A jet of water having a velocity of 20 m/s strikes a curved vane which is moving with a velocity of 9 m/s. The vane is symmetrical and is so shaped that the jet is deflected through 120°. Find the angle of the jet at inlet of the vane so that there is no shock. What is the absolute velocity of the jet at outlet in magnitude and direction and the work done per second per unit weight of water striking?

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Velocity Triangle Analysis for a Moving Vane

Problem Statement

A jet of water having a velocity of 20 m/s strikes a curved vane which is moving with a velocity of 9 m/s. The vane is symmetrical and is so shaped that the jet is deflected through 120°. Find the angle of the jet at inlet of the vane so that there is no shock. What is the absolute velocity of the jet at outlet in magnitude and direction and the work done per second per unit weight of water striking?

Given Data & Constants

  • Velocity of jet, \(V_1 = 20 \, \text{m/s}\)
  • Velocity of vane, \(u = 9 \, \text{m/s}\)
  • Jet deflection angle = 120°
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Determine Vane Angles (\(\theta, \phi\))

For a symmetrical vane, the inlet and outlet angles are equal (\(\theta = \phi\)). The deflection angle of the jet relative to the vane is \(180^\circ - (\theta + \phi)\).

$$ 120^\circ = 180^\circ - 2\theta $$ $$ 2\theta = 180^\circ - 120^\circ = 60^\circ $$ $$ \theta = \phi = 30^\circ $$

2. Inlet Velocity Triangle Analysis

For no shock at inlet, we find the required jet angle \(\alpha\).

$$ \tan\theta = \frac{V_{f1}}{V_{w1} - u} = \frac{V_1 \sin\alpha}{V_1 \cos\alpha - u} $$ $$ \tan(30^\circ) = \frac{20 \sin\alpha}{20 \cos\alpha - 9} $$ $$ 0.577(20 \cos\alpha - 9) = 20 \sin\alpha $$ $$ 11.547 \cos\alpha - 5.196 = 20 \sin\alpha $$ $$ \text{Solving this trigonometric equation gives } \alpha \approx 17^\circ $$

3. Outlet Velocity Triangle Analysis

First, find the relative velocity \(V_{r1}\). Assuming a smooth vane, \(V_{r2} = V_{r1}\).

$$ V_{w1} = 20 \cos(17^\circ) \approx 19.13 \, \text{m/s} $$ $$ V_{f1} = 20 \sin(17^\circ) \approx 5.85 \, \text{m/s} $$ $$ V_{r1}^2 = V_{f1}^2 + (V_{w1} - u)^2 = 5.85^2 + (19.13 - 9)^2 = 34.22 + 102.62 = 136.84 $$ $$ V_{r1} \approx 11.70 \, \text{m/s} \implies V_{r2} = 11.70 \, \text{m/s} $$

Now find the absolute velocity at the outlet (\(V_2\)) and its direction (\(\beta\)).

$$ V_{w2} = u - V_{r2}\cos\phi = 9 - 11.70 \cos(30^\circ) = 9 - 10.13 = -1.13 \, \text{m/s} $$ $$ V_{f2} = V_{r2}\sin\phi = 11.70 \sin(30^\circ) = 5.85 \, \text{m/s} $$ $$ V_2 = \sqrt{V_{w2}^2 + V_{f2}^2} = \sqrt{(-1.13)^2 + 5.85^2} \approx \sqrt{1.28 + 34.22} \approx 5.96 \, \text{m/s} $$ $$ \tan(180^\circ - \beta) = \frac{V_{f2}}{|V_{w2}|} = \frac{5.85}{1.13} \approx 5.177 \implies 180^\circ - \beta \approx 79.06^\circ $$ $$ \beta \approx 100.94^\circ $$

4. Work Done per Unit Weight of Water

$$ \text{Work Done} = \frac{1}{g} (V_{w1} + V_{w2})u \quad (\text{Note: + sign because } V_{w2} \text{ is in opposite direction}) $$ $$ \text{Work Done} = \frac{1}{9.81} (19.13 + 1.13) \times 9 = \frac{182.34}{9.81} \approx 18.59 \, \frac{\text{N-m}}{\text{N}} $$
Final Results:

Angle of the jet at inlet: \( \alpha \approx 17^\circ \)

Absolute velocity at outlet: \( V_2 \approx 5.96 \, \text{m/s} \) at an angle \( \beta \approx 100.9^\circ \) from the direction of motion (or 79.1° from the reverse direction).

Work done per unit weight of water: \( \approx 18.59 \, \text{m} \)

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