---
title: "Power delivered from sending end data, nominal-π"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:nominal-pi-power-delivered"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-pi-power-delivered
site: BCT Engineering
---

# Power delivered from sending end data, nominal-π

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2068 Bhadra, 6+2 marks
- Video: [PS25 Numerical on Medium Transmission Lines](https://www.youtube.com/watch?v=Dt4Y5BlylMo) — Lectures in Electrical Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/nominal-pi-power-delivered
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

A 3-phase line is represented by nominal-π model. The series impedance of the line is 12 + j65 Ohms/phase and shunt admittance is j2.5 × 10⁻⁴ S/phase. Compute the power delivered to a load at the receiving end and construct the phasor diagram for the line if the sending end voltage and current are: VS,L = 138.5∠15.2° kV, IS,L = 225∠18.3° A.

## Short answer

Working backward from sending end: receiving voltage V_R = 136.45 kV, current I_R = 223.0 A. Power delivered = 52.08 MW (with 8.04 Mvar supplied by the load side, power factor 0.988 leading).

## Answer

**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).

```text
 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°.

```text
          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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Given the sending-end quantities, what is the first step to find the receiving-end?
   A. Convert VS,L to phase voltage and subtract the sending shunt current Y/2 VS from IS to get IL
   B. Add Y VS to IS
   C. Multiply IS by Z
   D. Use VR = VS
   Answer: A. VS = 138.5/sqrt3 per phase, IC1 = (Y/2)VS, and IL = IS - IC1 is the current in the series arm. (Key point: Working backwards)

2. How is VR obtained?
   A. VR = VS - IL Z
   B. VR = VS + IL Z
   C. VR = VS - IS Z
   D. VR = VS/Z
   Answer: A. VR = VS - IL(12 + j65) = 78.78∠4.50° kV per phase. (Key point: Receiving voltage)

3. How is IR found after VR?
   A. IR = IL + (Y/2)VR
   B. IR = IL - (Y/2)VR
   C. IR = IS
   D. IR = IL - Y VR
   Answer: B. The receiving-end capacitor takes current IC2 = (Y/2)VR from the series current: IR = IL - IC2 = 223.0∠13.27° A. (Key point: Receiving current)

4. How is the power delivered calculated?
   A. S = 3 VR IR* (complex conjugate), whose real part is P
   B. S = 3 VR IR
   C. P = VR IR
   D. P = 3 IL^2 Z
   Answer: A. S = 3 VR IR* = 52.08 - j8.04 MVA, so P = 52.08 MW. (Key point: Power delivered)

## Related in Power System Analysis I

- [Nominal-π phasor, sending end, regulation, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-pi-phasor-regulation-efficiency) — Power System Analysis I, chapter 5
- [Sending and receiving power, 140∠10° and 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-140kv-132kv-a099) — Power System Analysis I, chapter 6
- [300 km line: short line vs nominal-π](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A300km-short-vs-nominal-pi) — Power System Analysis I, chapter 5, asked 2×
- [Power output with 46.85 kV sending, 33 kV receiving](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-output-46-85kv-33kv) — Power System Analysis I, chapter 6
- [Receiving and sending end powers from ABCD, 132/134 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-power-132kv-134kv) — Power System Analysis I, chapter 6
- [Model validity and nominal-π ABCD with phasor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-model-validity-nominal-pi) — Power System Analysis I, chapter 5

## Related in other subjects

- [Expression for receiving-end shunt compensation](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Ashunt-compensation-xp-expression) — High Voltage Engineering, chapter 3
- [Ybus with nominal π model, line lengths](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aybus-km-lines-pi-model) — Power System Analysis II, chapter 1, asked 2×
- [Double-stub shunt tuner for admittance load](https://www.bctengineer.com/question.md?key=semester_7%2Frf_and_microwave_engineering%3Adouble-stub-admittance-load) — RF and Microwave Engineering, chapter 2
- [Three-phase connections and series RC phasor](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Athree-phase-connections-series-rc) — Fundamental of Electrical and Electronics Engineering, chapter 3