- #1
- 3,487
- 1,163
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
Click For Summary
SUMMARY
The discussion focuses on the mechanics of a cumulative cascade connection of two induction motors, specifically addressing the reduction in net speed and the direction of torque. The first motor operates at 50 Hz with a synchronous speed of 3000 RPM, while the second motor, with four poles, has a synchronous speed of 1500 RPM. Both motors are mechanically coupled, sharing the load torque according to their torque-speed curves, with the slip being a critical factor in determining performance. The relationship between torque, slip, and speed is mathematically defined, illustrating how increased torque results in increased slip and reduced speed.
PREREQUISITES- Understanding of induction motor principles
- Knowledge of synchronous speed calculations
- Familiarity with torque-speed curves
- Basic electrical engineering concepts related to EMF and slip
- Study the mathematical relationships in induction motors, focusing on slip and torque equations
- Explore the implications of mechanical coupling in multi-motor systems
- Investigate the effects of varying supply frequencies on motor performance
- Learn about the design and operation of different types of induction motors, including squirrel cage and wound rotor types
Electrical engineers, motor control specialists, and anyone involved in the design or optimization of multi-motor systems will benefit from this discussion.
- #2
- 11,326
- 8,754
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
- 3,487
- 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
- 8,754
I never heard of an induction motor pair connected like that. I don't know.
- #5
- 3,487
- 1,163
anorlunda said:
It looks like this.
- #6
Babadag
- 611
- 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.
Attachments
- #7
Similar threads
- · Replies 5 ·
- · Replies 7 ·
- · Replies 18 ·
- · Replies 5 ·
- · Replies 13 ·
- · Replies 3 ·