The power losses in a transformer are of two types:
- Core or Iron Losses (Pi)
- Copper Losses (Pc)
(i) Core or Iron Losses (Pi):
Iron or core losses consist of: Pi
$$ P_i = \text{Hysteresis loss} + \text{Eddy current loss}
\quad = \text{Constant losses} $$
These can be determine by open circuit test .
- These occur in the Transformer core due to the alternating flux.
Hysteresis Loss:
- Hysteresis Loss: $$ Khf \mathrm{B}_{m}^{1.6} watt/m^{3} = Khf\mathrm{B}_{m}^{1.6}V $$
where:
- Kh = Hysteresis coefficient or Steinmetz constant
- f= frequency or no of cycles of magnetizing
- Bm = maximum flux density in the magnetic material
- V = volume of magnetic material
Eaddy Current Loss:
- Eaddy Current Loss:
$$ Khf ^{2}\mathrm{B}_{m}^{2}t^{2}watt/m^{3} = Khf ^{2}\mathrm{B}_{m}^{2}t^{2}V $$
where:
- Ke =coefficient of the eddy current
- f= frequency of reversal of the magnetic field
- Bm = maximum flux density in the magnetic material
- V = volume of magnetic material
- T=Thickness of material / Lamination
* The hysteresis loss can minimized by using Still of high silicon content , Where as eddy current loses can reduce by using core of thin Laminations
Minimization of Hysteresis Loss and Eddy Current Loss:
- Hysteresis loss → reduced by using steel with high silicon content
- Eddy current loss → reduced by using thin laminations in the core
(ii) Copper Losses (Pc):
These occur in both the primary and secondary windings due to ohmic resistance.
They are variable losses and can be determined by the short-circuit test.
Total Copper Loss
$$ P_c = I_1^2 R_1 + I_2^2 R_2
\ $$
$$
= I^2 R_{eq} \quad \text{(referred to secondary)} $$
Total Loss in a Transformer: –
$$ P_{i}+P_{c} = Constant losses + Variable Losses $$
Efficiency of a Transformer
$$ Efficiency (\eta) =\frac{Output Power}{Input Power}$$
$$ =\frac{Output Power}{output Power Losses} $$
$$=\frac{Output Power}{output Power + Copper loss+core loss}$$
$$=\frac{Output Power}{output Power +P_{c}+P_{i}}$$
Full load Efficiency =
$$ (\eta)_{FL} = \frac{Full Load V_{a}\ast P_{f}}{(Full Load V_{a}\ast P_{f})+P_{c}+P_{i}}$$
Condition for Maximum Efficiency:
$$ Output Power = V_{2}I_{2}cos\phi_{2} $$
If R02 is the total resistance of the transformer reffed to secondary then,
Total copper loss (Pc)
$$ P_c = I_2^2 R_{02}
\ $$
$$Total Loss =P_{i}+P_{c}$$
Therfore,
Transformer efficiency (ɳ)
$$ =\frac{Output Power}{Input Power} $$
$$= \frac{V_{2}I_{2}cos\phi_{2}}{V_{2}I_{2}cos\phi_{2}+P_{c}+P_{i}} $$
$$ = \frac{V_{2}I_{2}cos\phi_{2}}{V_{2}I_{2}cos\phi_{2}+I_2^2 R_{02}+P_{i}} $$
$$ \frac{\frac{V_{2}I_{2}cos\phi_{2}}{I_{2}}}{\frac{V_{2}I_{2}cos\phi_{2}+I_2^2 R_{02}+P_{i}}{I_{2}}} $$
$$ \frac{V_{2}I_{2}cos\phi_{2}}{V_{2}I_{2}cos\phi_{2}+\frac{P_{i}}{I_{2}}+I_{2}R_{02}} $$
V2 si is approximately constant.for two loads, of give power factor, efficiency depends upon load current I2. So, numerator is constant and the efficiency to be maximum, the denominator should be minimum.
$$ \frac{d(V_{2}cos\phi_{2}+\frac{P_{i}}{I_{2}}+I_{2}R_{02})}{dI_{2}} $$
$$\Rightarrow
\frac{0+(-\frac{P_{i}}{I_{2}})+1.R_{02}}{1}=0 $$
$$\Rightarrow
-\frac{P_{i}}{I_{2}^{2}}=-R_{02}$$
$$\Rightarrow
P_c = I_2^2 R_{02}
\ $$
Core loss = copper loss (Proved).
Hence,
efficiency of a transformer will be maximum when copper losses are equal to constant core losses.
Output KVA for maximum efficiency =$$ Full Load KVA \sqrt{\frac{Iron Loss}{Full Load Copper Loss}}$$
Types of Losses in Transformer
1. Hysteresis Loss (Ph):
Ph = Kh f Bm1.6 Ve
2. Eddy Current Loss (Pe):
Pe = Ke f2 Bm2 t2 V
3. Copper Loss (Pc):
Pc = I12 R1 + I22 R2
or (referred to secondary):
Pc = I12 R01 + I22 R02
4. Total Losses:
Total Losses = Pi + Pc
Total Losses = Constant Losses + Variable Losses