Saddle heat transfer calculation

[abehong]

Hello:

We have a very hot vessel (96”) diameter pipe (800 deg F) supported by two saddles. Is there a way to determine the heat dissipation or how long the supporting steel has to be to avoid over heating Teflon sliding plate installed underneath floating saddle?

I did a research on this forum. Someone proosed to calculate heat dissipation calculation using conduction heat transfer theory , but I am not sure how to determine conduction heat loss rate for carbon steel . It should be a somekind standard conduction heat loss rate .

Thanks

Abe

 

[gr2vessels]

Abe, it's not a simple thing to calculate the heat transfer from the shell to the wrapper, then to the web, baseplate and then to the teflon pad. You need to consider the conductivity of the steel, air between the shell and wrapper, assume a number of contact points between the shell and the wrapper for conductive heat transfer, then the heat transfer between the wrapper and the web. From there on, you need to assume the heat loss to the environment (mostly radiation) from wrapper and the web. By the time you get to the baseplate, the heat is all but gone. Make sure you don't insulate the web and anything below the wrapper. You might use a reflective plate to protect the baseplate from the radiated heat.
As far as the calculation goes, there are so many variables, that you might end up far away from reality. If you insist on the calculations, assuming the saddle is very low, try the above indications. You might have to insulate the baseplate from heat radiation.

 

[chicopee]

About showing a picture or a sketch of the saddles with the Teflon sliding plates so that it can be visualize because your description"... how long the supporting steel has to be..." confuses me; Also teflon cold flows under pressure and with a vessel @800dF would make the sliding plates extremely thin under the tank.

 

[prex]

I would solve it this way.
Consider the saddle web as a bar heated at one end and composed of two zones: the part covered by the insulation, where the temperature profile is linear, and the rest that releases heat in the air.
Let's define some symbols: T[sub]1[/sub]=pipe temp.,T[sub]2[/sub]=temperature at the interface between the two zones, T[sub]a[/sub]=room temp., k=th.conductivity of saddle mat'l,A=area of the cross section of web (assumed constant,equal to width x thickness), L[sub]1[/sub]=length of first zone (insulation thickness), h=heat exchange coeff.to air (take 10 w/m[sup]2[/sup]/degC), p=perimeter of web exposed to air (two times width+thickness).
Assuming the part exposed to air has infinite length (safe assumption), the thermal flux exchanged is:
q=kAn(T[sub]2[/sub]-T[sub]a[/sub]) where n=[√](hp/kA)
The same heat flux flows in the first zone:
q=(T[sub]1[/sub]-T[sub]2[/sub])kA/L[sub]1[/sub]
The temperature distribution in the second zone is:
T-T[sub]a[/sub]=(T[sub]2[/sub]-T[sub]a[/sub])e[sup]-nx[/sup] where x is the distance along zone 2.
Now by equating the two thermal fluxes you can eliminate T[sub]2[/sub] and from the last equation calculate the x where you reach a temperature acceptable for the teflon pad: if you have enough length of web exposed to air you are OK.
Of course all this is based on very rough assumptions, but you get a safe result and normally a few centimeters of exposed web could suffice, so this could be what you are looking for.

prex
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