264 replies to this topic

[PingPong]

I checked the calculations from your message #26, found no calculation errors, and have only a few remarks.

 

First of all it is important to note (and always include in your calculation report) on what reference state pressure P^o the data from your sources are based. See note 2 at the bottom of Table C.4 of your textbook. The K that you calculate for the reforming reaction is then valid for that P^o. For the WGS reaction it does not matter whether P^o is 1 bar or 1 atm because the number of moles does not change, but for the reforming reaction it does..

 

Secondly it is important to realize that due to the exponential character of K the slightest inaccuracy (or error) in input data will have a noticeable effect on the calculated K. For example:

 

You use 298 K as the reference temperature, resulting in K_ref = 386.5 and K_wgs = 0.925 , however when using the correct 298.15 K you would have found K_ref = 380.9 and K_wgs = 0.927

 

In your older calculation in message #22 you rounded the calculated ΔH_f^o to 206000 which then had an impact of 4 % on the calculated K.

Edited by PingPong, 05 January 2018 - 06:00 AM.

[MurtazaHakim]

I have prepared the atom balance sheet as suggested. I have considered 20 Kmol/hr of natural gas feed in my case. I have not taken into consideration the 5 mol% (1 mol% in my case) H₂ recycle. Kindly go through the attached images. The doubt is how do I solve three equations with five unknowns ?

 

 

there are the 2 equilibrium equations (equilibrium quotient equals the K value), the one for reforming, and the one for WGS

Which equations am I supposed to obtain ?

 

Attached Files

•  IMG-1.jpg   149.52KB   4 downloads
•  IMG-2.JPG   168.11KB   3 downloads
•  IMG-3.JPG   143.55KB   2 downloads

[PingPong]

You should also add on the last page: Let N_T be x₇

and then rewrite the equation for that in terms of all x's.

 

Which equations am I supposed to obtain ?

You calculated the numeric values for K_ref and K_wgs so now you write 2 equations to equate each of them to their reaction quotient. See equation (13.25) in your textbook.

In case of ideal gases the f^{^}_i of each component is equal to its partial pressure p_i

Again I advice you to read message #14 once more.

 

Then you have a total 7 equations for 7 unknowns. In this case N_N2 is easily solved as there are no other N-components except N₂ so that leaves you still with 6 equations with 6 unknowns.

[MurtazaHakim]

Should we consider the equilibrium mixture as an ideal gas since the reformer outlet pressure is 20 bar ?

If not, do we need to calculate the fugacity coefficients of all the reacting species and use equation 13.27 of the previously mentioned textbook ?

[PingPong]

At 850 ^oC the reformer effluent can be assumed to be a (nearly) ideal gas, even at 20 bar.

 

In any case it would be extremely difficult to calculate fugacities or fugacity coefficients by hand. A process simulator would solve an equation of state to obtain them.

 

For your manual calculations use (13.25) with f^{^}_i replaced by p_i or use (13.28) which is equivalent as p_i = y_i.P

Edited by PingPong, 08 January 2018 - 07:27 AM.

[MurtazaHakim]

I have considered the equilibrium mixture as an ideal gas as you have suggested. Kindly have a look into the following attached files and verify whether I have made the correct calculations. 

 

Can the two equilibrium constants be equated to the extent of reaction correlations ?

 

 

 

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•  IMG 2.jpg   127.69KB   2 downloads
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[PingPong]

To be clear, you need to solve following 6 equations with 6 unknowns:

 

equation #1: x₁ + x₂ + x₃ = 20.49

equation #2: x₂ + 2x₃ + x₄ = 61.67

equation #3: 4x₁ + 2x₄ + 2x₅ = 201.52

equation #4: x₁ + x₂ + x₃ + x₄ + x₅+ x₆ - x₇ = 0

equation #5: 0.95225·x₁·x₄·x₇² -x₂·x₅³ = 0

equation #6: 0.927·x₂·x₄ - x₃·x₅ = 0

 

In which x₆ = 1 (as N₂ in effluent = N₂ in feed = 1 kmol/h)

 

Solving that set can only be done by iteration. To do that manually one could try successive substitution.

Problem with that method however is to figure out which variables to calculate in which order from which equation, and which variables to give which value to start with to finally get a converged solution. That is for each set of equations always a time consuming exercise.

 

EDIT: typing error in equation #3 corrected.

Edited by PingPong, 09 January 2018 - 09:37 AM.

[MurtazaHakim]

 

 

equation #3: 4x₁ + 2x₂ + 2x₅ = 201.52

 I think the equation is 4x₁ + 2x₄+ 2x₅ = 201.52 . Please recheck and verify.

[PingPong]

You're right, I made a typing error.

I have corrected it in my post.

[MurtazaHakim]

After having done the calculations in the MS Excel (Solver) the variables obtained are as follows (in Kmol/hr)

N_CH4  (x₁) = 3.61703

N_CO    (x₂) = 10.28624

N_CO2  (x₃) = 6.58672

N_H2O  (x₄) = 38.21031

N_H2    (x₅) = 55.31562

N_N2    (x₆) = 1.00000

N_T      (x₇) = 115.01593

 

We have manually verified the above values by plugging them into the equations obtained and found them to be in good agreement with the R.H.S of the equations.

 

Now the following part is Water gas shift reactor. Which type (and how many) of water gas shift reactor(s) should be employed ? How do we further proceed into WGS ? I have been advised to have only one WGS reactor due to the low capacity of H₂ production.

Edited by MurtazaHakim, 10 January 2018 - 04:41 AM.

[PingPong]

You found the correct effluent composition.

 

It is difficult to predict what kind of process design licensors would propose for your small capacity, but most steam reformers for hydrogen production use only a High Temperature Shift (HTS) reactor with an inlet temperature such that the outlet temperature is around 420 ^oC. In this case the design ATE for the WGS equilibrium is usually in the order of 20 degrees, so K_wgs is then to be calculated at 420 + 20 = 440 ^oC. No others equilibria need to be considered (CH₄ will not react) so effluent composition can easily be calculated manually by incorporating a fractional conversion for CO in the formulas.

[MurtazaHakim]

The inlet temperature in most of the cases (from the literature survey) employing HTS is 350 degrees Celsius. Will there be any other reaction having considerable conversion apart from the CO + H₂O <==> CO₂ + H₂ ?

 

 

It is difficult to predict what kind of process design licensors would propose for your small capacity,

 

I think it is an open-art unit with Johnson Matthey as the catalyst supplier.

 

Should the former procedure be followed for obtaining the WGSR outlet composition and quantity i.e preparing atomic balance and calculating the K_WGS at the WGSR outlet temperature of say 440 Degrees Celsius ?

[PingPong]

The inlet temperature in most of the cases (from the literature survey) employing HTS is 350 degrees Celsius.

If you fix the HTS inlet temperature at 350 ^oC then you will not know what its outlet temperature will be and so you cannot calculate Kwgs. Therefor I suggest you fix the outlet temperature at say 420 ^oC and later calculate from the HTS heat balance what inlet temperature corresponds with outlet 420 ^oC.

Will there be any other reaction having considerable conversion apart from the CO + H₂O <==> CO₂ + H₂ ?

No.

Should the former procedure be followed for obtaining the WGSR outlet composition and quantity i.e preparing atomic balance and calculating the K_WGS at the WGSR outlet temperature of say 440 Degrees Celsius ?

Yes you can.

 

But once you have calculated K_wgs you could also simply define a conversion for CO and solve it by hand from the equation Reaction Quotient = K_wgs

[MurtazaHakim]

After having exercised the former procedure I have come out with the following.Please look into the attached files and verify the obtained results.

 

The downstream unit to the WGSR is PSA. What procedure should be followed for the PSA unit?

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•  IMG_20180112_160156#04.jpg   115.74KB   3 downloads

Edited by MurtazaHakim, 12 January 2018 - 06:14 AM.

[PingPong]

Your calculated shift effluent composition is correct.

 

The material balance around a PSA unit is very simple:

 

The hydrogen product is nearly pure H₂. For refinery applications a purity of 99.9 mol% is usual. It is possible to produce an even higher purity but at the cost of decreased recovery and/or higher investment cost.

 

The hydrogen recovery depends on:

- composition and hydrogen content of feed gas

- operating pressure of psa

- operating temperature of psa

- number of adsorption vessels in psa

- pressure of tail gas product

- number of operating steps in the psa sequence control system

 

Usually the psa is designed by a specialized vendor for a hydrogen recovery in the range of 80 - 90 %. It is an economical choice. More adsorption vessels mean more recovery but at a higher investment cost.

 

For your small unit the optimal recovery is likely at the lower end of the indicated range.

 

Unless your application needs a very high purity I suggest you base your material balance on a hydrogen product purity of 99.9 mol% (with 0.1 % CH₄ plus N₂ plus only traces of CO and CO₂) and a recovery of say 85 %.

[MurtazaHakim]

I think 99.9% purity would suffice our objective.How many vessels do we require for the flow rates obtained above ?

The inlet temperature of PSA in most of the cases encountered in the literature survey is 40 Degrees Celsius. 

We are unaware of the operating pressure of PSA.

The Hydrogen content in the feed gas is 54.22 mole %

We intend to use the tail gas for burning in the reformer furnace.(We do not know the tail gas product pressure)

 

Moreover we have around 31 kmol/hr (27 mole %) of steam in the feed gas to the PSA. What happens to the steam ? (It should condense in reaching 40 Degrees Celsius from 440 Degrees Celsius which is WGSR outlet temperature)

 

Now coming to the material balance, Should we just take the product of H₂ content in the feed gas and recovery efficiency (say 0.8*62.369) as our final product to the storage ?

Edited by MurtazaHakim, 12 January 2018 - 10:09 AM.

[PingPong]

The number of vessels is selected by the vendor to obtain a certain recovery. Capacity determines the size (L & D) of the vessels. More vessels (and therefor also more piping and valving) increase investment cost. For a small unit a few percent more recovery means not much more hydrogen product and therefor the additional investment cost may not be worth it.

 

I merely indicated that there are many factors that determine the recovery and all we can do now is assume an average number, say 85 % for a psa unit with not many vessels (maybe 4 or 5).

What happens to the steam ? (It should condense in reaching 40 Degrees Celsius from 440 Degrees Celsius which is WGSR outlet temperature)

The steam is mostly condensed at 40 ^oC and the condensate is separated from the syngas before it reaches the psa unit. The remaining steam ends up in the psa tail gas, together with the other non-hydrogen molecules and the not recovered hydrogen.

Should we just take the product of H₂ content in the feed gas and recovery efficiency (say 0.8*62.369) as our final product to the storage ?

Yes, if you assume 80 % recovery.

 

What do you mean by storage?

Edited by PingPong, 12 January 2018 - 11:32 AM.

[MurtazaHakim]

How much condensate would be produced before the shifted gas reaches the PSA unit ? The shift reactor effluent carries around 31 kmol/hr of unreacted steam. The amount of condensate further determines the tail gas product composition and flowrate.
The storage herein refers to may be surge vessel.

[PingPong]

You can simply use a steam table to obtain water vapor pressure at 40 ^oC and use that to estimate the remaining amount of water vapor at 40 ^oC & 18 bar in the PSA feed gas.

[MurtazaHakim]

Can you please elaborate on message number #44 ?

An additional question is, will any of the other components of shifted gas liquefy while reaching the inlet of PSA unit ?

Edited by MurtazaHakim, 17 January 2018 - 02:47 AM.

[PingPong]

PSA feed gas is water saturated as it leaves a separator. Steam table gives saturated vapor pressure so that is then the H₂O partial pressure in PSA feed gas, which gives you mol fraction assuming an absolute pressure of say 18 bar (usually 2 - 2.5 bar lower than reformer outlet).

 

The condensate from the separator will contain some dissolved gases. If you really want to calculate that you need to find solubility data or Henry constants for the gas components in the shifted syngas.

[MurtazaHakim]

Kindly look into the attached file and suggest any modifications if required.

 

We are unaware of the operating condition of the condenser ? The Henry's law constants are available at 298 K, but the temperature of the condenser seems to be more than 298 K.

Attached Files

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[ravindra@096]

I am getting multiple sources for the values of Henry's constant with different values for the same component. What reference should I take ?

Out of the values obtained only CO2 has considerable solubility so should I neglect the solubility of other components ?

[PingPong]

Murtaza,

your 0.47 kmol/h steam in PSA inlet gas is wrong.

Rethink to what you should apply that molfraction 0.004097

 

Remember what I wrote before: "If you really want to calculate that you need to find solubility data or Henry constants for the gas components in the shifted syngas.".

 

If you both feel that taking into account the gas that dissolves into the condensate will enhance the quality of your study report then you will have to do some effort. I am sure that Henry constants or solubility data must be available at other temperatures than 298 K. It is just a matter of searching, searching, searching. I don't know which sources you consulted up till now but I suggest you start in Handbook of Chemistry & Physics.

[MurtazaHakim]

I made a mistake by not taking n_T as 31.15686 kmol/hr which is the total number of moles of steam exiting the WGSR.

So taking n_T = 31.15686 kmol/hr

n_H2O to PSA inlet is 0.004097*31.15686 = 0.12764 kmol/hr. I hope this calculation is correct.