HW 9 - SP4 - Kinetic Analysis of a Chromatographic Column - 6 pts
Racemic amino acids can be separated using l-proline attached with silanes to the surface of silica gel. In one set of experiments, aspartame isomers gave the following results.
d-aspartame: t0 = 62 min , σt0 = 3 min
l-aspartame: t0 = 71 min , σt0 = 6 min
These results were obtained with a 25 cm column, 0.41 cm in diameter, filled with 45 μm silica gel spheres. The void fraction of the bed was 0.62. The flow in the column was 2.0 mL/min. Find the apparent rate constants for this separation. Compare these rate constants with those predicted from the following mass transfer correlation.
mass transfer correlation
where: d = packing particle diameter, v = superficial solvent velocity, ν = solution kinematic viscosity, D = diffusion coefficient of the solute (7.0 x 10^-6 cm2/s).
d-aspartame: t0 = 62 min , σt0 = 3 min
l-aspartame: t0 = 71 min , σt0 = 6 min
These results were obtained with a 25 cm column, 0.41 cm in diameter, filled with 45 μm silica gel spheres. The void fraction of the bed was 0.62. The flow in the column was 2.0 mL/min. Find the apparent rate constants for this separation. Compare these rate constants with those predicted from the following mass transfer correlation.
mass transfer correlation
where: d = packing particle diameter, v = superficial solvent velocity, ν = solution kinematic viscosity, D = diffusion coefficient of the solute (7.0 x 10^-6 cm2/s).
posted by Dr. B at 8:43 AM
19 Comments:
So you said that a=6/d(1-e). Isnt a the area? How does this give you area units?
blind:
"a" is the ratio of the surface area for MT to the volume of the bed. It has units of length^-1.
Oops.
Please use a void fraction of 0.38.
That way, you will get the answers that I got :)
How do we get the kinematic viscosity? I see the Re and Sc equations, but since we dont know the density or viscosity I dont see how those are of any help.
blinder:
Assume the solution is dilute and use the kinematic viscosity of water: 0.01 cm^2/s.
How do we use the data that you gave us? I know that to(min) is the time at which the peak occurs. I'm not sure what sigmato(min) is. NTU=1/sigma^2. What is sigma? Is sigma 3/62 and 6/71? Also, why can't we use Re and Sc to get the viscosity? Wouldn't it just involve setting two quantitues equal to each other, then solving for viscosity?
Basically, we're given a dimensional form of the standard deviation. How do I convert that to a dimensionless form? May seem like a silly question, but I don't know what to do...
I got a different K than you but when I used the diffusion coefficient in the a equation I got your answer. This seems wrong.
grrr :
Sigma*t0 is the dimensional form of the standard deviation. It can be approximated as half the width of the peak at half the height of the peak. Sigma is the dimensionless form of the standard deviation. Yes, sigma would be 3/72 and 6/71 for d- and l-aspartame, respectively.
No, you cannot use Re and Sc to determine the kinematic viscosity. The confusion here was caused by a typo in my PPT from Mon, 12/4. Sc = kinematic viscosity / diffusivity, period. The other part of that equation in slide 14 was a copy-and-paste error on my part. I apologize.
tired:
Not silly. Here you go.
Sigma = Sigma*t0 / t0
weird:
I don't think I understand your question.
Maybe you meant to post this question for problem SP3 ?
when calculating Ky it seems that the unit doesn't cancel out, we have velocity in mL/min in the numerator and Ky is in cm/s do we need to divide by volume? please help..
When we solve for the k values with the equation given in the problem statement, won't k be the same for both d- and l-aspartame since particle diameter, superficial velocity, kinematic viscosity, and Diffusion coefficients are the same for each compound? That is, unless the diffusion coefficients are somehow not equal...
Thanks
Whey you calculate the k from the correlation, I think you forgot to multiply by the velocity of the liquid. k/v = 1.7e-3 (unitless) where v is the velocity of the liquid (cm/s). I think your answer should be higher which will make the correlation bad vs your Ky.
Also, is k from the correlations on the same magnitude as Ky from the NTU calcs? Mine aren't which worry me.
Anon:
The units on Ky are cm/s. The units of v are cm/s. The units of F are cm^3/s. Perhaps it is your error on the units for velocity that is causing the problem.
hmm:
Yes, when you use the correlation given in the problm statement, you get just one value of Ky. This correlation cannot tell the difference between the two isomers. The correlation predicts a value of Ky that is very similar to the observed Ky for d-asp but it is surprisingly far from the observed value for l-asp.
Bill Nye:
I think my answer is correct. I did multiply by velcotiy and I got Ky = 1.71e-3 for l-asp. This value is about triple the value predicted by the correlation, so either the data is bad or the correlation just doesn't work well for some reason.
Yes, Ky is on the same order as the Ky values from the data. In fact, for d-asp, the agreement is remarkable.
Darn. Well I don't know what I did wrong then cause I think this should be easy.
Bill Nye:
I am sorry we couldn't work this out through the blog. Bring your work and or your Excel and we can slug through it in my office hours tomorrow.
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