Net thermal radiation from an object outside at night

• sian130
In summary, the conversation discusses a formula for calculating the net thermal radiation from an object placed outside at night, taking into consideration the emissivity and absorptivity of the object at 8-14μm wavelength. The formula also includes the emissivity of the sky and the ambient air temperature. However, the origin and derivation of the formula is unknown and there may be some inconsistencies with the units and assumptions used. It is suggested to find a different formula or approach, or consult a book on the subject for further clarification.

sian130

I am trying to model the cooling of an object (for example, a sheet of glass) placed outside at night. At the moment I am only considering heat loss by radiation.

I know that the net radiation from the object will be:

Rnet = Robj - Rsky

where:
Rnet = the net radiation from the object
Robj = the total thermal radiation from the object
Rsky = the thermal downwelling radiation from the sky

I have come across a formula which does the above, but also takes into consideration the absorptivity of the sheet of glass at the 8-14μm wavelength (I have only considered emissivity of the glass in this wavelength as I'm only concerned with thermal radiation heat loss to the sky).

The formula is:

Rnet = A((εobj2obj2)σTobj4 - εskyσTamb4)

where:
εobj = the emissivity of the object at 8-14μm
αobj = the absorptivity of the object at 8-14μm
εsky = the emissivity of the sky
Tobj = the temperature of the object
Tamb4 = the ambient air temperature
A = the area of the object

What I would like to know is where does the εobj2obj2 bit come from? I know if I wasn't considering absorptivity then it would just be εobj, but why are εobj and αobj now squared? I have since lost where I saw it, and I'm pretty sure there wasn't an explanation there anyway. I have searched all over for a derivation of it but have had no luck.

Can anyone help?

I am not familiar with this, but I would agree that something is off here:

Since the eqn is Rnet= Robj - Rsky

these terms should have equal forms.

But as you pointed out there is a squared ratio only on one term.
My guess is that it may be simplified, but who knows.

So I can't really answer your question, but I can say that I agree with it.
Perhaps try to find a different formula / approach to use?

I did find this though - a program for modelling thermal radiation. Looks like it is free to use, and mentions finding radiation from any given objects of given temperature. Maybe it will help. Here's the link http://www.fire-engineering-software.com/tra.html

good luck

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Without knowing how the terms are defined, and what units they have, it's impossible to say what is going on - "I found it on the internet somewhere" isn't very helpful. What units does each term in this equation have, and does the result ("net radiation" is not precise either) make sense?

Thanks for the link, elegysix.

JeffKoch - my problem in finding the source of the equation stems from the fact that it was found in a Solar Energy journal somewhere but I can no longer gain access to Solar Energy journals via my university account. Hence why I'm trying to find it by other means. I thought perhaps it may have been used elsewhere but now I'm beginning to realize that perhaps the assumptions and simplifications made in the particular journal article are more important than I'd first thought!

As for the units, they are all SI units, i.e. T in Kelvin, Stefan-Boltzmann in W(m^−2)(K−4). Emissivity and absorptivity are dimensionless. Net radiation is therefore in units of W. So the equation is dimensionally correct as far as I can tell.

What do you mean about "net radiation" not being precise? By net radiation, I mean the net thermal radiation (i.e. primarily in the 8-14micro m wavelength band) leaving the surface of the glass, as opposed to the thermal radiation leaving the glass before one has taken into account the downwelling thermal radiation from the sky.

The formula is:

Rnet = A((εobj2/αobj2)σTobj4 - εskyσTamb4)

where:
εobj = the emissivity of the object at 8-14μm
αobj = the absorptivity of the object at 8-14μm
εsky = the emissivity of the sky
Tobj = the temperature of the object
Tamb4 = the ambient air temperature
A = the area of the object

Your equation is most likely derived for a particluar situation since it has a ratio of emissivity/ absortivity, which is what I have no idea, and the ambiant air temperature rather than the temperature of the sky ( which is different than the sky temperature ),

sian130 said:
What do you mean about "net radiation" not being precise? By net radiation, I mean the net thermal radiation (i.e. primarily in the 8-14micro m wavelength band) leaving the surface of the glass, as opposed to the thermal radiation leaving the glass before one has taken into account the downwelling thermal radiation from the sky.

Think, man. Net thermal radiation per unit time? Per unit area? Per Hz? Per Sr? Integrated over any of these quantities? Photons or joules? What units do each of the terms have, and does the result make sense? Find a book on this stuff, there are many - for example the very useful one by Rybicki and Lightman.

What is net thermal radiation?

Net thermal radiation is the difference between the amount of thermal radiation emitted by an object and the amount of thermal radiation absorbed by that same object. It is a measure of the overall heat transfer of an object.

How is net thermal radiation calculated?

Net thermal radiation is calculated by subtracting the amount of thermal radiation absorbed by an object from the amount of thermal radiation emitted by that same object. This can be done using mathematical equations or through experimental measurements.

Why is net thermal radiation important to study?

Net thermal radiation is important to study because it plays a crucial role in the heat transfer of objects. Understanding net thermal radiation can help scientists and engineers design and improve insulation, energy efficient buildings, and other thermal systems.

How does net thermal radiation change at night?

At night, there is typically a decrease in the amount of thermal radiation emitted by an object due to cooler temperatures. However, the amount of thermal radiation absorbed by an object may remain the same, resulting in a higher net thermal radiation at night compared to during the day.

What factors affect net thermal radiation from an object outside at night?

The factors that affect net thermal radiation from an object outside at night include the object's temperature, surface properties, surrounding temperature, humidity, and wind speed. These factors can all impact the amount of thermal radiation emitted and absorbed by the object, thereby affecting the net thermal radiation.