Another Rotary Phase Converter

Blackjackjacques said:
Nice - when redoing your door, remember to leave about at least an inch of clearance between the panel door and any current carrying part. I cannot for certain figure out what is going on with your pf and the meter readings. The pf corrected circuit should normally read substantially less current than uncorrected. From your photos, the L1 L2 meter shows 11.8 A before pf correction and about 21A when corrected -- if I am reading your meters correctly and assuming that L1L2 is your single phase input and T1 T2 T3 is your derived 3 phase output. It would be helpful if you can provide a simple one-line schematic showing the connection of the motors, including the load motors and how everything is wired.
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It is a bit confusing. The L1-L2 meter at the top is showing PF corrected incoming power and the CT is on the input leg. From the panel T1, T2, T3, go to the motor, the lathe is paralleled off of the motor leads at a J-box on the wall. There is a CT on each leg in the panel. Once the two power contactors pull in, the PF correction caps are across T1(L1)-T2(L2), and the balance correction caps are across T1-T3, and T2-T3.

RPC As Built.png

Actually I have just figured out how to build an over unity machine, puts out more power than it is supplied with :cautious::grin:
 
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JimDawson said:
It is a bit confusing. The L1-L2 meter at the top is showing PF corrected incoming power and the CT is on the input leg. From the panel T1, T2, T3, go to the motor, the lathe is paralleled off of the motor leads at a J-box on the wall. There is a CT on each leg in the panel. Once the two power contactors pull in, the PF correction caps are across T1(L1)-T2(L2), and the balance correction caps are across T1-T3, and T2-T3.

View attachment 251419

Actually I have just figured out how to build an over unity machine, puts out more power than it is supplied with :cautious::grin:
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For over unity it is called a synchronous condenser, or just a synchronous motor to provide leading power factor. If you are making more power than put in, then Elon Musk and the Nobel committee would like to talk with you. :)

In the old days, engineers used to mix, where they could, synchronous motors with induction motors to balance pf where the motors are on at the same time, etc. However, having too much leading VARs can be as bad as having too much lagging VARs. I'm confused as to what the meters are showing. I assume that the Idler and Lathe motors are both wired Delta. If you can provide the particulars for these two motors, I can run a quick analysis for you to model and predict the actual parameter values, VA, VAR, I, etc. Motor nameplate data will do, but if you have actual resistance/inductance for the windings, I give you a precise picture.
 
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JimDawson said:
The idler is a standard 9 lead, Y connected. http://www.baldor.com/catalog/CEM2333T#tab="specs"

The lathe 1 hp hydraulic pump is also a 9 lead, Y connected, but the other motors are all AC servos.
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The following analysis may help your efforts to validate the design and if you are getting the targeted performance of your system. Double check my math, but it seems viable.

On the 15 hp Baldor - since you are running 2 Y (low voltage)or two identical Y windings in parallel, the impedance of the windings would be half than if you ran it in the 1 Y configuration. The 0.83 power factor rating for the motor does not indicate if that is for the motor running 1Y or 2Y, but my expectation is that the winding configuration would impact power factor. This is where having the actual measured impedance of the motor windings would have been more valuable – and especially when we are configuring this 15hp motor to be placed in parallel with yet another (load) motor winding system. But let's go with the manufacturer’s .83 rating for now:

Looking at the 15 hp motor alone and to take the pf from 0.83 to 0.95

AT.83 PF
Watt(P) (input) to motor = HP x 746 / eff. = 12.11 kW (bottom of the pftriangle)
VA(S) =P/pf = 14.6 kVA
VAR(Q) = VA Sin Θ= 14.6 kVA Sin 33.9 degrees = 8.14 kVAR
I= VA/V = (14.6 kVA/(230 x 1.73)) = ~36.7 A

AT.95 PF
Watt(P) (input) to motor = HP x 746 / eff. = 12.11 kW (bottom of the pftriangle)
VA(S) =P/pf = 12.7 kVA
VAR(Q) = VA Sin Θ= 12.7kVA Sin 18.2degrees = 3.97 kVAR
I= VA/V = (12.7 kVA/(230 x 1.73)) = ~32A

So 8.14 - 3.97 kVAR = 4.17 kVAR leading are needed or C=kVAR/ώV2=4.17 kVAR/2*Π*f*V2=209mfd. Connecting the capacitors in DELTA across the 2Y connections =209 mfd/3 = 70 mfd across each line.

So this approach improves the power factor for the 15 horse motor from0.83 to 0.95 and reduces the current from 36.7 A to 32 A.

Looking at the 1 hp motor alone and using typical NEMA values for eff. = 82.5% and pf = 0.8:

AT.80 PF
Watt(P) (input) to motor = HP x 746 / eff. = 0.9 kW (bottom of the pftriangle)
VA(S) =P/pf = 1.13 kVA
VAR(Q) = VA Sin Θ= 1.13kVA Sin 36.9degrees = 0.678 kVAR
I= VA/V = (1.13 kVA/(230 x 1.73)) = ~2.8 A

AT.95 PF
Watt(P) (input) to motor = HP x 746 / eff. = 0.9 kW (bottom of the pftriangle)
VA(S) =P/pf = 0.95 kVA
VAR(Q) = VA Sin Θ= 0.95kVA Sin 18.2degrees = 0.297 kVAR
I= VA/V = (0.95 kVA/(230 x 1.73)) = ~2.4 A

So 0.678 – 0.297 kVAR = 0.381kVAR leading are needed or C=kVAR/ώV2=0.381 kVAR/2*Π*f*V2=19mfd. Connecting the capacitors in DELTA across the Y connection = 19mfd/3 = 6.3mfd across each line.

So this approach improves the power factor for the 1 horse motor from0.82to 0.95 and reduces the current from 2.8A to 2.4A

Therefore, with the 15 hp generator motor in parallel with the 1 hp load motor, the total capacitance across-the-lines would, in theory, be 70 + 6.3mfd = ~75 mfd to bring the system pf to 0.95 and the total corrected rms current you should see on your Fluke should be about 34.5 A.

What is missing in the analysis is the fact that the 15 hp drive motor is being driven with one broken leg, so its shaft hp would be reduced at least one third and the inrush power requirements will increase correspondingly, and its anybody's guess how the power factor will present until the motor comes to steady-state speed.

However, the analysis is valid if the drive motor was operated in a true motor-generator configuration.
 
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