IM slip and starting current from torque ratios

Given: T_st = 1.25 T_fl, T_max = 2.5 T_fl. Stator resistance and rotational losses neglected, R₂ constant.

Using T / T_max = 2 s s_m / (s² + s_m²), where s_m = R₂/X₂.

(i) Slip at maximum torque

At starting s = 1:

T_st / T_max = 1.25 / 2.5 = 0.5 = 2 s_m / (1 + s_m²)

 s_m² - 4 s_m + 1 = 0
 s_m = (4 - √12)/2 = 2 - √3 = 0.268

(The root 3.73 is rejected, since s_m is below 1 for a practical motor.)

Slip at maximum torque = 0.268.

(ii) Slip at full load

T_fl / T_max = 1 / 2.5 = 0.4 = 2 s s_m / (s² + s_m²)

 0.4 s² - 2 s_m s + 0.4 s_m² = 0
 s² - 5 s_m s + s_m² = 0
 s = s_m (5 - √21)/2 = 0.268 × 0.2087 = 0.0559

(The other root, 0.268 × 4.79 = 1.28, is above 1 and rejected.)

Full-load slip = 0.0559 (5.59 %).

(iii) Starting current in per unit of full-load current

Neglecting stator impedance and magnetising current, I ∝ E₂ / √((R₂/s)² + X₂²) with R₂ = s_m X₂:

I ∝ 1 / √((s_m/s)² + 1) (X₂ common)

 Starting:   s = 1      : 1 / √(s_m² + 1)     = 1 / 1.0353 = 0.9659
 Full load:  s = 0.0559 : 1 / √((s_m/s)² + 1) = 1 / √(4.791² + 1)
                                              = 1 / 4.894   = 0.2043
 I_st / I_fl = 0.9659 / 0.2043 = 4.73

Starting current = 4.73 times full-load current (4.73 pu).