A Francis turbine with an overall efficiency of 70% is required to produce 147.15 kW. It is working under a head of 8 m. The wheel runs at 200 r.p.m. and the hydraulic losses in the turbine are 20% of the available energy. Assume radial discharge, determine : (i) The guide blade angle, (ii) The wheel vane angle at inlet, (iii) Diameter of the wheel at inlet, and (iv) Width of the wheel at inlet.

Leave a Comment / Fluid Mechanics, Numerical (Water) / By
Francis Turbine Design Calculation

Problem Statement

A Francis turbine with an overall efficiency of 70% is required to produce 147.15 kW. It is working under a head of 8 m. The peripheral velocity = \(0.30\sqrt{2gH}\) and the radial velocity of flow at inlet is \(0.96\sqrt{2gH}\). The wheel runs at 200 r.p.m. and the hydraulic losses in the turbine are 20% of the available energy. Assume radial discharge, determine : (i) The guide blade angle, (ii) The wheel vane angle at inlet, (iii) Diameter of the wheel at inlet, and (iv) Width of the wheel at inlet.

Given Data & Constants

  • Overall efficiency, \(\eta_o = 70\% = 0.70\)
  • Shaft Power, \(P_s = 147.15 \, \text{kW} = 147150 \, \text{W}\)
  • Head, \(H = 8 \, \text{m}\)
  • Peripheral velocity, \(u_1 = 0.30\sqrt{2gH}\)
  • Velocity of flow, \(V_{f1} = 0.96\sqrt{2gH}\)
  • Speed, \(N = 200 \, \text{r.p.m.}\)
  • Hydraulic efficiency, \(\eta_h = 1 - 0.20 = 0.80\)
  • 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 Key Velocities

First, we calculate the base velocity term from the head.

$$ \sqrt{2gH} = \sqrt{2 \times 9.81 \times 8} \approx 12.53 \, \text{m/s} $$ $$ \text{Peripheral Velocity, } u_1 = 0.30 \times 12.53 = 3.759 \, \text{m/s} $$ $$ \text{Velocity of Flow, } V_{f1} = 0.96 \times 12.53 = 12.029 \, \text{m/s} $$

(iii) Diameter of the Wheel at Inlet (\(D_1\))

The diameter is calculated from the peripheral velocity and the rotational speed.

$$ u_1 = \frac{\pi D_1 N}{60} \implies D_1 = \frac{60 u_1}{\pi N} $$ $$ D_1 = \frac{60 \times 3.759}{\pi \times 200} \approx 0.359 \, \text{m} $$

2. Determine Whirl Velocity at Inlet (\(V_{w1}\))

Using the hydraulic efficiency and the Euler turbine equation (with \(V_{w2}=0\)).

$$ \eta_h = \frac{\text{Work Done per Unit Weight}}{H} = \frac{(V_{w1} u_1)/g}{H} $$ $$ V_{w1} = \frac{\eta_h \cdot g \cdot H}{u_1} $$ $$ V_{w1} = \frac{0.80 \times 9.81 \times 8}{3.759} \approx 16.71 \, \text{m/s} $$

(i) The Guide Blade Angle (\(\alpha\))

This is found from the components of the absolute velocity at the inlet.

$$ \tan(\alpha) = \frac{V_{f1}}{V_{w1}} = \frac{12.029}{16.71} \approx 0.7199 $$ $$ \alpha = \arctan(0.7199) \approx 35.75^\circ $$

(ii) The Wheel Vane Angle at Inlet (\(\theta\))

This is found from the components of the relative velocity at the inlet.

$$ \tan(\theta) = \frac{V_{f1}}{V_{w1} - u_1} $$ $$ \tan(\theta) = \frac{12.029}{16.71 - 3.759} = \frac{12.029}{12.951} \approx 0.9288 $$ $$ \theta = \arctan(0.9288) \approx 42.89^\circ $$

(iv) Width of the Wheel at Inlet (\(b_1\))

First, calculate the required discharge (Q) using the overall efficiency and shaft power.

$$ P_w = \frac{P_s}{\eta_o} = \frac{147150}{0.70} = 210214 \, \text{W} $$ $$ Q = \frac{P_w}{\rho g H} = \frac{210214}{1000 \times 9.81 \times 8} \approx 2.678 \, \text{m}^3/\text{s} $$ $$ \text{Now, find the width from the discharge formula:} $$ $$ Q = \pi D_1 b_1 V_{f1} \implies b_1 = \frac{Q}{\pi D_1 V_{f1}} $$ $$ b_1 = \frac{2.678}{\pi \times 0.359 \times 12.029} \approx 0.197 \, \text{m} $$
Final Design Parameters:

(i) Guide blade angle: \( \alpha \approx 35.75^\circ \)

(ii) Wheel vane angle at inlet: \( \theta \approx 42.89^\circ \)

(iii) Diameter of the wheel at inlet: \( D_1 \approx 0.359 \, \text{m} \) or \(359 \, \text{mm}\)

(iv) Width of the wheel at inlet: \( b_1 \approx 0.197 \, \text{m} \) or \(197 \, \text{mm}\)

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top