Explanation Given: The diameter of the pipe ( d ) is 0.25 ft . The volumetric flow of the pump ( Q ) is 0.3 ft 3 /s . The friction loss is equal to 5 V 2 g . Calculation: The following assumptions are made in the problem: The water from A to B is considered as the control volume. The point A and B are open to atmosphere so the pressure at point A and B are zero. p A = p B = 0 The velocity of the pond A is negligible when the reservoir is large. V A = V B = 0 The datum line passes through the point A , so the datum at A is zero. z A = 0 and z B = ( 6 ft + 10 ft ) Calculate the area of the pipe ( A ). A = π d 2 4 A = π ( 0.25 ) 2 4 = 0.04908 ft 2 Calculate the velocity of water at B ( V A ) . Q = V B A B 0.3 ft 3 /s = V A × 0.049 ft 2 V B = 6.112 ft / s Refer Appendix A, “Physical Properties of Water versus Temperature (FPS Units)”, The density of water ( ρ w ) at 32 ° F as 1.94 slug / ft 3 . Calculate the unit weight of water ( γ ) . γ = ρ w g γ = 1.94 slug / ft 3 × 32.2 ft/s 2 = 62.468 lb / ft 3 Calculate the friction loss ( h L )

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ISBN: 2819480256061

Author: HIBBELER

Publisher: INTER PEAR

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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 5, Problem 99P

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The power of the pump supplies to the water in hp.

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Chapter 5, Problem 98P

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