A jet of water having a velocity of 30 m/s strikes a curved vane, which is moving with a velocity of 15 m/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 : (i) Vane angles, if the water enters and leaves the vane without shock, (ii) Work done per second per unit weight of water striking the vanes per second.

Analysis of a Jet on a Moving Curved Vane

Problem Statement

Given Data & Constants

Velocity of jet, $V_1 = 30 \, \text{m/s}$
Velocity of vane, $u = 15 \, \text{m/s}$
Inlet angle of jet, $\alpha = 30^\circ$
Outlet angle of jet, $\beta = 120^\circ$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Inlet Velocity Triangle Analysis

First, we resolve the absolute inlet velocity ($V_1$) into its whirl ($V_{w1}$) and flow ($V_{f1}$) components.

$$ V_{w1} = V_1 \cos(\alpha) = 30 \times \cos(30^\circ) \approx 25.98 \, \text{m/s} $$ $$ V_{f1} = V_1 \sin(\alpha) = 30 \times \sin(30^\circ) = 15.0 \, \text{m/s} $$

The inlet vane angle ($\theta$) is determined by the condition that water enters without shock.

$$ \tan(\theta) = \frac{V_{f1}}{V_{w1} - u} = \frac{15.0}{25.98 - 15} = \frac{15.0}{10.98} \approx 1.366 $$ $$ \theta = \arctan(1.366) \approx 53.79^\circ $$

2. Outlet Velocity Triangle Analysis

First, we find the magnitude of the relative velocity at the inlet ($V_{r1}$). Assuming a smooth vane, this will be equal to the relative velocity at the outlet ($V_{r2}$).

$$ V_{r1} = \frac{V_{f1}}{\sin(\theta)} = \frac{15.0}{\sin(53.79^\circ)} \approx 18.59 \, \text{m/s} $$ $$ V_{r2} = V_{r1} = 18.59 \, \text{m/s} $$

Now we use the outlet velocity triangle to find the outlet vane angle ($\phi$). The absolute velocity at the outlet ($V_2$) leaves at an angle $\beta=120^\circ$. The whirl component $V_{w2}$ is $u - V_{r2}\cos\phi$.

$$ \tan(180^\circ - \beta) = \frac{V_{f2}}{u - V_{w2}} $$ $$ \text{From the geometry, } V_{w2} = u - V_{r2}\cos\phi \text{ and } V_{f2} = V_{r2}\sin\phi $$ $$ \tan(180^\circ - 120^\circ) = \tan(60^\circ) = \frac{V_{r2}\sin\phi}{-(u - V_{r2}\cos\phi)} $$ $$ \text{This seems complex. Let's use components directly.} $$ $$ V_{w2} = u + V_{r2}\cos\phi \text{ (since } \phi \text{ is obtuse)} $$ $$ \text{Let's use the Law of Cosines on the outlet triangle with sides } u, V_{r2}, V_2 \text{ and angle } \phi. $$ $$ \text{A simpler way is to use components. The horizontal component of absolute velocity } V_2 \text{ is } V_{w2}. $$ $$ V_{w2} = V_2 \cos(120^\circ) $$ $$ \text{Also, } V_{w2} = u + V_{r2}\cos(180-\phi) = u - V_{r2}\cos\phi $$ $$ \text{And } V_{f2} = V_2 \sin(120^\circ) = V_{r2}\sin(180-\phi) = V_{r2}\sin\phi $$ $$ \text{From the last part: } V_2 = \frac{V_{r2}\sin\phi}{\sin(120^\circ)} = \frac{18.59\sin\phi}{0.866} = 21.46\sin\phi $$ $$ \text{Substituting into the whirl equation: } (21.46\sin\phi)\cos(120^\circ) = 15 - 18.59\cos\phi $$ $$ -10.73\sin\phi = 15 - 18.59\cos\phi \implies 18.59\cos\phi - 10.73\sin\phi = 15 $$ $$ \text{Solving this gives } \phi \approx 15.65^\circ $$

(i) Vane Angles

The calculated vane angles for shockless entry and exit are:

$$ \text{Inlet Vane Angle, } \theta \approx 53.8^\circ $$ $$ \text{Outlet Vane Angle, } \phi \approx 15.7^\circ $$

(ii) Work Done per Unit Weight of Water

First, we find the whirl velocity at the outlet, $V_{w2}$.

$$ V_{w2} = u - V_{r2}\cos\phi = 15 - 18.59\cos(15.65^\circ) \approx 15 - 17.9 = -2.9 \, \text{m/s} $$

The work done is calculated from the change in whirl velocity.

$$ \text{Work Done} = \frac{(V_{w1} - V_{w2})u}{g} $$ $$ \text{Work Done} = \frac{(25.98 - (-2.9)) \times 15}{9.81} = \frac{28.88 \times 15}{9.81} $$ $$ \text{Work Done} \approx 44.16 \, \frac{\text{N-m}}{\text{N}} $$

Final Results:

(i) Vane Angles: Inlet $ \theta \approx 53.8^\circ $, Outlet $ \phi \approx 15.7^\circ $

(ii) Work done per unit weight of water: $ \approx 44.16 \, \text{m} $