A 75mm diameter jet having velocity of 30m/s strikes a flat plate. This is inclined at 45 degree with axis of jet. Find normal force on plate if- a) When plate is stationary. b) When plate is moving with velocity of 15m/s & away from jet.
Answers
Answer:
Fn = 2811.51 N
η = 12.5%
Explanation:
Given data,
Diameter of the jet, d = 75 mm = 0.075 m
Absolute velocity of the jet, V = 30 m/s
Angle between jet and plate, θ = 450
Velocity of the plate, u = 15 m/s
When the plate is stationary, normal force on the plate,
Fn = ρaV2 sinθ
= 1000 x Π/4 x (0.075)2 x 302 x sin45
= 2811.51 N
When the plate is moving with a velocity 15 m/s and away from the jet, the normal force on the plate,
Fn = ρa(V-u)2 sinθ
= 1000 x Π/4 x (0.075)2 x (30 - 15)2 x sin45
= 702.88 N
Force in the direction of the jet, Fx = Fn sinθ
= 702.88 x sin45
= 497 N
Workdone per second, W = Fx x u = 497 x 15 = 7455 Nm/s
Power, P = 7455 W = 7.5 kW
Efficiency, = 2 (V-u)2 sin2θ u
V3
= 2 x (300-15)2 sin245 x 15
303
= 12.5 %