 I do not think you can see it, but you know what I mean exercises. Come to page 4 and we have F 2 properties of fluids and the first law set 2. F 2.0 is about plotting the various things which I leave to you ourselves. Let us look at 2.1. This is the simplest of the lot and the first ability we need to gain and the ability which we want our students to learn is to given some 2 properties find out 2 other properties. Now, here we have F 2.1 it is given that we have 1 kg of water substance. Water substance is ordinary water substance in a closed system and some 2 properties are given or a property and some specification is given and we have to determine H v x and S as appropriate. We will soon define x, but remember we first need to look at the D and E part. A and B it is very clear we are given saturated liquid at 5 bar and saturated vapor if otherwise mentioned unless otherwise mentioned saturated vapor is dry saturated vapor at 10 bar. So, you can read out these properties at from the steam table. Now, look at D and E here we have a P not given and a T not given and we have to determine first whether it is sub cool liquid the superheated steam and then read out the other properties. So, when P not and T not is given we should do first the following. If P not is greater than P critical or T not is greater than T critical then we have simply super critical steam horrible here otherwise we proceed as follows either A determine T sat at P not T not is greater than T sat at P not then what do we have we have super heated steam. If T not is less than T sat at P not then we have sub cooled determine that means read off P sat corresponding to T not here the equality change. If P not is less than P sat at T not then we have super heated steam and if P not is greater than sorry this should be P not P sat at T not then you have sub cooled liquid this is very clear from our phase diagram or P T diagram P not T not is given in the first case we are calculating the saturation pressure at P not suppose this is P not then this is sorry it go back suppose this is P not draw this line this line then this is T sat at P not. Then if the given temperature T not is higher than this we are in super heated steam zone if it is lower than this we are in the sub cooled liquid zone or if T not is given then we go like this this is P sat at T not then we look at the pressure if the pressure is lower than the P sat at T not then we are in the super heated vapor zone if it is higher than P sat at T not then we are in the sub cooled liquid zone. Now the question arises is what happens if we find that we are exactly on this line somewhere that is the saturation pressure and temperature sorry P not and T not correspond exactly to the saturation line and that means we have liquid and vapor together as we have seen earlier liquid and vapor in equilibrium this is the thermodynamic name for this the technical name for this is wet vapor and let us look at it like that when we have wet vapor liquid plus vapor in equilibrium and that means P and T are not independent. So we must select either P or T any one not both and some other property to completely define the state of the system. Let us draw two figures say pressure temperature our normal phase plot and let us say that our state lies exactly on this line then we do not know whether it is liquid or it is vapor in which we case now we say that the phase rule says pressure and temperature are not independent only one of them can be used. So let us use pressure say so instead of temperature now we use some other property. Let us say that we decide to use volume then on the volume plane the states look like this if this is the critical point the this is not to scale if I plot it to scale it will be pretty wide here we have the sub cooled liquid zone here we have the superheated vapor zone and this point here gets stretched from the saturated liquid f state to the dry saturated vapor g state and the state the actual state of the system could be anything from here which is saturated liquid to here dry saturated liquid or anything else in between. And now you can see that the pressure could be this but depending on the volume the state could be if the volume is this much the state could be saturated liquid if the volume is Vg the state could be dry saturated vapor if the volume is anything in between here here here the state would be from here to here wet vapor. Now I have selected volume one can select enthalpy one can select entropy one can select internal energy the characteristic is going to look very similar they are going to be a liquid point they are going to be a vapor point and there is going to be the wet vapor in between this is the critical point where we have seen that the distinction between liquid and vapor vanishes. So, the liquid point will be at the top of this on this curve when you plot it like this this side from the critical point containing all the saturated liquid states is known as the saturated liquid line. And similarly these states from the critical point downwards containing states of dry saturated vapor is known as the dry sat vapor. Now remember that we have selected P as between P and T, but I have here selected V someone may select V someone may select U someone may select H someone may select S and there will be some confusion. So, we decide on a common parameter which is really of convenience to us and for that we define what is known as a dryness fraction the symbol given to this is H. And the idea is like this whenever you have liquid and vapor in equilibrium it could be almost all liquid no vapor that will be saturated liquid or all vapor hardly any liquid that will be dry saturated vapor. Let us say the mass of the vapor part is m g mass of the liquid part is m f then the total mass of the system is m g plus m f and we define the dryness fraction as the mass fraction of the vapor mass of the vapor divided by mass of liquid plus mass sorry m f plus m g. What does this mean? This means that for the saturated liquid state x f will be 0 because there will be no vapor everything will be liquid and for dry saturated vapor state x g will be 1 and for the in between states wet vapor x will be between 0 and 1. This is the legal range 0 less than or equal to x less than or equal to 1 on our p and either v or u or s or h diagram. This is the saturated liquid line this is the dry saturated vapor line and if you take a pressure this is the saturated liquid state point this is the dry saturated vapor state point. So, this will be x equal to 0 this will be x equal to 1 and somewhere here all through this is 0 less than x less than 1 this is wet vapor. Now, in terms of x it is not difficult to determine other properties and it will be very clear. Now, suppose we want to determine the volume properties of now we know volume is an extensive property. So, if a system contains a liquid and vapor the total volume of the system will be made up of the liquid volume and the vapor volume mass will be mass of the liquid part plus mass of the vapor part and what about the volume the liquid have a volume liquid will be in the saturated liquid state because liquid and vapor are in equilibrium. So, the liquid volume will be mass of the liquid part into specific volume of the liquid plus mass of the vapor part g into specific volume of the vapor and this will be the total volume m into specific volume of the whole system and what is this m f is nothing but 1 minus x into m into v f plus m g is x into m into v g and that gives you this formula v will be 1 minus x into v f plus x into v g and similar formulae can be derived for u h and s you can write out this formulae you can turn this around and transpose this to solve for x we know that specific volume is 1 minus x into v f plus x into v g for wet vapor and. So, if you solve for x you will get v minus v f divided by v g minus v f and from similar formulae which I have asked you to write for u h and s you will get u minus u f u g minus u f h minus h f h g minus h f s minus s f s g minus and now you notice that it is here that these properties v g minus v f u g minus u f h g minus h f and h g minus h f the differences between the g properties and f properties turn up and authors of the steam tables have decided to tabulate at least h g minus s f and h g minus s f as h f g and s f g respectively these will be these are in our steam tables not all steam tables may have it. Now let us come to f 2.2 with our understanding of the dryness fraction you can tackle f 2.1 part c wet steam of quality thus only thing one has to remember that quality is another name for dryness fraction that is it. Now in f 2.2 the characteristic of this question is two properties are given and we have to determine whether it is subcooled liquid superheated steam wet steam dry saturated steam or dry or saturated liquid etcetera if you look at some of these options take for example, option a or option b we straight away have pressure and temperature specified we know how to tackle those cases. For example, if you come to option a 2 point f 2.2 a it is given to be a pressure of 2 bar temperature of 150 degree c you can come to say table 2 2 bar saturation temperature is 120.2 degree c oh sorry first one is 1 bar 1 bar saturation temperature is 99.6 degree c the specified temperature 150 degree c is higher than 99.6. So, we are in the superheated vapor zone you can even look at it the other way come to table 1 and look up the entry for 150 degree c at 150 degree c the saturation pressure is 4.758 bar the specified pressure is less than 4.78 bar. So, it is superheated steam in a similar way you can tackle b the last item t 100 degree c x 0.8 the item i it is very obvious that since the dryness fraction is 0.8 it is wet vapor and if you want to determine properties all that you have to do is go to 100 degree c and read off the saturated liquid values and trisaturated vapor values and use the formulae which we have just derived formulae like this or formulae like this I hope you have written all these four formulas now v f v g u f u g can be read off from our table and use the dryness fraction value provided 0.8 to compute out all the properties. So, we have so far looked at now among the other options notice the options c d g and h you will notice that of the two properties one property is either pressure or temperature in c d and h the pressure is specified 2 bar 2 bar and 0.5 bar respectively whereas, in h temperature is specified. So, one of the property is p or t and the other property is either h s or v here in c it is s in d it is v in h and g and h it is h. So, this is since it is 1 k g of water substance I can easily convert them to specific volume and other specific property v u h or s we do not have any illustration of u, but that could also be added. Now, when it is of this form this is what one should remember let me go to the next page this is a qualitative diagram. If you plot on one axis p or t and on the other axis this is a property phi which could be v or u or h or s you end up with an inverted dome this line could be straight down or this line could be a shallow like this or this line sometimes could even come in that. So, this could be very shallow as in case of specific volume case of enthalpy it will go down like this, but that is that is a minor variation at the top is the critical point. This is subcooled liquid this is superheated vapor and this is waxed steam and if you take any pressure or any temperature and as you traverse around this line you will notice that any one of these properties satisfy this characteristic that phi a property of subcooled liquid will be less than the property of saturated liquid will be less than the property of wet steam which will be less than the property of dry saturated vapor which will be less than the same property of superheated steam. This is what one should remember and you can look up the steam table may be plot the graphs you will be confirming this. So, long as you are below critical pressure or critical temperature this inequality or this qualitative relation will hold for any one of these four properties specific volume specific internal energy specific enthalpy or specific entropy. And if you remember this it is easy to solve any one of these four sub problems in F 2.2 take for example, F 2.2 let us take C P is 2 bar S because it is 1 kg I can write S at 6.2 kilo joule per kilogram kelvin I am just dividing the value by. Now, if you go to our table 2 the pressure based saturation table that is where one should go at 2 bar look at the entropy value at 2 bar at 2 bar at S F is how much was it at 2 bar S F is 1.530 S G is 7.127 1.530 S G is 7.127 right for 217 7.127. So, at 2 bar we read off from table 2 we S F is 1.530 kilo joule per kilogram kelvin S G is 7.127 kilo joule per kilogram kelvin get into a habit and enforce this habit on our students that whenever you write a quantity you must write the units with it do not even ever leave a quantity hanging without its unit. Of course, dryness fraction does not have any units and there may be a few other properties or parameters which may not have units, but most of the properties will have units and the numerical value of a property does not mean anything unless you have units associated with them. But, anyway after noticing this you will notice that S F is less than S is less than S G. Hence, what we have is wet steam once you have wet steam now you can calculate the dryness fraction you want to X will be S minus S F S G minus S F and once you get X you can determine all other properties U V H etcetera. I will leave it to you to calculate the dry out D G on your own. Now, that leaves E and F what do we do with this because both of these are of the H S kind and we do not have H as an independent variable nor do we have S as an independent variable anywhere. So, how do we proceed? We have to proceed in a graphical manner or we have to do what I call Sherlock Holmes Geary. Just look at our saturation table look at the pressure table and look at this is something of a should know because we are going to use steam tables for the rest of our lives and many of our students will also use the steam table for a significant part of their life. Look at table 2, notice that the specific enthalpy of dry saturated vapor at triple point is 2501.4 kilo joule per kilogram. It goes on increasing 2600, 2700, 2700, and then you see you notice that it is not increasing that fast 2763, 2771, 2780, 2799, 2800. Oh it started going back and now you will notice that there is a maximum of somewhere here 2804.2 at about 30 bar. So, the specific enthalpy of dry saturated vapor is 2804.2 at 30 bar and that is the maximum specific enthalpy of any vapor. If the specific enthalpy of any state is higher than 2804.2 kilo joule per kilogram well it has to be either superheated or supercritical steam. So, when you look at e specific enthalpy is 2900 kilo joule per kilogram which is definitely higher than 2804.2 kilo joule per kilogram and that means e is this e is going to be superheated steam superheated or supercritical. There is not much difference in them as we have seen during our tabulation. The f part is going to be more difficult because we cannot use the same trick which we used in H. It is 2500 kilo joule per kilogram specific enthalpy and specific entropy is 8.8 kilo joule per kilogram. Either you will have to do trial and error or interpolation, but then most of our tables also will have a chart associated with them and although we have been so far sketching charts with p or t on one axis y axis and v u h or s on the h axis. It turns out that when it comes to open thermodynamic systems particularly like turbine compressors and pumps. Another diagram becomes very useful and that is the diagram in which enthalpy and entropy are plus plotted as independent variables and this is known as the Mollier diagram. Our steam tables and you should open out the back of this see this diagram. What I have done is rather than I tried to scan it, but it does not look nice on the l m o it is too cluttered for you to see. So, I have obtained I went to the web which was the diagram which I came up I think it is here. They can see this. So, this is from a web site engineering tool box dot com you just put Mollier or H as diagram for steam. Do not put Mollier diagram because the word Mollier diagram is used for many other properties even what we call psychrometric chart is quite often known as. So, look at this as the qualitative version of the diagram which you have at the end of your chart. So, I am going to project this, but I want all of you to look at your diagram. Now, notice the line the red line which is going like a big wave across this. Can you show my paper for a time being the show the idea back? Oh, either see. If we plot generally we have been plotting things from the triple point to the critical point and belong beyond. So, if you take the same range on the H s diagram and start from the triple point. The triple point pressure line goes like this and the dry sorry saturated liquid line goes like this with a critical point here and the dry saturated vapor line goes like this. The critical point is here and all this part is very cluttered and is of no use to us and that is why the diagram which is plotted is this part of the diagram. And that is why you will notice that in our diagram the entropy goes from 5.5 to 9 kilo joule per kilogram Kelvin. Lower values are not shown and enthalpy goes from 2000 to nearly 4000 kilo joule per kilogram. So, this is a clipped version of the diagram. Now, coming to the Mollier diagram which you are seeing. I can do it from here. See this they can see the mouse. Sorry lower value I should not. This line is the red line is the dry saturated vapor line. The various isobars are these slanted blue lines. In our diagram the various isobars are slanted everything is black that is good. There are no color confusion here, but you can see the isobars shown there for various pressures. They are all slanted lines and then this is the zone of superheated vapor where you will see isotherms. In this figure the isotherm of say 400 degree c is seen here going to the left and then going down like this. You should be able to notice isotherms also in your Mollier diagram. And unfortunately in this distance mode it is very difficult for us to have a feedback, but since most of you would have seen this earlier spend time and be comfortable with it. Now, see the advantage of this and I would like you to look at it only qualitatively. When you come to exercise 2.2 part F, we have H specific enthalpy of 2500 kilo joule per kilogram. So, you take the 2500 line and 8.8 kilo joule per kilogram Kelvin entropy line and you will find that the point lies in the wet steam zone that is below that wavy line. And if you look at the line surrounding it the pressure is between 0.01 bar and 0.02 bar if you visually interpolate may be you can say 0.012 bar. And the other characteristic is you can see that the dryness fraction is approximately 0.99 because the that line almost goes through that point. But do not read of these values except as approximate values the exact values. Now, you will have to knowing that it is wet vapor knowing that it is approximately 0.012 bar go to table 1 and 2 and interpolate and obtain a more exact answer. The chart gives you a very nice visual description in the part of the state space, but that is what it is good for it is not good for reading values except approximately. The table is more precise the table is more precise, but it has discrete values unlike the chart which is continuous. So, you will have to interpolate and so my thing is use the chart for visual appreciation and use the table if necessary with interpolation for actual calculations. Now, I notice that it is a few minutes past 1 o clock. So, let us break for lunch and we will meet again after lunch.