Demonstrative Video
Power Stages in an Induction Motor
Stator iron loss (eddy+hysteresis losses) depends on the supply frequency and the flux density in the iron core, is practically constant
The rotor iron loss is negligible because frequency of rotor current under normal condition is always small
Total rotor Cu loss = \(3I_2^2R_2\)
Gross Torque developed by the motor \[\begin{aligned} T_{g} & =\dfrac{P_{2}}{\omega_{s}}=\dfrac{P_{2}}{2\pi N_{s}}\cdots\cdots\mbox{in terms of the rotor input}\\ & =\dfrac{P_{m}}{\omega}=\dfrac{P_{m}}{2\pi N}\cdots\cdots\mbox{in terms of rotor output} \end{aligned}\]
Shaft Torque \[T_{sh}=\dfrac{P_{out}}{\omega}=\dfrac{P_{out}}{2\pi N}\]
\(N\) and \(N_s\) are in r.p.s. If they are taken in r.p.m, then \[\begin{aligned} T_{g} & =\dfrac{P_{2}}{2\pi N_{s}/60}=\dfrac{60}{2\pi}\cdot\dfrac{P_{2}}{N_{s}}=9.55\dfrac{P_{2}}{N_{s}}\\ & =\dfrac{P_{m}}{2\pi N/60}=\dfrac{60}{2\pi}\cdot\dfrac{P_{m}}{N}=9.55\dfrac{P_{m}}{N}\\ T_{sh} & =\dfrac{P_{out}}{2\pi N/60}=\dfrac{60}{2\pi}\cdot\dfrac{P_{out}}{N}=9.55\dfrac{P_{out}}{N} \end{aligned}\]
Relations: \[\begin{aligned} \mbox{Stator input}~P_{1} & =\mbox{stator output+stator losses}\\ \mbox{Rotor input}~P_{2} & =\mbox{stator output}\\ \mbox{Rotor gross output}~P_{m} & =\mbox{rotor input}~P_{2}-\mbox{rotor Cu losses} \end{aligned}\]
If \(N\) in r.p.s be the actual speed of the rotor and if \(T_g\) is in N-m, then for rotor gross output \[\begin{aligned} T_{g}\times2\pi N & =P_{m}\\ \Rightarrow T_{g} & =\dfrac{P_{m}}{2\pi N} \end{aligned}\]
If there is no Cu losses in the rotor, then rotor output will equal rotor input and rotor will run at synchronous speed \[T_g=\dfrac{P_2}{2\pi N_s}\]
Relations: \[\begin{aligned} \mbox{Rotor gross output }~P_{m} & =T_{g}\omega=T_{g}\times2\pi N\\ \mbox{Rotor input }~P_{2} & =T_{g}\omega_{s}=T_{g}\times2\pi N_{s}\\ \therefore\mbox{Rotor Cu loss} & =P_{2}-P_{m}=T_{g}\times2\pi\left(N_{s}-N\right)\\ \Rightarrow\dfrac{\mbox{Rotor Cu loss}}{\mbox{Rotor input}} & =\dfrac{N_{s}-N}{N_{s}}=s \end{aligned}\]
Relations: \[\begin{aligned} \mbox{Rotor Gross Output}~P_{m} & =\mbox{Input}-\mbox{Rotor Cu loss}\\ P_{m} & =P_{2}-sP_{2}\\ P_{m} & =\left(1-s\right)P_{2}\\ \Rightarrow\dfrac{\mbox{Rotor Gross Output}~P_{m}}{\mbox{Rotor Input}~P_{2}}&=\left(1-s\right)=\dfrac{N}{N_{s}}\\ \Rightarrow\mbox{Rotor Efficiency}&=\dfrac{N}{N_{s}}\\ \Rightarrow\dfrac{\mbox{Rotor Cu loss}}{\mbox{Rotor Gross Output}}&=\dfrac{s}{1-s} \end{aligned}\]
\[P_{2}:P_{m}:P_{cr}::1:\left(1-s\right):s\]
Induction Motor Torque Equation
\[\begin{aligned} T_{g} & =P_{2}/2\pi N_{s}\cdots\cdots N_{s}~\mbox{in r.p.s}\\ & =60P_{2}/2\pi N_{s}=9.55P_{2}/N_{s}\cdots\cdots N_{s}~\mbox{in r.p.m}\\ P_{2} & =3I_{2}^{2}R_{2}/s\\ & =3\times\left[\dfrac{sE_{2}}{\sqrt{R_{2}^{2}+\left(sX_{2}\right)^{2}}}\right]^{2}\times\dfrac{R_{2}}{s}\\ & =\dfrac{3sE_{2}^{2}R_{2}}{R_{2}^{2}+\left(sX_{2}\right)^{2}} \end{aligned}\]
\[\begin{aligned} \therefore T_{g} & =\dfrac{P_{2}}{2\pi N_{s}}\\ & =\dfrac{3}{2\pi N_{s}}\times\dfrac{sE_{2}^{2}R_{2}}{\left[R_{2}^{2}+\left(sX_{2}\right)^{2}\right]}\cdots\cdots\mbox{in terms of }E_{2}\\ & =\dfrac{3}{2\pi N_{s}}\times\dfrac{sK^{2}E_{1}^{2}R_{2}}{\left[R_{2}^{2}+\left(sX_{2}\right)^{2}\right]}\cdots\cdots\mbox{in terms of }E_{1}\\ k & =\dfrac{3K^{2}}{2\pi N_{s}}=\mbox{constant of the given machine}\\ \Longrightarrow T_{g} & =k\dfrac{sE_{1}^{2}R_{2}}{R_{2}^{2}+\left(sX_{2}\right)^{2}}\cdots\cdots\mbox{in trems of }E_{1} \end{aligned}\]
Synchronous Watt
Synchronous watt is that torque which, at the synchronous speed of the machine under consideration, would develop a power of 1 watt \[\begin{aligned} T_{sw} & =\dfrac{P_{2}}{2\pi N_{s}}\\ & =\dfrac{1}{\omega_{s}}\cdot\dfrac{N_{s}}{N}\cdot P_{g}=\dfrac{1}{\omega_{s}}\cdot\dfrac{N_{s}}{N}\cdot P_{m} \end{aligned}\]
Torque in synchronous watt \[\mbox{=rotor input=}\dfrac{\mbox{rotor Cu loss}}{s}=\dfrac{P_{m}}{\left(1-s\right)}\]