How do you compare electric motor power to diesel power?

Discussion in 'All Things Boats & Boating' started by Yobarnacle, Dec 23, 2011.

  1. hoytedow
    Joined: Sep 2009
    Posts: 5,857
    Likes: 400, Points: 93, Legacy Rep: 2489
    Location: Control Group

    hoytedow Carbon Based Life Form

    Yo B, Just send phone photos to your email address, then move them from your inbox to pictures file.
     
  2. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    my phone doesnt have internet. It's a cheapo! :) Is there away to send to my email as a text message adress?
     
  3. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Wow. I just did search howto send fotos from phone to email. just type in email. Wow! I didn't know you could do that. Thanks Hoyt
     
  4. hoytedow
    Joined: Sep 2009
    Posts: 5,857
    Likes: 400, Points: 93, Legacy Rep: 2489
    Location: Control Group

    hoytedow Carbon Based Life Form

    (Pretend there is an upside down exclamation point here.)De nada!
     
  5. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Two scientists are standing in front of a blackboard covered in symbols and equations. One says to the author, "I follow your reasoning up to where you wrote in the formula, A miracle happens!"

    Well, I'd like to be open minded, but I don't believe I'm entitled to a miracle on my boat propulsion. And it would't likely be repeatable.

    So. Engineer Cockey gave me power formulas to calculate thrust, ect.
    There are a lot of unknown quantities in my rig. I don't even have a tach so even RPM is a mystery. I'm trying to reverse engineer a system that works, so I'll understand how and why it works, and can then explain it rationally to others.

    If this thread was dead, well, maybe as I post my math and formulas I discover, you'all will be interested in commenting. Merry Christmas!
     
  6. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    The first argument (not argumentative but a hypothesis defense) is my boat has a more easily driven hull than when she was diesel powered.
    I lengthened her from 25 to 30 ft. Length of waterline lengthened from 22 ft 2 inches to something near 26 ft +/-. Original empty weight (no fuel) was 1.8 tons. While she does have about 700 lbs of batteries not carried before, she is minus the former 600 lbs of the diesel engine. I ripped out most of the old accomodation, so lightened her. The new stern weighs only maybe 200 lbs. I lifted it in my arms when installing it. So the 4500 lbs I said she displaced is wrong. More like 3800 lbs displacement now.
    Since her beam remains the same, her L/B ratio is closer to 4:1, an ideal in this size by my thinking. Her block and prismatic co-efficients are both lower because of being longer and also she is now double ended. All making a much easier propelled hull.
     
  7. philSweet
    Joined: May 2008
    Posts: 2,682
    Likes: 451, Points: 83, Legacy Rep: 1082
    Location: Beaufort, SC and H'ville, NC

    philSweet Senior Member

    You can spend a lifetime doing that, Yo, and it will be well spent. Best wishes and Merry Christmas.

    The first time I pissed off a retired tug driver it didn't end well; and I've tried hard to prevent a repeat performance. Sorry if I opened a can of worms. But sometimes, you just gotta deal with worms.
     
  8. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    I found this formula:
    power = k times RPM cubed times prop diameter to 4th power times pitch.
    K is a co-efficient of prop type and unknown at this time.

    Deriving from the formula the following, because the unknown k is constant, learned this: Double the power and keep pitch and diameter the same will only increase RPM 1.26 times. 1.26 is the cube root of 2.
    Extrapolating reverse: RPM only decreases by 0.7936507936507937 or 79% when you cut imput power by half.
    So the 1400 max RPM of the troll motors @ 116 amps, turn 1111 RPM at half power or 58 amps, or 882 RPM @ 29 amps, or 700 RPM @ 15 amps.

    My test runs were done at 12 amps, 30 amps and 60 amps. I didn't exceed 60 amps because the motor was designed to withstand max power of 116 amps only for a few minutes, according to Minn Kota specs.
     
  9. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Well, I'm a retired tug driver, but nothing you have posted pissed me off. Not yet anyway. :)
     
  10. DCockey
    Joined: Oct 2009
    Posts: 5,229
    Likes: 634, Points: 113, Legacy Rep: 1485
    Location: Midcoast Maine

    DCockey Senior Member

    I did not give you formulas to calculate thrust, etc. I provided the relationships between torque, power and rotational speed for a shaft. Absolutely nothing about estimating thrust or any other quantity.
     
  11. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    yes sir. I get confused in my memory at times. Senior moments. :)
     
  12. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Mr Cockey,
    Do you have information on the value of k in the above formulae? The material i read said formula was for both air and water props, but only gave a value k for air.
     
  13. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    I discovered more in the past 24 hours. 2 bladed props are more efficient than multiblades because the addittional power required for same RPM is double for 4 blades than what's needed for 2. Thrust is much increased but not an equal amount as power. 4 blades won't produce double the thrust of 2 because of blade drag, more weight to spin, and resistance caused by tip vortex is multiplied with more tips.
    So, my 5 blade prop will give me an increase less than 2.5 times the thrust of the 2 blade prop (if blades of equal area), but cost more than 2.5 times increase in power to rotate. Because of increased resistances and weight.
    The 5 blade hydrofan has nearly square tips, and blades are 2 inches wide entire lenth. the plastic weed cutter OEM blade s are scimiter shaped with hooks and bends to brush off weeds. The weeder blade is admitted less efficient than a normal 2 blade prop.
    The hydrofan is working inside a nozzle, which it's nearly square tipped blades are well suited for. The shroud largely negates tip vortexes.
    The more efficient design and greater surface area of it's individual blades could offset the other losses in thrust, so hypothetically the 5 blade prop could be 2.5 times the thrust.
     
  14. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    The hydrofan is 11 inches diameter compared to OEM 10.5 inches.
    In the formula a few posts back, the difference in diameter affects the power reqired by power of 4. So 11 divided by 10.5 then multiplied by itself 4 times (actually should be 3 times but I was generous)equals 1.2618. Add power factor of 2.5 for 2.5 times extra blades, gives 3.76 times more power required to maintain RPM. Assuming weight or other encumberences I'll call it 4 times more power needed.
    So 60 amps would probably only spin this prop or props 700RPM.
    The 116 amps was for 2 motors each turning 1400 RPM. I just treated them together. So I have 2 props originally capable of 80 lbs thrust each but now 2.5 times or 200 lbs thrust each, total 400 lbs thrust at 700 RPM.
     

  15. Yobarnacle
    Joined: Nov 2011
    Posts: 1,746
    Likes: 130, Points: 63, Legacy Rep: 851
    Location: Mexico, Florida

    Yobarnacle Senior Member holding true course

    Can these motors @ 60 amps 24 volts turn the apparently much bigger props than the OEMs at 700 RPM? Need some more figuring. Have lots of unknowns. I only found one reference regarding pitch of the plastic props, that said 10 inch pitch. This is the same as the average pitch of the 5 blade hydrofans. However, the hydrofans are compound pitch and the tip of the blades are only 8.7 inch pitch best I can measure. I'm thinking that eases the strain at max diameter but an unknown amount. 8.7 inch pitch at 700 RPM is a tad more than 5 kts. Zero slip doesn't seem credible. Using the 10 inch average pitch claimed by mfg of these props @ 700 RPM = 5.8 kts, so actual 5 kts of theoretical 5.8 kts is 85% efficiency or 15% slip. Again hardly believable. If the RPM was greater than hypothesized, then 5 kts at a rational slip might be posible.
     
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.