MP Board · Class 10 · Science · Sources of EnergyA hydroelectric power plant station is situated at a height of 200 m above the water level. If 2 × 10^6 kg of water flows down per minute into the turbine generators, calculate the total power generated by the power plant in megawatts (MW), assuming 50% conversion efficiency of the plant. (Take g = 9.8 m/s²)

Given data:

• Height of the fall ($h$) = 200 m
• Mass of water flowing per minute ($m$) = $2 \times 10^6$ kg
• Time ($t$) = 1 minute = 60 seconds
• Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$
• Efficiency of the plant ($\eta$) = 50% = 0.50 \nStep 1: Calculate the potential energy of the water flowing per second.\nThe potential energy ($PE$) of water is given by the formula: $$PE = m \cdot g \cdot h$$\nSince mass is given per minute, let's find the mass of water per second ($m'$): $$m' = \frac{2 \times 10^6 \text{ kg}}{60 \text{ s}} = \frac{200,000}{6} \text{ kg/s} = 33,333.33 \text{ kg/s}$| \nStep 2: Calculate the total input power available from the falling water. $$\text{Input Power } (P_{\text{in}}) = \frac{PE}{t} = \frac{m \cdot g \cdot h}{t}$$\nSubstituting the values: $$P_{\text{in}} = \frac{2 \times 10^6 \text{ kg} \times 9.8 \text{ m/s}^2 \times 200 \text{ m}}{60 \text{ s}}$$ $$P_{\text{in}} = \frac{3.92 \times 10^9}{60}$$ $$P_{\text{in}} = 65,333,333.33 \text{ Watts (W)}$$ \nStep 3: Calculate the output power generated considering the 50% efficiency. $$\text{Output Power } (P_{\text{out}}) = P_{\text{in}} \times \text{Efficiency}$$ $$P_{\text{out}} = 65,333,333.33 \times 0.50$$ $$P_{\text{out}} = 32,666,666.67 \text{ W}$| \nStep 4: Convert the output power into Megawatts (MW). $$\text{Power in MW} = \frac{32,666,666.67}{10^6} \text{ MW} = 32.67 \text{ MW}$$ \nAnswer:\nThe total power generated by the hydroelectric power plant is 32.67 MW.