Question

(a) What is the effect on the field flux of an alternator current in the synchronous motor that leads the terminal voltage? (6)

(b) A 1,000-KVA, 11,000-V, 3-ϕ, star-connected synchronous motor has an armature resistance and reactance per phase of 3.5 Ω and 40 Ω respectively. Determine the inducted e.m.f. and angular retardation of the rotor when fully loaded at (10)

(a) Unity p.f.

(b) 0.8 p.f. lagging

(c) 0.8 p.f. leading

Answer 1

Part (a)

Effect on the field flux of an armature current in the synchronous motor that leads the terminal voltage:

In a synchronous motor, the armature current produces a magnetomotive force (armature reaction) which reacts with the field flux.
When the armature current leads the terminal voltage (i.e. the motor is over-excited, operating at a leading power factor), the armature reaction is magnetising - it assists (strengthens) the main field flux. The motor is said to be over-excited and draws a leading current, which helps to improve the power factor of the system.
When the current lags (under-excited), the armature reaction is demagnetising and weakens the field flux.
So a leading armature current increases (strengthens) the field flux, and the motor behaves as a capacitor, supplying leading reactive power to the system.

Part (b)

1000 kVA, 11,000 V, 3-phase, star-connected synchronous motor, armature resistance 3.5 ohm and reactance 40 ohm per phase.

Phase voltage Vph = 11000/root 3 = 6350.9 V.
Full-load current I = S/(root 3 x V) = 1,000,000/(1.732 x 11000) = 1,000,000/19052 = 52.49 A.
The induced e.m.f. E = V + I(Ra + j Xs), and the angular retardation (load angle) delta is the angle between E and V.

Part (a)

Unity p.f.:

I = 52.49 A at phi = 0. E = sqrt[(V + I Ra)^2 + (I Xs)^2] = sqrt[(6350.9 + 52.49 x 3.5)^2 + (52.49 x 40)^2] = sqrt[(6350.9 + 183.7)^2 + 2099.6^2] = sqrt[6534.6^2 + 2099.6^2] = sqrt[42,700,000 + 4,408,000] = sqrt[47,108,000] = 6864 V per phase.
Line value = 6864 x 1.732 = 11,889 V.
delta = atan(I Xs/(V + I Ra)) = atan(2099.6/6534.6) = atan(0.3214) = 17.8 degrees.

Part (b)

0.8 p.f. lagging:

I = 52.49(0.8 - j0.6) = 41.99 - j31.49.
I Ra = (41.99 - j31.49) x 3.5 = 146.97 - j110.2.
j I Xs = (41.99 - j31.49) x j40 = j1679.6 + 1259.6.
E = 6350.9 + 146.97 - j110.2 + j1679.6 + 1259.6 = 7757.5 + j1569.4.
|E| = sqrt(7757.5^2 + 1569.4^2) = sqrt(60,178,000 + 2,463,000) = sqrt(62,641,000) = 7914 V per phase.
Line value = 7914 x 1.732 = 13,708 V.
delta = atan(1569.4/7757.5) = atan(0.2023) = 11.4 degrees.

Part (c)

0.8 p.f. leading:

I = 52.49(0.8 + j0.6) = 41.99 + j31.49.
I Ra = 146.97 + j110.2.
j I Xs = (41.99 + j31.49) x j40 = j1679.6 - 1259.6.
E = 6350.9 + 146.97 + j110.2 + j1679.6 - 1259.6 = 5238.3 + j1789.8.
|E| = sqrt(5238.3^2 + 1789.8^2) = sqrt(27,440,000 + 3,203,000) = sqrt(30,643,000) = 5535 V per phase.
Line value = 5535 x 1.732 = 9587 V.
delta = atan(1789.8/5238.3) = atan(0.3417) = 18.9 degrees.

So: unity p.f. E = 11.89 kV line, delta = 17.8 deg; 0.8 lag E = 13.71 kV line, delta = 11.4 deg; 0.8 lead E = 9.59 kV line, delta = 18.9 deg.

✓ Verified Model Answer (Text Solution)

Effect on the field flux of an armature current in the synchronous motor that leads the terminal voltage:

1000 kVA, 11,000 V, 3-phase, star-connected synchronous motor, armature resistance 3.5 ohm and reactance 40 ohm per phase.

Unity p.f.:

0.8 p.f. lagging:

0.8 p.f. leading: