Does anyone have a calcultor or formulas for tank heating? I have a steel storage tank 80 ft dia.by 85 ft tall and need to maintain a skin temperature of 60 deg. F with a design temp of -1 deg. F. I need to know how many BTU's to supply. Thanks for any help.

csaw325:

Does anyone have a calcultor or formulas for tank heating? I have a steel storage tank 80 ft dia.by 85 ft tall and need to maintain a skin temperature of 60 deg. F with a design temp of -1 deg. F. I need to know how many BTU's to supply. Thanks for any help.

A buttload? /snark

Seriously, though - something here might help:

http://www.wisconsinperspective.com/key-water-heating-charts/

So - you would have to convert your tank size into gallons, then use one of the formulas on that page (for heating water - I'm not certain whether other liquids would need different formulas). Of course, I think those forumulas are for heating up the entire amount of liquid, and not just for the surface (which I would think would also need to involve ambient temperature at the surface, too).

But that should get you started, at least.

The heat needed is eventually the heat radiated away.

The heat radiated away depends on

- size of the surface
- is the tank closed on all sides or open on top? (diff heat loss)
- on the amount of liquid in the tank (the lower part will loose more heat)
- the delta with the outside temperature. (yeah it is a differential equation in the end)

So proper insulation might help keeping costs (and CO_{2} emission) low

If the tank is full the contents is 1/4 * 80 ft * 80ft * PI * 85 ft ~ **3,200,000 gallons**

That is quite large, is it for wine or ?

The outside surface is 80 ft *PI * 85 ft + 2 * 1/4 * 80 ft * 80ft * PI = **31,146 sqr ft**

assuming top is closed

There's also Newton's Law of Cooling: the rate at which the tank will cool off is proportional to its temperature vs ambient temperature.