Calculating thrust

[drmiller100]

I'm trying to calculate thrust from a jet.
If the pressure inside of the bowl is 125 psi, how fast is the water coming out of the nozzle?

Next, if the nozzle is 3 inches in diameter, I should be able to calculate GPM. Suggestions on where to find the formula?

[robherc]

Well, you'll have about 883 lbs of thrust from your pressure differential, but I'm not sure on the flow rate/velocity measurements...maybe you could find formulas for them if you "google it."
(3" dia circle = 7.069in^2 area x 125lbs/in^2 = 883.625lbs force)

[drmiller100]

oh wow.
is it really that simple???

thrust = nozzle area * PSI

[robherc]

That gives you the pressure imbalance that you're creating in the jet chamber.
There are 7.069in^2 feeling 125psi pressure on the front of the "bowl" in your jet motor that are NOT balanced by any area at the rear.

That said, I THINK there may be some effect on that number by the mass & velocity of the discharged water. That might just be another measurement that'll give you the same number though, I'm not sure. It seems to me that it'd make sense that way, but I'm just guessing now. ;)

[drmiller100]

here is another thought. do you use the diameter of the nozzle, or the diameter of the impeller???


i found this site.
http://www.grc.nasa.gov/WWW/K-12/airplane/propth.html

[drmiller100]

here is another equation
thrust = rho x pi x D x D x .25 x Vjb x (Vjb-Vbw)
where
D is nozzle diameter
rho is water density
Vjb is water jet exit speed
Vbw is speed of boat

This was provided by a guy a lot smarter then me from this list.

[Olav]

You can calculate the thrust from a waterjet drive using the principle of linear momentum.

There is a mass flow m' = ρ * AJ * uJ, with

ρ = water density
AJ = nozzle outlet cross sectional area
uJ = waterjet velocity

which gives the thrust force when multiplied by the velocity difference between advance velocity uA (at the jet intake) and the aforementioned waterjet velocity uJ at the nozzle:

T = m' * (uJ - uA).


Aside from "real" physics there is a regression formula (see attached Excel spreadsheet) developed by DENNY and FELLER which calculates the thrust of a waterjet for a given engine power, inlet diameter, and boatspeed. I checked some calculated values against actual waterjet performance curves and found them to be quite accurate.

Edit: Oops, too slow...

[robherc]

here is another thought. do you use the diameter of the nozzle, or the diameter of the impeller???
No, the impeller diameter is meaningless to thrust if you know the internal pressure of the chamber. The impeller has the same pressure against it from inside the chamber as do the chamber walls...the only real differential pressure is at the outlet nozzle.
That said, if you DO NOT know the pressure inside the "bowl," then your mass & velocity equations are necessary.

[drmiller100]

Aside from "real" physics there is a regression formula (see attached Excel spreadsheet) developed by DENNY and FELLER which calculates the thrust of a waterjet for a given engine power, inlet diameter, and boatspeed. I checked some calculated values against actual waterjet performance curves and found them to be quite accurate.

Edit: Oops, too slow...

Olav
Thank you! I tried to find the document mentioned online, and can't.

Where it says "Inlet Diameter", perhaps they really mean "Outlet Diameter", or "Nozzle Diameter?"
What is interesting is you enter the horsepower (which makes sense), and "inlet diameter", and it calculates thrust velocity. But, thrust velocity should be a function of Output diameter (nozzle) and NOT inlet area.

Thoughts?

Other then that, it is a really cool spreadsheet, which I agree shows very similar trends to the real world.

[robherc]

2 notes:

1. What I told you was based on theory, and I do not have mathematical formulas, or test results to prove it, so please take it for what it's worth.

2. I believe the intake diameter WILL have some factor in figuring throughput of a jet based on engine power....though I am still VERY sure that outlet diameter will be important as well.

[Olav]

drmiller,

admittedly I don't have DENNY and FELLER's paper either, but the formula is cited in a paper by Donald L. BLOUNT and Robert J. BARTEE called "Design of Propulsion Systems for High-Speed Craft" (1996).

From the drawing presented there (a longitudinal section view of a waterjet) the term "inlet diameter" refers to the cross section in the plane prior to the impeller, therefore inlet diameter is approximately equal to the impeller diameter. English is not my native language, but maybe there is a difference between "inlet" (= the entrance of the impeller chamber) and "intake" (= the opening in the hull bottom)?

You are right that the nozzle diameter is an important parameter for the jet velocity; however it's actually the ratio between inlet (as per definition from above) to nozzle diameter - I guess the formula assumes a fixed ratio to calculate the acceleration of the water.

[drmiller100]

2 notes:

1. What I told you was based on theory, and I do not have mathematical formulas, or test results to prove it, so please take it for what it's worth.

2. I believe the intake diameter WILL have some factor in figuring throughput of a jet based on engine power....though I am still VERY sure that outlet diameter will be important as well.

No worries, and thanks for offering your thoughts!
The jetpump is the simplest thing in the whole world to look at and operate, but it sure is complicated to derive the physics and math behind it.

[speedboats]

As was stated, inertia is a product of mass x velocity (or p=mv)

Remember that you can get the same inertia by reducing the mass and increasing the velocity, (and vice versa), but doing either will dramatically change the performance of the vessel.

For example, our boats are relatively light weight so we can reduce the nozzle diameter and flow less mass but higher output velocity. Now as you cannot go faster in one direction than the mass going in the other (Newton was a clever fellow), a higher output velocity will give the vessel more speed. This is however at the loss of low end 'grunt' as you havn't the mass to get things accelerating in a hurry.

There is a balancing act here, once you fail to flow enough mass even an increase in output velocity will not overcome specific drag. This drag is different for every boat, boat loads etc, so it is something that is developed in boat trials...

[drmiller100]

I spent a few hours playing with the excel spreadsheet provided by Olav. The spreadsheet does a great job with at least magnitudes and trends. I don't know if it describes actual numbers, but I don't think there is any way to validate it anyway.

I am pretty well convinced based upon this and the various thrust equations that the formula is not a simple matter. The simple formulas do not even come close to describing or explaining what happens in the real world.

In the real world, thrust from a jet pump goes down as you go faster. This is described in the formula I posted, and even better with the excel spreadsheet.
In the real world, for a given horsepower, the larger the nozzle the more thrust at less then 10 mph. Again, predicted in the spreadsheet.
In the real world, smaller nozzles go faster (above 70 mph) then big nozzles, predicted in the spreadsheet.

Does anyone have a copy of the Blount or Denny article they could send me?

Thanks,

[Olav]

The aforementioned paper by BLOUNT and BARTEE can be downloaded from the Donald L. Blount and Associates' web site (as well as lots of other interesting publications): PDF, 567kB (http://www.dlba-inc.com/photos/pubs/22.pdf).

[drmiller100]

I have spent many hours playing with your spreadsheet OLAV. In my opinion, the spreadsheet closely follows observable behaviors. I can't say the exact numbers are right, but it sure predicts well.

as an example, set your speed to about 80 feet per second, 350 horsepower, and start playing with nozzle size. If you make your nozzle size too large, you end up with negative thrust in a big hurry, which follows the real world.

Further, if leave nozzle at .25 feet (realistic high speed nozzle), change speed down to 20 mph or so, and note the thrust. Then change your nozzle size to 1 foot, and note the HUGE change.

Different nozzle sizes make a HUGE difference in pump characteristics.

I am currently reading frantically about propellers. I think there might be some efficiencies to be found in them.

[Doc Nozzle]

I've decided to take a crack at thrust formula and how/why a jetboat or PWC moves through the water. You can view or download my tech handbook on jetpump thrust at: www.mdotfx.com

I welcome any and all thoughts, questions, challenges, et cetera. I don't profess to know anything - I just profess to want to try!

Cheers! :)

[drmiller100]

I can't get it to download!!!

[Doc Nozzle]

Sorry that the site is so basic! I think you just have to right click on the funny little thing with the red arrow above it next to the file name: Jetpump Technical Handbook Vol1 vA.

When you right click, you can select "Open Link in New Tab" (or window). That should get it displayed for you and then you can print or save.

Hope that helps!

[Rick Willoughby]

I've decided to take a crack at thrust formula and how/why a jetboat or PWC moves through the water. You can view or download my tech handbook on jetpump thrust at: www.mdotfx.com

I welcome any and all thoughts, questions, challenges, et cetera. I don't profess to know anything - I just profess to want to try!

Cheers! :)

There is a simple formula back at post 6 that will get you a much more accurate answer than the 6 tonnes of thrust you arrive at for the jetboat.

I will work with metric using the formula because it does not have the confusion of units that has created your error. (I will let you work it out)

The 2360gpm through the 3.125" (0.079m) nozzle produces a flow of 30m/s. This is all you need to know to get the thrust. From the previous post #6:
thrust = rho x pi x D x D x .25 x Vjb x (Vjb-Vbw)

We are considering static thrust so Vbw is zero.

thrust = 1000*pi*0.079^2*.25*30^2 = 4411N (990lbf)

I think you will find this much more realistic than 13,293lbs. That sort of thrust would push the transom through the bow. With a boat weight of say 1000lb you would pull 13g - not plausible.

Rick W