\[E_{2} \propto\phi\propto V \Rightarrow T \propto sV^{2}\]
\(T\) at any \(N_r\) is proportional to \(V^2\)
If \(V\) decreases by 10%, then \(T\) decreases by 20%
Changes in \(V\) not only affect\(T_{st}\) but torque under running conditions also
If \(V\) decreases, then \(T\) also decreases
Hence, for maintaining same \(T\), \(s\) increases i.e. \(N_r\) falls
Let \(V \rightarrow V^{'}\), \(s \rightarrow s^{'}\) and \(T \rightarrow T^{'}\) \[\dfrac{T}{T'}=\dfrac{sV^{2}}{s^{'}V^{'2}}\]
Hardly any important changes in \(f\) take place on a large distribution system except during a major disturbance
However, large frequency changes often take place on isolated, low-power systems in which electrical energy is generated by means of diesel engine or gas turbines
Example of such systems are: emergency supply in hospital and the electrical system on a ship etc.
The major effect of change in \(f\) is on motor speed
If \(f\) drops by 10% then \(N_r\) also drops by 10%
A 50-Hz motor connected to 60-Hz supply then motor runs \((60-50)\times 100/50 = 20\%\) faster than normal
A 50-Hz motor operates well on a 60-Hz line provided its terminal voltage is raised to \(60/50=120\%\) of the name plate rating
In that case, the new breakdown torque becomes equal to the original breakdown torque and the starting torque is only slightly reduced
However, power factor, efficiency, and temperature rise remain satisfactory
Similarly, a 60-Hz motor can operate satisfactorily on 50-Hz supply provided its terminal voltage is reduced to \(5/6=80\%\) of its name-plate rating
Let \(s_f\) be the slip corresponding to the full-load torque \[\begin{aligned} T_{f} & \propto\dfrac{s_{f}R_{2}}{R_{2}^{2}+\left(s_{f}X_{2}\right)^{2}}~\mbox{and}~T_{max}\propto\dfrac{1}{2X_{2}}\\ \therefore\dfrac{T_{f}}{T_{max}} & =\dfrac{2s_{f}R_{2}X_{2}}{R_{2}^{2}+\left(s_{f}X_{2}\right)^{2}}\\ \Rightarrow & \dfrac{T_{f}}{T_{max}}=\dfrac{2s_{f}\cdot R_{2}/X_{2}}{\left(R_{2}/X_{2}\right)^{2}+s_{f}^{2}}=\dfrac{2s_{m}s_{f}}{s_{m}^{2}+s_{f}^{2}} \end{aligned}\]
In general, \[\dfrac{\mbox{operating torque at any slip s}}{\mbox{maximum torque}} = \dfrac{2s_ms}{s_m^2+s_f^2}\]
\[\begin{aligned} T_{st} & \propto\dfrac{R_{2}}{R_{2}^{2}+X_{2}^{2}}\\ T_{max} & \propto\dfrac{1}{2X_{2}}\\ \therefore\dfrac{T_{st}}{T_{max}} & =\dfrac{2R_{2}X_{2}}{R_{2}^{2}+X_{2}^{2}}\\ & =\dfrac{2R_{2}/X_{2}}{1+\left(R_{2}/X_{2}\right)^{2}}\\ & =\dfrac{2s_{m}}{1+s_{m}^{2}} \end{aligned}\]