Calculate Net Calorific Value at 25°C

[GeorgeP1]

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

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The Attempt at a Solution

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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

 

[Chestermiller]

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?

 

[GeorgeP1]

Hi chestermiller

yes flue gas is 1 bar.

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

 

[Chestermiller]

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.

 

[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?

 

[Chestermiller]

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?

 

[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

 

[Chestermiller]

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?

 

[GeorgeP1]

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

 

[Chestermiller]

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?

 

[GeorgeP1]

Is it 1.203 moles

 

[Chestermiller]

Is it 1.203 moles

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

 

[GeorgeP1]

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

 

[Chestermiller]

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?

 

[GeorgeP1]

great...

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

 

[Chestermiller]

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?

 

[GeorgeP1]

I make it 0.02206 M^3 for butane

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

 

[Chestermiller]

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?

 

[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

 

[Chestermiller]

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.

 

[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?

 

[Chestermiller]

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.

 

[GeorgeP1]

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

 

[GeorgeP1]

:-)