Question

Two three phase 415 V alternators supply a ship’s load comprising:

lighting totalling 800 kW at unity power factor; and
motors totalling 1700 kW at power factor 0.7 lag.

One alternator supplies 1400 kW at power factor 0.75 lag.

(a) Calculate EACH of the following for the other alternator: (16)

(i) the KVA output;

(ii) the power factor;

(iii) the line output current.

Answer 1

Two 415 V alternators supply a ship's load: lighting 800 kW at unity p.f.; motors 1700 kW at 0.7 p.f. lag. One alternator supplies 1400 kW at 0.75 p.f. lag.

Total kW = 800 + 1700 = 2500 kW.
Reactive power of motors: kVA = 1700/0.7 = 2428.6 kVA; sin phi = sqrt(1 - 0.49) = sqrt(0.51) = 0.714; kVAr = 2428.6 x 0.714 = 1734.7 kVAr (lagging). Lighting contributes no reactive power.
Total kVAr = 1734.7 kVAr (lagging).
Alternator 1: 1400 kW at 0.75 lag. kVA = 1400/0.75 = 1866.7 kVA; sin phi = sqrt(1 - 0.5625) = 0.661; kVAr = 1866.7 x 0.661 = 1233.9 kVAr.
Other alternator (alternator 2):
(i) kW on alternator 2 = 2500 - 1400 = 1100 kW.
kVAr on alternator 2 = 1734.7 - 1233.9 = 500.8 kVAr (lagging).
kVA output = sqrt(1100^2 + 500.8^2) = sqrt(1,210,000 + 250,800) = sqrt(1,460,800) = 1208.6 kVA.
(ii) Power factor = 1100/1208.6 = 0.910 lagging.
(iii) Line output current: I = S / (root 3 x V) = 1,208,600 / (1.732 x 415) = 1,208,600 / 718.8 = 1681.5 A.

So the other alternator supplies 1100 kW at 0.91 p.f. lagging, kVA output 1208.6 kVA, line current 1681.5 A.

✓ Verified Model Answer (Text Solution)