Efficiency of DC Generator and Condition for Maximum Efficiency with Examples

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Efficiency of DC Generator

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

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

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,

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

$$\mathrm{\text{Copper Losses } \: = \: B \: - \: C}$$

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

Mechanical Efficiency −

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

Electrical Efficiency −

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

Commercial Efficiency − (always consider this unless stated otherwise)

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

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,

I_L = load current

V = terminal voltage

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

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

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

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

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

Where,

I_a²R_a = Variable Losses = Copper Losses

W_c = Constant Losses = Iron Losses + Mechanical Losses

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

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

Hence, the efficiency of DC generator will be,

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

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

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 (I_L in this case) and equate it to zero, i.e.

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

$$\mathrm{\Rightarrow\:0\:+\:\frac{R_{a}}{V}\:-\:\frac{W_{c}}{VI_{L}^{2}}\:=\:0}$$

$$\mathrm{\Rightarrow\:\frac{R_{a}}{V}\:=\:\frac{W_{c}}{VI_{L}^{2}}}$$

$$\mathrm{\Rightarrow\:I_{L}^{2}R_{a} \:= \:W_{c}}$$

$$\mathrm{\Rightarrow\: \text{Variable Losses = Constant Losses}}$$

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 −

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

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

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

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

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

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

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

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

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

Therefore, the input power will be

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

$$\mathrm{\therefore\: Efficiency,\:\eta\:=\:\frac{P_{o}}{P_{i}}\:=\:\frac{22800}{29000.2}\:\times\:100\%\:=\:78.62\%}$$

The Load Current Corresponding to Maximum Efficiency −

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

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

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