RPC amps load question

[Ringo]

I built myself a RPC from a kit. I also used an instructions to install balance capacitors in it. I have 20a 220v 1ph plug, with a 5hp 3ph motor as the idler. With much fussing about with caps, I got the no-load 1ph amps down to 2.7a as read on a amp-clamp.
Now, my aircompressor tank died it rusted out. I need something to replace it.
By looking at electric motor HP charts & amps, and with 20a 1 ph available (less 2.7a overhead) that leaves me a 17.3amp availability? correct?
The charts telling me 3hp 1ph is 18amp and should work straight up 1ph.
The charts telling me 3hp 3ph is 10amp, for total 12.7amp with RPC (so, I can run 3hp 3ph for 5amp less, correct?)
The charts telling me 5hp 3ph is 15.2amp, plus RPC is 17.9 amp. (so, I can run 5hp 3ph for same amp as 3hp 1ph???)

The charts & amps tells me it will work, but can I really run a 5hp motor on a 5hp RPC?
I've looked at RPC with amp clamp and watch my mill & grinder ramp up the amps, it is the same baseline 2.7a plus the nameplateamp on the motors. this much math does add up. mill an grinder are 1hp an .5hp respectively.

[Doozer]

I have a 6hp RPC and it will start an 8hp motor on my lathe with NO load.
So it is more load dependent. If it is a real deal air compressor, and it has
an unloader that only cuts in when the compressor is up to speed (like oil
pressure based), them maybe the RPC will start it. But you really need
to try and see. Too many variables.

-D

[Rich Carlstedt]

Doozer Nailed it !
+1
Rich

[Ringo]

People been telling me a electric motor uses nameplate amps whether its loaded or not., and that part I don't fully understand. even if its running no load, it still uses the amps so they say.
and yes, the compressor I'm looking at has the unloaders, it starts no load and loads up as oil pressure comes up. the compressor unit tech says I can run it 3hp pulley reduction down, or 5hp higher speed reduction (rpm). its a real deal Quincy.

[chipmaker4130]

I don't think you can just assume 'no load' amps on your idler. It will load up as demand on the generated leg increases. I was told you should wire for idler full-load amps plus max expected attached load. Not that you'd regularly be at those levels of course.

[J Tiers]

People been telling me a electric motor uses nameplate amps whether its loaded or not., and that part I don't fully understand. even if its running no load, it still uses the amps so they say.
and yes, the compressor I'm looking at has the unloaders, it starts no load and loads up as oil pressure comes up. the compressor unit tech says I can run it 3hp pulley reduction down, or 5hp higher speed reduction (rpm). its a real deal Quincy.

"People" are telling you wrong.....

An unloaded motor uses usually about 40% of nameplate current, at a really low power factor, so the actual power involved is quite low compared to rated power.

Any motor that pulls nameplate maximum amps at no load has a problem and likely will heat up fairly fast, if it does not fail first.

[Ohio Mike]

Unfortunately its far more complex as noted above. Please note I an not an electrician, standard disclaimers apply.

By looking at electric motor HP charts & amps, and with 20a 1 ph available (less 2.7a overhead) that leaves me a 17.3amp availability? correct?

Mathematically yes, functionally no. You get close to that 20 amps and you will likely nuisance trip the breaker. This will depend on the load, length of the wire and frequency of motor cycles. National Electric Code actually states for continuous (non motor) loads you shouldn't exceed 80% of the rated capacity on the circuit. For motor loads they provide tables for proper circuit sizing. However the end result is conductors should be of a ampacity of not less than 125% of motor full load current.

All of this is a side bar from the original meat of the question which is can the RPC start the compressor. That will be impacted by the RPC itself, the load being started and the length and gauge of the wire running to the RPC. If I was betting and I guess I am I'd say with the setup you have can probably start the 3HP but not come close to starting the 5HP motor load. America Rotary suggests 10 gauge wire on a 30 amp breaker to feed their 5HP rotary, if the feed line is longer than 50 feet up the wire one gauge.

[J Tiers]

The general rule for "running" a motor is to oversize the RPC by 1 1/2 x. So a 10 HP load > 15 HP RPC. If the load is a hard starting load, then 2x is generally suggested.

That is what works for most. Maybe for you it is more than "needed", but generally, rather than trying to pare down the requirement to the absolute minimum-minimum size "needed" to "just barely start and run the load", it is better to go with what just plain works.

A compressor with an unloader is an easier start load, but you also have to RUN the load, and as the compressor comes up to full pressure, the load is increasing. Again, better to go with what works, rather than messing around trying to somehow squeeze 15 lb into the 5lb bag.

If you do not have enough "ampacity" on the circuit, rather than re-writing the laws of physics, put in what you need.

[challenger]

Should we even talk about "developed" HP? This "value" is something being shoved down the consumers throat as if it's a real thing. Put a 20lb flywheel on a 1hp electric motor and spin it up as slowly as necessary to get it up to the rated rpm and the hp may be whatever the marketing guru says it is.....
Everything is being dumbed down except the average consumer. They don't need dumbing down IYKWIM.

[Ringo]

I guess my OP point is the usage of the capacitors in the RPC. The capacitors made a HUGE difference in the idle amps of the RPC.
With no capacitors at all, the 1ph amps was whole lot higher, even at idle.
As I added in capacitors I saw incoming amps going down, way way down.
When I got done fussing with the capacitors, the incoming amps was way way way below the idler nameplate amps.
All said & done, I have varying size caps on all 3 legs. and the 5hp idler is idling at <3amps
As you load up the RPC, does that 3amp overhead remain a constant?
As you load up the RPC does overhead go up expotentially steep in amps?

[garyhlucas]

What you are seeing is the effect of power factor. An ideal power factor of 1 would be where the 60 cycle wave form of current and the one for voltage were exactly in synch. That is maximum voltage at maximum current and power is volts x amps under that condition. When a motor has no load the current and voltage go out of synch so minimal voltage times maximum current = very little power. So you still see the current with no load. What the capacitors do is supply the magnetizing current and correct the power factor so now you only see the actual current needed to magnetize the armature which is very little.

[Ringo]

What you are seeing is the effect of power factor. An ideal power factor of 1 would be where the 60 cycle wave form of current and the one for voltage were exactly in synch. That is maximum voltage at maximum current and power is volts x amps under that condition. When a motor has no load the current and voltage go out of synch so minimal voltage times maximum current = very little power. So you still see the current with no load. What the capacitors do is supply the magnetizing current and correct the power factor so now you only see the actual current needed to magnetize the armature which is very little.

I never quite understood power factor, but you indicating I corrected it.?
Does correcting power factor allow you to run a larger motor per given size RPC?
I fiddled with the caps until I reached the minimal amps possible idling. the additional caps lowered amps, until, I reached a point where additional mFd began increasing amps. I stopped at the low point.

[J Tiers]

The motor continues to draw what it always draws, there is no change in that. That current is required for magnetizing, and essentially does not change. The out-of phase condition of magnetizing current also does not change at all, it is essentially a constant.

It does not change when you put capacitors on it, and it does not change much when you put a load on the motor (the magnetizing current actually decreases a small amount*).

When you put a load on the motor, you increase the "in-phase" current draw, and the sum of the two currents (the in-phase power current and the out-of -phase magnetizing current) comes out to a larger current that is more in-phase, with a higher net power factor. The out of phase magnetizing current is still present, it just gets "diluted" by all the in-phase current.

When you add the capacitors, the motor again still draws the same current. What the capacitors do is to draw current when the inductance is not drawing current, and they then supply that current when the inductance does draw it. Net effect is that less current is drawn from the power line. But the same, or slightly more, actual "power" is drawn from the line. You trade a slight increase in net power requirement to get a net DECREASE in current required from the line (and thus less current in the breaker).



* The magnetizing current decreases because total current goes up, so the drops in resistance, both in the motor and in the wiring, consume more voltage, leaving the inductance with less net voltage across it, which means less current through it.)

[4miller1]

People been telling me a electric motor uses nameplate amps whether its loaded or not., and that part I don't fully understand. even if its running no load, it still uses the amps so they say.
and yes, the compressor I'm looking at has the unloaders, it starts no load and loads up as oil pressure comes up. the compressor unit tech says I can run it 3hp pulley reduction down, or 5hp higher speed reduction (rpm). its a real deal Quincy.

Another option is to add a small electric nc valve to the pipe between the compressor and tank.
Use a momentary switch to open the valve when starting the compressor releasing it when the motor is up to speed.
In effect a unloading valve.
I also put the same type valve on the drain and use a 2 position toggle switch so that one direction opens the drain for a moment
then the other direction to assist the compressor starting.

Jeff

[garyhlucas]

You really need a wattmeter to measure true power not amps. If you put a wattmeter on an unloaded motor you’ll see it using very little power, overcoming internal losses in wiring, the armature, and bearing drag. Back in the 70s during the oil embargo there were lots of scam artists selling devices that were just capacitors. They went around with a clamp on ammeter and small ac motor. They’d show you how their device ‘saved’ a huge amount of power when all they were doing was correcting power factor and reducing current. There are good reasons for correcting power factor. Resistive losses in motors and wiring are proportional to amperage not power so higher amperage due to poor power factor in increases those losses and reduces the total capacity of your system. I was industrial electrician back then and had to guide my customers around all kinds of power scams.