 Welcome back. So, in this snippet we discuss another small interesting device as an open system. Let me read out the problem first and then let us discuss the device. So, we have wet steam at 10 bar is passed through a throttling calorimeter. This is the device that we are going to discuss. The state of steam after throttling is 0.75 bar 100 degrees centigrade. What is the dryness fraction of steam before throttling? So, this is the problem. Let us first discuss the device first. So, normally when we are trying to find out the state of any system, we try to get two of its properties and we notice that in many cases it is the pressure and temperature which can be rather easily measured. For example, we can have a pressure gauge and we can have either a thermometer or some other form of thermocouple to measure temperatures and once we get these two properties, it becomes easy to get the state of the system. Two is enough. So, what happens in a phase chain zone? So, the same P and T can apply to let us say the saturated liquid part, the same P and T can apply to the saturated vapor part or anywhere in between where the dryness fraction changes continuously between 0 and 1. So, if we really want the state of the system, we need either P and the dryness fraction or T and the dryness fraction and that would be sufficient. But how do we get the dryness fraction? Suppose we know P, then we know along which isobar we are going. If we want to know the dryness fraction X, then we would want either U or H or S. If we know this, if we can use U in the liquid state and U in the vapor state for that particular pressure and get our X, because we can just interpolate. The same holds for H and S, but it is difficult to do this and one of the methods to go ahead doing this is a throttling calorimeter. So, for example, if you have a flow here and U somehow measured your P and T and you notice that they correspond to the same saturated line. I mean, so T is T sat for the P that you have measured. So, that means we are not sure what the dryness fraction of the flow is. What do we do in such a situation? We extract a small portion of the flow and we throttle it. So, what do we do while we are throttling? We reduce its pressure. So, this means the pressure is high here. We reduce it to a lower pressure. For example, this is what you do if you are opening a tap partially. The outside pressure is at atmosphere and the pressure in the pipe represents the head that is there between the tap and whichever overhead tank is stored in. Or you could have it in refrigerators. There is thin long tube across which the pressure reduces from a higher pressure to a lower pressure. The higher pressure is what is available in the condenser and the lower pressure is there in the evaporator. We have not discussed these things, but this is what you normally see. For example, if there is a small valve which lets a high pressure gas escape, that process is throttling. This is what we do here. We extract a small portion that does not change your conditions inside wherever you are measuring. The velocities here are negligible. If we expand it to a much lower pressure, you see what happens. If I have an HS diagram here and if this is in the wet zone, if I start going to a lower pressure, I will essentially expand it heavily. Of course, if I had kept the same tube diameter here, now what happens is that when the pressure drops down, the density also drops down. In steady state, if we want to maintain the same volumetric flow rate, the velocity would have increased if the tube diameter would have remained the same. Normally, it is given a much wider cross section as you start dropping down. There is the pressure reduction device, which can look like either a small tube or some kind of a valve. What happens is because you have provided a reasonably large cross section, even in this range, the velocity is much low and you do not have to deal with high velocities. You can assume that this is well insulated. There is no Q dot or Q dot is negligible. There is no work transfer. If I look at how this looks on the HS diagram, this is the high pressure line here. This is the low pressure line here. You are at this point and if I write the first log Q dot minus the Q dot, we realize that there is no real change in the height. This is negligible. We have already said that we have ensured to our best that this is negligible, that the velocities themselves are small. The Q dot is 0. It is adiabatic. There is no W dot. What are we left with now? We are just left with h e minus h i is equal to 0 or h e is equal to h i. What does that mean here? It means that we have traversed here and come to a state on the lower pressure line. So, this is p lower and this is p higher. So, the enthalpy of this state is the same as the enthalpy of this state, but what we have ensured is that we have come into the superheated zone here because this is the saturation line here. So, this is what we actually are trying to ensure that we come into a zone where we can have a separate p and t because it is in the superheated zone. It is a vapor. So, if I measure p and t, then I can pinpoint the h here. If I can pinpoint the h here because the h here is the same as the h here, I have got one of the quantities apart from p and I can now determine my x. So, that is the whole principle here. So, this is what happens in a throttling calorimeter and now the problem becomes reasonably simple. What one of course has to realize is that this is what one needs to do. If we reduce to a pressure where let us say we are still within the dome, this is of no use to us because here we still cannot get two separate quantities which are distinct. So, what we really need to do is to get into this zone and that is the whole goal of the throttling calorimeter. So, what happens now? We can find out h e because all we do is go to the superheated tables and find out what h is at the given conditions. This is given as 0.75 bar which is 0.075 mega Pascal and 100 degrees centigrade. You will notice that this is in the superheated zone as far as 0.075 mega Pascal go. So, we find out h e and since h e is equal to h i and I know what is p i, all I need to do is look up h f and h g at p equal to this particular pressure. You will notice that this h i should fall in between these two values and once this is known, we can get our state and hence we can get our dryness fraction. So, once we realize what the open system does, it is a reasonably straightforward calculation. Thank you.