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# Heat Balance For A Steam Reformer Unit

264 replies to this topic
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### #1 MurtazaHakim

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Posted 26 December 2017 - 04:32 AM

How do we calculate the heat energy required to raise the temperature of the inlet mixture of natural gas feedstock (35 degrees Celsius at BL) and steam (550 degrees Celsius in the Methane Steam mixer) to 850 degrees Celsius (temperature of the reformer tubes) ?

### #2 PingPong

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Posted 26 December 2017 - 05:21 AM

At 850 oC most of the natural gas will be converted to H2 , CO and CO2 and part of the steam is also converted.  You have to take into account the equilibrium constants of all reactions involved. The overall effect of combined reactions is endothermic.

Normally you would use a process simulator to do this calculation. In any case you need the exact composition of the natural gas, the steam/carbon ratio and the outlet pressure of the reformer heater.

Edited by PingPong, 26 December 2017 - 05:33 AM.

### #3 MurtazaHakim

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Posted 26 December 2017 - 06:10 AM

The following composition of natural gas is taken into assumption.(mole % basis)

Methane - 89

Ethane-    3.9

Propane-  1.25

i-butane-   0.1

n-butane-  0.25

CO2 -        0.5

N2-            5

The S/C ratio is 2 and the outlet pressure of reformer heater is 20 bar.

### #4 PingPong

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Posted 26 December 2017 - 02:22 PM

Your question looks very much like that of ravindra@096

https://www.cheresou...hane-reforming/

You both use natural gas with 89 % methane, an S/C ratio of 2 (which is unrealistically low) and a reformer outlet pressure of 20 bar, except that you use a more realistic reformer outlet temperature of 850 oC.

Normally SMR units operate on an S/C ratio of 2.5 - 4

Too low an S/C ratio can result in coke formation on the reformer catalyst.

### #5 MurtazaHakim

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Posted 27 December 2017 - 03:46 AM

Thank you very much for your useful feedback.

So considering a reformer outlet temperature of 850 degrees Celsius and S/C ratio of 3,how do I proceed further for energy balance for the reformer ?

### #6 breizh

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Posted 27 December 2017 - 04:08 AM

Did you check Google or similar engine ? key words steam reformer heat balance

a lot of stuff there.

Breizh

Edited by breizh, 27 December 2017 - 04:32 AM.

### #7 MurtazaHakim

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Posted 27 December 2017 - 06:22 AM

### #8 PingPong

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Posted 27 December 2017 - 06:36 AM

I looked at the first youtube clip called "Steam Reformer Material and Energy Balance" (by LearnChemE) but that was a big disappointment. Steam reforming of propane will result in a reformer effluent containing CH4 but there is no mention of that. Also there is no mention of the two equilibrium constants that are required to calculate the effluent composition. The person doing the presentation clearly does not know what he is talking about.

Sofar I did not see any other youtube clip that might be useful.

So considering a reformer outlet temperature of 850 degrees Celsius and S/C ratio of 3,how do I proceed further for energy balance for the reformer ?
Let's do this step by step.

First assume an arbitrary amount of natural gas feed, let's say 100 kmol, and based on the S/C ratio calculate the amount of steam that is to be added to that.

### #9 MurtazaHakim

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Posted 27 December 2017 - 06:53 AM

Based on the S/C ratio of 3 and the natural gas flowrate of 100 Kmol/hr the steam flow rate should be around 300 Kmol/hr. what should be done after that ?

### #10 PingPong

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Posted 27 December 2017 - 07:32 AM

No, that is not correct.

You need to take into account the actual composition of the natural gas and the number of C-atoms in each molecule.

### #11 MurtazaHakim

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Posted 28 December 2017 - 03:54 AM

So considering the below composition of natural gas what should be the steam flow rate in Kmol/hr ?

Methane - 89

Ethane-    3.9

Propane-  1.25

i-butane-   0.1

n-butane-  0.25

CO2 -        0.5

N2-            5

### #12 PingPong

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Posted 28 December 2017 - 06:05 AM

You need to look at the number of C-atoms in each molecule.

For 100 kmol of your natural gas and an S/C = 3 the required amount of steam is:

3 * (89*1 + 3.9*2 + .....................) = ............ kmol

Edited by PingPong, 28 December 2017 - 06:06 AM.

### #13 MurtazaHakim

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Posted 28 December 2017 - 07:15 AM

It comes out that we need 316 Kmol/hr considering 0.89,0.039,0.0125,0.0025,0.001 moles of C1, C2, C3, n-C4, i-C4 respectively, for a feed flow rate of 100 Kmol/hr. Now, the doubt is how much conversion of C1,C2,C3,C4 should we consider in the reformer?

### #14 PingPong

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Posted 28 December 2017 - 08:55 AM

It comes out that we need 316 Kmol/hr

I calculate 307 kmol steam, also taking the C of CO2 into account.

Apart from natural gas and steam there will also be a hydrogen feed to the unit to serve the HDS reactor. The hydrogen is usually taken from the unit product line (downstream the PSA unit) and recycled via a compressor to the unit feed line. For a reformer operating on natural gas this hydrogen quantity is usually in the order of 5 mol%. For a heavier feed (LPG or naphtha) the required hydrogen recycle would be much higher.

So in this case the combined feeds to the unit are 100 kmol natural gas + 307 kmol steam + 5 kmol H2.

All atoms in those feeds will end up in the reformer effluent. So there is a atom balance over the reformer for each element.

For example, the O balance: number of O-atoms out is equal to number of O-atoms in:

NH2O + NCO + 2*NCO2 = nH2O + 2*nCO2 =  307 + 2*0.5

NH2O + NCO + 2*NCO2 = 308

in which N refers to the number of moles of subscript molecule in the reformer effluent and n refers to the number of moles of subscript molecule in the reformer feed.

Now write down the C , H and N balances in a similar way. Note that the only hydrocarbon in the effluent will be CH4

Total number of moles in the effluent is: NT = NH2O + NCO + NCO2 + NCH4 + NH2 + NN2

In the effluent each component i has a partial pressure Pi, equal to their mole fraction times the total pressure P.

For example: PCH4 = P * NCH4/NT

Also you need to consider two equilibrium reactions at the reformer outlet:

CH4 + H2O <==> CO + 3 H2

CO + H2O <==> CO2 + H2  (watergas shift reaction)

For each reaction you need to find, or calculate, the equilibrium constant Kp at the reformer outlet conditions. In reality however equilibrium will not be reached as that would require an infinite long reactor. There is a so called Approach To Equilibrium (ATE) to be used.

The first reaction is endothermic and its ATE in a reformer is usually in the order of 10 - 15 oC, lets say 13 oC. That means that in this case you need its Kp at a temperature of 850 - 13 = 837 oC = 1110 K.

The second reaction is exothermic and its ATE in a reformer is usually only 0 - 5 oC, lets say 2 oC. That means that in this case you need its Kp at a temperature of 850 + 2 = 852 oC = 1125 K.

All the above together should give you a set of equations (element balances and equilibrium relations) and unknowns (the N's of all components) that can mathematically be solved. After that you have the exact reformer effluent composition and quantity.

Edited by PingPong, 28 December 2017 - 11:02 AM.

### #15 farid.k

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Posted 30 December 2017 - 02:01 AM

Dear ping pong:

I interested to know further on this.  What book that can i refer to?  I prefer to have own study.  I am interested in the ethane steam cracking process

### #16 MurtazaHakim

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Posted 30 December 2017 - 06:47 AM

Kindly look into the following calculations and suggest if any modifications are required.

### #17 Satyajit

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Posted 30 December 2017 - 09:37 AM

Hi Murtaza,

Very interesting topic. Is it a syngas plant you are talking about. The easiest way to find out answer to your reply is to run aspen plus with the conditions you have mentioned. Based on quick run of the model of SMR, I have following for you:

Basis:

NG feed flow = 200 kmol/hr ( 3,5 T/h), S/C= 2. reformer exit temperature = 850oC, 20 bar pressure

1. Steam flow = 5,3 t/hr

1. NG preheat duty                             = 0,74 Gcal/Hr

2. Mixed Feed coil duty                       =  2,83 Gcal/hr

3. Reforming duty including heat loss =  9,61 Gcal/hr ( 9,43 Gcal/hr Gcal/hr without heat loss).

Kind regards,

Satyajit

### #18 MurtazaHakim

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Posted 30 December 2017 - 10:14 AM

We are actually looking to calculate the conversion of Methane in the reformer using the thermodynamic calculations manually so that we can have the reformer outlet composition and quantity.Simulation is part of our projectwork but we are advised to first caalculate the parameters manually and then verify our calculations by using a process simulator.

### #19 Satyajit

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Posted 30 December 2017 - 04:25 PM

Dear Murtaza,

Thank you for your reply with your intention. Good luck ! It's a great platform to exchange knowledge. If you need anything, please feel free to ask the great people we have in this forum.

Wish you and other members a very Happy New Year 2018 !

Kind regards,

Satyajit

### #20 PingPong

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Posted 31 December 2017 - 04:29 AM

Dear ping pong. I interested to know further on this. What book that can i refer to? I prefer to have own study. I interested to ethane steam cracking process

Steam cracking to produce ethylene (and higher olefins) is a totally different process.

### #21 PingPong

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Posted 31 December 2017 - 05:54 AM

Kindly look into the following calculations and suggest if any modifications are required.

The actual Kp of the reforming reaction at 1110 K is about a factor 10 higher than you calculate.

Problem with the van 't Hoff equation is that it only works when the enthalpy of reaction ΔHr is constant over the range from T1 to T2 but that is not the case here where T1 and T2 are 800 degrees apart.

Therefor to use it properly you also need to take into account the difference in specific heats of the various reactants and products.

Look in your thermodynamics textbook for formulas that incorporate a ΔCp term.

By the way: which thermodynamics textbook are you using?

Edited by PingPong, 31 December 2017 - 05:55 AM.

### #22 MurtazaHakim

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Posted 01 January 2018 - 06:17 AM

We have tried to calculate the value of Kp at 1110 K using the following equations. It appears that the value of Kp at 1110 K comes out to be about 11.5 times higher than the value obtained at 298 K. Kindly look into the attached files and suggest any modifications required.

The book we are referring is CHEMICAL ENGINEERING THERMODYNAMICS by  J M Smith,H C Van Ness,M M Abbott.

Another question is that should we consider all the reactions or only the main reaction for calculating the fractional conversion of CH4 in the reformer ?

### #23 PingPong

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Posted 01 January 2018 - 09:32 AM

You now calculate for the reforming reaction that K = 363 @ 1110 K which is still a somewhat lower than the value of about 390 that I would expect. Your calculation description is not detailed enough for me to spot possible mistakes.

There are nevertheless some general remarks I would like to make:

The values that you now use for ΔH298 and Ko are the same as in your previous calculation, so that suggests that you are using the same ΔHfo and ΔGfo data as in the table in your previous calculation. These values are however different from the values in the appendix of Smith, van Ness & Abbott. Apparently you are using multiple sources.

Problem is that different sources report somewhat different values for ΔHfo and ΔGfo and moreover some sources report them for a standard reference pressure Po of 1 atm and others for Po = 1 bar and others don't bother to mention Po because some authors assume that "everbody knows that Po = ......", and some don't even realize that there is a difference. It's a bit of a mess really.

One should realize that when calculating K with data for Po = 1 atm the resulting K is slightly different from the K calculated from data at Po = 1 bar, for reactions that result in a change in number of molecules. One K has to be used for (partial) pressures in atm and the other for (partial) pressures in bar. Such difference is the case for the reforming reaction but not for the WGS reaction.

And then there is also the problem that different sources report somewhat different values for the coefficients for calculation of Cpo.

Another question is that should we consider all the reactions or only the main reaction for calculating the fractional conversion of CH4 in the reformer ?
You should consider the two equilibrium reactions that I mentioned before in message #14.

The effluent will only contain H2 + H2O + CO + CO2 + CH4 + N2

Edited by PingPong, 01 January 2018 - 09:37 AM.

### #24 MurtazaHakim

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Posted 04 January 2018 - 05:04 AM

We have calculated the values for both the reactions mentioned by you. The data used are taken from CHEMICAL ENGINEERING THERMODYNAMICS by  J M Smith,H C Van Ness,M M Abbott.

Kindly go through it.

The next question is how do we calculate the fractional conversion of components in the reformer ?

### #25 PingPong

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Posted 04 January 2018 - 02:51 PM

I will have a look at your calculations later.

The next question is how do we calculate the fractional conversion of components in the reformer ?

You need to write down the 4 atom-balances for C, H, N and O. I already did the one for O as an example in #14.

There is also the equation for the total number of moles NT in the effluent, see also #14.

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

That gives you a total of 7 equations for 7 unknowns (6 effluent components plus NT), so that can be solved.

It won't be easy by hand but I suppose that Matlab (or equivalent) can do it.