# Homework Help: Net calorific value

1. Oct 11, 2016

### GeorgeP1

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

Net calorific value (MJ m–3) at 25°C of:
Butane (C4H10) = 111.7 MJ m^3
Butene (C4H8) = 105.2 MJ m^3
Propane (C3H8) = 85.8 MJ m^3

I need help on how to calculate: -

the net calorific value (CV) per m^3 of the fuel/air mix at 25°C ?

the net calorific value (CV) per kmol of the fuel/air mix at 25°C ?

2. Relevant equations

3. The attempt at a solution

Net cv per m3 = 58.65 mJ / m^3

Net cv per kmol = 1466.25 mJ / kmol

These attempts are incorrect but I don't know where I have gone wrong. any help is much appreciated.

Last edited: Oct 11, 2016
2. Oct 11, 2016

### Staff: Mentor

Presumably, the flue gas is at 1 bar. How many moles of flue gas are there in 1 m^3 of the gas at 25 C? How many moles of butane, propane, and butene are there?

3. Oct 12, 2016

### GeorgeP1

Hi chestermiller

yes flue gas is 1 bar.

I make it a total of 35.35 moles with 10% excess air.

Last edited: Oct 12, 2016
4. Oct 12, 2016

### Staff: Mentor

I meant to ask how many moles each of butane, propane, butene, oxygen, and nitrogen are in 1 cubic meter of the air/fuel mixture.

Last edited: Oct 12, 2016
5. Oct 13, 2016

### GeorgeP1

I make it :
butane - 0.75mol
propane - 0.1mol
butene - 0.15mol
oxygen - 6.3mol
10% excess air = 0.63 + 6.3 = 6.93mol
nitrogen = 6.93 x 3.76 = 26.07mol

is this correct?

Last edited: Oct 13, 2016
6. Oct 13, 2016

### Staff: Mentor

That's the correct number of moles in the mixture, if you start with 1 mole of pure fuel. What is the total number of moles here, and, using the ideal gas law, what volume does this total number of moles of gas occupy?

7. Oct 13, 2016

### GeorgeP1

total number of moles is 34 moles.

n = pV / RT
n = no. of moles
p = pressure
v = volume
R = constant (8.314)
T = temperature

n = (101300 x 1) / (8.314 x 298)

n = 40.89 moles

8. Oct 13, 2016

### Staff: Mentor

So, if 1 m^3 of the gas contains 40.9 moles, and the total volume you calculated for the case of 1 mole of pure fuel is 34 m^3, how many moles of each species in the fuel mixture is present in 1 m^3 of mixture?

9. Oct 13, 2016

### GeorgeP1

V = (n/p)RT
The volume in the ideal gas equation for 34 moles of gas is 0.83 M^3

Last edited: Oct 13, 2016
10. Oct 13, 2016

### Staff: Mentor

So, I ask again, how many moles of each species in the fuel mixture is present in 1 m^3 of the mixture?

11. Oct 13, 2016

### GeorgeP1

Is it 1.203 moles

12. Oct 13, 2016

### Staff: Mentor

Yes, but I want each species individually, including the oxygen and nitrogen.

13. Oct 13, 2016

### GeorgeP1

butane - 0.902 mol
propane - 0.12 mol
butene - 0.181 mol
O2 - 8.33 mol
N2 - 31.36 mol

14. Oct 13, 2016

### Staff: Mentor

This agrees with what I got.

Now, I assume that those net caloric values that they give you is for burning 1 m^3 of each of the pure three species (i.e., for burning the number of moles of each species that occupy 1 m^3 at 25 C and 1 bar). Is that correct?

15. Oct 13, 2016

### GeorgeP1

great...

yes the net cv is for 1 m^3 of each gas.

16. Oct 14, 2016

### Staff: Mentor

If 40.89 moles of any gas would occupy 1 m^3 under these conditions, how many m^3 would 0.902 moles of pure butane occupy? What would be net caloric value of burning this amount of butane?

17. Oct 14, 2016

### GeorgeP1

I make it 0.02206 M^3 for butane

does that make the net CV for butane 2.46 Mj / M^3

Last edited: Oct 14, 2016
18. Oct 14, 2016

### Staff: Mentor

OK. What about the other two species, and what about their combined net CV?

19. Oct 14, 2016

### GeorgeP1

propane = 0.003 m^3
= 0.2574 MJ /M^3

Butene = 0.0044 M^3
= 0.463 Mj / M^3

Total = 3.1804 Mj /M^3

20. Oct 14, 2016

### Staff: Mentor

Well, If we did the arithmetic correctly, I think we have our answer.

21. Oct 15, 2016

### GeorgeP1

Excellent. Thanks chestermiller
My previous attempt was a country mile out then.
what about the net calorific value (CV) per kmol of the fuel/air mix at 25°C ?
Any ideas?

22. Oct 15, 2016

### Staff: Mentor

That's the total you already determined.

23. Oct 15, 2016

### GeorgeP1

so per kmol is 40.89 mols / 3.1804 = 0.013 Mj / Kmol
does this make sense??

24. Oct 15, 2016

:-)