How to work out thermal diffusivity using Angstrom's method?

[charlie mallet]

Hi guys,
Just looking for some help with the theory on a project I've been set at university.
The experiment involved heating a long thin bar from one end with a sinusoidal heat wave. Thermocouples were then placed every 7 cm (4 in total) and temperatures were recorded. The experiment brief sheet stressed that we work out thermal diffusivity from this and then thermal conductivity from that.

Any help would be greatly appreciated

 

[Chestermiller]

You are expected to show some effort at doping this out. The homework template asks you to provide whatever equations that you deem relevant, and then to discuss what you have done so far. So please proceed.

 

[charlie mallet]

OK sorry hadn't read that bit :)

I did some research and stumbled on a paper that was doing similar but with carbon fibre strands. Using the heat equation they had derived an expression whose thermal diffusivity was equal to (l^2)/(2tln(T1/T2)), where l is the distance between 2 thermocouples, t is the time between the two peak temperatures at each thermocouple and T is peak temp at each thermocouple. I tried using my figures yet it was slightly off. The radius of the bar I used was about 5mm. Do I need to adapt this equation or can my bar be treated at 1 dimensional?

 

[Chestermiller]

Your bar can be treated as 1D, but it should be insulated. What do you mean "it was slightly off?" I would not analyze the data this way. What was the equation they gave for the peak temperature as a function of distance?

 

[charlie mallet]

The bar was isolated using quite a primitive foam box. The value I found online for the thermal diffusivity of copper was 1.11E-4 M^2/S. However I calculated a value of 6.34e-4 M^2/S. The equation given was (dT/dt)=D(d^2T/dx^2) - BDT where B was a heat loss parameter

 

[Chestermiller]

The bar was isolated using quite a primitive foam box. The value I found online for the thermal diffusivity of copper was 1.11E-4 M^2/S. However I calculated a value of 6.34e-4 M^2/S. The equation given was (dT/dt)=D(d^2T/dx^2) - BDT where B was a heat loss parameter

I thought you said there was no heat loss. Also, what I'm asking for is their final solution for T(x,t). Also, what do you get if you make a semi-log plot of the amplitude of the temperature variation as a function of x (i.e., log (delta T) vs x)?

 

[charlie mallet]

sorry should have written insulated not isolated. I can't see that they've got any other solutions. I've copied the link if you want to have a look. I just tried plotting a semi log graph but it was not linear.
http://acs.omnibooksonline.com/data/papers/1999_178.pdf

 

[Chestermiller]

I solved this problem on my own, and obtained the following result:
$$A(z)=A_0e^{-\sqrt{\frac{\omega}{2\alpha}}z}$$where A(z) is the amplitude of the temperature variation at distance z from the driven end of the rod, $A_0$ is the amplitude of the temperature variation at z = 0, $\omega$ is the frequency of the oscillation (radians per second), and $\alpha$ is the thermal diffusivity. So, if you plot log A as a function of z, you should get a straight line. Please show me the data you obtained for A vs z.

 

[charlie mallet]

Thank you very much chester.
Did you derive this from the heat equation?
I've attached a graph of one set of the results I've obtained, please ignore the lack of formatting as this is still a process undergoing!
The image shows our results from an aluminium sample with the heating element working at a frequency of 1mHz. The thermocouples were placed every 4cm with the first one 4cm from the heating element.
I rearranged your equation for thermal diffusivity and obtained a result of 9.5e-3 but the true value is 9.7e-5. This makes me think I've made a calculation error somewhere?
With regards to the graph, each trend line represents the values from one of the thermocouples, each one with lower amplitude representing an extra 4cm along the bar.
I very much took estimates from the graph for temperature and time. If you haven't received it in a readable format please say and I will try to send again in a different type of file.

 

[Chestermiller]

Thank you very much chester.
Did you derive this from the heat equation?

Yes, this is my solution to the transient heat conduction equation, but I omitted the heat loss term. I can see from your data that the heat loss term is significant. The equation I derived describes the behavior at long times, when the average temperatures have leveled off. Do you have any data at such longer times? Analyzing the results during this transient period (before dynamic steady state has been achieved) requires a much more elaborate solution and game plan for comparison. In my judgment, you were supposed to have waited long enough for dynamic steady state to be achieved.

It looks from your graph that the period of the oscillation is about 10 minutes. How does this relate to your 1 mHz forcing. Is mHz milli-Hertz?

 

[Chestermiller]

I have another question. The heat flux at z = 0: It is constant, plus a superimposed sinusoidal variation?

 

[charlie mallet]

Hi Chester,
Using you solution to the heat equation I've actually got some correct answers. I think you are most certainly right that we were suppose to wait for it to level off. However to counter for this, when taking amplitudes for each "line", I have taken the zero point at different places; where the relative amplitude is zero.

Would you mind showing me a bit more on how you derived that solution from the heat equation?
Best
Charlie

 

[Chestermiller]

After I solved it on my own, I also found the solution fully worked out in Transport Phenomena by Bird, Stewart, and Lightfoot. They solved it for a sinusoidal heat flux, but the answer is the same for a sinusoidal temperature variation.