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Enthalpy Of Mixture Ethane-Ethylene

enthalpy

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#1 panoskagiou

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Posted 24 August 2018 - 02:47 PM

Hello engineers! I want to ask you for certain pressure and temperature if we have a mixture of ethylene-ethane/ Why by increasing the composition of ethylene in the mixture, why the enthalpy mixture increases substaintialy?



#2 PaoloPemi

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Posted 25 August 2018 - 03:25 AM

you should specify the conditions (or provide more details),
it can depend from several factors, for example different equilibria between vapor and liquid phases,
for gaseous mixtures there are limited variations,
for example (% mol) P = ATM T = 288.15 K model PRX (available in free edition of Prode)
C2H8 1.0 C2H4 0.0 -> cp 1.73 Kj/Kg-K
C2H8 0.9 C2H4 0.1 -> cp 1.71
C2H8 0.1 C2H4 0.9 -> cp 1.53
C2H8 0.0 C2H4 1.0 -> cp 1.51

if you integrate cp(T1->T2) the resulting values are not too different



#3 panoskagiou

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Posted 25 August 2018 - 04:22 AM   Best Answer

The results that I found have big difference in the enthalpy of vapor mixture.



#4 PaoloPemi

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Posted 25 August 2018 - 04:37 AM

for this mixture I expect  std. models to be quite accurate in a reasonable T, P range,

you may review your calc's...



#5 panoskagiou

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Posted 25 August 2018 - 11:36 AM

These are calculations of ASPEN PLUS. Paricularly, when I have conditions of pressure 35bar and temperature 20C, then if I go from 60% ethylene to 70% ethylene, then the mass enthalpy increases almost 5times. And I cannot understand why. The model that I use is Peng Robinson.



#6 PaoloPemi

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Posted 25 August 2018 - 01:39 PM

you should verify that your operatng point is not close to a critical point ,

PRODE PROPERTIES with standard PENG ROBINSON, VDW mixing rules, Kij 0.01 composition C2H6 0.4 C2H4 0.6  calculates  a critical pressure about 50 Bar.a (much higher than 35 Bar.a),

accordingly  at 20C, 35 Bar.a cp is about 2.48 Kj/Kg-K ,

but if you  increase pressure, for example moving operating point to 20C, 50Bar.a (close to critical point) cp increases to about 10 Kj/Kg-K



#7 panoskagiou

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Posted 25 August 2018 - 04:01 PM

Yes the operating pressure is below the critical point. The only change is in the composition not in the pressure or temp.



#8 PingPong

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Posted 29 August 2018 - 09:50 AM

You must have made an input error somehow.

As a result there could, for example, be partial liquid in one of the cases causing big impact on the calculated enthalpy.

 

So I suggest you check your input again and again and again until you find the mistake(s).



#9 panoskagiou

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Posted 01 September 2018 - 01:39 AM

It is vapor. The reason is the way that the enthalpy is calculated. Also the enthalpy of ethylene should be much higher than the enthalpy pf ethane. That's why I think.



#10 breizh

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Posted 01 September 2018 - 04:45 AM

Hi ,

You may use the calculator on line to support your work.

http://www.questcons...properties.html

 

hope this is helping you.

 

Breizh



#11 panoskagiou

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Posted 01 September 2018 - 03:34 PM

I am so happy to be part of this group. I enjoy your replies, because I learn a lot. Thank you everyone!



#12 panoskagiou

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Posted 01 September 2018 - 03:37 PM

Hi Breiz! How I can know the equations that are used for this site?

Thank you in advance!



#13 breizh

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Posted 01 September 2018 - 11:11 PM

Hello ,

Sorry I cannot offer more , you may contact the provider .

 

good luck,

Breizh



#14 PingPong

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Posted 02 September 2018 - 05:38 AM

The absolute value of enthalpy (and entropy) has no meaning.

Its reported value depends on the reference state that is used to calculate it. Every source (software, website, book, table, graph, et cetera) uses a different reference state, and it is even possible that a source uses a different reference state for different components.

 

What only matters in simulations and other calculations is the enthalpy difference. If a gas (mixture) is changed from one T and p to another T and p its enthalpy will change and this enthalpy change (difference) has a meaning, for example a heat exchanger duty, or compressor or turbine power, the absolute enthalpy values have not.

 

I have no idea what reference state(s) Aspen Plus uses for enthalpy, and you should not care as it does not matter at all.

 

Read also what I wrote here: https://www.cheresou...lpy-calulation/

 

 

If you are convinced that you did not make an input error then you can ignore the absolute enthalpy values and only focus on the calculated duties and powers.



#15 panoskagiou

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Posted 02 September 2018 - 10:51 AM

Hi PingPong! I am sure that there is not mistake in my calculations, because I just use a stream in ASPEN PLUS in which I specify the compositions of the two components, the temperature and pressure and a random flow rate. So it cannot be a mistake by my side. But I wanted to understand why the enthalpy is so strongly affected by the composition for the same T and P. 



#16 PaoloPemi

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Posted 03 September 2018 - 01:56 AM

you shouldn't obtain very different values for (spcific) enthalpy except for some cases as discussed in posts #2, and #6 ,

for (stream) enthalpy make sure to correct for different flows,

finally, enthalpy (and entropy) base values are constant and influence all (entropy and enthalpy) values which means that you can compare two different conditions,

some software (see for example https://www.cheresou...nces-in-prode/)allow to redefine base values but differences as H(t1,p1) - H(t2,p2) remain the same....






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