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### Topic: Solid expansion  (Read 4415 times)

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#### shafaifer

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##### Solid expansion
« on: January 08, 2015, 08:06:00 AM »
Let's say you have a metal composition of 88 % iron and 12 % aluminium and you wish to calculate the general volumetric thermal expansion coefficient. Would this be:

αcompositionironaluminium=(αFe-(0.12*αFe)) + (αAl-(0.88*αAl))

I want to know this in order to find out if a given metal composition will expand volumetrically when exposed to a temperature change of 8 K (in this case from 285.15 K to 293.15 K). The metal blank will be exposed to 293.15 K for 1 hour and then taken back to the temperature of 285.15 K.

#### Borek

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##### Re: Solid expansion
« Reply #1 on: January 08, 2015, 09:14:30 AM »
I don't know exact answer, but if there is one thing I have learned about properties of mixtures, it is - linear approximation may, or may not work. If you need really accurate numbers - it doesn't.
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#### Corribus

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##### Re: Solid expansion
« Reply #2 on: January 08, 2015, 10:07:15 AM »
It only works as an approximation if there is little to no interaction between the two components. In an alloy, this is almost certainly not the case.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### shafaifer

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##### Re: Solid expansion
« Reply #3 on: January 08, 2015, 10:39:02 AM »
Thank you very much, greatly. I am sorry I forgot to specify the composition as an alloy. So Borek, you say there will be no noticeable expansion of the metal?

Corribus, how should I calculate this coefficient of an alloy of 88 % Fe and 12 % Al, if what I suggested was incorrect?

#### Corribus

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##### Re: Solid expansion
« Reply #4 on: January 08, 2015, 12:01:41 PM »
I don't think Borek was saying that. An alloy is essentially a solution of one metal in another. So you can think of it in terms of an ideal solution versus a non-ideal solution.  In an ideal mixture, there is no or minimal interaction between the solute and solvent, and many physical properties can be approximated as a mass- or mole-fraction-weighted average between the values for the two "pure" substances. (E.g., vapor pressure, Raoult's Law.) In a non-ideal mixture, this is not the case, because the interaction of the two substances is not-negligible. In these cases, it is not a simple matter to calculate what the final observable value will be... particularly for something as complicated as thermal expansion coefficient.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Borek

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##### Re: Solid expansion
« Reply #5 on: January 08, 2015, 03:42:31 PM »
So Borek, you say there will be no noticeable expansion of the metal?

No. What I was saying was that a simple linear interpolation will most likely NOT give a reliable value of the expansion coefficient.
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#### shafaifer

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##### Re: Solid expansion
« Reply #6 on: January 09, 2015, 06:01:41 AM »
Thank you.

Just an application if it is okay for you to help me again.

The thermal expansion coefficient for C45 steel is given by 11.5 10-6 0C-1 at 20 - 100 0C. Is it credible to assume that the coefficient has the same value for the interval 120C - 1000C? If assumed, a volume change for isotropic materials (http://en.wikipedia.org/wiki/Thermal_expansion) can be calculated by:

ΔV = (3V-aLΔT) - V

#### mjc123

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##### Re: Solid expansion
« Reply #7 on: January 09, 2015, 07:19:07 AM »
I think you mean
ΔV = 3VaLΔT

#### Enthalpy

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##### Re: Solid expansion
« Reply #8 on: January 13, 2015, 04:09:30 PM »
I would not use any sort of interpolation by composition for the expansion coefficient, even less so for steel.

First, steel is extremely complicated (the worst alloy). Expect the unexpected with it. For instance, common alloys can be martensitic (12ppm/K), austenitic (17ppm/K for usual stainless steel), ferritic (13ppm/K), ledeburitic (12ppm/K) and many more. Alloying elements, in amounts far smaller than your 12% (1% for C, 2% for Si...), decide the crystal structure which changes stepwise the expansion coefficient.

From an uncertain source I found >2% aluminium to be alphagenic, so your 12% Al would make a ferrite or martensite, beginning near 12% rather than 17% - and then Al would supposedly increase the thermal expansion. Maybe - or maybe not.

Because then, you habe abnormal behaviours. 36%Ni in Fe make the reknown Invar alloy, with zero expansion around room temperature, despite Fe contributes 12% and Ni slightly more. Said to be a consequence of magnetism - essentially the same alloy is also called Permalloy, with a huge magnetic permeability.

Other alloys change their crystal structure near room temperature, and then the expansion coefficient can be extreme, with no relation with the constituents. For instance shape memory alloys do that.

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I know no steel with just 12% Al. AlNiCo contains 12% Al but also much Ni and Co. Is it a permanent magnet, a metal glass...? What uses?

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I'd use the same expansion coefficient between 12-100°C and 20-100°C. For "normal" steel (C45 is, as far as steel can be normal) the expansion coefficient just increases slowly with temperature, and at 100°C it's but bigger than at RT, maybe 1-2ppm/K more. You can compare the temperature range with the melting point or with the temperature of crystal transformation (austenitization 700-900°C for steel), for "normal" alloys.

Wiki states that the volume increases three times faster than the linear dimensions, nice, no worry with that. But I'm not sure of your derivation, and also, remember this is for small temperature variations, nothing more than some calculus. It is not a working physical model that predicts a value over a wide range from basic properties.

More generally, alloys are not quite predictible. Don't believe theories.