I can take you a little way along with this (homework?) assignment.
You would like to convert the available shaft power to maximal thrust.

Thrust is another term for propulsive force.
Power = force times velocity.
(watts = newtons X meter/sec)
So for a given power, force (that means thrust) is maximized for the
smallest speed over the water.
The usual Newtonian equations give
force = mass times acceleration or mass flow rate times velocity
or change of momentum, in the physicists' favorite equation form.

( N = kg X m/s^2 or N = kg/sec X m/s or N = dm.v/dt )

Thrust = massflow rate times velocity is the equation on which to
focus some attention.
( N = kg/sec X m/sec )

You presumably know your target velocity and your available shaft
power and shaft speed, or you know the hull drag for given speeds.
The available thrust must equal that drag so with a little paper and
pencil work you can can reasonably work out the desired mass flow
per revolution of the prop.
The prop will be pitched to move a longer column of water than that
given in the previous sentence. The difference expresses the slip and
some other losses.

Clear as mud?
Sorry.

Brian W

On 21 Dec 2003 13:55:00 -0800, (MBS) wrote:

I am doing a project that involves using a small electric motor to
power a small boat. Basically, my question involves the propeller
design. I have read that more efficiency can be realized by utilyzing
a lower rpm and a larger propeller. However, a large propeller is out
of the question and the motor that I will be using has continous
torque rating at around 2500 rpm. Therefore, I would appreciate some
expert help with a few questions.
1) In order to stay within efficiency, what is the maximum rpm range
for a small propeller setup using a Kort nozzle? Is the rpm above too
high?
2) When we are talking about efficiency, are we talking about
propeller or overall energy efficiency (I have read that they aren't
the same)?
3) How does blade area relate to efficiency (it seems like the larger,
the less slip)?

Thanks!
Bailey--