Efficiency of DC Generator & Condition for Maximum Efficiency with Examples

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,

• 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,

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

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 (I_L 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 −

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

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

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