A Pelton wheel is having a mean bucket diameter of 0.8 m and is running at 1000 r.p.m. The net head on the Pelton wheel is 400 m. If the side clearance angle is 15° and discharge through nozzle is 150 litres/s, find : (i) Power available at the nozzle, and (ii) Hydraulic efficiency of the turbine.

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Pelton Wheel Efficiency Calculation

Problem Statement

A Pelton wheel is having a mean bucket diameter of 0.8 m and is running at 1000 r.p.m. The net head on the Pelton wheel is 400 m. If the side clearance angle is 15° and discharge through nozzle is 150 litres/s, find : (i) Power available at the nozzle, and (ii) Hydraulic efficiency of the turbine.

Given Data & Constants

  • Mean bucket diameter, \(D = 0.8 \, \text{m}\)
  • Speed, \(N = 1000 \, \text{r.p.m.}\)
  • Net Head, \(H = 400 \, \text{m}\)
  • Side clearance angle (outlet blade angle), \(\phi = 15^\circ\)
  • Discharge, \(Q = 150 \, \text{L/s} = 0.15 \, \text{m}^3/\text{s}\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

(i) Power Available at the Nozzle (Water Power)

This is the total potential energy of the water entering the turbine per second.

$$ \text{Water Power, } P_w = \rho g Q H $$ $$ P_w = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.15 \, \text{m}^3/\text{s} \times 400 \, \text{m} $$ $$ P_w = 588600 \, \text{W} $$

(ii) Hydraulic Efficiency of the Turbine (\(\eta_h\))

First, we need to calculate the power delivered to the runner. For this, we need the jet velocity (\(V_1\)) and the bucket velocity (\(u\)).

$$ \text{Jet Velocity, } V_1 = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 400} \approx 88.59 \, \text{m/s} \quad (\text{Assuming } C_v=1) $$ $$ \text{Bucket Velocity, } u = \frac{\pi D N}{60} = \frac{\pi \times 0.8 \times 1000}{60} \approx 41.89 \, \text{m/s} $$

Now, calculate the power delivered to the runner.

$$ \text{Runner Power, } P_r = \rho Q (V_1 - u)(1 + \cos\phi)u $$ $$ P_r = 1000 \times 0.15 \times (88.59 - 41.89)(1 + \cos(15^\circ)) \times 41.89 $$ $$ P_r = 150 \times 46.7 \times (1 + 0.9659) \times 41.89 $$ $$ P_r = 150 \times 46.7 \times 1.9659 \times 41.89 $$ $$ P_r \approx 576430 \, \text{W} $$

Finally, the hydraulic efficiency is the ratio of runner power to water power.

$$ \eta_h = \frac{\text{Runner Power}}{\text{Water Power}} = \frac{P_r}{P_w} $$ $$ \eta_h = \frac{576430}{588600} \approx 0.9793 $$
Final Results:

(i) Power available at the nozzle: \( 588.6 \, \text{kW} \)

(ii) Hydraulic efficiency of the turbine: \( \approx 97.9\% \)

Explanation of Pelton Wheel Performance

  • Power at Nozzle (Water Power): This represents the total energy available from the water source. It's the theoretical maximum power that can be extracted, based on the flow rate and the height (head) the water is falling from.
  • Runner Power: This is the actual mechanical power transferred from the water jet to the rotating wheel (the runner). It is calculated by analyzing the change in the water's momentum as it strikes the buckets and is deflected.
  • Hydraulic Efficiency: This crucial metric tells us how effectively the turbine converts the available water power into mechanical power at the runner. An efficiency of 97.9% is very high, indicating an excellent design where minimal energy is lost due to fluid dynamic effects like splash and friction as the water interacts with the buckets.

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