 3. Flow or pressure energy Relates to the pressure of the flow FE = wp/
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If you consider these three forms of energies The total energy at any point for fluid flow in a pipe can be given as E = FE + PE + KE
E = wp/ + wz + wv2/2g
***Principle of conservation of energy If no energy is added or removed, the total energy at two points is equal E1 = E2 Substituting and dividing by w on both sides
p1/ + z1 + v12/2g = p2/ + z2 + v22/2g
referred to as Bernoullis Equation******
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Each term of the equation refers to a form of energy in the fluid per unit weight p/ pressure head z elevation head v2/2g velocity head units of all three terms = energy per unit wt. = N*m/N = m since each term is in length units, the terms can be plotted as heights --- see Figure 6.6
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Bernoullis Equation accounts for the changes elevation, velocity, and pressure head between two points in the system. Assumption no energy losses or addition between the two points.
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Key conditions for Bernoullis equation 1. Valid for incompressible fluids only. Specific weight cannot change between the two points. 2. No mechanical devices between the two sections that might add or remove energy. 3. No heat transfer to or from the fluid. 4. No energy loss due to friction. These assumptions will be relaxed later by addition of more terms to the equation.
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Procedure for application of Bernoullis Equation: 1. determine what is know and what is to be found 2. identify the two locations typically we know data for Pt. 1 and need to find out for Pt. 2 3. equation is written in direction of flow; flow must proceed from LHS to RHS of equation 4. draw a sketch, label the points, determine the reference level 5. simplify the equation cancel zero or terms that are equal on both sides 6. use consistent units!
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Problem 6.9 Fluid water at 10C; = 9.81 kN/m3 Flow from Section 1 Section 1: D1 = 25 mm p1 = 345 kPa v1 = 3.0 m/s Section 2: D2 = 50 mm Z2-z1 = 2.0 m (pt. 2 is 2m above pt.1) g = 9.81 m/s2 v2 = ??? Compute p2.????????? How will we solve this????? What are the steps???? Section 2
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First, we need to find out V2. Apply the continuity equation between 1 and 2 A1*V1 = A2*V2 A1 = 491 mm2 A2 = 1963 mm2 Therefore V2 = (A1*V1)/A2 = (491*3.0)/(1963) = 0.75 m/s Now apply the Bernoulli equation between 1 and 2 Set up the equations for points 1 and 2 p1/ + z1 + v12 / 2g = p2/ + z2 + v22 / 2g we need to find p2, so re-arrange the equation so that p2/ = p1/ + z1 + v12 / 2g - z2 + v22 / 2g p2/ = p1/ + (z1 - z2) + (v12 - v22 )/2g p2/ = (345/9.81) 2.0 + (3.0*3.0 0.75*0.75)/(2*9.81) 20
p2/ = 33.59 p2 = 33.59*9.81 = 329.6 kPa
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Applying Bernoullis Equation to tank with a siphon and nozzle.
Key Observations: When fluid is exposed to atmosphere, the pressure is zero (gage pressure), and the pressure head term can be cancelled [e.g., points A and F] Velocity head at the surface of a tank or reservoir is considered to be zero and can be cancelled. [e.g., at point A]
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When two points are inside a pipe of the same diameter, the velocity head terms for the two points cancel each other. [e.g., points B, C, D] When the two points are the same elevation, the elevation head terms can be cancelled [e.g., points A, B, D].
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Problem 6.10:
Compute the volume flow rate Q through the siphon Determine pressures at points B through E.
First, lets calculate the volume flow rate Q for the problem We will choose points A and F. what is known and unknown about these points? pA = 0 = pF 24
vA = 0 ( we assume a very tank, large so the velocity at A will be very small and can be assumed as zero) zA and zF are known. Set up the equation Points A & F
pA/ + zA + vA2 / 2g = pF/ + zF + vF2 / 2g pA// + zA + vA2 // 2g = pF// + zF + vF2 / 2g pA vA2 2g pF zA = zF + vF2 / 2g 25
vF = [(zA - zF)*2g]1/2 vF = [(3.0)*2*9.81]1/2 vF = 7.67 m/s QF = AF * vF QF = 3.77 x 10-3 m3/s The same Q will apply at points B, C, D, E!
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Now lets find the pressures at the various points. Set up the equation Points A & B
pA/ + zA + vA2 / 2g = pB/ + zB + vB2 / 2g pA// + zA + vA2 // 2g = pB/ + zB + vB2 / 2g pA vA 2 2 g pB = [ (zA - zB) - vB2 / 2g] what is (zA - zB)???? what is vB ? vB = Q/AB
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vB = 3.77 x 10-3/0.001257 = 3.0 m/s Therefore pressure at B pB = [ (zA - zB) - vB2 / 2g] pB = 9.81* [ 0 (3)2 / 2*9.81] pB = -4.50 kPa *** Hmmmmm pressure at B is negative!!!!! A and B are at the same level in the fluid why arent the pressures the same?????????? Didnt we learn that in the previous lecture???????? SO WHAT IS DIFFERENT HERE?????
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Set up the equation Points A & C
pA/ + zA + vA2 / 2g = pC/ + zC + vC2 / 2g pA// + zA + vA2 // 2g = pC/ + zC + vC2 / 2g pA vA 2 2 g pC = [ (zA - zC) vC2 / 2g] what is (zA - zC)???? = -1.2 m what is vC ?????????
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vC = Q/Ac = ???? Pressure at C pC = [ (zA - zC) vC2 / 2g] pC = 9.81* [ -1.2 -0.459] pC = -16.27 kPa What about the pressure at D?????????????
Any guesses???????????????? And WHY???? 30
pD = Set up the equation Points A & E
pA/ + zA + vA2 / 2g = pE/ + zE + vE2 / 2g pA// + zA + vA2 // 2g = pE/ + zE + vE2 / 2g vA 2 2 g pA pE = [ (zA - zE) vE2 / 2g] what is (zA - zE)???? = 3.0 m what is vE ????????? 31
vE = 3.0 m/s pressure at E pE = [ (zA - zE) vE2 / 2g] pE = 9.81* [ 3.0 3.02 / 2*9.81] pE = 24.93 kPa
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Important take-home points from this problem
The velocity of the siphon and the flow rate out of the siphon depends on the elevation difference between free surface of fluid and level of siphon. The pressure at B is negative (below atmospheric) even though it is at the same level as A in the fluid! The decreased pressure energy (negative) at B got converted to what?????? The velocity of flow is the same at all points if the pipe size does not change (B, C, D, E) Pressure at C is the lowest, since its at the highest elevation
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Pressure at D and B are the same since the elevation and velocity heads for these two locations are the same (the other two terms in the Bernoullis equation) Pressure at point E is the highest in the system since it is at the lowest position.
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Torricellis Theorum (year - 1645): If Bernoullis equation is applied to a tank with an orifice.
p1/ + z1 + v12 / 2g = p2/ + z2 + v22 / 2g p1// + z1 + v12 // 2g = p2// + z2 + v22 / 2g p1 v12 2g p2 v2 = [2g (z1 - z2)] 1/2 v2 = [2gh] 1/2 VELOCITY at ORIFICE dependent of fluid elevation! 35
Velocity will change as h changes! Velocity and flow rate will decrease with time as the tank drains! why because the value of h decreases! Problem 6.13:
Compute the flow rate Q from the nozzle if: h starts at 3.0 m and for increments of 0.5 m. diameter of nozzle 50mm area of nozzle = 0.001963 m2 for h = 3.0 m v = [2* 9.81* 3.0] 1/2 v = 7.67 m/s
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therefore Q = Av = 0.001963 * 7.67 = 0.0151 m3/s compute Q for decreasing 0.5 m increments and plot Excel spreadsheet computations h 3 2.5 2 1.5 1 0.5 0 v 7.672 7.004 6.264 5.425 4.429 3.132 0.000 A 0.001963 0.001963 0.001963 0.001963 0.001963 0.001963 0.001963 Q 0.015 0.014 0.012 0.011 0.009 0.006 0.000
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9.000 8.000 7.000 6.000 5.000
0.016 0.014 0.012 0.010 0.008
discharge Q
velocity (v)
v Q
4.000 0.006 3.000 2.000 1.000 0.000 0 1 2 h (m) 3 4 0.004 0.002 0.000
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Another case of Torricellis theor...