 I myself Sachin Rathod working as assistant professor in mechanical engineering department from Wall Street Stop Technology, Solapur. Today we will discuss about the topic that is the selection of the belt part 7 from the course Machine Design 1. So, the learning outcome of this session is the learner will able to design the flat belt from the manufacturing catalog. So, we will just discuss how to select the proper belt for the particular applications by considering one case study. So, this is a statement by referring this statement we have to select the proper type of the belt which will run without failure. The statement is that it is required to select the flat belt drive for a compressor running at 720 rpm which is driven by 25 kW power having the speed of 1440 rpm motor. The space available for a central distance of 3 meter the belt is open type. So, through this problem statement you will get the given data, so this is a given data. So, in that step number 1 calculate the pitch diameter of the pulley that the we are having the two pulley one is a smaller pulley one is a larger pulley for the smaller pulley we are using the letters small d and for the larger pulley we are using the letter capital d. So, just consider the belt velocity is 18 meter per second and calculate the small d. So, we are knowing the formula for finding the velocity or for finding the diameter that the relation between the velocity and the speed is v is equal to pi d n1 by 60 rpm put that the value of the n1 you will get the diameter of the smaller pulley. So, we are getting the diameter of the smaller pulley is 238.73 mm. Next we have to select the pitch diameter small d of the smaller pulley for that purpose we have to refer the standard design data book. So, these are the diameter of the pulley from the standard data book. So, we have to select the diameter from this table. So, as we are getting the diameter of the smaller pulley is 238.73 mm we have to check out whether it is in this table or not. So, here we can select the diameter of the smaller pulley is 250 mm. So, from the standard data manufacturing data take d is equal to 250 mm. Now, we have to find out the value of capital D diameter of the larger pulley that the relation we are knowing n1 by n2 is equal to capital D by small d that the speed is inversely proportional to the diameter. So, we are getting this equation. So, from this equations we have to calculate the value of capital D. So, you will get the value of the capital D is 500 mm. So, it is in the table that is the 500. Now, in the step number 2 calculate the corrected velocity of the bed. So, we are having the formula v corrected is equal to pi d n1 by 60. So, just put the value that the small d is 250 mm and the n1 speed is 1440 rpm. You will get the corrected velocity as 18.85 m per second this is a corrected velocity. Now, the step number 3 calculate the load correction factor FA from the manufacturing data. So, the already in the question they had given us the value of the load correction factor FA that is the 1.3. If they had not given us the value of the FA you have to refer this table which is from the manufacturing data book and in that we have to check out. So, for which applications we are using the bell drive. So, in the question or in the problem statement they had given us we have to use for the running the compressor and for the compressor the correction factor is 1.3. So, we have to take the load correction factor is 1.3. So, in the step number 4 calculate the angle of contact for the smaller pulley. So, angle of contact for the or the wrap angle we are calculating by using this equation. So, just put the value of capital D small d and the center distance you will get the wrap angle for the smaller pulley. So, in the step number 5 calculate the arc of contact factor FD from the manufacturing data book. So, this is a table from the manufacturing data book you have to calculate the wrap angle alpha s. So, you have the alpha s value is 175.23 degree. So, from this tables you have to calculate the value of alpha s. So, here the alpha s is 170 and 180 the two data we are knowing and the value we have to find out that is a 175.23 the factor we have to find out that will which will lie in between 1.04 and 1. So, it is lying in between these two value. So, you have to use the interpolation formula and calculate that factor that by using the interpolation formula we are getting FD is equal to 1.019. Next in the step number 6 calculate the design power. So, as we are calculated the factors you will get the design power. So, design power is calculated for the flat bell drive is P D is equal to P into F A into F D. So, just put the value of load correction factor and the arc factor is 1.3 and 1.019 that we have calculated from the previous slide just put that value you will get the design power. So, here the design power is 33.12 kilo Watt. So, you can think about this just pause the video and think what is the difference between design power and you are the actual power. So, why the design power is more than that of the actual power. So, you can pause this video and you can think about this. So, in the next step step number 7 calculate the power rating of the belt. The power rating is nothing but you are the power transmitting capacity of the belt. So, you can calculate by using this equation the PR is equal to 0.0118 divided by 5.08 into v corrected. So, this is a formula we are getting for the high speed belt. So, if you put the corrected velocity that we had already find out that is 18.85 you will get the power rating of the belt is equal to 0.0476 kilo Watt. In the step number 8 calculate the width of the belt. So, the width of the belt is calculated by using this equation that the width of the belt W is equal to design power divided by rated power into the number of the ply. So, this is a belt suppose this is a open flat belt in which if you observe these are the number of the layers 1, 2, 3, 4 so this layer is called as a ply. So, there are the 4 ply is having the total width W. So, you have to find out the number of the ply required for the belt and the total width. So, by referring these tables you have to find out the width as well as the number of the ply for the belt. So, the PD value we are knowing the design power and the rated power we are knowing. So, just consider if you put the number of the ply as a 3 in this equation calculate the W. So, if you put the value of number of the ply is equal to 3 you will get the W value that is the width is 252.28 mm. Now, we have to check the first row for the 3 ply the width is ranging from 25 to 76. So, as we are getting the width of the ply or the width of the belt is 252.28 it is not ranging in the 3 ply. So, you cannot take the 3 ply. Now, put the number of the ply is equal to 4 in this equation calculate the W. So, the width we are getting 189.04 mm. So, we have to go for the second row for the 4 ply this value is not ranging that is up to 152 you can consider the width. So, 189 is not ranging in the second row. So, you cannot consider the 4 ply. Now, if you put the number of the ply is equal to 5 calculate the W. So, W value we are getting 151.23 mm. Now, I will check the third row. In the third row the value yes it is ranging up to 152 that is comes. So, we are getting the value of the W is 151.23 so, we can consider the value as a 152. So that we are having the value the width of the belt is 152 mm wide and we can use the number of the ply as 5. So, in the step number 8 we had calculated the number of the ply and the width of the belt. In the step number 9 calculate the length of the belt. So, already we are knowing the equation for the open belt drive just put the value of capital D small d and c in this equation you will get the length of the belt. So, by this way we are calculating the length of the belt. So, by the above steps the final design solutions will be obtained as so, first we had calculated the diameter of the smaller pulley at the small d is equal to 250 mm. The diameter of the larger pulley is 500 mm. The width of the belt is 152 mm. Number of the ply is required or the number of the layers are 5 and the length of the belt is 7183.31 mm. So, this will be the your last specification of the belt which is 7.5 meter length of 152 mm width and 5 ply high speed belting. So, this specification of the belt is required to identify the belt which we are going to mount over the pulley. So, we will get the working of that belt drive, open belt drive without the failure. So, I have taken a reference for this PPT from the design of machine element book which is written by VB Vandari.