Dear Jimmy

Here I enclose the answer, which is a little approximated, as it does not take into account serial parameters, which are completly insgnificant durin the idle run.

 Piotr Wach, Opole, Poland

Dear Mr. Wach,

I think you got my question wrong. I want to calculate the parameters mathematically. The values W0 and I0 are obtained from the open circuit test. I am looking for W0 and I0 to be calculated mathematically.

I provided these two values just for verification between calculated and measured value.

Dear Jimmy

Now I understand that You want to obtain these parameters from the contruction data of the transformer. It is posible but more difficult and You will need much more data.

The iron losses You can calculate having the total mass of the iron core, average induction B in the core and a frequency (50 Hz or 60 Hz). There are respactable formulae for magnetizing losses and eddy current losses (which depend on a thickness of iron sheets).

To calculate magnetizing current Im You need more data. You have  to calculate a magentomotive force  in all pieces of the magnetic circuit along the way of the flux. The total magnetomotive force calculated in such way You have to devide by the No of turns of a respective windindg and finally You will get magnetizing current if this paricular winding .

So You will need dimensions of core and characteristics of ferromagnetic material of thet core and also number of turns of the windings.

with kind regards  - Piotr Wach

Dear Mr. Wach,

I am using a Ring Transformer. Core dimensions are 46x80x30 (IDxODxH) mm. The primary number of windings are Np=1262

Secondary windings Ns=150.

The core material is Fe M5 graded steel.

Now is it possible?

Dear Jimmy

Now we are much closer to the solution. I don't have the characteristics of that steel and  Yoy will have to make calculations by yiourself.

For the average induction in the core (I understant that it is 1.57 T)  You will find the losses per 1 kg of mass and You will multiply that total mass of the core. So You will get iron losses to calculate iron losses current  IFe.

Then You will calculate the average length of the magnetic flux path in the core ( in fact the length of the middle path of the flux)  and You will multiply that length by a data for electromotive force (in Am - Ampermeters) respective to B = 1.57 average induction, given in the Table (or characteristic) for a unit length (cm or m ). The result of this miltiplication is the total magnetomotive force to magnetize Your core to 1.57 T.  This result You will devide by the number of turns ( 2162 or 150) depending of which side of transformer You want to receive magnetizing current.

For IFe current Yoy should use the nominal voltage of the same side.

So finalize Your task, I hope successfuly.

   Piotr Wach

Dear Mr. Wach,

I am using this transformer.

http://www.sedlbauer.de/media/ringkerntrafo_datenblatt_826029.pdf

The weight provided here is the total weight with the windings.

If I take loss per kg=1 and Weight of core=1.04kg

so core loss= 1.04W. Is this correct?

Now for mean magnetic length for ring/toroid transformer I found this formula

https://www.cwsbytemark.com/CatalogSheets/MPP%20PDF%20files/13.pdf

So I get Le=19.30cm. But I dont have any data for electromotive force.

 Hello,

I also have a doubt. When we do short circuit test, from the measurement test we always take R1=R2'(referred to primary). In this case it was around 25ohm.

When I actually measure the primary and secondary resistance I get R1= 24ohm(i.e okay). However, R2'=38ohm.

I have two secondaries so with multimeter I get R21=R22=0.9 ohm and then calculate equivalent resistance in parallel and finally finally multiply by the square of turns ratio (As referred to primary).

Why so much difference?

Dear Jimmy

The multiplier is  kN=1262/150 = 8.413 

So R2*kN2 = 63.7  Ohm

Having two of secondaries, You can say that they are equivalent to 0.5*63.7 = 31.9 Ohm

In conclusion the difference to  R1 = 24 Ohm is not dramatic and it happens so, because the primary winding takes also the magetizing current and it should have little lower resistance. 

As for magnetizing current Im  I will try to get some data and to see how it is.

The length of way of magnetic flux You can calculate, as it is a toroid and You have the diameter of outside and inside of the toroid.

Nice weekend, Piotr Wach

Dear Jimmy

Now we can finish the task. We can find approximate material data for the M5 steel:

7.55 g/cm2  - steel density;   1.3 W/kg (for 1.7 T) - for iron losses;  800 A/m (for 1.8 T)

magnetizing MMF.

I understand that induction in the core of Your transformer is 1.57 T.

So the approximations of the specific will be:  1.1 W/kg - for specific losses and

about 400 A/m for specific magnetization. 

The length of the mediom flux path is:  Lav = Pi*(ID+OD)/2 = 0.198 m.

The mass of the core is:  M = Lav *(OD-ID)/2 *H* density = 

= 19.8 cm * 1.7 cm * 3 cm *  7.55 g/cm3 = 762 g =0.762 kg

So the iron losses are:  PFe = 1,1 W/kg * 0.762 kg = 0.84 W     (OK)

Magnetizig current is:

MMF =Lav *400 A/m = 0.198 m *400 A/m = 79.2 A

and the Im current in the promary winding is

Im1= MMF/1262 turns = 0.063 A  , that is not much more than 5 mA You have from

measuremets. It is quite good because 400 A/m was rough approximation.

I hope that this is a finish of this problem.

  Piotr Wach, Opole, Poland

I am appalled that people are asking such questions rather than googling "Core and copper losses in transformers" and developing their basic knowledge.  Is this person trying to be a technician or an engineer?

If they do not have or cannot develop the necessary  fundamental understanding of transformers then they are a danger to themselves, their employer and society.

Plugging numbers into formulae may produce an answer but do they have any ability to recognise when the result is wrong?  Even scaling errors between mm cm and meter or worse cubic millimeters and cubic meters are a serious risk let alone if they are converting from imperial to metric.  Ask NASA, they flew a spacecraft into Mars because navigation data in imperial units was not converted to metric as used by the spacecraft.  

Dear Mr. Brown,

I think you have understood my question completely wrong. I have searched everything on google. However, there are some things which are difficult for us to understand as a student. This is why this platform is developed where we can get some knowledge form great researchers and professors.

I know all these parameters and their meanings. I am just trying to understand the relation between them from pre and post measurements. Nowadays, there are softwares also which makes our task much easier for designing any model. My aim is to develop an algorithm which can be useful and a faster way to design a transformer and analysing all the factors beforehand to increase the efficiency.

You are a senior and a much experienced professor and I completely respect your statement. In case I have offended you, I completely apologize for it.

Firstly to develop the no-load equivalent circuit and then  apply voltage balance equation . if voltage and all equivalent circuit parameters are known . Then, current value will come out after applying Ohm's law (taking impedance into consideration). After this multiplication of resistance (responsible for core loss) .to the no-load current squared , will equal to no-load losses.

Dear Mr. Wach,

Can you send me the link where you found the material data for this core material?

Jimmy Lalani