Question

Please Solve this question as soon as possible

Please Solve this question as soon as possible

(Power)

A three-phane lune has an impedance of Z-04 + /250 per phase The line feeds two balanced three-phave loads that are connected in parallel as shown in Fug 01 The first loads absorbing 560 IVA ato power factor lagging The second load shorts 132 KW ty pona factor The lune-to-me voltage at the load end of the line is 3810.5 V (see the Fig 01) The lune-to-coutral voltage at the load can be taken as a reference Z-0.4-0.30 132 kW 560.1 kVA -0.707 lagring 1 CS Scanned with Canl banner Fig. 21

Answer 1

Drawing per phase circuit Z2=0.4 -10.5 =V Voltage across load = 3810.5 For Loadi as Si =[]17,*4 560.1 X103 = 3810.5 x15 / 17.1= 146.99A I = 146-99 -45° Ą ! tot load 2 P = V Iz loso 132 x1000 = 3810.5 x 12 xI I2 = 34.64 Iz = 34.64L0A. To tal current cussent (I) = I + Iz = 146.994-4** +34.64 I = 173-22 /-3687° A Power factor of total load = Cos 36.87 = 0.8 lagging

Total complex power absorbeel by load = 3V, I* = 3 x3810.5 x 178-22/36.87 SL = 1980164.43/36.87 VA 12 or [S2 = 1980.14/36.84 kVA KUA and SL = 1584.13 +11188.10 b) Cussent drawn by total Load oy total Load 'I = 173.22 _-36.87 A c) across Voltage across I = IXZ. Vz) = 173.22 ) -36.87 *0•4-70.5) = 110.921-88.2 Vello d) k, = 3.46-j10.86 Sy = VzL I* = 110.92 /_882 x 173-22 36.87 = (12 -J15) K =p. +JQ active Power loss =P = 12 kW Geactive Power = Q = - 15 KVAR