newbie question-drag and power calculation

Discussion in 'Boat Design' started by boatofhoha, Nov 18, 2013.

  1. boatofhoha
    Joined: Nov 2013
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    boatofhoha Junior Member

    hello guys,

    I am student in fourth year of mechanical engeneering from the middle-east.
    I am part of small team in project of desighining autonomic sailboat, we just started, some of us including me havn't studied specialy in naval arhetecture so we basecly trying to learn as much as possible fast.

    I need to do the calculation that will lead to choosing back up motor,propelor, and other motors that will be needed.

    My problem is that I dont know realy where to start from my calculation, the boat should be small model-1.5 meter length, slow speed-something like 5 knots speed realy no importand, and long time in water-24 hours, calm water.
    now I thought about starting from calculation generaly the drag dorce, as I don't know yet all te final sizes.

    my questions:
    -is it right to use the general drag equation, and if so how to calculate the drag coeficent?
    -how to use the thrust to calculate the sizes of my propellor and choosing the right motor(electric), what things need to be considered?

    first real project and I feel abit lost right now, would be glad if you can help me out alittle bit.

    thanks.
     
  2. kerosene
    Joined: Jul 2006
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    kerosene Senior Member

    for that scale you might want to ask rc-groups.com (radio controlled boats).
    Gerr's nature of boats is a good book for general introduction to boats. For your boat 5 knotts is not slow as wave resistance is largely dependent on speeds relation to boat (waterline) length. Short boats will not go as fast as long ones (generally speaking).

    Someone will hopefully fill with a simple rough ideas how to estimate the power use. You can probably browse the propulsion subforum and do searches there for a formula for displacement speed (vs. planing speed) resistance calculations.
    The pedal powered boat thread is probably a very useful one too.

    Factors that need to be known are the displacement (weight of the boat with all gear), length at waterline and the hull shape. I assume efficient power use is crucial so it should be something like a canoe or kayak shaped.
     
  3. rwatson
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    rwatson Senior Member

    My 1.75 M model boat can do 5.4 mph easily with this size electric motor. I hope to get it to ~ 9 mph for a scale speed of ~20 mph once I sort out the propeller situation.

    It has a weight of 11 kilos in total ( ballast, motor, hull etc )

    The trick is to know how to calculate the performance - I used a GPS camera mounted inside.
     

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    Last edited: Nov 21, 2013
  4. boatofhoha
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    boatofhoha Junior Member

    hey,
    thanks for replys.

    We decided that the boat hull shape will be simple v shape and not catamaran.
    I am in the pair that doing the propeller and motor calculation, it was hard for me to find from where to start, but in the end I did simple calculation based on the drag dorce formula with two assamptions:1.drag coeeficent=0.04(streamline body) and some wetted area formula:0.85*LOA*B in feets, in the end I got 34 newton of drag and 80 watt of power from the propeller.

    The resulets seem to me quiet low, so I wanted to ask you few extra questions:
    1.can you give me and idea if my assomption that I found in the net are no too far, and if yes where I can find better formulas for drag coeeficent and area.

    2.Is there any way of knowing the requeired speed in rpm and tourqe needed from my system before choosing some examples?, becouse after looking at the problem is seem to me that I only can know the power requred from motor and propeller.

    3.what ussualy come first, choosin the right propeller or motor?, form what I read it seems that propeller diameter is decided from the speed of the motor.
    and if not how do you choose your diameter and pitch.

    4.about the boat you mentioned rwatson, We realy dont focus on speed, its a back up motor, the main idea is to use the wind for as much time as possible and in the best way.

    thanks.
     
  5. Baltic Bandit

    Baltic Bandit Previous Member

    Well the simplest thing to do is to actually calculate the drag of your hull. Mock it up out of cardboard, spray paint it with some waterproof enamel, and tow it behind a real boat using a well calibrated scale to measure the load on the tow

    that way you can put all the instrumentation in the tow boat (which can be as simple as a canoe except for the scale.

    the trick on props is that the more lift (power) you generate out of one, the more you start to get to the limit of cavitation. there are a variety of torque/rpm/pitch/shape tables out there that you can look up a decent first go at it. but if you are going for really precise stuff, that's where the fiddly biits come in.


    http://www.offshoreelectrics.com/propeller-chart.php
     
  6. Baltic Bandit

    Baltic Bandit Previous Member

    BTW by looking at the pitch and the diameter of a prop, you can calculate how much water it moves in a single rotation (obviously making a lossless flow assumption) http://www.boatingmag.com/maintenance/understanding-propeller-pitch

    So from the chart in the previous post a Prather Prop with a 3" pitch (the 230) moves 3" of linear water each rotation.

    it has a 1.9" diameter which means its radius is 0.95"

    2piR = Area so 1.9" x pi x 3" ===> 17.9 ci of water ==> 290 cc of water.

    so at 600 rpm, it is moving 2,900 cc of water per second. IE 2.9 liters of water per second.

    A liter is 2.2kg so roughly 6.4kg/sec

    now a Watt is in units of kg-m/s so at 600 RPM ==> 10 Rps and 1 Rev = 3" so 30"/s = 0.76 meters/sec

    so that prop is moving 6.4 x 0.76 kg-m/s ==> 48.5 watts of power being generated.

    And you can set up a spreadsheet with that calculation and just plug in the pitch and diameter numbers from the linked table and figure out how much power the prop is generating in the ideal. and figure some amount of loss.

    then you look at the power curve of the motor and figure out its optimal rotation speed

    So then at its optimal rotation speed you also know its power consumption in watts.

    So you then match the power consumption watts to the output watts and you have the prop you need
     
  7. rwatson
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    rwatson Senior Member

    Thanks BB for that rough outline. It is of interest to me as well, as I am still developing my scale model.



    Did I detect a typo in your notes ?

    " it has a 1.9" diameter which means its radius is 0.95"

    2piR = Area so 1.9" x pi x 3" ===> 17.9 ci of water ==> 290 cc of water."


    Should that last line be
    2piR = Area so .95" x pi x 3" ===>



    I have developed a spreadsheet as you suggested, with figures more appropriate for a ~2 meter scale model and props. I also got rid of imperial figures, as they are just a nuisance.

    It resulted in about 8 watts of power needed.

    I should be able to verify these figures if I put a meter on the motor wires, and calculated the Amps being generated.
    http://www.rapidtables.com/calc/electric/Volt_to_Watt_Calculator.htm

    shows I would need to draw 2 Amps at 3.1 Volts to operate at 2000 RPM




    It would be good if you and anyone else interested could check it out for errors, and to use themselves if it is useful.

    Many thanks for the info again.
     

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  8. rwatson
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    rwatson Senior Member

    XLS Version 2

    Hmm, already found an error I think.

    The Wattage ( Power ) calculations brought the power needed way down to .07 Watts.

    This is the 'fixed' version'
     

    Attached Files:

  9. Baltic Bandit

    Baltic Bandit Previous Member

    Another error in the Spreadsheet. Prop Efficiency is something you DIVIDE your lossless power output by, not multiply.


    FActual = FLossless * (EfficiencyLossLess/EfficiencyLossy)

    Except that 'Efficiency' is typically (EfficiencyLossy/EfficiencyLossLess)

    so you have to multiply by 1/Efficiency

    IOW if you want 0.07 watts of output power and you have 70% efficiency, you need to 0.1 watts of power into the prop.
     
  10. Baltic Bandit

    Baltic Bandit Previous Member

    BTW I had some trolls go through and ding my "rep" so if I've been helpful, I'd appreciate a bump on my rep points.
     
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  11. rwatson
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    rwatson Senior Member

    Useful info points added. :)

    That's a good point about the efficiency, but I was thinking that no prop can push 100% of the water.

    A lot of losses of volume due to tip flow, propeller shape, some vortex influence etc.

    So, maybe if a percentage of water was not 'pushed', this would lead to a reduction in power required, not an increase ?

    If the 'water pusher' was an enclosed screw, then friction losses etc would add easy 30% to the motor load for sure.

    If I was after a boat performance speed, then the theoretical power required would have to be increased to allow for prop efficiency loss, but in my mind, I see the prop doing a lot of 'spin' without 'push', therefore not requiring the full power to shift the theoretical water volume.

    In other words, does the 'Prop Slip' factor reduce or increase power requirements ?
    Would a Prop with a slip factor of .1 require more power to spin at a given RPM than one with a slip factor of .9 ?

    Is a prop spinning at 2000 RPM with little pitch ( thus a lot of slip ) using more power than another prop spinning at 2000 rpm with correct pitch ( and little slip ).

    Any thoughts ?

    Found an interesting prop calculator at
    http://www.csgnetwork.com/marinepropcalc.html
     
  12. rwatson
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    rwatson Senior Member

    Last edited: Nov 22, 2013
  13. Baltic Bandit

    Baltic Bandit Previous Member

    No because the loss of "water pushed" is due to turbulence and lateral flow off the tips and tip vortices and such - all of which increase the drag on the prop and hence rob it of power at the prop. So you need more energy to move the same amount of water.
     
  14. rwatson
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    rwatson Senior Member

    Yes, I agree - to move the theoretical amount of water would need more power.

    But, what I was talking about though, is that the calculated power required to turn the prop at a set RPM, is less because the theoretical amount of water actually pushed is less.

    The calculations assume that the 'magic propeller' is pushing every atom of water within the diameter of the prop - thus requiring X amount of power to do it.

    In reality - it can only push a percentage of that weight of water, so the X amount of calculated power required to spin at the set RPM is also reduced.
     

  15. Baltic Bandit

    Baltic Bandit Previous Member

    Well except that you can assume that the reduced efficiency is not gained back from pushing less water - intstead it gets wasted in the things I mentioned: tip vortices, surface drag, induced drag etc.

    think of it this way. imagine if instead of water we were talking about really really thick oil: Now as you generate lift from the prop (that's what you are doing essentially you are "flying" the "wing" that is the prop in a circle and "lifting" that wing forwards ) the viscosity of the oil is such that the theoretical amount of lift you could get is reduced so you more less oil than expected.

    Well ok, but you are having to fight the viscosity, and some of the energy goes into heat energy in the viscous bonding and some of it goes into heat energy of surface friction in that dense material etc.

    Now you could calculate out each of these factors and add them back into your power costs. Much simpler to just assume that you put in 100% power theoretically needed, but get out less work - with the rest lost to other effects
     
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