In a 25 KVA, 3300/233 V, single phase transformer, the iron and
full-load Cu. losses are respectively 350 and 400 watts. Calculate the
efficiency at half-full load 0.8 power factor.
If you are noticing some error in problems kindly comment below.Thanks
Given
KVA = 25
V1 = 3300
V2 = 233
Iron loss = 350 W
F.L Cu loss = 400 w
Cos Φ = 0.8
To find
1.Half load efficiency
2.Full load efficiency
Solution
1. Full load output at 0.8 pf = 25 x 0.8
= 20 KW
At half load output = 20 / 2
= 10 KW
Cu loss at half load is 1/4th of that full load.
At half load Cu loss = 400 / 4
= 100 W
= 0.1 KW
Iron loss remains same = 0.35 KW
Efficiency = Output / (Output + Cu loss + Iron loss)
Half load efficiency = H.L output / (H.L output + H.L Cu loss + Iron loss)
= 10 / (10 + 0.1 + 0.35 )
Half load efficiency = 95.69 %
2. Full load efficiency = F.L output / (F.L output + F.L Cu loss + Iron loss)
= 20 / (20 + 0.4 + 0.35)
Full load efficiency = 96.38 %
If you are noticing some error in problems kindly comment below.Thanks
KVA = 25
V1 = 3300
V2 = 233
Iron loss = 350 W
F.L Cu loss = 400 w
Cos Φ = 0.8
To find
1.Half load efficiency
2.Full load efficiency
Solution
1. Full load output at 0.8 pf = 25 x 0.8
= 20 KW
At half load output = 20 / 2
= 10 KW
Cu loss at half load is 1/4th of that full load.
At half load Cu loss = 400 / 4
= 100 W
= 0.1 KW
Iron loss remains same = 0.35 KW
Efficiency = Output / (Output + Cu loss + Iron loss)
Half load efficiency = H.L output / (H.L output + H.L Cu loss + Iron loss)
= 10 / (10 + 0.1 + 0.35 )
Half load efficiency = 95.69 %
2. Full load efficiency = F.L output / (F.L output + F.L Cu loss + Iron loss)
= 20 / (20 + 0.4 + 0.35)
Full load efficiency = 96.38 %
Why Cu loss is multiplied with ¼.?
ReplyDeleteWhen u calculate efficiency at half load..!
Where it is mentioned this.?
You can use formula directly
ReplyDeleteEfficiency = (X x KVA x COS¢) ÷ { (X x KVA x COS¢) + FE loss + (X^2 x Cu Loss)}
Where X=1 when efficiency@Full load.
Where X=½ when efficiency@Half load.
Direct method.
Cu loss directly proportional to square of current. So load is halved then cu poss will be 1 by 4th.
ReplyDeleteefficiency formula is output/input know??
ReplyDeleteyes correct. here in the problem input = output + F.L Cu loss + Iron loss. both are same.
Delete