Calculating thrust

Discussion in 'Jet Drives' started by drmiller100, Feb 15, 2009.

  1. drmiller100
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    drmiller100 Junior Member

    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?
     
  2. robherc
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    robherc Designer/Hobbyist

    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)
     
  3. drmiller100
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    drmiller100 Junior Member

    oh wow.
    is it really that simple???

    thrust = nozzle area * PSI
     
  4. robherc
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    robherc Designer/Hobbyist

    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. ;)
     
  5. drmiller100
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    drmiller100 Junior Member

  6. drmiller100
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    drmiller100 Junior Member

    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.
     
  7. Olav
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    Olav naval architect

    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...
     

    Attached Files:

  8. robherc
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    robherc Designer/Hobbyist

    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.
     
  9. drmiller100
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    drmiller100 Junior Member

    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.
     
  10. robherc
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    robherc Designer/Hobbyist

    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.
     
  11. Olav
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    Olav naval architect

    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.
     
  12. drmiller100
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    drmiller100 Junior Member

    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.
     
  13. speedboats
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    speedboats Senior Member

    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...
     
  14. drmiller100
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    drmiller100 Junior Member

    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,
     

  15. Olav
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    Olav naval architect

    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.
     
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