# Transformer efficiency | Efficiency calculation

Every device has this functionality or efficiency Depending on how effective the device is Similarly, a transformer is an electrical device. The transformer also has a skill We will know better how to diagnose this skill.

## transformer efficiency or Efficiency calculation

In each case, the power given as input to the machine is not found as output. Less power is available at the output than the input, so the ratio of output to input power in power loss is called efficiency.

The efficiency of the transformer is much higher than other machines, the amount can be up to 95% - 99%. Because the earth is only a core and a copper loss.

Which can be greatly reduced through quantity and proper use Skills are denoted by the Greek letter (η) and are expressed as a percentage.

\eta\%=\frac{Output}{Input}\times100

\eta\%=\frac{Output}{Output+Losses}\times100

\eta\%=\frac{v_sI_sP.f\times100}{V_sI_sP.f+CUloss+CORELoss}

\eta\%=\frac{Input-Losses}{Input}

\eta\%=1-\frac{Losses}{Input}\times100

### transformer core loss and other considerations that affect copper loss: -

1.Transformer Core Loss: - Consists of hysteresis loss and eddy current loss. Loss changes 1% - 3% in no-load condition.

It is affected by various factors: -

1. Power factor: - The core loss varies proportionally with the power factor. That is, if it increases, the value of the power factor increases

Core Loss =W_0=V_0I_0Cos\theta

2. Voltage: - Core Loss changes when voltage changes Such as: - Core Loss \propto V^2 That is, the square ratio of the voltage changes

3. Frequency: - Its value depends on the frequency. As the frequency increases, the loss also increases. It also depends on the type of iron in the core loss core

2. Copper Loss: -

It is affected by various factors Such as: -

1. Power factor: - Copper loss changes in inverse proportion with power factor. That is, copper loss \propto\frac1{Cos\theta} . If the value of the power factor increases, copper loss decreases

2. Voltage: - Copper loss \propto V^2 But if you increase the voltage to get constant KVA, the copper loss will decrease.

3. Frequency: - Copper loss does not depend on the frequency of the distribution transformer.

4. Load: - Depending on the note Copper loss decreases as load and unload decreases

### The equation for determining maximum efficiency

We know,

Cupper loss=> W_{cu}={I^2}_pR_e'\;W\;(or,={I^2}_pR_e")

Core loss => W_{core}=Hysterisis+Eddy\;Current\;loss

W_{core}=W_h+W_e

Primary Input= I_p=V_pI_pCos\theta_p

\eta=\frac{V_pI_pCoss\theta_p-Losses}{V_pI_pCos\theta_p}

\eta=\frac{V_pI_pCoss\theta_p-{I^2}_pR_e'-W_{CORE}}{V_pI_pCos\theta_p}

\eta=1-\frac{{I^2}_pR_e'}{V_pI_pCoss\theta_p}-\frac{W_{CORE}}{V_pI_pCos\theta_p}

\eta=1-\frac{I_pR_e'}{V_pCoss\theta_p}-\frac{W_{CORE}}{V_pI_pCoss\theta_p}

I_p=Differentiating both sides w.r to I_p

\frac{d\eta}{dI_p}=0-\frac{R_e'}{V_pCos\theta_p}+\frac{W_{CORE}}{V_pI_pCos\theta_p}

When,\frac{d\eta}{dI_p}=0 maximum efficiency

\Rightarrow0-\frac{-R_e'}{V_pCos\theta_p}+\frac{W_{CORE}}{V_p{I^2}_pCos\theta_p}

\frac{R_e'}{V_pCos\theta_p}=\frac{W_{CORE}}{V_p{I^2}_pCos\theta_p}

R_e'=\frac{W_{CORE}}{{I^2}_p}

{I^2}_pR_e'=W_{CORE}

Cupper Loss = Core Loss

Therefore it is seen that The copper loss will work at the maximum efficiency of the transformer when it is equal to the core loss,

{I^2}_pR_e'=W_{CORE}

I_p=\sqrt{\frac{W_{CORE}}{R_e'}}

I_s=\sqrt{\frac{W_{CORE}}{R_e"}}  =\frac{I_{FL}}{I_{FL}}\sqrt{\frac{W_{CORE}}{R_e"}}

I_s=\sqrt{\frac{W_{CORE}}{{I^2}_{FL}R_e"}}\times I_{FL}

I_{FL=\sqrt{\frac{W_{CORE}}{F.L.CUloss}}}

#### Change of efficiency with power factor

Percentage efficiency of a transformer η
\eta=\frac{OUTPUT}{INPUT}

\eta=\frac{INPUT-LOSSES}{INPUT}

\eta=1-\frac{LOSSES}{INPUT}

\eta=\{1-\frac{LOSSES}{(V_sI_sCos\theta+Losses)}\}

\eta=\{1-\frac{\frac{Losses}{V_sI_s}}{(Cos\theta+\frac{Losses}{V_sI_s})}\}

### All-day efficiency of transformer formula.

The primary site of the distribution transformer is always connected to the supply line. A full voltage transformer is connected to this side 24 hours a day. As a result, its core loss is always equal.

Only when the voltage of this line is low-high then core loss can be low-high However, such incidents are more or less rare That is why core loss is always parallel.

But since the secondary is directly connected to the load of the customer, the copper loss (-) is less and more with the amount of load of the customer.

Since the load is not equal in 24 hours, its copper loss is not equal Therefore, it is wrong to determine the efficiency of such transformers with full load copper loss If you want to find out the real skill, it is better to find out the skill of the whole day.

It translates the energy received from the transformer line throughout the day with the energy consumed by the customers throughout the day (24 hours).

All-day efficiency

=\frac{24\;Hours\;KVA\;Output}{24\;Hours\;KVA\;Input}

=\frac{24\;Hours\;KVA\;Output}{24\;Hours\;KVA\;Input+Hours\;Losses}

=\frac{ENERGY\;Output\times24}{ENERGY\;Output\times24+CORE\;Loss\times24+CUPPER\;Loss}

All-day efficiency must be less than the maximum efficiency.