Liquid Air Cycle Rocket Equation by Steven S. Pietrobon (19 Dec 1994)
--------------------------------
Let us assume that we have a Liquid Air Cycle Engine (LACE) where the fuel is
provided by the rocket and the oxidiser is from surrounding air. To simplify
our derivation we assume that the rocket is frictionless (except for engine
drag). For a normal rocket which carries its own fuel and oxidiser we have
from Newton's third law that
dv(t) F
a(t) = ----- = ------------------------------------
dt m_c + m_s + m_f + m_o - (r_f + r_o)t
where
a(t) = acceleration varying with time (m/s^2)
v(t) = the speed varying with time (m/s)
F = engine thrust (N)
m_c = cargo mass (kg)
m_s = structure mass (kg)
m_f = fuel mass (kg)
m_o = oxidiser mass (kg)
r_f = fuel rate (kg/s)
r_o = oxidiser rate (kg/s)
We have T = m_f/r_f = m_o/r_o is the firing duration of the rocket.
Integrating and letting t = T we have the standard rocket equation
F (m_c + m_s + m_f + m_o)
v(T) = --------- ln (---------------------)
r_f + r_o ( m_c + m_s )
= v_e ln (1 + (m_f + m_o)/(m_c + m_s))
where
v_e = F/(r_f + r_o) = effective exhaust speed of the engine (m/s)
For a LACE m_o is not carried aboard and must be accelerated to the speed of
the rocket. This induces an effective drag on the vehicle equal to v(t)r_o.
We derive this from the fact that the effective thrust of a rocket equation
is that obtained by accelerating the propellant mass continuously entering the
engine up to some exhaust speed v_e (the energy being obtained from the
combustion of a fuel and an oxidiser). It does not matter how the mass was
accelerated. Only the final speed is important. Conversely, to bring the
oxidiser up to the same speed of the rocket before entering the combustion
chambers requires a force equal to v(t)r_o. The acceleration equation then
becomes
dv(t) v_e(r_f + r_o) - v(t) r_o
----- = -------------------------
dt m_c + m_s + m_f - r_f t
For a positive acceleration we must have
v_e(r_f + r_o) - v(t) r_o > 0
or
v(t) < v_e (1 + r_f/r_o)
The upper speed obtained by a LACE only powered vehicle is therefore fixed.
For hydrogen/oxygen propellants v_e = 4000 m/s (the space shuttle main engine
has v_e = 3543 m/s at sea level and 4441 m/s in a vacuum). Air has only about
25% oxygen by mass (a rough guess since I could only find out that 21% of air
by volume is oxygen and oxygen is denser than nitrogen which makes up most of
the rest of air). This means that the combustion energy is distributed over
about four times the mass resulting in an exhaust speed of about half (since
exhaust energy is proportional to v_e^2). Thus v_e = 2000 m/s. Now to make up
for the fact that nearly 75% of air is nitrogen with a mass 14/(16+2) = 7/9
times smaller than water we have v_e = 4000 / sqrt(1+3x7/9) = 2190 m/s. So
lets say v_e = 2200 m/s.
For the SKYLON LACE engine 2/3 of the hydrogen is used to cool the incoming
air and is then dumped overboard. A normal hydrogen/oxygen engine has a mixture
ratio of O/F = 6:1. For LACE we have (1+3x14/16)x6:(1+2) = 7.25:1. Thus the
maximum speed that we can expect for a hydrogen powered vehicle starting from
rest is about 2500 m/s which is far short of the orbital speed of about 8000
m/s (and that's ignoring gravity and air drag losses).
We can solve the LACE differential equation (I used the Laplace transform)
and get (assuming starting at rest)
r_o/r_f
[ ( m_c + m_s ) ]
v(T) = v_e (1 + r_f/r_o) [1 - (---------------) ]
[ (m_c + m_s + m_f) ]
This is the LACE Rocket Equation (probably discovered yonks ago, but what the
heck, it was fun deriving it). For the LACE the air powered portion of the
flight has
m_c + m_s = 224 t
m_c + m_s + m_f = 262 t
Thus v(T) = 1700 m/s. The quoted final speed for SKYLON was 1657 m/s. Ways
to increase the performance of the LACE is to not cool the air so that the
thermal energy can be used in the rocket engine. In fact the SKYLON SABRE
engine does not liquify the air which increases the performance of the system.
SKYLON then uses normal rocket powered flight to reach orbit. We have
m_c = 10 t
m_s = 45.7 t
m_f + m_o = 168.3 t
Using the rocket equation and v_e = 4400 m/s we have v(T) = 6123 m/s. SKYLON
receives an initial rocket powered boost requiring 13 t of propellant. This
gives an initial v(T) of 169 m/s assuming v_e = 3500 m/s. The total deltaV
is thus about 8000 m/s. Hmm, might be a bit low, but then again SKYLON does
take off horizontally which should save on gravity losses (since it follows
the Earth's velocity vector, unlike a rocket which takes off perpendicular
to the velocity vector).
In case you're wondering, this is the LACE equation if you start with an
initial speed of v_i
r_o/r_f
( m_c + m_s )
v(T) = v_e(1 + r_f/r_o) + (v_i - v_e(1 + r_f/r_o)) (---------------)
(m_c + m_s + m_f)
Food for thought. How do you handle SCRAM (supersonic combustion RAM) jets?
The air is decelerated to some supersonic value (say v_s) and combustion takes
place. The drag is now (v-v_s)r_o. Now I can imagine incoming oxygen molecules
smashing into the hydrogen molecules with an almighty whack. The air velocity
vector has been partially converted to thermal energy. You should be able to
get this energy back again in the engine exhaust (with a suitable expansion
nozzle) plus the energy of combusting hydrogen with oxygen. As you can imagine
the drag increases with speed and so the thrust must increase with it. This is
the reverse of rocket engines where you usually decrease the thrust the faster
you go. A SCRAM powered vehicle would have three different types of operation:
zero to the supersonic speed where the SCRAM kicks in (using a LACE engine?),
supersonic to as fast as you can go, and then a normal rocket engine to put you
into orbit. Plus the fact that a SCRAM engine has such a big mouth (and a
matching rear end) that you have to design your vehicle around it.