# Heat flow through rubber sample

1. Feb 18, 2014

### examorph

I have a rubber sample here, basically it is a slice with a diameter of 45mm and a depth of 3mm. If this sample was at 25 degrees Celsius and I applied 180 degrees temperature all around it (top, bottom faces and outside diameter) how long would it take for the centre point (last point to get heated) to reach the 180 degrees Celsius temperature?

The materials thermal properties are:

Thermal Conductivity: 0.36 W/m*K
Specific Heat Capacity: 2 kJ/Kg K

I am really stuck with this and would appreciate any guidance at all. Thank you.

2. Feb 18, 2014

### .Scott

It will never happen - but it will get very, very close.

3. Feb 18, 2014

### examorph

Thank you for the reply, could you please explain why it would never happen and also how long would it take to get very close?

4. Feb 18, 2014

### Staff: Mentor

Scott is right. It will approach 180 asymptotically. However, on a practical basis, it will approach 180 very closely in a finite amount of time. This is a transient heat conduction problem to a slab. The half-thickness of the slab is b=0.15 cm, and the thermal diffusivity of the rubber (based on the data you provided and a density of 1 gm/cc) is about α=0.0018 cm2/sec. The center of the slab should closely approach steady state when

$$\frac{αt}{b^2}=1.2$$

I leave it to you to calculate t.

Chet

5. Feb 18, 2014

### examorph

Thank you very much, the results were similar to what I got from my simulations also. Could you please help me understand what this equation is?
I searched for it online and kept on running into the energy equation: https://imageshack.com/i/0k6mvlp
Is this where the equation came from and if so, it looks very familiar, it reminds me of the Navier-Stokes equation used in fluid dynamics, does it have a similar derivation?

Also, could you explain how you got the value 1.2 to me please.

6. Feb 18, 2014

### Staff: Mentor

See Transport Phenomena by Bird, Stewart, and Lightfoot. There is a chapter on Temperature Variations with More Than One Independent Variables. In the equation you showed, the velocity is equal to zero, and the heat generation rate is equal to zero. This gives the 1D transient heat conduction equation. Bird et al presents a set of 3 graphs showing transient temperature profiles for a slab, a cylinder, and a sphere. In the graph for a slab, a Fourier number of 1.2 looked like it would be about adequate for the center temperature to approach the wall temperature.

chet

7. Feb 18, 2014

### examorph

Thank you very much for your help, it is greatly appreciated!

When I start looking at thicker, larger cross section 1D problems will I need to use different equations or would the above equation still apply?

Also, if I have a large diameter hollow cylinder such as that shown below, with a temperature applied to the top face could I apply the 1D formula on the 2D cross section (slice) of the cylinder to determine how long it would take for the heat to transfer??
http://www.onlinemathlearning.com/image-files/surfaceareahollowcylinder.gif

8. Feb 18, 2014

### Staff: Mentor

The same equation would still apply.
The same equation would apply if the cylindrical surfaces are insulated, or the cylinder is very short. Otherwise, there can be heat loss out the sides.

9. Feb 19, 2014

### Staff: Mentor

This is ridiculous.

The entire sample of rubber will reach thermal equilibirum in a finite time. That's an observable fact. If you have an equation that says otherwise, either that equation is wrong or an approximation to reality.

10. Feb 19, 2014

### examorph

Thanks for the replies, what if I was trying to find out how long it would take a cylinder such as that shown in my pervious post to reach the wall temperature. The conditions are that there would be a temperature applied to the outside diameter surface, inside diameter surface and top/bottom. Would this now become a 2D problem and require a 2D version of the energy transportation equation? If so, could you please give me an example?
Also, I would imagine that something bigger than a slab would not reach equilibrium in a finite time therefore, the results should be similar to what occurs in reality, is this true?

11. Feb 19, 2014

### Staff: Mentor

If the annular thickness is small compared to the cylinder diameter, then you can neglect the curvature, and the result will still be the same as a flat slab. You can also neglect the 2D feature if you are willing to accept the infinitely-wide slab result as an upper bound to the true amount of time required for the annulus.
I don't understand this question.

Chet

12. Feb 19, 2014

### Staff: Mentor

If it reaches equilibrium in a finite amount of time, please tell us exactly what your equation for that finite amount of time is (in terms of the thickness of the slab and the physical properties of the material). Also tell us how you determine observationally the precise finite moment that equilibrium has been reached.

Chet

Last edited: Feb 19, 2014
13. Feb 19, 2014

### Staff: Mentor

I never said that I had an equation. I just found the statement "it will never happen" to be completely unhelpful, and actually wrong, as thermal equilibirum is reached in practice.

The heat equation might say that equilibrium is only reached asymptotically, but the model neglects both the discrete (molecular) nature of the substance and the fact that the temperature of the environment need only be reached within the range of fluctuation of said temperature (or within the precision of the thermometer in the real world).

Just to be clear, I am not criticising the usefulness of the heat equation, just the usefulness of short, unhelpful answers that pop up on PF from time to time.

14. Feb 19, 2014

### Staff: Mentor

Thanks DrClaude. I can certainly see what you are driving at. I was just a little concerned that your earlier response might possibly have detracted a bit (for the uninitiated) from the validity of the practical result I presented for calculating the time at which the sample effectively reaches equilibrium. On second thought, probably not.

Chet

15. Feb 19, 2014

### examorph

Thank you again for your help.

Is there some sort of rough ratio which I could use to determine whether or not the component that I am dealing can be considered as a flat slab or not?

For example I am currently dealing with a component which has an outer diameter of 120mm a width of 5mm (therefore a 110mm inner diameter) and a thickness of 10mm. Another component I am looking at has an outer diameter of about 300mm with all other dimensions being the same as above. Would the 1D energy transportation formula you described still apply?

16. Feb 19, 2014

### Staff: Mentor

Actually, I think you mean that the thickness is 5 mm.
Actually, in my judgement, the 1D approximation would be adequate in both these cases.

chet

17. Feb 20, 2014

### examorph

Thank you for all of the wonderful help it has really helped allot!

There is one thing that is still confusing me however. How is this equation independent of the temperature difference and contact area as I thought that these two variables played a large role in how heat traveled through a component?

18. Feb 20, 2014

### Staff: Mentor

The equation tells you when the temperature is reduced to a certain small fraction of the original temperature difference. So it's a percentage thing. The contact area determines the total amount of heat transferred, but this analysis focuses on the heat flow per unit area.

Chet

19. Feb 21, 2014

### examorph

So, based on this equation higher temperatures at the surface don't result in heat travelling faster through the component but rather the same distance when compared to a lower applied temperature but in larger quantities?

Also, you mentioned that a rough ratio of 10:1 could be used to determine whether or not to use the 1D equation,if this ratio was not met would I have to move onto the 2D equation? Is there any other cases in which I would have to leave the 1D equation and instead use 2D? Also, does the 2D equation also compare the temperature difference as a percentage?

Thank you.

20. Feb 21, 2014

### Staff: Mentor

yes
No. it's still 1D radial, but the more exact solution involves the ratio of the diameters. The two solutions merge when the ratio of the diameter to the thickness becomes large. You can probably find the solution in Carslaw and Jaeger, Conduction of Heat in Solids.
It gives the absolute temperature difference as a direct proportion to the initial temperature difference. You can divide both sides of the equation by the initial temperature difference to get the percentage, which is independent of the initial temperature difference.

Chet