SCRAM Rocket Equation by Steven S. Pietrobon (20 Dec 1994)
---------------------

We showed in http://www.sworld.com.au/steven/space/lace.txt the rocket
equation for a liquid air cycle engine as

                                          r_o/r_f
                         (   m_c + m_s   )
v(T) = v_m + (v_i - v_m) (---------------)
                         (m_c + m_s + m_f)

where

v(T) = final speed (m/s)
v_m  = v_e(1 + r_f/r_o) = maximum theoretical speed (m/s)
v_e  = exhaust speed of engine (m/s)
v_i  = initital speed of vehicle
m_c  = cargo mass (kg)
m_s  = structure mass (kg)
m_f  = fuel mass (kg)
r_f  = fuel rate (kg/s)
r_o  = oxidiser rate (kg/s)

For a SCRAM powered rocket incoming air is slowed down to some supersonic speed
v_s to allow the fuel enough time to burn with the oxygen in the air. Thus the
drag will be (v(t)-v_s)r_o. This is equivalent to converting velocity energy
to thermal energy. We can regain this energy by thermally expanding the heated
air in a rocket nozzle with some efficiency e (e=1 is 100% efficient). Since
the thermal energy provided by combusting the fuel uses the same nozzle we have
the differential equation

dv(t)   e v_e(r_f + r_o) - (1-e) (v(t) - v_s) r_o
----- = -----------------------------------------
 dt              m_c + m_s + m_f - r_f t

Similar to the derivation of the LACE rocket equation we have

                                          (1-e)r_o/r_f
                         (   m_c + m_s   )
v(T) = v_m + (v_i - v_m) (---------------)
                         (m_c + m_s + m_f)

where

      v_e(1+r_f/r_o)
v_m = -------------- + v_s
         1/e - 1

If we assume an efficiency of e = 0.9, v_e = 2200 m/s, r_o/r_f = 3x7.25 = 21.75 
(since we're not dumping 2/3 of the hydrogen overboard) and v_s = 1000 m/s
(about Mach 3) we have v_m = 21,710 m/s. 

For an actual vehicle let us start at v_i = 1657 m/s (Mach 5). For the masses 
let us use the same values as SKYLON with m_c = 10 t, m_s = 45.7 t. To calulate 
m_f we'll replace the oxygen with hydrogen. Assume a 5:1 O/F ratio for the 
rocket powered flight (a lower ratio is used for vacuum operation to get better 
v_e, whereas 6:1 is used for sea level operations where higher thrust is 
desired). We thus have m_o = 5/6 x 168.3 t = 140.25 t. The equivalent hydrogen 
mass is 0.071/1.14 x 140.25 = 8.74 t (0.071 and 1.14 kg/m^3 is the density of 
liquid hydrogen and oxygen, respectively). The total hydrogen mass is then m_f 
= 8.74 + 1/6 x 168.3 = 36.79 t. Lets round it up to m_f = 36.8 t. Plugging in 
the numbers we get v(T) = 15056 m/s (!). So the potential appears to be there
to obtain very high performance. Of course, when vehicle drag losses are taken
into account, this may be a different story. Only very little of these losses
can be recovered which will severly affect performance (it won't take much
to counteract the net thrust produced by the engine).

If we assume that rocket powered 300 m/s is required to go into and out of
orbit (150 m/s each way) then using the rocket equation we require

m_p = (m_c + m_s) (exp (v_d/v_e) - 1)
    = (10 + 44.7) (exp (300/4400) - 1)
    = 3.9 t (propellant mass)

The SCRAM powered v(T) is then 14664 m/s, only slightly less than before. If
the efficiency drops to e = 0.8 then v_m = 10205 m/s and v(T) = 9149 m/s.