- #1
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!
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!
