A jet of water of diameter 50 mm, having a velocity of 30 m/s strikes a curved vane which is moving with a velocity of 15 m/s in the direction of the jet. The jet leaves the vane at an angle of 60° to the direction of motion of vanes at outlet. Determine : (i) the force exerted by the jet on the vane in the direction of motion, (ii) work done per second by the jet.

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Analysis of a Jet on a Moving Curved Vane

Problem Statement

A jet of water of diameter 50 mm, having a velocity of 30 m/s strikes a curved vane which is moving with a velocity of 15 m/s in the direction of the jet. The jet leaves the vane at an angle of 60° to the direction of motion of vanes at outlet. Determine : (i) the force exerted by the jet on the vane in the direction of motion, (ii) work done per second by the jet.

Given Data & Constants

  • Diameter of jet, \(d = 50 \, \text{mm} = 0.05 \, \text{m}\)
  • Velocity of jet, \(V_1 = 30 \, \text{m/s}\)
  • Velocity of vane, \(u = 15 \, \text{m/s}\)
  • Absolute outlet angle of jet, \(\beta = 60^\circ\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)

Solution

1. Analyze Velocity Triangles

First, we determine the relative velocity at the inlet. Since the jet strikes in the direction of motion, the inlet velocity triangle is a straight line.

$$ \text{Relative velocity at inlet, } V_{r1} = V_1 - u = 30 - 15 = 15 \, \text{m/s} $$

Assuming the vane is smooth, the relative velocity at the outlet equals the relative velocity at the inlet.

$$ V_{r2} = V_{r1} = 15 \, \text{m/s} $$

Now, we find the whirl velocity at the outlet (\(V_{w2}\)) using the outlet velocity triangle. We know \(V_{r2}\), \(u\), and the absolute outlet angle \(\beta\).

$$ V_{r2}^2 = V_{f2}^2 + (V_{w2} - u)^2 \quad \text{and} \quad \tan(\beta) = \frac{V_{f2}}{V_{w2}} $$ $$ \text{Substituting } V_{f2} = V_{w2} \tan(\beta): $$ $$ V_{r2}^2 = (V_{w2} \tan\beta)^2 + (V_{w2} - u)^2 $$ $$ 15^2 = (V_{w2} \tan 60^\circ)^2 + (V_{w2} - 15)^2 $$ $$ 225 = 3 V_{w2}^2 + V_{w2}^2 - 30V_{w2} + 225 $$ $$ 0 = 4V_{w2}^2 - 30V_{w2} \implies 0 = V_{w2}(4V_{w2} - 30) $$ $$ \text{This gives } V_{w2} = \frac{30}{4} = 7.5 \, \text{m/s} $$

(i) Force Exerted by the Jet (\(F_x\))

The force is the rate of change of momentum in the direction of motion. The mass flow rate striking the vane is based on the relative velocity.

$$ \text{Area of jet, } A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.05)^2 \approx 0.001963 \, \text{m}^2 $$ $$ \text{Mass flow rate, } \dot{m} = \rho A V_{r1} = 1000 \times 0.001963 \times 15 \approx 29.45 \, \text{kg/s} $$ $$ \text{Whirl velocity at inlet, } V_{w1} = V_1 = 30 \, \text{m/s} $$ $$ F_x = \dot{m} (V_{w1} - V_{w2}) $$ $$ F_x = 29.45 \times (30 - 7.5) = 29.45 \times 22.5 \approx 662.6 \, \text{N} $$

(ii) Work Done per Second by the Jet (Power)

Work done per second is the force exerted on the vane multiplied by the velocity of the vane.

$$ \text{Work Done per second} = F_x \times u $$ $$ \text{Work Done} = 662.6 \, \text{N} \times 15 \, \text{m/s} \approx 9939 \, \text{J/s} \text{ or } \text{W} $$
Final Results:

(i) Force exerted by the jet: \( \approx 662.6 \, \text{N} \)

(ii) Work done per second: \( \approx 9939 \, \text{W} \) (or 9.94 kW)

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