 We're now going to work an example problem involving an adiabatic air compressor and an adiabatic steam turbine and we'll look at both the first law as well as a calculation involving entropy for this combined unit. So there's a description of the example problem that we're going to work. What we have are two different systems. We have a compressor and we have a turbine. So the air compressor and it's being driven by an adiabatic. Notice that it's adiabatic. So you got to look for keywords whenever you're solving these problems. It's being driven by an adiabatic steam turbine. So here the keyword is adiabatic and adiabatic. Looking for properties, we have information here and the rate and it exits at 10 kPa and a quality of 0.92. That's for the steam and then the air. The air enters at 98 kPa, 295 at a rate of 10 kilograms per second and exits at 1 mPa and 550 kPa. So there we have the information. We're looking for two things. One is what is the net power to the generator. We have a coupled compressor and turbine and also what is the entropy generation for this process. So what I'm going to do, I'm going to draw it a little diagram that will help keep things a little straighter because it's a rather complex problem. So it's not a difficult problem. It's just a little complex. So here we have our air compressor and what it is being fed by is 98 kPa, 295 Kelvin, mass flow rate 10 kilograms per second and what I'm going to do, I'm going to write state information on this diagram here. It's adiabatic. That tells us Q dot is equal to 0 and then the air leaves and I'll call this state 2 at 1 mPa and 550 Kelvin. Now we have a different system that the air compressor is hooked up to. It's mechanically coupled to a steam turbine and the steam turbine is also mechanically coupled to a generator. So what the steam turbine is doing is it is taking some of its work that is being performed and some of it is going there and some of it is going there. And so what we need to do is find out the difference or the quantities of those and and then we can estimate one of the things that they want us to look for. But here what we have, this is a steam turbine and we have air coming in. I'll call this state 3. Be careful here because this is a different working fluid from what we looked at with the air flow. So don't get them mixed up. 500 degrees C and we're told that it exits a little on the wet side so that means that we have water droplets starting to form because our quality is 0.92. I'll call this state 4. 10 kPa pressure and the quality is 0.92. Now I'm not going to draw out the TS diagram for this or a PV diagram because we have two different working fluids so really we get a little on the complex side. Instead what I'm going to do, I'm going to dive straight into the tables in the back of your textbook and I'm going to write out the property data. And when I do this what I'm going to try to do is get as much property information as I can mainly looking for things like enthalpy and entropy and I'm going to get them at both the inlet and exit state for both the air compressor and the steam turbine. So beginning let's look at air. So looking at tables at the back of any thermodynamics book and I'm going to pull out the enthalpies. Now they also asked us to calculate the entropy generation. So what I'm going to do is pull out information, oops sorry that should be a 4, information on entropy and if you recall when we were deriving properties for an ideal gas we said that if you use exact specific heats you need to use this relationship which is entropy integrated with respect to the zero state which is the zero Kelvin state. And then for steam we go into the steam tables and we're beginning they say that at the end of the process we have a quality of 0.92 so that would tell me that we're starting off. Let's see we would be super heated and you can figure that out when you look on the tables but anyways H3 and then H4 we're in the two phase region so we have to use the quality and our relationship for evaluating values that are in the mid region or in the two phase region and then S4 out of the steam tables again we're in the two phase region. Okay so there we have all of the different properties that we'll probably need to solve this problem. What I'm next going to do is I'm going to apply the first law to the turbine and then to the air compressor. So we're writing out the first law and what we need to do now well we said it was adiabatic so heat transfer disappears and we're going to neglect both kinetic energy and potential energy and we end up with this which we can plug in our values and evaluate the work out of the turbine and given it's a work producing device it's positive through the first law and we can also say it's out. Now we're going to look at first law for the compressor again you know writing out the first law we can cancel out adiabatic there's no kinetic energy and no potential energy we're left with now when we plug in the values for enthalpy we get a negative number indicating that we're doing work on the system which makes sense because we're dealing with a compressor so you could also say that's twenty six oh five point seven kilowatts in now the network on the turbine is going to equal the work out of the turbine minus the work that we have to put into the compressor and with that we end up with this value so that is the answer to the first part of the problem the second thing they ask us to look at is entropy generation now we have not come out with or I haven't presented to you yet the entropy generation equation but it's not all that complex I'll write it out here sometimes the most complex part is figuring out where the heat flow is going in this last term however we're told that both the turbine and the air compressor are adiabatic so that kind of simplifies our lives on this problem that term disappears then all we have is mass flow rate times entropy change for individual fluids so entropy generation is going to be mass flow rate of air times the change in entropy for that fluid stream plus mass flow of steam times that entropy change there now for air it's an ideal gas and so what we need to do we need to use the relationship remember I said that we pulled out the entropy integrated with respect to absolute zero and that enables us to account for the fact that specific heats are changing in the process and when you plug the numbers in you get minus point zero three three seven that's the change of the air you combine the two together so you throw in the value for the steam and when you do all of that you get the entropy generation rate being twenty six point eight seven oh five kilowatts per Kelvin so that is the answer to the second part of the problem and that concludes this particular example problem as well as this lecture thank you very much