Ch 4. Particle Momentum and Impulse Multimedia Engineering Dynamics Impulse & Momentum Consv. Linear Momentum Impact AngularMomentum MassFlow
 Chapter - Particle - 1. General Motion 2. Force & Accel. 3. Energy 4. Momentum - Rigid Body - 5. General Motion 6. Force & Accel. 7. Energy 8. Momentum 9. 3-D Motion 10. Vibrations Appendix Basic Math Units Basic Equations Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll

 DYNAMICS - EXAMPLE Water Flow Direction Change Example A 4 ft diameter pipe is supported by a block as it makes a 90o turn. What is the force in the y-direction on the block due to the water direction change? Assume the specific weight, γw, of the water is 62.4 lb/ft3. Solution Forces Due to Flow Momentum Change To determine the change in momentum, fluid velocity needs to be determined for the out flow. Using the conservation of mass equation,      vAAA = vBAB      (2) [(4/2)2π] = vB [(1.5/2)2π]      8 = vB 0.752      vB = 14.22 ft/s The mass flow rate can be calculated using the 4 ft or the 1.5 ft pipe,     dm/dt =ρvAAA (= ρvBAB)        = (γ/g) [(4/2)2π]        = (62.4/32.2) 12.57        = 48.70 slug/s The vertical reaction force must equal the change of momentum in the vertical direction,       Fy = dm/dt (vBy - vAy)            = 48.70 (14.22 - 0) Fy = 692.7 lb

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