Joe wrote:

I'm hoping someone might have a reference site for free wheeling prop
torque produced by a prop around the size of a 20X20 sailing at hull
speed. (9kts)

It's a 200 HP perm magnet motor... They need 420 volt DC using
two banks of batterys (35 batteries each bank) and will require 75 LBS
or torque to generate electricity.
Now I'm pretty sure the torque on a prop around the 22" size will be
way over 75 lbs but do not know exactly. Im hoping it's twice that at
least for gear reduction.

A SWAG for you: the MOST efficient a fluid mill gets, is to extract
the kinetic energy represented by halving the fluid's speed.


The energy in question is given by prop area (diam 20 inches)
and effective speed 4.5 kts.

Now: reduce the units to easy SI units:

4.5 kts = 2.3 m/s
diameter of water column: 20 ins = 0.51 m
volume of water/sec = 2.3 X pi X 0.51^2 / 4 = 0.47 cu meters
density of water 1050 kg /m^3 (for salt water say...)
Mass flow rate: 0.47 X 1050 kg/sec = 489 kg/sec

Force times speed = power
So power = 489 kg/sec X 9.8 N/kg = 4797 watts.
Let's suppose 50% transfer efficiency and we end up with
2.4 kW to play with, at hull speed.

Finding a prop speed 10 % less than the no slip speed of a 20 /20
prop, and knowing the available power from above, leads easily to
the torque available...

Brian W

On Oct 31, 12:10*pm, brian whatcott wrote:
Joe wrote:
I'm hoping someone might have a reference site for free wheeling prop
torque produced by a prop around the size of a 20X20 sailing at hull
speed. (9kts)
It's a 200 HP perm magnet motor... They need 420 volt DC using
two banks of batterys (35 batteries each bank) and will require 75 LBS
or torque to generate electricity.
Now I'm pretty sure the torque on a prop around the 22" size will be
way over 75 lbs but do not know exactly. Im hoping it's twice that at
least for gear reduction.

A SWAG for you: the MOST efficient a fluid mill gets, is to extract
the kinetic energy represented by halving the fluid's speed.

The energy in question is given by prop area (diam 20 inches)
and effective speed 4.5 kts.

Now: reduce the units to easy SI units:

4.5 kts *= 2.3 m/s
diameter of water column: 20 ins = 0.51 m
volume of water/sec = 2.3 X pi X 0.51^2 / 4 = 0.47 cu meters
density of water *1050 kg /m^3 (for *salt water say...)
Mass flow rate: 0.47 X 1050 kg/sec = 489 kg/sec

Force times speed = power
So power = 489 kg/sec X 9.8 N/kg = 4797 watts.
Let's suppose 50% transfer efficiency and we end up with
2.4 kW to play with, at hull speed.

Finding a prop speed 10 % less than the no slip speed of a 20 /20
prop, and knowing the available power from above, leads easily to
the torque available...

Brian W

Thanks Brian,

I can use this data and go back to the company and talk to an
applications engineer.

Joe