Induction motor: torque -slip characteristics explained with numerical

Basic Terms

Slip = (Ns - Nr)/Ns

Torque (T) is the turning force produced by the motor shaft

Characteristic Curve Overview

The torque vs. slip curve has 3 regions:

1. Low Slip Region

 (Normal running condition, small slip ~0 to 0.05)

Motor runs near synchronous speed

Torque increases almost linearly with slip

Torque ≈ proportional to slip (T ∝ S)

2. Medium Slip Region

 (Moderate load increase)

Torque increases non-linearly

Approaches maximum torque (breakdown torque)

3.  High Slip Region 

(Startup or overload)

Torque starts decreasing with increase in slip

Rotor resistance dominates

Torque 

∝ 1/S (inversely proportional to slip)

Parameter	Description
Starting Torque (S = 1)	Non-zero torque, depends on rotor resistance.
Maximum Torque	Also called Breakdown Torque or Pull-out Torque.
Slip at Max Torque	S= R2/X2  independent of supply voltage.
Low Slip Region	Torque is directly proportional to slip (T ∝ S).
High Slip Region	Torque is inversely proportional to slip (T ∝ 1/S).

Numerical on induction motor 

Induction Motor Solved Numericals

 Slip and Rotor Frequency

Q: A 4-pole, 3-phase, 50 Hz induction motor runs at 1440 rpm. Find:

• Synchronous speed
• Slip
• Rotor current frequency

Solution:

Ns = (120 × f) / P = (120 × 50) / 4 = 1500 rpm

S = (Ns - Nr) / Ns = (1500 - 1440) / 1500 = 0.04 or 4%

fr = S × f = 0.04 × 50 = 2 Hz
  
 Torque Developed

Q: A 3-phase induction motor delivers 25 kW at 1470 rpm. Calculate the torque developed.

Solution:

T = (P × 60) / (2πN) = (25000 × 60) / (2π × 1470) ≈ 162.16 Nm

  

 Rotor Copper Loss

Q: A motor has rotor input of 12 kW and slip of 5%. Find rotor copper loss and mechanical power developed.

Solution:

Rotor Cu loss = S × Rotor Input = 0.05 × 12000 = 600 W

Mechanical Power = 12000 - 600 = 11400 W

  

 Rotor Input Power

Q: Motor output power = 18.5 kW, slip = 3%. Find rotor input power.

Solution:

Rotor Input = Output / (1 - S) = 18500 / 0.97 ≈ 19072.16 W

  

Starting Current vs Full Load

Q: Starting torque is 1.5 times full-load torque. Slip at full load is 4%. Find the ratio of starting current to full-load current.

Solution:

T ∝ I² × R₂ / s

1.5 = (Istart / Ifl)² × 0.04

(Istart / Ifl)² = 37.5

Istart / Ifl = √37.5 ≈ 6.12

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Back emf and torque in dc motor