To determine The input power required with and without conical expansion. Explanation Given information: A 70 % efficient pump delivers water from one reservoir to another 20 f t higher. 60 f t of galvanized 2 i n iron pipe is used in the piping system. The flow rate is 0.4 f t 3 / s The power of the pump can be defined as below, P o w e r = ρ g Q h p η In above equation, ρ - Density of the fluid Q - Volumetric flow rate h p - Head increase across the pump The volumetric flow rate Q can be defined as below, Q = V A In above equation, V - Velocity of the flow A -Area Reynolds’s number can be defined as below, R e = ρ V d μ μ - Dynamic viscosity of the fluid d - Diameter of the pipe Assume that the water at 20 ° C will have, ρ = 1.94 s l u g / f t 3 μ = 2.09 × 10 − 5 s l u g / f t s Assume that the galvanized iron will have a roughness of, ε = 0.0005 f t Calculation: First of all, the roughness ratio will be, ε d = 0.0005 f t 2 12 f t = 3 × 10 − 3 Assume that the loss co-efficient K without the 6 ° cone as below, K r e e n t r a n t = 1.0 K e l b o w s = 0.82 K g a t e v a l v e = 0.16 K s h a r p e x i t = 1.0 To evaluate the velocity of the flow, V = Q A = 0.4 f t 3 / s π ( 1 12 f t ) 2 = 18.334 f t / s To find the relevant Reynolds’s number, R e = ρ V d μ = ( 1.94 s l u g / f t 3 ) ( 18.334 f t / s ) ( 2 12 f t ) 2.09 × 10 − 5 s l u g / f t s = 283636.04 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 Therefore, h p = z 2 − z 1 + h f + ∑ h m = Δ z + V 2 2 g ( f L d + ∑ K ) = 20 f t + ( 18

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ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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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 6, Problem 6.102P

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The input power required with and without conical expansion.

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Chapter 6, Problem 6.101P

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