Star-delta starting torque ratio and full-load slip

Given: $T_{max}/T_{fl} = 2.5$, $R_2 = 0.4\ \Omega$, $X_2 = 4\ \Omega$ at standstill.

Slip at maximum torque

$$s_m = \frac{R_2}{X_2} = \frac{0.4}{4} = 0.1$$

Starting torque with direct switching

$$\frac{T_{st}}{T_{max}} = \frac{2s_m}{1+s_m^2} = \frac{0.2}{1.01} = 0.198$$ $$\frac{T_{st}}{T_{fl}} = 0.198\times 2.5 = 0.495$$

Star-delta starting

In star, the phase voltage is $1/\sqrt3$ of the line voltage, and torque $\propto V^2$, so the starting torque is one third of the direct (delta) value:

$$\frac{T_{st(Y\Delta)}}{T_{fl}} = \frac{0.495}{3} = 0.165$$

Full-load slip

$$\frac{T_{fl}}{T_{max}} = 0.4 = \frac{2 s s_m}{s^2+s_m^2} \;\Rightarrow\; s^2 - 0.5 s + 0.01 = 0$$ $$s = \frac{0.5 \pm\sqrt{0.25-0.04}}{2} = 0.0209 \ \text{(stable root)}$$

Answer: starting torque with star-delta starter $= 0.165\,T_{fl}$; full-load slip $= 0.0209$ (2.09 %).