Q178 / 180 · COMEDK 2024 (Evening Shift) · Physics
When water falls from a height of $80 \mathrm{~m}$ at the rate of $20 \mathrm{~kg} \mathrm{~s}^{-1}$ to operate a turbine the losses due to frictional force are $20 \%$ of input energy. How much power is generated by the turbine?
- A25.6 KW
- Correct answerB12.8 KW
- C62.5 KW
- D21.6 KW
Solution
The potential energy of the water falling per unit time is the input power, given by $P_{in} = \dfrac{mgh}{t} = \left(\dfrac{m}{t}\right)gh$.
Given $\dfrac{m}{t} = 20 \text{ kg s}^{-1}$, $g = 10 \text{ m s}^{-2}$, and $h = 80 \text{ m}$, we have $P_{in} = 20 \times 10 \times 80 = 16000 \text{ W} = 16 \text{ kW}$.
The losses due to frictional force are $20 \%$ of the input energy, so the efficiency of the turbine is $100 \% - 20 \% = 80 \%$.
The power generated by the turbine is $P_{out} = 80 \% \text{ of } P_{in} = 0.8 \times 16 \text{ kW} = 12.8 \text{ kW}$.
Answer: 12.8 KW
Given $\dfrac{m}{t} = 20 \text{ kg s}^{-1}$, $g = 10 \text{ m s}^{-2}$, and $h = 80 \text{ m}$, we have $P_{in} = 20 \times 10 \times 80 = 16000 \text{ W} = 16 \text{ kW}$.
The losses due to frictional force are $20 \%$ of the input energy, so the efficiency of the turbine is $100 \% - 20 \% = 80 \%$.
The power generated by the turbine is $P_{out} = 80 \% \text{ of } P_{in} = 0.8 \times 16 \text{ kW} = 12.8 \text{ kW}$.
Answer: 12.8 KW
Hence, the correct answer is option: 'B'.