 Welcome back, we take up another problem in open systems and we discuss a device called the venturimeter. It is possible that you have seen this device described in fluid mechanics. Let us read the problem now. The problem is as follows. Water flows through a horizontal venturimeter at a steady state of 600 kilograms per minute. The inlet and throat diameters of the venturi are 6 and 3 centimeters respectively. If there is no transfer of heat or work and no change in internal energy and if the density remains constant at 1000 kilogram per meter cube, what will be the pressure drop between the inlet and throat in bar? The last answer could of course be in Pascal or any other units for pressure. So let us just go ahead. If you do not know what a venturi is, let me just draw it for you here, it is an open system and often described in fluid mechanics, but let us just start over here. So it is a device where the flow passes through a constriction like this. So it is a converging and then a diverging passage and it is often used to measure the flow rates, in which case one measures the pressure drop across the inlet part, inlet port here. So this is where the flow is going in, let me put it I and this is the throat region and normally one measures the pressures here and then tries to reduce the flow rate. This device is often used for mixing too and if you are a mechanical engineer, you may have heard at least in cars which were around a decade ago, there used to be a carburetor. Maybe there is still in some a few cars around and it is used for mixing the regular fuel with air. So this is normally the principle of a carburetor. What happens is, flow comes in from an end with a bigger cross section and as it passes through the throat, the cross section is smaller and hence the velocity increases. This leads to a reduction in pressure and that is the point where you try to suck in another fluid and mix with the first fluid. This is also used in some chemical mixers and but here we are just purely looking at the flow through the system and we are not bothered about mixing here. We are just looking at the pressure drop. As far as we are concerned, we will put the boundaries of our system here. This is close to the inlet and at the throat, we will put our exit because this is all that we are investigating in the present problem. We know what the mass flow rate is. It is given as 600 kilogram per minute. Of course, I divide by 60 to get it in kilograms per second. It is very easy. I just put it 10 kilograms per second. The diameter here is given d i and the diameter here is given d e. It is given as d throat, which means that a i can be calculated as pi by 4 d i square and this must be converted to meters. If it is 6 centimeters, this is 0.06 meters, I would not bother about the calculation here. Similarly, a e can be also calculated. E represents throat here. What do we have? We write down the first law again. It is a steady state system. It is what we assume. For further ado, we just write its m dot is constant h e minus h i plus v e square by 2 minus v i square by 2 plus g z e minus z i. You will notice that we have already mentioned that there is negligible heat transfer. It is an adiabatic system. There is no work transfer and it is a horizontal venturimeter. Some venturimeters can be arranged at an angle with a difference in heights. At that time, the difference in heights is significant right now because it is horizontal. The height at the inlet here and the height at the exit here is the same and this difference is as far as we are concerned is 0. Now you are left with only these terms. You realize that the m dot no longer appears in the equation because the left hand side is entirely 0. There is no more term on the right side and hence, we will just have h e minus h i plus v e square by 2 minus v i square by 2 is equal to 0. In this case, we can write h e as u e plus p e v e. Similarly, h i is u i plus p i p i. Now notice what else is given? It is given that the temperature does not change and which means that we can assume that u e is equal to u i. When we subtract these two here, the use go off. Similarly, the density does not change which means that the specific volume does not change. These two quantities are equal which means that I can write h e minus h i is equal to some specific volume p e minus p i or we can write it as p e minus p i upon rho and p e minus p i is nothing but the delta p which is expected and hence, we have finally p e minus p i upon rho plus v e square by 2 minus v i square by 2 is equal to 0 or p e minus p i is equal to v i square by 2 minus v e square by 2 multiplied by rho. Now you realize that v i and v e, these are the velocities can be obtained from mass flow rate. This is because mass flow rate is equal to rho i a i v i is the same as rho e a e v e and since a i is given because d i is given rho i is given this is 1000 kilogram per meter cube and this is the same as rho e. We can find out v i as m dot upon rho i a i. Similarly, v e would be m dot upon rho e a e of course, rho i and rho e are the same. Once we have v i and v e, we substitute it back here, we know rho we can get our delta p and that would finish our answer here. So, what one should realize is that the inlet velocity would be lesser than the exit velocity because as you go to lower cross sections v e would be higher and hence v i squared minus v e squared would be a negative quantity and hence p e minus p i would be a negative quantity which means the pressure at the throat would be lower than the pressure at the inlet and that is what we had mentioned all along right from the beginning. Thank you.