[TRNSYS-users] Heat losses in horizontal storage tank

[thornton at tess-inc.com]

On 2022-12-14 12:37, Adrian Riebel Brummer via TRNSYS-users wrote:
> I would like to ask a question about Type 533: Horizontal Cylindrical
> Storage Tank With Immersed Heat Exchanger.
> 
> My question is: how are the heat losses computed, given that the
> difference between "top", "edge" and "bottom" are not so clear in a
> horizontal tank, in contrast with a vertical tank? The documentation
> is not very clear in this regard. Actually, the information about heat
> losses is exactly the same for vertical and horizontal tanks.

Type 533 has been replaced with Type 1533 in the latest version of the 
libraries and the documentation updated.  The Type 533 and 1533 models 
have left and right surfaces (circular geometry Pi*D^2/4), and edge 
losses (this is the surface with total area = Pi*Diameter*Length).  For 
all tanks, the individual surface losses are based on the loss 
temperature for that node/surface, the tank nodal temperature, the edge 
loss coefficient for that node, and a calculation of the surface area 
for each surface for that horizontal tank node using a horizontal 
cylinder geometry:

Qloss = U * SA * (Tnode - Tenv)

The surface area for each horizontal, iso-volumetric tank node, is 
calculated based on the tank geometry and the number of nodes.

Jeff


-- 
----
Jeff Thornton
President
Thermal Energy System Specialists (TESS)
3 N. Pinckney Street - Suite 202
Madison WI 53703
(608) 274-2577
www.tess-inc.com