[Matt@RFR]

I know there's a few guys here that have their machines running on phase converters. My utility bill each month is between $500-$600 and I only run the machine after work and on the weekends. Maybe 40 hours max or so per week. There's been many weeks that I didn't even start the thing too. The machine is an '07 VF-2ss running on a 50hp American Rotary phase converter.

The majority of my work is prototyping and very small runs (max of 10 parts on average) so the machine is sitting idle quite a bit, and the biggest cut I take for the most part has the spindle at 50%.

I took some amp readings today and will share the results. I'm no electrician, so hopefully I did everything correctly. The way I measured amperage is with a clamp-on amp probe clamped onto one of the wires coming from the main panel into the phase converter.

At idle, it pulls 14A per leg.
Starting spindle from 0 to 12k rpm pulls 140A per leg (200% spindle load).
12k rpm steady state pulls 18A per leg (3% spindle load).
12k rpm, 50% spindle load cut pulls 27A (50% spindle load).

I used this formula to come up with Killowatts:
Amps per leg x 2 legs x voltage (240) /1000 = killowatts.

We get charged roughly $0.20 per kwh. So worst case scenario, if I roughed something out at 54A for 40 hours/week that would cost me $415 over and above our regular house usage. That works out to taking up $11 of my hourly rate. Not bad, but I don't use the mill like that. A big chunk of its time is sitting idle while messing with programs, changing offsets, etc. That, and the majority of the tooling I run is 1/4" and smaller, down to very small endmills. I'll see 3-5% spindle loads for hours on end.

What are your thoughts? Is this in line with what everyone is seeing?

[gar]

080323-1604 EST USA

Matt@RFR:

First, you doubled your current, and second you do not know your power factor. Also you are assuming the current is a sinewave.

Your cost sounds high.

Let us assume your $0.20 cost per KWH, then $600 = 3000 KWH. Assume your house base average load is 1.5 KW and a 30 day month, then 30 * 24 * 1.5 = 720 + 360 = 1080 KWH. This leaves 1920 for your shop. Divide by 160 and you have an average load of 12 KW or 12000/746 = 16 HP. Does not seem to compute.

Even if you assumed the idle load of the phase converter was resistive that is a load of 14 * 240 = 3.36 KVA . Actually it is probably about 2 KW of power, or slightly under 3 HP.

Use your service entrance watt-hour meter to measure the actual power input at idle load and then at your 50% machine load. Also try to determine the average base load for everything except your shop.

.

[Donkey Hotey]

I second the idea about checking the meter. In fact, I tracked usage on my house many years ago. I read the meter daily and tracked it in Excel. It was an interesting exercise in learning just how much power a refrigerator costs per month (!!!) or a waterbed heater (!!!!).

Take a reading before you leave in the morning, mark the time. Take another reading when you get home and mark the time. You can figure out roughly what your home's parasitic load looks like from that. Then fire up the shop and do a 'typical' evening of work. That'll give you some indication of the real power consumption.

Somewhere in there, you're running a good sized compressor to keep up with the spindle. With that kind of money tied up, you probably treated yourself to a bunch of big florescent lights. Do you have a TV running out there while you're working? A gas furnace (fan)? An electric heater? They all sound insignificant but they add up quickly. Especially if you've got a family and the rest of your house is still running 'normally' while you're outside (other TVs, lights, cooking, etc).

After all of that, your bill does still sound high. How often do they really come out and read your meter? Has this been a couple of months? One of the reasons I started tracking usage was they misread one of the dials on my bill and sent me a huge bill one month, then didn't read it for two more (estimated). It caught up on the fourth month but that didn't matter: I had to pay it.

[davereagan]

50hp sounds like major overkill. I don't think you often use more than 15hp on that mill. For reference, 15 hp will cut over 20 cubic inches per minute of mild steel with a good face mill. Example: .15" depth of cut, 2" width of cut, 65 inches per minute.

[gar]

Another perspective.

If you think your phase converter is the cause of high electric bills, then it has to get very hot.

A phase converter is a relatively small package and suppose from my above assumptions that you take 1/2 of the 12 KW figure or 6 KW and dissipate this within your converter package and imagine how hot it will get. You can judge this by what would happen if you put 6 1 KW heaters in side the converter package.

A phase converter only supplies 1/3 of the power to the 3 phase load, the rest comes directly from the single phase lines. View the load as a three phase Y connected resistor bank and you can see why. If a 50 HP converter is only supplying 50/3 HP and if the converter is 75% efficient, then the losses in the converter at full load would be about 50/(3*4) or about 4.16 HP or 3.1 KW. But you will never run at full load. A converter will not be as efficient at light load as full load, but my guess is that you might expect that a 1 KW average load from power loss in your converter. Thus 160 hours is 160 KWH or $32 per month for your converter. These are wild guesses but illustrate an analysis method.

Cross checking this with your 14 A at idle and 240 V we have 3.36 KVA and I do not think my 1 KW figure for average dissipation is too far off.

(edit) Get a wattmeter and an appropriate current transfomer and measure actual power input to your phase converter. Maybe better get an electronic wattmeter that uses a clamp on Hall device current sensor. Note: ordinary transformer type current transfomers are a serious problem if the secondary is not connected to a low impedance. With the secondary unloaded it can produce a destructive voltage. (end edit)

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[bob1112]

When we tested converters, it seemed they pulled between .2-.3A/HP of the slave motor. This could be 10-15A just sitting there on the single phase side. Rotary converters are pretty efficient "when" loaded between 20-90%. The SS can easily pull 10A of three phase just sitting from the servos being fired up and all the fans and such running. The SS also does not have a counter balance and has to hold more power on the Z servo to keep it in position. I can't remember what they do under power down.

Our other shop has 2 of them and they love power but even with 2 of those and three others, the bill does not exceed 600. I would guess the reason you have a 50HP is because of the amperage?? I know the breaker in the SS is 100A! They really do not need it. You could sneak by with a 25HP converter on that machine for what you are doing.

We almost had to run the SS machines with converters so this is interesting. The only thing I can recommend if you only need control functions, you can add a circuit to only power the 120V control of the machine until you are ready to fire up the rest of it. You may also try testing with the estop on to see if that can drop consumption a bit.

If you wanted to back up to a 25HP converter, just find a slave motor and retune the caps a bit. My self or others here could walk you through it. I am interested to see what happens here because this could be a scenario for me soon. I have a lower HP Haas on a converter right now but need to get an SS.

[gar]

080324-1906 EST USA

Our 20 HP VF-0E uses the Z servo for counter balance on Z.

With servos off the input lines read about 1 A for the pair supplying the single phase load and the wild leg reads near 0. With servos on and no motion the said 2 legs are about 1.2 to 1.3 A and the wild leg is about 0.8 A.

Not very much power consumption.

.

[bob1112]

Wow, that is much lower than ours. That might mean he has a problem with the converter slave motor or mill. Maybe a meter issue??? Never heard of it but 600 bucks sure seems high. I did notice he said 20cents/kw. I think ours is around 9 cents here.


Gar, just curious what meter you used for that test? Hell, my old machine has three cooling fans in it that pull more than that! Now I have to go test my VF1.

[Matt@RFR]

Thanks for the help guys. What I'm going to do is track my electicity useage via the meter and see if that brings any light to the subject. I think that will take a couple weeks to get a good average, so I'll update this thread when I feel I have something decent to post.

The reason I have a 50HP phase converter is simple: When I bought this mill, I had literally never touched a CNC machine before and didn't know exactly what I was getting into. I had no customers (it was bought for another business venture that failed) so I didn't know how I was giong to use it, and there's allways the possibility of getting a CNC lathe at some point. It made sense at the time, anyway.

And just to clarify, I never refered to all this as a problem. I just wanted to get an idea of whether my electicity bill was inline with what others were paying in similar situations. Ofcourse, if there IS a problem I'd love to get it solved because these electricity bills are a much bigger portion of my shop rate than I initially figured.

Our other shop has 2 of them and they love power but even with 2 of those and three others, the bill does not exceed 600.

Do these machines run on a phase converter?

[SORCHEROR]

the pahse converter is running even if your machine isnt,50 hp rotary uses alot of electricity,even unloaded,you are paying a higher rate in CA then i do in NY,when i ran my 50hp converter my bill was always higher,alot higher

[gar]

080326-0913 EST USA

If you are going to make usefull comparisons, then you need some quantative data.

As a residential customer what you buy from the power company is energy, not current. A current input measurement to a device may not be a good predictor of power consumption. Note energy is the integral of power over time.

Consider the 50 HP converter that has been discussed. Unloaded this is nothing more than an unload motor. The primary power consumption of an unloaded motor is core losses, friction, and windage (really part of friction). All input power in this case goes into heat. It is not unreasonable to assume these losses represent 10% of the motor rating. Thus, 50 HP would be 5 HP = 3.73 KW. If this motor idled for 40 hours per month that is 149.2 KWH and at 0.10/KWH this is $14.92 per month, or double that at 0.20/KWH. This is not a large figure in comparison with a normal residential bill.

What you need to do is put a wattmeter at the input of the unload RPC and see what is the actual power input.

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[bbrown01]

This is a very interesting topic. I just installed a 25hp rotary converter for a yet to be installed VF-2SS. The company claims it has a starting capacity of 25hp but a running capacity of 50hp. Unloaded, the converter draws 55 amps at 248v. That's a whopping 13,640 watts without the machine connected! All the literature on the product claimed the converter didn't consume any power. I feel like I've been duped.

This particular model has circuitry that measures the voltage and current of T3 and switches in and out a set of capacitors to maintain the voltage of T3 to within 5% of L1 and L2. This point is settable and it is called low and high modes. The unloaded current is 55 amps in low mode and 35 amps in high mode. High mode is what the converter should be running in while SS is cutting.

The problem is (other than my future electric bill) just having the SS powered up is not enough to switch the converter from low to high even at it's lowest setting. It requires something like 20% of the converter's rating to switch it and the set point is adjustable after that. It takes the spindle to be turned on to switch from low to high. I'm not sure what the effects (current spikes) of switching low to high and back will have on the machine.

Should I be concerned? I don't want to risk damaging the machine. Even though I just spent $5k plus shipping on this I may rip it out and install a Phase Perfect.

[davereagan]

How are you measuring the 55 amp draw? That is enormous. It sounds just as high as the FLA rating on the nameplate. If the motor is not doing any work, that all turns to heat. What happens when you run the converter for even a half hour unloaded? I'd be worried about it burning up. Have you called the company and asked for guidance or an explanation?

Dave

[bbrown01]

Hi Dave,

I measured the current with a Fluke 322 clamp meter. My electrician's clamp meter had the same value. I have only run the unit long enough to check the voltages and current. Maybe 15 min. Doesn't seem to get warm. The company claims it can be run continuously without load. I have been in touch with the mfg. and they said this is normal operation. I also have a test report that came with the unit with similar results.

-Brian

[slatern44]

I know I have posted this before but call your power company and just asked them if you can get 3-phase power. I was running on a rotary converter for about a month when one of my customers asked me why don’t you call PGE and ask if you can get 3 phase. So I did and 30 days later they put it in for FREE, I had to pay to re-wire my shop and the meter base but that’s it. Now two more shops down the road from me that were also running rotary converter also call PGE and they go 3-phase for free also.

[bbrown01]

Slatern44,

Thanks for the suggestion but I already tried that route. I applied for 3-phase from PG&E on Jan. 12, 2006. Not sure what it's like in Yamhill, but here in Santa Cruz county they are worse than the planning department. I've heard almost nothing (except for personnel change notices) from them since my application. I think I'm stuck using a phase converter.

[slatern44]

I understand, I also got told no the first time that why I got the rotary converter. If you do try it again make sure you talk with someone in the service and design department because they are the one that will draw up the plans and get them approved.

[gar]

080406-1832 EST USA

bbrown01:

Did you read and in any way understand what I said in post numbered #2?

Current and power are not the same thing. Instantaneous power is the product of the instantaneous current and voltage of a load. If you multiply sin b * cos b and average this over one full cycle what is the result? Suppose v = Vp * sin a, and i = Ip * cos a, and you let p = v * i and average this over one cycle to determine average power what is the result?

Go find yourself a wattmeter and measure your input.

.

[Andre' B]

Probably the converter has a really bad power factor when at idle. Which would mean the current you were measuring was not real but reactive current. Over simplified reactive current cycles back and forth between the device and the power grid and does not transfer any energy.

http://en.wikipedia.org/wiki/Reactiv...apparent_power

If that thing was pulling that much real power at idle it would melt down in a short time.

[larrycoyle]

I have two rotary phase converters running in my shop and they pretty much run close to twelve hours a day at least five days a week. One is a 40HP CNC phase converter that runs my 5-axis FIDIA and the other is a 30HP that runs my 15HP Ingersoll Rotary Air compressor. That along with the air conditioner that runs constantly and my highest electric bill is around $350.00. The phase converters are GWM and are made in Missouri. They are damn nice too. It sounds as though you have something else going on in the woodpile.