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some test results.

some test results.

Here is what I did today.
Setup is the same. Ted 1001 installed. Clamp ons on service entrance conductors before main breakers , transmitter on 20A dedicated single pole , Power Saver1200 on double pole dedicated 20A.
First step: All branch circuit breakers are off except Ted. Reading is 0.000
TED is working properly.

2nd :I turn PS1200 breaker on. Ted shows 100 Watt load, but the meter outside does not turn.
Amp reading on PS1200 wires , both 6.25A
Looks like PS pumping some reactive power back to the source.

3rd. Oven test. PS 1200 is off , I turn the oven on.
Ted reads 3.870 KW load ,steady.

I turn PS1200 on.
Ted reads 3.780 KW steady.
This is a nearly 100% resistance load, reading is 90W less!

4th step. Mixed load , single phase, my heat gun.

Oven is off , PS1200 is off, heat gun plugged in,
transmitter -heat gun on the same leg.
Ted reads 1350 watt.

Ps1200 is on
Ted reads 1260 watt.

5th step. Mostly inductive , motor load. Split A/C , 3ton. 240V
All other breakers are off. A/C in cooling mode.

A/H , compressor is running, without PS1200
Ted reads 3700 watt
I turn PS1200 on Ted reads 3600 watt.

That's what I've got. I know very well that there are other factors ,accuracy,
all sort of errors. I'd like to point out that we are Compering Data in this case.
Accuracy is more or less irrelevant.
The results are consistent. Dan
 
2nd :I turn PS1200 breaker on. Ted shows 100 Watt load, but the meter outside does not turn.
Click to expand...

Right there sends up a red flag with me.

Amp reading on PS1200 wires , both 6.25A
Click to expand...

6.25 A x 120 V = 750 VA. So this device draws 100 W + 650 kVAR and the meter doesn't move?


Looks like PS pumping some reactive power back to the source.
Click to expand...

I think there is more to this story. Interesting data so far, I have to admit.
 
Oven test. PS 1200 is off , I turn the oven on.
Ted reads 3.870 KW load ,steady.

I turn PS1200 on.
Ted reads 3.780 KW steady.
This is a nearly 100% resistance load, reading is 90W less!
Click to expand...
Uless it can be chalked up as measurment error, this to me proves this is smoke and mirrors with the measuring device. There is nothing that can make a purely resistive load consume less wattage, short of reducing the voltage getting to the resistive element. Given that this device is in parallel, it cannot be doing that. So what it is doing is probably some sort of pulsed leading power factor trick that is making the meter read less watts. There have been/are illegal devices out there that do this to make a meter run backwards, maybe these guys are backing off the full potential to just make a meter falsely read less watts and thus not attract the attention of the PoCo computerized fraud alert systems. So I say again, this is NOT saving energy, at best it is a (barely?) legal way of not paying for what you use. It may even be found to be illegal if a PoCo takes time to investigate them.

Warning: The Bondee B4-88 device in the link I posted above IS ILLEGAL and if caught using it, PoCos prosecute to the full extent. They know that these devices exist, they know how to spot them in use.
 
090509-1830 EST

K8MHZ:

The TED is not UL listed. What difference does that make?


ttdezo:

The TED unit has an instrumentation error that occurs on a highly reactive load. I do not know how TED determines power.

I suspect the 100 W reading you saw with only the PowerSaver connected is probably a negative 100 W. They may use a zero crossing detector in their power algorithm and there might be a slight phase shift here. Lacking any knowledge of how the device works one is left to run various experiments to see if one can find some relationship.

Some suggestions:

In each of these do the no load test with and without the PowerSaver. Then repeat with the oven load.

1. Swap the two hot leads to the PowerSaver. Do we get an addition of the 100 W?
2. Supply the voltage input to TED from the opposite phase.
3. Interchange the two current transformers but keep the same orientation relative to the source.

.
 
090509-1847 EST

Brian:

By running the no load test with and without the PowerSaver (capacitor) with any power measuring instrument is a useful reference type of test.

Today with my 75 W full scale Simpson two coil electrodynamometer wattmeter I tried to run an experiment with a capacitor loads. The capacitor was an electrolytic motor starter or run 21-25 mfd unit. It has considerable leakage and is not satisfactory for my test. I will try some oil filled later and report back.

In M. B. Stout's book on "Basic Electrical Measurements" he mentions special wattmeters for low power factor measurements. So even with standard two coil wattmeters there are potential errors with low power factor.

More later.

.
 
090509-1925 EST

K8MHZ:

Negative power is not a joke. It simply indicates the direction of power flow. If you connect a wattmeter incorrectly you will get a negative reading. Try it sometime.

.
 
brian john said:
I am not sure what benefit you think UL brings to this discussion?
Click to expand...

Well, it WOULD add a little credibility to both devices. It would also provide for some more information, especially testing done by an independent lab.

These devices are scams and may even be illegal, on more than one level.

FIRST, consult an AHJ and see if they would sign off on one.

Next, consult the POCO and see what their policies are.

Then, consult an insurance company and get their view on people connecting capacitors across the 120 volt electrical lines in order to reduce their electric bills. UL listing would come in handy if one of these babies caught fire and burnt the place down. I'll bet an adjuster would see this as an easy way out.

Sorry to sound like a spoil sport....
 
gar said:
090509-1925 EST

K8MHZ:

Negative power is not a joke. It simply indicates the direction of power flow. If you connect a wattmeter incorrectly you will get a negative reading. Try it sometime.

.
Click to expand...

I didn't say it was a joke. Just improperly used in AC. Think about it, don't we get half of our AC from positive watts and the other from negative? Yes, we do as the direction of the current changes, the volts change from positive to negative, hence positive and negative watts. So, why don't they cancel each other out and we get nothing???

Hmm....

You may be thinking of REMF.

Sorry if it sounds like I am jerking your chain but this isn't the first time I have seen people get sucked into scams.

I know your intentions are true. Perhaps I have been overly risque in my responses. I apolize. :cool:
 
K8MHZ said:
Well, it WOULD add a little credibility to both devices. It would also provide for some more information, especially testing done by an independent lab.

....
Click to expand...

It does not have to work to be UL listed just meet the criteria UL requires for listing.

FIRST, consult an AHJ and see if they would sign off on one.
Click to expand...

And what would they add to the discussion, if the install meets NEC where is their beef, maybe the listing?
 
Last edited:
quote=K8MHZ;1049394---improperly used in AC. Think about it, don't we get half of our AC from positive watts and the other from negative?
Nope. or not exactly. As I heard, right here, sometime back, Watts is a pseudo vector. (I think it was Bob Anderson) Watts has no real direction - you can't point in the direction that watts is moving like you can with the wind. So a positive direction is assigned. In this case, gar picked: to us = positive; to utility = negative. So everything we get is positive.

Yes, we do as the direction of the current changes, the volts change from positive to negative, hence positive and negative watts.
not exactly. For a resistive load, with the voltage positive the current is positive and the product (watts) is positive. for the other half cycle, the voltage is negative, and the currrent is negative, and the product (watts again) positive.

So, why don't they cancel each other out and we get nothing?
And I'm thinking you already knew all of this
---/quote


And if the load is reactive things change a bit, there we do trade power back and forth. When one connects a capacitor across the line, the current to charge the e-field has to go all the way to the generator and back, twice each cycle.

and I thinking you knew that too.

cf
 
090509-2008 EST

K8MHZ:

As Cold Fusion has correctly pointed out you do not get negative power on a resistive load on the voltage negative half cycle.

For any load, including non-linear, the instantaneous power is the product of the voltage and current at that instant. Product meaning you multiply.

Average power is the instantaneous power averaged over some time period.

If you look at instantaneous power to a resistive load from a sine wave excitation you get (sin (x) )^2 = 1/2 - (cos 2x)/2. Thus, there is an average value of 1/2 with a double frequency cos term subtracted from the constant. Instantaneous power varies from 0 to 1 to 0 to 1 to 0 for each full cycle of excitation.

If you multiply sin x * cos x you get (sin 2x) / 2. This has an average value of 0 over an integral number of cycles and a double frequency component that cycles between + and -. So in a purely reactive load the power flows in in 1/2 cycle and out in the other half.

.
 
090509-2034 EST

Brian:

My experiment with a good capacitor. The capacitor is a GE 45F277, 12.5 ufd. The capacitive reactance at 60 Hz is 10^6/(2*Pi*60*12.5) = 212 ohms. At 120 v a current of 0.566 A. A 60 W bulb at 120 V is 0.5 A.

With my 75 W Simpson wattmeter the following results were obtained:

AC = 80 V
Capacitor only --- maybe 0.25 W
Bulb only -------- 30 W
Bulb + cap ------- about 30.25 W

AC = 128 V
Capacitor only --- about 1 W
Bulb only -------- 60 W
Bulb + cap ------- about 61 W

These results are what I would expect. The power consumed by the capacitor represents its internal losses, and correspond to a shunt resistor of
about 16380 ohms at 120 V, and
about 25600 ohms at 80 V.

This does not mean there is really this difference because of voltage. There is in all likelihood a significant error in my small power measurements. That does not change the theory or proof of what this thread is about.

.
 
brian john said:
It does not have to work to be UL listed just meet the criteria UL requires for listing.



And what would they add to the discussion, if the install meets NEC where is their beef, maybe the listing?
Click to expand...

As an inspector my concern would be having my signature related in any way to the installation of non-listed devices of a known controversial nature, no matter what that device or installation may entail.

So far, we haven't even agreed on the effectiveness of the device. What about the safety of it's use?

By all means, I am in favor of carrying on with the experiments, but I am NOT in favor of promoting or installing such devices.

I prefer to err on the side of safety.

"It didn't catch fire 'till I left".

Catchy, isn't it? :grin:
 
Cold Fusion said:
quote=K8MHZ;1049394---improperly used in AC. Think about it, don't we get half of our AC from positive watts and the other from negative?
Nope. or not exactly. As I heard, right here, sometime back, Watts is a pseudo vector. (I think it was Bob Anderson) Watts has no real direction - you can't point in the direction that watts is moving like you can with the wind. So a positive direction is assigned. In this case, gar picked: to us = positive; to utility = negative. So everything we get is positive.

Yes, we do as the direction of the current changes, the volts change from positive to negative, hence positive and negative watts.
not exactly. For a resistive load, with the voltage positive the current is positive and the product (watts) is positive. for the other half cycle, the voltage is negative, and the currrent is negative, and the product (watts again) positive.

So, why don't they cancel each other out and we get nothing?
And I'm thinking you already knew all of this
---/quote


And if the load is reactive things change a bit, there we do trade power back and forth. When one connects a capacitor across the line, the current to charge the e-field has to go all the way to the generator and back, twice each cycle.

and I thinking you knew that too.

cf
Click to expand...

So much for negative horsepower, eh? :grin:
 
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