[apex]

Hi,

I am a student and one of the things I have to do for my endreport is to design a MDEA unit. I have some question that hopefully someone can answer me.

The system I have to design is practically always lean end pinched. I added some phosphoric acid to lower the steam consumption. I have read that the phosphoric acid increases the protonated amine, and at low HS- concentrations like for instance at the reboiler helps the regenerating process because of the much higher protonated amine concentration.

I did some runs on the simulator, I expected that the system would stay lean end pinched. But it didn't. The phosphoric acid concentration has an optimum. By increasing the R2NH+, you actually decrease the amine concentration, so where phosphoric acid helps the regenerating process, it has a negative effect on absorption explainin the optimum concentration.

What I also saw, and this is what I don't get, the lean loading corresponding with the lean approach in % before phosphoric acid addition, do not match the lean approach % after addtion. This can only be explained that the equilibrium H2S vapor composition has changed. So H3PO4 has an effect on the equilibrium curve of H2S. Can someone explain this?

Also what is the difference between plates/ random packed/ structured packing. I do not have CO2 in the gas, only H2S, so residence time is not that important for me.

Why is the highest MDEA concentration set to 50 wt%?

How can foaming occur besides hydrocarbon presence in the absorber?

Thanks in advance,

apex

[eugenedakin]

Hi apex,

Wow, you have a ton of questions...good news... I have 'some' answers.

I have been working with sweetening units for over 10 years, and I am guessing that your term 'lean end pinched' means that the lean end of the process is the most inefficient portion of the system.

Yes, adding an acid will increase the protonated amine concentration...but.. why would you want to do this? The H2S protonates the amine also, and adding something else will tie up the MDEA with a phosphorous group, instead of tying up H2S?

Addition of phosphoric acid and amine form a new chemical... one which is used for corrosion inhibition, and neutralizes its H2S capabilities.  In other words, adding phosphoric acid lowers the available MDEA for H2S removal.

As a general rule of thumb, random packed is most efficient for contact, structured packing second (it depends on design) and plates are least efficient for contact.

MDEA is at 50% concentration due to freeze point depression.

Yes, a wide variety of other chemicals can cause foaming .. such as corrosion inhibitors ... 

I hope this helps,

Sincerely,

Eugene

[apex]

Hi eugenedakin,

Thanks for replying. You're right, by the term lean end pinched I mean that the bottom plates do most of the work, and the lean amine has the minimum purity required to achieve 50 ppmv overhead (what I need).

But adding H3PO4 will bring the steam consumption down substantially. You're also right by saying it lowers the capacity of the amine, that's why there is an optimum concentration. Here is how H3PO4 works:

the K value of the reaction of amine with H2S is:

K=(R2NH+ . HS-)/(R2NH . H2S)

In the reboiler both R2NH+ and HS- concentration are low, but by adding H3PO4 you increase R2NH+ significantly, and since the K-value is constant HS- goes down alot, thus achieving a much leaner amine. The concentration of H3PO4 doesn't have to be high to achieve this, mostly it's around 1 wt% to 2 wt%, depending on the amine circulation. So you actually add some heat stable salts to the amine, it sounds contradictory, but I have read several articles about it, this doesn't work for DEA and MEA. The HSS concentration needs to be followed closely. An interesting article that describes this effect is:

http://ogtrt.com/pdf/AIChE_Austin.pdf

Also, could you give me the temperature of the freezing point depression?

Best regards,

apex

[eugenedakin]

Hello apex,

Thanks for your reply and the interesting article.

I would be quite hesitant on partially neutralizing any amine with phosphoric acid for a few reasons: 1) yes, addition of phosphoric acid will increase the hydrogen concentration at the reboiler, but the purpose of the amine is to prevent the hydrogen (via an elevated pH) from the contactor to the reboiler.  If more hydrogen is what is needed, instead of using a 50% concentration of MDEA, use a 40% concentration, this will achieve the same effect. 2) increasing heat stable salt increase foaming. Elevated foaming decreases the contact of fluids and gas in the contactor, making the whole process much more inefficient. 3) adding phosphoric acid neutralized some free amine, decreasing the efficiency of the transfer of acid gases.

MDEA (pure) has a freezing point of -21 C (-5.8 F), addition of 5-50% water lowers the freeze point below -25 C (-13C), and also increases the free water in the system. Free water is very important, as the H2S and CO2 reaction occur in the water phase only. H2S and CO2 must dissolve into the water, then the majority of the acid-base reaction occurs. Also, the viscosity decreases signifacently, allowing a faster exchange rate (in both contactor and reboiler).

Here is a chart where concentration data is available for MDEA:

http://www.coastalchem.com/PDFs/GAS_SPEC/MDEA.pdf

Lowering steam consumption is typically a minor problem compared with removing all of the acid gases. Purchasing a cube of water compared to 10 days of downtime due to addition of phosphoric acid is inexpensive.

I used to spend much of my time working with Dow and Union Carbide on sweetening.  Much work has been completed to make the systems work efficiently.  I would be VERY cautious about adding phosphoric acid to the system, as many known and unknown reactions are occuring. 

I hope this helps.

Sincerely,

Eugene

[apex]

Hi eugenedakin,

I agree absolutely, the simulation program just calculates the outcome with thermodynamical packages and does not take actual problems like foaming into account. So here is where your expertise come in handy! Thanks for the MDEA properties list. This advice will help me for sure!

Best regards,

apex

[apex]

eugenedakin,

Do you know something about the design of a static mixer? The simulation program calculated the amineflow to equilibrium. Bot you need an endless long mixer to achieve this point. So how can you decide a good geometry for the mixer?
Also you said this in your last post in this topic:

"If more hydrogen is what is needed, instead of using a 50% concentration of MDEA, use a 40% concentration, this will achieve the same effect."

I don't believe that will be the case, because you won't influence the equilibrium of H2S since there is no common ion effect.

Best regards,

apex

[eugenedakin]

Hello apex,

I am guessing that you are talking about the equilibrium in the reboiler (mixing).  There is enough steam being generated that no mixing is required, the bubbles from steam provide the agitation.

After re-reading the phosphorous information you provided, I am still not sure how the phosphoric acid willwork.  Do you happen to know the mechanism by which this works? Could you explain to me what your interpretation of the 'common ion' effect is?

Thank you,

Eugene

[apex]

Hi eugenedakin,

Concerning the common ion effect, maybe this text helps:

In Kohl and Nielsen’s Gas Purification, they say this concerning the chemical equilibrium explanation as to why the presence of another acidic component (such as CO2 or H3PO4) increases the vapor pressure of H2S over an amine solution. They use [H2S][RNH2]/[RNH3+][HS-] = K  and go on to say “Since the equilibrium partial pressure of H2S in the gas is proportional to the concentration of molecular H2S in the solution (Henry’s Law), the equation indicates that the vapor pressure of H2S increases with both H2S and CO2 [or other acid] loadings in the solution. The enhanced stripping of H2S in the presence of dissolved CO2 [or other acid] can, therefore, be attributed to the common ion RNH3+ produced by both acid gases in reacting with an amine.”

Concerning the static mixer, I didn't explain myself, sorry. There will be a static mixer placed before the absorber, because of reasons concerning a rapid change in feed and concentration. There will be an aminesolution sent to the mixer, and a concentrated H2S stream. I found the best amineflow to the mixer with a simulation program. The problem however is that the program just calculates by using a simple equilibrium/flash calculation, and to achieve this equilibrium point you need an endless long mixer. So the problem is what length should I take. Is there something I can relate the amount of H2S removed to the length of a mixer.

Best regards,

apex

[eugenedakin]

Hi Apex,

Although I agree with the theory of common ion effect, I am still in disagreement with the addition of phosphoric acid for a couple of reasons: 1) lowering of available 'free' amine, 2) yes, the addition of phosphoric acid will lower the vaport pressure of H2S, and will contaminate the MDEA by forming heat stable salts. Since production streams are rarely clean, there will be water which can have ions such as calcium and magnesium dissolved into solution. In particular, Calcium forms a calcium phosphate salt, which can be almost insoluble.  The added heat stable salts will cause an elevated risk of foaming, which decreases efficiency of adsorption and removal of acid gases. 3) The tolerance of salts will be decreased.  With the insoluble nature of phosphate salts, and that phosphates are a common detergent (variations), this tends to promote emulsification and foaming tendencies (through different mechanisms than Heat Stable Salts).

I agree that a mixer-container be placed before the absorber be used due to changes in feeds and concentrations. This will help the system loadings become more predictable, and also provide a reservoir in case some amine fluid is accidently removed from the system.  As far as the length is concerned, one length doesn't immediately appear. From a 'surging' perspective, the more fluid reserve you have, the better.  I hope this helps.

Sincerely,

Eugene

[apex]

Eugenedakin,

I have a question about the overhead condensor of the stripper. I set the temperature to 40°C after the condensor. But does the type of condensation matter if you want to calculate the amount of water that has condensed?

You have integral and differential condensation. Differential is for instance when the H2S/ water mixture passes through the shell and the condensor is placed vertical,the condensed material and material in vaporphase stay in contact then.  Integral is for instance when the H2S/water mixture flows through the shell, and the condensor is placed horizontal.

Does the heat required to achieve 40°C differ between integral and differential condensation?

Best regards,

apex

[eugenedakin]

Hello apex,

Theoretically speaking, if the temperature is 40 C and both condensors are able to keep the temperature at 40 C, there should be no difference between the amount of water condensed - with everything remaining equal.

The only aspect which I can see being different is the concentration of H₂S that remains in the condensed water.  A vertical condensor may maximise the exposure time of H₂S and water making the reflux drum contain slightly elevated concentrations of acid gases over that of the horizontal condensor.  This difference will only be slight.

Great questions!

Sincerely,

Eugene

[apex]

Hi eugenedakin,

Let's say you want to condense H2S and water completely.

If you would have a vertical condensor, you would just have to cool down to the bubble point of the water/H2S mixture coming into the condensor.

If you would have a horizontal condensor, you would have to go to the lowest dew point to condense the last bit of H2S, since near the end of the condensor you (almost) only have H2S left to condense.

So the temperature profile is very different I think. So that's why I thought you would condense a different amount of water at 40°C, if you had a vertical or horizontal condensor. So you are saying this difference is very small here?

Do you happen to know where I can find a T,x diagram of H2S and water?

Best regards,

apex

[eugenedakin]

Hello apex,

I may not have explained it clearly.  The temperature of the fluid will ultimately be 40 C (by your suggestion), the only difference that I can see is that there is more contact time for the H2S with the water (appears to look like a reflux).  Either condensor (as long as all of the gas has cooled to 40C, and has sufficient time to coalese, will condense.  The only aspect that I can guess with our theoretical thoughts is an increased in contact time of H2S and water in the vertical condensor by some slightly increased time. Since H2S is a gas at 40 C, it will eventually dissipate out of the condensed water (which is at 40C).

If you have a high velocity of heated gas and a condensor which is too small (either type), you will not condense the water efficiently.

I hope that I cleared up any misunderstanding.

Sincerely,

Eugene

[apex]

Hi eugenedakin,

I have yet another question! Maybe a weird one.

It's about the addtion of H3PO4 again. If you add H3PO4, you get a heat stable salt R3NH+H2PO4-. Alot of anions like sulfates and phosphates can be used as an inhibitor by forming an insoluble compound with a metal ion and therefore covering the anode side of the corroding metal. But if you wanna add these inhibitors you always add them as H2SO4 or H3PO4 or not? Can the H2PO4- in the heat stable salt react with a metal ion? Does the H2PO4- ion always have to be in the vicinity of the R3NH+? Or do they just come together and form a salt when you evaporate the solution for instance.

Also, could you tell me something about your experiences with corrosion in your MDEA unit? I am guessing corrosion occurs in the stripper, absorbers, and places with pressure drops, like heat exchangers or control valves? Stripping of H2S will occur with higher temperatures and so a higher partial and pressure and hence an increased H2S concentration (H+) in the solution. But is there really alot of corrosion at pressure drop places? I have simulated that the rich loading of the amine doesnt change much after a pressure drop, I think places with higher temperatures are more dangerous. Can also some kind of passivation occur? Is stainless steel (AISI 316) good enough?

Best regards,

apex

[eugenedakin]

Hello apex,

When you add H₃PO₄, the R₃NH⁺H₂PO₄^- is only one of the possible heat stable salts formed. There could be easily 10 or hundereds of variations of heat stable salts formed.  The most common derivatives are insoluble metal salts. 

When you mention the word 'inhibition', I am guessing that you mean corrosion inhibition, right?  Inhibition can be present in many forms. I am just making sure that we are both talking about the same topic 

There are some huge assumptions that I will be making when talking about corrosion in these units: 1) all metals are the same corrosion resistant material, 2) proper insulation (conducitivity) measures have been implemented, 3) the system is working properly (no upsets recently), 4) the system has fresh MDEA (proper strength), 5) there is little to no heat stable salts, 6) the knockout drum is working properly, 7) no surges in pressure or rate are occuring, 8.) pumps, valves, and absorber or regeneration areas are unobstructed and free of all debris without constrictions, I am sure there is more, but this is a good start 

As long as there is sufficient excess of MDEA, there should be little/tolerable levels of corrosion in the MDEA system.  The most likely areas of attack exist in the rich amine stream from the absorber to the regenerator, and occur at the top of the regenerator when acid gases exit the reflux drum.  This is due to a lower pH, higher concentration of sulphur, and lower concentration of MDEA. Areas where pressure drops exist can be a problem if there is low concentrations of MDEA and elevated levels of heat stable salts.

Passivation can definately occur. The best protection of passivation occurs with very low velocities. Accelerated corrosion will occur during rapid flow.

I hope this helps

Sincerely,

Eugene