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Posted: 8/31/2004 11:00:45 AM EST
[Last Edit: 9/1/2004 12:37:31 PM EST by scottryan]
I need help with these two problems for my Metals 360 college class.....


Question 3-16 reads:

Alpha-Mn has a cubic structure with side length a=0.8931 nm and a density of 7.47 g/cm^3. Beta-Mn has a different cubic structure with side length a=0.6326 nm and a density of 7.26 g/cm^3. The atomic weight of manganese is 54.938 g/mol and the atomic radius is 0.112 nm. Determine the percent volume change that would occur if Alpha-Mn transforms to Beta-Mn.

When the problem says "cubic structure", that is on the atomic level with how the atoms are arranged.

Question 3-50 reads:

You would like to sort iron specimens, some of with are face centered cubic and others body centered cubic. Design an x-ray diffraction method by which this can be accomplished.



thanks
Scott
Link Posted: 8/31/2004 11:03:29 AM EST

USNA EE Final Exam:

"Using this paperclip, this piece of string, the battery from your calculator, and your working knowledge of electrical engineering, construct a working model of the Aegis SPY-1A radar."

Sorry, dude, but I almost failed Materials, and I don't have my books handy (yes, I kept them) otherwise I'd be happy to help.....
Link Posted: 8/31/2004 11:04:09 AM EST
thats tough, I'm a chemist and I think you need different help than I can give
Link Posted: 8/31/2004 11:06:23 AM EST
UHHH 2+2 = 10 YEAH!

This is why I am POLS major, Izz stupidzz in mathzz!
Link Posted: 8/31/2004 11:17:42 AM EST
3-16 is just dimensional analysis. Mass is conserved, therefore 1gram A = 1 gram B, use the densities to calculate volume change.

It;'s been too long for me to know the xray solution off the top of my head
Link Posted: 8/31/2004 11:18:53 AM EST
yeah 3-16 is sort of a trick question in that they give you way too much info.
Link Posted: 8/31/2004 11:25:19 AM EST
Volume of Alpha = 0.8931 ^ 3 = 0.7124

Volume of Beta = 0.6326 ^ 3 = 0.2532

Difference = 0.4592

% difference is -64.5%

That said, I have no idea how Alpha would convert to Beta, or even if it can.....
Link Posted: 8/31/2004 12:39:35 PM EST

Originally Posted By ROKIT88:
3-16 is just dimensional analysis. Mass is conserved, therefore 1gram A = 1 gram B, use the densities to calculate volume change.

It;'s been too long for me to know the xray solution off the top of my head



Could you explain this further? I am not quite following how you would calculate the volume change from the densities?
Link Posted: 8/31/2004 12:45:11 PM EST
[Last Edit: 8/31/2004 12:46:03 PM EST by ajm1911]
sorry i was wrong
Link Posted: 8/31/2004 1:54:00 PM EST
ROKIT88's solution yields the mass change, that's why you don't understand.

Can't help with the second question, except with a smart ass answer.
Link Posted: 8/31/2004 2:38:56 PM EST
scottryan,

The volume would change by 2.9% (7.47/7.26 = 1.029).
Link Posted: 8/31/2004 2:45:01 PM EST

Originally Posted By Mike_Mills:
scottryan,

The volume would change by 2.9% (7.47/7.26 = 1.029).




How?
Link Posted: 8/31/2004 2:50:30 PM EST
scottryan:
On 3-16, assume you have 1 gram of Alpha.
Then Alpha converts to Beta, you still have 1 gram of material, but now in Beta form.
You dont need the side lengths, its just the ratio of the densities.

Kharn
Link Posted: 8/31/2004 2:57:31 PM EST
Man, with the Collective knowledge we have on this site, we could rule the world!
Link Posted: 8/31/2004 3:27:29 PM EST
Well, if both Alpha and Beta have cubic structures, and the sides are given, then wouldn't the volume reduction be calculated as I did above, irrespective of density?

I don't think I made any arithmetic mistakes.....
Link Posted: 8/31/2004 4:10:07 PM EST

You would like to sort iron specimens, some of with are face centered cubic and others body centered cubic. Design an x-ray diffraction method by which this can be accomplished.


You can determine the angle between the lattice points for fcc and bcc Fe by using the Bergers vector. By directing the x-rays at a certain angle, you can cause x-rays to refract at different angles, depending on which structure the sample is.
Link Posted: 8/31/2004 4:13:45 PM EST
Mary Ann +1 all the way. Unless of course you are making the WKRP calculation then its Bailey +1 all the way.
Link Posted: 8/31/2004 4:14:37 PM EST

Originally Posted By Zaphod:
Well, if both Alpha and Beta have cubic structures, and the sides are given, then wouldn't the volume reduction be calculated as I did above, irrespective of density?

I don't think I made any arithmetic mistakes.....



Your calculation relies on two unstated assumptions: (1) each bcc and fcc crystal is a cube and (2) that each bcc crystal has the same mass as a fcc crystal. The second assumption is not generally true. A single fcc crystal generally has more mass than a single bcc crystal due to a tighter "packing factor." Because mass is conserved, the number of crystals changes after the transition, which throws off your calculation.

X-ray diffraction question - I'm guessing, but I think that there are only a limited number of possible bragg angles for bcc and fcc crystals. If you know the bragg angle, you can tell by looking at a table whether it's bcc or fcc. Doesn't your textbook or class notes tell you something on the subject?
Link Posted: 8/31/2004 4:37:12 PM EST

Originally Posted By AeroE:
ROKIT88's solution yields the mass change, that's why you don't understand.

Can't help with the second question, except with a smart ass answer.



The inverse of the density is the specific volume.

Fractional volume change A -> B:

(1/DensityB - 1/DensityA)/(1/DensityA)

Multiply by 100 for %


Link Posted: 8/31/2004 5:42:46 PM EST

Originally Posted By danno-in-michigan:

Originally Posted By Zaphod:
Well, if both Alpha and Beta have cubic structures, and the sides are given, then wouldn't the volume reduction be calculated as I did above, irrespective of density?

I don't think I made any arithmetic mistakes.....



Your calculation relies on two unstated assumptions: (1) each bcc and fcc crystal is a cube and (2) that each bcc crystal has the same mass as a fcc crystal. The second assumption is not generally true. A single fcc crystal generally has more mass than a single bcc crystal due to a tighter "packing factor." Because mass is conserved, the number of crystals changes after the transition, which throws off your calculation.

X-ray diffraction question - I'm guessing, but I think that there are only a limited number of possible bragg angles for bcc and fcc crystals. If you know the bragg angle, you can tell by looking at a table whether it's bcc or fcc. Doesn't your textbook or class notes tell you something on the subject?



My classes are taught by foreigners. I had 8 problems in this assignment. I am lucky I could solve 6 of them on my own.
Link Posted: 8/31/2004 5:49:47 PM EST

Originally Posted By atomicferret:
Man, with the Collective knowledge we have on this site, we could rule the world!



+1, let's go for it.
Link Posted: 8/31/2004 8:47:10 PM EST


--------------------------------------------------------------------------------
Originally Posted By Mike_Mills:
scottryan,

The volume would change by 2.9% (7.47/7.26 = 1.029).
--------------------------------------------------------------------------------


How?






scottyryan,

The exam question does not ask how, hence, in my answer I did not address the mechanism whereby the volumetric change occurs - neither should you. It asks only for the percent volumetric change. This, I gave.
Link Posted: 8/31/2004 9:37:40 PM EST
MikeMills has the right answer as does Weise. If you are still having trouble seeing it, here is a long form explanation that may help illustrate it for you:

Start with any known volume of A (1cc works nicely)
Multiply by the density of A to eliminate the original dimension (volume)
You are now left with grams of A
grams A = grams B (no material is being added or removed from the system/sample)
Now divide (or multiply by the inverse) by the density of B
You are now left with the volume of B
Vol.A subtracted from Vol.B = delta V
divide delta V by Vol. A
multiply by 100%

voila

Write it all out with the units, watch the units cancel. Get comfortable with the units and how to convert them backwards in your sleep because it sounds like you will be doing a lot of this in your class.Length, mass, heat, energy, time, velocity, and the coefficient of pie...


Link Posted: 8/31/2004 10:02:49 PM EST
[Last Edit: 8/31/2004 10:03:06 PM EST by Mike_Mills]
.
Link Posted: 8/31/2004 10:15:12 PM EST
Wow.....I'm just a CS major, but hanging around with a Metallurgical Engineering PHD student all the time musta rubbed off on me. I coulda solved the first one myself, and actually understood the answers for the 2nd one, even though I probably wouldn't have come up with them on my own.
Link Posted: 9/1/2004 8:54:31 AM EST
Thanks for helping me guys. If someone would like to explain this problem further, it would be appreciated. I need to know how these problems work, not just how to get the answer.
Link Posted: 9/1/2004 12:28:54 PM EST
[Last Edit: 9/1/2004 12:34:30 PM EST by scottryan]
I don't think anyone has the correct answer on problem 3-16 yet. My teaching assistant told me I needed to find the number of atoms per unit cell and the mass does matter.

When the problem says "cubic structure", that is on the atomic level.
Link Posted: 9/1/2004 3:41:46 PM EST
I don't think the answer will be any different, but here is another method which combines all of the given info.

start with the density of A
Divide by weight of Mn now you have moles/cc of A
convert moles to Atoms
convert cubic centimeters to cubic Angstroms
You now have atoms per angstrom^3
multiply by the cubic structure volume (cube the given length)
Now you have the Number of atoms in a unit cell

Do the same for B

Now that you have the number of atoms in each cube

Take the cubic structure volume of B multiplied by the number of atoms in structure A and divide by the number of atoms in structure B (this gives you the volume of B that can be created from one cubic structure unit of A)

Subtract this volume from the Cubic Structure of A and divide that number by the Volume of the cubic structure of A multiplied by 100%

The answer should be the same.
Find a nice cast iron frying pan to demonstrate its metallurgical properties on your TA.
Link Posted: 9/1/2004 3:48:52 PM EST
As far as I can tell it has somethign to do with 3.1415926
Link Posted: 9/1/2004 6:30:23 PM EST
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