Categories: Electrical Engineering

Effect Of Change In Rotor Resistance On Torque

It is shown that in slip ring induction motor, externally resistance can be added in the rotor. Let us see the effect of change in rotor resistance on the torque produced.

Let R₂= Rotor resistance per phase

Corresponding torque, T α (s E₂² R₂)/√(R₂² +(s X₂)²)

Now externally resistance is added in each phase of rotor through slip rings.

Let R₂‘ = New rotor resistance per phase

Corresponding torque T‘ α (s E₂² R₂‘ )/√(R₂‘² +(s X₂)²)

Similarly the starting torque at s = 1 for R₂and R₂‘ can be written as

T_st α (E₂² R₂ )/√(R₂² +(X₂)²)

and T‘st α (E₂² R‘₂ )/√(R‘₂² +(X₂)²)

Maximum torque T_m α (E₂²)/(2X₂)

Key Point : It can be observed that T_m is independent of R₂ hence whatever may be the rotor resistance, maximum torque produced never change but the slip and speed at which it occurs depends on

R₂.
For R₂, s_m = R₂/X₂ where T_m occurs
For R₂‘, s_m‘ = R₂‘/X₂‘ where same T_m occurs

As R₂‘ > R₂, the slip s_m‘ > s_m. Due to this, we get a new torque-slip characteristics for rotor resistance . This new characteristics is parallel to the characteristics for with same but T_m occurring at s_m‘. The effect of change in rotor resistance on torque-slip characteristics shown in the Fig. 1.

It can be seen that the starting torque T‘st for R₂‘ is more than Tst for R₂. Thus by changing rotor resistance the starting torque can be controlled.

If now resistance is further added to rotor to get resistance as R₂‘ and so on, it can be seen that T_m remains same but slip at which it occurs increases to s_m‘ and so on. Similarly starting torque also increases to T‘st and so on.

Fig. 1 Effect of rotor resistance on torque-slip curves

If maximum torque Tm is required at start then s_m = 1 as at start slip is always unity, so
s_m= R₂/X₂ = 1
R₂ = X₂ Condition for getting Tst = T_m
Key Point : Thus by adding external resistance to rotor till it becomes equal to X₂, the maximum torque can be achieved at start.

It is represented by point A in the Fig. 1.

If such high resistance is kept permanently in the circuit, there will be large copper losses (I² R) and hence efficiency of the motor will be very poor. Hence such added resistance is cut-off gradually and finally removed from the rotor circuit, in the normal running condition of the motor. So this method is used in practice to achieve higher starting torque hence resistance in rotor is added only at start.

Thus good performance at start and in the running condition is ensured.

Key Point : This is possible only in case of slip type of induction motor as in squirrel cage due to short circuited rotor, extra rotor resistance can not be added.

Example : Rotor resistance and standstill reactance per phase of a 3 phase induction motor are 0.04 Ω and 0.2 Ω respectively. What should be the external resistance required at start in rotor circuit to obtain.
i) maximum torque at start ii) 50% of maximum torque at start.

Solution :

R₂ = 0.04 Ω,      X₂ = 0.2 Ω
i)    For T_m = T_st ,     s_m = R₂‘/X₂ = 1
.^..        R₂‘ = X₂ = 0.2
Let R_ex = external resistance required in rotor.

R₂‘ = R₂ + R_ex
.^..      R_ex = R₂‘ – R₂ = 0.2 – 0.04 = 0.16 Ω per phase
ii)     For       T_st = 0.5 T_m,
Now       T_m= (k E₂²)/(2 X₂) and
T_st = (k E₂² R₂)/(R₂² + X₂²)

But at start, external resistance R_ex is added. So new value of rotor resistance is say R₂‘.
R₂‘ = R₂ + R_ex
.^.. T_st = (k E₂² R₂‘)/(R₂‘² + X₂²) with added resistance
but T_st = 0.5T_m required.

Substituting expressions of T_st and T_m, we get
(k E₂² R₂‘)/(R₂‘² + X₂²) = 0.5 (k E₂²)/ (2X₂)
.^..                   4 R₂‘ X₂= (R₂‘² + X₂²)
.^..                 (R₂‘²) – 4 x 0.2 x R₂‘ + 0.2² = 0
.^..                    (R₂‘²) – 0.8 R₂‘ + 0.04 = 0
.^..                    R₂‘ = {0.8 + √(0.8² – 4 x 0.04)}/2
.^..                  R₂‘ = 0.0535 , 0.7464 Ω