How to simulate a water-saturated dry-basis HC gas mixture ?

[KernOily]

Good morning. Here's a dumb mech engr question for you fine fellers.

Suppose I have the following compositional analysis of a gas. This was sampled at a compressor discharge. This is an ASTM-whatever-method dry gas analysis, done in a lab, at STP:

[CO2] = 0.4
[C1] = 0.4
[C2] = 0.02
[H2S] = 0.002
[C6+] remainder

Now suppose I am simulating a process that uses this gas with my trusty process simulator. Problem is, this gas exists in the process in a water-saturated form. So the gas is not in fact dry at all but is water-saturated. In real life the process runs at 190° F and 0.1 psig, according to my infallible field PIs and TIs ;-).

To properly do the simulation of the whole process, then, I need to know what mole fraction water (vapor) will exist in equilibrium with this particular 'dry gas' mixture. So to saturate the gas, at the front end of the process, I made my simulator to feed this dry gas stream at 190°, along with a liquid water stream also at 0.1. psig and 190°, into a 'mixer' and then on to a fake separator to do the flash and to determine the water vapor fraction. As long as I have liquid water coming off the bottom of the fake separator, I know I have an equilibrium saturated gas. The overheads off this fake separator is then the saturated 'real gas' that will be used to model the real process. The real process, then, actually starts just downstream of this 'saturation module'.

Problem is, when I do this, the predicted temperature of the now-saturated gas from the overheads of the fake separator is nowhere near the 190° that I need, that the real life TI says, but it's like 176° or so. I need the real 190 for my HEX calcs. No matter what I do in the way of adjustments to the front of the model, I can't get the water-saturated gas to exist at 190 and 0.1. psig.

So I am thinking I am forgetting something fundamental here. I need to hear from you experts on phase equilibrium. I am thinking that at 0.1 psig, this particular saturated gas mixture can't exist at 190° F. The 'dry gas' can exist at anything I want, but add the water and it's no more than 176 or so due to the species present and the physics of the k-values. Yes? Art - I bet you know the answer. Thanks!!! Pete

PS - I think this points up two issues (1) I have forgotten the fundamentals and am relying on a simulator to do my work - Not Good (2) it's not possible to get real life to match a process imulator, and I would do well to remember that. Your comments? What was it John Campbell said - "Knowledge of the fundamentals is the required foundation for successful professional practice" - or some such.




Thanks!
Pete

 

[Montemayor]

Pete:

This is a challenging problem for various reasons. I’m going to have to do some presumptions, so bear with me since we don’t want to fill up the thread with responses back-and-forth and not give the cleverer and smarter guys on the forum a chance to jump in on this.

1. Your gas is essentially pure n-Hexane (99.18% vol. C6) or n-Hexane & heaviers (although I doubt it). I’m going to assume the remaining light gases don’t contribute much effect.
2. At 190 oF Hexane has a vapor pressure of 25 psia. It has a normal boiling point of 156 oF at atmospheric pressure (which is essentially where you find yourself – 14.796 psia). That means you have approximately 30 oF of superheat in your Hexane gas stream. This is what is keeping it in the gas phase.
3. At 190 oF, water has a vapor pressure of 9.33 psia.; the liquid phase has not started to boil yet.
4. I’m going to assume the C6 doesn’t form hydrates.

I believe you’ve got the right routine to simulate the saturated conditions. However, I don’t know what went awry with your mixer, resulting in a lower temperature of 176 oF. But that can be fixed by introducing a heat exchanger, prior to the mixer, to heat the mixture to a temperature above the 190 oF (205 oF?) such that it will result in an overhead saturated product of 190 oF. This, I would hope, should give you the saturated mixture at the conditions you want. Mind that you are working in an area that is relatively close to the saturated vapor line of the n-Hexane. Visualize, if you will, a mental picture of the n-Hexane T-S diagram. We used to call the light hydrocarbon phase envelopes (or “domes”) “witch’s tit” because of the manner that they tend to droop and hang over to the right-hand side of the graph. What I mean by this is that if you refer to a similar T-S diagram (like the GPSA’s diagram for Pentane) you will see that you are probably in an area that is confined by the saturated vapor curve being on the left side and directly overhead. Depending on how you handle the simulation program, it could interpret the mixture as being a saturated vapor and not superheated anymore – which then throws out the criteria for having phase equilibria in the gas phase and nullifies Dalton’s Law of partial pressures – which is what I would depend on to enter into play.

If your hydrocarbon mixture is truly in the gas phase and it does not react to form hydrates, then I don’t believe you have reason to fear that it won’t get saturated with water vapor in the superheated region. This would be a normal, physical event to be expected – as I’m sure you are expecting it to occur. Rest assured that if you maintain a superheated gas and expose it to water at the same temperature, it will get saturated if left to establish an equilibrium condition. Your initial results lead me to believe that the simulation is not set up right or that there is a hydrate or other reaction taking place – which I don’t know about. But bottom line, I’m in agreement with you in expecting that the gas get saturated when exposed sufficiently to excess water – of course, with the caveat that it is in the superheated gas region (and stays there). I trust that is the situation because otherwise the compression stage wouldn’t make any sense at this point. I’m going to close this contribution and eagerly wait for the next responses which I anticipate will be from such smart and knowledgeable guys like 25362 and others.

Good Luck.


Art Montemayor
Spring, TX

 

[KernOily]

Guys thanks for your replies. Art - actually the component fractions I provided above are mole (volume) fractions, so the gas is mostly CO2 and mostly methane with a tinky bit of C6+. And to be totally correct I should have presented that particular fraction as C3+.

Sorry about the confusion - everybody represents fractions a little differently. Thanks! Pete

Thanks!
Pete

 

[hacksaw]

at the risk of a stupid comment, since when is a saturation process isothermal?

 

[sailoday28]

For the non-chemical engineers:
What is C1 ,C2 C6+ C3+ ??

 

[Montemayor]

sailoday28:

Sorry for the Chem jargon. Usually one uses the symbols to mean the "saturated" hydrocarbons such as:

C1 = Methane
C2 = Ethane
C3 = Propane
C6+ = Hexane + higher hydrocarbons (Heptane, etc.)

Sometimes un-saturated hydrocarbons are included (such as acetylene, ethylene, propylene, benzene, etc.), but I have not seen this to be the case when listing a gas composition.

Hope this clears things up.

Art Montemayor
Spring, TX

 

[EGT01]

74Elsinore,

It might help to know what simulator you are using and the settings you've chosen for your unit operations.

The simulator I use at work is Aspen. It would allow elimination of the mixer operation and you could just introduce the water and hydrocarbon streams as two separate feeds into a separator (flash) operation. As Hacksaw points out, if you set your flash operation to zero heat duty, then water vaporizing will cause a temperature decrease. If you set the flash operation temperature to what you desire then you shouldn't get a temperature decrease.

Have you tried performing the flash at a specified temperature and pressure? Seems like you could start with a small flow of water and gradually increase it until you start seeing liquid from the flash operation. Depending on the thermo property method you use you probably need to watch the liquid composition to see if hydrocarbons go out with the water which would tend to distort your dry gas composition if it does.

 

[siretb]

I think parts of your problems comes from the way you define your mixer
If you just specify you mix 190°F dry gas with 190°F water, it will do an adiabatic mix. Since some heat is needed to vaporize the water, temperatture will drop. And the temperature is close to what we call the "saturation temperature", the one you reach when you quench a gas.
I have to deal with your problem very often. Unless I misunderstand you, the proper way is
Not to use a mixer,
But use a Flash , with specified pressure and temperature (190°F 0.1 psig) and to ensure that some liquid water is left over.
Be careful about the thermodynamical package you use, you have water and hydrocarbons

 

[engzeidan]

Hi 74Elsinore,

To get your stream do the following (assuming you are using HYSYS)

1.Create 3 streams: stream1, stream2, stream3

2. Define the hydrocarbons composition in stream1 on molar basis. Define pressure as 0.1 psig and any temperature (such as 190F). Define an arbitrary flow value (such as 100 lbmole/hr)

3. Define water as composition in stream2. Define pressure as 0.1 psig. DO NOT DEFINE TEMPERATURE.

3. Use a mixer unit to mix stream1 and stream2 into stream3.

4. For stream3, define the dew point (vapor fraction = 1). This represent the saturation point.

5. HYSYS will calculate stream3 temperature as 179F. Manipulate stream2 flow rate to get stream3 temperature of 190F. In this case, it will be 180 lbmole/hr. Stream3 is then saturated with water at 0.1 psig and 190F.

Note that the above results could be produced in other software tools but would require more work. HYSYS has a functionality of back calculating which is not availble in other softwares.

Do not use a heat exchanger to superheat the flow. This will give you incorrect superheated stream with wrong water content.




A.Zeidan

 

[25362]

A simple ideal gas calculation shows that if at 190 F the water vapor pressure is 9.34 psia, and the total pressure is 14.696+0.1=15.696 psia, the mol fraction of water in the saturated gas (water dew point) would be:

9.34/15.696=0.595​

Please, anybody tell me what is my error.