Electrical Machines · Induction Motors

Induction Motor Load: Torque-Speed Insights

Dr. Mithun Mondal BITS Pilani, Hyderabad Campus Electrical Machines

Demonstrative Video

SECTION 01

Effect of parameters on Torque and speed

\[E_{2} \propto\phi\propto V \Rightarrow T \propto sV^{2}\]

  • \(T\) 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

  • \[\dfrac{T}{T'}=\dfrac{sV^{2}}{s^{'}V^{'2}}\]
    and , Let
Key Concepts

  • 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 runsfaster than normal

  • 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

SECTION 02

Full-load Torque and Maximum Torque

  • \[\begin{aligned} T_{f} & \propto\dfrac{s_{f}R_{2}}{R_{2}^{2}+\left(s_{f}X_{2}\right)^{2}}~\mbox{and}~T_{max}\propto\dfrac{1}{2X_{2}}\\ \therefore\dfrac{T_{f}}{T_{max}} & =\dfrac{2s_{f}R_{2}X_{2}}{R_{2}^{2}+\left(s_{f}X_{2}\right)^{2}}\\ \Rightarrow & \dfrac{T_{f}}{T_{max}}=\dfrac{2s_{f}\cdot R_{2}/X_{2}}{\left(R_{2}/X_{2}\right)^{2}+s_{f}^{2}}=\dfrac{2s_{m}s_{f}}{s_{m}^{2}+s_{f}^{2}} \end{aligned}\]
    be the slip corresponding to the full-load torque Let
  • \[\dfrac{\mbox{operating torque at any slip s}}{\mbox{maximum torque}} = \dfrac{2s_ms}{s_m^2+s_f^2}\]
    In general,
SECTION 03

Starting Torque and Maximum Torque

\[\begin{aligned} T_{st} & \propto\dfrac{R_{2}}{R_{2}^{2}+X_{2}^{2}}\\ T_{max} & \propto\dfrac{1}{2X_{2}}\\ \therefore\dfrac{T_{st}}{T_{max}} & =\dfrac{2R_{2}X_{2}}{R_{2}^{2}+X_{2}^{2}}\\ & =\dfrac{2R_{2}/X_{2}}{1+\left(R_{2}/X_{2}\right)^{2}}\\ & =\dfrac{2s_{m}}{1+s_{m}^{2}} \end{aligned}\]