ElectricalMachinesmediumFE ExamCourseworkPE / Advanced

Slip of an Induction Motor

A 4-pole three-phase induction motor runs from a $60\ \text{Hz}$ supply. Its rotor turns at $1750\ \text{rpm}$ .

Find the per-unit slip.

Given

p	4 -	Number of poles
f	60 Hz	Supply frequency
Nr	1750 rpm	Rotor speed

Hint 1

Find synchronous speed first — the speed of the rotating stator field.

Hint 2

N_s = 120f/p, where p is the number of poles, not pole-pairs.

Hint 3

s = (N_s − N_r)/N_s, normalised by synchronous speed.

Worked solution — try the problem first

Synchronous speed:

$N_s = \frac{120f}{p} = \frac{120 \times 60}{4} = 1800\ \text{rpm}$

The 120 is a unit constant (60 s/min × 2 poles per pole-pair). Note $p$ is the number of poles, not pole-pairs.

Slip:

$s = \frac{N_s - N_r}{N_s} = \frac{1800 - 1750}{1800} = \frac{50}{1800} = 0.0278$

$\boxed{s = 0.0278 = 2.78\%}$

Why slip must be nonzero. An induction motor develops torque only when the rotor cuts the rotating field — at exactly synchronous speed there is no relative motion, no induced rotor current, and no torque. Typical full-load slip is 2-5%, so 2.78% is entirely normal.

Concepts:Induction motorsSlipSynchronous speed

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