Question

The infinite bus in the Figure operates at 480 V. Load 1 consumes 100 kW at 0.9 Pflagging and load 2 consumes 80 KVA O Plagne load is synchronous motor whose real power consumption is 74.6 kW at 0.8 PF leading (a) If the switch was initially open what is the equivalent power factor of the system (b) When the switch is finally closed, what is the new equivalent power factor of the system Bus 1 Load 1 100 KW 0.9 PF laging Load 2 80 KVA 08 Flag 40 V Y connected M

Answer 1

Ang PLI Su= - SL 1 = [Cosg=PF=0.9] Load -15- 100 KW at 0.9PF lagging Real power absorbed - 100kW apparent power of Load 1 = Piz cos, 100 kW 0.9 SL2 = 111.11 KVA lagging nowo, Since the 'of' is lagging, it absorbs reactive power Reactive power absorbed is Qu= vsi PLI 48.43 KVAR 2 PL1 - 100kW, QL1 = 48-43KVAR, SL1 - 11). 11kVA load-2- Lagging SOKVA at 0.8 PF lagging Reactive power is absorbed =) SlQ - 80KVA Real power absorbed= Pla= Slax cosda (cos2=PF= 0.8 80 x 0.8 64 kW Reactive power absorbed - Qla- vscé-Puz 48 KVAR PL2=64kW Pla -48KVAR, S12= 80KVA

3 total Real power absorbed PL+Pla 100+ 64 -) PL - PL = 164 kW total Reactive power absorbed : QL+QL2 AL - 48.434 48 QL 96.43 KVAR apparent power = Sl= VP ²+0 2 = V1642 +96.432 190.25 kVA -0.862 powerfactor of the system = PL 164 SL 190.25 Since 'Al is absorbed pf is lag power factor = 0.862 lagging when Scoitch is closed :- Iload 1 म G IL Iload 2 M 74.6kW Ofpf lead Load3 100kW => 100kW, 18.43 KVAR 0.9 pflag Sokva => UKW, 48 KVAR 1o8pf lag absorbs both real and reactive power 74.6KW -55.95KVAR Labsorbs real power but delivers reactive power : Reactive power has '-' sign.

total Real power absorbed - P4+Plat PL3 -100+ 60+ 74.6 PL = 238-6 KW total Reactive power absorbed = Qur+Ql2+Q13 = 48.43+48 - 55.95 = + 40.48 KVAR Bince 'QL is the positive => Reactive power is absorbed pfis lagging and apparent power 242 KVA Sl= VP22+Q2² S4=242 KVA power factor PF = PL = 0.986 lag SL |: powerfactor of the system: 0.986 lag