Transformer Efficiency

Transformer Efficiency

Transformer Efficiency is defined as the ratio of output power to the input power. It is denoted as η.

    \[ \eta = \frac{Output \, Power}{Input \, Power} \]

    \[ \eta = \frac{Output \, Power}{Output \, Power + Copper \, loss + Iron \, loss } \]

    \[ Output \, Power = V_2 I_2 cos \phi_2 \]

    \[ Copper \, loss = I_2^2 R_{e2} \]

    \[ Iron \, loss = P_i \]

Therefore, per-unit efficiency at load current I2 and power factor cosф2 is;

    \[ \eta = \frac{V_2I_2Cos\phi_2}{V_2I_2Cos\phi_2 + I_2^2R_{e2} + P_i} \]

Per-unit efficiency at full load;

    \[ \eta_{fl} = \frac{V_2I_{2fl}Cos\phi_2}{V_2I_{2fl}Cos\phi_2 + I_{2fl}^2R_{e2} + P_i} \]

The transformer is a static device means there is no rotating parts. So, in transformer, the friction and windage loss is zero.

Therefore, the efficiency of a transformer is much higher than other electrical machines. For example, a transformer’s well-maintained and good quality of parts has efficiency as high 99%.

Condition for Maximum Efficiency (ηmax)

The per-unit efficiency at load current I2 and power factor cosф2 is;

    \[ \eta = \frac{V_2I_2Cos\phi_2}{V_2I_2Cos\phi_2 + I_2^2R_{e2} + P_i} \]

    \[ \eta = \frac{V_2Cos\phi_2}{V_2Cos\phi_2 + I_2R_{e2} + \frac{P_i}{I_2}} \]

In transformer, how the efficiency varies with change in load is described by the efficiency curve. The load curve is a plot of efficiency vs load current.

The below figure shows the efficiency curve for the transformer.

Transformer Efficiency
Transformer Efficiency

The above figure shows that the efficiency is low at a small load and maximum at a specific load. After that, the efficiency decreases as the load increases.

Therefore, it is desired to find under which condition and at which value the maximum efficiency. So, we can operate the transformer at maximum efficiency.

For maximum efficiency,

    \[ \frac{d \eta}{d I_2} = 0 \; and \; \frac{d^2 \eta}{d I_2^2} > 0\]

For a given load condition, the power factor cosф2 and output voltage V2 remains constant. Hence, the efficiency is maximum when the denominator is minimum.

    \[ D = V_2Cos\phi_2 + I_2R_{e2} + \frac{P_i}{I_2} \]

For minimum value of denominator;

    \[ \frac{dD}{d I_2} = 0 \; and \; \frac{d^2 D}{d I_2^2} > 0\]

    \[ \frac{dD}{d I_2} = \frac{d}{d I_2} (V_2Cos\phi_2 + I_2R_{e2} + \frac{P_i}{I_2}) \]

    \[ \frac{dD}{d I_2} = 0 + R_{e2} - \frac{P_i}{I_2^2} \]

Hence,

    \[ R_{e2} - \frac{P_i}{I_2^2} = 0 \]

    \[ R_{e2} = \frac{P_i}{I_2^2} \]

    \[ I_2^2 R_{e2} = P_i \]

Now,

    \[ \frac{d^2 D}{d I_2^2} = \frac{d}{d I_2} (R_{e2} - \frac{P_i}{I_2^2}) = 0 + \frac{2P_i}{I_2^3} > 0 \]

Therefore, \frac{d^2 D}{d I_2^2} is positive.

And the condition for maximum efficiency is;

    \[ I_2^2 R_{e2} = P_i \]

So, the efficiency of transformer at given power factor is maximum when the variable copper loss is equal to the constant iron loss (core loss).

Transformer Losses

 345 total views,  1 views today

Leave a Reply

Your email address will not be published.