 What we'll do now is solve an example problem that enables us to use some of the tables that are in the back of the textbook, and specifically we'll look at the steam table. So what we're going to consider is a piston-cylinder device. So this is an example problem. So we have a piston-cylinder device. So we have a piston-cylinder device that contains 0.1 meters cubed of liquid water and 0.9 meters cubed of water vapor in equilibrium at 800 kPa. We're then told that heat is transferred at constant pressure until the temperature reaches 350 degrees C. And so a schematic of what we're looking at, it's a piston-cylinder device whereby the piston is free to move up or down depending upon the pressure within the system. And we have water, and it's both in the vapor and in the liquid state. Water is 800 kPa, and so we have a little bit in the liquid. Some of it is in the liquid state, and the rest of it is as a vapor. So that describes the problem, and what it wants us to do is find a number of things. It wants us to find the initial temperature of the water, determine the total mass of the water, calculate the final volume, and show the process on a PV diagram with respect to the saturation line. So writing those out, find, okay, so that's the problem that we are going to solve. So when you have a problem like this, the best thing to do is to begin by writing out the information that you know, and the unknowns are the things that we have to solve for there. So let's begin by writing out the information that we know. So we know that we have a piston cylinder device. We know that the volume of the liquid is 0.1 meters cubed, and the volume of the gas phase, or the vapor phase, is 0.9 meters cubed. We know that the pressure, we're told it's a constant pressure process, and consequently P1 and P2 will be the same. So it is 800 kPa, and we also know that we have some process. We go from state 1 to state 2, and that consists of heat transfer until we get to state 2, where we know the temperature. We're told it is 350 degrees C, and P1 equals P2 equals a constant in this process. So it's a constant pressure process. So what we need to do is find the four variables, being initial temperature, total mass, final volume, and showing the process on a PV diagram. Now whatever you're faced with these problems, another thing that is very handy to do is to write out a schematic of the process in terms of what you understand, what is going on. And I will write it in terms of a TV, although they want a PV diagram. In the end, I prefer to look at things in terms of the TV diagram. So there's our two phase region, a constant pressure line coming along. So this would be P equals 800 kPa. So we start off with a mixture of both vapor and liquid, meaning that we're somewhere in the two phase region where we don't know. So I'll just denote that with one, and it's someplace in the middle of the two phase region there. And then with our process, what we're doing is we're adding heat, and we're heating up the fluid, or the system, and eventually we're going to move up to some point here on our superheated region, and that would be at a temperature of 350 degrees C. So that's what we know about the process in terms of the schematic. So let's go into the analysis of this problem. So for analysis, what we know is the pressure one is 800 kPa, which is also 0.8 mPa. Now what we can do, remember I said that we were in the two phase region at state one as shown on the schematic. So we can go into our tables at the back of the book, and in this particular case what we want to do is look at the saturated water pressure table. And the reason for that is we know the pressure. We don't know the temperature, we know the pressure. And so if we look there, and we look for a pressure of 800 kPa, we will find a value for that, and it turns out there's a saturation temperature. So from the table we can determine that the saturation temperature is 170.41 degrees C. So that's one piece of information we have. What that also tells us is that, and it's as I sketched on the figure earlier, at T2 equals 350 degrees C, that's higher than 170, and that means that we are superheated. Which makes sense because we start off in the two phase region, we add heat, there's only one place we're going to go and that's into superheated vapor. Okay, so the first question asked us to determine the initial temperature T1. Well we just obtained that by getting the saturation temperature. So we can write T1 equals 170.41 degrees Celsius. Now the second thing they want us to determine is the total mass of the system. So again what we'll do is state 1, we go back to our saturated water pressure table and we look up the values for the specific volume of the liquid. And we also know the total volume of the liquid, it was 0.1 meters cubed. And we know of the vapor phase, the specific volume. And we also know that from the problem statement itself is 0.9 meters cubed. Therefore we can determine the mass of the liquid. So doing this, we obtain that the total mass is equal to 93.4298 kilograms. So that answers the second part of the problem. So there we've solved part one and part two. The next thing they ask us to do is to determine the final volume when we're at state two. So final volume. So what do we know there? We know that the pressure is 800 kPa. And we know that the temperature there is 350 degrees C. Well remember earlier we said that we were in the superheated region. So we go back into our tables and we look under superheated water. And from that we can then determine, we'll find a table at a pressure of 0.8 MPa. And then we look for 350 degrees Celsius. And from there we can obtain the specific volume. So we get the specific volume at state two. And from that, given that we already know the mass, we can determine the total volume at state two. And that's going to be the total mass multiplied by the specific volume at state two. And from that we get 33.111 meters cubed. So that solves the third part of the problem. And finally what they wanted us to do is they wanted us to sketch the process on a PV diagram. So for that what we can do, we read out our pressure, specific volume. And this is what the lines would look like. That constant temperature on a PV diagram. So what I've drawn here are two different isotherms. One is at 170.43 degrees C. And the other one is an isotherm at 350 degrees C. And remember we started off at saturated liquid at 800 kPa. The pressure does not change, but we're somewhere in here, state one. And then with our process we move to the right, to the right, until we hit the 350 degrees C line or isotherm. So that's what our process would look like on a PV diagram. And that concludes the solution to that particular problem.