A Kaplan turbine working under a head of 15 m develops 7357.5 kW shaft power. The outer diameter of the runner is 4 m and hub diameter is 2 m. The guide blade angle at the extreme edge of the runner is 30°. The hydraulic and overall efficiencies of the turbine are 90% and 85% respectively. If the velocity of whirl is zero at outlet, determine : (i) runner vane angles at inlet and outlet at the extreme edge of the runner and (ii) speed of the turbine.

Kaplan Turbine Design Calculation

Problem Statement

Given Data & Constants

Head, $H = 15 \, \text{m}$
Shaft Power, $P_s = 7357.5 \, \text{kW} = 7357500 \, \text{W}$
Outer diameter, $D_o = 4 \, \text{m}$
Hub diameter, $D_h = 2 \, \text{m}$
Guide blade angle, $\alpha = 30^\circ$
Hydraulic efficiency, $\eta_h = 90\% = 0.90$
Overall efficiency, $\eta_o = 85\% = 0.85$
Radial discharge: $V_{w2} = 0$
Density of water, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Calculate Discharge (Q) and Velocity of Flow ($V_{f1}$)

First, find the required water power and the corresponding discharge.

$$ P_w = \frac{P_s}{\eta_o} = \frac{7357500}{0.85} \approx 8655882 \, \text{W} $$ $$ Q = \frac{P_w}{\rho g H} = \frac{8655882}{1000 \times 9.81 \times 15} \approx 58.82 \, \text{m}^3/\text{s} $$

Now, calculate the velocity of flow through the runner.

$$ \text{Area of flow, } A_f = \frac{\pi}{4} (D_o^2 - D_h^2) = \frac{\pi}{4} (4^2 - 2^2) = \frac{\pi}{4}(12) \approx 9.425 \, \text{m}^2 $$ $$ V_{f1} = \frac{Q}{A_f} = \frac{58.82}{9.425} \approx 6.24 \, \text{m/s} $$

2. Analyze Inlet Velocity Triangle

First, find the whirl velocity ($V_{w1}$) from the guide blade angle.

$$ \tan(\alpha) = \frac{V_{f1}}{V_{w1}} \implies V_{w1} = \frac{V_{f1}}{\tan(\alpha)} $$ $$ V_{w1} = \frac{6.24}{\tan(30^\circ)} = \frac{6.24}{0.5773} \approx 10.81 \, \text{m/s} $$

Now, use the hydraulic efficiency to find the peripheral velocity of the runner at the outer edge ($u_1$).

$$ \eta_h = \frac{V_{w1} u_1}{gH} \quad (\text{since } V_{w2}=0) $$ $$ u_1 = \frac{\eta_h g H}{V_{w1}} = \frac{0.90 \times 9.81 \times 15}{10.81} \approx 12.25 \, \text{m/s} $$

(ii) Speed of the Turbine (N)

The rotational speed is calculated from the peripheral velocity at the outer diameter.

$$ u_1 = \frac{\pi D_o N}{60} \implies N = \frac{60 u_1}{\pi D_o} $$ $$ N = \frac{60 \times 12.25}{\pi \times 4} \approx 58.48 \, \text{r.p.m.} $$

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

The analysis is done at the extreme edge ($D_o$), so $u_1 = u_2 = 12.25$ m/s and $V_{f1} = V_{f2} = 6.24$ m/s.

$$ \tan(\theta) = \frac{V_{f1}}{V_{w1} - u_1} = \frac{6.24}{10.81 - 12.25} = \frac{6.24}{-1.44} \approx -4.333 $$ $$ \theta = \arctan(-4.333) \approx -77.0^\circ \implies \theta \approx 180^\circ - 77.0^\circ = 103.0^\circ $$

For the outlet vane angle, since discharge is radial ($V_{w2}=0$):

$$ \tan(\phi) = \frac{V_{f2}}{u_2} = \frac{6.24}{12.25} \approx 0.509 $$ $$ \phi = \arctan(0.509) \approx 27.0^\circ $$

Final Results:

(i) Runner Vane Angles: Inlet $ \theta \approx 103.0^\circ $, Outlet $ \phi \approx 27.0^\circ $

(ii) Speed of the turbine: $ N \approx 58.5 \, \text{r.p.m.} $

Problem Statement

Given Data & Constants

Solution

1. Calculate Discharge (Q) and Velocity of Flow (\(V_{f1}\))

2. Analyze Inlet Velocity Triangle

(ii) Speed of the Turbine (N)

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

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Problem Statement

Given Data & Constants

Solution

1. Calculate Discharge (Q) and Velocity of Flow (\(V_{f1}\))

2. Analyze Inlet Velocity Triangle

(ii) Speed of the Turbine (N)

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

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