Heat Loss of the Vessel Skirt 1

[ivana24]

Based on IEEE515 I caculated Heat Loss of the vertical Vessel (natural convection, forced convection, radiation etc.)
My problem is vertical vessel skirt which is as high as vessel, where high heat losses can be expected.
My skirt is a large, cylindrical skirt with large surface area.
how to calculate these heat losses or approximate ( not given by IEEE 515).
Thanks.

 

[Flareman]

Ivana24
If you are doing a complex heat loss calculation from a vessl, you are probably already using the thermal conductivity of something (insulation ?) in an interative manner to "home in" on the temperature profiles on the horizontal plane through the complex wall. You simply have to build in a third dimension, down, starting with the stable equilibrium temperature at the bottom of the section you already know about. The complicated bit is that the temperature along the skirt will decrease incrementally as the heat flowing through the shell thickness at the upper end reaches the lower elevations and is lost in some relationship to the temperature which it achieves. I haven't ever needed to do this calculation so don't have a handy quick solution, but if you treat it like a shell and tube heat exchanger you can probably come up with a "log mean temperature difference" approach, remembering that the energy input is confined to that which can flow downward through the metal wall at the top end of the skirt.

If you're lucky, somebody already did this and will be nice enough to send you a spreadsheet.

David

 

[gr2vessels]

ivana24,
Is the 24 related to someone's age?
Are you sure that you need to calculate the heat loss of the supporting skirt, welded to the vertical pressure vessel? Is this related to some IEEE 515 heat tracing of the vessel? Are you are worried about the heat loss from vessel to the skirt, through the few millimiters of skirt plate welded to the bottom of the vessel?
Sorry, this remindes me of the day when I was sent to bring some elbow grease to the MIG welder....
gr2vessels

 

[ivana24]

gr2vessels,
vessel skirt ( cilindrical is welded to vessel, but skirt has surface area almost the same as as that for the vessels
are you sure that that heat loss can be neglected.
Sorry, this remindes me that gr2vessels is making joke with my question.
(i am senior enginer not of thermodynamics, 56 yrs old not 24))

 

[gr2vessels]

My apologies again Ivana,
From the short description of your vertical vessel, I can only assume that the heat transfer area between the vessel and skirt is the thickness of the weld. That heat loss to the skirt is probably insignificant, compared with other heat losses from the vertical shell (through the insulation, for instance).
Cheers,
gr2vessels

 

[jte]

Ivana-

I presume that the skirt has fireproofing at least on the outside. As gr2vessels has implied, the bottleneck is the skirt to head/shell weld. I'll presume that you're worried about loss of heat from the process fluid. Without getting an exact solution, you can easily get an upper bound on the problem by figuring out how much heat you can possibly pass through the weld. Limited by cross sectional are of the weld (no more than the thickness of the skirt x circumference), conductivity of the steel, and delta T no more than the process fluid temp minus a reasonably low ambient temp for the neighborhood.

Or, more simply, neglect it. You'll lose far more from the shell. Fireproofing is a reasonably good insulator...

My rule of thumb when lacking other info: You'll drop 100 deg F per vertical foot of skirt for the first few feet.

jt

 

[prex]

The way of attacking this problem is by considering the skirt as a unidimensional bar heated at one extremity, with constant cross section and infinite length.
Under such conditions the thermal flux (equivalent to what is calculated as hA[Δ]T elsewhere) is (from any heat transfer book):
q=kAn[Δ]T
where
k=thermal conductivity of skirt material
A=2[π]Dt=cross section of skirt
D=skirt diameter
t=skirt thickness
[Δ]T=temperature difference between vessel and ambient
n=[√](hP/kA)
P=4[π]D=exposed perimeter (considering both faces of skirt as effectively exchanging)
h=heat exchange coefficient from skirt surface to ambient
The key factor is of course h: this should account for a higher than ambient temperature at inside of skirt, and for any fireproofing or insulation covering the outer face (the initial portion of skirt covered by vessel insulation may be neglected, as essentially at uniform temperature).
I agree with the comments above that the heat loss through the skirt should be comparatively small if the skirt is fireproofed, however the length of the skirt plays a role in this case, and it all depends on what's considered as small by the owner...

prex

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[prex]

Oops! Of course 2[π]Dt and 4[π]D above are respectively [π]Dt and 2[π]D

prex

http://www.xcalcs.com

: Online tools for structural design

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads