Calculating CV of Fuel/Air Mixture @ 25°C

  • Thread starter Thread starter Andy86
  • Start date Start date
  • Tags Tags
    Cv Mixture
Click For Summary
SUMMARY

The discussion focuses on calculating the net calorific value (CV) of a fuel gas mixture consisting of 75% butane (C4H10), 10% propane (C3H8), and 15% butene (C4H8) at 25°C with 10% excess air. The net CV per cubic meter of the fuel/air mix is determined to be 103.2 MJ/m³, while the net CV per kilomole is calculated to be 0.0104 MJ/kmol. Key calculations involve the stoichiometry of combustion and the heat released from the combustion of each component in the mixture.

PREREQUISITES
  • Understanding of combustion chemistry and stoichiometry
  • Familiarity with calorific value (CV) calculations
  • Knowledge of the ideal gas law (PV = nRT)
  • Basic thermodynamics principles related to heat transfer
NEXT STEPS
  • Research the calculation of calorific values for different fuel mixtures
  • Learn about the effects of excess air on combustion efficiency
  • Explore the stoichiometric calculations for combustion reactions
  • Study the thermodynamic properties of butane, propane, and butene
USEFUL FOR

Engineers, chemists, and energy analysts involved in fuel combustion analysis, as well as students studying thermodynamics and combustion processes.

  • #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:
Physics news on Phys.org
  • #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

??

 
Last edited:
  • #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

?
 
Last edited:
  • #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
 
  • Like
Likes   Reactions: Chestermiller

Similar threads

Calculate Net Calorific Value at 25°C
  • · Replies 23 ·
Replies
23
Views
7K
Net calorific value (CV) per kmol of an fuel/air mix
  • · Replies 5 ·
Replies
5
Views
4K
How Does Fuel Composition Affect Combustion Efficiency and Steam Generation?
  • · Replies 9 ·
Replies
9
Views
8K
Solving Combustion Q: Find Steam Raised Per Hour
  • · Replies 4 ·
Replies
4
Views
3K
Molar Gibbs Energy: 25{\circ} C Expansion Calculation
  • · Replies 1 ·
Replies
1
Views
2K