A jet of water having a velocity of 30 m/s, strikes a series of radial curved vanes mounted on a wheel which is rotating at 300 r.p.m. The jet makes an angle of 30° with the tangent to wheel at inlet and leaves the wheel with a velocity of 4 m/s at an angle of 120° to the tangent to the wheel at outlet. Water is flowing from outward in a radial direction. The outer and inner radii of the wheel are 0.6 m and 0.3 m respectively. Determine : (i) vane angles at inlet and outlet, (ii) work done per second per kg of water, and (iii) efficiency of the wheel.

Outward Flow Turbine Velocity Triangle Analysis

Problem Statement

Given Data & Constants

Absolute velocity at inlet, $V_1 = 30 \, \text{m/s}$
Speed, $N = 300 \, \text{r.p.m.}$
Jet angle at inlet, $\alpha = 30^\circ$
Absolute velocity at outlet, $V_2 = 4 \, \text{m/s}$
Jet angle at outlet, $\beta = 120^\circ$
Inner (inlet) radius, $r_1 = 0.3 \, \text{m}$
Outer (outlet) radius, $r_2 = 0.6 \, \text{m}$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Calculate Tangential Velocities ($u_1, u_2$)

For an outward flow turbine, the water enters at the inner radius and exits at the outer radius.

$$ u_1 = \frac{2 \pi N r_1}{60} = \frac{2 \pi \times 300 \times 0.3}{60} \approx 9.425 \, \text{m/s} $$ $$ u_2 = \frac{2 \pi N r_2}{60} = \frac{2 \pi \times 300 \times 0.6}{60} \approx 18.85 \, \text{m/s} $$

2. Analyze Inlet and Outlet Velocity Triangles

We resolve the absolute velocities into whirl ($V_w$) and flow ($V_f$) components.

$$ V_{w1} = V_1 \cos(\alpha) = 30 \times \cos(30^\circ) \approx 25.98 \, \text{m/s} $$ $$ V_{f1} = V_1 \sin(\alpha) = 30 \times \sin(30^\circ) = 15.0 \, \text{m/s} $$ $$ V_{w2} = V_2 \cos(180^\circ - \beta) = 4 \times \cos(60^\circ) = 2.0 \, \text{m/s} \quad (\text{Note: angle is measured from tangent, so we use } 180-\beta)$$ $$ V_{f2} = V_2 \sin(180^\circ - \beta) = 4 \times \sin(60^\circ) \approx 3.464 \, \text{m/s} $$

(i) Vane Angles at Inlet ($\theta$) and Outlet ($\phi$)

The vane angles are determined from the velocity triangle components.

$$ \tan(\theta) = \frac{V_{f1}}{V_{w1} - u_1} = \frac{15.0}{25.98 - 9.425} = \frac{15.0}{16.555} \approx 0.906 $$ $$ \theta = \arctan(0.906) \approx 42.18^\circ $$ $$ \tan(\phi) = \frac{V_{f2}}{u_2 - V_{w2}} = \frac{3.464}{18.85 - 2.0} = \frac{3.464}{16.85} \approx 0.2056 $$ $$ \phi = \arctan(0.2056) \approx 11.62^\circ $$

(ii) Work Done per Second per kg of Water

This is calculated using the Euler turbomachine equation for a turbine.

$$ \text{Work Done per kg} = (V_{w1} u_1 - V_{w2} u_2) $$ $$ \text{Work Done} = (25.98 \times 9.425) - (2.0 \times 18.85) $$ $$ \text{Work Done} = 244.88 - 37.7 = 207.18 \, \text{J/kg} $$

(iii) Efficiency of the Wheel

The efficiency is the ratio of the work done to the initial kinetic energy of the jet, per unit mass.

$$ \text{Initial Energy per kg} = \frac{V_1^2}{2} = \frac{30^2}{2} = 450 \, \text{J/kg} $$ $$ \eta = \frac{\text{Work Done per kg}}{\text{Initial Energy per kg}} = \frac{207.18}{450} \approx 0.4604 $$

Final Results:

(i) Vane Angles: Inlet $ \theta \approx 42.2^\circ $, Outlet $ \phi \approx 11.6^\circ $

(ii) Work done per second per kg of water: $ \approx 207.2 \, \text{J/kg} $

(iii) Efficiency of the wheel: $ \approx 46.0\% $

Problem Statement

Given Data & Constants

Solution

1. Calculate Tangential Velocities (\(u_1, u_2\))

2. Analyze Inlet and Outlet Velocity Triangles

(i) Vane Angles at Inlet (\(\theta\)) and Outlet (\(\phi\))

(ii) Work Done per Second per kg of Water

(iii) Efficiency of the Wheel

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Problem Statement

Given Data & Constants

Solution

1. Calculate Tangential Velocities (\(u_1, u_2\))

2. Analyze Inlet and Outlet Velocity Triangles

(i) Vane Angles at Inlet (\(\theta\)) and Outlet (\(\phi\))

(ii) Work Done per Second per kg of Water

(iii) Efficiency of the Wheel

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