The transformer on a utlity pole steps down voltage from 10,000 to 220 V and del
ID: 3278063 • Letter: T
Question
The transformer on a utlity pole steps down voltage from 10,000 to 220 V and deliver energy at the rate of 10 kW. a) The transformer is ideal, during that time, what are the primary and secondary currents of the transfomer? b) if the transformer is only 90% efficent (but still deliver electric power at 10kW) how does its input current compare to the ideal case? c)at what rate is heat lost in the nonideal tranformer? d) if you wanted to keep the transformer cool and to do this needed to disspiate half of the joule heating of part c using water cooling lines (the other half is taken care of air cooling) what should be the rate flow of liters per minute of water in the lines? Assume the input cool water is at 68 degrees F and the max allowable output water temperature is 98 degrees F.
Explanation / Answer
(A) Ip Vp = P
Ip (10,000) = 10 x 10^3
Ip = 1 A ..........Current in primary
Is Vs = P
Is (220) = 10 x 10^3
Is = 45.45 A ...............Current in secondary
(B) now Input power = 10kW/ 0.90 = 11.11 kW
Ip (10000) = (11.11 x 10^3)
Ip = 1.11 A
(C) heat lost = 11.11 - 10 = 1.11 kW
(D) Q = m C deltaT
Energy per minute will be, Q = 1.11 x 10^3 x 60 s = 66.7 x 10^3 J = 66.7 kJ
66.7 = m (4.186) ( 36.67 - 20)
{ 98 deg F = 36.67 deg C
68 deg F = 20 deg C }
m = 0.955 kg / min
{ 1 kg = 10^3 L}
m = 955 L / min
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