Efficiency of Transformer

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 pmitations, 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 −

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

$$mathrm{Rightarrow eta :=:frac{VA imes Power:Factor}{left ( VA imes Power:Factor ight ):+:Losses}}$$

Where,

$$mathrm{mathrm{Output:power}:=:VA imes Power:factor}$$

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.

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

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.

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

Condition for Maximum Efficiency

For a given transformer, we have,

$$mathrm{mathrm{Output:power}:=:mathit{V_{mathrm{2}}I_{mathrm{2}}cosphi _{mathrm{2}}}}$$

Let the transformer referred to secondary side, then R_o2 is the total resistance of the transformer. The total copper loss is given by,

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

Therefore, the transformer efficiency is given by,

$$mathrm{eta :=:frac{mathit{V_{mathrm{2}}}I_{mathrm{2}}cosphi _{mathrm{2}}}{mathit{V_{mathrm{2}}I_{mathrm{2}}cosphi _{mathrm{2}}}+mathit{P_{i}}+mathit{I_{mathrm{2}}^{mathrm{2}}}R_{omathrm{2}}}}$$

On rearranging the expression, we get,

$$mathrm{eta :=:frac{mathit{V_{mathrm{2}}}cosphi _{mathrm{2}}}{mathit{V_{mathrm{2}}cosphi _{mathrm{2}}}+left ( mathit{frac{P_{i}}{I_{mathrm{2}}}} ight )+mathit{I_{mathrm{2}}}R_{omathrm{2}}}:=:mathit{frac{V_{mathrm{2}}cosphi _{mathrm{2}}}{D}}:cdot cdot cdot (1)}$$

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.

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

$$mathrm{Rightarrowmathit{frac{d}{dI_{mathrm{2}}}}left [ mathit{V_{mathrm{2}}cosphi _{mathrm{2}}}+left ( mathit{frac{P_{i}}{I_{mathrm{2}}}} ight )+mathit{I_{mathrm{2}} R_{0mathrm{2}}} ight ]:=:0}$$

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

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

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

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,

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

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,P_i = 450 W

Copper loss,P_c = 900 W

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

Transformer efficiency at full-load −

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

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

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

$$mathrm{ herefore eta _{mathit{FL}}:=:0.9834 imes 100\%:=:98.34\%}$$

Maximum efficiency of the transformer −

For maximum efficiency,

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

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

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

$$mathrm{ herefore eta _{mathit{max}}:=:0.9888 imes 100\%:=:98.88\%}$$