A water-cooled engine produces 1589 W of power. Water enters the engine block at 14.61 °C and exits at 29.91 °C. The rate of water flow is 147.3 L/h. What is the engine’s efficiency?

Answer 1

You know that the mechanical power out is
W = 1589 J/s

The mass flow rate of coolant is
dm/dt = 147.3 l/h = (147.3L/h*10^3 g/L)/3600s/h
= 40.9 g/s

The heat energy (rate) out to the cooling water is
Qout = C*(dm/dt)*deltaT
= 4.18 J/(°C*g)*40.9 g/s *15.3 °C
= 2617 J/s

The heat energy rate in = Qout + Wout
= 2617 + 1589 = 4206 J/s

Efficiency = Wout/(Qin) = 1589/4206 = 37.8%