1st Part of the Course
Please post any questions relating to material that was covered on the 1st test as comments on this message.
posted by Dr. B at 8:28 PM
Friday, December 08, 2006
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24 Comments:
On slide 5 of your second lecture, you have Na= Ja +xa(Na + Nb). I know the first Na is the molar flux. Is the second Na the number of moles of a? Too many N's and K's in this course.
Billy Bob:
BOTH of the Na terms are the molar flux of species A.
I agree, too many ks. I think the Ns are not so bad.
Good to see you are studying already.
Best of luck on the final!
For problem 3.14 How do you know not to use the log mean in the Na eqn. I know they are both derived from each other but why would you use ln((1-xa2)/(1-xa1))it in this problem but the log mean in others?
Dimensionless diffusion equation when and why would you use it. I'm just confused in general about the whole concept.
For prob 3.40 How do you know that it is a non-dilute solution of water in air? By looking at the mol fraction of water in air it looks as though it's pretty dilute. Am I missing something here? Also would it be true to say that you have to use (yai-yab)lm whenever you can't neglect bulk flow caused by diffusion i.e. a non dilute solution??
One more question about related to 3.40, So would it be right to say if all of the mass xfer resistance were in the liquid phase than Kx=kx otherwise 1/Kx = 1/(m''ky) + 1/kx.
One question on your previous test was, "Write the form of Fick’s Law that is applicable to unimolecular diffusion of A through stagnant component B." What is the answer to this question?
what are the assupmtions on the kremser group method?
Why are expressions for overall mass transfer coefficients more complex than expressions for overall heat transfer coefficients?
I can't find it in the notes.
Why is the column efficiency in distillation higher than in absorption?
Where can we find the answer to this?
stewie:
Eqns 3-33 and 3-35 are equivalent. So, in some sense, we DID use the log-mean. These two eqns are just algebraic rearrangements of each other. In general I like to use the log-mean form.
Anon:
I am not sure when or where we made the diffusion equation dimensionless. Help me narrow it down a bit and I can give you some more information.
Rocy Balboa:
I think the outlet air is 8.8 mol%. This is more than 1 or 2 mol%, so it is not dilute.
(yai-yab) is a driving force for a stripper. When the driving force is not the same throughout the process, such as in a stripping column, you need to use some sort of average driving force. The right choice for the driving force is the log mean driving force: (yai-yab)lm.
Rocky ii:
Yes, exactly.
guy in the elevator:
See eqn 3-27.
Jonny:
The Kremser method is based on four key assumptions.
1- Equilibrium stage model applies
2- Constant temperature
3- The equilibrium curve is linear: y = K x
4- The process is well-described by an average absorption/stripping/extraction factor that attempts to account for the fact that A, S and E vary from stage to stage.
5- Feed flow rate and compositions must be specified (not really an assumption).
sick of MT:
Expressions for overall mass transfer coefficients are more complex than expressions for overall heat transfer coefficients because , although temperature is continuous across a phase interface, concentration is not. Consequently, overall mass transfer coefficients must take into account the discontinuity of concentration at the interface. This is accomplished by assuming the two phases are in equilibrium at the interface and using an equilibrium equation such as Henry's Law or the Modified Raoult's Law.
nerd:
The key to tray efficiency is the liquid viscosity. In distillation the liquid is hot...at its boiling point. In absorption the liquid is cold so it will absorb gas better. The hot liquid has a lower viscosity and if you check the correlations for tray efficiency, you will find that lower liquid viscosity leads to higher tray efficiency. Higher tray efficiency results in higher column efficiency.
What is the breakthrough time for an adsorption column?
One of the old tests asks "For mass transfer to or from the wall of a tube through which a fluid flows, is the appropriate driving force an arithmetic or log-mean average ?" Is log mean because the concentration through the tube does not vary linearly?
On the final from 2005, how do you go about starting, doing and finishing problem number 4? Have we learned how to do this? Please Help...
Taz:
See slides 10 & 11 from lecture 9 on 10/16.
We called it break point instead of breakthrough time.
When you begin charging an adsorption column at time t = 0, the break point is the time that elapses before the concentration at the outlet rises to some specified level. A typical threshold value might be 5% of the solute concentration in the feed.
graceful:
Imagine a fluid flowing through a tube in which some solute comes out of the wall and dissolves in the fluid. The fluid at the wall surface is saturated. The fluid in the bulk has a low solute concentration at the inlet and a high solute concentration at the outlet. The driving force is the difference between the saturated fluid at the wall and the bulk fluid. The driving force is large at the inlet and smaller at the outlet. We clearly need to use some sort of average driving force. The correct average driving force turns out to be the log-mean driving force.
For more details see slides 6-8 from the PPT of lecture 5 on 10/6.
sleepless:
Good news, my friend, this is a drying problem and we did NOT cover drying this year ! So, you do not need to be able to solve problems like this on the final.
This year, I decided to spend more time on membrane processes and chromatography as well as introducing electrophoresis. I cut back on distillation and eliminated humidification and drying.
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