Finding optimal propeller dimension based on ships speed and hull resistance

Discussion in 'Hydrodynamics and Aerodynamics' started by Simme_swede, Apr 2, 2024.

  1. Simme_swede
    Joined: Apr 2024
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    Simme_swede New Member

    Hi


    I’m researching for my thesis on how to find the optimal propeller for an electrified version of a current boat. What I have: Hull resistance, weight, speeds of the vessel.


    My thoughts of process:

    I know the resistance of the boat, therefore I also know at least how much thrust I need to propel the boat. But how does torque come into play? How can I calculate the required torque needed to go 5 and 8 knots? '


    Any help is appreciated

    Simon & Fredrik
     
  2. hashtag_laeuft
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    hashtag_laeuft Junior Member

    Hi Simon & Fredrik,

    the torque highly depends on the size of the propeller and if it is designed for high or low RPM. As you know from your bicycle you need more torque to climb a hill with a high gear as compared to small gear.
    So very important is also the specification of your engine.

    To have a good starting point for your propeller design I would recommend you to have a look into the Wageningen B Series.
    There is also an onlinge propeller generator -> Home https://www.wageningen-b-series-propeller.com/

    This might give you a good idea of which information are required to find a suitable propeller .. at least a first shot.

    Best regards
    Nicolas
     
  3. Simme_swede
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    Simme_swede New Member

    Thank you for the fast answer!


    A part of our question for the thesis is to also select a suitable DC engine. We are aiming to preserve as much current as possible but we are limited to stay under 60v. Preferably low RPM, since voltages are often linear with the rpm.



    upload_2024-4-2_11-17-7.png


    We are somewhat using this model as an example, excluding the gearbox.

    Would it be a good starting point to find the minimum power needed for the propeller to achieve 8 knots?

    We have a datasheet of a current combustion engine that is in the boat right now.

    upload_2024-4-2_11-26-10.png


    The boat guy that gave us the job says that it achieves 5 knots on 2-2.5 k RPM and 8 knots around 3 k RPM.

    Is it a good aiming point for us to go for the same torque used by the diesel engine or does it change due to the fact that we have a DC engine?

    The DC engine can have lower RPM with just as much torque, therefore my thoughts are that the propeller can be shaped quite differently compared to the combustion engine.
     
  4. hashtag_laeuft
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    hashtag_laeuft Junior Member

    Hi ..

    it all depends on the size of the propeller. The torque of a propeller is related to the diameter with the power of FIVE.
    So, if the geometry of the boat, position / inclination of the shaft & the tip clearance (min distance between propeller tip and the hull) would allow a bigger propeller, you could aim for a slightly bigger one, in order to reduce the rpm.

    The question is, if you are looking for a theoretical specification of a new propeller to include that in your thesis, if you would have the possibility to manufacture a newly designed propeller or if you need to know best suitable specifications to choose the right stock propeller.

    As I said, as a starting point you can check the wageningen b series, in order to get a first impression of what would be possible. Or similar propeller series.
    Please find attached a document with the Kt, Kq and Efficiency curves of the b series propellers.

    Kt (thrust coefficient), Kq (torque coefficient) and J (advance coefficient) are your best friends in order to find a suitable propeller for your project. They are explained in the introduction on page 13 of the document.


    Best regards
    Nicolas
     

    Attached Files:

  5. jmwoodring
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    jmwoodring Junior Member

  6. jehardiman
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    jehardiman Senior Member

    Welcome to the Forums.

    As Nicolas implies, all the data you have shown so far really doesn't show the total picture. Is the existing diesel grared to the propeller shaft? What is the existing propeller? If the diesel is not geared, the data indicates that the existing propeller is small, shaft speed is high, and therefore the slip is large. This is inefficient. Most electric launches in the speed range you ask for have historically had large slow turning propellers (typically something like a B 2.30 or B 3.30 from the referenced paper). Therefore you should enter this exercise with the largest diameter you can fit into the hull form.

    Additionally, you need to understand how power is transferred from the motor to the water. It is a function of how a boat draws power, a propeller delivers it, and a motor provides it.
    (NAs out there, let us keep it simple ...OK)

    Boats need horsepower as thrust (T) at speed (V). This is effective horsepower required (ehp =(T*V)/550 with T in lbs, V in ft/sec)
    Propellers produce T as a function of blade area (A), diameter (D) and pitch (P) times propeller rps (n) when there is slip (s), i.e. slip ratio s=1-(V/n*P). Think of a large s as a large angle of attack of the propeller blade.

    Propeller absorb horsepower as a function of Torque (Q) and n, where Q is a function of A, D, and s. This is the delivered horsepower (dhp = (Q*n*2*pi)/550)
    Motors provide horsepower as a function of motor torque (q) and rpm. This is shaft horsepower. (shp = (q*rpm)/5252)
    Because n and rpm are locked by gearing (if any), the torque (q/Q) relationship is the inverse of the gear ratio (60*n/rpm) = (q/Q).
    And typically shp>dhp>ehp where dhp = 0.95*shp and ehp = 0.75 dhp (or greater for larger D and lower n).

    So to produce an exact T at an exact V, there are many combinations of D, P, A, and n.
    However at any P, the n interacts with the V to generate a specific s, which sets Q....larger s, larger Q
    Given Q for that specific A, D, P, s and n then the motor q output for that combination of n and rpm needs to exactly match, or exceed the required Q, this is called the operating point.
    1) Too little motor torque, and the motor slows down (i.e. "lug" or overload the motor) until a) the Q is supportable, or b) you stall the motor.
    2) Too much motor torque and the propeller speeds up (i.e. the motor over speeds) until either a) the motor control design limits n, b) s and Q increase to match motor output, or c) the prop tears the water apart and the prop fully cavitates in which case Q falls to effectively zero and the motor runs away.
    3) Increase the A, D, or P, you increase the Q at a given n, see 1)
    4) Decrease the A, D, or P, you decrease the Q at a given n, see 2)

    Rather than use the Kt, Kq charts referenced by hashtag_laeuft; you might consider using the older Bp-delta form of the data which allows you to calculate optimum propeller diameter for a given propeller type and desired operational point. Some of these (a B 2.38 and 3.50) can be found in the 1966 edition of Principles of Naval Architecture or many more in the original Netherlands Ship Model Basin work (i.e. the Wageningen/Troost series "Open Water Test Series with Modern Propeller Forms").

    Edit to add: BP-delta Chart | PDF | Propeller | Aerospace Engineering https://www.scribd.com/document/489717128/BP-delta-Chart
     
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  7. Ad Hoc
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    Ad Hoc Naval Architect

    To add to JEH's excellent summary.

    Once you read JEH's post you'll understand that there is no such thing as "optimal", other than in the minds of managers or students.
    Why?..because it this "condition" (the one being seeked) can only be true at one point in time at one condition, one speed, one rpm etc etc.
    Once any of the multitude of variables is changed - which it shall - it is no longer working at its "optimal" design point - it has moved from said minima/maxima.
    Ergo, it is no longer at "optimal".

    That should really be the thrust (no pun intended) for your thesis.
    Demonstrating the number of variables that come into play and hence - what is optimal?
     
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  8. BlueBell
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    BlueBell . . . _ _ _ . . . _ _ _

    The "best" compromise, perhaps?
     
  9. Ad Hoc
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    Ad Hoc Naval Architect

    Do you mean..."best" or "compromise"..?

    Im now wholly sure either would be appropriate replacements.
    I just think such issues can be better framed than "optimal", is it gives a false sense of what is really happening and/or being achieved.
    Engineering and design across multiple disciplines cannot be boiled down to a one liner as a 'justification' or 'result'....
     
  10. baeckmo
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    baeckmo Hydrodynamics

    Simon/Fredrik, at what institution do you study, and what is the level of your thesis? I'm asking, because the answers to your quest depend on that. Although I concur with the recommendation to start with the "classical" Bp charts, there is a better way once you get the hang of it. As noted above, there are many factors to deal with, and you have to understand what happens at off-design conditions. I strongly recommend that you read the following article:

    "Small-Craft Power Prediction" by Donald L. Blount and David L. Fox. It was published in MARINE TECHNOLOGY Vol 13, No 1, Jan 1976, pp 14-45. Years ago, it was available from the library at Chalmers uni Gothenburg.

    By using the suggested format there (Kt/J^2), the dominant design driving factor; the diameter, is boiled down to a simple function of (thrust/speed^2). The efficiency can be plotted as function of Kt/J^2 for each blade area ratio ("BAR"), pitch/diameter ratio and cavitation condition, and you can select what level of compromize in efficiency you have to accept to cover the necessary operating conditions. For electrical propulsion with energy storage in batteries, the efficiency is the primary operational driving factor, and you will most probably select from low BAR, say ~0,5 to 0,65, and sigma ("cavitation number") atmospheric (or maybe ~2).

    That gives the diameter required for the conditions, and the efficiency. Since the effective power is (roughly, to keep it simple): thrust * speed of advance/efficiency; you have the shaft power. And since you have : Kt=T/(density*N^2*D^4), you have the shaft speed (N), and from the power/(shaft rotational speed) you get the required torque. Just remember using the correct dimensional system all the way!
     
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  11. rwatson
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    rwatson Senior Member

    Although this is a Model Building Exercise, it mentions several software tools that would be helpful

     
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