---
title: "20 MVA salient pole motor maximum power"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:salient-motor-20mva-max-power"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-20mva-max-power
site: BCT Engineering
---

# 20 MVA salient pole motor maximum power

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2071 Chaitra, 6 marks
- Video: [Lec 78 | Salient Pole Synchronous Machines | Power Angle Equation of Salient Pole SM](https://www.youtube.com/watch?v=1uQhWUxOtzM) — Kreatryx GATE - EE, ECE & IN by Unacademy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-20mva-max-power
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 20 MVA, 3 phase star connected 11 kV, 12 pole 50 Hz salient pole synchronous motor has a direct axis reactance of 50 ohm and quadrature axis reactance of 3 ohm per phase. The armature resistance being negligible. At rated load, unity power factor and rated voltage determine (i) excitation voltage (ii) The maximum value of power angle and corresponding power.

## Short answer

By two-reaction theory, excitation voltage Ef = 29007 V per phase (50.24 kV line), load angle δ = 26.38°. Maximum power occurs at δm = 50.36°, with Pmax = 27.14 MW.

## Answer

**Given:** 20 MVA, 11 kV, star, $X_d = 50\ \Omega$, $X_q = 3\ \Omega$, $R_a = 0$, rated load at unity pf and rated voltage.

$$
\begin{aligned}
V &= \frac{11000}{\sqrt3} = 6350.85\ \text{V per phase} \\
I_a &= \frac{20\times10^6}{\sqrt3 \times 11000} = 1049.73\ \text{A}\ (\text{in phase with } V)
\end{aligned}
$$

**Method (two-reaction theory, motor, $R_a = 0$):** $\vec V = \vec E_f + jX_d\vec I_d + jX_q\vec I_q$.

1. $\vec E' = \vec V - jX_q\vec I_a$ gives the direction of $E_f$ (the q-axis), so its angle is $-\delta$.
2. Angle between $I_a$ and the q-axis: $\psi = \delta - \phi$ ($\phi$ positive for lagging, negative for leading); $I_d = I_a\sin\psi$, $I_q = I_a\cos\psi$.
3. $E_f = E' + (X_d - X_q)I_d$.

#### (i) Excitation voltage

$$
\begin{aligned}
\vec E' &= V - jX_qI_a = 6350.85 - j(3 \times 1049.73) = 6350.85 - j3149.18 \\
&= 7088.8\angle -26.38^\circ\ \text{V} \quad \Rightarrow\ \delta = 26.38^\circ \\
\psi &= \delta - 0 = 26.38^\circ \ (\text{upf}) \\
I_d &= 1049.73\sin26.38^\circ = 466.34\ \text{A},\quad I_q = 940.45\ \text{A} \\
E_f &= E' + (X_d - X_q)I_d = 7088.8 + 47 \times 466.34 \\
&= 29007\ \text{V per phase} = 50.24\ \text{kV line}
\end{aligned}
$$

Check: $P$ at $\delta = 26.38^\circ$ from the power-angle equation $= 20.00$ MW $=$ rated input.

#### (ii) Maximum power and its power angle

$$
P = 3\left[\frac{E_fV}{X_d}\sin\delta + \frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta\right]
$$

Let $a = \dfrac{E_fV}{X_d} = 3.6844\times10^6$ and $b = V^2\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right) = 12.6378\times10^6$ (per phase).

$\dfrac{dP}{d\delta} = 0 \Rightarrow 2b\cos^2\delta + a\cos\delta - b = 0$:

$$
\begin{aligned}
\cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.6380 \\
\delta_m &= 50.36^\circ \\
P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m \\
&= 8.51 + 18.63 = 27.14\ \text{MW}
\end{aligned}
$$

The maximum occurs below $90^\circ$ because of the reluctance term.

**Answer:** (i) $E_f = 29007$ V per phase (50.24 kV line), load angle $26.38^\circ$; (ii) maximum power angle $\delta_m = 50.36^\circ$, $P_{max} = 27.14$ MW.

## Check yourself

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

1. What is the first step of the two-reaction method for this motor?
   A. Compute E' = V - jXq Ia to find the angle delta of the q-axis
   B. Compute Pmax
   C. Assume delta = 90 degrees
   D. Add Xd Ia to V
   Answer: A. E' lies along the q-axis, giving delta = 26.38 degrees, from which Id is found. (Key point: Method for Ef)

2. What is Ef per phase?
   A. 7089 V
   B. 29 007 V
   C. 50 240 V
   D. 6351 V
   Answer: B. Ef = 7088.8 + (50 - 3) x 466.34 = 29 007 V per phase (50.24 kV line). (Key point: Excitation voltage)

3. What are delta_m and Pmax?
   A. 26.38 degrees and 20 MW
   B. 90 degrees and 20 MW
   C. 50.36 degrees and 27.14 MW
   D. 50.36 degrees and 8.51 MW
   Answer: C. cos(delta_m) = 0.638 gives 50.36 degrees, Pmax = 8.51 + 18.63 = 27.14 MW; the large reluctance term moves the maximum below 90 degrees. (Key point: Maximum power)

## Close variants set in other exams

- [30 MVA salient pole motor maximum power](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-30mva-max-power) — Electric Machines II, chapter 2, asked 3×
- [25 MVA salient pole motor maximum power](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-25mva-max-power) — Electric Machines II, chapter 2, asked 3×

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