Recently asked questions in Kochi mmd and Class 2 Numerical solutions

Search

22 March 2019

39.MET March 2019 Q.7

March 22, 2019 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.

                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.

0 comments:

Post a Comment