What is the Relation between torque and rotor power factor ? -

In this article we will study about the Relation between torque and rotor power factor and also read different types of torque like starting torque, starting torque of a squirrel-cage motor, Condition for maximum starting torque in induction motor, Torque under running condition, Full load torque and maximum torque, Relation between torque and slip. You will get to study all this in this article.

The relation between torque and rotor power factor

In case of induction motor the torque is proportional to the product of flux per stator pole and the rotor current. and power factor of the rotor. Therefore, T is proportional to ΦI₂ cosΦ₂ or T = kΦI₂ cosΦ_2.

Where, I₂ = rotor current at standstill, Φ₂ = is the angle between rotor E.M.F. and rotor current, k = is a constant. E.M.F. at standstill is denoting by E_2.

At standstill condition E₂ is proportional to Φ, therefore T ∝ E₂I₂ cosΦ₂ or T = k₁E₂I₂cosΦ_2. The effect of rotor power factor on rotor torque is given for various value of Φ_2.

From the above relation it is conclude that if I₂ increase cos Φ₂ decrease and also the torque decrease and vice versa.

( 1 ) If the Rotor is Assumed to the Non-inductive or Φ₂ = 0

In this case, the rotor current I₂ is in phase with E.M.F. induce in the rotor E_2.The power factor cos Φ = become 1 hence, the torque is always +ve and that is unidirectional. and this +ve torque is know as forward torque.

( 2 ) Rotor Assumed to the Inductive

Φ₂ = tan^-1 X₂/R₂ , In this case rotor current I₂ lags behind E₂ by an angle Φ₂ = tan^-1 X₂/R_2.In this case a portion of torque of -ve that is it is revers. The -ve torque is also know as backward torque.

Hence, the total torque is the difference of forward torque and backward torque. In this case the torque is bidirectional.

Φ₂ = 90° in this case the total torque is zero. Because the backward torque and forward torque becomes equal and opposite.

What is the Relation between torque and rotor power factor ?

Starting torque of induction motor

The starting torque is that torque produce in the motor or develop by the motor at the instant of starting. It is denoted by the letter T_st .

Let E₂ = rotor e.m.f. at standstill per phase

I₂ = Rotor current at standstill

R₂ = rotor resistance at standstill

X₂ = rotor reactance at standstill

Z₂ = rotor impedance at standstill

Φ₂ = Phase angle between E₂ and I₂

$Z_{2}=\sqrt{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}} What is the Relation between torque and rotor power factor ?$

$I_{2}= \frac{V_{2}}{Z_{2}}=\frac{E_{2}}{Z^{2}} What is the Relation between torque and rotor power factor ?$

$I_{2}= \frac{E_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}} What is the Relation between torque and rotor power factor ?$

So the Standstill or starting torque is T_st = K₁ E₂ I₂ Cos Φ₂

$T_{st}=K_{1}E_{2}\times \frac{E_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}}Cos\Phi _{2}$

$Cos \Phi _{2}= \frac{R_{2}}{Z_{2}}=\frac{R_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}}$

$T_{st}= K_{1}E_{2}\times \frac{E_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}}\times \frac{R_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}}$

$T_{st}= K_{1}\frac{\left ( E_{2} \right )^{2}R_{2}}{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}} What is the Relation between torque and rotor power factor ?$

We know , K₁ = 3 / 2𝛑N_s

Therefore,

$T_{st}= \frac{3}{2\pi N_{s}}\times \frac{\left ( E_{2} \right )^{2}R_{2}}{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}} What is the Relation between torque and rotor power factor ?$

Read also this vector diagram of loaded alternator

Some numerical of starting torque – Relation between torque and rotor power factor

Q-1. A 3-phase induction motor having a star-connected rotor has an induced E.M.F. of 80 volts between slip-ring at standstill on open-circuit. and The rotor has a resistance and reactance per phase of 1 Ω and 4 Ω respectively. Calculate current per phase and power factor when (i) slip rings are short-circuited (ii) slip-rings are connected to a star-connected rheostat of 3 Ω per phase.

Solution :

( a ) Standstill e.m.f. / rotor phase ( E₂ ) = 80/√3 = 46.18 volt

Impedance phase ( Z₂ ) = $\sqrt{R_{2}^{2}+X_{2}^{2}}$

= $\sqrt{1^{2}+4^{2}}$

= 4.12 ohm

Current per phase ( I₂ ) = e.m.f. per phase / Impedance per phase

= 46.18 / 4.12

= 11.2 A

Power factor ( Cos Φ₂ )

$\frac{R_{2}}{\sqrt{R_{2}^{2}+X_{2}^{2}}}= \frac{1}{\sqrt{1^{2}+4^{2}}}$

= 0.242

( b ) Rotor resistance / phase = 1+3 = 4 ohm

Rotor impedance / phase = $\sqrt{R_{2}^{2}+X_{2}^{2}} What is the Relation between torque and rotor power factor ?$

= $\sqrt{4^{2}+4^{2}}$

= 5.65 ohm

Current per phase ( I₂ ) = e.m.f. per phase / Impedance per phase

= 46.18 / 5.65

= 8.17 A

Power factor ( Cos Φ₂ )

$\frac{R_{2}}{\sqrt{R_{2}^{2}+X_{2}^{2}}}=\frac{4}{\sqrt{4^{2}+4^{2}}}$

= 0.70

Starting torque of a squirrel cage induction motor

The resistance of 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 current is equal to the supply frequency.

Hence, the starting current I₂ of the rotor though very large in magnitude, lags by very large angle behind E_2, with the result 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 motor are not useful where the motor has to start against heavy load.

Starting torque of slip ring induction motor.(What is the Relation between torque and rotor power factor ?)

Starting torque of slip ring induction motor

The starting torque of slip ring induction motor is increase by improving power factor by addition resistance in the rotor circuit from the star-connected the rheostat, the rheostat resistance being progressively cut out as the motor gather speed.

Addition of external resistance increase the rotor impedance and so reduce the rotor current. The effect of improve power factor predominates the current decreasing effect of impedance .Hence, starting torque is increased.

Condition for maximum starting torque in induction motor

T_st = k₂R₂ / (R₂)² + (X₂)²

$\frac{dT_{st}}{dR_{2}}=k_{2}\frac{d}{dR_{2}}\left ( \frac{R_{2}}{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}} \right )$

$\frac{dT_{st}}{dR_{2}}=k_{2}\left [ \frac{1}{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}-\frac{R_{2}\left ( 2R_{2} \right )}{\left ( \left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2} \right )^{2}} \right ]$

So for maximum condition dT_st / dR₂ = 0

$\frac{1}{\left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2}}=\frac{R_{2}\left ( 2R_{2} \right )}{\left ( \left ( R_{2} \right )^{2}+\left ( X_{2} \right )^{2} \right )^{2}}$

( R₂)² + ( X₂)² = 2( R₂)²

( X₂)² = ( R₂)²

Click for what is a 3 phase induction motor

Rotor E.M.F. and reactance under running condition

Let E₂ = Standstill rotor induced E.M.F. per phase

X₂ = Standstill rotor reactance per phase

f₂ = Rotor current frequency at standstill

When rotor is stationary that is s = 1, the frequency of rotor E.M.F. is the same as that of the stator supply frequency the value of E.M.F. induced in the rotor at standstill is maximum.

Because the relative speed between the rotor and the revolving stator flux is maximum. In fact, the motor is equivalent to a 3-Φ transformer with short circuit rotating secondary.

When rotor start running the relative speed between it and the rotating stator flux is decrease. Hence, the rotor induce E.M.F. which is directly proportional to this relative speed is also decrease.

Hence, for a slip ‘s’ the rotor induced E.M.F. will be ‘s’ time the induced E.M.F. at standstill. Therefore, under running condition E_r = sE_2.The frequency of the induced E.M.F. will be f_r = sf_2.

Due to decrease in frequency of the rotor E.M.F. the rotor reactor also decrease X_r = sX₂. Where E_r and X_r are rotor E.M.F. and the reactance under running condition.

Torque under running condition – Relation between torque and rotor power factor

T_r ∝ E_r I_r CosΦ₂ or T ∝ Φ I_r Cos Φ₂

Where, E_r = is the rotor e.m.f. per phase under running condition

I_r = is the rotor current per phase under running condition

E_r = sE₂

$I_{r}=\frac{E_{r}}{Z_{r}}=\frac{sE_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( sX_{2} \right )^{2}}}$

$Cos \Phi _{2}= \frac{R_{2}}{Z_{2}}=\frac{R_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( sX_{2} \right )^{2}}}$

T_r ∝ E_r I_r Cos Φ₂

$T_{r}\propto sE_{2}\times \frac{sE_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( sX_{2} \right )^{2}}}\times \frac{R_{2}}{\sqrt{\left ( R_{2} \right )^{2}+\left ( sX_{2} \right )^{2}}}$

So we know that k₁ is constant and its value can be prove = 3 / 2𝛑N_s . Hence in that case the expression for torque is

$T_{r}= \frac{3}{2\pi N_{s}}\times \frac{sE_{2}^{2}R_{2}}{\left ( R_{2} \right )^{2}+\left ( sX_{2} \right )^{2}}$

$T_{r}= \frac{3}{2\pi N_{s}}\times \frac{sE_{2}^{2}R_{2}}{Z_{r^{2}}}$

At standstill when s = 1

$T_{st}= \frac{3}{2\pi N_{s}}\times\frac{E_{2}^{2}R_{2}}{R_{2}^{2}+X_{2}^{2}} What is the Relation between torque and rotor power factor ?$

Relation between torque and slip in induction motor

What is the Relation between torque and rotor power factor ?

So , T = kΦsE₂R₂ / (R₂)² + (sX₂)²

When s = 0 and T = 0

Normal speed is directly proportional to slip and the curve start from the origin 0. At normal speed, close to synchronism the term ‘ sX₂ ‘ is small and hence, negligible with respect to R_2.Therefore T ∝ s/R₂ or T = s [ If R₂ is constant].

Hence, for low value of slip the torque /slip curve is approximately is straight line. As slip increases the torque also increases and becomes maximum when s = R₂/X_2.

This torque is also know as pull out torque or maximum torque or breakdown torque. denoted by T_b or stalling torque. As the slip further increases with further increase in motor load then R₂ becomes negligible as compared to sX_2.

High speed condition

At high speed or at last speed R₂ is small and hence negligible with respect to sX_2.Therefore T ∝ 1/sX₂ if X₂ is constant. therefore T ∝ 1/s. ( Torque inversely proportional to slip.)

Hence, the torque slip curve is a rectangular hyperbola. So, we see that beyond the point of maximum torque any further increase in motor load results in decrease of torque developed by the motor.

The result is that motor slow down and eventually stop the circuit breaker will be tripped open if the circuit has been protected.

Click for what is resistance? in electrical

Full load torque and maximum torque

T_f ∝ S_f R₂ / ( R₂ )² + ( s_f X₂ )² ———- (1)

T_max ∝ 1 / 2×X₂ ———- (2)

$\frac{T_{f}}{T_{max}}= \frac{\frac{s_{f}R_{2}}{\left ( R_{2} \right )^{2}+\left ( s_{f}X_{2} \right )^{2}}}{\frac{1}{2\times X_{2}}}=\frac{2s_{f}R_{2}X_{2}}{\left ( R_{2} \right )^{2}+\left ( s_{f}X_{2} \right )^{2}}$

Now dividing both numerator and denominator by ( X₂)² then we get

$\frac{T_{f}}{T_{max}}= \frac{2s_{f}R_{2}X_{2}/X_{2}^{2}}{R_{2}^{2}+(s_{f}X_{2})^{2}/X_{2}^{2}}=\frac{2s_{f}R_{2}/X_{2}}{\left ( R_{2}/X_{2} \right )^{2}+s^{2}_{f}}$

Where, R₂ / X₂ = a

and ‘a’ is the resistance per reactance per standstill

So , T_f / T_max = 2s_fa / a² + s²_f

Starting torque and maximum torque – Relation between torque and rotor power factor

T_st ∝ R₂ / ( R₂ )² +( X₂ )²

T_max ∝ 1 / 2X₂

So , T_st / T_max = 2R₂ X₂ / ( R₂ )² + ( X₂ )²

Now dividing both numerator and denominator by ( X₂ )²

$\frac{T_{st}}{T_{max}}= \frac{2R_{2}X_{2}/X_{2}^{2}}{R_{2}^{2}+X_{2}^{2}/X_{2}^{2}}=\frac{2R_{2}/X_{2}}{\left ( R_{2}/X_{2} \right )^{2}+1}$

Where , a = R₂ / X₂

‘a’ is the rotor resistance per standstill per phase

So , T_st / T_max = 2a / a² + 1