A total load of 8000 KW at 0.8 power factor is supplied by two alternators in parallel. One alternator supplies 6000KW at 0.9 power factor. Find the KVA rating of the other alternator and the power factor.
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Given
Total power PT = 8000 KW
Total pf Cos øT = 0.8
Generator 1 power P1 = 6000 KW
Generator 1 pf Cos ø1 = 0.9
To find
1. Generator 2 apparent power, KVA2
2. Generator 2 pf, Cos ø2
Solution
Apparent power total, KVAT = KWT / Cos øT
= 8000 / 0.8
= 10000 KVA
Reactive power total, KVArT = Sin øT X KVAT
= 0.6 X 10000
= 6000 KVArT
Apparent power Generator 1, KVA1 = KW1 / Cos ø1
= 6000 / 0.9
= 6667 KVA
Reactive power Generator 1, KVAr1 = Sin ø1 X KVA1
= 0.4357 X 6667
= 2905 KVAr
Power of generator 2, KW2 = KWT – KW1
= 8000 - 6000
= 2000 KW
Reactive power of generator 2, KVAr2 = KVArT – KVAr1
Reactive power of generator 2, KVAr2 = KVArT – KVAr1
= 6000 - 2905
= 3095 KVAr
1. Apparent power of generator 2, KVA22 = KW22 + KVAr22
KVA2 = √ (KW22 + KVAr22 )
= √ (20002 + 30952)
= 3684 KVA
2. Power factor of generator 2, Cosø2 = KW2 / KVA2
= 2000 / 3684
= 0.543
In 1st answer 2nd step how kvar2 came..Its KVA2 know??
ReplyDeleteThanks for the info. Same updated.
Deletehow sinPhi t value is calculated from given value
ReplyDeleteCos øT = 0.8
DeleteøT = cos-1 (0.8) {cos inverse 0.8}
øT = 0.6435011 (use scientific calculator)
So sin øT = Sin (0.6435011)
Sin øT = 0.6
power generator2 KV2 answer unit is KW
ReplyDeleteThanks for the info. Updated.
Delete