Power factor correction experiment

[gar]

111128-1000 EST

Last night I removed the power factor correction capacitor, also temperature was colder at about 35 F. The results were TED 8.0 kWh, and power company 7.0 kWh, ratio = 1.14 from 2230 until 0700. Shorter time period than the previous night. The ratio for the previous night with PF capacitor in place was 0.84 .

There is potential for a lot of error in these short time measurements. This points out the difficulty for an ordinary homeowner to determine whether installation of a PF capacitor will save any money.

Using the measured amount of energy over some time period with a highly variable load profile is a very difficult means to evaluate a product such as a PF capacitor. An instantaneous power measurement is a much better method. But this means that the power measuring instrument must perform with adequate accuracy for the intended purpose. I think the TED 1000 system can not.

Just looking at TED 1 hour data for 0000 to 0900 and 11-23-2011 thru 11-28-2011 you can see the effect of the whole house capacitorr at the main panel. Even though there is weather variation ity is clear that TED has a problem. Days 25, 26, 27 are the only ones with the capacitor present.

[FONT=courier new]
      no cap  no cap     cap     cap     cap  no cap
Hour     kWh     kWh[FONT=Courier New]     kWh     kWh[FONT=Courier New]     kWh     kWh
[/FONT][/FONT]
Day---->  23      24      25      26      27      28

0000   1.114   1.004   1.353   1.116   1.160   1.106
0100   1.056    .965    .792    .752    .616   1.015
0200   1.091    .942    .835    .731    .643   1.034
0300   1.064    .987    .828    .726    .672    .976
0400   1.056   1.042    .745    .823    .663   1.007
0500   1.045   1.054    .825    .744    .695   1.077
0600    .976   1.195    .590    .762    .694   1.090
0700   1.486   1.120    .872   1.106   1.424   1.834
0800   2.446   1.578   2.752   1.610   2.149   2.364  
[/FONT]

The capacitor was not connected until after midday on the 24th. Very clear the TED readings were substantially lower during the nights when the power factor correction capacitor was present.
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[mivey]

Think it's called moving the goalposts.

A relative of grasping, flailing, spinning. But it all depends on the apparent direction of spin imparted on the charge carriers by the gradient force across the Lorentz thrust-plate motivator.

 

[iwire]

A relative of grasping, flailing, spinning. But it all depends on the apparent direction of spin imparted on the charge carriers by the gradient force across the Lorentz thrust-plate motivator.

I think I just sprained my kanfler trying to read that.

 

[mivey]

111128-1000 EST

...But this means that the power measuring instrument must perform with adequate accuracy for the intended purpose. I think the TED 1000 system can not.

I had toyed with the idea of unbalanced loading (causing an issue with accurate metering), DC or harmonics filtering, etc, but the numbers seemed way off for these to even begin to address the difference. It seems the TED is just not adequate for the task.

It does seem odd about the Cirrus chip. Not sure what is different about the utility meter circuitry. I know we don't anticipate fractional low power factors, but the meters are much more accurate than the numbers you seem to be getting from the TED.

 

[gar]

111128-1312 EST

mivey:

Just for reference my outside meter is a spinning disk.

I really believe that the Cirrus chip in a utility meter provides very good performance all across the power factor range. The chip's specifications are quite good.

There are various problems with the TED design, both software and hardware.

I may stop at DTE today and see if they will sell a spinning disk cheap. DTE is in some areas switching to digital meters but are getting resistance because of the RF link. Customer worry about about RF is total nonsense, but it is still occuring.

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[mivey]

111128-1312 EST
Just for reference my outside meter is a spinning disk.

I actually like the spinning disk better with small loads and estimating fractional disk turns. A lot easier sometimes to estimate the fractional turns when you can see the whole disk through the glass as opposed to the electronic disk simulator on the meter face of the digital meter.

 

[mivey]

111128-1312 EST

I may stop at DTE today and see if they will sell a spinning disk cheap.

Ever thought about buying a PQ meter? $3k-$5k is the problem if you can't find a good business use.

 

[gar]

111128-1514 EST

mivey:

I do not want to spend that amount of money for that function.

But more important I need to work with and understand the problems of devices that the average consumer may be and is using.

For instance some person on the TED forum mentioned that he was monitoring his home with TED and added a power factor correction capacitor, and TED data indicated a saving of about 10% on his bill.

I will get around to putting on line several of my plots of data for my home usage. There is wild variation from day to day, but looking at the data over a monthly period it smooths out, and one can see some change that is attributable to seasonal variations. One either has to use rather long averages or do controlled instantaneous measurements to get useful information the PF capacitor scam.

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[GeorgeB]

Think it's called moving the goalposts. Maybe that's just a UK expression.

Nah, common expression; our goalposts just look different than yours.<g>

Of course, we colonists are BEGINNING to understand that game with the round ball that's black and white ... non-American football<bg> and the goalposts are easier to move than ours.

 

[T.M.Haja Sahib]

The curves I gave you are for a real life motor on a centrifugal load.

But that diagram in your post #170 is not showing how the motor will operate on unstable region,when the motor voltage is reduced much more beyond 80%rated voltage and you are giving a wrong interpretation for that!Again see your graph at post #170 .Take the motor characteristic at rated voltage.You can see that the motor develops the same torque at two speeds one at 'unstable region' and another at 'stable region'.The current the motor takes when its torque equals load torque at stable region is still a multiple of full load current.But this starting current then reduces to match with the full load on the motor.The same happens when the motor operates on unstable region with fan type load.Even though the current is 3.4 p.u at 0.55 p.u voltage at unstable region for fan type load (your post #170),this starting current can only drop to match with the load at that point,because there is not any more need for the motor to accelerate further.

 

[Besoeker]

But that diagram in your post #170 is not showing how the motor will operate on unstable region,when the motor voltage is reduced much more beyond 80%rated voltage and you are giving a wrong interpretation for that!

Oh dear! I've been chastised for being wrong yet again.
My bad......but I'll get over it somehow. And when I do, I'll be good as new when I get over....(apologies to Miss C Gayle)

Again see your graph at post #170 .Take the motor characteristic at rated voltage.You can see that the motor develops the same torque at two speeds one at 'unstable region' and another at 'stable region'.

Yep. That's what you get when there is a hump in the curve. Well spotted!

The current the motor takes when its torque equals load torque at stable region is still a multiple of full load current.

Again, not so. As you can see from the curves I presented in post #170, the load torque and the motor torque are equal at about 0.996 pu speed (4-pole motor at 1494 rpm) and 0.64 pu current.

But this starting current then reduces to match with the full load on the motor.

Not so. See above.

The same happens when the motor operates on unstable region with fan type load.Even though the current is 3.4 p.u at 0.55 p.u voltage at unstable region for fan type load (your post #170),this starting current can only drop to match with the load at that point,because there is not any more need for the motor to accelerate further.

You obviously didn't understand the understand the figures and explanation I gave in post #209. Not a criticism - it's just not your field.
The 3.4 pu current at 0.6 pu speed is not starting current. It's the current at that slip at 0.55pu voltage.
You need 0.55 pu voltage and thus about 3.4 pu current are the conditions at 0.6 speed to just match the required load torque.
Not viable.

 

[T.M.Haja Sahib]

Again, not so. As you can see from the curves I presented in post #170, the load torque and the motor torque are equal at about 0.996 pu speed (4-pole motor at 1494 rpm) and 0.64 pu current

But see the diagram,The red line is the load torque.It intersects the rated voltage motor characteristic at a point corresponding to approximately 1.75 p.u motor current.How you arrive it o.64 p.u current?

The 3.4 p.u current at 0.6 p.u speed is not starting current.

You are self-contradicting,You arrived at the above values from the starting current for rated voltage.Then the value you are deriving for reduced voltage should be starting current for that reduced voltage only.
Do not think that motor starting current exist only at the instant of switching on the motor.

 

[mivey]

But see the diagram,The red line is the load torque.It intersects the rated voltage motor characteristic at a point corresponding to approximately 1.75 p.u motor current.How you arrive it o.64 p.u current?

The red line is there, but you might not be able to see it because everything is scrunched together.

You are self-contradicting,You arrived at the above values from the starting current for rated voltage.Then the value you are deriving for reduced voltage should be starting current for that reduced voltage only.
Do not think that motor starting current exist only at the instant of switching on the motor.

While not "instanstaneous", I thought "starting" was kind of self-explanatory.

 

[T.M.Haja Sahib]

Again, not so. As you can see from the curves I presented in post #170, the load torque and the motor torque are equal at about 0.996 pu speed (4-pole motor at 1494 rpm) and 0.64 pu current.

Congratulations :thumbsup:,you got one point lead! I committed a blunder by stating

''But see the diagram,The red line is the load torque.It intersects the rated voltage motor characteristic at a point corresponding to approximately 1.75 p.u motor current.''

The load current is,as you stated,0.64 p.u current only.

 

[T.M.Haja Sahib]

While not "instanstaneous", I thought "starting" was kind of self-explanatory.

Hi mivey,
Don't you think Besoeker committed a blunder by assuming the starting current of motor at the reduced voltage as its running current,because he derived it proportionately from the starting current of motor at rated voltage?

 

[mivey]

Hi mivey,
Don't you think Besoeker committed a blunder by assuming the starting current of motor at the reduced voltage as its running current,because he derived it proportionately from the starting current of motor at rated voltage?

I think Beseoker's approximations were based on sound reason and I have no problem with them. Do you have a different formula/method you are proposing for approximating the voltage and current?

 

[Besoeker]

You are self-contradicting,You arrived at the above values from the starting current for rated voltage.Then the value you are deriving for reduced voltage should be starting current for that reduced voltage only.

It is the motor current at 0.6 pu speed.
You can then compute what it be approximately at reduced voltages at that speed.
There is no contradiction.

 

[T.M.Haja Sahib]

I think Beseoker's approximations were based on sound reason and I have no problem with them. Do you have a different formula/method you are proposing for approximating the voltage and current?

Yes.At 60% speed,the power required by load is proportional to cube of o.6 i.e 0.22 of that at rated speed only.Let us take an example.Let KW of load be 10 kW at rated speed.If its speed reduces to 60%,its power requirement is 2.2kW only.Let it be driven by a 10kwmotor.The current corresponding to 2.2 kW can be calculated taking into account the reduced efficiency and power factor of the motor for this reduced load by using the formula for input power to motor.Can you demonstrate any suitable values for this reduced efficiency and power factor to be used in that formula?

 

[mivey]

Yes.At 60% speed,the power required by load is proportional to cube of o.6 i.e 0.22 of that at rated speed only

For a centrifugal machine, the power required is a function of the square of the speed, not the cube.

 

[mivey]

For a centrifugal machine, the power required is a function of the square of the speed, not the cube.

Actually, that is not right. When comparing throttle control to VSD, the power needed compared to throttling is a factor of the square of the speed.