An inward flow reaction turbine has an external diameter of 1 m and its breadth at inlet is 200 mm. If the velocity of flow at inlet is 1.5 m/s, find the mass of water passing through the turbine per second. Assume 15% of the area of flow is blocked by blade thickness. If the speed of the runner is 200 r.p.m. and guide blades make an angle of 15° to the wheel tangent, draw the inlet velocity triangle and find: (i) The runner vane angle at inlet (ii) Velocity of wheel at inlet, (iii) The absolute velocity of water leaving the guide vanes, and (iv) The relative velocity of water entering the runner blade.

Inward Flow Reaction Turbine - Inlet Velocity Triangle

Problem Statement

Given Data & Constants

External (inlet) diameter, $D_1 = 1 \, \text{m}$
Breadth at inlet, $b_1 = 200 \, \text{mm} = 0.2 \, \text{m}$
Velocity of flow at inlet, $V_{f1} = 1.5 \, \text{m/s}$
Area blockage = 15% (Effective Area = 85%)
Speed, $N = 200 \, \text{r.p.m.}$
Guide blade angle, $\alpha = 15^\circ$
Density of water, $\rho = 1000 \, \text{kg/m}^3$

Solution

1. Mass of Water Passing Through Turbine per Second

First, calculate the effective flow area, then the volume flow rate (discharge), and finally the mass flow rate.

$$ \text{Gross Area, } A_g = \pi D_1 b_1 = \pi \times 1 \times 0.2 = 0.6283 \, \text{m}^2 $$ $$ \text{Effective Area, } A_e = A_g \times (1 - 0.15) = 0.6283 \times 0.85 \approx 0.534 \, \text{m}^2 $$ $$ \text{Discharge, } Q = A_e \times V_{f1} = 0.534 \times 1.5 \approx 0.801 \, \text{m}^3/\text{s} $$ $$ \text{Mass Flow Rate, } \dot{m} = \rho \times Q = 1000 \times 0.801 = 801 \, \text{kg/s} $$

(ii) Velocity of Wheel at Inlet ($u_1$)

$$ u_1 = \frac{\pi D_1 N}{60} = \frac{\pi \times 1 \times 200}{60} \approx 10.47 \, \text{m/s} $$

(iii) Absolute Velocity of Water ($V_1$)

From the inlet velocity triangle, we can find the absolute velocity using the guide blade angle and the velocity of flow.

$$ \sin(\alpha) = \frac{V_{f1}}{V_1} \implies V_1 = \frac{V_{f1}}{\sin(\alpha)} $$ $$ V_1 = \frac{1.5}{\sin(15^\circ)} = \frac{1.5}{0.2588} \approx 5.796 \, \text{m/s} $$

Inlet Velocity Triangle Analysis

First, we need the whirl velocity component ($V_{w1}$) to solve for the remaining parts.

$$ \tan(\alpha) = \frac{V_{f1}}{V_{w1}} \implies V_{w1} = \frac{V_{f1}}{\tan(\alpha)} $$ $$ V_{w1} = \frac{1.5}{\tan(15^\circ)} = \frac{1.5}{0.2679} \approx 5.599 \, \text{m/s} $$

(i) The Runner Vane Angle at Inlet ($\theta$)

$$ \tan(\theta) = \frac{V_{f1}}{u_1 - V_{w1}} $$ $$ \tan(\theta) = \frac{1.5}{10.47 - 5.599} = \frac{1.5}{4.871} \approx 0.3079 $$ $$ \theta = \arctan(0.3079) \approx 17.1^\circ $$

(iv) The Relative Velocity of Water ($V_{r1}$)

$$ \sin(\theta) = \frac{V_{f1}}{V_{r1}} \implies V_{r1} = \frac{V_{f1}}{\sin(\theta)} $$ $$ V_{r1} = \frac{1.5}{\sin(17.1^\circ)} = \frac{1.5}{0.294} \approx 5.10 \, \text{m/s} $$

Final Results:

Mass of water passing per second: $ \approx 801 \, \text{kg/s} $

(i) Runner vane angle at inlet: $ \theta \approx 17.1^\circ $

(ii) Velocity of wheel at inlet: $ u_1 \approx 10.47 \, \text{m/s} $

(iii) Absolute velocity of water: $ V_1 \approx 5.80 \, \text{m/s} $

(iv) Relative velocity of water: $ V_{r1} \approx 5.10 \, \text{m/s} $

Problem Statement

Given Data & Constants

Solution

1. Mass of Water Passing Through Turbine per Second

(ii) Velocity of Wheel at Inlet (\(u_1\))

(iii) Absolute Velocity of Water (\(V_1\))

Inlet Velocity Triangle Analysis

(i) The Runner Vane Angle at Inlet (\(\theta\))

(iv) The Relative Velocity of Water (\(V_{r1}\))

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Problem Statement

Given Data & Constants

Solution

1. Mass of Water Passing Through Turbine per Second

(ii) Velocity of Wheel at Inlet (\(u_1\))

(iii) Absolute Velocity of Water (\(V_1\))

Inlet Velocity Triangle Analysis

(i) The Runner Vane Angle at Inlet (\(\theta\))

(iv) The Relative Velocity of Water (\(V_{r1}\))

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