Rotary phase converter

Hi,I have a rotary phase converter with a three horse motor which I use to run a boxford 280 lathe which has a two horse motor can anyone tell me even roughly how much this set up costs to run?

If you can get a clamp-style ammeter around the line or neutral of the supply to the converter that will tell you the current, then V x I x time/1000 to give you an idea of kWh. It would be interesting to see the difference between when the converter is idling and when the Boxford is running under a bit of load.

Edited By Steve Pavey on 25/08/2020 16:57:57

Andy

You need to measure the input current to the converter and multiply that by the supply voltage, divide by 1000 to convert to kW so you can multiply the kW by your kWh rate cost per unit to determine running cost per hour.

Small plug in meters are available that will do all the calcs for you, just set it to kWh and input your unit cost, you plug the appliance into the meter unit.

Emgee

Assuming the lathe works flat out:

2HP = 1500W

and the lathe's motor is only 80% efficient = 1800W input

and rotary converter is 60% efficient = 2520W input

Average price UK electricity is 18.75p per unit, so running the machines flat out would cost a bit under 50p per hour.

In practice the lathe is unlikely to work anything like flat out so the bill would be less.

However, rotary converters waste power continually. If keeping a 3HP converter spinning consumed 250W then it will uselessly burn 4.5p per hour. If the workshop runs 8 hours a day for 200 days a year, the rotary converter wastes about £75 per annum, or £750 over ten years. In comparison, a VFD consumes almost nothing until the lathe demands power. It's another reason VFDs are preferred to rotary converters.

My lathe has a 1500W motor. In practice, it has to be worked exceptionally hard to pull more than 1200W from the mains, and that only in short bursts when the lathe takes a deep cut. In my workshop the lighting (6 x 58W tubes) consumes more power than the machines. It's because the lights are always on and I spend more time thinking and setting up than cutting metal.

Also assuming I did the maths correctly - don't bet the farm on it!

Dave

There ia a plethora of power meters on the interweb. Well worth the investment.

Thanks for all the replies I will get a power meter as well.So I hooked up a clamp on ammeter the inverter while idling is drawing 2.6 amps.When I turn the lathe on it draws 5.6amps and when I do a heavy cut its 6.4 amps

Flat out that's about 1.5kWh, SOD needs to change supplier and get a reduction on his kWh charge.

Emgee

Edited By Emgee on 25/08/2020 17:51:13

I quite often worked in lift rooms before retiring as an Electrician. Many had huge rotary coverters. These were AC input & DC output. I often used to look at the control panels , watching the load current when the lift was in motion & when static. AC draw.

When static the load was very low. Obviously when being used the current went up. I cannot remember any figures, only that the draw was minimal when not loaded. Point being i am not sure about some comments. If it was running all day of which it would only be when you required to run the machine. The current draw would only be at peak when being drawn from.

Steve.

Steve, the machines you are referring to would be Ward-Leonard sets. They use an AC motor, driving a DC generator, to produce DC power for the DC motor driving the load. The reason for all this is that by controlling the field currents of the DC generator and motor, you can get very smooth control of the speed of the load. They were very popular before solid state controls came along, being pretty much the only way to get the job done. They can also be used with an IC motor driving the DC generator, for instance for a locomotive.

The rotary converters that the other guys were discussing convert single phase AC into 3 phase, using a mixture of capacitors, inductors, and a three phase motor running as an idler. They produce better results than a simple static converter with no idler motor, but obviously cost more. These days, I wouldn't bother with either, since a VFD will do the same job and give speed control as well. Also with no power consumption when the load is not running.

John

Nah, I quoted the UK average price for a Unit of Electricity, which is 18.75p per kilowatt hour. Andy's figures are interesting, and I submit he isn't running his lathe flat-out! Based on the energy consumption, his 'heavy-cut' is much less than his lathe's motor could do.

Andy's numbers, assuming 240vac:

• Rotary Converter alone, 2.6A. That's 624W wasted, and worse than my estimate of 250W.
• Lathe Idling, 5.6-2.6=3.0A. That's 720W wasted turning the lathe, power gone in the belts, gearing and bearings without doing any useful work.
• Total current while cutting, 6.4-5.6 = 0.8A, or 192W of useful work.

Andy's Rotary Converter and Lathe are both more inefficient than my estimates. Rotary convertor is worse than I guestimated but they're doomed to be inefficienct! The lathe may be worth looking at. because it's consuming a lot of power uselessly. My Warco 280 takes less than 250W to spin the chuck and turn the lead-screw, maybe because it has roller bearings and a simple gearbox. If I were Andy, I'd let the lathe spin for 20 minutes and see if the current drops as the machine warms up (normal), and check the bearings for overheating (bad news).

On the face of it. Andy's "heavy cut" consumed 192W compared with the 1000W cuts I've pushed my Chinese machine up to. I'd expect Andy's Boxford to perform similarly to mine in terms of cutting power, whilst getting better finish and accuracy because his Boxford is more rigid than my hobby lathe.

By a heavy cut, I mean power traverse set to 0.28mm per revolution, chuck at 2400rpm, and a 5mm deep cut with a square carbide insert into mild-steel. Not what I normally do!

Dave

Guys....Do not forget about the power factor in AC circuits !

Independent measurement of line current and line voltage only gives real power in kW assuming the load is purely resistive eg an electric fire element. In this case the power factor, which is a measure of the phase relationship between voltage and current is close to 1.

For any equipment with coils and windings (inductance) eg motors, solenoids and traditional ballast type fluorescent fittings the power factor is less than 1 and it is more complex to measure the real power that the utility meter will measure.

Assuming that the rotary converter is driven by an induction motor, this may operate at a power factor of say 0.8 at full load. So in this case, using round figures if the line current is measured at 4A with the clamp meter and the voltage is 250V this gives an apparent power of 1kVA.

If the power factor is 0.8 this gives a real power of 1kVA x 0.8 = 0.8kW. A residential utility electricity meter only measures and aggregates to kWh the real kW = 0.8kW whereas you might incorrectly think this is 1kW by just measuring the line current and voltage.

For an extreme case of a purely inductive load the power factor will be close to zero. So even though 4A is flowing in the line, the electricity meter will read close to zero kW. All the current is "wattless" and just heats up the cables and utility distribution transformers without doing anything useful. The very reason why utility companies don't like poor power factor and many industrial companies use capacitive type power factor correction devices to avoid being penalised !

An additional problem is that the power factor for an induction motor gets worse as the mechanical load decreases. So a motor may take 30 to 50% of it's full load current even with no mechanical load connected and the current is mainly used to magnetise the motor. So with no load on the rotary converter you might incorrectly think that significant real power is being consumed just by measuring current and volts.

The only way is to measure the rotary converter power input is with a power meter, making sure that this measures both line current and voltage and can take account of the prevailing power factor.

Beware though that there have been a number of low cost domestic power meters which were only provided with a current probe and no voltage measurement. There was a big hoo ha with over reading of these meters in dwellings using traditional ballast type fluorescent fittings simply due to the low power factor.

I would have thought that the low cost "plug in" type power meters that can be obtained these days ought to include the voltage measurement but there is no guarantee of this though so careful reading of the specification is required to make sure that power factor is accounted for.

Lots of info out there on the interweb about power factor if there is interest eg. https://en.wikipedia.org/wiki/Power_factor

Variable speed drives are better in the fact that the "wattless" current just sloshes around between the drive and the motor and is isolated from the AC input supply by the internal DC link within the drive. However there are other supply issues introduced by VSD's such as harmonics and high frequency emissions but that is another story !

Sorry for my ramblings and hope this helps.

Adrian

Adrian and SOD

The question was regarding cost to run the gear, given the flat out input current my "about 1.5kWh" was based on 4.2A = 1kW

There is no need to take account of any efficiency of equipment and in the normal domestic kWh meter PF can be ignored, it would have to be taken account of if metering was by maximum demand kVA type metering as installed in industrial and some commercial properties.

Emgee

Have I missed a trick? I'm sure there is a reason but just can't help wonder why anyone would want to run a rotary converter in this day and age.

Emgee.

What you don't want to do though is to take an electrical measurement that grossly inflates the cost of running the gear and that must be the point.

It's a tough call Brian if you already own the rotary converter and it powers several machines. It might take many years of running the rotary converter to pay for the VSD's. Unless there are other reasons for doing so eg. "I'd quite like variable speed operation".

Adrian

The figure quoted doesn't exaggerate the cost if the lathe is running flat out at all times of use, i'm sure Andy would know that having quoted "off load" and "on load" currents.

Emgee

The problem is if the rotary converter is running light at low power factor (which it probably is most of the time) then the current and volts measurement of apparent power will be much higher than the utility meter will measure. So will provide a grossly inflated view of running costs.

You can see this from the 2.6A running light measurements which are getting on for 50% of the on load values.

There's no way you could push 624W into a small rotary converter with no output load. This would all end up as heat and overheat the converter. This does not happen due to the power factor when the real power absorbed (and measured at the utility meter) may be a couple of hundred watts, most of which will end up as heat in the converter motor.

All solved though if a correct power measurement is taken similar to the utility meter.

Adrian

Adrian

Did you see this message:

Thanks for all the replies I will get a power meter as well.So I hooked up a clamp on ammeter the inverter while idling is drawing 2.6 amps.When I turn the lathe on it draws 5.6amps and when I do a heavy cut its 6.4 amps

Emgee

Brian

Main reason for a rotary converter right now is if you need 440 V three phase for either permanently star connected motors or 440 V delta motors. (The latter being why my Hydrovane HV4 still hasn't been hooked up after a decade in storage.)

Until recently relative costs of VFD and rotary for multiple machines made the rotary converter attractive if you had several machines. Especially if they might need to be run at the same time. How to share VFD boxes used to be a very live topic! For anything we are likely to use that gap has more than closed. I'd be at about break even on 8 machines with individual drives against a 10 or 15 HP rotary.

But I use one of Drives Directs 10 HP modified VFD "whole shop" boxes. Around £2,500 and a bit today.

Couple of interesting things about motors and rotary converters. This one, written from the motor repairers perspective, is about idle, unloaded, current and why 1/3 rd rated is a decent rule of thumb **LINK** . This one, from another forum, is a discussion mostly about why about balancing the capcitors makes a big difference to current draw **LINK** tails off a bit with no final resolution but still interesting.

Clive

Emgee

Yes I saw this and the stator current figures seem reasonable for an induction motor no load to top end load.

Just beware that low cost power meters might just measure line current and not take PF into account.
Careful checking of the specs required !

Adrian

Au contraire, I suggest we do forget about Power Factor. Reason: we don't know what it is and it doesn't alter the price of fish much! (Maybe worth considering later if more facts come to light.)

Years ago I found a second-hand book and foolishly decided not to buy it. Regretted ever since and now I can't trace it! I think it was called 'The Weight of a Round Cow', though that may have been a Chapter heading. Anyhoo, the book was about the importance of making reasonable assumptions rather than bogging down in detail whilst estimating. Believe it or not the weight of a cow can be estimated without measuring an actual animal, let alone worrying about if it's full of grass or has been milked recently. Not an accurate answer, but in certainly in the right ball-park.

In the same way it's possible to suggest an answer to Andy's question. My estimate is of the worst case cost, ie assuming his converter and lathe are run flat out. As the boundary assumptions are stated, anyone can improve the estimate by injecting new facts. However, at this stage, none of us knows how hard Andy runs his workshop or for how long per day. Power factor is trivial compared with these unknowns.

Andy might torpedo my guess of an 8 hour 200 day year by revealing he's a shift-working slave driver. Not likely he's a running a sweat shop I feel. If he's a hobbyist a few hours light work per day are more probable, say 5% or less of my worst case number. My own workshop is considerably less loaded below capacity at the moment: I only used my lathe for about 10 minutes last week to skim some Aluminium, and I don't recall using it at all the week before. Only Andy knows how hard he works his machines, but he can get an idea of the cost by working back from my worst case numbers. In the context of the guesstimate, it's not worth worrying about power factor. The best Andy can expect is an order of magnitude indication, not precision.

Richard Feynman was once asked during a lecture if he'd given the surface temperature of the sun in degrees Celsius or degrees Kelvin. He replied: 'I don't care'. Not because he was a grumpy sod, but because a few hundreds don't matter when discussing several thousands with 10% uncertainty in the measurement.

Adrian's point becomes interesting if Andy nails down major uncertainties by measuring his set-up's consumption of power over a representative period of real work. As the worst-case is £75pa, it may not be worth the effort.

Dave