Demonstrative Video
- Gross-Mechanical Power Developed:
\[\begin{aligned} P_{m} & =P_{2}-P_{c}\\ & =\dfrac{3\left(I_{2r}^{'}\right)^{2}R_{2}^{'}}{s}-3\left(I_{2r}^{'}\right)^{2}R_{2}^{'}\\ & =3\left(I_{2r}^{'}\right)^{2}R_{2}^{'}\left(\dfrac{1-s}{s}\right) \end{aligned}\]
- Torque Developed:
\[\begin{aligned} T & =\dfrac{P_{m}}{\omega}=\dfrac{3\left(I_{2r}^{'}\right)^{2}R_{2}^{'}\left(\dfrac{1-s}{s}\right)}{\dfrac{2\pi N}{60}}\\ & =\dfrac{3\left(I_{2r}^{'}\right)^{2}R_{2}^{'}\left(\dfrac{1-s}{s}\right)}{\dfrac{2\pi N_{s}\left(1-s\right)}{60}} \\ & =9.55\times\dfrac{\dfrac{3\left(I_{2r}^{'}\right)^{2}R_{2}^{'}}{s}}{N_{s}} \end{aligned}\]
- Maximum Power Output:
On further simplifying the approximate equivalent circuit
exciting circuit is omitted i.e. \(I_0\) is neglected
\(K\) is assumed unity
Gross-power is given by \[\begin{aligned} P_{g} & =3I_{1}^{2}R_{L}\\ & =3\left[\dfrac{V_{1}}{\sqrt{\left(R_{01}+R_{L}\right)^{2}+X_{01}^{2}}}\right]^{2}R_{L}\\ & =\dfrac{3V_{1}^{2}R_{L}}{\left(R_{01}+R_{L}\right)^{2}+X_{01}^{2}} \end{aligned}\]
The condition for maximum power output is obtained by differentiating and equating with zero, we get \[\begin{aligned} R_{L}^{2} & =R_{01}^{2}+X_{01}^{2}\\ & =Z_{01}^{2}\\ \Rightarrow R_{L} & =Z_{01} \end{aligned}\]
Hence, power is maximum when the equivalent load resistance is equal to the standstill leakage impedance of the motor
Corresponding slip:
\[\begin{aligned} Z_{01} & =R_{L}=R_{2}\left[\left(1/s\right)-1\right]\\ & \Rightarrow s=\dfrac{R_{2}}{R_{2}+Z_{01}}\\ P_{m,max} & =\dfrac{3V_{1}^{2}Z_{01}}{\left(R_{01}+Z_{01}\right)^{2}+X_{01}^{2}}\\ & =\dfrac{3V_{1}^{2}}{2\left(R_{01}+Z_{01}\right)} \end{aligned}\] NOTE: \(V_1\) is volt/phase and \(K\) is unity