|Apr14-07, 03:23 PM||#1|
Hello, I have the following problem and am unsure how to solve it, although I believe the heat of dissolution to be -71.76kJ/mol, but I can only see how this would help me if my acid was infinetly dissoluted which is not the case so I need to know how the calculate a temperature rise when this is not the case.
My acid starts off at 98% wt. total wieght 115.985041 kg, (1158.9 moles H2SO4, and 128.8 moles H2O).
The acid ends up at 75% wt. total weight 151.55284 kg, (1158.9 moles H2SO4, and 2103.1 moles H20).
Both the acid and water initial temps can be taken as 30oC,
So basically 115.98kg of 98% wt. H2SO4 at 30oC, with 35.5kg water at 30oC
|Apr14-07, 03:52 PM||#2|
the water is also in vapour phase at the beggining, so will I also have to consider the heat of condensation of the water going from the vapour phase to the liquid phase? thankyou.
|Apr16-07, 03:43 AM||#3|
I vote no to the second question, but I'm not sure. It's not like it's steam? If the pressure is 1 atm of course... They are little particles of water at 30 °C.
And, the solution of H2SO4 seems 18 M so, I think you may use the formular of:
Which you posted yourself.
So if all H2SO4 dissociates it will give: 83170 kJ of energy to the solution.
For the warmth capacity I'd say approximately:
H2SO4: 3,5 J/gC . 113000 g = 400 kJ/C
H2O: 4,18 J/gC . 35,3 kg = 150 kJ/C
So dQ = (cp1m1 + cp2m2) dT
dT = 152 ° C
Oh boy, there must be something wrong with the calculation??? Hey I'm not a professional... yet.
Your reactor is going to explode... do you have some kind of cooling medium?
The boiling point of H2SO4 was something at 350°C so maybe, just maybe your solution won't evaporate at 180 °C. Otherwise, you're in a big trouble. I don't know how your installation looks like, so that info would be interesting too.
I'm curious myself how you're going to handle this calculation! Please post everything you got.
|Apr16-07, 05:23 AM||#4|
Hi can I just say first thanks loads for your reply, I will be using a cooling loop in the system, sorry but I gave you all the information I got, but it going to 150oC sounds correct. So do you recon that I would be able to half or maybe even third the heating due to it not beening infinet dilution woukld you be able to hazard a guess with a vague explanation, if not it doesn't matter too much all I need to do is put in my report that I presume that it is infinet dissolution, its just that I dont want to be using more cooling than I sholud as I have to do a detailed costing on the system.
One other thing I also found the heat of dissolution to be 800kJ/mol, but that was on wiapedia.com, or however it's spelt, so it's probaly not reliable.
|Apr16-07, 05:30 AM||#5|
I'm just looking for a basic temperature rise, the whole system is an absorption tower, a wet gas feed at the bottom, and the acid feed at top, so somehow I'm also going to have to come up with a compromise of the two phases increasing with temperature due to the dissolution and the general heat transfere between the two counter flows as both streams will rise in temperature but flow past the inlet streams as they exit which will be at the original lower temperature, but I'm probaly just going to say and presume there is no heat transfer from counter flows but that both the acid and gas are heated together to the same temperature, ha lazy....!
|Apr16-07, 01:32 PM||#6|
hi, would you not also have to take into consideration the heating of the gas via the acid (heat transfer), I have a spread sheet set up which considers this and as the gas flow rate is many time the liquid flow rate, and whole system only reaches 55.65oC, and this is good as the column works at optimum absorption at 55-60oC. This is assuming that the heat is shared equally let me know what you think of this if you think its a bad idea then I scrap it, thanks
|Apr16-07, 07:06 PM||#7|
So you spray HS2O4 above and you have some kind of gas at the bottom? Or is it air with water?
As for the 800 kJ/mol, that seems much! Meaning the reactor will even blow more up than I predicted. You should know exactly what you are doing here!!!
The thing is, I calculated that it's 18 M and the internet site also uses 18 M solution as an example. So the heat of reaction they use there should be correct. Maybe I read something wrong there.
I think -800 kJ/mol is the formation of HSO4- from pure H2SO4 (solid) with H2O and -70 kJ/mol takes in account the solution of H2SO4/H2O.
I'm just a student don't take me so seriously! I make mistakes. :)
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