Fourier's Law: please tell me where I am in error

I'm not specialist of this discipline, but to my mind here's two controversial things that I've noticed at once skimming your post.

First: isn't pod thickness too thin? I would supposing that one should be at least 10 times greater.

Second: Fourier's law equation is just differential one. So it should be integrated properly. You have initial condition for difference of temperatures, this temperature difference shall decrease gradually to direction of heating stream. So it will be take less power consumption every time moment.

Third: to be more precisious you should take in account ambient air conditions as media of heat spreading as well unless you cooking in vacuum.

Hope it helps. Hope someone here with more experience at heat transfer field will deliver you more profound answer.

I double-checked the figure that I used for pot thickness; I don't have calipers, so I had assumed thickness of 1/16", which I think is reasonable because this is a very light pot, and after converting it to meters, the 0.0015875 figure is correct.

On your second point, I am not trying to figure out how long it takes for a pot to cool off after being removed from the burner -- although that would be a handy formula to have for when I am cooling the wort -- so the temperature differential never really changes; i.e., the contents of the pot remain at 212F during the 90 minutes that it is boiling, and the ambient temperature outside the pot remains the same, as well.

I don't understand your third point unless what you mean is what a few others have already corrected me about; i.e., that a layer of warm air forms around the pot, and that I should be using a formula other than Fourier's. I'm still trying to figure out the other replies I have received which appear to address that issue. Naturally, the biggest mistake I made was in not converting Fahrenheit to Kelvin before subtracting the two temps to get the temperature delta.

I am astounded by how helpful this group is; many thanks to you and everyone else.

Cheers.

Bill Velek

No - you are correct in using that Fourier's equation - for ONE part of the puzzle. Note that there are actually three different heat problems here.

For a good fundamental reference on thermodynamics - download the thermo handbooks from the DOE website.

What you are particularly interested in is explained (fundamentally) in Volume 2 page 20.

Your other question regarding the cooling of the pot off the burner is a transient heat problem and is too advanced for those handbooks.

Yes you're right --- thickness 1.5 mm is likely enough [I was wrong with mine miscalculations at deep night time getting thought it has foil thickness some 0.15mm].

For first approach you could disregarded influence of air as well. Formula is the same.

I just get home. I hope cr4 community has helped you already.

In my first post it should be read: "...this temperature difference shall decrease gradually opposite to direction of heating stream.

I do agreed with Guest comment regarding spec of equation for hollow cylinder.

Differential Equation is [what i could find]

dT=-P/[2pi*l*k]dr/r,

where r-radius; l- cylinder's length, P - what you assigned as J/sec at your post.

In agreement with Guest #1 above, your analysis is somewhat correct, somewhat incorrect (more towards the incorrect side as you have realized).

The equation you used is for heat conduction through a material with a driving force of temperatures - and in your analysis you used 70 F as the lower temperature, which is wrong.

In your situation, the outside temperature of the aluminum will be almost 212 F - with the main driving force of heat flow being convection of air against the wall.

Without doing a full thermal analysis (I am at work and I don't have time) - here is a rough idea for you:

Taken from my trusty piping handbook for heat losses from bare surfaces - a hot surface temperature of 93 C and ambient temp of 21 C (still air conditions) - you will loose approximately 930 W/square meter. Note that your hot surface temperature with such a low wall thickness may be higher.

Therefore - you are going to loose approximately 930 * 0.2919 = 272 watts due to conduction and convection (and perhaps a little radiation).

If you want a more detailed answer - just reply and I will see if I can provide better numebrs tonight at home.

I do believe that the heat transfer co-efficient is quite different from the Thermal conductivity coefficient, and as such the division by Th is no longer necessary as it has already been usually accounted for in the heat transfer coefficient. Also because your pot is not very high the axial variation of temperature on the outside surface of the pot that would follow due to the natural convection on which the 'k' is based can be ignored - meaning dT/dz along the pot: from bottom to top; is zero, and all heat losses are radial.

So try Q = kA(212 -70) making sure that the units are consistent; and A is πDh.

Good catch - there is a big difference between heat transfer coefficient and thermal conductivity. The overall heat transfer coefficient takes into account the convection, thickness, and thermal conductivity.

So therefore, the question to the original poster is -

Where did the value of 109 come from - in your post you mention both heat transfer coefficient and 'k' - note they are different. And most importantly what are the units of that 109 value? W/m²•K or W/m•K

If the value of 109 is TRULY an overall heat transfer coefficient (I would have to ask based on what fluid convection properties, thermal conductivity, and thickness) then the equation Q= UA(ΔT) is the right equation. NOTE that in pure technical terms - that 'k' should be noted as U for heat transfer coefficient.

billvelek:

More simply stated, your problem is in assuming that the temperature gradient occurs through the aluminum pot wall. That gradient will be near zero because of high aluminum conductivity. The real one, to be analyzed, will be through the adjacent air, not metal.

Both sides of the pot wall are at nearly the same temperature (212 degrees) because they are closely coupled by the thin, highly conductive aluminum.

Best regards.

DickL

I do agree with some of the posts before... the main problem is that the "model" you are using is not fitting the real situation...

In your case you can model using sum of two different layers:

the first one where the conduction occurs, having the thickness of the pot and applying Fourier Law; Q = k ( Tinside - Twall ) / z Note that k is W/m K

the second one where the convection occurs, having a Q = h ( Twall - Tair ) A

Note that h is W/ m2 K

Now, considering that k will be really high for metals and z is low (your pot is really thick), I would assume that Twall is more or less equal to Tinside... so resistance to conductivity maybe could be negligible... it depends on how accurate you need!

Then you can consider just convection Q = h (Tinside - Tair ) A

Again, this is just estimation because you have not a flat surface... be careful in h estimation!!! With temperatures higher than 300°C is not wrong to consider even radiation...

I believe your error is trying to address this problem as a straight conduction one rather than a coupled heat transfer problem of conduction and convection (occuring on the outside of the lid). I would suggest you look at "Process Heat Transfer" by Donald Q. Kern or a similar text in chemical engineering on Heat Transfer (eg: McAdams text)..