Propulsion Efficiency

[BretCahill]

In a propulsion engine that accelerates the fluid through which it is moving,
propulsion efficiency is defined as the thrust times the velocity divided by
the total kinetic energy in the exhaust jet, Vj. Since the thrust will drop
off as the velocity of the vessel or craft increases due to Newton's 2nd Law,
there will be an optimum speed, 1/2 Vj, that gives maximum propulsion
efficiency. A lot of kinetic energy necessarily remains in the exhaust jet and
is wasted, lowering the maximum efficiency.

Since KE ~ velocity squared, the KE remaining in the jet of a vessel or craft
moving at 1/2 Vj should be 1/4 of the KE of the exhaust of a jet sitting on the
runway or a ship tied to a dock with the engine[s] wide open. The maximum
propulsion efficiency would, therefore, be 75% at 1/2 Vj.

The formula in the books give an efficiency of 2/3rds or 67% at 1/2 Vj.

This is an example of:

1. confusing reference points in KE calculations

2. different efficiencies

3. ambigous exhaust velocities

4. worrying about 8.33% which ain't jack when you are talking about trillions
of dollars in fuel.

Bret Cahill

All conservatism is based on censorship of
economic information.
-- Bret Cahill