# Static phase converter

• posted

I have a rotary phase converter, the good thing about which is that it works very well and is 17.5 HP idler capacity.

What bugs me, however, is a question of how I can get full power out of a 3 phase motor, using some sort of a static phase converter scheme.

As far as I understand, in a static phase converter, there is a starting circuit using a start capacitor and a relay that gets the motor started, and then the motor runs at 2/3 of power from single phase.

What I do not understand is, why is it not possible to permanently connect a properly rated run capacitor to legs 1-3 (legs 1 and 2 being connected to single phase input). The start capacitor would still be relay controlled. I thought this would shift the phase on leg 3 and allow the motor develop full power during running. There must be something that I am missing.

• posted

Forget the question, I think that I understand now.

i
• posted

A VFD is a type of static phase converter and provides full power. Static means no moving parts like an idler motor, not no electronics.

• posted

You can switch and run a 3 phase motor in the way you describe but it's a bit unkind to the switching contacts.

If you separately switch in start and run capacitors the maximum peak switching current is only the normal current drawn by the motor windings.

If you switch the start capacitor in parallel with the already connected run capacitor you are instantaneously paralleling two capacitors each of which may be charged to a different voltage. The peak current that then flows is only limited by the internal series resistance of the capacitors and is typically MANY times steady state winding current.

Jim

• posted

Well, let's set this question aside for a minute. Do you think that I could get full power out of the motor?

Sure. But that is a question that can be addressed. The question that I wanted to find out was, could you get the motor to run at full nameplate power.

• posted

Think about the phase relationship, will each phase of the motor draw the same voltage and current when using capacitors as the same motor running on utility supplied 3-phase ? How does the phase relationship change when a load is applied to the motor ? Can you ever get 120 degree phase to phase angle using capacitors and maintain that relationship as the motor load varies?

Best Regards Tom.

• posted

Assume constant load for a first try.

This is a compressor application, by the way.

• posted

Your going to have to take some measurements and do the math to determine the best combination. The two web pages should give you an idea of what needs to be done (caution some heavy math involved) there is no rule of thumb when you want to get name plate rating you will have to do the math. It probobaly will be faster, easier and less expensive to use a VFD unless this is a learning experience.

These sould help if you want konw whats realy going on in AC LC circuits.

Best Regards Tom.

• posted

developing or is capable of developing unless you test it on a dynamometer. I believe that a properly sized run capacitor will permit the motor to adequately run your compressor on single phase power. It seems to me that the requirements are to start the compressor and to run the compressor at full pressure without destructively overheating for as long as necessary. Of course this is largely a function of how conservatively the motor was sized and coupled to the pump in the first place.

In my opinion, nothing beats trying it if there is a reasonable chance of success and I think there is. Keep in mind that the nameplate rating is the manufacturer's stated power that the motor can safely develop for long periods of time under the specified conditions and does not tell how much power it can develop under other conditions or for shorter time periods.

I have seen some test results for this but have not seen complete tests monitoring developed dynamotor horsepower and long term temperature rise. It would be an interesting experiment.

Don Young

• posted

What is helpful is that every time, the horsepower in use depends only on the pressure in the compressor tank. So at any time, I will know that the same certain pressure implies the same horsepower. Also, the pump in question has hydraulic unloaders that only make it start pumping when lubricating oil pressure rises sufficiently, which means RPM is reached, so starts are easy.

I will see what I can do, I want to do something that is cheap, easy and robust, which is not a very promising combination, but nothing beats trying.

• posted

You might want to talk to this company:

Looks like they have done what you want to do, maybe they have some litrature that would be helpfull in your experiments.

Best Regards Tom.

• posted

Then you *might* be able to do that -- but consider the cost. Motor starting capacitors are high capacitance, but will only work for a short duration of current. Keep them switched on for longer and they will spew their gut all over the shop. They *are* relatively inexpensive, however.

For run capacitors of the same value, however, you need much more expensive and physically much larger oil filled capacitors -- and likely a bunch of them connected in parallel to reach the needed value.

That is not a constant load. The motor has an easy job starting out, and has to work harder and harder as the pressure in the tank increases.

Assuming a disconnector which allows the motor to free wheel the range will change even more between the free wheeling and the full tank pressure points.

You might be able to tune it to something semi reasonable for the cycling between the low-pressure switch on and the higher pressure swtich off, but it would be very poorly balanced for when the motor is idling if you do have a load disconnector.

Enjoy, DoN.

• posted

It would appear that yes, they have dome something with great similarities to what I am trying to do. Very interesting. I am reading their white paper right now.

• posted

Upon reading the article, no, this is not what I was considering.

• posted

If the run capacitor is selected for current balance (more accurate than voltage balance) at the nameplate motor load the short answer is yes. Under these conditions the supply to the motor is as well balanced or better than the same motor operating from the more elaborate rotary converter/idler setup.

It's true that static converters (start and run capacitor systems with no idler) can deliver the full rated power of the motor for surprisingly long periods but that is not the whole story.

A converter of this type is basically a capacitor/inductor phase shift system which produces an open vee 3 phase system. This phase shifter is a series resonant circuit and when it is set up to give the 60 deg phase shift it is working a long way below its natural resonant frequency. 60 deg is of course the correct phase angle between the two legs of an open vee system.

The motor(s) is the inductor in the system and unfortunately the apparent inductance of the motor changes with rotor speed. For any particular rotor speed greater than about

90% of synchronous speed (the lower limit varies a bit with motor type) it is possible to choose a capacitor combination which produces a pretty close approximation to balanced 3 phase at the motor terminals.

For near the full load rated speed of the motor, large run capacitance is needed with most or all of it as a single capacitor feeding the phantom phase from supply live. At light load the impedance of the rotor rises and if the capacitor value is chosen to achieve the right phase angle the phantom phase voltage will be excessive. This could be corrected by feeding the capacitor from a lower voltage single phase source but this would mean feeding it from an auto transformer across the supply.

It is much simpler (and of course everybody does this) to use two capacitors arranged as a voltage divider to simultaneously achieve the correct phase angle and phase voltage.

The effective capacitanceof the two capacitors connected in series across the supply is the sum of the capacitances because the source impedance of the supply is zero and this effectively parallels the two capacitors.

Because the they also act as a voltage divider, this sum capacitance is effectively fed from a voltage of supply voltage times C1/(C1+C2) where C1 is the top capacitor and C2 is connected phantom phase to neutral.

Because it looks nicely symmetrical there seems to be a tendency to believe that C1 and C2 should be equal and any inequality in their optimum value must result from some strange second order effect. This is NOT true. There is nothing magic about equal C1 and C2. It simply results in a capacitor of value C1+C2 fed from half the supply voltage. At this low effective supply voltage it is only possible to get close to balanced operation at no load or light loads which enable the rotor to operate close to synchronous speed. As the load increases with consequent slowing of the rotor speed the total capacitance needs to increase with both more in C1 and less in C2. By the time full load is reached the optimum value for C2 is usually zero.

These effects are very noticeable if you're using a single motor on a variable load up to near rated full load power and some compromise necessary. The saving grace is that industrial motors are surprisingly tolerant of reasonable overvoltage when operating at light loads. The trick is to size the capacitors for at or near full load and to accept some overvoltaqe at light loads. This increases the motor losses at light load but the total motor losses still remain below the losses at rated full load so temperature rise is acceptable.

Summing up - if you need to cope with heavy loads on a static converter throw away the bottom capacitor and be sure to choose C1 for operation near full load.

None of this helps with starting torque - this is inherently poor with the static converter arrangement however large the starting capacitor. This is because correct low speed phasing requires the capacitor to be fed fed from a voltage many times the supply voltage.

Jim

• posted

If the run capacitor is selected for current balance (more accurate than voltage balance) at the nameplate motor load the short answer is yes. Under these conditions the supply to the motor is as well balanced or better than the same motor operating from the more elaborate rotary converter/idler setup.

It's true that static converters (start and run capacitor systems with no idler) can deliver the full rated power of the motor for surprisingly long periods but that is not the whole story.

A converter of this type is basically a capacitor/inductor phase shift system which produces an open vee 3 phase system. This phase shifter is a series resonant circuit and when it is set up to give the 60 deg phase shift it is working a long way below its natural resonant frequency. 60 deg is of course the correct phase angle between the two legs of an open vee system.

The motor(s) is the inductor in the system and unfortunately the apparent inductance of the motor changes with rotor speed. For any particular rotor speed greater than about

90% of synchronous speed (the lower limit varies a bit with motor type) it is possible to choose a capacitor combination which produces a pretty close approximation to balanced 3 phase at the motor terminals.

For near the full load rated speed of the motor, large run capacitance is needed with most or all of it as a single capacitor feeding the phantom phase from supply live. At light load the impedance of the rotor rises and if the capacitor value is chosen to achieve the right phase angle the phantom phase voltage will be excessive. This could be corrected by feeding the capacitor from a lower voltage single phase source but this would mean feeding it from an auto transformer across the supply.

It is much simpler (and of course everybody does this) to use two capacitors arranged as a voltage divider to simultaneously achieve the correct phase angle and phase voltage.

The effective capacitanceof the two capacitors connected in series across the supply is the sum of the capacitances because the source impedance of the supply is zero and this effectively parallels the two capacitors.

Because the they also act as a voltage divider, this sum capacitance is effectively fed from a voltage of supply voltage times C1/(C1+C2) where C1 is the top capacitor and C2 is connected phantom phase to neutral.

Because it looks nicely symmetrical there seems to be a tendency to believe that C1 and C2 should be equal and any inequality in their optimum value must result from some strange second order effect. This is NOT true. There is nothing magic about equal C1 and C2. It simply results in a capacitor of value C1+C2 fed from half the supply voltage. At this low effective supply voltage it is only possible to get close to balanced operation at no load or light loads which enable the rotor to operate close to synchronous speed. As the load increases with consequent slowing of the rotor speed the total capacitance needs to increase with both more in C1 and less in C2. By the time full load is reached the optimum value for C2 is usually zero.

These effects are very noticeable if you're using a single motor on a variable load up to near rated full load power and some compromise necessary. The saving grace is that industrial motors are surprisingly tolerant of reasonable overvoltage when operating at light loads. The trick is to size the capacitors for at or near full load and to accept some overvoltaqe at light loads. This increases the motor losses at light load but the total motor losses still remain below the losses at rated full load so temperature rise is acceptable.

Summing up - if you need to cope with heavy loads on a static converter throw away the bottom capacitor and be sure to choose C1 for operation near full load.

None of this helps with starting torque - this is inherently poor with the static converter arrangement however large the starting capacitor. This is because correct low speed phasing requires the capacitor to be fed fed from a voltage many times the supply voltage.

Jim

• posted

Jim, You are one of the reasons I read this newsgroup. I usually get a bit more insight everytime I read one of your posts.

Dan

• posted

DITTO

• posted

Pentagrid, this is very interesting indeed. Thank you.

I have some questions in this regard.

This is for a compressor duty application. The compressor is a Quincy QR-25, model 340 pump based compressor, 80 gallon tank, that I bought the other day for \$200.

I have several options for running it.

1. Use a 5 HP, 1725 single phase RPM motor that I have. The plus is that it is easy, the minus is that it is only 5 HP, 13 CFM at 175 PSI. Maybe I can get more CFM is I run it slightly faster, but regulate to be between 110-140 PSI. I will call Quincy today about it.
2. Try to make a 10HP VFD which I have, run a 10HP 3ph motor that I have, but I would need to try to supply its DC bus so that the drive does not "phase fault".
3. Run from phase converter. (will work fine but painful!)
4. Try to develop some sort of starting and running circuitry for the
10 HP motor that we are discussing in this thread.
1. Buy a 7.5 or 10 HP single phase motor (very expensive! for a real motor)

For this particular compressor, the motor starts, and spins up the pulley. Then unloader valves make the pump start pumping, only when oil pressure reaches a certain number. This way, the motor has a chance to spin up the pump unloaded.

After that, depending on initial tank pressure, and air use, the motor load would gradually increase until the tank is almost full and the motor reaches its maximum power output.

So, here are the q u e s t i o n s for option 4.

1. Suppose that I found out the capacitance needed to run the motor at full nameplate horsepower. How would the motor behave, if it was connected to the same capacitance, but at lower power output (and therefore slightly higher RPM).
2. Practically, what would be a range of capacitance for a 10 HP 1725 RPM motor. I would need to buy a set of run caps to make up the capacitance I need, and it would be good to know the cost beforehand.
3. How would I find this capacitance. Is that a function of hertz and industance? If so, it would seem that I would need to find out the inductance of the motor at full load first. So how would I do it? If I run the motor from real three phase, I can connect a capacitor between two legs and compute inductance from knowing the current through the capacitor, frequency (60 Hz), and the capacitance. Right?
4. Under the less favorable assumption that capacitance needs to be adjusted as tank pressure rises, would it be possible, as it seems, to use pressure switches based on tank pressure, and zero crossing SSRs.

This means that if, say, the motor is 1740 RPM, as is my 10 HP motor, then it runs at 96.6% of synchronous speed at highest output power, meaning that it is even faster (closer to 100%) at lower outputs. So, it would appear that it is a good case for giving it a phase shifted third leg with a capacitor.

Do you have any idea how large, in uF per HP?

You are referring to feeding the capacitor with voltage, but I thought that it would be simply plugged between legs 1 and 3, without any voltage connected to leg 3?

How fine would these changes need to be? Would one change be enough, for example, as the motor increases its output due to pressure rise?

This sounds good. The 10 HP motor is a great Reliance motor that should be as good as any motor.

Sounds easy enough.

Starting is not a problem, I am not worried, this is something that can be solved very easily with A/C starting caps. The motor would start at low torque anyway.

I appreciate your very thoughtful reply. I bought and read some motor books a while ago, and I believe that one was yours.

• posted

SNIP

If I've understood you correctly you have a 5 HP and a 10 P motor and you're looking for the simplest and cheapest method of using one of them to drive your compressor. In this case there's no point in worrying about the precision of the balance achieved over the load range. Use the bigger motor (the bigger the motor in relation to the load the less critical the setup) and choose your run capacitor for best balance near or at full compressor load. When the compressor is running light the phantom phase voltage will will be well above normal. Motors are pretty tolerant to this. The yard stick is simple - check the phantom phase current - so long as it remains below rated full load current it's OK.

In practically every case the residual mechanical load is sufficient and this is no problem. If you should be unlucky and the phantom phase current is marginally high, reduce the run capacitance a bit - with 10HP drive you don't neeed the last ounce of torque.

For 230V 60 Hz motors 24uF per HP is typical but since ypu're not fully loading your 10HP motor about 80% of this is probably enough. At least 3 times this for the starting capacitor.

I think this probably also answers your comments on the longer part of my post. A possible exception is on he use of an autotransformer. This is really only relevant to the purist who is trying to arrive at the closest possible balance. The interdependence of both the relative and the total value of C1 and C2 can make optimisation protracted and frustrating. If a single capacitor is used fed from the slider of a variac connected across the supply, the variac setting independently controls the phantom phase voltage and the choice of capacitor the phase angle so it's much easier to find the optimum setting.

Not sure which of my books you've got but the recent 2nd edition of Electric Motors Workshop Practice Series no 16 includes a lot more information on 3 phase systems and VFDs.

Jim

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