- 2072 Magh
IM equivalent circuit; starting/running torque
Short answer
At standstill (s=1), rotor circuit has E2, R2, X2. Starting torque: T_st = (3/ω_s) E2² R2/(R2²+X2²). At any slip s, rotor emf is sE2, reactance sX2; running torque: T = (3/ω_s) s E2² R2/(R2²+s²X2²). Maximum torque occurs at s_m = R2/X2, T_max = 3E2²/(2ω_s X2).
Draw equivalent circuit of 3 phase induction motor at stand still and running conditions. Derive the expression for starting torque and running torque.
A three-phase induction motor is like a transformer with a rotating secondary. Its per-phase equivalent circuit is drawn with all rotor quantities referred to the stator.
Rotor circuit at standstill (s = 1) and running (slip s)
Rotor emf per phase at standstill = E₂, reactance X₂. At slip s: emf = sE₂ and reactance = sX₂ (rotor frequency f₂ = sf).
Standstill (s = 1) Running (slip s)
R2 X2 R2 sX2
--/\/\/--(XXX)--+ --/\/\/--(XXX)--+
E2 -> | sE2 -> |
(short) (short)
I2 = E2/√(R2²+X2²) I2 = sE2/√(R2²+s²X2²)
Dividing the running circuit by s gives the same current with a constant-frequency circuit: I₂ = E₂ / √((R₂/s)² + X₂²). Here R₂/s = R₂ + R₂(1−s)/s, where R₂(1−s)/s is the load resistance that stands for the mechanical output.
Complete per-phase equivalent circuit (referred to stator)
I1 R1 X1 X2' R2'
-->--/\/\/--(XXX)--+-------+--(XXX)--/\/\/--+
| | |
V1 (phase) Rc| Xm| | R2'(1-s)/s
| | | (load)
--------------------+-------+---------------+
Rc and Xm form the exciting branch (core loss and magnetising reactance). At standstill the load resistance R₂'(1−s)/s = 0 (s = 1). The motor behaves like a short-circuited transformer.
Starting torque
At start s = 1. Rotor current I₂ = E₂ / √(R₂² + X₂²), rotor power factor = R₂ / √(R₂² + X₂²).
Rotor input (air-gap power) P₂ = 3 I₂² R₂ / s. Torque T = P₂ / ω_s. At s = 1:
T_st = (3/ω_s) · E₂² R₂ / (R₂² + X₂²)
(Neglecting stator impedance, E₂ is proportional to the supply voltage, so T_st ∝ V².)
Running torque
For slip s, I₂ = sE₂ / √(R₂² + s²X₂²). Rotor input P₂ = 3 I₂² R₂ / s. Hence
T = P₂/ω_s = (3/ω_s) · s E₂² R₂ / (R₂² + s²X₂²)
Mechanical power P_m = (1−s)P₂ and T = P_m/ω_r = P₂/ω_s. The maximum running torque occurs at s_m = R₂/X₂ and equals T_max = 3E₂² / (2ω_s X₂).
Relations
- Rotor copper loss = s · P₂
- Mechanical power P_m = (1 − s) · P₂
- P₂ : P_cu : P_m = 1 : s : (1 − s)
✓ Answer checked by Claude Opus 5.5 (high effort)
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