Taming the Speed: Solved Problems on DC Motor Speed Control Techniques

Problem-1

A 250V, dc shunt motor has shunt field resistance of 250 \(\Omega\) and an armature resistance of 0.25 \(\Omega\). For a given load torque and no additional resistance included in the shunt field circuit, the motor runs at 1500 rpm, drawing an armature current of 20 A. If a resistance of 250 \(\Omega\) is inserted in series with the field, the load torque remaining the same, find out the new speed and armature current

Solution-1

\[\begin{aligned} \dfrac{N_{2}}{N_{1}} & =\dfrac{E_{b2}}{E_{b1}}\times\dfrac{\phi_{1}}{\phi_{2}}\\ \Rightarrow & =\dfrac{E_{b2}}{E_{b1}}\times\dfrac{I_{sh1}}{I_{sh2}}\\ E_{b1} & =V-I_{a1}R_{a}\\ E_{b2} & =V-I_{a2}R_{a} \end{aligned}\]
Motor speed changed by changing flux
\[\begin{aligned} T_{a} & \propto\phi_{1}I_{a1}\propto\phi_{2}I_{a2} \quad \Rightarrow\phi_{1}I_{a1} =\phi_{2}I_{a2}\\ I_{a2} & =I_{a1}\times\dfrac{\phi_{1}}{\phi_{2}}=I_{a1}\times\dfrac{I_{sh1}}{I_{sh2}}\\ I_{sh1} & =250/250=1\mathrm{A}\\ I_{sh2} & =250/\left(250+250\right)=1/2\mathrm{A} \end{aligned}\]
Since load torque remains same

\[\begin{aligned} I_{a2} & =20\times\dfrac{1}{1/2}=40\mathrm{A}\\ E_{b2} & =250-\left(40\times0.25\right)=240\mathrm{V}\\ E_{b1} & =250-\left(20\times0.25\right)=245\mathrm{V}\\ \dfrac{N_{2}}{1500} & =\dfrac{240}{245}\times\dfrac{1}{1/2}\\ \Rightarrow N_{2} & =2930\mathrm{RPM} \end{aligned}\]

Problem-2

A 220 V shunt motor has an armature resistance of 0.5 \(\Omega\) and takes a current of 40 A on full load. By how much must the main flux be reduced to raise the speed by 50% if the developed torque is constant?

Solution-2

Torque remains constant

\[\begin{aligned} T_{a} & \propto\phi_{1}I_{a1}\propto\phi_{2}I_{a2}\\ \Rightarrow\phi_{1}I_{a1} & =\phi_{2}I_{a2}\\ I_{a2} & =I_{a1}\times\dfrac{\phi_{1}}{\phi_{2}}=40x~\text{where}~x=\dfrac{\phi_{1}}{\phi_{2}}\\ E_{b1} & =220-\left(40\times0.5\right)=200\mathrm{V}\\ E_{b2} & =220-\left(40x\times0.5\right)=\left(220-20x\right)\mathrm{V}\\ \dfrac{N_{2}}{N_{1}} & =\dfrac{E_{b2}}{E_{b1}}\times\dfrac{\phi_{1}}{\phi_{2}}\\ \Rightarrow\dfrac{3}{2} & =\dfrac{\left(220-20x\right)}{200}x\\ \Rightarrow & x^{2}-11x+15 = 0\\ \Rightarrow x & =9.4^{*}~\text{or}~1.6 \end{aligned}\]

\[\begin{aligned} \dfrac{\phi_{1}}{\phi_{2}} & =1.6~\text{or \ensuremath{\dfrac{\phi_{2}}{\phi_{1}}=\dfrac{1}{1.6}}}\\ \Rightarrow\dfrac{\phi_{1}-\phi_{2}}{\phi_{1}} & =\dfrac{1.6-1}{1.6}=\dfrac{3}{8}\\ \Rightarrow\%\text{change in flux} & =\dfrac{3}{8}\times100=37.5\% \end{aligned}\]

Problem-3

A 220 V shunt motor with an armature resistance of 0.5 \(\Omega\) is excited to give constant main field. At full load the motor runs at 500 rpm and takes an armature current of 30 A. If a resistance of 1 \(\Omega\) is placed in the armature circuit, find the speed at

full-load torque
double full-load torque

Solution-3

Since flux remains constant, \(N_2/N_1 = E_{b2}/E_{b1}\)

\[\begin{aligned} & \text{With no additional resistance in armature circuit}\\ N_{1} & =500\mathrm{rpm}\\ I_{a1} & =30\mathrm{A}\\ E_{b1} & =220-30\times0.5=205\mathrm{V}\\ T & \propto I_{a}\left(\because\phi=\text{constant}\right)\\ \Rightarrow\dfrac{T_{2}}{T_{1}} & =\dfrac{I_{a2}}{I_{a1}}\quad \text{Since}T_{2}=T_{1}; \quad I_{a2}=I_{a1}=30\mathrm{A}\\ & \text{When additional resistance of 1}\Omega\\ E_{b2} & =220-30\left(1+0.5\right)=175\mathrm{V}\\ \dfrac{N_{2}}{500} & =\dfrac{175}{205}\Rightarrow N_{2}=427\mathrm{rpm} \end{aligned}\]
Full-load torque:

Double Full-load torque:

\[\begin{aligned} \frac{T_{2}}{T_{1}} &=\frac{I_{a 2}}{I_{a \mathrm{l}}} \quad \text { or } \quad \frac{2 T_{1}}{T_{1}}=\frac{l_{a 2}}{30} \\ I_{a 2}=60 \mathrm{~A} \\ E_{b 2} &=220-60(1+0.5)=130 \mathrm{~V} \\ \frac{N_{2}}{500} &=\frac{130}{205} \\ N_{2}=317 \mathrm{r.p.m} \end{aligned}\]

Problem-4

A dc series motor drives a load, the torque of which varies as the square of the speed. Assuming the magnetic circuit to be remain unsaturated and the motor resistance to be negligible, estimate the percentage reduction in the motor terminal voltage which will reduce the motor speed to half the value it has on full voltage. What is then the percentage fall in the motor current ? Stray losses of the motor may be ignored.

Solution-4

\[\begin{aligned} T_{a}& \propto \Phi I_{a} \propto I_{a}^{2} \\ T_{a}& \propto N^{2} \\ N^{2} & \propto I_{a}^{2}~ \text{or}~ N \propto I_{a} \\ N_{1} & \propto I_{a 1} ~ \text{and}~ N_{2} \propto I_{\alpha 2} \\ N_{2} / N_{1}&=I_{a 2} / I_{a 1} \\ N_{2} / N_{1} & =1 / 2 \\ I_{a 2} / I_{a1} & =1 / 2 \text { or } I_{a 2}=I_{a 1} / 2 \end{aligned}\]

be the voltages across the motor in the two cases. Since motor resistance is negligible, and Let

\[\begin{aligned} \frac{N_{2}}{N_{1}} &=\frac{E_{b 2}}{E_{b 1}} \times \frac{\Phi_{1}}{\Phi_{2}} \\ \frac{1}{2}&=\frac{V_{2}}{V_{1}} \times 2 \\ \frac{V_{2}}{V_{1}}&=\frac{1}{4} \\ \therefore \quad \frac{V_{1}-V_{2}}{V_{1}} &=\frac{4-1}{4}=0.75 \\ \therefore \% \text { reduction in voltage }&=\frac{V_{1}-V_{2}}{V_{1}} \times 100=0.75 \times 100=75 \% \\ \%\text { change in motor current }&=\frac{I_{a 1}-I_{a 2}}{I_{a 1}} \times 100\\ &=\frac{I_{a1}-I_{a 1} / 2}{I_{a1}} \times 100=50 \% \end{aligned}\]