pump thrust

[Corey7320]

i am a structural guy so i am not super familiar with pumps but have some bakground knowledge from my civil engineering curriculum. Here is my question. I am trying to devise a propulsion system using a small gas powered water pump but can not convince myself it will work. I am trying to move a small sailing catamaran that weighs about 700# max with crew when the wind dies. i think i need somewhere between 30-50 lbs of thrust based on what electric trolling motors provide. I have been looking at small gas powered water pumps that you can buy pretty cheap that are from 0.8hp up. The info they give for these pumps vary but for example one i am looking at is a 0.8 hp pump with 1.5in dia output port, 114ft overhead lift, 26ft suction height, 65 gal per min volume. what can i use to calculate the thrust? Also, if i change the output port is there an optimal configuration for thrust? Your help is appreciated.

 

[BigInch]

I'm fuzzy on thrust, so I'm stretching it here,

At 65 gpm you have about 9 lbs/sec of water
through a 1.25" id +/- => 17.6 fps velocity

Discharge water kinetic energy
KE = 1/2 mv^2 = 1450 lbs-ft2/sec2

assume boat velocity = 5 fps

power = 1450/5 = 290 ft-lbs/sec
= 0.53 Hp (at 5 fps)
accounting for pump efficiency (0.65)
then 0.53/0.65 = 0.8 HP, so looks about right at 5 fps (3.4) kn speed.
Thrust = 290 ft-lbs/sec / 17.6 fps / 5 fps
= 16.5 lbs thrust

Now the question is, can you get 5 fps out of your hull with 16.5 lbs thrust?

Reducing the outlet diameter, increases velocity and thereby KE increases by the square of velocity,
BUT that would also increase backpressure on the pump and reducing flow, which has a linear affect on KE.
That seems the way to go is low mass-high velocity, but lowering flow with a smaller discharge nozzle decreases pump efficiency, if the pump is not designed for the new flowrate.
Better to buy a pump with low flow, high velocity BEP and not change the outlet diameter.

The problem I think I see is the 26 ft of head needed for suction. You'll need to have the pump hanging 26 ft below the waterline to get that. Hope you're a blue water sailor.

Now my advice is to wait and see if anybody wants to critique my fuzzy physics.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[JRLAKE]

I spent some time designing jet ski propulsion systems in grad school.

The question marks with this setup, as Big Inch implied, is the inlet conditions. You need to account for the effect of your vessel's velocity at the inlet of the pump. The inlet nozzle design would therefore have a significant effect on the thrust, so would drag. I would assume drag is small at the low speed you wish to achieve on a catamaran, but I have no idea how low. Since you assumed 50 lbs of thrust is necessary to accelerate you to your desired speed, I would stick with that since your logic with that number is sound.

Unfortunately, an end suction top discharge centrifugal pump's inherent design is not ideal for this application. And since most small gas powered water pumps are of this design, I assume this is what you have. I would consider and axial flow design pump, but I realize this is just for giggles.

Therefore I would simply assume a simple F=ma solution with V1 = 0.

 

[BigInch]

I assumed there is some reason that prohibits using an axial flow pump.

The 26 ft inlet head could be provided by scouping up water at 20 mph, but that would require a more powerful pump to reach that velocity.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[JRLAKE]

Yea I was thinking that too. I was also thinking any such scoop from the fluid stream to the pump, after friction losses, elevation change and additional drag, would make any incoming energy insignificant to the control space at such low velocities.

 

[Corey7320]

Thanks for you thoughts all. The type of pump i was trying to use is pretty cheap and i assume its made to move water efficiently from one location to another. i figured the 26 ft of suction was how far the pump could pull water vertically (if you wanted to move water from a low tank to a higher tank and needed the pump on the high side) am i mistaken about that pump stat? i found one source that gave the simple equation of (fluid density [kg/m^3])*(mass flow rate [m^3/s])^2/(area [m^2])= force [N]. going through the unit conversions using that formula i got 3.31 lbf of thrust which is not much from a 0.8hp motor.

 

[JRLAKE]

My guess is the 26' suction height is the pump's ability to prime. Is it a self priming pump?

 

[Corey7320]

Yes these pumps are typically self priming. see website if interested.

http://www.gasoline-engine-generator.com/water-pump/gasoline-water-pump-gwp033-25.htm

or

http://www.duropower.com/item.asp?PID=22&FID=8&level=1

 

[Corey7320]

i found a website that gives a pump curve. The engine is much bigger 2.5 hp

http://www.steadypower.com/catalog/honda_wx15.php

 

[JRLAKE]

Why are you not considering a small outboard motor? Just curious. That's probably too easy, I understand completely

I am pretty sure the suction head number is a reference to the pump's self priming ability. It sounds a bit high to me. The other likely possibility is that they are giving you a design point, "With a 26' suction lift, you will get X GPM and Y feet of additional head beyond that." I think marketing people do this for people that are scared of curves.

As far as your discharge nozzle arrangment, you should ask yourself which is more important, top speed or acceleration. The larger the discharge nozzle, the higher the acceleration and the slower the top speed. And vice versa for smaller diameter nozzles. I would keep it on the wide side.

For simplicity I would estimate this pump being half as efficient as a propeller. Therefore I would double the HP of the pump as compared to a propeller design. I would spend more time enjoying the sunshine and the family.

 

[JRLAKE]

...and I think Big Inch is right on the money with his original thoughts. I would go high flow and low head pump (aka axial flow impeller type pump). It's basically a classic boat propeller setup.

 

[hydtools]

Sir Hamilton's discovery that resulted in his orginal centrifugal pump working as a propulsion unit was that the discharge had to be above the waterline.
Here is a link to some of Hamilton Jet papers on how jet pumps work. Thrust is directly related to mass flow rate and the change in velocity from intake to outlet.

http://www.hamiltonjet.co.nz/hamiltonjet_waterjet/jettorque

Ted

 

[JRLAKE]

Sir Newton may be offended that Sir Hamilton is taking credit for F=ma.

 

[hydtools]

I don't think Hamilton is taking credit for F=ma. He just realized that his water jet propulsion worked better when the nozzle was above the water level rather than when submerged.
Sir Newton would be pleased with the application of action and reaction.

Ted

 

[hydtools]

Let's make some assumptions. The pump performance is 65gpm at minimum head, flow right out the 1.5" port, and submerged intake.

Thrust is = mass flow rate * exit velocity. Intake velocity is 0 fps. F = dm/dt * (v2 -v1), v1 = 0, v2 is water exit velocity. dm/dt is mass flow rate, 9.04/32.2
Thrust is then (9.04/32.2) * 11.77 = 3.3 lbs. The pump will not give you 30 - 50 lbs of thrust.

The force formula cooks down to (gpm^2/dia^2) * 0.0018. dia is the outlet diameter in inches. Reducing the discharge diameter will increase thrust if the pump can deliver 65gpm through the smaller area. To get 30 lbs or 10 times the thrust you must increase the gpm 3.2 times (sqrt of 10). Or reduce the diameter by 0.316. I dought the pump will push 65 gpm through a 1/2-inch nozzle.

Mess with gpm and diameter to find what pump performance you need to get the thrust you want.

BigInch you need to divide by g, 32.2, to get mass in your KE equation.

Ted

 

[BigInch]

Then I'd have slug-ft2/sec2.
Mass flowrate = 9 lbm/sec
Imagine the boat is in outer space where there is no gravity.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[hydtools]

No, then you would have ft-lbs, the units for energy.

9 lbs is not mass, it is weight, a force. You determined this weight rate by using water weight density, not mass density. You must divide by g to get mass units.

lbs-ft^2/sec^2 are not energy units. Divide by g, ft/sec^2, and your units become proper energy units, ft-lbs.

Ted

 

[BigInch]

Yes I agree I missed the 32.2. Thanks for picking up my mistake.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain