Chemical Forums
Specialty Chemistry Forums => Chemical Engineering Forum => Topic started by: yaserbm on September 16, 2011, 03:57:22 PM
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hi all,
we have stream of gas mixture ( CH4 + C2H2 + C2H6 + H2 ) and it has temperature and pressure of -50C and 12.8 barg respectively, we would like to install pressuer controller to drop the pressure from 12.8 barg to 5.88 barg in this situation how much the temperature going to be, any body can give me a method to calculate the resulted temperature?!! :-\.
the stream has a mass flow rate of 32650 t/h of the above mentioned components, and it has molar flow of 1413.19 kmol/hr as well.
please help me in solving this issue... :-[
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http://www.wolframalpha.com/input/?i=adiabatic+expansion
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Any help please, i need one help the attached link make me Little pit confuse
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32650 t/h and at the same time there are 1413.19 kmol/h ... if you devide the mass with the mol you get the average molar mass . but how could it be like 23103,758 g/mol :)
But it don't matter. if you have 1413.19 kmols with the volume of 2046.9m^3 and expand it from 12.8 bar to 5.88bar you get 3706.24m^3 if you have a heat capacity ratio of 4/3 . Then you get the temperature of -87.51°C (wolfram alpha gave me -90°C ) ... ideal gases!!
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32650 t/h and at the same time there are 1413.19 kmol/h ... if you devide the mass with the mol you get the average molar mass . but how could it be like 23103,758 g/mol
But it don't matter. if you have 1413.19 kmols with the volume of 2046.9m^3 and expand it from 12.8 bar to 5.88bar you get 3706.24m^3 if you have a heat capacity ratio of 4/3 . Then you get the temperature of -87.51°C (wolfram alpha gave me -90°C ) ... ideal gases!!
thanks dude for the effort you done, i really appreciate it :)
but, i would like to ask you from where you obtain the volume of the stream, did obtain this figure from the relation of mass flow rate and the density??!!, because i tried to calculate it and it was showing 1945.74 m^3. also from where you get the volume of the stream after dropping the pressure to 5.88. and the last question is based on what you are chosing the heat capacity.
i'll be so glad if you answer my question.
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sorry there were a mistake, it is not 32650 t/h, actullay it is 32650 kg/hr
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I got the volume from the ideal gas law. pV = nRT
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Oh and the final volume i got from adiabatic expansion equasion p1*V1^k = p2*V2^k ... k = heat capacity ratio ( Cp/Cv ) ... the 4/3 ratio is for polyatomic gases ... didnt do the math tho ... just remeberd it because we mentioned it in class.
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i tried to find the voulme using the ideal gas equation, and i end up with value which is around 1.9 m^3
i think if you make all the unit consistent you will end up with such value
i used 8.314 (Pa.m^3/mol.k) and i convert the pressure from barg to Pa also the tempreature will be subjected to change to K instead of C, all of these done to make the unit consistent.
please can you write down your solution step in finding the intial voulme, to make sure that the figures obtained above are right.
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I used 1413,19 x 10^3 mol
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i really appricate your coopreation,
i'll try to do it my self from A to Z and i '' let you know my approach.
by the way you are right as you said that the heat capcity for polyatomic gases such as CO2, SO2, NH3 and CH4 are 1.3=4/3
but what if i have a mixture of gases in one stream like in my cases, can i use 1.3 in the adiabtic expansion equation for the stream of gas mixture.
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I am not shure. Maybie if you had more info about the mixture you could calculate the average heat capacities and devide the two to get the ratio.
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By combining the gas law and the adiabatic expression, you will arrive at following relationship:
T2 = T1 [P2/P1]^m
Where T and P are in absolute temperature and pressure and m is given by
m = (k-1)/k
Where k = the specific heat ratio, Cp/Cv