Question

#3. For the 3-phase power system shown in Fig. 3, the load is 30 kVA at a p.f. of 0.94 lagging. If the phasor load voltage, V. = 480/0 Compute: (i)The phasor line current, I. Ia Fig. 3 (ii) The phasor voltage Vec of the source. A . Contjo. 6). Ba [Zunet (0.1+30.6). 30 kVA hase/p. f. 0.94 las I ho o.1tjo. 652 ů 2

Answer 1

Solution:

solution : 3 Tune to a 31 Cool+J0.61 | 30 KVA Voi hicho phose pe 0.94 logging 7 I b Load 5 Load Cooltjo.621 Cooltjo.662 IC (a)(i) the phasoo current, Ia. Nou Ipb = Vph & I = 3x Iph Iph 3 xPf x Taload X Vab Pload = 30 x103 = 3 x 0.94 x Ice X 480 60° - Р the phason line, Current Iph ca) 3 x pfx Vab 30x10² 3 x 0.94 X 480 600 = 38.3876 Amps page- Iph (a)

I = Iph x3 = 115. 1629 Amps.tube 1h B-bus the phasor Voltage Vic of the source, Now from us we have observed that for every 3-phase line there is an equal live impedance of z line = (0.1 +30.61 Zlep = 0.6082 180:53° & it is driven by a lood of 30KVA at 0.94 lagging Pf Now we can consider the current per phase as → 38.3876 Amps & 0.038 38 k Amy By the above condition we get H Bc = BC CLine & I = 0.6082 80.530 X 38.3976 V = BC 23.3473 90.530 V 1 page 2)

Note:

Be careful about the 3- phase load and 3- phase power as shown in the above figure. We can observe that there are represented with capital letters that are V_BC (phase source voltage) and as given in question phasor load voltage V_ab. I want to say that phases/terminals of 3-phase load are represented with a,b,c and those of 3- phase source are represented with A, B, C and N(neutral).

As both are the 3- phases only and delta connected. So, as a result the formulas in the solution are used, dont get confused. I hope it is helpful for a better understanding.