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slip ring motor resistance


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#1 tahira

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Posted 27 October 2002 - 03:08 PM

Hi sir

I have seen your effective calculations for selection of slip ring motor resistance
I have still some confusion in my mind as I have following relation to calculate the resistance
Rotor coefficient is
K = U (lock rotor voltage) /I (rated current of rotor at rated torque). Root 3

in (ohm /phase)

And rotor resistance is then R = S. K (ohm /phase)

Slip = S =(Ns N / Ns)

Therefore for motor rotor current = 224 A
Rotor V = 579v with rmp = 1477, 4 pole

The value is

K = 1.5

S = (1500-1477/1500) = 0.0153

R = 0.0153 * 1.5 = 0.022 ohm per phase

Plz help me in this regard


An advance thanks to all senior who will guided me.

#2 marke

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Posted 27 October 2002 - 11:00 PM

Hello tahira

Yes you are on the correct track, except that you would not normally select resistors for such a small value of slip. Shorting the rotor will still give some slip due to the resistance of the rotor circuit.
You can use that method of calculation to calculate resistance values for different values of slip delivering maximum torque. i.e. if you want a torqu peak to ocure at half speed during start, then use slip as 0.5
Generally, starting a slip ring motor is achieved by have a number of steps with the resistors selected to provide a maximum torque at different speeds. If you are starting a difficult load, you may have the first resistor stage with a maximum torque at say 20% speed, followed by a stage at 40% speed etc. This will give an average high torque across the whole speed range during start. The greater the number of steps, the smoother the effective speed torque curve is.
Always arrange for the final resistor stage to be close to full speed, i.e. about 10% slip to prevent a large curent transition when shorting out the rotor. - do not short out the rotor at less than 90% speed.

Best regards,

#3 RalphChristie

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Posted 08 November 2002 - 08:18 AM

Hello all

To marke:

I had already asked a similar question in an another forum (eng-tips), and you had also responded to that - but with not enough information (or say not enough info that I can help myself)
Look if my calculculations is correct ( obtained from tahira's post) and correct if neccessary.

Vrotor = 288V
Irotor = 113A
Speed = 575rpm

k = Vrotor / Irotor * root3
= 288 / 113 * root3
= 1.471 Ohm/ph (Is this internal resistance of rotor?)

Let say we want a speed of 20%

slip = (Ns - N) / Ns
= (575 - 0.2*575) / 575
= 0.8

Rtotal = k*slip
= 1.471*0.8
= 1.177 Ohm/ph

Is this last value the value of the External resistors?

Like I have said in my eng-tips' post, The problem is not on "full speed", but to obtain slower speed control. If I am just sure how to determine the size of the external resistors (also to obtain a certain speed), it would help me a lot.

Thanks
Ralph

#4 marke

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Posted 08 November 2002 - 09:38 AM

Hello Ralph

These slip ring starters really cause some major head achs for some people and there are many misconceptions about them.
Firstly, the rotor resistors determine the speed or slip where the maximum torque occurs. This does not actually equate to the operating speed of the motor.
If you consider the torque curve of a standard cage motor, you have a maximum torque at say 5% slip and it may be 250% FLT. These figures vary from motor to motor. It we take that curve and say that that is equivilent to our slip ring motor with the rings shorted, then as we add resistance, we are shifting that maximium torque point to lower speeds.
If we wish to achieve maximum torque from zero speed to full speed for a difficult load, then we would use a number of stages of rotor resistance. The first stage may have a torque maximum at say 10% speed (90% slip) If we apply voltage and load to the motor, it will accelerate until the torque developed by the motor equals the load torque. If we had a maximum torque of 250% and a load torque of 125% (loaded conveyor) then the motor may accelerate to about 50% speed before the motor ran out of torque. At that time, it will continue to run at that speed until either the load torque changes, or the motor torque changes. Let us now reduce the resistance in the rotor to provide a maximum torque at say 70% speed. The motor will accelerate again until it runs out of surplus torque. This may be at about 90% speed. We now switch in the last resistance stage with a torque peak at 90% and the motor accelerates through to full speed and we short out the rotor.
The important point is that the rotor resistance determines the speed at which the maximum torque occurs. The running speed is determined by the intersect of the motor torque curve and the load torque curve. If either change, the motor speed will change also.

The rotor resistance will also reduce the starting current of the slip ring motor. If you apply full voltage to a slip ring motor with shorted rings, the Locked Rotor Current is very high, far higher than a standard cage motor. It is important that you short the rings only when the motor is very close to full speed. It is normall for the final stage resistance to be designe for a torque maximum at about 10% slip to prevent a very high current step when shorting the rings.
A three stage, high torque starter may be designed for maximum torque to be at say 20% speed, 60 % speed and 90% speed. You can plot out the curves and determine exactly what will happen, how the speeds will be affected by changing resistances.

Does this help?
Perhaps I need to draw some diagrams to further illustrate the point. - don't be afraid to ask further questions,ther are others with the same questions.
Best regards,

#5 Joe

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Posted 15 January 2003 - 06:43 AM

Dear Mark,

So glad to find this website.

I have a question;


Please allow me to explain my problem:

I have a slip ring motor with the following specs:
Primary: 380Vac 675A
Secondary: 600Vac 380A
Power Factor: 0.89
Speed: 735 RPM
Power: 370kWatt


I need to design a 4 stage start up circuit for the rotor.
My resistor values are as follows:
Stage : Start-up Resistance per Phase: I take all my resistors per phase and total them = 0.75 ohm
Stage 1 Contactor engaged: 0.35 ohm
Stage 2 Contactor engaged: 0.15 ohm
Stage 3 Contactor engaged: 0.05 ohm
Stage 4 Contactor engaged: Rotor Shorted

How do I calculate the current per phase in the rotor for each stage? I need to select cable and contactor size for each stage.(Cable that runs between the contactors and resistors)

Please help me.

Awaiting reply.

Joe
;p;

#6 marke

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Posted 15 January 2003 - 06:01 PM

Hello Johan
The actual current flowing is a function of the stator / Rotor turns ratio, the rotor resistance and the rotor speed.
Typically, the rotor current is going to be in the order of twice the full load rotor current during start, however in the case of the resistors, this is for a short period of time for the intermediate stages. I would normally size the cable for the full load full speed current, i.e.380A and you could undersize the cable on the intermediate stages by an amount depending on the period of time each stage was carrying current. This is of course dependant on the load inertia etc.
Someone out there may have better rule of thumb than that, but you will be safe using that rating.

Best regards,




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