Calculate Heating Time of Object with Radiation

In summary, the equation can be solved analytically, but is more accurate if the initial temperature and final temperature are known.
  • #1
tjosan
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2
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
$$mC_p\frac{dT}{dt}=\sigma A_1(T_1^4-T^4)$$
subject to some initial condition on T
 
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  • #3
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
russ_watters said:
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
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
russ_watters said:
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
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
Thank you for the answers!
Chestermiller said:
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
tjosan said:
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
 

Related to Calculate Heating Time of Object with Radiation

What is the formula for calculating the heating time of an object with radiation?

The formula for calculating the heating time of an object with radiation is t = (Q/m)(1/σAε(T^4 - T0^4)), where t is the heating time, Q is the energy transferred through radiation, m is the mass of the object, σ is the Stefan-Boltzmann constant, A is the surface area of the object, ε is the emissivity of the object, T is the final temperature, and T0 is the initial temperature.

How does the surface area of an object affect its heating time with radiation?

The surface area of an object directly affects its heating time with radiation. The larger the surface area, the more energy can be transferred through radiation, resulting in a shorter heating time. This is because a larger surface area allows for more contact with the surrounding radiation, while a smaller surface area has less contact and therefore takes longer to heat up.

What is the significance of the emissivity value in calculating the heating time of an object with radiation?

The emissivity value represents the efficiency of an object in emitting thermal radiation. A higher emissivity value means that the object is more efficient in emitting thermal radiation, resulting in a shorter heating time. On the other hand, a lower emissivity value means that the object is less efficient in emitting thermal radiation, resulting in a longer heating time.

How does the initial and final temperatures of an object affect its heating time with radiation?

The initial and final temperatures of an object play a significant role in determining its heating time with radiation. The larger the difference between the two temperatures, the longer the heating time will be. This is because a larger temperature difference means that more energy is needed to be transferred through radiation, resulting in a longer heating time.

What are the limitations of using the formula for calculating the heating time of an object with radiation?

The formula for calculating the heating time of an object with radiation assumes that the object is in a vacuum and that there are no other heat transfer mechanisms at play. In reality, there may be other factors such as conduction and convection that can affect the heating time. Additionally, the formula does not take into account any changes in the emissivity of the object as its temperature increases. Therefore, the calculated heating time may not be entirely accurate in real-world scenarios.

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