Question

Q3. from a 240 V, 50 Hz supply as shown in the Figure 3. 24020 Vrms Inductive load with 0.6 lagging Load current 50 A Ohle Capacitor to be connected for power factor improvement Figure 3: a) Assuming that the capacitor shown in the figure is disconnected, calculate the load real (PL), reactive(Q1), and apparent power(SL) (5 marks) b) Draw the load power triangle and indicate real (P.) reactive (Q.), apparent (SL) power, and power factor angle (@.) (5 marks) c) Suppose that the capacitor shown in the diagram is connected to improve the power factor to unity (p.f=1), Calculate the reactive power (Qc) the capacitor should inject to the system (5 marks) d) Calculate the current (Ic) drawn by the connected capacitor (5 marks) e) Calculate the value of capacitance (C) required (5 marks)

Answer 1

SOLUTO ON: GIVEN: VSUPPLY = 240V; f = 50HZ, I lowp=50A POWER FACTOR CPS = COS OY = 0.6 .. THE CIRCUITS AS FOLLOWING: 2NDUCTIVE LOAD WTH 0,6 LACCAG LOAD CURRENT 501 CAPACTOR TO BE CONTTECTED FOR POWER PACTOR 2MPROVEMENT 240209 W REAL POWER SUPPLIED TO LOAD CPU)= VSUPPLY ZLOAD COSO, =2200w = 7.2kW REACTIVE POWER SUPPL2ED TO LOAD(Q2= Vs UPPLY ZLOAD Song, = 240*50* -10.6)2 = 960OVAR = 9.6kVAR. APPARENT POWER SUPPLIED TO LOAD (5.)= VSUPPLYILDADI = 240 * 50 = 12000 VA = 12 kVA. 20, P,=702kW, Q₂ = 9.6kVAR, SL=12 kVA. LANS). b) LOAD POWER TRIA NALE 2S AS FOLLOWING? HERE, PL=ACTIVE POWER(2NK WINNIT) ankua 12 Q₂ = REACTIVE POWER (2N KVAR UNN) KVAR) AND, SL = APPARENT POWER (2N VA UNIT) PL (21 kw) Se vaata AND, Cose;- 5 J2102 CN . PL 2 so, ;=cos 7.2x103 12x105 = 53.13° POWER FACTOR ANGLE(0;)23 33.13°(IN DEAREE), CANS)

9 REACTIVE POWER (Qc) THAT IS NEEDED TO BE INJECTED BY THE CAPAQTOR 25 = Q, = 9.6KVAR. THE REACTIVE POWER (Q) THE CAPACITOR SHOULD ANJECT TO THE SYSTEM IS 9.6 k VAR. CANS.) Qe d THE CURRENT DRAWN BY CAPA (2TOR()2) = - SUPPLY 9.68103 = 40AMP SO CURRENT DRAWN BY CAPACITOR 23 40AMP. CANS) e) WE KNOW THAT VOLTAGE ACROSS CAPAC2 TOROS, Ve = Usupply = al 40 SOC= - -=0.053 NF - 2nf * VSUPPLY 2* 3.14*50*240 20 so, VALUE OF CAPACITOR SHOULD BE 0.053NF. (ANS)