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RLine XLine AC Voltage Source Load Z Line = 2.4Ω + j 15Ω Fill in the...

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

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
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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 Iline = 288 ∠0 Amp. This will be same for source and line since all are connected in series

We know

S = Vsource Iline*

Then load voltage is Vsource = S / Iline* = 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 Vload = Vsource - Iline Zline = 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

Sload = Vload Iline*

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

Line power calculations:

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

Pline = I2 R = 2882 x 2.4 = 199.065 kW

Qline = I2 X = 2882 x 15 = 1244.160 kVAR

Sline = 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
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