 Welcome back, up until now we were looking at heat transfer devices as open systems. Then we looked at work transfer devices as open systems where we looked at turbines and compressors, we looked at their H's diagrams and here we assumed that the open system was adiabatic that is there is no heat transfer. Then we looked at an adiabatic nozzle where the main purpose was to increase the velocity of the fluid. We look at another adiabatic device now where we do not expect either heat transfer, work transfer or an increase in velocity. We just expect the fluid to flow through a duct and you realize that a duct is a very common system in many engineering systems. We have ducts going from the boiler to the turbine, we have ducts which are connecting compressors to condensers and etc. So we can have a simple open system as a duct where we have an inlet and exit. Of course we expect that whenever we have a duct if it is well insulated the q dot is 0, we do not expect any work transfer from it and it is not a nozzle where we expect the velocity to increase. And hence we will write the first law as follows just writing q dot minus w dot s is equal to m dot h e minus h i plus v e square by 2 minus v i square by 2 plus g z e minus z i and we draw a schematic of the duct. Let us say this is the system, there is an inlet, the duct probably went out like this, we do not know. There is a change in z, there could be a change, so there is a change in z, there could be a change in the velocity and there could be a change in h e. But overall we expect that it is well insulated and this is nearly equal to 0. We definitely do not expect shaft work out, there is no work output expected, so this we put as 0. So what do we get as first law, we get h i plus v i square by 2 plus g z i is equal to h e plus v e square by 2 plus g z e. So this is what we get from this situation. Now if we think of a situation where the temperature does not change and it is a fluid like an ideal gas, the u also does not change and since h is equal to u plus p v, we say if, so this is an if here, if u i equal to u e probably because the temperature does not change, then instead of h we can remove the u and write only p v here and since p v can be written as p by rho, we would have an equation instead of h i, we would write p i by rho i, the remaining terms would be the same and this would of course on the right side be equal to p e upon rho e plus v e square by 2 plus g z e. Now this is a reasonably familiar equation, if one has studied fluid mechanics, this is nothing but the Bernoulli equation. One should of course notice that in fluid mechanics, this equation was derived from the momentum equation along a streamline. Here we have used the energy equation, the first law for open systems derived. So, one can look at how the procedure was to derive this equation in fluid mechanics and here, but this is what we get for a simple fluid flowing through a duct and this is of course an adiabatic system. We could write down the second law in this case and since q dot is 0, we would get that s e actual is probably greater than or equal to s i, which is what it will always turn out for an adiabatic system and an adiabatic open system I should say. Thank you.