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Re: [TRNSYS-users] Heat losses in horizontal storage tank



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


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Jeff Thornton
President
Thermal Energy System Specialists (TESS)
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