# Calculating CV of Fuel/Air Mixture @ 25°C

• Andy86
In summary, the 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.
Andy86

## 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

## 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 = 0.0104 MJ / KMOL

Last edited:
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.

Like I said I am struggling with the question, a bit of guidance would be appreciated.

Thanks

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.

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?

What are the heats of combustion in kJ/mole of the three species at 25 C and 1 atm? (lower heating value)

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

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?

Butane = 0.75 * - 2660 = -1995 kj/mol
Propane = 0.10 * - 2046 = -204 kj/mol
Butene = 0.15 * - CANT FIND VALUE = -? kj/mol

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.

Chestermiller said:
It's right in the table under monoolefins.
Thats MJ/KG, I am looking for KG/MOL

Are you saying that you don't know how to convert between the two? What is the molecular weight of Butene?

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

?

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.

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

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

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

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

2580 / 26.002 = 99.2 KJ / MOL

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

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!

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

LOST!

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.

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:
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

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:
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?

Andy86 said:
Any thoughts?
In post #25, you got 99.2 kJ/mole of fuel mixture. This looked correct to me.

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