# Homework Help: CV calculation

1. Feb 11, 2017

### Andy86

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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: Feb 11, 2017
2. Feb 11, 2017

### Staff: Mentor

What does CV stand for? Is this happening at a specific pressure?

3. Feb 11, 2017

### Andy86

Hi Chester, thanks for the reply, iv had a breakthrough- ill just type it up now! 2 secs

4. Feb 11, 2017

### Andy86

Amended my original effort, I now have an answer! How does it look?

5. Feb 11, 2017

### BvU

It asks for the CV of the fuel/air mixture in part i). Where's the air ?

6. Feb 11, 2017

### Andy86

I havent got one unfortunately!

7. Feb 11, 2017

### Andy86

Anyone????

8. Feb 12, 2017

### Andy86

CV - Calorific Value

9. Feb 12, 2017

### BvU

You haven't got one - what ?

You calculate CV for the fuel mixture, not for the fuel/air mixture as the exercise asks.

10. Feb 12, 2017

### Andy86

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

Thanks

11. Feb 12, 2017

### Staff: Mentor

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. Feb 13, 2017

### Andy86

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. Feb 13, 2017

### Staff: Mentor

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

14. Feb 13, 2017

### Andy86

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. Feb 13, 2017

### Staff: Mentor

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. Feb 13, 2017

### Andy86

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

17. Feb 13, 2017

### Staff: Mentor

It's right in the table under monoolefins.

18. Feb 13, 2017

### Andy86

Thats MJ/KG, im looking for KG/MOL

19. Feb 13, 2017

### Staff: Mentor

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

20. Feb 13, 2017

### Andy86

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

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