Heat loss from thermos

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Homework Statement



A thermos consists of two layers (see fig). The core contains a liquid with temperature [tex]T_1[/tex]. The outer layer contains air at low pressure with temperature [tex]T_2[/tex]. The surrounding air has temperature [tex]T_3[/tex]. At [tex]t=0[/tex], [tex]T_2=T_3[/tex]. Assume that [tex]T_3[/tex] does not change with time, and that temperature is conserved between the core and outer layer. What is the magnitude of [tex]T_2[/tex] after [tex]t[/tex] seconds?

http://img141.imageshack.us/img141/3128/termos.png [Broken]

Homework Equations



Heat equation: [tex]\frac{dT}{dt}=k\Delta T[/tex]

The Attempt at a Solution



First I write down the differential equations for each temperature and layer:

[tex]\frac{dT_1}{dt}=k_1(T_2-T_1)[/tex]

[tex]\frac{dT_2}{dt}_1=k_1(T_1-T_2)[/tex]

[tex]\frac{dT_2}{dt}_2=k_2(T_3-T_2)[/tex]

[tex]\frac{dT_2}{dt}=k_1(T_1-T_2)+k_2(T_3-T_2)[/tex]

[tex]\frac{dT_3}{dt}=0[/tex]

Since I have two functions to deal with, [tex]T_1[/tex] and [tex]T_2[/tex], I set up a set of differential equations with two functions:

[tex]\frac{dT_2}{dt}=-T_2(k_1+k_2)+k_1T_1+k_2T_3[/tex]

[tex]\frac{dT_1}{dt}=k_1(T_2-T_1)[/tex]

How do I solve this set of differential equations. I dont think I can solve them one by one, since neither [tex]T_1[/tex] or [tex]T_2[/tex] is constant.
 
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Answers and Replies

  • #2
kuruman
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You have a system of coupled differential equations that you need to decouple. To do this,

1. Solve the second equation for T2 to get T2= (blah, blah, blah).
2. Replace T2 with (blah, blah, blah) on the right side of the first equation.
3. Replace the left side of the first equation with the time derivative of (blah, blah, blah).

This leaves you with a second order differential equation in T1.
 
  • #3
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Okay, thanks for the help!

I solve the second equation for [tex]T_2[/tex], obtaining

[tex]T_2=T_1+\frac{1}{k_1}\frac{dT_1}{dt}[/tex]

and

[tex]\frac{dT_2}{dt}=\frac{dT_1}{dt}+\frac{1}{k_1}\frac{d^2T_1}{dt^2}[/tex]

Putting it into the second equation gives me

[tex]\frac{dT_1}{dt}+\frac{1}{k_1}\frac{d^2T_1}{dt^2}=-(k_1+k_2)(T_1+\frac{1}{k_1}\frac{dT_1}{dt})+k_1T_1+k_2T_3[/tex]

leaving me with

[tex]\frac{d^2T_1}{dt^2}+(2k_1+k_2)\frac{dT_1}{dt}+k_1k_2T_1-k_1k_2T_3=0[/tex]

Solving this yields

[tex]T_1=C_1e^{-\frac{1}{2}t(\sqrt{4k_1^2+k_2^2}-2k_1-k_2)}+C_2e^{\frac{1}{2}t(\sqrt{4k_1^2+k_2^2}-2k_1-k_2)}+T_3[/tex]

I wonder if this expression is an accurate description of how heat really transfers out of a thermos. For positive values of [tex]C_2[/tex], the function increases. That can't be, because [tex]T_1>T_2\geq T_3[/tex]. For negative values, it quickly drops to 0 K, absolute zero. That's illogical, because the function should approximate to [tex]T_3[/tex] after a long time. The only way for that to happen is if [tex]C_2=0[/tex]. Do you agree?

[tex]T_1=(T_1)_0e^{-\frac{1}{2}t(\sqrt{4k_1^2+k_2^2}-2k_1-k_2)}+T_3[/tex]

I replaced [tex]C_1[/tex] with [tex](T_1)_0[/tex] because it is the value of [tex]T_1[/tex] at [tex]t=0[/tex].

[tex]\frac{dT_1}{dt}=-(T_1)_0e^{-\frac{1}{2}t(\sqrt{4k_1^2+k_2^2}-2k_1-k_2)}(\frac{1}{2}(\sqrt{4k_1^2+k_2^2}-2k_1-k_2))[/tex]
 
  • #4
Integral
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Your model is not correct. You completely neglect the spatial aspect. The temperature in the region you have labeled T2 is not and will not be constant. It will be T1 adjacent to the inner boundary and T3 at the outer boundary. If you assume pure conduction (a bad assumption by the way!) then there will be continuous gradient in the T2 region.

You have also neglected material properties of the liquid in the center, the walls of the container and the air in the gap.

The wiki article on the http://en.wikipedia.org/wiki/Heat_equation" [Broken]



Edit:
I should also mention that what you are calling the heat equation is simply not correct. See the wiki artical on the heat equation.
 
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