Combustion EQ of liquid Hydrogen and liquid Oxygen

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Discussion Overview

The discussion revolves around the balanced chemical equation for the combustion of liquid hydrogen (LH2) and liquid oxygen (LOx) in the context of a Space Shuttle Main Engine (SSME). Participants explore the complexities of the reaction, including the effects of stoichiometry, excess reactants, and byproducts, while addressing the challenges of achieving a balanced equation without considering dissociation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially proposes the equation H2 + 0.5O2 > H2O but is informed by their professor that this is not the correct representation for the SSME scenario.
  • Another participant suggests that the reaction mixture is not stoichiometric and that thermal decomposition of water may occur due to high flame temperatures.
  • Contamination of the reaction by air is questioned by one participant, indicating potential complications in the combustion process.
  • A participant later claims to have figured out the equation as 2H2 + O2 > 2H2O + H2, suggesting that hydrogen is produced as a byproduct.
  • Another participant challenges the balancing of the equation, noting that it leads to a coefficient of zero for hydrogen on the product side, reverting to the original equation proposed.
  • One participant mentions research indicating the presence of excess energy in the form of hydrogen during combustion, with a promise to provide references.
  • A participant emphasizes the importance of the air-fuel (AF) ratio of 6 for determining the complete equation.
  • Concerns are raised about the potential for multiple solutions when balancing the equation without the AF ratio, highlighting the complexity of the problem.
  • Another participant distinguishes between mass balance and balanced reaction equations when excess reagents and inert compounds are included in the analysis.

Areas of Agreement / Disagreement

Participants express differing views on the correct balanced equation and the implications of excess reactants and byproducts. The discussion remains unresolved, with multiple competing perspectives on how to accurately represent the combustion process.

Contextual Notes

Participants note that the balancing of the equation is complicated by the presence of excess hydrogen and the need for specific AF ratios, which may affect the overall reaction representation.

MechStudent123
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Im working on a problem regarding a SSME and need to figure out the balanced eq for LH2 and LOx.

I thought it was simply.. H2 +0.5O2 > H2O however my professor has confirmed that it is not the case in this situation. I am stumped as to what it could possibly be, maybe excess H2? And am not supposed to include dissociation for the reaction.
 
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No idea what else it could be, this is the correct overall reaction equation. The mixture is definitely not exactly stoichiometric during the reaction, and the flame temperature is so high some of the water produced undergoes thermal decomposition (or, more precisely, the mixture never reacts to the end till it cools down a bit outside of the engine and the output flame), but in the end it is just a water synthesis.
 
Do you have contamination of the reaction by air?

BoB
 
rbelli1 said:
Do you have contamination of the reaction by air?

BoB
I ended up figuring it out

2H2 + O2 >2 H2O + H2 then baa fed for the AF of 6
 
I believe this was the eq I ended up using, not looking at it right now. *balanced
 
MechStudent123 said:
2H2 + O2 >2 H2O + H2 then baa fed for the AF of 6

Huh?
 
Then balanced the equation. I guess the ssme produces h2 as a byproduct
 
If you balance

H2 + O2 → H2O + H2

you will find no hydrogen on the right (coefficient of zero) and you will end with exactly the same equation you listed in your very first post.
 
I have researched a few articles about the combustion in a ssme. There is excess 'energy' in the form of monotomic or diatomic hydrogen. Will link the reference soon.
 
  • #10
If you know there is a AF ratio of 6 on a mass basis then you will get the complete eq
 
  • #11
Balancing the eq you stated above could result in an infinite number of solutions without the AF ratio..then again I could be wrong, haven't received my grade back
 
  • #12
Perhaps that's a bit of nitpicking on my side, but when you include excess reagents (together with the inert compounds) you get a mass balance for the process, not the balanced reaction equation. In some ways they are similar, in other ways they are distinct.
 

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