HT model of insulated Cylinder in ambient air

[UBbaja]

Hi everyone. I am trying to model my design and could use some help.

I have a cylinder that has a exothermic chemical reaction occurring within it. On the ID of the cylinder there are 3 layers of insulation. The OD is exposed to amb air. I have started to model it but I am having difficulty understanding everything with respect to time regarding the convective HT on the OD and the exothermic reaction occurring on the inside. I can assume the exothermic reaction is always producing a given temp, then model the conduction through the insulation and then the cylinder steel wall. At that point then the radiative and convective HT kicks in. With all of this happening within the realm of time, I am having trouble setting up the equations. Any advice?

need more info?

any ideas on how to find the "h" HT coeficient provided by the reaction?

 

[IRstuff]

You need to provide more detail as to what trouble you are having. Fundamentally, you are solving

heat.cond = heat.conv+heat.rad

for the temperature that balances the equation.

TTFN

FAQ731-376

 

[UBbaja]

Sorry for my delayed response. On the outside of my cylinder I know the surface temp as well as the surrounding environment temp. Using these I can calculate q, the rate of HT. Therefore I am looking at the boundary conditions of the outside environment and the cylinder surface. Between these two mediums heat is being transferred via radiation and natural convection.

Now I would like to look at my next boundary layer, the cylinder ID to an interior liner, both having radii, r3 and r2 respectively. I assume the cylinder wall to be very thin, so the temp on the cylinder ID = temp at cylinder OD. Across this air gap between r2 and r3, I assume no convection because its so small, I am only considering radiation from r2 to r3 and conduction across the air gap. My knowns are

r2,r3, Temp at r3, k air, emissivities of both r2 and r3. My unknown is the temp of r2. To solve this can I use the q I found earlier? Can I continue to use this q to eventually find the temperature at my inside most radius, r1? Between r1 and r2 is a layer of insulation where only conduction occurs. At r1 is where the heat is generated. I do not know the local "h" associated with r1. I did the calcs assuming the q i achieved from the outside boundary layer throughout the layers. I assume q to be constant across all layers becuase I assumed zero energy storage within the layers, as I am looking at this at an instant in time. Is this a correct assumption?

Thanks for the insights!

 

[zekeman]

Three questions
1) why do you need temperature -time response (transient)and
2) where doers this air gap come from?
Is it part of a design or your idea of the imperfect attachment of insulator to wall?
3) how long does this chemical reaction last and what temperature are you assuming at the inner wall?

 

[UBbaja]

Three questions
1) why do you need temperature -time response (transient)-->I don't think that it is appropriate to assume 2D steady state. The reaction begins at one and continues until the end, much like a strip of paper burning.
2) where doers this air gap come from? -->The air gap comes from a .003 cylinder of nickle. Based upon the 2D analysis it is better to have this nickle cylinder and therefor the air gap instead of just a layer of vermiculite insulation.
^^Is it part of a design or your idea of the imperfect attachment of insulator to wall?-->uhhh....what?
3) how long does this chemical reaction last and what temperature are you assuming at the inner wall? The reaction lasts about 14 minutes. I am assuming ~1900 DEG F on the inner wall. I haven't had a chance to play with the equations more, I know from experiments that the inner temp is more like 1200 DEG F, thus telling me my assumptions like the k values and e values are a bit off, the only question is which ones to tweak to understand it better.

 

[zekeman]

If you are looking for the maximum outside temperature, you have an essentially steady state problem in one dimension and your book gives you the answer.
The reason is that radially and longitudinally it takes the order of seconds to reach steady state and the reaction is going on minutes
If you think .003" nickel gives you an air gap, it doesn't.

So to solve the problem you have.


The solution is
To+(Ti-T0)*(1/2pi*h*R)/[(sum of insulator resistances+1/(2Pi*h*R)]
where
R= outside radius of insulated cylinder
To= ambient temp
Ti=inner wall temp
h- convective film conductance, outside surface-to ambient, (about 2.5 BTU/HrFT^2deg F

 

[UBbaja]

Thanks Zekeman, I think that's the same conclusion I came to. What I am trying to understand is what is going on in 3D. Conduction longitudinally doesn't happen in a matter of seconds. For example, at t=30s the outside diameter temperature is up around 450 DEG F at z=1" (longitudinally along the axis of 6.5" x 2.75" cylinder). At this point in time, t=30s, the OD temperature at z=4" is only 100 DEG F. While the conductive HT happens rather quickly due to the coefficient of conductivity of stainless steel, the effect of this mode of HT is moot because as it conducts down the cylinder it is also being transmitted off the surface due to natural convection as well radiation. This happens due to the small effective surface area that the energy can transfer thorugh (the wall is only .008"). As time progresses and the reaction moves down the length of the z axis (longitudinally) the "hot spot" on teh OD doesn't really move much because while the reaction moves, the earlier points on the z axis are cooling off because the hot reaction is over with at that point.

Our limit is 500 DEG F, we currently fail this ~30% of the time. We know these failures are therefore due to variation. We brainstorm that our variation can come from 3 places; mechanical design tolerances, temperature measurement variances, or variation from sample to sample of the reaction taking place.

I am running tests to quantify the variation which is coming from the measurement apparatus/process. It looks like thus far the variation due to this is +/- 36 DEG F, 42% of overall variation (our overall variation ranges between +/- 86 DEG F overall). We found that by changing our method of measurement we can reduce this variation due to measurement down to +/- 7 DEG F. This should therefore should reduce the overall variation down to +/- 50 DEG F, with variation due to measure being 14%.

An option also is to try and find a design change that will reduce this variation "mechanically." My overall goal is to reduce the OD surface temperature in this case. My thought process, based on my 2D analysis, is to effectively lower overall surface temperature. Based on the analysis, by lowering emissivity of the ID it will significantly lower the outside shell temperature (by ~30% all else held constant). I am thinking about doing this by a copper liner that would interface directly with the cylinder ID. Will this also eliminate "hot spots" better than just the regular stainless steel due to the copper's thermal conductivity coeff which is 4x greater?

The thrid source of variation, the reaction, is currently undergoing tests to understand the difference in energy output from sample to sample.

After understanding the potential variation from these three areas, we can look at which method of reducing the variation would get us the "biggest bang for the buck."