Receiving and sending end powers from ABCD, 132/134 kV

Use the power-circle equations (taking $D = A$). Let $A = |A|\angle\alpha$, $B = |B|\angle\beta$, $V_S$ leading $V_R$ by $\delta$. With line voltages in kV, powers come out directly in three-phase MW/MVAr.

Data: $|A| = 0.98$, $\alpha = 0.8^\circ$, $|B| = 45\ \Omega$, $\beta = 74^\circ$, $|V_S| = 132$ kV, $|V_R| = 134$ kV, $\delta = 10^\circ$.

Useful terms:

$$\begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{132\times134}{45} = 393.07 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.98\times134^2}{45} = 391.04 \\ \frac{|A||V_S|^2}{|B|} &= \frac{0.98\times132^2}{45} = 379.46 \end{aligned}$$

Angles: $\beta-\delta = 64^\circ$, $\beta-\alpha = 73.2^\circ$, $\beta+\delta = 84^\circ$.

Receiving end

$$\begin{aligned} P_R &= 393.07\cos64^\circ - 391.04\cos73.2^\circ = 59.29\ \text{MW} \\ Q_R &= 393.07\sin64^\circ - 391.04\sin73.2^\circ = -21.07\ \text{MVAr} \\ \text{pf}_R &= \frac{59.29}{\sqrt{59.29^2 + 21.07^2}} = 0.942 \end{aligned}$$

$Q_R$ is negative, so reactive power flows from the receiving end into the line: the receiving-end pf is 0.942 leading. This is expected because $|V_R| > |V_S|$.

Sending end

$$\begin{aligned} P_S &= 379.46\cos73.2^\circ - 393.07\cos84^\circ = 68.59\ \text{MW} \\ Q_S &= 379.46\sin73.2^\circ - 393.07\sin84^\circ = -27.65\ \text{MVAr} \\ \text{pf}_S &= \frac{68.59}{\sqrt{68.59^2 + 27.65^2}} = 0.927\ \text{(leading)} \end{aligned}$$

Line losses

$$\begin{aligned} P_{loss} &= P_S - P_R = 68.59 - 59.29 = 9.30\ \text{MW} \\ Q_S - Q_R &= -27.65 - (-21.07) = -6.59\ \text{MVAr} \end{aligned}$$

The negative reactive difference means the line, on net, generates about 6.59 MVAr (its charging exceeds its $I^2X$ absorption).

Voltage regulation

No-load receiving voltage $= |V_S|/|A| = 132/0.98 = 134.69$ kV.

$$\%VR = \frac{134.69 - 134}{134}\times100 = 0.52\%$$

Answer: $P_R$ = 59.29 MW, $Q_R$ = −21.07 MVAr, pf 0.942 lead; $P_S$ = 68.59 MW, $Q_S$ = −27.65 MVAr, pf 0.927 lead; loss 9.30 MW; regulation 0.52 %.