3 phase, 2 speed motor load
- Thread starter Doops
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What do you think it would be?
Can’t figure out if this is a real question or a test question.
kwired
Electron manager
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- NE Nebraska
should be 1.4 kW at high speed and 1.1 kW at low speed.
Efficiency should change input power though. Motor nameplate power rating should be output rating. Amps and voltage ratings however are what it is expecting for input to product the output rating
Okay more explanation.
Trying to size a single to 3phase converter for this motor. Supplier says to add the two speed loads together and double it to get the size of converter needed. = 7HP rated converter. Intuitively this is seeming like a very large converter for that small a motor.
I have two identical motors ( same model no and same rating from two different manufacturers) .
The one installed in the mixer has a nameplate amperage of 4.5 A at 1.4 kw and 5.7 A at 1.1 kw; cos*.75 and .67 respectively. This works out by the formula.
However, the overload set from the factory is only at 9 amps. Also because the the higher speed 1.4 kw is less amperage than the lower speed, I am guessing the two loads are not added together, but the windings are connected in a different sequence. this would mean the supplier of the phase converter is wrong. I asked the supplier of the mixer and the tech told me it cannot be connected to a phase converter. No explanation. There is nothing unusual about the controls that would indicate this. ( a 208/24volt transformer runs everything).
But I have a spare motor also 1.4kw/1.1 kw 208 v, cos 0.99 /0.93 same speeds that has amperage stamped at13.5/ 9.6 .
Something doesn't compute for me. I need to size the phase converter correctly before ordering.
I have no opportunity to measure this load with proper 3 phase power.
Can you make sense of this? Thanks.
Trying to size a single to 3phase converter for this motor. Supplier says to add the two speed loads together and double it to get the size of converter needed. = 7HP rated converter. Intuitively this is seeming like a very large converter for that small a motor.
I have two identical motors ( same model no and same rating from two different manufacturers) .
The one installed in the mixer has a nameplate amperage of 4.5 A at 1.4 kw and 5.7 A at 1.1 kw; cos*.75 and .67 respectively. This works out by the formula.
However, the overload set from the factory is only at 9 amps. Also because the the higher speed 1.4 kw is less amperage than the lower speed, I am guessing the two loads are not added together, but the windings are connected in a different sequence. this would mean the supplier of the phase converter is wrong. I asked the supplier of the mixer and the tech told me it cannot be connected to a phase converter. No explanation. There is nothing unusual about the controls that would indicate this. ( a 208/24volt transformer runs everything).
But I have a spare motor also 1.4kw/1.1 kw 208 v, cos 0.99 /0.93 same speeds that has amperage stamped at13.5/ 9.6 .
Something doesn't compute for me. I need to size the phase converter correctly before ordering.
I have no opportunity to measure this load with proper 3 phase power.
Can you make sense of this? Thanks.
kwired
Electron manager
- Location
- NE Nebraska
What is the driven load?
What type of phase converter?
"Static" phase converters (the ones that just use capacitors to create a phase shift) are best suited for a constant load type of application. If the load varies they don't work as well at balancing currents in the motor. They often have a boost function for starting the motor, but once running like to see a load that doesn't change much and generally only can be used for a single motor.
Rotary converters work good when you have multiple motors being driven. They still tend to unbalance current in the motor though, and often is good idea to derate the motor, especially if it will run at rated load very much, that imbalance current will overheat it if running at full rating, but then proper overload protection should still shut it down also, but the damage can add up over time.
Use of a VFD for phase converter is probably the best and most energy efficient option most cases. But if only using two input lines instead of all three you must derate the drive, general rule is to double the drive capacity over the motor capacity, though techically it kind of should really need to be 1.732 times the motor current rating.
What type of phase converter?
"Static" phase converters (the ones that just use capacitors to create a phase shift) are best suited for a constant load type of application. If the load varies they don't work as well at balancing currents in the motor. They often have a boost function for starting the motor, but once running like to see a load that doesn't change much and generally only can be used for a single motor.
Rotary converters work good when you have multiple motors being driven. They still tend to unbalance current in the motor though, and often is good idea to derate the motor, especially if it will run at rated load very much, that imbalance current will overheat it if running at full rating, but then proper overload protection should still shut it down also, but the damage can add up over time.
Use of a VFD for phase converter is probably the best and most energy efficient option most cases. But if only using two input lines instead of all three you must derate the drive, general rule is to double the drive capacity over the motor capacity, though techically it kind of should really need to be 1.732 times the motor current rating.
Thanks. Sorry for not being more concise.
It is a rotary converter, the load is a dough mixer, 2 speed. 208v 3 ph. Made in Italy. Running at variable loads but perhaps 30 % of the time at full load.
We have come to the conclusions you have indicated, but the bottom line question is what is the actual load of this motor.
The argument is whether the rated loads of 5.7 amps on speed one and 4.5 amps on speed two are added for a full load of 10.2 amps or is 5.7 the highest amperage this mixer should draw under normal use.
The secondary question is which motor would draw less power, ( there are two, with identical power ratings), the one with S.F. cos .75/ .67 or the one with S.F cos .91/.93 ?
Thanks in advance for your response.
It is a rotary converter, the load is a dough mixer, 2 speed. 208v 3 ph. Made in Italy. Running at variable loads but perhaps 30 % of the time at full load.
We have come to the conclusions you have indicated, but the bottom line question is what is the actual load of this motor.
The argument is whether the rated loads of 5.7 amps on speed one and 4.5 amps on speed two are added for a full load of 10.2 amps or is 5.7 the highest amperage this mixer should draw under normal use.
The secondary question is which motor would draw less power, ( there are two, with identical power ratings), the one with S.F. cos .75/ .67 or the one with S.F cos .91/.93 ?
Thanks in advance for your response.
kwired
Electron manager
- Location
- NE Nebraska
It only runs at one speed or the other, so worst case is going to be the 5.7 amps.
If you try to run it at both speeds simultaneously you will be letting the smoke out of winding insulation if you don't have proper overload protection settings to shut it down in fairly short time.
Add you probably need to have separate overload protection on each speed, otherwise if single device set for higher value then it will allow overloading at the lower speed without tripping.
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