 Hello students, myself, Siddhaswara Mithuljapure, Associate Professor, Department of Mechanical Engineering, Balchan Institute of Technology, Sulapur. So in this session, we are going to cover the numerical and maximum fluctuation of energy in flywheel, the learning outcome. At the end of this session, students will be able to solve numerical concerned with the maximum fluctuation of energy in a flywheel. Now the numerical which we are going to solve is given here, while reading the numerical, let us note the data which is given by the author. So in case of this, firstly the turning moment diagram for a petrol engine is drawn to the following scales, means the scale has been given here, so for the turning moment it is 1 mm is equal to 5 Newton meter, so it is given data, so it is 1 mm is equal to 5 Newton meter, then one more it is given data which is crank angle, so scale for crank angle it is 1 mm is equal to 1 degree. Let us have the conversion of this one into the radians, it is pi divided by it is 180 radians. Then the turning moment diagram repeats itself at every half revolution of the engine and the areas above and below, see the areas which are given these are above and below and the below the mean turning moment lines and then in the order, means this is the order we have to follow. Let us see how many areas are there, it is 1, 2, 3, 4, 5 and it is 6, total 6 areas are there. Let us draw the turning moment diagram here only, so this one it is representing it is crank angle and here it is torque or the turning moment and let us say that it is t it is mean and 6 areas are there in the order and these are above and below means 3 areas are to be drawn on the upper side and 3 areas are to be drawn on the lower side. So we can observe it is 1, 2, 3 these are above the mean torque line and 1, 2, 3 these are below the mean torque line. So now let us give the values to these areas, it is 295 mm square, then the lower side it is 685, then it is 40, then it is 340, then it is 960, then it is 270, then the rotating parts are equivalent to mass, mass has been given, so it is mass equal to m equal to it is 36 kg, then radius of variation of 150 mm, k radius of variation is equal to it is 150 mm is equal to 0.15 meters. Determine coefficient of fluctuation of speed when the engine runs at 1800 rpm that is engine speed they have given, so this is nothing but it is only one speed it is given means it is mean, so it is 1800 rpm. So in case of this, this can be converted to the angular velocity that is omega mean is equal to 1800 into 2 pi divided by it is 60, so you can cancel this 630, then comes out to be 65, it is radius per second, now in case of this, so we are interested in the value of the coefficient of fluctuation of speed, so what it is? It is to be determined, are you able to recall any formula which is containing coefficient of fluctuation of speed, then to some extent energy which is fluctuating, then it is mass, then it is kinetic, then it is radius of variation, then angular velocity etc. We are having one formula that is delta E max is equal to we are going to have i omega mean square it is Cs, i containing mk square and omega mean, then it is Cs means I will be able to determine from the given data, omega mean already knowing, so the value of coefficient of fluctuation of speed it is as and delta E it is to be determined, then only we will be able to determine coefficient of fluctuation of speed, now in case of this, let us give names to different points, it is A, B, then it is C, then it is D, E, then it is F and lastly it is G, so energy at A like it be equal to EA, then energy at B it is EB is equal to EA plus 295, then energy at C we are interested, energy at C will be equal to energy at B minus 685, so this we can write it as EA plus it is 295 minus it is 685, this comes out to be EA minus it is 390, so then it is ED, ED is equal to EC plus 40, so this EC is EA minus 390 it is plus 40, so this comes out to be EA minus it is 350, then it is energy at E, this one is equal to ED, then minus it is 340, so this energy at D we are knowing it is EA minus it is 350, minus it is 340, so this comes out to be EA minus it is 690, then we will go for energy at F it is energy at F is equal to energy at E plus it is 960, so this one will be energy at A it is E, it is EA minus it is 690 plus it is 960, then it comes out to be EA plus it is 270, so energy at lastly the point it is G is there, energy at G is equal to EF minus it is 270, E is equal to it is EA plus it is 270 minus 270, so these two will get cancelled we have got EA and this is EG, so see that EA is equal to EG, that is at the end of this cycle, we are going to get the energy which is at the start of this cycle, because after this one this cycle it is going to be repeated, so this tells you that the calculations which we have done it is correct, now further we are going to have the calculation of delta E max, delta E max is equal to E max minus E minimum, in case of this one E maximum we have to observe, so let us see how much it is EA plus or EM minus everywhere, so EA plus it is 295, so where it is 270 is there, so this one it is going to be maximum, so it is EA plus it is 295 minus EA minimum we can observe, so EA minus, E minimum we can observe, it is minus 390 minus it is 690 is there, so what we can have is we can have EA minus it is 690, so this will come out to be, EA will get cancelled actually, EA it is plus 295 minus EA plus it is 690, so this comes out to be, so this comes out to be minus it is 985, it is mm square, now this is in terms of the mm square, so we can have the conversion of this into, so it is 1 mm is equal to 5 Newton meter and the second one with reference to crank angle it is pi by 180, so it is 5 into pi upon 180 we can do, this comes out to be near about 86 joules, then we can put the value in this equation, equation number 1, so this 86 will be equal to m k square it is 36 into k square it is 0.15 square into omega mean it is 60 pi, so it is 60 pi bracket square into C s, so lastly after calculating this one we will observe that the value of C s is 0.003, so these are the references which are used for this session, thank you.