Homework 2

[Josh Duckett]

Have you gotten anywhere with Homework 2? This stuff isn't really
like anything we've been doing in class.

For 5.1, the boundary layer would just serve to make the available
diameter smaller right? So the number density would increase for the
mean free path since the usable volume would get a bit smaller in
diameter by two times the boundary layer thickness delta?

I haven't gone through 5.2 yet but its seems fairly straight forward
from following the energy and momentum transport sections.

For 6.1 how did you calculate the diameter to use with equation 6.15b
or is that even what you did? I calculated the diameter based on
inverse of number density which would give me volume/molecule - then I
used spherical molecule assumption to calculate the diameter based on
that volume/molecule? I didn't get the answer they had calculated in
the text - I got a smaller answer of 1.64e19 /cm^3sec. Have you
gotten this one worked out?

Do you understand what the difference is for Problem 6.2 from what
they did in the text to get equation 6.15b? It seems like we are just
supposed to simplify 6.15b for single species in part A - that what
you are thinking? Total rate for part b would just be collisions of
like particles plus collisions of unlike particles?

Josh

[Josh Duckett]

I emailed Hassan about 5.1. He said it has to do with viscosity but
I'm still not sure how that links to Mean Free Path. Have you had any
luck with that one.

I got 5.2 and 6.2 but I'm getting a smaller answer for 6.1. Have you
gotten the answer they had in the text?

Josh

[Bryan Susi]

Josh, I was able to get the answer in the text for 6.1. I used the
relation of ch.1 eq. 5.5 , ch.2 eq.5.12, and the mfp relation cited
below ch.2 eq. 6.16. These should provide the necessary substitutions
to solve for d^2 which then goes into the Zab relationship. For 6.2,
is it as simple as Ztotal = Zab + Zaa + Zbb? It seems that you would
end up with way too many collisions...

[Josh Duckett]

Thanks for the info for 6.1. I got 6.2 worked out and to be the right number of collisions the same way you did it. For the proof in part c just set Ntotal = Na + Nb and substitute in...when u do all the algebra it will work out the same as part A...let me know if that doesn't make sense

Josh
Sent from my Verizon Wireless BlackBerry

[Josh Duckett]

Bryan

Have you gotten 5.1 worked out?

Josh

[Josh Duckett]

Did you get the answer in the book for 6.1? I worked through the
equations you gave and got 1.696e25 /cm^3*s. What did you get for d -
I got .02278cm.

josh

[Bryan]

I was able to get the answer in the book, but I never solved for d^2 directly, instead I just subbed in the expression for d^2 into the Zab expression. (Since we're dealing with huge/tiny numbers, I kept things as symbolic as possible until the end) 

The d you've calculated seems very, very high. Some of the other values I did solve for were mAr=6.6412*10^-23 and nAr=3.0115*10^19, maybe they help locate your error? 

[Josh Duckett]

Yea I know its large...haven't found the error yet

Sent from my Verizon Wireless BlackBerry

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From: Bryan <sprtn...@gmail.com>

Date: Tue, 16 Feb 2010 23:04:32 -0500

To: <mae...@googlegroups.com>

Subject: Re: Homework 2