A
72 KVA transformer supplies (a) a heating and lighting load of 12 KW at
unity power factor and a motor load of 70 kVA at 0.766 (lagging) power
factor; Calculate the minimum rating of the power-factor improvement
capacitors which must be connected in the circuit the ensure that the
transformer does not become overloaded.
If you are noticing some error in problems kindly comment below.Thanks
Given
Transformer KVA = 72
Motor KVA = 70
Motor PF = 0.766
Active power = 12 KW
To find
minimum rating of the power-factor improvement capacitors
Solution
cos fi = 0.766, fi = 40 deg, sin fi = 0.6428
Motor active power KW = KVA x Cos ø
= 70 x 0.766
= 53.62 KW
Total Active power = 53.62 + 12
= 65.62 KW
From triangle ,
Motor reactive power KVAr = KVA x Sin ø
= 70 x 0.6428
= 44.996 KVAr
Transformer reactive power KVAr = Square root of (KVA2 – KW2)
= Square root of (722 – 65.622)
= 29.63 KVAr
Therefore reduction in reactive load = Motor reactive load - Transformer reactive load
= 44.996 - 29.63
= 15.366 KVAr
The minimum rating of capacitor will be providing 15.366 KVAr at leading power factor.
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