In the mining industry I have encountered several 575 volt motors connected to 480 volt systems. I was wandering what precautions should be observed in this practice? I have not seen any adverse effects. No burned up motors or overload issues at all. How would the actual output of the motor horsepower and current load compare to its nameplate rating?
Run 575 volt motors on 480 ?
- Thread starter dorymill
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Operating this 575V motor at a reduced voltage at 480V will reduce the designed V/Hz ratio of the motor. With a reduced V/Hz ratio the motor will not be able to produce the rated flux and therefore will have difficulty producing the rated torque.
In an attempt to produce the rated torque the motor will draw more current, so you will see an increase in current.
In an attempt to produce the rated torque the motor will draw more current, so you will see an increase in current.
Cold Fusion
Senior Member
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I don't have any references available right now, so I am going from recollection. The available torque is porportional to the square of the voltage. So a 100 hp, 575V motor is only good for 64 hp at 460V.
(460^2)/(575^2) = .64
As long as the load is kept down, they should work fine.
cf
(460^2)/(575^2) = .64
As long as the load is kept down, they should work fine.
cf
Assuming we are discussing a 100hp motor as a reference, my comments were I guess geared toward what would happen if you tried to run a 100hp load with this motor. The current would increase, however I'm not sure if this is a function of the current trying to supplement for the reduced flux to create the desired torque, or the fact that the motor torque curve is shifted vertically downward and therefore this new torque point intersects with a higher current?
So assuming that the torque reduces at the square of the voltage then at 64hp we would expect to see full rated current load on the motor? Is this correct?
Is there a way to corrolate reduced voltage values to an increased current value when the motor is running? For example if the motor voltage was 3% lower than nameplate, can a linear interperatation be made to say the the current will increase by 3%?
I had a situation the other week where I was measuring 470V at the motor starter. I know the standard is that motors withstand a 10% voltage dip however the motor would see increased current. Enen though this 470V was below the rated 480V of the circuit the motor nameplate was rated at 460V. With that being said, I wouuld think that this 470V was still above the 460V rating of the motor so this would be o.k. We would have more avaliable toruqe and would see less current for a given HP? Am I on the right track here?
Cold Fusion
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Yes - That's what I figured you were meaning. From the OP's comment that there were no adverse effects, I figured they had to be running lightly loaded.
Philly -
With all due respect, you are way out of my area of expertise. I have some knowledge on the basic physics/math model, but that is about it.
I think so. My thinking is the loading (required torque) is reduced to where the speed is within the normal 3% - 5% slip and the current will be at nameplate FLA.
Probably, but I couldn't without a bunch of research. I attached a chart that has been hanging around for 20 years or so. I'm not sure where it came from. I think it was Nema MG-1, but it is not in the current version of MG-1.
cf
Some background is here http://www.baldor.com/pdf/manuals/PR2525.pdf
I think graph above is on page 60 but there is a lot of additional info if want to dig a little.
I think graph above is on page 60 but there is a lot of additional info if want to dig a little.
Run a 575v motor on 480v
Run a 575v motor on 480v
Thanks for the the discussion.
I asked my boss and this is what he said.
"The output HP of the motor will be a product of the square of the ratio of the name plate voltage to the applied voltage" times the rated horsepower.
We are working with a 200 HP motor on a high pressure hydraulic pump. The motor is rated 575/995V on the nameplate.
He says take 480/575=.83 Now square this .83
.83^2=.6889 Now multiply this by the Horsepower Rating
.6889 x 200 =137.78 HP
So the 200HP ends up being around 137 HP on 480V.
I'll get a chance to check the current and monitor this motor in the next few weeks. They have been using this set-up for a long time. I was rebuilding the starter box when I discovered the motor rating was not 480 but 575. I will size the starter,overload, and wire for the 200 rating and keep an eye on the performance.
Run a 575v motor on 480v
Thanks for the the discussion.
I asked my boss and this is what he said.
"The output HP of the motor will be a product of the square of the ratio of the name plate voltage to the applied voltage" times the rated horsepower.
We are working with a 200 HP motor on a high pressure hydraulic pump. The motor is rated 575/995V on the nameplate.
He says take 480/575=.83 Now square this .83
.83^2=.6889 Now multiply this by the Horsepower Rating
.6889 x 200 =137.78 HP
So the 200HP ends up being around 137 HP on 480V.
I'll get a chance to check the current and monitor this motor in the next few weeks. They have been using this set-up for a long time. I was rebuilding the starter box when I discovered the motor rating was not 480 but 575. I will size the starter,overload, and wire for the 200 rating and keep an eye on the performance.
Run a 575v motor on 480v
Run a 575v motor on 480v
Thanks for the the discussion.
I asked my boss and this is what he said.
"The output HP of the motor will be a product of the square of the ratio of the name plate voltage to the applied voltage" times the rated horsepower.
We are working with a 200 HP motor on a high pressure hydraulic pump. The motor is rated 575/995V on the nameplate.
He says take 480/575=.83 Now square this .83
.83^2=.6889 Now multiply this by the Horsepower Rating
.6889 x 200 =137.78 HP
So the 200HP ends up being around 137 HP on 480V.
I'll get a chance to check the current and monitor this motor in the next few weeks. They have been using this set-up for a long time. I was rebuilding the starter box when I discovered the motor rating was not 480 but 575. I will size the starter,overload, and wire for the 200 rating and keep an eye on the performance.
Run a 575v motor on 480v
Thanks for the the discussion.
I asked my boss and this is what he said.
"The output HP of the motor will be a product of the square of the ratio of the name plate voltage to the applied voltage" times the rated horsepower.
We are working with a 200 HP motor on a high pressure hydraulic pump. The motor is rated 575/995V on the nameplate.
He says take 480/575=.83 Now square this .83
.83^2=.6889 Now multiply this by the Horsepower Rating
.6889 x 200 =137.78 HP
So the 200HP ends up being around 137 HP on 480V.
I'll get a chance to check the current and monitor this motor in the next few weeks. They have been using this set-up for a long time. I was rebuilding the starter box when I discovered the motor rating was not 480 but 575. I will size the starter,overload, and wire for the 200 rating and keep an eye on the performance.
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
Keeping in mind that approximations abound in motor theory, and that these approximations will fall apart more or less quickly depending upon how complex they are, and also keeping in mind that I am working from memory here:
The torque versus speed curve will scale with the square of the applied voltage. If you multiply the applied voltage by 0.5, then the torque at any given speed will be multiplied by 0.25. The approximation is that the over-all shape of the torque curve does not change; this falls apart because of saturation effects.
The current versus speed curve will scale linearly with the applied voltage. This means that at a given fixed speed if you halve the applied voltage you will halve the current that the motor draws. Again this approximation ignores saturation effects.
The motor speed is determined by the intersection of the motor torque/speed curve and the load torque/speed curve. When you reduce the voltage the motor torque/speed curve will drop as above, and the operating speed will change.
Okay, my guesses:
If you take a 100 Hp motor and operate it at 80% voltage, then the available peak torque and starting torque will look like that of a 64 Hp motor. If you then operate this motor with a 64 Hp mechanical load, the operating speed would be the same as expected with a 100 Hp mechanical load at full voltage. Thus I would expect the current to be 80% of full load current. As an independent check: 80% applied voltage * 80% current = 64% input power. Assuming similar efficiency, I would expect 64% input power to get 64% output power.
-Jon
I agree with that.
For my own interest, I checked estimated performance on a motor model, one that I know to be sound and, to within two significant digits, it give the figures you did.
Well, there is no simple answer that I know of.
It depends on loading, the extent to which the voltage is reduced and, of course, motor design.
For one specific motor...
Take the 3% reduction in voltage you suggested. To get full load torque (FLT), the stator current would increase by a little under 3%. For 0.5 FLT, the increase is about 1.3%. At a quarter load, the reduced voltage makes the current lower than at full voltage.
If you go for a greater reduction, say 20% (which is where this thread started) the 20% reduction in voltage results in nearer a 30% increase in current at FLT but, at 0.25 FLT, there is an even greater reduction when compared to full voltage operation.
PowerQualityDoctor
Senior Member
- Location
- Israel
Experiemental data
Experiemental data
All the calculations in this thread neglected one issue - the power factor. The motor's power factor is changed with the load. If we have 50% load on 575 motor and we provide only 480 volt, this means we reduce the voltage by 16.5% or the power by 30%. Now, the motor is loaded by 71% (50%/70%). The result is three issues happening at the same time:
1. Reduced voltage -> increased current
2. Improved power factor -> reduced current
3. Improved motor efficiency -> reduced kW -> reduced current.
The result is that for 50% load, reduction of voltage by 16.5% would reduce the current.
http://www.powersines.com
Experiemental data
All the calculations in this thread neglected one issue - the power factor. The motor's power factor is changed with the load. If we have 50% load on 575 motor and we provide only 480 volt, this means we reduce the voltage by 16.5% or the power by 30%. Now, the motor is loaded by 71% (50%/70%). The result is three issues happening at the same time:
1. Reduced voltage -> increased current
2. Improved power factor -> reduced current
3. Improved motor efficiency -> reduced kW -> reduced current.
The result is that for 50% load, reduction of voltage by 16.5% would reduce the current.
http://www.powersines.com
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Incorrect.
That is just garbage.
I can explain why but I suspect it would be a wasted effort.
But, if you can convince me otherwise.....
PowerQualityDoctor
Senior Member
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- Israel
I learn new things every day
I learn new things every day
Today I have learned that some people think that if they use certain words, this means they know better :roll:
You are more than welcome to comment seriously to my post. I am willing to learn and/or debate any of my statements.
I learn new things every day
Today I have learned that some people think that if they use certain words, this means they know better :roll:
You are more than welcome to comment seriously to my post. I am willing to learn and/or debate any of my statements.
My comments were serious, succinct, and accurate.
But if you want to debate your statements, let's do just that.
For sure, motor power factor varies with loading. It's a consequence, not a driving factor. And it was not neglected in my calculations.
Reduce the voltage and the speed at 50% load drops very slightly. By around 0.3% depending on motor design. So output power (torque times speed) at 50% load, would drop by a similar amount and even less for a load with a positive speed torque curve. Most loads have a positive speed torque curve.
So, output power is substantially unchanged.
Input power is output power plus losses. Efficiency is little different. There is no 30% power loss to be saved.
That's the easy stuff.
skeshesh
Senior Member
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- Los Angeles, Ca
I'm gonno have to agree with Besoeker. Also I'm wondering about about your avatar nomeclature PowerQualityDoctor - are you perchance involved in the magic capacitor power reduction industry?
PowerQualityDoctor
Senior Member
- Location
- Israel
Some clarifications
Some clarifications
The idea is to reduce the voltage, but not reducing the speed. In this way the motor effective power is reduced by 30%, not the power consumption. This is exactly what happens in the situation described in this post - you supply 480 volts to 525 volt motor.
Another example is 6 wire motor - if you connect it in Delta with 480v, the power is 100% but if you connect it in Star, where the coils receive 277v, the power is 33% (the voltage reduction squared).
After we agree on this (hopefully), we have to see what happens when the full power is reduced. If the load is less than 70% (assuming 16% voltage reduction), the motor will continue to work as it can provide the required counter torque. The different motor loading will have effect on the motor efficiency and power factor. If the load is 50% (measured in power on the motor shaft), this means the efficiency will be improved and the motor will consume less active power. Of course, the saving depends on the wasted energy and the motor's original efficiency. In mining applications, some loads are very lightly loaded (less than 30%), which can get 15% or even better consumption reduction.
Some clarifications
The idea is to reduce the voltage, but not reducing the speed. In this way the motor effective power is reduced by 30%, not the power consumption. This is exactly what happens in the situation described in this post - you supply 480 volts to 525 volt motor.
Another example is 6 wire motor - if you connect it in Delta with 480v, the power is 100% but if you connect it in Star, where the coils receive 277v, the power is 33% (the voltage reduction squared).
After we agree on this (hopefully), we have to see what happens when the full power is reduced. If the load is less than 70% (assuming 16% voltage reduction), the motor will continue to work as it can provide the required counter torque. The different motor loading will have effect on the motor efficiency and power factor. If the load is 50% (measured in power on the motor shaft), this means the efficiency will be improved and the motor will consume less active power. Of course, the saving depends on the wasted energy and the motor's original efficiency. In mining applications, some loads are very lightly loaded (less than 30%), which can get 15% or even better consumption reduction.
I assume the 525V is a typo and that you are still referring to the 480V on a 575V motor you gave in post #13.
I don't agree.
Again, from post your #13
For the motor on which did the calculations, the reduced voltage (480/575) resulted in an increase in supply current from 366A to 404A at half load. True, the efficiency did improve - by one quarter of one percent (of running load). Of course, that might easily be completely negated by the increased losses in the supply cabling and distribution resulting from the increased supply current.
At one quarter load on the same motor, the current was not significantly different between full voltage and reduced voltage. Again, there was a very small improvement in efficiency, this time about 0.8% of running load. Or 0.2% of motor rating.
Cold Fusion
Senior Member
- Location
- way north
I'm having trouble following your physics. There are no models known to me that support you thinking - perhaps you could help me out.
Quotes are from post 18:
"After we agree on this (hopefully), we have to see what happens when the full power is reduced. If the load is less than 70% (assuming 16% voltage reduction), the motor will continue to work as it can provide the required counter torque."
This has already shown in post 3. I got that - I'm pretty sure everyone else has as well. - cf
"The different motor loading will have effect on the motor efficiency and power factor."
This is the part you will need to show the model you are using. - cf
If the load is 50% (measured in power on the motor shaft), this means the efficiency will be improved and the motor will consume less active power.
50% of what? The motor nameplate rating? Or the reduced rating of available power caused by the reduced voltage. I can't tell which you are talking about. -cf
Now lets look at the 'efficiency will be improved" part. It appears that you are saying that loading a motor less than full load improves the efficiency? If so that is contrary to every motor model I've ever seen, and contrary to every motor measurement I've ever made. Here's why I say that:
As I recall, the magnetizing current is porportional to the applied voltage. (Motor whizes are welcome to jump in here and help out). So the reduced voltage requires less magnetizing current. But that does not appear to help out the power factor because the real power is down and the motor current is down.
Now let's look at real power consumption:
The motor is turning about the same speed, so the windage and friction are about the same, but wait, the motor is loaded less than nameplate - the shaft output is down. So the losses are about the same, but the shaft output is down. That certainly sounds like the efficiency went down. -cf
"Of course, the saving depends on the wasted energy and the motor's original efficiency."
What is this wasted energy? Where is it going? This is anoter statement that appears unsupported.- cf
In mining applications, some loads are very lightly loaded (less than 30%), which can get 15% or even better consumption reduction.
This makes no sense at all.
PQD -
If you have a motor model that supports this conjecture I'd like to see it.
If you have emperical data (test data) that shows your conjecture is true - I'd like to see that as well.
cf
edit
I didn't see besoeker's post until I was done. And I don't have access to that computer generated motor model. I am supprised the efficiency went up at all - I would have expedcted it to go down.
Bes - makes me wonder if your example isn't pushing the limits of your model. I'm sure you are right - just surprised.
Quotes are from post 18:
"After we agree on this (hopefully), we have to see what happens when the full power is reduced. If the load is less than 70% (assuming 16% voltage reduction), the motor will continue to work as it can provide the required counter torque."
This has already shown in post 3. I got that - I'm pretty sure everyone else has as well. - cf
"The different motor loading will have effect on the motor efficiency and power factor."
This is the part you will need to show the model you are using. - cf
If the load is 50% (measured in power on the motor shaft), this means the efficiency will be improved and the motor will consume less active power.
50% of what? The motor nameplate rating? Or the reduced rating of available power caused by the reduced voltage. I can't tell which you are talking about. -cf
Now lets look at the 'efficiency will be improved" part. It appears that you are saying that loading a motor less than full load improves the efficiency? If so that is contrary to every motor model I've ever seen, and contrary to every motor measurement I've ever made. Here's why I say that:
As I recall, the magnetizing current is porportional to the applied voltage. (Motor whizes are welcome to jump in here and help out). So the reduced voltage requires less magnetizing current. But that does not appear to help out the power factor because the real power is down and the motor current is down.
Now let's look at real power consumption:
The motor is turning about the same speed, so the windage and friction are about the same, but wait, the motor is loaded less than nameplate - the shaft output is down. So the losses are about the same, but the shaft output is down. That certainly sounds like the efficiency went down. -cf
"Of course, the saving depends on the wasted energy and the motor's original efficiency."
What is this wasted energy? Where is it going? This is anoter statement that appears unsupported.- cf
In mining applications, some loads are very lightly loaded (less than 30%), which can get 15% or even better consumption reduction.
This makes no sense at all.
PQD -
If you have a motor model that supports this conjecture I'd like to see it.
If you have emperical data (test data) that shows your conjecture is true - I'd like to see that as well.
cf
edit
I didn't see besoeker's post until I was done. And I don't have access to that computer generated motor model. I am supprised the efficiency went up at all - I would have expedcted it to go down.
Bes - makes me wonder if your example isn't pushing the limits of your model. I'm sure you are right - just surprised.
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