Calculating Hydrogen-Oxygen Ratios at the Space Center

[flatmaster]

I went to visit the space center in my new home of Huntsville AL the other day. I figured i'd do a quick calculation from tank capacity as read from the plaque for the second and third liquid hytrogen stages. Assuming an ideal reaction, there should be a mass ratio of oxyogen to hydrogen of 4:1. This calculation comes from the straight foreward chemistry of the reaction and the respective molar masses of hydrogen and oxygen.

The third stage was close to this ratio. However, the second stage was closer to 5:1. Could a difference this big be atribuable to the engineering of a particular geometry of the tank to fit within the structure?

 

[flatmaster]

Also, I was sure to read the fuel masses rather than tank volumes.

 

[D H]

The stoichiometric ratio is 8:1 (H₂O=1₁₆O+2₁H, 16:2), not 4:1. Both the second and third stages used a very rich fuel mixture. The reason: While running rich does slows the exhaust, it does so just a little bit. You can think of it as if those extra hydrogen ions in the exhaust stream as hitching a free ride. The vehicle gets a corresponding free ride the other direction.

 

[flatmaster]

Ok. I see my mistake now. I understand how running fuel rich would provide extra mass for some more momentum transfer rearwards, resulting in more thrust. However, what would be the difference between running fuel rich rather than oxidizer rich? Expense of LOX? storage temperature? Reaction rate? Can I assume that the extra reaction force from a heavier oxygen would be the same as from the lighter hydrogen?

 

[D H]

Can I assume that the extra reaction force from a heavier oxygen would be the same as from the lighter hydrogen?

No. A lean mixture would essentially act as a quench.

A *very* old paper: http://digidoc.larc.nasa.gov/report/tn/19/NACA-TN-4219.PDF