Inductance and Capacitance as the Secondary Load

[jim hardy]

[QUOTE="Jimmy Lalani, post: 5825807, member: 628145"T]he turns ratio =Vp/Vs= 230/24=9.5833.[/QUOTE]
Published numbers are for operation at rated load. They include voltage drop across Rp Xp Rs and Xs.

Turns ratio looks to me more like 230 / 27.39 = 8.397

That's why you run open circuit test, to eliminate or at least minimize the voltage drop across Rp Xp Rs and Xs

@Jimmy Lalani Try your calculations using observed turns ratio at open circuit
because that is when voltage drops across those impedances are negligible.. You'll come a lot closer to the actual turns ratio.

 

[Jimmy Lalani]

Hello Hardy Sir,
1) Is it possible to calculate the no load voltage mathematically? According to my observations it is always between 10-20% more than the rated voltage. However I am interested in calculating it mathematically.

2)For core losses I applied the following method:
The length of the medium flux path is: Lav = Pi*(ID+OD)/2 = 0.198 m.

The mass of the core is: M = Lav *(OD-ID)/2 *H* density =
= 19.8 cm * 1.7 cm * 3 cm * 7.55 g/cm3 = 762 g =0.762 kg
So the iron losses are: PFe = 1,1 W/kg * 0.762 kg = 0.84 W

3)And no load current= Pfe/230= 3.67mA (without considering the voltage drop across primary impedence)
Is it correct?Jimmy

 

[jim hardy]

1) Is it possible to calculate the no load voltage mathematically? According to my observations it is always between 10-20% more than the rated voltage. However I am interested in calculating it mathematically.

Sure it's possible if you know the actual turns ratio. The voltage ratio they give you is not the turns ratio it's the voltage you should expect at normal load.

2)For core losses I applied the following method:
The length of the medium flux path is: Lav = Pi*(ID+OD)/2 = 0.198 m.

Geometrically that looks fine and is probably close enough for experimenting at home..
Some folks correct for uneven flux density inside the core, see worked example here
http://pe2bz.philpem.me.uk/Comm01/- - Parts-NonActive/- Inductor/MakeCoils/terms.html
you'll find this approximation is used as well

The mass of the core is: M = Lav *(OD-ID)/2 *H* density =
= 19.8 cm * 1.7 cm * 3 cm * 7.55 g/cm3 = 762 g =0.762 kg
So the iron losses are: PFe = 1,1 W/kg * 0.762 kg = 0.84 W

The result of that will be as good as the 1.1 w/kg number. From where did that come ?

3)And no load current= Pfe/230= 3.67mA (without considering the voltage drop across primary impedence)
Is it correct?

?? i thought you measured no load current as 5 milliamps. See your 'observation ' table.
Which instrument do you prefer to believe , the wattmeter or the ammeter ? If the ammeter is digital and reading only on its least significant digit it's not very accurate.

 

[Jimmy Lalani]

Hello Hardy Sir,
1) I do not know the actual turns ratio. Is it still possible to calculate the No load voltage? Does the calculation differ for a frequency of 400Hz?
2) W/kg is an assumed value. However, I found that it does not work for all VA rating. I think this W/kg depends on the VA rating and the frequency considered, because when I performed the test for 200VA 50Hz I got core loss of around 2 W and for 600VA, 400Hz it was around 9W.
So, I searched on internet and I found that either I should find the W/kg for the given rating of transformer or the Core loss density at that frequency and Bm(Maximum Induction). How can I move on from here?

 

[jim hardy]

1) I do not know the actual turns ratio.

That's why you measure open circuit voltage ratio.

Is it still possible to calculate the No load voltage?

Why ? You measured it.. Your questions are circular . What information do you have about the transformer that you would use to calculate its open circuit voltage?

Does the calculation differ for a frequency of 400Hz?

? Open circuit voltage? To what accuracy? For all practical purposes, no the calculation does not differ. Of course core losses and leakage reactance will have more effect at higher frequency.. . but at no load they should still be small.
Have you studied the wikipedia article on transformer model? It's at
https://en.wikipedia.org/wiki/Transformer .

2) W/kg is an assumed value. However, I found that it does not work for all VA rating. I think this W/kg depends on the VA rating and the frequency considered, because when I performed the test for 200VA 50Hz I got core loss of around 2 W and for 600VA, 400Hz it was around 9W.

Look up Steinmetz formula for iron loss.

So, I searched on internet and I found that either I should find the W/kg for the given rating of transformer or the Core loss density at that frequency and Bm(Maximum Induction). How can I move on from here?

Take a look at this old old thread. It addresses similar questions.
https://www.physicsforums.com/threa...ses-hysteresis-eddy-current-constants.850475/

 

[Jimmy Lalani]

Hello Hardy Sir,
1) I looked at the steinmetz equation. However, it shows the constant value for frequencies above 1Khz. I am looking for 50Hz and 400hz.
2) Now Hardy Sir with your help I have managed to calculate the transformer parameters mathematically. However, I was just thinking that how are the cores selected? I mean how are the dimensions of the core selected? I thought about the maximum VA capacity but was not satisfied with it as each time I put in values of higher core dimensions the VA capacity is more, so it cannot be the only reason. Is there anything else from which I could decide the core dimension for a particular transformer?

Jimmy

 

[jim hardy]

Here's a "How To" from a manufacturere of cores
see page 16
you'll have to zoom into make it legible

https://issuu.com/magnetics/docs/2016_magnetics_tape_wound_cores_cat/1?ff=true&e=3217463/32125843

a snip from top of page

 

[jim hardy]

I thought about the maximum VA capacity but was not satisfied with it as each time I put in values of higher core dimensions the VA capacity is more, so it cannot be the only reason. Is there anything else from which I could decide the core dimension for a particular transformer?

Well think about simple geometry and building blocks.

You have to have a window large enough to encircle however many turns of wire of whatever gauge you need to deliver the power.
You have to surround that window with a core having enough cross sectional area to carry the flux it takes to make your needed volts per turn.
The higher the amps the bigger the wire.
The bigger the wire the bigger the window for it to pass through.
The bigger the window the longer the core to surround it.
The higher the volts per turn more the flux.
The more the flux the thicker and wider the core to carry it..

each time I put in values of higher core dimensions

You must be playing with a core calculator program of some sort ?
That's the danger of not starting with the basics you get really confused.

Why do you think VA rating goes up as the core gets bigger ?

 

[Jimmy Lalani]

Hello Hardy Sir,
Thank you for the pdf file you provided me. I have 600VA, Vp=230V, Vs=36V, f=400Hz Toroid transformer. B=1T(I assumed this value as at 400 Hz B is between 1-1.2) Is it correct? Or should I also calculate Bm?

1) From this I found data such as Ip=2.608A, wire diameter=1.12mm, Area=0.9858mm^2 b) Is=16.67A, wire diameter=2.85mm, Area= 6.1575mm^2

2) From the pdf file you provided I calculated WaAc Product and compared it with my table for cores and I found the core 70x42x40mm. Is it correct?

I have a doubt how do we decide IDmin, ODmax, Hmax? Can I directly calculate WaAc product from the normal core provided and then compare it with the value from above formula?

3) Number of turns N= V/(4.44BfA)

4) length of one turn = 2h+(OD-ID)

 

[jim hardy]

B=1T(I assumed this value as at 400 Hz B is between 1-1.2) Is it correct? Or should I also calculate Bm?

Your core datasheet will tell you if it can operate at 1T. Sounds reasonable for decent transformer steel.

2) From the pdf file you provided I calculated WaAc Product and compared it with my table for cores and I found the core 70x42x40mm. Is it correct?

You do realize you're asking me to evaluate a table I've never seen , don't you ?

I don't want to check your arithmetic - surely you have a co-worker who will.

 

[Jimmy Lalani]

Hello Hardy Sir,
Ohh I am sorry. I asked the question wrongly. Actually from the formula I get WaAc= 70 and from the core given I found this core WaAc product as 77 which was the nearest as compared to other core. So, I chose this core.

 

[Jimmy Lalani]

Hello Hardy Sir,
This is how I calculated IDmin: Np=231, Ns=36
(Np x Primary wire area)+(Ns x Secondary wire area)= Total area occupied by the turns=4.49cm^2
Inner diameter area= 13.85cm^2.
So IDmin Area= (13.85- 4.49)=9.36cm^2.
So IDmin= 3.4cm. Is it correct?

 

[Jimmy Lalani]

hello Hardy Sir,

From the figure provided above I have managed to calculate all the parameters mathematically. However, I am confused in calculating the value of Xm. Can you please help me.
my data was f=100Hz, Vp=100V, Vs=10V. Core weight =0.615kg
Therefore, Core loss Pi =Specific no load loss (1.1 W/kg)x Core weight = 0.6765 W
Rc=Vp^2/(Core loss)= 14782 W

No load current= Pi/Vp= 6.765 mA ?( is this correct way to calculate as I have not included power factor)

How do I calculate Xm?

 

[jim hardy]

Rc=Vp^2/(Core loss)= 14782 W

Using your numbers I got for core loss 1.1w/kg X 0.6765 kg = 0.74415 Watts
and if your 1.1 W/kg is the right number for whatever is your flux, that should be right.

Neglecting small drop across Rp and Xp, Vp^2 /(that core loss) = ~100^2 /0.74415 = 13438 Ω (ohms symbol is in the sigma sign atop posting area)
and for 100 volts across that many ohms i get 7.4415 ma

No load current= Pi/Vp= 6.765 mA ?( is this correct way to calculate as I have not included power factor)

No load current is the sum of I_C and I_M . What we just calculated is I_C

How do I calculate Xm?

To find I_M you will have to go back to your wattmeter and current readings.
Excitation current that you measured with ammeter in open circuit test, I₀ , is I_C - j I_m .
However - i am skeptical about resolution of a wattmeter at such low power. It might be advantageous to measure current and its phase angle with an oscilloscope. Then you'll have current and power factor at no load from the same instrument.

 

[Jimmy Lalani]

Hello Hardy Sir,
Thank you for the reply. However, I am interested in calculating Xm mathematically without using any meters, in the same way as I found Rc. Now can you help me?

 

[jim hardy]

C'mon, you already know the answer. Use what you know and what facts you have at hand and figure out what steps would lead you to the answer you seek.

What is cross section of core ?
What is mean path length of core?
What is permeability of core material ?
What flux is required to support 100 volts(or your operating voltage) of counter-emf in your number of primary turns? Maybe you prefer to use volts per turn..
How many amp-turns are required to push that much flux through your core, given its dimensions and permeability? That result does not include iron losses.
How many amps then flow at no load to magnetize the core? Isn't Xm equal to applied volts / that many amps ?

 

[Jimmy Lalani]

Hello Hardy Sir,
I used the same process you quoted above. However,the problem is I do not know the permeability of the given core. I could only calculate the following parameters:
1. The core selected is 60x38x50mm, cross section Area=550 mm^2. at Bm=1 T, f=100Hz, Vp=100V
2. Mean path length of core lc=15.13cm
3. Np=431
4. Volt/turn=0.232

 

[jim hardy]

Hello Hardy Sir,
I used the same process you quoted above. However,the problem is I do not know the permeability of the given core. I could only calculate the following parameters:
1. The core selected is 60x38x50mm, cross section Area=550 mm^2. at Bm=1 T, f=100Hz, Vp=100V
2. Mean path length of core lc=15.13cm
3. Np=431
4. Volt/turn=0.232

Then what is the question?
Are you still trying to calculate Xm ?

I am interested in calculating Xm mathematically without using any meters, in the same way as I found Rc.

And how did you find Rc ?
Did you not use published watt/kg loss?
Is there not a published μ ?
Or a published AL value?
http://www.encyclopedia-magnetica.com/doku.php/al_value
http://www.pocomagnetic.com/html/2014/03/06/2014030605592122540691.html

 

[Jimmy Lalani]

Hello Hardy Sir,
Yes I am still trying to find Xm.
I did use published watt/kg loss. However, there is no published µ and AL value given.
I have tried these formulas which you have provided in the links but the solution was not as desired. Therefore, I have asked for your help.
I am really sorry that I am bothering you so much.

 

[jim hardy]

Well you have to use either properties as given in datasheet or empirical measurements..

If you know volts per turn you know flux, e = n dΦ/dt
and is you know flux and area you know flux density B
and if you know flux density , magnetic path length, and magnetizing current's amplitude and phase , then you can separate magnetizing current into inductive and real components
the inductive component makes the flux and the real component makes the heat
solve the familiar
B = μNI/Length
for permeability μ

but you keep insisting that i tell you how to calculate something from nothing.

I did use published watt/kg loss. However, there is no published µ and AL value given.

Well, where is it? What IS published ?

 

[Jimmy Lalani]

Hello Hardy Sir,
I ahve attached a file below regarding the watt/kg value which I have taken of the material specified. It shoes only for 50 and 60 Hz. looking at this values I ahve taken 1.1 for 100 Hz. unfortunately there was no µ and Al value.

 

[jim hardy]

Well here's their list of datasheets
https://cogent-power.com/downloads
though i don't see M097-30N

here's a snip from one at https://cogent-power.com/cms-data/downloads/Hi-Lite_NO30.pdf
it seems to have B vs H tabulated
sometimes those part numbers are cryptic

if you can't find your core material on their site
i guess you'll have to take them up on this offer

and ask for help finding the right datasheet.

Does that "Polarization" in right hand column work out to μ_relative about 1900 ?

 

[Jimmy Lalani]

Hello Hardy Sir,

Does that "Polarization" in right hand column work out to μ_relative about 1900 ?

B=µH, µ=µ0*µr.
µr= B/( µ0*H), µ0=4*pi*10^(-7), B=1T, H=800A/m
from this I get
µr=995 Is this correct?
However with this permeability I get I am very large which should not be. I am totally confused right now and mixing things here and there. Please help me.

 

[jim hardy]

µr=995 Is this correct?

It's correct for 1T

but they show 1.88 to 1.9 Tand that looks reasonable , here's a similar datasheet from another manufacturer .
http://www.aksteel.com/pdf/markets_products/electrical/AK%20CARLITE%20Lite%20042413.pdf

10 Oersteds is approximately 800 A/M(795.7) and 18 kilogauss is 1.8 Teslas

However with this permeability I get I am very large which should not be. I am totally confused right now and mixing things here and there.

I cannot see your work.
I have to think that
were it orderly and neat
with one calculation per line,
columns arranged so it's easy to read,
with notes interspersed explaining what is purpose of each calculation

you would not be confused .

However with this permeability I get Im very large

In the font on my screen lowercase letters "l"(ell) and uppercase "I"(eye) are indistinguishable
so i do not know which you mean, inductance or current, is very large.

 

[Tom.G]

Try these. They were the first two hits of a google search for the core material.

https://www.stanz-und-lasertechnik.de/en/leistungen/elektroblechqualitaeten-uebersicht.html

https://books.google.com/books?id=s...DAB#v=onepage&q=thyssenkrupp M097-30N&f=false

 

[jim hardy]

Fantastic find there @Tom.G

Never dawned on me there was a method to those cryptic numbers .

I learn something every day here .
From your first link

I think that was @Jimmy Lalani 's missing piece.