#### aelectricalman

##### Senior Member

- Location
- KY

- Thread starter aelectricalman
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- Location
- KY

I can do the calculations on a case by case basis.

But I suspect that's not quite what you want.

- Location
- Placerville, CA, USA

But I am betting that you want a more accurate answer than that.

Tapatalk!

- Location
- KY

Well, I can figure out the Ohm's law part. I'm really looking for a savings calculator or a good start where I can create my own. I created a really nice one for kVA (that I have no problem sharing) but I want to start looking into what savings to expect when utility companies don't bill on kVA.I can do the calculations on a case by case basis.

But I suspect that's not quite what you want.

OK. Utility companies here generally charge on kWh. Industrial users may have a maximum kVA demand limit beyond which they are hit with stiff penalties.Well, I can figure out the Ohm's law part. I'm really looking for a savings calculator or a good start where I can create my own. I created a really nice one for kVA (that I have no problem sharing) but I want to start looking into what savings to expect when utility companies don't bill on kVA.

But PFC doesn't do a lot for kW reduction. Fitted close to the load they can somewhat reduce conductor currents and consequently losses in those.

But that's usually not a lot.

Or, at the installation phase, they can allow smaller conductors to be used.

Not much help, I know. I do it on a case by case basis.

- Location
- KY

I am aware of this but thank you for your input.OK. Utility companies here generally charge on kWh. Industrial users may have a maximum kVA demand limit beyond which they are hit with stiff penalties.

But PFC doesn't do a lot for kW reduction. Fitted close to the load they can somewhat reduce conductor currents and consequently losses in those.

But that's usually not a lot.

Or, at the installation phase, they can allow smaller conductors to be used.

Not much help, I know. I do it on a case by case basis.

- Location
- Placerville, CA, USA

Knowing the two VD figures will give you the difference in the (V^2)/R losses. That is close enough to the entire difference in billed power to serve your needs.

I think the biggest problem will be in inputting all of the wire lengths and transformer impedances as well as the motor loads with power factor. Especially since the motor PF will vary with load rather than being a constant for the motor design.

Tapatalk!

- Location
- San Francisco Bay Area, CA, USA

- Occupation
- Electrical Engineer

Hmmm... lots of little holes in this concept and I have some time off today...

*
*OK, no. First off I have NEVER seen a utility that bills solely on kVA demand. They bill on kWh, then in SOME cases there is a **PENALTY for exceeding an allotted kVA demand**. But they do not "bill on" kVA demand. Secondly, "marked savings" is the kind of nebulous fluff terminology you see in a marketing pitch, I differ with what it implies.*
*Hmmm... I'd say that value is debatable. If the conductors are properly sized to code requirements and adjusted for voltage drop under full load, then the losses would be very low to start with. I believe that 1-2% would be a stretch.*
* *ONLY *if you are being penalized for exceeding your allotted kVA demand . If you are not, the savings are ZERO.*
* That would be your first purely true statement. The reason they are not installed to reduce kW is because... doing so doesn't save enough kWh to make it worth doing! *
* Sort of. Technically they STORE (not "add") the VARs for use at a different point in the cycle to reduce the current necessary to make the motor into a motor. But the VARs are still coming from the line source, that's why they do not save kW.*
* "Those kW savings"? What kW savings? We just established they are not there for worthy consideration, or are you trying to hang your hat on the I[SUP]2[/SUP]R loss reductions? If so and they are minimal to start with on a properly designed system, then in my opinion, you are clutching at straws. Do they exist? Yes. Are they worth the cost of installing PFC capacitors *JUST* for the purpose of attaining that miniscule reduction in losses? **No.**

So are you trying to quantify the SIDE BENEFIT of reducing the I[SUP]2[/SUP]R losses from having installed PFC caps to reduce PF penalties? Because THAT is, in my opinion, the only reason to indulge this exercise.

So here are your column headings, solve for each circuit that has PFC caps:

A | Wire size

B | Wire resistance in ohms / ft.

C | Total ckt length in feet from feeder to PFC cap. ***

D | Total circuit ohms

E | Delta I-Load vs I-load wit PFC caps applied

F | I[SUP]2[/SUP]R Delta

G | Heat reduction in kW

H | Cost /kWh

I | Operating hours

J | kWh $ saved per circuit

B comes from wire mfr. chart using column A

D = C * B

E = Empirical data from current measurements

F = E[SUP]2[/SUP] * D

G = F x Line Voltage

H = Utility data

I = Empirical data

J = I * H * G

After last circuit Row: sum all of column J

Next row, column J: $ value of your time to do all this, including recording and collecting empirical data

Last row, column J: [Sum of Column J] - [your time cost] = net savings.

*** You have to assume that above the feeder, circuit conductor size increases to the point where the conductor heating delta from the reduction in load becomes so negligible that it cannot be calculated.

Ok, so its well known that you can reap marked savings from PF correction equipment when you are being billed on KVA demand.

There are savings associated with I^2R loss reduction that actually gives you some (1%-2%) savings when installing capacitors on a system to reduce kW

...but KVA savings can yield up to around 25%.

We all know however that PF capacitors are not installed to reduce kW.

They are installed to reduce kVA, by adding Vars to the system.

So, with this being said, does anyone know of or have an excel sheet or program that will let one guestimate, taking into consideration distance from motor, impedance, age of equipment, current, etc... I'll gladly pay for it or accept it as a token of friendship.lolthose kW savings

So are you trying to quantify the SIDE BENEFIT of reducing the I[SUP]2[/SUP]R losses from having installed PFC caps to reduce PF penalties? Because THAT is, in my opinion, the only reason to indulge this exercise.

So here are your column headings, solve for each circuit that has PFC caps:

A | Wire size

B | Wire resistance in ohms / ft.

C | Total ckt length in feet from feeder to PFC cap. ***

D | Total circuit ohms

E | Delta I-Load vs I-load wit PFC caps applied

F | I[SUP]2[/SUP]R Delta

G | Heat reduction in kW

H | Cost /kWh

I | Operating hours

J | kWh $ saved per circuit

B comes from wire mfr. chart using column A

D = C * B

E = Empirical data from current measurements

F = E[SUP]2[/SUP] * D

G = F x Line Voltage

H = Utility data

I = Empirical data

J = I * H * G

After last circuit Row: sum all of column J

Next row, column J: $ value of your time to do all this, including recording and collecting empirical data

Last row, column J: [Sum of Column J] - [your time cost] = net savings.

*** You have to assume that above the feeder, circuit conductor size increases to the point where the conductor heating delta from the reduction in load becomes so negligible that it cannot be calculated.

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If you're looking at the i[SUP]2[/SUP]R losses, the savings are not really going to be considerable.

- Location
- Illinois

Can you expand on that? I don't understand how the available fault current has anything to do with this.

- Location
- Placerville, CA, USA

The change in available fault current when you get to the branch circuit may be useful, but you probably are more likely to know the VD directly at the branch level than to know the fault current.

Tapatalk!

http://www.samwha.com/fc/tech/tech/tech4.pdf

As for the derivation of equation describing the relationship between voltage drop and fault level given in the above paper, I am sure Bes will help us.

Interesting. A capacitor company making a case for the installation of capacitors..............

http://www.samwha.com/fc/tech/tech/tech4.pdf

As for the derivation of equation describing the relationship between voltage drop and fault level given in the above paper, I am sure Bes will help us.

- Location
- Placerville, CA, USA

Tapatalk!

Reduction of voltage drop rate e by power factor improvement can be calculated to

e = Qc / RC x 100%

Qc : Input capacitor capacitance[kvar]

RC : Short capacity[kVA] of bus-bar (source)

To derive the above equation, proceed as follows.

If Ixl is the load inductive current and Ixc is the capacitor current, the capacitor connected across the load, then the size of the capacitor is so chosen that Ixl=Ixc.

So the percentage reduction in voltage drop e=(Ixc*R/V)*100% , where V is the source voltage and R is the resistance of wire lengths from source to load.

e =(Ixc*R/V)*100 %

=(Ixc/V/R)*100 %

=(Ixc/Is)*100% , where Is is the short circuit current

=(1.732*V/1.732*V)* /Is)*100 %

=(1.732*V*Ixc/1.732*V*Is)*100 %

= Qc / RC x 100%.

e = Qc / RC x 100%

Qc : Input capacitor capacitance[kvar]

RC : Short capacity[kVA] of bus-bar (source)

To derive the above equation, proceed as follows.

If Ixl is the load inductive current and Ixc is the capacitor current, the capacitor connected across the load, then the size of the capacitor is so chosen that Ixl=Ixc.

So the percentage reduction in voltage drop e=(Ixc*R/V)*100% , where V is the source voltage and R is the resistance of wire lengths from source to load.

e =(Ixc*R/V)*100 %

=(Ixc/V/R)*100 %

=(Ixc/Is)*100% , where Is is the short circuit current

=(1.732*V/1.732*V)* /Is)*100 %

=(1.732*V*Ixc/1.732*V*Is)*100 %

= Qc / RC x 100%.

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- Location
- London, England

It might be worthwhile in the case of a long hour load of low power factor on a long run of marginaly sized wire.

Take as an example, a low power factor load that uses 100 amps single phase, with voltage drop in the wires of 6 volts. The losses are clearly 600 watts.

Now add capacitors such that the current drops to 66.6 amps, the voltage drop will now be 4 volts giving a watts loss of 266 watts.

Clearly 334 watts have been saved, and for a long hour load that could amount to over 2,000KWH a year, worth perhaps a few hundred $

That is however an extreme case, smaller loads, or more generously sized wires, or better original power factor would give far less favourable results.

No great accuracy can be claimed since the load may be slightly more or less efficient at the higher voltage, the temperature and therefore resistance of the wires will alter slightly.

I've seen it done, and done it myself for two main reasons.Saving in Kw by installing capacitors at load ends is a worthwhile effort in facilities covering large areas such as paper mills.

It reduces maximum demand in terms of reduced kVA. Exceeding maximum demand is extremely costly in this country.

The other application is reducing the size of conductors and/or supply capacity. We currently have an installation for pumps where the customer decided to change the motor from a synchronous induction motor with unity power factor to cage induction motors with 0.76 p.f. One serious consequence of this was that the unit transformer supplying the original motor was not rated for the increased kVA. PFC fixed this by getting the pf up to about 0.97.

Energy saving? Not really. As is our usual practice, we fitted detuning chokes and the losses in these, though not a lot, was about the same as the reduction cable losses resulting from the reduced current.

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