6 replies to this topic

[alfirin]

Dear all, I am trying to calculate the wt% hydrogen in coke, coke yield and catalyst to oil ratio using some formulas and basic calculations. This is a theoretical problem. I have attached a pdf file with the problem and my solution. I seem to get a hydrogen content of around 13.6%, but I'm not quite sure if I am doing the right calculations for the catalyst to oil ratio. The first question involves partial combustion. The second one assumes full combustion and I am doing some new calculations. Again, I am not sure if my methodology is correct. I would be really glad if someone could have a quick look and comment on my work. Regards

Edited by alfirin, 20 May 2011 - 08:35 AM.

[pavanayi]

Alfirin,
There are errors in this calculation as well. I expect a masters student to know the difference between FLUE GAS and FUEL GAS. I am sure this calculation pertains to flue gas, but you have used 'fuel gas' throughout instead!!

1. If in the flue gas,
amount of O₂ = 6.76 moles
amount of CO₂ = 40.58 moles
amount of CO = 81.16 moles,

Then the total moles of O₂ that contributed to the above results should be
6.76+40.58+ (81.16/2) = 87.92 moles

That will affect calculations that follow to find hydrogen%.


2. The catalyst to oil ratio:

Amount of carbon = 1460.9 TPH

This is the carbon that has been released from the catalysts through combustion. If so,
amount of catalyst = 1460.9/(0.016-0.0006) = 94865 TPH
Catalyst to oil ratio = 94865/37500 = 2.53

[Art Montemayor]

Alfirin:

Like pavanayi, I also have a lot of trouble following these calculations – plus a headache.

A combustion calculation is really simple and should be easy to develop in a logical, step-wise, explained algorithm. The submitted work doesn’t do that. It leaves the reader to struggle and figure out what is the planned and logical methodology being used.

Besides the misspellings and typos – which make it difficult to follow what is presented, there is no logical explanation for the algorithm being employed. This makes it very difficult to justify following calculations that may -or not – be applicable. I cite just some examples:

• Dry air has a known molecular weight of 28.97 and all engineers use it as such. But this is dry air – without humidity, and with argon and CO₂. My point here is that the calculation should make known that it is an ESTIMATE based on estimated values and not real, actual values. Nowhere are these bases even mentioned – even though the “ideal” air molecular weight is calculated. No mention is made as to what it really represents.
• Are we checking numbers written in the European style (commas instead of decimal points?) – or are the numbers expressed in USA style (decimal points instead of commas?). I am confused because I see the molecular weight of air expressed as 28.84 (gr/grmol) and not as 28,84. However, that number is divided into 20000 and the answer is expressed as 693.481 Mmoles/hr (which I have to assume means 693,481,276 grmols/hr in the USA style – which is the correct math). If you are going to use European, then stick to it; if you are going to use USA, then stick to that. Mixing both only makes for confusion and mistakes.
• If you are going to use symbology, then IDENTIFY IT. What do you mean by n_FG? You make the mistake of calling it Fuel Gas. Failure to identify what you mean by symbols confuses your reader – and you as well!! The proof of this is your mistakes.
• How is it that you can state “H₂ + 1/2O₂ = H₂O which means one mole of Oxygen will give one mole of water” and then later state “C + O₂ = CO₂ the mols of C, CO₂, and O₂ will be equal”????
• I don’t agree with this basic Chemistry interpretation. In my day ½ mol of O₂ combined with one mol of H₂ to make one mol of water.

There are more lapses and errors that would be far easier and more convenient to both yourself and your checker if you did the engineering calculations on an engineering format – like an EXCEL SPREADSHEET – and protected it without a password to allow the checker to make corrections, comments, add sketches and side calculations, etc. to your work. That would be more constructive, faster, easier, more convenient, more accurate, and more efficient. I have been preaching this for years and years on our Forums for all students to heed and follow – but then what do I know?

I am sorry that I can’t make any comments directly on your .pdf document and I hope that the above comments are taken positively and you can improve on your engineering work output. I regret that I would have to reject the submitted calculations if I were your supervisor and I hope that this serves as a learning step. I am glad that you are striving to have your calculations work checked and I encourage you to continue this practice. This is a good and excellent practice that is a requisite step in industry. However, be aware that you must present your work in a polished, professional manner if you expect professionals to give you valued comments and checks.

[alfirin]

Thank you both for your comments. I can clearly see my mistakes now. Yes, I meant flue gas, it was a bad mistyping! I have done some recalculations. My apologies if I didn't prepare an excel spreadsheet, but this was an exam question and I tried to solve it in the same way that I will have to if I'm asked to do it this year.

I now get a 13.6% Hydrogen in coke. Could I ask, how can we interpret this high hydrogen content in real engineering? I mean how does this affect the performance of the FCC unit and is it possible to be reduced through operational or design modifications? I know that an acceptable content in the industry is about 5-6%.

I am also having trouble doing this problem. Where can i download the pdf you have attached to the post? I cant find a link of some sort directing me towards it.
If anyone is willing to help ourt. I will state the problem again:

For the following conditions calculate [a] wt% hydrogen in coke, [b] coke yield, [c] catalyst to oil ratio

carbon on spent catalyst: 1.50 wt%
carbon on regenerated catalyst: 0.80 wt%
Air from blower: 155,000 lb/hr
hydrocarbon feed to reactor 295,000 lb/hr

Fuel gas analysis, vol%
CO: 12%
CO2: 6.0%
O2: 0.7%
N2: 81.3%

Edited by rxh5015, 27 October 2011 - 10:20 PM.

[pavanayi]

rxh5015,
Start a new thread, and attach the calculations you have done till now(preferably in an excel file).
State clearly what part of the calculation you are not able to understand. Then we will be able to comment/direct you.

[Profe]

For your problem from: Handbook of Chemical Engineering Calculations, third Ed, Nicolas Chopey

Select a basis (100 kg· mol dry flue gas), a tie component (N₂ passes through the system unreacted), and write out the relevant equations involved.
Dry flue gas is the basis, containing CO₂ and CO, the relevant reactions and the quantities per 100 mol flue gas are:

C + O₂ = CO₂

12 kg· mol 12 kg· mol =12 kg ·mol

C + 1/2O₂ = CO

6 kg· mol 3 kg· mol = 6 kg ·mol

The N₂ in the entering air must equal the moles in the flue gas, from input data = 81.3 kg· mol.
The air has 79.02% N₂ and 20.98% O₂. Then O₂ contents will be (20.98/79.02)(81.3) = 21.59 kg · mol.

Calculate the O₂ in the flue gas as H₂O. The kg· moles of O₂ in the flue gas should be the same as in the entering air, that is = 21.59 kg · mol of O₂. Therefore, the oxygen not accounted in the analysis of the flue gas is the oxygen converted to water.
The flue gas analysis yields 12 kg· mol of O₂ as CO2, 3 kg· mol of O₂ as CO, and 0.7 kg· mol of O₂ as unreacted O₂.
The O₂ for the H₂O leaving the system in the flue gas will be (21.59 − 12 − 3 − 0.7) = 5.89 kg· mol of O₂
Since 2 mol of H₂O is produced per mole of O₂ reacted, the amount of H₂O in the wet flue gas will be 2(5.89) = 11.78 kg· mol. This amount of water contains 11.78 kg· mol H₂ or (11.78)[2.016 kg/(kg · mol)] = 23.75 kg H₂.
Then, the carbon associated with this H₂ is the 12 kg· mol that reacted to CO2 plus the 6 mol that reacted to CO or (12 + 6)[12.011 kg/(kg· mol)] = 216.20 kg carbon. Therefore, the weight percent of H₂ in the coke will be (100)(23.75)/(23.75 + 216.0) = 9.91 %.

Source: Handbook of Chemical Engineering Calculations, third Ed, Nicolas Chopey, p 2.5 Stoichometry, 2.4 USE OF A TIE ELEMENT IN MATERIAL-BALANCE CALCULATIONS