 Hello students, myself, Shri Siddeshwar B. Thulzapure, Associate Professor, Department of Mechanical Engineering, Balchen Institute of Technology, Sulapur. So, today, we are going to deal with the topic from the section Fluid Dynamics and the topic is application of Bernoulli's theorem, that is ORI-5 meter, the learning outcome. At the end of this session, students will be able to explain construction and working of ORI-5 meter. ORI-5 meter, we are knowing that it is used for the measurement of the discharge and it is out of the three applications, that is, venture meter is also used for the measurement of discharge and one more is there, that is, spit out tube used for the measurement of velocity. Now, let us see the construction of the ORI-5 meter. So, as compared to the construction of the venture meter, we are having the simplicity in case of the ORI-5 meter. So, in case of the ORI-5 meter, we have to cut the pipe at one of the sections, then we are going to introduce this ORI-5 split. So, this yellow colored part, it is shown as ORI-5 split here. So, it is having to some extent the sharp edge and then we are going to have the measurement of the difference in the pressure with the help of the manometer. Then we are having the flow from the left-hand side to the right-hand side. Now, we can observe that we are having the flow as to some extent it is with the diameter of that flow, fluid flow it is getting reduced. So, initially at this particular section, the fluid it is stretching to the amount is continuously and here in case of the ORI-5 split, here the sharp edge is there. So, because of the sharp edge of the ORI-5, we are going to have the reduction in the energy of the fluid. So, due to the loss of that energy due to the sharp edge of the ORI-5, the fluid flow it is getting contracted and we are having at this particular section the reduced diameter of the fluid flow it is and it is the smallest cross section in the area and which have been named as Vena contracta and the diameter of this one it is called as diameter of Vena contracta or Vena contracta diameter. So, this is one section where we are going to attach one of the tapings of the manometer and one more section is there that is before the ORI-5 split where the fluid flow is there and then these two tapings we are going to have and we are going to measure the pressure drop in between these two with the help of the differential manometer. Now fluid flow it is from the left hand side to the right hand side and we are going to consider the section one and the section two at the Vena contracta portion and we are having the ORI-5 split in between these two. Now, let us start with the derivation. So, in case of the discharge through the ORI-5 split, we are going to have this derivation. So, applying Bernoulli's theorem at section one and section two. Section one it is a pipe only and section two belongs to the Vena contracta. In case of this one say the Bernoulli's theorem it is P1 by rho g plus V1 square by 2g plus z1 is equal to it is P2 by rho g plus V2 square by 2g plus z2. So, here z1 is equal to z2 then rearranging the terms it is P1 by rho g minus P2 by rho g is equal to V2 square by 2g minus V1 square by 2g. So, let us say that the difference in the pressure head is equal to it is H. So, small h we can indicate and this one is equal to V2 square minus V1 square divided by it is 2g. So, we can rearrange the terms it is V2 square minus V1 square is equal to we can write it as 2g H. Now, in this case from continuity equation from continuity equation A1 V1 is equal to A2 V2. Now, in case of this one we are going to introduce one parameter called as coefficient of contraction it is coefficient of contraction is equal to Cc. So, we observed in case of the construction that the diameter of the flow it is getting reduced. So, there the diameter of the orifice is there one and next to that one the diameter of the flow at the Vena contracta portion that is the Vena contracta diameter. So, the ratios of these areas at the two places can be taken and here we are going to have the diameter considering the diameter of the Vena contracta portion the area can be calculated it is A2 and the area of the orifice we can have which is equal to AO we are writing. So, in this case it is A2 can be written as Cc into it is AO. So, now replacing this one. So, in case of the equation it is A1 V1 is equal to it is Cc into it is AO into it is V2 then V1 is equal to we can write it is Cc that is coefficient of contraction area of orifice then it is V2 divided by it is A1. Then the value of V1 we have got in terms of the V2. So, let us put this value in this equation where we are having V2 and V1 and now both can be put in terms of the V2 only. So, now it is V2 square minus see this one it is V1 square. So, here it will be Cc square AO square then it is V2 square divided by it will be A1 square will be equal to 2gh. Now can have the V2 square common. So, it is V2 square into bracket it is 1 minus Cc square AO square divided by it is A1 square and this one is equal to say it is 2gh. So, let us simplify the bracket it is V2 square into bracket it will be A1 square minus Cc square AO square. So, then it is divided by it is A1 square is equal to it is 2gh then it is V2 square is equal to we can write. So, this one A1 square will be taken to the right hand side A1 square into it is 2gh divided by it is A1 square minus Cc square and multiplied by AO square. Then it is V2 we can write say the square root of both the sides we have taken. So, this one will be A1 then it is under root 2gh then it is A1 square minus Cc square and AO square the whole thing will be under root. So, from equation it is discharge Q is equal to is equal to say we are having this one as A2 V2 then this Q is equal to this A2 into it is A1. So, value of V2 we are going to put it is under root 2gh divided by it is under root it is A1 square minus Cc square and AO square. Now, this one can be written as Q is equal to we will take this here it is A1 A2 it is under root 2gh divided by. So, it is under root A1 square minus Cc square and it is AO square. So, now, this can be written in some another form. So, what we can have is we can have replacement of some of the parts and then instead of going for it is A2 we can go for AO which is the area of the orifice and then next to that one we are going to have it is say A1 square then again Cc square is there then AO square is there instead of AO square also we can go for say A2 square it is. So, in case of this one AO square we will keep it as it is, but Cc square we are going to convert it. So, now the same can be written as say now here in case of this one A2 divided by under root it is A1 square minus Cc square into it is AO square can be written as just say it is Cd into it is under root A1 square minus AO square. So, now in this case we are going to have the in the earlier equation it is A1 and then it is A2. A2 we cannot measure. So, because of which we are going to have the replacement. Replacement will be in terms of the area of the orifice say it is area of the orifice we will write. So, the terms which we are changing you can observe it is A2 is replaced by AO A1 square is kept as it is and then this one we are going to change. So, this one is going to simplify the thing. So, Q is equal to it will be Cd into say it is A0 A1 it is under root 2gh divided by it is under root A1 square minus it is AO square. So, this is the equation which we are going to make use and this Cd it is nothing but it is coefficient of discharge. These are the references. Thank you.