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Topic: Steady State mass balance for separation of cyclopentadiene  (Read 5481 times)

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Offline Woopy

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Steady State mass balance for separation of cyclopentadiene
« on: October 20, 2013, 12:54:31 AM »
Hello,

Here is a homework problem that I have been working on. I am wondering if for the mass balance of substance A, do I include the terms in the reaction rate that involve species R and B? Also, wouldn't A and B start reacting in the pipe before they even go into the CSTR? (I guess put in 2 different pipes then!).

Also, I wasn't sure if my mass balance for the PFR is right, I'm pretty sure it isn't. The trouble is I don't know how to put it in terms of volumes when given moles for the PFR, so I just did a volume flow rate (or rather just a velocity I suppose, in terms of z, a length).
« Last Edit: October 20, 2013, 01:37:38 AM by Woopy »

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #1 on: October 20, 2013, 01:07:34 AM »
Can you post as images. I hate to open pdfs to see problems.

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #2 on: October 20, 2013, 01:08:12 AM »
Link moles n volume by gas law or density.

Offline Woopy

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #3 on: October 20, 2013, 01:12:20 AM »
My attempt is too large a file to post as a image on here

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #4 on: October 20, 2013, 01:22:31 AM »
Quote
I am wondering if for the mass balance of substance A, do I include the terms in the reaction rate that involve species R and B?

Yes.


Quote
Also, wouldn't A and B start reacting in the pipe before they even go into the CSTR? (I guess put in 2 different pipes then!).

If they react so easily you need a pipe-tee not a reactor. Typically a reactor has heat / catalyst / mixing something a pipe wouldn't offer. Or just residence time.

For hyper fast gas reactions a pipe is indeed all you need. Maybe cooling.

Offline Woopy

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #5 on: October 20, 2013, 01:27:00 AM »
Okay, so what am I missing in my solution? By the way, part C is for a batch stirred tank reactor

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #6 on: October 20, 2013, 01:27:49 AM »
Okay, so what am I missing in my solution? By the way, part C is for a batch stirred tank reactor

Try reducing image size n posting. Cant read a pdf.

Offline Woopy

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #7 on: October 20, 2013, 01:40:02 AM »
okay, I have reduced the size

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #8 on: October 20, 2013, 02:14:51 AM »
First, re your CSTR: why no B in outlet for B-balance?

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #9 on: October 20, 2013, 02:16:06 AM »
In your A-balance signs seem screwed up for rate expression.

Offline Woopy

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #10 on: October 20, 2013, 02:24:25 AM »
For the CSTR, I said no B comes out because the reaction went to completion. My A-balance for which part is screwed up? I changed the signs for all of them because it is depletion so I am subtracting

Edit: Or rather, my intent was for all the signs to be opposite of what they are in the problem statement!

Offline curiouscat

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Re: Steady State mass balance for separation of cyclopentadiene
« Reply #11 on: October 20, 2013, 02:57:24 AM »
For the CSTR, I said no B comes out because the reaction went to completion. My A-balance for which part is screwed up? I changed the signs for all of them because it is depletion so I am subtracting

Edit: Or rather, my intent was for all the signs to be opposite of what they are in the problem statement!

Yeah look again. Not all are changed.

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