Hi, I have some questions about using Aluminum as a rocket fuel with(adsbygoogle = window.adsbygoogle || []).push({});

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 exaust be a

liquid? Who ever heard of a liquid exaust 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 exaust will be liquid, molten Al_2_O_3!

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# Aluminum Rocket Questions

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