Power Factor Correction for a small 60W AC Induction motor

[DjMadness]

Added:

I detached the fan, and saw a built-in capacitor of 2uF. As you assumed. This helps the motor at its initial to start.

If the attached circuit is not justified, how can we compensate the presence of 2uF with another 2uF externals? Without the 2uF the fan would not start by it self automatically (this is one reason of existence of this inner capacitor,) how can we manipulate it to maximize pf or to reduce ampere drawing?

May main concentration be on the attached schematic. it serves as an inductive load, that directly reply and react to cap compensation

thank you

 

[jim hardy]

Without the 2uF the fan would not start by it self automatically (this is one reason of existence of this inner capacitor,) how can we manipulate it to maximize pf or to reduce ampere drawing?

aha ! So it's not the "shaded pole" motor i thought it was.

Chances are that capacitor is not switched out after start.
If you snip it out the fan will not self start
but most of them just hum but will start after you give the blade a spin by hand.
Try it, it'll run either way i'll bet. My neighbor's air conditioner fan did that when its capacitor went out.
Without that cap you will find the power factor lags and you can correct it with an external capacitor as you initially thought.

It looks to me as if with that capacitor installed your fan has pf very near unity and you can't improve on that.
Start winding current would lag too but it gets its phase gets shifted by the series capacitor
they sized things to get a good power factor.

worth removing it temporarily to gain understanding ? depends on how valuable is the fan.

Please look at the attached schematic and give me your opinions if it would serve as power factor behavior monitor. If this is applied, I can simplify several amperes load, to this single small board, fed by a 220/50VAC transformer, that results would look obvious on PIC circuit, and hence on the clamp-meter. The highlighted compensating cap, is there to be installed and removed and to see equivalent circuit (pf) behavior.

if i understand , you intend this to be a test load for your pic power factor application?
Should work fine provided R1's 10M means ten milli-ohms not megohms..
If you move R1 down into the ground leg it'll be a current sampling resistor that you can observe safely with a grounded oscilloscope.

old jim

 

[DjMadness]

Dear prof Jim

I already tested removing/switching caps from within the fan.

At total removal, the fan don't start unless supplied by an arm force exercised upon the shaft. When you manually give the shaft a starting turn, it would continue rotating but not the same speed when the cap is on. But at removal, you CANNOT replace the inner cap with an external shunt on, because it has a specific wiring connection, that I wasn't able to access it, because it was so tiny and deep.

I tried decreasing the 2uF to 1uF and saw a bit increase (few tens milli-amps) in load dissipation. With the 1uF I tried to shunt another 1uF in parallel, but also no results were found.

For the circuit, I have calculated the C to be around 20uF. This circuit can be described as a huge inductive load, with small circuit, it is way better than buying a 2HP water pump or equivalent, and it *should* be more effective than the fan. Could you please confirm the value of C for pf correction, and state whether I see a reduction in current dissipated or not?

For a normal domestic inductive load, math was done to demonstrate that at best cases, using domestic loads, the best ampere reduction won't exceed 0.1A. I believe the microcircuit I attached previously, would make more ampere reduction and visible.

Please tell me your advises/opinions

 

[jim hardy]

For the circuit, I have calculated the C to be around 20uF. This circuit can be described as a huge inductive load, with small circuit, it is way better than buying a 2HP water pump or equivalent, and it *should* be more effective than the fan. Could you please confirm the value of C for pf correction, and state whether I see a reduction in current dissipated or not?

In your test circuit ?
I'll assume that 10M is ten milliohms. Makes no difference if it's more, that'd just reduce voltage across the reactive branches.

X_L of 50 mh at 50 hz = 2πfL = 15.7 ohms
i see 4.7 ohms of resistance in series with 50 mh of inductance,, 4.7 + j15.7 ohms = 16.4∠73.2°
At 50 volts ohm's law says that'll pass 3.05 amps(call it 3) at 73.2° lagging
real component of current is 3cos73.2 = 0.87 amp
imaginary is 3sin73.2 = 2.87 amps
Of course you'll measure and add in the DC resistance of your inductor and repeat calculation... And make sure your inductor has enough volt-seconds to handle 50 volts...

to get to unity pf your compensating capacitor needs to draw 2.87 amps
at 50 volts that'd be 50/2.87 = 17.4 ohms of X_C
C = 1/(2πfX_C) = 183uf

i showed you my arithmetic
show me yours ?

 

[DjMadness]

Dear prof Jim,

Hello,

please check the above attachment to see cap size (very identical to 183uF). From another part, and before beginning with this circuit as "load" what are you concerns about neglecting the 10m resistor? It's hard to fit in size, and takes too much space to parallely add res to achieve 10m

 

[DjMadness]

And make sure your inductor has enough volt-seconds to handle 50 volts...

Could you provide rated power of inductor? I have a 2W 50mH coil. Do I need more?

Using simple math, P = V^2 / R , and at 50Hz resistance of inductance has a value of 15.7Ohms.

So: P(inductor) = 50^2 / 15.7 = 160W that is large enough.

Please state your thoughts.

 

[jim hardy]

Could you provide rated power of inductor? I have a 2W 50mH coil. Do I need more?

Using simple math, P = V^2 / R , and at 50Hz resistance of inductance has a value of 15.7Ohms.

So: P(inductor) = 50^2 / 15.7 = 160W that is large enough.

Please state your thoughts.

realizing English may not be your first language
and there are short-hand notations on some specialty datasheets

let me ask do you understand the difference between inductive reactance and resistance ?

I have no idea what is meant by "2W Inductor"
can you post the datasheet from which you got that specification ?

beginning with this circuit as "load" what are you concerns about neglecting the 10m resistor?

i did neglect the 10m resistor.
You still haven't said whether it's ten milli ohms , 0.01 Ω
or it's ten meg ohms, 10,000,000 Ω .

Your image seems to agree with my arithmetic though it's so small i can't read most of the numbers.

Do the arithmetic by paper and pencil for yourself , that's the only way you learn to recognize when your computer is producing gibberish. To be dependent on the computer renders one helpless. See my signature line.

old jim

 

[DjMadness]

Dear prof Jim,

I have no access for uploading any image at the moment unfortunately. The resistor in the supply leads (0.01R) is neglected. It is only concerned for the simulator to work with.

I need the correct power of the 4.7R resistor to maintain the 3Amps drawn by this circuit. Also I need the 50mH to withstand a 3Amps current drawing.

Plus thickness of wire should also withstand 3Amps drawing; in other words, no simple 0.8mm soldering wire using soldering gun could interconnect this circuit. Right? At least I need a 3mm wire to interconnect between R, L and added C, or?

 

[DjMadness]

let me ask do you understand the difference between inductive reactance and resistance ?

Simply, a resistance does not get affected by frequency. Inductive reactance could be understood as a frequency dependent resistance. There are other large explanations about reactance and equivalent phasor representation but could minimally understood by previous phrase.

 

[jim hardy]

Simply, a resistance does not get affected by frequency. Inductive reactance could be understood as a frequency dependent resistance.

Good. It's just terminology then...
We use the umbrella term "Impedance" to include both resistance and reactance.
Impedance impedes flow of current and is the ohms we calcuate as ratio of voltage / current.

Impedance comes in two flavors - resistance and reactance.
The difference is current flowing through reactance does not make heat as it does when flowing through resistance.
Of course reactance also has two flavors, inductive and capacitive , and they have their associated time relations between voltage and current.

So when you say 2W(att?) inductor i don't know what you mean. Inductance doesn't make heat

The resistor in the supply leads (0.01R) is neglected. It is only concerned for the simulator to work with.

good. i thought it was a current sampling resistor, but it's more likely your generous allowance for the resistance of your interconnectig wires.

I need the correct power of the 4.7R resistor to maintain the 3Amps drawn by this circuit.

I think you already know the answer: P = I²R

Also I need the 50mH to withstand a 3Amps current drawing.

it needs to be rated for that much current.
You'd like it to have low resistance, though you could reduce your 4.7R by whatever resistance your inductor does have.
Have you picked out an inductor or are you planning to wind one ?

Plus thickness of wire should also withstand 3Amps drawing; in other words, no simple 0.8mm soldering wire using soldering gun could interconnect this circuit. Right? At least I need a 3mm wire to interconnect between R, L and added C, or?

If i read this right http://www.powerstream.com/Wire_Size.htm
#20 wire(.81mm) is 0.033 ohms/meter , #16(1,3mm) is 0.013 ohms/meter .
Either should do fine for hookup and for winding the inductor.old jim

 

[DjMadness]

Dear old Jim,

We are in trouble! I think that my previous posted circuit makes no sense! Once fully charged, an inductor plays the role of a wire, that means when I shunt connect it to 50VAC voltage source, this would be a form of short-circuit.

On the other hand, how come a single inductor with no function (does not turn shaft of rotor, does not transform Vin to Vout,...,does not do any thing, only placed in circuit) would spend 3 Amps?

The purpose of this circuit was to monitor power factor behavior, since R, and L were similar to ones for a motor. This was to let go of motors, especially to save their space/cost

Please I didn't have found any inductor but this small one (at the right of the resistor, the purpose of putting a resistor in this image is to let conclude smallness of inductor).

I got the 50VAC from a step-down transformer 220/50VAC.

I think either I buy a several horse power motor to see its amperage rating, then apply shunt capacitance across it, then see pf behavior, or I`m wasting my time.

What do you think? It feels so down man

 

[DjMadness]

This is the image of the 50mH inductor:

 

[DjMadness]

This for sure cannot withstand a 50VAC.

This is the image of the 50mH inductor:View attachment 100307

This for sure cannot withstand a 50VAC.

 

[jim hardy]

What is the voltage across 50mh for 3 amps at 50 hz?

Are you familiar with inductor ratings? Volt-seconds ?
http://ecee.colorado.edu/copec/book/slides/Ch15slides.pdf

Your little inductor does not look big enough to handle the required flux.
Let's just guess that for a 50 volt 50 hz sinewave the gray area in that graph would be

λ_{i =} 50volts X √2_{(rms to peak)} X 0.636_{(peak to average)} X .01sec_{(half-cycle at 50hz)} = 0.45 volt-second

that's going to be a physically big inductor

What would be λ_i for your 50 mh at 3 amps 50 hz ?

i think you should try your fan motor. Remove the blades and it'll probably run (after a hand start) without that internal capacitor.

 

[DjMadness]

Dear old Jim,

Things got very confused to me.

What do you mean by "try your fan motor"? You mean use it as the choke? Or do the power factor measurement on it?

Suddenly I`m not able to understand anything I don't know why...Too many theories, too many real circuit problems.

Can you advice me what to do please?

 

[DjMadness]

Dear old Jim,

Hi

Following is the secondary of a 220/50VAC transformer:

This is the intended choke to be used as a load, when fed by a 50VAC source. The problem is that I don't know its Henry's value, and no inductance function exists in my multi-meter.

How can I know its L[H]?? Beside impedance changing, what are the consequences if its L is too below or too large than 50mH?

What are you notes about this? Should this translate uploaded circuit to a real connection? Is this true?

Thanks

 

[jim hardy]

What do you mean by "try your fan motor"? You mean use it as the choke?

Yes, that's what i meant - sorry.

Suddenly I`m not able to understand anything I don't know why...Too many theories, too many real circuit problems.

Can you advice me what to do please?

Don't worry, a lot of people don't understand practical inductors very well.
See an old thread "how the power transfers across an ideal transformer. We did a lot of basic work there.

The problem is that I don't know its Henry's value, and no inductance function exists in my multi-meter.

How can I know its L[H]?? Beside impedance changing, what are the consequences if its L is too below or too large than 50mH?

It's easy to get flustered when confronted with what they rushed past in class.
You know the answer already
apply some AC volts, measure the AC current, per Ohm's Law divide the two and result is the impedance Z in ohms.
Next measure the DC resistance of the windings. That's the "real " part R of the impedance .
"Imaginary" inductive reactance part of the Z is , as you know from basic AC circuits ,

X_L = √(Z² - R²)
and L = X_L / 2πf

might as well measure inductance of both primary and secondary windings...

This is why labs are so good for students.

Do you have access to an oscilloscope ?

Okay,
awaiting your measurement of transformer's inductance.

old jim

 

[DjMadness]

Dear old Jim,

I did all this, but on the primary side to the transformer, because I have no access to a variac at the moment, and therefore I cannot supply the secondary with 50VAC.

Using your methodology, I've found that primary inductance is of 8H which is tooo big. Even thought I`m not concerned with primary inductance, but if its has this large value, does this have to mean that the secondary also have a large value of tens of H instead of mH?

If so, this experiment will not work, because it would result in a very small (few mA) current draw, that won't let me show effect of current reduction using caps.

What can you tell me prof?

 

[jim hardy]

Using your methodology, I've found that primary inductance is of 8H which is tooo big. Even thought I`m not concerned with primary inductance, but if its has this large value, does this have to mean that the secondary also have a large value of tens of H instead of mH?

Well how about that !
You reckon they design transformers to have large amount of inductance so the magnetizing current they draw will be not a whole lot ?

If so, this experiment will not work, because it would result in a very small (few mA) current draw, that won't let me show effect of current reduction using caps.

What can you tell me

Hmmm

firstly as to inductance of secondary

Formula for inductance windings around a closed magnetic path like a transformer core
is L = μμ₀N²A_{rea of core} / l_{ength around path}

what do you know about your transformer ? Its L . And its turns ratio,, 220/55 .

Since μ, μ₀, A_{rea of core} and l_{ength around path} are physical characteristics of the transformer

you could lump them all in one constant , call it K_onstant
and write L_primary = N²_primary X K_onstant,

Knowing that , can you figure L_secondary ?
L_secondary = N²_secondary X K_onstant
knowing N_secondary = N_primary X 55/220 ?

See how close you come to 2H . EDIT See post 51 - you shouldn't come very close..What to do ?
I think i'd remove the blades from that fan. and its internal capacitor,
see if it'll run at 50 volts without the blades after a hand start.
If it will, measure its inductance as before.
It has an airgap so should have less inductance.
If it won't run,
think of something else...
...I wonder what would be the inductance of its stator with the rotor out? need low voltage to test it...
Now - remember i said that'd be a big inductor ? It needs a big core to handle the flux required for almost a half volt second with reasonable number of turns.
check these
http://www.newark.com/hammond/195r10/dc-filter-choke-0-05h-10a-15/dp/24B8753/false
http://www.newark.com/hammond/195p5/dc-filter-choke-0-03h-5a-15/dp/24B8752/false

here's a 30 mh 5 amp that weighs only 6½ pounds...
http://www.mouser.com/ds/2/177/5c0034-44711.pdf
http://www.mouser.com/_/?Keyword=195P5

 

[DjMadness]

Dear old Jim,

Since the total resistance in the choke "load" is too high, and could result in several milliamps drawing, can I make its resistance lesser, by using parallel resistors with the choke? In this case 50/diminished R would have a significant value, to be noticed especially when adding cap bank across.

What are (if any) the methods to still use the ~2H secondary as the choke, in the same time drawing several amps?

I will do test on fan once returnedThanks

 

[Tom.G]

Post 49, above:

Lsecondary = N²secondary X Konstant
knowing Nsecondary = Nprimary X 55/220 ?

Looks like a Square got dropped. I get 0.5H.

L_PRI = 8H
L is proportional to turns²
220/55 = 4 (turns ratio)
4² = 16
8H/16 = 0.5H

 

[jim hardy]

Tom's right of course, my mistake, haste makes waste... Thanks Tom !Look at that inductance equation.
A gap in the magnetic path adds a section of l_ength with term μ of only 1 instead of the few thousand that's typical of transformer steel.
So it knocks inductance down smartly.
If you can find a transformer with say a winding good for 48volts at 3 amps
and add to it an air gap
that should do.
Sometimes transformers that look like this

have underneath that sheet metal wrapper a core that looks like this,

http://www.rfcafe.com/references/po...nsformer-october-1960-popular-electronics.htm
(an interesting little article there, by the way)

you can see how easy it would be to place a thin spacer between the E and I shaped pieces and measure resulting inductance change..

You could wind your own coil, too.

old jim

 

[DjMadness]

Dear old Jim,

Your references, and explanations are vociferous!

Let me ask you at the moment about a switch, driven by PIC, that connects/disconnects caps according to code.

Popular one, is about Open-Relay Switch, but it is cored on Relay, that in turn have mechanical moving part.

I want to get rid of motion in switching, and referring to a solid-state device. the ULN2003 is a solid-state device, but its problem is that it doesn't connect load at full cycle.

What I want is a solid device, functions just like an open-relay without moving parts.

Any suggestions?

Following is a schematic for open-relay switch:

 

[jim hardy]

you're after a "solid state relay"

there are thousands to choose from
http://www.crydom.com/en/tech/newsletters/solid statements - ssrs switching types.pdf

 

[jim hardy]

Your references, and explanations are vociferous!

Webster
Simple Definition of vociferous
: expressing feelings or opinions in a very loud or forceful way : expressed in a very loud or forceful way

I've been called worse...

when you read that Crydom link
look at the 'zero crossing' variety of SSR's. They'll be gentle on your caps.

 

[DjMadness]

Dear old Jim,

It was a metaphor or equivalent.

Please tell me how to add images via private messaging. <Beside adding image URL, and if having image on the hard disk>

Thank you