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

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

wrote:
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


To repeat my earlier post, my 30 lb Endura at 30 amp/12 volt pushes my
16' skiff or 14' Hobie at about 3.5 mph. My 50 lb Endura at 42 amp/12
volt pushes the same skiff at 4 mph. The force required as hull speed is
approached (for a displacement hull) increases astronomically. 28 pounds
might snuggle a 25 ft hull up to the dock, but it wouldn't do much in
open water.

BS

wrote:

Keith Said:

snip

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...

OK, I'm confused. Are you talking about *just* a bow thruster operation?
If not, why would you add another mechanically lossy system instead of
just using the marine engine? No matter what system you bolt onto the
engine, it will be less than 100% efficient at energy conversion, so
you'll just lose power in the process.

For bowthruster operation, should be easily doable.

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.

My understanding is that they are considerably more efficient than
typical outboards, since the "ring" around the prop eliminates a
significant amount of prop slip relative to having an open prop.

Keith Hughes

Jets (be they axial like real waterjets or centrifugal like a bilge
pump) are more efficient than larger diameter propellers ONLY when the
boat is moving fast enough that the drag from propeller
strut/shaft/and rudder becomes a significant part of the drag values.

In the real world that works out to be around 25-30 knots. A bilge
pump will never beat a small trolling motor propeller at displacement
speeds.

I've seen Cal 20's pushed by a big trolling motor for a sailing
school. Works o.k. in calm water but in any sort of breeze it's not
enough thrust. The Cal 20 with a 3.5 HP outboard was a LOT faster and
would end up towing the ones with a trolling motor.

For a Thunderbird, a typical 5-8 HP long shaft outboard is the only
solution that makes sense.

Evan Gatehouse

"Evan Gatehouse2" wrote in message
...
Jets (be they axial like real waterjets or centrifugal like a bilge pump)
are more efficient than larger diameter propellers ONLY when the boat is
moving fast enough that the drag from propeller strut/shaft/and rudder
becomes a significant part of the drag values.

In the real world that works out to be around 25-30 knots. A bilge pump
will never beat a small trolling motor propeller at displacement speeds.

I've seen Cal 20's pushed by a big trolling motor for a sailing school.
Works o.k. in calm water but in any sort of breeze it's not enough thrust.
The Cal 20 with a 3.5 HP outboard was a LOT faster and would end up towing
the ones with a trolling motor.

For a Thunderbird, a typical 5-8 HP long shaft outboard is the only
solution that makes sense.

Evan Gatehouse

hi Evan,
there may have been some crossed wires here.... the bilge pump/trolling
motor solution was for a 14 foot beach cat.

Shaun

Keith Said:
OK, I'm confused. Are you talking about *just* a bow thruster operation?

**Yes, ONLY Bowthruster, maybe also piped to SternThruster ?? Just for
smooth docking/undocking or dead-slow movement in calm water.. I can
and do bring my 22' Inboard CuddyCabin to shore / rock points with the
canoe paddle. But having a fully-controllable couple of horsepower
should allow total maneuverability. The marine engine would be in
neutral, just running the pump...

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.

My understanding is that they are considerably more efficient than
typical outboards, since the "ring" around the prop eliminates a
significant amount of prop slip relative to having an open prop.

Keith Hughes

Right! Apparently this is a significant factor in dead-slow
operations. There are huge (Kort Nozzles ?) on big tugboat propeller
installations. Interesting idea...

We've had some excellent information in several above posts. What I'm
getting from all this is:

- Inboard pumps to various outlet nozzles can be effective for
maneuvering and very slow speeds.

- Efficient forward-motion propulsion up to 3 or 4 knots is much more
efficient using external propeller (such as 'trolling motor' types).

My idea is a different one: I have conventional marine propulsion for
running underway, but I wish I could have good fine-control
maneuvering / docking / dead-slow sightseeing based on an inboard pump
driven from my regular marine engine. Especially with a conventional
fixed-propeller / rudder type boat that steers for c*** in reverse,
and is very difficult to turn in a short radius, this would be "Nice".

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

wrote:
Keith Said:


We've had some excellent information in several above posts. What I'm
getting from all this is:

- Inboard pumps to various outlet nozzles can be effective for
maneuvering and very slow speeds.

- Efficient forward-motion propulsion up to 3 or 4 knots is much more
efficient using external propeller (such as 'trolling motor' types).

I'd say that covers it.

My idea is a different one: I have conventional marine propulsion for
running underway, but I wish I could have good fine-control
maneuvering / docking / dead-slow sightseeing based on an inboard pump
driven from my regular marine engine. Especially with a conventional
fixed-propeller / rudder type boat that steers for c*** in reverse,
and is very difficult to turn in a short radius, this would be "Nice".

I certainly agree with that. Our Catalina 30 is in a very tight slip,
and we get a lot of prop-walk with our 3-blade prop. A couple of
thrusters would certainly be handy at times. Cheers,

Keith