🚀 go-pugleaf

>>Flowrates:
>>   Total                 1,130 lb/s     22,557 gpm
>>   Hydrogen                160 lb/s     16,436 gpm
>>   Oxygen                  970 lb/s      6,121 gpm
>
>Does the above flowrates indicate tha actual combustion mixture present at the
>combustion chamber? If so then why is there a greater ratio than 2 parts H to
>one part O? I would think that the extra hydrogen would be unconsumed making
>the engine less efficient than it could be... What am I missing?
>
>Phil Biehl

While the temperature of the products will be lower because of the energy
used to heat the excess hydrogen, the SSME is not the most efficient at
a 2 to 1 mixture ratio.  Not getting involved in discussions on units, the
performance of a rocket is measured by its specific impulse.

                                  
                             F       Vexit
                      I  =  ---  ~  -------
                       s     .         g
                             w
                                              .
Where F is thrust, Is is specific impulse, w is the mass flow rate, and
Vexit is the nozzle exit velocity of the gases.  Trusting me that exit velocity
is proportional to the square root of combustion temperature over the 
molecualr weight of the exhaust gas, by lowering combustion temperature 
you are also lowering the molecular weight (Water = 18, Hydrogen = 2, Both
mixed < 18) and at some point is the ideal mixture ratio.  If you really
want the equation for Vexit, let me know.  It is rather larger than the
one above :-)


Bob Campbell                Some times I wish that I could stop you from 
Hewlett Packard             talking, when I hear the silly things you say.
hplabs!hpda!campbelr                         - Elvis Costello