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smaerd
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I'd like to see some math! Or a picture explaining the problem. AAHA and you could be looking at two different problems.
Anyways this really sounds like a job for finite element simulation. If I was better with ELMER FEM a numerical solution for this could probably be
found very quickly.
[Edited on 10-5-2015 by smaerd]
Edit - http://www.syvum.com/cgi/online/serve.cgi/heat/heat1003.html Does this have the temperature profile you are looking for?
[Edited on 10-5-2015 by smaerd]
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blogfast25
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Quote: Originally posted by smaerd  | I'd like to see some math! Or a picture explaining the problem. AAHA and you could be looking at two different problems.
Anyways this really sounds like a job for finite element simulation. If I was better with ELMER FEM a numerical solution for this could probably be
found very quickly.
[Edited on 10-5-2015 by smaerd]
Edit - http://www.syvum.com/cgi/online/serve.cgi/heat/heat1003.html Does this have the temperature profile you are looking for?
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The geometry is simple enough not to require finite element analysis, although a quick numerical solution would be useful, of course.
I think AAHD and I are now on the same wavelength re. the geometry. It's the geometry outlined in your link (thanks for that!)
That page's approach is slightly different from mine and at first glance correct. Damned Bessel functions though!
[Edited on 10-5-2015 by blogfast25]
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annaandherdad
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You did.
| Quote: Originally posted by blogfast25 |
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)
[Edited on 10-5-2015 by blogfast25] |
Your solution is valid for a long rectangular fin attached to a wall. I get the same solution for this geometry. If I assume a circular geometry,
the solution involves Bessel functions. Bessel functions are normal in circular geometry.
I didn't say it was. I said my formula was equivalent to yours. Note that my formula involves cosh a(R1-x), not cosh a(2R1-x). Factor out 2 exp aR1
from your numerator and denominator, and it becomes my formula. I don't claim that this is a big deal. I'm agreeing with your solution completely,
but only for rectangular geometry.
| Quote: Originally posted by blogfast25 |
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.
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You described the problem verbally as involving circular geometry, but you did not show your derivation, only the solution. Your solution applies in
rectangular geometry, not circular.
I will write this up as a pdf and send it. In the meantime, I have a challenge for you, if you think you solved the problem in circular geometry.
Work out the problem in rectangular geometry. Then either show me how the solution is different in rectangular geometry, or else claim to me that the
geometry makes no difference.
Any other SF Bay chemists?
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annaandherdad
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| Quote: Originally posted by smaerd | | I'd like to see some math! Or a picture explaining the problem. AAHA and you could be looking at two different problems. |
[Edited on 10-5-2015 by smaerd]
I was looking at a different problem at the beginning. I didn't understand the physical model b25 was using. Now I do understand it. He is using a
phenomenological model for the convection from the surface, which assumes that the local heat flux (W/m^2) due to convection is proportional to the
difference between the ambient temperature and the local temperature of the fin. It's a reasonable model to begin with for a fin in quiet air.
[Edited on 10-5-2015 by annaandherdad]
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blogfast25
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Ok, mystery solved.
Although my derivation did take circular geometry into account, due to an error in setting up the infinitesimal I obtained a DE of the general type:
d<sup>2</sup>y/dx<sup>2</sup> – a<sup>2</sup>y = 0
And not of the type:
d<sup>2</sup>y/dx<sup>2</sup> + (1/x)dy/dx – c<sup>2</sup>y = 0
As coincidence would have it, the first type leads to a solution similar of that for a rectangular fin (but my definition of a is
different, so not exactly the same either).
So yes, modified Bessel functions pop up. Pity!
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annaandherdad
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If you mean that your "c" in this post is different from your "a" in your earlier post, then I disagree. I get c=a=sqrt(2h/kt), same as your
definition of a. This is shown in my earlier post in which I gave the solution in terms of the modified bessel functions I_0(ar) and K_0(ar).
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annaandherdad
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A strange feature of this model and solution is that it assumes that the heat flux is purely in the radial direction (or x-direction, in the
rectangular geometry), which would make it tangent to the surface of the fin. However, heat is radiating away from the fin, so by conservation of
energy, the heat flux must have a component normal to the surface, just below the surface, which must equal the heat flux transferred convectively
away from the surface. Evidently this normal component must be small compared to the component tangential to the surface, in order for the model to
be valid.
Also, since the heat flux inside the fin is normal to the contour surfaces of temperature T, the contours surfaces inside the fin must be bowed
outward slightly (in the x-direction) in the middle of the fin, that is, the temperature is slightly higher in the middle of the fin. This is logical
physically, but in order to account carefully for conservation of energy it is interesting to see how much the contours are bowed.
Analyze this in rectangular geometry, for simplicity. Center the fin on the x-axis, with the y-axis vertical and orthogonal to the surface of the
fin. The fin lies in the x-z plane. A solution of the Laplace equation inside the fin (|y| <= t/2) is
T(x,y) - T_infty = cos(ay) [A exp(ax) + B exp(-ax)],
for constants A,B. Only cosine terms are kept in y because of symmetry we must have T(x,y) = T(x,-y). This is a solution of the Laplace equation for
all values of a, but it turns out that for the solution we want, a=sqrt(2h/kt), as defined by b25, so I'm not redefining symbols here. But this has
to be shown.
The flux vector is given by J = -k grad T, or J_y = -k partial T/partial y. Evaluating this at y=t/2, you get
J_y(x,t/2) = ka sin(at/2)[A exp(ax) + B exp(-ax)].
But by the model for the conductive layer, this must equal
J_y(x,t/2) = h (T-T_infty).
Combining equations then gives
h cos(at/2) = ka sin(at/2),
or,
a tan(at/2) = h/k.
Graphically it is easy to see that this equation has an infinite number of solutions in a, but we are interested only in the smallest one. If we
assume that at << 1, then this solution is given by approximating tan x approx = x, or
a^2 = 2h/kt,
same as earlier definition. The condition at << 1 is inherent in the model used here. It is physically reasonable if you think about it, but
it doesn't come out in an obvious way in a simplified derivation in which you just balance energy flows into and out of an element of the fin.
If at/2 << 1, then ay/2 << 1 for all y inside the fin, and the sin and cos of ay/2 can be approximated. This gives
T(x,y) - T_infty = [1-(1/2)(ay)^2] [A exp(ax) + B exp(-ax)]
showing the bulge in the middle of the temperature profile. The temperature contour is roughly parabolic, and the center is hotter than the surface
by an amount,
T(center) = T(surface) + [T(surface)-T_infty] (at)^2/8.
The same formulas for the shape of the temperature contours inside the fin are obtained in circular geometry.
In this way conservation of energy is accounted for.
[Edited on 11-5-2015 by annaandherdad]
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aga
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Theory and Maths are wonderful.
The maths are frankly beyond my means.
Experimentation however would prove/disprove the mathematically derived claims, and measuring Heat is pretty much within our grasp.
Make a heatsink (bit of metal), heat it (candle), measure the temperatures at various points and test if the maths work as expected.
Now the results of that'd be interesting.
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blogfast25
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| Quote: Originally posted by aga | Experimentation however would prove/disprove the mathematically derived claims, and measuring Heat is pretty much within our grasp.
Now the results of that'd be interesting. |
The math is applied to physics that's as safe as houses and equally well established.
The math itself is routine (advanced calculus).
Some assumptions we know are not 100 % correct:
1. h and k may not be very accurately known and not completely independent of temperature. But you'd use a different set up to measure that.
2. Convection at end of fin is small but not zero.
3. Radiation was neglected. Only works at reasonably low T. Model could be modified for that though.
Experimentation (not w/o problems either) is very unlikely to unearth something new/wrong here.
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aga
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Experimentation would at the very least provide some Factual Foundation to the maths being spouted.
A complex Conduction/Convection/Radiation system following a hyperbolic cosine function almost perfectly ?
Reality seldom does that.
Proof please, if anyone would be so kind.
[Edited on 11-5-2015 by aga]
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Magpie
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Aga, what tolerance would you allow to consider the experimental results to verify the math?
I didn't realize that this was just an academic exercise (mental masturbation*). I thought blog25 needed this result for practical purposes.
*(I indulge in this occasionaly myself.  )
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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| Quote: Originally posted by aga | Experimentation would at the very least provide some Factual Foundation to the maths being spouted.
A complex Conduction/Convection/Radiation system following a hyperbolic cosine function almost perfectly ?
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As Magpie also hinted, you're somewhat pissing the point.
Experimental verification of the temperature profile has measuring error. It's an estimate of the 'true' curve, requiring several replica measurements
and some nifty math statistics to obtain it. Uncertainty on it is inherent.
Where a model is based on established science and math, experimental verification is not required, only verification of HOW the model has been arrived
at. Some of its weaknesses we already know.
I did. It's seems I probably don't anymore though.
[Edited on 12-5-2015 by blogfast25]
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annaandherdad
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| Quote: Originally posted by smaerd | Anyways this really sounds like a job for finite element simulation. If I was better with ELMER FEM a numerical solution for this could probably be
found very quickly.
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Hi, smaerd, you would use a finite element computer code to solve PDE's such as the Laplace equation in an irregular geometry. The temperature
satisfies the Laplace equation Del^2 T = 0 in equilibrium in any homogeneous medium. But blogfast's model is simplified, and doesn't require the
solution of the Laplace equation. The simplification is possible because the fin is thin, so you just assume that the temperature is constant across
the thin direction of the fin, and that it only depends on the radius (for the circular fin). However, I did solve the Laplace equation in my
previous note, in order to understand better the flow of heat inside the fin and to resolve an apparent paradox. I did not use a computer code to do
this because the Laplace equation is separable in rectangular geometry and can be solved analytically.
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smaerd
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Edit -
AnnaandHerDad - Woops misunderstood what you were saying. Yea there is some radial symmetry which doesn't necessitate FEM, but I'm lazy.
[Edited on 13-5-2015 by smaerd]
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