Calculate the power delivered by water to the runner of a Pelton wheel turbine and the hydraulic efficiency of the turbine. The Pelton wheel has a bucket speed of 10 m/s with a jet of water flowing at a rate of 700 L/s under a head of 30 meters. The bucket deflects the jet to an angle of 160 degrees, and the coefficient of velocity ( C v C v ​ ) is 0.298.

Problem Statement

Calculate the power delivered by water to the runner of a Pelton wheel turbine and the hydraulic efficiency of the turbine. The Pelton wheel has the following parameters:

• Bucket speed, $u=10\text{m/s}$

• Flow rate, $Q=700\text{L/s}=0.7{\text{m}}^3\mathrm{/}\text{s}$

• Head, $H=30\text{m}$

• Jet deflection angle, $\theta =160^{\circ }$

• Coefficient of velocity, $C_v=0.98$ (corrected from the unrealistic $0.298$)

• Acceleration due to gravity, $g=9.81{\text{m/s}}^2$

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Step 1: Calculate Jet Velocity ($V_1$)

The velocity of the water jet is given by:

$$V_1=C_v\sqrt{2gH}$$

Substituting the values:

$$V_1=0.98\times \sqrt{2\times 9.81\times 30}$$$$V_1=0.98\times \sqrt{588.6}$$$$V_1=0.98\times 24.26$$$$V_1\approx 23.77\text{m/s}$$

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Step 2: Calculate Power Delivered to the Runner ($P$)

The power delivered to the runner of a Pelton wheel is given by:

$$P=\rho Qu(V_1-u)(1-\cos \theta )$$

Where:

• $\rho =1000{\text{kg/m}}^3$ (density of water)

• $Q=0.7{\text{m}}^3\mathrm{/}\text{s}$

• $u=10\text{m/s}$

• $V_1=23.77\text{m/s}$

• $\theta =160^{\circ }$, so $\cos 160^{\circ }=-0.9397$, and $1-\cos \theta =1.9397$

Substituting the values:

$$P=1000\times 0.7\times 10\times (23.77-10)\times 1.9397$$$$P=1000\times 0.7\times 10\times 13.77\times 1.9397$$$$P\approx 184.6\text{kW}$$

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Step 3: Calculate Hydraulic Efficiency (${\eta }_h$)

The hydraulic efficiency is given by:

$${\eta }_h=\frac{\text{Power Delivered to Runner}}{\text{Water Power}}\times 100$$

First, calculate the water power ($P_w$):

$$P_w=\rho gQH$$

Substituting the values:

$$P_w=1000\times 9.81\times 0.7\times 30$$$$P_w=206.01\text{kW}$$

Now, calculate the efficiency:

$${\eta }_h=\frac{184.6}{206.01}\times 100$$$${\eta }_h\approx 89.6\mathrm{\%}$$

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Final Answer

• Power Delivered to Runner: $184.6\text{kW}$

• Hydraulic Efficiency: $89.6\mathrm{\%}$

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Note

The original problem had unrealistic values for the coefficient of velocity ($C_v=0.298$) and bucket speed ($u=10\text{m/s}$), which would result in invalid calculations. The solution above assumes a corrected $C_v=0.98$, which is typical for Pelton wheels. Always ensure that the bucket speed ($u$) is less than the jet velocity ($V_1$) for valid results.