A single-phase \(400\;V,\;50\;Hz\) transformer has an iron loss of \(5000\;W\) at the rated condition. When operated at \(200\;V,\;25\;Hz\), the iron loss is \(2000\;W\). When operated at \(416\;V,\;52\;Hz\), the value of the hysteresis loss divided by the eddy current loss is ______.
Concept:
Hysteresis loss:
It is due to the reversal of magnetization of the transformer core whenever it is subjected to the alternating nature of the magnetizing force.
The power consumed by the magnitude domains to change their orientation after every half cycle whenever core is subjected to alternating nature of magnetizing force is called as hysteresis loss.
Hysteresis loss can be determined by using the Steinmetz formula given by
\({W_h} = \eta B_{max}^2fV\)
Where
x is the Steinmetz constant, Bm = maximum flux density
f = frequency of magnetization or supply frequency, V = volume of the core
Eddy current losses:
Eddy current loss is basically I2R loss present in the core due to the production of eddy currents in the core, because of its conductivity.
Eddy current losses are directly proportional to the conductivity of the core.
Eddy current loss \({W_e} = K{f^2}B_m^2{t^2}\)
Where
K - coefficient of eddy current., Bm - maximum value of flux density
t - thickness of lamination in meters, f - frequency of eddy current
If (V / f) ratio is constant:
As we know,
Bmax ∝ (V / f)
⇒ Bmax = constant
⇒ Hysteresis loss Wh ∝ f
And, eddy current loss We ∝ f2
If V/f is not constant:
Hysteresis losses
\( \Rightarrow {W_h} \propto V_1^{1.6}{f^{ - 0.6}}\)
Eddy current losses
\( \Rightarrow {W_e} \propto V_1^2\)
Explanation:
At \(400\;V,\;50\;Hz\), iron loss \({P_i}\; = \;5000\) Watt
At \(200\;V,\;25\;Hz\) ⇒ \({P_i}\; = \;2000\) Watt
\(\begin{array}{l} {P_i}\; = \;{P_h}\; + \;{P_e}\\ {P_h}\;\propto \;fB_m^x\;\\ {P_e}\;\propto \;{f^2}B_m^2 \end{array}\)
We have, \(\frac{V}{f} = \frac{{400}}{{50}} = \frac{{200}}{{25}} = \frac{{416}}{{52}} = 8 = constant\)
Hence,\({P_h}\; = \;Af,\;{P_e}\; = \;B{f^2}\)
Form given data
\(2000\; = \;A\;\left( {25} \right)\; + \;B{\left( {25} \right)^2}\) …(i)
\(5000\; = \;A\;\left( {50} \right)\; + \;B{\left( {50} \right)^2}\)…(ii)
Solving (i) and (ii)
\(A\; = \;60,\;B\; = \;0.8\)
\(\begin{array}{l} \left( {{P_h}} \right)\; = \;Af\; = \;60\; \times \;52 = 3120\;W\\ \left( {{P_e}} \right) = \;Bf\; = \;0.8\; \times \;{\left( {52} \right)^2} = 2163.2\;W\\ \frac{{{{\left( P \right)}_h}}}{{{{\left( P \right)}_e}}} = \frac{{3120}}{{2163.2}} = 1.4423 \end{array}\)
Important Questions from Transformer Copper Losses
What is the condition at which a transformer gives maximum efficiency?
- Which of the following losses remains variable during normal operation of transformer?
For a single phase transformer, the maximum efficiency occurs at 75% of the load, then \(\rm \frac{iron \ loss \ at\ full \ load }{copper \ loss \ at\ full \ load}=?\)
In a single-phase transformer, the ratio of transformation is 2 and the secondary resistance is 0.24 Ω. Find the resistance of secondary in terms of primary.
The full load copper loss and iron loss of a transformer are 6400 W and 5000 W respectively the copper loss and iron loss at half load will be respectively.
