Calculate Net Calorific Value at 25°C

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Discussion Overview

The discussion revolves around calculating the net calorific value (CV) of a fuel gas mixture consisting of butane, propane, and butene at 25°C, specifically focusing on the CV per cubic meter and per kilomole of the mixture. The context includes combustion processes and the use of flue gases for steam generation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants discuss the composition of the fuel gas mixture and the need to calculate the net calorific value per m³ and per kmol.
  • One participant provides initial calculations for the net CV per m³ and per kmol, which they later identify as incorrect.
  • There is a focus on determining the number of moles of each component in the mixture, including butane, propane, butene, oxygen, and nitrogen.
  • Several participants calculate the total number of moles in the mixture and explore the implications of using the ideal gas law.
  • Participants arrive at specific mole fractions for each gas in the mixture and discuss the net calorific values for burning these gases.
  • Further calculations are made to determine the volume occupied by each gas and their respective contributions to the total net CV.
  • There is a query regarding the net CV per kmol of the fuel/air mix, with participants attempting to derive this value based on previous calculations.

Areas of Agreement / Disagreement

Participants generally agree on the calculations of moles and the approach to determining the net calorific values, but there is no consensus on the correctness of the initial attempts at calculating the net CV. The discussion remains unresolved regarding the final values for the net CV per kmol.

Contextual Notes

Some calculations depend on the assumptions made about the ideal gas behavior and the specific conditions of temperature and pressure. There are unresolved steps in the calculations that could affect the final results.

GeorgeP1
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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

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 ?

Homework Equations


[/B]

The Attempt at a Solution


[/B]
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.

thanks in advance
 
Last edited:
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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?
 
Hi chestermiller

yes flue gas is 1 bar.

I make it a total of 35.35 moles with 10% excess air.
 
Last edited:
GeorgeP1 said:
Hi chestermiller

yes flue gas is 1 bar.

I make it a total of 35.35 moles with 10% excess air.
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:
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:
GeorgeP1 said:
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?
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?
 
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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
 
GeorgeP1 said:
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
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?
 
V = (n/p)RT
The volume in the ideal gas equation for 34 moles of gas is 0.83 M^3
 
Last edited:
  • #10
GeorgeP1 said:
V = (n/p)RT
The volume in the ideal gas equation for 34 moles of gas is 0.83 M^3
So, I ask again, how many moles of each species in the fuel mixture is present in 1 m^3 of the mixture?
 
  • #11
Is it 1.203 moles
 
  • #12
GeorgeP1 said:
Is it 1.203 moles
Yes, but I want each species individually, including the oxygen and nitrogen.
 
  • #13
butane - 0.902 mol
propane - 0.12 mol
butene - 0.181 mol
O2 - 8.33 mol
N2 - 31.36 mol
 
  • #14
GeorgeP1 said:
butane - 0.902 mol
propane - 0.12 mol
butene - 0.181 mol
O2 - 8.33 mol
N2 - 31.36 mol
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
great...

yes the net cv is for 1 m^3 of each gas.
 
  • #16
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
I make it 0.02206 M^3 for butane

does that make the net CV for butane 2.46 Mj / M^3
 
Last edited:
  • #18
GeorgeP1 said:
I make it 0.02206 M^3 for butane

does that make the net CV for butane 2.46 Mj / M^3
OK. What about the other two species, and what about their combined net CV?
 
  • #19
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
GeorgeP1 said:
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
Well, If we did the arithmetic correctly, I think we have our answer.
 
  • #21
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
GeorgeP1 said:
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?
That's the total you already determined.
 
  • #23
so per kmol is 40.89 mols / 3.1804 = 0.013 Mj / Kmol
does this make sense??
 
  • #24
:-)
 

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