How to Calculate Heat of Combustion and Heat Transfer in a Furnace?

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SUMMARY

The discussion focuses on calculating the standard heat of combustion for a natural-gas fuel comprising 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen at 25°C, with water vapor (H2O(g)) as a product. The heat of combustion is determined using Hess's Law, where the enthalpy changes of methane and ethane are combined, while nitrogen remains unreacted. Additionally, the heat transfer in a furnace with 50% excess air, where products exit at 600°C, is calculated by considering the sensible heat changes from the combustion process.

PREREQUISITES
  • Understanding of Hess's Law for thermochemical calculations
  • Knowledge of standard heat of formation values for methane and ethane
  • Familiarity with combustion reactions and stoichiometry
  • Basic principles of heat transfer in chemical processes
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  • Research standard heat of formation values for methane and ethane
  • Learn about the application of Hess's Law in complex thermochemical calculations
  • Study the principles of heat transfer in furnaces and reactors
  • Explore the impact of excess air on combustion efficiency and heat transfer
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Chemical engineers, thermodynamics students, and professionals involved in combustion analysis and heat transfer optimization in industrial processes.

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Homework Statement


A natural-gas fuel contains 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen.
a) What is the standard heat of combustion (kJ mol-1) of the fuel at 25°C with H2O(g) as a
product?
b) The fuel is supplied to a furnace with 50% excess air, both entering at 25°C. The products
leave at 600°C. If combustion is complete and if no side reactions occur, how much heat
(kJ mol-1 of fuel) is transferred in the furnace?


Homework Equations





The Attempt at a Solution


For this problem, I am unsure if in general, the exiting temperature is the reaction temperature. I mean, initially the gases are at 25 C, but when they exit they are 600C. That doesn't mean it reacted at 600 C, the result of the reaction just happened to be that temperature increase, right?
 

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Maylis said:

Homework Statement


A natural-gas fuel contains 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen.
a) What is the standard heat of combustion (kJ mol-1) of the fuel at 25°C with H2O(g) as a
product?
b) The fuel is supplied to a furnace with 50% excess air, both entering at 25°C. The products
leave at 600°C. If combustion is complete and if no side reactions occur, how much heat
(kJ mol-1 of fuel) is transferred in the furnace?


Homework Equations





The Attempt at a Solution


For this problem, I am unsure if in general, the exiting temperature is the reaction temperature. I mean, initially the gases are at 25 C, but when they exit they are 600C. That doesn't mean it reacted at 600 C, the result of the reaction just happened to be that temperature increase, right?

Yes. You have the right idea. I didn't check every last detail of what you did, but you certainly knew to use Hess's Law. The heat given off by the reaction at 25C minus the heat loss from the reactor is enough to raise the temperature of the products from 25C to 600C.

chet
 
Does nitrogen combust with oxygen?

Also, is my part (a) correct? A friend of mine is trying it my doing .85ΔHf,methane + .10ΔHf,ethane + .05ΔHf,N2 - ΔHf, CO2 - ΔHf,H2O(g)
 
Last edited:
Maylis said:
Does nitrogen combust with oxygen?

Also, is my part (a) correct? A friend of mine is trying it my doing .85ΔHf,methane + .10ΔHf,ethane + .05ΔHf,N2 - ΔHf, CO2 - ΔHf,H2O(g)
The N2 does not react.

I don't follow what your friend did, but in part (a) you should have taken 0.85 times the heat of reaction 1 plus 0.10 times the heat of reaction 2. There is no change in the enthalpy of the N2 at 25C.

In part (b), I really like the way you did it. That's what I would have done.

Chet
 
Yes, I did that. Once you get the answer to the heat of combustion to part (a), you just add up the sensible heat changes. However, I actually changed my basis from 1 mol in part (a) to 100 moles in part (b), so I went back, changed the basis to 1 mol, and multiplied all the values I got for the sensible heats of the species by the number of moles coming out of the reactor.

It gives the right answer, but I was a little shaky why I had to do that. It seems like it has to do with the wording based on the fact that its per mole of fuel.
 
Maylis said:
Yes, I did that. Once you get the answer to the heat of combustion to part (a), you just add up the sensible heat changes. However, I actually changed my basis from 1 mol in part (a) to 100 moles in part (b), so I went back, changed the basis to 1 mol, and multiplied all the values I got for the sensible heats of the species by the number of moles coming out of the reactor.

It gives the right answer, but I was a little shaky why I had to do that. It seems like it has to do with the wording based on the fact that its per mole of fuel.
Yes. That's correct.

Chet
 

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