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I am trying to flash a mixture of Ethylene, ethanol and Water. I have given the outlet pressure as 500KPa and heat as zero. The vapor stream is has 9 percent liquid. I repeated this calculation using a pressure reduction valve prior to flashing. Still the flash unit shows liquid in the vapor stream. I have downloaded your examples in my pc. The example show expected results. Please help.

I am not sure what your question is.

If pressure is reduced to about 18.627 kPa (with the selected thermodynamic model), the liquid disappears. Not so far from the 12.35 kPa at 50 C for pure water (using the steam tables), considering the mixture has 60% water.

Normally one would flash at a vapor fraction of 1 and given temperature to find the dew point pressure. It is also listed in the overall properties of the stream.

Aha - I see what you are saying now, the stream out of the flash contains liquid. Something appears to go wrong with the exit flash. Normally the TP flash option should not be selected. But something fails without this option, looking into it.

First please run CUP to update COUSCOUS, to resolve a string allocation bug in the Gibbs reactor that causes problems elsewhere in the simulation.

Next, the problem you are seeing is caused by the fact that the actual system (given the selected thermodynamic model, which surely is not accurate; a cubic equation of state is selected without binary interaction parameters to describe the non-ideal behaviour of the compounds involved) has a phase equilibrium at 500 kPa and 28.626775 C (which are the solution conditions of the adiabatic flash) with 2 liquid phases.

TEA does not support two liquid phases. So it finds an incorrect phase equilibrium where it splits off a single liquid phase from the vapor; please note that this is an incorrect solution; at the actual solution a second liquid phase takes part in the equilibrium, changing the compositions and fugacities of all phases involved. A split off vapor would normally be in equilibrium with a liquid at the same conditions as the liquid that was split off, with zero liquid phase fraction. However, another liquid phase exists that is in equilibrium with this vapor phase, which is found in the exit flash of the vapor stream, causing the 9% liquid to appear.

Using TEA I do not think there is a correct solution for this problem. To treat this problem correctly, thermodynamics need to be used that support 2 liquid phases. There are many such thermodynamic servers around, but TEA is not one of them. TEA is restricted to systems that form a single liquid phase.

For completeness, below is the TEA solution compared to the actual solution (given Peng Robinson without binary interaction parameters) of stream3, flashed at 500 kPa, adiabatically:

[table]
[tr] [th] [/th] [th] TEA [/th] [th] Actual solution [/th] [th] unit [/th] [/tr]
[tr] [td] Overall pressure [/td] [td] 500 [/td] [td] 500 [/td] [td] kPa [/td] [/tr]
[tr] [td] Overall temperature [/td] [td] 28.626775 [/td] [td] 28.626775 [/td] [td] °C [/td] [/tr]
[tr] [td] Overall mole fraction [Ethylene] [/td] [td] 35.921661 [/td] [td] 35.921661 [/td] [td] % [/td] [/tr]
[tr] [td] Overall mole fraction [Water] [/td] [td] 59.047568 [/td] [td] 59.047568 [/td] [td] % [/td] [/tr]
[tr] [td] Overall mole fraction [Ethanol] [/td] [td] 5.0307707 [/td] [td] 5.0307707 [/td] [td] % [/td] [/tr]
[tr] [td] Overall flow [/td] [td] 1574.9004 [/td] [td] 1574.9004 [/td] [td] kmol / s [/td] [/tr]
[tr] [td] Overall MW [/td] [td] 0.023032508 [/td] [td] 0.023032508 [/td] [td] kg / mol [/td] [/tr]
[tr] [td] Vapor mole fraction [Ethylene] [/td] [td] 88.633487 [/td] [td] 97.410387 [/td] [td] % [/td] [/tr]
[tr] [td] Vapor mole fraction [Water] [/td] [td] 0.71590845 [/td] [td] 0.71941453 [/td] [td] % [/td] [/tr]
[tr] [td] Vapor mole fraction [Ethanol] [/td] [td] 10.650604 [/td] [td] 1.8701983 [/td] [td] % [/td] [/tr]
[tr] [td] Liquid mole fraction [Ethylene] [/td] [td] 0.001119511 [/td] [td] 0.000842379 [/td] [td] % [/td] [/tr]
[tr] [td] Liquid mole fraction [Water] [/td] [td] 98.797753 [/td] [td] 99.854636 [/td] [td] % [/td] [/tr]
[tr] [td] Liquid mole fraction [Ethanol] [/td] [td] 1.2011278 [/td] [td] 0.14452166 [/td] [td] % [/td] [/tr]
[tr] [td] Liquid 2 mole fraction [Ethylene] [/td] [td] N/A [/td] [td] 2.6717359 [/td] [td] % [/td] [/tr]
[tr] [td] Liquid 2 mole fraction [Water] [/td] [td] N/A [/td] [td] 14.630337 [/td] [td] % [/td] [/tr]
[tr] [td] Liquid 2 mole fraction [Ethanol] [/td] [td] N/A [/td] [td] 82.697927 [/td] [td] % [/td] [/tr]
[/table]

Note on the "correct" solution: the TEA solution is the adiabatic one; the solution that it is compared to is taken at the same T & P of the TEA solution. The adiabatic solution would be at different T, due to the difference in phase equilibrium.

The flash uo should be designed to separate two phases in equilibrium. This is not happening in flash. I feel that it has something to do with the algorithm of the flash uo. I would have always tried to code the flash uo assuming that we have two phases in equilibrium. Incase there is a phase seperation then we would have three phases. I feel it has nothing to do with the TEA package. Anyways I will try to study the papers that you have referred to in for design of the flash uo. I even tried to uses chemsep uo as flash, I got the same result.

The flash UO does little more than invoke a flash calculation on the underlying thermo.

The problem is that this is a VLL system, which TEA does not support.

TEA returns a VL solution where there should be a VLL solution. A second liquid phase splits off the resulting vapor phase. Note that this VL "vapor" product is not in phase equilibrium with the liquid product, so VL + L does not make VLL.