CAPE-OPEN

Hi,
ln the paper attached, we are given two reactions and l assumed they occur in series as cumene has to be a product in reaction 1, before it can be consumed as a reactant in reaction 2. The paper gives us the conversion, was wondering if this is the overall conversion of propylene consumption in two reactions. Also we are given selectivity, how do l implement selectivity in COCO?
https://cbe.statler.wvu.edu/files/d/595 ... mene-a.pdf

Selectivity can be implemented easiest using the fixed conversion reactor.

Ok, so l have to do hand calculation to convert selectivity to conversion?

Well, if the reactions are in series, this is straight forward. If there is 99:1 cumene:DIPB, then the follow-up reaction occurs to the extent of 1%, so the conversion of the follow-up reaction is 1%.

Hi Jasper,
Thanks for the reply, l really appreciate it. l noticed the reaction is neither a series or a parallel reaction after reading around a bit, but it falls under complex reaction category (see attachment). l then did some hand calculations using the extent method and obtained that extent of reaction 1 is 107.17 and that of reaction to is 3.36. I then assumed the reaction is in series and computed the conversion thereafter, but using these conversions l could not obtain the required selectivity given in the spec. My question is, is there a way to handle complex reactions in COCO, or assumptions that can be made to be able to handle it since the series reaction failed. if not, if l have inputs and outputs to a reactor is there any unit operation l can use to represent my reactor. Thank you

I have some trouble decyphering your notes. But from what I can see the concept of the reactions-in-series still holds w.r.t. the conversion of cumene. To check, I set the conversion of the second reaction to 1% Cumene, as discussed below. Note that the conversion should be that of Cumene, not of DIPB. Then the ratio of cumene : DIPB is indeed 99.

Hi Jasper,
Stream 18 contain the following substances with their boiling points:
propylene = -47.6 C
propane = -42 C
benzene = 80.1 C
cumene = 152.4 C
DIPB = 204 C
l wanted to vaporise the liquid so that only propane and propylene are condensed and eliminated as vapour in the separator. I then entered 0.0259 as this is the mole flow fraction of pentane and propylene divided by the total stream flow of stream 18. But the heavy components become condensed too, see stream 19.

As you can see from the fugacities of stream 19 and 20, this is where (given the thermodynamic model) the phase equilibrium lies. If you want to separate further you will need more than a flash.

Thank you for your vast knowledge and time, l really appreciate it. l am lost as far as your comment on Fugacities, l went back to check and the fugacities were equal in the vapour and liquid phases, since the two states are in equilibrium. Can we make any deductions from the magnitude of the fugacity value. Also, l was wondering since l have a vapour at a particular pressure and temperature, can l not cool it down just enough to condense all the heavy components, and leave the non-condensables i.e. propane and propylene since they have negative boiling point temperatures. When l try to do so, l lose propane to the liquid phase, but its a non-condensable. Attached is a PFD from the problem statement, showing how they eliminated the non-condensables by flashing before doing distillation on the liquid phase.



REF: https://cbe.statler.wvu.edu/files/d/595 ... mene-a.pdf

Hydrocarbons tend to really like each other. This means that hydrocarbons dissolved in other hydrocarbons feel 'comfortable', reflected by a lower partial fugacity (or 'effective partial pressure', or 'tendency to want to get out of that phase') then what you would get for the pure fluid.

You could run a parametric study to check the effect of temperature on separation by a single flash.