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.

"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

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.

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

"Keith Hughes" wrote in message
...

snip

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.

If i understand what you're saying here, it sounds prettymuch
counterintuitive. I may be mis-using some of the terms? let me give an
example just to be sure that i understand what you're saying here, and bear
in mind that of course the numbers im going to use are entirely made up in
my head, so they'd be wrong....

lets say that you have two identical boats with the same pump on each one,
running at whatever flow you like, say 5,000GPH. Boat A has a huge
outlet... say 5 inches in diameter. for arguements sake, because i dont
know how to calculate the speed of the water for that given outlet, lets say
the speed of the water coming out the back is slow. i dont know how slow,
but lets say it comes out at 3 knots.

Now boat B has the same pump, but the outlet is so small, that even though
its using the same pump, the water is coming out at a speed of 20 knots.
what you're saying is that both boats because they have the same amount of
energy put into them, and the same total force.... they'd go the same
speed? is there no relationship between the speed the water comes out and
the speed of the boat ie. it seems pretty obvious boat A could never go
faster than 3 knots, so boat B would never go faster than 3 knots either?

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.

sorry, i think i was just continuing on from something i was writing in a
previous post... i was meaning to say iff you had two pumps in parallel as
opposed to two pumps in series... two pumps in parallel would give you
double the GPH flowing, but having two in series would allow you to have a
higher velocity through a smaller outlet right? you can probably see where
im going with this, but it really does hinge on the question i was just
asking about the relationship between the speed of the flow and the speed of
the boat....

if a higher flow speed allows a higher boat speed, then it would seem
logical to me that you might get more boat speed by running two pumps in
series as opposed to parallel because you could then have theoretically a
much smaller outlet diameter than you could with parallel pumps, and
therefore a higher speed of water being pumped....

you're probably feeling like you're banging your head against a wall here...
but im sure ill get what you're saying pretty soon... all the math that
you're giving me looks right, i think there may be just some basic concept
that im misunderstanding?

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)

ok, wait i should have read this first and thought about it more.... so
there is a direct relationship between water velocity and boat speed, if you
can maintain the same volume of water flowing.... right? so the sweet spot
would be just before the pump starts to be slowed by the backpressure? now
this may be pure conjecture on all our behalfs, but assuming you could get
double the pressure (which you probably cant) at the same flow rate by
having pumps in series as opposed to parallel, and for the same current
draw, which boat do you think would go faster, the boat with double the flow
and half the speed, or the boat with double the speed and half the flow?
the total numbers add up the same right, but wouldnt the boat with higher
speed water jet go faster?

Shaun

snip Keith Hughes

Shaun

snip

If i understand what you're saying here, it sounds prettymuch
counterintuitive. I may be mis-using some of the terms? let me give an
example just to be sure that i understand what you're saying here, and bear
in mind that of course the numbers im going to use are entirely made up in
my head, so they'd be wrong....

It seems counterintuitive if you're thinking in terms of the discharge
stream "pushing against the water".

lets say that you have two identical boats with the same pump on each one,
running at whatever flow you like, say 5,000GPH. Boat A has a huge
outlet... say 5 inches in diameter. for arguements sake, because i dont
know how to calculate the speed of the water for that given outlet, lets say
the speed of the water coming out the back is slow. i dont know how slow,
but lets say it comes out at 3 knots.

Now boat B has the same pump, but the outlet is so small, that even though
its using the same pump, the water is coming out at a speed of 20 knots.
what you're saying is that both boats because they have the same amount of
energy put into them, and the same total force.... they'd go the same
speed?

Yes, that is the case. It's no different than if you used your hands,
and applied the same force to boat A, using only one finger, as you
apply to boat B, using the whole palm of your hand. It would "seem"
that you're pushing much harder with the finger than you have to with
the hand, but you're really not. With the finger, you have to apply a
much greater force *per unit of area* than with your palm (which has a
much larger surface area). For example, if you apply 10 lbf/sq.in. to
one square inch on Boat A, and 1 lbf/sq.in. to 10 square inches on Boat
B, the total force applied to both is the same, and the resulting
acceleration would be the same (assuming the same time interval of force
application).

is there no relationship between the speed the water comes out and
the speed of the boat ie. it seems pretty obvious boat A could never go
faster than 3 knots,

It may seem obvious, but that is incorrect. The discharge water is being
*accelerated* to 3 knots faster than the intake water. So there is a
constant force being applied that is totally independent of boat speed.
If there were no friction (and bow waves, etc.) the boat would
continue to accelerate indefinitely (well...see below). This is for an
axial system, where water comes in the bow, leaves the stern, the only
practical way to do it. As I said earlier, if your intake isn't pointed
forward, then you have Bernoulli effects, and you have pump cavitation
problems that reduce the flowrate. In the axial configuration, the boat
speed increases system efficiency by pressurizing the suction line and
overcoming the intake line pressure drop. At some speed, you'll reach a
point where the pump cannot maintain an acceleration of 3 knots (outlet
vs inlet), and your thrust will drop from that point on.

snip


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)


ok, wait i should have read this first and thought about it more.... so
there is a direct relationship between water velocity and boat speed, if you
can maintain the same volume of water flowing.... right?

Exactly. It takes more pressure (force per unit area) to get that
higher velocity, so you have to do more work on the system (by the
pump). That additional 'work' is then available in the form of thrust.

so the sweet spot
would be just before the pump starts to be slowed by the backpressure?

Just before the discharge *volume* (mass flow techically) decreases.
Keep in mind that shrinking the nozzle is not "free", since that creates
higher pressure requirements, and thus higher load on the pump (i.e.
more amp draw).

now
this may be pure conjecture on all our behalfs, but assuming you could get
double the pressure (which you probably cant) at the same flow rate by
having pumps in series as opposed to parallel, and for the same current
draw, which boat do you think would go faster, the boat with double the flow
and half the speed, or the boat with double the speed and half the flow?
the total numbers add up the same right, but wouldnt the boat with higher
speed water jet go faster?

Nope, 'cause it's the accelerated mass of water that supplies the
thrust, not the discharge water 'pushing against' anything. Does the
boat you push with your finger (above) go faster than the one you push
with your palm because your fingertip is applying so much more pressure
(over a small area) than your palm is? No, if the total force applied
is the same. It doesn't matter whether it's great force on a small area
(fingertip, small water stream) or a lesser force over a greater area
(palm, large water stream) if the total force is the same, the thrust is
the same.

Keith Hughes

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

Keith Said:
.... 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.

THAT makes a lot of sense, (a 3 or 4 to 1 ratio) and gives us some
real-world numbers to think about... And the following implies that a
decent experiment could be done by using a maximum-outlet-diameter
smooth hose to the outlet, and then fitting different experimental
nozzles:

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.

Let me try an approximation based on the above, looking at at my idea
of running a large? inboard pump connected to my existing marine
engine thru an air-conditioning compressor clutch, and piping it thru
a control valve to vary bow thrust port-to-starboard.

30 Amps at 12 V gave maybe 7 pounds thrust. That's about 1/2
horsepower. Say I can use 2 HP (What I understand a car air-
conditioner uses) to a pump with the same losses as the example Keith
showed. So maybe that's 28 pounds thrust. That sounds like plenty
for a 25 foot boat...

Question: How much thrust do typical electric bow-thrusters give in
the smaller sizes?? (We'd expect them to be more efficient).. BTW,
they are expected to be used at close-to-zero hull speed, so the
thrust measurement is reasonable here.

Maybe I can try some of this out this Summer on Lake Champlain
(Vermont) . (Now I'm boatless :-( on the Med this year, but moving to
the shore of the South China Sea for the next 2 or 3 years where I
WILL Mess With Boats!).

Interesting discussion!

Regards, Terry King ...On The Mediterranean in Carthage

"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