What is the ideal mixture ratio of oxygen to aluminum for a rocket fuel?

[foulbr3@yahoo.com]

Hi, I have some questions about using Aluminum as a rocket fuel with
oxygen.
According to my calculuations, V_ex = sqrt(2*Ed) where V_ex = exaust
velocity at 100% efficiency and Ed = energy density of the propellant
in J/kg. This is proportional to sqrt(T/m_w) where T is temperature and
m_w is molecular weight.

Let's try this with LH2-LOX first: H2O has a molecular weight of
18.02e-3 kg/mol and enthalpy of formation of 242e3 J/mol. V_ex =
sqrt(2*242e3/18.02e-3) = 5182.57 m/s.
Dividing by 9.8 we get 528 s so at 88% efficiency we get 465 s which is
the actual specific impulse.

Next let's try this with Al-LOX. V_ex = sqrt(2*1675.7e3/101.96e-3) =
5733.2 m/s.
Dividing by 9.8 we get 585 s, so it beats the space shuttle main
engines! At 88% efficiency we get 514 s. This is enough to reach space
in 2 stages very easilly.

We can only conclude from the above that the combustion temperature of
Al-LOX is very, very hot. How can I find the temperature exactly, to
determine if the Al_2_O_3 product will be in gasseus form or not?

My questions are:
1. How can I find the ideal mixture ratio of oxygen to aluminum? Why
don't LH2-LOX rockets just have twice as much H_2 as O_2? Why 4 times
the O_2 as H_2 instead? It seems to me this reduces the combustion
temperature, thus increasing the mass ratio at the expense of the
exaust velocity (specific impulse). Will an molten Aluminum-fueled
rocket need excess LOX too?? How can I find the ideal ratio? I want to
compute the volume of tanks but to do this I need to know the
Oxygen-Aluminum ratio.

2. After passing through the nozzle, will the Al_2_O_3 exhaust be a
liquid? Who ever heard of a liquid exhaust rocket, except
water-compressed-air rockets! But my understanding is that as the
exaust passes through the nozzle it cools and speeds up, like a heat
engine. I imagine then the exhaust will be liquid, molten Al_2_O_3!

 

[dav2008]

https://www.physicsforums.com/showthread.php?t=122759

 

[sir-pinski]

The ideal mixture ratio of oxygen to aluminum for a rocket fuel depends on several factors, including the efficiency of the rocket engine, the desired specific impulse, and the temperature at which the fuel will burn. Generally, a higher oxygen to aluminum ratio will result in a higher specific impulse, but it may also increase the combustion temperature, which can have negative effects on the rocket engine.

To find the ideal mixture ratio, you can use equations and calculations similar to the ones you have already done for LH2-LOX and Al-LOX. However, it is important to note that the ideal mixture ratio may vary for different rocket engines and designs, so it is best to consult with experts or conduct thorough testing to determine the most efficient ratio for your specific rocket.

In terms of the combustion temperature and whether the Al_2_O_3 exhaust will be in liquid form, it is difficult to determine without specific information about the rocket engine design and temperature. The exhaust temperature can vary depending on the efficiency of the engine and the amount of excess oxygen used. It is possible that the Al_2_O_3 exhaust may be in liquid form, but it could also be in a gaseous or solid form depending on the temperature and pressure.

Overall, the ideal mixture ratio of oxygen to aluminum for a rocket fuel will depend on several factors and may need to be determined through testing and experimentation. It is important to consider the specific design and goals of the rocket when determining the ideal ratio.