Which equation gives the receiving-end active power?

P_R = |Vs||Vr|cos(beta-delta)/|B| - |A||Vr|^2 cos(beta-alpha)/|B|

P_R = |Vs||Vr|sin(delta)/|B| only

P_R = |A||Vs|^2 cos(beta-alpha)/|B| - |Vs||Vr|cos(beta+delta)/|B|

P_R = |Vs||Vr|cos(beta+delta)/|B|

What angles are used in the two cosine terms?

beta - delta = 60.6 and beta - alpha = 78.4

beta + delta = 97 and beta + alpha = 79.2

beta - delta = 78.8 and beta - alpha = 0.4

2068 Magh · 5 marks

Receiving end power, 140.2∠18.2° kV and 132 kV

Short answer

Using P_R = (|V_S||V_R|/|B|) cos(β−δ) − (|A||V_R|²/|B|) cos(β−α) with given ABCD parameters yields receiving-end active power of 90.2 MW.

For a 3-phase line the following data is available at a particular instant: VS (line) = 140.2∠18.2° kV, VR (line) = 132 kV∠0°, A = 0.94∠0.4°, B = 64.2∠78.8° Ohms. Compute the receiving end active power.

Use the receiving-end power-circle equation. With line kV, the result is three-phase MW:

P_R = \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)

Data: $|V_S| = 140.2$ kV, $\delta = 18.2^\circ$ , $|V_R| = 132$ kV, $|A| = 0.94$ , $\alpha = 0.4^\circ$ , $|B| = 64.2\ \Omega$ , $\beta = 78.8^\circ$ .

Substitution

\begin{aligned} \frac{|V_S||V_R|}{|B|} &= \frac{140.2\times132}{64.2} = 288.26 \\ \frac{|A||V_R|^2}{|B|} &= \frac{0.94\times132^2}{64.2} = 255.12 \\ \beta-\delta &= 60.6^\circ,\quad \beta-\alpha = 78.4^\circ \end{aligned}

\begin{aligned} P_R &= 288.26\cos60.6^\circ - 255.12\cos78.4^\circ \\ &= 141.51 - 51.30 = 90.21\ \text{MW} \end{aligned}

Answer: Receiving-end active power ≈ 90.2 MW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Receiving end power, 140.2∠18.2° kV and 132 kV

Substitution

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