Explanation Given: Temperature of water, T w = 20° C Length, L = 5 m Diameter, d = 3 cm = 0.03 m Flow rate, Q · = 11 m 3 /h = 11 3600 m 3 /s = 0.003055 m 3 /s. Calculation: We know, Reynolds Number, Re = ρ v d μ Where, ρ = density of water at 20° C = 998 kg/ m 3 μ =viscosity of the water at 20° C = 0.001 kg/ms V = velocity of flow = Q · A , where A is the cross-sectional area of the pipe ⇒ A = π d 2 4 = π ( 0.03 ) 2 4 = 0.00071 m 2 . ⇒ V = 0.003055 0.00071 = 4.303 m/s Therefore, Re = 998 × 4.303 × 0.03 0.001 = 1.28831 × 10 5 which is a turbulent flow. Now, applying Bernoulli’s equation for real fluid: P 1 + ρ V 1 2 2 + ρ g z 1 = P 2 + ρ V 2 2 2 + ρ g z 2 + h L , Here P 1 = P 2 = atmospheric pressure h L = head loss or energy loss ⇒ V 1 2 2 g + z 1 = V 2 2 2 g + z 2 + h L ⇒ h L = ( z 1 − z 2 ) − ( V 2 2 − V 1 2 ) 2 g ⇒ h L = 4 − ( 4

8th Edition

ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Chapter 6, Problem 6.49P

To determine

The maximum roughness height.

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Chapter 6 Solutions

Fluid Mechanics

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