# Frequency appears to have to effect on hysteresis

Tags:
1. Jul 9, 2016

### entity314

This question is primarily regarding transformers, particularly the impact frequency has on hysteresis loss.

Given hysteresis loss = η×Bmax×f×V

and Bmax = (V ×〖10〗^ 8)/(√2 π×f×N×A)

This means that hysteresis loss

= η (V ×〖10〗^8)/(√2 π×f×N×A) fV

= (η ×〖10〗^8)/(√2 π×N×A) V^2

And as such frequency has no impact on hysteresis loss.

This seems to me to be completely incorrect, as my understanding of hysteresis is that it's the energy input to change the direction of electron motion, and thus greater frequency means greater energy input. Is this incorrect, or is there a fault in the equations?

In addition, I found that if the core in a transformer is parallel to the lines of flux, then power loss to eddy currents is negligible. Is also correct? Why/why not?

2. Jul 9, 2016

One major part of hysteresis loss is due to Faraday EMF's that occur in the core material of the transformers and the resulting currents from them. The transformer works because of Faraday EMF's but besides generating voltages in the secondary coils, you also generate voltages in the iron of the transformer from the changing magnetic flux. Laminations (layering the iron with a thin insulating layer between them) in the transformer material can be used to reduce the eddy currents that result from the EMF's. The frequency comes into play here because the EMF depends upon the amplitude of dB/dt. The higher the frequency the larger the higher the dB/dt. There may be other factors for the hysteresis loss, but hopefully this helps.

3. Jul 9, 2016

### jim hardy

Last edited by a moderator: Jul 11, 2016
4. Jul 10, 2016

### Staff: Mentor

Wow, @jim hardy is quoting my hero, Steinmetz. We'll make an analyst of you yet Jim.

But once again I am so impressed by Jim's skill at researching source materials for PF answers.

5. Jul 10, 2016

### jim hardy

There's some higher math disconnect here that i don't quite see yet..

Energy per cycle X cycles per second gives power

but rearranging this equation from post #1
Bmax = (V ×〖10〗^ 8)/(√2 π×f×N×A)
into
Bmax X A(rea) = V(olts) X K(onstant ) / N , where K(onstant ) includes 10^8, √2, N and pi

Since B X A = Flux Φ

it means
Φ = K X V / F)
let's discard the K so we focus on physics not algebra , and replace the = sign with proportional to sign α just for rigor in our thinking.
Φ α v/F , which we know to be true it's the familiar volts per hertz ratio that's the basis of overflux.protection

But there's a catch - that's based on assumption we have sinusoidal voltage
really flux is proportional to time integral of voltage, we get away with not integrating because integral of sine gives cosine
which has same shape

we all know also
dΦ/dt = V (per turn)
Φ = ∫Vdt ,
OOPS gotta fix this and rethink

integrating voltage isn't same as multiplying it by frequency , isn't it closer to dividing by frequency?
Indeed integrating voltage divides it by frequency
∫sin(ωt) = -$\frac{1}{ω}$cos(wt)

i thought i had something but it hasnt worked out

back later

old jim

Last edited: Jul 10, 2016
6. Jul 10, 2016

### jim hardy

Thanks anorlunda for the kind words

Steinmetz is indeed a giant.

His biography "Modern Jupiter" is quite a good read. To our shame, it was published by ASME not IEEE.

7. Jul 10, 2016

### entity314

The input on from the article is a massive help, thanks.

In regards to the voltage, that was just a formula misread, but doesn't effect the frequency issue.
However, the formula on there is energy loss per cycle, and thus multiplying by frequency gives energy loss per second.

Furthermore, as I go reading through the article, I've found that substituting the B equation into the Hysteresis equation (per second; frequency accounted for) cancels frequency when the 1.6 isn't accounted for, and when the 1.6 is accounted for reducing the hysteresis loss.

This brings me back to my original problem: My understanding of hysteresis loss is that it's the energy input to change the direction of the electrons, and thus the more frequently this occurs, the more energy is lost to hysteresis. However, as I bring the B value in, it shows the frequency either having no effect, or an effect of reducing the hysteresis.

In addition, I'm a bit lost on page 199 (page 3 on the site) as to how a power of 1.6 came into it.

Last edited: Jul 10, 2016
8. Jul 10, 2016

### jim hardy

That's the curve fit to Steinmetz's empirical data.

When i went through school they taught it's between 1.4 and 1.6

mine is that it's tugging the iron atoms around in their crystal lattice into alignment with the mmf , just like aligning the polar atoms of a dielectric in an E-field, both processes do some mechanical work on the medium that becomes heat.
Look up "Barkhausen effect" for a good word picture of domains snapping into alignment. It leads me to a vivid mental picture of the grains distorting, just like cracking your knuckles. So i think of whole atoms cogging over . Mental analogies , pictures if you will, to help me believe the formulas.

There's also eddy current loss in the iron mentioned a few slides later in that colorado link.
So i save my electrons for eddy current and use the nucleus for hysteresis, though your mental model probably works as well for you. We constantly refine our modelsas needed so they lead to the right formula.

Right now i'm same place as you - frequency shouldn't cancel but it appears to
which says i probably misunderstand or have misused a term in the formula.

gotta get away from this keyboard and use a pencil

old jim

9. Jul 10, 2016

A very good reference @jim hardy . I would also like to run across a good reference that explains why some types of iron and other materials make good transformers with minimal hysteresis loss and also don't go into a permanent magnet mode while other materials make good permanent magnets whose permanent magnetism is very difficult to disrupt.

10. Jul 11, 2016

### jim hardy

i think our trouble is we're equating flux to energy

volts per hertz tells what is fux
and units are even correct, a weber is a volt-second
http://www.physics.nist.gov/cuu/Units/units.html

but webers is not energy
so perhaps Steinmetz's 'numerical coefficient' η has units associated with it ? Were one of those units time it might cure the conundrum.

It's a riddle.I
haven't abandoned you, just plodding along.

old jim

11. Jul 12, 2016

### Staff: Mentor

THANK YOU Jim. Hadn't heard of that. I just ordered a copy.

Edit: a bit of trivia. I have a check signed by Steinmetz. He was a local celebrity while living in Schenectady. Local merchants accepted his checks for goods and services but instead of cashing them, they sold them on the collectibles market. My check is one of those. Schenectady recently erected bronze statues of Steinmetz and Edison on Erie Boulevard. Only city I know that honors EEs so much.

Last edited: Jul 12, 2016
12. Jul 12, 2016

### jim hardy

hmmm let's try this approach

i usually start from a mechanistic description of what's happening and figure out the formula instead of starting with a formula and going the other way

@sophiecentaur implores us "do the maths" but i am a child of the lesser gods of math . So i struggle to believe formulas, have tp figure them out mechanically.

That said,
I think the trouble lies in algebra not in physics.
a Triad is a set of three related things
Force Mass and Acceleration are a triad related by simple ratio, if you know any two you know the third
same for Distance Rate and Time

so it is not necessary that all three terms of a triad appear in a formula describing some phenomenon.
and any one term of a triad can be replaced by the other two or vice versa

for example f = ma, so a = f/m
substituting f/m for a into the familiar high school physics distance formula s= ½at2 + v0t gives
s= ½ $\frac{f}{m}$t2 + v0t,
which might tempt the uninitiated to think distance traversed by a falling object is independent of acceleration.

do you mind if i swap to SI to get rid of that pesky 10^8

Steinmetz figured out that hysteresis loss per cycle is ηB1.6
and η has value about 0.0024 if you have flux density in lines per cm2 (Maxwells in my day) and you want ergs per cc per cycle
actually he found various η's for various materials because it's a property of the material.
We could figure a new η for SI units, maybe i will another day ..
Point is, η X flux gives us energy

i'll remove that mistaken V per your post #7
Given hysteresis loss = η×Bmax×f
SI units are EDIT(thanks Mr Entity)
Joules per second = η X Teslas X Hertz
may i use ω instead of hz and lump the 2π in with η ? Call it n' ? (I'll probably forget to type the ' someplace)
Hysteresis loss = η' X Bmax X ω

now to the next formula
Bmax = (V ×〖10〗^ 8)/(√2 π×f×N×A)
That's for Gauss area in cm2, SI uses Teslas and m2
Bmax = V / (√2πf X N X A)
Bmax X A = V/)√2πf X N)
Now let's simplify a bit more
the √2 is there because Steinmetz used RMS volts but peak flux
so if i say we'll use peak both places i can drop the max from B and the √2 from the denominator, just to reduce the clutter
and if i say we'll use a single turn i can drop the N, or if you prefer say we're using volts per turn
B X A = V/2πf
B X A is flux Φ and 2πf is angular velocity ω
Φ = V/ω flux in Webers and volts in volts per turn
Do you agree this formula is functionally equivalent to the second one one of your post #1 ?
how about that - it's a Triad that relates flux Φ, volts per turn V , and frequency ω
Which means a formula without an explicit frequency term in it may well include frequency , it might appear as the other two terms of the triad, flux and voltage
Let me put it back in terms of B by re-introducing area, we'll need it that way in a minute
divide both sides by A,, and since B = Φ/A
B = V/Aω and we'll call that eq(1)
Webers = Volts X seconds/Area

You substituted the second formula into the first, so let me repeat that using my decluttered versions
Hysteresis loss = η' X Bmax X ω
let me change from Bmax to B by lumping also into η a and √2 ,, call it η" now

Hysteresis loss = η" X B X ω , and we'll call that eq(2)
joules per sec = n" X Webers per second

Substituting eq1 into eq2
Hysteresis loss = η" X V/Aω X ω
so ω appears to disappear
eq(3) Hysteresis loss = η" X V/A sure enough no ω
But did ω really disappear ? Or did a Triad fool us ?

Since, from that link in post # 10
Weber = Volt X second ? then
Volt = Weber X sec-1
it follows that a Volt is a Weber per second, which would be weber X sec-1 which is Weber X Hertz

and your formula has volts in it and volts has Hertz in it
so frequency hasn't disappeared it's just hiding in a triad !

i have said often i struggle with math
so must apologize for this 'sow's ear ' of a presentation
but this is how my alleged brain plods along,

I could clean the presentation up in another evening or two
but have a lot of time in it already
and for me it'll be way more productive to toss it out now for a 'child of the greater math gods' to show me my error, or to point me how to clean it up.

Lavoisier says 'Science is just language well arranged' and the above is my awkward first attempt Doubtless fraught with typos.

Is it plausible ?

Corrections welcome

old jim

PS of course we'll have to put Steinmetz's B exponent of 1.6 back in there
i simply forgot to type it in early and carried that omission the rest of the way

Last edited: Jul 14, 2016
13. Jul 12, 2016

### sophiecentaur

I'm a worshiper of the Maths but hardly a consultant. LOL
t-1 is rate of change and, although it has the same unit that frequency has, it is not necessarily the same. Does that resolve some of the question? We have 'slew rate' in op amps which means that the frequency response would be amplitude dependent for a sinusoid. We are talking non-linear here.

14. Jul 12, 2016

### jim hardy

I keep coming back to

it's AC so voltage is of form Asinwt
so the voltage - flux - frequency triad is a derivative or integral one but we often treat it as if it were algebra not calculus

meaning whenever we express flux as a function of voltage it appears to sneak ω back in

Last edited: Jul 14, 2016
15. Jul 13, 2016

### entity314

I'm not sure whether this impacts the final result of the frequency being hidden in the voltage, but you said that
That's incorrect; the unit is
Joules = η X Teslas
as can be seen from the Steinmetz article:

Unrelated, but thank-you for the slideshow, that had everything on eddy currents that I needed.

16. Jul 13, 2016

### jim hardy

thanks - indeed i included hz so it should be watts not joules

correction is appreciated , i fixed it up above

don't get old !

jim

Last edited: Jul 14, 2016