Paddle Wheel Design Project

Discussion in 'Propulsion' started by drortiz1, Sep 20, 2012.

  1. drortiz1
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    drortiz1 Junior Member

    Good afternoon ladies / gentleman,
    I'm currently working on a reduction gear design to drive a "decorative" casino-boat paddle wheel. In order to effectively design the gear-box I need to size an electric motor that will be big enough to rotate the paddle wheel which has a mass of 11850 lbs at a constant 1 RPM. Here are some details of the paddle wheel:
    data: weight paddle wheel = 11850 lbs, diameter 15' , velocity 1 rpm,
    this paddle will WILL NOT impulse a boat it is only a decoration it will sit on top of two low friction bearings and it will be submerged 2ft in a small pool where two paddles will be inside the water at the same time. the area of each paddle is 22ft^2
    Initial approach: calculate the torque required to accelerate the paddle wheel from 0 rpm to 1 rpm in 60 sec. I decided to exclude the friction in the bearings and also the drag of the water in the paddle since the speed of the wheel is very slow ( is it correct to exclude the water drag from the calculation?)
    torque =( wheel moment inertia )* (angular acceleration)
    torque = (1/2)*(11850/32.2)*7.5^2* (2pi/3600) --> covert RPM to0647 radians per second^2
    torque = 18.0647 lbs * ft
    HP = torque* RPM/ 5252
    HP = 18.0647 * 900/5252 = 3.09 HP * 1.6 (Gear box service factor )
    HP = 5hp (notice that my gear box will reduce the rpm from 900:1)
    Is this calculation correct? I'm missing something ? I will greatly appreciate any advice and opinion this is the first time I'm doing something like this and I not sure if I'm doing it the right way.

    Thank you for your time.
  2. upchurchmr
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    upchurchmr Senior Member

    If this is decorative, reduce the weight of the paddle wheel to 200# max, make the blades of plastic thin enough to bend when they are in the water.
    No load, no issues.

    11,000# decorative wheel? Not very sensible.
  3. Jim_Hbar
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    Jim_Hbar Junior Member

    Where to start.........

    Firstly, I'll apologize for being blunt - but as they say, facts are friendly..

    Let's try the most obvious math error.

    In your hp calc, you've calculated the torque to accelerate the wheel to 1 rpm, yet used 900 rpm in the hp calc.
    Yet the motor will likely be running at 1800 rpm, so really the overall gear reduction will really be 1800:1..... But that doesn't matter either, continue reading.

    So calculate the torque to accelerate to 1 rpm, and then calculate the hp from that - Hp is a scaler quantity, so gear ratio doesn't enter into it (other than the inherent efficiency issues that result from high gear ratios)

    If you really do the numbers, you will find that the rotor of the motor will take significantly more energy to accelerate than your wheel!

    It's all a function of the change in angular velocity squared while accelerating, and 1800 squared is a much larger number than 1 squared.

    Yes, the Hp required from the motor to accelerate the wheel will be negligible !

    Next - One logical error.

    Start that motor across the line, and it and the wheel will be at speed in approximately a second +/- - none of this 60 second stuff. The motor will accelerate the inertia as fast as it's torque curve allows. The rim speed of the wheel, at 47 ft/min is slow.. very slow.. so it has very little stored energy.

    But the torque required to accelerate the mechanical system is a red herring.

    Next - Second logical error.

    The force to push 22 sq.ft. of blade through the water will be very significant, depending on what you do with the moving water, and how you let the water flow - accelerating the mass of the water to speed will take significant power, and a likely a significant ramp time.

    BTW, gear box service factors have nothing to do with the size of the motor. They have everything to do with the size of the gearbox selected, given the size of the motor, type of prime mover, the shock of the application, duration of operation, required reliability, etc...

    Just some quick points.

    Hope that helps!

  4. drortiz1
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    drortiz1 Junior Member

    Thank you both for your replies,
    We can’t build a decorative wheel that weights less because the owner wants to go back to original design, when the boat was in circulation back in the day.
    Thank you for your help,
    My previous post was not that great, please let me give you more details and if you don’t mind I would like to know your opinion. I have a 5 HP electric motor rated at 900 RPM which works with a sprocket gear reduction with a 900:1 ratio and its service factor is 1.6.
    A better question will be: Is my electric engine going to be able to move that wheel?
    Reading your response I modified my calculation completely, this is what I did:
    I know that my max motor torque will be T= HP*5252/900 = +/- 29.18 ft*lb
    So I asked myself the question what torque will I need to deliver to my wheel in order to make it move.
    I know I need torque to break the static state (accelerate the wheel) + I need torque to push the water away; so my new formula is as follow:
    T = (moment inertia wheel)*(angular acceleration) + (the force of the water against the paddle)*(radius of the wheel)
    Assuming 1 second as my time to accelerate the wheel and using fluid dynamics to get the force on the water pushing against the paddle I get:
    T= ½ * (11850/32.2)*(7.5^2)*(2pi/60) + ∫_1^3▒〖62.4(1-y)11 dy *7.5〗  my wheel is going to be submerge on a 3 ft deep pool but the but the blade is only 2ft wide, and 11 ft long
    T = 17450+ 10626 = 28076 lb * ft
    This is a huge amount of torque I know this I’m making a huge mistake here.
    I know this has nothing to do with all of the above but my 6.5 HP winch can pull up to 12000 lbs faster than 1 rpm, and I know the faster it moves the more power it needs.
  5. Submarine Tom

    Submarine Tom Previous Member

    So, I am to understand you cannot use a propeller and have the paddle wheel just idle along looking the part?
  6. drortiz1
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    drortiz1 Junior Member

    yes, it will be nice to do that but the boat is not moving, it is one of those old paddle wheel river cruise boats that are reconstructed and use as a Casino. It is going to sit on the side of a lake. If it was a going to be place at a river I will use the current of the river to move my wheel but a lake doesn’t have any current so I need a motor to do the job.
  7. Petros
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    Petros Senior Member

    you are leaving out two very important considerations. trying to acceleration a 12,000 lb wheel, even if not in the water, will take more than 5 hp. You have angular acceleration, all that rotational inertia takes power to get it moving up to speed, even if you have enough power to hold it as steady state 1 rpm, you will need to get it up to speed. And of course the faster it turns, the more water resistance will come into effect.

    The others is a 900/1 reduction drive will not be with out friction and system loss as well. Even the best will take a hp or two to get moving and keep moving by itself.

    YOu have a complex equation to solve for with several unknowns, you have to know what the rotational inertia of the wheel is, what the friction of the reduction drive is, what the water resistance will be at each rpm. This has to all be reduced to equations that are related to input rpm, and than use calculus to interrelate the changing condition as the wheel accelerates. For each given hp what you will find it will accelerate the wheel at different speeds up to the final steady state condition, which of course will be different for each size engine you use.

    You can short cut this by just determining what the hp is required to hold it at a stead state condition will be of 1 rpm, and use a motor rated at about 1.5 times this power requirement. than take the acceleration up to speed as it comes, though it is possible when you flip it on you overload the motor and it burns up before it can get the whole 12,000 lb contraption (and gear box inertia) up to speed. If you help accelerated it by hand each time you start it up likely you can get by with a smaller motor, but that is risky. Of course if you burn up your first motor, get a bigger one, and keep doing that until you have one that does not over heat on start up. but that is the costly way to do it.

    good luck with that.
  8. Jim_Hbar
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    Jim_Hbar Junior Member

    Sorry, Wrong...

    Lets look at it another way - in a frictionless linear system HP= FV/33000, so if 5hp is available, and the rim speed is 47fpm, the force available is about 3500 lbf.

    Since the mass is 12000 lbm, and the force is 3500 lbf the acceleration will be .3 g's.. So, when g=32.2 ft/sec^2, the acceleration rate will be about 10 ft/sec^2..
    When we are trying to accelerate to about .8 ft/sec, it's over in .08 seconds.. (In our very simple model!)
    But the motor can't accelerate that fast. That's less than 5 cycles of the AC current!
    Like I said initially, the inertia of the rotor will be have more significance in the acceleration time than that 12000 lb wheel.
    I'll let you satisfy yourself that using a linear model errs on the conservative side.
    It's not a bandmill with a 6 to 10,000 ft/min rim speed - it only has a 47 ft/min rim speed.

    BTW, I designed industrial machinery for 30+ years, and in that period of time I've selected literally thousands of AC electric drives, and a fair number of dynamic closed-loop drives. (Servo, Flux-vector, hydraulic, etc.)
    If you doubt my math (or qualifications), PM me.


    Okay, the maximum torque at the motor may be 29 ft.lbs, but the torque available at the wheel is 900 times that, so call it 26,000 ft-lbs at the wheel, so it is right in the ball-park of where you think you need it to be...
    But if I read your equation correctly, you're calculating the torque required to accelerate the wheel at 1g.

    But steady state, the torque to accelerate the wheel doesn't enter into the problem, so drop that term in your equation.

    Also, you note that the service factor of the motor is 1.6 - so that motor will pull 8hp before it is over-loaded, steady state. (Sounds to me like you have your hands on an old U-frame.)
    On top of that, the breakdown torque and the locked rotor torque are essentially available to accelerate the load, and they can be 200% to as high as perhaps 400% of the full load torque. All of which depend on the individual motor design, and are not known to us here.

    Like I said originally, the inertia of the wheel is a red-herring - the moving of the water is the problem, and I can't help you with that. Site conditions will determine how much energy you will need to pump into it.
  9. Petros
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    Petros Senior Member

    I do not think so, your over simplified model neglects rotational inertia, which is WAY larger than just the mass of the rotor. Go look up the equations, they are quite complex, you need to know the weight distribution of the wheel, for example if the 12,000 lbs were distributed around the rim it would have hundreds of times more rotational inertia than if it were at the hub (related to a square of the radius of the wheel).

    I do not have time to do the calculations for you, but you can find them on the net and do them yourself and see. I have done them before, they are not insignificant, and not directly related to just the mass of the wheel, you need to know the mass distribution on it to determine rotational inertia, when you spin things up the rules are very different. You need to educate yourself on this stuff before you make sweeping assumptions with over simplifications.
  10. Jim_Hbar
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    Jim_Hbar Junior Member


    The problem with typed communication is 90% of the message is lost, depending on the readers frame of mind and view-point when reading the message. I apologize if you've taken offence..


    My simplified model assumes ALL of the mass is at the circumference, and therefore is conservative, if the paddle wheel has any significant mass other than exactly at the circumference... I thought this was obvious, and didn't state it.
    My simplified model will give the same answer as if one assumes the mass is at the circumference, without getting into all of the fog caused by confusing the issue with PI, omega, and all the units mess.

    You are missing the point - the speed is so slow, that the torque required from the motor, to accelerate the wheel, is insignificant. Run the numbers in your internet calculator, and you'll see that I am correct.

    I've done similar calculations many, many times, for machinery that has sold for millions of dollars, and my mechanical engineering degree didn't come from the internet.

    - I posted to the OP that the inertia of the wheel was not significant to his problem, and then you posted a totally opposite opinion.
    - I provided a simplified proof, and a glimpse at my qualifications.
    - You're stated my simplified calculation is wrong, but haven't provided anything to empirically show that I am incorrect.
    - You haven't stated what your background is to lend credence to support your argument.

    The ball is actually in your court... A polite response would show me the error of my ways, or apologize and thank me for showing a different way to look at such a problem.

    But, I'll help you a bit - The simple way to prove this, is to calculate the kinetic energy of my linear systems mass and compare it to the energy in the rotary wheel with the mass lumped at the circumference.

    If you go look in Wikipedia, here, you can see where they derive it for you. (Wiki isn't the best engineering reference, but it was the first Google came up with)

    Hell, this is too easy to not do it - forget the easy energy approach, let's play "Angular Acceleration"..

    Assuming 100% of rated torque of the motor is available to accelerate the paddle-wheel, calculate the time to speed.
    Assume the mass of the wheel is a thin shell at 7.5 ft. radius.

    Given: Hp = T*N/5252 where T is in Ft-lbs and N is Rpm.
    Given: 5hp and 1Rpm therefore T= 26260 Ft-lbf.

    I = mr^2 and m=12000 lbm and r=7.5 ft. therefore I = 675000 lbm-ft^2

    And T = I alpha or, alpha = T/I = (26260 ft-lbf)(32.2 ft/sec^2)/(675000 lbm-ft^2) = 1.25 rad/sec^2

    1rpm = 2pi radians/min = 2pi/60 rad/sec = .105 rad/sec

    To go from 0 rad/sec to .105 rad/sec at a accel rate of 1.25 rad/sec^2 will take (.105 rad/sec)/(1.25 rad/sec^2) = .08 sec.

    Funny how it's still the same number as before............

    And I quote:
    Like I said before, the inertia of the wheel is a red-herring, and isn't part of the problem.
    Last edited: Sep 22, 2012
    2 people like this.
  11. alan white
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    alan white Senior Member

    I'd bet the wheel could be gently started by having hidden water spigots fill the cavities at the top of the wheel. Once run up to speed an electric motor could maintain a steady rpm. I mighty have suggested simply using water to run the wheel, just like an old time mill on a stream.
    While the water "drive" would be less efficient by itself, it would be very cheap to install and run.
  12. Frosty

    Frosty Previous Member

    All this for a visual attraction. Just do away with it and make the entrance fee cheaper.

    This 12000 ilb wheel will have 6000 coming down and 6000 going up. and acceleration is not linear.

    Once moving and if not in the water it would need minimal Hp to rotate
  13. drortiz1
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    drortiz1 Junior Member

    Thank you very much for your help, now I have a better understanding of how things really work. Just to summarized my motor will work close to its MAX torque for .3g's in order to accelerate the wheel. ones the wheel is moving I only have to worry about the water and since the water is only 10626 lbs. ft (assuming I did my integration the right way) a 5 HP should be enough.
    we did a job a few years ago using an air hose hidden in the water. The air created a current that pushed the wheel and also it gave the illusion that the boat was actually moving. we may do the same thing with this one but it is always nice to learn new things and look for new options.
    "All this for a visual attraction. Just do away with it and make the entrance fee cheaper."
    I like that idea!
  14. gonzo
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    gonzo Senior Member

    None of these calculations are adding the power needed to move the huge amount of water the paddles are going to be pushing. That also will depend on whether the water is in a free flowing or closed system.

  15. DCockey
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    DCockey Senior Member

    Simple approximation of the force on a blade.

    Force = Cd * 0.5 * water density * linear velocity squared * frontal area

    Density of fresh water = 62.4 lbm / ft^3 = 1.94 slugs / ft^3

    Velocity = 47 ft/min = 0.785 ft/sec

    A guess for Cd of a single paddle is 2.0 assuming the width of the paddle is much greater than the height. This does not take into account presence of the bottom, sides or free surface.

    Put these values into equation above and a coarse estimate for the force on a single paddle is 26 lbf without considering the sides, bottom or free surface. Also there are 2 paddles submerged at a time. I'll use 70 lbf as a relatively conservative initial guess for the resistance force of a pair of submerged paddles.

    Power = Force * Velocity

    Power = 70 lbf * 47 ft/min = 3290 ft * lbf / min = 0.1 HP

    Anyone see an error in the math or calculations, or disagrees with the assumptions?

    By the way, the rim speed is less than 0.5 knots.
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