Answer:
4.1%
Explanation:
Given that water specific heat c = 4186 J/kgC. If water is flowing with mass rate of 200kg/s and temperature is dropping from 200C to 25C. Then the thermal energy rate from the water should be
[tex]P_i = mc\Delta t = 200*4186*(200 - 25) = 146510000 J/s[/tex] or 146510kW
Since the turbine is only able to extract 6000kW of power, then the thermal efficiency is:
[tex]Eff = \frac{P_o}{P_i} = \frac{6000}{146510} = 0.041[/tex] or 4.1%