Conductive heat transfer through a finite cylinder with generation

[jparrish]

I'm having trouble developing a formula to calculate hotspot temperature. I'm an EE so I haven't had much background in thermodynamics. I found the following equations for an infinitely long cylinder but am having trouble getting the right value for generation.

1/r*d/dr(k*r*dt/dr)*q'''=0
solved to find:
t(r)=q'''*ro^2/(4*k)*[1-(r/ro)^2]+to
to get hotspot temperature, set r=0
t(hotspot)=q'''*ro^2/(4*k)+to

When evaluating the units, it looks like the generation q''' needs to be in W/(length unit) but I'm having some difficulty arriving at a value with these units. Its easy to find out how many watts of heat the cylinder as a whole should be dissipating but I can't figure out where to get the "/in". At first I thought it would come from Power*l/(pi*r^2) but the equation above is one dimensional so I shouldn't be using the length of the cylinder right?

Also I know heat is transferred to the ends of the cylinder as well so I am already missing that portion of the formula. Any help would be greatly appreciated.

 

[jparrish]

The full equation would be:

dx*dy*dz
T = Ta + Tr + ----------------------------------------
k(dx*dy*D*L+pi*r^2*(dx+dy)*dz)

Where
T = Hotspot temp
Ta = Ambient temp
Tr = Calculated surface temperature rise

Tr is another approximated equation that calculates the temperature rise of the surface of the capacitor winding. This is where the generation is built in. So essentially what I'm currently doing is calculating the temperature rise at the surface and using the thermal resistance of the dielectric and foil to sort-of reverse calculate what the hotspot would be. Like I said earlier, my original thought was to try to find the cylindrical equivalent of this method to get a more accurate result. Below is the equation we're using for Tr.

Tr = K*P/A

Where
P = Power dissipated
A = Surface area of winding
K = Convection coefficient (176 in^2*C/W for air)

This assumes the winding is in non moving free air so when it's not, I'm using a thermocouple attached to the surface to get the heat rise.

For the 1D equation being off, it was low. But again, the length can quite often range from 8*diameter to 1/2*diameter so I'm petty sure we need to take it into account.

Now for the construction:
The windings are made up of 2 layers of dielectric and 2 layers of foil rolled up like a Twinkie. One layer of foil would extend out one side just a little bit and the other would extend out the other side just a little bit. These extensions are then sprayed with a molten metal (zinc) that turns the end of the winding into a solid metal plate. A typical dielectric thickness would be around 8 microns and a typical foil thickness would be around 6 microns. The metal spray on then ends is about .025" thick. If a winding were completely unwound it would quite often be over 10,000in long.

When I was trying to come up with a better way of doing the heat transfer, I tried to put it in terms that I could understand (being an electrical engineer). Here's what I've come up with so far:

1) Look in the radial direction and combine all the dielectric layers into one resistance and all the foil layers into another resistance.

2) Look at these resistances as series resistors.

3) Then in the lengthwise direction do the same thing only look at the resistive materials like parallel resistors.

4) Since the center of the winding would be the hotspot, treat it's temperature like one voltage potential.

5) Since the outside of the winding would be generally uniform in temperature, treat the outside like another voltage potential.

6) Again, since the outside of the winding is generally uniform in temperature, connect the "series" resistances in parallel with the "parallel" resistances.

7) The power dissipated is known and the resistance is known so the temperature or "Voltage potential" between the two should be V=sqrt(P*R).

The only thing that's really left to be found is how the cylindrical construction effects the equivalent resistances in #2 above.
What are your thoughts on approaching the problem this way?

 

[Dave41A]

OK. That is much clearer. A few comments on your terminology and your computations:

1) "K" is typically reserved for thermal conductivity (W/mK or W/mC--they are the same since the K or C int he denominator is actually a temperature difference), while "h" is used for a convection coefficient (W/m^2K). Thus, your "K" that appears in your temperature rise equation is actually a 1/h. It is in weird units (mixed inches/metric), but is actually the reciprocal of a convection coefficient.

2) Converting to metric units (I am more accustomed), means that your "K" of 176 in^2/W equates to an "h" of about 8.8 W/m^2K, which is about right for "free convection" to otherwise still air (i.e. circulation due to buoyancy forces only).

Question: given your known power dissipation (in watts) does this value produce a surface temperature that agrees with your experimental results? "h" can be anywhere between 5 and 50 W/mK, depending upon the configuration, so you may need to adjust your value of 1/h (or "K" as you put it) accordingly ("h" is very configuration dependent and is best measured directly if possible).

3) Based on your description of the capacitor, it does not sound like you can assume homogeneous composition due to the layered nature of the capacitor, unless the thermal conductivities of the film and dielectric are very close. Therefore, I do not recommend the infinite cylinder approximation, nor the bessel-function based finite cylinder approximation. Instead, I recommend a numerical approach--not too different from what you are proposing--but with a couple of modifications:

a) The heat generated in each layer of the film can actually move in 3 directions: (1) axially along the film, (2) radially through the dielectric and into the other film, and (3) "spirally" along the film until it gets all the way to the outside surface of the capacitor. How much flows in each direction depends upon the relative thermal conductivities of the component materials. I would then suggest modeling each layer of the capacitor as a node with three resistors connected to it: one connected to the end (axial conduction), one connected to the dielectric, and one connected (spirally) to the next layer of the film. You would basically be treating the spiral construction as a series of annular rings (like rings of a tree), but where each ring is joined to the ring two rings further out by a resistor (two rings out because of the other layer of film).

b) Be sure to take into account the increased size of each ring, which will increase linearly with radius from the center. The area over which the heat is passing is increasing (A=2*pi*r*L, where L is the length of the capacitor).

c) At each node, add in a "current source" that reflects the amount of heat added at that layer of the capacitor. Assuming the heat is generated in a uniform volumetric manner within each layer, this will also increase radially within the capacitor.

d) Rather than try to model this as (literally) 1000's of rings, I suggest maybe 5 rings or so as a first approximation. Recall that the radial resistance of a ring is actually give by ln(r_2/r_1)/(2*pi*k*L), where "k" is the thermal conductivity of the material (here a weighted average of the film and dielectric conductivities), and L is the length of the capacitor (axial direction). Increase the number of rings to 10 or even 20. The results should start converging to a reasonable value. Beyond a point you should not see appreciable change in the results.

The differences between what I am proposing and what you are doing are that (1) the proposed method takes the internal generation into account, (2) the proposed method takes the spiral conduction into account and (3) the proposed method takes the changing radius into account. Of the three changes, the 2nd is probably the least important (I don't anticipate a lot of "spiral" heat transfer), so you could probably skip it and still work out a reasonably accurate answer.

Hope that all makes sense. Good luck,

Dave

 

[jparrish]

Dave,
Thanks a lot for your extensive reply.
First to answer your question, yes it comes pretty close to what we usually observe for outer surface heat rise for individual windings in an open environment. In cases where the winding is embedded in a potting material, we generally get this number using a thermocouple (or as you said, measured directly).
Secondly I follow what you're saying pretty well. I think I'm going to try (test) both approaches but I have a couple questions about the formulas.

1) I've come across this equation -> ln(r_2/r_1)/(2*pi*k*L) various times but have not been able to implement it because I have an "r_1" of zero. I have looked at making it really really small but depending on how small I make it, it returns a drastically different result (appears to converge at infinity).
Could you please explain in a little more detail how to get an equation for the radial direction?

2) You state using a "current source" at the nodes as the generation. I guess, I would agree except I don't know what value or function to apply as the current source.

3) You said that your proposed method takes the changing radius into account. That's what I was trying to do with my new method as well with the 2D r,z equation. I planned on the resistance in the r direction being a function of the outer radius. My main problem is that I don't have this function and really don't know how to go about deriving it. As I stated in #1, I've looked at using the ln(r2/r1) but really don't understand how to apply it since I can't use r1=0 and don't trust using a really really small r1.

I think we're on the right track and thanks again for bearing with me. I guess I just need a short lesson in radial math & geometry.

 

[zekeman]

"Based on your description of the capacitor, it does not sound like you can assume homogeneous composition due to the layered nature of the capacitor, unless the thermal conductivities of the film and dielectric are very close. Therefore, I do not recommend the infinite cylinder approximation, nor the bessel-function based finite cylinder approximation."

Dave,

Not exactly. You can get the equivalent K =Keq from the well known equation for radial flow

(L1+L2)/Keq=L1/K1+L2/K2

Thus for the long cylinder , you can use Keq as the homogeneous value of conductivity.

Also, given the thicknesses of each layer in microns you would need an enormous number of rings to get a result.

I think the OP should at least use the solution to the infinite cylinder to get a first look at the result for the long cylinder case, namely

T=Q"'*R^2/(4*Keq)+T0

T0= surface temperature
R= radius of cyliner

and do the numerical solution as you suggested for the short cap, but with larger layers. The axial flow would suggest using an equivalent K there.
Keq=(L1K1+L2K2)/(L1+L2)

 

[jparrish]

zekeman,
I have been using the same formula as you to get the Keq. I used that value in the 1D equation you mention but that's the one I said is undershooting what the actual hotspot temperature is. The 3D equation I mentioned above is getting much closer to the test results so I'm thinking we still have something missing. I have not attempted the bessel function version yet because I would like to try the other two methods first. However I still need the equation for the radial direction. Unless you're saying the 1D equation mentioned above is the radial portion of the 2D equation I'm suggesting. I thought it would be different though because when you add another dimension to the mix, the differential equation solution would be different.

 

[zekeman]

"I have been using the same formula as you to get the Keq. I used that value in the 1D equation you mention but that's the one I said is undershooting what the actual hotspot temperature is. "


There is something wrong , since the 1-D equation yields the hottest temperature possible, basically shutting down axial heat flow.
Have you used the 1-D solution:

T=Q"'*R^2/(4*Keq)+T0

Or,Possibly, T0 or Keq may be not known accurately.

 

[jparrish]

Actually if anything's off I'd think it would be Q'''. The formula I think we settled on was:

Q'''=Power/Length/(PI()*Radius^2)

However since we're considering the problem as an infinitely long cylinder, I'm not convinced we should be using Length.

I do agree that it does not make sense to get a lower than measured hot-spot when ignoring heat transfer in one axis.

 

[jparrish]

It also seems strange to me that when using the 3D (x,y,z) formula the hot-spot is higher. The heat has 3 directions to go, whereas with the 1D (r) formula only has one.

Granted, the two formulas take generation into account in very different ways so this may not be a very good comparison. But it was just another observation.

It almost seems like you should be able to approach the 1D (r) formula the same way as the 3D though. This would mean using the generation to estimate the surface heat rise (from ambient). As shown earlier, I already have this equation and it has "proven itself" pretty accurate. Then, using the thermal resistance in the r direction we should be able to get the hotspot. That's all I did in the 3D equation. Again this leads us to the ever elusive radial resistance function. This would be the same function I need for the other 2 approaches Dave and I mentioned above.

 

[Dave41A]

zekeman:

You originally wrote:

"Not exactly. You can get the equivalent K =Keq from the well known equation for radial flow(L1+L2)/Keq=L1/K1+L2/K2Thus for the long cylinder , you can use Keq as the homogeneous value of conductivity.Also, given the thicknesses of each layer in microns you would need an enormous number of rings to get a result."

I believe you are incorrect on the first part. Let me illustrate why.

For the axial conduction, the heat flows along the foil film and along the thin layers of dielectric. This is parallel flow through thermal resistors, and the appropriate equivalent thermal resistance is:

L1*L2/(K1*K2*A1*A2)
Req = R1*R2/(R1+R2) = -----------------------
L1/(K1*A1) + L2/(K2*A2)

In the special case where the lengths L1 and L2 are the same and the thickness of the layers of foil are 3/4 the thickness of the dielectic (i.e. the case here), this reduces to (where K1 is the thermal conductivity of the dielectric and K2 the thermal conductivity of the foil, and A is the area in the axial direction):
4L
Req = --------------
4A*K1 + 3A*K2

In contrast, the thermal resistances in the radial direction act in series, so they are simply added:
ln(r3/r2) ln(r2/r1)
Req = -------------------- + ------------------
2*pi*K1*L1 2*pi*K2*L2

Now in this case, the thickness of the layers is tiny compared to the radius (except at the very center), so the "large radius" (flat wall) approximation will hold here as being over 99% accurate. Therefore, the above expression may be approximated to a high degree of precision with:

r3-r2 r2-r1
Req = ----------- + ------------
2K1*pi*r2*L 2K2*pi*r1*L

Anyway, if this expression (above) is extended to an arbitrary radial thickness, and converted to an equivalent average conductivity (Keq = (r_outer-r_inner)/(2*Req*pi*r_average*L)), you should be able to see that the conductive properties in the radial direction are different from those in the axial direction. This can be easily verified for a variety of materials.

In general, when speaking of laminated materials, the lamination results in a material with overall properties that are neither homogeneous nor isotropic. Hence, the Bessel function solution is invalid.



On the second part, the method I proposed (5 rings, then 10, etc.) should eventually start to converge to a value. jparrish can use as many layers as he needs to reach the desired precision required.

Good luck,

Dave

 

[jparrish]

Dave,

What are your comments on what I said about the inner radius being zero? I know in all actuality that it's not really zero and that there really is an inner radius but it just seems strange to me how much the result changes the closer you get to zero.

 

[zekeman]

"For the axial conduction, the heat flows along the foil film and along the thin layers of dielectric. This is parallel flow through thermal resistors, and the appropriate equivalent thermal resistance is:

L1*L2/(K1*K2*A1*A2)
Req = R1*R2/(R1+R2) = -----------------------

L1/(K1*A1) + L2/(K2*A2)"

Dave,

By my analysis L1=L2=L; L1,L2, L your nomenclature
or
1/Req=K1A1/L+K2A2/L which is precisely yours

Keq(A!+A2)/L=K1A1/L+K2A2/L
Keq(A1+A2)=K1A1+k2A2
I use A1=2pirL1 and A2=2pirL2
L1 and L2 are MY THICKNESSES
so after substitutuion and elimination of 2pir
Keq(L1+L2)=K1L1+K2L2

Now for your radial answer,

r3-r2 r2-r1
Req = ----------- + ------------
2K1*pi*r2*L 2K2*pi*r1*L

My L1=r3-r2 and L2=r2-r1 and the r1 and r2 in the denominators are essentially equal as you mention except near 0
So your answer becomes
Req=1/2pirL*(l1/K1+L2/K2)
and since Req=(L1+L2)/Keq2pirL I get after substitution
1/2pirL*(L1+L2)/Keq=1/2pirL*(l1/K1+L2/K2)
and
(L1+L2)/Keq=L1/K1+L2/K2
the same as you would get if you carried it through.

I agree that the Bessel solution would be invalid for the finite cylinder but the analytic solution for the infinite cylinder should be quite accurate using the radial Keq.

And as the OP knows, it should yield the highest temperatures.
If his intent is to get a worst case central temperature, I recommend the simple analytic solution and save a lot of unnecessary work using numerical methods.

 

[jparrish]

So you're not using:
ln(r3/r2) ln(r2/r1)
Req = -------------------- + ------------------
2*pi*K1*L1 2*pi*K2*L2
?
Instead you're saying to use r3-r2 and r2-r1 in the numerators?
This would solve the center radius of 0 problem but I'm not sure how ln(x/y) = x-y

 

[Dave41A]

jparrish:

To answer your earlier question, the radial heat flow near the center of the cylinder will be essentially zero (at the very center, it will be zero, due to the symmetry). To this end, I would draw the thermal circuit so that "resistor" in the radial direction was "open circuit" as almost all the heat flow will be axial and "spiral." For the surrounding rings, you will have appreciable temperature gradient in the radial direction, and hence, heat flow will be in all three directions.

On the confusion over the appropriate expression to use for thermal resistance, the difference between a radial resistance (R=ln(r2/21)/(2piLk) ) and a flat wall resistance (R=L/kA, where L=r_outer-r_inner) only becomes significant for wall thicknesses that are on the same order as the inner radius or thicker. Once the inner radius is greater than the wall thickness, the error drops. For thin-walled vessels, the difference if essentially nil (even at r_outer = 2*r_inner, the error between flat wall and radial is only about 4%). The flat wall approximation uses a slightly simpler expression for thermal resistance, that's all, and was introduced to simplify the accompanying discussion.

zekeman: Forgive me if I mis-understood you, but the point I was trying to make was that the Keq in the radial direction will not be the same as in the axial or spiral directions (spiral and axial will be very, very close). One condition for the application of almost all analytic solutions is that the material be homogenous and isotropic...conditions that don't hold here. Homogeneity is pretty closely met (as you observe, due to the thin layers), but the isotropic condition doesn't hold at all--possibly (probably?) far from it.

In all fairness, I don't have the reference cited, so if that particular solution is valid for non-isotropic materials, then I stand corrected. However, considering that only one "k" appears in the solution (and not "k_axial" and "k_radial"), I would be surprised if it did.

You do raise an important point regarding the cumbersome nature of numerical solutions. However, here it seems that the leading analytic contender is an infinite series solution. Hence, using it to obtain a solution with any accuracy will require a lot of terms, or the inevitable truncation of the series to some manageable number. So the trade off is between accuracy and the number of terms to keep in the solution...a trade off that is also present with any numerical solution. For me, however, the clincher is that the analytic solution assumes an isotropic material, while the numerical does not. That alone is enough to tip the balance for me, but I can see a good argument being made for the analytic solution if the isotropic condition is at least approximately met. I don't know the exact material properties of this metallic foil and the dielectric used, but their conductivites should be pretty close to each other (within 50% or so I would guess) before I would believe the analytic solution. Hope that makes sense.

Good luck to all,

Dave

 

[zekeman]

Dave,

If you look at the axial flow everywhere within the infinite cylinder, you get

Kax*dT/dz=0

So it makes no difference what the axial conductivity since no flow takes place and accordingly, the condition for the validity of the analytic solution depend on the constancy of Keq in the radial direction only.

Just think of an arbitrary slice of this cylinder and you can see it.

 

[jparrish]

Okay, thanks Dave and Zekeman. So you're saying not to really worry about the center since the heat transfer is negligible there?

With that being the case would you recommend using an inner radius that is oh say... 5% of the total radius?

Now, sorry to repeat earlier portions of the thread but in this case I feel it is necessary. Earlier you stated the following (in a nutshell):

a) The heat generated can move in 3 directions:
- axially
- radially
- "spirally" <-- negligible
Model each <of the 4 layers> as a node with <two> resistors connected to it (radial and axial).

c) At each node, add a "current source" that reflects the amount of heat added for that node.
Generation is uniform at all points for a given radius.
Generation increases with radius.

Question: What would the formula for generation be then?

d) Rather than model 1000's of rings, use 5 to 20.
Radial resistance is given by:
ln(r_2/r_1)/(2*pi*k*L)
k = film & dielectric weighted average thermal conductivity
L = length in axial direction

As the # of rings increase, "The results should start converging to a reasonable value. Beyond a point you should not see appreciable change in the results."

Question: Are you saying that beyond a certain # of rings, the # of rings no longer matters? It seems to me that this is like saying that beyond a certain radius, the radius no longer matters... Please clarify.

 

[Dave41A]

jparrish:

For the amount of "current" to add at each node, the amount is determined by the volume of the ring and the volumetric generation rate (the q''' from the way back).

I would take the radius of the cylindrical capacitor and divide it into, say, five "rings" (like rings in a tree) and work out the solution that way. Then I would repeat with the same total radius divided into 10 rings, etc. As the number of rings goes to infinity, you will have a numerical approximation to the actual temperature distribution within the cylindrical capacitor. I doubt you'll have to exceed 20 rings (each of thickness equal to 1/20 the total radius) before you stop seeing appreciable change.

zekeman: No argument from me on the use of Keq for the infinite cylinder. My original comments on the use of Keq were directed towards the finite cylinder case. For infinite cylinder--absolutely--assume Keq and procede with the well-known analytic solution. For the finite, a numerical solution is best, due to the non-isotropic nature of the material.

I think we agree that the capacitor should not be modeled as an isotropic finite cylinder (with the analytic, Bessel-function based solution from Carslaw and Jaeger--this was the "analytic soluion" I was criticizing--i.e. that only one "k" appears in the Bessel-function solution).

For a first approximation, the infinite cylinder approximation might indeed be a good choice if the radial Keq is such that radial heat transfer dominates. Likewise, an axial-only heat transfer model might even be appropriate if the axial conduction dominates (solution also well known). A comparison of the relative values of the two "k's" would determine which (if either) is applicable. In both cases, the model will over-predict the "hot spot" temperature (as jparrish and you have both observed).

Good luck,
Dave

 

[sailoday28]

Dave41A (Mechanical) 19 Nov 10 14:35 States:
jparrish:To answer your earlier question, the radial heat flow near the center of the cylinder will be essentially zero (at the very center, it will be zero, due to the symmetry).

What is the basis for radial heat flow near center near zero?

 

[Dave41A]

jparrish, sailoday28:

Heat flow in the radial direction will be governed by fourier's law, so the heat rate will be proportional to the temperature gradient in the various directions. At the very center of the cylinder (assuming symmetry of boundary conditions and construction of the capacitor), the temperature will be at a maximum. Hence, the derivative of the temperature with respect to radius will be zero ("first derivative rule"--the slope is zero at a max or min). This only holds true at the very center (a geometric point), but in close vicinity, the heat rate will be very small, even if not exactly zero.

That said, I want to change my earlier recommendation to model the first radial resistor as "open circuit." Upon further reflection, I recommend imposing the condition that the temperature at the inner edge of the first "ring" be the same as that of the center cylinder. By making these temperatures the same, the "delta T" between them will be zero, which will reflect the "zero heat rate" condition properly. I would then add a current source at the inner edge of the first ring that reflects the heat "current" generated by the central cylinder. The second ring out will then be connected to the first ring by a resistor (radial resistance) and itself be "fed" by a current source that reflects the amount of heat generated in the first ring. The third ring is then fed by a current source that reflects the heat generated in the second ring, etc.

The resulting resistor diagram will end up looking something like this (with current added at T1, T2, etc as described above not shown; "Tend" is the temperature at the end of the capacitor):

Tend Tend Tend
| | |
T0&T1-----T2----T3----- etc.
| | |
Tend Tend Tend

This will not be exact, but should be reasonably accurate. I suggest Incropera. et al (Wiley) for more detailed discussion of finite difference methods, as the above approximation is very crude. Hope this all works out for you. We certainly spent enough time on it.

Good luck and happy Thanksgiving,

Dave