---
title: "440 V IM with 50 kW rotor input"
subject: "Electrical Machine"
chapter: 5
key: "semester_4/electrical_machine:im-440v-rotor-input-50kw"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-440v-rotor-input-50kw
site: BCT Engineering
---

# 440 V IM with 50 kW rotor input

Electrical Machine · Chapter 5: Three Phase Induction Machines

- Asked in: 2074 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-440v-rotor-input-50kw
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5

## Question

The power input to the rotor of a 440 V, 50 Hz, 3-phase, 6-pole, induction motor is 50 kW. The rotor emf makes 120 cycles per minute. Friction and windage losses are 2 kW. Calculate (i) slip (ii) rotor speed (iii) rotor copper losses (iv) mechanical power developed (v) output power (vi) output torque.

## Short answer

Slip s = f₂/f = 2/50 = 0.04. Rotor speed N = 1000(1−0.04) = 960 rpm. Rotor copper loss = s×50 kW = 2 kW. Mechanical power = 48 kW, output power = 46 kW, output torque = 46000/(2π×960/60) = 457.6 N·m.

## Answer

**Given:** 440 V, 50 Hz, 3-phase, 6-pole; rotor input P₂ = 50 kW; rotor emf frequency = 120 cycles/min = 2 Hz; friction and windage loss = 2 kW.

**Synchronous speed:** N_s = 120f/P = 120 × 50 / 6 = 1000 rpm (ω_s = 104.72 rad/s).

#### (i) Slip
Rotor frequency f₂ = s·f, with f₂ = 120/60 = 2 Hz:

s = f₂ / f = 2 / 50 = **0.04 (4 %)**

#### (ii) Rotor speed
N = N_s(1 − s) = 1000 × (1 − 0.04) = **960 rpm** (ω = 100.53 rad/s)

#### (iii) Rotor copper loss
P_cu = s × P₂ = 0.04 × 50 kW = **2 kW**

#### (iv) Mechanical power developed
P_m = P₂ − P_cu = (1 − s)P₂ = 50 − 2 = **48 kW**

#### (v) Output power
P_out = P_m − friction and windage = 48 − 2 = **46 kW**

#### (vi) Output torque
T_out = P_out / ω = 46 000 / (2π × 960/60) = 46 000 / 100.53 = **457.6 N·m**

(For reference, the gross torque is P₂/ω_s = 50 000/104.72 = 477.5 N·m.)

| Quantity | Value |
|---|---|
| Slip | 0.04 |
| Rotor speed | 960 rpm |
| Rotor copper loss | 2 kW |
| Mechanical power | 48 kW |
| Output power | 46 kW |
| Output torque | 457.6 N·m |

## Check yourself

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

1. Rotor emf makes 120 cycles per minute on a 50 Hz motor. What is the slip?
   A. 0.04
   B. 0.4
   C. 0.02
   D. 0.4 %
   Answer: A. f2 = 120/60 = 2 Hz, so s = f2/f = 2/50 = 0.04. (Key point: Slip from rotor frequency)

2. How is rotor copper loss obtained from the rotor input?
   A. Pcu = (1 - s) x P2
   B. Pcu = s x P2 = 2 kW
   C. Pcu = P2 / s
   D. Pcu = P2 - friction loss
   Answer: B. Rotor copper loss is slip times rotor input: 0.04 x 50 = 2 kW. (Key point: Rotor copper loss)

3. Which step gives the output power here?
   A. Output = rotor input minus copper loss only
   B. Output = mechanical power plus friction and windage
   C. Output = mechanical power (48 kW) minus friction and windage (2 kW) = 46 kW
   D. Output = 50 kW x (1 - s) x 2
   Answer: C. Pm = 48 kW; subtracting friction and windage of 2 kW leaves 46 kW. (Key point: Output power)

4. Output torque is calculated by dividing output power by which speed?
   A. Synchronous angular speed (104.72 rad/s)
   B. Rotor angular speed (100.53 rad/s)
   C. Rotor frequency 2 Hz
   D. Speed 1000 rpm in rpm units directly
   Answer: B. Output torque = 46000/100.53 = 457.6 N·m. Dividing by synchronous speed gives gross torque. (Key point: Output torque)

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