264 replies to this topic

[PingPong]

No, is not correct.

 

The correct calculation of steam in PSA feed gas is:

 

(115.0159 - 31.15686) * 0.004097 / (1 - 0.004097) = ..............kmol/h

[MurtazaHakim]

We referred to the sources suggested by you and found that some of the data had temperature constraints so we searched from other sources.

 

We have obtained the values of Henry's constant from NIST website and have used the temperature dependence correlation to get the values of those constants at other temperatures.

 

Kindly have a look at the attached file and suggest the further treatment required.

 

 

Attached Files

•  Henry constants.PNG   28.38KB   1 downloads

[ravindra@096]

 

 

(115.0159 - 31.15686) * 0.004097 / (1 - 0.004097) = ..............kmol/h

 

  Can you please explain how did you arrive at this formula ?

Edited by ravindra@096, 20 January 2018 - 05:28 AM.

[PingPong]

Murtaza, your calculated values are correct but note that the units of measurement for k_H on NIST website are mol/kg.bar not mol/l.atm as you are reporting. Difference is small but nevertheless.

 

Note also that the third column heading in your table should read -Δ_solH/R

 

Ravindra, the formule simply takes the dry gas part, divides it by its mole fraction (to give total PSA gas feed), and multiplies that by water vapor mole fraction.

Edited by PingPong, 21 January 2018 - 06:11 AM.

[breizh]

Hi ,

 Hope this document is going to help you .

Breizh

[MurtazaHakim]

According to Henry’s law, the solubility of a gas in the solvent is equal to the product of Henry’s law constant and the partial pressure of the gas in the system. Having known the Henry constants of all the constituent gases, we require the partial pressure of all the gases to compute their solubility.

 

Considering the Dalton’s law of partial pressure, we need the total pressure of the system to calculate the partial pressure of each gas since we have the mole fractions of individual gases in the system.

 

What is the temperature and pressure of the system(condenser) ?

[PingPong]

The temperature is whatever you choose it to be, Depends on the temperature of the cooling medium(s) available.

 

The pressure I already indicated before as typically 2 - 2.5 bar lower than the reformer furnace outlet pressure.

[MurtazaHakim]

I have calculated the amount of shifted gas components dissolved in the condensate considering the operating temperature and pressure of the condenser to be 45 degrees Celsius and 18 bar.Kindly have a look at an image attached below.

Attached Files

•  condenser calculations.JPG   120.71KB   0 downloads

[PingPong]

I can't see from the printout what is calculated from what and how, as no formulas are visible.

 

However I do notice that your partial pressure of steam at 45 ^oC is wrong. Check with steam table.

 

And I also notice that your T(K) = 343.15 is wrong.

[MurtazaHakim]

I am attaching an excel spreadsheet. I have tried to correct the mistakes made in the former message.

I have calculated the Henry constants at 45 Degrees Celsius and used them to calculate the solubility of each component.

The mole fractions are calculated by dividing individual flow rate by total wet gas in the condenser.

The mole fraction is then multiplied by the total pressure in the condenser to compute the partial pressure of each gas.

Finally the product of partial pressure and Henry constants give the solubility in mol/kg which is further multiplied by the flow rate of steam condensed in kg/hr and divided by 1000 to get the solubility in kmol/hr

Attached Files

•  condenser calc..xlsx   25.8KB   74 downloads

[MurtazaHakim]

The tail gas consists of CH₄,CO,CO₂,N₂ and unrecovered H₂. Which of these gases are to be used in the reformer furnace ?

After completion of mass balance we are required to head toward the heat balance. How to begin with the heat balance ?

[PingPong]

The whole tail gas stream is used as fuel gas in the reformer furnace.

 

I want to remind you what I already mentioned in message #14: there will be a hydrogen recycle over the unit of about 5 % on natural gas feed to service the HDS reactor. So far you chose not to take that into account for the material balance (see your message #27) for reasons unknown to me, although it does have some negative impact on the equilibrium calculations.

It will have an impact on the heat balance as it has to be heated up and cooled down in the unit.

 

For the heat balance I suggest you start with the radiant coil.

(1) Outlet temperature and composition are known so you can calculate the effluent enthalpy relative to 25 ^oC.

(2) For radiant coil inlet assume a temperature of say 550 ^oC so you can calculate the combined feed mixture enthalpy relative to 25 ^oC.

(3) Calculate enthalpy of reaction at 25 ^oC for actual conversion of feed to effluent.

Calculate radiant coil absorbed duty using (1) , (2) and (3).

[MurtazaHakim]

The reason behind my ignorance of H₂ recycle is mere the simplicity in designing the plant. I was advised to assume that the natural gas utilised as the feed stock is fairly free of impurities (mainly sulphur) and does not require any HDS. If any impurities are assumed to be present,the task of designing the hydrogenation reactor followed by ZnO scrubbers becomes inevitable. Is it possible to obtain pre-treated gas in order to refrain from the pre-treatment section design ?

 

Kindly comment on the former attached excel spreadsheet.

[PingPong]

Every SMR has an HDS reactor and ZnO beds to remove any sulfur.

Every natural gas contains some sulfur compounds (mainly mercaptans) even if they are not mentioned in the composition.

 

You can't buy pretreated natural gas for SMR use. It is much cheaper to treat the natural gas inside the SMR as the natural gas is preheated anyway in the convection section of the furnace, so it is just a matter of taking it out of the convection bank at the right temperature, treat it, and send it back to the convection bank for further preheating to say 550 ^oC.

 

What you agree with your supervisor is between you two. If he does not find it necessary to include pretreating in your project than that makes your life a little easier. I merely wanted to remind you (and other interested readers) that in the real world there will be a hydrogen recycle.

 

I'll look at the spreadsheet later. You need not wait for that as it does not affect your heat balance calculation.

[ravindra@096]

After considering the 5 mol% (1 kmol in our case) H₂ recycle in the feed, the reformer effluent and WGSR effluent changes slightly. We have calculated the outlet flow rates taking into account the 5 mol% H₂ recycle and obtained the following.The reported flow rates in the image attached are in kmol/hr.We have now agreed upon the consideration of the HDS reactor and ZnO beds into our design.

 

Should we consider the mass balance around the HDS reactor and ZnO beds since impurities removed are of ppm level ?

Attached Files

•  New Flowrates.JPG   26.49KB   3 downloads

Edited by ravindra@096, 25 January 2018 - 04:55 AM.

[PingPong]

Spreadsheet result looks correct.

 

HDS & ZnO have no impact on material balance of the unit.

 

Updated Reformer and WGS effluent compositions look correct for 5 % hydrogen recycle.

[MurtazaHakim]

The reformer operates at 20 bar. Do we need to take the effect of pressure into account for the values of specific heat of the components in the mixture ?

[PingPong]

To take that effect into account in a manual calculation would be difficult to do as it would require the use of an equation of state like SRK or PR or ......

 

In this case the gases will be (nearly) ideal gases even at 20 bar so I suggest you do not bother with that.

[MurtazaHakim]

(1)The effluent enthalpy relative to 25 ̊C comes out to be 1009514145 KJ/hr

(2)The feed mixture enthalpy relative to 25 ̊C is calculated to be 4565191477 KJ/hr

(3)The enthalpy of reaction for actual conversion of feed to effluent is 13578438.09 KJ/hr 

   (calculated by adding standard heats of reaction of two equilibrium reactions considered and then multiplying it to the no.of moles of feed)

 

The heat capacities at temperatures 850  ̊C and 550  ̊C are calculated from the correlations provided in the book CHEMICAL ENGINEERING THERMODYNAMICS by  J M Smith,H C Van Ness,M M Abbott (App. C).Please have a look at the attached spreadsheet.

Attached Files

•  SMR Mass and Heat .xlsx   56.94KB   107 downloads

[PingPong]

I am sure those numbers are not calculated correctly as they are simply far to huge for such a small flowrate.

In any case use kW instead of kJ/hr.

 

To calculate each gas mixture enthalpy properly you need to calculate the integral of C_p.dT between its T and 298.15 K

 

I do not have the time to check spreadsheets for bugs so from now on I will simply compare what you calculate with result of a process simulator:

Reaction enthalpy @ 25 ^oC should be roughly 855 kW.

Enthalpy of feed mix @ 550 ^oC should be roughly 480 kW

Enthalpy of effluent @ 850 ^oC should be roughly 915 kW

So total absorbed radiant coil duty should be roughly 1290 kW

 

Keep trying until you find similar numbers.

Edited by PingPong, 27 January 2018 - 10:28 AM.

[MurtazaHakim]

I have corrected the spreadsheet according to the formula suggested by you and have obtained the values of 915.2735 KW and 478.5397 KW for reformer effluent and mixed feed enthalpy respectively.Please have a look at attached images.

 

I have two doubts,

(1) How to calculate the reaction enthalpy at 25 degrees Celsius for actual conversion of feed to effluent ?

 

(2) The mixed feed enthalpy calculation should include the sensible heat of liquid water from 25 degrees to                         100 degrees Celsius and latent heat of water at 100 degrees Celsius.Please comment.

Attached Files

•  Heat balance 1.JPG   96.12KB   0 downloads
•  Heat balance 2.JPG   117.24KB   0 downloads

Edited by MurtazaHakim, 28 January 2018 - 11:29 AM.

[ShreeJeeth]

Alright.... Now I am new to this discussion and I have just started my project on designing a SMR. So my aim is to get 50 tpd methanol. So can anyone share the process flow diagram of this process... So that I can verify it with mine...

[PingPong]

I have corrected the spreadsheet according to the formula suggested by you and have obtained the values of 915.2735 KW and 478.5397 KW for reformer effluent and mixed feed enthalpy respectively.Please have a look at attached images.

Those calculated numbers are correct.

 

However I have difficulty with the last term in the integral formula that you write: D/T_o(τ - 1/τ)

Is that a typing error and did you actually use (D/T_o)(1 - 1/τ)

 

Personally I don't like the use of τ in those formulas. Smith et al seem to be fond of it but it makes the formulas only more complicated and increases the risk of mistakes.

 

(1) How to calculate the reaction enthalpy at 25 degrees Celsius for actual conversion of feed to effluent ?

By comparing feed and effluent compositions you can see which components have reacted to form which products, so calculating reaction enthalpy is straightforward.

 

(2) The mixed feed enthalpy calculation should include the sensible heat of liquid water from 25 degrees to 100 degrees Celsius and latent heat of water at 100 degrees Celsius.Please comment.

No.

Remember Hess' Law: when going from one state (feed) to another (effluent) the enthalpy change is independent of the route taken. In case of a chemical reaction the reaction enthalpy is easiest calculated at 25 ^oC as standard enthalpies of formation are available for that temperature. So your simplest route goes from actual feed temperature to ideal (feed) gas at 25 ^oC, reaction at 25 ^oC, and from ideal (effluent) gas at 25 ^oC to actual effluent temperature.

Therefor the calculations become simplest by defining the enthalpy of ideal gas at 25 ^oC as the zero point for the enthalpy of all streams.

[RaviGaanth]

Hey guys....Can u say where the mass balance started in this discussion? I am really liking to get into this discussion...so please help me out.

[MurtazaHakim]

I am stuck in the process of evaluating enthalpy of reaction at 25 Degrees Celsius.I have taken the first reaction (which is endothermic in nature) for calculation but I am not sure how to evaluate the second reaction since the second reaction has CO as the reactant which is not initially present but is produced during the reaction.Kindly see the attached image and provide your suggestion for the same.

 

We have the composition of the tail gas to be used in the reformer furnace. How do we evaluate the amount of duty fulfilled by the tail gas to determine the remaining duty to be fulfilled by the other source (electrical or natural gas) ?

Attached Files

•  heat of reaction.jpg   117.94KB   1 downloads