Cascade connection of induction motors

[cnh1995]

In cumulative cascade connection of two induction motors, why does the net speed reduce? Also, how are the torques of the two motors in the same direction?

 

[anorlunda]

Explain more about your cascade.

Are the cascaded electrically or mechanically on the same shaft?

Are two motors connected to provide torque in the same direction, or opposing connections?

How much mechanical load is applied relative to the ratings of the motor?

Edit: a drawing would be very helpful.

 

[cnh1995]

Explain more about your cascade.

Are the cascaded electrically or mechanically on the same shaft?

Are two motors connected to provide torque in the same direction, or opposing connections?

How much mechanical load is applied relative to the ratings of the motor?

Edit: a drawing would be very helpful.

They are mechanically coupled, so both run at the same speed . One motor is driven by 50 Hz mains supply and its rotor emf is given to the stator of the other motor. How does this assembly work? How are the connections made to provide torques in the same direction and in the opposite direction? There are mathematical formulae about this in my book but I want to know what's happening in there physically.

 

[anorlunda]

I never heard of an induction motor pair connected like that. I don't know.

 

[cnh1995]

I never heard of an induction motor pair connected like that. I don't know.

It looks like this.

 

[Babadag]

If the rotor rotates synchronously with stator rotating magnetic field no EMF is induced.

The voltage [EMF] and the frequency in rotor are directly proportional with the slip.

The slip it is the difference between synchronous velocity and the actual velocity ,divided by synchronous speed.

The slip increases linearly [approximate] with the motor torque up to maximum torque.

More torque= more slip= less speed.

At start, in order to accelerate, the motor torque is more than load torque. At steady state motor torque is equal with load torque. Then Tqload=Tq1+Tq2.

In my opinion,the load torque is shared by the two motors according to their torque-speed curve and the supply voltage[see the fig.no.1].

In this example I took motor A of two poles and motor B of 4 poles.

Since p*nsyn/120=f where p=no.of poles, nsyn=synchronous rpm, f=frequency[Hz]

the first motor synchronous speed is 3000 rpm[50 Hz] and for the second 1500 rpm.

The first motor stator supply voltage is constant. The second motor stator voltage depends on speed.The both rotors are rotating at the same speed but both stators are standing.

Curve 1 and 2 represent the torque as speed function for rated voltage and frequency.

Motor A-the brushes are up and the slip rings are short-circuited.

Motor B is squirrel cage rotor.

Curve 3 it is for motor A when the slip rings is connected with an impedance equal with

Zext [impedance equivalent with of stator and rotor of motor B –see fig.no.2]

Curve 4 for motor B when the supply voltage is V2=kr1*s1*V1 where:

kr1=Vstator/Vrotor for motor A at rated supply ;s1=slip of motor A at actual speed.

V1 =the motor A rated voltage and the frequency will be f2= s1.frated [1/3 in this case].

In assemble the synchronous speed will be nsyna*p2=120 *f2=120*s1*frated.

If nsyna[actual]=x and nsyn1=n and 120*frated=p1*n then

x*p2=p1*n*s1 s1=(n-x)/n x*p2=p1*(n-x)

x(p2+p1)=p1*n

nsyna=p1*nsynch1*p1/(p1+p2)=2/(2+4)*nsynch1=nsynch1/3.

 

[Babadag]

Sorry! wrong scale! If the power rated is the same for both motors the less velocity motor

presents more torque. In this case Tq2=~2*Tq1 at rated voltage and frequency.