Reducing Energy Use with Power Factor Correction (PFC)

The calculation you have made are for watt-hrs rather than kilowatt hrs. It appears that you need to take losses formula and add another 1000 in the dividend making the resulting savings $1.232 and $0.536

I'm not sure about that. he had the 1000 dividend already.

What I question is the basic math at the beginning. The "spice they slipped into the soup" here is the 51A derived from "a test" without any further qualification. They are (subtly) making you accept that at face value. But do the math at that point; 51A at a .99 PF (which is also questionable) makes the kW calculation come out to only 40.3kW. Where did the other 1.8kW in motor power go to? Funny thing is, that alone accounts for $540.00 in their $705.00 savings scenario. In other words, they introduced an error that can account for 3/4 of the purported savings without even coming close to using real empirical test data provided by a reputable independent testing authority.

So what I see is the same BS sandwich, served on different bread.

The calculation is correct and it is based on KWH. The resistance is 0.75 Ohms per 1000 feet. In the calculation, it is used 0.00075 Ohm (0.00075 Ohm per foot). So, the denominator 1000 is used to convert it to KW.

The calculated savings is theoretically correct but in reality, the PFI itself consumes some real power, especially when it is old. I had an experience in this regards. I was hired to investigate the excessive billing of one of my clients. While investing their system, I noticed that the energy meter was rotating while all other loads are off. Finally, I figured it out that the PFI was in service and it was consuming the energy. I turned the PFI off, then the energy meter stopped. I suggested keeping it off and the client found the energy bill was ok since then.

- MS

PFC equipment, when properly installed according to NEC Article 460, should be switched with the motor load. Reactive power should not be feeding back into the system when there is no demand for it from the motor.

That being said, reactive power does not turn a meter. Reactive power doesn't do any work. It is true that if a capacitor is being fed on a isolated circuit (when the motor load is off and the capacitor is not installed so that it switches on and off with the motor) there is a measureable current draw, but that current draw will not spin the meter because of the fact that it is only reactive power.

As a mathematical convenience for engineering design, the nontransfer of power is described as the product of a physically nonexistent (imaginary) current wave that is a quarter out of phase with the actual voltage wave. This nonexistent current is called reactive current because it is normally caused by loads that react to voltage by storing current. If the load is purely reactive, then the voltage and current are 90 degrees out of phase and there is no net power flow over an entire cycle.

A real capacitor bank has power losses. There are about 1-2% of the KVAR in KW losses. It can be even more if detuning reactors and dampening resistors are used in high harmonics installations.

Another problem from hard switched capacitor banks is the voltage spikes that are produced. These can cause sensitive equipment to fail after a while (light ballasts, PLC cards...). The only way around this is to use semiconductors capacitor bank switches but the higher cost will increase the payback.

In general, small capacitor banks in parallel with the loads spread around the plant work well but it should not be overdone. Looking for a PF of 0.95 and above is asking for trouble as the risk of parallel resonance becomes non negligible.

Actually. by improving the Power factor from 0.6 to 0.9 DOES save energy cost. Most of the time, we always base on the load (KW) to do the energy cost calculation, as a simple estimation calculation for the energy cost.

For example, if the load or a motor is rated at 300KW, we will just take the 300 KW and multiply with the hour and we get KWh. Actually, the actual correct power consume by the load or the motor is not 300KW/hr.

Let use the formula to calculate it.

Motor/ Load = 300KW

Voltage = 415V

Power Factor = 0.85

Current =

Current (I) =KW/( V*1.732*PF. )

Current (I) = 300,000 / (415*1.732*0.85)

Current (I) =491.1 Amps – Full Load Current of the motor/load

Actual running current is always lower than the Full Load Current. Assume that the running current is 80% of the full load current.

Actual Running Current = 491.1 x 80% = 392.86 – Actual running current consume.

Therefore, for the actual power consumption in KWh is ;

KW = V*I (actual running current)*1.732* Power Factor

KW= 415 * 392.86 *1.732* 0.85

KW = 240KW * 1hr = 240 KWh (For the motor / load running for 1 hr)

If the power factor improved again to higher to 0.95 or 0.9 you will get better saving of money for your electricity bill. Our electricity meter are running base on the current flow. The more current you use the faster it rotate or you have to pay your bill more.

If you use recording power meter to measure the power consumption after the power factor had been improved, it does not show the actual values or data of the saving. Whatever, it print out will still show the same KW (because the meter use the P=VI√3 cos ø to do the KW calculation) but you can see that the consumption of the current is reduce very much as compare before the improved power factor.

The calculations are more or less correct. So you do get a saving on the I²R front (and this has been already covered in all the other threads)

But this is realisable only when the pf correction device is near the load.

However the other part - the motor size, efficiency and the cable sizes - you don't get the benefit. The load still draws the reactive power, only now it is not supplied by your power generator but by the capacitor bank.

So in isolation, motor current, pf, loss, efficiency etc are still same as earlier.

The saving = (2.33KW (I²R Loss) - Losses in the PFC device) - time adjusted cost of the PFC device

JR

You are correct. I added 1-1000 too many. But I am not following you comment. It seems to me that both estimates of savings are calculated values and not tested values.

Quote :What I question is the basic math at the beginning. The "spice they slipped into the soup" here is the 51A derived from "a test" without any further qualification. They are (subtly) making you accept that at face value. But do the math at that point; 51A at a .99 PF (which is also questionable) makes the kW calculation come out to only 40.3kW.

480 volts x 51 amps x 1.73 = 42.3 kva x .99 = 41.9 kw

480 volts x 78 amps x 1.73 = 64.8 kva x .65 = 42.1 kw

As you said a PF of 0.99 is questionable.

Huh. that's what I get for using the i-Phone as a calculator...

Must have missed something.

Why is a PF of .99 questionable? We correct PF's to .99 all the time. We have the ability to take the capacitiance into the field and "size" the inductive loads to correct the power factor to as close to unity as possible. I can correct PF to .999 alot of time.

Additionally, on rare occasions, we see a drop in kW right at the point of installation of the PFC equipment. This only happens when we initially see low PF's but it does happen. Normally we are only looking for a drop in kW at the panel but on rare occasions we see small (1%-3%) drop in kW at the point of installation.

Pete

The reason I said that 0.99 pf was questionable was because you may not have the exact size capacitor to get the pf to = 0.99. Maybe a std size capacitor only get it to 0.98.

Gotcha.

And there are times when we can't get the PF that high because of three phase current imbalances as well as phase PF imbalance.

The only time I have ever achieved .99pf was with a synchronous condenser and a very good closed loop pf feedback system, the cost of which, if you had no other need for a large synchronous motor to be running, would far out strip any potential reduction in cable losses. The general rule of thumb is to achieve at least a .95pf. That is the value generally proscribed by utilities to avoid their penalties; they use that because it is attainable without too much difficulty. So to use a relatively unattainable value such as .99 in an argument regarding the efficacy of saving on energy losses is to cheapen the argument by using unrealistic goals, further making me question the other values cited.

The other issue is using a value of 6000 operating hours per year. That equates to a machine running 24 hours per day NON-STOP, 5 days per week, 50 weeks per year. Most "common" applications, and CERTAINLY any application in a residential installation (for which most of these PF "energy savers" are targeted) will not operate anywhere near these values. So again, hyperbole generated by using an extreme example.

I disagree that the .99 PF is unattainable. We do it all the time. The other thing I would say here is that the example given is exactly what we look to do for our cleints. That is we target large loads with long runtimes and /or lightly loaded or underloaded motors (i.e. large conveyors, escalators, etc.). We look for clients who have loads that run 24/7/365. We are in no way interested in selling PFC to the residential market. Our target market are very large commercial, institutional, municipal and industrial clients with extremely large electric bills. Those type of clients can and do benefit from the reduction of system current through PFC.

I agree with you wholeheartedly about small commercial and residential clients. Only in the most extreme cases do they see any benefit from PFC. I think we are coming from the same place but your opinion is biased from doing battle with fools.

"I think we are coming from the same place but your opinion is biased from doing battle with fools."

Fair statement. I suppose I have ben a bit too harsh because these types of threads usually descend into promoting the "little magic boxes" being sold all over the internet now by people making wild clams of "25-40% reduction in your energy useage" and all they do is connect a capacitor t the main line terminals. I do understand there are many benefits of PFC in industrial environments and I'm affraid my battle weariness has got the best of me.

Pereciliberto, regarding to your systems where you do it all the time. What is the resolution of those capacitor banks giving this 0.99 PF? (% step size) Can you maintain it consistently with a varying load without going leading? What is the topology?

One thing that has not being discussed is how to implement the PF regulator. If you use a 10 steps bank, your resolution is 10% of the total KVARs in a fix size system and down to 0.1% with a binary size scheme. This might allow for a good regulation but the cost becomes much more than a simple bank and contactor approach. The large step number system is only cost effective at the plant feeder.

As in every engineering problem, the cost ans performances can vary considerably depending on the scheme used. One must evaluate the real savings of the system proposed by the vendor. This means using your real life numbers and the capacitor bank(s) system that will be required to achieve the performances needed.

IT DEPENDS ON THE TARIFF!

Dear peteciliberto,

The ohms/1000ft equates to #8AWG copper. For the loads stated at the distance stated this wire would be causing quite a voltage drop, (not to mention performance and heat problems due to greatly undersized conductors). This does not show up in the stated measured values, why?

The values stated to have been measured appear to be calculated instead, since the values of the power correction results also do not indicate a voltage drop, and the values are not in line with physical realities. Correction factors resulting in .99 PF at full load, which you show, will result in dangerous over voltages in the capacitors and motor for static correction when the motor spins down to stop.

Proper PFC values are set around 80% of the charging current only, which is 20-60% of the full load current.

If the correction is not static (connected at the individual load), then the cost of PF-monitoring style correction will be hugely prohibitive or loaded up with non-sinusoidal values, in which case all measurements in the nature you have suggested are invalid.

The .99 PF values shown in your example are ideal "on paper" calculations, and would never be attainable as an average over 6000 hours of usage by any induction motor.

Simply parallelling another set of #8 AWG copper wires (900' total as you stated) would in effect cut the resistance by 1/2.

$350 dollars/year under extremely ideal 'on paper only' conditions does not add up to the expense of PFC over proper circuit correction.

I know we have explained this at length here on CR4, and there seems to be no end of folks willing to believe beyond reason and practicality that PFC can be cost effective over proper circuit design, unless there are tariff incentives or it is not possible to make corrections to the design, so I will include this link to a very concise and highly technical explanation, " unbiased, intelligent comments and input" as you requested, including links to the many related energy savers, devices and theories.

Glad to be of service, CJM

Friends,

First, a few general observations on this thread:

1. CJMcGill in #17 pointed out the incorrect wire resistance value in the original post's calculations for wire losses.
2. Both Masamud #4 and Simon Wan #2 were figuring metering consumption on the KVA without any adjustments for power factor. I believe this is correct, although at least one other post suggested that the consumption was based on real power.
3. The original post didn't include any savings of consumption because of the use of PFC's, just the over-stated savings of line losses (due to the unrealistic use of 0.75Ω/1000') for wire resistance.
4. SB had a spreadsheet with rather little explanation of the entries, but it included the corrected values of line losses and the consumption savings, as well as a factor for the internal losses of the capacitor bank. I fail to see why it was off-topic.

Therefore, back to the original questions:

• The annual costs should be corrected to show a metered usage of 78A before PFC and 51A after PFC. These two numbers, entered into the formula with 6000 hours per year, will give energy costs before and after correction.
• Assuming that these were measured current flows, the line losses are included in these two numbers. Assuming a 5% maximum voltage drop the resisteance is 0.27Ω/1000' and line losses are $427 and $182 per year, respectively. They are already included in the energy savings and are of significance only if they occur within an environmentally cooled space (therefore adding to the cooling load).
• The additional internal losses within the capacitor bank are also included in the metered bills.
• At an original power factor of 0.80, I am surprised that the utility was not separately metering the KVARs. The post didn't mention this, so I will ignore it.

The decision to add PFC to this customer's equipment should be made on the energy savings of 27A for 6000 hrs per year, or $6734 per year (using the rates in the original post). Other factors in making this decision would be the presence or absence of KVAR metering and the need for local energy conservation.

J Raef, your posts were excellent examples of politeness on CR4. Thanks.

--John M.

Dear jmueller,

Those amp differences are the most extreme possibilities, and will not be sustainable for more than a brief period in actual practice.

Extending the snapshot of .99 PF and 27A reduction over a 6000 hour usage period is unsupportable logic in the least and outright fraud in the worst case. Those results are not physically possible to maintain as the average over that period of time, even if it is possible to demonstrate them for a brief moment of time.

The reality check is that 6000 hours is 2760 hours short of a full year, not continuous operation, and the start-up and stopping process at that level of PFC will cross the frequency (in the resonant circuit with the capacitor and motor) that produces extremely high and damaging voltages every time it is operated.

PF correction to that level is only recommended and appropriate for an aggregate load such as a distribution system. It is not applied to a single load since at some common values of operation there is serious probability of damage due to over-correction. Loads do not stay at full load value, except on paper.

Have you been to the link I provided that details this?

In short, it is not $6734 per year upon which a decision for PFC will be made, unless the decision -maker is unaware of reality and ill advised.

CJMcGill,

Your points are all good. I didn't take the time to look at the link you had posted. My practice is to correct only to 0.95 PF and do so at the major individual loads. Thanks--

JMM

Most savings come when the company is penalized for not correcting power factor I believe.