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Working Of Flywheel

  1. Jun 10, 2010 #1
    Hi guys,
    I have a very basic question which I have never been able to completely grasp the concept behind.

    This is regarding the Working Of A Flywheel.

    My question is:

    The Flywheel is said to store the kinetic energy from the excessive power produced during the power stroke and return it in the idle stoke to run the crank upwards to the T.D.C.

    And in the course it is said to SMOOTH OUT the fluctuation in the power produced in the power stroke.

    It is said that the flywheel absorbs energy by accelerating and supplies /gives back energy by decelerating.

    Now the purpose of the flywheel is said to be SMOOTHING THE FLUCTUATION in the power> i.e. To regulate the variation in the speed (angular) of the crankshaft due to variable power produced throughout the power stroke > i.e. to regulate and maintain a nearly uniform angularvelocity/rotational speed of the crankshaft/ultimately the jerking linear speed of the vehicle. (had the flywheel not been there)

    But what I couldnt really grasp was that:

    1. How actually does the extra power surge (in the first 90 degree of crank angle) gets absorbed by the flywheel (being connected to the crankshaft)?

    Wouldnt the flywheel simply get turned with a variable angular velocity throughout the power stroke as the the torque produced at the crank is continously varying due to varying force at the crank. i.e the the flywheel will rotate at a varying angular speed (increasing continously throughout the stroke)

    And inturn wouldnt the crankshaft itself rotate at the same (varying - increasing) speed as that of the flywheel (due to flywheel being mounted on the crankshaft so they both will have same angular velocity at any instant)

    2. And if this is the case How come the flywheel serve the purpose of Smoothing Out the fluctuation in speed?

    Wouldnt it keep on accelerating during the power stroke and then keep on decelrating in the next stroke and so would do the cankshaft resulting in a faster - slower type of speed at wheels (assume no transmissions for simplicity of theory)

    3. HOW ACTUALLY IS THE ENERGY ABSORBED and HOW ACTUALLY IS THE SPEED/POWER FLUCTUATION SMOOTHED OUT?

    4. Does really a flywheel be needed to give motion for the exhaust stroke as crank wheel will already have some inertial motion at the end of 90 degree crank angle which will drive the crank beyond the first half circle rotation?

    Please answer the 3rd question in maximum detail because my main intent is to build the clear concept of this.

    Thanks.
     
    Last edited: Jun 10, 2010
  2. jcsd
  3. Jun 10, 2010 #2
    Hey people,

    No one with a reply to my query??

    Come on guys, I need to understand this very clearly and as quickly as it can be. My further course in learning more of automobiles depends on this a lot.

    So please answer this question, and the more detailed or made easy the better, it can be more interactive so helps in discussion and geting the concept cleared and doubts solved.

    So please go ahead.
     
  4. Jun 10, 2010 #3
    The flywheel's "smoothing" out function is most applicable at idle. It simply decreases the "rate" of angular acceleration (when speeding up or down) during the cycle of an engine, "dampens" velocity fluctuations. When the car is in gear, the rest of the drivetrain is part of the "flywheel" system, so the actual flywheel is less important. The flywheel also make the car easier to drive if it is a manual, less likely to stall.

    I think your understanding sounds pretty complete. For #3, think of the flywheel as a drag on the system when it tries to change speed, but offers no drag at steady velocity. The flywheel actually slows the acceleration of the vehicle, that is fly racecars have the minimum mass flywheel possible, but they idle rough and are hard to drive from a stop.
     
  5. Jun 10, 2010 #4
    http://en.wikipedia.org/wiki/Internal_combustion_engine#Flywheels"
     
    Last edited by a moderator: Apr 25, 2017
  6. Jun 10, 2010 #5

    jack action

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    If you push on a block sliding on the ground with a force lower than static friction, the block will not move. So as soon as you will release the force, the block will stay where it is. Simple.

    If you push on the same block but with a force high enough to start motion then the force will be equal to the friction force plus the mass times the acceleration. The energy you are storing will be equal to ½mv². When you will release the force, the block will be at a velocity v, so it will not stay where it was like in the previous case. It will go on, because the energy stored will be released. It will go on for a distance x equals to the energy stored divided by the friction force. If m is very small, it will stop faster (less energy stored).

    Imagine that you are giving a series of pulses to a block such that you have an average speed of v for the block. If m is very large, the difference between vmax and vmin will be less than if m is very small. With a large mass, it will be very difficult to attained that velocity v, but once there, it will also be very difficult to slow down. And when you'll stop applying the pulses, all the initial energy you used to get to velocity v will be returned to you and the block won't stop easily.
     
  7. Jun 10, 2010 #6
    MRFMengineer, thanks a lot for the reply.

    But my actual doubt remains sort of unaddressed.

    My question is that suppose the engine be idle (no load and no transmissions drivetrains - for the sake of simplicity) then for a particular Power produced per cycle (by keeping a fixed amount of A/F mixture) when there is variation in presssure exerted (to be precise a very high pressure blast in the beginning of combustion nearer to T.D.C. and continually reducing pressure on the piston head thereafter till the end of power stroke).

    Then as a sresult of such pressure variation the torque produced at the crank will also vary in the same fashion (i.e. from a very high to continually lower). Thus this would naturally produce a very high and sudden high angular velocity at the beginning and then further down as there would be still some more force acting on the crank (though in small quantities as pressure would though reduce yet it would still add to the previous thrust on the piston) an increasing angular velocity of the flywheel and crankshaft (both being same assembly of shaft).

    Thus the angular velocity of the flywheel - crankshaft should go on increasing progressively in a power stroke. Then if we consider our assumption of no transmissions and the crankshaft being the direct shaft to drive wheels (for simplicity) - shouldnt the drive wheels also also rotate a a varying - infact increasing velocity during the power stroke.

    Then how do we say that the flywheel 'Dampens' or 'regulates' the fluctuation in the power / speed of the vehicle. Wouldnt the crankshaft rotate at a variable rate (angular velocity)

    This is the part which I could not understand clearly.

    Secondly wouldnt also the inertia of the crankshaft (even without a flywheel) during the power stroke continue to move it forward or rotate the shaft to start the exhaust stroke.

    Please answer these two questions in detail particularly the firrst part.

    All members are invited to reply. Guys get ahead and get me going.
     
  8. Jun 10, 2010 #7
    No.
    When force is along the radius rod,torque is zero, no matter how it is force. (upper dead point).
    From this point of view, the torque is maximum when the force is perpendicular to the radius rod.And then decreases up to zero. (Lower dead point) Is a sinusoidal variation.

    Adds force variation you described earlier, and get a complex variation of torque.
     
  9. Jun 10, 2010 #8
    emiltr has it right, you have to remember the crankshat/rod/piston relationship, the piston cannot exert any torque on the crankshaft at TDC or BDC, with the max torgue applied at half stoke (rod journal at 90 deg).

    Think of the flywheel this way, if you had a one cylinder engine with no rotational mass, it would fire at TDC go to BDC on the power stoke, then stop and not rotate anymore. The "flywheel" aka the rotational intertial of the entire system absorbs energy on the power stroke (speeds up) and then dissipates (slows down) on the other portions of the cycle. Energy is continually being dissipating to friction and the drivetrain, but is only being "charged" during the power stoke. An engine with very little flywheel interia would be very jerky and rough (racecar style). Once a system is at high speed these effects are not notciable by the driver, but are still happening, just hundreds of time per second. And with an engine with 8 cylinders, these effects are even more averaged out as compared to a 1 cylinder.

    Two answer your 2 questions:
    1. Yes (it has to), the craftshaft rotates at variable velocity, it is just a question if it is noticable or not to the drive/sensor being used to observe it.

    2. Yes, the interia of the crankshaft might be enough to keep it going, F1 cars barely have a flywheel at all, it is just the clutch more or less. An F1 car would be pretty hard to drive in stop and go traffic, high idle speed and jerky response, stalls, etc... A "flywheel" just adds interia to the system, the flywheel could be part of the crank for that matter. Sports car/motorcycles have light flywheels, trucks/bulldozers etc have heavy ones. It is just a question of response.
     
  10. Jun 11, 2010 #9
    Thanks a lot guys for answering back.

    And MRFMengineer this was exactly the point I was looking to for a conformation and My comcept is clear now.

    Thanks again for the responses and looking forward to more cooperation like this in the future from you.
     
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