Explanation The figure below shows the arrangement of tanks and pipes. Figure-(1) Write the energy equation for fluid flowing from tank A to tank B . Z A = Z B + f 1 l 1 V 1 2 2 g d 1 + f 2 l 2 V 2 2 2 g d 2 (I) Here, the acceleration due to gravity is g , the elevation of tank A is Z A , the elevation of tank B is Z B , the friction factor for the pipe 1 is f 1 , the length of pipe 1 is l 1 , the velocity in pipe 1 is V 1 , the diameter of the pipe 1 is d 1 , friction factor for the pipe 2 is f 2 , the length of pipe 2 is l 2 , the velocity in pipe 2 is V 2 and the diameter of the pipe 2 is d 2 . Write the continuity equation for the given arrangement. A 1 V 1 = A 2 V 2 + A 3 V 3 (II) Here, the velocity in pipe 2 is V 3 . Write the expression to calculate the area of pipe 1. A 1 = π 4 D 1 2 Here, the diameter of pipe 1 is D 1 . Write the expression to calculate the area of pipe 2. A 2 = π 4 D 2 2 Here, the diameter of pipe 2 is D 2 . Write the expression to calculate the area of pipe 3. A 3 = π 4 D 3 2 Here, the diameter of pipe 3 is D 3 . Write the energy equation for fluid flowing from tank A to tank C . Z A = Z C + f 1 l 1 V 1 2 2 g d 1 + f 3 l 3 V 3 2 2 g d 3 (III) Here, the pressure in Tank C is P C , velocity of fluid in tank C is V C , the elevation of tank C is Z C , friction factor for the pipe 3 is f 3 , the length of pipe 3 is l 3 , the velocity in pipe 3 is V 3 and the diameter of the pipe 3 is d 3 . Write the expression for discharge from pipe 1. Q 1 = A 1 V 1 (IV) Write the expression for discharge from pipe 2. Q 2 = A 2 V 2 (V) Write the expression for discharge from pipe 3. Q 3 = A 3 V 3 (VI) Conclusion: Substitute π 4 D 1 2 for A 1 , π 4 D 2 2 for A 2 and π 4 D 3 2 for A 3 in Equation (III). π 4 D 1 2 V 1 = π 4 D 2 2 V 2 + π 4 D 3 2 V 3 D 1 2 V 1 = D 2 2 V 2 + D 3 2 V 3 (VII) Substitute, 0.1 m for D 1 , 0.08 m for D 2 and 0.08 m for D 3 in Equation (VII). ( 0.1 m ) 2 V 1 = ( 0.08 m ) 2 V 2 + ( 0.08 m ) 2 V 3 0.01 V 1 m 2 = 0.0064 V 2 m 2 + 0.0064 V 3 m 2 V 1 = 0.64 V 2 + 0.64 V 3 (VIII) Substitute 0.1 m for D 1 , 0.08 m for D 2 , 60 m for Z A , 20 m for Z B , 0.015 for f 1 , 0.020 for f 2 , 200 m for l 1 , 200 m for l 2 and 9.81 m / s 2 for g in Equation (I). 60 m = [ 20 m + 0.015 × 200 V 1 2 m ( 0.10 m ) ( 2 × 9.81 m / s 2 ) + 0.020 × 200 V 2 2 m ( 0.08 m ) ( 2 × 9.81 m / s 2 ) ] 40 m = 3 V 1 2 1.962 m / s 2 + 4 V 2 2 m 1.5696 m / s 2 40 m = 1.529 V 1 2 m − 1 ⋅ s 2 + 2.584 V 2 2 m − 1 ⋅ s 2 (IX) Substitute 0.1 m for D 1 , 0.08 m for D 3 , 0 m for Z C , 0.015 for f 1 , 0.020 for f 3 , 200 m for l 1 , 400 m for l 3 and 9.81 m / s 2 for g in Equation (III). 60 m = [ 0 m + 0.015 × 200 V 1 2 m ( 0.10 m ) ( 2 × 9.81 m / s 2 ) + 0.020 × 400 V 3 2 m ( 0.08 m ) ( 2 × 9.81 m / s 2 ) ] 60 m = 3 V 1 2 1.962 m / s 2 + 8 V 3 2 1

Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, Binder Ready Version

8th Edition

ISBN: 9781119080701

Author: Philip M. Gerhart, Andrew L. Gerhart, John I. Hochstein

Publisher: WILEY

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Applied Fluid Mechanics

A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

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Chapter 8.5, Problem 119P

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The flow rate in each pipe.

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Chapter 8.5, Problem 118P

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