Calculating CV of Fuel/Air Mixture @ 25°C

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The discussion focuses on calculating the net calorific value (CV) of a fuel gas mixture consisting of butane, propane, and butene, with excess air at 25°C. The net CV per cubic meter is calculated to be approximately 105.599 MJ/m³, while the net CV per kilomole is determined to be around 0.0104 MJ/kmol. Participants clarify the distinction between calculating the CV for the fuel mixture versus the fuel/air mixture, emphasizing the importance of including air in the calculations. There are also discussions about the heats of combustion for each gas and the need to balance chemical equations correctly. Overall, the calculations and methodologies for determining the CV are refined through collaborative input.
  • #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
 
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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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