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Transformer Losses:
Two types of losses exist in the transformer which is as follows:
- When the process of mutual induction is occurring in the transformer, its core is subjected to alternating flux due to which core losses occur.
- Copper losses happen to take place when the transformer is loaded and both the windings are carrying currents.
Core losses:
The transformer undergoes magnetization and demagnetization due to alternating flux set up in its magnetic core. Due to the hysteresis effect, there is the loss of energy in this process which is called hysteresis loss.
It is given by,
Hysteresis loss = Kh*Bm1.67*f*v Watts
Where,
Kh = hysteresis constant depends on materials
Bm = maximum flux density
f = frequency
v = volume of the core
The induced emf in the core tries to initiate the eddy currents in the core and it is hence responsible for the eddy current losses.
Basically, Eddy currents are the leakage currents caused by the induced emf because when emf is induced in the primary and then it causes mutual induction in the secondary, the secondary winding sets up its own emf and this emf creates its own driving current on both windings and hence these currents are the eddy currents.
The eddy current loss is given by,
Eddy current loss = Ke*Bm2*f2*t2 Watts/unit volume
Where,
Ke = eddy current constant
t = thickness of the core
As the flux generated in the core is always almost constant as supply voltage at rated frequency f is always constant due to which Bm in the core is constant and hence, all the core losses are constant at all the loads. So, the core losses are also known as the constant losses and are denoted Pi.
Pre-caution to minimize the core losses: The core losses can be minimized by using high-grade silicon steel of having a very low hysteresis loop and the manufacturing process of it in the form of laminations.
Copper Losses:
Copper losses also known as heat losses are caused by power wastage in the form I2R loss due to the resistance of the primary and secondary windings of the transformer. Actually, the copper loss depends on the magnitude of the currents flowing through the primary and secondary windings which are I1 & I2, respectively and the corresponding resistances are R1 & R2.
Total Cu loss=
= I12*R1 + I22*R2 = I12*(R1 + (I22*R2/ I12)) = I22*((I!2*R1/ I22) + R2)
= I12*(R1 + (I22*R2/ I12)) = I22*((I!2*R1/ I22) + R2)
where V2/V1 = I1/I2 = K
= I12*(R1 + (R2/ K2)) = I22*(K 2*R1 + R2) = I12*(R1 + R2’) = I22*(R1’ + R2)
= I12*R1e = I22*R2e
The copper losses are denoted by PCu. If the transformer is bearing the full load then both the primary and secondary windings are bearing the full load current and so does the copper losses are at full load. If a transformer is subjected to the half load, then both the windings currents are half-load currents and the copper losses subject to half load. Thus, the copper losses are variable losses. The transformer rating depends on V1*I1 or V2*I2. As V1 is constant, so the copper losses are totally proportional to the square of the kVA rating of the transformer.
So,
PCu ∝ I2 ∝ (kVA)2
Thus for a transformer,
Total losses = Core losses + Copper losses = Pi + PCu
Transformer Efficiency:
The output power of the transformer is always less than the input power supplied because of losses in the transformer.
Therefore,
Output Power = Input Power – Total Losses
Input Power = Output Power + Total Losses = Output Power + Pi + PCu
The ratio of output power to the input power is known as the efficiency of any device. So, for the transformer efficiency can be expressed as,
η = Output Power / Input Power
η = Output Power / Output Power + Pi + PCu
Now,
Output Power = V2*I2*cosФ
Where,
cosФ = Power Factor of load
Now,
PCu = Copper loss on full load = I22*R2e
Therefore,
η = V2*I2*cosФ / V2*I2*cosФ + Pi + I22*R2e
V2*I2 = VA Rating of the transformer
η = (VA Rating)*cosФ / (VA Rating)*cosФ + Pi + I22*R2e
In percentage form, we can write efficiency as follow:
% η = [(VA Rating)*cosФ / (VA Rating)*cosФ + Pi + I22*R2e]*100
This is the full load efficiency with,
I2 = full load secondary current
But there is a point to ponder which is what if the transformer is subjected to fractional load. Then the efficiency will subject to the ratio of actual load to full load.
Let
n = fraction by which load is less than full load = actual load / full load
When load changes, the load current changes by the same proportion.
Therefore,
new I2 = n(I2) Full load
So, due to it the power also reduces by the same fraction. Thus, the fraction of VA rating is also available at the output.
Similarly, copper losses are proportional to the square of the current then,
new PCu = n2(PCu) Full load
So Copper losses get reduced by n2.
Therefore, in general, efficiency can be written as
% η = [n*(VA Rating)*cosФ / n*(VA Rating)*cosФ + Pi + n2*I22*R2e]*100
Conditions for Maximum Efficiency:
The transformer efficiency varies with the load working with constant voltage and frequency. With the load increases, the efficiency increases till at a certain point which is the maximum efficiency. After this point, the efficiency of the transformer decreases. The load current at maximum efficiency is denoted by I2m and the maximum efficiency can be denoted as ηmax.
- For Maximum Efficiency:
Copper losses = Iron Losses
- Load Current I2m at ηmax:
I2m = (I2) Full load*(Pi/(PCu) Full load)1/2
- kVA Rating supplied at ηmax
kVA at ηmax = (V2*I2) Full load*(Pi/(PCu) Full load)1/2
kVA at ηmax = (kVA Rating) Full load*(Pi/(PCu) Full load)1/2
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