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Unformatted text preview: Chapter 6 Momentum Principle Problem 6.1 Water at 20 o C is discharged from a nozzle onto a plate as shown. The g32 ow rate of the water is 0.001 m 3 g64 s, and the diameter of the nozzle outlet is 0.5 cm. Find the force necessary to hold the plate in place. Solution This is a onedimensional, steady g32 ow. Since the system is not accelerating, the ve locities with respect to the nozzle and plate are inertial velocities. The momentum equation in the g123direction (horizontal direction) is X g73 g123 = X g98 g112 g114 g121 g114 g123 g3 X g98 g112 g108 g121 g108 g123 Draw a control volume with the associated force and momentum diagrams. 41 42 CHAPTER 6. MOMENTUM PRINCIPLE From the force diagram X g73 g123 = g3 g85 From the continuity equation, the mass g32 ow in is equal to the mass g32 ow out so g98 g112 g114 = g98 g112 g108 = g98 g112 The velocity at the inlet is g89 . The component of velocity in the g123direction at the outlet is zero, so the momentum g32 ux is X g98 g112 g114 g121 g114 g123 g3 X g98 g112 g108 g121 g108 g123 = g3 g98 g112g89 Equating the forces and momentum g32 ux g3 g85 = g3 g98 g112g89 or g85 = g98 g112g89 The volume g32 ow rate is 0.01 m 3 g64 s, so the mass g32 ow rate is g98 g112 = g29g84 = 1000 × g61 001 = 1 kg/s. The velocity is g89 = g84 g68 = g61 001 g28 4 (0 g61 005) 2 = 50 g61 9 m/s The restraining force is g85 = 50 g61 9 × 1 = 50 g61 9 N 43 Problem 6.2 A water jet with a velocity of 30 m/s impacts on a splitter plate so that 1 4 of the water is de g32 ected toward the bottom and 3 4 toward the top. The angle of the plate is 45 g114 . Find the force required to hold the plate stationary. Neglect the weight of the plate and water, and neglect viscous e g30 ects. Solution The pressure is constant on the free surface of the water. Because frictional e g30 ects are neglected, the Bernoulli equation is applicable. Without gravitational e g30 ects, the Bernoulli equation becomes g115 + 1 2 g29g89 2 = constant Since pressure is constant, the velocity will be constant. Therefore, each exit velocity is equal to the inlet velocity. Momentum and force diagrams for this problem are The forces acting on the control surface are F = g73 g123 i + g73 g124 j 44 CHAPTER 6. MOMENTUM PRINCIPLE The momentum g32 ux from the momentum diagram is X g102g118 g98 g112 g114 v g114 g3 X g102g118 g98 g112 g108 v g108 = 3 4 g98 g112 g108 (30 i cos 45 + 30 j sin 45) + 1 4 g98 g112 g108 ( g3 30 i cos 45 g3 30 j sin 45) g3 g98 g112 g108 30 i Equating the force and momentum g32 ux g73 g123 i + g73 g124 j = g98 g112 g108 ( g3 19 g61 4 i +10 g61 6 j ) The inlet mass g32 ow rate is g98 g112 g108 = g29g68g89 = 1000 × g61 1 × 30 = 3000 kg/s The force vector evaluates to g73 g123 i + g73 g124 j = g3 5 g61 82 × 10 4 i + 3 g61 18 × 10 4 j (N) Thus g73 g123 = g3 5 g61 82 × 10 4 N g73 g124 = 3 g61 18 × 10 4 N Problem 6.3 A 12in. horizontal pipe is connected to a reducer to a 6in. pipe. Crude oil g32 ows through the pipe at a rate of 10 cfs. The pressure at the inlet to the reducerows through the pipe at a rate of 10 cfs....
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This note was uploaded on 12/18/2011 for the course ECON 102 taught by Professor Drfcvvv during the Spring '11 term at Politechnika Wrocławska.
 Spring '11
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