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andy_fl
12-13-2002, 09:26 AM
So how do they do it without setting up convectional heat transfer at the same time ?

I think maybe they do it in small films or something. Of course space (zero g) might be a good place to do this. But

Duck Duck Goose
12-13-2002, 09:31 AM
Does this help?

http://www.thermo.com/eThermo/CDA/Industry/Industry_Technology_Listing/0,1223,11859-1-1000000009918,00.html

andy_fl
12-13-2002, 02:45 PM
Thanks Duck Duck Goose, but it does'nt help. Thats just a way of comparing the thermal conductivity with a reference.

It does'nt measure absolute thermal conductivity.

Napier
12-13-2002, 07:37 PM
You put the hot object on top and the cold object on the bottom.

andy_fl
12-13-2002, 07:43 PM
Sorry Napier, but that does'nt stop convection currents. You have to consider the diffusivity of gases still.

Dr_Paprika
12-13-2002, 10:17 PM
When a temperature gradient exists in a body, there is an energy transfer from the high temperature regoin to the low temperature region. The heat transfer rate (q) per unit area (A) is proportional to the normal temperature gradient. Using the proportionality constant k, it follows that

q=-kA*dT/dx

where dT/dx is the temperaqture gradient in the direction of heat flow and k is called the thermal conductivity. The minus sign is inserted to satisfy the Second Law of thermodynamics, heat flows downhill on the temperature scale.

The situation is more complex if the temperature changes with time or a heat source is located in the body. Using a one-dimensional cross section, energy conducted in the left face (at x)of a cross section of elemental volume plus the heat generated in the element equals the change in internal energy plus the energy conducted out the right face (at x+dx), ie:

q(left face)=-kA*dT/dx
energy in element=QAdx (where Q is energy per unit volume)
change in internal energy = pcA*dT/dt*dx
(p=density, c=specific heat)
q(right face)=-kA(dT/dx) at x+dx
=-A{k*dT/dx = d/dx *(k*dT/dx)dx}

hence, in one dimension, d/dx(k*dT/dx)+Q=pc(dT/dt)

In three dimensions and rectangular coordinates, using element volume limited by x, y, z and x+dx, y+dy, z+dz,

d/dx(k*dT/dx)+d/dy(k+dT/dy)+d/dz(k*dT/dz)+Q=pc(dT/dt)

or, if the thermal conductivity is constant, and a=k/pc=(thermal diffusivity),

d2T/dx2 + d2T/dy2 + d2T/dz2 = Q/k = (dT/dt)/a

Experimental models to determine k can be made as a result of the first equation for gases at low tempertures. The mechanism of gas conduction is simple; kinetic energy and velocity depend on temperature. Molecules exchange momentum and energy when they collide with each other, and gas molecules move randomly and collide regardless of whether a temperature gradient exists. Thus, kinetic energy is transported to the lower temperature parts of the system and thermal conductivity is strongly temperature dependent.

A hot plate of metal cools faster when a fan is beside it (rather than still air). This convection heat transfer would seem to depend on the velocity of the air that passes over the plate and would probably differ if cool water (rather than air) was used. For a plate of temperature Tw and fluid (gas is a fluid) temperature Tf, the velocity of the fluid layer at the wall is zero due to viscosity and heat is only transfered by conduction at this point. The heat transfer at the wall can be computed only with the temperature gradient and thermal conductivity at the wall. But the temperature gradient depends on the rate at which the heat is carried away by the fluid; a fan with high air velocity produces a larger temperature gradient above the hot plate. To express the effect of convection, Newton's Law of Cooling states

q=hA(Tw-Tf)

The heat transfer rate is related to the overall temperature diffeence between the surface area A and the wall and fluid. h is called the convection heat transfer coefficient.

Heat can also be lost through radiation (heat may be transferred through a perfect vacuum), this radiated heat q can be found using a modified Stefan-Boltzmann law where

Let's ignore this.

Combining all modalities, heat conducted through a plate which is removed via convection and radiation would yield

-kA(dT/dy) at the wall = hA(Tw-Tf) + q(radiation)

But in practice q(radiation) can often be ignored. And convection is minimal if (as at the wall) tw is close to Tf.

Anyway, the first equation listed gives good values for gases at moderately low temperatures. And the equations for both are relatively easy to combine and solve for using boundary conditions. I'd refer you to J.P. Holman's Heat Transfer for more details.

andy_fl
12-13-2002, 11:16 PM
Originally posted by Dr_Paprika

q(left face)=-kA*dT/dx

Not correct for gases since conduction is not the only way heat can be transferred at left. Also you are assuming that no mass enters or leaves the area inside dx,dy,dz which is wrong.

Desmostylus
12-14-2002, 12:01 AM
andy_fl:

1. Google "absolute measurement of thermal conductivity of gases". Read it yourself.

2. Don't respond rudely to posters like Dr_Paprika who are trying to answer your question for you.

3. I don't know why you start these threads in the first place. You usually end up with an ass-kicking.

andy_fl
12-14-2002, 09:30 AM
Originally posted by Desmostylus
andy_fl:

[QUOTE]1. Google "absolute measurement of thermal conductivity of gases". Read it yourself.

Done that - the search in quotes give no results. Searched without quotes too, did'nt find any detailed explanation.

2. Don't respond rudely to posters like [b]Dr_Paprika who are trying to answer your question for you.

Apart from disagreeing with Dr_Paprika, I don't see how I am being rude. Is it wrong to disagree on the SDMB ? Is there any rule that you cannot disagree with some of the "posters" ?

3. I don't know why you start these threads in the first place. You usually end up with an ass-kicking.

Thats your opinion and guess who is being rude now. I am asking an question with an factual answer and as far as I know, I'm not violating any forum rules. If I am, let me know.

Squink
12-14-2002, 10:58 AM
"thermal conductivity of gases" or "thermal conductivity of freon" give some fairly detailed links. The measurement makes a good lab exercise for aspiring p-chemists.

Dr_Paprika
12-14-2002, 11:18 PM
The given equation defines thermal conductivity. It is perfectly valid for gases. I believe I stated heat loss also occurs due to convection and radiation. I am sure you agree that if a closed body containing a gas is given a specific amount of energy, and that this body is located in a fluid at roughly the same temperature as the body, convection is minimal.

But don't take my word for it. "The equation q=-kA(dT/dt) is the defining equation for thermal conductivity. On the basis of this definition, experimental measurements may be made to determine the thermal conductivity of different materials. For gases at moderately low temperatures (addendum: hence a small Tw-Tf), analytic treatments in the kinetic theory of gases may be used to accurately predict the experiementally observed values." (Heat Transfer, 7th ed., JP Holman, p.6).

Dr_Paprika, Bachelor Mechanical Engineering (Hons)
Gold Medal Recipient