Question

(a) Show how the power that is transferred across the air gap of the three-phase induction motor is represented. Explain the terms. What portion of this is useful power? (6)

(b) A 440 V load of 400 kW at 0.8 (lagging) power factor is jointly supplied by two alternators A and B. The kW load on A is 150 kW and the KVAr load on B is 150 KVAr (lagging). Determine the kW load on B, the KVAr load on A, the power factor of operation on each machine and the current loading of each machine. (10)

Answer 1

Part (a)

Power transferred across the air gap of a three-phase induction motor:

The stator input power P1 is the electrical power drawn from the supply.
Stator losses (stator copper loss and iron/core loss) are subtracted to give the air-gap power Pg (also called the rotor input power), which is the power transferred across the air gap to the rotor by electromagnetic induction.
Pg = P1 - stator losses.
The air-gap power is divided into two parts: the rotor copper loss (I2^2 R2) and the mechanical power developed (gross mechanical power Pm).
Pg = rotor copper loss + gross mechanical power.
Rotor copper loss = s x Pg (where s is the slip), and gross mechanical power = (1 - s) x Pg.
The useful (shaft) power is the gross mechanical power minus the rotational losses (friction, windage and iron losses in the rotor). So the useful power = Pg(1 - s) - rotational losses.
The useful power is the portion that appears as mechanical output at the shaft.

Part (b)

Two alternators A and B supplying a 440 V load of 400 kW at 0.8 p.f. lagging:

Total load: kW = 400 kW. Total kVA = 400/0.8 = 500 kVA. Total kVAr (lagging) = 500 x 0.6 = 300 kVAr (since sin phi = 0.6).
Given: kW on A = 150 kW; kVAr on B = 150 kVAr (lagging).
kW on B = 400 - 150 = 250 kW.
kVAr on A = 300 - 150 = 150 kVAr (lagging).
Machine A: kVA = sqrt(150^2 + 150^2) = sqrt(45000) = 212.1 kVA. p.f. = 150/212.1 = 0.707 lagging.
Machine B: kVA = sqrt(250^2 + 150^2) = sqrt(85000) = 291.5 kVA. p.f. = 250/291.5 = 0.858 lagging.
Current loading: I = S / (root 3 x V).
I_A = 212100 / (1.732 x 440) = 212100 / 762.1 = 278.3 A.
I_B = 291500 / 762.1 = 382.5 A.

So A supplies 150 kW at 0.707 p.f. lagging, current 278 A; B supplies 250 kW at 0.858 p.f. lagging, current 382 A.

✓ Verified Model Answer (Text Solution)

Power transferred across the air gap of a three-phase induction motor:

Two alternators A and B supplying a 440 V load of 400 kW at 0.8 p.f. lagging: