Mechanical Engineering Programme of Study


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1 Mechanical Engineering Programme of Study Fluid Mechanics Instructor: Marios M. Fyrillas SOLVED EXAMPLES ON VISCOUS FLOW 1. Consider steady, laminar flow between two fixed parallel plates due to a pressure gradient. Using a control volume of unit depth, height y, and width x (centred at y 0 ) obtain an expression for the velocity profile. a. By integrating the velocity profile obtain an expression for the volumetric flow rate and the mean velocity. b. Obtain an expression for the dimensionless pressure loss as a function of the Reynolds number. w x y Conservation of Momentum of the control volume Consider xmomentum conservation M M F 0 (steadystate so net momentum flux is zero) out in x The forces acting are: i. right surface: pa ii. left surface: pa iii. top surface: A iv. bottom surface: A l t r l t r b b
2 Balance of forces:  pa r r pa l l ( tat bab) Because of symmetry: t b Areas are given by: Ar Al 1 y, At Ab 1 x y( p  p ) x l r du (Newtonian fluid) dy dp constant (pressure increases linearly) dx du dp du yp ( p p) x y dy dx dy Integrate above expression y dp y dp y dp du dy u dy constant dx dx dx To find the constant use the boundary conditions, i.e. at y h and y  h the velocity is zero (u 0) d d u( y h) 0 constant constant= dx dx h p h p y dp h dp h dp y so u=  1 dx dx dx h To find the volumetric flow rate: h h dp y h dp4h Q u d Au 1dy 1 d dx y h dx 3 h 3 h dp 3 dx (mean velocity) u m 3 h dp Q h dp 3 dx A h 3 dx b. Dimensionless pressure drop 3um p 3um 6 1 p h 1 u umh h umh R m umh h e
3 . Working in a similar fashion as for the case of a horizontal cylinder, obtain the velocity profile of Poiseuille s law in an inclined pipe using the control volume suggested in the figure. Conservation of Momentum of the control volume Consider xmomentum conservation M M F 0 (steadystate so net momentum flux is zero) out in x The force balance can be written as: ( pp) r pr r mgsin 0 m r pgsin r du pgsin r du dr dr pgsin pgsin r u rdr cons tant Evaluate the constant using the boundary conditions: pgsin ur R R 4 ( ) 0 0 consta pgsin constant= R 4 nt
4 ( pgsin ) R r u 1 4 R R ( pgsin ) R R r Q uda u rdr 1 r r 0 0 d R ( pgsin ) R R ( pgsin ) R An oil with a viscosity of 0.4 N s/m and density 900 kg/m 3 flows in a pipe of diameter D 0. m. (a) What pressure drop p 1 p, is needed to 5 3 produce a flowrate of Q.010 m /s if the pipe is horizontal with x1 0 and x 10 m? (b) How steep a hill,, must the pipe be on if the oil is to flow through the pipe at the same rate as in part (a), but with p1 p? (c) For the conditions of part (b), if p1 00 kpa, what is the pressure at section x3 5 m where x is measured along the pipe?
5 4. Consider steady, laminar flow in a circular pipe due to a pressure gradient. Using a control volume of length and radius r obtain an expression for the velocity profile. Follow the steps below: a. Consider the control volume below (Figure 1) and indicate the forces exerted on the control volume. Give a physical explanation. Control Volume Figure 1: Laminar flow in a circular pipe. a. Doing a force balance show that the momentum equation can be simplified to: p. r c. Assuming laminar flow of a Newtonian fluid and applying an appropriate boundary condition obtain that the velocity profile is: pd r u 1 16 D d. Integrate above expression to find the volumetric flow rate.. The forces acting on the control volume are the shear forces acting on the perimetric area r, and pressure forces acting on the fore and aft crosssectional areas pr and ( p p) r, respectively. p By doing a force balance pr r ( p p) r r
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9 5. Determine the head loss for a sudden expansion. Consider the control volume shown on the figure below and use conservation of mass and conservation of momentum. Mass Conservation VA VA m density is constant Momentum Conservation ( pa pa pa) pa M M a a b b c c 3 3 out in M M m V m V m V V V A V V out in out out in in ( 3 1) 3 3( 3 1) Assume that pa pb pc pa pa VA( V V )
10 Energy Equation (Bernoulli's equation) p1 V1 p3 V3 h g g g g L From momentum equation: p p V ( V V ) Substitute above in energy equation V1 V3 V3( V3 V1) ghl V1 V 3 Solve above for ghl V3 V3 V1 VA 1 1 From mass conservation: V3 A3 Substitute V 3 gh L VA V V A A3 A3 gh 1 L A 1 A A 1 A 1 1 A V1 A3 A3 A3 A3 A3 hl gh L A 1 The loss coefficient KL 1 V V 1 1 A3 g
11 6. Calculate the power supplied to the pump shown in Figure 3 if its efficiency 34 is 76%. Methyl alcohol ( 790 kg/m, Pa s ) is flowing at the rate 3 of 54 m /hr. The suction line is a standard 4 in steel pipe, 15 m long. The total length of in steel pipe in the discharge line is 00 m. Assume that the entrance from reservoir 1 is through a squarededged inlet and that the elbows are standard. The valve is a fully open globe valve. The roughness of the pipe is є= mm. Figure 3: Pump/pipeline configuration Consider a streamline joining the points 1 and. Applying the energy equation we obtain 1 W pump 1 p1 u1 gz1 p u gz ghl. Q p = p = p. If we take as the datum the point 1 then z 0 and z 10 m. 1 atm 1 If we further assume that u1 0 and u 0 the energy equation simpilfies to W Q gz gh. pump L
12 Given: Q D suction suction in= m m /hr m /s=0.015 m /s 15 m D discharge discharge in= m 00 m = 790 kg/m Pa s g= 9.81 m/s z 10 m 3 The only uknown in the equation for W pump V V V V V hl f f + KL + KL + KL D g D g g g g major losses major losses minor losses minor losses fully minor losses of minor losses suction discharge pipe entrance open globe valve the standard elbows pipe exit KL0.5 KL10 KL0.3 KL1 is + V KL g The loss coefficients can be obtained from a table, and the velocities from V = Q/( D / 4) / 3.14 / m/s suction discharge suction V = Q/( D / 4) / 3.14 / m/s discharge To find the major losses we need to find the Reynolds number and the relative roughness Re D f suction suction suction V suction D suction from Moody chart Re D f discharge discharge discharge 4 discharge discharge V D from Moody chart Substitute all above information in the equation for hl,calculate hl and finally substitue in equation for W pump
13 7. For the system shown in Figure 4, compute the power delivered by the 3 o pump to the water to pump m /s of water at 15 C to the tank. The air in the tank is at 76 kpa gauge pressure. Consider the friction loss in the 5ftlong discharge pipe, but neglect other losses. Then, redesign the system by using a larger pipe size to reduce the energy loss and reduce the power required to no more than 379 W. The roughness of the pipe is 4 є= and 1 in=0.054 m. Figure 4: Pump/pipeline configuration
14 8. In the turbulent region the friction factor associated with pipe flow is approximated by the formula: 0.5 f 5.74 log D Re Find an expression for the friction factor f for large Re number. For large Reynolds number (Re) above expression simplifies to f log because lim 0. Re 0.9 Re 3.7 D Liquid with specific gravity 3 g 10 kn/m is flowing in a vertical pipe. If the diameter of the pipe is D 15 cm and the viscosity of 3 the fluid is 310 Nm/s determine the direction of the flow and the mean velocity if the pipe relative roughness is / D The pressures shown are static pressures. Hint: Assume a high Reynolds number and verify. Energy Equation (Bernoulli's equation) p1 V1 p V z1 z h g g g g um where the losses are estimated using hl f D g and we have assumed that the flow is directed upwards. L
15 Using mass conservation and assuming uniform flow V V u 1 m. So Bernoulli's equation simplies to um um 0 10hL g g h h 1 L L Hence, our original assumption was wrong and the flow is directed downwards, i.e. p V p V g g g g h L 1 1 z1hl z 1 If we assume that the flow has a high Reynolds number then f log10 log D 3.7 h L 10 u 0.15 g m 0.1 um 1 um.89 m/s ud Verify Reynolds number Re=
16 9. Estimate the elevation required in the upper reservoir to produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?
17 10. Water flows from a reservoir through a pipe 150mm diameter and 180m long to a point below the surface of the reservoir where it branches into two pipes, each 100mm in diameter (see Figure ). One of the pipes is 48m long discharging to atmosphere at a point below reservoir level and the other 60m long discharging to atmosphere 4m below reservoir level. Assuming that f 0.03 calculate the discharge from each pipe, neglecting all loses other than friction. 18m 180m 48m 60m 4m Figure : Reservoir pipeline configuration
18 11. The three waterfilled tanks shown in the figure (Figure P8.10 in textbook) are connected by pipes as indicated in the figure. If minor losses are neglected determine the flowrate in each pipe.
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20 1. Water is to be pumped from one large, open tank to a second large, open tank as shown in the figure. The pipe diameter throughout is 15 cm and the total length of the pipe between the pipe entrance and exit is 61 m. Minor loss coefficients for the entrance, exit, and the elbow are shown on the figure, and the friction factor for the pipe can be assumed constant and equal to 0.0. A certain centrifugal pump having the performance characteristics shown in the figure is suggested as a good pump for this flow system. With this pump, what would be the flowrate between the tanks? Do you think this pump would be a good choice?
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