Single Phase current draw for a 3 phase output VFD: technical discussion

[Besoeker]

You're saying it's probably not a lot different... then you're saying it may very well be different.

It's simple physics If you put 10.5kW 3Ø in and you 10kW out, the process has a 95% efficiency (i.e. 5% loss). Then everyone (but Jraef) is saying it'll likely take significantly more 1Ø power in to get the same 10kW 3Ø power out. Logic dictates a lesser efficiency (greater loss). How can this be "not a lot different"?

No, I'm not saying that at all.
In both cases, the DC current conducts through two diodes at any one time. There's just more of them taking part when it's three phase.
The issue is how hard you can push the input rectifier when not using all of its capacity. The efficiency is a bit of a red herring.

 

[Besoeker]

If I can't tell them "1.732x the motor current, because you are feeding the same kW load from a single phase source", then what DO I say?

Consult the manufacturer.

 

[gar]

170517-1556 EDT

There seem to be two different discussions that have been going on.

One was whether an existing system designed for 3 phase input could be used with a single phase source.

The other seems to have something to do with input to output power, and input current related to something.

I am going to address current and power for the moment. Assume the three phase source is a 208 wye. The load is a balanced wye resistor. For example each leg of the load consists of two 20 ohm resistors in parallel making the resistance of each leg = 10 ohms.

Each 3 phase line current is 12 A. Power into each load leg is 1440 W. Total power is 3 * 1440 = 4320 W. You can do this with DC or AC supplies.

Now take all 6 resistors and put them in parallel and apply 120 V. Each resistor is worth 6 A and 720 W, or a total of 36 A and 4320 W. The current is 3 times the current on one 3 phase line. As expected.

Next organize the 6 resistors to be balanced across 240 V. Now the line current is 6+6+6 = 18 A, or 1.5 times the 3 phase line current. Total power is 18 * 240 = 4320 W.

.

 

[Jraef]

OK, but again, look at the current formulae for three phase vs single phase that I posted in the first place.
How does that relate to what you just said?

So as a form of backup, take a 10HP 230V 3 phase motor, FLA = 28A
Now look at a comparable 10HP 230V single phase motor, FLA = 50A
50/28 = 1.78; efficiency would be different, but it is a lot closer to 1.732 than it is to 1.5.

 

[Besoeker]

OK, but again, look at the current formulae for three phase vs single phase that I posted in the first place.
How does that relate to what you just said?

So as a form of backup, take a 10HP 230V 3 phase motor, FLA = 28A
Now look at a comparable 10HP 230V single phase motor, FLA = 50A
50/28 = 1.78; efficiency would be different, but it is a lot closer to 1.732 than it is to 1.5.

But the topic is not about single phase v three phase motors.
It is about 3-phase VFD output powered from either a single phase or three phase input.
Can we stick to that?

 

[Jraef]

But the topic is not about single phase v three phase motors.
It is about 3-phase VFD output powered from either a single phase or three phase input.
Can we stick to that?

Ok sure.

So if i send 10kW of power out of my VFD to a 3 phase motor, the input power cannot be LESS than 10kW, correct?

So if my input power is 10kW, what is the current that is represented within that value, an how it that NOT represented by the formula:

Three phase; I = W/V x PF x 1.732 ergo I = 10,000/230 x .95 x 1.732 = 26.42A

Single phase; I = W/V x PF ergo I = 10,000/230 x .95 = 45.76A

45.76/26.42 = 1.732

???

If I_sp = 1.5 x I_tp as purported, then is the PF the difference? So is the PF for the single phase input only .6 then?

 

[Ingenieur]

Much of this has to do with standard sizing
usually that will give you some headroon over the 1.732
in addition the motor is seldom loaded to 100%
most mfgs now give the 1 ph ratings anyways

sanity check I use
calc 3 ph fla at sf, 1.15 for example
mult x 1.732
find a 3 ph drive rated for that i, it will usually be larger than the calculated
slop
sf
assumed 100%+ loaded
margin to next standard size
usually 2x plus

 

[gar]

170517-2006 EDT

Jraef:

In response to your post #24. You are comparing three phase line current to three phase line to line load voltage for your three phase motor input, and not three phase motor winding curtrent (a voltage to current phase angle problem), to the single phase winding voltage and its winding current (no phase angle problem).

Change your three phase thinking to a wye motor. Now the voltage to current phase angle problem is corrected by using the voltagre magnitude from line to neutral.

Line to neutral is 230/1.732 = 132.8 V. Then 123.8*28 = 3466.4 VA per winding or 10,399 VA for the entire motor.

For the single phase motor we have 230*50 = 11,500 VA.

The two values are adequatly close to indicate a close comparison.

.

 

[mike_kilroy]

Gar, you are mixing up heating in diodes with power thru them. The current on 1ph is exactly 1.73 x higher on 1ph into the vfd compared to 3ph into it for the same load.

Your 1a vs 50a scenarios are not realistic to this discussion.

Your idea that the diode drop is constant is no different than the 208v input voltage - it too is constant.

You cannot supply the same 3ph load from anything less than exactly 1.73x more current. And don't bring in efficiency or pf, neither is changing enough to discuss.

If you want to argue how much heat the diode junction will see on 1ph vs 3ph load, be it resistor or motor, ok, and your 1.5 vs avg vs rms discussion has merit. But not regarding 1.73x more current for the same load, be it resistor or motor.


Sent from my SM-G900V using Tapatalk

 

[topgone]

Consult the manufacturer.

There is a simpler way--> conduct a load test!

 

[gar]

170517-23349 EDT

mike_kilroy:

A 208 V sine wave voltage is not an instantaneous constant voltage source. It varies from a zero value to a positive peak, back to zero, to an equal negative peak, and back to zero once per cycle. At no time is this anything like a constant value. If in some way you process this sine wave signal, average it, and assume it is a stationary statistical process, then you can get a resultant signal that looks constant. This is what a Simpson 260/270 and other meters do. Put a low enough frequency sine wave into a Simpson and you will see the needle follow the instantaneous value of the waveform.

By contrast the forward voltage drop across the silicone diode I referenced is moderately constant for a rather large change of forward current. About 0.5 V at 1 A and 1 V at 50 A.

I am sorry if you can not see and understand the difference.

.

 

[Besoeker]

Ok sure.

So if i send 10kW of power out of my VFD to a 3 phase motor, the input power cannot be LESS than 10kW, correct?

So if my input power is 10kW, what is the current that is represented within that value, an how it that NOT represented by the formula:

Again, very simple.
You are using linear relationships in a non-linear application.
Look at the current waveforms both gar and I have presented.

 

[junkhound]

quick illustrative analysis FWIW: 2X is good enough rule of thumb for electricians, eh?

Typical VFD front end circuits and waveforms with drive and motor represented as simple resistor.

One hint to lessen increased diode stress is lower RH schematic, e.g tie 2 phase inputs together for single phase drive


too lazy this morning (or other paying jobs to do <G>) to type out complete explanations

 

[gar]

170518-2130 EDT

From where does the number 1.732 derive, and when is it applicable?

First see https://en.wikipedia.org/wiki/Phasor . At GE https://en.wikipedia.org/wiki/Charles_Proteus_Steinmetz developed the idea that analysis of steady state AC circuits with linear elements and sine waves could be performed with algerbraic equations and vectors (phasors) instead of more difficult analysis using differential equations.

On to 1732 that is the year of George Washington's birth under the Julian calendar, and the square root of 3. An easy way to remember both, high school physics.

If you create a right triangle with a hypotenuse of 1 unit length and one leg of 0.5 unit length, then what is the length of the other leg? h^2 = a^2 + b^2 (pythagorean theorm). b^2 = 1 - 0.5*0.5 = 1 - 0.25 = 0.75 . The approximate square root of 0.75 = 0.8660 , and 2 times this is 1.7321 . And what is the sq-root of 3, again from my HP32S = 1.7321 . What is the sine of 30 deg? It is 0.5000 . Put this all together and you can see a derivable reason for the use of 1.732 in sine wave linear circuits.

In any non-linear application the use o 1.732 is just someone pulling an approximate random number out of the hat.

.

 

[Besoeker]

One hint to lessen increased diode stress is lower RH schematic, e.g tie 2 phase inputs together for single phase drive

Paralleling semiconductors like that could help but their forward characteristics would have to be pretty well matched. It's something I've seen done in other applications but it required the devices to be specially selected. Even then, it doesn't help one leg that isn't paralleled. That would still be a limiting factor.

 

[electrofelon]

170518-2130 EDT



In any non-linear application the use o 1.732 is just someone pulling an approximate random number out of the hat.

.

But doesnt the concept of RMS values make it always work regardless of the waveform or no?

 

[Besoeker]

But doesnt the concept of RMS values make it always work regardless of the waveform or no?

Yes but......but
For a sinewave you can give fixed relationships between peak, average and RMS.
But once you get into the square wave, triangular, trapeziodal, discontinuous, asymmetic it's a whole new ball of wax.

 

[gar]

170518-1135 EDT

electrofelon:

If you have a linear invariant resistor, and send a stationary (statistical definition) RMS current of I of arbitrary waveform thru the resistor, then the average power dissipated in the resistor can be calculated from I^2*R.

If the resistor is replaced with an ideal diode of non-zero voltage drop, or a real world diode this equation does not apply. What is the value of R to use?

Suppose an ideal diode of 1 V forward drop at any current, then what is its apparent resistance? At 1 A it is 1 ohm, at 10 A it is 0.1 ohm, etc.

.

 

[Besoeker]

170518-1135 EDT

electrofelon:

If you have a linear invariant resistor, and send a stationary (statistical definition) RMS current of I of arbitrary waveform thru the resistor, then the average power dissipated in the resistor can be calculated from I^2*R.

If the resistor is replaced with an ideal diode of non-zero voltage drop, or a real world diode this equation does not apply. What is the value of R to use?

Suppose an ideal diode of 1 V forward drop at any current, then what is its apparent resistance? At 1 A it is 1 ohm, at 10 A it is 0.1 ohm, etc.

.

I admire your fortitude.

 

[Smart $]

But doesnt the concept of RMS values make it always work regardless of the waveform or no?

Yes but......but
For a sinewave you can give fixed relationships between peak, average and RMS.
But once you get into the square wave, triangular, trapeziodal, discontinuous, asymmetic it's a whole new ball of wax.

Say what?!!!

I believe the distinction would be periodic vs. non-periodic. And even with non-periodic, the RMS value through a specified time window is still the effective value through that period of time.

https://meettechniek.info/compendium/average-effective.html