Calculating core losses in transformers

  • #1
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If i have a transformer, and i need to calculate the core losses using this approximate equivalent circuit, according to the lectures ( taking an introduction to electrical machines course in college) the Pc ( power lost in core) = (V1)^2/Gc, but the professor clearly stated that the core losses are always the same, independent of the load, but V1= V2'+I2'*(R1+R'2+(X1+X'2)J) and I2' depends on the load, how come ? when i thought about it i was like maybe he meant V1 at full load, but there is no way to make sure if i am correct and i cant seem to find any answers on the internet
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  • #2
anorlunda
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When you have a load, both V2 and I2 depend on the load, V1 is independent of load and those core losses depend on V1.
 
  • #3
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When you have a load, both V2 and I2 depend on the load, V1 is independent of load and those core losses depend on V1.
yea but how come V1 doesnt depend on the load and it depends on I'2 ?
 
  • #4
anorlunda
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You are assuming that V2 is given and constant. It isn't.
 
  • #5
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You are assuming that V2 is given and constant. It isn't.
that's the problem, as far as the course goes ( because it's very basic) if we have for example a 2400/240 transformer, V2 is always 240
 
  • #6
anorlunda
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No, V2 is 240 only with no load, when I2=0.

Put a load resistor R across V2, then solve for both I2 and V2. With a short circuit, R=0 and V2=0.
 
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  • #7
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No, V2 is 240 only with no load, when I2=0.

Put a load resistor R across V2, then solve for both I2 and V2. With a short circuit, R=0 and V2=0.
can i show you a solved example from my textbook?
 
  • #8
jim hardy
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If i have a transformer, and i need to calculate the core losses using this approximate equivalent circuit, according to the lectures ( taking an introduction to electrical machines course in college) (this is orange)the Pc ( power lost in core) = (V1)^2/Gc, (this should be yellow but yellow is impossible to read)the professor clearly stated that the core losses are always the same, independent of the load, but V1= V2'+I2'*(R1+R'2+(X1+X'2)J) and I2' depends on the load, how come ?
Do you realize what a run-on disjointed sentence that is?
Six separate thoughts.
Video games teach us to rush without taking time to think , we just react .
I color coded your six separate thoughts, match each of mine to yours of the same color:
Your circuit says "Referred to primary side" . That's where V1 is connected.

This drawing i guess is from your textbook?
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(orange)Power in the core is V12 / Gc . You know that from basic Ohm's Law and P=VXI Bm is inductance and dissipates no power. Those other R's and X's are in the windings. And X's dissipate no power.

(yellow again) V12 / Gc is core loss, and there's no load term in that

V1= V2'+I2'*(R1+R'2+(X1+X'2)J) that's Ohm's Law, isn't it ?

I2' depends on the load, how come ? Ohm's Law ? Isn't I2' = V2'/Zload ?

yea but how come V1 doesnt depend on the load and it depends on I'2 ?
I think they intend for you to re-arrange the equation and solve for V2.

Anorlunda has answered twice .
Usually we do transformer calculations forward to find out how much loss we get from input to output
but one can calculate backward as you are trying to do, to figure out how much voltage we'd have to apply at input to get nominal voltage at output.


When Professor said "core loss is constant", that's a giveaway that V1 is constant.
Ask him about that
and ask him about this possible mistake in his model ----
I submit that R1 and Xl1 are shown on wrong side of core because magnetizing current flows through primary windings.

Any help ?


Part of teaching is figuring out what are your students' prior misconceptions.
Yours was logical enough, you took the equation as it was written instead of solving it for V2 as f(V1 ) and the circuit elements .

Teachers can be astonished to learn that, despite their best efforts, students do not grasp fundamental ideas covered in class. Even some of the best students give the right answers, but are only using correctly-memorized words. When questioned more closely, these students reveal their failure to understand the underlying concepts fully. Students are often able to use algorithms to solve numerical problems without completely understanding the underlying scientific concepts. Mazur (1997) reported that students in his physics class had memorized equations and problem-solving skills, but performed poorly on tests of conceptual understanding. Nakhleh and Mitchell (1993) studied 60 students in an introductory course for chemistry majors. In an exam which paired an algorithmic problem with a conceptual question on the same topic, only 49 percent of those students classified as having high algorithmic ability were able to answer the parallel conceptual question.
Always relate the equation to a physical picture.

I too await a worked example.

old jim
 
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  • #9
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so...first about the model, the professor said the current flowing to the Rc and Xm is very small about 2%, so he would neglect it and say that I1 is = I2 and thats why they are series, we usually do a lot of things wrong for the sake of being easy because it's an introductory course after all
secondly about V1 being constant, like you said we're doing things backwards, we're calculating the V1 that will give us constant V2, which means that if a 2300/230 transformer is given we set V2 as 230 and start calculating V1
about the I2' , we usually calculate it by X*Srated/V2' where X is the load ratio because about 80% of the questions he never mentions a Zl
and i just found this solved example in the textbook and i hope if you can help me understand why he did that

he had this transformer then he did the OC and SC tests to figure out the Req Xeq Rc and Xm
then we drew the circuit, just like the one above, this was a 2300/230 transformer, he said he will refer it to secondary, then he said that V2= 230 and then he said he wanted to get V1' at full load and Pf of 1 and 0.8 , to get them he first got I2 by dividing Srated by 230 (V2) then he said that V1'= V2+I2*Zeq , but of course because the Pf is changing the angle of I2 is changing hence V1' is changing so he got for Pf =1 V1'= 232 V and for pf= 0.8 V1'= 234 V
then he asked for the efficiency at pf = 0.8 and full load so what he did was simply 234^2/Rc
and thats why i am asking this question, somethings like the Pf and the load ratio (X) can and will change V1 which will then change the core loss which i know is constant, but the textbook author just uses whichever V1' suitable for the conditions given in the question which is the V1' at Pf=0.8 and X=1

but i think you do have a point, maybe this whole misunderstanding comes from the fact that we do things backwards, consider V2 constant and calculate V1, not the other way around. ( i tried to ask the professor, sent him emails but he wouldn't answer, and the final is on Thursday so i am kinda stuck)
 
  • #10
jim hardy
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he said he will refer it to secondary,
Aha ! He decided to work it backward. I wonder why ?
I would not have done that to undergrads.

he had this transformer then he did the OC and SC tests to figure out the Req Xeq Rc and Xm
Aha !
I'll wager he worked the problem same way he tested the transformer,,,
because he didn't have a 2300 volt source at the bench but did have a 230 one ?
And he had a meter that'd safely measure high side voltage when he energized it from low side secondary ?
Of course i'm guessing. But when faced with one of these conundrums i always ask myself "What would make a rational man do that?" . And don't judge him until i've walked in his shoes.
Were I in his shoes with no 2300 volt source, i'd have worked it out backwards too. But i'd have told the students "I'm working this backwards because that's how we tested it ...., and then we're going work it again but forward. "

but i think you do have a point, maybe this whole misunderstanding comes from the fact that we do things backwards, consider V2 constant and calculate V1, not the other way around. ( i tried to ask the professor, sent him emails but he wouldn't answer, and the final is on Thursday so i am kinda stuck)
Just be sure you can work it either direction. Then you understand the principles , and from the principles you figure out the formulas instead of memorizing them. It's a LOT easier that way.

Try my technique for finals:
Practice it both ways. Before Wednesday.
Wednesday nite , have a small steak for dinner, no booze, and go to bed early ,
Walk into that test confident and well rested and mentally sharp , no 'beer fog'

Trust me - it beats cramming.

old jim
 
  • #11
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Surprisingly enough he did use a 2300 V source And energized the high voltage side :D no idea why tho.
And I try soooo hard to understand any formula thts why I made this thread, all my friends are like " dude, just memorized it"
But I don't think engineers are supposed do that.
Anyways
Thanks so much for ur help
@anorlunda @jim hardy
 
  • #12
jim hardy
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all my friends are like " u don't have to understand everything, just memorized it" but I don't think engineers are supposed to do so.
Hmm he had 2300 volts on the bench ? Wow !
Okay. I stand corrected, and thank you.

If the test asks you to work one frontwards
then he's testing to see whether you grasped the concept or just memorized the formula.
As you see, the formula and the transformer model are parallel constructs - you can fix either end, V1 or V2, and solve for the other

Next paragraph of that teaching link i quoted:
Besides offering students information and helpful examples, we must show them the reasoning processes that lead to algorithms and conceptual generalizations. Inclusion of conceptual questions on tests is another way to emphasize the importance of this aspect of problem solving.
Seriously - i wish you good luck in this course. I found i like machinery as much i do electronics and had a really interesting career working on the electronics in a power plant full of huge machinery. Industry needs folks who can do cross-over work because computers are moving into power in a big way.

Best wishes for Thursday -
 
  • #13
jim hardy
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he said he will refer it to secondary, then he said that V2= 230 and then he said he wanted to get V1' at full load and Pf of 1 and 0.8 , to get them he first got I2 by dividing Srated by 230 (V2) then he said that V1'= V2+I2*Zeq , but of course because the Pf is changing the angle of I2 is changing hence V1' is changing so he got for Pf =1 V1'= 232 V and for pf= 0.8 V1'= 234 V
Try thinking in one thought per line, like a computer program.

Referred to secondary is a hint he started at output side.
note z's get multiplied by square of turns ratio ,
so be careful when figuring your Zeq,,,
If R' and X' are referred to secondary, secondary ones are as measured but primary ones are X1/100 ...
with 234 volts at V1, core loss is more than with 230 volts at V1 ,
it's really that simple. But it conflicts with statement "core loss is constant" .
Define your model and be rigorous in applying laws of Ohm and Kirchoff.

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then he asked for the efficiency at pf = 0.8 and full load so what he did was simply 234^2/Rc
Efficiency ?
What's Rc ?
234^2/Rc is core loss, or core plus R1 & R2 losses ?
Power input minus losses = power output, and efficiency is Pout/Pin.


and thats why i am asking this question, somethings like the Pf and the load ratio (X) can and will change V1 which will then change the core loss which i know is constant,
IF you're forcing output side voltage V2 to be constant , the math tells you V1 must change with load..
But to achieve that constant V2 you must apply a V1 that you can adjust when load changes ,
you'll see V1 change from 230 volts unloaded (as anorlunda said) to 234 volts as in your worked example
and core loss at V1=230 volts IS NOT THE SAME as at V1=234 volts !
(Did professor plant that prior misconception?)

If professor insists that it is, show him this post. Print a copy for your backpack.



but the textbook author just uses whichever V1' suitable for the conditions given in the question which is the V1' at Pf=0.8 and X=1
Aha ! Textbook author shows that you can adjust V1 to get desired V2.
Transformers typically have taps in the windings so we can tweak the turns ratio slightly.
We select the tap that gives us comfortable voltage at the point of use.
He might be leading up to that subject.
Our main transformer had a 20.9 kv winding connected to a 22KV generator
the mismatch was to account for voltage loss across internal impedances of the transformer and still give 230KV on high side, which was " the grid ".
We adjusted generator voltage as required to deliver proper voltage to the grid.

I'm sensing too much haste in presenting this material.

Stick to your basics. You are obviously bright and will do fine if you keep your thinking to one simple step at a time.

@abdo799 Arrive at that test rested.
 
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  • #14
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about the efficiency thing, sorry that's my fault i meant core losses, i was awake for 26 hours straight at that point. About the professor, most of the students are the "memorizing" type, so he doesn't bother and explain in the first place, and i didn't have the chance to really think about it back then, well, because i thought the the V1 will be the constant because that makes more sense because it's the source (it's like assuming the voltage drop across a resistor is constant and then i figure out the voltage of a battery ) . I think i should go to the test after at least a 10 hour sleep :D .
 
  • #15
jim hardy
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I think i should go to the test after at least a 10 hour sleep :D .
I hope you'll let us know how it goes .

Good Luck to you, sir !
 
  • #16
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@jim hardy
hey...so the test was today...despite being the hardest question, i did solve the transformer correctly..so yay
i think i got a 0 though in the dc motor ( my friends are insisting the torque and rpm are directly proportional just because the question said they were"proportional" ) anyways...thanks for the support
 
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  • #17
jim hardy
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That's great news !

my friends are insisting the torque and rpm are directly proportional just because the question said they were"proportional"
There's a constant K for any DC motor with constant excitation
Counter-Emf = Konstant X Flux X RPM
Torque = 7.04 Konstant X Flux X Iarmature
we establish Konstant X Flux empirically by open circuit test, spinning at known RPM and excitation
first formula gives you volts
second gives foot-pounds, it's easy enough to convert those to Newton-meters for you new guys....
of course Flux is proportional to excitation (corrected for armature reaction probably later in your studies).


so , your friends say torque and RPM are "proportional" to what ?
A torque vs speed curve at constant excitation will be pretty linear from stall to no load RPM
as applied voltage shifts from IR drop in armature(wasting energy as heat) to counter-emf(doing work) .

I hope you aced the DC motor part too.
DC motors was the only exam where i ever got 100% .

Did you walk in rested ?
 
  • #18
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That's great news !



There's a constant K for any DC motor with constant excitation
Counter-Emf = Konstant X Flux X RPM
Torque = 7.04 Konstant X Flux X Iarmature
we establish Konstant X Flux empirically by open circuit test, spinning at known RPM and excitation
first formula gives you volts
second gives foot-pounds, it's easy enough to convert those to Newton-meters for you new guys....
of course Flux is proportional to excitation (corrected for armature reaction probably later in your studies).


so , your friends say torque and RPM are "proportional" to what ?
A torque vs speed curve at constant excitation will be pretty linear from stall to no load RPM
as applied voltage shifts from IR drop in armature(wasting energy as heat) to counter-emf(doing work) .

I hope you aced the DC motor part too.
DC motors was the only exam where i ever got 100% .

Did you walk in rested ?
well i slept about 9 hours before the exam, the Kconstant in our course= 2p/2a * Z/60
his exact words were " given that the torque is proportional to the speed squared" he didnt specify inverse or direct, but i know that in case of a series dc motor, they are pretty much always inversely
 
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