Blackbody Heat Shields/Metal Sheets

[sweetdreams12]

A “blackbody heat shield” is used to reduce radiant heat loss in situations where standard low conductivity insulation (i.e. normal insulation) cannot be used. The shield is constructed using a set of parallel metal sheets, each with emissivity 1.0.

Explain how/why this “heat shield” works and determine how many sheets of metal are
required to reduce the heat loss to less than 20% of the unshielded value. (Hint: think
about the radiation absorbed by each sheet and how this energy will be re-radiated.)

Diagram:

I really need help with this xD I don't even know where to begin.

 

[mfb]

Here are some ideas:
- assume that the whole problem is 1-dimensional
- begin with 1 heat shield: Where does it receive/emit radiation?
- assuming that the environment has a temperature of ~0, calculate the temperature of that heat shield in equilibrium

 

[sweetdreams12]

ummmm I still don't get it xD

do I use the equation:

Q = mC deltaT?

 

[mfb]

No, that looks like regular conduction.
$P_{emitted}=\sigma A T^4$ (Stefan-Boltzmann)

 

[sardar]

I would first like to clarify the concept of emissivity. Emissivity is a material's ability to emit thermal radiation. A material with an emissivity of 1.0 is considered a perfect emitter, meaning it can radiate all of the heat that it absorbs.

Now, let's consider the construction of the blackbody heat shield. The shield is made up of a set of parallel metal sheets, each with an emissivity of 1.0. This means that each sheet can absorb and radiate all of the heat that it receives. When placed between a heat source and the surrounding environment, the metal sheets act as a barrier, preventing the heat from escaping through radiation.

To determine the number of sheets required to reduce heat loss to less than 20% of the unshielded value, we need to consider the radiation absorbed and re-radiated by each sheet. As the heat passes through each sheet, it is absorbed and re-radiated in all directions. However, since the sheets are parallel to each other, some of the radiation will be directed back towards the heat source, effectively trapping the heat within the shield.

The amount of heat that is trapped and re-radiated back towards the heat source decreases with each additional sheet. This is because the first sheet absorbs and re-radiates a certain amount of heat, and the subsequent sheets will only absorb and re-radiate the remaining heat that was not trapped by the previous sheets.

To reduce the heat loss to less than 20%, we would need enough sheets to trap at least 80% of the heat. This can be achieved with approximately 5 sheets. However, the exact number of sheets required may vary depending on the specific materials and conditions. Other factors such as the thickness and spacing of the sheets may also impact the effectiveness of the shield.

In conclusion, the blackbody heat shield works by using multiple parallel metal sheets with high emissivity to trap and re-radiate heat, preventing it from escaping through radiation. The number of sheets required to reduce heat loss to less than 20% will depend on various factors, but it can be estimated at approximately 5 sheets.