Propeller Selection using Effective Power

Discussion in 'Props' started by Mitch1990, May 27, 2020.

  1. Mitch1990
    Joined: Feb 2020
    Posts: 71
    Likes: 0, Points: 6
    Location: Maine, USA

    Mitch1990 Junior Member

    Please delete
     
    Last edited: May 29, 2020
  2. jehardiman
    Joined: Aug 2004
    Posts: 2,798
    Likes: 363, Points: 83, Legacy Rep: 2040
    Location: Port Orchard, Washington, USA

    jehardiman Senior Member

    Actually, lets not and with referencing the other thread (How to find Delivered Power https://www.boatdesign.net/threads/how-to-find-delivered-power.63949/) I will walk you through how to get a prop and prime mover out of a known effective hull power.

    So, by some means (testing, simulation, or standard series) you have an effective hull power of 600kW at 14kn.
    In this case Power is thrust*V and actually what we are interested in right now is the thrust of the prop so the tow resistance (Rt) = 600/(14*0.514) = 83.4 N
    Since you are twin screwed, we need a prop that delivers 41.7 N (83.4/2) of thrust at 14 knots. Whether that is a lot of a little depends on the propeller diameter. Since you have a hull shape, the next question is what is the propeller diameter you want to fit.
    In the other thread you say:
    You never say what the propeller diameter (D) is or what n (rps) you are using....but yet you selected a J? This is what Ad Hoc was getting at.

    Anyway......Instead of stopping here for your input, I will just assume a set based upon a 70m x 700 Tn twin screw CPP.
    Reaching into the hat...superyacht!....SYCARA V Yacht for Sale | Fraser https://www.fraseryachts.com/en/yacht-for-sale/sycara-v/ (I would use a different set for CODAG PC)
    So lets say D=1.5m
    J=Va/nD and Va = V*(1-w) so
    Va =14*.88 = 12.3 knts = 6.33 m/s
    n = Va/JD = 6.2 rps = 377 rpm (which I think is a little high FWIW).
    Now lets check your selections so far
    T=Kt*rho*n^2D^4 =38.6 N
    So you are low by 10%...not only that we have to consider the thrust deduction (t)
    T required =Rt/(1-t) = 41.7/(1-0.15) = 49.0 N
    So realistically you are 21% short on thrust
    Anyway...lets continue to calculate power required.
    Q= Kq*rho*n^2*D^5 = 9.78 Nm
    Shaft power delivered is
    Ps = Qn *n*2pi = 381 W = 0.381 kW
    This low of Ps means that J is too large for the selected P/D or P/D is too small for the required J

    This could be done on Bp delta charts (or the more modern Bp1 or Bp2 1/J charts) The Wageningen B-Screw Series - PDF Free Download https://docplayer.net/48370380-The-wageningen-b-screw-series.html
    I don't have a Bp delta for B4 60 immediately on hand (it's in a box somewhere) but I could do it for a B4 55 or B4 70 as an example.
     
    Mitch1990 and Ad Hoc like this.
  3. Mitch1990
    Joined: Feb 2020
    Posts: 71
    Likes: 0, Points: 6
    Location: Maine, USA

    Mitch1990 Junior Member

    Sorry, I missed the propeller diameter of 1.7m which I understand is key. As you would have read I was provided with a spreadsheet that has a macro that I am told works for most Wag B Props. What I find comforting is what you described is what I did. The below is the table generated by the macro for a wag B BAR of 0.6 and Z = 4. If you are happy to provide an example on a chart that would be amazing

    upload_2020-5-29_22-16-24.png
     
    Last edited: May 29, 2020
  4. Mitch1990
    Joined: Feb 2020
    Posts: 71
    Likes: 0, Points: 6
    Location: Maine, USA

    Mitch1990 Junior Member

    I have a B4-60 and I have plotted the ships resistance to overlay on the prop chart in excel, I just don't know how to overlay them.
     

    Attached Files:

  5. jehardiman
    Joined: Aug 2004
    Posts: 2,798
    Likes: 363, Points: 83, Legacy Rep: 2040
    Location: Port Orchard, Washington, USA

    jehardiman Senior Member

    If that is your real Kt Kq chart, WTF, made you decide on a J = 0.67 and a P/D =1 ? and FWIW why would the macro miss peak efficiency by that much? And G*D!, the axis selection on that chart sucks! 0.04 for the sub-grid?
    Calm down John, you must remember that skills now are different from skills then...
    But why make something hard to to interpret! He setting himself up for failure!
    John..put the lead holder and proportional dividers down...What are you going to do what that duck?....NOOOOOO...that's a CRT!


    BTW, I F'ing hate a Kt Kq chart laid out that way. It is sooooo stupid to do it that way because it is so prone to errors in reading. Harvard Graphics did such a disservice to the engineering world when it set up auto axis like that. Really..The axis should be Kt,10Kq on the left and eta on the right....makes a much clearer chart.
     
    Eric ruttan likes this.
  6. Mitch1990
    Joined: Feb 2020
    Posts: 71
    Likes: 0, Points: 6
    Location: Maine, USA

    Mitch1990 Junior Member

    I should mention that I chose the J and P/D based on the spreadsheet not the chart. This is what we were given, sorry if its rubbish, but I don't know any better. As I am sure you can tell...

    Also, for good measure and to assist in calming you down... i had to google proportional dividers... PS. They would be excellent as a weapon. I imagine they still teach it to undergrads for naval architecture, I vaguelly remember isometric drawing.
     
  7. jehardiman
    Joined: Aug 2004
    Posts: 2,798
    Likes: 363, Points: 83, Legacy Rep: 2040
    Location: Port Orchard, Washington, USA

    jehardiman Senior Member

    Do you know what the macro does? And why? Normally when using an effective powering datum determined by testing, simulation, or standard series, you add 20% to the value for 'sea service' which is power in your hip pocket for fouling and seaway issues. Additionally, the prime mover is normally de-rated ~20% for sustained speed.
    So...
    What is the 14 knots?....and what is the requirement?
    Right now, the data being spit out of the macro makes no sense, even with a CCP, which is important for prime mover selection. It is not an optimum nor is there any other reason to select that particular point in the propeller curve.
    I know this may be a homework problem...but if you want us to point you in the right direction, we need to know what the expected result is.
     
  8. Eric ruttan
    Joined: Jul 2018
    Posts: 162
    Likes: 17, Points: 18
    Location: usa

    Eric ruttan Senior Member

    One like is not enough

    .
     
  9. Mitch1990
    Joined: Feb 2020
    Posts: 71
    Likes: 0, Points: 6
    Location: Maine, USA

    Mitch1990 Junior Member

    The macro takes the inputs of effective hull power, ship velocity, thrust deduction and wake factor, diameter, BAR and number of blades and produces the table that I provided above along with a Jadvance vs Kt plot for the ship. I would post it here but it doesn't belong to me, I would be happy to send it too you separately but that sort of defeats the point of posting in a forum. The effective power was given as 600kw at 14 knots. From here I believe I need to plot the ships resistance and overlay it on the prop chart to find the correct prop stats to achieve the thrust required for that speed and effective power.

    The overall aim is to match a prop to an engine through the prop cubic law which I am fairly comfortable with.
     
  10. jehardiman
    Joined: Aug 2004
    Posts: 2,798
    Likes: 363, Points: 83, Legacy Rep: 2040
    Location: Port Orchard, Washington, USA

    jehardiman Senior Member

    Whom ever owns it, you should tell them it is not working, at least for your application.

    Ok....since you don't seem to have a Bp delta drawing, we'll do it the classic Kt Kq chart method.
    1) Select your prop chart (i.e. B4.60)
    2) on that chart draw the line of maximum efficency (i.e. strike a fair line through the max eta O's for all P/Ds)
    3) select an open water efficiency (eta O) about in the middle of the chart
    4) calculate eta D where eta D = ((1-t)/(1-w)) * eta O *eta R, where eta R is generally between 0.95 and 1.0 for twin screw ships. Use 0.95 because in real applications this will guarantee the required power is overestimated (a good thing)
    5) determine required power delivered (Pd) where Pd = Rt/eta D
    6) determine Va remembering to include wake fraction (w)
    7) Select an n (not a prop D! you will see why) If you have a D in mind, use that D to calculate n based on the J dropped from the selected eta O on the prop chart. Also ensure that your n is supported by your prime mover, gearbox choice and a cavitation check.
    8) calculate the torque (Q) for the Pd. Q in Nm and Pd kW; Q = 1000*Pd/ (2*pi*n). We now have shaft torque that the propeller needs to absorb to meet the required power
    9) Calculate Kq for various D's. Kq = Q/(rho*n^2*D^5) = Constant 1/D^5
    10) for the same D's, calculate J = Va/n*D = Constant 2/D
    11) plot the curve of D based Kq's to J's of the prop chart
    12) where the D based curve intersects the maximum efficiency curve, read J and interpolate D.
    13) if the eta O determined in 12) is significantly different than eta O in 3), revise it slightly then repeat steps 3) through 12 until convergence
    14) once you are satisfied with eta O, n, and D; determine Kt and back check thrust against Rt.
    15) also remember to check propeller clearances, cavitation, prime mover,...etc.
     

  11. Mitch1990
    Joined: Feb 2020
    Posts: 71
    Likes: 0, Points: 6
    Location: Maine, USA

    Mitch1990 Junior Member

    I have a maximum D which I have used as the limit for D, thanks for you help. I have attached my final workings spreadsheet for those who are interested and am working through the graphical at the moment
     

    Attached Files:

Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.