A 50 Hz, 3-phase induction motor has a rated voltage V_{1}. The motor’s breakdown torque at rated voltage and frequency occurs at a slip of 0.2. The motor is instead run from a 60 Hz supply of voltage V_{2}. The stator impedance can be neglected.
(a) If V_{2}=V_{1}, find the ratio of currents and torques at starting. Also find the ratio of maximum torques.
(b) Find the ratio V2 /V1 such that the motor has the same values of starting current and torque at 50 and 60 Hz.
(a) s_{\max , T}=\frac{R_{2}^{\prime}}{X_{2}^{\prime}}=0.2
Where X_{2}^{\prime}= = standstill 50 Hz rotor reactance
At rated voltage ( V_{1} ) and frequency (50 Hz)
I_{s}(1)=I_{2}^{\prime}(1)=\frac{V_{1}}{\sqrt{R_{2}^{\prime 2}+X_{2}^{\prime 2}}} (i)
T_{s}(1)=\frac{V_{1}^{2} R_{2}^{\prime}}{R_{2}^{\prime 2}+X_{2}^{\prime 2}} (ii)
T_{\max }(1)=\frac{3}{\omega_{s}} \cdot \frac{0.5 V_{1}^{2}}{X_{2}^{\prime}} (iii)
At voltage V_{2} and frequency 60 Hz
I_{2}(2)=I_{2}^{\prime}(2)=\frac{V_{2}}{\sqrt{R_{2}^{\prime 2}+\left(\frac{6}{5}\right)^{2} X_{2}^{\prime 2}}} (iv)
T_{s}(2)=\frac{V_{2}^{2} R_{2}^{\prime}}{R_{2}^{\prime 2}+\left(\frac{6}{5}\right)^{2} X_{2}^{\prime 2}} (v)
T_{\max }(2)=\frac{3}{\left(\frac{6}{5}\right) \omega_{s}} \frac{0.5 V_{2}^{2}}{\left(\frac{6}{5}\right) X_{2}^{\prime}} (vi)
Dividing Eqs (iv), (v) and (vi) respectively by Eqs (i), (ii) and (iii)
\frac{I_{s}(2)}{I_{s}(1)}=\frac{V_{2}}{V_{1}} \cdot \sqrt{\frac{R_{2}^{\prime 2}+X_{2}^{\prime 2}}{R_{2}^{\prime 2}+\left(\frac{6}{5}\right)^{2} X_{2}^{\prime 2}}}=\frac{V_{2}}{V_{1}} \cdot \sqrt{\frac{s_{\max , T}^{2}+1}{s_{\max , T}^{2}+\left(\frac{6}{5}\right)^{2}}}
=1 \times \sqrt{\frac{(0.2)^{2}+1}{(0.2)^{2}+\left(\frac{6}{5}\right)^{2}}}=0.838
\frac{T_{s}(2)}{T_{s}(1)}=\frac{V_{2}^{2}}{V_{1}^{2}} \cdot \frac{s_{\max , T}^{2}+1}{s_{\max , T}^{2}+\left(\frac{6}{5}\right)^{2}}
=1 \times \frac{(0.2)^{2}+1}{(0.2)^{2}+\left(\frac{6}{5}\right)^{2}}=0.703
\frac{T_{\max }(2)}{T_{\max }(1)}=\frac{V_{2}^{2}}{V_{1}^{2}}\left(\frac{5}{6}\right)^{2}=0.694
(b) \frac{V_{2}^{2}}{V_{1}^{2}} \cdot \sqrt{\frac{(0.2)^{2}+1}{(0.2)^{2}+\left(\frac{6}{5}\right)^{2}}}=1
\frac{V_{2}}{V_{1}}=1.19
This ratio will also give equal staring torques.