Heat Transfer in Batch Reactors

[RJB32482]

I performed an experiment to determine the heat transfer in a batch reactor. The test was performed at normal agitation for production. We cooled 270,000 lbs of water from 122.5 to 104 degrees F in 78 minutes. I calculated a heat transfer rate of 65,959.6 BTU/min. Now we are looking at adding an exothermic reaction that produces 300 BTU/lb of material in the reactor. The amount of reactant is 90,000 lbs. The reaction usually occurs around 200 degrees F and lasts about 6 hours holding at this temperature.

My question is this enough heat transfer to perform this reaction at the specific conditions? Do I need more information to conclude this? I believe it will be tough to remove all the heat. Please comment.

Thanks.

 

[Latexman]

It depends. You have to use the fundamentals you were taught at the university to figure it out. No one in their right mind would give you a yes or no answer based on the limited information you gave.

During the entire reaction process, does the reactor contents have similar physical properties to water at 104 to 122.5 F? If not, I'm not sure it was worth the effort. If yes, I hope you got enough data to calculate the overall heat transfer coefficient, U. You are going to need it to answer your questions.

Good luck,
Latexman

 

[RJB32482]

So if I have the heat transfer area, I could calculate the overall heat transfer coefficient on the reactor wall side. I will look at the heat capacity data for the reactor contents. If they look constant to water, could it be assumed that the heat transfer rate needs to be more than the total heat generated/time of reaction?

Except the above information, what other info do I need to determine this question?

Thanks.

 

[StoneCold]

Give us more information.
How are you doing the reaction? Feeding one component? Use that to control the heat load on the reactor.

If you need to control the reactor at 200F for six hours after the reaction is completed and you are using cooling water in the jacket you are going to have a hard time.
270,000 lbs of water is roughly 32,400 gallons of water. This is a very big reactor to be fooling around with.

Are you heating the reactor to 200F and then starting the feed or are you dumping in both components then heating them to 200F and then the reaction initiates and you get an exothermic reaction that you have to try to hold down to 200F?

If you explain things better we can probably give you better advice.

Regards

StoneCold

 

[RJB32482]

The component in the reactor is added before the heat up to 200F. The reactor is batch, so no cold feed can be added to the reactor.

The reactor is 30,000 gallons, so my information must be wrong from the test. The charge of water for the test is less than that. The reactor has a half-pipe jacket with a 30% propylene glycol solution. It goes through a plate heat exchanger while full cooling with cooling water as the cooling medium. The reaction is polymerization of styrene in a 50/50 weight charge with water.

Here was my thoughts on a test I could do:
1. Charge the reactor just with water (x lbs)
2. Heat the reactor up to 50 degrees C
3. Cool the reactor to 40C
4. Find the amount of heat transfered (Q=m*Cp*delta T)
5. Find the heat exchange surface area of the half-pipe from the drawings.
6. Calculate U= (Q/A*delta T)
7. Average the proposed amount of styrene added and calculate the heat needed to be removed from the reactor
(lbs styrene*300BTU/lb)/360 minutes=heat transfer rate needed
8. If heat removal at full cool>needed heat transfer rate its good, if not need more heat transfer.

Also agitation would be the same for both the test and reaction.

Please add any insight on this test method.

Thanks.

 

[Latexman]

Is the viscosity of the reactor contents going to build to a viscosity higher than water? What's the end viscosity? Also, the heat capacity, thermal conductivity, and density of the reactor contents will be different than water AND they will change as the reaction conversion increases.

Are you charging initiator batchwise up front or feeding it over time?

Some of your numbers don't add up. 90,000 lbs of reactant (styrene) is not a 50% solids batch in a 30,000 gallon reactor. It's more like 33%.

Good luck,
Latexman

 

[StoneCold]

RJB32482
One problem with your reaction logic is that the rate of reaction is not linear, especially the way you are doing it. This dump and run method is going to cause about 2/3 of your reactants to react in the first couple of minutes. Then the reaction will slow down and taper off to zero at your six hour number. (Be advised that these numbers are only to illustrate my point not to design the reactor).

So you need a much higher heat transfer rate to overcome the rate of reaction in the beginning.
So you really need
Q= reaction rate* heat of reaction < heat removal capablility of the reactor.


Equation 6 above U=Q/A*delta T. You realize that delta T here is the temperature difference between the batch water and the cooling water right? So Delta T is changing with time as the batch cools, and the best way to get your U value will be to integrate the equation with respect to time. I would just solve it for UA and forget about finding just the U. It takes one unknown off the table.

I hope that I am helping you here and not just making things worse.

Regards
StoneCold

 

[djack77494]

RJB32482,
I agree with StoneCold. Be careful. You have an unsteady state heat transfer problem complicated with reaction kinetics. The needed heat transfer will undoubtedly be much higher than any steady state analysis would indicate. You will probably need to charge the reactor, then heat the contents until you get "ignition". Then you must quickly shift to cooling mode to control the maximum reactor temperature. Maximum heat release and removal will be at the beginning of the cooling phase.
Doug