Explanation Given information: The length of the pipe is 4500 m and the diameter is 4 c m Δ z = 100 m The power delivered by the pump is equal to 3 k W The power of the pump can be defined as below, P o w e r = ρ g Q h p In above equation, ρ - Density of the fluid Q - Volumetric flow rate h p - Head increase across the pump The steady flow equation between the reservoir 1 & 2 will be, p 1 ρ g + V 1 2 2 g + z 1 = p 2 ρ g + V 2 2 2 g + z 2 + h f + ∑ h m − h p We know that, p 1 = p 2 and V 1 = V 2 ≈ 0 , therefore the above equation can be rewritten as, h p = z 2 − z 1 + h f + ∑ h m Calculation: To find the flow rate, the head loss needs to be found, for that h p = z 2 − z 1 + h f + ∑ h m Assume, f = 0.02 K e n t r a n c e = 0.5 K e x i t = 1.0 K o p e n f l a n g e d g l o b e = 10.0 Therefore, h p = z 2 − z 1 + h f + ∑ h m = Δ z + V 2 2 g f L d + ∑ K = 100 m + Q π 0

8th Edition

ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Chapter 6, Problem 6.100P

To determine

The predicted flow rate.

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Chapter 6, Problem 6.99P

Chapter 6, Problem 6.101P

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Chapter 6 Solutions

Fluid Mechanics

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