Power delivered from sending end data, nominal-π

Given: $Z = 12 + j65\ \Omega$, $Y = j2.5\times10^{-4}$ S (so $Y/2 = j1.25\times10^{-4}$ S); $V_{S,L} = 138.5\angle 15.2^\circ$ kV, $I_S = 225\angle 18.3^\circ$ A. Working backwards from the sending end through the Π circuit (per phase).

 IS       IL    Z = R + jX        IR
 o-->--+-----[ R ]--[ jX ]-----+-->--o
       |                       |
 VS  [Y/2] Ic1           Ic2 [Y/2]  VR
       |                       |
 o-----+-----------------------+-----o

Step 1: Sending-end phase voltage

$$V_S = \frac{138.5}{\sqrt3}\angle 15.2^\circ = 79.96\angle 15.2^\circ = 77{,}166 + j20{,}965\ \text{V}$$

Step 2: Sending-end shunt current and series current

$$\begin{aligned} I_{C1} &= \frac{Y}{2}V_S = j1.25\times10^{-4}\times79{,}963\angle 15.2^\circ = 9.995\angle 105.2^\circ = -2.62 + j9.65\ \text{A} \\ I_S &= 225\angle 18.3^\circ = 213.62 + j70.65\ \text{A} \\ I_L &= I_S - I_{C1} = 216.24 + j61.00 = 224.7\angle 15.75^\circ\ \text{A} \end{aligned}$$

Step 3: Receiving-end voltage

$$\begin{aligned} I_LZ &= (216.24 + j61.00)(12 + j65) = -1370 + j14{,}787\ \text{V} \\ V_R &= V_S - I_LZ = 78{,}536 + j6178 = 78.78\angle 4.50^\circ\ \text{kV/phase} \\ V_{R(L)} &= \sqrt3\times78.78 = 136.45\ \text{kV} \end{aligned}$$

Step 4: Receiving-end current

$$\begin{aligned} I_{C2} &= \frac{Y}{2}V_R = 9.847\angle 94.50^\circ = -0.77 + j9.82\ \text{A} \\ I_R &= I_L - I_{C2} = 217.01 + j51.19 = 223.0\angle 13.27^\circ\ \text{A} \end{aligned}$$

Step 5: Power delivered

$$\begin{aligned} S_R &= 3V_RI_R^* = 3\times78.78\angle 4.50^\circ\times223.0\angle -13.27^\circ\ \text{kVA} \\ &= 52.08 - j8.04\ \text{MVA} \end{aligned}$$

So the load receives $P_R = 52.08$ MW, and the receiving-end p.f. is $\cos 8.77^\circ = 0.988$ leading ($I_R$ leads $V_R$); the load side supplies 8.04 Mvar. Sending power is $53.90$ MW, so the line loss is $3|I_L|^2R = 3\times224.7^2\times12 = 1.82$ MW.

Phasor diagram ($V_R$ near reference)

• $V_R = 78.78\angle 4.5^\circ$ kV, $I_R = 223\angle 13.27^\circ$ A (leading $V_R$ by 8.77°).
• $I_{C2}$ is 90° ahead of $V_R$; $I_L = I_R + I_{C2}$ at 15.75°.
• $V_S = V_R + I_LR + jI_LX$, at 15.2°.
• $I_{C1}$ is 90° ahead of $V_S$; $I_S = I_L + I_{C1}$ at 18.3°.

          IC1 ^  IC2 ^
               \     |      VS (15.2 deg)
                \    |    /  ---+ IL.X
          IS,IL  \   |  /       |
           ^ ^    \  |/  . IL.R |
            \ \    \ /.---------+
   O---------+-----------> VR (4.5 deg)
      IR (13.3 deg) near IL

Answer: Power delivered to the load $= 52.08$ MW (with 8.04 Mvar supplied by the load side, p.f. 0.988 leading), $V_R = 136.45$ kV, $I_R = 223.0$ A.