Explanation Given information: The surface elevation is equal to 40 m The diameter of the steel pipe is equal to 8 c m The desired flow rate is equal to 200 N / s The volumetric flow rate Q can be defined as below, Q = V A In above equation, V - Velocity of the flow A -Area The energy equation for this problem can be defined as, p 1 ρ g + V 1 2 2 g + z 1 = p 2 ρ g + V 2 2 2 g + z 2 + h f Where head loss h f is equal to, h f = f L D V 2 2 g The Reynolds’s number is defined as, R e = D V ν D - Diameter of the pipe V - Velocity of the flow ν - Kinematic viscosity The friction factor can be determined by, 1 f 1 / 2 ≈ − 1.8 log 6.9 R e + ∈ / d 3.7 1.11 Calculation: To find the velocity of the flow, V = Q A = 200 N s 9.81 m / s 2 1000 π 0.04 m 2 = 4.056 m / s To find the Reynolds’s number, R e = V d ν = 4.056 m / s 0.08 m 1.01 × 10 − 6 m 2 / s = 321267.32 To find the roughness ratio, ε d = 0.05 × 10 − 3 0.08 = 6.25 × 10 − 4 According to the above obtained results, the friction factor will be equal to 1 f 1 / 2 ≈ − 1.8 log 6.9 R e + ∈ / d 3

8th Edition

ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Chapter 6, Problem 6.74P

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The proper length of the pipe

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