A squirrel-cage induction motor has a slip of 4% at full load. Its starting current is five times the full-load current. The stator impedance and magnetizing current may be neglected; the rotor resistance is assumed constant.
(a) Calculate the maximum torque and the slip at which it would occur.
(b) Calculate the starting torque.
Express torques in pu of the full-load torque.
(a) I_{s}^{2}=\frac{V^{2}}{R_{2}^{\prime 2}+X_{2}^{2}} ;(s=1) (i)
I_{f l}^{2}=\frac{V^{2}}{\left(R_{2}^{\prime 2} / s_{f l}\right)^{2}+X_{2}^{\prime 2}} (ii)
Dividing Eq. (i) by (ii)
\left(\frac{I_{S}}{I_{f l}}\right)^{2}=\frac{\left(R_{2}^{\prime} / s_{f l}\right)^{2}+X_{2}^{\prime 2}}{R_{2}^{\prime 2}+X_{2}^{\prime 2}}=\frac{s_{\max , T}^{2}+s_{f l}^{2}}{s_{f l}^{2}\left(s_{\max , T}^{2}+1\right)} (iii)
Substituting the values 25=\frac{s_{\max , T}^{2}+(0.04)^{2}}{(0.04)^{2}\left(s_{\max , T}^{2}+1\right)}
Or s_{\max , T}=0.2 \quad \text { or } \quad 20 \%
T_{\max }=\frac{3}{\omega_{s}} \cdot \frac{0.5 V^{2}}{X_{2}^{\prime 2}} (iv)
T_{f 1}=\frac{3}{\omega_{s}} \cdot \frac{V^{2}\left(R_{2}^{\prime 2} / s_{f l}\right)}{\left(R_{2}^{\prime 2} / s_{f l}\right)^{2}+X_{2}^{\prime 2}} (v)
Dividing Eq. (iv) by (v)
\frac{T_{\max }}{T_{f l}}=0.5 \times \frac{R_{2}^{2}+s_{f l}^{2} X_{2}^{\prime 2}}{R_{2}^{\prime} X_{2}^{\prime} s_{f l}}=0.5 \times \frac{s_{\max , T}^{2}+s_{f l}^{2}}{s_{\max , T} s_{f l}}=0.5 \times \frac{(0.2)^{2}+(0.04)^{2}}{0.2 \times 0.04}=2.6
Or T_{\max }=2.6 pu
(b) As per Eq. (9.60)
\frac{T_{s}}{T_{f l}}=\left(\frac{I_{s}}{I_{f l}}\right)^{2} s_{f 1}=(5)^{2} \times 0.04=1Or T_{s}=1 pu