Combustion EQ of liquid Hydrogen and liquid Oxygen

  • #1
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. Im stumped as to what it could possibly be, maybe excess H2? And am not supposed to include dissociation for the reaction.
 

Answers and Replies

  • #2
Borek
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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.
 
  • #3
rbelli1
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Do you have contamination of the reaction by air?

BoB
 
  • #4
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
 
  • #5
I believe this was the eq I ended up using, not looking at it right now. *balanced
 
  • #7
Then balanced the equation. I guess the ssme produces h2 as a byproduct
 
  • #8
Borek
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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.
 
  • #9
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
Borek
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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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