Question

QUESTION 4: AC POWER - PF CORRECTION points) The voltage and current for a load Z are given by Vrms = 120290° and IRMS = 40230 Volts and Amps respectively; F-60 Hz. Do the following: • Determine the load Z6 • Determine the PF • State whether this is a leading/lagging PF. • Determine the Power delivered to the Load (Watts) • Sketch the power triangle and include all relevant information (quantities, units etc.) • What type of element must be added in parallel with the load in order to achieve a PF of 1.0? • Calculate the value of the element in part (f)

Answer 1

biven, VRms - 1202900 and I Rms - 40 (30° 10 = VRMs = 120 < 90 = 3460 Irms 40 30 ③ power factor = cos (60°) - 0.5 3 When Q (= 60°) > 0, is said to then be poker factor Lagging. ③ power de Livered to the load, P - 1x rm x Im cos (Qv - Qi). Ix 120 x 40 cas (go-300) 1 x 120 x 40 x(0.5) 1200 Watt Agong s Q lzl (Lagging Pf) Where, P = Q = o Real power measured in watt. (W) Reactive power measured in volt ampere Reactive (VAR) complex power measured in volt – ampere (VA) y s =

Since given power Hence to improve a capacitive element with Load. factor is Lagging, power Factor to unity is added in parallel 2222 For unity power factor, the total reactive power is zero. Assuming VAR of Capacitor bank - Qc he Qu + Q c = 0 Qc = – Q - Reactive power of Load - – I x Vm x Imx sin (Qu-Qi). =-x 120 x 40 x sin (90°– 30") 1x 120 x 40 x 0.866 the =- 2078.4 VAR. Relation, Use Qca y Vm = 7 Xc - 3.469 (120)2 2X 2078. 4 RXQc and xc = jwe he jwe C = ASAS Xcx Raf - - 3.464 x 24 x 60 7. 657810-4 - C - 765 MF &