roto phase converters

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hardworkingstiff

Senior Member
Location
Wilmington, NC
The manufacturer says their 10KVA unit will run 10 HP worth of motor so long as the largest starting motor in not greater than 5 HP.

OK, so I ordered that unit for a drill press that has (8) 1 HP motors. The customer said he can control the starting of the motors so only 1 is started at a time.

I picked up the unit today and the output is rated 22-amps.

The motors are rated 3.6-3.8 amps (208-230V). That's close to 30-amps.

3.6*(208*1.732)=1296*8=10368 VA.

I called the manufacturer and they confirmed it will run it OK.

Checking the web I found a HP = 746 watts. So, let's say 8*746=5968 W.

5968/10368=57.5%

So does this mean that the motors have a 57.5% power factor?

Am I wrong to be suspicious about the unit being cable of handling these motors?

Thanks
 

billyzee

Member
Efficiency

Efficiency

You have to make sure that you get efficiency included in the calc.
Motor HP is output shaft HP. So you need to divide by the efficiency to get input HP. Then go through the rest of the calc
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
101030-0753 EDT

hardworkingstiff:

A simple calculation may not produce a useful answer.

An example: My son had a sawing operation a number of years ago for several years. The part was an extruded aluminum shape. Roughly every 7 seconds a cut was made. It took between 1 and 2 seconds for the cut. The motor was rated at 5 HP and my measurements were that peak input power to the motor during the cut was about 7.5 HP. The motor was not overloaded from an average power perspective. Approximately 600,000 cuts per year were performed. Never a motor problem.

Different conditions. On our various CNC machines, all rated about 20 HP, the load is seldom run near full load, even on peaks.

In your application the important question is what is the peak simultaneous KVA requirement? Do all motors get loaded to a peak value simultaneously? If not can you determine the maximum peak KVA input? How long is the maximum peak? How does this peak relate to the converter capability? Is this application peak limited or average limited? You need to ask the customer questions, and also make some measurements.

.
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
The drill press drills 3/16" holes in wood deck boards (pine at 1.5" thick or Brazilian hardwoods at 1" thick).

The basic operation is the table pushes up with air powered hydraulics into the drill bits. I don't think the motors will be loaded severely since it's wood and small holes.

I just didn't understand how the motor nameplate loads could be so high compared to the 746 watts per HP.
 

Rick Christopherson

Senior Member
The reason why the total output can exceed the motor rating of the Idler motor is because each running tool motor partially serves as an idler motor to the whole system.

Make sure you size your L1 and L2 phase for the total load. They don't need to pass through the converter.

An idling motor will have a very low powerfactor, so a good rule of thumb is that it will draw 1/2 the nameplate FLA.

For most motors (especially for woodworking equipment) I use the following to approximate loading:

HP = I x V x pf x eff, where
pf = 0.85 to 0.9 at full load
eff = 0.8

Don't forget sqrt-3 for 3-phase calcs.

Also, when sizing the incoming single-phase power, take the total 3-phase load and divide by sqrt-3, and then add the single-phase load of the RPC.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
101030-1223 EDT

hardworkingstiff:

I looked at a 2.4 HP motor, 1.8 KW, 230 V, and it listed current at 6.74 A. Scaling this to 1 HP produces a current of 2.8 A. The ratio of HP to KVA for this motor is 0.67 .

An experiment with a Rockwell 3/8 drill with a 3/8 bit in aged Douglas fir and pushing fairly hard required about 300 W input power.

With an appropriate feed rate I doubt you will get to 1/2 HP load in pine with the smaller bit. However, you may have a higher RPM than the about 1000 for the Rockwell. You need to run an experiment in the hardwood.

What supplies the air? And what is air hydraulics? Is this air over oil, an air to hydraulic cylinder, or a hydraulic feed control?

.
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
101030-1223 EDT


With an appropriate feed rate I doubt you will get to 1/2 HP load in pine with the smaller bit. However, you may have a higher RPM than the about 1000 for the Rockwell. You need to run an experiment in the hardwood.
The motors are 1750 RPM and drive the heads via belts, I didn't look at whether their was a pulley diameter difference.
What supplies the air? And what is air hydraulics? Is this air over oil, an air to hydraulic cylinder, or a hydraulic feed control?
I didn't look too closely, the customer just said the air worked the table and the table is run by hydraulic cylinders. I just assumed it worked similar to the in the ground lifts that used to be popular in gas stations years ago, air over hydraulic fluid. I know how to spell ass u me. :grin:
 

broadgage

Senior Member
Location
London, England
It should be fine, the total VA use of the motors is in excess of the converter rating, but only by a very small amount.
It appears unlikely that all the motors will be fully loaded.
It also appears unlikely that the motors will be continualy loaded, the drilling of holes is inherently an intermitant load, unlike say a continually running pump or fan.
 

mcclary's electrical

Senior Member
Location
VA
The bad thing about rotarty phase converters, is they need to be oversized to work properly. I was involved in one for a compressor, and the converter was engineered really borderlined. On cold mornings, or when it starts under load, the thing would not start the compressor. They really should not be designed "borderlined".
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
The bad thing about rotarty phase converters, is they need to be oversized to work properly. I was involved in one for a compressor, and the converter was engineered really borderlined. On cold mornings, or when it starts under load, the thing would not start the compressor. They really should not be designed "borderlined".

Was there only 1 motor connected to the converter?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
101031-1509 EDT

Some basic comments.

A rotary converter using a standard three phase induction motor and capacitors is not a very good three phase power generator. However, they do work and serve a useful purpose.

First, a standard induction motor is not a very good generator. So the phase converters may have capacitors added in the run mode that differ from just what is required to start the phase converter. But the choice of the capacitors is load dependent.

Second, one phase passes thru directly to the load. So the phase converter only has to supply power to the other two phases. The power to these other two phases comes from power to the first phase and the capacitors.

Third, if not heavily loaded a three phase motor can produce mechanical output power with a single input phase once the motor is up to speed.

Fourth, a rotary converter with a single large load may not be able to start that load because of the large inrush current required by the load.

Fifth, if the load is made up of a number of small loads where each load has its own large inrush, but in proportion to its steady state value, then the total of these loads can be started and run by only starting one at a time even though these total the one large load of "Fourth". Basically the approach that was stated in the first post.

Sixth, I believe that if a synchronous motor with an adjustable field was used in the phase converter that much better performance could be obtained. I do not have such a motor and have never run the experiment.

.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Fifth, if the load is made up of a number of small loads where each load has its own large inrush, but in proportion to its steady state value, then the total of these loads can be started and run by only starting one at a time even though these total the one large load of "Fourth".
Don't these other, running motors in fact assist in the starting (and running) of additional ones?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
101031-1537 EDT

Larry:

Yes.

But, induction motors are still not good generators.

Before the drilling operation the ones already started help in starting the next one in the sequence. Once into the drilling operation it is somewhat like adding some loosely coupled mechanical inertia.

.
 

iwire

Moderator
Staff member
Location
Massachusetts
There is theory and there is what actually works:), I always think of this thread on another forum when this subject comes up. :cool:



With my 200 amp single phase service I run a 40 hp on my sawmill, a 20 hp on my band resaw a 10 hp on my dust blower, Two 5hp and a 7.5hp on my molder, 5hp on my planer and a 5hp on my blower which pulls the shavings from around the molding and planing heads. All of these motors are 3 phase. To make the 3rd leg (manufactured leg) I use a 25hp 3 phase motor.

http://www.electrical-contractor.net/forums/ubbthreads.php/topics/128478/1.html
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
I read the entire thread. (it wasn't easy!)

What the OP is missing is that, when you use any type of transformer to raise voltage (without reducing the load current), the primary current will increase. The extra energy has to come from somewhere.

The only advantage of doubling a service voltage to supply a dual-voltage motor is to be able to use smaller conductors and OCP, but the voltage ratings may have to be increased, so it could be a financial wash.
 
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