A jet of water having a velocity of 30m/s strikes a curved vane which is moving with a velocity of 15m/s.The jet makes an angle of 30° with the direction of motion of vane at inlet and leaves at an angle of 120° to the direction of motion of vane at outlet .calculate.1)vane angles so that the water enters and leaves the vane without shock.2)work done per second per unit weight of water striking the vane per second.
Answers
Answer:
37.876°, 6.565°, 20.24 N-m/N
Step-by-step explanation:
Vane angles so that the water enters is and the vane angle so that the water leaves the vane without shock are Θ = 52.33° and ∅ = 18.84°.
Work done per second per unit weight of water striking the vane per second is 2.94 Nm/N weight of water.
Given:
Jet velocity v2 = 30m/s
initial velocity v1 = 15m/s
vane velocity = 20m/s
Jet Inlet angle, α = 30°
Jet outlet angle, β = 120°
Vr₁ = Vr₂ (Because of no shock)
To Find:
Vane angle at the inlet, Θ =?
Vane angle at the outlet, ∅ =?
Work done / N weight of the water =?
Solution:
Using V1 and α
= Vω₁ / V₁
Vω₁ = V₁ =
Vω₁ = 21.65 m/s
Now,
= V&₁/V₁
V&₁ = V₁ =
V&₁ = 12.5 m/s
Similarly, tanΘ = V&₁ / (Vω₁ - u)
Θ =
Θ = 52.33°
Now,
SinΘ = V&₁/ Vr₁
Vr₁ = V&₁ / SinΘ
Vr₁ = 12.5 / Sin 52.33°
Vr₁ = Vr₂ = 15.79 m/s
By using the Sin rule,
Vr₂ / Sin(180 - 60) = u₂ / Sin(60 - ∅)
15.79 / Sin(180 - 60) = 12/ Sin(60 - ∅)
Sin(60 - ∅) = 12 × Sin(180 - 60) / 15.79
60 - ∅ = Sin⁻¹ (0.658)
∅ = 60 - 41.159
∅ = 18.84°
Cos ∅ = u₂ + Vω₂ / Vr₂
Vω₂ = Vr₂ × Cos ∅ - u₂
Vω₂ = 2.944 m/s
We know that work done is given by the formula :
Work done = 1/g × ( Vω₁ + Vω₂)
Work done = 1/9.8 × ( 21.65 + 2.944) × 12
Work done = 30.084 Nm
Hence, Vane angles so that the water enters and leaves the vane without shock are Θ = 52.33° and ∅ = 18.84°.
Work done is 30.084 Nm weight of water.
For more examples of curved vane sums, refer to these answers:
https://brainly.in/question/31307784
https://brainly.in/question/11201393
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