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Plug Flow Reactors In Series
#1
Posted 04 May 2009 - 06:39 AM
Compound A undergoes a reversible isomerization reaction. Under pertinant conditions, A and B are liquid, miscible, and nearly identical in density; the equilibrium constant for the reaction (in concentration units) is 5.8. In the Plug Flow Reactor, feed of pure A undergoes a net conversion of 55% to B . The reaction is elementary. If a second, identical, Plug Fow Reactor is placed downstream from the first what overall conversion of A would you expect if:
a) The reactors are connected in series?
b) The products from the first reactor are separated and only unconverted compound A is fed to the second reactor?
#2
Posted 04 May 2009 - 06:14 PM
Look if you are getting someone here to do your homework, this is not the place to be and definitely not the place to ask.
Go through your reaction engineering class notes, and you can solve it quite simply.
Now go and do your own homework, and when you have done some work to show us that you really don't understand, then we will help.
#3
Posted 05 May 2009 - 02:00 AM
#4
Posted 05 May 2009 - 02:25 AM
If I understand it right you feed A to a plug flow reactor and get 55% conversion.
With the second identical PFR you feed again pure A into it (since the formed B has been separated). Now it seems to me that all conditions in this second reactor are identical to those in the first reactor, so what would you think the conversion would be in this 2nd reactor?
Now you know the conversion in the 1st and in the 2nd reactor, so what would be the overall conversion?
#5
Posted 05 May 2009 - 02:41 AM
#6
Posted 05 May 2009 - 05:11 AM
#7
Posted 05 May 2009 - 05:27 AM
part a: the reactors are in series
part b: the products from the first reactor are seperated and only the unconverted A is fed to the second reactor.
It is P4-13 from Fogler's Elements of Chemical Reaction Engineering if you happen to have the book.
#8
Posted 05 May 2009 - 06:51 AM
Part (a): I believe you should not have any problems with this.
Part (b): Please look at the schematic I have provided. I also advise you to always do the same whenever you solve problems, and whenever you ask any questions here in this forum.
cheers,
clarence
Attached Files
#9
Posted 05 May 2009 - 07:32 AM
#10
Posted 05 May 2009 - 06:06 PM
i don't think you quite understood me, you said you have done some work, but you do not upload it for any of us here to help you. we have already thrown out the lifebuoy, but you are not wanting to hang onto it.
please, i remind you, to upload your work first before you want us to find out any errors.
cheers,
clarence
#11
Posted 07 May 2009 - 01:26 AM
The rate law for A:
-rA = k1Ca - k2Cb
At equilibrium, -rA = 0
Therefore:
Cb/Ca = K = k1/k2
Where K is the equilibrium constant and Cb and Ca are the concentrations of A and B at equilibrium.
Rate law then becomes:
-rA = k1[Ca - (Cb/K)]
Stoichiometry:
Ca = Cao(1-X) , Cb = CaoX
Substituting into rate law:
-rA = k1Cao[1-(1+1/K)X]
For the first PFR:
V = Cao.Q (integral from 0 to X) dx / k1Cao[1-(1+1/K)X]
= [-Q/ k1(1+1/K)] . ln[1-(1+1/K)X]
Can solve for lumped parameter which is the operation damkoler number:
k1.V/Q = k1T = -ln[1-(1+1/K)X] / (1+1/K)
= -ln[1-(1+1/5.8)0.55] / (1+1/5.8)
= 0.883
For 2nd PFR reactor:
V = Cao.Q (integral from X1 to X2) dx / k1Cao[1-(1+1/K)X]
= [Q/ k1(1+1/K)] . {ln[1-(1+1/K)X2] + ln[1-(1+1/K)X1]}
Rearranging to solve for X2:
ln {[1-(1+1/K)X2] / [1-(1+1/K)X1]} = k1T(1+1/K)
= 0.883(1+1/5.8)
= -1.035
Subbing in values and rearranging to solve for X2 I find that X2 = 0.745
Still not sure how to do part b. My thinking was that because the products get separated and only A goes into the second reactor it is much like doing the calculations for the 1st PFR reactor above. But I have been unable to figure out how to take into account the fact that it is only the unconverted A and not 100% pure A.
#12
Posted 08 May 2009 - 02:51 AM
#13
Posted 08 May 2009 - 04:32 AM
But I have been unable to figure out how to take into account the fact that it is only the unconverted A and not 100% pure A.
After the separation step I would say you have again 100% pure A that is now fed to the 2nd PFR. The fact that it is "unconverted A" does not mean that it cannot be 100% pure...
#14
Posted 08 May 2009 - 05:37 AM
#15
Posted 09 May 2009 - 08:01 AM
i don't know if i am helping you or just doing your homework for you. you already know the volume of the PFR, you know all the other conditions, so why can't you find the conversion of A coming out of the 2nd PFR?
please do not take my comments as a direct/personal attack, but i seriously think you are not studying your notes enough.
#16
Posted 17 January 2012 - 04:49 AM
I was searching for plug flow reactors in series and i find this topic. this question is exactly my question , so can anyone help me too? I read all the posts but still i have problem.
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