How the power transfers across the Ideal Transformer

[tim9000]

tells me current will reverse in center leg wrt outers ? We'll arrive at some voltage across windings as required to force that current?

When you say the current reverse, I don't really get what you mean, do you mean like the voltages are both halfs that are opposite, so if the control was twice the turns of the outers, then the MMFs'd be equal and opposite?

with R zero, center leg current is aided by outer legs
i think it'll really saturate

Thats what I wanted to get varified, that the bigger the control current means the more back emf, which means the more voltage on the control meaning the more back emf and so on, so it's that sort of feed back.Pretty fascinating.So would the dΦ/dt of the circuit in post #266 be the same as a dΦ/dt and inductance of a regular saturatable reactor that wasn't rectifiying AC?Thanks!

 

[tim9000]

Hey Jim, I was hoping I could grab you attention if possible.
Back to another previous point:
I gather when you're designing an inductor (putting an amount of turns on it) the amount of current through it is equally as important as the amount of V.s on it, however I recall you saying you saw no purpose for designing an inductor to operate at μ_max, and I see that if you have a set V.s and you want a big inductance, than you may as well operate down the bottom of the BH curve. But/And I raised the point that if you were opperating at a specific current through the inductor, than wouldn't μ_max be a good point to be operating at, because that way you're getting current through it, you're getting your inductance, and you're getting value out of your steel. I was wondering what your response was to that? (as well as the top posts)
Kind Regards

 

[jim hardy]

Of course Tim , i apologize... i haven't applied the requisite concentration to this thread...

from 260
Could you please elaborate on how low flux and mag current, preserves the current ratio?
Remember that statement was for a current transformer.
Go back to your transformer model. If flux is low then there's not much magnetizing current. So primary current all goes into making secondary current , in accordance with turns ratio.
Or if you prefer, because voltage is low there's hardly any voltage across X_M

So Ideally in a perfect world you'd be operating down the bottom of the BH curve for any magnetic core?
You operate current transformers at low flux but not necessarily low MMF. Primary and secondary amp-turns ideally would cancel .

A transformer intended to move lots of power at substantial voltage you'd operate at higher flux so as to keep core physically small enough to be affordable.

from 260
At first I thought you meant he wound the primary inside, then was like 'oh crap, there's not enough space left to wind the secondary, so then the flared the secondary out the sides, around it like a rectangle. But then you said 'one winding extends int to the air' which winding? do you mean by that? because both harlfs seem to be symmetrically flaring out the sides into the air.

Look closely at that picture.

He's filled the passageways in his cores completely. Those passageways for the windings are called "windows"...Indeed that winding marked by the red stripe extends way out into the air on both sides of the core, so any flux out there is leakage flux.

from 260
Hmm. Why is it most efficient when copper loss in a VT is equal to the core loss?

it's maybe not most perfectly energy efficient but it assures you're using neither excess copper nor excess iron for the power you're moving

from 271
When you say the current reverse, I don't really get what you mean, do you mean like the voltages are both halfs that are opposite, so if the control was twice the turns of the outers, then the MMFs'd be equal and opposite?

. look at that picture(264 & 268)

With R = zero, we have opposing mmf's because all 3 windings are trying to push flux down. So total current will be set by Rload.
How will current divide between the center coil and the outer ones ? I'm not sure.
With your postulated turns numbers, center and series combination of outer legs would have same volts per turn wouldn't they ? Inferring same flux? Seems to me current would have to reverse in center leg... Since primary current is set by Rload it's operating as a current transformer.brain overload - back tonite..

 

[tim9000]

Hey Jim, Thanks for the reply. That's ok. But I'm hoping I can get the last few curiosities tied up soon because time is starting to run out.

That's a good answer to illustrate the importance of flux in preserving the current ratio (of a CT), my mind is still a bit like a sieve in this field sometimes. How important is low flux (small magnetising current) though in a VT? Let me see if I've got this right/straight:
I remember from our previous discussion that if the magnetising current is high (such as from saturation) than the increasee in current will be on the larger voltage drop on the resistance of the coil, and so less induced voltage on the primary and secondary. So is it fair to say that while the impedence of the magnetising branch is linear, the amount of reduced voltage on the primary and secondary will be linear due to it will be primary resistance. While it's not what you'd call good, it is atleast predictable, so not a big deal...Actually hang on, if You want a more efficient transformer, have less resitance in the copper and less reluctance in the core, for less flux, because the lower the reluctance the lower Rc but the higher Xm, so bigger inductance means less current.
And/But with a CT the current ratio is parramount so you don't want to be having to account for magnetising current (as Ipri = I_in - I_M) in your equation, that would be unprofessional for a manufacturer, so you have to a big core (less reluctance) for bigger Xm, for less flux.
So the MMF might not be low, but the Net MMF will be low.
How am I travelling?

Indeed that winding marked by the red stripe extends way out into the air on both sides of the core, so any flux out there is leakage flux.

I think the issue was I was misinterpretting what you meant by winding, you mean't like primary or secondary, I was thinking you meant physical side of the TX, my bad. So he wasn't trying to get more turns on the thing by flaring it out the sides? Like trying to be a smart-arse geometrically, or you could say engineering a solution to the small window problem? Instead you think he was experementing with leakage flux?

it's maybe not most perfectly energy efficient but it assures you're using neither excess copper nor excess iron for the power you're moving

Hmm, that still begs the question, how is it indicating the design is equally ustilising copper and steel for the job?

At the risk of sounding like a broken record I'm going to put this to you again:

I gather when you're designing an inductor (putting an amount of turns on it) the amount of current through it is equally as important as the amount of V.s on it, however I recall you saying you saw no purpose for designing an inductor to operate at μmax, and I see that if you have a set V.s and you want a big inductance, than you may as well operate down the bottom of the BH curve. But/And I raised the point that if you were opperating at a specific current through the inductor, than wouldn't μ_max be a good point to be operating at, because that way you're getting current through it, you're getting your inductance, and you're getting value out of your steel. I was wondering what your response was to that? (as well as the top posts)

Because you previously said you saw no problem in, if one wanted a bigger inductor, they could just wind more turns around it, and I can see that would be fine provided you were designing it for V.s, but if you were designing it for a Current, than doesn't that luxury evaporate? -> Say you've hit the wall as far as increasing the side of your core goes. You've got the current going through it, and it's hunky dorry, down the bottom of the BH curve. You think, "ok room to play with", and you wind more turns on it, B goes up, then isn't μ_max the last port of call, it will give you the best inductance and you got more turns on it for the same current?

With R = zero, we have opposing mmf's because all 3 windings are trying to push flux down. So total current will be set by Rload.
How will current divide between the center coil and the outer ones ? I'm not sure.
With your postulated turns numbers, center and series combination of outer legs would have same volts per turn wouldn't they ? Inferring same flux? Seems to me current would have to reverse in center leg... Since primary current is set by Rload it's operating as a current transformer.

I still don't see why the control current would reverse direction?
With the rheostat at zero, Wouldn't it act like a normal VT with a SC secondary? They're both desaturating each other and so the dΦ/dt will be really small or zero?

- back tonite..

Look forward to it!
Thanks

 

[The Electrician]

Hmm. Why is it most efficient when copper loss in a VT is equal to the core loss?

Have a look at this chapter on transformers from the famous Radiotron Designers Handbook: http://headfonz.rutgers.edu/RDH4/CHAPTR05.PDF

At the bottom of page 205 it says: "A particular transformer reaches its maximum efficiency when the copper losses have become equal to the core losses (proof given in Ref. A10)"

Reference A10 is found at the end of the chapter on page 252. It refers to a 1950 book which is probably not easy to find.

The fact that maximum efficiency of a transformer occurs when iron losses equal copper losses is often stated but a proof is not often found.

You might also be rewarded by obtaining a copy of the out-of-print, but fairly easy to find book, "Magnetic Circuits and Transformers", by the staff of MIT's EE department, published in 1943.

 

[tim9000]

Have a look at this chapter on transformers from the famous Radiotron Designers Handbook: http://headfonz.rutgers.edu/RDH4/CHAPTR05.PDF

At the bottom of page 205 it says: "A particular transformer reaches its maximum efficiency when the copper losses have become equal to the core losses (proof given in Ref. A10)"

Reference A10 is found at the end of the chapter on page 252. It refers to a 1950 book which is probably not easy to find.

The fact that maximum efficiency of a transformer occurs when iron losses equal copper losses is often stated but a proof is not often found.

You might also be rewarded by obtaining a copy of the out-of-print, but fairly easy to find book, "Magnetic Circuits and Transformers", by the staff of MIT's EE department, published in 1943.

Right, so it isn't about 'a most affective use of steel and copper for a design of a TX, minimising materials for construction' The benifite is as an efficient in energy use, like some sort of 'maximum power transfer' thing?
Ok, I'll follow up your lead, thanks a lot.

 

[tim9000]

Have a look at this chapter on transformers from the famous Radiotron Designers Handbook: http://headfonz.rutgers.edu/RDH4/CHAPTR05.PDF

At the bottom of page 205 it says: "A particular transformer reaches its maximum efficiency when the copper losses have become equal to the core losses (proof given in Ref. A10)"

Reference A10 is found at the end of the chapter on page 252. It refers to a 1950 book which is probably not easy to find.

The fact that maximum efficiency of a transformer occurs when iron losses equal copper losses is often stated but a proof is not often found.

You might also be rewarded by obtaining a copy of the out-of-print, but fairly easy to find book, "Magnetic Circuits and Transformers", by the staff of MIT's EE department, published in 1943.

You were right, I couldn't get an ebook of 'transformers by F. C Connelley, 1950, and I can't afford to pay $100 dollars for one on ebay.

So would "Magnetic Circuits and Transformers", by the staff of MIT, also have a proof do you think?

Edit: actually much the same story there, I'm curious but not USD $75 curious.

 

[The Electrician]

It means that if you have a given core, the copper windings should be such (number of turns, wire diameter) that the losses in the copper are the same as the losses in the core, for a given power level. If a transformer is used at full rated power all the time, the windings will be different than if the transformer is only used at, say, 50% of rated power 80% of the time, and 100% of rated power for 20% of the time,

This makes it an interesting problem to design a transformer that will be used at varying power levels throughout the day, such as a distribution transformer supplying your home.

In the evening when you're cooking supper, heating the house in winter, etc., heavily loading the transformer, the total losses in the transformer will greater than during the day when you're away at work. The transformer designer has to find the optimum so that the initial cost of the transformer, plus the cost of supplying the losses during the life of the transformer is minimized. The designer has to make a guess about how the residential loading on the transformer may go up as the homeowners start using more electrical appliances, or the loading may go down if the homeowners buy more efficient appliances. Not a simple problem.

 

[The Electrician]

So would "Magnetic Circuits and Transformers", by the staff of MIT, also have a proof do you think?

No it does not, but it does have an extensive discussion of the many factors involved in the problem of designing a transformer for varying load conditions like I discussed in the previous post.

 

[tim9000]

No it does not, but it does have an extensive discussion of the many factors involved in the problem of designing a transformer for varying load conditions like I discussed in the previous post.

hah, ok, it's a bit of a mystery proof.
That does sound like a useful book though.

It means that if you have a given core, the copper windings should be such (number of turns, wire diameter) that the losses in the copper are the same as the losses in the core, for a given power level. If a transformer is used at full rated power all the time, the windings will be different than if the transformer is only used at, say, 50% of rated power 80% of the time, and 100% of rated power for 20% of the time,

So an under-used transformer will have most of the losses in the core, and a fully loaded transformer will have most of the losses in the copper. So are you saying the designer of a distribution transformer will try and average out where the losses in the TX are, based on how loaded the TX is? For instance, say it was at rated power 100% of the time, THEN you'd want copper and iron losses to be equal, but say it was at rated power 20% of the time and over rated at 80%, then you'd have a preference to minimising copper losses? Conversely if it was at rated power 20% of the time and under rated power 80%, then you'd have a preference to design it to minimise core losses?
Is that how I should be interpretting your statement?

 

[The Electrician]

hah, ok, it's a bit of a mystery proof.
That does sound like a useful book though.So an under-used transformer will have most of the losses in the core, and a fully loaded transformer will have most of the losses in the copper. So are you saying the designer of a distribution transformer will try and average out where the losses in the TX are, based on how loaded the TX is? For instance, say it was at rated power 100% of the time, THEN you'd want copper and iron losses to be equal, but say it was at rated power 20% of the time and over rated at 80%, then you'd have a preference to minimising copper losses? Conversely if it was at rated power 20% of the time and under rated power 80%, then you'd have a preference to design it to minimise core losses?
Is that how I should be interpretting your statement?

Not quite. It wouldn't be a good idea to run a transformer over its rating 80% of the time.

If it's running at less than full rated power most of the time, which is typical for distribution transformers, you would use less copper, so the copper losses would be higher at full rated power.

There can be other considerations that determine the allocation of copper and core losses.

For example, here's a transformer from a 4000 watt sine wave inverter:

This transformer has 12 watts of core loss at no load, and at full load. The copper loss at full load is about 200 watts. The reason for this disparity is that customers seldom run their inverters at full load, and they want very low no load and light load losses, core and copper combined. The no load losses determine how fast your battery will run down when only the nite lite is being powered.

 

[tim9000]

Not quite. It wouldn't be a good idea to run a transformer over its rating 80% of the time.

If it's running at less than full rated power most of the time, which is typical for distribution transformers, you would use less copper, so the copper losses would be higher at full rated power.

There can be other considerations that determine the allocation of copper and core losses.

For example, here's a transformer from a 4000 watt sine wave inverter:
View attachment 88865

This transformer has 12 watts of core loss at no load, and at full load. The copper loss at full load is about 200 watts. The reason for this disparity is that customers seldom run their inverters at full load, and they want very low no load and light load losses, core and copper combined. The no load losses determine how fast your battery will run down when only the nite lite is being powered.

Interesting.
Yeah I see at 80% would be a bad idea, I was just going for proportions. So the main point is that equal copper and Iron is only ideal if the thing is actually going to be used at rated, i.e. that rule is FOR rated efficiency. You never want to run it over rated, or atleast not for long, and If it is under rated a lot of the time, it makes sense ot have better efficiency at under rated, than actually AT rated.
Makes sense
Cheers!

 

[tim9000]

I think I have a better way to pose my question about whether the dΦ/dt of a rectified mag amp or one that just sees both parts of the AC supply makes any difference:

I know rectifying it is better from the stand-point of no magneto-striction, but I was wondering if there were any magnetic down-sides? So to speak.

 

[jim hardy]

Great input from Electrician !

{o me o my, c'mon alleged brain work on one thing at a time.........
got the baseplate level and nailed in for first row of siding just before dark.

I've wanted for weeks to post a "how to wire up a washing machine motor" , and lo- somebody asked.}

Okay back to inductors.

I thought you had it pretty well surrounded in this old post 107

your saturable reactor is well saturated by 2 vdc across the control winding
i don't know how many amp-turns that is though.

I've printed this from post 180 several times, but am unable to relate the numbers to the graph.

Column V mag Amp stays less than 4V but graph goes to near 70.

You asked where do you sit on the BH curve as you adjust DC.

From Post 187

With extreme DC control current, there is never enough load current to push the outer legs out of saturation. So you're way out on the "wings" sweeping current back and forth but not changing flux appreciably(hence hardly any voltage appears)
With zero DC control current you're on the steep part of the curve sweeping current back and forth and changing flux appreciably (hence voltage appears).

But in that snip from post 180 , columns Excitation, V L and I (A) are all practically constant.

so i have not folowed some of your thought process since then and have probably answered what i thought were your questions but probably weren'tnow on to more recent posts

 

[jim hardy]

now to this one , i rather muffed

. look at that picture(264 & 268)

With R = zero, we have opposing mmf's because all 3 windings are trying to push flux down. So total current will be set by Rload.
How will current divide between the center coil and the outer ones ? I'm not sure.
With your postulated turns numbers, center and series combination of outer legs would have same volts per turn wouldn't they ? Inferring same flux? Seems to me current would have to reverse in center leg... Since primary current is set by Rload it's operating as a current transformer.

Aha , yes of course
rectified AC has AC fundamental at twice frequency, and DC offset.

Flux will flow in direction of dominant amp-turns so may not be in direction of the arrows . Arrows show direction of MMF.
If we disallow current in center leg by setting R high
that thing should saturate itself because DC component of load current pushes flux down outer and up center legs.
If we allow current in center leg, its MMF opposes outer leg MMF ?
Like any good self-saturator, control winding MMF current opposes load current MMF. But usually the control source is independent of load. This one's more like a feedback winding?

I think i have that one right now. Sanity check ? Tim ? Electrician ?

 

[jim hardy]

From post 274

Hey Jim, Thanks for the reply. That's ok. But I'm hoping I can get the last few curiosities tied up soon because time is starting to run out.

That's a good answer to illustrate the importance of flux in preserving the current ratio (of a CT), my mind is still a bit like a sieve in this field sometimes. it's okay i struggle too

How important is low flux (small magnetising current) though in a VT? in a plain power transformer not very , it just wastes some heat to push the core out past the knee twice per cycle... and makes it draw harmonic current. In a metering transformer where you want accurate reproduction of Vp you'd want to operate the transformer further below its knee. That'll reduce magnetizing current but the objective was mainly to keep transformer in its linear range.

Let me see if I've got this right/straight:
I remember from our previous discussion that if the magnetising current is high (such as from saturation) than the increasee in current will be on the ?? Verb alert ? maybe "show up as" ? larger voltage drop on the resistance of the coil, and so less induced voltage on the primary and secondary.

So is it fair to say that while the impedence of the magnetising branch is linear, the amount of reduced voltage on the primary and secondary will be linear due to it will be primary resistance. While it's not what you'd call good, it is atleast predictable, so not a big deal...Actually hang on, if You want a more efficient transformer, have less resitance in the copper and less reluctance in the core, for less flux, because the lower the reluctance the lower Rc but the higher Xm, so bigger inductance means less current.
You have things moving in the right directions...
And/But with a CT the current ratio is parramount so you don't want to be having to account for magnetising current (as Ipri = Iin - IM) in your equation, that would be unprofessional for a manufacturer, take a look at phasor diagrams for a CT...
so you have to a big core (less reluctance) for bigger Xm, for less flux.
So the MMF might not be low, but the Net MMF will be low. you got it
How am I travelling?

You're doing well. A CT is just a very good PT that's operated with its secondary shorted, as near as is practical.
Primary mmf is canceled by secondary mmf.

Magnetizing current causes phase shift between primary and secondary so they want it small which means like you said, a good core.
http://www.crmagnetics.com/assets/technical-references/analysis_of_ct_error.pdf

 

[tim9000]

i've printed this from post 180 several times, but am unable to relate the numbers to the graph.
Column V mag Amp stays less than 4V but graph goes to near 70.

So the vertical axis is V3, and the horizontal is Vdc. But the table only has the saturation data points, the data points while it was amplifying were from Vdc = 0 to 0.9V but I didn't include them...probably due to space limitations.

Flux will flow in direction of dominant amp-turns so may not be in direction of the arrows . Arrows show direction of MMF.
If we disallow current in center leg by setting R high
that thing should saturate itself because DC component of load current pushes flux down outer and up center legs.
If we allow current in center leg, its MMF opposes outer leg MMF ?
Like any good self-saturator, control winding MMF current opposes load current MMF. But usually the control source is independent of load. This one's more like a feedback winding?

Yeah I agree with that, I'd like to point out that it's a ficticious design. I was trying to replicate Steiner's parallel Amp, rectification, but for a Series Amp, so you think this will operate similarly to that?

rectified AC has AC fundamental at twice frequency, and DC offset.

Does it?? I can imagine that it would have a DC, but the frequency surprises me, Are you addressing/refering to my picture in #283? So are you saying if you supply a mag amp with rectfied ac, the fundamental will make it's reactance twice as large??

 

[tim9000]

You're doing well. A CT is just a very good PT that's operated with its secondary shorted, as near as is practical.
Primary mmf is canceled by secondary mmf.

The way I rationalise "If you open circuit the secondary of a CT since there's heaps of turns on it, it's like a big inductor that wants to keep it's current flowing so it induces a big voltage and might arc." As in I personify the inductance like it has a disire. I remember ages ago you said something like 'since the other flux opposing it disapears...' so I assume the way you rationalise it is that 'if the secondary flux suddenly dissapears the flux from the primary can increase very fast causing a big dΦ/dt'
But I'm assuming the same ISN'T true if you open circuit the primary passing through it? because ampere's law around the primary doesn't quite work out the same.One thing that bothered me was that when you increase the cross sectional area of a TX and the reluctance drops, so Rc drops, I'm assuming the core power loss drops also? Ic²*Rc = core really power loss
Because even though the magnetising current I am <90^O has dropped, decreasing the reluctance by increasing the cross sectional area has increased the magnetising current Ic < 0^O through Rc
Or might the core loss actually go up if you decrease the reluctance?

 

[tim9000]

Again, why was the 840 turn curve different from the 400 and 200 ones:

??

Also I reitterate the question priorities of designing for current and dΦ/dt or inductance when rectified:

Rectified AC has a fundamental twice the frequency? Does it?? I can imagine that it would have a DC, but the frequency surprises me, Are you addressing/refering to my picture in #283? So are you saying if you supply a mag amp with rectfied ac, the fundamental will make it's reactance twice as large??

I think I have a better way to pose my question about whether the dΦ/dt of a rectified mag amp or one that just sees both parts of the AC supply makes any difference:

I know rectifying it is better from the stand-point of no magneto-striction, but I was wondering if there were any magnetic down-sides? So to speak.

At the risk of sounding like a broken record I'm going to put this to you again:

I gather when you're designing an inductor (putting an amount of turns on it) the amount of current through it is equally as important as the amount of V.s on it, however I recall you saying you saw no purpose for designing an inductor to operate at μmax, and I see that if you have a set V.s and you want a big inductance, than you may as well operate down the bottom of the BH curve. But/And I raised the point that if you were opperating at a specific current through the inductor, than wouldn't μmax be a good point to be operating at, because that way you're getting current through it, you're getting your inductance, and you're getting value out of your steel. I was wondering what your response was to that? (as well as the top posts)

Because you previously said you saw no problem in, if one wanted a bigger inductor, they could just wind more turns around it, and I can see that would be fine provided you were designing it for V.s, but if you were designing it for a Current, than doesn't that luxury evaporate? -> Say you've hit the wall as far as increasing the side of your core goes. You've got the current going through it, and it's hunky dorry, down the bottom of the BH curve. You think, "ok room to play with", and you wind more turns on it, B goes up, then isn't μmax the last port of call, it will give you the best inductance and you got more turns on it for the same current?

 

[jim hardy]

I think I have a better way to pose my question about whether the dΦ/dt of a rectified mag amp or one that just sees both parts of the AC supply makes any difference:
View attachment 88880
I know rectifying it is better from the stand-point of no magneto-striction, but I was wondering if there were any magnetic down-sides? So to speak.

Rectifying just flips the negative half cycles
so slope at every instant has same absolute magnitude in both waves
from 0 to pi radians the waves are identical and have same slope
but note from pi to 2pi, slopes have same magnitude but opposite sign

so would it be rigorous use of language to say dΦ/dt was same for both waves?
RMS of both waves would be same, presumably any function that squares would give same result for both waves

but i'd be very cautious about claiming "same" dΦ/dt for them.

 

[jim hardy]

So are you saying if you supply a mag amp with rectfied ac, the fundamental will make it's reactance twice as large??

be careful with that thought.
Look at Fourier of full wave rectified sinewave

http://www.reed.edu/physics/courses/Physics331.f08/pdf/Fourier.pdf
rectifying your sinewave it changes it from a plain single frequency to a summation of even harmonics
and the line frequency fundamental disappears !

So your AC meters measuring V and I are going to see that summation...

Impedance at each harmonic is jωL at that frequency
but remember a real core has frequency dependent losses(eddy currents) , so L itself if a f(frequency)

so Z will be larger but 2X larger would be a fluke.

Not meaning to be difficult or contrary, just be careful about taking too big leaps of thought.

 

[jim hardy]

One thing that bothered me was that when you increase the cross sectional area of a TX and the reluctance drops, so Rc drops, I'm assuming the core power loss drops also? Ic2*Rc = core really power loss
Because even though the magnetising current I am <90O has dropped, decreasing the reluctance by increasing the cross sectional area has increased the magnetising current Ic < 0O through Rc
Or might the core loss actually go up if you decrease the reluctance?

Two components to core loss, hysteresis and eddy currents.

Usually it's all lumped and ascribed to Steinmetz, fB^1.6

Steinmetz's equation, sometimes called the Power equation,[1] is a physics equation used to calculate the core loss of magnetic materials due to magnetic hysteresis. The equation is named after Charles Steinmetz, who proposed a similar equation in 1892.[2] The equation is as follows:

[PLAIN]https://upload.wikimedia.org/math/b/b/5/bb544bd4abdacdf8ce85f1cb2e3ea3de.png[/QUOTE]
https://en.wikipedia.org/wiki/Steinmetz's_equation (i was taught fB^1.4, maybe his 1892 number?)
That's loss per unit volume; volume and B are both proportional to area so i think losses will go down because of the 1.6 exponent

NASA separates them here
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660001049.pdf

and

So does Rc get larger or smaller ?

Seems to me if the core gets less lossy, R_C will increase and I₀ decrease , ?
If the core becomes easier to magnetize, X_M will increase and I_M decrease, ?
It'd be approaching ideal...

 

[jim hardy]

Again, why was the 840 turn curve different from the 400 and 200 ones:

??

How far back was that post ?

i thought it was length of magnetic paths

but only 147.5 when driven from center ?

 

[tim9000]

How far back was that post ?

i thought it was length of magnetic paths

View attachment 89063

but only 147.5 when driven from center ?

I think you might be onto something, I can totally see that it takes less amps to get the same flux when there are more turns, but just looking at the term how does magnetic length effect 'V.s/N'? Because there's not 'length' in that term, also as far as I can see those curves are just flux Vs. Amps.

 

[tim9000]

Thanks for the reply Jim!

Rectifying just flips the negative half cycles
so slope at every instant has same absolute magnitude in both waves
from 0 to pi radians the waves are identical and have same slope
but note from pi to 2pi, slopes have same magnitude but opposite sign

so would it be rigorous use of language to say dΦ/dt was same for both waves?
RMS of both waves would be same, presumably any function that squares would give same result for both waves

but i'd be very cautious about claiming "same" dΦ/dt for them.

So then when you rectify it, the supply becomes the sum of the even harmonics, so how does that change the dΦ/dt for each hump? I imagine it would increase it?

Yeah I can see they'd have the same RMS. I'm really interested in 'dΦ/dt' because of the implications on Inductance.
So from pi dΦ/dt is negative, and from 2pi dΦ/dt is positive, so for instane in the equation E = NdΦ/dt
what implications does this positive only (due to rectification) have on E?
Thanks

 

[tim9000]

H'mm I'll have to get my head around Steinmetz, thanks.

Seems to me if the core gets less lossy, RC will increase and I0 decrease , ?
If the core becomes easier to magnetize, XM will increase and IM decrease, ?
It'd be approaching ideal...

Is that simply to say as the Reluctance of the core DECREASES then the Xm and Rc will BOTH INCREASE?

 

[jim hardy]

I think you might be onto something, I can totally see that it takes less amps to get the same flux when there are more turns, but just looking at the term how does magnetic length effect 'V.s/N'?

?? Isn't V.s flux ? V = -N dΦ/dt ; Vdt = -N dΦ ; ∫Vdt = -N ∫dΦ ; Vt -= NΦ ; for t in seconds s , V.s / N = Φ

Φ = μ μ₀ N I A_rea / L_ength
V.s / N = μ μ₀ N I A_rea / L_ength

V.s /n = 1/L_ength X (N I μ μ₀ A_rea )

check my thinking ? And my algebra ?

 

[tim9000]

?? Isn't V.s flux ? V = -N dΦ/dt ; Vdt = -N dΦ ; ∫Vdt = -N ∫dΦ ; Vt -= NΦ ; for t in seconds s , V.s / N = Φ

Φ = μ μ₀ N I A_rea / L_ength
V.s / N = μ μ₀ N I A_rea / L_ength

V.s /n = 1/L_ength X (N I μ μ₀ A_rea )

check my thinking ? And my algebra ?

Yeah I see where you're coming from:
I = Φ * Length / ( μ μ₀ .N. Area)
But...OOOH, ok, what I was confused about was, why the two bottom curves were flattening off at a lower flux, but that's because their cross sectional area is half! Does that sound right?
So they need more current to get to the same flux because the length is longer, but also since Area is half, it requires half the flux to saturate.

That finally makes sense, thanks.

what did you make of my question in #295 and my assessment in #296?

 

[jim hardy]

But...OOOH, ok, what I was confused about was, why the two bottom curves were flattening off at a lower flux, but that's because their cross sectional area is half! Does that sound right?

Sure does ! How'd i miss that ?

So then when you rectify it, the supply becomes the sum of the even harmonics, so how does that change the dΦ/dt for each hump? I imagine it would increase it?

i guess i was looking at slope of voltage...
dΦ/dt IS voltage (per turn).
so dΦ/dt also becomes the sum of even harmonics

what does that series evaluate to ? I think it'll be less than 1 so dΦ/dt decreases a little ?

Yeah I can see they'd have the same RMS. I'm really interested in 'dΦ/dt' because of the implications on Inductance.
So from pi dΦ/dt is negative, and from 2pi dΦ/dt is positive, so for instane in the equation E = NdΦ/dt
what implications does this positive only (due to rectification) have on E?
Thanks

E an dΦ/dt are one and the same, scaled by N_{Turns ?}

 

[jim hardy]

bedtime here