Solved Problems on DC Motors Fundamentals

Problem-1

A 25-kW, 250-V, dc shunt machine has armature and field resistances of 0.06 \(\Omega\) and 100 \(\Omega\) respectively. Determine the total armature power developed when working

as a generator delivering 25 kW output and
as a motor taking 25 kW input.

Solution-1

\[\begin{aligned} \text { Output current } &=25,000 / 250=100 \mathrm{~A} \\ I_{s h}&=250 / 100=2.5 \mathrm{~A} \\ I_{a}&=102.5 \mathrm{~A} \\ E_g&=250+I_{a} R_{a}=250+102.5 \times 0.06\\ &=256.15 \mathrm{~V} \\ \text { Power developed in armature } &=E_{b} I_{a}=\dfrac{256.15 \times 102.5}{1000}=26.25 \mathrm{k} \mathrm{W} \end{aligned}\]

As Generator

Problem-2

A 4-pole, 32 conductor, lap-wound dc shunt generator with terminal voltage of 200 volts delivering 12 A to the load has \(R_a = 2 ~\Omega\) and \(R_f=200~ \Omega\), it is driven at 1000 rpm.

Calculate the flux per pole in the machine.
If the machine has to be run as a motor with the same terminal voltage and drawing 5 A from the mains, maintaining the same magnetic field, find the speed of the machine.

Solution-2

\[\begin{aligned} E_{g}&=200+13 \times 2=226 \mathrm{~V}\\ 226&=\phi \dfrac{Z N}{60} \times \dfrac{P}{a}\\ P &=a \\ \phi &=\dfrac{226 \times 60}{1000 \times 32}=0.42375 \mathrm{wb} \end{aligned}\]

Problem-3

A d.c. motor takes an armature current of 110 A at 480 V. The armature circuit resistance is 0.2 \(\Omega\). The machine has 6-poles and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate

the speed and
the gross torque developed by the armature.

Solution-3

\[\begin{aligned} E_{b}&=480-110 \times 0.2=458 \mathrm{~V} \\ \Phi&=0.05 \mathrm{~W}\\ Z&=864 \\ E_{b}&=\dfrac{\Phi Z N}{60}\left(\dfrac{P}{A}\right) \\ \rightarrow 458&=\dfrac{0.05 \times 864 \times N}{60} \times\left(\dfrac{6}{6}\right) \\ \rightarrow N&=636~ \mathrm{rpm} \\ T_{a}&=0.159 \times 0.05 \times 864 \times 110(6 / 6)=756.3 \mathrm{~N}-\mathrm{m} \end{aligned}\]

Problem-4

Determine the developed torque and the shaft torque of 220-V, 4-pole series motor with 800 conductors wave-connected supplying a load of 8.2 kW by taking 45 A from the mains. The flux per pole is 25 mWb and its armature circuit resistance is 0.6 \(\Omega\)

Solution-4

\[\begin{aligned} T_{a} &=0.159 \Phi Z A(P / \mathrm{A}) \\ &=0.159 \times 25 \times 10^{-3} \times 800 \times 45(4 / 2)\\ &=286.2 \mathrm{~N}-\mathrm{m} \\ E_{b} &=V-I_{a} R_{a}=220-45 \times 0.6=193 \mathrm{~V} \\ E_{b} &=\Phi Z N(P / A) \\ \rightarrow 193&=25 \times 10^{-3} \times 800 \times N \pi \times(4 / 2) \\ N&=4.825 ~\mathrm{r.p.s}\\ 2 \pi N T_{s h}&= \text{output} \\ 2 \pi \times 4.825 T_{sh}&=8200 \\ \rightarrow T_{s h}&=270.5 \mathrm{~N} \cdot \mathrm{m} \end{aligned}\]

Developed or gross torque is the same thing as armature torque.

Problem-5

A 500-V dc shunt motor draws a line-current of 5 A on light-load. If armature resistance is 0.15 \(\Omega\) and the field resistance is 200 \(\Omega\), determine the efficiency of the machine running as a generator delivering a load current of 40 A

Solution-5

\[\begin{aligned} \text { Input Power } &=500 \times 5=2500 \text { watts } \\ \text { Field copper-loss } &=500 \times 2.5=1250 \text { watts } \end{aligned}\]
No Load, running as a motor:
Neglecting armature copper-loss at no load ( \(2.5^{2} \times 0.15=1\) watt), the balance of 1250 watts goes towards no load losses of the machine running at rated speed.
These losses are mainly the no load mechanical losses and the core-loss.

\[\text { Output delivered }=500 \times 40 \times 10^{-3}=20 \mathrm{~kW}\]
As a Generator, delivering 40 A to load:
\[\begin{aligned} \text{Field copper-loss} &=1250~\text{watts} \\ \text{Armature copper-loss} & =42.5^{2} \times 0.15=271~ \text{watts}\\ \text{No load losses} &=1250~ \text{watts} \\ \text{Total losses} &=2.771 \mathrm{~kW}\\ \text{Generator Efficiency} &=(20 / 22.771) \times 100 \%=87.83 \% \end{aligned}\]
Losses :

Problem-6

A dc series motor operates at 800 rpm with a line current of 100 A from 230-V mains. Its armature circuit resistance is 0.15 \(\Omega\) and its field resistance 0.1 \(\Omega\). Find the speed at which the motor runs at a line current of 25 A, assuming that the flux at this current is 45% of the flux at 100 A

Solution-6

\[\begin{aligned} \dfrac{N_{2}}{N_{1}}&=\dfrac{E_{b 2}}{E_{b 1}} \times \dfrac{\Phi_{1}}{\Phi_{2}} \\ \Phi_{2}&=0.45 \Phi_{1} \\ \rightarrow \dfrac{\Phi_{1}}{\Phi_{2}}&=\dfrac{1}{0.45} \\ E_{b 1} &=230-(0.15+0.1) \times 100=205 \mathrm{~V} \\ E_{b 2}&=230-25 \times 0.25=223.75 \mathrm{~V} \\ \dfrac{N_{2}}{800} &=\dfrac{223.75}{205} \times \dfrac{1}{0.45}\\ N_{2}&=1940~ \mathrm{r.p.m} \end{aligned}\]