Q10 (10 Marks) Electrical Circuits & Calculations
MET • Written Exam

With reference to Synchronous Motors:

(a) Draw and Explain the principle of operation of Synchronous Motors. (6)

(b) Find the synchronous impedance and reactance of an alternator in which a given field current produces an armature current of 200 A on short circuit and a generated e.m.f. of 50V on open-circuit. The armature resistance is 0.1 ohm. To what induced voltage must the alternator be excited if it is to deliver a load of 100 A at a p.f. of 0.8 lagging, with a terminal voltage of 200 V. (10)

Appeared In: Dec 2018

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Principle of operation of a synchronous motor:

  • A synchronous motor has a three-phase stator winding (like an induction motor) and a rotor with a d.c. field winding (or permanent magnets). The stator produces a rotating magnetic field at synchronous speed Ns = 120 f/P.
  • The rotor is excited with d.c., producing a fixed magnetic field. The rotor locks in step with the rotating stator field and rotates at exactly synchronous speed (no slip).
  • The motor is not self-starting: the rotor must be brought up to near synchronous speed (by a starting winding or by an external drive) before the d.c. field is applied, so that the rotor poles can lock onto the rotating field.
  • Once running, the motor maintains synchronous speed regardless of load (up to the pull-out torque). The load angle (torque angle) delta increases with load. By varying the d.c. field excitation, the motor can operate at unity, lagging, or leading power factor (over-excitation gives a leading power factor, useful for power-factor correction).
  • The sketch shows the stator winding, the rotor field winding, the d.c. excitation supply, and the rotating field.
Part (b)

Synchronous impedance and reactance of an alternator:

  • Synchronous impedance Zs = open-circuit e.m.f./short-circuit current = 50/200 = 0.25 ohm.
  • Synchronous reactance Xs = sqrt(Zs^2 - Ra^2) = sqrt(0.25^2 - 0.1^2) = sqrt(0.0525) = 0.229 ohm.
  • Induced voltage to deliver 100 A at 0.8 p.f. lagging with terminal voltage 200 V:
  • Assume star-connected. Phase voltage Vph = 200/root 3 = 115.5 V. I = 100 A. cos phi = 0.8, sin phi = 0.6.
  • E = sqrt[(Vph cos phi + I Ra)^2 + (Vph sin phi + I Xs)^2]
  • = sqrt[(115.5 x 0.8 + 100 x 0.1)^2 + (115.5 x 0.6 + 100 x 0.229)^2]
  • = sqrt[(92.4 + 10)^2 + (69.3 + 22.9)^2] = sqrt[102.4^2 + 92.2^2] = sqrt[18987] = 137.8 V per phase.
  • Line value = 137.8 x root 3 = 238.6 V.

So the alternator must be excited to give about 137.8 V per phase (238.6 V line).

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