A 25 kVa signal phase transformer 2200:200V has a primary and secondary resistance of 1Ω and 0.01 Ω respectively. Find the equivalent secondary resistance and full load efficiency at 0.8pf lagging, if the iron losses of the transformer are 80% of the full load copper losses.
If you are noticing some error in problems kindly comment below.Thanks
Given
KVA = 25000
V1 = 2200
V2 = 200
R1 = 1 ohm
R2 = 0.01 ohm
PF = 0.8
Losses = 80%
To find
1. Secondary resistance
2. Full load efficiency
Solution
Voltage transformation ratio, K = V2 / V1
= 200 / 2200
= 0.09
1. Secondary resistance, R02 = R2 + (K2 x R1)
= 0.01 + (0.092 x 1 )
= 0.018 ohm
Secondary Full load current I2 = KVA / V2
= 25000 / 200
= 125 A
Full load Cu loss = I22 x R02
= 1252 x 0.018
= 281.25 W
Iron loss = 80 % of full load Cu loss
= 0.8 x 281.25
= 225 W
Total loss = 281.25 + 225
= 506.25 W
Full load output = 25000 x 0.8 = 20000 W
2. Full load efficiency = 20000 / (20000 + 506.25)
= 97.53 %
If you are noticing some error in problems kindly comment below.Thanks
Given
KVA = 25000
V1 = 2200
V2 = 200
R1 = 1 ohm
R2 = 0.01 ohm
PF = 0.8
Losses = 80%
To find
1. Secondary resistance
2. Full load efficiency
Solution
Voltage transformation ratio, K = V2 / V1
= 200 / 2200
= 0.09
1. Secondary resistance, R02 = R2 + (K2 x R1)
= 0.01 + (0.092 x 1 )
= 0.018 ohm
Secondary Full load current I2 = KVA / V2
= 25000 / 200
= 125 A
Full load Cu loss = I22 x R02
= 1252 x 0.018
= 281.25 W
Iron loss = 80 % of full load Cu loss
= 0.8 x 281.25
= 225 W
Total loss = 281.25 + 225
= 506.25 W
Full load output = 25000 x 0.8 = 20000 W
2. Full load efficiency = 20000 / (20000 + 506.25)
= 97.53 %
What about the copper loss in primary? Isn't it necessary to take it into account?
ReplyDeleteIts already counted. When we calculated the equvalent resistance reffered to secondary.
ReplyDelete