An electric motor operating at steady state draws a current

An electric motor operating at steady state draws a current of 20 amp at a voltage of 110 V. The power developed by the output shaft is 2.05 kW. The heat transfer rate is related to the average temperature at the surface of the motor, Tsurf, and the temperature of the surroundings, Tsurr = 25oC, by the expression for convective heat transfer in Eq. 2.34. In this case, hA = 4.2 W/K. For a closed system that encloses just the motor, determine the rate of entropy production, in W/K. Repeat for an enlarged system boundary such that the heat transfer occurs at the temperature of the surroundings, Tsurr. Discuss.

Solution

Energy input = VI = 110*20 = 2200 Watts

Energy output = 2.05 kW = 2050 Watts

Net energy in = 2200 - 2050 = 150 Watts

Heat transfer from surface Q = hA*(Tsurf - Tsurr)

150 = 4.2*(Tsurf - 25)

Tsurf = 60.7 deg C = 333.7 K

Rate of Entropy production = Q/Tsurf

= 150 / 333.7

= 0.4495 W/K

For enlarged system boundary, we\'ll have Rate of entropy production = Q / Tsurr

= 150 / (25+273)

= 0.503 W/K

With enlarged boundary, the rate of entropy production has increased due to lower temperature at which heat transfer occurs.