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energy saving through current balancing

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COMAPRO

Industrial
May 17, 2003
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Anybody there can help me finding out percentage wise how much can one save in KW when currents are balance in a 3 phase system with an example ?

I read somwhere that if in a electrical 3 phase motor phase currents are unbalanced , one can try changing the phase cables at the motor terminals , pushing one to the next one and so on , in order to prevent motor turn rotation and measure again to see if they are not any more unbalanced . is it true ? where can I read about it ?
finally if in the secondary side of a bank transformer one measures 210 volts and the 3 phase motor are rated 230 volts , if I raise the voltage through the taps to 230volts , would I save energy ? if yes , how much , 2% or what ?
Thank you for your coming help on this matter.
 
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Hello CAMAPRO,

Will not attempt to answer all your questions here but can provide some comment to the questions raised in the second paragraph of your post.

Yes, the level of phase imbalance can in some cases be reduced by 'rolling the phases'. This is particularly true in cases where variations in supply voltage and motor impedance work together (add) creating a larger phase imbalance. The principle behind rolling the phases is to have these variations subracting from each other, thereby minimising imbalance.

I have tried 'rolling the phases' on several occassions now and have been pleasantly surprised with the outcome each time. This is certainly an Eng-tip worth recommending to others.

Regards,
GGOSS
 
GGOSS and Stevekw , thank you very much for promtly answers. -that is one of the things I like from this forum but there is no way in a formula to calculate the percentage of savings if one balance currents and or raise the voltage to the required one from the motor plate ?
 

Phase rolling is worth a try, but the influence of various combinations of loads on the same bus bears a degree of scrutiny. The procedure is most effective in one-motor/one-transformer situations.
 
Suggestion: The raised voltages to 230V will mean higher losses in the magnetizing branches of motors and transformers. Also, various loads, e.g. coils, heaters, etc. will consume more electricity. However, there will be some savings in the power distribution conductors RI**2 watt loss since the current I will be lower. The increased Watt consumption of some loads can surpass the RI**2 Watt loss savings. For example, a heater will have the Watt increase according to:
Pincrease=Pwatt230V-Pwatt210=(230**2 - 210**2)/R=8800/R, Watts
Now, if R=10Ohms then
Pwatt210=4410 Watts
Pwatt230=5290 Watts
Pincrease=880 Watts
 
jbartos , I am very confused with your statement "The raised voltages to 230V will mean higher losses in the magnetizing branches of motors and transformers."because I always thought that one should calibrate the taps of the transformer in such a way that the secondary voltage would be at least 230 volts , which is the voltage needed for the motors , if the voltage is less the current will be higher and the power losses will be also be higher as you mentioned it.that heaters "will consume more electricity" I do not see how because current value will be lower.
 
Hi,
*Voltage increasing raises losses in core due to material histeresis. This part of losses is quite small and not dependable from motor load.
*Voltage decreases puts down torque of asyncronous motor
in power 2.
*If load has M=const not function of speed (conveyor),
then rotor slips and current increases dramatically (active and inductive).
*If load has Vent shape - low voltage may bring motor to unstable part of mechanical characteristics - really bad.
*Non symmetric voltage bribgs odd harmonics to magnetic field and creates heat, decreases useful torque.

Thanks,

Vladimir.
 
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