To determine The horizontal and vertical force components, and the moment exerted on the support D . Explanation Given: Velocity of water flowing through the pipe ( V ) is 4 m / s . The total mass of bend and the water within it ( M ) is 20 kg . Pressure acting at point A ( p A ) is 50 kPa . Calculations: The flow of the fluid is assumed to be steady, incompressible, and inviscid. So, the density of ideal fluid water is constant and average velocity can be used for analysis. Consider the fixed control volume consists of elbow, support and water it contains. Draw the free body diagram for the control volume as in Figure (1). Determine the discharge of water from the pipe ( Q ) . Q = V A = ( 4 m / s ) [ π ( 0.05 m ) 2 ] = 0.01 π m 3 / s From the Bernoulli’s equation for negligible elevation difference, the pressure at B ( p B ) can be determined as p B = p A = 50 kPa Determine the pressure forces acting on the surfaces A ( F A ) and B ( F B ) . F A = F B = [ 50 ( 10 3 ) N / m 2 ] [ π ( 0.05 m ) 2 ] = 125 π N 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 A ) x ρ w ( − V A A A ) + ( V B ) x ρ w ( V B A B ) D x − F A − F B cos 60 ° = ( V A ) x ρ w ( − V A A A ) + ( V B ) x ρ w ( V B A B ) [ D x − 125 π N − 125 π cos 60 ° N ] = [ ( − 4 m / s ) ( 1000 kg / m 3 ) ( − 0.01 π m 3 / s ) + ( 6 cos 60 ° m / s ) ( 1000 kg / m 3 ) ( 0.01 π m 3 / s ) ] D x = 777.54 N Thus, horizontal component developed at support D is 777.54 N _ . For the steady and incompressible flow, the linear momentum equation in y direction can be applied as ∑ F = ∂ ∂ t ∫ cv V ρ d V + ∫ es V ρ V ⋅ d A ∑ F y = 0 + ( V B ) y ρ w ( V B A B ) D y − M g − F B sin 60 ° = ( V B ) y ρ w ( Q ) [ D y − ( 20 × 9

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Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 6, Problem 55P

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Chapter 6, Problem 54P

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