An electric adaptor for a notebook computer converting 110 v

An electric adaptor for a notebook computer (converting 110 volts to 21 volts) operates 11 degree C warmer than the surroundings, which is at 18 degree C. The output current is measured at 5 amps and heat is lost from the adapter at a rate of 11 W. Determine the energetic efficiency of the device. Express your answer to three significant figures. Determine the rate of internal entropy generation. Express your answer to three significant figures. Determine the rate of external entropy generation. Express your answer to three significant figures.

Solution

Q_lost = 11 W

V_in = 110 v

V_out = 21 v

T_c = (11+18) = 29 ^oC

T_s = 18 ^oC

Q_out = VI = 21*5 =105 watts

Q_in = Q_out + Q_lost

Q_in = 105+11 =116 watts

a) energetic effeciancy (n) = Q_out / Q_in = 105 / 116

n = 0.9052 or 90.52 %

b) rate of internal entropy genration = Q_lost / T_c

S _internal = -11 / (29+273 ) = - 0.036 W/K

c) rate of external entropy genration = Q_lost / T_s

S _external = 11 / (18+273 ) = 0.0378 W/K