phase converter power use

dear sirs

how much power does a rotary phase converter consume?

if I had a 10 hp idler RPC, would the run capacitors consume much power? how many amps should I expect it to run while idling?

let's say I use this 10 hp RPC to start and power a 5 horsepower mill. how much power will the RPC unit use during startup of the mill? how much power will it use while the mill is running? how much power would the phase converter use if I was running 5 mills?

I don't want to give you guys some kind of assignment, I'm asking these questions to get a conceptual understanding, and I'm sure anyone responding to this post will easily do that.

The 10 hp RPC I build typically draw 3 to 3.5 amps while "idling", that is, with no load motor running. How much current one draws depends upon the amount and placement of run capacitors added. However, I have seen posts where guys (or gals) stating 10 to 15 amp draw for a 10 hp. As for running load motors, figure the increase in the 240 1ph current (what you are paying the utility for) to increase about 1.73 times the load motor's rated current.

As an example, assume you have a 10 hp RPC which draws 10 amp while idling, and a 5 hp load motor (assume full load of 15 amp @ 240 v). The load motor will increase the current draw of your 1 ph input to about 26 amps (15 X 1.73 = 25.95). This added to the 10 amps the converter draws equals about 36 amps. However, your actual current draw should be less, because you are probably not drawing the full 5 hp from the machine you are running.

The above example is approximate, but should give you a indication of what to expect.

d

i'm not quite clear on this subject.

the amperage of the phase converters you describe are amazing. I'm interested in that level of tuning.

my project is on a 400 amp 240 service and a phase generator system comprised of 2 50 horse baldor idlers (1.15) to drive what draws 350 amps single phase.

the phase converter setup right now draws about 50 amps with the two motors running.

my main concern is about the draw of these idlers.

let's say the two 50 horse idlers draw 50 amps with nothing else going on. how much do the 50 horse idlers draw when the shop is in full swing and our service is at 400 amps?

let's say the machinery really would draw 202 amps off a utility three phase service, and that's the draw of normal use. this is about 350 amps converted to single phase, am I right?

so we can say that 350 amps single phase are going to be consumed by the machinery. we can also say that 50 amps will be consumed by the dual motor phase converter at idle. how many amps will the phase converter system draw with this full load going on?

As I understand your statement, you have two 50 hp motors, running in parallel (which is electrically equavilent to a single 100 hp converter motor) and combined, they draw 50 amps. As you start other motors, the current of your 240 1 ph service will increase about 1.73 times the data plate amperage of each load motor. The extra current that will be drawn by the converter motor as load motors are started is included in that calculation. So, if all the load motors data plate 3 ph current totals to 202 amps, then you can expect the 1 ph power draw to increase by 350 amps. This, plus the 50 amps drawn by the two 50's will total 400 amps. As stated earlier, unless all load motors are drawing their full horsepower (which is usually not the case), you should not see the full 350 amp increase.

Remember that most of the current drawn by the load motors is directly from the 240 v, single phase incoming lines. The converter motor is generating voltage in the third leg only. The increase in current draw by the converter for this "third leg generation" is included in the 1.73 multiplier.

Hope this helps.

heh

ok ok. I'm up to speed I THINK.

the way our converters are wired and the potential problems this poses are too much to recount right here.

can I ask you how much you would expect a dual 50 horse converter unit to draw, if you were building it?

The biggest I have built have been 40 hp and the highest current draw of these was 15 amp.
The 30's have all been less then 10 amp, and the 20's have all been less than 8 amps. Based upon the trend, I would expect a 50 to draw about 20 to 25 amps (that's idle current, not starting). Of course, two 50's (or a 100hp) would then draw 40 to 50 amps.

one question on this old topic:

Let's say a 10 horse idler- rotary phase converter draws 5 amps while idling. I'm assuming most of these 5 amps is being used by the idler motor just to rotate at speed. Therefore, if I took an ampere reading on one of the legs of the idler motor, I would see about 5 amps.

Let's say a 10 horsepower three phase motor is started on the RPC's output... and then ran at full load.

if you took an amperage measurement directly off the idler motor at this full load condition, would the idler be drawing more than it did at idle?

Agree with above in general. The converters I've built draw about 12% of theoretical maximum at idle, which is in line with above.

Two points:

1) Idle current is strongly influenced by power factor of the converter. Sometimes this is even more true under load. Balancing the converter with run caps goes a long way to reduce idle current. I have never had to use power factor correction caps (across phases A and C) if the converter is well balanced.

2) How you wire the shop setup will influence the amperage you measure. If all your load current on phases A and C goes through the wire (bus) feeding the idler, the amperage will obviously be much greater under load than if the only thing on that wire is the idler itself. This is really not a good wiring setup. A much better setup is to feed the three phase load machines from a subpanel which gets phases A and C directly from the single phase service panel, and phase B from the idler. This is what I call an integrated panel design. It makes current measurements and overcurrent protection easy compared to the single wire shop bus system.

atetsade, to answer your last question, use the conservation of power method.

The 10 hp motor at full load will draw 7500 watts. At three phase (balanced), each leg will draw 7500 watts/240 volts/1.732 = 18 amps.

If the shop uses a panel bus I described above, the idler will have its own single phase feeder. The current definitely will increase under the load condition. By how much???

The load motor is getting 18 amps single phase from phases A and C. That's 18 amps X 240 volts = 4320 watts. The other 3180 watts must come from the idler for conservation of power to hold. For the single phase idler, the current increase above idle current will be 3180 watts/240 volts = 13.25 amps. So your idler will go from drawing 5 amps at idle to 18.25 amps under load.

As a check on the power conservation principle, you can view your load motor as getting all its power from single phase. That would be 18 amps + 13.25 amps = 31.25 amps single phase.

Single phase power is 31.25 amps X 240 volts = 7500 watts.

450 amps total

-60 overhead (literally)

-22 both phase converters idling

= 368 amps 240v single phase consumed by the machinery

except instead of 22A both phase converters idling now I have 60 and 25, or 85 amps running the phase converters.

the phase converter idlers seem to hold an equal balance on the three legs, and they'd be the only three phase motors in the shop that do. but as a correlary this means that they supply only 85 amps of third leg power.

Overall the shop machinery pulls 305 on phases A and C. Phase B, if I'm to believe this, is 85 amps.

Am I to believe in this much of an imbalance? I have seen imbalance... but I don't believe the machines are only seeing 85 amps of generated leg power.

Is this possible?

Is there a net effect rotary phase generation occurring in each three phase motor we have running right now, bolstering the whole system, or did I just prove that there is precisely 50hp* of phase converter idlers in those two units and these machines are running horribly imbalanced...

*50 x 750 \1.732 \240 = 90.2

Are you taking into account the fact that the single phase current supplying the idlers will increase dramatically under load(compared to idling under no load)?

Is there any way you can measure the B phase current actually going to your load machines?

Using the logic from my last post, your B phase current should also equal 305 amps if shop power were perfectly balanced to the load machines.

The increase in idler current on the single phase input to the converters should be 223 amps, for a total idler consumption of 223 + 85 = 308 amps.

Since your service can only supply 400 amps, something does not add up correctly (305 + 308 = 613 amps), and it is very possible your converter IS horribly unbalanced, causing your load machines to operate essentially single phase on legs A and C.

The above does not apply if you are including the single phase converter supply amps in the 305 amps. In that case you have a common bus, and different math must be used in applying the conservation of power priciple.

bnelson

because of the goofy way this thing is cobbled together, I'm looking at current readings directly off the idler leads.

line to line 215, 235, 225. 350 single phase on the main. phase converter leads 50A, 55A(*generated), 55A.

The other one is at 25A(*generated).

This is telling me that a total of about 80 amps is being generated by the RPC's.

Unless there is a net system effect of 25 machines, all with three phase motors contributing to the idler effect, there is a problem.

Let's turn off the phase converters and see if these screw machines run by themselves eh?

They will, once they're started!

The real question you have is how well-balanced your shop power is.

OK, let's do the math assuming a common bus. There's 350 amps total right off the service feeders. That's 240 volts X 350 amps = 84,000 watts.

If both your idlers together are generating 80 amps under load, assuming perfect balance, your combined screw machines should be drawing 240 volts X 80 amps X 1.732 = 33,254 watts, which equals 44.6 hp.

That's not even close to 84,000 watts (112.6 hp).

Are you REALLY measuring 350 amps right off your service feeders when the whole shop is running???

If so, your shop is essentially relying on the multiple load machines to supply B phase current on the common bus. I would strongly suspect if you took current measurements directly on the lines feeding each load machine, you'd find a high degree of current imbalance (> 50%).

bnelson:

500 amps at times. steady, for maybe 1/2 hour at a time. 450, 500 steady no problem. imagine 400 amp slo-blow fuses in a 400 amp Challenger box (should only be at 75%), it drops 1-2 volts across the fuses at those levels.

When they blow, they need to be removed with channel-locks, as they are about 300 degrees farenheit.

even at 500 amps single phase draw the phase converter idlers will show 55 amps and 25 amps respectively on the manufactured phase. 60 and 30 tops. If we are at 260 amps on the service then I will see 50 and 23. I think this is telling me that if I can get 90 amps out of them they are combined 50 horsepower.

I have a 100 hp unit ready to go, I just need a few capacitors and some other little stuff to hook it up.

But if I do hook it up do I suddenly see that the machines really use 500 amps and add on 50 for the lights and another 100 for the phase converter and now even a 600 amp service won't do??

hahaha. that would make it a hard sell in a place when things have been running like this for 25 years. This is difficult because the current (!) situation is beyond my control and even imagination. Everything works and has for a long time but Jesus.

I'll take some more measurements today and tell you about line-to-line and current balance under various conditions. thanks for the insight so far. it is very much appreciated and quite timely if you know what I mean.

To give you an idea of the cap values needed for balance, figure 15 uf/hp across phases C and B, and 9 uf/hp across phases A and B. That works out to 1500 and 900 uf respecitvely.

They make 100 uf run caps:
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&category=4662&item=3825060493

bnelson

Lord only knows what else I've ignored. Here are some of the latest readings:

taken at 450 amp draw on the service, 237.1 volts above the fuses, 236.4 below

three phase power voltage at 210, 219, 236 line to line. 118, 178, 118 line to ground.

RPC idlers amperage at 48, 54, 56*(generated); and 28.3, 29.2, 25.5*(generated).

I took readings on a machine this morning before anyone was around, 15 horse brown & sharpe ultramatic. amperage on phases was 7, 17.8*(generated leg), 17.

Same machine now 18, 3.5, 17.

There is no doubt.

Now my question is whoa, what if I had the phase perfect running this brownie... would it be 18, 18, 18.... 12, 12, 12.... or what?

if I had balanced three phase would it draw more current and put me in an even worse situation with the service entrance melting? almost certainly.. I feel very alive.

Very useful data. No doubt that you have an exceptionally weak B phase. The B phase to ground volts should be 208.

Essentially most of the machines in your shop are likely running close to single phase. This means they consume MORE current and put out LESS mechanical power, with consequent overheating of everything except the B phase.

In spite of what you might think, it will cost you very little to fix this situation. Probably all that's needed is lots of run capacitors. You don't need a new converter, unless you want to consider adding a third idler (not a bad idea).

After all is tweaked into balance, your shop overall will consume LESS power and machine performance will improve (slightly more mechanical power, smoother operation with less vibration and noise).

Basically there are two ways to do it:

1) Place all your run caps on the two (or three) idlers, using a balanced design. For a shop your size, this is what I'd do.

2) Place run caps ONLY across phases C and B on the idlers, and at each load motor, place only enough capacitance across phases A and B to get current balance below 35% for that single motor.

Either way, you'll see your shop power bill decrease. You'll probably stop blowing fuses. It'll be a quieter place to work.

Proper balancing NEVER raises power consumed unless the balancing was so off to begin with that the machines were only putting out about 50% of rated hp. Usually, proper balancing causes a noticeable DECREASE in power consumed.

To find out the real power draw of an
idling rotary phase converter, one needs
to measure *only* the in-phase (real)
power consumed by the device.

The state of tuning will affect the
total power draw - real plus reactive
power, but the real power consumed will
be the same for a tuned vs an untuned
converter.

If one does not have a meter that allows
separate meaurement of real and reactive
components, the best way is to use a
floating oscilliscope with a small sampling
resistor in the power feed of the converter.

This allows a simultaneous meaurement of
the total current, the voltage, and the
phase angle between the current and voltage.

As a point of reference, my five hp untuned
converter draws about two hundred watts of
of real power when running unloaded. All
of that supposedly goes into small bearing
losses and windage loss on the rotor.

A simple amp-clamp measurment shows around
ten amps being drawn at 240 volts - but
this is nearly all reactive current.

Jim

bnelson:

according to our discussion... I've given it a rest, I think I overloaded myself trying to fix things in this shop.

there is a great resistance against doing anything to advance the shop I work at. it's a lot of negative *****ing and carping about everything that goes on. mainly about things they don't control because they don't know anything about them... but these improvements have really been something they can control and strangle.

that's the way I see it.

I guess I'm interested in your last post because it is an academic-test situation. I started writing down a lot of the stuff we've been talking about here. I'm also coming to grips slowly with more and more of the interesting stuff.

especially this phase converter technology. people don't like it.. but they're stupid.

I'm stupid too, but I have a couple of questions.

bnelson:

from our previous discussion I had assumed that the two arco roto-phases I was looking at were about a 30 hp and 15 hp. this could jive if only because of the 55 amps per leg on the big idler and 25 amps per leg on the little guy... and I'm assuming full load because they reach these currents at the idler when the shop is in half swing.

then everything above that is just starving the B phase... the more machines we run the worse it gets.

another calculation: 30 + 15 = 44 hp or 33 kilowatts. 33kw at unity pf is 79.4 amps 240VAC three phase. 80 amps three phase is what those idling motors add up to.

I don't think that's right though--I suspect they're bigger motors that never reach full load and have bad efficiency. but now I'm confused because all these motors are parallel, the idling unloaded ones should they exist only provide a little boost to the B phase.

you think it's possible to balance out this situation with a bunch of capacitors? equal capacitance A-B and B-C would bolster this B phase enough to balance the whole shop?

these motors are running with about a 80% current imbalance on one leg, basically single phase. assuming the motors load out to ~100 horsepower I'd need as you say 2.4 millifarads of capacitance but is this really going to balance out the system?

or is it necessary to provide more horsepower in the unloaded idler motors to catch up... using capacitance to fine tune and shed pf losses...

I guess the answer would be... could I turn off the idler motors after everything is started running and add a grip of capacitance into the unpowered B leg and still achieve balance?

seems to me capacitance is cheaper than horsepower why not use it alone to create three phase?

jim rozen:

did you just describe a way to measure real power with an oscilloscope, obtaining vars and watts?

could you please explain the procedure for this if it's possible? I don't have a current clamp transducer type thing for an oscilloscope but I do have a 2 channel kikusui... what did you mean by "floating oscilloscope" and where exactly is the placement of a small resistor in the power feed of the converter?

[This message has been edited by atetsade (edited 07-15-2004).]

these motors are running with about a 80% current imbalance on one leg, basically single phase. assuming the motors load out to ~100 horsepower I'd need as you say 2.4 millifarads of capacitance but is this really going to balance out the system?

or is it necessary to provide more horsepower in the unloaded idler motors to catch up... using capacitance to fine tune and shed pf losses...

Click to expand...

Capacitance alone should do it. For EACH of the two 50 hp idlers, place 750 uf between phases B and C, and 450 uf between A and B.

Certainly do that before adding another idler.