Design a Pelton wheel for a head of 80 m and speed 300 r.p.m. The Pelton wheel develops 103 kW S.P. Take C_v = 0.98, speed ratio = 0.45 and overall efficiency = 0.80.

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Pelton Wheel Design Calculation

Problem Statement

Design a Pelton wheel for a head of 80 m and speed 300 r.p.m. The Pelton wheel develops 103 kW S.P. Take C_v = 0.98, speed ratio = 0.45 and overall efficiency = 0.80.

Given Data & Constants

  • Head, \(H = 80 \, \text{m}\)
  • Speed, \(N = 300 \, \text{r.p.m.}\)
  • Shaft Power, \(P_s = 103 \, \text{kW} = 103000 \, \text{W}\)
  • Co-efficient of velocity, \(C_v = 0.98\)
  • Speed ratio, \(K_u = 0.45\)
  • Overall efficiency, \(\eta_o = 0.80\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate Jet Velocity (\(V_1\)) and Bucket Velocity (u)

The theoretical velocity is found from the head, then adjusted by the speed ratio and velocity coefficient.

$$ \text{Theoretical Velocity, } V_{th} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 80} \approx 39.62 \, \text{m/s} $$ $$ \text{Jet Velocity, } V_1 = C_v \times V_{th} = 0.98 \times 39.62 \approx 38.83 \, \text{m/s} $$ $$ \text{Bucket Velocity, } u = K_u \times V_{th} = 0.45 \times 39.62 \approx 17.83 \, \text{m/s} $$

2. Determine the Wheel Diameter (D)

The mean diameter of the wheel is calculated from the bucket velocity and rotational speed.

$$ u = \frac{\pi D N}{60} \implies D = \frac{60 u}{\pi N} $$ $$ D = \frac{60 \times 17.83}{\pi \times 300} \approx 1.135 \, \text{m} $$

3. Calculate Total Discharge Required (Q)

First, find the input water power required using the overall efficiency, then find the discharge.

$$ \text{Water Power, } P_w = \frac{\text{Shaft Power}}{\eta_o} = \frac{103000}{0.80} = 128750 \, \text{W} $$ $$ P_w = \rho g Q H \implies Q = \frac{P_w}{\rho g H} $$ $$ Q = \frac{128750}{1000 \times 9.81 \times 80} \approx 0.164 \, \text{m}^3/\text{s} $$

4. Determine the Diameter of the Jet (d)

The total discharge is a function of the jet area and jet velocity. Assuming a single jet.

$$ Q = \text{Area of jet} \times V_1 = a \times V_1 $$ $$ a = \frac{Q}{V_1} = \frac{0.164}{38.83} \approx 0.004223 \, \text{m}^2 $$ $$ a = \frac{\pi}{4} d^2 \implies d = \sqrt{\frac{4a}{\pi}} $$ $$ d = \sqrt{\frac{4 \times 0.004223}{\pi}} \approx 0.0733 \, \text{m} $$
Final Design Parameters:

Wheel diameter: \( D \approx 1.135 \, \text{m} \) or \(1135 \, \text{mm}\)

Diameter of the jet: \( d \approx 0.0733 \, \text{m} \) or \(73.3 \, \text{mm}\)

Total discharge required: \( Q \approx 0.164 \, \text{m}^3/\text{s} \) or \(164 \, \text{L/s}\)

Explanation of Design Parameters

  • Velocities: The head determines the theoretical maximum velocity of the water. The speed ratio (\(K_u\)) is an empirical value that defines the optimal bucket speed relative to this velocity for maximum efficiency. The coefficient of velocity (\(C_v\)) accounts for minor friction losses in the nozzle.
  • Wheel Diameter: This is a direct consequence of the required bucket speed and the rotational speed (RPM). A specific bucket speed at a given RPM dictates a unique wheel diameter.
  • Discharge and Jet Diameter: The overall efficiency tells us how much water power is needed to produce the required shaft power. From this required water power and the known head, we calculate the necessary flow rate (discharge). The diameter of the jet is then sized to deliver this exact flow rate at the calculated jet velocity.
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