Answer MET Question 11

Question: List the factors that determine the starting torque of the three-phase induction motor. How does this torque generally compare with the value of the rated torque?

Answer: Torque of an induction motor is proportional to the stator flux, rotor current and rotor power factor. thus,
$\displaystyle \small \mathrm{T\ \alpha \ \phi I_2cos\phi_2}$
at stand still
$\displaystyle \small \mathrm{E_2\ \alpha\ \phi }$
thus,
$\displaystyle \small \mathrm{T\ \alpha \ E_2 I_2cos\phi_2}$
The Rotor Circuit:
let, Rotor Resistance/ph = $\displaystyle \small \mathrm{R_2}$
Rotor Reactance/ph = $\displaystyle \small \mathrm{SX_2}$
Rotor Impedence/ph = $\displaystyle \small \mathrm{Z_2}$
= $\displaystyle \small \mathrm{Z_2=\sqrt{R_2^2+(SX_2)^2}}$
Rotor Induced emf/ph at slip 'S' = $\displaystyle \small \mathrm{SE_2}$
Rotor Current/ph = $\displaystyle \small \mathrm{I_2}$
= $\displaystyle \small \mathrm{\frac{SE_2}{Z_2}=\frac{SE_2}{\sqrt{R_2^2+(SX_2)^2}}}$
Rotor power factor = $\displaystyle \small \mathrm{cos\phi _2}$
= $\displaystyle \small \mathrm{cos\phi _2= \frac{R_2}{Z_2}}$
= $\displaystyle \small \mathrm{cos\phi _2= \frac{R_2}{\sqrt{R_2^2+(SX_2)^2}}}$

thus,
form: $\displaystyle \small \mathrm{T\ \alpha \ E_2 I_2cos\phi_2}$
$\displaystyle \small \mathrm{T\ \alpha \ E_2\times \frac{SE_2}{\sqrt{{R_2^2+(SX_2)^2}}}\times \frac{R_2}{\sqrt{{R_2^2+(SX_2)^2}}}}$
or, $\displaystyle \small \mathrm{T\ =K_1 \times \frac{SE_2^2R_2}{{R_2^2+(SX_2)^2}}}$
where, $\displaystyle \small \mathrm{K_1 =\frac{3}{2\pi N_s}}$
This equation is to find torque at any slip.

From the above expression, it is evident, that
i. Torque is zero when slip s = 0 (i.e., speed is synchronous).
ii. When slip 'S' is say low the value of the term $\displaystyle \small \mathrm{sX_2 }$ is very small and is negligible in comparison with $\displaystyle \small \mathrm{R_2 }$, therefore torque T is approximately proportional to slip 'S' if rotor resistance $\displaystyle \small \mathrm{R_2 }$ is constant. This means that at speeds near to synchronous speed the torque.
iii. Speed and torque-slip curves are approximately straight lines.
iv. When the slip 'S' increases (i.e. as the speed decreases with increase in load) torque increases and reaches its maximum value when$\displaystyle \small \mathrm{R_2=sX_2 }$.

The maximum torque is also known as 'pull-out' or 'break-down' torque. normal range operation of the squirrel-cage induction motor. Even under normal operation, however, a suddenly applied load may require the temporary development of this maximum torque, after which the motor will speed up to its full-load value. Thus, in a manner similar to the shunt motor, the induction motor inherently adjusts itself to the applied external torque.

Maximum Torque at any time:
from the torque equation we can see $\displaystyle \small \mathrm{ \frac{SR_2}{{R_2^2+(SX_2)^2}}}$ should be maximum.
or $\displaystyle \small \mathrm{ \frac{R_2}{{(\frac{R_2}{\sqrt{s}}-\sqrt{s}X_2)^2+2R_2X_2}}}$ should be maximum.
or $\displaystyle \small \mathrm{\frac{R_2}{\sqrt{s}}-\sqrt{s}X_2 =0}$
or $\displaystyle \small \mathrm{R_2 = sX_2}$
Thus the Torque will be maximum when, $\displaystyle \small \mathrm{R_2 = sX_2}$ which is $\displaystyle \small \mathrm{T_{max}= \frac{KE_2^2}{2X_2} }$

From the above expression, the following conclusions can be drawn
i. Maximum torque is independent of rotor circuit resistance.
ii. Maximum torque varies inversely as standstill reactance of the rotor. Therefore to have maximum torque, standstill reactance (i.e. inductance) of the rotor should be kept as small as possible.
iii. The slip at which the maximum torque occurs depends upon the resistance of the rotor.

Starting torque (S=1):
$\displaystyle \small \mathrm{T\ =K_1 \times \frac{E_2^2R_2}{{R_2^2+(X_2)^2}}}$
$\displaystyle \small \mathrm{T_{st}\ =K_2 \times \frac{R_2}{{R_2^2+(X_2)^2}}}$
where, $\displaystyle \small \mathrm{K_2= K_1 \times E_2^2 }$
Maximum Starting Torque:
$\displaystyle \small \mathrm{\frac{dT_{st}}{dR_2}\ =K_2 \times\left ( \frac{1}{{R_2^2+(X_2)^2}}-\frac{R_2(2R_2)}{{(R_2^2+X_2^2)^2}} \right )=0 }$
$\displaystyle \small \mathrm{R_2=X_2}$
Thus it is clear the maximum starting torque will be when, $\displaystyle \small \mathrm{R_2=X_2}$

Starting torque of a squirrel-cage motor.
The squirrel cage rotor resistance is fixed and small as compared to its reactance which is very large especially at start (because at standstill the frequency of rotor current is equal to that of supply frequency). Hence, the starting current $\displaystyle \small \mathrm{I_2 }$  of the rotor, though very large in magnitude, lags by a very large angle behind $\displaystyle \small \mathrm{E_2 }$  ; consequently 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. Thus such motors are not suitable for applications where these have to be started against heavy loads. Starting torque of a slip ring motor.
In a slip ring motor the torque is increased by improving its power factor by adding external resistance in the rotor circuit from the star. connected rheostat ; as the motor gains speed the rheostat resistance is gradually cut out. This additional resistance, however, increases the rotor impedance and so reduces the rotor current. At first, the effect of improved power factor predominates the current-decreasing effect of impedance, hence 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.