V 1 22 ms v 2 21 ms q 0 15 m 3 s s 0 9 find

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v 1 = 22 m/s, v 2 = 21 m/s. Q = 0 . 15 m 3 /s, S = 0 . 9 . Find : Components of force to hold vane stationary (kN). Sketch : PLAN Apply the momentum equation. SOLUTION Force and momentum diagrams Mass fl ow rate ˙ m = ρQ = 0 . 9 × 1000 kg / m 3 × 0 . 15 m 3 / s = 135 kg/s Momentum equation ( x -direction) X F x = ˙ m ( v o ) x ˙ m ( v i ) x F x = ˙ m ( v 2 cos 30) ˙ mv 1 F x = 135 kg / s(21 m / s cos 30 + 22 m / s) F x = 5 . 43 kN (acts to the left) 34
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Momentum equation ( y -direction) X F y = ˙ m ( v o ) y ˙ m ( v i ) y F y = ˙ m ( v 2 sin 30) = 135 kg / s ( 21 m / s sin 30) = 1 . 42 kN F y = 1 . 42 kN (acts downward) 35
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6.24: PROBLEM DEFINITION Situation : A fi xed vane in the horizontal plane. v 1 = 70 ft/s, v 2 = 65 ft/s. Q = 1 . 5 cfs, S = 0 . 9 . Find : Components of force to hold vane stationary (lbf). Sketch : PLAN Apply the momentum equation. SOLUTION Force and momentum diagrams Mass fl ow rate ˙ m = ρQ = 0 . 9 × 1 . 94 slug / ft 3 × 1 . 5 ft 3 / s = 2 . 62 slug/s Momentum equation ( x -direction) X F x = ˙ m ( v o ) x ˙ m ( v i ) x F x = ˙ m ( v 2 cos 30) ˙ mv 1 F x = 2 . 62 slug / s(65 ft / s cos 30 + 70 ft / s) F x = 331 lbf (acts to the left) y -direction 36
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X F y = ˙ m ( v o ) y ˙ m ( v i ) y F y = ˙ m ( v 2 sin 30) = 2 . 62 slug / s ( 65 ft / s sin 30) = 85 lbf F y = 85 lbf (acts downward) 37
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6.25: PROBLEM DEFINITION Situation : A horizontal, two-dimensional water jet de fl ected by a fi xed vane. v 1 = 40 ft/s, w 2 = 0 . 2 ft, w 3 = 0 . 1 ft. Find : Components of force, per foot of width, to hold the vane stationary (lbf/ft). Assumptions : Neglect elevation changes. Neglect viscous e ff ects. Properties : Water, Table A.5: ρ = 1 . 94 slug/ft 3 . PLAN Apply the Bernoulli equation, the continuity equation, and fi nally the momentum equation. SOLUTION Force and momentum diagrams Bernoulli equation v 1 = v 2 = v 3 = v = 40 ft/s Continuity equation w 1 v 1 = w 2 v 2 + w 3 v 3 w 1 = w 2 + w 3 = (0 . 2 + 0 . 1) = 0 . 3 ft Momentum equation ( x -direction) X F x = X ˙ m o ( v o ) x ˙ m i ( v i ) x F x = ˙ m 2 v cos 60 + ˙ m 3 ( v cos 30) ˙ m 1 v F x = ρv 2 ( A 2 cos 60 + A 3 cos 30 + A 1 ) F x = 1 . 94 slug / ft 3 × (40 ft / s) 2 × ( 0 . 2 ft cos 60 + 0 . 1 ft cos 30 + 0 . 3 ft) F x = 890 lbf/ft (acts to the left) 38
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Momentum equation ( y -direction) X F y = X ˙ m o ( v o ) y F y = ˙ m 2 v sin 60 + ˙ m 3 ( v sin 30) = ρv 2 ( A 2 sin 60 A 3 sin 30) = 1 . 94 slug / ft 3 × (40 ft / s) 2 × (0 . 2 ft sin 60 0 . 1 ft sin 30) F y = 382 lbf/ft (acts upward) 39
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6.26: PROBLEM DEFINITION Situation : A water jet is de fl ected by a fi xed vane. v 1 = 30 ft/s, ˙ m = 35 lbm/s = 1.086 slug/s. Find : Force of the water on the vane (lbf). Sketch : PLAN Apply the Bernoulli equation, and then the momentum equation. SOLUTION Force and momentum diagrams Bernoulli equation v 1 = v 2 = v = 30 ft/s Momentum equation ( x -direction) X F x = ˙ m o ( v o ) x ˙ m i ( v i ) x F x = ˙ mv cos 30 ˙ mv F x = ˙ mv (1 cos 30) = 1 . 086 slug / s × 30 ft / s × (1 cos 30) F x = 4 . 36 lbf to the lef y -direction X F y = ˙ m o ( v o ) y F y = ˙ m ( v cos 60) = 1 . 086 slug / s × 30 ft / s × sin 30 F y = 16 . 29 lbf downward Since the forces acting on the vane represent a state of equilibrium, the force of water on the vane is equal in magnitude & opposite in direction.
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  • Winter '15
  • SteveTarrent
  • Force, y-direction, momentum equation

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