# The rocket motor

As noted on page 527 the rocket motor is the only current example of aeronautical interest in Class II of propulsive systems. Since it does not work by accelerating atmospheric air, it cannot be treated by Froude’s momentum theory. It is unique among current aircraft power plants in that it can operate independently of air from the atmosphere. The consequences of this are:

(i) it can operate in a rarefied atmosphere, or an atmosphere of inert gas

(ii) its maximum speed is not limited by the thermal barrier set up by the high ram – compression of the air in all air-breathing engines.

In a rocket, some form of chemical is converted in the combustion chamber into gas at high temperature and pressure, which is then exhausted at supersonic speed through a nozzle. Suppose a rocket to be travelling at a speed of V, and let the gas leave the nozzle with a speed of v relative to the rocket. Let the rate of mass flow of gas be m[77] This gas is produced by the consumption, at the same rate, of the chemicals in the rocket fuel tanks (or solid charge). Whilst in the tanks the mean m of fuel has a forward momentum of mV. After discharge from the nozzle the gas has a rearward momentum of m(v — V). Thus the rate of increase of rearward momentum of the fuel/gas is

m(v — V) — (—mV) — mv

and this rate of change of momentum is equal to the thrust on the rocket. Thus the thrust depends only on the rate of fuel consumption and the velocity of discharge relative to the rocket. The thrust does not depend on the speed of the rocket itself. In particular, the possibility exists that the speed of the rocket V can exceed the speed of the gas relative to both the rocket, v, and relative to the axes of reference, v — V.

When in the form of fuel in the rocket, the mass m of the fuel has a kinetic energy of mV2. After discharge it has a kinetic energy of m(v — V)2. Thus the rate of change of kinetic energy is

^ = ^[(v – V)2 – V2] = т(^ – 2vV)

the units being Watts.

Useful work is done at the rate TV, where T = mv is the thrust. Thus the propulsive efficiency of the rocket is

rate of useful work

rate of increase of KE of fuel 2vV 2

v2 — 2vV (v/V) -2

Now suppose v/V = 4. Then

7/p = = 1 or 100%

If v/V < 4, i. e. V > v/4, the propulsive efficiency exceeds 100%.

This derivation of the efficiency, while theoretically sound, is not normally accepted, since the engineer is unaccustomed to efficiencies in excess of 100%. Accordingly an alternative measure of the efficiency is used. In this the energy input is taken to be the energy liberated in the jet, plus the initial kinetic energy of the fuel while in the tanks. The total energy input is then

giving for the efficiency

2(v/V)

{v/V)2 + 1

By differentiating with respect to v/V, this is seen to be a maximum when v/V = 1, the propulsive efficiency then being 100%. Thus the definition of efficiency leads to a maximum efficiency of 100% when the speed of the rocket equals the speed of the exhaust gas relative to the rocket, i. e. when the exhaust is at rest relative to an observer past whom the rocket has the speed V.

If the speed of the rocket V is small compared with the exhaust speed v, as is the case for most aircraft applications, V2 may be ignored compared with v2 giving

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