EECE 370 Problems Power Factor Correction Example, Summer 2011

LEARNING EXAMPLE . Required :pf 0.94 leading | 34 5. |, 60 = = = kV rms Hz V f line ⎫ f S P jQ P S | cosθ |= + = MVA Q leading pf MW P new new old 8.6 94 .0 18.72 = − ⇒ ⎭ ⎬ ⎫ = = MVA Q 21.82 15.02 8.6 = − − = − Δ 2 1 tan pf pf f − = θ f f f pf S Q θ θ cos |sin | | | = = f P Q tanθ = MVA Q capacitor = − .7 273 per kV rms V Y 3 = 34 5. ⇒ − capacitor connection .0 626 1 sin cos 2 = − = ⇒ = pf pf f f θ θ > 0 ⇒ Qold lagging 2 3 6 3 10 34 5. 60 2 .7 273 10 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × × × = − × − C π F C 6 48 MW P MVA Q old old .72 18 | 15.02 | = = F C = 48 6. μ