2 replies to this topic

[Maistral]

So I wanted to create an enthalpy-concentration diagram generator in Excel using Dalton's Law and Raoult's Law (ideal stuff).

The problem is with the formulation. I studied and used this .pdf file form a university in HK:
http://ihome.ust.hk/...3210/221-05.pdf

Seeing as it also agrees with this paper in SA:
http://faculty.ksu.e.../CHE411/CH3.pdf

I then decided to give it a roll. I managed to do their computations in an Excel spreadsheet with the reference temperature equal to the boiling point of the lighter component.

The problem began to appear when I changed the reference temperature. Changing the reference temperature totally changed the values of the intermediate enthalpies. If I am correct, the difference in enthalpies for the vapor lines and the liquid lines must be constant whatever the reference temperature is. I then backtracked to the previous example by the .pdf files, the computations seem to be correct. The changes in enthalpy for (0,0) and (1,1) of the VLE were held constant, though.

I managed to whip up an Excel spreadsheet for it. I'm hoping you guys could pinpoint what I did wrong, or did I miss something else. And yes, this again wasn't discussed in our undergraduate studies (lol). Attached is the spreadsheet.

 

(Note: This isn't an assignment, I'm trying to do these kinds of things just for the sake of knowing them)

Attached Files

•  23373291.xlsx   10.13KB   25 downloads

[Art Montemayor]

Allan:

 

What you have taken on is a commendable task.  By diving directly into the heart of a distillation unit operation's basic theory of separation, you are showing the "right stuff".  I really like your aggressive attitude and enthusiasm in going after this material.

 

Yes, you are correct.  Enthalpy is a state function - and as such, the reference state is what sets the numerical values.  The value of enthalpy does not depend on the PATH taken to achieve it.  That is why it is called a "point" or "state" function.  The differences in enthalpy at the same initial and final conditions between two different referenced temperatures should be identical - regardless of the path taken.  There should be no difference between the delta enthalpies.  You are working on good, solid thermodynamic ground in establishing your enthalpy differences.

 

However, it is difficult to follow your calculations because you are presuming that our members will know your mental organization of the calculation algorithm you use.  You are not explaining the various steps and showing what principles you apply in each of your set of tabular calculations.  That would make it much more easier to check your application of the fundamentals.  --- What are "HK" and "SA"?

 

Keep up the good work.

[Maistral]

Thanks. I'm trying to do my best. :D

So yeah, This is how it went. Given numerical values are (the units are consistent):
CPL (component A) = 138.2 (kJ/kmol-K)
CPL (component B) = 167.5 (kJ/kmol-K)
CPV (component A) = 96.3 (kJ/kmol-K)
CPV (component B) = 138.2 (kJ/kmol-K)
LATENT (component A) = 30820 (kJ/kmol)
LATENT (component B) = 33330 (kJ/kmol)
BOILING POINT (component A) = 353.25 (K)
BOILING POINT (component B) = 383.75 (K)

Let:
Tref = reference temperature
Tb = boiling temperature of mixture (the temperature, given x or y (Txy))
Td = boiling temperature of mixture (the temperature, given x or y (Txy))
CPLA = heat capacity component A (liquid)
CPLB = heat capacity component B (vapor)
CPVA = heat capacity component A (liquid)
CPVB = heat capacity component B (vapor)
LATENTA = latent heat of vaporization (component A)
LATENTB = latent heat of vaporization (component B)
TbA = Boiling point of pure A
TbB = Boiling point of pure B

I'm given the T-x-y data for components A and B:
T         X        Y
353.25    1        1
358.15    0.78     0.9
363.15    0.581    0.777
368.15    0.411    0.632
373.15    0.258    0.456
378.15    0.13     0.261


First, assuming that the boiling point of pure A is the reference temperature. Tref = 353.25K

Say for example, xA = 0;

hL = xA*hLA + xB*hLB
HV = yA*HVA + yB*HVB

-------------------------
hL = xA*CPLA*(Tb-Tref) + xB*CPLB*(Tb-Tref)
hL = xA*CPLA*(Tb-Tref) + (1-xA)*CPLB*(Tb-Tref)

hL = 0*138.2*(353.25-353.25) + (1-0)*167.5*(353.25-353.25) = 0 kJ/kmol

HVA = CPLA*(TbA-Tref) + LATENTA - CPVA*(TbA-Tref)
HVB = CPLB*(TbB-Tref) + LATENTB - CPVB*(TbB-Tref)

HVA = 138.2*(353.25-353.25) + 30820 - 96.3*(353.25-353.25) = 30820 kJ/kmol
HVB = 167.5*(383.75-353.25) + 33330 - 138.2*(383.75-353.25) = 34223.65 kJ/kmol

HV = yA*(HVA + CPVA*(Td-Tref)) + (1-yA)*(HVB + CPVB*(Td-Tref))

HV = 0*(30820 + 96.3*(353.25-353.25)) + (1-0)*(34223.65 + 138.2*(353.25-353.25)) = 30820 kJ/kmol
-------------------------
dH = change in enthalpy = 30820 - 0 = 30820 kJ/kmol**

Assuming one of the intermediate points; say xA = 0.78

-------------------------
hL = 0.78*138.2*(358.15-353.25) + (1-0.78)*167.5*(358.15-353.25) = 708.7654 kJ/kmol
HVA = 138.2*(353.25-353.25) + 30820 - 96.3*(353.25-353.25) = 30820 kJ/kmol
HVB = 167.5*(383.75-353.25) + 33330 - 138.2*(383.75-353.25) = 34223.65 kJ/kmol
HV = 0.9*(30820 + 96.3*(358.15-353.25)) + (1-0.9)*(34223.65 + 138.2*(358.15-353.25)) = 31652.766 kJ/kmol
-------------------------
dH = change in enthalpy = 31652.766 - 708.7654 = 30944.0006 kJ/kmol***

Now, I'll be assuming a different reference temperature. Say Tref = 450K.

Say for example again, xA = 0;
-------------------------
hL = 0*138.2*(353.25-450) + (1-0)*167.5*(353.25-450) = -13370.85 kJ/kmol
HVA = 138.2*(353.25-450) + 30820 - 96.3*(353.25-450) = 26766.175 kJ/kmol
HVB = 167.5*(383.75-450) + 33330 - 138.2*(383.75-450) = 31388.875 kJ/kmol
HV = 0*(26766.175 + 96.3*(358.15-450)) + (1-0)*(31388.875 + 138.2*(358.15-450)) = 17449.15 kJ/kmol
-------------------------
dH = change in enthalpy = 17449.15 - (-13370.85) = 30820 kJ/kmol** <<-- hooray it agrees. But not quite...

Assuming one of the intermediate points; say xA = 0.78, with the reference at 450K.

-------------------------
hL = 0.78*138.2*(358.15-450) + (1-0.78)*167.5*(358.15-450) = -13285.735 kJ/kmol
HVA = 138.2*(353.25-450) + 30820 - 96.3*(353.25-450) = 26766.175 kJ/kmol
HVB = 167.5*(383.75-450) + 33330 - 138.2*(383.75-450) = 31388.875 kJ/kmol
HV = 0.9*(26766.175 + 96.3*(358.15-450)) + (1-0.9)*(31388.875 + 138.2*(358.15-450)) = 17998.4385 kJ/kmol
-------------------------
dH = change in enthalpy = 17998.4385 - (-13285.735) = 31284.1736 kJ/kmol*** <<-- It's totally different o_o

Recall that at the reference temperature, the change in enthalpy is equal to 30944.0006 kJ/kmol**. I have no idea what's wrong...



 

Edited by Maistral, 22 March 2014 - 11:47 AM.