Question

A delta connected source delivers power to a delta connected load as shown in the figure below. The line-to-line voltage at the source is 494 Vrms. The load impedance is ZA = 9+ 3;12. The line impedance is Zi = 1+2j12.

a) Find the magnitude of the line current IaA (A rms)
b) Find the phase of the line current IaA. (Phase deg)
c) Find the magnitude of VAB the line-to-line load voltage (V rms)
d) Find the phase of VAB (deg)
e) Find the magnitude of the load phase current IAB (A rms)
f) Find the phase of IAB (deg)
g) Find the total power delivered to the load (kW)
h) Find the power lost in the line. (kW)

Answer 1

#. for the given problem per phase analysis would be easy. To perform per phase calculation we can Convert dela connected system to star connected. 14 tjar mm 14 jan b ugu Loo ? 494L-120 G 484 (20 10 jan mm

i Z della = 3 2 star cii, Phase - Vine (for Star 13 comected syslem) (1) Phase voltage lags by 30 to line voltage Per phase cct - © 205.21 20 3 (3+j1) L (phase) applying KUL 205-2340 = Jive ( 1+2j) +1 (3+j1) 205. 2120 205.2120 4+3j 1 +2j+ 3tj I line = 57.04 L-36.86 ) la magnitude of Line current II Line = I aa l = 57.04 Amp (b) phase of the line current 0 = -36-06° Vphane) = I line ( 20 ) = (57.04 L-36-86)X( 3+12) Vulphone) = 180.38 (-10-42 -ui)

ce for star conne of phase Holtage and er star connected system Line voltage is 3 times phase Moltage and leads by 30 (line voltage of star connected system will be equal to della Vlcine) = Vee = 13X180-38 (-18.42730; 2connected System Vae = 312.42 11.52° O magnitude of line to line voltage IVB) = 312-42 volt, (d) phase of VAB o = 11.52% // As Previously calculated fline = 57.04 (-36-86 current nes of a ge is at for delta connected phase current and system line lags by 30" Theo - Iphone D = 57:04_C3686+30) = 32:934-6-06 Amp. (e) magnitude of load phase current (FABI = 32.93 Amp A phase of IAB = -6:06)

Per phase delivered compelex power 3 (18) = I line x Je phase) Slcao) = (57.04 2+36-86) X(180.30-18.42) 52(40) = 10200.88 10.44 as 30 system is noting but 3 times 10 system SL (30) = 351(10) = 30.67 /18.44 KVA 51 (30) – (29-26 +j 9-76 ) KUA 3. =P tją (9) P = 29.28 kW, 8L(30) = 9-76 KVAR "L(30) Total power dilivered to load PL (38) = 29.20 Kwatt (active power) (h) Power disciputed in line Plossid) = 3 | Itine 1 x 1 cactive power) Ploss (30) = 3x(57-04)2x1 PLOSS (30) = 9760.69 watt OR PLOSS (38) = 9.76 kW 1