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Efficiency of a DC Motor – Condition for Maximum Efficiency
Efficiency of a DC Motor
The efficiency of a DC motor is defined as the ratio of output power to the input power.
Mathematically,
$$\mathrm{Efficiency,\eta \:=\:\frac{Power\:Output}{Power\:Intput}\times 100 \%\:=\:\frac{P_{out}}{P_{in}}\times100\%\:...(1)}$$
Since,
$$\mathrm{Power\:input,\:P_{in} = Power\:Output(P_{out}) + Losses}$$
Therefore,
$$\mathrm{Efficiency,\eta \:=\:\frac{P_{out}}{P_{out}\:+\:losses}\times100\%\:....(2)}$$
Condition for Maximum Efficiency of DC Motor
The efficiency of a DC motor is not constant but varies with the load. Consider a shunt motor (as shown in the figure) drawing an armature current of Ia amperes and the back emf is Eb.
The mechanical power developed by the motor is (neglecting the mechanical losses),
$$\mathrm{P_{m}=P_{out}=E_{b}I_{a}}$$
The input power to the motor is,
$$\mathrm{P_{in}=P_{out}\:+\:variable\:losses\:+\:constant\:losses}$$
$$\mathrm{Efficiency,\eta\:=\:\frac{P_{out}}{P_{in}}\:=\:\frac{E_{b}I_{a}}{P_{out}\: +\:variable\:losses\:+\:constant\:losses}}$$
$$\mathrm{⇒\eta\:=\:\frac{E_{b}I_{a}}{E_{b}I_{a}\:+\:I_a^2R_{a}\:+\:W_{c}}}$$
$$\mathrm{⇒\eta\:=\:\frac{1}{1+(\frac{I_{a}R_{a}}{E_{b}})+\frac{W_{c}}{E_{b}I_{a}}}\:....(3)}$$
The efficiency of the motor will be maximum, if the denominator of the eqn. (3) is minimum. Hence, to determine the condition for maximum efficiency, differentiating the denominator of the eqn. (3) with respect to armature current and equating it to zero, i.e.
$$\mathrm{\frac{d}{dt}[1+(\frac{I_{a}R_{a}}{E_{b}})+(\frac{W_{c}}{cd})]\:=\:0}$$
$$\mathrm{⇒\:(\frac{R_{a}}{E_{b}})\:-\:(\frac{W_{c}}{E_{b}I_a^2})\:=\:0}$$
$$\mathrm{⇒\:\frac{R_{a}}{E_{b}}\:=\:\frac{W_{c}}{E_{b}I_a^2}}$$
$$\mathrm{⇒\:I_a^2R_{a}\:=\:W_{c}}$$
$$\mathrm{⇒\:Variable\:Losses\:=\:Constant\:Losses … (4)}$$
Therefore, the maximum efficiency occurs in a DC motor when the variable losses become equal to the constant losses for the motor.
The armature current corresponding to maximum efficiency is given by,
$$\mathrm{I_{a}=\sqrt\frac{W_{c}}{R_{a}}\:\:......(5)}$$
Numerical Example
A 10 kW DC shunt motor has the following losses at full load
Mechanical Losses = 300 W
Iron Losses = 400 W
Shunt field Cu loss = 100 W
Armature cu loss = 500 W
Calculate the full-load efficiency of the motor and armature current corresponding to maximum efficiency if the armature resistance is 0.25 Ω.
Solution
Here,
$$\mathrm{Constant\:Losses,\:W_{c}\:=\:300 + 400 + 100\:=\:800 W}$$
$$\mathrm{Variable\:losses\:=\:Armature\:cu\:loss\:=\:500 W}$$
$$\mathrm{Total\:Losses\:=\:800 + 500\:=\:1300 W}$$
Therefore,
$$\mathrm{Power\:input,P_{in} = 10000 + 1300 = 11300 W}$$
$$\mathrm{\therefore \:Efficiency,\eta \:=\:\frac{P_{out}}{P_{in}}\times100\:=\:\frac{10000}{11300}\times 100 \:}$$
$$\mathrm{⇒\:\eta\:=\:88.49\%}$$
The armature current corresponding to maximum efficiency,
$$\mathrm{I_{a}\:=\:\sqrt\frac{W_{c}}{R_{a}}\:=\:\sqrt\frac{800}{0.25}\:=\:56.57A}$$
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