- #1
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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?
Cascade connection of induction motors
- Thread starter cnh1995
- Start date
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Discussion Overview
The discussion revolves around the cumulative cascade connection of two induction motors, focusing on the mechanics of their operation, the effects on speed and torque, and the physical principles involved. Participants explore theoretical and practical aspects of this configuration, including electrical and mechanical coupling, torque direction, and load considerations.
Discussion Character
- Technical explanation
- Conceptual clarification
- Debate/contested
Main Points Raised
- Some participants inquire about the reasons for reduced net speed in a cascade connection of induction motors and how the torques are aligned in the same direction.
- Questions are raised regarding whether the motors are electrically or mechanically coupled, and how they provide torque in the same or opposing directions.
- One participant describes a scenario where one motor is driven by a 50 Hz supply, and its rotor emf is fed to the stator of another motor, seeking clarification on the physical workings of this setup.
- Another participant expresses unfamiliarity with the concept of induction motors connected in this manner, indicating a lack of prior knowledge on the topic.
- A detailed explanation is provided regarding the relationship between rotor speed, slip, and torque, including mathematical relationships and torque-speed curves for different motor configurations.
- It is noted that if both motors have the same power rating, the motor with lower speed will exhibit higher torque under certain conditions.
Areas of Agreement / Disagreement
Participants express varying levels of understanding and familiarity with the cascade connection of induction motors, leading to a mix of exploratory questions and technical explanations. No consensus is reached on the mechanics or implications of the setup, and multiple viewpoints remain regarding the operation and effects of the connection.
Contextual Notes
Some discussions involve complex mathematical relationships and assumptions regarding motor ratings, slip, and torque-sharing, which may not be fully resolved or universally accepted among participants.
- #2
- 11,326
- 8,755
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.
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.
- #3
- 3,489
- 1,163
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 said:
- #4
- 11,326
- 8,755
I never heard of an induction motor pair connected like that. I don't know.
- #5
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- 1,163
anorlunda said:
It looks like this.
- #6
Babadag
- 614
- 162
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.
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.
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