Definitions of jet thrusts

Discussion in 'Jet Drives' started by Ben Land, Apr 1, 2015.

  1. Ben Land
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    Ben Land Junior Member

    I am having hard times understanding which of the following equals boat total drag at certain speed (required thrust): net thrust of gross thrust. I am aware of that there are different definitions for these thrusts; Terwisga: Tnet=Tgross*(1-t). And Allison:T gross= mass flow* fluid velocity in jet, Tnet=mass flow*(fluid velocity in jet-velocity of vessel).

    In Allison version the net thrust and the gross thrust seem to have rather big difference, unlike in Terwisga version where they are almost equal. Which one is commonly accepted definition?

    My actual question here is that if I know the boat drag at certain speed which thrust in Allison definition ;net or gross should i consider equal to the total drag of the boat ( or required thrust) in the name of estimating the forces coming out of the nozzle?

    Thanks in advance

    Link to Allison paper below:

    http://academic.amc.edu.au/~psahoo/Research/Int.Workshop-AMC-2003/Gottaschalk/Allison.pdf
     
  2. drmiller100
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    drmiller100 Junior Member

    I'm trying to understand it as well, and take this as my interpretation.
    Gross thrust is nozzle area times PSI in the pump. It tells how much thrust the pump as a system makes at 0 speed.

    As your speed picks up, your gross thrust wont' change for a given RPM no matter how fast you go.
    HOWEVER, your NET thrust falls significantly. Net trust subtracts the inlet pressure and the work done to accelerate the water up to the boat's speed. You will reach a velocity where the net thrust is zero. In general, the higher the PSI of the pump the higher that velocity will be.

    BTW, thank you for the link to the Allison paper. For my limited understanding, that is the best resource I have run across to date regarding what is really going on with a jet pump and how it works.
     
  3. Ben Land
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    Ben Land Junior Member

    Great to know that there are other people struggling with the same question! :) I have noticed this phenomenon too, that the net thrust falls down when the speed is accelerated. But I still don't understand the relation between the two and the so called required thrust (which would be bit more than the boat drag) that determines the effective horsepower.

    I almost thought that the net thrust would be the equivalent to required thrust but required thrust should increase as the boat speed increases, right? So is it so that none of these are the req. thrust but instead it would be the third thrust value? The more I think, the more....
     
  4. drmiller100
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    drmiller100 Junior Member

    cant' tell you what "required" thrust is in relation to net.

    Here is how i look at it this week.
    You are cruising across a lake at 30 mph. You pick up a bucket of water, and throw it out teh back of the boat.

    If you throw the bucket of water out the back at 45 mph, you have thrust keeping you going.

    If you throw the bucket out teh back at 25 mph, you are movnig water across the lake in the direction you are headed, and slow down in a big hurry.

    So, now, you are sitting still in the lake. You can throw 5 gallons out the back at 45 mph. Or you can throw 10 gallons out the back at 30 mph.

    The second case will give you more thrust at standstill, but the first case may very well have a higher top speed depending on hull drag.
     
  5. Barry
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    Barry Senior Member

    In the thirty odd years of building jet boats, I have never ran across the term gross thrust and wonder what it is you are seeking
    To simplify the process,
    Thrust is generated when you take a mass and accelerate it
    For continued force,

    The formula would be F= M (mass flow rate) x acceleration

    So you can change either increase the amount of mass that you have flowing, for a specific acceleration ( which is a change in velocity) to increase thrust or make the change in velocity larger ( a higher rate of acceleration) with a fixed mass flow rate to increase thrust

    Not sure where you get the idea that thrust falls down as you go faster
    Boats require more thrust as their speed increases. (ignoring hump speed issues)

    When the boat is going as fast as it can, the thrust of the pump in pounds is equal to the all forms of the drag of the boat. ( assuming that the direction of the nozzle is horizontal)

    Ben your statement
    net or gross should I consider equal to the total drag of the boat ( or required thrust) in the name of estimating the forces coming out of the nozzle?

    This is not correct, the force coming out of the nozzle. The action of the impeller pushing the water, and the stators changing direction is what gives the water the change in velocity, not thrust coming out of the nozzle

    Say the nozzle and the impeller were the same diameter, you would still generate thrust, like a propeller. When you restrict the size of the nozzle, the pressure builds inside the bowl which increases the change in velocity, ie acceleration. But there is a limit as higher pressures require higher horsepower so you could not just reduce a 12 inch impeller/bowl to say one inch in diameter and expect to go anywhere.

    There are three basic types of pumps.
    Low volume, high pressure, centrifugal, which I don't think anyone uses this as a jet pump for a boat. (Hamiltons first entry into building jets was a centrifugal pump which was not effective)

    High volume, low pressure, axial flow pumps, the older Hamilton three stages pump is a good example

    Medium volume, medium pressure, mixed flow pumps, American Turbine, Berkley

    So if you just take the amount of water coming out the back of the pump, the mass flow rate, and take the exit velocity ( as the inlet velocity with respect to the lake is zero) ( horizontally, plug this into the F=ma equation, you will get the drag of your boat.

    This is simplified and omits a lot of complicated issues such as induced turbulence, flow vectors, curl, impeller loading, etc. This is a complex issue
     
  6. Ben Land
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    Ben Land Junior Member

    Thank you Barry, this was very informative! In the Allison paper I posted before is mentioned Gross thrust as well as net thrust. Also in one older Hamilton manual of 402 jet the thrusts are determined as followed

    Tnet:Tgross-Td
    Tgross:Q*rho*Vjet
    Td:Q*rho*Vboat, all results in newtons.

    As I understand it, the Tnet decreases as the boat speed increases where its plot cuts the plot of boat total drag and that point where they cross determines the boat max speed. And this chart would be drag (or thrust) versus boat speed.

    My guess is that the Gross thust is the power measured coming out of the nozzle, which as you said is not actually the thrust that moves the boat forward. But this is the force that gives the steering force when making a turn
     
  7. daiquiri
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    daiquiri Engineering and Design

    The nature knows nothing about our definitions. The ship will go as fast as the power input will allow it to go. The terms "net thrust" and "gross thrust" are just products of someone's necessity to simplify the calculations and find a way to apply the results from towing tanks measurements to full-size ships.

    Imagine a streamlined hull which is moving through the water at a speed V, pulled by a tow line. For the moment, let's ignore the wave drag and consider the remaining components of the ship's resistance. There will be a turbulent wake left behind the hull, created by the adverse pressure gradient in the stern area of the hull. This adverse (increasing) pressure gradient forces the water streamlines to detach from the hull and to create a wake. A dynamometer at the towing line will measure a certain resistance Rt, and we can calculate the required power as
    Pt = Rt x V (Eq. 1).​

    Now detach the towing line and attach a propeller (or a waterjet) at the stern. Things will change, and numbers will change too. The working propulsor will change the pressure field in the stern area of the hull, and will create a more favorable pressure gradient. It sucks the water in at the inlet and ejects it at the outlet. Hence, there will be a low-pressure area around the inlet, and the water streamlines will tend to converge to the inlet rather than to detach from the hull.

    The apparent paradox here is that the resistance of the hull alone will increase by adding a propulsor, because of the pressure decrease at the stern, but the overall propulsive efficiency will likely increase. This is how it happens.

    The increase of resistance of a self-propelled hull is expressed through the formula:
    Rt = (1-t) x T​
    where Rt is the resistance of the towed hull, and T is the resistance of the propelled hull (which is equal to the thrust - hence the symbol T).
    "t" is called "thrust deduction coefficient", and it's typical value is somewhere around 0.1-0.2 (depending on hull and propulsion layout).

    By the way, by comparing this to the formula in your last post we see that what you call "gross thrust" is actually the thrust effectively required by the self-propelled ship and what you call "net thrust" is the resistance of the ship towed by an external towing line.

    Now, what happens to the propeller (or a waterjet), once it has been placed in the wake of the ship? The water inflow to the inlet will be reduced (when compared to the open-water condition), because it works in the ship's wake. The average water speed seen by the propeller is calculated by the formula:

    Vp = (1-w) x V​
    where V is the ship speed (equal to the towing speed), and w is the so-called "Taylor wake fraction" (typical values between 0.05-0.3).

    Considering that the propulsive power of the propeller is calculated as:
    P = T x Vp (Eq. 2).​
    we can combine the eqs. 1) and 2) and obtain the following relationship:
    Pt / P = (Rt/T) x (V/Vp) = (1-t) / (1-w)​

    the last term is called "hull efficiency", and it tells us how advantageous will be the interaction of the propulsor with the hull.
    eff,h = (1-t) / (1-w)
    Pt = eff,h x P​
    If we take typical values of t=0.2 and w=0.3 (displacement vessels), we get:
    eff,h = (1-0.2) / (1-0.3) = 1.14​
    If t=0.1 and w=0.05 (fast displacement or planing vessels), we get:
    eff,h = (1-0.1) / (1-0.05) = 0.95​

    indicating that a towed hull can typically (but not necessarily) require more power than a self-propelled hull.

    Don't know it this is the explanation you were looking for, but I hope it will help.
     
    Last edited: Apr 2, 2015
  8. Ben Land
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    Ben Land Junior Member

    Now I got it! Thanks to you all and have a nice easter in case you celebrate it!
     
  9. daiquiri
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    daiquiri Engineering and Design

    I would also suggest you to read this nice and easy-to-read presentation, which explains the same thing, more in-depth and with some explanatory graphics: http://research.ncl.ac.uk/cavitation/archive/MAR2010 - prop hull.pdf

    It appears to be more oriented towards displacement ships, while this one (for example): http://www.dtic.mil/dtic/tr/fulltext/u2/a265818.pdf will give you some typical values of thrust deduction and wake coefficients for high-speed vessels. However, neither of them talks explicitly about waterjets. You will have to dig some specific info on your own. ;)

    Cheers
     
  10. daiquiri
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    daiquiri Engineering and Design

  11. drmiller100
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    drmiller100 Junior Member


    Interesting. Every other engineer, physicist, and math guy calculates thrust as

    F=MV and treats it as a momentum exercise. The OP provided excellent literature along those lines.
     
  12. drmiller100
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    drmiller100 Junior Member

    No.

    The article the OP posted clearly describes Net Thrust, Gross thrust, and the relatively trivial effects of the propulsion method on the hull.

    In fact, the article even goes into how the jet and prop have very different effects on the hull, and how the jet will have some beneficial sideeffects.

    If you do real number calculations, the difference between a self propelled hull vs towed hull is relatively small especially above planing speed.

    However, net thrust vs gross thrust is a huge effect, and net thrust approaches zero as speed increases as per my description above.
     
  13. daiquiri
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    daiquiri Engineering and Design

    Ok, thanks for the clarification. Actually, I did not read the article (my fault) and was only replying to the direct question asked by Ben Land, and was commenting the formula written by him. My fault, I should have checked first if the formulas in the OP were transcribed and described correctly.

    Cheers



    P.S.:

    Just read the article. Very good reading. At the page 282 it explains the formula Rt=(1-t)T in exactly the same terms as I did. And yes, the formula doesn't describe the Gross and the Net Thrust.

    The gross thrust Tg, as given by the formula in the Allison paper, is simply the maximum thrust obtainable by accelerating a mass of water from zero to a speed Vj (the exit speed of the jet). It is also a thrust at zero boat speed (bollard pull).

    The net thrust Tn takes account of the fact that the water is not getting accelerated from zero speed (relative to the boat) but from the speed Vs (the ship speed).

    Since the maximum Vj is limited (by material strength, by cavitation etc.), the difference (Vj - Vs) becomes smaller as the ship speed increases. Hence the net thrust (given by the difference between momentums of the mass of water at the nozzle and at the inlet) has to decrease with increasing Vs. When the net thrust equals the vessel's drag, the equilibrium speed Vs (and the final and non-zero value of Vj-Vs) is reached. As simple as that.
     
  14. drmiller100
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    drmiller100 Junior Member

    I had an idea.

    Take a propeller powered boat.
    Let's say the boat has very small drag, 10 pounds of drag at 30 mph.
    Let's say the propeller makes 100 pounds of thrust at zero mph.

    We turn the boat loose, and let it accelerate across the water? How fast does the boat go? At some point, the pitch and RPM of he propeller will mean the boat goes no faster on its own. Therefore, the THRUST of the propeller decreases at some speed, and the thrust of the propeller approaches zero.

    If we were to put a rope on the boat, and tow the boat even faster, the propeller would actually create DRAG - the boat is going faster than the propeller rpm and pitch will support.

    One way to look at it is the cone of water behind the propeller has a velocity. if the velocity of the water is going the opposite direction of the boat, RELATIVE to the calm water, the propeller is producing thrust. If the propeller is simply moving through the water, and no water is being moved, the propeller is not producing thrust.

    Likewise a jet. If the water out the back of the jet is not going "backwards" in relation to the free standing water, the jet is not producing thrust.

    The VELOCITY of the water out the back of the nozzle times the mass moved determines thrust as a function of boat speed.
     

  15. cmckesson
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    cmckesson Naval Architect

    Note that the Q and Vj values for the jet (and thus the gross thrust) are normally found by experiment or calculation for the jet operating by itself....with no boat in front of it.

    We then find that, with a boat upstream of the jet, the jet actually produces slightly more thrust than it does in the alone condition. For this reason the thrust deduction "t" is usually a negative number, say -0.05.

    KaMeWa's Rolf Svensson has published on this point, you might try Google Scholar or Compenex using Rolf's name.

    Chris
     
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