Question 6 of 6: DC Motor

Question 6 — DC Motor

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 30 hp, 240 V, 1150 rpm DC shunt motor at rated load, efficiency 88.5%, with $R_a=0.064\ \Omega$ and $R_f=93.6\ \Omega$.

Find. (a) the meaning of $E_a$; (b)(i) rotational-loss fraction of total losses; (ii) series armature resistance to cap starting current at 175% of rated; (iii) the new speed with 10% less flux at rated armature current.

Approach. $E_a$ is the back-emf. From $\eta$ and rated output get $P_{in}$ and line current; split into field and armature currents; find $E_a$; separate copper losses from the total loss to get rotational losses; use the standstill ($E_a=0$) condition for the starting resistor; and use $E_a=k\Phi\,n$ with constant $E_a$ (armature current held) for the new speed.

(a) Meaning of $E_a$. $E_a$ is the internal generated voltage — the back-emf (counter-emf) induced in the rotating armature conductors, $E_a=k\Phi\omega$. It opposes the applied voltage, so the net voltage driving armature current is $V_t-E_a=I_aR_a$; it is the term through which electrical input becomes mechanical power ($P_{dev}=E_aI_a$).

1. Input power and currents. $P_{out}=30(746)=22{,}380\ \text{W}$, so $P_{in}=P_{out}/\eta=22{,}380/0.885=25{,}288\ \text{W}$ and $I_L=P_{in}/V_t=105.4\ \text{A}$. The field draws $I_f=V_t/R_f=240/93.6=2.56\ \text{A}$, so the rated armature current is $I_a=I_L-I_f=102.8\ \text{A}$ and $E_a=V_t-I_aR_a=240-6.58=233.4\ \text{V}$.
2. (i) Loss split. Total losses $=P_{in}-P_{out}=2908\ \text{W}$. Field copper loss $=V_tI_f=615\ \text{W}$; armature copper loss $=I_a^2R_a=(102.8)^2(0.064)=676\ \text{W}$. Hence rotational (friction, windage and core) losses $=2908-615-676=1616\ \text{W}$, and $$\frac{P_{rot}}{P_{loss}}=\frac{1616}{2908}=\boxed{55.6\%}$$
3. (ii) Starting resistor. At standstill $E_a=0$, so the armature circuit must limit the current to $1.75I_a=1.75(102.8)=179.9\ \text{A}$: $$R_a+R_{ext}=\frac{V_t}{1.75\,I_a}=\frac{240}{179.9}=1.334\ \Omega \;\Rightarrow\; R_{ext}=1.334-0.064=\boxed{1.27\ \Omega}$$
4. (iii) New speed with weakened field. Holding $I_a$ at its rated value keeps $E_a=V_t-I_aR_a=233.4\ \text{V}$ unchanged. Since $E_a=k\Phi\,n$, $$\frac{n_2}{n_1}=\frac{E_{a2}/\Phi_2}{E_{a1}/\Phi_1}=\frac{\Phi_1}{0.9\,\Phi_1} \;\Rightarrow\; n_2=\frac{1150}{0.9}=\boxed{1278\ \text{rpm}}$$ Weakening the field 10% raises the speed about 11%.