How to calculate iron losses in a steel ring around a conductor?

In summary, AC current is never passed through a steel ring without an equal and opposite current through the same ring that will cancel the magnetic field. That is required and specified in every wiring code, which is why you cannot find examples in the literature.
  • #1
OSR
17
4
TL;DR Summary
How to calculate iron losses in steel ring?
Hi

I'm looking for calculation example how to calculate iron losses in steel ring around current carrying conductor.
I have tried to find examples online, but without a suitable result.
1690649869065.png
 
Engineering news on Phys.org
  • #2
Welcome to PF.

AC current is never passed through a steel ring, or a hole in a steel plate, without an equal and opposite current through the same ring that will cancel the magnetic field. That is required and specified in every wiring code, which is why you cannot find examples in the literature.

As a magnetic transformer core, there is no air gap in the ring to control or limit the inductance of the single conductor primary.

What is your application of such an AC conductor and magnetic topology ?
 
  • #3
Thanks, my goal is to determine how close a ferromagnetic loop can be to a current-carrying conductor with a high AC current so that there is practically no heating effect. What causes the loop to heat up if it is not due to eddy currents and hysteresis losses?
 
  • #4
OSR said:
Thanks, my goal is to determine how close a ferromagnetic loop can be to a current-carrying conductor with a high AC current so that there is practically no heating effect.
If you have only one side of the AC electric circuit passing through the ring, there will never be zero heating. What path will be taken by the return circuit?

OSR said:
What causes the loop to heat up if it is not due to eddy currents and hysteresis losses?
It is the orientation of those eddy currents that is important. The magnetic field of the AC conductor follows the big circumference of the toroidal ring, a magnetic circuit. The electric eddy currents that heat the ring flow around the small circumference of the toroidal ring, through the hole.

The electric current path is short = 2⋅π⋅r , compared to the longer magnetic path length = 2⋅π⋅R , so the resistance is very low. Skin effect at 50 Hz will limit the depth of that conductive path in the surface of the iron.
 
  • #5
Return path is outside of the ring. Theoretically yes, there will be always some losses regardless of the size of the ring, but I would assume that in practice, the heating effect on a 50 mm diameter loop and a 500 mm diameter loop would be different.
 
  • #6
OSR said:
... but I would assume that in practice, the heating effect on a 50 mm diameter loop and a 500 mm diameter loop would be different.
Let's call the electric wire a loop, and the iron a ring. Those two circuits, one electric, the other magnetic, are like two interlocked chain links. As you increase the size of one, so the other must become bigger. Scale and separation are important.

The sectional area of the ring, will determine the volume of the loop's B field that will cut the iron ring. That B field, will come from all elements of the electric circuit loop you have not specified.

In effect, you have three circuits, one is the primary electric loop, the second is the magnetic circuit in the ring, the third is the long sheet of short eddy current, flowing in the surface of the iron ring.
 
  • #7
Ok, here is more information of my test setup.
1690707038878.png
 
  • #8
All you have stated so far are some dimensions and low-C Steel.

Steel is a metal alloy much like Curry is a mixed bag of spices. There is a wide range of materials under the same name, not just Fe and C.

"Mild" steel is mostly iron, Fe arranged in random atomic crystals with low carbon, C which can yield high permeability but with a 100:1 wide range on conduction losses from high resistance due to lack of molecular grain orientation of conductor electrons. Alloys Cr, Ni, Mb, Cu, O2, and other trace elements, affect the crystal lattice structures. Also, processing affects electrical loss greatly with how it was annealed, cold rolled or printed as well as the conductor thickness/skin depth ratio.

Each variable affects the mechanical, electrical, magnetic, thermal, and corrosive properties widely.

Although you are only interested in the magnetic, conductive and inductive properties of this 40mm thin ring, keep in mind the wide range of permeabilities, and conductance losses of steel make this impossible to predict without these metallurgical values.
 
  • #9
How would you apply the formulae to estimate these losses?
Starting with ballpark numbers for ρ = 1e-6 [Ω-m]
R = (ρ • L) / A [Ω] with L=π×D for ring diameter D, cross-sectional area A
A = π • (OD² −ID²) [m]

This calculation doesn't account for eddy current losses, skin effect, permeability, mutual coupling or other factors that could affect the overall thermal losses. For a more detailed and accurate analysis, specialized software or simulations may be required.

Assume relative permeability μr = 1000 and a low mutual coupling factor M due to the PVC gap.

Then temp rise of the iron ring depends on the thermal resistance of the PVC from the ring junction to the ambient or Rja and the power dissipation P= I²R
Mutual coupling is typically calculated using numerical methods like finite element analysis (FEA) or finite difference time domain (FDTD) simulations.

In short ( no pun intended) although the mutual coupling which affects the impedance ratio and current might be 1% ballpark while the conductor resistance of the 125 mm loop might be 5 times the path length compare to the length of conductor = ring width=25mm but perhaps 100 times the electrical resistance and maybe 10~100 times the thermal resistance of PVC compared to the busbar with convection cooling and continuous conductor conduction cooling.

So it could be easily be hotter than the wire.
 
Last edited:
  • #10
TonyStewart said:
All you have stated so far are some dimensions and low-C Steel.

Steel is a metal alloy much like Curry is a mixed bag of spices. There is a wide range of materials under the same name, not just Fe and C.

"Mild" steel is mostly iron, Fe arranged in random atomic crystals with low carbon, C which can yield high permeability but with a 100:1 wide range on conduction losses from high resistance due to lack of molecular grain orientation of conductor electrons. Alloys Cr, Ni, Mb, Cu, O2, and other trace elements, affect the crystal lattice structures. Also, processing affects electrical loss greatly with how it was annealed, cold rolled or printed as well as the conductor thickness/skin depth ratio.

Each variable affects the mechanical, electrical, magnetic, thermal, and corrosive properties widely.

Although you are only interested in the magnetic, conductive and inductive properties of this 40mm thin ring, keep in mind the wide range of permeabilities, and conductance losses of steel make this impossible to predict without these metallurgical values.
Yes that is true, that steel ring is some unknown piece of water pipe. Is there some good book or source where i can find metallurgical values for some known material?

TonyStewart said:
How would you apply the formulae to estimate these losses?
Starting with ballpark numbers for ρ = 1e-6 [Ω-m]
R = (ρ • L) / A [Ω] with L=π×D for ring diameter D, cross-sectional area A
A = π • (OD² −ID²) [m]

This calculation doesn't account for eddy current losses, skin effect, permeability, mutual coupling or other factors that could affect the overall thermal losses. For a more detailed and accurate analysis, specialized software or simulations may be required.

Assume relative permeability μr = 1000 and a low mutual coupling factor M due to the PVC gap.

Then temp rise of the iron ring depends on the thermal resistance of the PVC from the ring junction to the ambient or Rja and the power dissipation P= I²R
If i understand it right estimated loss is regarding to this is about 0.6 W = 250A^2 * 1.031*10^-5

I tried to estimate the required heating power by measuring the temperature as a function of time.
The temperature change was about 67,8 degrees Celsius over about 20 minutes.

1690896677485.png

If Q = c·m·ΔT and c=0,46 kJ/(K·kg) then if I'm correct, heating power has to be at least about 1.6 W ?Are these formulas good even estimating eddy current and hysteresis losses in this kind of situation, because i'm getting about 10-20 mW total power loss or is it just that my guess of resistivity and hysteresis co-efficient factor are too off. I'm not also sure did i calculate Bmax correctly, i use Bmax =0.075 T.

1690898503923.png


1690898909477.png


I can use Ansys and Maxwell to simulate this, but I want to better understand the theory and get some estimated values before attempting the simulation.
 
  • #11
If you have the instrumentation set up, repeat the experiment up to some max T and then record the cooldown as well. From this you can easilly calculate the final steady state temperature directly I believe.
Does a log plot of that data give a good straight line?
 
  • #12
hutchphd said:
If you have the instrumentation set up, repeat the experiment up to some max T and then record the cooldown as well. From this you can easilly calculate the final steady state temperature directly I believe.
Does a log plot of that data give a good straight line?
I created this plot by manually collecting temperature readings every 30 seconds. I need to make some kind of data logger to monitor the cooling process so that I don't have to watch it myself continuously.
 
  • #13
Yes. But are you really only interested in the steady state T rise in the ring? Then I think that is the easiest route to a good number without any physical modeling. Worth a look.
 
  • #14
I'm not just looking for a answer, i want to know how iron losses can be calculated or estimated through calculations.
 
  • Like
Likes hutchphd
  • #15
hutchphd said:
If you have the instrumentation set up, repeat the experiment up to some max T and then record the cooldown as well. From this you can easilly calculate the final steady state temperature directly I believe.
Does a log plot of that data give a good straight line?
I made this plot manually by collecting temperature points in every 30 s, i need to make some datalogger to record also cooling.
 
  • #16
I made test with a bit lower 150 A current and recorded also cooling. If we ignore conductor heating effect and look only what plot says, then if i'm right steel ring heating power is about 0.5 W. I would like to verify how the maximum magnetic flux density is calculated in this case, because I get very small values for the total iron losses compared to measured heating power?

1691153890490.png
 
Last edited:
  • #17
If we don't know all the details, for an approximation, we may improvise.
I take the ring as a short pipe of 42 out/37 in 25 mm long.
The average diameter is (42+37)/2=39.2 mm. Since the busbar current is 250 A, I take field intensity H= 250/(PI()*3.95)=20.15 A/cm=2015 A/m and for a mild steel the flux density B=0.55 Wb/m^2.
If the losses for 1.5 T are 3-90 W/kg and the weight of this ring it is Vol[dm^3]*7.85[kg/dm^3]
Vol=pi()*(0.42^2-0.37^2)*0.25/4= 0.007756 dm^3. The weight= 0.007756*7.85= 0.061 kg.
Let’s take the maximum W/kg=90.
Magnetic losses [max.]=90*(0.55/1.5)^2*0.061=0.7381 W
The eddy current is proportional with B^2 and hysteresis with B^1.6.
However, the aluminum busbar evacuates some losses from the part included in the ring.
The resistance of 25mm bar- at 90oC, including skin effect -it is
2.43E-6 ohm. Then the losses are 250^2*2.43/10^6=0.15 W.
Total losses=0.15+0.7381=0.8881 W
The temperature drop through outside pvc tube of 5 mm thick is 0.6 oC and calculating Qconv+Qrad =0.881 W the ring temperature will be 60 oC [if the ambient air temperature is 40oC] or 52oC [if Ta=30oC].
 
  • Like
Likes OSR
  • #18
Ok i found the problem why i was getting so low values, problem was that i multiplied current with vacuum permeability for no reason when was calculating field intensity. Now I'm getting similar results what I get from previous heat energy calculations, if I adjust steel ring relative permeability between 300-500. I might try to measure hysteresis loop and calculate relative permeability for this steel ring according to this instruction http://info.ee.surrey.ac.uk/Workshop/advice/coils/BHCkt/index.html#top

I did not understand what are you calculating here, can you explain it more:
Babadag said:
If the losses for 1.5 T are 3-90 W/kg

Babadag said:
Magnetic losses [max.]=90*(0.55/1.5)^2*0.061=0.7381 W
 

1. How do I calculate the iron losses in a steel ring around a conductor?

To calculate the iron losses in a steel ring around a conductor, you will need to use the formula: P = (K x f x Bmax^2 x t^2) / R, where P is the power loss in watts, K is a constant value dependent on the material and geometry of the ring, f is the frequency in hertz, Bmax is the maximum flux density in tesla, t is the thickness of the ring in meters, and R is the resistivity of the material in ohm-meters.

2. What is the constant value K in the iron loss calculation?

The constant value K is a material and geometry dependent factor that takes into account the specific properties of the steel ring and the geometry of the conductor. It is typically provided by the manufacturer or can be calculated using the material's permeability and the ring's dimensions.

3. How does the frequency affect iron losses in a steel ring?

The frequency plays a significant role in determining the iron losses in a steel ring. As the frequency increases, the rate of change of the magnetic field also increases, leading to higher iron losses. Therefore, higher frequencies result in higher iron losses in the steel ring.

4. Can I use the same formula to calculate iron losses in different types of steel rings?

While the basic formula for calculating iron losses in a steel ring is the same, the value of the constant K may vary for different types of steel rings. It is essential to use the correct value of K for the specific material and geometry of the ring to get an accurate calculation of iron losses.

5. How can I reduce iron losses in a steel ring around a conductor?

There are a few ways to reduce iron losses in a steel ring. One way is to use a material with a lower resistivity, which will decrease the value of R in the formula. Another way is to use a thinner ring, as the thickness is squared in the formula. Additionally, minimizing the maximum flux density in the ring can also help reduce iron losses.

Similar threads

Replies
10
Views
1K
  • Electrical Engineering
Replies
6
Views
876
  • Electrical Engineering
2
Replies
38
Views
5K
Replies
21
Views
2K
  • Electrical Engineering
Replies
8
Views
2K
Replies
6
Views
1K
Replies
1
Views
1K
  • Electrical Engineering
Replies
1
Views
798
Replies
19
Views
2K
Replies
22
Views
2K
Back
Top