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 = \(3I_2^2R_2\)
- \[\begin{aligned} T_{g} & =\dfrac{P_{2}}{\omega_{s}}=\dfrac{P_{2}}{2\pi N_{s}}\cdots\cdots\mbox{in terms of the rotor input}\\ & =\dfrac{P_{m}}{\omega}=\dfrac{P_{m}}{2\pi N}\cdots\cdots\mbox{in terms of rotor output} \end{aligned}\]Gross Torque developed by the motor
- \[T_{sh}=\dfrac{P_{out}}{\omega}=\dfrac{P_{out}}{2\pi N}\]Shaft Torque
- \[\begin{aligned} T_{g} & =\dfrac{P_{2}}{2\pi N_{s}/60}=\dfrac{60}{2\pi}\cdot\dfrac{P_{2}}{N_{s}}=9.55\dfrac{P_{2}}{N_{s}}\\ & =\dfrac{P_{m}}{2\pi N/60}=\dfrac{60}{2\pi}\cdot\dfrac{P_{m}}{N}=9.55\dfrac{P_{m}}{N}\\ T_{sh} & =\dfrac{P_{out}}{2\pi N/60}=\dfrac{60}{2\pi}\cdot\dfrac{P_{out}}{N}=9.55\dfrac{P_{out}}{N} \end{aligned}\]\(N_s\)\(N\)
- \[\begin{aligned} \mbox{Stator input}~P_{1} & =\mbox{stator output+stator losses}\\ \mbox{Rotor input}~P_{2} & =\mbox{stator output}\\ \mbox{Rotor gross output}~P_{m} & =\mbox{rotor input}~P_{2}-\mbox{rotor Cu losses} \end{aligned}\]Relations:
- \[\begin{aligned} T_{g}\times2\pi N & =P_{m}\\ \Rightarrow T_{g} & =\dfrac{P_{m}}{2\pi N} \end{aligned}\]is in N-m, then for rotor gross output in r.p.s be the actual speed of the rotor and if If
- \[T_g=\dfrac{P_2}{2\pi N_s}\]If there is no Cu losses in the rotor, then rotor output will equal rotor input and rotor will run at synchronous speed
- \[\begin{aligned} \mbox{Rotor gross output }~P_{m} & =T_{g}\omega=T_{g}\times2\pi N\\ \mbox{Rotor input }~P_{2} & =T_{g}\omega_{s}=T_{g}\times2\pi N_{s}\\ \therefore\mbox{Rotor Cu loss} & =P_{2}-P_{m}=T_{g}\times2\pi\left(N_{s}-N\right)\\ \Rightarrow\dfrac{\mbox{Rotor Cu loss}}{\mbox{Rotor input}} & =\dfrac{N_{s}-N}{N_{s}}=s \end{aligned}\]Relations:
- \[\begin{aligned} \mbox{Rotor Gross Output}~P_{m} & =\mbox{Input}-\mbox{Rotor Cu loss}\\ P_{m} & =P_{2}-sP_{2}\\ P_{m} & =\left(1-s\right)P_{2}\\ \Rightarrow\dfrac{\mbox{Rotor Gross Output}~P_{m}}{\mbox{Rotor Input}~P_{2}}&=\left(1-s\right)=\dfrac{N}{N_{s}}\\ \Rightarrow\mbox{Rotor Efficiency}&=\dfrac{N}{N_{s}}\\ \Rightarrow\dfrac{\mbox{Rotor Cu loss}}{\mbox{Rotor Gross Output}}&=\dfrac{s}{1-s} \end{aligned}\]Relations:
\[P_{2}:P_{m}:P_{cr}::1:\left(1-s\right):s\]
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
- \[\begin{aligned} T_{sw} & =\dfrac{P_{2}}{2\pi N_{s}}\\ & =\dfrac{1}{\omega_{s}}\cdot\dfrac{N_{s}}{N}\cdot P_{g}=\dfrac{1}{\omega_{s}}\cdot\dfrac{N_{s}}{N}\cdot P_{m} \end{aligned}\]Synchronous watt is that torque which, at the synchronous speed of the machine under consideration, would develop a power of 1 watt
- \[\mbox{=rotor input=}\dfrac{\mbox{rotor Cu loss}}{s}=\dfrac{P_{m}}{\left(1-s\right)}\]Torque in synchronous watt