In this topic, you study Torque Equation of Induction Motor.
The exact mathematical expression for the torque developed by an induction motor can be derived from the relationship between the rotor input, rotor copper loss, total mechanical power developed by the rotor and slip seen in the previous section. Let T be the total torque (in newton-metres) developed by an induction motor while operating at a speed of N r.p.m. and let S be the corresponding slip.
The rotor copper loss is given as
\[\text{Rotor copper loss}=\text{3I}_{\text{2S}}^{\text{2}}{{\text{R}}_{\text{2}}}….(1)\]
The rotor current per phase under running condition (at slip S),
\[{{\text{I}}_{\text{2S}}}=\frac{\text{S}{{\text{E}}_{\text{2}}}}{\sqrt{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}}….(2)\]
The Rotor input P2 is given as
\[{{\text{P}}_{\text{2}}}=\frac{\text{Rotor copper loss}}{\text{S}}\]
Using Equations (1) and (2), the Rotor input P2 will be
\[{{\text{P}}_{\text{2}}}=\frac{\text{3}}{\text{S}}\text{ }{{\left[ \frac{\text{S}{{\text{E}}_{\text{2}}}}{\sqrt{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}} \right]}^{2}}\times {{\text{R}}_{\text{2}}}\]
\[=\frac{\text{3SE}_{\text{2}}^{\text{2}}{{\text{R}}_{\text{2}}}}{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}\text{ watts}\]
∴ Total mechanical power developed by the motor,
\[{{\text{P}}_{\text{m}}}=(1-\text{S})(\text{Rotor input})\]
\[=(1-\text{S})\times \frac{\text{3SE}_{\text{2}}^{\text{2}}{{\text{R}}_{\text{2}}}}{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}\text{ watts}\]
But total mechanical power developed by the motor is also given by,
\[{{\text{P}}_{\text{m}}}=\text{T}\times \frac{\text{2 }\!\!\pi\!\!\text{ N}}{60}\text{ watts}\]
∴
\[\text{T}\times \frac{\text{2 }\!\!\pi\!\!\text{ N}}{60}=(1-\text{S})\times \frac{\text{3SE}_{\text{2}}^{\text{2}}{{\text{R}}_{\text{2}}}}{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}\]
or,
\[\text{T}=\frac{60\text{ }(1-\text{S})}{\text{2 }\!\!\pi\!\!\text{ N}}\times \frac{\text{3SE}_{\text{2}}^{\text{2}}{{\text{R}}_{\text{2}}}}{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}\]
However, we know that
\[\text{N}={{\text{N}}_{\text{S}}}(1-\text{S})\]
Substituting this in the above equation, we have
\[\text{T}=\frac{60}{\text{2 }\!\!\pi\!\!\text{ }{{\text{N}}_{\text{S}}}}\times \frac{\text{3SE}_{\text{2}}^{\text{2}}{{\text{R}}_{\text{2}}}}{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}\]
or,
\[\text{T}=\frac{3}{\text{2 }\text{ }\!\!\pi\!\!\text{ }\text{ }{{\text{n}}_{\text{S}}}}\times \frac{\text{SE}_{\text{2}}^{\text{2}}{{\text{R}}_{\text{2}}}}{\text{R}_{\text{2}}^{\text{2}}+{{\text{(S}{{\text{X}}_{\text{2}}}\text{)}}^{\text{2}}}}\text{ newton-metres}\]
….(3)
where nS = Synchronous speed of the motor in r.p.s.
The above equation (3) is known as torque equation of an induction motor.
TORQUE IN SYNCHRONOUS WATTS
The total torque developed by the induction motor is directly proportional to the rotor power input, regardless of actual speed of the motor. For this reason, it is common practice to express the torque in terms of synchronous watts. The synchronous watt is defined as that torque which, al the synchronous speed of the motor, would develop a power of 1 W. Hence, Total torque will be
\[\text{T}=\frac{{{\text{P}}_{\text{m}}}}{\text{2 }\!\!\pi\!\!\text{ N/60}}=\frac{60\text{ }(1-\text{S}){{\text{P}}_{\text{2}}}}{\text{2 }\!\!\pi\!\!\text{ }{{\text{N}}_{\text{S}}}(1-\text{S})}\]
\[=\frac{60\text{ }{{\text{P}}_{\text{2}}}}{\text{2 }\!\!\pi\!\!\text{ }{{\text{N}}_{\text{S}}}}\text{ newton-metres}\]
\[={{\text{P}}_{\text{2}}}\text{ synchronous watts}\]
Thus the torque in synchronous watts equals the rotor input in watts. It is important to note here that the two motors having different synchronous speeds, may have identical values for the total torque developed when it is expressed in synchronous watts but will have different values when the same torque is expressed in newton-metres.