Calculating Pitch, DAR etc when diameter is known

Discussion in 'Props' started by Mat-C, Oct 3, 2012.

  1. Mat-C
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    Mat-C Senior Member

    I've been running through the calcs layed out in Gerr's Prop Handbook and am sure the answer is there in simple terms, but my teeny mind must have missed it....

    I have a little timber launch that is 22' on the waterline. Max Loaded displacement is approx 2000kg. I have come across an electric motor that can supply 4.7 kW @ 1400rpm max and suposed 700rpm cruise (though why you have cruise reves on an electric motor I am yet to understand, but that's a side issue...). There is no reduction gear.

    Running through the Bp method, I wind up with a 3-bladed prop of 12" diameter and I think something like 10" pitch. But my setup allows for a 16" diameter prop. Bigger is almost always better, I understand...

    So - the question... how do I go about calculating the best pitch, DAR etc?
     
  2. BertKu
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    BertKu Senior Member

    Yes , I have also been struggling to find certain details out in Dave Gerr's book. Tremendous information, but sometimes for the insider and not for people like you and me. It depends what electro motor you have, a brushless, or a brushed permanent magnet motor or a normal DC motor. You will have losses which shows up into heat dissipation. The temperature at 4.7 KW may create such a development of heat, that you can only run that for 10 minutes. While at 700 rpm, you will dissipate minimal heat over long periods of time.

    Download "propcalc" propeller calculation program from the site “Castle Marine”. (Google “ Castle Marine and it will give you the correct URL) It is free, but you cannot ask questions. Except if you order a prop.
     
  3. johneck
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    johneck Senior Member

    Bigger is usually better, but there are limits. And generally as RPM goes up the optimal diameter goes down. In this case, the optimal diameter is probably about 10" because of the high RPM.

    In order to calculate pitch you need to have an idea of how fast the boat will be going. From that you can calculate the advance coefficient J and using Wageningen B series data in the form of Bp-Delta or regression data you can calculate pitch. If I assume 6 kts, then the pitch for a 10" prop would be 8.5"

    DAR is based on the area required to avoid cavitation problems. Sometimes due to loading, you will have to increase diameter to provide sufficient blade area. In this case, due to the low power, a DAR of 0.5 - 0.60 would be enough to work fine (assuming a reasonable inflow and low shaft inclination).
     
  4. FAST FRED
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    FAST FRED Senior Member

    Skene's Elements of Yacht Design

    Better book to use , at your library.
     
  5. jehardiman
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    jehardiman Senior Member

    As johneck said, get a Bp-delta chart and pull the point off the maximum efficiency curve (correcting Va for the wake fraction of course). But on the other hand, do you want maximum efficiency for propulsion or are you looking for maneuvering response? For maximum efficicency you want a 2 bladed prop if you can go bigger and swing it slower.
     
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  6. Mat-C
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    Mat-C Senior Member

    Hmm... so the old bigger is always better mantra ain't necessarily so....

    With so little power, my calcs suggest that 6 knots would be the most I could hope for, so yes, I am aiming for maximum efficiency... but wanted to avoid the chance of any 'drumming' caused by a 2-blade prop, hence the choice of a 3-blade.

    So - what you are saying is that ideally there would be a reduction gearbox involved to get rpm down to more sensible numbers, then I could swing a larger diameter, more efficient prop...

    compromises, compromises...:rolleyes:
     
  7. BertKu
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    BertKu Senior Member

    Mat, can't you power your motor with PWM (Pulse width modulation) Normally with a permanent magnet motor you have the full torgue at 1 rev per minute already. Thus running at 700 revs and a big prop should be in my view not a problem. Don't put a reduction gearbox, you waste money and efficiency.
    P.S. with a PWM, at any time you can decrease or increase the revs/min and have full 4,7 KW available for an emergency.
     
    Last edited: Oct 7, 2012
  8. johneck
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    johneck Senior Member

    With 2:1 reduction, a 16" prop is just about right. It will be more efficient to go bigger and slower. With the RPM at 1700 and diam 10" you are looking at open water efficiency of about 0.55, at 850 RPM and 16", efficiency is about 0.68, which is a good number.
    A two bladed prop will be slightly more efficient and with the loading and tip speed of the propeller so low, vibration should not be an issue, assuming a decent inflow.
     
  9. Mat-C
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    Mat-C Senior Member

    The motor is a Mastervolt like this
    I don't quite understand though.... if I have full torque at 1 rpm, what's the point in running at anything faster than the minimum revs that will allow me to swing the max size (16") prop? Obviously I'm not drawing the full 3.5kW at say 700 rpm, but if I have full torque, doesn't that mean I could size the prop to run at 700 rpm and still get the same (approx 6knot ) top speed?
    Is there any advantage to be had by 'revving' any higher?
     
  10. BertKu
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    BertKu Senior Member

    I cannot answer that question, as I myself, have made a propulsion system and have to figure out what the differences are between a diesel which gives you torque at about 1600 - 2000 revs and brushless electro motors, which has the full torque from the beginning to the highest rev. What it means is, if your motor is NOT strong enough and you have a bigger prop, your motor will slow down and creates more heat if you turn up the voltage on the motor to get the right revs. The speed from the brushless motor depends on RPM per Volt of the windings multiplied by the Voltage applied to the motor. If you pulse it, for that short period of time, you have the full voltage and the full current for that short period of time. I think you will be OK, but let see what the others say. Bert
    P.S. If your pulse is 10 millisecond , every 1 second , your full torque is applied to your prop for 10 mS, your approx 4,7 KW of torque. However, it takes time to start turning your prop, thus your prop will turn very slowly. This is different for the diesel. As there, the engine will probably stall and you need a clutch to have first the engine running on speed before it has the torque to start the prop. The mass of the flywheel and the slip will bring your prop at the speed required. This is totally different from an electric propulsion system. In that case, if your prop is so big, that the power required at full speed needs more than the 4,7 KW, all what means is that your motor will not reach the RPM as indicated by the manufacturer i.e. 1400. It will turn slower and generate more heat. If you can get rid of the heat by water cooling or oil cooling and the internal temperature stays within manufacturers specifications, it is not ideal, but it will work.
    Put your info in the program “propcalc” and it will give you some answers you are looking for.
     
    Last edited: Oct 8, 2012
  11. BertKu
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    BertKu Senior Member

    I put some data in for you into the Propcalc program. Although some people think it is not accurate at low power, it is not confirmed that this is the case. However it gives you still an indication. At a displacement hull of 22 feet, 2000 kg, single engine, outboard, 1400 revs and 6 horsepower you will achieve 5,53 Knots. But I have no idea, whether at smooth water surface or at waves of 50 cm. Your prop should be 11.73 inch, pitch ratio 0.87

    At 700 revs , your average power will be 3 hp and you will do 4.41 knots, but your prop should be 15,47 Inch and your Pitch ration is 1.24

    The problem is, this program "propcalc" is made for diesel engines, thus for electro motors, you have to go on your gut feeling and make a prop which is large enough, but made so that you can grind something away. Well that is my plan for my prop. The problem is, maybe only Jeremy Harris (or John) can answer our question. He has experience with electro propulsion.

    Because your power is 4,7 KW at 700 revs, but for a short period of time (1/2) your usage power is thus 2,35 KW But your torque at 700 revs is the full torque. What propeller therefore to use is in my view not accurate and there for I am also going for the biggest prop my material is allowing me.
    I would say go for it and experiment.
    Bert
     
  12. baeckmo
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    baeckmo Hydrodynamics

    Don't bother about the torque, you have to understand the basics first. The vessel has a thrust versus speed characteristics. Thrust times speed equals "hull power needed"; ergo you have to supply an amount of power that equals the "hull power needed" plus losses, i.e "available power".

    The "available power" from your motor is a function of rpms. Thus you have two variables to compare (hull velocity and motor rpms), but the common factor to care about is POWER.

    A rough, but reasonable equation for a preliminary estimate of prop dia is as follows for BAR ~0,5-0,7:

    D ~ 630 * P^0,2 / N^0,6; where D is prop dia in inches, P is power in hp, and N is shaft rpms.

    Now, the losses are a function of prop dia and power input, and as you can see in the equation, to keep input power constant, the dia has to match the rpms, so if you gear down, the dia goes up and losses are reduced. But there are limitations to this; bending stress at blade roots beeing one, added power losses in transmission another.

    At maximum torque, you obviously do NOT have the full power available, and of course you then cannot deliver the "power necessary" to drive your hull at a speed that requires the full power! Listen to johneck here, he is pointing you in the right direction!
     
  13. BertKu
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    BertKu Senior Member

    That is my problem. All calculations and formula's are based for diesel type of "powerhouse". With electro motors I believe that the maximum power is continuous there, but for intermittent periods and short periods. While with a normal diesel you have the power related to the engine revolutions and is lower by lower revolutions, but continuous. Until somebody can definite put his head on the block that the formula's for both type of "powerhouses" are the same, I am quite willing to accept any formula. But so far my gut feeling is that we need for Brushless permanent magnet electro motors different formula's.
    Prove it to me and Mat.
    Bert
     
  14. BertKu
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    BertKu Senior Member

    Here I agree with you. Because the full power is there for a very short period, the avarage power is lower. Therefore the boat will move slower. But what Mat and I don't know, what is the relation of peak power for a short period of time versus prop size and if one has lower average power, do we need really need larger props or can we do from the start with a large prop for all speeds. I hope that you understand our dilemma and what I try to get accross.
    bert
     

  15. BertKu
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    BertKu Senior Member

    Let me try to put it in a different way. We don't have a gear down box. Our reduction box is electronically PWM mode (pulse width modulation) whereby we don't turn down the maximum power, that stays the same, just the average power is reduced. For the short period of time the full power is applied, you need a certain size prop. But our power is always maximum, always the same, just not the average power. Why do we need than to change the size of the prop? We just have one size prop. While if I take 5 different reduction boxes, I need 5 different size props, when using a diesel engine keeping the same speed. Or, do we have to calculate with average power for a permanent magnet motor and ignore the fact that we always have peak power during whatever pulse time period.
    For that reason, I think we need a different formula for pulsed brushless magnet motors. If I look at the props used on Torqeedo, that may give us an indication. Love to know whether we talking a lot of crap, or is there some truth in the above.
     

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