Question

At a hydroelectric power plant, the water pressure head is at a height of $300m$ and the water flow available is $100m^3{s}^{-1}$. If the turbine generator efficiency is $60\mathrm{\%}$, estimate the electric power available from the plant. $($Given $g=9.8{ms}^{-2})$

• (A) $176.4MW$
• (B) $170.4MW$
• (C) $172.4MW$
• (D) $160.4MW$

Answer 1

The Correct Option is A

To estimate the electric power available from the hydroelectric power plant, we use the formula for power generated by water flow through a turbine:

P = \eta \cdot \rho \cdot g \cdot h \cdot Q

where:

• \eta is the efficiency of the turbine generator (60% or 0.60).
• \rho is the density of water (approx. 1000 \, \text{kg/m}^3).
• g is the acceleration due to gravity (9.8 \, \text{m/s}^2).
• h is the height of the water pressure head (300 m).
• Q is the water flow rate (100 \, \text{m}^3/\text{s}).

Substituting the values into the formula:

P = 0.60 \times 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 300 \, \text{m} \times 100 \, \text{m}^3/\text{s}

Calculating the result:

P = 0.60 \times 1000 \times 9.8 \times 300 \times 100

= 0.60 \times 294,000,000 \, \text{W}

= 176,400,000 \, \text{W}

Since 1 MW (megawatt) is equal to 10^6 watts, we convert the result to megawatts:

P = 176.4 \, \text{MW}

Hence, the electric power available from the plant is 176.4 MW.

The correct answer is:

(A) 176.4 \, \text{MW}