Explanation Given: Velocity of flow through the pipe ( d ) at A is 6 m / s . Diameter of pipe A ( d A ) is 100 mm . Diameter of pipes at B and C ( d ) is 60 mm . Calculations: The flow of the fluid is assumed to be steady, incompressible, and inviscid. So, the density of water is constant. Consider the fixed control volume consists of pipes and the water it contains. Draw the free body diagram for the control volume as in Figure (1). The flow is exposed to atmospheric condition at section B and section C . Thus, pressures are expressed as p B = p C = p a t m = 0 Apply the continuity equation for the flow in simplified form. ∂ ∂ t ∫ cv ρ d V + ∫ es ρ v ⋅ d A = 0 0 − V A A A + V B A B + V C A C = 0 − ( 6 m / s ) [ π ( 0.05 m ) 2 ] + V B [ π ( 0.03 m ) 2 ] + V C [ π ( 0.03 m ) 2 ] = 0 V B + V C = 16.67 (I) Apply the Bernoulli’s equation between sections A and B . p A ρ w + V A 2 2 + g z A = p B ρ w + V B 2 2 + g z B p A 1000 kg / m 3 + ( 6 m / s ) 2 2 + g z = 0 + V B 2 2 + g z V B 2 = 0.002 p A + 36 (II) Apply the Bernoulli’s equation between sections A and C . p A ρ w + V A 2 2 + g z A = p B ρ w + V B 2 2 + g z C p A 1000 kg / m 3 + ( 6 m / s ) 2 2 + g z = 0 + V C 2 2 + g z V C 2 = 0.002 p A + 36 (III) From Equation (II) and (III), V B = V C Thus, velocities at sections B and C from Equation (I) are V B = V C = 8.333 m / s Determine the pressure at section A from Equation (III). ( 8.333 m / s ) = 0.002 p A + 36 p A = 16.72 ( 10 3 ) Pa For the steady and incompressible flow, the linear momentum equation in x direction can be applied as ∑ F = ∂ ∂ t ∫ cv V ρ d V + ∫ es V ρ V ⋅ d A ∑ F x = 0 + ( V B ) x ρ w ( V B A B ) + ( V C ) x ρ w ( V C A C ) F x = ( V B ) x ρ w ( V B A B ) + ( V C ) x ρ w ( V C A C ) F x = [ − ( 8

2nd Edition

ISBN: 9780134626055

Author: HIBBELER

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 6, Problem 34P

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Chapter 6, Problem 33P

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