Emissivity, radiation and heat transfer

In summary, the temperature of the inner wall of a space station in outer space can be found by equating the power flowing out of the wall via conduction to the power dispersed by emission to the cold environment. The emissivity of the wall, its thickness, and its thermal conductivity can be used to solve for the temperature difference between the inner and outer walls, which can then be added to the outside temperature to find the inner wall's temperature.
  • #1
CeeCoops
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Homework Statement


A space station in outer space (far from the sun) has a total surface area of 580 m^2 with emissivity of .62. The temperature of the outside surface is 156 K. the walls are .25 m thick with an average thermal conductivity of .038 Wm^-1K^-1. Find the temperature of the inner wall of the space station, assuming that the outside surface radiates into an environment that is very cold - essentially at 0.0 K.

Homework Equations



P= σA(T^4 - To^4) and P= kA ((Th-Tc)/L)

The Attempt at a Solution

Just not too sure where to start...does the outer surface of the wall lose some of its heat to the surroundings and then we try to determine how much heat will be transferred to the inner wall? thanks for any help :)
 
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  • #2
CeeCoops said:

Homework Statement


A space station in outer space (far from the sun) has a total surface area of 580 m^2 with emissivity of .62. The temperature of the outside surface is 156 K. the walls are .25 m thick with an average thermal conductivity of .038 Wm^-1K^-1. Find the temperature of the inner wall of the space station, assuming that the outside surface radiates into an environment that is very cold - essentially at 0.0 K.


Homework Equations



P= σA(T^4 - To^4) and P= kA ((Th-Tc)/L)



The Attempt at a Solution

Just not too sure where to start...does the outer surface of the wall lose some of its' heat to the surroundings and then we try to determine how much heat will be transferred to the inner wall? thanks for any help :)

Heat obviously travels from the inside to the outside wall. The space station is generating heat which flows thru the wall to deep space at 0K.

Your emissivity equation is for a black body (α=1). If α < 1 how should that equation go?

So all that's left is to realize that power flowing out of the platform by conduction = power dispersed by emission to cold, cold space!
 
Last edited:

1. What is emissivity and how does it affect heat transfer?

Emissivity is a measure of an object's ability to emit thermal radiation. It is a dimensionless value between 0 and 1, where 1 represents a perfect emitter (known as a blackbody) and 0 represents a perfect reflector. The higher the emissivity of an object, the more efficiently it can transfer heat through radiation.

2. How does radiation differ from conduction and convection in terms of heat transfer?

Radiation is a form of heat transfer that does not require a medium, whereas conduction and convection both require a medium (such as a solid or fluid) to transfer heat. Radiation occurs when electromagnetic waves are emitted by a hot object and absorbed by a cooler object, resulting in heat transfer.

3. Can emissivity be changed or controlled?

Yes, emissivity can be changed by altering the surface properties of an object. For example, by adding a coating or painting a surface, the emissivity can be increased or decreased. However, the emissivity of a material is typically constant at a given temperature and wavelength.

4. How does the Stefan-Boltzmann law relate to emissivity and heat transfer?

The Stefan-Boltzmann law states that the total amount of thermal radiation emitted by a blackbody is proportional to the fourth power of its absolute temperature. This law can be used to calculate the amount of heat transferred through radiation between two objects, taking into account their emissivity values.

5. What is the role of emissivity in thermographic imaging?

Emissivity plays a crucial role in thermographic imaging, as it affects the accuracy of temperature measurements. Infrared cameras used for thermographic imaging rely on the emissivity of an object to determine its temperature. Incorrect emissivity values can lead to inaccurate temperature readings and potential errors in thermal analysis.

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