Are these calculatable? (is that a word)

The 3 main differences in dry type transformers, are copper-clad, copper windings, and copper windings energy effiecent.

What I am trying to figure out is whether or not cores losses increase porpotionally with load?
I think they should......

Can some one help me with that formula?




Dnk....

Core losses of a transformer are considered fixed. I have never seen a calcualtion that varied the core loss in realtion to the loading.

Where did you come up with only three differences in type dry type transformers? The construction features most likely to impact core losses that I am familiar with are:
Aluminum or copper conductor
Shell or core core construction
Butt, wound or mitered core
Standard, high, or energy efficient design
150, 115, or 80C temperature rise

Three types= Cutler hammer supplier information

That wrong?


Isn't effiency atribbuted somehow to core losses?

The more losses, the less efficient.

Not so?

Dnk....

Transformers have many different losses, each caused by different factors. In some cases, the things that you do to reduce one loss will increase others.

The greater the losses, the lower the efficiency.

Under full load, the biggest loss term will be conduction losses; the current moving through the coils causes voltage drop and heat generation. The lower the resistance of the conductors, the smaller this loss term.

But under no load at all there are still losses. Most of these get lumped together as 'core loss', but since I think motor design I tend to separate them out.

You have 'hysteresis losses'. This is the energy lost in continually changing the magnetic flux through the core. This loss depends upon the flux density (how much magnetic 'current' you have), the frequency (how quickly the flux is changing), and the mass of the core. These losses increase slowing with core saturation, since the flux itself isn't increasing once the core is saturated.

You have 'eddy current losses'. These are electric currents in the core, caused by transformer coupling between the coils and the core. You reduce this loss by increasing the electrical resistance of the core, and my laminating the core.

You have magnetization losses. This is the energy lost in primary current flowing to maintain the magnetic field in the core. This loss will increase drastically once the core starts saturating, since magnetizing current starts to go through the roof.

The above three losses are pretty much _constant_ for a fixed primary voltage, and will mean that the transformer uses power even with no load at all connected. In fact, if you increase the load, because of voltage drop on the primary side, these core losses will actually go down slightly.

The load dependant I^2R losses will increase quite rapidly as current increases, and these will make the transformer less efficient at higher loads.

You can decrease your conduction losses by decreasing the resistance of your coils. You can do this by decreasing the resistivity of the coil materials (copper rather than aluminium). You can pack more wire into the same space (square wire rather than round wire, more difficult winding techniques). You can reduce the temperature of the conductor, lowering its resistance. Or you can increase the space for windings. This latter approach means more iron in the core, and thus more iron losses.

You can decrease your saturation losses by increasing the core cross section, thus reducing the flux density. By taking the iron out of saturation you greatly reduce the magnetizing current...but the cost of a larger core is longer conductors, meaning greater conduction losses.

It just keeps being traded; going in circles.

If someone tells you that a copper transformer is more efficient than an aluminium transformer, call bull. _All other things being equal_, meaning same size conductors, same core, same insulation, etc., if you replace the aluminium with copper, then the machine will be more efficient. But all other things won't be equal. The conductors will be made smaller, the core will be made smaller, etc, to balance out the losses; the net result is that the copper transformer _might_ be more efficient.

Also one of the important design criteria is just what loading should be the point of greatest efficiency. Since to some extent you can trade off core losses and conduction losses, if a transformer is expected to operate with low loading, you will want a machine with low core losses; but for high loading you want low conduction losses. I _believe_ but have not confirmed that current practice is to design dry type local transformers to have highest efficiency at about 35% loading.

-Jon

When you see someone answer a question with such dexterity, eloquence, and accuracy, it's like watching a master artist paint the perfect painting.

Nice work Jon!

M.

Jon, you are toooooo cool.

Thanks..........

Dnk...

I'd take a bow, but cannot because I forgot a few things.

The big thing that I left out is that my description was incomplete. There are loads of other details, and the above is just supposed to whet your appetite for more

-Jon

Are you bringing the full course later?

Jon, you mentioned awhile ago about doing a paper on harmonics, did you ever complete that?

Dnk.........

Core losses {generally iron losses or no-load losses} are not so much calculated, but measured with a 3ø power analyzer in terms of watts and vars. They are highly reactive, so are of very low [lagging] power factor.

If part of the specs and purchase order, they are published in a certified test report along with load losses.

Scott35, maybe you can comment on the legal aspects of [drytype] transformer losses in California? ;`\

bjarney (Scott, welcome back).

The California requirements for energy effcient transformer design is going national. The recently passed US energy bill is requiring all new transformers (2008?) to be designed with maximum energy effciency at 35% loading.

Guys, I appreciate the responses.........

However, I need a little more help.

A customer has (5) 75kva-112.5Kva general purpose dry transformers, feeding (5) machines in one area of production. they are 480X208.
(4) come from one 400a 480v panelboard, and the other one is it's own feed. Both feeds are about 200' from the main service MDP.


How would someone calculate the savings on this?
How much better would one transformer save vs 5 or 6 smaller "general purpose".

What would you do? Any suggestions?


Dnk....

One more very important piece of infomation is needed, the load profile. Published loss data is usually only provided at 100% loading so the actual loading is important.

Core loss is present 100% of the time a transformer is energized. Conductor loss is only present when the transformer is loaded. This loss varies with the square of the loading, at 90% load there will only be 81% of the full load conductor losses.

So if your transformer is fully loaded 24hx7d (like a production line) you would want a design that minimizes conductor losses. But if your transformer was only loaded 80% for 16hx5d and 30% for 8hx5d + 24hx2d (like an office building) you might want a design that minimizes core losses.

For a not-completely-unrelated question, what would the in-rush be if a typical 7200/240VAC pole transformer were to be back energized? Say, by a clueless homeowner supplying the grid? I tried to calculate this last night but couldn't find the right xfmr specs to work it out

Dnkldorf: Personally? I'd keep the 5 transformers, but I'd bus them on both ends with ACBs, so that if I lost any of them, I could isolate it and use the others to assume the load. If I was designing new, I'd run the numbers first, but I'd probably just use 2, each capable of carrying the full load in case of failure. But then, efficiency savings from the transformers is very low on my priority list

[This message has been edited by SteveFehr (edited 12-04-2005).]

Just to add further complexity to winnie's excellent description, the losses are also temperature dependant, increasing by about 0.4% per degree C over ambient. And you have to factor in the quality of the iron sheets used in the core laminations, because 'impurities', [carbon, phosphorus, sulphur, silicon, manganese, copper, etc.], affect permeability. Historically, many of these impurities could not be commercially totally removed from the iron in smelting, so pure ores were used; the best being pure red haematites from Sweden smelted with charcoal- because coal and derived cokes contaminated the metal. Hence the term Swedish Iron. Absolutely no idea what iron is used today though.

Alan

math error

[This message has been edited by Alan Belson (edited 12-05-2005).]

I put in a 75kva zig zag today. 480X208Y.

Anyway, I remembered to take readings before and after....

75kva that I took out, Dry transformer..general purpose, yada,yada...

With no load, primary current was 7.5-8 amps/phase...

New transformer...1 amp/phase, no load....

The old transformer had 6.5kw of core losses and magentizing currents?


Dnk......

Dnk,
That sounds rather worrying for a transformer of that size.

This may sound extremely odd - and even down right incorrect, but the idling current on the original ("Old") Transformer may have been drawing mostly Reactive Power (AKA "VARs"), and very little True Power (AKA "Watts"), within the complete Apparent Power (AKA "Volt-Amps") package.

Simply put, the measured 7.5 Amps on the Primary side would result in 6.2 KVA - the complete Apparent Power Package.
Since the Transformer is at idle - and the only True Power carried within the input KVA would be the resultant of internal losses of the Transformer (see "Losses" below for more info.), the "remainder" of the KVA input would be Reactive Power.

Example figures (just for fun!):

<OL TYPE=A>

[*] Input Apparent Power "package" (KVA): 6200 VA (+/- 7,5 Amps @ 480V 3Ø),


[*] Input True Power (Wattage): 416 Watts (0.5 Amps @ 480V 3Ø),


[*] Input Reactive Power (VARs): 6186 VARs (7.435 Amps @ 480V 3Ø)
</OL>

The True Power would be creating _Most_ of the heat that could be felt at the Transformer [makes sense... True Power = Heat energy...in a simple way ...], directly from "Resistive-Like" losses,

... BUT:

The Reactive Power would do nothing more than flow between the Poco's Transformer and this Transformer - however, due to it's Magnetising abilities, it will result in exibiting "Resistance-Like" losses; and thus will create a scenario where heat energy is released, and therefore, an additional True Power load on the Primary side.

What does all this mean?

Well, not very much!

Seriously though, when taking an Amp Reading on an idling Transformer (and even some Induction Motors running unloaded), the measured readings may not directly reflect the entire KW drawn.

One big clue to the amount of True Power (KW) drawn is the amount of heat "Pouring Out" of the Transformer's Enclosure.

Say you took an Amp reading on the Primary side, and found the idling Transformer to have a "sort-of equal" reading of 3.5 Amps on all 3 conductors, you could "Guesstimate" the consumed KW by:

If there is just a small amount pouring out - like the amount of heat that would be felt from a 100 Watt Incandescent Lamp, the input KW would be small (possibly 100 Watts!)

However, if the amount of heat pouring out is large - like what comes out of a 1500 Watt Portable Heater, the input KW would be much higher (possibly 1500 Watts!).

*** Losses ***
(and some other stuff)

Losses at the Transformer may be from the Windings' Resistance (most responsible), effects in the core, and other stuff.
But what causes a large VAR draw, and why would a Transformer be designed to draw such a large VAR?

The VAR input is Magnetising power, and it's a resultant of the core design.
The VARs are stored in the core/windings assembly, and having them "readily available" makes the Transformer react better when there is a large draw on the Secondary side (such as a loaded Motor starting).

Another item is the Power Factor of the idling Transformer's Primary Windings.
At idle, the Power Factor is very low - only True Power coming in is heating the surrounding air, and maybe passed on to the charging of the circuitry + windings on the Secondary side. The rest of the power is Reactive.

When the Transformer is in operation, the Power Factor changes.
Off hand, I think the "changing points" are at 35% and 65% of the Full-Load ratings.

With heavy draws across the core (from winding to winding), the losses may become higher - especially with cores made from more reluctant materials - or using "El-Cheapo" brand of Silicon Steel (more impurities than the "Non El-Cheapo" brand of Silicon Steel).

Some Transformers are designed to idle "high", others aren't. Some are designed with larger air gaps on the core (which, even with low permeable materials, increases the core's reluctance drammatically), some designed with EMI / RFI sheilding, some designed with "Impedance Protection", yadda-yadda-yadda.
All these things effect the "lossy-ness" of the Transformer.

Here in California, Transformers need to conform to the State's Energy Commission guidelines (AKA: "TP-1").
Losses, which result in excessive input KW draws, are frowned upon bigtime!

I wish there was enough time to include as much Transformer information as should be, but the Dinner Bell is ringing - and I am STARVING!!!

So see ya later everyone!

Let me know if there are way too many inconsistancies (sp???) in this message.

Scott35

edited to fix spelleeng airrerereerrrszzz

[This message has been edited by Scott35 (edited 12-10-2005).]

Thanks everyone for your input..........

Now then, let's say the transformer in question here has a 6.5kw loss at idle.

Does this loss increase with load by a percentage?

At no load, we have a 6.5kw loss, what could we expect if the transformer was loaded at 75%. Would the losses increase porportionally to the load, or not?

Would more magnetizing currents and eddy currents be present, therfore increasing the total amount of losses, or is it the other way around. Whereby loading the transformer more, decreases the idle losses?


Dnk.....

[This message has been edited by Dnkldorf (edited 12-11-2005).]

In almost all cases _total_ losses will increase as the load increases.

However the various internal losses will change in relative proportion.

My gut feeling is that the 'core' losses (magnetizing and eddy current) would _decrease_, since the input voltage for magnetizing would drop slightly. But this is from experience with induction motor design; the various loss aspects mentioned above might have the opposite effect on core loss.

At the same time the load conduction losses would increase in proportion to the square of the current. I expect lower losses in the steel and higher losses in the copper as the load increases.

-Jon

I should of checked PF before.....

Didn't think of that one for some reason..

And I own the meter, DUH.....


Thanks guys...


Dnk......

Hello everyboby!

Core losses (mostly hysteresis and eddy currents [Foucault currents]) are not going to change (theoretically and with the usual simplifications) with the load increase. But load losses (I2R) will.

About the 6.5 KW you mention, Scott35 wrote an excellent answer (and previously Bjarney have said something about it) explaining that those are not KW. Some part or it, the lesser one, are KVA, and the other, the bigger one are KVAR (I’ll bet my shoes that the ratio is about 1 to 5).

So, when load is starting to be added to the transformer, the transformer imput impedance will start gain resistance and loose reactance. This means that the Q factor of the transformer equivalent circuit will start to decrease and the “stored energy” over “dissipated energy” ratio will go down.

For a 75 KVA transformer, the total losses at a load level of 40 % should be around 0.8 KW. If the transformer you mention is in good working condition I think you should try to measure this value (let’s make science!).

Joe, go ahead. I'm listening.....


Dnk....

That's it! My brain is empty.

Oh.., if you want you can load the transformer up to 40% and measure (with your Fluke 2060) real power on both sides (input and output) and calculate the total loss (substracting). You can also use that instrument to measure the KVA in open circuit (the 6500 you were referring) to see the real magnitude of the core losses (no-load losses).

But if you are like me, without a penny in your pocket for a Fluke 2060 or similar, you can always use a normal True-RMS clamp meter and make the measures of voltages and currents to calculate KVAs, KWs and KVARs with pencil and eraser. If need some help with that just let me know.

Regards,

Joe.-