Heating time radiation

  • Thread starter tjosan
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  • #1
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Hello.

I want to calculate how long time it takes to heat an object with radiation. The object is the inner cylinder of two concentric cylinders. The inner cylinder is not hollow. If assuming black body:

[itex]\dot{Q}_E=\sigma A_1(T_1^4-T(t)^4)[/itex] [1]

Energy emitted must the same as the energy absorbed for the inner cylinder. I want the inner cylinder to reach temperature [itex]T_2[/itex] [itex](T_2<T_1)[/itex].
So the total energy needed for for inner cylinder is given by:

[itex]q=C_p(T_2-T(t))m[/itex] [2]

However, as can be seen, as the temperature of the inner cylinder increase, [itex]\dot{Q_E}[/itex] ([2]) will decrease.

So how do I calculate how long time it will take for it to reach [itex]T_2[/itex]?

Assumptions:
Black body
Inner cylinder has uniform temperature, i.e. surface temperature = core temperature at all times.
 

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  • #2
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$$mC_p\frac{dT}{dt}=\sigma A_1(T_1^4-T^4)$$
subject to some initial condition on T
 
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  • #3
russ_watters
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To actually solve for the answer though, I'd use a numerical method in a spreadsheet to avoid dealing with the differential equation directly.
 
  • #4
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To actually solve for the answer though, I'd use a numerical method in a spreadsheet to avoid dealing with the differential equation directly.
I think what you're saying is that, because the equation is non-linear in T, one needs to integrate the differential equation numerically. Am I interpreting what you're saying correctly?
 
  • #5
russ_watters
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I'm not sure if one "needs to", I'm just saying I would: the equation looks fine, I'm just not sure if the OP is going to know what to do with it. I vaguely recall solving such equations for T or t back in college (this is a modified version of Newton's Law of Cooling, which was a specific example used), but would avoid it today.
 
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  • #6
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I'm not sure if one "needs to", I'm just saying I would: the equation looks fine, I'm just not sure if the OP is going to know what to do with it. I vaguely recall solving such equations for T or t back in college (this is a modified version of Newton's Law of Cooling, which was a specific example used), but would avoid it today.
During my career, this kind of thing was my "bread and butter."

There are some good approximations that can be made to obtain an analytic estimate, depending on the initial temperature T0 and the final temperature T2. If (T2-T0) is small compared to both (T2+T0)/2 and to T1-T2, then the right hand side of the equation can be evaluated at Tave = (T2+T0)/2. The heating time then becomes:
$$t=\frac{\sigma A_1(T_1^4-T_{ave}^4)}{mC_p(T_2-T_0)}$$
A better approximation is obtained by first factoring the right hand side of the differential equation:
$$mC_p\frac{dT}{dt}=\sigma A_1(T_1^2+T^2)(T_1+T)(T_1-T)$$
If we then substitute Tave for T in the first two parentheses, we obtain:
$$mC_p\frac{dT}{dt}=h_{ave} A_1(T_1-T)\tag{1}$$
where have is the average heat transfer coefficient, given by:
$$h_{ave}=\sigma (T_1^2+T_{ave}^2)(T_1+T_{ave})$$
Eqn. 1 is the standard Newton's cooling linear heat transfer equation that is readily solved analytically:
$$\frac{T_1-T_2}{T_1-T_0}=e^{-\frac{h_{ave}A_1}{mC_p}t}$$

Chet
 
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  • #7
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The exact analytic solution to this problem can also be obtained by grouping the temperature terms on one side of the equation and using partial fractions. I'm just too lazy to do it.

Chet
 
  • #8
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Thank you for the answers!


The exact analytic solution to this problem can also be obtained by grouping the temperature terms on one side of the equation and using partial fractions. I'm just too lazy to do it.
Chet
I have another question! I also found that you can use this aproach: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html
When and Why would you do it this way?
 
  • #9
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Thank you for the answers!
I have another question! I also found that you can use this aproach: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html
When and Why would you do it this way?
That link inherently assumes that it is a monoatomic gas that is cooling (because of the expression used for the heat capacity), and it assumes that the temperature of the body it is radiating to is at 0 K (absolute zero).

Chet
 

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