Explanation Consider the free surface of the water is section (1) and outlet of the pipe is section (2). Write the expression for the energy balance equation for the given system including the major losses. p 1 γ + V 1 2 2 g + z 1 + h P = p 2 γ + V 2 2 2 g + z 2 + f l V 2 2 g D (I) Here, the pressure at point 1 is p 1 , the velocity at point 1 is V 1 , the elevation at point 1 is z 1 , the pump head is h p , the pressure at point 2 is p 2 , the velocity at point 2 is V 2 , the elevation at point 2 is z 2 , loss coefficient is K L , friction factor is f , and length of the pipe is l and the diameter of the pipe is D . Write the expression to calculate the area of pipe. A = π 4 D 2 (II) Here, the diameter of pipe is D Write the expression for the velocity using continuity equation. V = Q A (III) Here, discharge is Q . Write the expression for the discharge. Q = A V (IV) Write the continuity equation for the pipe and the nozzle. A V = A 2 V 2 V 2 = ( A A 2 ) V (V) Here, area of the pipe is A and area of the nozzle is V 2 . Substitute equation (II) in equation (V). V 2 = ( D D 2 ) 2 V (VI) Write the relationship between the power and head. h P = P γ Q (VII) Power input is P and discharge is Q . Conclusion: Substitute 40 mm for D 2 in equation (II). A 2 = π 4 ( 40 mm ) 2 A 2 = π 4 ( 40 mm × 1 m 1000 mm ) 2 = 0 .00126 m 2 Substitute 60 mm for D in equation (II). A = π 4 ( 60 mm ) 2 A = π 4 ( 60 mm × 1 m 1000 mm ) 2 = 0 .00283 m 2 Substitute 0.04 m 3 /s for Q and 0 .00126 m 2 for A 2 in equation (III). V 2 = 0.04 m 3 /s 0 .00126 m 2 = 31.83 m/s Substitute 0.04 m 3 /s for Q and 0 .00283 m 2 for A in equation (III). V = 0.04 m 3 /s 0 .00283 m 2 = 14.15 m/s Substitute 25 kN for P , 62.4 lb/ft 3 for γ , and 0.04 m 3 /s for Q in equation (VII). h P = 25 kN ( 62.4 lb/ft 3 ) ( 0.04 m 3 /s ) = 25 kN × 1000 N 1 kN ( 62.4 lb/ft 3 ) ( 0.04 m 3 /s ) = 63.8 m Substitute 0 for p 1 , 0 for V 1 , h for z 1 , 63

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 89P

To determine

The flow rate of the system without pump.

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Chapter 8.5, Problem 88P

Chapter 8.5, Problem 90P

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

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