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.

 

[btrueblood]

jp,

Heat is generated due to current flowing against electrical resistance thru the wire cross section, so the power term must depend on pi*r^2 (cross section area).

Heat is dissipated on the periphery of the wire due to radiation or convection (or conduction), so the loss term must depend on 2*pi*r .

You need both terms to fully address hot spot development. It can get a bit tricky if you are considering both convective heat transfer and radiation, but it's still solvable numerically.

 

[ione]

The solution you have posted is valid for 1-dimensional, steady state with uniform heat generation (anyway your differential equation is not correct at all, as in the first term you should have + q’’’ and not*q’’’)

q’” is the heat flux, unit of measure is W/m^2 (or any other consistent) and not W/m

 

[jparrish]

ione,
Sorry, that was a typo. I did start with +q'''
Also, I thought q'' was heat flux, and q''' was... something else (the derivative of of heat flux? so... the rate of change in heat flux?)

Following is my new check to find what units q''' needs to be (which is totally different than what I originally calculated):
t(hotspot)=q'''*ro^2/(4*k)+to
where:
t => deg. C
ro => in
k => W/(in*deg. C)
to => deg. C

so we have:
C=?*in^2/(W/(in*C))+C
C=?*in^3*C/W+C
?=W/(in^3*C)

so q''' needs to be W/(in^3*C)


btrueblood,
This is a slightly more complicated problem than a wire. It is similar but there are various materials that make up the cylinder (it's a round film capacitor). I'm calculating power by using the specified current and the capacitor winding ESR: I^2*ESR

 

[btrueblood]

Err, then I'm not sure you can use the equation for an infinite cylinder...won't the current flow (charge/time) be non-uniformly distributed across the capacitor plate faces/edges? Maybe I am not envisioning what you are modelling correctly...or maybe you aren't.

 

[Dave41A]

Your derivation is correct. q''' (triple prime) is the volumetric heat generation rate (W/m^3). If you are thinking of heat generation in a wire (due to current passing through), this will be the watts of heat generated per unit length divided by the cross sectional area.

Some texts (e.g. Incropera, et al) use "q dot" for volumetric rate (W/m^3) which I don't like because "dot" normally means d/dt. q dot is also inconsistent with q'' for flux (W/m^2). However, it is still a popular notation, so become accustomed to seeing it.

I agree with btrueblood that the infinite cylinder is not applicable here. If this is for a capacitor, I suggest either a lumped-parameter model (capacitor all a uniform temperature) if you want an analytic solution (check the "Biot Number"). If that is not valid due to the construction of the capacitor, I suggest a numerical approach.

ione's observation that the posted equation is for 1-D (radial) steady-state conditions with uniform generation and constant thermal conductivity is also correct. I don't think you have this situation.

Good luck,

Dave

 

[jparrish]

Thanks Dave,
I just figured this 1D equation for an infinite cylinder would be similar to a 2D equation for a finite cylinder. I'm also afraid I can't assume uniform heat since the goal is to try and calculate the hotspot temperature based on surface temp and the power dissipated in the cap.

I don't suppose you would have that similar 2D equation for a finite cylinder? That would be exactly what I'm looking for.

What I've posted already is as much as I could get out of my old college textbook and a good number of google searches.

 

[zekeman]

Your solution

"t(r)=q'''*ro^2/(4*k)*[1-(r/ro)^2]+to

is missing the finite heat transfer across the surface.

It should be

t(r)=q'''*ro^2/(4*k)*[1-(r/ro)^2+2/(ro*h)]+to

Noting the addition of the term 2/(ro*h).

where h includes convection plus the linearized radiation term. The units for Q"' are for the volumetric power generated W/m^3

The solution is conservative for the finite cylinder and may even be fairly accurate for the finite cylinder if the length is long next to the radius.

For your purposes I think this solution is good if the answer is less than the max hot spot allowed. Otherwise you have a messy 2 dimensional problem whose solution I can point you to if needed.

 

[ione]

I have checked my previous post and discovered a typo q''' is in W/m^3 as already noticed by others.

I think the right solution for 1-dimensional, steady state with uniform heat generation plus convective + radiative boundaries should be:

T(r) = T0 + q'''*[(r0^2-r^2)/(4k) + r0/(2h)]

with h in W/(m^2 K) (Zekeman please check)

 

[jparrish]

Thanks again guys.

Dave,
When you say q''' would be "watts of heat generated per unit length divided by the cross sectional area", that would effectively be P/(L*pi*(D/2)^2) correct?

Zekeman and Ione,
You mention convective heat transfer which I wasn't real sure whether to include or not. This gets tricky because the capacitor is quite often potted in an epoxy or urethane with other capacitors and materials, which would be another conduction transfer before the convection. However I was thinking we could side-step that by placing a thermocouple against the surface of the cap to get its OD temp reading.

Also, Zekeman, I think we may need the 2D equation after all since quite often the capacitor length can be pretty close to the radius... as much as this may complicate things, I would like to try and get the best approximation of hotspot. So if you wouldn't mind pointing me to that "messy 2 dimensional problem solution" I would be very grateful.

 

[zekeman]

Here goes [from Carslaw and Jaeger, Oxford Clarendon Press,1959(adjusted for surface temperature constant=To)]

T=To+q"'z(L-z)/2k+ 4L^2q"'/[k(3.14)^3]*infinite sum from n=0 to n= infinity:
Io[(2n+1)*3.14r/L]*sin[(2n+1)*3.14z/L]/{(2n+1)^3Io[(2n+1)*3.14a/L]

finite cylinder
radius a
length L
T0 surface temperature
0>z<L
Io Bessel function

Hot spot at z=.5L, r=0

Hot spot solution:

T=To+q"'z(L-z)/2k+ 4L^2q"'/[k(3.14)^3]*infinite sum from n=0 to n= infinity:
sin([n+.5)*3.14]/{(2n+1)^3Io[(2n+1)*3.14a/L]

Good luck




















































.

 

[jparrish]

Thanks. And yeah that's pretty complicated.
Just a couple of questions:

1) Is the attached equation the form of the bessel function you're using? It's been a while since I've worked with bessel functions (college calc 2 I think).

2) I'm a little lost in this portion of it:
sin([n+.5)*3.14]/{(2n+1)^3Io[(2n+1)*3.14a/L]

did you mean

sin(n+.5)*pi
--------------- * Io[(2n+1)*pi*a/L]
(2n+1)^3

or

sin(n+.5)*pi
--------------------------------
(2n+1)^3*Io[(2n+1)*pi*a/L]

 

[jparrish]

Sorry, the equation image didn't attach.

 

[zekeman]

1) No, Io is another solution to the cylindrical equation modified due to the finite length and boundaries . But don't sweat it . Just get the values of Io(x) for the various values from mathematical tables ( maybe google it). You will have to sum several terms of the series before it converges and it should not be too difficult to use the hot spot formula
below.

2) The second form
sin(n+.5)*pi
--------------------------------
(2n+1)^3*Io[(2n+1)*pi*a/L]

 

[zekeman]

Oops, forgot the first part,
To+q"'z(L-z)/2k+ 4L^2q"'/[k(3.14)^3]* the sum of the infinite series
sin(n+.5)*pi
--------------------------------
(2n+1)^3*Io[(2n+1)*pi*a/L]

n=0,1,2,3... to infinity

 

[jparrish]

Now that I look at it again is the pi supposed to be inside or outside the sine function?

sin[(n+.5)*pi]
--------------------------------
(2n+1)^3*Io[(2n+1)*pi*a/L]

 

[jparrish]

Also, wouldn't it simplify down even more when substituting .5L for z?

To + q'''L²/8k + 4L²q'''/[k*pi³] *

? sin[(n+.5)*pi]
? [--------------------------------]
n=0 (2n+1)³*Io[(2n+1)*pi*a/L]

 

[Dave41A]

jparrish:

To answer your question, yes, q''' is simply heat (in watts) divided by volume (which is what you have).

I am not so sure that the Bessel function route is the way to go. The assumption in such a solution is that the cylinder is of a homogeneous material of constant properties. If this is a "can style" capacitor, the inside is going to be a wound assembly of the charge-holding material and the dielectric. If the thermal conductivities of these materials are significantly different, then the assumptions inherent in the Bessel-function solution are not met, and hence the solution itself is invalid.

If you're looking for a "hot spot" in such a device, I suggest a numerical appoach.

Good luck,

Dave

 

[jparrish]

The cap is not the "can" form. It is just the winding element. Also the the thickness of the current carrying element and the dielectric is extremely thin in comparison to the total diameter, so I think the homogeneous equation should give a pretty good approximation.

Also I've got some results from the previous 1D equation.
Using the q''' of P/(L*pi*(D/2)^2) gets a hotspot result nowhere close to what I was seeing in tests.

So far the closest calculation (to the test results) is to treat the winding as a rectangular box and use the following formula to get the hotspot-to-surface temperature difference:

dx*dy*dz
----------------------------------------
k(dx*dy*D*L+pi*r^2*(dx+dy)*dz)

Where
dx & dy = pi*D/8
dz = L/2
L = length
D = diameter
r = radius

Then add the surface temp (To) to get the hotspot temp.

I realize that this assumes the surface temperature to be sourced entirely by the capacitor element.

My original thought was to get a better approximation by taking a similar approach but using cylindrical coordinates.
What are everyone's thoughts on this? I thought it would help to simplify things since generation would not have to be bothered with.

 

[Dave41A]

jparrish:

I have to admit I don't really understand what you're doing above. For starters, the formula you have has no equals sign...what does it produce?

I would caution against trying different equations until you find one that seems to be close. "A stopped clock is right twice a day" as the saying goes. Any similarity between the actual temperature and the calculation will be overwhelmingly (if not entirely) due to luck. A solution that has no generation in it (such as the one you proposed)is clearly wrong and any similarity to real life will be entirely due to luck.

Could you describe the construction of this capacitor in greater detail? I could then help you find the best approximate solution technique. Also, for the 1D radial solution that was "nowhere close" to the test results, which way (high or low) was it off?

Good luck,

Dave