Question 3 of 5: DC Motors

Question 3: DC Motors (20 marks)

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.

Part (a) — Three speed-control methods for a DC shunt motor

The speed of a DC shunt motor follows \(n=\dfrac{V_t-I_aR_a}{k\,\phi}\), so speed can be changed by adjusting the armature voltage \(V_t\), the field flux \(\phi\), or the effective armature-circuit resistance. This gives the three classical methods:

1. Armature-voltage control. The voltage applied to the armature is varied (formerly by a Ward-Leonard set, today by a controlled rectifier or DC chopper) while the field is held constant. Because flux is constant, torque capability is maintained, giving smooth, efficient, constant-torque control over a wide range below base speed. Its disadvantage is that it requires a variable-voltage supply and, on its own, cannot raise the speed above the base (rated-voltage) value.

2. Field (flux) control. A rheostat in the shunt-field circuit reduces the field current and hence the flux, which raises the speed above base. It is simple, cheap and efficient because the field carries only a small current, so little power is wasted. Its disadvantages are that it only gives speeds above base, the available torque falls as flux is weakened (constant-power region), and very weak fields cause poor commutation and can lead to dangerous overspeed if the field is lost.

3. Armature-resistance control. A resistor is inserted in series with the armature, dropping voltage and reducing speed below base. It is simple and needs no special supply, but it is very inefficient because the added resistor dissipates \(I_a^2R\) as heat, and it gives poor speed regulation because the speed then varies strongly with load. It is used only for intermittent or low-power duty.

Part (b) — DC shunt-motor performance

Given. DC shunt motor rated 20 hp, 230 V, 1250 rpm; \(R_a=0.18\ \Omega\); at rated speed and load \(I_a=78\text{ A}\), \(I_f=1.8\text{ A}\).

Find. (i) induced armature voltage \(E_a\); (ii) output torque; (iii) mechanical (rotational) losses; (iv) efficiency; (v) starting current; (vi) no-load speed.

Approach. Use the armature KVL for \(E_a\); shaft output power over mechanical speed for torque; the gap between developed power \(E_aI_a\) and shaft output for rotational losses; input \(V_tI_L\) for efficiency; \(V_t/R_a\) at standstill for starting current; and the proportionality \(E_a\propto n\) at constant flux for no-load speed.

1. Induced armature voltage. From the armature loop \(V_t=E_a+I_aR_a\), $$E_a=V_t-I_aR_a=230-78(0.18)=216.0\text{ V}.$$
2. Output torque. The rated shaft output is \(P_{out}=20\times746=14\,920\text{ W}\) at \(\omega_m=1250\cdot\dfrac{2\pi}{60}=130.9\text{ rad/s}\), so $$T_{out}=\frac{P_{out}}{\omega_m}=\frac{14\,920}{130.9}=114.0\ \text{N}\cdot\text{m}.$$
3. Mechanical (rotational) losses. The developed (electromagnetic) power is \(P_{dev}=E_aI_a=216.0(78)=16\,845\text{ W}\); the rotational losses are the part not delivered to the shaft, $$P_{rot}=P_{dev}-P_{out}=16\,845-14\,920=1925\text{ W}.$$
4. Efficiency. The input current is \(I_L=I_a+I_f=78+1.8=79.8\text{ A}\), so \(P_{in}=V_tI_L=230(79.8)=18\,354\text{ W}\) and $$\eta=\frac{P_{out}}{P_{in}}=\frac{14\,920}{18\,354}=81.3\%.$$ (Check: armature copper \(I_a^2R_a=1095\) W + field \(V_tI_f=414\) W + rotational \(1925\) W \(=3434\) W \(=P_{in}-P_{out}\) ✓.)
5. Starting current. At standstill \(n=0\Rightarrow E_a=0\), so the armature is limited only by \(R_a\): $$I_{a,\text{start}}=\frac{V_t}{R_a}=\frac{230}{0.18}=1278\text{ A}.$$ This is about 16 times rated armature current, which is why a starting resistor or controlled supply is required.
6. No-load speed. The shunt flux is constant (fixed \(I_f\)), so \(E_a\propto n\). At no load the armature current is small, so \(E_{a,\text{nl}}\approx V_t=230\text{ V}\) and $$n_{nl}=n_{rated}\,\frac{E_{a,\text{nl}}}{E_{a,\text{rated}}} =1250\,\frac{230}{216.0}=1331\text{ rpm}.$$