 Hello students. Myself, Siddhaswara B. Turzapure, Associate Professor, Department of Mechanical Engineering, Balchen Institute of Technology, Sulapur. So today, we are going to deal with the topic maximum fluctuation of energy in a flywheel. The learning outcome. At the end of this session, students will be able to derive a formula to determine maximum fluctuation of energy in a flywheel. Now, during the earlier sessions, we observed the turning movement diagram and the flywheel. So now in case of the maximum fluctuation of energy, again we are referring to the turning movement diagram. Let us say that this one it is representing crank angle and the y axis it is representing torque or turning movement. Then we are going to have the mean torque line is T mean. Then we are going to have the fluctuation of the energy like this. Now suppose we are having this area as A1. Let us say this area is A2, this one is A3, this one is A4 and this one is suppose A5. Now let us name the different points. So this one let it be A, this one it is B, this one it is C, this one it is D and say it is E and then it is F. Now we will calculate the energies at these different points. Let energy at A is equal to EA. Then we will have the notations as E for energy and then suffix it is going to indicate the energy at that particular point. So it is energy at B is equal to. Now here we can observe that in case of the turning movement diagram we are having T mean and the areas corresponding to the torque at different positions of the crank angle we are having the areas either on the upper side or on the lower side. So if the areas on the upper side we have to add the energy which is going to be say the excess energy than the required and when we are going to have the areas below the mean torque line we are going to deduct it from the available energy in the earlier stage. Now energy at point B is equal to, so it is energy at A then from A to B we are having the area on the upper side so it is going to be EA plus it is A1. Then energy at point C is equal to we are having this one as energy at B then from B to C we are moving. So the area is on the lower side it is energy at B minus it is A2. So this can be further written as energy at B we are knowing already this is energy at A plus A1 minus it is A2. Then energy at D we can calculate so energy at D is equal to from C to D we are moving so energy at C and from C to D the area is on the upper side of the mean torque line so it is plus it is A3 but energy at C we are knowing so it is energy at C is nothing but it is this one it is energy at A plus A1 minus A2 then it will be plus A3. So like this the energy is at different points we can calculate. Now in case of ours now suppose from the point A to the point F so A2F on the x axis now if it is representing one cycle now think of the energy at F so without calculating at D, E etc. So what will be the energy at F now see we have said that A to F now it is one cycle means after this again it is going to be the repetition of this pattern with reference to this torque and that crank angle that is F is going to be nothing but it is point A only that is so it is point A for the next cycle so it is energy at F will be equal to energy at A it is so this one you should remember so when you are going to calculate in case of the numerical etc so at the end what you should get is energy at the last point of the cycle should be equal to the energy at the first point of the cycle. Now in case of this one again let us introduce two parameters it is one it is omega mean so we are having the angular velocity of the crankshaft or the flywheel which is mounted on that one it is maximum and minimum. So it is we are going to have the ratio of these two it is omega max and omega mean so divided by 2 that is this one it is nothing but the average angular velocity of the flywheel or crankshaft. Then we are going to have say it is coefficient of fluctuation of speed this one is equal to say it is fluctuation referring to that one it is omega max and it is omega minimum and divided by it is omega mean so this one and this one is represented by coefficient of fluctuation of it is speed C suffix it is yes. Now we are interested in delta E that is the maximum fluctuation of energy so delta E it is max. So energy is continuously fluctuating we should focus for the maximum value of the energy and minimum value of the energy so this is going to be E max minus E minimum. Now we can have the formula it is but energy E can be represented as say it is I into half into it is omega square where I is this mass moment of inertia of flywheel and omega is angular velocity of flywheel omega it is omega mean so in case of the normal one now suppose we are going to have this one as say it is delta E max wheel height is equal to corresponding to maximum this one will be half I omega max square minus half I omega mean square. So you can take this one half I common then it is omega max square minus it is omega minimum square then we can rearrange the terms it is half I then we are going to have this one as omega max plus it is omega minimum and then multiplied by it is omega max minus it is omega minimum then we can think of this term it is omega max plus omega minimum divided by it is 2. So it is nothing but it is the mean or the average angular velocity so this one will be now I into omega it is mean now and then omega max minus omega minimum is there it is omega max minus omega minimum then we can think of again what we will have is we will have the rearrangement of terms. If you concentrate on the coefficient of fluctuation of speed it is C s is given by omega max minus omega minimum divided by omega mean so here we are having already omega max minus omega minimum then we can have the denominator side omega mean and one more omega mean we can have on the numerator side then see here it is I it is omega mean so it is 1 omega mean multiplied by second omega mean so it is I omega mean square and this one is nothing but it is coefficient of fluctuation of speed so it is delta E comes out to be I that is the mass moment of inertia of the flywheel and then omega mean square into it is C s then the same can be written as here it is I is equal to mass of the flywheel into the radius of gyration of the flywheel and in case of the flywheel we are going to have the radius of gyration can be written as K is equal to it is R it is mean radius of now this one will be I is equal to it is M R square and then delta E is equal to M R square and it is omega mean square C s this can be written as delta E is equal to R omega is nothing but it is linear velocity it is mean so it is mean square then it is C s this is another form of this fluctuation of energy which is going to be the maximum so these are the references which are used for the preparation of this session thank you