THREE PHASE A.C.CIRCUITS Expression for power in a 3 phase delta connected load

Power  = Power in R phase + Power in Y phase + Power in B Phase             =VRY .IR'Y' cos Φ + VYB IY'B' cos Φ + VBR IB'R' cos Φ But              VRY = VYB = VBR = Vph as line voltage = phase voltage in case of delta              IR'Y' = IY'B' = IB'R' = Iph              P = Vph.Iph cos Φ  + Vph. Iph. cos Φ + Vph. Iph.cos Φ P = 3ph.Iph. cos Φ  watts ( in terms of phase quantities )  Also Vph = VL          Iph = IL/√3 and cos Φ = R/Z                    IL P = 3 VL --------- cos Φ                   √3 P = √3 VL VL cos Φ watts (Power in terms of line quantities) 3 phase reactive power Q Q = √3 VL IL sin Φ VAR Three phase apparent power S =  √3 VL IL     volt amp Worked Examples: Three equal impedances each having a resistance of 25Ω and reactance of 40Ω are connected in star to a 400 V, 3-phase, 50 Hz system. Calculate: The line current Power factor, and Power consumed Solution: Resistance per phase,                 Rph  = 25Ω Reactance per phase,                  Xph  = 40Ω Line voltage,                                 EL   = 400Ω To find:  Line current, IL: Power factor, cos θ: Power consumed, P Impedance per phase,          Zph  = √Rph2  +Xph2 Therefore                              Zph= Phase voltage,                      EL  =   400     volts                                             -----     -------     =   231 V                                              √3       √3 Phase current,                                 Eph        231                                              Iph = ------- = ---------  =   4.9 A (app.)                                                         Zph      47.17 Line current,                        IL = Iph  as it is star connected                                             IL = 4.9 A  (Ans) Power factor,                                     Rph          95                                             cos Φ= ------  =  -------- = 0.53(lag).                                                           Zph        47.17 Power consumed,            P= √ 3  ELIL  cos Φ  =  √3 *  400 *  4.9 * 0.53                                           = 1800 W (app.)                                   (or P = 3Iph2 Rph =3 * 4.92 * 25 =1800 W) Three identical coils are connected in a star to a 400V (line voltage) 3-phase A.C supply and each coil takes 300W. If the power factor is 0.8(lagging). Calculate: The line current Impedance and Resistance and inductance of each coil Solution. Line voltage,     EL= 400 V Power taken by each coil, Pph = 300W Power factor,                 cos Φ= 0.8 (lagging)  IL  ; Z ; Rph ; Lph : Phase voltage,                      EL       400                                   Eph = ------ = ------  V                                              √3       √3 Also                            Pph  = E ph Iph   cos Φ                                              400                                   300 = ------ *  Iph * 0.8                                               √3                                               300* √3                                   Iph = --------------- = 1.62 A                                               400*0.8 Line current,               IL = phase current, Iph  as it is star connected                                    IL = 1.62 A                                   Eph=     400  = 230.94 V                                              -------                                                √3     Coil impedance,            Eph      230.94                           Zph= ------- = ----------   =   142.5 Ω                                      Iph        1.62                                    ZC= 142.5 Ω. (Ans)                                    Rph= Zphcos Φ = 142.5 * 0.8 = 114 Ω Coil reactance,           Xph= Zphsin Φ = 142.5 * 0.6 = 85.5 Ω But,                            Xph= 2 π f Lph                                             Xph          85.5                                   Lph = -----  =  ------------   = 0.272H                                             2πf         2π * 50 hence,                        Rph  =  114 Ω.  (Ans)                                   Lph   = 0.272 H.   (Ans) A star- connected, 6000V, 3-phase alternator is supplying 4000kW at a power factor of 0.8. Calculate the active and reactive components of current in each phase. Solution.  Line voltage,                  EL  = 6000 V                 Power supplied,                p = 4000kW                 Power factor,              cos Φ= 0.8 Active and reactive components of current: We know that ,                     P =  √3  EL IL   cos Φ                              4000*1000 = √3  * 6000 * IL   * 0.8                                                       4000 * 1000                                              IL= --------------------- = 481A                                                      √3 * 6000 *0.8                                                  Iph= IL= 481 Active component                  = Iphcos Φ = 481 * 0.8 = 384.8 A Reactive component              = Iphsin Φ  = 481*0.6 = 288.6 A A balanced 3- phase star connected load of 100 kW takes an leading current of 80 A, when connected across a 3-phase, 1100 V, 50 Hz supply .Find the circuit constant of the load per phase.Solution. Given: P = 100 kW; Iph(=IL) = 80; EL = 1100 V; f = 50 Hz Circuit constants of the load per phase, R, C: As the – 3 θ load is balanced and star connected, line or phase current                                                     P                                100 * 103                     IL= ( Iph) = --------------------- =   --------------------------------                                            103 EL cos Φ                  √3* 1100 * cos θ                                            100 * 103 or,                 cos Φ  =   --------------------- =   0.656                                       √3 * 1100* 80                                     Eph        (1100/√3)                              Z = -------- = ---------------- = 7.94 Ω                                     Iph               80                              R = Z cos Φ  7.94 * 0.656 = 5.2 Ω (Ans)                                         1 Now                    XC = --------- as the current given is leading current                                     2 π f C                                         1                              C = -----------                                     2 π f XC  But                      XC = Z sin Φ = 7.94 * 0.755 * 5.99 Ω                                                 1                              C = ---------------------- F                                      2π * 50 * 5.99                                              106                                 = --------------------- μF = 531.4 μF. (Ans)                                      2π * 50 * 5.99