V oy x cs m i v iy w m v m v cos 45 o w mv 1 cos 45 o

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v oy X cs ˙ m i v iy W = ˙ m ( v ) ˙ m ( v ) cos 45 o W = ˙ mv (1 cos 45 o ) REVIEW Thus, weight provides the force needed to increase y-momentum fl ow. This weight is produced by the fl uid swirling up to form the shape show in the above sketches. 47
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6.31: PROBLEM DEFINITION Situation : A cone is supported by a vertical jet of water. W = 30 N , V 1 = 15 m / s . d 1 = 2 cm , θ = 60 . Find : Height to which cone will rise (m). Assumptions : Speed of the fl uid as it passes by the cone is constant ( V 2 = V 3 ) . PLAN Apply the Bernoulli equation and the momentum equation. SOLUTION c.s. V 1 1 2 3 60 o Bernoulli equation V 2 1 2 g + 0 = V 2 2 2 g + h V 2 2 = ( V 1 ) 2 2 gh V 2 2 = 225 19 . 62 h Momentum equation ( y -direction). Select a control volume surrounding the cone. X F y = ˙ m o v oy ˙ m i v iy W = ˙ m ( v 3 y v 2 ) 30 N = 1000 kg / m 3 × 15 m / s × π × (0 . 01 m) 2 ( V 2 sin 30 V 2 ) Solve for the V 2 V 2 = 12 . 73 m/s 48
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Complete the Bernoulli equation calculation V 2 2 = 225 19 . 62 h (12 . 73 m / s) 2 = 225 19 . 62 h h = 3 . 21 m 49
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6.32: PROBLEM DEFINITION Situation : A fl uid jet strikes a vane that is moving at a speed. v 1 = 20 m / s , v v = 7 m/s. D 1 = 6 cm. Find : Force of the water on the vane. Sketch : 45 o x y v 1 v v v 2 SOLUTION Force and momentum diagrams Select a control volume surrounding and moving with the vane. Select a reference frame attached to the moving vane. v 1 = 20 m/s v 2 v v = 7 m/s Momentum equation ( x -direction) X F x = ˙ mv 2 x ˙ mv 1 x F x = ˙ mv 2 cos 45 ˙ mv 1 Momentum equation ( y -direction) X F y = ˙ mv 2 y ˙ mv 1 y F y = ˙ mv 2 sin 45 Velocity analysis v 1 is relative to the reference frame = (20 7) = 13 m / s . 50
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in the term ˙ m = ρAv use v which is relative to the control surface. In this case v = (20 7) = 13 m/s v 2 is relative to the reference frame v 2 = v 1 = 13 m/s Mass fl ow rate ˙ m = ρAv = (1 , 000 kg) ( π/ 4 × (0 . 06 m) 2 )(13 m / s) = 36 . 76 kg/s Evaluate forces F x = ˙ mv 1 (1 + cos 45 ) = 36 . 76 kg / s × 13 m / s(1 + cos 45 ) = 816 N which is in the negative x direction. F y = ˙ mv 2 sin 45 = 36 . 76 kg / s × 13 m / s sin 45 = 338 N The force of the water on the vane is the negative of the force of the vane on the water. Thus the force of the water on the vane is F = (816 i 338 j ) N 51
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6.33: PROBLEM DEFINITION Situation : Cart, moving with a steady speed of 3m/s. Jet, V = 50 m/s, D = 15 cm, is being sprayed from behind the cart, and the jet is divided and de fl ected by a vane situated on the cart. Speed of jet as it impacts the vane is 47 m/s, as shown. Find : Force exerted by the vane on the jet: F PLAN Apply the momentum equation. SOLUTION Make the fl ow steady by referencing all velocities to the moving vane and let the c.v. move with the vane as shown in the fi gure above. Momentum equation ( x -direction) F x = ˙ m 2 v 2 x ˙ m 1 v 1 ˙ m = ρAV = 1000 kg / m 3 × ( π/ 4) × (0 . 15 m) 2 × 47 m / s = 830 . 1 kg/s F x = ( ˙ m 2 v cos 45 o ˙ mv ) = ˙ mv ( cos 45 o 2 1) = 830 . 1 kg/s × 47 m/s × (0 . 3535 1) = 25200 N 52
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Momentum equation ( y - direction) F y = ˙ m 2 v 2 y ˙ mv 3 y = ˙ m 2 v sin 45 o ˙ m 2 v = ˙ m 2 v (sin 45 o 1) = 830 . 1 kg/s 2 × 47 m/s × (0 . 707 1) = 5720 N F ( water on vane ) = (25200 i + 5720 j ) N 53
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6.34: PROBLEM DEFINITION
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  • Fall '09
  • Cotel
  • Force, y-direction, momentum equation