Posted by linda on Saturday, January 7, 2012 at 11:04am.
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%
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