MP Board · Class 10 · Science · Sources of EnergyA hydroelectric power plant generates electrical energy by utilizing the potential energy of stored water. If a waterfall has a height of 50 m and water is falling at a rate of 1000 kg/s, calculate the total potential energy available per second. Assuming that 40% of this potential energy is successfully converted into electrical energy, determine the total electrical power generated by the plant. (Take acceleration due to gravity, g = 9.8 m/s²).

Step-by-Step Solution:

Given Data:

• Height of the waterfall ($h$) = $50\text{ m}$
• Rate of mass flow of water ($\frac{m}{t}$) = $1000\text{ kg/s}$
• Acceleration due to gravity ($g$) = $9.8\text{ m/s}^2$
• Efficiency of energy conversion ($\eta$) = $40% = 0.40$

Formula:

1. Potential Energy ($PE$) = $m \times g \times h$
2. Power available per second ($P_{available}$) = $\frac{PE}{t} = \left(\frac{m}{t}\right) \times g \times h$
3. Electrical Power generated ($P_{electrical}$) = $\eta \times P_{available}$

Step 1: Calculate the total potential energy available per second. $$P_{available} = \left(\frac{m}{t}\right) \times g \times h$$ $$P_{available} = 1000 \text{ kg/s} \times 9.8 \text{ m/s}^2 \times 50 \text{ m}$$ $$P_{available} = 1000 \times 490$$ $$P_{available} = 490,000 \text{ Watts} = 490 \text{ kW}$|

Step 2: Calculate the electrical power generated. $$P_{electrical} = 40% \text{ of } 490,000 \text{ W}$$ $$P_{electrical} = 0.40 \times 490,000 \text{ W}$$ $$P_{electrical} = 196,000 \text{ W} = 196 \text{ kW}$$

Final Answer:\nThe total potential energy available per second is $490 \text{ kW}$ and the total electrical power generated by the plant is $196 \text{ kW}$ (or $0.196 \text{ MW}$).