Induction Motor Mastery: Starting, Running, & Max Torquee

Demonstrative Video

Starting Torque

The torque developed by the motor at the instant of starting is called starting torque
In some cases, it is greater than the normal running torque, whereas other cases it is somewhat less

$\begin{aligned} E_{2} & =\mbox{rotor e.m.f per phase} ~{\color{magenta}{\text{at standstill}}}\\ R_{2} & =\mbox{rotor resistance/phase}\\ X_{2} & =\mbox{rotor reactance/phase} ~{\color{magenta}{\text{at standstill}}}\\ \therefore Z_{2} & =\sqrt{\left(R_{2}^{2}+X_{2}^{2}\right)}=\mbox{rotor impedance/phase } ~{\color{magenta}{\text{at standstill}}}\\ \Rightarrow I_{2} & =\dfrac{E_{2}}{Z_{2}}=\dfrac{E_{2}}{\sqrt{\left(R_{2}^{2}+X_{2}^{2}\right)}}\\ \Rightarrow\cos\phi_{2} & =\dfrac{R_{2}}{Z_{2}}=\dfrac{R_{2}}{\sqrt{\left(R_{2}^{2}+X_{2}^{2}\right)}} \end{aligned}$

Standstill or starting torque $T_{st} = K_1E_2I_2\cos\phi_2$ $\begin{aligned} T_{st} & =K_{1}E_{2}\cdot\dfrac{E_{2}}{\sqrt{\left(R_{2}^{2}+X_{2}^{2}\right)}}\times\dfrac{R_{2}}{\sqrt{\left(R_{2}^{2}+X_{2}^{2}\right)}}\\ \Rightarrow T_{st} & =\dfrac{K_{1}E_{2}^{2}R_{2}}{\left(R_{2}^{2}+X_{2}^{2}\right)} \end{aligned}$
If $V$ is constant, then $\phi$ and hence $E_2$ both are constant $\therefore T_{st} = K_2 \cdot \dfrac{R_2}{\left(R_2^2+X_2^2\right)} = K_2 \dfrac{R_2}{Z_2^2}$
Now, $K_1=\dfrac{3}{2\pi N_s}$

$T_{st}=\dfrac{3}{2\pi N_{s}}\cdot\dfrac{E_{2}^{2}R_{2}}{R_{2}^{2}+X_{2}^{2}}$ where, $N_s \rightarrow$ synchronous speed in r.p.s

Starting torque of a squirrel-cage motor

The resistance of the Squirrel-cage motor is fixed and small as compared to its reactance which is very large especially at the start because at standstill the frequency of the rotor currents equal the supply frequency
Hnece, the strating current $I_2$ of the rotor though very large in magnitude, lags by a very large angle behind $E_2$ , with the result that the starting torque per ampere is very poor
It is roughly 1.5 times the full-load torque although the starting current is 5 to 7 times the full-load current
Hence such motors are not ver useful where the motor has to start against heavy loads

Starting torque of a slip-ring motor

The starting torque of such motor is increased by improving its power factor by adding external resistance in the rotor circuit from the star-connected rheostat
The rheostat resistance being progressively cut out as the motor gathers speed
Addition of external resistance, increase the rotor impedance and so reduce the rotor current
At first the effect of improved power factor predominates the current - decreasing effect of impedance. Hence, the starting torque is increased
But after a certain point, the effect of increased impedance predominates the effect of improved power factor and so the torque starts decreasing

Condition for maximum Starting torque

$\begin{aligned} T_{st} & =\dfrac{K_{2}R_{2}}{R_{2}^{2}+X_{2}^{2}}\\ \therefore\dfrac{dT_{st}}{dR_{2}} & =K_{2}\left[\dfrac{1}{R_{2}^{2}+X_{2}^{2}}-\dfrac{R_{2}\left(2R_{2}\right)}{\left(R_{2}^{2}+X_{2}^{2}\right)^{2}}\right]=0\\ & \Rightarrow R_{2}^{2}+X_{2}^{2}=2R_{2}^{2}\\ \Longrightarrow & R_{2}=X_{2} \end{aligned}$

Thus, $T_{st}$ is maximum when $R_2=X_2$

Effect of change in supply voltage on Starting torque

$\begin{aligned} E_{2} & \propto V\\ \therefore T_{st} & =\dfrac{K_{1}E_{2}^{2}R_{2}}{R_{2}^{2}+X_{2}^{2}}=\dfrac{K_{1}V^{2}R_{2}}{R_{2}^{2}+X_{2}^{2}}=\dfrac{K_{3}V^{2}R_{2}}{Z_{2}^{2}}\\ \Longrightarrow T_{st} & \propto V^{2} \end{aligned}$

Thus, torque is very sensitive to change in supply voltage
A change in 5% in $V$ will produce a change of approx. 10% in the rotor torque

Rotor E.M.F and Reactance under Running Conditions

At standstill, $s=1$ and $f_r=f$
$E_2$ at standstill is maximum because relative speed between the rotor and the revolving stator flux is maximum
In fact, IM is equivalent to 3-phase TF with short-circuited rotating secondary
Under running condition, the relative speed decreases, $E_2 \propto$ relative speed also decreases

Hence under running condition $\begin{aligned} E_r & = sE_2\\ f_r & = sf_2\\ X_r & = sX_2 \end{aligned}$ where $E_2, X_2,$ and $f_2$ are rotor quantities under standstill conditions
Due to decrease in the frequency of $E_2$ , $X_r$ also decreases

Torque Under Running Conditions

$\begin{aligned} T_{r} & \propto\phi I_{r}\cos\phi_{2}\\ \Rightarrow T_{r} & \propto E_{2}\dfrac{E_{r}}{Z_{r}}\cdot\dfrac{R_{2}}{Z_{r}}~\left(\because E_{2}\propto\phi\right)\\ \Rightarrow T_{r} & \propto\left(E_{2}\right)\left(\dfrac{sE_{2}}{\sqrt{R_{2}^{2}+\left(sX_{2}\right)^{2}}}\right)\left(\dfrac{R_{2}}{\sqrt{R_{2}^{2}+\left(sX_{2}\right)^{2}}}\right)\\ \Rightarrow T_{r} & \propto\dfrac{sE_{2}^{2}R_{2}}{R_{2}^{2}+\left(sX_{2}\right)^{2}}\\ \Rightarrow T_{r} & =\dfrac{3}{2\pi N_{s}}\left[\dfrac{sE_{2}^{2}R_{2}}{R_{2}^{2}+\left(sX_{2}\right)^{2}}\right] \\ & \Rightarrow \text{Substitute}~ s=1~\text{ to get}~ T_{st} \end{aligned}$

Condition for maximum torque under running condition

The torque under running condition is given by $T_r = K_1 \cdot \dfrac{sE_2^2R_2}{R_2^2+(sX_2)^2}$
To make calculation simple take $Y=1/T_r$

Then, to obtain the condition of maximum torque $\begin{aligned} \dfrac{dY}{ds} & =\dfrac{d}{ds}\left[\dfrac{R_{2}^{2}+\left(sX_{2}\right)^{2}}{K_{1}sE_{2}^{2}R_{2}}\right]\\ &=\dfrac{d}{ds}\left[\dfrac{R_{2}}{K_{1}sE_{2}^{2}}+\dfrac{sX_{2}^{2}}{K_{1}E_{2}^{2}R_{2}}\right]\\ \Rightarrow\dfrac{dY}{ds} & =\dfrac{-R_{2}}{K_{1}s^{2}E_{2}^{2}}+\dfrac{X_{2}^{2}}{K_{1}E_{2}^{2}R_{2}}=0\\ \Rightarrow\dfrac{R_{2}}{K_{1}s^{2}E_{2}^{2}} & =\dfrac{X_{2}^{2}}{K_{1}E_{2}^{2}R_{2}}\\ \Longrightarrow R_{2} & =sX_{2} \end{aligned}$

Thus, torque under running condition is maximum when $R_2=sX_2$ , and thus the maximum torque is $\begin{aligned} T_{max} & =K_{1}\dfrac{sE_{2}^{2}\left(sX_{2}\right)}{\left(sX_{2}\right)^{2}+\left(sX_{2}\right)^{2}}\\ & =K_{1}\dfrac{sE_{2}^{2}\left(sX_{2}\right)}{2\left(sX_{2}\right)^{2}}\\ & =K_{1}\dfrac{E_{2}^{2}}{2X_{2}}\\ \Rightarrow T_{max} & =\dfrac{3}{2\pi N_{s}}\cdot\dfrac{E_{2}^{2}}{2X_{2}} \end{aligned}$

$T_{max}$ is independent of the $R_2$
However, the speed or slip at which $T_{max}$ occurs is determined by $R_2$
By varying $R_2$ (possible only with slip-ring motors) $T_{max}$ can be made to occur at any desired slip (or motor speed)
$T_{max} \propto 1/X_2$ Hence, it should be kept as small as possible
$T_{max} \propto V^2$
For obtaining $T_{max}$ at starting ( $s=1$ ), $R_2=X_2$

Rotor Torque and Breakdown Torque

The rotor torque at any slip $s$ can be expressed in terms of the maximum (or breakdown) torque $T_b$ by the following equation $T=T_{b}\left[\dfrac{2}{\left(s_{b}/s\right)+\left(s/s_{b}\right)}\right]$ where $s_b$ is the breakdown or pull-out slip