Calculating CV of Fuel/Air Mixture @ 25°C

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Discussion Overview

The discussion revolves around calculating the calorific value (CV) of a fuel/air mixture consisting of butane, propane, and butene at 25°C. Participants explore the implications of excess air in combustion and the resulting flue gases, with calculations for both net and gross CV per cubic meter and per kilomole of the mixture.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the gross CV of the fuel mixture and proposes a method for determining the net CV by accounting for water vapor produced during combustion.
  • Another participant questions the absence of air in the initial CV calculation, suggesting that the exercise specifically asks for the fuel/air mixture.
  • Multiple participants seek clarification on what CV stands for and whether the calculations are being done at a specific pressure.
  • Some participants discuss the heats of combustion for each fuel component, with differing values presented and concerns raised about the accuracy of the data used.
  • A participant suggests a different approach to the problem, recommending that part ii be calculated first to simplify part i.
  • There are ongoing discussions about the conversion between different units of heating values and the molecular weight of butene.
  • Several participants express confusion about the calculations and seek guidance on how to proceed with the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the CV of the fuel/air mixture, with multiple competing views and methods presented. There is also disagreement regarding the values for heats of combustion and the interpretation of the problem statement.

Contextual Notes

Some calculations depend on assumptions about the combustion process and the specific conditions under which the CV is being calculated. There are unresolved discrepancies in the heating values cited by participants, and the discussion reflects varying interpretations of the problem requirements.

Who May Find This Useful

This discussion may be useful for students and professionals interested in combustion engineering, thermodynamics, and chemical engineering, particularly those dealing with fuel properties and energy calculations.

Andy86
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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 25°C, 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 90°C

i) Calculate the net CV per M^3 of the fuel/air mix @ 25°C

ii) Calculate the net CV per kmol of the fuel/air mix @ 25°C

Homework Equations

The Attempt at a Solution


[/B]
i) 1m^3 @ 25°C

Butane = 111.7 MJ m^-3
Propane = 85.8 MJ m^-3
Butene = 105.2 MJ m^-3

CV in M^3

(0.75*111.7)+(0.1*85.8)+(0.15*105.2)

ANS= 108.135 MJ M^-3

----------------------------------------------------------------------------------

ii)

Butane @ 75%
C4H10 + 4.5O2 → 4CO2 + 5H2O
Stoichiometric value 1 MOL 4.5 MOL
Actual Value 0.75 MOL 3.375 MOL

Propane @ 10%
C3H8 + 5O2 → 3CO2 + 4H2O
Stoichiometric value 1 MOL 5 MOL
Actual Value 0.10 MOL 0.5 MOL

Butene @ 15%
C4H8 + 6O2 → 4CO2 + 4H2O
Stoichiometric value 1 MOL 6 MOL
Actual Value 0.15 MOL 0.9 MOL

TOTAL MOLES in 1M^3

Butane = 0.75 MOL
Propane = 0.1 MOL
Butene = 0.15 MOL
O2 = (3.375+0.5+0.9) * 1.1 (excess air) = 5.253 MOL
N2 = 5.253*3.76 = 19.751 MOL
------------------------------------------------------
TOT = 26.004 MOL
------------------------------------------------------

n = PV / RT
=(101300*1) / (8.314*298.15)
=40.86
=40.9 MOL

26/40.9
=0.64M^3

Butane = 0.75 / 0.64 = 1.17MOL
Propane= 0.10 / 0.64 = 0.15MOL
Butene = 0.15 / 0.64 = 0.23MOL
Oxygen = 5.253 / 0.64 = 8.20MOL
Nitrogen = 19.751 / 0.64 = 30.86MOL

V= (n*RT) / P

Butane = 1.17 * (8.314*298.15) / 101300 = 0.028 MJ / M^3
Propane = 0.15 * (8.314*298.15) / 101300 = 0.003 MJ / M^3
Butene = 0.23 * (8.314*298.15) / 101300 = 0.005 MJ / M^3

Net CV

Butane = 0.028 * 111.7 = 3.127 MJ M^3
Propane = 0.003 * 85.8 = 0.257 MJ M^3
Butene = 0.005 * 105.2 = 0.526 MJ M^3
-----------------------------------------------------
TOT = 3.910 MJ M^3
-----------------------------------------------------

ANSWER = 40.9/3.910

ANSWER = 0.0104 MJ / KMOL
 
Last edited:
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What does CV stand for? Is this happening at a specific pressure?
 
Chestermiller said:
What does CV stand for? Is this happening at a specific pressure?

Hi Chester, thanks for the reply, iv had a breakthrough- ill just type it up now! 2 secs
 
Amended my original effort, I now have an answer! How does it look?
 
It asks for the CV of the fuel/air mixture in part i). Where's the air ?
 
BvU said:
It asks for the CV of the fuel/air mixture in part i). Where's the air ?

I haven't got one unfortunately!
 
Anyone?
 
Chestermiller said:
What does CV stand for? Is this happening at a specific pressure?

CV - Calorific Value
 
Andy86 said:
I haven't got one unfortunately!
You haven't got one - what ?

You calculate CV for the fuel mixture, not for the fuel/air mixture as the exercise asks.
 
  • #10
Like I said I am struggling with the question, a bit of guidance would be appreciated.

Thanks
 
  • #11
I would do this differently. I would do part ii first, and then part i is easy.

I would start out with:

0.75 moles of Butane
0.10 moles of Propane
0.15 moles of Butene

Then determine the heat for each. Then do the stoichiometry to find the number of moles of air. Then I would get the heat per mole of mixture.
 
  • #12
Butane = 111.7 MJ m^-3
Propane = 85.8 MJ m^-3
Butene = 105.2 MJ m^-3

Gross CV in M^3 (not accounting for water vapour produced)

(0.75*111.7)+(0.1*85.8)+(0.15*105.2)

ANS= 108.135 MJ M^-3

Net CV

1. Calculate number of H2O moles produced by each gas

Butane @ 75%
C4H10 + 4.5O2 → 4CO2 + 5H2O
0.75*5 = 3.75mol of H2O

Propane @ 10%
C3H8 + 5O2 → 3CO2 + 4H2O
0.10*4 = 0.4mol of H2O

Butene @ 15%
C4H8 + 6O2 → 4CO2 + 4H2O
0.15*4 = 0.6mol of H20

n = 100,000 * (3.75+0.4+0.6) / (8.314*288) = 57.4mol = 0.0574 Kmol

KG = Kmol * Mol Mass
= 0.0574 * 18
= 1.033kg of water are produced

Heat released = (mass * latent heat) + ( mass*specific heat capacity*fall in temp)
= (1.033 * 2258 ) + (1.033*4.20*85)
= 2701KJ
= 2.701MJ

Heat in the water vapour = 1.033 * 1.88 * 85 = 165KJ

TOTAL DIFFERENCE = 2701KJ - 165KJ
=2536 KJ
=2.536 MJ

NET CV = GROSS CV - HEAT LOST DUE TO VAPORISATION
= 108.135MJ - 2.536 MJ
= 105.599 MJ / m^-3
= 0.105 MJ / M^3

Is this any better?
 
  • #13
What are the heats of combustion in kJ/mole of the three species at 25 C and 1 atm? (lower heating value)
 
  • #14
Chestermiller said:
What are the heats of combustion in kJ/mole of the three species at 25 C and 1 atm? (lower heating value)
Butane = 0.75 * - 2874 = -2155.5 kj/mol
Propane = 0.10 * - 2220 = -222 kj/mol
Butene = 0.15 * - 2718 = -407.7 kj/mol

TOTAL = - 2785.2 KJ/MOL
 
  • #15
Andy86 said:
Butane = 0.75 * - 2874 = -2155.5 kj/mol
Propane = 0.10 * - 2220 = -222 kj/mol
Butene = 0.15 * - 2718 = -407.7 kj/mol

TOTAL = - 2785.2 KJ/MOL
The table of lower heating values in Wiki does not agree with your input data: https://en.wikipedia.org/wiki/Heat_of_combustion

Are you sure you are not using higher heating values?
 
  • #16
Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - CANT FIND VALUE = -? kj/mol
 
  • #17
Andy86 said:
Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - CANT FIND VALUE = -? kj/mol
It's right in the table under monoolefins.
 
  • #18
Chestermiller said:
It's right in the table under monoolefins.
Thats MJ/KG, I am looking for KG/MOL
 
  • #19
Are you saying that you don't know how to convert between the two? What is the molecular weight of Butene?
 
  • #20
Butene LLV = MJ/KJ * Mol (w) = 45.334 * 56.106 = 2543

Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - 2543 = -381 kj/mol

?
 
  • #21
Andy86 said:
Butene LLV = MJ/KJ * Mol (w) = 45.334 * 56.106 = 2543

Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - 2543 = -381 kj/mol

?
Looks good. What is the total heating value per mole of flue gas? What are the total number of moles of air required per mole of flue gas? What are the total number of moles of mixture gas per mole of flue gas? What is the total heat value per mole of mixture.
 
  • #22
Chestermiller said:
Looks good. What is the total heating value per mole of flue gas? What are the total number of moles of air required per mole of flue gas? What are the total number of moles of mixture gas per mole of flue gas? What is the total heat value per mole of mixture.

TOTAL HEATING VALUE PER MOLE OF FLUE GAS

Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - 2543 = -381 kj/mol
-------------------------------------------------
TOT = 2580 kj / mol
-------------------------------------------------

TOTAL MOLES in 1M^3

Butane = 0.75 MOL
Propane = 0.1 MOL
Butene = 0.15 MOL
O2 = (3.375+0.5+0.9) * 1.1 (excess air) = 5.253 MOL
N2 = 5.253*3.76 = 19.751 MOL
H2O= 4.75 MOL
Co2 = 3.9 MOL

LOST! I am really in the dark here
 
  • #23
Andy86 said:
TOTAL HEATING VALUE PER MOLE OF FLUE GAS

Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - 2543 = -381 kj/mol
-------------------------------------------------
TOT = 2580 kj / mol
-------------------------------------------------

TOTAL MOLES in 1M^3

The following are not the number of moles in 1 m^3. They are the number of moles of mixture per mole of pure fuel. We are using 1 mole of fuel as the basis of our calculation.
Butane = 0.75 MOL
Propane = 0.1 MOL
Butene = 0.15 MOL
O2 = (3.375+0.5+0.9) * 1.1 (excess air) = 5.253 MOL
N2 = 5.253*3.76 = 19.751 MOL
H2O= 4.75 MOL
Co2 = 3.9 MOL
We are only looking at the fuel/air mixture, which does not contain water and CO2. So how many moles of air-fuel mixture do we have (per mole of pure fuel)?

Chetr
 
  • #24
Chestermiller said:
The following are not the number of moles in 1 m^3. They are the number of moles of mixture per mole of pure fuel. We are using 1 mole of fuel as the basis of our calculation.

We are only looking at the fuel/air mixture, which does not contain water and CO2. So how many moles of air-fuel mixture do we have (per mole of pure fuel)?

Chetr

Sorry Chester, you asked for "flue gas". Here is my calculation of moles per moles for air fuel mix;

TOTAL MOLES in 1MOL or pure fuel

Butane = 0.75 MOL
Propane = 0.1 MOL
Butene = 0.15 MOL
O2 = (3.375+0.5+0.9) * 1.1 (excess air) = 5.2525 MOL
N2 = 5.253*3.76 = 19.7494 MOL
------------------------------------------------------
TOT = 26.002 MOL
------------------------------------------------------
 
  • #25
NET CV BY MASS

TOTAL HEATING VALUE PER MOLE OF FLUE GAS

Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - 2543 = -381 kj/mol
-------------------------------------------------
TOT = 2580 kj / mol
-------------------------------------------------
TOTAL MOLES = 26.002
-------------------------------------------------

ANSWER

2580 / 26.002 = 99.2 KJ / MOL
 
  • #26
Good. Now all you have to do is to find the volume of 26.002 moles at 25 C and 1 atm to get the answer to part i in kJ/m^3
 
  • #27
Chestermiller said:
Good. Now all you have to do is to find the volume of 26.002 moles at 25 C and 1 atm to get the answer to part i in kJ/m^3

My first equation for butane doesn't balance Chester :-( ill repost my latest effort!
 
  • #28
NET CV BY MASS

TOTAL HEATING VALUE PER MOLE OF FLUE GAS
Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - 2543 = -381 kj/mol
-------------------------------------------------
TOT = 2580 kj / mol
-------------------------------------------------
TOTAL MOLES = 33.8 (including O2 and N2)
-------------------------------------------------
ANSWER 2580 / 33.8 = 76.3 KJ / MOL

PV = N * R * T

Rearranged to make V the subject

V = N * R * T / 100,000
= 33.8 * 8.314 * 298 / 100,000
= 0.837 m^3

I don't know what to do from here
 
  • #29
LOST!
 
  • #30
You have the number of kJ per mole and, if you calculate the number of m^3 per mole, you can divide to get the number of kJ per m^3.
 

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