## What do you mean by Maximum Efficiency Criteria?

In every field of engineering, there is a tremendous use of the machine and all those machines are equipped for their popular work efficiency so it very much important for operation engineers to monitor the efficiency of the machine, planning engineers to check out the efficiency of the machine before installing the machine and design engineers to design machine for higher efficiency than and then the utility will procure their products that will ultimately lead to profit and loss of the company. It indicates the importance of efficiency right from the initial stage as manufacturing units, intermediate stage as planning coordinators, and end-users stage as a utility.

## Efficiency

It is the ratio of output to the input of any Machine or Apparatus or Device. The unit of Output and Input should be the same.

${\text{\eta}}_{\text{eff}}\text{=}\frac{\text{Output}}{\text{Input}}$$$

## Maximum Efficiency Criteria

They are the condition at which the device has maximum efficiency. To understand the maximum efficiency criteria, let's take the example of the transformer. The efficiency of the transformer can be written as,

${\text{\eta}}_{\text{eff}}\text{=}\frac{\text{Output}}{\text{Input}}$

There are two types of losses in the transformer. The first one is copper loss related to the winding of the transformer which depended on the current. The second is core loss occurs in the core of the transformer also known as magnetic losses. Core loss comprises two components, Hysteresis loss and Eddy current loss which depend upon the magnetic material of the core. The core loss does not depend upon the current so this component of the loss remains constant so it is also known as fixed losses. It is very much easy to compute when all the parameters are known but if one or more parameters are then so for that different equations relating to output should be known and they are as follows:

$Input=Output+losses$

The output power can be obtained by just subtracting losses from the input that can be represented by a single line diagram. We can highlight power flow from input toward output by a single line diagram called power flow diagram as shown below figure.

$Losses={P}_{i}+{P}_{c}$

${\eta}_{eff}=\frac{Output}{Output+Losses}$

${\eta}_{eff}=\frac{Output}{Output+{P}_{i}+{P}_{c}}$

${\eta}_{eff}=\frac{{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}}{{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}+{P}_{i}+{P}_{c}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{.....}\left(1\right)$

Here,

${P}_{i}=\text{FullloadIronlossesofthetransformer}$

${P}_{c}=\text{FullloadCore}\left(\text{magnetic}\right)\text{\hspace{0.17em}}\text{lossesoftransformer}={I}_{2}^{2}{R}_{es}$

$Output={V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}$

${I}_{2}=\text{Fullloadsecondarycurrent}$

${V}_{2}=\text{Secondaryterminalvoltage}$

$\mathrm{cos}{\varphi}_{2}=\text{Powerfactoroftheload}$

Now let consider the x is the fraction of the full load. The transformer efficiency equation can represent in the form fraction of full load the equation can be rewritten as,

${\eta}_{eff}=\frac{x{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}}{x{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}+{P}_{i}+{x}^{2}{P}_{c}}$

${\eta}_{eff}=\frac{x{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}}{x{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}+{P}_{i}+{x}^{2}{I}_{2}^{2}{R}_{es}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{....}\left(2\right)$

Where R_{es} is the resistance. It is much clear that copper loss varies according to the fraction of full load. Copper loss is the electric power loss that varies as the square of the current in each winding. From the above equation, if the efficiency of the transformer to be maximum denominator should be as small as possible. The numerator contains firstly the terminal voltage V_{2} which approximately constant, secondly the power factor which depends on load current I_{2}. So, it can be said that the numerator is constant.

The transformer efficiency will be a maximum if the denominator becomes small. To make the denominator small we need to differentiate the denominator with respect to I_{2} and equate it to zero.

$\begin{array}{c}\frac{d}{d{I}_{2}}\left({V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}+{P}_{i}+{I}_{2}^{2}{R}_{es}\right)=0\\ \frac{d}{d{I}_{2}}\left({V}_{2}\mathrm{cos}{\varphi}_{2}+\frac{{P}_{i}}{{I}_{2}}+{I}_{2}{R}_{es}\right)=0\\ 0-\frac{{P}_{i}}{{I}_{2}^{2}}+{R}_{es}=0\\ -\frac{{P}_{i}}{{I}_{2}^{2}}+{R}_{es}=0\\ \frac{{P}_{i}}{{I}_{2}^{2}}={R}_{es}\\ {P}_{i}={I}_{2}^{2}{R}_{es}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{.....}\left(3\right)\\ {P}_{i}={P}_{c}\end{array}$

$\text{FullloadIronlossesofthetransformer}=\text{FullloadCorelossesofthetransformer}$

Hence, the transformer will give the maximum efficiency when their copper losses are equal to Iron losses. It means variable (copper losses) losses are equal to constant (core losses) losses.

Using the above equation no. 3 in equation no. 1 we get,

$\begin{array}{l}{\eta}_{\mathrm{max}}=\frac{{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}}{{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}+{P}_{i}+{P}_{i}}\\ {\eta}_{\mathrm{max}}=\frac{{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}}{{V}_{2}{I}_{2}\mathrm{cos}{\varphi}_{2}+2{P}_{i}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{.....}\left(4\right)\end{array}$

From equation no. 3 the value of secondary current I_{2} at which the transformer efficiency will be maximum is given as,

${I}_{2}=\sqrt{\frac{{P}_{i}}{{R}_{es}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{......}\left(5\right)$

In other words, it can be said that a transformer operates at maximum efficiency when the copper loss curve intersects with the iron loss curve as shown below figure.

To obtain efficiency at maximum KVA ratings of the transformer multiply and divide by I_{2} on the right-hand side of the equation we get,

${I}_{2}={I}_{2}\sqrt{\frac{{P}_{i}}{{I}_{2}^{2}{R}_{es}}}$

Multiply both the side by secondary terminal voltage V_{2} in the above equation we get,

$\begin{array}{c}{V}_{2}{I}_{2}={V}_{2}{I}_{2}\sqrt{\frac{{P}_{i}}{{I}_{2}^{2}{R}_{es}}}\\ {\left.VA\right|}_{\mathrm{max}.eff}={\left.VA\right|}_{rated}\sqrt{\frac{{P}_{i}}{{P}_{c}}}\end{array}$

If x is the fraction of full load KVA then the copper loss is equal to Copper loss =xP_{c} (where P_{c} is the full load copper loss) the equation no. 3 can be written as,

${P}_{i}={x}^{2}{P}_{c}$

Therefore

$x=\sqrt{\frac{{P}_{i}}{{P}_{c}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{....}\left(6\right)$

Thus, the output KVA of the transformer corresponding to maximum efficiency is

${\eta}_{\mathrm{max}}=x*\text{FullloadKVA}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{......}\left(7\right)$

Equating both the equating no. 6 & 7

$\begin{array}{c}{\eta}_{\mathrm{max}}=\sqrt{\frac{{P}_{i}}{{P}_{c}}}*\text{FullloadKVA}\\ {\eta}_{\mathrm{max}}=\text{FullloadKVA*}\sqrt{\frac{\text{Ironlosses}}{\text{copperlossesatfullload}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{.....}\left(8\right)\end{array}$

The above equation (8) is the maximum efficiency condition of the transformer. The actual KVA rating of the transformer can be obtained by performing various tests on transformers

## Common Mistakes

When the student or researcher thought about the efficiency, he or she forgets about the data fetch for computing the maximum efficiency operating point for the machine so it is necessary to take real practical data while analyzing.

## Context and Application

Maximum efficiency criteria are used to obtain the condition at which the device or apparatus like DC generator, Transformer, induction motor, and any machine. It is studied in

- Bachelors in Technology (Electrical Engineering)
- Bachelors in Science (Physics)
- Masters in Science (Physics)

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