Max Flame Temp of C4H10 + 4.5O2 -> 4Co2 + 5H2O

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SUMMARY

The discussion centers on calculating the maximum flame temperature for a fuel gas mixture consisting of 75% butane (C4H10), 10% propane (C3H8), and 15% butene (C4H8) when combusted with 10% excess air. The calculated heat content of the fuel is 2598.6 kJ/mol, but discrepancies arise when comparing this value to enthalpy tables, which only extend to 2200°C. Participants discuss the importance of accounting for the enthalpy of vaporization of water and the dissociation of combustion products at high temperatures, with a consensus that the maximum flame temperature is likely around 2000°C.

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C4H10 + 4.5O2 -> 4Co2 + 5H2O1. Homework Statement

A fuel gas consists of 75% butane (C4 H10 ), 10% propane (C3 H8 ) and butene (C4 H8 ) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºwhere 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 90ºC.

Determine the maximum flame temperature.

Homework Equations


Actual oxygen supplied = theoretical x excess air
Heat content = enthalpy x no of moles

The Attempt at a Solution



lower valvue for propane 2046 KJ/mol, butane 2660 KJ/mol

(.75*2660)+(.1*2046)+(.15*2660)
1995+204.6+399=2598.6KJ = 2598.6KJ/mol fuel

C4H10 + 4.5O2 -> 4Co2 + 5H2O
1mol 4.5mol 4 mol 5 mol
.75 mol 3.375mol 3mol 3.75mol

C3H8 + 5O2 -> 3Co2 + 4H2O
1mol 5mol 3 mol 4 mol
.1 mol .5mol .3mol .4mol

C4H8 + 6O2 -> 4Co2 + 4H2O
1mol 6mol 4 mol 4 mol
.15 mol .9mol .6mol .6mol

4.775 O2 Req
actual O2 = theo x excess air
4.775 x 1.1 = 5.2525 mol
N2 = 5.2525 x 3.76=19.749 moles in flue gas (3.76 is 79%/21% conversion factor from O2 to air to N2 )

the problem is the next bit once i work out the heat content the enthalpy table goes up to 2200 Deg C however when i work out the heat content it it below what i worked out in the first steps (2598.6KJ/mol fuel) could someone check my calcs above to ensure there are correct as i believe it should be somewhat less to tie into the enthalpy table to give me a max flame temp.

thanks
 
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Mitch1 said:
N2 = 5.2525 x 3.76=19.749 moles in flue gas (3.76 is 79%/21% conversion factor from O2 to air to N2 )

Plus combustion products?
 
Bystander said:
Plus combustion products?
Hi bystander
I'm not sure what you mean, do you need to add more values to the nitrogen?
Thanks
 
What is the composition of the flue gas?
 
O2=0.4775
N2=19.749
H2O=4.75
CO2=3.9moles
 
And the heat capacity?
 
Bystander said:
And the heat capacity?
I believe it to be 4.18 KJ/kg/k
Obtained from steam tables
 
That's for liquid water --- we're still in the firebox rather than the boiler. What's the heat capacity of the flue gas?
 
Bystander said:
That's for liquid water --- we're still in the firebox rather than the boiler. What's the heat capacity of the flue gas?
I'm not sure how to obtain this value
With the flue gas being a gas I'm assuming this is going to be a really low value?
 
  • #10
Just checked through your original post, and the heat of combustion for butene is NOT the same as that of butane, it's about 95% that of butane, which reduces your total enthalpy of combustion by 1-2%.

That enough to fit your table? And we'll forget I started trying to get you to do things the hard way.
 
  • #11
No problem, the lecturer has stated we can use the same value for both butane and butene although these are slightly different. 1-2% would not change my final value so it would fit that of the table.
Is it possible it is above 2200 deg C or is this too high ?
The question stated max temp so I'm assusung it is below 2200?
 
  • #12
What I've been trying to do is enthalpy divided by heat capacity for del T. This ignores the fact that CO2 and H2O are significantly dissociated at high temperatures. If you hit "flame temperature" on Wiki, it comes up around 2000 C, give or take. Whereas, total enthalpy of combustion which is measured by cooling products to 25 C ------- and all of a sudden the golden age brain death becomes transparent!

Total enthalpy is as you've calculated MINUS enthalpy of vaporization of water for the 4.75 moles of water since they are not condensed in the flue gas --- that'll be around 200 kJ you can take from the total heat of combustion since it has not been recovered by condensation.

Sorry, Man --- I'm up too late.
 
  • #13
Bystander said:
What I've been trying to do is enthalpy divided by heat capacity for del T. This ignores the fact that CO2 and H2O are significantly dissociated at high temperatures. If you hit "flame temperature" on Wiki, it comes up around 2000 C, give or take. Whereas, total enthalpy of combustion which is measured by cooling products to 25 C ------- and all of a sudden the golden age brain death becomes transparent!

Total enthalpy is as you've calculated MINUS enthalpy of vaporization of water for the 4.75 moles of water since they are not condensed in the flue gas --- that'll be around 200 kJ you can take from the total heat of combustion since it has not been recovered by condensation.

Sorry, Man --- I'm up too late.
Ah I get this now I will re do this and hopefully it will work out- thanks for your help
Much appreciated
 
  • #14
Hi I was wondering if you would know what equation to use for this ...

  1. (i) If 5% of the heat available for steam production is lost to the atmosphere, determine the amount of steam raised per hour when the total flow of flue gases is 1400 kmol h–1.

    Can I just deduct 5% from the heat available then input that into an equation to give the amount of steam raised per hr?
    Thanks mate
 
  • #15
Should do it.
 
  • #16
Mitch1 said:
Hi I was wondering if you would know what equation to use for this ...

  1. (i) If 5% of the heat available for steam production is lost to the atmosphere, determine the amount of steam raised per hour when the total flow of flue gases is 1400 kmol h–1.

    Can I just deduct 5% from the heat available then input that into an equation to give the amount of steam raised per hr?
    Thanks mate

Mitch, could you help me out by letting me know what equation you are referring to here?
 
  • #17
SeaofEnergy said:
Mitch, could you help me out by letting me know what equation you are referring to here?
Hi
I will dig out my notes and let you know. I am not 100% off the top of my head
Regards
 
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  • #18
appreciate that!cheers
 

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