Question

(a) Derive an expression for the e.m.f induced in an a.c. generator. (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 percent reduction in speed. The torque for both conditions can be assumed to remain constant. (10)

Answer 1

Part (a)

Derivation of the e.m.f. induced in an a.c. generator:

Consider a coil of N turns rotating in a uniform magnetic field of flux density B. The flux linking the coil is phi = B A cos(wt), where A is the coil area and w the angular velocity.
By Faraday's law, the induced e.m.f. is e = -N d(phi)/dt = -N d(B A cos wt)/dt = N B A w sin(wt) = Em sin(wt).
The maximum e.m.f. Em = N B A w = N phi_m w, where phi_m = B A is the maximum flux linking the coil.
For a machine with Z conductors, P poles, flux per pole phi, speed N rev/min, the generated e.m.f. is:

E = (P x phi x Z x N)/(60 x A) volts, where A is the number of parallel paths (A = 2 for wave winding, A = P for lap winding).

The r.m.s. value per phase for a distributed winding is E = 4.44 f phi T k_w, where T is the turns per phase, f the frequency, and k_w the winding factor.

Part (b)

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

Back e.m.f. E1 = V - Ia Ra = 220 - 40 x 0.5 = 220 - 20 = 200 V.
Speed proportional to E/phi. For 50% speed reduction, N2 = 0.5 N1, so E2/phi2 = 0.5 E1/phi1, i.e. E2 = 100 (phi2/phi1).
Torque constant: phi1 Ia1 = phi2 Ia2, so Ia2 = 40 (phi1/phi2).
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 -> 5 r^2 - 11 r + 1 = 0.

r = [11 +/- sqrt(101)]/10 = [11 +/- 10.05]/10. Valid root r = 0.095.
So phi2 = 0.095 phi1, i.e. flux reduced to about 9.5% of original. Percentage reduction = 90.5%.
New armature current Ia2 = 40/0.095 = 421 A.

So the flux must be reduced by about 90.5% for a 50% speed reduction at constant torque.

✓ Verified Model Answer (Text Solution)

Derivation of the e.m.f. induced in an a.c. generator:

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