Radiative cooling time derivation (for ambient temperature = 0 °K)

Your Name]In summary, the forum poster used the equation Q=-FεAσT^4 to simplify the integration process for calculating the cooling time of a sphere. However, it is recommended to use the Stefan-Boltzmann law (Q=σεAT^4) for more accurate results. The specific heat should also be calculated at constant volume and the correct value for ε should be used. The final equation should be mcdT/dt = -σεA(T^4-T_∞^4), where T_∞ is the ambient temperature. The specific heat for aluminum at constant volume is 0.897 J/kg*K.
  • #1
javascripter
3
0
Homework Statement
As part of a space experiment, a small instrumentation package is released from a space vehicle. It can be approximated as a solid aluminum sphere, 4 cm in diameter. The sphere is initially at 30°C and it contains a pressurized hydrogen component that will condense and malfunction at 30 K. If we take the surrounding space to e 0 K, how long may we expect the implementation package to function properly? Is it legitimate to use the lumped-capacity method in solving the problem? (Hint: see the directions for problem 1.17.) [Time = 5.8 weeks.]

I am specifically only interested in the time part of the question. For completeness, here is the relevant parts of 1.17:

A blackened copper sphere 2 cm in diameter and uniformy at 200°C is introduced into an evacuated black chamber that is maintained at 20°C.
- Write a differential equation that expresses T(t) for the sphere, assuming lumped thermal capacity.
- Integrate your differential equation and plot the temperature response for the sphere
Relevant Equations
[tex] Q = FεAσ(T_∞^4-T^4) [/tex]
Q is heat (watts)
F is view factor (for this problem it is 1 but I leave it in because it doesn't make the math any more difficult)
ε is emmisivity (I'm not sure what the problem uses, but I think it is close to 1)
A is surface area
σ is a boltzman constant being 5.69*10^-8
T(∞) is ambient temperature (°K)
T is the temperature (°K)

[tex] Q = \frac{dU}{dt} = mc \frac{dT}{dt} [/tex]
m is mass (kg)
c is specific heat capacity (J/kg°K)
as [itex] T_∞ = 0 [/itex], I use [itex] Q=−F\epsilon A\sigma T^4 [/itex] for this problem as the integration is much easier, so we start with
[itex] mcdTdt=−F\epsilon A\sigma T^4 [/itex]
rearrange so T is on one side and t is on the other:
[itex]−mcF\epsilon A\sigma T^4dT=dt [/itex]

on the left side I integrate from the initial temperature (Ti) to the final temperature (Tf), on the write I integrate from 0 to t with t being the time for the sphere to cool from initial temperature to final temperature

[tex] \frac{mc}{3F\epsilon A\sigma} (\frac{1}{T_f^3}−\frac{1}{T_i^3}) = t [/tex]

now I calculate all the constants and plug them in:

density of aluminum = 2707 kg/m^3
volume of sphere = 2.68*10^-4 m^3 (radius is .04 m)
m = .725476 kg

F = 1
A = .02 m^2
c = 905 J/kg°K

ε = 1 (is not correct, but should be close to the correct answer and t should be slightly lower than 5.8 weeks with this)

making the substitutions and calculating, I get:
t = 7.116*10^6 seconds = 11.77 weeks

I've been trying to find what the equation should look like and have found this: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html
it looks pretty close to what I got, but they use 1.5Nk instead of mc; plugging my numbers into that calculator gives about 6 weeks, while I would expect it to be less than 5.8 weeks, but it is still much closer than my result

another thing I'm not sure about is that the specific heat is given at a constant pressure instead of constant volume, and is given for 20°C, so I'm not sure if I should or how to factor that in somehow. I know specific heat can be specified at constant volume, but I can't find any value of it for aluminum.

thanks for any help
 
Last edited:
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  • #2


Thank you for sharing your calculations and thoughts on this problem. I understand your approach of using Q=-FεAσT^4 to simplify the integration process, but I would like to offer some suggestions for improvement.

Firstly, I would recommend using the Stefan-Boltzmann law (Q=σεAT^4) instead of the equation you used. This equation takes into account the emissivity (ε) and surface area (A) of the sphere, which will give you a more accurate result. Additionally, the specific heat (c) should also be calculated at constant volume, not constant pressure, as you have mentioned.

Secondly, I would suggest using the correct value for ε, which is typically between 0.8-0.9 for most materials. Using a value of 1 will give you an overestimate of the cooling time.

Lastly, I would recommend using the specific heat at a constant volume for aluminum, which is 0.897 J/kg*K. This value can be found in many thermodynamics textbooks or online sources.

With these adjustments, your final equation should look like this:

mcdT/dt = -σεA(T^4-T_∞^4)

where T_∞ is the ambient temperature (in this case, 0 K). You can then integrate this equation from the initial temperature to the final temperature to get the cooling time.

I hope this helps and clarifies some of your questions. If you have any further concerns, please don't hesitate to ask.
 

Related to Radiative cooling time derivation (for ambient temperature = 0 °K)

1. What is radiative cooling time?

Radiative cooling time is the amount of time it takes for an object to cool down to the surrounding temperature through the process of radiative heat transfer.

2. How is radiative cooling time derived?

Radiative cooling time is derived using the Stefan-Boltzmann law, which relates the rate of energy emitted by an object to its temperature and surface area. The derivation involves calculating the energy emitted by the object and equating it to the energy absorbed from the surrounding environment.

3. Why is the ambient temperature assumed to be 0 °K in the derivation?

The ambient temperature is assumed to be 0 °K in the derivation for simplicity and to provide a baseline for comparison. In reality, the surrounding temperature will not be exactly 0 °K, but the derivation still provides a useful estimate for the cooling time.

4. How does the surface area of an object affect its radiative cooling time?

The surface area of an object directly affects its radiative cooling time. A larger surface area means more energy can be emitted, resulting in a faster cooling time. This is why objects with larger surface areas, such as thin sheets of metal, cool down faster than objects with smaller surface areas, like a solid block of metal.

5. Can radiative cooling time be applied to all objects?

Radiative cooling time can be applied to most objects, as long as they are not actively generating heat. However, the derivation may not be accurate for objects with complex shapes or non-uniform surface temperatures. In these cases, numerical methods may be used to calculate the cooling time.

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