Doug,

Thanks for the derivation - I was too lazy to look up the volumetric
flow/velocity relationship. Looking at the RULE site, their largest
bilge pump is 8000gph, or 133gpm. That calculates to 7.8 Lbf thrust,
with a 31 amp draw at 12VDC. Comparing that to a Minn Kota Endura 30,
with 30 Lbf thrust, at 30A/12VDC gives a good comparison of the relative
efficiencies.

Nice site, BTW. Looks like someone's got a lot of time on their
hands...or a buttload more motivation than I have :-)

Keith Hughes

Doug J wrote:
The following was posted by Cliff on the psubs.org group. Personal
submarine and ROV builders consider bilge pumps frequently. They are
often used in ROV's because they are easy, but bilge pumps or jet
propulsion is rarely used in Subs because they are inefficient. If
you do go with a pump, be sure to place the discharge just above the
water line to increase the efficiently. --Doug www.submarineboat.com

Below is a derivation of thrust that can be developed from a axial
flow pump
in terms of volumetric flowrate.
The thrust due to accelerating fluid through a pump can be written as
F=M(V1-V0)
Where M is the mass flow rate, V0 is the free stream velocity upstream
of
the pump and V1 is the velocity exiting the pump.
But the mass flow rate M can be related to the volumetric flow rate Q
as
M=Density*Q
Substituting, the thrust in terms of volumetric flow rate is
F=Density*Q(V1-V0)
But the volumetric flow rate Q is related to velocity in the pump duct
ID as
Q=V1*A=V1*Pi*D^2/4
Where D is the duct ID.
Solving for V1, and substituting, the thrust can be written as
F=Density*Q(Q/(Pi*D^2)-V0)
For a thruster oriented approximately normal to the direction of flow,
the
inlet velocity can be assumed to be zero. The thrust then reduces to
F = 4*Density*Q^2/(Pi*D^2)
Or
F= 0.001766*(q/d)^2
for freshwater where,
F = Thrust, lbf
q = pump volumetric flow rate in gpm
d = pump outlet duct inside diameter in inches
As an example, a pump with a capacity of 200 gpm flowing through a 2"
duct
would develop 17.7 lbf of thrust.