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Fin tube heat absorbtion

Q: I am designing a containment structure (box) approximately 5 ft (H) x 3 ft (W) x 8 in (L) that will be exposed to direct sunlight. I would like to internally include a heat exchanger built with fin tubes that would remove/absorb the heat in the container in the box itself. I have seen calculations from fin tube suppliers that can relate the amount of Btus being dissipated in a foot of fin tube, but not a calculation for the amount of heat that could be absorbed in a contained environment. When I see that a foot of fin tube might dissipate 900/1,100 Btu hours, is there a simple calculation or factor that could be applied to determine what that same foot of fin tube would absorb? In other words, if a foot dissipates 900 Btus, absorbtion might be 80% or 90% of the 900 Btu's in that closed container. I would appreciate any information you can provide.

A: The amount of heat which can be removed in a finned tube is dependent upon the summation of resistances to heat transfer and the temperature difference between hot and cold fluids. Typically for a finned tube in this application, there is some fluid (refrigerant perhaps) that flows through the inside of the tube and a gas that is either forced to circulate across the tubes or flows across the tubes due to natural convection. Most of the time, it is the heat transfer on the outside of the tubes that represents the greatest resistance and is controlling the dissipation or absorption of heat.

In your application, it appears that there is no forced flow of gas across the finned tubes, thus, resulting in a natural convection flow of gas across the finned bundle. The rate at which heat is absorbed by the coolant is measured in Btu per hour and would be calculated by using the following equation:

Q (Btu/hr) = U x (Tb-Tc) x A

Where U is the heat transfer coefficient in Btu/hrft²°F; Tb is the enclosure temperature; Tc is the coolant temperature (flowing through the tubes) and A is the surface area of the finned tube. For natural convection cases, the heat transfer coefficient is usually quite low and may fall in the range of 0.5 – 5.0 Btu/hrft²°F. As an example, consider a 5/8 in o.d. tube with 1/4-in high fins spaced at 8 fins per inch. This style of tube would have 6.7 ft² of finned surface per feet-square of bare tube, resulting in a total of about 1 ft² of surface area per foot of tube length. The amount of heat absorbed per foot of tube could therefore be estimated as follows assuming a mid-range value of U (2.5 Btu/hrft²°F):

2.5 Btu/hrft²°F x (Tb-Tc) x 1 ft² = 2.5 (Tb-Tc) Btu/hr

If the coolant is at 40°F and the enclosure is at 140°F then the heat removal rate would be on the order of 250 Btu/hr. Significantly higher heat transfer rates can be accomplished by reducing the temperature of the coolant or blowing the air in the enclosure across the finned bundle. The design of many types of heat exchangers is covered by Hewitt, Shires & Bott in “Process Heat Transfer.”