Power factor correction experiment

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T.M.Haja Sahib

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It is classic trolling without a doubt.
Trolling means deceiving.If you think I posted false information with an intention to deceive,you may list them for all others to see.
 

Besoeker

Senior Member
Location
UK
I would have been beaten senseless, grammar and spelling has always been a weak point for me.
Or it might have improved before it got to that parlous state of affairs.:D

College composition and English courses wrecked my GPA. I blew a 4.0. It happens.
GPA? 4.0? Remember I'm a :? Brit.

Back on topic: What meter are you using? Your readings are impressive.
Thank you kindly.
I usually use a digital storage oscilloscope.
I routinely carry around a Tektronix TDS 220 with an RS232 serial port so that I can download waveforms directly to a PC - provided the PC has a serial port of course.
More commonly for the measurements of the sort I've given in this thread I use this:



It has removable storage media which can store several hundred waveforms in *.csv format. These can be downloaded into a spreadsheet for calculations, made into pretty graphs etc.
Voltage is measured directly or sometimes via a VT depending on magnitude and I use a clamp on CT with a resistive burden for current.


Are UK residential customers charged for PF?
No.

Are all UK residences derived from a 3 phase source?
I think the great majority are typically with a single fairly fairly large 11kV/400V (Dyn) transformer feeding a residential area.
But I have seen a few pole-mounted single-phase transformers in rural locations.
 

jumper

Senior Member
Thank you kindly.
I usually use a digital storage oscilloscope.
I routinely carry around a Tektronix TDS 220 with an RS232 serial port so that I can download waveforms directly to a PC - provided the PC has a serial port of course.
More commonly for the measurements of the sort I've given in this thread I use this
My simulation software has this. It is the same as I used in classes, but ours were 2 channel only.

 

Besoeker

Senior Member
Location
UK
If h.p,power factor,and efficiency of the pump motor are known,it is possible to calculate the increased current X at reduced voltage without the capacitor and the reduced current Y at increased voltage with the capacitor across the terminals of the motor.Once the values X and Y are known,further calculations can be made to ascertain the saving that could be achieved.
The above is based on the assumption that the motor current will increase at reduced voltage. That is by no means a certainty.
 

Strife

Senior Member
All I can say is WOWWWWWWWWWWWWWWW!
But to throw my two cents in, if it walks like a duck, and quacks like a duck, it is a duck.
Don't have any of those tests you guys have, but the moment I heard about these "energy saving devices" it quacked like a duck to me. I mean scam.
 
T

T.M.Haja Sahib

Guest
All I can say is WOWWWWWWWWWWWWWWW!
But to throw my two cents in, if it walks like a duck, and quacks like a duck, it is a duck.
Don't have any of those tests you guys have, but the moment I heard about these "energy saving devices" it quacked like a duck to me. I mean scam.
I want to tell you one incident from history.The famous scientist Sir Issac Newton was wrongly blamed for his interest in Alchemy.The real intention of Newton was to discover whether there were any scientific principles behind the quest of alchemists for gold that could be beneficial to the humanity as a whole.The same is going on here.
 

gar

Senior Member
111115-2117 EDT

T.M.Haja Sahib:

The first two paragraphs of my post #1 are:
I am working on a set of notes on Electrical Energy Measurement, Conservation, & Methods to Reduce Your Electric Bill. To try to illustrate to readers that power factor correction black boxes are probably useless to the average residential customer I have run the experiment that follows. This is based on the type of demonstration shown in the fraudulent advertisements.

It is probably correct to say that virtually no residential customer in the US pays a penalty for poor power factor. Thus, power factor correction is of no value to the customer.
These two paragraphs set the stage for what I am trying to illustrate to the average homeowner interested in trying to reduce their electric bill, reduce energy consumption, and "carbon footprint".

Since you may not be familiar with the fraudulent ads, and what they describe may be the reason you drifted to unrelated points. All of these ads describe installation of the power factor correction box at the service entrance main circuit breaker panel. I believe that in most homes this means only a few feet of wire from the meter to the panel. In my case it is probably 6 to 8 feet of 0000 copper.

I do thank you for reminding me of the low dissipation factor of polypropylene. At the time I ran the experiment I had not actually measured the dissipation factor of the capacitors I used. Since then I have made the measurement with a General Radio 1650-A bridge at 1 kHz. The 30 ufd capacitors measure about 0.002 for D and the 12.5 ufd about 0.02. Thus, the capacitor power dissipation should not be as large as I measured. This implies there are inaccuracies in the instrumentation I am using that prevent high accuracy in the result. But the accuracy is sufficiently good to dispute the claims of the advertisers.

You need to do an Internet Google search for Power Saver 1200 and then look at some of the sales video material to get an understanding of my goal. No way can any homeowner reduce their energy consumption by the implied 25%, or even the guaranteed 6%. One place told me they use two 40 ufd capacitors, one for each side of neutral. X[SUB]C[/SUB] is about 66 ohms at 60 Hz, and at 120 V is about 1.8 A. This current always flows whether inductive loads are present or not. But this does nothing for I[SUP]2[/SUP]*R losses in the house.

I need to re-describe the results of my experiment to indicate that within the accuracy of my measuring equipment there is no energy saving for the homeowner by the use of one of these power factor correcting devices at the main panel.

To see a typical day of my load power see photo P26 at
http://beta-a2.com/EE-photos.html

.
 
T

T.M.Haja Sahib

Guest
No, a pump motor may behave that way but not always.
Yes.The current drops with drop in voltage across the pump motor,if the water output of the pump also drops at the same time.This is an abnormal condition not to be taken into account for the present study.
 
T

T.M.Haja Sahib

Guest
111115-2117 EDT

T.M.Haja Sahib:

The first two paragraphs of my post #1 are:


These two paragraphs set the stage for what I am trying to illustrate to the average homeowner interested in trying to reduce their electric bill, reduce energy consumption, and "carbon footprint".

Since you may not be familiar with the fraudulent ads, and what they describe may be the reason you drifted to unrelated points. All of these ads describe installation of the power factor correction box at the service entrance main circuit breaker panel. I believe that in most homes this means only a few feet of wire from the meter to the panel. In my case it is probably 6 to 8 feet of 0000 copper.

I do thank you for reminding me of the low dissipation factor of polypropylene. At the time I ran the experiment I had not actually measured the dissipation factor of the capacitors I used. Since then I have made the measurement with a General Radio 1650-A bridge at 1 kHz. The 30 ufd capacitors measure about 0.002 for D and the 12.5 ufd about 0.02. Thus, the capacitor power dissipation should not be as large as I measured. This implies there are inaccuracies in the instrumentation I am using that prevent high accuracy in the result. But the accuracy is sufficiently good to dispute the claims of the advertisers.

You need to do an Internet Google search for Power Saver 1200 and then look at some of the sales video material to get an understanding of my goal. No way can any homeowner reduce their energy consumption by the implied 25%, or even the guaranteed 6%. One place told me they use two 40 ufd capacitors, one for each side of neutral. X[SUB]C[/SUB] is about 66 ohms at 60 Hz, and at 120 V is about 1.8 A. This current always flows whether inductive loads are present or not. But this does nothing for I[SUP]2[/SUP]*R losses in the house.

I need to re-describe the results of my experiment to indicate that within the accuracy of my measuring equipment there is no energy saving for the homeowner by the use of one of these power factor correcting devices at the main panel.

To see a typical day of my load power see photo P26 at
http://beta-a2.com/EE-photos.html

.
Your study is proven one beyond doubt.My endeavor is to see whether any thing can be salvaged from that scam,just as the great Newton tried to do for Alchemy.
 
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iwire

Moderator
Staff member
Location
Massachusetts
Yes.The current drops with drop in voltage across the pump motor,if the water output of the pump also drops at the same time.
That can be true.

But in many cases the volume of water would remain constant.

Typically an electric motor is not fully loaded, when it is not fully loaded the efficiency drops.

If the motor is underloaded, as most centrifugal pump motors would be, then reducing the voltage will also reduce the current without a reduction in water flow.

Reducing voltage, or 'voltage optimization' is a common technique for reducing power consumption.

It should go without saying that there are limits to how much you can reduce the voltage.

This is an abnormal condition not to be taken into account for the present study.
There is nothing abnormal about reducing current through voltage reductions.
 

Besoeker

Senior Member
Location
UK
Yes.The current drops with drop in voltage across the pump motor,if the water output of the pump also drops at the same time.This is an abnormal condition not to be taken into account for the present study.
Fixed frequency cage motors are typical of this application.
If the motor is not fully loaded, reducing the voltage will not appreciably affect the motor speed. Slip will increase but by very little.
This will have little impact on the pump head and flow.
 
T

T.M.Haja Sahib

Guest
If the motor is underloaded, as most centrifugal pump motors would be, then reducing the voltage will also reduce the current without a reduction in water flow.
Do you know pump affinity laws? Have you checked your statement with them?
 
T

T.M.Haja Sahib

Guest
If you actually read the article you will see that it resoundingly contradicts your assertion.
Look at the tables and you will see that the current in save mode is significantly reduced than in bypass.
And look at the title:
"Reduced Motor Voltage to Mud Pump Motor Load."
Please look again at tables.The current in save mode is less but the voltage more;the current in the bypass mode is more but the voltage is less.This is what I am asserting.
 
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