Hi Shaun,

Ive been reading up a bit on pumps, but some of the math is beyond me. i do
know that its possilb eo hook must pumps up either in series, or in
parallel. in parallel you quite logically get a doubling of flow in gallons
per hour or whatever, while in series you combine the 'heads' whatever that
means. i think it means head pressure?

Yep, head is pressure. Basically, you have one pound/sq.inch for each
27.68" of water column (height).

i know a lot of the losses in small pumps are from pumping 'up'.

That's kind of a misconception regarding 'head'. Pumping up, down, or
horizontal, the flowrate is dependent on the total backpressure on the
discharge line (but of course, 10' of vertical pipe does have more total
backpressure than 10' of horizontal pipe - of the same size).

most small
pumps are rated by how high they can pump water, and the rating for flow
goes down as the height increases. installed in a boat, i would try to keep
the whole thing on the level with the shortest hose runs possible. on a
beach cat, i would have a thru hull on the side of the hull with maybe 6
inches of hose going to the pump, then another foot of hose going to the
outlet.

Keeping the tubing runs as short as possible is certainly the right
approach to reduce frictional losses. One problem with the inlet on the
side of the hull (or any hull surface tangential to the water flow) is
that you get Bernoulli effects as the boat speed increases, that tends
to create a vacuum in the suction line (the same concept that makes
paint sprayers - the kind that use air hoses - or end-of-hose garden
sprayers work. The high speed stream across the diptube end creates
suction to raise the paint/roundup into the discharge stream).

i think youd have to start with two pumps in each hull, both running off a
common larger diameter inlest, and through a Y joiner to a common outlet.
this would give you some options. you could run the pumps in parallel, or
in series. then you would have to experiement with various reductions in
the outlet to see what the smallest diameter nozzle you could use without
losing flow would be. this is probably how you would use 'gearing'.

if you used too large of a diameter nozzle, you really wouldnt get any force
at all.

Don't confuse "velocity" with "Force". Just like with a garden hose
where you have, say 80psig, you can pinch the end to get a higher
velocity stream, but you get less flow (i.e. less mass). Since the
force = mass x acceleration, the force however is the same (you only
have 80psig to start with). The same is true for pumps, as you note
above, when you create more backpressure (pinching the hose), the
flowrate goes down. If you move 100gpm of water through the system, the
force is the same whether the discharge is 1" or 3", only the velocity
of the dischage changes. Remember, PSI is pounds per square inch (i.e.
force per unit area), so the 1" discharge stream may be at 10 times the
pressure of the 3" stream, but the 3" stream has 10 times the
cross-sectional area of the 1" stream.


Keith Hughes

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.

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.

On Feb 16, 1:24 pm, Keith Hughes wrote:
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.


I actually saw this being done once:

Get a large cordless electric drill, mount a long shaft in it and put
a trolling prop on the shaft. I saw a guy pushing an 18' canoe once
this way and I nearly fell overboard watching it.
For that matter, you could attach leads to power it from your 12V
battery.

does having the outlet above the water line really increase the efficiency?
All the jet boats I''ve seen have the outlet below the waterline, but i
could be wrong... is this what all the RC boat builders do?

Ill have to have a beer or two before i try to get my head around the
numbers, but thanks for the information!

Shaun

"Doug J" wrote in message
ups.com...
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.

On Feb 16, 2:43 pm, "Shaun Van Poecke"
wrote:
does having the outlet above the water line really increase the efficiency?
All the jet boats I''ve seen have the outlet below the waterline, but i
could be wrong... is this what all the RC boat builders do?

I was afraid someone would ask that. So I'll admit that I am going
on what I have been told by jet boat and jet ski people. If anyone
knows better about jet pumps, please correct me.

The efficiency of the jet pump is based on the mass of water it
discharges. Any back pressure and turbulence at the outlet only
reduces the velocity of the flow and therefor the rate of the flow.
The discharge ports are below the water line but only when the craft
is not yet up to speed. I think there is a benefit to having a higher
outlet pressure during start up or the "hole shot". Jet pumps also
depend on the design of their intake ports, because at top speed the
forward motion of the craft and the shape of the intake actually
assist in directing the water flow into the pump, much like an air
intake scoop on a dragster. You might milk another 2 or 3 oz of
thrust from that bilge pump if you put a scoop on it. Then again
that would really jack with the drag on a sail boat hull.

I'd go with an old used cheep trolling motor with a busted speed
controller. Clean it, replace the brushes and mount it on one of the
transoms with a hinge that lets it flip down into the water and then
steer with the rudders. Add a simple on/off switch and avoid the
variable speed controller or any other electronics that will just
present other points of potential failure.

Best Regards
Doug
www.submarineboat.com

"Doug J" wrote in message
ups.com...
On Feb 16, 2:43 pm, "Shaun Van Poecke"
I'd go with an old used cheep trolling motor with a busted speed
controller. Clean it, replace the brushes and mount it on one of the
transoms with a hinge that lets it flip down into the water and then
steer with the rudders. Add a simple on/off switch and avoid the
variable speed controller or any other electronics that will just
present other points of potential failure.

Best Regards
Doug
www.submarineboat.com

im with you there doug, that would suit my budget and my temperament
perfectly. even used trolling motors on ebay in australia attract quite a
premium... Ive seen second hand 40lb motors going for up to AU$250! since
ill probably chop it anyway, buying new is not a big concern of mine, and
ill only want full speed, so drect wiring seems the way to go.

what is the general thought on the life of a trolling motor? are
replacement bushes readily available?

Shaun

Doug J wrote:

On Feb 16, 2:43 pm, "Shaun Van Poecke"
wrote:

does having the outlet above the water line really increase the efficiency?
All the jet boats I''ve seen have the outlet below the waterline, but i
could be wrong... is this what all the RC boat builders do?


I was afraid someone would ask that. So I'll admit that I am going
on what I have been told by jet boat and jet ski people. If anyone
knows better about jet pumps, please correct me.

The efficiency of the jet pump is based on the mass of water it
discharges. Any back pressure and turbulence at the outlet only
reduces the velocity of the flow and therefor the rate of the flow.

Yep, the discharge stream has to displace water already behind the boat,
and that requires work (seen as higher backpressure at the discharge
nozzle, reducing the mass flow rate).

Where you could gain efficiency would be in having a directionally
adjustable discharge nozzle (primarily with a planing hull) so you could
optimize the discharge vector, for maximum thrust in the direction of
boat travel, for different bow angles. For e.g., as the bow rises, the
discharge angle, for a fixed nozzle, rotates downward. The thrust is
now a vector addition of the upward + forward thrusts, with the upward
thrust component being wasted energy. Seems that angling the stream
back to horizontal (and letting the hull do the lifting) would increase
the forward thrust.

The discharge ports are below the water line but only when the craft
is not yet up to speed. I think there is a benefit to having a higher
outlet pressure during start up or the "hole shot". Jet pumps also
depend on the design of their intake ports, because at top speed the
forward motion of the craft and the shape of the intake actually
assist in directing the water flow into the pump, much like an air
intake scoop on a dragster. You might milk another 2 or 3 oz of
thrust from that bilge pump if you put a scoop on it. Then again
that would really jack with the drag on a sail boat hull.

Details, details... :-)

I'd go with an old used cheep trolling motor with a busted speed
controller. Clean it, replace the brushes and mount it on one of the
transoms with a hinge that lets it flip down into the water and then
steer with the rudders. Add a simple on/off switch and avoid the
variable speed controller or any other electronics that will just
present other points of potential failure.

Sounds like the best plan to me.

Keith Hughes

"Keith Hughes" wrote in message
...
Hi Shaun,

Ive been reading up a bit on pumps, but some of the math is beyond me. i
do know that its possilb eo hook must pumps up either in series, or in
parallel. in parallel you quite logically get a doubling of flow in
gallons per hour or whatever, while in series you combine the 'heads'
whatever that means. i think it means head pressure?

Yep, head is pressure. Basically, you have one pound/sq.inch for each
27.68" of water column (height).

i know a lot of the losses in small pumps are from pumping 'up'.

That's kind of a misconception regarding 'head'. Pumping up, down, or
horizontal, the flowrate is dependent on the total backpressure on the
discharge line (but of course, 10' of vertical pipe does have more total
backpressure than 10' of horizontal pipe - of the same size).

most small pumps are rated by how high they can pump water, and the
rating for flow goes down as the height increases. installed in a boat,
i would try to keep the whole thing on the level with the shortest hose
runs possible. on a beach cat, i would have a thru hull on the side of
the hull with maybe 6 inches of hose going to the pump, then another foot
of hose going to the outlet.

Keeping the tubing runs as short as possible is certainly the right
approach to reduce frictional losses. One problem with the inlet on the
side of the hull (or any hull surface tangential to the water flow) is
that you get Bernoulli effects as the boat speed increases, that tends to
create a vacuum in the suction line (the same concept that makes paint
sprayers - the kind that use air hoses - or end-of-hose garden sprayers
work. The high speed stream across the diptube end creates suction to
raise the paint/roundup into the discharge stream).

i think youd have to start with two pumps in each hull, both running off
a common larger diameter inlest, and through a Y joiner to a common
outlet. this would give you some options. you could run the pumps in
parallel, or in series. then you would have to experiement with various
reductions in the outlet to see what the smallest diameter nozzle you
could use without losing flow would be. this is probably how you would
use 'gearing'.

if you used too large of a diameter nozzle, you really wouldnt get any
force at all.

Don't confuse "velocity" with "Force". Just like with a garden hose where
you have, say 80psig, you can pinch the end to get a higher velocity
stream, but you get less flow (i.e. less mass). Since the force = mass x
acceleration, the force however is the same (you only have 80psig to start
with). The same is true for pumps, as you note above, when you create
more backpressure (pinching the hose), the flowrate goes down. If you move
100gpm of water through the system, the force is the same whether the
discharge is 1" or 3", only the velocity of the dischage changes.
Remember, PSI is pounds per square inch (i.e. force per unit area), so the
1" discharge stream may be at 10 times the pressure of the 3" stream, but
the 3" stream has 10 times the cross-sectional area of the 1" stream.


Keith Hughes


in the very simples sense though, if i had the same volume of water flowing
through both a very large and a very small outlet, the speed would be much
greater for the smaller outlet right? this seems like a way to achieve some
sort of gearing to me, despite whatever losses are incurred from
backpressure. runing pumps in series would allow you to have a smaller
outlet and still maintain the same volume of flow right?

While there would obviously be a sweet spot for any given pump, having more
velocity at the outlet seems like it would probably result in more real
world 'thrust'. I was reading a page by an RC boat builder who use a bilge
pump for drive on his boat. he used a fishing scale to measure the trust
produced by the boat, and found that making the nozzle on the outlet
increased thust, but only to a certain point.

Shaun

Shaun,

in the very simples sense though, if i had the same volume of water flowing
through both a very large and a very small outlet, the speed would be much
greater for the smaller outlet right?

The velocity (speed) of the water stream would be greater from the
smaller outlet. The resulting force, however, would be the same since
you're moving the same volume of water per unit time.

this seems like a way to achieve some
sort of gearing to me, despite whatever losses are incurred from
backpressure.

It's not a matter of backpressure, it's a matter of reaction mass. It
is Newtons second law of motion, paraphrased; for every action, there is
an equal and opposite reaction. The 'little' stream puts a lot of force
over a small area, whereas the 'big' stream puts a small amount of force
over a big area. In each case, the "force/unit area x area" quantity
(total Force) is the same. As long as the volume remains constant,
every increase in velocity will be offset by a proportional decrease in
the area over which it is applied.

It's not a matter of the water stream "pushing" against the water behind
the boat. Its just like how rocket thrusters work in a vacuum; you shoot
out 10kg of gas at 10m/s over a 10 second period, and you'll get exactly
that much "thrust" in the opposite direction. To be sure, there are
lots of hydrodynamic losses and effects for the boat, but the basic
properties of thrust are the same.

runing pumps in series would allow you to have a smaller
outlet and still maintain the same volume of flow right?

The same volume as what, a single pump with larger outlet? If you mean
use a second series pump to overcome all the frictional losses to
maintain flowrate, sure...but you're now powering 2 pumps. The cost of
the higher velocity, at the same volume, is all the additional power you
burn up in the second pump.

While there would obviously be a sweet spot for any given pump, having more
velocity at the outlet seems like it would probably result in more real
world 'thrust'.

The higher the velocity *at a given volumetric flow rate* the higher the
thrust. It's Newtons formula:

F = m x a

Where F = Force
m = mass (proportional to the volumetric flow rate)
a = acceleration (proportional to the velocity of the water leaving the
pump versus velocity entering the pump)

I was reading a page by an RC boat builder who use a bilge
pump for drive on his boat. he used a fishing scale to measure the trust
produced by the boat, and found that making the nozzle on the outlet
increased thust, but only to a certain point.

Yes, and that certain point is where the flowrate begins to decrease as
a result of the additional head pressure caused by restricting the
outlet. There are other issues that arise when the outlet is
sufficiently large that it represents a significant percentage of the
width of the boat, which you can do with an RC boat, that just don't
arise in 'real' boat applications.

Keith Hughes