Calculating Heat Loss via Radiation Transfer

In summary, the conversation discusses calculating heat loss through radiation heat transfer and the factors that affect it, such as temperature of surroundings and emissivity. It is important to consider variables like weather conditions and the properties of different materials in order to accurately calculate heat loss. The use of Stefan's law and Planck's law are mentioned as possible methods, but it is noted that experimental results may be more reliable. The conversation ends with a question about the accuracy of the calculations and the potential for heat to be distributed to different components of the surroundings.
  • #1
chetanladha
59
0
Hi.

I want to calculate the heat which will be lost from a body at temperature 't' to the atmosphere through radiation heat transfer. I can use Stephen's Boltzmann equation, but what should i take the temperature of surroundings as?

Should it be the temperature of air or space?

Thanks.
 
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  • #2


chetanladha said:
Should it be the temperature of air or space?
It depends on weather conditions and surroundings.
Have you ever parked your car wintertime (clear night) on the open area or under tree or under open sched? It gets more frosted on the open area, as it loses heat to the sky, while under sched you should take the temperatur of sched roof.
Similar effect you may find comparing temperature on clear and cloudy nights - clear nights are much colder, because you lose heat to the sky, why on cloudy nights most of the radiation is reflected.

BTW - using Stefan's law - be careful - most of the objects cannot be treated as black-body. Even if some object is visually black, don't believe it is also 'black' outside visual spectrum.
 
  • #3


That was a good description.
I am giving the value of emmissivity in Stefan's law, so i guess that shouldn't be a problem.

I am calculating the radiation heat loss from a steel plate in open atmosphere clear sky. So, now what should i take the temperature of sky as?
 
  • #4


chetanladha said:
I am giving the value of emmissivity in Stefan's law, so i guess that shouldn't be a problem.
Not be that sure. For most substances emissivity depends on a wavelength. So you should rather integrate Planck's law over wavelength, than assuming it is uniform. For many materials results may dramatically differ. Snow is a perfect example - having very small emissivity in visual spectrum, while pretty large in infrared - that's why the weather changes to much colder as soon as first snow covers ground.

I am calculating the radiation heat loss from a steel plate in open atmosphere clear sky. So, now what should i take the temperature of sky as?
As the rough approximation you may just forget about sky temp and assume it is 0K. Error you make will be still much less probably than by assuming uniform emissivity of of your plate.
 
  • #5


xts said:
Not be that sure. For most substances emissivity depends on a wavelength. So you should rather integrate Planck's law over wavelength, than assuming it is uniform. For many materials results may dramatically differ. Snow is a perfect example - having very small emissivity in visual spectrum, while pretty large in infrared - that's why the weather changes to much colder as soon as first snow covers ground.

Thanks a lot. It was really helpful.
 
  • #6


Hey.. Sorry but 1 more small q.

I calculated the heat flux from sun on my plate to be around 51kW. (assuming solar intensity 1000kW/m2).
The stephen law gives me heat lost to the surroundings in order of 72kW (using constant value of emissivity).

Apart of not using constant value of emissivity, can i assume part of heat is going to air at 30 C and rest to 0 K.
Also m really afraid of integrating Plank's law, unless its a must..

What do u think..??
 
  • #7


Oh, I see you must live in some place in the world where my examples with car getting frosted in the night and snow emissivity are not so common experience :redface:

Probably your steel has lower emissivity in infrared than in visual spectrum. Your result is against my intuition: it absorbs less from Sun than emits. It would mean it is getting colder than surroundings. My intuition tells me that steel (maybe except very shiny polished acidproof steels) gets hot in the sunlight faster than soil, concrete, bricks, etc., so something had to be wrong in your calculations.

Anyway - it is not that easy to model the heatflow with reasonable accuracy. Even if you integrate Planck's law over whole spectrum...
In most cases the experimental approach is used to determine the values used for practical calculations.
 

Related to Calculating Heat Loss via Radiation Transfer

1. What is radiation transfer and how does it contribute to heat loss?

Radiation transfer is the transfer of thermal energy through electromagnetic waves. It contributes to heat loss by allowing heat to escape from a warmer object to a cooler one without the need for direct contact.

2. How does the Stefan-Boltzmann law relate to calculating heat loss via radiation transfer?

The Stefan-Boltzmann law states that the rate of radiation emitted by an object is proportional to the fourth power of its absolute temperature. This means that as the temperature of an object increases, the amount of heat it radiates also increases, resulting in a higher rate of heat loss.

3. What factors influence the rate of heat loss via radiation transfer?

The rate of heat loss via radiation transfer is influenced by several factors, including the temperature difference between the two objects, the surface area of the objects, and the emissivity of the objects. Objects with a larger temperature difference, larger surface area, and higher emissivity will experience a higher rate of heat loss through radiation transfer.

4. How can the rate of heat loss via radiation transfer be calculated?

The rate of heat loss via radiation transfer can be calculated using the Stefan-Boltzmann law and the emissivity of the objects involved. The formula is Q = εσAT⁴, where Q is the rate of heat loss, ε is the emissivity, σ is the Stefan-Boltzmann constant, A is the surface area, and T is the absolute temperature.

5. How can the rate of heat loss via radiation transfer be reduced?

The rate of heat loss via radiation transfer can be reduced by increasing the temperature of the warmer object, decreasing the temperature of the cooler object, decreasing the surface area of the objects, and using materials with lower emissivity. Insulation can also be used to reduce the rate of heat loss via radiation transfer by creating a barrier between the two objects.

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