Transformer Design

Hi all,

I need to design the transformer core calculations:

Area of the core
Length of the core

I mean the dimensions of the core w.r.t to power.

Input voltage of the primary side of the transformer is 565V and 175A

Total power is for 100KW...

Quite a simple question!
The 100Kw bit makes the problem interesting; like the thing probably will weigh 500 Kg and be oil cooled.

Actually I am doing the simulations for IBE Robots resistive welding machines at 1000 Hz frequency. The transformer design should be step down where the current should be more than 10KA at secondary side.. (Iron Powder Core, Amorphous core, Powerlite C core or nano crystaline core)..
I am trying to calculate the area and length of the core.. So can ant please help me with some example formulas ..

Thanking you in advance...

rumpfy said:

Quite a simple question!
The 100Kw bit makes the problem interesting; like the thing probably will weigh 500 Kg and be oil cooled.

Click to expand...

From power and time(frequency) you can get energy, and since energy is a function of L and I
out of it all you can get what it's inductance should be.
Now inductance is related to core dimensions...

Edit:

Number of turns (important to inductance) you get from Faraday's law,
for sine wave: N=U/(4.44*S*f*B)
(depending on core material B should be about 1T ; S is area ; f is frequency )

maheshyedla said:

Actually I am doing the simulations for IBE Robots resistive welding machines at 1000 Hz frequency. The transformer design should be step down where the current should be more than 10KA at secondary side.. (Iron Powder Core, Amorphous core, Powerlite C core or nano crystaline core)..
I am trying to calculate the area and length of the core.. So can ant please help me with some example formulas ..

Thanking you in advance...

Click to expand...

1000Hz is not high enough to help that much... is there any reason not to go to 100KHz?

the length is immaterial, aside for fitting in more turns. That said the higher the frequency the lower the cost, but at 100KW it will be expensive regardless.

Dan

Mahesh,
Vlad is almost helpful, but his contribution highlights the need to establish B.
This is the saturation flux density of the core material.
I can do the design for you, but you will learn nothing.
So, the first thing you must do is to select a core material.
At 1000 Hz, this frequency is low enough that eddy current loss wont be too great if you select a silicon steel. For a transformer of this size, it is unlikely that a ceramic type material will be available. With ceramic material one has to have a very big mould. Its possible that large ceramic cores are made but are probably not readily available. With sheet steel, the core can be laminated to a very large size.
You need to examine manufacturers data for their transformer steels and you will be looking for the power loss versus flux density. It is probable that data will be available for 50/60 Hz and not 1000. You need to establish the factor for power loss increase versus frequency. This will be at least 1000/50 times greater than the 50 Hz value. When you have the material selected, then come back to me.

Hi Dan,

At present in industries for the Resistive welding Transformers they were using mostly 50Hz and 400Hz transformers. Now they are increasing the frequency of the transformers. So, my task is to increasing the frequency from 1000Hz to 10KHz and checking for the efficiency.

Mostly and at present in an IBE Robot the welding machine efficiency is 44%

Hi,

Thanks for the feedback.

My prffesor suggested me to use Iron Powder core and also I selected Powerlite_c_core and Amorphous cores. These cores are having more than 1.7T of flux density. And few days back I have visited the company and I have seen the transformer of the Welding Machine and I have measures the length of the transformer including lamination its about 24Cm.
V = 565 and I = 175 A , F= 1000Hz and Power = 100Kw
I have calculated the Inductance and Energy its about
V=I*Z
Assuming that the load is 100% inductive, then Z = 2*pi*f*L
substituting,
L=V/(I*2*pi*f) = 514uH
Energy Storage (J) = 0.5*L*I^2 = 0.5*514e-6*175^2 = 7.87J

rumpfy said:

Mahesh,
Vlad is almost helpful, but his contribution highlights the need to establish B.
This is the saturation flux density of the core material.
I can do the design for you, but you will learn nothing.
So, the first thing you must do is to select a core material.
At 1000 Hz, this frequency is low enough that eddy current loss wont be too great if you select a silicon steel. For a transformer of this size, it is unlikely that a ceramic type material will be available. With ceramic material one has to have a very big mould. Its possible that large ceramic cores are made but are probably not readily available. With sheet steel, the core can be laminated to a very large size.
You need to examine manufacturers data for their transformer steels and you will be looking for the power loss versus flux density. It is probable that data will be available for 50/60 Hz and not 1000. You need to establish the factor for power loss increase versus frequency. This will be at least 1000/50 times greater than the 50 Hz value. When you have the material selected, then come back to me.

Click to expand...

Hi,


Perhaps what you are looking for is this equation, sometimes called "The Transformer Equation"...

B=E*1e8/(4.44*F*A*N)

where
B is the flux density in Gausses,
E is the RMS voltage in volts,
F is the frequency in Hertz,
A is the core cross sectional area in square centimeters, and
N is the number of turns.

From this we can see right away that:
If we start with a frequency of 100Hz and a voltage of 10 volts RMS and end up with a flux density of 2000G, if we increase the frequency by 10 times (to 1000Hz) we lower the flux density by 10 times (to 200G for this example), so that allows us to increase the voltage by ten times to 100 volts. Thus, when we increase the frequency by ten times we can use ten times the primary voltage.
Looking at it from the standpoint of the core area, if we increase the frequency by ten times we can decrease the core area by a factor of 10 also. So if we started with 2000G at 100Hz and a core area of 10 square inches (about 3.2 inches by 3.2 inches), then increased the frequency to 1000Hz, we could use a core that has only 1 square inch cross sectional area.
There are other things to think about too though when thinking about an increase in frequency. Even at 60Hz there is some skin effect to consider, so at 1000Hz it would be more to worry about, and if you go beyond that you might really have to rethink this project. For example, it would not be uncommon to use tubular 'wiring' rather than actual heavy copper wire because the skin effect makes the heavy copper wire look like a tubular conductor anyway, so the extra copper expense and weight is saved. There also might be coolant pumped through the center of the tubular windings.

I think you should stick to transformer equation, rewritten this is what it says:

N*S=U/(4.44*f*B)

In your case:

N*S=565V/(4.44*1000Hz*1.7T)

https://www.scribd.com/doc/27073343/Transformer-Engineering

As for the first part of my previous post, it may be misleading and not useful in this case
but here is what I get.

Your power is not reactive, your load is resistive.
Average P=100kW so E=P/(2*f) for one semi-cycle.
1/(2*f) is time for one semi-cycle.
L=E/(I^2)
L=1.63 mH

maheshyedla said:

Hi Dan,

At present in industries for the Resistive welding Transformers they were using mostly 50Hz and 400Hz transformers. Now they are increasing the frequency of the transformers. So, my task is to increasing the frequency from 1000Hz to 10KHz and checking for the efficiency.

Mostly and at present in an IBE Robot the welding machine efficiency is 44%

Click to expand...

This sounds like somebody sitting in a room came up with the notion that all they have to do is crank up the frequency to get "more efficient" and save money and size on the transformer.

It is WAYYYYY more complicated than that as losses in the core also increase with frequency.

I can say without fear of contradiction: you need to get with a transformer manufacturer to see what options are available and what the size and cost criteria will be if you go to increased frequency. There probably are core materials that can achieve it and they will be more expensive than the laminated steel used at low frequencies.

Hi again,

https://en.wikipedia.org/wiki/User:Msiddalingaiah/Transformer_design

Scroll down a little, check out the first equation under the section titled "Equations".
Solve that equation for B and you get the same formula i had posted in post #9 of this thread.
This has been known for a long time and is usually referred to as the "Transformer Equation", but unfortunately there are other equations that want to go by that name too.

Usually a given core will work up to a certain frequency so the core area can be reduced as frequency is increased, but then it reaches a point where after a certain frequency the core area has to be increased.

Hi ,

Thanks and its help full.

I have small doubt, that can we assume the number of turns for calculating the area of the core.
Suppose for example Let N is 75:1, then the area of the core is
A = U/ (4*F*N*B) = 565/(4*1000*75*1.7) = 11Cm^2 is this way is correct, if it is okay. So please can I now how can I find the length of the core and Windows area of the core. Please help with this.

Thanking you in advance.

vlad777 said:

I think you should stick to transformer equation, rewritten this is what it says:

N*S=U/(4.44*f*B)

In your case:

N*S=565V/(4.44*1000Hz*1.7T)

https://www.scribd.com/doc/27073343/Transformer-Engineering

As for the first part of my previous post, it may be misleading and not useful in this case
but here is what I get.

Your power is not reactive, your load is resistive.
Average P=100kW so E=P/(2*f) for one semi-cycle.
1/(2*f) is time for one semi-cycle.
L=E/(I^2)
L=1.63 mH

Click to expand...

Ok,
Mr al says the transformer equation is;
N=E x 10^8 /(4.4 x F x A x Bsat)
Bsat you say is at least 17000 gauss(1.7T). Frankly I think this is high because it might be for the material, but in a real core could be less. Nevertheless, you can play with these numbers. Now you need to play with the core cross section area (in cm 2) and N. You will need to determine the cross section area of the wire. What current density in the primary wire will you assume? For starters I usually take 2000 A /per square inch.
The loading on this transformer is likely to be hard to predict, and is likely to vary through the weld cycle The load is effectively a short circuit and the connecting cables will probably be a significant influence on the final performance. This is all about refining the design. You might want to develop an equivalent circuit using the 'perfect transformer' to get the final secondary turns.

Hi again,

A flux density of 17 thousand Gauss is not unreasonable for standard transformer EI laminations which as the grain oriented steel laminations usually used in power transformers that operate directly off of the line voltage. The frequency of operation though is usually 50 or 60 Hz, with thinner laminations being used for 400Hz. At 1000Hz i would think some concessions would have to be made in order to use these same laminations, such as reducing the max flux density. This would be done mainly to reduce core heating.

Other types of cores are available of course, but they may already have more limited max flux densities so it is kind of a trade off.

Before building this though it would be a good idea to consult with a professional in the field of (high) power transformers, someone with a lot of experience in building these higher power transformers. There are a lot of little catches that come up such as thermal management and winding insulation as well as audible noise management. If anything about it does not work right it could easily overheat and require constant stopping to allow cool down, a feature which they may not want in their end product. This is not the same as building a little wall wart to power a transistor radio alarm clock

Hi,

With the help of Inductance value i have calculated the area of the core is this is correct

W.r.t current = 175A, voltage= 565V and frequency= 1000Hz I have calculated inducatnce L = 514UH

From the inductance area of the core Ae = (L*I)/(N*B)

Where I = 175, N = 75 , B= 1.56T
Ae = 7.6 Cm^2.

Now I have the doubt that how can I calculate the length of the core and area of the product.
or
Can I select the core dimensions with respect to the calculated area of the core.

Can you help me please. I was strucked with this thing I have to finish the calculations by the end of this month and its the dead line. so Could any one help me with this.

Thanking you,

Best Regards,
Mahesh

rumpfy said:

Ok,
Mr al says the transformer equation is;
N=E x 10^8 /(4.4 x F x A x Bsat)
Bsat you say is at least 17000 gauss(1.7T). Frankly I think this is high because it might be for the material, but in a real core could be less. Nevertheless, you can play with these numbers. Now you need to play with the core cross section area (in cm 2) and N. You will need to determine the cross section area of the wire. What current density in the primary wire will you assume? For starters I usually take 2000 A /per square inch.
The loading on this transformer is likely to be hard to predict, and is likely to vary through the weld cycle The load is effectively a short circuit and the connecting cables will probably be a significant influence on the final performance. This is all about refining the design. You might want to develop an equivalent circuit using the 'perfect transformer' to get the final secondary turns.

Click to expand...

Hi,

When i use E=565 and F=1000 and B=15600G and N=75 i get Area=10.9 square centimeters.

Mahesh,
you've had a good bit of help and now YOU have to lay the egg. The consenus is that the crossection is at lest 10 square cm. You have information on 'C' cores. You have information on current density in the copper. You now have to put it together. Take the winding space and fill half of it with primary winding and half of it with secondary winding. Will you use one 'C' core and two coils or or 2 'C' cores and one winding. Make the winding space as big as you need to and if you have to increase the cross section of the core to make it all fit, then drop the number of turns on the primary a bit. Just keep on iterating till you get something close. That's how its done. Theory supports the practice.
Hope you get to your dead line but dont wait much longer; over to you and out from me.

Hi,

Thanks for your feedback..

To calculate the current density I am referring this article is it good way of calculating the current desity..

https://www.electro-tech-online.com/custompdfs/2012/06/1263932428_1721_FT82593_hvhf_transformer.pdf

or

Can you give me some suggestion or formula to find out the current density please.


Thanking you,


Best Regards,
Mahesh

rumpfy said:

Mahesh,
you've had a good bit of help and now YOU have to lay the egg. The consenus is that the crossection is at lest 10 square cm. You have information on 'C' cores. You have information on current density in the copper. You now have to put it together. Take the winding space and fill half of it with primary winding and half of it with secondary winding. Will you use one 'C' core and two coils or or 2 'C' cores and one winding. Make the winding space as big as you need to and if you have to increase the cross section of the core to make it all fit, then drop the number of turns on the primary a bit. Just keep on iterating till you get something close. That's how its done. Theory supports the practice.
Hope you get to your dead line but dont wait much longer; over to you and out from me.

Click to expand...

Hi

Thanks for the feedback.. But I am assuming the number of turns in this case. Is this is the right way of considering the number turns to calculate the area of the core.

Or

Is it have any other formulas to calculate the area of the core with out assuming the number of turns .

MrAl said:

Hi,

When i use E=565 and F=1000 and B=15600G and N=75 i get Area=10.9 square centimeters.

Click to expand...