---
title: "Synchronous motor: pf when input rises to 1500 kW"
subject: "Electrical Machines"
chapter: 6
key: "semester_4/electrical_machines:sync-motor-pf-input-1500kw"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/sync-motor-pf-input-1500kw
site: BCT Engineering
---

# Synchronous motor: pf when input rises to 1500 kW

Electrical Machines · Chapter 6: Three Phase Synchronous Machines

- Asked in: 2067 Mangsir (old course), 8 marks
- Video: [PEO solved examples (ELEC A6): Synchronous Machines no. 3](https://www.youtube.com/watch?v=6VMQvjgEBy0) — Ayman El-Hag - Electrical Engineering for Everyone
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/sync-motor-pf-input-1500kw
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-6

## Question

A 6.6 kV star connected, 3 phase synchronous motor works at constant voltage and constant excitation. Its synchronous reactance is $20\ \Omega$/phase, when the input is 100 kW and power factor is 0.8 leading. Find the power factor when the input is increased to 1500 kW.

## Short answer

The power factor becomes 0.95 lagging. The motor shifts from leading to lagging because excitation is fixed while load increases.

## Answer

**Given:** 6.6 kV, star, $X_s = 20\ \Omega$ per phase (resistance neglected), constant $V$ and constant excitation. Condition 1: input 100 kW at 0.8 leading. Find the power factor at 1500 kW input.

$$
V_{ph} = \frac{6600}{\sqrt3} = 3810.5\ \text{V}
$$

#### Step 1: excitation emf at 100 kW, 0.8 leading
$$
I_1 = \frac{100\times10^3}{\sqrt3\times6600\times0.8} = 10.93\ \text{A}, \qquad \vec I_1 = 10.93\angle+36.87^\circ
$$
For a motor, $\vec E = \vec V - j\vec I X_s$:
$$
\vec E = 3810.5 - j20\,(8.748 + j6.561) = (3810.5 + 131.2) - j174.95 = 3941.7 - j174.95\ \text{V}
$$
$$
E = 3945.6\ \text{V per phase}\quad (\delta_1 = 2.54^\circ)
$$

#### Step 2: at 1500 kW (E unchanged)
Power per phase $= 500$ kW:
$$
P_{ph} = \frac{VE}{X_s}\sin\delta \;\Rightarrow\; \sin\delta = \frac{500\times10^3\times20}{3810.5\times3945.6} = 0.6651, \quad \delta = 41.69^\circ
$$
(maximum power $= 3VE/X_s = 2255$ kW, so 1500 kW is below the pull-out limit.)

Current:
$$
\vec I = \frac{\vec V - \vec E}{jX_s} = \frac{3810.5 - 3945.6\angle-41.69^\circ}{j20}
$$
$$
|I| = 138.15\ \text{A}, \qquad \angle I = -18.23^\circ
$$
$$
\cos\phi = \cos 18.23^\circ = 0.950 \ \text{(lagging)}
$$
Check: $\sqrt3\times6600\times138.15\times0.9498 = 1500$ kW.

**Answer:** power factor at 1500 kW input $\approx 0.95$ **lagging** (current 138.15 A). The motor moved from leading to lagging because the excitation was fixed while the load increased.

## Check yourself

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

1. Which equation gives the excitation emf of a synchronous motor?
   A. E = V + jI Xs
   B. E = V - jI Xs
   C. E = V + I Xs with I in phase with V
   D. E = V - I Xs with no phase
   Answer: B. For a motor the current flows in and E = V - jI Xs. (Key point: Motor emf equation)

2. What is E at 100 kW, 0.8 leading, 6.6 kV, Xs = 20 ohm?
   A. About 3810 V
   B. About 3946 V
   C. About 6600 V
   D. About 4500 V
   Answer: B. I = 10.93 A at +36.87 degrees gives E = 3941.7 - j174.95, magnitude 3945.6 V per phase. (Key point: Excitation emf)

3. How is the load angle found at 1500 kW with E constant?
   A. From sin(delta) = P Xs/(V E) per phase, giving delta = 41.69 degrees
   B. From cos(delta) = P Xs/(V E)
   C. From delta = P/E
   D. It stays at 2.54 degrees
   Answer: A. P per phase = (VE/Xs) sin(delta) with P = 500 kW gives sin(delta) = 0.6651. (Key point: Power angle at 1500 kW)

4. What is the power factor at 1500 kW?
   A. 0.8 leading
   B. 1.0
   C. 0.95 lagging
   D. 0.95 leading
   Answer: C. I = (V - E angle -delta)/(jXs) = 138.15 A at -18.23 degrees, so pf = 0.95 lagging; it moves from leading to lagging because excitation is fixed. (Key point: Final power factor)

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