Explanation The figure below represents the arrangement of the tanks. Figure-(1) Write the Bernoulli’s Equation for tank A and Tank B . P A ρ g + V A 2 2 g + Z A = P B ρ g + V B 2 2 g + 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 pressure on the tank A is P A , the velocity of fluid in tank A is V A the pressure on the tank B is P B , the velocity of fluid in tank B is V B 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 expression for flow rate from cylindrical pipes. Q x = A x V x (II) Here, the area of the pipe is A x and the velocity of flow is V x . Write the expression to calculate the area of pipe. A = π 4 d 2 (III) Here the diameter of the pipe is d . Write the expression for flowrate after addition of pipe. Q 1 = 1.3 Q (IV) Write the expression for discharge from pipe 3. Q 3 = Q 1 − Q 2 (V) Write the Bernoulli’s Equation for tank A and Tank B . P A ρ g + V A 2 2 g + Z A = P C ρ g + V C 2 2 g + Z C + f 1 l 1 V 1 2 2 g d 1 + f 3 l 3 V 3 2 2 g d 3 (VI) Here, the pressure on the tank C is P C , the velocity of fluid in tank C is V C the acceleration due to gravity is g , the elevation of tank C is, 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 . Conclusion: Substitute P a t m for P A , P a t m for P B , 0 for V A , 0 for V B , 15 ft for Z A , 0 for Z B , 0.02 for f 1 , 0.02 for f 2 , 600 ft for l 1 , 500 ft for l 2 , 32.2 ft / s 2 for g , 6 in for d 1 , 6 in for d 2 , V for V 1 and V for V 2 in Equation (1) to calculate the velocity of flow without second pipe. P a t m ρ g + 0 + 15 ft = [ P a t m ρ g + 0 + 0 + 0.02 ( 600 V 2 ft 2 × 32.2 ft / s 2 × 6 in ) + 0.02 ( 500 V 2 ft 2 × 32.2 ft / s 2 × 6 in ) ] 15 ft = 0.02 ( 600 V 2 ft 2 × 32.2 ft / s 2 × 6 in ( 1 ft 12 in ) ) + 0.02 ( 500 V 2 ft 2 × 32.2 ft / s 2 × 6 in ( 1 ft 12 in ) ) V 2 = 21.95 ft 2 / s 2 V = 4.685 ft / s Substitute π 4 d 1 2 for A 1 and 4.685 ft / s for V 1 to calculate the flow rate without addition of second pipe. Q = π 4 d 1 2 × 4.685 ft / s = 3.67 d 1 2 ft / s (VII) Substitute 6 in for d 1 in Equation (VII). Q = 3.67 × ( 6 in ) 2 = 3.67 × ( 6 in ( 1 ft 12 in ) ) 2 = 0.9199 ft 3 / s Substitute 0 .9399 ft 3 / s for Q in Equation (IV) to calculate the flow after the addition of second pipe. Q 1 = 1.3 × 0 .9399 ft 3 / s = 1.195 ft 3 / s Substitute 1.195 ft 3 / s for Q 1 and π 4 d 1 2 for A 1 in Equation (II). 1.195 ft 3 / s = π 4 d 1 2 V 1 (VIII) Substitute 6 in for d 1 in Equation (VIII). 1.195 ft 3 / s = π 4 ( 6 in ) 2 V 1 V 1 = 4 × 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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Chapter 8.5, Problem 115P

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The diameter of pipe 3.

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

Ch. 8.1 - Prob. 1P Ch. 8.1 - Rainwater runoff from a parking lot flows through...Ch. 8.1 - Blue and yellow streams of paint at 60 °F (each...Ch. 8.1 - Air at 200 °F flows at standard atmospheric...Ch. 8.1 - To cool a given room it is necessary to supply 4...Ch. 8.1 - Prob. 6P Ch. 8.1 - Prob. 7P Ch. 8.1 - (See The Wide World of Fluids article titled...Ch. 8.2 - For fully developed laminar pipe flow in a...Ch. 8.2 - Prob. 10P

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