Question

1. Water enters the rotor shown to the right at a rate of 5 liters/s and enters along the axis of rotation. The cross-sectional area of each of the three nozzle exits is 18 mm. Nozzle exit area normal to relative velocity 18 mm How large is the resisting torque required to hold the rotor stationary if: (a) = 0°, (b) 0 = 30°, (c) = 60° ? How fast will the rotor spin steadily if the resisting torque is reduced to zero and: (d) 0 -0°, (e) 0 = 30°, (1) 0 = 60° ? <o bliters's

Answer 1

Sol Using moment of momentum Egn- Tome = Ishat moment of momentum flux moment of a momentum flux out - O = m Tout vout coso n = mass flow rate = PQ • m = 100 kg x 510 = 5 kg Vent = 0.5m 92.59 8 m/s Vout = 5X1D - 3x18x10 92.59 5x0.5 x * coso 231.48 uso = co=0 For < 0 = 30° → lo -60 - Tshaft = [231.48 Nm Tshatr 231.48 6930° = [ 200 N.ml Tames 231.48 Con6° = 115-74 N.ro/

© let c" be angular veloáty of noktion - V = Abs. Tangential velocity of jet. Tu - wir = un 92.59 -0.5 Similarly, Vtz = V43 = 92.590.5o. харе moment of momentum flux leaving the sprinkler, S T = PQ, Vt, 41 * 3 - Since external Troque as zero 0 - 3* 1000x5wo’x 0.5 (42.59coseco 0.56) =) 0 715 ( 92159 code - 05W) to =) 694.425 40 = 3.75W =) W = 185.18 Cosol For o=0°, rw = 185.18 Jad/ For 8 = 300 W = 185-18 Cos 30° - [160 gids For 0= 60° w = 185:18 costo = 192.59 rad/ w = 185.18 coso.