---
title: "25 MVA salient pole motor maximum power"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:salient-motor-25mva-max-power"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-25mva-max-power
site: BCT Engineering
---

# 25 MVA salient pole motor maximum power

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2075 Asoj, 7 marks
- Video: [Two Reaction Theory | Theory of Salient-Pole Synchronous Machine](https://www.youtube.com/watch?v=uItlycioiyg) — EduNeev Solutions
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-25mva-max-power
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

A 25 MVA, 3-phase star connected 11 kV, 12 poles, 50 Hz salient pole synchronous motor has direct axis reactance of 48 ohm and quadrature axis reactance of 3.5 ohm per phase. The armature resistance being negligible. At rated load, unity power factor and rated voltage, Determine: (i) Excitation Voltage (ii) Maximum value of power angle and corresponding power

## Short answer

With V=6350.85 V/phase, Ia=1312.16 A at unity pf, two-reaction theory gives excitation voltage Ef = 42.05 kV/phase (72.84 kV line) at load angle 35.87°. Maximum power occurs at δm=53.93°, Pmax=28.75 MW.

## Answer

**Given:** 25 MVA, 11 kV, star, $X_d = 48\ \Omega$, $X_q = 3.5\ \Omega$, $R_a = 0$, rated load at unity pf and rated voltage.

$$
\begin{aligned}
V &= \frac{11000}{\sqrt3} = 6350.85\ \text{V per phase} \\
I_a &= \frac{25\times10^6}{\sqrt3 \times 11000} = 1312.16\ \text{A}\ (\text{in phase with } V)
\end{aligned}
$$

**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

$$
\begin{aligned}
\vec E' &= V - jX_qI_a = 6350.85 - j(3.5 \times 1312.16) = 6350.85 - j4592.56 \\
&= 7837.4\angle -35.87^\circ\ \text{V} \quad \Rightarrow\ \delta = 35.87^\circ \\
\psi &= \delta - 0 = 35.87^\circ \ (\text{upf}) \\
I_d &= 1312.16\sin35.87^\circ = 768.90\ \text{A},\quad I_q = 1063.28\ \text{A} \\
E_f &= E' + (X_d - X_q)I_d = 7837.4 + 44.5 \times 768.90 \\
&= 42053\ \text{V per phase} = 72.84\ \text{kV line}
\end{aligned}
$$

Check: $P$ at $\delta = 35.87^\circ$ from the power-angle equation $= 25.00$ 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 $a = \dfrac{E_fV}{X_d} = 5.5641\times10^6$ and $b = V^2\left(\dfrac{1}{X_q} - \dfrac{1}{X_d}\right) = 10.6835\times10^6$ (per phase).

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

$$
\begin{aligned}
\cos\delta_m &= \frac{-a + \sqrt{a^2 + 8b^2}}{4b} = 0.5888 \\
\delta_m &= 53.93^\circ \\
P_{max} &= 3a\sin\delta_m + \tfrac{3b}{2}\sin2\delta_m \\
&= 13.49 + 15.25 = 28.75\ \text{MW}
\end{aligned}
$$

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

**Answer:** (i) $E_f = 42053$ V per phase (72.84 kV line), load angle $35.87^\circ$; (ii) maximum power angle $\delta_m = 53.93^\circ$, $P_{max} = 28.75$ MW.

## Check yourself

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

1. For the 25 MVA, 11 kV motor at rated load, how is Ia found and what is its phase?
   A. Ia = S/(sqrt3 VL) = 1312 A, in phase with V
   B. Ia = S/(3 VL) = 758 A, lagging V
   C. Ia = S/(sqrt3 VL) = 1312 A, 90 degrees leading V
   D. Ia = P/V per phase, lagging
   Answer: A. At unity pf the current is in phase with V, and Ia = 25e6/(sqrt3 x 11000) = 1312.16 A. (Key point: Armature current at unity pf)

2. What is Ef per phase at rated unity pf load?
   A. 7837 V
   B. 6351 V
   C. About 42 053 V
   D. About 72 840 V
   Answer: C. E' = 7837 V at -35.87 degrees, Id = 768.9 A, so Ef = 7837 + 44.5 x 768.9 = 42 053 V per phase (72.84 kV is the line value). (Key point: Excitation voltage)

3. Why is the maximum power angle of a salient pole motor less than 90 degrees?
   A. Because Ra is not zero
   B. Because of the reluctance (sin 2delta) term from Xd differing from Xq
   C. Because the speed drops at high load
   D. Because Ef is too small
   Answer: B. The reluctance power term adds to the excitation term, shifting the maximum below 90 degrees. (Key point: Why Pmax occurs before 90 degrees)

4. Which pair gives the maximum power angle and power for this motor?
   A. 90 degrees and 25 MW
   B. 35.87 degrees and 25 MW
   C. 53.93 degrees and 28.75 MW
   D. 53.93 degrees and 13.49 MW
   Answer: C. Solving dP/d(delta) = 0 gives cos(delta_m) = 0.5888, delta_m = 53.93 degrees, and Pmax = 13.49 + 15.25 = 28.75 MW. (Key point: Maximum power result)

## 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×
- [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×

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