Question

A 240-Volt rms, 60-Hz source is connected in parallel with two loads. Compute the total reactive power in KVAR) supplied by the source and the overall power factor. Then, select the nearest answer. Load 1 12 kW Load 2 18 kW Ws = 24020° V rms pf = 0.7 lagging pf=0.8 leading

Answer 1

Solution : I lead 1 12 kW Load 2 18 kW PF - 6-7 10 PF -6.8 Vs=24010'v hms dمفا | Load : P = 12 kw , Pf, = 0.7 lag Reactive power consumed ty lood ů Q, = P, Tand, • P, Tan (605.190.] = 12x 1.02 KVAR = 12:24 KVAR Coso, = 0.7 Load 2 P₂ = 18kw, Pf₂ = 0.8 lead Reactive power connumed by lead 2 is los qz = 0.8 Q₂ = - P₂ Tam P2 [-ve because leading load , = -18 Tan/Cos'.87 so realtive poweň is gonerated] = - 18X0.75 KVAR = -13.5 KVAR

11 Total Total reactive power supplied by source ů Q = Q,+ Q2 = -1.26 KVAR Total actave power supplied PEP+ P, = 30 kW :: Pt = cos(Tamil ) = cos Tom. 1726 = 0.999 corect option is Q=-1-3, p. f = 0.999