Composition of flue gases by volume on a wet basis and dry basis

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Discussion Overview

The discussion revolves around calculating the composition of flue gases by volume on both a wet and dry basis, specifically focusing on the amounts of carbon dioxide (CO2), water (H2O), oxygen (O2), and nitrogen (N2) produced from the combustion of butane, propane, and butene. Participants explore the relationships between these gases and the calculations needed to derive the nitrogen content from the given oxygen values.

Discussion Character

  • Technical explanation
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the amounts of CO2, H2O, and O2 produced from the combustion of butane, propane, and butene based on their respective molar amounts.
  • Another participant notes the need to find the nitrogen value and references a previously seen value of 26.07mol, expressing confusion about how this was derived.
  • A later reply explains that since air consists of 21% O2 and 79% N2, the nitrogen can be calculated by dividing the oxygen amount (6.93mol) by 21% to find the total air provided, then multiplying by 79% to find the nitrogen content, arriving at 26.07mol.
  • Another participant reiterates the calculation method for nitrogen, emphasizing the same approach and welcoming the original poster to the forum.
  • One participant reminds others that they should not do the student's work for them but can provide hints or ask probing questions instead.

Areas of Agreement / Disagreement

Participants generally agree on the method to calculate nitrogen from the oxygen value, but there is no consensus on the appropriateness of providing direct solutions versus guiding the student through the problem.

Contextual Notes

There is a lack of clarity on the assumptions made regarding the combustion process and the specific conditions under which the calculations are performed. The discussion does not resolve the confusion expressed by the original poster regarding the nitrogen calculation.

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Homework Statement
A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºC, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90ºC

(e) Determine the composition of the flue gases by volume (assuming the inlet air is dry) :
(i) on a wet basis
(ii) on a dry basis.

I'm aware that there is a similar question already posted from 6+ years ago. I have read that particular forum and seem to still struggle.
Relevant Equations
Balanced reactions
Butane C4H10 + 6.5O2 = 4CO2 + 5H20
Propane C3H8 + 5O2 = 3C02 + 4H2O
Butene C4H8 + 6O2 = 4CO2 + 4H20
Butane
0.75mol of C4H10

0.75 x C4 = 3mol of CO2
0.75 x H5 = 3.75mol of H2O
0.75 x 6.5O2 = 4.875mol of O2

Propane
0.10mol of C3H8

0.10 x C3 = 0.3mol of CO2
0.10 x H4 = 0.4mol of H2O
0.10 x 5O2 = 0.5mol of O2

Butene
0.15mol of C4H8

0.15 x C4 = 0.6mol of CO2
0.15 x H4 = 0.6mol of H2O
0.15 x 6O2 = 0.19mol of O2

We know, Oxygen = 6.3mol + 10% excess air = 1.1 x 6.3mol = 6.93mol.

I know I need to find the value of Nitrogen. I've seen on other threads, that they have got a value of 26.07mol...but I'm struggling to understand how they've achieved this.

This is as far as I can manage to get...Please help me with some advise!!
 
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Don't know if you still need this but...

Air consists of 21% O2 and 79% N2, so 6.93mol is 21% of the total air provided.

To get the N2, divide 6.93 by 21 to get 1% then multiply by 79, which gives 26.07mol :)
 
rc404 said:
Don't know if you still need this but...

Air consists of 21% O2 and 79% N2, so 6.93mol is 21% of the total air provided.

To get the N2, divide 6.93 by 21 to get 1% then multiply by 79, which gives 26.07mol :)
Welcome to PF.

We are not allowed to do the student's work for them here at PF. Instead, ask probing questions, give hints, find mistakes, etc. The student must do the bulk of the work on their schoolwork questions.

Having said that, this thread is several months old so presumably the student no longer has this problem to solve. So in cases like this, it is okay to post a solution or an alternate solution versus the one the student came up with.
 
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