……… (1) Here S in the volt ampere rating of the transformer (V2, I2 at full load), Pconst being Constant losses and Pvar the variable losses at full load.
Pvar = (IFL)2 R
If I is the current at any load, then we can define the fractional load “x” as under:
x = I/ IFL
Efficiency is maximum, when constant losses are equal to variable losses,
I2 R = Pconst (x. IFL)2 R = Pconst x2 Pvar = Pconst
Finally, this equation gives the fractional load at which the efficiency is maximum. Then the value of x and other variables are used in equation 1 to get the efficiency. Moreover, equation 1 is useful whenever we need to calculate efficiency at any value of load. change the value of x and get the efficiency at desired load.
EEN340 Energy Conversion Efficiency of Transformers
Problem 1: In a 25 KVA, 2000/200 V power transformer the Iron and full load copper losses are350 W and 400 W respectively. Calculate the Efficiency at unity power factor at i) ii) Full Load Half Load
(Solution: η= 97.087 % & 96.525 % respectively) Problem 2: A 220/400 V, 10 KVA, 50Hz single phase Transformer has at full load, a copper loss of 120 W. If it has an efficiency of 98% at full load, unity power factor, determine the iron losses. What would be the efficiency of the transformer at half load at 0.8 p.f. lagging? (Solution: Pi=84.08 W & η= 97.23%) Problem 3: The efficiency of a 400 KVA, single phase transformer is 98.77 % when delivering full load at 0.8 power factor and 99.13% at half load and unity power factor. Calculate i) ii) The Iron losses The full load Copper losses (Solution: Pi = 1.012 W & Pc = 2.973 W)...