WEEKS 14-15 Dynamics of Machinery References Theory of Machines and Mechanisms, J.J.Uicker, G.R.Pennock ve J.E. Shigley, 2003 DESIGN OF MACHINERY, Robert L, Norton. 1999 DYNAMICS OF MACHINERY, Prof.Dr. Sadettin Kapucu Theory of Machines, Khurmi, R. et al.; 2005 Prof.Dr.Hasan ÖZTÜRK Dr.H.ÖZTÜRK-2010

FLYWHEEL A flywheel is an energy storage device. It absorbs mechanical energy by increasing its angular velocity and delivers energy by decreasing its angular velocity. Commonly, the flywheel is used to smooth the flow of energy between a power source and its load. If the load happens to be a punch press, the actual punching operation requires energy for only a fraction of its motion cycle. If the power source happens to be a two-cylinder four-cycle engine, the engine delivers energy during only about half of its motion cycle. More recent applications under investigation involve using a flywheel to absorb braking energy and deliver accelerating energy for an automobile and to act as energy-smoothing devices for electric utilities as well as solar and wind-power generating facilities. Electric railways have long used regenerative braking by feeding braking energy back into power lines, but newer and stronger materials now make the flywheel more feasible for such purposes Prof.Dr.Hasan ÖZTÜRK

FLYWHEEL There are two general situations which we will be faced with in mechanical systems non smooth operation due to fluctuation of speed. Output= Average of input Input= Average of output Input Output Punching Machine IC Engine Electrical Motor Generator Prof.Dr.Hasan ÖZTÜRK

FLYWHEEL ENERGY: The near figure shows a flywheel designed as a flat circular disk, attached to a motor shaft which might also be the derive shaft for the crank of our linkage. The motor supplied a torque magnitude Ti which we would like to be as constant as possible, i.e., to be equal to the average torque Tavg . The load, on the other side of the flywheel, demands a torque To which is time varying as shown in the below figure. The kinetic energy in a rotating system is: where I is the moment of inertia of all rotating mass on the shaft. This includes the I of the motor rotor and of the linkage crank plus that of the flywheel, We want to determine how much I we need to add in the form of a flywheel to reduce the speed variation of the shaft, to an acceptable level. We begin by writing Newton's law for the free-body diagram in the above Figure. Prof.Dr.Hasan ÖZTÜRK

Prof.Dr.Hasan ÖZTÜRK

OR Prof.Dr.Hasan ÖZTÜRK

Prof.Dr.Hasan ÖZTÜRK

Prof.Dr.Hasan ÖZTÜRK

the torque-time diagram …….1 The left side or this expression represents the change in energy E between the maximum and minimum shaft ω's and is equal to the area under the torque-time diagram between those extreme values of ω. The right side of equation (2) is the change in energy stored in the flywheel. The only way we can extract energy from the flywheel is to slow it down as shown in equation (1). Adding energy will speed it up. Thus it is impossible to obtain exactly constant shaft velocity in the face of changing energy demands by the load. The best we can do is to minimize the speed variation (ωmax- ωmin) by providing a flywheel with sufficiently large I. ……..2 the torque-time diagram Prof.Dr.Hasan ÖZTÜRK

Find the total energy variation over one cycle Example: An input torque-time function which varies over its cycle. The torque is varying during the 3600 cycle about its average value. Nm Find the total energy variation over one cycle Calculate the average value of the torque-time function over one cycle. which in this case is 70.2 Nm. Note that the integration on the left side of equation (4) is done with respect to the average line of the torque function, not with respect to the  axis. Prof.Dr.Hasan ÖZTÜRK Nm-rad

Prof.Dr.Hasan ÖZTÜRK

Coefficient of fluctuation of speed System Permissible k Pumps, Shearing machines 1/5 – 1/30 Machine Tools, Textile Machines 1/40 – 1/50 Generators 1/100 – 1/300 Automobiles 1/200 – 1/300 Aircraft Engines 1/1000 – 1/2000 Prof.Dr.Hasan ÖZTÜRK

1 mm2 on turning moment diagram =600*p/60=31.42Nm Example: The turning moment diagram for a multicylinder engine has been drawn to a scale 1 mm = 600N-m vertically and 1 mm = 3o horizontally. The intercepted areas between the output torque curve and the mean resistance line, taken in order from one end, are as follows: +52, -124, +92, 140, +85, -72 and +107 mm2, when the engine is running at a speed of 600 rpm. If the total fluctuation speed is not to exceed -+1.5 % of the mean, find the necessary mass of the flywheel of radius of gyration 0.5 m. Given: Since the total fluctuation of speed is not exceed +- 1.5 % of the mean speed, therefore Since the turning moment scale is 1 mm=600N-m and crank angle scale is 1 mm=3o=3op/180=p/60, therefore 1 mm2 on turning moment diagram =600*p/60=31.42Nm Prof.Dr.Hasan ÖZTÜRK

Given: 1 mm2 =31.42Nm Maximum Energy Minimum Energy Prof.Dr.Hasan ÖZTÜRK