Two alternators working in the parallel supplies a unity...

Two alternators working in the parallel supplies a unity power factor load of 2500 kW together with a motor load of 4000 kVA having a lagging power factor of 0.707. One alternator is loaded 4000 kW, 0.80 leading. Calculate the kVA load and pf of the other alternator. The stator of a 3-phase, 20 pole alternator has 120 slots and there are 4 conductors per slot accommodated in two layers. If the speed of the alternator is 360 rpm, determine the emf induced between terminals. Air gap flux is 60 mWb per pole and the coil span is 150 electrical degrees.

Transcript

00:01 In this question, first let us solve the total apparent power.

00:04 So total apparent, so total apparent power is given by total load divided by power factor.

00:16 So this is equal to, so total load is given as 2500 kilowatt divided by 1.

00:22 Therefore, total apparent power is 2500 kva.

00:26 Then we have to calculate the apparent power of motor load.

00:32 So motor load, it is given by the formula.

00:37 So it is given by motor load divided by power factor, which is equal to, so this is equal to, it is 4000 kva divided by 0 .707.

00:55 So apparent power of the motor load is 4000, so 5657, so 5657 .3 kva.

01:05 Then we have to calculate the, so we have to calculate the apparent power of loaded alternator, alternator.

01:15 So this is given by the formula.

01:17 It is the, so it is the loaded alternator divided by power factor.

01:25 So this is on substitution.

01:26 It is equal to 4000 divided by 0 .8, which is equal to 5000 kva.

01:33 Then we have to calculate the apparent power of the another alternator, alternator 2.

01:43 So this is given by alternator 2.

01:46 So this is given by total apparent power minus motor apparent power minus apparent power of the loaded alternator, that is loaded alternator apparent power.

01:59 So this is equal to, it is 2500 minus 5657 .3 minus 5000.

02:08 So apparent power of the another alternator is equal to minus 8157 .3 kva.

02:16 So negative sign indicates that loaded alternator's power is leading.

02:21 So it is leading.

02:23 Therefore, so we can calculate the power factor of the other alternator, power factor of other alternator.

02:33 So it is given by the formula.

02:37 It is loaded alternate...