---
title: "Salient pole motor excitation and reluctance power"
subject: "Electric Machines II"
chapter: 2
key: "semester_5/electric_machines_ii:salient-motor-50mva-pu"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-50mva-pu
site: BCT Engineering
---

# Salient pole motor excitation and reluctance power

Electric Machines II · Chapter 2: Three Phase Synchronous Motor

- Asked in: 2078 Bhadra, 7 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/salient-motor-50mva-pu
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2

## Question

Salient pole synchronous motor has reactance Xd = 0.8 pu, Xq = 0.4 pu. It is a 3-phase, 50 MVA, 11 kV, 50 Hz. Motor draws rated current at a supply power factor of 0.8 lagging. Rotational losses are 0.15 pu and armature resistance losses are neglected.
a) Calculate the excitation voltage.
b) Find the power due to field excitation and that due to the salience of the machine.
c) If the field current is zero, will the machine stay in synchronism, explain why?

## Short answer

Per-unit: V=1, Ia=1∠–36.87°. E' = V – jXq Ia gives δ=22.83°, then Ef = 0.728 pu (8.0 kV line). Excitation power 0.353 pu (17.65 MW), saliency power 0.447 pu (22.35 MW). With zero field, max reluctance power 0.625 pu < required 0.8 pu, so motor falls out of step.

## Answer

Work in per unit: $V = 1\angle0^\circ$ pu, rated current $I_a = 1$ pu at 0.8 lagging, so $\vec I_a = 1\angle -36.87^\circ = 0.8 - j0.6$ pu.

**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. $E_f = E' + (X_d - X_q)I_d$.

#### a) Excitation voltage

$$
\begin{aligned}
\vec E' &= 1 - j0.4(0.8 - j0.6) = 1 - 0.24 - j0.32 \\
&= 0.76 - j0.32 = 0.8246\angle -22.83^\circ\ \text{pu} \\
\delta &= 22.83^\circ,\quad \psi = 22.83^\circ - 36.87^\circ = -14.04^\circ \\
I_d &= 1 \times \sin(-14.04^\circ) = -0.2425\ \text{pu},\quad I_q = 0.9701\ \text{pu} \\
E_f &= 0.8246 + (0.8 - 0.4)(-0.2425) = 0.7276\ \text{pu}
\end{aligned}
$$

$E_f = 0.7276 \times 11 = 8.00$ kV (line), i.e. 4.62 kV per phase. $E_f < V$, as expected for an under-excited (lagging) motor.

#### b) Excitation power and reluctance power

$$
P = \underbrace{\frac{E_fV}{X_d}\sin\delta}_{\text{excitation}} + \underbrace{\frac{V^2}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin2\delta}_{\text{saliency}}
$$

$$
\begin{aligned}
P_{exc} &= \frac{0.7276 \times 1}{0.8}\sin22.83^\circ = 0.3529\ \text{pu} = 17.65\ \text{MW} \\
P_{rel} &= \frac{1}{2}(2.5 - 1.25)\sin45.67^\circ = 0.4471\ \text{pu} = 22.35\ \text{MW} \\
P &= 0.3529 + 0.4471 = 0.8\ \text{pu} = 40\ \text{MW}
\end{aligned}
$$

Check: input $= VI\cos\phi = 1 \times 1 \times 0.8 = 0.8$ pu. Shaft output $= 0.8 - 0.15 = 0.65$ pu $= 32.5$ MW.

#### c) Field current reduced to zero

With $E_f = 0$ only reluctance power remains:

$$
P_{rel,max} = \frac{V^2}{2}\left(\frac{1}{X_q} - \frac{1}{X_d}\right) = \frac12(2.5 - 1.25) = 0.625\ \text{pu} \ (\text{at } \delta = 45^\circ)
$$

The motor must take $0.8$ pu (0.65 pu load + 0.15 pu rotational loss). Since $0.625 < 0.8$, the reluctance torque cannot carry the load, so **the machine falls out of synchronism**. It would stay in step as a reluctance motor only if the shaft load were reduced so that total input is below 0.625 pu (load below 0.475 pu).

**Answer:** (a) $E_f = 0.728$ pu (8.0 kV line), $\delta = 22.8^\circ$; (b) excitation power 0.353 pu (17.65 MW), saliency power 0.447 pu (22.35 MW); (c) No, it loses synchronism.

## Check yourself

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

1. In the two-reaction method, what does E' = V - jXq Ia give?
   A. The magnitude of Ef directly
   B. The direction of Ef (q-axis), hence the load angle
   C. The d-axis current
   D. The reluctance power
   Answer: B. E' lies along the q-axis, so its angle is -delta; Ef is then E' + (Xd - Xq) Id. (Key point: Direction of Ef)

2. What is Ef in pu for rated current at 0.8 lagging?
   A. 0.7276 pu
   B. 1.0 pu
   C. 0.8246 pu
   D. 1.685 pu
   Answer: A. E' = 0.8246, Id = -0.2425, so Ef = 0.8246 + 0.4(-0.2425) = 0.7276 pu, less than V, as an under-excited motor. (Key point: Excitation voltage)

3. Which statement about the power components in (b) is correct?
   A. Excitation power 0.353 pu and saliency power 0.447 pu, summing to 0.8 pu
   B. Excitation power 0.8 pu and saliency power 0
   C. Saliency power is zero because Ef < V
   D. Excitation power 0.447 pu and saliency power 0.353 pu
   Answer: A. Pexc = (Ef V/Xd) sin(delta) = 0.353 pu and Prel = 0.5(2.5 - 1.25) sin(45.67 degrees) = 0.447 pu, totalling the input 0.8 pu. (Key point: Power split)

4. With zero field current, why does the machine fall out of synchronism here?
   A. Maximum reluctance power 0.625 pu is below the 0.8 pu input needed
   B. There is no stator field
   C. Reluctance power is zero
   D. The speed doubles
   Answer: A. With Ef = 0 only reluctance power (max 0.625 pu) remains, which cannot supply the 0.65 pu load plus 0.15 pu losses. (Key point: Zero field current)

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