# kW reductions from capacitors

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#### aelectricalman

##### Senior Member
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 those kW savings, 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.lol

#### Besoeker

##### Senior Member
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 those kW savings, 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.lol
I can do the calculations on a case by case basis.
But I suspect that's not quite what you want.

#### GoldDigger

##### Moderator
Staff member
I can write you a spreadsheet which accepts all of those inputs and returns a value of zero as a first approximation.
But I am betting that you want a more accurate answer than that.

Tapatalk!

#### aelectricalman

##### Senior Member
I can do the calculations on a case by case basis.
But I suspect that's not quite what you want.
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.

#### Besoeker

##### Senior Member
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.
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.

#### aelectricalman

##### Senior Member
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.
I am aware of this but thank you for your input.

#### GoldDigger

##### Moderator
Staff member
One step in the right direction would simply be to run a standard voltage drop calculation for both the full current and the resistive component of the current. The simplifying assumption that the resistive component will not be changed by the slight increase in voltage to the motor should be valid.
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!

#### Jraef

##### Moderator
Staff member
Hmmm... lots of little holes in this concept and I have some time off today...
Ok, so its well known that you can reap marked savings from PF correction equipment when you are being billed on KVA demand.
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.
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
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.
...but KVA savings can yield up to around 25%.
ONLY if you are being penalized for exceeding your allotted kVA demand . If you are not, the savings are ZERO.
We all know however that PF capacitors are not installed to reduce kW.
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!
They are installed to reduce kVA, by adding Vars to the system.
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.
So, with this being said, does anyone know of or have an excel sheet or program that will let one guestimate those kW savings, 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.lol
"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.

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#### Sahib

##### Senior Member
One important factor for KW saving in wire lengths from source to capacitor installed equipment is the fault level in MVA at source. Lower is the fault level, higher is the KW saving. So if fault level is high, there is practically no saving by installing capacitor at equipment. But if the fault level is low, there may be considerable saving in installing capacitor at equipment.

#### Besoeker

##### Senior Member
One important factor for KW saving in wire lengths from source to capacitor installed equipment is the fault level in MVA at source. Lower is the fault level, higher is the KW saving. So if fault level is high, there is practically no saving by installing capacitor at equipment. But if the fault level is low, there may be considerable saving in installing capacitor at equipment.
If you're looking at the i[SUP]2[/SUP]R losses, the savings are not really going to be considerable.

#### Sahib

##### Senior Member
If you only know how to express wire length voltage drop in terms of source fault level, you would not say that. Do you know how to do it?

#### don_resqcapt19

##### Moderator
Staff member
One important factor for KW saving in wire lengths from source to capacitor installed equipment is the fault level in MVA at source. Lower is the fault level, higher is the KW saving. So if fault level is high, there is practically no saving by installing capacitor at equipment. But if the fault level is low, there may be considerable saving in installing capacitor at equipment.
Can you expand on that? I don't understand how the available fault current has anything to do with this.

#### GoldDigger

##### Moderator
Staff member
Since the fault current available is most strongly influenced by the impedance of the service transformer, and all power losses in the transformer are upstream of the meter, I do not see the point at all for service and feeder level fault current.
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!

#### Sahib

##### Senior Member
Knowing the fault level and how it relates to voltage drop makes it unnecessary to do actual voltage drop measurment to ascertain how much percentage reduction in voltage drop and so in power loss in wire lengths takes place due to installation capacitor at the equipment. It is described in the paper below.
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.

#### Besoeker

##### Senior Member
Knowing the fault level and how it relates to voltage drop makes it unnecessary to do actual voltage drop measurment to ascertain how much percentage reduction in voltage drop and so in power loss in wire lengths takes place due to installation capacitor at the equipment. It is described in the paper below.
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..............

#### GoldDigger

##### Moderator
Staff member
Another interesting factor is that since resistive voltage drop with the capacitors in place is no longer in phase with the motor current, the effect of the reduced voltage drop on motor performance is not completely straightforward.

Tapatalk!

#### Sahib

##### Senior Member
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%.

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##### Senior Member
As others post, the KW saving by adding capacitors is normaly very low, it can be calculated approximatly by determining the load current with capacitors and without, and then calculating the voltage drop for each case.

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.

#### Sahib

##### Senior Member
Saving in Kw by installing capacitors at load ends is a worthwhile effort in facilities covering large areas such as paper mills.

#### Besoeker

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