miki83 Posted May 28, 2011 Report Share Posted May 28, 2011 Hi! Need some help again if anyone has some spare time (and knowledge of course:) Few days ago I was installing a couple of new control cabinets with wye-delta starters on two 30-40 years old piston compressors, one of them having a squirrel cage motor, and the other one fitted with a slip-ring one (both motors 45 kW, completely equal compressors). The deal was that the customer is to replace the slip-ring motor with a squirrel cage, but instead they only bridged out the rotor. Now the first compressor with the squirrel cage motor started without any trouble. The second one had the wrong direction during first start, we switched the phases, started again, but this time the motor only hums, draws around 200 Amps and does not move at all. Now i know that in this case the slip-ring motor will have reduced starting torque, but does anyone know if this starting torque is even smaller than the one on the squirrel cage motor? Or generally why the motor won't start? All the phases are ok (the cabinet has a phase control relay). Thanks in advance! Miki. Link to comment Share on other sites More sharing options...
marke Posted May 28, 2011 Report Share Posted May 28, 2011 When you short out the slip rings on a slip ring starter and then try to start it, you will have problems because the locked rotor current will be very high and the locked rotor torque will be very low.see http://www.lmphotonics.com/slipring.htm and http://www.lmphotonics.com/star_delta.htm The start torque will be much less than a standard induction motor and the locked rotor current will be much higher.If you leave the final stage resistors from the secondary resistance starter, in the rotor circuit while starting, you may get it going, then use a timer to control a contactor bridging out those resistors. Best regards,Mark Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
miki83 Posted May 29, 2011 Author Report Share Posted May 29, 2011 Hi Mark. Thank You very much for the reply and for the links. I saw the articles already and suspected that much, but wasn't 100% sure that a bridged-out slip ring motor has much lower locked-rotor torque than a standard one. That pretty much explains it. Thanks again and best regards, Miki. Link to comment Share on other sites More sharing options...
marke Posted May 29, 2011 Report Share Posted May 29, 2011 Hello Miki The torque produced at any speed is a function of the power dissipated in the rotor circuit at that speed.If the resistance is zero, there will be no torque.The current in the rotor circuit is dependent on the impedance of the rotor circuit which changes with speed as the frequency of the current flowing depends on the slip.In analysis, that major difference between the slip ring motor and a squirrel cage motor, is the turns ratio. This impacts on the equivilent rotor resitance. Best regards,Mark. Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
miki83 Posted May 30, 2011 Author Report Share Posted May 30, 2011 Hi Mark. Thank You very much for yet another clarification. I've been going through Your first reply for some time now. As I said the motor drew around 200-250 (which is about 200-300% of the rated current) Amps and wasn't moving, and based on Your first post and the links the current should have been much higher if I'm not mistaking. Do You have any idea why the current stayed relatively low in this case? Sorry for bothering You again with (I suppose) trivial questions Best Regards, Miki. Link to comment Share on other sites More sharing options...
marke Posted May 30, 2011 Report Share Posted May 30, 2011 Hi Miki the start current is limited by the effective impedance of the motor at different speeds and this is made up of the supply impedance, the stator impedance and the reflected rotor impedance. Normally, it would be the rotor impedance that is dominant, but if there is a high leakage reactance between rotor and stator, that can have a major impact.The torque developed is a function of the power dissipated in the rotor circuit. If the rotor resistance is low compared to the other reactances, then the current can be low and the torque also low. Did you check that the start current was equal on each phase? Best regards,Mark. Mark Empson | administratorSkype Contact = markempson | phone +64 274 363 067LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters Link to comment Share on other sites More sharing options...
miki83 Posted May 30, 2011 Author Report Share Posted May 30, 2011 Hi Mark. I didn't have the time to check the current on each of the phases. The cabinet has a single ammeter measuring the current on one of the phases. We had a fire in the junction box during our first start in the right direction, which took about 5 sec. from the moment the contactors switched in to ignite (the motor didn't rotate), so we didn't want to risk afterwards . We tried the motor three more times after the burnt cables in the junction box have been fixed, but only for a couple of seconds before we switch it off once we notice the rotor's not turning. The windings seem fine (they checked the resistance). Thanks for Your time. It seems I'll just have to wait and see what they'll come up with. Hopefully they won't destroy anything in the process Best regards, Miki. Link to comment Share on other sites More sharing options...
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