Efficiency of Transformer

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Transformer Efficiency

The ratio of the output power to the input power in a transformer is known as efficiency of transformer. The transformer efficiency is represented by Greek letter Eta ($\eta $).

$$\mathrm{\mathrm{Efficiency,}\eta \:=\:\frac{Output\:Power}{Input\:Power}}$$

From this definition, it appears that we can determined the efficiency of a transformer by directly loading the transformer and measuring the input power and output power. Although, this method of efficiency determination has the following disadvantages −

  • In practice, the efficiency of a transformer is very high, and a very small error (let say 1%) in input and output wattmeters may give ridiculous results. Consequently, this method may give efficiency more than 100%.

  • In this method, the transformer is loaded, hence a considerable amount of power is wasted. Therefore, this method becomes uneconomical for large transformers.

  • It is very difficult to find a load which is capable of absorbing all of the output power.

  • This method does not provide any information about losses in the transformer.

Thus, due to these limitations, the direct-loading method is rarely used to determine the efficiency of a transformer. In practice, we use open-circuit and short-circuit tests to find out the transformer efficiency.

For a practical transformer, the input power is given by,

$$\mathrm{\mathrm{Input\:power}\:=\:\mathrm{Output\:power\:+\:Losses}}$$

Therefore, the transformer efficiency can also be calculated using the following expression −

$$\mathrm{\eta \:=\:\frac{Output\:power}{Output\:power\:+\:Losses}}$$

$$\mathrm{\Rightarrow \eta \:=\:\frac{VA\times Power\:Factor}{\left ( VA\times Power\:Factor \right )\:+\:Losses}}$$

Where,

$$\mathrm{\mathrm{Output\:power}\:=\:VA\times Power\:factor}$$

And, losses can be determined by transformer tests.

Efficiency from Transformer Tests

When we perform transformer tests, the following results are obtained −

  • From open-circuit test −

$$\mathrm{\mathrm{Full\:load\:iron\:loss}\:=\:\mathit{P_{i}}}$$

  • From short-circuit test −

$$\mathrm{\mathrm{Full\:load\:copper\:loss}\:=\:\mathit{P_{c}}}$$

Therefore, the total losses at full load in a transformer are

$$\mathrm{\mathrm{Total\:FL\:losses}\:=\:\mathit{P_{i}+\:P_{c}}}$$

Now, we are able to determine the full-load efficiency of the transformer at any power factor without actual loading the transformer.

$$\mathrm{\mathit{n_{FL}}\:=\:\frac{(VA)_{\mathit{FL}}\times Power\:factor}{[(VA)_{\mathit{FL}}\times Power\:factor]+\:\mathit{P_{i}}+\mathit{P_{c}}}}$$

Also, the transformer efficiency at any load equal to x full load. Where, x is the fraction of loading. In this case, the total losses corresponding to the given load are,

$$\mathrm{(Total\:losses)_{x}\:=\:\mathit{P_{i}+\:x^{\mathrm{2}}\mathit{P_{c}}}}$$

It is because, the iron loss ($\mathit{P_{i}}$) is the constant loss and hence remains the same at all loads, while the copper loss is proportional to the square of the load current.

$$\mathrm{\therefore\eta _{x}\:=\: \frac{\mathit{x}\times (VA)_{\mathit{FL}}\times Power\:factor}{[\mathit{x}\times (VA)_{\mathit{FL}}\times Power\:factor]+\:\mathit{P_{i}}+\:x^{\mathrm{2}}\mathit{P_{c}}}}$$

Condition for Maximum Efficiency

For a given transformer, we have,

$$\mathrm{\mathrm{Output\:power}\:=\:\mathit{V_{\mathrm{2}}I_{\mathrm{2}}cos\phi _{\mathrm{2}}}}$$

Let the transformer referred to secondary side, then Ro2 is the total resistance of the transformer. The total copper loss is given by,

$$\mathrm{\mathit{P_{c}}\:=\:\mathit{I_{\mathrm{2}}^{\mathrm{2}}\mathit{R_{o\mathrm{2}}}}}$$

Therefore, the transformer efficiency is given by,

$$\mathrm{\eta \:=\:\frac{\mathit{V_{\mathrm{2}}}I_{\mathrm{2}}cos\phi _{\mathrm{2}}}{\mathit{V_{\mathrm{2}}I_{\mathrm{2}}cos\phi _{\mathrm{2}}}+\mathit{P_{i}}+\mathit{I_{\mathrm{2}}^{\mathrm{2}}}R_{o\mathrm{2}}}}$$

On rearranging the expression, we get,

$$\mathrm{\eta \:=\:\frac{\mathit{V_{\mathrm{2}}}cos\phi _{\mathrm{2}}}{\mathit{V_{\mathrm{2}}cos\phi _{\mathrm{2}}}+\left ( \mathit{\frac{P_{i}}{I_{\mathrm{2}}}} \right )+\mathit{I_{\mathrm{2}}}R_{o\mathrm{2}}}\:=\:\mathit{\frac{V_{\mathrm{2}}cos\phi _{\mathrm{2}}}{D}}\:\cdot \cdot \cdot (1)}$$

In practice, the secondary voltage V2 is approximately constant. Hence, for a load of given power factor, the transformer efficiency depends upon the load current (I2). From the equation (1), we can see that the numerator is constant and for the efficiency to be maximum, the denominator (D) should be minimum, i.e.

$$\mathrm{\mathit{\frac{d(D)}{dI_{\mathrm{2}}}}\:=\:0}$$

$$\mathrm{\Rightarrow\mathit{\frac{d}{dI_{\mathrm{2}}}}\left [ \mathit{V_{\mathrm{2}}cos\phi _{\mathrm{2}}}+\left ( \mathit{\frac{P_{i}}{I_{\mathrm{2}}}}\right )+\mathit{I_{\mathrm{2}} R_{0\mathrm{2}}} \right ]\:=\:0}$$

$$\mathrm{\Rightarrow 0-\left ( \mathit{\frac{P_{i}}{I_{\mathrm{2}}}} \right )+\mathit{R_{o\mathrm{2}}}\:=\:0}$$

$$\mathrm{\Rightarrow \mathit{P_{i}}\:=\:\mathit{I_{\mathrm{2}}^{\mathrm{2}}R_{o\mathrm{2}}}}$$

$$\mathrm{\Rightarrow \mathrm{Iron\:loss}\:=\:Copper\:loss}$$

Therefore, the transformer efficiency for a given power factor will be maximum when the constant iron loss is equal to the variable copper loss.

The maximum efficiency at any load is given by,

$$\mathrm{\mathit{\eta _{max}}\:=\:\frac{\mathit{x\times (VA)_{\mathit{FL}}\times \mathrm{Power\:factor}}}{[\mathit{x\times (VA)_{\mathit{FL}}}\times Power\:fctor]+\:2\mathit{P_{i}}}}$$

Also, the load current (I2) corresponding to the maximum efficiency of transformer is,

$$\mathrm{\mathit{I_{\mathrm{2}}}\:=\:\sqrt{\frac{\mathit{P_{i}}}{R_{o2}}}}$$

Numerical Example

In a 100 kVA transformer, the iron loss is 450 W and full-load copper loss is 900 W. Find the transformer efficiency at full load and the maximum efficiency of the transformer, where the load power factor is 0.8 lagging.

Solution

Given data,

  • Full load VA = 100 kVA = 100 1000 VA

  • Iron loss,Pi = 450 W

  • Copper loss,Pc = 900 W

  • cos$\mathit{\phi _{\mathrm{2}}}$ = 0.8

Transformer efficiency at full-load −

$$\mathrm{\mathrm{Total\:losses}\:=\:450\:+\:900\:=\:1350\:W}$$

$$\mathrm{\mathit{\eta _{\mathit{FL}}}\:=\:\frac{(VA)_{\mathit{FL}}\times Power\:factor}{[(VA)_{\mathit{FL}}\times Power\:factor]+\:Total\:losses}}$$

$$\mathrm{\Rightarrow \mathit{\eta _{\mathit{FL}}}\:=\:\frac{100\times 1000\times 0.8}{(100\times 1000\times 0.8)+1350}\:=\:\frac{80000}{81350}\:=\:0.9834}$$

$$\mathrm{\therefore \eta _{\mathit{FL}}\:=\:0.9834\times 100\%\:=\:98.34\%}$$

Maximum efficiency of the transformer −

For maximum efficiency,

$$\mathrm{\mathrm{Iron\:loss}\:=\:Copper\:Loss}$$

$$\mathrm{\therefore \eta _{\mathit{max}}\:=\:\frac{(VA)_{\mathit{FL}}\times Power\:factor}{[(VA)_{\mathit{FL}}\times Power\:factor]+2\mathit{P_{i}}}}$$

$$\mathrm{\Rightarrow \eta _{\mathit{max}}\:=\:\frac{100\times 1000\times 0.8}{(100\times 1000\times 0.8)+(2\times 450)}\:=\:0.9888}$$

$$\mathrm{\therefore \eta _{\mathit{max}}\:=\:0.9888\times 100\%\:=\:98.88\%}$$

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