Prop tip velocity

Discussion in 'Props' started by michaeljc, Jun 10, 2013.

  1. michaeljc
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    michaeljc Senior Member

    Can anyone out there give me some idea of max tip velocity on a 700 mm displacement trawler-type prop (cruise 5 knt)

    Thanks

    Michael C

    Floatainer Systems
     
  2. DCockey
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    DCockey Senior Member

    Tip speed in m/sec

    = Prop diameter (m) / 2 * Shaft speed (rpm) * 6.28 radians / 1 revolution * 1 min / 60 seconds

    = Prop diameter (m) * Shaft speed (rpm) * 0.0523 (radians/revolution*min/sec)

    This ignores the speed of the vessel. To add in the speed of the vessel use

    Tip speed with speed of advance =

    [(Tip speed without speed of advance)^2 + (Vessel speed)^2]^1/2

    with vessel speed in same units as the tip speed
     
    Last edited: Jun 10, 2013
  3. michaeljc
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    michaeljc Senior Member

    :) Wow, thanks!

    Actually not so wow - on looking at this it is giving me tip speed based on RPM. I am after typical tip speed for this type of prop from which I can establish RPM and reduction ratios from a 1800 rpm motor. I am looking for industry norms.

    Thanks anyway

    MC
     
  4. DCockey
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    DCockey Senior Member

    Note that the formulas above ignore the shaft angle, and also are the tip speed relative to an inertial frame of reference. For tip speed relative to the water flowing into the prop an adjustment is needed.
     
  5. michaeljc
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    michaeljc Senior Member

    I am looking for rotational tip speed around its axis - not speed through the water
     
  6. DCockey
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    DCockey Senior Member

    The tip is rotating at the same speed that the rest of the propeller and shaft are rotating.

    The "linear" speed of the tip relative to the boat is the first formula above.

    The second formula is the "linear" speed relative to the undisturbed water.

    What do you need the "rotational tip speed" for?
     
  7. baeckmo
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    baeckmo Hydrodynamics

    There are three basic factors that determine the tip speed; propeller diameter, rotational speed and power.

    The first approximation of their relationship is:
    "D = 630 * power^0.2/rpm^0.6" ; where "power" is hp, "rpm" = shaft revs per minute, "D" = prop dia in inches.

    In order to proceed from your 700 mm dia (27.6 in.) you have to know the shaft power in order to calculate the missing factor, i.e. the rpms. That done, you follow Davids instructions.

    Generally speaking, the propeller disc area is determined by the thrust requirements of the vessel, in terms of a limiting allowable pressure on the blading, which in its turn is linked to the cavitation limit and from there to the tip speed.

    An example: Your 700 mm prop is mated to an engine that feeds 250 hp into the prop shaft. This means that the shaft speed should be 1157 rpm. Translate the rpm into radians/second and you have 121.2 rad/s. Multiply with prop radius (0.35 m), and you have your tip speed = 42.2 m/s. Corrected for advance speed 5 knots, you have a relative tip inflow velocity of 42.5 m/s.

    The dynamic pressure head corresponding to this speed is 92 meter water column. In order to avoid cavitation, the minimum pressure must not be reduced below vapour pressure of water at the propeller. With water at +20 centigrades, the limiting head is close to 10 m. The pressure constant at the tip leading edge must then be less than 10/92 = 0.11, which is a low value for ordinary blade profiles, telling me that this propeller will probably have cavitation issues with 250 hp. If we add one meter of water above the propeller disc, the pressure constant becomes 11/92 = 0.12; still too low.

    This is roughly the procedure, admittedly with a few intermediate steps excluded for clarity.
     
  8. daiquiri
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    daiquiri Engineering and Design

    Yes it will, and I know it from experience. :rolleyes:
     
  9. michaeljc
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    michaeljc Senior Member

    'Your 700 mm prop is mated to an engine that feeds 250 hp into the prop shaft. This means that the shaft speed should be 1157 rpm'

    This is an assumption right? I don’t know yet, but I am expecting craft that maximise thrust at low hull speed will have greater reduction ratios than 2:1

    An alternative to calculation is to take data from industry norms. There is a very good book on boat design written by an American which outlines variables in existing designs based around an average. He covers everything like scantling spacing, beam dimension, hull thickness, B:L ratio etc etc in the form of tables and graphs. It is most useful because when designing as one can compare the design with the norms and establish if we are pushing the envelope too much. As in all data relating to real-world examples there is a bell curve. I have this book but have mislaid it for the moment. I will find its title and post it here.

    Anyway I am taking this approach to establish reduction ratios for low hull speed, high thrust props. I would be very surprised if there are not graphs out there somewhere that gives acceptable tip velocity for a range of pitch and diameters. I know the BHP and the typical diameter of props being used in my application.

    I was in India recently where there were hundreds of trawlers using 150 HP diesels and props ranging from 500 to 750 mm diameter. I will shortly get info on what gear ratios they are normally running. I am picking it will be 3 – 4:1. There are a range of props being cast. By now hundreds of operator’s experience will have established what works best. We can expect that the designs will be conservative, staying well away from a cavitation risk, and that: as diameter increases tip velocity will decrease to avoid vibration.

    Thanks for the formula. They are useful.

    Thanks

    MC
     
  10. michaeljc
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    michaeljc Senior Member

    This is the book. It is available in the book store on this site

    The Elements of Boat Strength: For Builders, Designers, and Owners
    by: Dave Gerr
     
  11. daiquiri
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    daiquiri Engineering and Design

    Gerr's book has been widely covered in this forum. Use the "search" function and you will find the relevant topics. Ok for a rapid initial scantling and weight estimate, but soon after that you'll have to switch to ISO or Class rules, if you intend to commercialize your design.

    Anyways, I don't remember that book saying anything about propellers, perhaps you have confused it with the other one by Gerr, "The Propeller Handbook"?
    Cheers
     
  12. baeckmo
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    baeckmo Hydrodynamics

    The power figure is an example to show you how to use the formula, no more, no less. The purpose is to demonstrate that the tip speed per se is not a factor in the basic dimensioning. Generally speaking, a bigger prop (in terms of propeller area per horsepower) will produce higher thrust and higher efficiency (=less losses).

    Now, if you put the figures you mention (150 hp) into the equation, you will get a much lower shaft speed with a corresponding reduction in tip speed and increase in the pressure constant. This then, results in less danger for cavitation.
     
  13. michaeljc
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    michaeljc Senior Member

    I don't have time right now but I do want to continue this interesting discussion e.g. if in practice you have a 250 hp motor, then you will not be using its full output for any length of time. It will probably work at its most efficient at around 180 HP. This is an example of how we have to be careful using pure theory (IMO)

    M
     
  14. daiquiri
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    daiquiri Engineering and Design

    The point in choosing a right prop for cruising is to find the type, diameter and pitch such to get the best efficiency at engine's optimum working point, with the constraint of being able to get an acceptable efficiency and cavitation level at engine's WOT.

    So installing the 250 HP engine with intention to make it work at 180 HP, and choosing the prop for just the latter condition is not a good strategy. The very likely cavitation at higher blade loadings will not allow the engine to give 250 HP when needed and you will end up with useless engine weight on board.

    Cheers
     

  15. michaeljc
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    michaeljc Senior Member

    Yes, I get your point, but working an engine at close to max power reduces working life. And - the 250 hp engine may use less fuel at say 75% max power output than a 180 hp working at max. Only race engines work at max - and not for long. Many modern motors have the same components (weight) but various options on output. If you are buying for a work boat the manufacturer will de-tune to give longer working life. If you are buying for racing or showing off they will up-tune. Modern trucks are doing it on the fly.

    Most engines work at their most efficient at RPM around half way between max torque and max BHP

    However this does not relate directly to the formula above on tip velocity. I will comment later.

    PS I referred to Gerr's book as an example of reverse engineering which is exactly what I am attempting to do with tip velocity. We will see if I can get enough data to prove my point.

    There will be a trend, of that I am sure.

    Cheers

    M
     
    Last edited: Jun 11, 2013
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