---
title: "30 MVA salient pole motor maximum power"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:salient-motor-30mva-max-power"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-30mva-max-power
site: BCT Engineering
---

# 30 MVA salient pole motor maximum power

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2073 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-30mva-max-power
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 30 MVA, 3 phase star connected 11 kV, 12 pole 50 Hz salient pole synchronous motor has a direct axis reactance of 55 ohm and quadrature axis reactance of 4 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

At unity pf, V=6350.85 V/phase, Ia=1574.59 A, excitation voltage Ef = 65.49 kV/phase (113.44 kV line) at δ=44.76°. Maximum power angle is 57.77° with Pmax=31.85 MW.

## Answer

**Given:** 30 MVA, 11 kV, star, $X_d = 55\ \Omega$, $X_q = 4\ \Omega$, $R_a = 0$, rated load at unity pf and rated voltage.

$$
\begin{aligned}
V &= \frac{11000}{\sqrt3} = 6350.85\ \text{V per phase} \\
I_a &= \frac{30\times10^6}{\sqrt3 \times 11000} = 1574.59\ \text{A}\ (\text{in phase with } V)
\end{aligned}
$$

**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

$$
\begin{aligned}
\vec E' &= V - jX_qI_a = 6350.85 - j(4 \times 1574.59) = 6350.85 - j6298.37 \\
&= 8944.4\angle -44.76^\circ\ \text{V} \quad \Rightarrow\ \delta = 44.76^\circ \\
\psi &= \delta - 0 = 44.76^\circ \ (\text{upf}) \\
I_d &= 1574.59\sin44.76^\circ = 1108.77\ \text{A},\quad I_q = 1118.01\ \text{A} \\
E_f &= E' + (X_d - X_q)I_d = 8944.4 + 51 \times 1108.77 \\
&= 65492\ \text{V per phase} = 113.44\ \text{kV line}
\end{aligned}
$$

Check: $P$ at $\delta = 44.76^\circ$ from the power-angle equation $= 30.00$ 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 $a = \dfrac{E_fV}{X_d} = 7.5624\times10^6$ and $b = V^2\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right) = 9.3500\times10^6$ (per phase).

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

$$
\begin{aligned}
\cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.5332 \\
\delta_m &= 57.77^\circ \\
P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m \\
&= 19.19 + 12.65 = 31.85\ \text{MW}
\end{aligned}
$$

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

**Answer:** (i) $E_f = 65492$ V per phase (113.44 kV line), load angle $44.76^\circ$; (ii) maximum power angle $\delta_m = 57.77^\circ$, $P_{max} = 31.85$ MW.

## Check yourself

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

1. What is the rated armature current for the 30 MVA, 11 kV motor and its phase relative to V?
   A. 1574.6 A, in phase with V
   B. 909 A, in phase with V
   C. 1574.6 A, lagging V by 90 degrees
   D. 2727 A, leading V
   Answer: A. Ia = 30e6/(sqrt3 x 11000) = 1574.59 A, and at unity pf it is in phase with V. (Key point: Rated current)

2. Which is the correct value of Ef per phase?
   A. 8944 V
   B. 65 492 V
   C. 113 440 V
   D. 6351 V
   Answer: B. Ef = E' + (Xd - Xq)Id = 8944.4 + 51 x 1108.77 = 65 492 V per phase (113.44 kV is the line value). (Key point: Excitation voltage)

3. What are the maximum power angle and maximum power?
   A. 90 degrees and 30 MW
   B. 44.76 degrees and 30 MW
   C. 57.77 degrees and 31.85 MW
   D. 57.77 degrees and 19.19 MW
   Answer: C. From dP/d(delta) = 0, cos(delta_m) = 0.5332, delta_m = 57.77 degrees, and Pmax = 19.19 + 12.65 = 31.85 MW. (Key point: Maximum power)

## Close variants set in other exams

- [20 MVA salient pole motor maximum power](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-20mva-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×

## Related in Electric Machines II

- [Salient pole motor excitation and reluctance power](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-50mva-pu) — Electric Machines II, chapter 2
- [Salient pole generator phasor diagram in per unit](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-generator-phasor-pu) — Electric Machines II, chapter 1
- [Internal emf and load angle of salient pole motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-200kw-load-angle) — Electric Machines II, chapter 2
- [Excitation emf of salient pole generator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-generator-excitation-500v) — Electric Machines II, chapter 1
- [Excitation emf and maximum power of 4 kVA motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-4kva-max-power) — Electric Machines II, chapter 2
- [Excitation emf of 6600 V salient pole motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asalient-motor-6600v-excitation) — Electric Machines II, chapter 2

## Related in other subjects

- [Reactance diagram: 30 MVA, 6.6 kV base, three lines](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-30mva-g1-motor) — Power System Analysis I, chapter 3
- [Reactance diagram: 20 MVA generator, 15 kV/25 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-20mva-15kv-25mva) — Power System Analysis I, chapter 3
- [Reactance diagram: three generators, 30 MVA 6.6 kV base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-30mva-three-generators) — Power System Analysis I, chapter 3
- [Reactance diagram: 50 MVA, 13.8 kV base, three generators](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-50mva-13-8kv) — Power System Analysis I, chapter 3