# Power-Save 1200

1. May 2, 2008

### Ivan Seeking

Staff Emeritus
http://www.savepoweramerica.com/store/1562424/product/PS-1200

I linked the site above because the manufacturer's site was having problems and the activity seemed suspicious - lots of data transfer but no screen update - but here is their claim.
http://www.power-save1200.com/1200?gclid=CNDo_I2qiJMCFQY_agodenhEgg

Last edited: May 2, 2008
2. May 2, 2008

### Staff: Mentor

The first quote is obviously completely bogus. The second is less bold, but still bogus. Typically these types of devices act to reduce power factor, but homeowners (in the US, anyway) don't pay for bad power factor anyway. Here's what it says in the FAQ:
Bad power factor increases your reactive power and thus your apparent power, but your power meter measures you true power, thus the above claim is a straightforward lie. Here's a description of the power factor concept: http://www.allaboutcircuits.com/vol_2/chpt_11/2.html

Businesses, however, benefit greatly from power factor correction. The utility has a separate device attached to the meter to measure it (the meter itself still measures only "true power"). But I wouldn't buy one of these devices, I'd buy one from a reputable manufacturer. Ie: Square-D

Last edited: May 2, 2008
3. Jul 8, 2008

### WCWally

The big simple answer is NO. Although power factor is an issue in large commercial and industrial applications, there is no U.S. utility I know of that asses a penalty on residential customers for low power factor. The utility meters measure watt hours not KVA hours, so if the power factor is low, the meter doesn't see it. Correcting power factor in a residential application will save exactly zero dollars.

4. Jul 9, 2008

### Ivan Seeking

Staff Emeritus
Well, there may be a slight gain by eliminating I2R losses in the wiring that result from the increased currents, but that probably amounts to something in the order of 10 watts [assuming 0.00118 ohms per foot of wiring at an average of 50 feet, and an average of 10 amps reactive, which is generous], which would suggest something around 7 KWHrs, or about 70 cents per month in savings. So at $300, we might expect the unit to pay back in about 30 or 40 years. 5. Jul 9, 2008 ### WCWally The problem with these devices is that the makers spin around with the I2R losses, distinct from the KWHr, as if they matter. They are real enough but the big point is that the residential utility meter will not see them nor will the utility bill directly for them. Corrections to power factor will help the utility, which is not a bad thing, but it is important to know that your 400 bucks are going to a noble global cooling cause and not coming directly back to your wallet. 6. Jul 9, 2008 ### LowlyPion FWIW I came across a diagram for an electrical meter that might be found in a residential location. http://www.themeterguy.com/Theory/watthour_meter.htm#The Meter Diagram Such meters do read true power I see. Anecdotally I use a table saw that when it switches on causes a noticeable surge in the circuit. I have thought that balancing that might be useful, but the few times a month I may use it, I couldn't imagine much of a savings. After stumbling across this thread and I now see the way the power is measured I'm just as glad that it was one of those things I never bothered with even looking further into. I haven't seen the Power Save 1200 itself, so I was wondering if the device was just a simple array of passive capacitors or whether there are any active elements to it that might switch in response to inductive loading? The real test of whether it could ever be a cost benefit is that if the device gets warm at all in normal continuous operation. If so then whatever heat the device itself generates surely would result in added cost to the bill and not any savings, even if the meter in the box wasn't measuring true power. A little knowledge can be a dangerous thing in the hands of an infomercial sales guy. (Are you following this camera guy?) 7. Jul 9, 2008 ### kokain Maybe if I tell my motor to say please when it “asks” the power company for more power. Current is current. When your motor has a 20A inrush, it uses 20A. If you run it through a cap first, you still use 20A. There is no way to damage wire in a house built to code. A breaker’s sole purpose is to protect the wire. Wire rated for 20A on a 20A breaker is safe even if you inrush 50A. The breaker trips and protects the wire. A short is theoretically infinite current and the wire is still safe thanks to the breaker. Fear makes people dumb. The desire to saving energy makes people dumb. 8. Jul 9, 2008 ### Ivan Seeking Staff Emeritus There are real reactive currents flowing through the wiring, and the resulting I2R losses are real and would be measured by the meter, but they are insignificant. Last edited: Jul 9, 2008 9. Jul 9, 2008 ### WCWally Yes, the losses are real, no they would not be measured by the watt-hour meter. The reason utilities penalize large users for low power factor is that they would otherwise not get paid for it. In the case of residential users with low power factor they don't directly charge for it. It's only relatively recently that the utilities have had real time pf measurement on large users. They used to come out and test once in a while and assess a penalty based on the test until the next test. Large users that installed correction capacitors, or in some cases, synchronous motors, had to be re-tested to get rid of the penalties. The overall point is that these power factor correcting device claims of savings for residential users are basically fradulent. They play on something true but completely mis-applied. 10. Jul 9, 2008 ### Ivan Seeking Staff Emeritus There is a difference between charging for the power factor, which determines how much extra current must be carried by the transmission lines, and metering losses resulting from wiring in the house. The power meter would measure these losses just as it would the energy used by a heater. If work is done, the meter will measure it. The power company has nothing to do with it. They don't charge industry for I2R losses resulting from VAR because those losses are measured by their meter. They are charging for the added load on the xmission lines. Last edited: Jul 9, 2008 11. Jul 9, 2008 ### WCWally The utilities attempt to charge the cost of delivering whatever they deliver plus whatever profit the markets and the public utility commissions will stand for. I2R losses are a component of KWHr as well. Standard watt hour meters do not pick up the difference between KWHr and KVAHr and there is no residential tariff in the US (that I am aware of) that contains a provision for power factor adjustments for residential service. The utility eats it case by case and just tries to adjust the overall rates to compensate in general. Here, for example, are excerpts from the current PECO tariff. "power factor - As used herein, power factor is, in a single-phase circuit, the ratio of the watts to the voltamperes, and in a polyphase circuit, is the ratio of the total watts to the vector sum of the voltamperes in the several phases." "billing demand - The calculated or measured demand after correction, if any, for power factor; except that the billing demand may be limited to a minimum figure." They correct what their meter sees for the power factor, which it doesn't see. http://www.exeloncorp.com/NR/rdonly...8D7-7AB2FCE4CA2A/4375/s77_complete_011009.pdf 12. Jul 9, 2008 ### LowlyPion Here is a picture of the product Apparently it requires a separate 20A dedicated circuit. (Another cost sink.) It is apparently only a bank of capacitors. (I do see a pilot light. I wonder if that constant energy cost eats up any possible marginal savings.) I see prices between$250 to $315. (There goes the economic stimulus check.) 13. Jul 9, 2008 ### russ_watters ### Staff: Mentor Wally, what Ivan is basically saying is that the IR^2 losses lower the efficiency of the motor: the motor generates more heat without power factor correction than with power factor correction. 14. Jul 10, 2008 ### Ivan Seeking Staff Emeritus I was also thinking of the wiring in the house. We get slightly greater losses in the house wiring because of the reactive currents from motors and other inductive loads supplied by those wires. But now that I think about, if this device is mounted at the panel and not near the appliance, the losses would still occur as the charge cycles through the wires between the capacitors and the inductive loads. Last edited: Jul 10, 2008 15. Jul 12, 2008 ### WCWally It's true whether I agree or not, but I completely agree that with a low power factor there will be more heat generated by any resistance that the additional current passes through. My point is that the electric meter doesn't see it, so the residential customer isn't charged for it, so eliminating it will not save any money. This illustrates my basic complaint about the marketing. The discussion of power factor and I2R losses is perfectly legitimate. There is a slick video showing a motor drawing significantly less amperage with the device (capacitors) connected. All true, all worthless to the residential power consumer. It's the money savings angle that is that is mis-stated. Truth is, you buy this thing, install it, improve the power factor by some amount (they don't state the KVAR value of the device) and save exactly no dollars. Not 5%, not 10%, not "up to 25%!!" .. no dollars. That's my whole point. Find me a utility in the US that charges residential customers for low power factor and I'll be right on board for customers of that utility. 16. Jul 13, 2008 ### Ivan Seeking Staff Emeritus VARs do not do any work on the inductor, but they do work on the intervening wiring. If it does work, it is measured by the meter. The reason VARS do not do work on the inductor is that the energy stored in magnetic fields is returned to the supply. And a perfect inductor returns all of the stored energy. But, if there is work done on any intervening resistance, then the meter will see it. It will see it because not quite all of the energy is returned to the supply, and we see a slight voltage drop relative to what we would see in perfect wires. If we had perfect inductors and perfect wires [ie no resistance], then the VARs would not do work, and the meter would never see it. However, if this device is mounted at the panel, it will not reduce I2R losses because the stored energy oscillates through the house wires between the inductors and capacitors. As this happens, current flows, and losses occur. Last edited: Jul 13, 2008 17. Jul 13, 2008 ### Averagesupernova There is one place that this little gizmo could actually do something useful that I don't believe is mentioned. Ivan, you are 100% correct in stating that it will NOT reduce current between the Power-Save 1200 and the motor. That wiring is part of a tank circuit and will pass circulating currents as you stated. However, imagine you have a hobby type workshop in an unattached garage that is fed by a sub-panel from your house. Suppose it's an older garage with a not-very-large service. What could happen if the Power-Save 1200 is installed at the service in the garage is that current would be lowered between the house and the garage. So without the gizmo the current used in the garage could approach the limits of the service in this hypothetical case. Add the 1200 and it will be reduced as it claims to. Might not save you money on power useage but it could certainly improve voltage drop on the run between the house and garage. - However, this could then also allow any inductive load in the house to cause circulating current on the wires between the house and the garage. Power factor correcting devices should always be installed at the load in my opinion. Technically your neighbor could have large inductive loads and the gizmo you have installed in your house would be helping correct them. Last edited: Jul 13, 2008 18. Dec 4, 2008 ### turin I'm surprised no one mentioned this, out of confusion or otherwise. I just thought I would throw it in because it gave me such a headache once-upon-a-time. I was trying to figure out why my electric bill was so high. It turned out to be due to the crappy appartments I was living in, but I learned from my investigation that the electric company was charging for a "power factor". I was curious how they were doing this, because, as has been pointed out by many of you, the meter cannot determine this. I had a frustrating conversation with a very unintelligent "technician" who kept telling me to "multiply by the power factor", which made absolutely no sense to me. Then, he told me how to determine my "power factor". I don't remember the details, but it had something to do with timing the actual motion of the wheel that spins around. After much skepticism I finally realized that their so-called "power factor" was basically a rate adjustment to increase the billing rate for increased monthly consumption, and it had nothing to do with the real/reactive power factor. Just thought someone might find that amusing. 19. Dec 6, 2008 ### pp_314156 Energy Loss due to low power factor in a residence Here are the assumptions: 1) Residential power meters measure real power. The Oscillating power (the reactive component) that is generated by inductive motors travels through the wires between the motor and the utility grid, passing back and forth through the residential meter. 2) Capacitors can be used to intercept the reactive power from inductive motors, and return it to the source on the next cycle. However, if the capacitors are at the service entrance or utility panel, they will do nothing to reduce I2R losses in the wiring between the panel and the motor. In order to eliminate line losses, the power factor correction device must be mounted at the inductive load. 3) If the resistance of the wiring was zero, then the power losses due to the reactive components would also be zero. However, residential wire does have a resistance as we will calculate below. Now for some tests with a top fill washing machine on a regular cycle: Make Whirlpool Model LCR7244JQ3 Rated 9.8A 120 Volts Age: 3 years First the fill cycle: Real power: 3W Current: 0.4A Voltage: 121.3VAC Apparent power: 5VA Power factor: 0.68 Length of fill 5 min Next the wash cycle: Real power 630W Current: 9.54A Voltage: 120.2 VAC apparent power: 1144VA Power factor: 0.53 Length of wash: 25 minutes Finally the spin cycle while water still draining: Real power: 689W Current: 9.11A Apparent power: 1106VA Power factor: 0.58 And the spin when it gets up to full speed: Real power 513W Current 8.94A Apparent power: 1081VA Power factor: 0.47 The complete cycle took 41 minutes. Total real energy used (the part we pay for) 0.23 kWh. From the data above, we can do the following: Ignore the fill cycle since it is 200 time smaller than the wash and spin. Lump the wash and spin together since they both use a similar amount of energy (another reason to get a front loader. They use about 22 watts on the wash cycle because they don't have to slosh around gallons of water). We will be conservative and use the worst numbers, those of the spin cycle: That leaves us with: 30 minutes Power Factor: 0.47 Current: 8.94A Real power: 513W Apparent power: 1081VA From this we can calculate the following: Impedance phase angle is arcos(True power / apparent power) = 62 degrees Reactive power = Real power * atan(phase angle) = 423 VAR The above is easier to conceptualize as a power triangle, a right angle triangle where the apparent power is the hypotenuse, and the other two sides are the reactive and real powers. The reactive part does not do work. The real part only does work. The apparent power is what you get when you measure the voltage and current as measured with a meter. Now, we will assume that 513W of real work can wash and spin the clothes. Had the power factor been 1, we would have needed 513W/110V = 4.66A. However, because of the poor power factor, we drew an extra (8.94 - 4.66) = 4.27A Stage Two: We need to calculate the energy loss between the washing machine and the service panel. Lets assume the following: 1) The distance between the machine and the panel is less than 40 feet. I think that is generous in most cases. 2) 110V machines are usually wired with 14AWG copper. Looking on the net, we can find some values: 2.524 Ohms/1000 feet http://www.mwswire.com/barecu5.htm 2.525 Ohms/1000 feet http://en.wikipedia.org/wiki/American_wire_gauge 2.6 Ohms/1000 feet http://www.engineersedge.com/copper_wire.htm Let's take the highest value. Using this, we can calculate the resistance of a 40 length of wire to be 40ft * (2.6Ohms/1000ft) = 0.104 Ohm. There are also connections, a screw clamp on the circuit breaker, possible a few spring connectors if the wall boxes are daisy chained, and a screw connection on the wall plug. So lets calculate it directly. When the washer moved from fill to wash, the voltage dropped (121.3 - 120.2) = 1.1 volts. This occurred at a load increase of 9.41A. V = I/R so R = V/I = 1.1v/9.41 = 0.11 Ohms. The washer above is located on the 3rd floor, the breaker panel is on the garage. So, it appears that both methods of calculation give us similar results for resistance. Let's use the higher value, 0.11 Ohms. Stage Three: We want to know the energy loss by heating the wires between the washing machine and the utility power meter. We really don't care what happens on the other side of the meter, although the power company does. Power = I2R = 4.27A * 4.27A * 0.11Ohms = 2 Watts. Energy = (2W/(1000W/Kwh)) * 1/2 hour = 0.001 kWh Cost of this power: 0.001 * 12 cents/kWh = 0.012 of a penny Finally, lets assume you run a wash every day of the year (perhaps true if you have children). Annual power cost:$0.00438

Stage Four the Economics:

1) The unit must be installed at the source, not the panel or we defeat the purpose of eliminating the I2R losses.

2) We will use an interest rate of 7%, the inflation adjusted average growth of the stock market over the last 200 years.

3) Life of capacitors is 15 years. After this, you will discard the power factor unit, or perhaps it will just become useless.

Let's calculate the present value of a hypothetical power saving device:

PV(7%,15years,$0.00438) =$0.04

Provided that MS Excel's PV calculation is still accurate with fractional cents, it looks like the investment is worth 4 cents.

Maybe.

I live in a very cold climate and heat with electricity because natural gas it not available and trees are scarce. All the wiring in the house aside from exterior lighting, is in heated spaces. Most of the year, the "wasted heat" is actually useful to me.

Hope this helps. If you see a mistake here, please let me know and I will correct it.

Regards,
Peter

20. Jan 27, 2009

### Heinz Rosen

Great analysis, Peter. Particularly your point about connecting the power factor correction capacitor(s) directly across the inductive motor so that they are only in the circuit when the motor is switched on. Now, multiply that exercise for every inductive motor in the home -the compressor motors for the ac and the refrigerator(s), dishwasher, hot water circulating pump(s), etc. Every one of these appliances needs its own capacitor. It thus becomes obvious why these devices are not even remotely cost effective for residential applications. And certainly not in the way the manufacturers recommend that they be installed, that is, permanently connecting them at the main panel. Doing that drags the power factor capacitive when the inductive motors are off and could create some real problems with ringing voltages.