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.

  • #31
Chestermiller said:
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.
Rearranged to make V the subject

V = N * R * T / 100,000
= 34 * 8.314 * 298 / 100,000
= 0.84 m^3

34 moles = 0.84m^3

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

Volume of 1 mol = 0.84m^3 / 34 = 0.025 m^3

1 / 0.025 m^3 = 40
2580kj / mol * 40 = 103200KJ / M^-3
NET CV BY VOL = 103.2MJ / M^-3
 
Last edited:
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  • #32
Andy86 said:
Rearranged to make V the subject

V = N * R * T / 100,000
= 34 * 8.314 * 298 / 100,000
= 0.84 m^3

34 moles = 0.84m^3

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

Volume of 1 mol = 0.84m^3 / 34 = 0.025 m^3

1 / 0.025 m^3 = 40
2580kj / mol * 40 = 103200KJ / M^-3
NET CV BY VOL = 103.2MJ / M^-3
This is correct. But, you didn't need to calculate the volume of 34 moles to get your answer. All you needed to do was to use the ideal gas law to find the volume of 1 mole: V = (8.314)(298)/100000
 
  • #33
Chestermiller said:
This is correct. But, you didn't need to calculate the volume of 34 moles to get your answer. All you needed to do was to use the ideal gas law to find the volume of 1 mole: V = (8.314)(298)/100000

EXCELLENT! how does this net CV by mass look...

NET CV BY MASS

n = PV / RT
= 101300 * 1 / 8.314 * 298
= 40.0 mol
= 0.0408 kmol

CV = CV in MJ M^-3 / kmol M^3
= 103.2 MJ M^-3 / 0.0408 kmol
= 2529.4 MJ kmol^-1 @ 25°C

??

 
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  • #34
Andy86 said:
EXCELLENT! how does this net CV by mass look...

NET CV BY MASS

n = PV / RT
= 101300 * 1 / 8.314 * 298
= 40.0 mol
= 0.0408 kmol

CV = CV in MJ M^-3 / kmol M^3
= 103.2 MJ M^-3 / 0.0408 kmol
= 2529.4 MJ kmol^-1 @ 25°C

??

Any thoughts?
 
  • #35
Andy86 said:
Any thoughts?
In post #25, you got 99.2 kJ/mole of fuel mixture. This looked correct to me.
 
  • #36
That's the incorrect number of total mols in the mix, total ,mols is 33.8 > 34 Mols
 
  • #37
OK. So 2580/33.8=76.3 kJ/mole

The volume per mole is 0.0245 m^3/mole

So, 76.3/0.0245 = 3115 kJ/m^3 of fuel mixture = 3.12 MJ/m^3 of fuel mixture
 
  • #38
OK. So 2580/33.8=76.3 kJ/mole

The volume per mole is 0.0245 m^3/mole

So, 76.3/0.0245 = 3115 kJ/m^3 of fuel mixture = 3.12 MJ/m^3 of fuel mixture
 
  • #39
Chestermiller said:
OK. So 2580/33.8=76.3 kJ/mole

The volume per mole is 0.0245 m^3/mole

So, 76.3/0.0245 = 3115 kJ/m^3 of fuel mixture = 3.12 MJ/m^3 of fuel mixture

It doesn't look right that the CV is 3.12 MJ / m^3 as the gross CV is 108.1 MJ M^3, my calculated NET CV of 103.2 MJ M^3 looks more realistic?
 
  • #40
Andy86 said:
It doesn't look right that the CV is 3.12 MJ / m^3 as the gross CV is 108.1 MJ M^3, my calculated NET CV of 103.2 MJ M^3 looks more realistic?
The stream is diluted by a factor of 33,9
 
  • #41
Chestermiller said:
The stream is diluted by a factor of 33,9

Meaning?
 
  • #42
Andy86 said:
Meaning?
You have 1 mole of actual fuel and 32,9 mioles of nitrogen and oxygen that dilute the heating value. Just take your 100MJ and divide by 33.9. What do you get?
 
  • #43
NET CV BY VOL = 103.2MJ / M^-3

103.2 / 33.8 = 3.05MJ / M^-3
 
  • #44
Andy86 said:
NET CV BY VOL = 103.2MJ / M^-3

103.2 / 33.8 = 3.05MJ / M^-3
Pretty close, huh?
 
  • #45
VERY! :-)

So.....

NET CV BY MASS

n = PV / RT = 100000 * 1 / 8.314 * 298 = 40.3 MOL = 0.0403 KMOL

CV = MK M^-3 / KMOL
= 3.05 MJ M^-3 / 0.0403 KMOL
= 75.6 MJ KMOL^-1

?
 
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  • #46
Andy86 said:
VERY! :-)

So.....

NET CV BY MASS

n = PV / RT = 100000 * 1 / 8.314 * 298 = 40.3 MOL = 0.0403 KMOL

CV = MK M^-3 / KMOL
= 3.05 MJ M^-3 / 0.0403 KMOL
= 75.6 MJ KMOL^-1

?
You're going in a circle. You already had the CV per mole.
 
  • #47
Andy86 said:
VERY! :-)

So.....

NET CV BY MASS

n = PV / RT = 100000 * 1 / 8.314 * 298 = 40.3 MOL = 0.0403 KMOL

CV = MK M^-3 / KMOL
= 3.05 MJ M^-3 / 0.0403 KMOL
= 75.6 MJ KMOL^-1

?
You're going in a circle. You already had the CV per mole.
 
  • #48
Ok Great, thanks for all your help Chester I really appreciate it.

Andy
 
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