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25 January 2021

101.MET November 2020 Q.6

January 25, 2021 Posted by AK No comments

 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.


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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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