Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine? (Take g = 10 m/s^2 )

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine? (Take \( g = 10 \, \text{m/s}^2 \))

(a) 12.3 kW

(b) 7.0 kW

(c) 8.1 kW

(d) 10.2 kW

Detailed Answer:

The power generated by the turbine is the rate at which the water loses potential energy. The potential energy \(PE\) of water falling from a height \(h\) is given by the formula: \[ PE = mgh \] Where: - \(m\) is the mass of water falling per second (rate of mass flow), - \(g\) is the acceleration due to gravity, - \(h\) is the height. The rate of energy transfer is the power generated. We are given: - Mass flow rate \( \dot{m} = 15 \, \text{kg/s} \), - Height \( h = 60 \, \text{m} \), - Gravitational acceleration \( g = 10 \, \text{m/s}^2 \). The total potential energy per second (which is the power before losses) is: \[ P_{\text{total}} = \dot{m}gh = 15 \times 10 \times 60 = 9000 \, \text{J/s} = 9000 \, \text{W} = 9 \, \text{kW}. \] Now, considering the energy losses due to friction (10% of energy): \[ \text{Power after losses} = P_{\text{total}} \times (1 - \text{Loss percentage}) \] \[ \text{Power after losses} = 9 \, \text{kW} \times (1 - 0.10) = 9 \, \text{kW} \times 0.90 = 8.1 \, \text{kW}. \] Therefore, the power generated by the turbine is: \[ \boxed{8.1 \, \text{kW}} \ (\text{Option (c)}). \]