HW 7 - p 13.11 - Batch Rectification of an Equimolar Mixture of Benzene and Toluene at Constant Reflux - 6 pts
Please post all of your questions about this problem as comments on this post. I will respond with my own comments.
posted by Dr. B at 5:47 PM
12 Comments:
In your hints you say "Then, choose a set of xD values that lie between the initial values of xD and the final xD value of 0.55 mol B/mol. I chose the initial xD and even steps of 0.05. This gave me 6 values of xD and therefore I made 6 McC-T Diagrams !" But if you go from .55 to .9 in increments of .05, isnt this 8 McC-T diagrams? Do we actually have to make the diagrams for each of the xd's? Or can we just do the steps numerically without a visual representation of it?
Are you sure of your solution for Wf? I get 450 for Wf if I use the equation ln(Wf/Wo)=integral, which is not what you have in the slides or is in the book. When I use the equation in the book/slides, ln(Wf/Wo)=int, I get ~2000.
Never mind, we got it.
stage fright:
Doh ! Yes, that makes 8 values of xD.
I would like you to submit just ONE of the 8 McC-T diagrams. If you can do 1, you could probably do the others. Try to use Excel to generate this McC-T plot instead of doing it by hand. For the other values of xD, just your Excel work will be sufficient.
madagascar:
I am glad you got this problem resolved. I wish you had explained the problem and how you resolved it because I cannot figure out what you did ! Oh, well. I am glad it all worked out.
We had the f and o locations reversed in our minds, so we basically reversed the bounds of the integral.
madagascar:
I kind of suspected that, but thanks for clearing it up for everyone here.
I am having the same problem that madagascar had but I don't know why the limits of integration don't correspond for getting a W/W0 of 2(ie a W of 2000). the equation I'm using the equation on slide 18 which is the same as equation 13-3 in the book. Don't you just sum up the area of the trapazoids and set that equal to ln(W/W0)? and then take the exp() of the sum of the areas to get the ratio? when does the switch occur if ln(W0/W) gets you the right answer?
how did you only used 6 McCT diagrams when you took .05 intervals for xD from xDinitial=.91 to xD=.55? wouldn't that be at least 8? only asking because i think this might be why i'm getting a Wf off by 20mol from you which will make the xD ave value quite a bit different.
thanks
well, off by 2% for the average composition, but would still like to know which xD values you used for your 6 McC-T diagrams
very, very:
Eqn 13-3 is the right choice and yes, you sum the areas of the trapezoids. Ln(Wfinal / Winit) = area. I am not sure whatis bugging you here. My guess is the sign issue. Because of the limits on the integral in eqn 13-3, the left hand side...the sum of the areas is NEGATIVE. xfinal < xinit ! That is OK since Winit > Wfinal, Wfinal/Winit < 1 therefore the right side of eqn 13-3 is ALSO NEGATIVE.
It's all good here ! Whoohoo !
karen:
Sorry for the confusion. I actually used 9 McC-T diagrams...{0.55, 0.60,...0.90} moakes 8 and I also did one for xD,init ~ 0.91.
But I do not think this could explain much of an error in the integral. But maybe it could explain 2%. I am not sure. The main thing is that you DO understand how to solve the problem.
You can check my solution tomorrow night.
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