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How to utilize the material properties inside Mathematica?

Posted 9 years ago

Hello everyone,

I have questions regarding properties stored inside Mathematica. I am a master student researching on heat transfer. As I found out that Mathematica itself have data related to the thermodynamic data for the materials, for example, steel, aluminium, water, ammonia and others. I would like to use the data for my calculations. But the problem is that, the properties are dependant on the temperatures of the material itself (though, for heat conductivity, does not change much in the range of 300 K to 400 K).

Now how do I extract data for the properties for different temperature? For example, thermal conductivity for aluminium for the range of 100 K to 1000 K.

What I have learned is how to find the properties of Air, for room temperature? ThermodynamicData["Air", "ThermalConductivity"]

Thank you very much for your help.

POSTED BY: Thai Kee Gan
9 Replies
Posted 9 years ago

Hello again, here I encounter 1 problem, Apparently when I wanted to have the thermal conductivity of water above 373 K, the data retrieved change into water vapour thermal conductivity. Is there anyway to get the data in liquid form?

POSTED BY: Thai Kee Gan

I don't know. What about changing the pressure to prevent the water from evaporating?

Say

ThermodynamicData["Water", "ThermalConductivity", {"Temperature" -> Quantity[375, "Kelvins"], "Pressure" -> Quantity[1114, "Hectopascals"]}]

(*Quantity[0.679663, ("Watts")/("KelvinsDifference" "Meters")]*)

vs the same at standard pressure:

ThermodynamicData["Water", "ThermalConductivity", {"Temperature" -> Quantity[375, "Kelvins"]}]

(*Quantity[0.0252191, ("Watts")/("KelvinsDifference" "Meters")]*)

Cheers,

M.

POSTED BY: Marco Thiel
Posted 9 years ago

Thank you

POSTED BY: Thai Kee Gan
Posted 9 years ago

Is there any way for me to store the data into my PC? As I need to iterate the information according to the temperature of the material.

POSTED BY: Thai Kee Gan

Well, you can use the Export command.

Export["~/Desktop/conductivitydata.csv", data]

would be one option (you obviously might need to adapt the directory).

Cheers,

M.

POSTED BY: Marco Thiel
Posted 9 years ago

So, the data is extracted from WolframAlpha, which is connected to the server and it does not stored in my PC?

POSTED BY: Thai Kee Gan
Posted 9 years ago

Oh thanks alot Marco. Your are right, that graph looks weird. Thanks for verifying it for me. I need to write the program to change the value of the properties of the materials according to the temperature given. Now that I know what to do, you really helped me :D

POSTED BY: Thai Kee Gan

Just to confirm one point. On this website:

http://www.efunda.com/materials/elements/TCTable.cfm?ElementID=Al

there is some data on Aluminium. I can use that data (Import does not work so this requires manual work):

dataurl = Partition[ToExpression@StringSplit[
     {"1 101325 4110 2 101325 8180 3 101325 12100 4 101325 15700 5 \
101325 18800 6 101325 21300 7 101325 22900 8 101325 23700 9 101325 \
23900 10 101325 23500 15 101325 17600 20 101325 11700 30 101325 4950 \
40 101325 2400 50 101325 1350 60 101325 850 70 101325 585 80 101325 \
432 90 101325 342 100 101325 302 150 101325 248 200 101325 237 250 \
101325 235 300 101325 237 350 101325 240 400 101325 240 500 101325 \
236 600 101325 231 800 101325 218"}, " "][[1]], 3]

If we plot everything together (red dots Mathematica and blue curve from website) we get:

Show[ListPlot[Transpose[{Range[100, 1000, 100], QuantityMagnitude@ data}], PlotRange -> All, PlotStyle -> Red], 
ListLinePlot[dataurl[[-10 ;;, {1, 3}]]]]

enter image description here

which shows good agreement.

Cheers,

Marco

POSTED BY: Marco Thiel

This is not elegant and a bit slow, but

data=Table[WolframAlpha["Thermal Conductivity of Aluminium at " <> ToString[h] <>  "K", {{"Result", 1}, "ComputableData"}], {h, 100, 1000, 100}]

appears to be giving results. Here is a plot

ListPlot[Transpose[{Range[100, 1000, 100], QuantityMagnitude@ data}]] 

This looks a bit strange to me, but there we go:

enter image description here

Cheers,

M.

I cannot verify the values, but some do not seem to be in total disagreement with, data, e.g. here.

POSTED BY: Marco Thiel
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