Heat loss through shell to atmosphere 1

[mielke]

anyone know a quick equation or rule of thumb to figure out how much heat can be loss through a shell and tube heat exchanger shell side fluid to the atmosphere?

 

[rmw]

I don't believe there is a quick equation. The rule of thumb would be that you have to know all your conditions, fluid temperatures, outside ambient temperature, outside air or wind velocity, know or make some reasonable assumptions with regards to fluid velocities and flow patterns inside the shell, shell materials and thicknesses, and then do some basic heat transfer calculations.

Without getting too complicated with internal flows, etc, as a quick indicator, I'd treat it like a length of equivalent sized pipe and use a heat trace calculator program to approximate the heat loss.

rmw

 

[Chance17]

For a crude estimate use 3 btu/hr-ft2 for stagnant air
or 20 btu/hr-ft2 on a windy day.
Of course Q = U * Area * Delta T

 

[IRstuff]

Bear in mind that any equation that does not specify temperature difference has an implicit temperature difference built-in, which may be different than your operating conditions.

For example, 3 BTU/hr-ft^2 might be the result of using a 5 W/m^2-K heat transfer coefficient with only a 2°C temperature difference.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[StewPad]

You can't use those numbers without knowing the dT

Check this old post:

http://www.eng-tips.com/viewthread.cfm?qid=276525&page=7

 

[mielke]

The average temperature of the shell side fluid is 595F and the ambient air between 30F.

 

[IRstuff]

That's a good start, but still insufficient for the purposes of analysis. I hope that you'll recognize that there is no "rule of humb" at work here, because each installation is unique. You'll need to know the view factor and emissivity, and you'll have to apply radiative loss as well.

You'll have some sort of conductive thermal resistance to the surface that's in contact with air. The resistance induces a temperature drop reflected in the surface temperature. That surface temperature is what you use for both the radiative and convective losses. You'll need to know the emissivity of the surface. The solution to the heat balance between heat getting to the surface from the interior of the exchanger and the heat leaving the surface through radiation and convection will give you the final surface temperature. That temperature, when applied to the heat balance expressions, will give you the heat lost.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize

 

[zekeman]

"Bear in mind that any equation that does not specify temperature difference has an implicit temperature difference built-in, which may be different than your operating conditions.

For example, 3 BTU/hr-ft^2 might be the result of using a 5 W/m^2-K heat transfer coefficient with only a 2°C temperature difference"


If you read further down his post ,Chance17 does have the correct equation that includes the delta T and his result is a very good "Rule of Thumb" that the OOP needs.

 

[IRstuff]

I did read it; it was inconsistent with the first part, hence the caveat.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize