Approximate Slip Ring %Starting Current?

[nightfox1925]

As a rule of the thumb, what is the approximate percent starting current that can gained out from a slip ring motor started by step resistances at the rotor circuit for a high inertial load?

GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata

 

[electricpete]

I'm not sure of the direct answer to your question in terms of a gain (reduction?) in starting current, but I think the big advantage will be in making the motor come up to speed quicker.

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[nightfox1925]

I am under the impression that starting a slip ring motor will provide a low starting current at a very high torque. I am currently sizing my power cables for a slip ring motor and I'm ok with my ampacity and the steady state voltage drop (limited at 3% in my case). What bothers my is the starting voltage drop limitation I have to set on my power cable size since I believe that the starting current I get for starting a slip ring motor is less compared to a normal induction motor started at DOL. I have a dilemna of how much in percent of full load current a slip ring motor typically draws during starting. Any idea?

GO PLACIDLY, AMIDST THE NOISE AND HASTE-Desiderata

 

[electricpete]

I am not very familiar with this subject.

As has recently been discussed, the value of slip ring resistance that produces max torque at a given slip is the resistance which matches the total inductive reactance at that slip.

So if the resistor were initially optimized for the standstill slip=1 case, then we have R=Xlrc
|Z| = sqrt(R^2+2Xlrc^2)=sqrt(2)*Xlrc
So it appears initial current would be a factor of sqrt(2) less than if there were no resistance at all.

Just my thoughts. You better check it with someone else.

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[LionelHutz]

If you pick the right steps and enough steps and control the steps correctly you can keep the current below about 150%. If you don't, then it will be higher.

The current is completely dependent on the resistance connected to the rotor. And, at each speed, the amount of resistance required to end up with a certain current changes. So, it's not possible to just give a number. It would require knowing the motor details, load details and the resistor control details.

 

[aolalde]

The induction wound rotor motor performs like a transformer with variable voltage and frequency on the rotor side.
The rotor voltage is proportional to the Stator (primary) Voltage, the ratio of winding turns rotor/stator (r) and the rotor slip (s) V2= r*V1*s while the frequency on the rotor (F2) changes with the rotor slip F2 = s*F1
the slip s=(RPMsynch – RPM rotor)/ RPMsynch
If the rotor is kept stand-still the slip s=1 and when the rotor is at synchronous speed s=0. This means the rotor voltage reduces as the rotor accelerates and the rotor frequency drops proportional to the increase in motor speed.

The external resistance, connected to the rotor circuit modifies (reduces) the rotor current and the phase angle with the induced voltage V2. For high slip (s= 1 to 0.5) the motor torque is almost proportional to the current. That means 100% current will develop 100% torque, 150% current will develop 150% torque, etc. Then, tailoring of the rotor external resistances will depend on your load application and torque requirements.

as a guideline the external resistance (Re) per phase required to develop 100% Torque at stand still and 100% locked current is around:
Re = 0.95V2n / 1.732 *I2n (This is for wye connection )
V2n = Rotor nameplate voltage, I2n Rotor nameplate current at full load (100% load)

 

[waross]

The normal procedure for squirrel-cage motors is to size motor conductors at 125% of full load current.
In the NEC a table is used to determine full load current while in Canada under the CEC the nameplate current is used.
Motor conductors are not generally sized for starting voltage drop.
Are you concerned with plant voltage drop on the service conductors?
Many applications of wound rotor motors such as cranes can run at much reduced speeds continually without exceeding rated current. wound rotor motors are occasionally used with permanently connected, unswitched, resistors to give the motor low starting current and high slip characteristics. This application may be found driving shears and punch presses that utilize large flywheels to provide the high instantaneous energy required.
The starting current in your instance will be much less than the normal 300% to 600% of a squirrel-cage induction motor.
The actual value will depend on the value of the starting resistors and the timing of the steps.
If this were my problem I would make every effort to determine the starting current from the supplier or manufacturer.
If I was unsuccessful I would use 200% as conservative and 125% to 150% as normally acceptable.
And remember, there is such a wide variety of conditions and applications for wound rotor motors that were I to be in your plant, viewing the installation first hand, my answer may be different.
respectfully