Question

(a) D.C. motors are used where very high torque and/or precise speed control is required. How does the control of magnetic field flux and armature current relate to the starting characteristics of a D.C. motor in such applications? (6)

(b) A 220 V, D.C. shunt motor has an armature resistance of 0.5 ohm and an armature current of 40 A on full load. Determine the reduction in flux necessary for a 50 per cent reduction in speed, the torque for both conditions can be assumed to remain constant. (10)

Answer 1

Part (a)

Control of magnetic field flux and armature current in relation to starting characteristics of a d.c. motor:

The torque of a d.c. motor is T = k phi Ia, and the speed is N = (V - Ia Ra)/(k phi). By controlling the field flux phi and the armature current Ia, both torque and speed can be controlled.
At starting, the armature current must be limited to a safe value (typically 1.5 to 2 times full-load current) because at standstill the back e.m.f. is zero and the armature would otherwise draw a very large current (V/Ra). This is done by inserting a starting resistance in series with the armature, or by applying a reduced voltage.
The field flux is kept at maximum (full field) during starting to give maximum torque per ampere of armature current, so that a high starting torque is obtained with a limited armature current. This is important for applications requiring very high starting torque.
For precise speed control, the field flux is weakened (field weakening) to increase speed above base speed, while the armature current (and hence torque) is controlled by the armature voltage. Below base speed, speed is controlled by armature voltage at full field; above base speed, by field weakening at constant power.
By controlling both flux and armature current, the motor can provide high torque at low speed and precise speed regulation, as required for deck machinery and traction.

Part (b)

220 V d.c. shunt motor, armature resistance 0.5 ohm, armature current 40 A on full load. Determine the reduction in flux for a 50% reduction in speed, torque constant.

Back e.m.f. at full load: E1 = V - Ia Ra = 220 - 40 x 0.5 = 220 - 20 = 200 V.
Speed is proportional to E/phi: N proportional to E/phi.
For 50% reduction in speed, N2 = 0.5 N1, so E2/phi2 = 0.5 (E1/phi1), i.e. E2 = 0.5 E1 (phi2/phi1) = 100 (phi2/phi1).
Torque constant: T = k phi Ia, so phi1 Ia1 = phi2 Ia2, giving Ia2 = Ia1 (phi1/phi2) = 40 (phi1/phi2).
Back e.m.f. at new condition: E2 = V - Ia2 Ra = 220 - 0.5 x 40 (phi1/phi2) = 220 - 20 (phi1/phi2).
Equating: 100 (phi2/phi1) = 220 - 20 (phi1/phi2). Let r = phi2/phi1.

100 r = 220 - 20/r -> 100 r^2 - 220 r + 20 = 0 -> 5 r^2 - 11 r + 1 = 0.

r = [11 +/- sqrt(121 - 20)]/10 = [11 +/- sqrt(101)]/10 = [11 +/- 10.05]/10.
Valid root: r = (11 - 10.05)/10 = 0.095 (the other root 2.1 is not physically valid for a speed increase).
So phi2 = 0.095 phi1, i.e. the flux is reduced to about 9.5% of its original value.
Percentage reduction in flux = (1 - 0.095) x 100 = 90.5%.
New armature current Ia2 = 40/0.095 = 421 A.

So the flux must be reduced by about 90.5% (to about 9.5% of the original) for a 50% speed reduction at constant torque.

✓ Verified Model Answer (Text Solution)

Control of magnetic field flux and armature current in relation to starting characteristics of a d.c. motor:

220 V d.c. shunt motor, armature resistance 0.5 ohm, armature current 40 A on full load. Determine the reduction in flux for a 50% reduction in speed, torque constant.