Improving power factor

Hello. I have a devise (an ice cream machine but just think of it as a big motor) which uses 1100 watts at 240 volts. So it should draw 4.6 amps if the power factor was good however the manufacturers seem to have built it in a way where the power factor is terrible. It actually draws closer to 8.5 amps and has a power factor of around 0.55

This high amp draw is causing me problems as I have one line powering three of these devises so the amp draw is around 24 amps while the line has a trip at I think 20 amps so it is tripping. As such I can only use two of these machines at once.

So I have two choices. Install an additional new line (cost and time) or try to improve the very poor PF of the devises.

So any ideas on how I can improve the PF without opening up the machines and soldering on an appropriate capacitor? I am not confident to do that as I am no electrician and only know a little about electronics from A-level classes a long time ago.

Is there any off the shelf solution which are plug and play for this type of problem?

Any help appreciated

 

I was thinking would a transformer help/work?

Could I get a 240 volt to 240 volt transformer and would that stop the phase shift?

So although the units would still have a poor power factor that will only be to the new transformer and the limited line behind the new transformer should have a PF of 1??

Does that sound correct?

As you can tell I am no expert I just know a little from school physics classes many years ago.

 

I don't think there is an off the shelf solution. Adding another transformer will only decrease the power factor.
I would say the best thing to do is to add another power line, or buy newer machines that use less power.

 

I think I may have found a solution

Some UPS are double conversion, so they convert the incomeing AC to DC and then to AC again so the UPS should draw near PF 1

The problem is then a question of the efficency loss being too big a problem or not as AC to DC to AC must suffer perhaps as much as 10-20% loss?

may need to try investigate this before digging up the floors for a new line

 

Hello. I have a devise (an ice cream machine but just think of it as a big motor) which uses 1100 watts at 240 volts. So it should draw 4.6 amps if the power factor was good however the manufacturers seem to have built it in a way where the power factor is terrible. It actually draws closer to 8.5 amps and has a power factor of around 0.55

what is the nameplate supply requirements.

Transformers approach a PF of 1 when they approach full load, but the opposite is also true.

Some UPS are double conversion, so they convert the incomeing AC to DC and then to AC again so the UPS should draw near PF 1

how does that alter the characteristics of your load?

 

http://www.eaton.com/Eaton/Products...ection/index.htm?wtredirect=www.eaton.com/PFC

 

what is the nameplate supply requirements.

240v
1100 watts is what it says on the back

Actual supply is about 250v PF 0.55 Current draw 8.5A

how does that alter the characteristics of your load?

It doesnt

But the line now sees the UPS as the load
And the UPS should draw close to PF 1.0

So althouogh the cable from the machine to the UPS is still carrying 8.5A PF 0.55 the cable from the UPS to the supply should be closer to 5A and PF 1.0

However there will be a constant efficency loss of perhaps 150 watts....

 

http://www.eaton.com/Eaton/Products...ection/index.htm?wtredirect=www.eaton.com/PFC

Thanks but I really need something that is plug and play, eg UPS would be something I can do, anything involving more than pluging in is largely out of the question

The alternative of a new line isnt that bad, will cost about £500 tops I think so that might be the route.

Just posting here as there may be a solution im not aware of

Of course the best thing would have been for the dam manufacturers to fit a capacitor across their motor so the machine has a better PF to begin with

 

Improving power factor


So any ideas on how I can improve the PF without opening up the machines and soldering on an appropriate capacitor?

Is there any off the shelf solution which are plug and play for this type of problem?

Any help appreciated

There is no need to open up the machine. The simplest way is to use PF Correction Capacitor Outside the Machine, Across the Input Lines. Take help from someone who can do this.

Ref: http://www.allaboutcircuits.com/vol_2/chpt_11/4.html

Ramesh

 

There is no need to open up the machine. The simplest way is to use PF Correction Capacitor Outside the Machine, Across the Input Lines. Take help from someone who can do this.

Ref: http://www.allaboutcircuits.com/vol_2/chpt_11/4.html

Ramesh

Hi

Can you please advise as to who can do this for me. I assume this is outside the scope of a normal electrician???

Are there any users here that could wire the appropriate capacitor to a small extention cord and mail me that so its effectively plug and play for me. Obviously I would pay for the service. However and this is critical the item needs to be absolutely safe. So presumably just a capacitor on a lead would not be a good idea as it would need a way to safely discharge the capacitor of the extention cord if the cord is ever unplugged

Also would such a capacitor on a lead system cause any problems with other devises. The line also powers a few other things which are resistive loads (a few lights and a resistive heater)


Having spent the best part of ten hours trying to research the issue it seems a correcting capacitor is the best and only realistic solution. I've looked into ups and power line conditioners but don't think they will help.

 

My first post here but yes I agree wit the CAP idea, if the PF is to inductive, (Voltage follows current)which it usually is with AC, then a CAP on the line will bring it closer to 1, theres a formula for PF correction I know. Thats getting into power dist, I had a few classes on that, not many.

ELI the ICE man knows.

 

You might also want to open the machine and see if the motor has a CAP built in or on. Most motors do. A bad CAP makes the motor run slow sluggish and pull to many amps.

 

You might also want to open the machine and see if the motor has a CAP built in or on. Most motors do. A bad CAP makes the motor run slow sluggish and pull to many amps.

I don't think they have a capacitor to try and reduce the impact of the inductance of the motor. I have multiple sets of these machines and they are all reading between 0.55 and 0.6 power factor so I can rule out a dud capacitor its just a popr design as far as I can tell

I've emailed the manufacturer telling them that their machines have veey poor power factor so hopefully they will try and improve on that for future versions

 

Can anyone advise as to companies or trades that can build an extention cord with a capacitor built in?

Presumably its a job for an electrical engineer rather than an electrician?

If so how does one find an electrical engineer that does odd jobs?

To be honest I'm surprised you can't buy extention cords with variable capacitors. Plug in and turn on your equipment and vary the capacitor until your meter reads over 0.95 power factor. I can't be the only one who could do with one

 

I would think that one way to do it would be to take a junction box and wire a plug into it (that you will plug into your power outlet) and an outlet (that you will plug your motor into). In side the box you have a run capacitor across the outlet.

For 250VAC, you probably want your caps rated for at least 400VAC.

Your apparent power is 8.5 A * 250 V = 2125 VA
With a power factor of 55%, your real power is 1169 W
That makes your reactive power sqrt[(2125 VA)^2 - (1169 W)^2] = 1775 VAR
That means you need a capacitive reactance of -(250 V)^2 / 1775 VAR = -35.2 Ω.

Since you talked about costs in pounds, I'm assuming you are someplace where the line frequency is 50 Hz. That means the capacitor would need to be

C = 1/(2pi*50Hz*35.2Ω) = 90μF

Put a bleeder resistor across it that will discharge it. I'm guessing something in the neighboorhood of 100 kΩ rated for 1W. I don't know if there is a code for what the time constant for such a bleed circuit needs to be.

That might serve at least as a starting point for you.

Another member of the forum, #12, does this kind of work for a living, so he can tell you whether this is a good/safe way to go and whether you should use a different way to size your cap. I'll shoot him a PM and see if he can take a look.

 

Hi

Can you please advise as to who can do this for me. I assume this is outside the scope of a normal electrician???

Are there any users here that could wire the appropriate capacitor to a small extention cord and mail me that so its effectively plug and play for me. Obviously I would pay for the service. However and this is critical the item needs to be absolutely safe. So presumably just a capacitor on a lead would not be a good idea as it would need a way to safely discharge the capacitor of the extention cord if the cord is ever unplugged

Also would such a capacitor on a lead system cause any problems with other devises. The line also powers a few other things which are resistive loads (a few lights and a resistive heater)


Having spent the best part of ten hours trying to research the issue it seems a correcting capacitor is the best and only realistic solution. I've looked into ups and power line conditioners but don't think they will help.

The PF Correction Capacitor bank will be something like this.

A normal electrician can wire it up - like wiring up a washing machine. It is Not connected with an Extension Cord
These normally have a discharge resistor.
It will not affect the working of any other device.
Can be disconnected with a switch though it may not be necessary.

Ramesh

 

If so how does one find an electrical engineer that does odd jobs?

Since you have given no idea of where in the world you are located, don't you think it would be hard for us to tell you how to go about something like this?

 

Since you have given no idea of where in the world you are located, don't you think it would be hard for us to tell you how to go about something like this?

Apologies I didn't even think about that.

I am in the UK near Birmingham

 

I think #12 might be away right now. He's been a very active poster the past few months but has gone pretty dark the last weekish or so. Hopefully he will come back and comment soon, as I think he can give you a good, reliable answer.

EDIT: Actually, I was thinking of someone else. #12 is still around and active, so hopefully he'll get a chance to swing by sometime.

 

[QUOTEahn;638817]I would think that one way to do it would be to take a junction box and wire a plug into it (that you will plug into your power outlet) and an outlet (that you will plug your motor into). In side the box you have a run capacitor across the outlet.

For 250VAC, you probably want your caps rated for at least 400VAC.

Your apparent power is 8.5 A * 250 V = 2125 VA
With a power factor of 55%, your real power is 1169 W
That makes your reactive power sqrt[(2125 VA)^2 - (1169 W)^2] = 1775 VAR
That means you need a capacitive reactance of -(250 V)^2 / 1775 VAR = -35.2 Ω.

Since you talked about costs in pounds, I'm assuming you are someplace where the line frequency is 50 Hz. That means the capacitor would need to be

C = 1/(2pi*50Hz*35.2Ω) = 90μF


Put a bleeder resistor across it that will discharge it. I'm guessing something in the neighboorhood of 100 kΩ rated for 1W. I don't know if there is a code for what the time constant for such a bleed circuit needs to be.

That might serve at least as a starting point for you.

Another member of the forum, #12, does this kind of work for a living, so he can tell you whether this is a good/safe way to go and whether you should use a different way to size your cap. I'll shoot him a PM and see if he can take a look.[/QUOTE]

Thank you v.much

I am wondering if this is something I could do. I've done a lot of soldering in the past so that shouldn't be a problem and I've got basic electronics knowledge and could confirm correct design here but I would still prefer to have someone more experienced and knowledgeable do this for me.

One more quick question if you guys don't mind. If I was to put a run capacitor directly across the motor presumably there would be no need for a bleed capacitor as the motor would discharge the capacitor fully and nearly instantly when the machine is turned off?

I think that would be the better design option but presents further challenges (the inside of the machine is fairly warm perhaps 50 centigrade and I've read capacitors don't fair well as ambient temperature increases) so the junction box idea may have to suffice