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Working out the angular acceleration of a flywheel

  1. Feb 9, 2016 #1
    I am trying to work out the angular acceleration of a small flywheel (with axle running through) when an attached piston (by means of a pin) with an outstroke force acts on it. I assumed I would need to work out the torque using the formula torque = force x radius, and find out the flywheel’s and axle’s moment of inertia using I = 1/2 mr^2 , since the flywheel is shaped as a hollow cylinder, in order to calculate the angular acceleration.

    Flywheel’s outer diameter = 30mm

    Flywheel’s inner radius = 10mm

    Axle diameter = 6mm

    Mass of flywheel = 0.075 kg

    Mass of axle = 0.00888 kg

    Distance from centre of wheel to pin = 11.25 mm

    Diameter of piston/cylinder bore = 10mm

    Cylinder air pressure = 85 PSI = 0.586 MPa

    This is what I did but I got an unrealistic answer. Ignoring frictional resistance

    Working out the piston outstroke force

    F=pa = (0.586x10^6) x π(5x10^-3)^2 = 46.0 N

    Working out the torque

    T=Fr = 46(11.25x10^-3) = 0.518 Nm

    Working out the moment of inertia for the flywheel

    I = 1/2*mr^2 = (1/2) (0.075) (15x10^-3)^2 = 8.4375x10^-6

    Working out the moment of inertia for the axle

    I = 1/2*mr^2 = (1/2) (0.00888) (3x10^-3)^2 = 39.96x10^-9

    Therefore, total moment of inertia is 8.4375x10^-6 + 39.96x10^-9 = 8.48x10^-6

    Using torque = moment of inertia x angular acceleration,

    Angular acceleration = 0.518/(8.48x10^-6) = 61103 rads^-2

    This value seems ridiculous. I must have gone wrong somewhere?
  2. jcsd
  3. Feb 9, 2016 #2


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    I haven't worked the problem completely through, but your torque calculation is in error. The radius should be 11.25x10^-2 m = 5.18 Nm which will increase the acceleration by a factor of 10.
  4. Feb 9, 2016 #3
    Unless you're using bearings, shouldn't your axle be approaching 10mm? Given this isn't a theoretical situation.
  5. Feb 10, 2016 #4
    Thanks for your reply, I am making use of brass bearings.
  6. Feb 10, 2016 #5
    Thanks for your input, perhaps I am mistaken, but wouldn't that mean the radius (distance from centre of wheel to pin) is in centimeters, not millimeters, since 11.25x10^-2 m = 11.25 cm = 112.5 mm, and the actual radius is 11.25 mm, not 11.25 cm?
  7. Feb 10, 2016 #6


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    For each move of the decimal to the left you divide by 10; so, if you start with 11.25 mm, the first move to the left is 1.125 cm and the next move to the left is .1125 m. As result, 11.25 mm = 11.25x10^-2 m
  8. Feb 11, 2016 #7
    I inputted the value of 0.1125 into a conversion calculator, and this was the result.


    0.1125 m is in fact equal to 11.25 cm, not 11.25 mm.
  9. Feb 11, 2016 #8


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    The accelerating torque on the flywheel is not constant .

    It is roughly a maximum at mid stroke of piston and zero at start and end .

    Also one single acting piston only applies force to crank and hence torque to flywheel for something less than half of each crankshaft rotation .
  10. Feb 11, 2016 #9
    Thanks for your reply, I have produced a basic diagram (below) to show how the piston is connected to the crank-wheel (which I am treating as a flywheel). All the dimensions stated in the original post still apply.

    What approach should I take to ultimately work out the angular velocity* of the axle, considering my calculations are incorrect?

    *I assumed I would need to work out the angular acceleration first in order to find the angular velocity
  11. Feb 11, 2016 #10


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    What you are basically describing is a simple oscillating cylinder engine - similar design to a toy steam engine .

    You need to start in a different place to make any headway in analyzing the performance of an engine .

    Basic reason for having a flywheel on an engine is to prevent speed variations . Most engine design is done by considering the crankshaft to be turning at a constant speed .

    The flywheel is then designed to even out the cyclic variations in driving torque from the cylinders (and sometimes in the reaction torque from the load) and keep the engine running smoothly .

    With one cylinder and one active side of piston there are long dead zones in the rotation of the crank where the cylinder is not generating any driving force and the flywheel has to perform the additional task of keeping the engine running until the cylinder produces the next power stroke .

    With the engine running at steady speed and a known load you can then work out the temperature/pressure/quantity of air or steam needed to make the engine work under those conditions and develop the practical design of the engine .

    Some ways of improving your design would be :

    To make cylinder double acting - ie have pressure alternately on each face of piston .
    Have two double acting cylinders acting on cranks at 90 deg to each other .

    Having more effective working strokes reduces the flywheel weight requirement considerably .

    There's a lot more to all this . Ask any questions you like .
  12. Feb 11, 2016 #11


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    One of the first things that I ever designed and built was an oscillating cylinder steam engine .

    This one is a bit cruder but it's fun to watch working :

  13. Feb 11, 2016 #12


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    I apologize for my error. I did the same verification check and found you are absolutely correct. At first, I couldn't figure out my error; and then I realized in the metric system there is no designation for "one tenth of a meter" so a 100 divisor is required to get from cm to m.
    Being someone that has spent my entire life in the US with our inch system the crossover to the metric system sometimes has its perils.
    The normal succession for our US decimal inch system is thousands/hundredths/tenths/inches, so the metric jump from cm to m without an intervening term just simply slid right by me when running my powers of ten conversion chain.
  14. Feb 17, 2016 #13
    Thank you for your input, sorry for the late reply. I appreciate your design improvement suggestions, however currently I am looking to analyse the engine's performance as it is now, and then moving on to improve its design (taking your suggestions on board!). In order to see the effects of the changing the design, I need to work out the current engine's axle velocity, and I'm not sure on how to do this (as my attempts have proved fruitless). The image below pictures the engine as it is now, where the piston acts on the crank-wheel which rotates the axle, which in turn rotates the flywheel.

    I wish to work out the velocity of the axle under a piston outstroke force of 46N, and I’m wondering how to go about this.

    Thanks again.


    Attached Files:

  15. Feb 18, 2016 #14


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    You can only work out the force generated by the pressure on the piston and then work out the (varying) axle torque generated by the piston force on the crank pin . The actual running speed will depend on many things but primarily on the load that the engine is driving .

    Assuming adequate supply of air/steam :

    No load at all - engine will run at high speed .

    Load within practical working range - engine will run at a moderate speed .

    Excessive load - engine will stall .
  16. Feb 21, 2016 #15
    Can the running speed be simply calculated when the engine is not driving a load? Would I need to work out the varying torque on the flywheel, or can I assume the torque is constant if I were to calculate the running speed theoretically?
  17. Feb 21, 2016 #16


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    The maximum no load speed will occur when all power developed by the engine is wasted in overcoming internal friction .

    There is usually no easy way to calculate this speed .

    Your second idea of specifying a load torque and then estimating running speed is possible and an easier calculation .
  18. Feb 21, 2016 #17
    If I were to say the applied torque generated by the piston force on the crank pin is 0.6 Nm, and this torque is constant, where would I go from here to estimate the running speed?
  19. Feb 24, 2016 #18


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    Output power = Area of piston * Length of stroke * Mean effective pressure * number of power strokes/rev * Number of revs/second .

    Output Torque = Output Power / ( Number of revs/second * 2 * Pi )

    Mean effective pressure can be taken as about 0.85 * supply pressure .

    Number of power strokes/rev is 1 for a single cylinder single acting engine .

    The above assumes that the practical arrangement of engine and flywheel allow it to work smoothly and continuously .
  20. Feb 24, 2016 #19


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    The answers for Output Power and Torque are averaged values for complete revs of the engine . There will be some cyclic variation as previously mentioned .
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