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blogfast25
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Radial Cooling Fin: temperature profile and heat loss
I had some difficulty finding (in books or online) the mathematical expressions for the temperature profile and heat loss for a radial cooling fin, so
I developed them from the relevant thermodynamics and differential equation.
Geometry:
A single radial cooling fin (circular plate, hot tube runs perpendicularly through centre) mounted on a tube with radius
R<sub>0</sub>. Outer radius of fin is R<sub>1</sub>, uniform fin thickness is
t. With x the radial distance from the centre of the tube.
Temperature and cooling:
Tube temperature is a uniform T<sub>0</sub>, the surrounding medium is at a uniform temperature
T<sub>∞</sub>. Radiative cooling losses are neglected.
Material constants:
h is the heat transfer coefficient (convective) in W/m<sup>2</sup>K
k is the heat conduction coefficient (Fourrier) in W/mK
Temperature profile:
T(x) – T<sub>∞</sub> = (T<sub>0</sub> – T<sub>∞</sub> ) [(e<sup>ax</sup> +
e<sup>a(2R1-x)</sup> / (e<sup>aRo</sup> +
e<sup>a(2R1-Ro)</sup>]
Overall heat loss (Q = dq/dt):
Q = - 2 π R<sub>0</sub> t √(2h/tk) [(e<sup>aRo</sup> - e<sup>a(2R1-Ro)</sup> / (e<sup>aRo</sup> + e<sup>a(2R1-Ro)</sup>] (T<sub>0</sub> – T<sub>∞</sub> )
With a = √(2h/tk) (√ is square root)
Does anybody here have these formulas at hand, for verification?
[Edited on 5-5-2015 by blogfast25]
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Sulaiman
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a quick 'rule of thumb' for temperature rise with convection cooling used in electronics (<200C) is;
(Temp rise in C) = (dissipation in mW) / (exposed surface area in cm2) all to the power 0.833
many assumptions and simplifications but it seems to give reasonably good estimations.
e.g. a piece of copper 4cm x 3cm vertical in free air dissipating 2 W
Temp rise = (2000mW / 2x 4x3cm2)^0.833=39.8. C above ambient.
For your configuration add up the disk and pipe area exposed as
n x 2 x pi(R1^2 - R0^2) + (length of pipe) x 2 x pi x R0 where n=number of disks
ignoring disk thickness as it is only an approximation.
if you design, build, test (or whatever) such a device and measure its performance
I would be interested to know how well this 'rule of thumb' works.
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blogfast25
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Thanks for your reply. What do you mean by 'temperature rise' here?
If that piece of copper, of whatever shape, acts as a fin, then thickness is impossible to ignore. The surface contact area between the object to be
cooled and the fin is the ‘bottleneck’ through which all conducted heat must flow:
Q = - A k (dT/dx)<sub>boundary</sub> (Fourrier) at the boundary between object and fin, with A that contact surface area. That’s why
tapered fins are often used: thicker at base promotes conduction and reduces ‘bottleneck’.
[Edited on 5-5-2015 by blogfast25]
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Sulaiman
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I was assuming a pipe with several (n) disks,
each thick enough only to minimise temperature gradients within the bulk of the metal
spaced far enough apart to consider each disk to be in free air.
the 'bottleneck' would be 'bypassed' by the method of fixing disks to pipe.
(weld/solder/glue/press-fit etc.)
'temperature rise' is the difference in temperature between ambient air and the bulk of the metal
if the thickness of the fins becomes significant (lots of metal/weight/cost) then
the 'rule of thumb' is probably too inaccurate for detailed cost analysis.
and probably 'forced air' would be economical?
[Edited on 5-5-2015 by Sulaiman]
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blogfast25
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You want to maximise the temperature along the fin, no temperature gradient means no cooling at all. In practice that means high
h (and high k). At h = 0 (insulated fin) the fin dissipates no heat whatsoever.
Temperature gradients across the fin are considered negligible (otherwise the math treatment becomes very complicated) and that is true for
thinish fins.
[Edited on 5-5-2015 by blogfast25]
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Magpie
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Have you checked "Transport Phenomena" by Bird, Stewart, & Lightfoot (1960)? They work an example using a rectangular fin extending from a wall
to give a temperature profile. Fin thickness is a parameter in the derivation.
A problem at the end of the chapter deals specifically with your situation, asking for a temperature profile and heat dissipation. Please do not ask
me to work this problem.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Sulaiman
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there are theoretically correct methods, or
I mentioned the 'rule of thumb' http://en.wikipedia.org/wiki/Rule_of_thumb as a quick 'sanity' check,
I can't remember where I got it from and I wish I'd never mentioned it !
the 'rule of thumb' assumes all of the metal is at the same temperature,
even though there must be a temperature gradient within the metal to transport heat,
it is assumed to be negligible compared to (bulk metal temperature) - (ambient temperature).
e.g. I think that 20 thin aluminium disks punched from a sheet, press-fit onto a tube
could out-perform a precision profiled solid copper disk, cost wise.
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blogfast25
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Quote: Originally posted by Magpie  | Have you checked "Transport Phenomena" by Bird, Stewart, & Lightfoot (1960)? They work an example using a rectangular fin extending from a wall
to give a temperature profile. Fin thickness is a parameter in the derivation.
A problem at the end of the chapter deals specifically with your situation, asking for a temperature profile and heat dissipation. Please do not ask
me to work this problem. |
You can find that derivation here:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/...
I adapted that for the radial fin. I just wanted to get some corroboration on the specific formulas, in case someone here has an engineering handbook
that has that solution in it. Mine doesn't treat that case.
I'm fairly sure of my case.
@sulaiman: there is absolutely nothing wrong with rules of thumb or with bringing one up. But that one doesn't fly: without taking into account the
interface through which all heat must flow such a model cannot possibly work. All other things being equal, a fin that's twice as thick will dissipate
twice the heat, for instance.
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aga
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The maths is beyond me.
Silicone paste is generally used at metal-metal interfaces to maximise heat transfer between separate metal parts for the porpoise of conduction.
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Sulaiman
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but two disks would have the same nett cross-sectional metal area (relative to direction of heat flow)
but would have (almost) twice the surface area exposed to ambient air etc.
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blogfast25
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| Quote: Originally posted by aga | The maths is beyond me.
Silicone paste is generally used at metal-metal interfaces to maximise heat transfer between separate metal parts for the porpoise of conduction.
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Hmmm... silicones are generally poor conductors of heat, you know? Mixing with metal powder would increase k, or so I'd imagine...
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blogfast25
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| Quote: Originally posted by Sulaiman |
but two disks would have the same nett cross-sectional metal area (relative to direction of heat flow)
but would have (almost) twice the surface area exposed to ambient air etc.
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Correct. Multiple thinner fins amounting to the same thickness as a single one are more effective.
[Edited on 5-5-2015 by blogfast25]
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aga
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I have a few 5ml syringes of a silicon based white pasty stuff that is sold as heat transfer enhancer when sticking heatsinks on chips.
What it is, i do not know.
It seems to work.
http://uk.farnell.com/electrolube/htc10s/heat-transfer-compo...
There's a data sheet on there too.
[Edited on 5-5-2015 by aga]
Surprised ! The data sheet says 'NO silicones used'
[Edited on 5-5-2015 by aga]
Back to normal : they have a silicon based product for High Temperature applications. Phew.
[Edited on 5-5-2015 by aga]
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Magpie
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I think for automobile radiator work some kind of solder is normally used. Silver solder should be good. Brazing might be too much heat for thin
fins.
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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| Quote: | | Conductivity: 0.9W/mk |
Hmm. Al is about 200 W/mK.
Of course you only need the thinnest layer of it.
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annaandherdad
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| Quote: Originally posted by blogfast25 |
You want to maximise the temperature along the fin, no temperature gradient means no cooling at all. In practice that means high
h (and high k). At h = 0 (insulated fin) the fin dissipates no heat whatsoever.
Temperature gradients across the fin are considered negligible (otherwise the math treatment becomes very complicated) and that is true for
thinish fins.
[Edited on 5-5-2015 by blogfast25] |
Hi, blogfast, I've been trying to understand the problem but I'm not sure I get the geometry. The disk is annular, with the central shaft going
through it, right? And the temperature along the wall at the inner radius is T0, while along the top, bottom and outer radius it is T_infty, right?
I'm guessing that these are the boundary conditions you want.
If this is right, I don't understand your statement, that the temperature gradient across the fin is negligible, because that would mean that the
temperature inside the disk was constant (namely T_infty). Of course the actual temperature, at fixed radius but moving across the fin (in the "z"
direction) would be a curve that bows upward in the middle, where the temperature would be highest.
Any other SF Bay chemists?
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blogfast25
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| Quote: Originally posted by annaandherdad | I'm guessing that these are the boundary conditions you want.
If this is right, I don't understand your statement, that the temperature gradient across the fin is negligible, because that would mean that the
temperature inside the disk was constant (namely T_infty). Of course the actual temperature, at fixed radius but moving across the fin (in the "z"
direction) would be a curve that bows upward in the middle, where the temperature would be highest.
|
It was perhaps clumsily explained on my part. A diagram would have helped. Where I wrote 'across the fin' the term 'in the fin' would have been
clearer.
Call any axis with origin in the centre of the tube and running along a radial line of the fin, the x-axis. Temperature gradually decreases along that
axis. I called that profile T(x). dT/dx < 0 (in the case of cooling).
The y-axis runs perpendicularly to the x-axis and along the centre line of the tube. It's assumed the temperature in the fin, at any x, in the y
direction is constant: dT/dy = 0. As you said, this is of course not fully accurate but I've never seen a treatment that takes the temperature profile
'in' the fin into account, if the fin is relatively thin compared to its radius.
| Quote: Originally posted by annaandherdad | | And the temperature along the wall at the inner radius is T0, while along the top, bottom and outer radius it is T_infty, right?
|
No. T<sub>∞</sub> is simply the ambient temperature. The outer edge of the fin would only reach T<sub>∞</sub> for x =
∞.
The differential equation requires a second boundary condition: (dT/dx)<sub>x=R1</sub> = 0.
The undetermined solution of the DE is:
T(x) - T<sub>∞</sub> = C<sub>1</sub>e<sup>ax</sup> + C<sub>2</sub>e<sup>-ax</sup>
Both constants are determined from T(R<sub>0</sub> =
T<sub>0</sub> and (dT/dx)<sub>x=R1</sub> = 0.
Does that clear it up, AAHD?
[Edited on 6-5-2015 by blogfast25]
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annaandherdad
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I have a lot of questions, but let me just start with one. Why are the boundary conditions at x=R1, dT/dx=0? If I have a slab with different
temperatures on each side, for example, the wall of your house in winter, then the temperature profile inside the slab is just a straight line going
from T0 at one side to T1 at the other (modelling the wall as if it were a uniform material). My point is that dT/dx is not zero at the interface.
The boundary conditions are given by the two temperatures, not the derivatives of temperatures.
Any other SF Bay chemists?
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blogfast25
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| Quote: Originally posted by annaandherdad | | I have a lot of questions, but let me just start with one. Why are the boundary conditions at x=R1, dT/dx=0? If I have a slab with different
temperatures on each side, for example, the wall of your house in winter, then the temperature profile inside the slab is just a straight line going
from T0 at one side to T1 at the other (modelling the wall as if it were a uniform material). My point is that dT/dx is not zero at the interface.
The boundary conditions are given by the two temperatures, not the derivatives of temperatures. |
Actually, with the temperature profile of a wall, from inside to outside, the outside temperature of the wall is not T1 either, see here:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/...
You start from the assumption that the edge of the fin is at T<sub>∞</sub> but that <i>simply isn’t true</i>.
If you stick one end of a poker (rod) into an open fire, do you expect the ‘cold’ end to be at room temperature? I sure as hell hope not
because otherwise you’re going to get a lot of burnt fingers!  The
edge temperature of the rod tends to T<sub>∞</sub> only for rods with length L = ∞ (all other things being equal). It's the
same for radial fins.
What IS true is that at the edge of the fin no more heat is conducted (because there is no more fin!) and by Fourrier that means that at x =
R<sub>1</sub>, the temperature gradient (dT/dx)<sub>x=R1</sub> = 0. The only error we’re making is that we neglect that
little bit of convection heat that comes off the edge of the fin, which would be 2 π R<sub>1</sub> h t (T<sub>1</sub> -
T<sub>∞</sub>. For very 'stubby' fins that would be a concern.
[Edited on 7-5-2015 by blogfast25]
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aga
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For any part of a heat conducting Thing to remain at ambient, all other parts will also be at ambient temperature.
Once you put any heat into one part, it creates a heat gradient to all other parts.
The reverse is true for a heat Insulator.
Yes, a metal rod stuck into a fire can glow red (700C+) on one end, yet the other will certainly not be at 700+ or ambient temperature (far from it).
Many fingers (including some of mine) bear witness to this fact.
[Edited on 7-5-2015 by aga]
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blogfast25
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Correct, aga. The fact that for a cooling fin the edge temperature is lower than the tube temperature is due to convective (and radiative but I didn't
take these into account) losses. Without these, for example an (absurd, of course) insulated fin would have no temperature gradient (dT/dx = 0, for
one dimensional or radially symmetrical geometries). In steady state conditions, of course.
[Edited on 7-5-2015 by blogfast25]
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annaandherdad
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| Quote: Originally posted by blogfast25 | | Quote: Originally posted by annaandherdad | | I have a lot of questions, but let me just start with one. Why are the boundary conditions at x=R1, dT/dx=0? If I have a slab with different
temperatures on each side, for example, the wall of your house in winter, then the temperature profile inside the slab is just a straight line going
from T0 at one side to T1 at the other (modelling the wall as if it were a uniform material). My point is that dT/dx is not zero at the interface.
The boundary conditions are given by the two temperatures, not the derivatives of temperatures. |
Actually, with the temperature profile of a wall, from inside to outside, the outside temperature of the wall is not T1 either, see here:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/...
You start from the assumption that the edge of the fin is at T<sub>∞</sub> but that <i>simply isn’t true</i>.
If you stick one end of a poker (rod) into an open fire, do you expect the ‘cold’ end to be at room temperature? I sure as hell hope not
because otherwise you’re going to get a lot of burnt fingers! The
edge temperature of the rod tends to T<sub>∞</sub> only for rods with length L = ∞ (all other things being equal). It's the
same for radial fins.
What IS true is that at the edge of the fin no more heat is conducted (because there is no more fin!) and by Fourrier that means that at x =
R<sub>1</sub>, the temperature gradient (dT/dx)<sub>x=R1</sub> = 0. The only error we’re making is that we neglect that
little bit of convection heat that comes off the edge of the fin, which would be 2 π R<sub>1</sub> h t (T<sub>1</sub> -
T<sub>∞</sub>. For very 'stubby' fins that would be a concern.
[Edited on 7-5-2015 by blogfast25] |
The MIT engineering diagram shows the linear profile of temperature between the two sides of the wall, which is what I was talking about in my
previous post. The reason the diagram shows the inside and outside temperatures as being different from the temperatures at the wall is that the
diagram is taking into account the layer of fluid near the wall over which there is a temperature gradient, for example, connecting T1 with Tw1 in the
diagram.
The thickness of this layer, and the temperatures on the two sides of it, depend on complicated factors such as the motion of the fluid (which may be
turbulent, etc). It also depends on the heat flux coming out from the wall.
The temperature profile inside the wall, however, is simple, being just a straight line. This is the solution of the Laplace equation for the
temperature in a slab, with boundary conditions in which the temperature is constant on each wall.
Another simple fact is that as the cooling at the surface becomes more efficient, by increasing the velocity of the cooling fluid, for example, or the
nature of the fluid (lower viscosity is better), the temperature T1 (in the diagram) and the temperature Tw1 approach one another. The same happens
when the heat flux goes to zero, for example, by making the wall thicker while holding T1 and T2 fixed. Thus, taking T1=Tw1 (and T2=Tw2) is an
idealization, being the condition for the maximum efficiency of cooling available, given T1 and T2.
This idealization has the advantage that it is mathematically simple to characterize. There is no fluid mechanics involved, just solving the Laplace
equation in the interior of the heat conductor, with a given temperature on the boundary. It may not be realistic in practice, it depends, but it's
a place to start in analyzing the problem.
I have seen engineers in the computer business solve this equation in an analog manner by using carbon paper, cut to the shape of the heat fin, with
fixed voltages clamped to the appropriate edges. By measuring the voltage in the interior with a probe, you get the temperature profile. Of course
the Laplace equation can be solved on numerically too, and in some cases it can be solved analytically (if the boundary conditions are nice).
More later.
Any other SF Bay chemists?
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blogfast25
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AAHD:
I've no gripe with what you write here. The boundary layer is indeed subject to some nifty fluid mechanics, if you really want to go there. But for a
fin in stationary air the simplified treatment is what ANY decent engineering textbook prescribes.
What that has to do with the second boundary condition is what I'd like to know though. So far you've only managed to dodge your erroneous judgement
that edge of the fin is supposed to be at T<sub>∞</sub>, which is manifest nonsense.
[Edited on 7-5-2015 by blogfast25]
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annaandherdad
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| Quote: Originally posted by blogfast25 | AAHD:
So far you've only managed to dodge your erroneous judgement that edge of the fin is supposed to be at T<sub>∞</sub>, which is manifest
nonsense.
[Edited on 7-5-2015 by blogfast25] |
Hey Blog, no need to get insulting, I was trying to understand the physical model you were using and at first I thought you were solving the Laplace
equation with fixed T boundary conditions, as I've seen in coursework and as I mentioned in my job in a computer company that was designing chips for
a main frame. And by the way your solution gives T=T_infty in the limit that R1>>1/a so it's not absurd even in your model.
Now I do understand your model. I worked out the equations and got the same results as you, except I'd write them in the form
T(x) = T_infty + (T0-T_infty) cosh a(R1-x) / cosh a(R1-R0)
and
J(x) = ak (T0-T_infty) sinh a(R1-x) / cosh a(R1-R0)
where J(x) is the heat flux (W/m^2).
Notice that if R1 >> 1/a, this simplifies,
T(x) = T_infty + (T0-T_infty) exp[-a(x-R0)]
J(x) = ak (T0-T_infty) exp[-a(x-R0)]
However, you did not take into account the circular geometry. As near as I can tell, that's what you're talking about in the exchange above. Your
derivation applies to a long, rectangular fin attached to a wall. For your actual circular fin, the solution is a linear combination of the modified
Bessel functions. I'll use r here instead of x for the radial coordinate. The solution is
T(r) = T_infty + A K_0(ar) + B I_0(ar)
J(r) = -ka [ A K'_0(ar) + B I'_0(ar)]
where constants A and B can be determined by the boundary conditions, T(R0) = T0, J(R1) = 0. Again, if R1 >> 1/a, there is a simplification,
in that the coefficient B=0 (the solution is purely in terms of K-type Bessel functions).
I'll bet the difference between the simpler exponential-type solution and the more complicated Bessel function solution would be important in
practice, that is, it's worth it to use the Bessel functions.
[Edited on 10-5-2015 by annaandherdad]
Any other SF Bay chemists?
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blogfast25
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@AAHA:
First you manage to misinterpret the problem (and think I’m insulting you when I point that out!)
Next you tell me you’ve arrived at the same conclusion as me but that simply isn’t true. (How could they be the same if I didn’t take the
circular geometry into account, huh? Think about that for 5 minutes)
e<sup>ax</sup> + e<sup>a(2R1-x)</sup> is not cosh(a(2R1-x), for instance.
Then you tell me I’ve not taken the circular geometry into account but I can assure you I have. I would not have formulated the problem as
“A single radial cooling fin (circular plate, hot tube runs perpendicularly through centre) mounted on a tube with radius R0. Outer radius
of fin is R1, uniform fin thickness is t. With x the radial distance from the centre of the tube”, if I hadn’t taken the geometry into
account, that should go without saying really.
In the setting up of the differential equation, which is the heat balance of a thermostatic infinitesimal element, the circular geometry is taken into
account. Due to symmetry it simplifies a lot, though. Symmetry is a mathematician’s best friend.
Bessel functions are not needed here. If t (fin thickness) was a function of x (say, t = t(x)) then they would pop up (first hand
experience, BTW). Very annoyingly too.
Post script :
I understand you want to help and much appreciate that. I therefore kindly ask you to post your derivation as an attachment (*.doc or *.pdf) and I
will do the same with mine. I think this is the only way forward.
I look forward to reading it.
[Edited on 10-5-2015 by blogfast25]
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