Thermos Heat Loss: Solving Differential Equations for Changing Temperatures

[espen180]

Homework Statement

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

http://img141.imageshack.us/img141/3128/termos.png

Homework Equations

Heat equation: \frac{dT}{dt}=k\Delta T

The Attempt at a Solution

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

\frac{dT_1}{dt}=k_1(T_2-T_1)

\frac{dT_2}{dt}_1=k_1(T_1-T_2)

\frac{dT_2}{dt}_2=k_2(T_3-T_2)

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

\frac{dT_3}{dt}=0

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

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

\frac{dT_1}{dt}=k_1(T_2-T_1)

How do I solve this set of differential equations. I don't think I can solve them one by one, since neither T_1 or T_2 is constant.

 

[kuruman]

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.

 

[espen180]

Okay, thanks for the help!

I solve the second equation for T_2, obtaining

T_2=T_1+\frac{1}{k_1}\frac{dT_1}{dt}

and

\frac{dT_2}{dt}=\frac{dT_1}{dt}+\frac{1}{k_1}\frac{d^2T_1}{dt^2}

Putting it into the second equation gives me

\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

leaving me with

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

Solving this yields

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

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

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

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

\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))

 

[Integral]

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"



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.