Thermodynamics(hard problem)try to help

1. May 21, 2007

gollum

Liquid helium is stored in containers fitted with 7 cm-thick "superinsulation" consisting of a large number of layers of very thin aluminized Mylar sheets. The rate of evaporation of liquid in a 200 L container is about 0.7 L per day. Assume that the container is spherical and that the external temperature is 20 C. The specific gravity of liquid helium is 0.125 and the latent heat of vaporization is 21kj/kg. Estimate the thermal conductivity of superinsulation.

This problem is very hard so we need all the help we can get. Even simple ideas in solving the problem will do. Thank you

2. May 21, 2007

lalbatros

This problem is extemely simple.
You simply need to use a few definitions, some unit conversions, and the first principle.

I did the calculation and found 3.1E-6 W/m/K .
I did the calculation with the simplified assumption that the temperature gradient is entirely within the mylar thickness. The helium tempreture is assumed to be uniform, as well as the surrounding air temperture.

You can see that at the liquid helium temperature, this is roughly consistent with the value above.
Unfortunately, the mylar heat conductivity varies by nearly a factor 3 fom 4K to 300K.

Therefore, this exercice is a very simplified version of the reality.

When you are ready with this exercice, you should try to calculate the evaporation rate on the basis of the real mylar heat conductivity that changes quite a lot with temperature. You could also plot the temperature profile within the mylar thickness.

It is also possible to refine the calculation to take into account the thermal resistance of the surrounding air. But I don't know how far you are supposed to go. Let's start with simplified version of this problem.

Tell me if you are supposed to go into more detail.

Last edited: May 21, 2007
3. May 22, 2007

Andrew Mason

How much heat required to evaporate .7 L of He? So how much heat must pass through the insulation in one day? What is that in Joules/sec or Watts? What does that tell you about the thermal conductivity?

AM