Q7 (16 Marks) Electric Machines (Motors & Generators) 🔥 Repeated 2x in exams
MET • Written Exam

(a) Derive an expression for the emf induced in an A.C. generator. (6)

(b) A 3000 KVA, 6-pole alternator runs at 1000 r.p.m. in parallel with other machines on 3300V bus-bars. The synchronous reactance is 25%. Calculate the synchronizing power for one mechanical degree of displacement and the corresponding synchronizing torque. (10)

Appeared In: Dec 2025Apr 2025

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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 average e.m.f. per conductor is E_av = 2 phi N P / 60, and 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 of the generated e.m.f. per phase for a distributed winding is E = 4.44 x f x phi x T x k_w, where T is the number of turns per phase, f the frequency, and k_w the winding factor.
Part (b)

3000 kVA, 6-pole alternator at 1000 rev/min in parallel on 3300 V busbars. Synchronous reactance 25%.

  • Synchronous speed Ns = 1000 rev/min (6-pole at 50 Hz). Angular speed (mechanical) w = 2 pi x 1000/60 = 104.72 rad/s.
  • Per-unit synchronous reactance Xs = 0.25 p.u. (based on 3000 kVA).
  • Synchronizing power: for a small angular displacement, the synchronizing power per phase = (Ef V / Xs) x (electrical angle in rad). At no load Ef = V.
  • In per unit, Ef V / Xs = 1 x 1 / 0.25 = 4 p.u. of the per-phase base power. Per-phase base power = 3000/3 = 1000 kW, so Ef V/Xs = 4 x 1000 = 4000 kW per phase; total for 3 phases = 12000 kW.
  • One mechanical degree = (P/2) electrical degrees = 3 electrical degrees = 3 x pi/180 = 0.05236 rad.
  • Synchronizing power Ps = 12000 x 0.05236 = 628.3 kW.
  • Synchronizing torque Ts = Ps / w = 628300 / 104.72 = 6000 N m.

So the synchronizing power for one mechanical degree of displacement is about 628 kW and the corresponding synchronizing torque is about 6000 N m.

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