What's new
What's new

capacitor/voltage relationship for RPC

challenger

Stainless
Joined
Mar 6, 2003
Location
Hampstead, NC-S.E. Coast
I've built a 10hp RPC. Everything went well and I had been running it for test purposes-w-only the A-C legs having capacitors on it of 120MF. The unloded voltages were AB-244 AC 241 & BC-255 (I may have the A & B voltage mixed up). I increased the A-C to 160 and got the exact same voltages on all 3 legs. In looking at the Fitch design I then started adding caps to B-C and got an imbalance of 20 volts between all three legs 244/251/261. This was still within the Fitch tolerance so I figured I would add Cpf. I put a 20mf on A-B and got over 270 volts on one leg (I forget which). I thought putting a load on would help but only had a 1/2hp motor to try and this actually made for a larger imbalance so I got pissed off and called it a night.
My question is, given I understand this is trial and error, what is the relationship between increasing the sizes of the capacitors and the resulting outcome? Do the caps raise or lower the voltage? In addition I wonder why we measure between legs and not just to ground? This seems to me like the best way to find out what each leg is doing.
Lastly when this design raises the input voltage (A/B-which I cannot for the life of me understand as it is coming form the power co. right?) isn't this increasing the voltage in the whole house?
Thanks - Howard
Hampstead, NC
 
"My question is, given I understand this is trial and error, what is the relationship between increasing the sizes of the capacitors and the resulting outcome? Do the caps raise or lower the voltage?"

Recall that an electric utility is an infinite source of power. An infinite source of neither lagging nor leading volt-amperes (real power). An infinite source of lagging volt-amperes reactive (imaginary power, inductive). An infinite source of leading volt-amperes reactive (imaginary power, capacitive).

If you add an inductive load, such as a motor, to a limited source of power, such as a residential feeder or branch circuit, there will be a lack of leading volt-amperes reactive. This will cause the voltage at that point to decrease, unless it is compensated for by the addition of a source of leading volt-amperes reactive, such as a capacitor.

If you add a capacitive load to a limited source of power, there will be an excess of leading volt-amperes reactive. This will cause the voltage at that point to increase.

So, all other things being equal, adding a capacitor causes the voltage to increase, whereas removing a capacitor (or adding an inductor, such as a motor) causes the voltage to decrease.

It is perhaps significant to mention that power in an ac distribution system is more affected by the relative phase difference, not by the relative voltage difference, hence why we use capacitors and inductors to control voltages. Capacitors add a +90 degree phase shift (a source of leading volt-amperes reactive); inductors add a -90 degree phase shift (a source of lagging volt-amperes reactive). Resistors cause no phase shift at all (resistors are said to be neither capacitive nor inductive, hence neither leading nor lagging).


"In addition I wonder why we measure between legs and not just to ground? This seems to me like the best way to find out what each leg is doing."

By convention, we are electing to avoid so-called "zero sequence" currents, which are those currents attributable to imbalance in a three-phase system. By eliminating the ground return from the circuit, we are avoiding the undesirable "zero sequence" currents. The so-called "positive sequence" and "negative sequence" currents are still present, however.


"Lastly when this design raises the input voltage (A/B-which I cannot for the life of me understand as it is coming form the power co. right?) isn't this increasing the voltage in the whole house?"

I would expect a small increase in L1-L2.

However, by convention, the manufactured phase is labeled B, whereas the incoming lines are, by convention, labeled A and C.

L1 is identical to A; L2 is identical to C.
 
I can't do better than Peter in description, but

Measure between lines and not to ground, because puting caps on one leg does effect the others. Measure and see for yourself.

I really like it when all the voltags are close when the machine is running! The RPC by it's self is less significant.

CalG
 
"This is the first time I've heard of L1+A and L2+C?"

It's been in the NEC for what seems like centuries. Decades, anyway.

The buses in panelboards and switchboards are required to be labeled: left to right or top to down, as A, B and C.

By convention, the "wild leg", also called the "high leg" is B.

Therefore L1, the left-most (top) bus is A while L2, the right-most (bottom) bus is C.

The manufactured phase is considered a "wild leg", and such legs must be "indicated" with orange tape.

One reason for labeling the manufactured leg as B is single-phase loads may never be powered by such a leg.
 
As it turns out (I hope) I got the best results-w-a lot less capacitance than I thought. I have 115mf on L1/generated Leg, 80 MF on the L2/Generated Leg and 15mf on L1/L2. This gives me voltage within 5 volts-w-the highest at 248vac. The current I read, and I don't know how this all works but, I have 9.7 amp draw on L1, 9.5 on L2 and the generated leg-w-a 1/2 hp motor running off the RPC is 3.1 amp??
Does all this sound proper? I spent a lot of time before I could get the balance-w-the lowest current draw. I think the voltage is high but I can't get it lower and the incoming is 247vac.
Thank all - Howard
 
Well said Peter. Basic AC circuit fundamentals.

I am also having a hard time balancing my 10hp, but I haven't spent a lot of time playing with it.
 
I

If you add a capacitive load to a limited source of power, there will be an excess of leading volt-amperes reactive. This will cause the voltage at that point to increase.

So, all other things being equal, adding a capacitor causes the voltage to increase, whereas removing a capacitor (or adding an inductor, such as a motor) causes the voltage to decrease.

It is perhaps significant to mention that power in an ac distribution system is more affected by the relative phase difference, not by the relative voltage difference, hence why we use capacitors and inductors to control voltages. Capacitors add a +90 degree phase shift (a source of leading volt-amperes reactive); inductors add a -90 degree phase shift (a source of lagging volt-amperes reactive). Resistors cause no phase shift at all (resistors are said to be neither capacitive nor inductive, hence neither leading nor lagging).

In the general case of an infinitely powerful source, adding a leading volt-ampere load won't change the voltage.

In the case of an all resistive limited power network, adding a leading volt ampere load will REDUCE the voltage, just as ANY load will.

The effect of leading volt amperes in RAISING voltage is because the general tendency of the distribution network is to have lots of lagging volt-amperes. Therefore, the addition of leading volt-ampere loads partly compensates for the lagging VA, and so reduces the drops.

it isn't because there is any magic effect of leading VA loads that increases voltage independently of any other effect.

Same for a SOURCE that is leading vs lagging in nature....... both will reduce the voltage for a resistive load..... The leading source will tend to reduce less (i.e. increase) voltage with a LAGGING load, while the lagging source will do the same with a LEADING load.

The powerco puts capacitors on the pole when they decide that they need to compensate a general lagging load in an area. You very seldom if ever see inductors inserted.......... the only ones I ever see are for reducing short circuit current.
 
Ouch - this makes my head hurt. From the little that I can take from it, and I may be way off, the Cpf act as the "leading" side and the capacitors for Cs and Cp are 'lagging"?
Either way I am still confused when the Fitch design is in the mix because he is seeking a happy medium with the voltage balance and the lowest current draw.
Does any of the below information suggest why I see approx 9.5Amps on the L1/L2 and 3.1 on the manufactured leg-w-a 1/2 hp motor on it?
Should I think of putting something in the RPC to make the voltage lower than 245?
Thanks for the study material and thehelp.
Howard

"In the general case of an infinitely powerful source, adding a leading volt-ampere load won't change the voltage.

In the case of an all resistive limited power network, adding a leading volt ampere load will REDUCE the voltage, just as ANY load will.

The effect of leading volt amperes in RAISING voltage is because the general tendency of the distribution network is to have lots of lagging volt-amperes. Therefore, the addition of leading volt-ampere loads partly compensates for the lagging VA, and so reduces the drops.

it isn't because there is any magic effect of leading VA loads that increases voltage independently of any other effect.

Same for a SOURCE that is leading vs lagging in nature....... both will reduce the voltage for a resistive load..... The leading source will tend to reduce less (i.e. increase) voltage with a LAGGING load, while the lagging source will do the same with a LEADING load.

The powerco puts capacitors on the pole when they decide that they need to compensate a general lagging load in an area. You very seldom if ever see inductors inserted.......... the only ones I ever see are for reducing short circuit current."
 
If you read fitch's comments carefully he makes one point that you should be
aware of:

Balance the converter for voltage balance from L1 to manufactured, and from L2 to
manufactured leg. Do this by using capacitors from L1 to Man. and L2 to man. first.
Do not look at the currents anywhere when doing this, and do it when the
load motor is running and if possible loaded.

Only *then* do you add a power factor correction capacitor from L1 to L2, and
this is done by looking at an amp-clamp on the incoming line (L1 or L2) and tune
for the minimum current draw.

(you are not really reducing the real current draw, you lowering the reactive currents
flowing into your setup)

The reason for the above set of procedures is, that adding capacitence from L1 to
Man. leg, and from L2 to Man. leg, will tend to overall reduce the reactive current
draw of the converter. Power factor correction will not change the converter
ballance, but changing the converters balance condition *will* affect the power
factor.

So you do the PF correction last.

Jim
 
Does any of the below information suggest why I see approx 9.5Amps on the L1/L2 and 3.1 on the manufactured leg-w-a 1/2 hp motor on it?
Should I think of putting something in the RPC to make the voltage lower than 245?
Thanks for the study material and thehelp.
Howard

Well, it suggests that you don't need the cap across the L1/L2. Lose it for now.

And it suggests that caps may raise OR lower, but probably in most real cases, will raise voltage, which you asked.

Quite a few people report good balance with NO capacitors, if the RPC is considerably larger than the loads. My Arco has ONE fairly small capacitor from l1 to manufactured, yet it produces good 3 phase power which smoothly runs the motors I use it with.

There is no LAW that says you must have capacitors hung on all over. Only if they improve performance are they needed. Most likely you WILL need at least one.

The "Fitch" info is one way to do it, certainly.
 
"In the general case of an infinitely powerful source, adding a leading volt-ampere load won't change the voltage.

"In the case of an all resistive limited power network, adding a leading volt ampere load will REDUCE the voltage, just as ANY load will."

But, no power company source is truly infinitely powerful, just as no other source is truly all resistive.

Power company sources tend to be exceptionally stiff at the generating station, but significantly less stiff elsewhere. And, this stiffness decreases rapidly as the distance away from the generating station increases.

It is still stiff enough, for most practical purposes, at a commercial site fed by a relatively high voltage feeder.

At the very large municipal utility where I was an EE in a former lifetime, we fed our large industrial customers from our 34.5 kV subtransmission system, and at least 1 MW (1,000 KVA) was possible with using a relatively compact pole-mounted distribution station (3 x 333 KVA transformers), and more if a transformer vault was available and a large three-phase transformer was installed therein.

We did not service any customer at all from our transmission system, which was, variously, 138, 230, 287.5 or 500 kV, and could supply perhaps 1,500 MW (1,500,000 KVA) on a single point-to-point interconnection.

As I recall, the smallest capacity line within our system was a single circuit 230 kV underground line from Scattergood Steam Plant (near LAX) to Receiving Station 26 (Bundy, in West L.A.) and its capacity was 315 MW (315,000 KVA).
 
Yes this is exactly how I tuned the RPC-w-the Cpf going on last and with a motor on it (although a small one).
My incoming L1,L2 read 244vac. I suspect in order to reduce this I would have to step it down-w-a transformer. I'm curious if a voltage of 245 to 248 is out of the proper range for a motor that is labeled for 230vac?

If you read fitch's comments carefully he makes one point that you should be
aware of:

Balance the converter for voltage balance from L1 to manufactured, and from L2 to
manufactured leg. Do this by using capacitors from L1 to Man. and L2 to man. first.
Do not look at the currents anywhere when doing this, and do it when the
load motor is running and if possible loaded.

Only *then* do you add a power factor correction capacitor from L1 to L2, and
this is done by looking at an amp-clamp on the incoming line (L1 or L2) and tune
for the minimum current draw.

(you are not really reducing the real current draw, you lowering the reactive currents
flowing into your setup)

The reason for the above set of procedures is, that adding capacitence from L1 to
Man. leg, and from L2 to Man. leg, will tend to overall reduce the reactive current
draw of the converter. Power factor correction will not change the converter
ballance, but changing the converters balance condition *will* affect the power
factor.

So you do the PF correction last.

Jim
 
I doubt your 148 volt incoming line is a problem for the motor. It will run a bit
cooler (yes, cooler, it will draw less current) and as long as the capacitors in your
converter are rated conservatively I doubt it will cause problems.

Jim
 








 
Back
Top