Thanks for having the time to read my question. A 50kVA, 1-phase, 60 Hz, 4400/20
ID: 2991177 • Letter: T
Question
Thanks for having the time to read my question.
A 50kVA, 1-phase, 60 Hz, 4400/200 V distribution transformer has a maximum efficiency at full load. The transformer gave the following test results
Open-circuit test: 4400 V applied to HT side, power taken 320 W.
Short-circuit test: Secondary short-circuited, power taken 320 W at full load.
Calculate the transformer energy efficiency (all-day efficiency) for the load cycle over 24 hour period given that the transformer works at full-load for 4 hours, at half load for 8 hours, at quarter load for 6 hours and at no-load for 6 assuming a unity power factor.
Explanation / Answer
Energy Output = kva*cos?*Kwh
%Rated Load
pf
x* kva cos?
KW
Hours
O/P Kwhs
1
1
1*1*50
50
4
200
1/2
1
0.5*1*50
25
8
200
1/4
1
.25*1*50
12.5
6
75
0
1
0
0
6
0
Copper loss at load = m*m*copper loss at any load
Copper loss at rated load
%Rated Load
m
KW
Hours
O/P Kwhs
1
1
50
4
1.28
1/2
1/2
25
8
1.28
1/4
1/4
12.5
6
0.48
0
0
0
6
3.04Kwh
Total Energy Output over 24 hr period excluding 6 hrs at no load is Wout = 475kwH.
Total Energy loss in the core for 24hrs including 6 hrs at no load ,
OC test = Pi*t = (320/1000)*24= 7.68Kwh.
Given Copper loss at rated load = 320Watts.
All day efficiency = Wout/( Wout+Wi+wC)= 475/485.72 = 97.792 Percent
Energy Output = kva*cos?*Kwh
%Rated Load
pf
x* kva cos?
KW
Hours
O/P Kwhs
1
1
1*1*50
50
4
200
1/2
1
0.5*1*50
25
8
200
1/4
1
.25*1*50
12.5
6
75
0
1
0
0
6
0
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