A Pelton wheel has a mean bucket speed of 35 m/s with a jet of water flowing at the rate of 1 m³/s under a head of 270 m. The buckets deflect the jet through an angle of 170°. Calculate the power delivered to the runner and the hydraulic efficiency of the turbine. Assume co-efficient of velocity as 0.98.

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Pelton Wheel Turbine Calculation

Problem Statement

A Pelton wheel has a mean bucket speed of 35 m/s with a jet of water flowing at the rate of 1 m³/s under a head of 270 m. The buckets deflect the jet through an angle of 170°. Calculate the power delivered to the runner and the hydraulic efficiency of the turbine. Assume co-efficient of velocity as 0.98.

Given Data & Constants

  • Mean bucket speed, \(u = 35 \, \text{m/s}\)
  • Discharge, \(Q = 1.0 \, \text{m}^3/\text{s}\)
  • Head, \(H = 270 \, \text{m}\)
  • Jet deflection angle = 170°
  • Outlet blade angle, \(\phi = 180^\circ - 170^\circ = 10^\circ\)
  • Co-efficient of velocity, \(C_v = 0.98\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate the Velocity of the Jet (\(V_1\))

The velocity of the water jet as it strikes the buckets is determined by the head and the co-efficient of velocity.

$$ V_1 = C_v \sqrt{2gH} $$ $$ V_1 = 0.98 \sqrt{2 \times 9.81 \times 270} = 0.98 \sqrt{5300.4} $$ $$ V_1 \approx 0.98 \times 72.80 \approx 71.34 \, \text{m/s} $$

2. Calculate the Power Delivered to the Runner

The power delivered to the runner (runner power) is calculated from the change in momentum of the water jet.

$$ \text{Power, } P = \rho Q (V_1 - u)(1 + \cos\phi)u $$ $$ P = 1000 \times 1.0 \times (71.34 - 35)(1 + \cos(10^\circ)) \times 35 $$ $$ P = 1000 \times 36.34 \times (1 + 0.9848) \times 35 $$ $$ P = 1000 \times 36.34 \times 1.9848 \times 35 $$ $$ P \approx 2523100 \, \text{W} $$

3. Calculate the Hydraulic Efficiency (\(\eta_h\))

First, we calculate the input power available from the water (water power). Then, we find the ratio of the runner power to the water power.

$$ \text{Water Power, } P_w = \rho g Q H $$ $$ P_w = 1000 \times 9.81 \times 1.0 \times 270 = 2648700 \, \text{W} $$ $$ \eta_h = \frac{\text{Runner Power}}{\text{Water Power}} = \frac{P}{P_w} $$ $$ \eta_h = \frac{2523100}{2648700} \approx 0.9526 $$
Final Results:

Power delivered to the runner: \( \approx 2523.1 \, \text{kW} \) (or 2.52 MW)

Hydraulic efficiency of the turbine: \( \approx 95.3\% \)

Explanation of Pelton Wheel Operation

A Pelton wheel is an impulse turbine, which means it extracts energy from the kinetic energy of a moving fluid, rather than from its pressure. The process involves a few key steps:

  • Nozzle: High-pressure water from a reservoir is converted into a high-velocity jet by a nozzle. The jet's velocity (\(V_1\)) is determined by the head (\(H\)).
  • Buckets: This high-velocity jet strikes a series of spoon-shaped buckets mounted on the rim of a wheel (the runner).
  • Momentum Transfer: The buckets are designed to turn the water jet through a large angle (170° here). This significant change in the direction of the water's momentum results in a large impulse force on the buckets, causing the runner to spin.
  • Power and Efficiency: The power delivered is the rate at which this impulse force does work. The hydraulic efficiency measures how effectively the kinetic energy of the jet is converted into mechanical work at the runner.

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