Calculating Propulsion Efficiency
BretCahill
not a rocket --, 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
Pete K.
"BretCahill" <bretc...@aol.com> wrote in message
news:20021022232002...@mb-fz.aol.com...
Or it could be that you're misinterpreting propulsive efficiency.
Pete
paradigm
Maybe it's just your interpretation of numbers is all wrong, last time I
checked 8.33% of just one trillion dollars is *only* 83.3 billion
dollars...try convincing any engine building company that 8.33% of
efficiency worth 83.3 billion dollars "ain't jack."
paradigm