Question

RLine XLine AC Voltage Source Load Z Line = 2.4Ω + j 15Ω Fill in the missing information in the table below pf | (lead/lag)| (NIVA) | (MW) | (MVAr)| (deg) (kV) | (A) Load Line 288 ource 3

Answer 1

Given

Apparent Power delivered to the load S = 3 MW +j2 MVAR = 3.606 ∠33.69 MVA

Line current is 288, lets assume it as reference phasor I_line = 288 ∠0 Amp. This will be same for source and line since all are connected in series

We know

S = V_source I_line*

Then load voltage is Vsource = S / I_line* = 3.606 ∠33.69 * 10⁶/288 ∠0 = 12.519 kV ∠ 33.69

The source voltage leads the current by 33.69 degrees hence the power factor is cos(33.69) = 0.832 lagging

And load voltage is V_load = V_source - I_line Z_line = 12.519 kV ∠33.69 - 288∠0 x (2.4+j15) /1000 kV = 10.07 ∠15.1 kV

The load voltage leads the current by 15.1 degrees hence the power factor is cos(15.1) = 0.965 lagging

The load power can be thus calculated by

S_load = V_load I_line*

==> S_load = 10.07 ∠15.1 kV x 288∠0 = 2.9 ∠15.1 MVA = 2.8 MW + j 0.755 MVAR

Line power calculations:

V_line = I (R+jX) = 288 (2.4+j15) = 4.375 ∠80.9 kV

P_line = I² R = 288² x 2.4 = 199.065 kW

Q_line = I² X = 288² x 15 = 1244.160 kVAR

S_line = 1260.27 ∠80.9 kVA = 1.26∠80.9 MVA

pf = cos 80.9 lag = 0.158 lag

Hence the completed table is :

	S (MVA)	P(MW)	Q(MVAR)	Φ(deg)	pf(lead/lag)	V(kV)	I(A)
Load	2.9	2.8	0.755	15.1	0.965 lag	10.07	288
Line	1.26	0.199	1.244	80.9	0.158 lag	4.375	288
Source	3.606	3	2	33.69	0.832 lag	12.519	288