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

In summary, the conversation covers the calculation of maximum flame temperature in a combustion chamber using a fuel gas mixture of butane, propane, and butene. The calculation takes into account the heat content of the fuel and the actual oxygen supplied. However, the heat capacity of the flue gas needs to be considered as well. Additionally, the conversation also discusses the deduction of heat lost to the atmosphere and its impact on the amount of steam raised per hour in a given flow of flue gases.
  • #1
Mitch1
55
1
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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  • #2
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?
 
  • #3
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
 
  • #4
What is the composition of the flue gas?
 
  • #5
O2=0.4775
N2=19.749
H2O=4.75
CO2=3.9moles
 
  • #6
And the heat capacity?
 
  • #7
Bystander said:
And the heat capacity?
I believe it to be 4.18 KJ/kg/k
Obtained from steam tables
 
  • #8
That's for liquid water --- we're still in the firebox rather than the boiler. What's the heat capacity of the flue gas?
 
  • #9
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
 
  • Like
Likes SeaofEnergy
  • #18
appreciate that!cheers
 

1. What is the balanced chemical equation for the reaction between C4H10 and O2?

The balanced chemical equation for the reaction between C4H10 and O2 is C4H10 + 4.5O2 -> 4CO2 + 5H2O. This means that for every molecule of C4H10, 4.5 molecules of O2 are required to produce 4 molecules of CO2 and 5 molecules of H2O.

2. What is the maximum flame temperature that can be achieved from this reaction?

The maximum flame temperature that can be achieved from this reaction is approximately 3,500 degrees Celsius. This is due to the high energy release from the combustion of C4H10 and O2, which results in the formation of the products CO2 and H2O.

3. How does the ratio of C4H10 to O2 affect the maximum flame temperature?

The ratio of C4H10 to O2 has a direct impact on the maximum flame temperature. This is because the reaction requires a specific amount of oxygen to completely combust the C4H10. If there is an excess of C4H10, the flame temperature will be lower as not all of the molecules will be able to react with the available oxygen. On the other hand, if there is an excess of O2, the flame temperature can increase as more C4H10 molecules can react and release energy.

4. Can the maximum flame temperature be affected by external factors?

Yes, the maximum flame temperature can be affected by external factors such as pressure, temperature, and the availability of other reactants or catalysts. These factors can alter the rate of the reaction and therefore impact the overall energy release and flame temperature.

5. What are the potential applications of this reaction and its maximum flame temperature?

The reaction between C4H10 and O2 has various applications in the energy and chemical industries. The maximum flame temperature can be harnessed for heating and energy production, as well as for the production of other chemicals through the combustion of C4H10. Additionally, this reaction can also be used in rocket propulsion and welding processes.

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