2 replies to this topic

[chemguy22]

What is a good way to estimate the heat loss due to radiation from a bare pipe exposed to the outdoors?

 

I currently am working with this equation:

 

Q.radiation = α*( e1 T.outside.pipe^4 - e2 T.ambient^4 )*A (radiation)

A = Area of Radiating surface (m2)
T.outside.pipe = Temperature or radiating body (K) << to find out
T.ambient = Temperature or Suroundings (K)
α = Stefan Boltzman constant = 5,673 x 10-8 W m-2 K-4
e 1 = Emissivity of Radiating surface
e 2 = Emissivity of Surroundings

 

But I am wondering whether the emissivity of surroundings (i.e. the atmosphere, there are no buildings/structures immediately beside the section of exposed pipe) is significant??

 

Will the heat loss due to radiation be insignificant if there are no significant structures adjacent to the exposed pipe to absorb the radiation??

 

Any help or input would be greatly appreciated!

[Steve Hall]

Your highest radiation loss is to interstellar space, if there are no structures or clouds in the way. This is why you can, theoretically, freeze water in the desert at temperatures approaching 20C. The siimplifying assumption is to set e2 = e1. Since you are raising the pipe temperature to the 4th power, if the temperature difference between pipe and environment is significant then the emissivity of the environment isn't very important. However, if the temperature difference is small then e2 can have a big influence on the calculation.

[chemguy22]

Great. Thank you very much Steve!