Explanation Draw the sketch of the system when the reservoir is full as shown in figure (1). Refer the figure (1), Apply the energy balance between the point (1) and (2). 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 + ∑ K L V 2 2 g 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 + [ n K L V 2 2 g + K L V 2 2 g ] (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 , loss coefficient is ∑ K L , and number of the elbows is n . The surface of the reservoir is atmospheric, hence the velocity and pressure at point (1) is zero. P 1 = 0 V 1 = 0 Similarly, the exit of the pipe is atmospheric pressure so the pressure at point (2) also zero. P 2 = 0 Substitute 0 for P 2 , 0 for P 1 , 0 for V 1 , and V for V 2 in equation (I). z 1 = z 2 + f l V 2 2 g D + [ n K L V 2 2 g + K L V 2 2 g ] ( z 1 − z 2 ) = f l V 2 2 g D + [ n K L V 2 2 g + K L V 2 2 g ] (II) Write the expression for the Reynolds number. Re = V D v (III) Here, kinematic viscosity is ν . Write the expression for the area of the pipe at section (2). A 2 = π 4 d 2 2 (IV) Write the expression for the continuity equation. Q 1 = Q 2 A 1 V 1 = A 2 V 2 (V) Here, area of the cross section (1) is A 1 and area of the cross section (2) is A 2 . Conclusion: Substitute 55 m for ( z 1 − z 2 ) , 1000 m for l , 9.81 m/s 2 for g , 7 cm for D , 5 for n , 1.5 for K e l b o w , 10 for K v a l v e in equation (II). ( 25 + 30 m ) = f × 1000 m × V 2 2 × 9.81 m/s 2 × 7 cm + [ 5 × 1.5 × V 2 2 × 9.81 m/s 2 + 10 × V 2 2 × 9.81 m/s 2 ] ( 25 + 30 m ) = f × 1000 m × V 2 2 × 9.81 m/s 2 × 7 cm × 1 m 100 cm + [ 0.3822 V 2 + 0.5097 V 2 ] f × 1000 × V 2 2 × 0.07 × 9.81 = 55 − ( 0.8919 + 0.05097 ) V 2 V = 55 728.12 f + 0.94287 (VI) From Table B.2, “Physical properties of water (SI units), obtain the following properties of water at 10 ° C . ν = 1.307 × 10 − 6 m 2 /s Substitute 7 cm for D and 1.307 × 10 − 6 m 2 /s for ν in equation (III). Re = V ( 7 cm ) 1.307 × 10 − 6 m 2 /s Re = V ( 7 cm × 1 m 100 cm ) 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 98P

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The time required for the water drop from 25 m to 5 m.

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Chapter 8 Solutions

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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