# Efficiency of DC Generator & Condition for Maximum Efficiency with Examples

Digital ElectronicsElectronics & ElectricalElectron

## Efficiency of DC Generator

The efficiency of a DC generator is defined as the ratio of mechanical input power to the output electrical power.

$$\mathrm{Efficiency,\:\eta\:=\frac{Electrical\:Power\:Output(P_{o})}{ Mechanical\:Power\:Input(P_{i})}}$$

## Explanation

Consider the power flow diagram of a DC generator (see the figure), here the power is represented in three stages as  By referring the power flow diagram,

$$\mathrm{Iron\:and\:Friction\:Losses\:=\:𝐴\:−\:𝐵}$$

$$\mathrm{Copper\:Losses\:=\:𝐵\:−\:𝐶}$$

Therefore, the efficiency of a DC generator can also be defined for the three stages as follows

• Mechanical Efficiency

$$\mathrm{\eta_{mech}\:=\frac{B}{A}=\:\frac{Power\:Developed\:in\:Armature\:(E_{g}I_{a})}{Mechanical\:Power\:Input\:(P_{i})}}$$

• Electrical Efficiency

$$\mathrm{\eta_{elect}\:=\frac{C}{B}=\:\frac{Electric\:Power\:Output\: (VI_{L})}{Power\:Developed\:in\:Armature\:(E_{g}I_{a})}}$$

• Commercial Efficiency − (always consider this unless stated otherwise)

$$\mathrm{\eta\:=\frac{C}{A}=\:\frac{Power\:Output\:(P_{o})}{Power\:input\:(P_{i})}}$$

## Condition for Maximum Efficiency

The efficiency of a DC generator is not constant but changes with the change in load.

Let, for a shunt generator,

$$\mathrm{I_{L}\:=\:load\:current}$$

$$\mathrm{V\:=\:terminal\:voltage}$$

Then, the output power of the DC generator is given by,

$$\mathrm{Output\:Power,=P_{o}=VI_{L}}$$

$$\mathrm{Total\:Input\:Power,\:P_{i} = P_{o} + Losses}$$

$$\mathrm{⇒P_{i}=VI_{L}+I_a^2R_{a}+W_{c}}$$

$$\mathrm{⇒P_{i}=VI_{L}+(I_{L}+I_{sh})^{2}R_{a}+W_{c}}$$

Where,

• Ia2Ra=variable losses = Copper losses
• Wc= Constant losses = Iron losses + Mechanical losses

Practically, the shunt field current (Ish) is very small as compared to load current (IL), hence it can be neglected. Therefore,

$$\mathrm{P_{i}=VI_{L}+I_L^2R_{a}+W_{c}}$$

Hence, the efficiency of DC generator will be,

$$\mathrm{\eta=\frac{P_{o}}{P_{i}}=\frac{VI_{L}}{VI_{L}+I_L^2R_{a}+W_{c}}}$$

$$\mathrm{\eta=\frac{1}{1+(\frac{I_{L}R_{a}}{V})+(\frac{W_{c}}{VI_{L}})}}$$

The efficiency will be maximum when the denominator of the above expression is minimum. In order to determine minimum value of denominator, differential it with respect to variable (IL in this case) and equate it to zero, i.e.

$$\mathrm{\frac{d}{dI_{L}}[1+(\frac{I_{L}R_{a}}{V})+(\frac{W_{c}}{VI_{L}})]\:=\:0}$$

$$\mathrm{⇒\:0+\frac{R_{a}}{V}-\frac{W_{c}}{VI_L^2}\:=\:0}$$

$$\mathrm{⇒\:\frac{R_{a}}{V}=\frac{W_{c}}{VI_L^2}}$$

$$\mathrm{⇒\:I_L^2R_{a}=W_{c}}$$

$$\mathrm{⇒\: Variable\:Losses = Constant\:Losses}$$

Hence, the efficiency of a DC generator is maximum when the load current is such that the variable losses are equal to the constant losses.

The load current corresponding to maximum efficiency is given by,

$$\mathrm{I_{L}=\sqrt{\frac{W_{c}}{R_{a}}}}$$

## Numerical Example

A shunt generator supplies 95 A at a terminal voltage of 240 V. The armature and shunt field resistances are 0.2 Ω and 60 Ω respectively. The iron and frictional losses are 2000 W. Find the efficiency of DC generator. Also determine the value of load current at which maximum efficiency occurs.

## Solution −

Efficiency of DC Generator −

$$\mathrm{Shunt\:field\: current,\:I_{sh}\:=\:\frac{V}{R_{sh}}\:=\:\frac{240}{60}\:=\:6A}$$

$$\mathrm{Armature \:current,\:I_{a} = I_{L} + I_{sh} = 95 + 6 = 101 A}$$

$$\mathrm{Armature \:cu \:losses\:=\:=\:I_a^2R_{a}\:=\:(101)^2 × 0.2 = 2040.2 W}$$

$$\mathrm{Shunt \:field \:cu \:losses\:= I_{sh}^{2}R_{sh}\:= \:(6)^2 × 60 = 2160 W}$$

$$\mathrm{\therefore \:Total \:cu \:losses\:=\:2040.2 + 2160 = 4200.2\: W}$$

The total stray losses are given equal to 2000 W. Thus,

$$\mathrm{Total \:losses \:= \:Stray \:losses \:+\: Cu \:losses}$$

$$\mathrm{⇒\:\:Total \:Losses\: = \:2000 + 4200.2\: = \:6200.2\: W}$$

$$\mathrm{Output \:Power,\:P_{o} \:=\: VI_{L}\: = \:240 × 95 \:= \:22800 W}$$

Therefore, the input power will be

$$\mathrm{Input \:Power, \:P_{i} \:=\: P_{o} + Losses \:= \:22800 + 6200.2\: = \:29000.2 W}$$

$$\mathrm{\therefore\:Efficiency,\:\eta\:=\:\frac{P_{o}}{P_{i}}\:=\:\frac{22800}{29000.2}\times100\%\:=\:78.62\%}$$

The Load Current Corresponding to Maximum Efficiency −

$$\mathrm{Constant \:Losses,\:W_{c}\: =\: Shunt \:field \:cu \:loss \:+\: Stray \:losses}$$

$$\mathrm{⇒W_{c}=\:2160 + 2000 \:= \:4160 W}$$

$$\mathrm{I_{L}=\sqrt\frac{W_{c}}{R_{a}}\:=\:\sqrt\frac{4160}{0.2}\:=\: 144.22 A}$$