Sunday, February 18, 2007

Test #2 - SVN Chapters 11 & 12

If you have any questions related to the first test, please post them here instead of using email.

Best of luck on Tuesday !

21 comments:

Dr. B said...

Thumbtack recently wrote :
"I am reviewing concepts and I don't understand how the summation of chemical potentials with respect to one another is equal to zero in phase equilibria or how the first two terms in the expression for total gibb's energy looked generalized to be applicable to both phases. "

Dr. B said...

Thumbtack:
I am not sure I understand your question about chemical potential. When phases A & B are in equilibrium, the chemical potential of each species in phase A is the same as the chemical potential of the same species in phase B. I think you are asking about section 11.2 on pgs 380 & 381. There SVN started with the Gibbs Eqn and simplified for constant T & P. Next, they used the fact that the system is closed. A molecule that leaves phase A must entr phase B. So, the change in moles of each species in phase A must be equal and opposite to the change in the number of moles of that species in phase B. This allows them to simplify Gibbs Eqn to the summation in the 2nd to last eqn on pg 380. Because the eqn holds for any change in moles in the phases, the only way for this sum to be zero is if the chemical potential for each species is the same in phase A as in phase B. That is the last eqn on this page.

I hope this helps. I am not sure I answered your question. Ask again if I missed the point.

Dr. B said...

Thumbtack:
In section 11.2, SVN have applied the Gibbs Eqn to a system that contains 2 phases, alpha and beta (I will call them A & B). The 1st three terms apply to the whole system. nG is the total Gibbs free energy. That is the total Gibbs free energy of all the species in all the phases in the system. nV and nS are similarly defined.

Does this help at all ?

Anonymous said...

Yeah, that helps. Is the final statement only true for a closed system?

Dr. B said...

Thumbtack:
Yes. In an open system, there are other ways to change properties like G, S, H, etc. That is by having inlet and outlet streams with different properties.

SKY said...

I just want to make sure what these symbols mean again...

wiggle: molar property
bar: partial molar property
hat: I am not sure
nothing: the property

What is partial molar property?

ahayles said...

Would you suggest we program van Laar and Margulas into our calculators in addition to Wilson for the test?

Dr. B said...

Jerry:
Yes. Hat means a specific property. H^ = H/m = specific enthalpy = enthalpy per kg = kJ/kg.

Dr. B said...

ahayles:
I would say that Margules, van Laar and Wilson are equally probable for the test. If you know how to do it and you have the time, go for Margules and van Laar. You may find these useful on the final even if they don't show up on Tuesday. Write yourself some notes about how to use them !

Dr. B said...

A student asked me...

"So I have my
old note sheet also the supplement from class #3 on jan 8th. This
suppliment has the equations to find the resedentual properties. And I am
also looking at the VLE using SRK equation of state supplement. Ok so i
thought that natural log of fugasity coeffcient was equal to Gr/RT. but the two supplements don't agree on how to caclulate that. the one from class has a term ln(z/(z+B)) and the on from VLE says ln((z+B)/z) basically they are resipricals of each other inside the log, which one is right, or if they are both right why are they different?

Dr. B said...

Here is my reply to that student:

Both are correct. The one from 1/8 is ...
+ (A/B) * Ln[Z/(Z+B)].

The one from VLE with SRK is... - (A/B) * Ln[(Z+B)/Z].

The minus sign flipped the Ln term. I should fix this. But the good news is that both are correct !

Good question.

Anonymous said...

In the solutions for HW5 (problem WB-3 and 4 specifically) you list the equation for ln(phi1) for a binary system with Wilsons. You list it as lnphi1=-ln(mess)+x1(mess), but the book changes that x1 to an x2 (pg 448) to get lnphi1=-ln(mess)+x2(mess).

Which of these is correct and the one we should use on the test (if Wilsons comes up)?

Dr. B said...

Katie:
This was a typo in Eqn 3 in my solution to to WB-3 and Eqn 4 in WB-4 on HW #5. The numerical results inthese solutions are correct.

The Wilson eqns in my class notes and in the book are correct.

I will email people about this to avoid any confusion.

Anonymous said...

Hi DR. I still do not understand about plotting the graph. For margules, as I know, it is straight forward to find ln gamma and GE/RT. Then, straight line represent (GE/x1x2RT) and the intersection shows A12 and A21 value. But, how to plot van laar? Is it same value for ln gamma and GE/RT? if so, whatis the difference betweem them? Thanks

Dr. B said...

Anonymous, 6:15 AM
I will be happy to help you, but I am not certain which problem you are asking about. If you can tell me which problem from my blog or which problem from your textbook, I should be able to help.

I am not sure I understand your question, but I will take my best shot anyway. I think you are talking about SVN problem 12.3.

It is not so simple to use GE/(x1 x2 R T) to determine the van Laar parameters. If you plot x1x2RT/GE, you should get something resembling a straight line. From the two intercepts, where x1=0 or x2=0, you will be able to determine the two van Laar parameters. But it will not be as simple as with the Margules Model.

I usually ask my students to use Solver in Excel to perform a non-linear regression to determine the model parameters for both Margules and van Laar. They "guess" the parameter values and let Solver adjust the parameter values until the root-mean-square error in the bubble point and dew point pressures in the data set are minimized. This gets pretty complicated. We spend a lot of time on this in class and in problem sessions.

Anonymous said...

I referred from Introduction for Chemical Engineering Thermodynamics(7th Ed) by McGraw Hill books. Chapter 12:solution thermodynamics:applications. One for the sample problems:

12.1 Given VLE data for methanol(1)/water(2)

(a) find parameter values for Margules equation. we just apply the equation of margules right?

(b) repeat (a) for the Van Laar equation. How to do this question? My lecturer also could not give answer that satisfy us. He just said we have to plot graph and the difference only on [x1x2/(GE/RT)].. thanks.

Dr. B said...

Anonymous 7:43 PM:
OK. Look at Eqn 12.16, the equation that relates GE/x1x2RT to A12, A21 and x1 and x2.

Invert both sides of the eqn and replace x2 with 1 - x1 on the right-hand side. You essentially have x1x2RT/GE = a x1 + b where
a = slope =(1/A21-1/A12) and
b = Yintercept (where x1=0) = 1/A12.

So, if you plot x1x2RT/GE vs x1, A12 = 1/Yintercept. Then, A21 = 1/(slope+1/A12).

So, plot x1x2RT/GE and do a linear regression to determine the slope and intercept and use them to evaluate the two van Laar parameters.

Best of luck to you !

Anonymous said...

I understand Dr. I got it. Thank you so much. You are very helpful. I will be tested on 25th March. Thanks.

Anonymous said...

Hi. DR.
I have one question regarding Solution Thermodynamic:Application.

The Van Laar question, i guessed it's done. This about Wilson Equation.
Is it true we do not have to plot the graph in order to find the wilson's parameter? According to my lecturer, we need to guess the parameter and substitute in the equation.
All my friends ask him, which x1, x2 and GE/RT are going to be used, and he answered choose any and make comparison. We do not understand. Hope you can help me. Thanks

Anonymous said...

Hi. DR.
I have one question regarding Solution Thermodynamic:Application.

The Van Laar question, i guessed it's done. This about Wilson Equation.
Is it true we do not have to plot the graph in order to find the wilson's parameter? According to my lecturer, we need to guess the parameter and substitute in the equation.
All my friends ask him, which x1, x2 and GE/RT are going to be used, and he answered choose any and make comparison. We do not understand. Hope you can help me. Thanks

Dr. B said...

Hi Nikshaz,

I am sorry my reply is late. I was out of town. Maybe my reply will be helpful anyway.

I do not think that the Wilson Parameters can be determined by linear regression of data based on a manipulation of equation 12.18 in Chemical Engineering Thermodynamics(7th Ed) from McGraw Hill. I do not think that Eqn 12.18 can be algebraically manipulated into a linear form. As a result, non-linear regression is required to determine the values of the Wilson parameters that best fit a data set consisting of x1, x2 and Gexcess.

You can do this using whichever software package you have available. I would use Solver in Excel.

In Excel, setup a cell for Lambda1 and another cell for Lambda2. Put initial guesstimates in these cells, say 1.04 for each. Use these parameter cells to compute the right-hand side of Eqn 12.18. I will call this cell RHS. Use the Gexcess data to compute the right-hand side of Eqn 12.18. I will call this cell LHS. Do this for EACH data point you have.

In a new cell, compute the %difference or % error between LHS and RHS. %Error = 100*(LHS-RHS)/LHS. Repeat this for each data point you have. Now, SQUARE each %Error and sum the result for all you data points. Finally, in a new cell, divide the sum of the squared %Errors by N-1, where N is the number of data points you have and take the SQUARE-ROOT for the result. This cell is the Percent Root-Mean-Squared error = %RMS. Now use Solver (Solver appears on the Tools menu. If it is missing from your Tools, you must install it before going on). In Solver, you want to MINIMIZE the %RMS cell BY CHANGING the LAMBDA1 and LAMBDA cells. For best results, click the Options tab in Solver and set the precision to 1e-15 and the convergence to 1e-12. Click OK to run Solver. It will change the values of Lambda1 and Lambda2 until it finds the best values based on your data set. You may need to run Solver a few times starting with different initial guesstimates of the parameters. Whichever result yields the smallest %RMS is the winner !

I hope you find this helpful even if your assignment has already been completed.