\[\%~\mbox{slip}~(s)=\dfrac{N_{s}-N_{r}}{N_{s}}\times100\]
Slip is ratio and have no units
\(N_s>N_r\)
\(N_{s}=N_{r}\Rightarrow\)relative speed = 0 \(\Rightarrow\) No rotor emf/current \(\Rightarrow\) No Torque
At standstill/blocked rotor \(N_r=0 \Rightarrow\) \(s=1\)
At synchronous speed \(N_s = N_r \Rightarrow\) \(s=0\)
When the rotor is stationary, the frequency of rotor current is the same as the supply frequency
When rotor starts revolving, then the frequency depends upon the relative speed or on slip speed
Let any slip-speed, the frequency of the rotor current be \(f_r\). Then \[\begin{aligned} N_{s}-N & =\dfrac{120f_r}{P}~\mbox{and}~N_{s}=\dfrac{120f}{P}\\ \Longrightarrow\dfrac{f_r}{f} & =\dfrac{N_{s}-N}{N_{s}}=s\\ \Rightarrow f_r & =sf \end{aligned}\]
Thus rotor currents have a frequency \(f_r=sf\) and when flowing through the individual phases of rotor winding, give rise to rotor magnetic fields
These individual rotor magnetic fields produce a combined rotating magnetic field, whose speed relative to rotor is \[=\dfrac{120f_{r}}{P}=\dfrac{120sf}{P}=sN_{s}\]
However, the rotor itself is running at speed \(N\) w.r.t space
Hence, the speed of rotor field in space = speed of rotor magnetic field relative to rotor + speed of rotor relative to space \[= sN_s+N=sN_s+N_s(1-s)=N_s\]
No matter what the vale of \(s\), rotor currents and stator current each produce a sinusoidally distributed magnetic field of constant magnitude and constant space speed \(N_s\)
In other words, both rotor and stator field rotate synchronously, which means that they are stationary w.r.t each other
These two synchronously rotating magnetic fields, in fact, superimpose on each other and give rise to the actually existing rotating field
Relation between Torque and Rotor Power factor
We know, \[\begin{aligned} T & \propto\phi I_{2}\cos\phi_{2}\\ \Rightarrow T & =K\phi I_{2}\cos\phi_{2} \end{aligned}\] where, \[\begin{aligned} I_{2} & =\mbox{rotor current at standstill}\\ \phi_{2} & =\mbox{angle between rotor e.m.f and rotor current}\\ K & =\mbox{a constant} \end{aligned}\]
Denoting rotor e.m.f at standstill by \(E_2\), we have \(E_2 \propto \phi\)
Therefore, \(T \propto E_{2}I_{2}\cos\phi_{2} \quad \Rightarrow T =K_{1}E_{2}I_{2}\cos\phi_{2}\)
Hence, \(\phi_{2}\uparrow\Rightarrow\cos\phi_{2}\downarrow\Rightarrow T\downarrow~\mbox{and vice-versa}\)