Sciencemadness Discussion Board - Radial Cooling Fin: temperature profile and heat loss

Thanks for your reply. What do you mean by 'temperature rise' here?
[Edited on 5-5-2015 by blogfast25]

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]

blogfast25 - 5-5-2015 at 10:35

Quote: Originally posted by Sulaiman

each thick enough only to minimise temperature gradients within the metal

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]

Magpie - 5-5-2015 at 10:36

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.

Sulaiman - 5-5-2015 at 11:01

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.

blogfast25 - 5-5-2015 at 11:43

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.

aga - 5-5-2015 at 12:04

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.

Sulaiman - 5-5-2015 at 12:15

@sulaiman: All other things being equal, a fin that's twice as thick will dissipate twice the heat, for instance.

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.

blogfast25 - 5-5-2015 at 12:15

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.

Hmmm... silicones are generally poor conductors of heat, you know? Mixing with metal powder would increase k, or so I'd imagine...

blogfast25 - 5-5-2015 at 12:22

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.

Correct. Multiple thinner fins amounting to the same thickness as a single one are more effective.

[Edited on 5-5-2015 by blogfast25]

aga - 5-5-2015 at 12:29

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]

Magpie - 5-5-2015 at 12:59

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.

blogfast25 - 5-5-2015 at 13:14

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.

Quote:

Conductivity: 0.9W/mk

Hmm. Al is about 200 W/mK.

Of course you only need the thinnest layer of it.

annaandherdad - 5-5-2015 at 14:08

Quote: Originally posted by Sulaiman

each thick enough only to minimise temperature gradients within the metal

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.

blogfast25 - 5-5-2015 at 17:28

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.

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∞ is simply the ambient temperature. The outer edge of the fin would only reach T∞ for x = ∞.

The differential equation requires a second boundary condition: (dT/dx)x=R1 = 0.

The undetermined solution of the DE is:

T(x) - T∞ = C1eax + C2e-ax

Both constants are determined from T(R0

= T0 and (dT/dx)x=R1 = 0.

Does that clear it up, AAHD?

[Edited on 6-5-2015 by blogfast25]

annaandherdad - 6-5-2015 at 11:45

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.

blogfast25 - 6-5-2015 at 12:40

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∞ but that simply isn’t true.

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∞ 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 = R1, the temperature gradient (dT/dx)x=R1 = 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 R1 h t (T1 - T∞

. For very 'stubby' fins that would be a concern.

[Edited on 7-5-2015 by blogfast25]

aga - 7-5-2015 at 10:09

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]

blogfast25 - 7-5-2015 at 10:19

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]

annaandherdad - 7-5-2015 at 13:21

. 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.

blogfast25 - 7-5-2015 at 13:40

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∞, which is manifest nonsense.

[Edited on 7-5-2015 by blogfast25]

annaandherdad - 9-5-2015 at 15:35

AAHD:
So far you've only managed to dodge your erroneous judgement that edge of the fin is supposed to be at T∞, 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]

blogfast25 - 10-5-2015 at 07:51

@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)

eax + ea(2R1-x) 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]

smaerd - 10-5-2015 at 08:11

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]

blogfast25 - 10-5-2015 at 09:37

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?

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]

annaandherdad - 10-5-2015 at 10:50

First you manage to misinterpret the problem (and think I’m insulting you when I point that out!)

You did.

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.

eax + ea(2R1-x) is not cosh(a(2R1-x), for instance.

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.

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.

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.

annaandherdad - 10-5-2015 at 11:03

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]

blogfast25 - 11-5-2015 at 09:00

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:

d2y/dx2 – a2y = 0

And not of the type:

d2y/dx2 + (1/x)dy/dx – c2y = 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!

annaandherdad - 11-5-2015 at 09:32

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).

annaandherdad - 11-5-2015 at 10:32

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]

aga - 11-5-2015 at 11:39

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.

blogfast25 - 11-5-2015 at 12:28

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.

aga - 11-5-2015 at 14:46

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]

Magpie - 11-5-2015 at 14:54

Proof please, if anyone would be so kind.

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.

)

blogfast25 - 11-5-2015 at 16:20