 So, we can write the steady flow energy equation in a general form like this, we use the differential form of the energy equation in deriving the expression for stagnation temperature. But if you use the normal form of the energy equation connecting two states, we may write it like this. In the general form because now here we are adding the effects of both heat addition removal and work addition removal. So, you can see from this that when we add heat to a flow, Q is positive and the stagnation temperature increases. If we remove heat from the flow, Q is negative and stagnation temperature decreases. Now, we can also bring about a change in stagnation state by actually adding or removing work. So, when we have a compressor where work is done, basically what we are saying is we can change the stagnation state in the absence of work interaction by heat interaction. Adding heat increases the stagnation temperature, removing heat decreases the stagnation temperature. In the absence of heat adiabatic flow, we can change the stagnation state by adding or extracting work from the flow. So, across an adiabatic compressor where work is done on the flow, stagnation pressure increases. Remember, W is negative in this case in our sign convention and across a turbine where work is extracted from the fluid, W is greater than 0, stagnation pressure decreases. So, any loss of stagnation pressure is undesirable because it is tantamount to a loss of work that is the most important part. So, any loss of stagnation pressure indicates that there are irreversibilities in the flow and as a result of irreversibilities, there is entropy generation in the universe, loss of exergy and loss of work. So, as I mentioned before, so we may have two states like this 1 and 2 and the corresponding stagnation state here would be T01 or T01 or let me just write it as stagnation state 0,1, this is stagnation state 0,2 and keep in mind that S02 equal to S2 and S01 equal to S1 because the deceleration process is isentropic. So, I can now apply TDS relationship between these two states and write this expression. So, if we have irreversibilities in an adiabatic flow field, if we have an adiabatic flow field that means T02 equal to T01 and if we have irreversibilities then S2 is greater than S1 which means that S2 is greater than S1 which then implies that P02 is less than P01. So, it is also stagnation pressure if there are any irreversibilities between state 1 and state 2, assuming that this is the flow direction. If there are any irreversibilities between state 1 and state 2, S2 is greater than S1, so P02 is less than P01, the flow is adiabatic. Now, if the flow itself is isentropic everywhere then the stagnation pressure and stagnation temperature will be the same everywhere. So, it is important to bear that also in mind. If the flow itself is isentropic everywhere then S2 is equal to S1, equal to S3 and so on and the stagnation pressure and stagnation temperature will remain the same between 0.123 and so on. So, in that case stagnation state becomes a global reference state because it is the same for the entire flow. So, stagnation state can be thought of as a local reference state as well as a global reference state. It is an excellent indication in case stagnation pressure changes from one point in the flow field to another, we can then directly infer that in the absence of any heat interaction that there must be an irreversibility in the flow field. So, this is why we said earlier that inferences, the advantages of defining reference state is that important physics in the flow can be brought out clearly by the use of this reference state. So, you can see how the stagnation state brings about inferences on irreversibility is present in the flow field or any work addition or removal in the flow field, heat addition or removal in the flow field. We customarily use TS diagram and PV diagram incompressible flows to indicate thermodynamic states and thermodynamic processes. Now, let us first start with the TS diagram. The TDS relation for a calorically perfect gas reads like this. So, we will discuss calorically perfect gas first, then I will make a mention about steam and refrigerant which are not calorically perfect. So, we may write this as CV dt minus PDV. So, constant specific volume lines or lines of constant specific volume have a slope which is given by this expression, T over CV. So, in the same manner from the alternative form of TDS relationship, we may show that the slope at any point of an isobar. So, this would be a slope of an isocore. So, this is the slope of an isobar. Slope of an isobar is given as T over CP. So, it is clear from here that the isocores are actually steeper than isobars because this T over CV, since CP is greater than CV, isocores have a higher slope than the isobar. So, and that is illustrated here. So, you can see that isocores are actually steeper than isobars and the slope is positive which is also indicated here and it increases with temperature for both of them as temperature increases the isobars and isocores become steeper and as temperature decreases isocores and isobars become shallower and the slope is positive. Most importantly, isocores are steeper than isobars. This information would be very useful later on when we start looking at actual flow fields. Now, in the same manner, notice that basically we are looking at two sets of coordinates T S or PV. So, in a T S diagram, we want to know how constant isocores and isobars are appearing in the T S diagram. So, basically we have two sets of coordinates T S and PV. So, on a T S diagram, we want to know the shape and orientation of isobars and isocores whereas, in a PV diagram, we want to know the shape and orientation of isotherms and isotropes. So, this would be an isotrope and this is an isotherm. So, notice that both the isotrope and the isotherm have a negative slope in the PV diagram which is why these lines look like this and once again the magnitude of the slope of the isotrope is actually greater than the slope of the isotherm and which is indicated here. So, S equal to constant lines are steeper than the T equal to constant of isotherms. So, you should remember this diagram, the appearance or rather the slope and orientation of isobars and isocores in a T S diagram and the shape or slope and orientation of isotherms and isotropes on a PV diagram. So, we mentioned that we used the T S and PV diagrams to locate states to illustrate states and also to illustrate thermodynamic processes. So, when we have a one-dimensional flow field that is one component of velocity and two thermodynamic properties, let us take them to be T 1 and S 1. It can be T 1 and P 1 and so on, but for the sake of simplicity here, we will take them to be T 1 and S 1. So, how do we locate the state in a T S or a PV diagram? That is the next question that we are going to ask. The reason why we ask this question is that it requires a little bit of careful thinking because T 1 and S 1 are thermodynamic properties. So, showing them, showing the state even a value of T V and S is trivial to do on a T S diagram. So, basically I draw a T S coordinates like this. So, I locate state 1, I am given T 1, I am given S 1. The difficulty is how do I show V 1? So, I may have many states all of which have same T 1 and S 1, but different velocities. How do I illustrate velocity, which is a fluid dynamic quantity? So, these are thermodynamic properties, whereas V 1 is a fluid dynamic property. How do we illustrate the velocity in this diagram? Velocity corresponding to state 1 in this diagram because state 1 is unique. Three property values are required and they are given P, S and V. A clue about this is available from the energy equation. Remember, the differential form of the energy equation looks like this. D of h plus V square over 2 equal to 0 and this quantity may be defined as a stagnation enthalpy. We defined it as stagnation enthalpy at 0 equal to h plus V square over 2 at a given point in the flow field. Now, for a calorically perfect gas, this expression may be rewritten like this. D of T plus V square over 2 Cp equal to 0 and the clue that we are looking for is provided here. So, the quantity V square over 2 Cp has the same unit as the static temperature and it is simply being added to the static temperature. So, that is the clue that we are looking for. So, if you go to a TS diagram like this, we ask the question, how do we show fluid dynamic quantity in this? So, basically what it seems is that we simply add an amount of let us say V 1 square over 2 Cp, as you can see from here, we simply add a quantity V 1 square over 2 Cp to state 1. And so, now this state is completely defined. So, the state here has as you can see the velocity added to this and that actually fixes I am sorry that allows us to show velocity also in a TS diagram. So, this was the given state. So, state 1 is shown here, T 1 was given, S 1 was given. So, on a PV diagram again, T 1 was given and S 1 is also given. So, the given state lies here and we may retrieve P 1 at the isobar corresponding to the static pressure looks like this and it looks like this here. It is shown as a straight line only for the sake of illustration. Now, we will add an amount equal to V 1 square over 2 Cp to this like this, excuse me. Again, remember we need to do this without changing the given value of specific entropy. T 1 and S 1 were given. So, this addition must take place without changing the value of S 1, which means that we should end up with the state which has the same S 1. And that is what I have shown here. So, I end up with the state which has the same S 1 as state 1. Similarly, here I travel along this isentrope like this and add an amount V 1 square over 2 Cp to the static temperature, which means that I end up at a new isotherm which is here and a new isotherm which is here in these two diagrams. And you know what that state is because T 1 plus V 1 square over T Cp is nothing but T 0 1. So, this isotherm just is equal to T 0 1 and same is shown on a PV diagram here. So, this is state 0 comma 1. Again, this is state 0 comma 1. So, the isobar corresponding to state 0 comma 1 is shown like this and T 0 1 is shown like this. So, every static state. So, notice that. So, the thermodynamic properties actually define the static state. But the thermodynamic and the fluid dynamic property together define the stagnation state. And the stagnation state is nothing but stagnation pressure, stagnation temperature, stagnation density which are all thermodynamic properties and they can be easily illustrated on the TS or PV diagram. That is the general idea. Velocity is a fluid dynamic property, but we convert it into a into a thermodynamic property by using the notion of adiabatic deceleration, which is why we insisted that we should travel along the same S equal to constant line. And similarly, here also we insisted that we should travel along the same S equal to constant line. So, this is nothing but the stagnation state T 0 1 and P 0 1. So, this is how we actually can depict state in the flow field at a point in the flow field on a TS or PV diagram. So, now you begin to understand or realize why reference states are so important because they play such a role such an important role in conveying actual information to us. Now, this is all we can do. Can we can we depict any more information? For instance, let us say that you know we ask the question, is the state 1 as shown here, can we possibly show information concerning speed of sound? For instance, or is this state subsonic or supersonic? Is state 1 subsonic or supersonic? Can we possibly indicate that information? So, if you want to indicate that information, then you must be able to show speed of sound on this diagram, which should not be too difficult to do because it involves only the temperature in the case of a calorically perfect gas. So, we should be able to do that. So, we can in fact do that in the following manner. Remember, if you go back to the expression involving the stagnation temperature simply by setting M equal to 1. So, in this expression, if I set M equal to 1, that means the local velocity is equal to the local speed of sound, which means the static temperature is actually the sonic state. So, by setting M equal to 1, I actually can get once I know T0, I can get T star by simply setting M equal to 1 like this. So, once I have this, I can illustrate again the sonic state and T star, which corresponds to M equal to 1. So, states that are above this line or subsonic states and states that are below this line M equal to 1 line or supersonic states. Why? Because this distance is nothing but A1 square over 2 Cp. So, any state which lies above this line must then have a velocity V to reach the same stagnation state. It has to have a velocity V, which is less than the speed of sound and any state on this side must have a velocity greater than the speed of sound in order to reach the same stagnation state. That is why this is M equal to 1 line, these are subsonic states and these are supersonic states. Of course, when I say states, I am using a plural here, I have to be careful because we are doing everything with respect to the given state 1. So, this allows us to decide whether the given state is a subsonic state or supersonic state. Now, in case the stagnation temperature remained the same throughout the flow field, then I can carry over this to all the other states in the flow field. So, any state in the flow field which is above this line has to be a subsonic state and any state below this has to be a supersonic state. So, in that case, the stagnation state becomes a global reference state. So, I can use the plural of state. Otherwise, I can only say that the given state is subsonic or supersonic. And in the same manner, you can see the isotherm corresponding to t equal to t star. So, this is the isotherm corresponding to t equal to t star. So, we can see that the yield has stated here, this is the sonic state. And states that lie along this s equal to s1 like this or subsonic states and states that lie along this s equal to s1 below this are supersonic states. So, these are subsonic states m less than 1 and these are supersonic states m greater than 1. So, you can see how much information can be about the flow field can be conveyed in a TS or PV diagram. And here we have looked only at one state. Now, in case you know the entire flow is isentropic, then the entire flow field information can be contained in just a single line s equal to s1. So, this is a very powerful way of illustrating the flow field number one. And what makes it possible or the inferences that we are drawing from the sonic state which was the one reference state that we one of the reference states that we discussed and the stagnation state which was the other reference state that we discussed. These are the two most important reference states in gas dynamics or compression. In case the given state instead of being a subsonic state had it been supersonic, it would have looked like this. So, one prime here is a supersonic state. So, notice that this is a1 square over 2 Cp and this is a1 prime square I am sorry. So, this is v1 prime square over 2 Cp and as you can see v1 prime is greater than a1 which is why it is a supersonic state. The same manner one prime is indicated and the same s equal to s1 line s equal to s1 isentrope in a PV diagram. Now, so we have done all this for a calorically perfect gas. Notice that most of these things would not be possible in the case of a real gas like steam or refrigerant. But the ideas and concepts carry over without any difficulty. So, basically what we what we do in the case of a real gas is the following. We do not define stagnation temperature but we actually use a stagnation enthalpy. So, we do not use TS diagrams in the case of real gases we use HS diagrams. This is still true. We have not made any assumption while deriving this and s0 equal to s is applicable regardless of whether the fluid is calorically perfect or not. So, those are the two important things. So, s equal to constant and in going from state 1 to the stagnation state and h0 equal to this are still relevant and applicable. So, we can use the HS diagram to illustrate this you know these ideas. We would not use TS but we will use HS and if you use HS then notice that instead of v square over 2 Cp we simply add v square over 2 to the static enthalpy. Notice that in that case this is called a static enthalpy. So, we add v square over 2 to a to the static enthalpy to arrive at the stagnation state. So, here we added v1 square over 2 Cp to the static temperature to arrive at the stagnation state. In the case of real gas we will add v square over 2 to the static enthalpy to arrive at the stagnation state. Stagnation state is still fixed because h0 is known s0 is equal to s only two properties are required to fix a state in a TS or PV diagram. So, in the case of real gas such as steam was represented we add v square over 2 to the static enthalpy and we will use HS coordinates for depicting states and processes.