# (20 Dec 1994)

We showed in Liquid Air Cycle Rocket Equation 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.

```Steven S. Pietrobon,  Australian Space Centre for Signal Processing
Signal Processing Research Institute, University of South Australia
The Levels, SA 5095, Australia.     steven@spri.levels.unisa.edu.au

```