 Because in the morning we discussed only one carrier equation to calculate the value of partial pressure of the water vapor that is C w. But there are some other empirical relations like F John and so on. So, if we apply those equations the value of partial pressure remain the same it will change. See these are all empirical equations and the values only differ by small amount and one cannot say with complete confidence that this equation is better than that or that equation is better than this. Carrier equation is by far one of the most used equations and hence we are recommending it otherwise you will not really suffer too much using the other equations it is just that very slight deviations are observed. So, you can use any of those other equations that people give we are just recommending carrier. So, that all answers are more or less the same, but they would not be too different if you use other equations. Thank you. My second question is regarding the properties of moisture. In the psychrometric chart there are various properties that the partial pressure w, v, t, d, v, t then d, p, t and then humidity and all some thermo line properties. Now, to calculate the w, v, t of the moisture how to apply the relations for this psychrometric theory? Can you calculate the value of wet bulb temperature with the help of the psychrometric table or not? Yes. So, I would say that if you know the partial pressure. So, let us say you know the specific humidity and the total pressure and the dry bulb temperature then so you know p v. Now, what I can do is I can always use the carrier equation in which there is wet bulb temperature and the saturation pressure at wet bulb temperature, but I cannot directly get it because I need to know both, but I can always do a trial and error. I can assume a value of wet bulb temperature from the steam tables, get the saturation at that wet bulb temperature and see if I get the correct p v. If not, I will change my guess for w, v, t and within two, three trial and errors I can easily get my wet bulb temperature knowing my p v. So, it is not difficult at all. So, what do you mean that w, v, t can be calculated only with the help of trial and error? Yes. This is, if you are using any of the empirical relations, yes, you can get it using trial and error, yes. No, see one way people would normally do it is, you know, calculate the adiabatic wet bulb temperature which you can get exactly, but the adiabatic wet bulb temperature as it turns out is not a perfect number for the empirical wet bulb temperature. So, what you can do is, use the adiabatic wet bulb temperature as the first guess for your trial and error and you will be there within two guesses. So, that would be it I would say. One more question. What is the difference between thermodynamic wet bulb temperature and adiabatic saturation temperature? See the thermodynamic wet bulb temperature is the same as what is called as an adiabatic wet bulb temperature. So, this is different from the practical method in which the wet bulb temperature is got. That is because there you do not have a perfect control volume or a control on what you are doing. It is just the temperature which is measured by the wet cloth of the thermometer. So, those two will be slightly different whereas what is called as the thermodynamic wet bulb temperature is purely based on an adiabatic process and those two are the same. Thank you. They have some relationship with the Levy's number. Yes. So, this is what is given that if the Lewis number is close to 1, then we say that the thermodynamic wet bulb temperature is same as the empirical wet bulb temperature and you can see what the situation is. If I draw an adiabatic control volume, so I come in with a particular temperature and a particular omega. There is a particular T star which is the same as the adiabatic or thermodynamic wet bulb temperature. If I keep water at this point, then I can hypothesize and a process which is adiabatic so that all the energy is taken from here and I come out with T star and W star where W star corresponds to saturation W. So, this is the adiabatic process. So, this is an entirely thermodynamic process that we can think of. Whereas, when we think of a thermometer like this and it is surrounded by a clock, we are actually talking of a quantity of air coming and a quantity of air going here and in steady state this has reached some temperature T star. This T star is not the same as this T star. In fact, this T star and the temperature here need not even be the same. Now, when what we mean by saying that the Lewis number is 1 is that let us say I could somehow draw a control volume here and all the energy see here what is happening is all the energy that is required to evaporate this is entirely coming from this air and hence its temperature is dropping down. Here you are not sure at all it is just some air which is flowing on top of this thing and not only that you are not sure whether this is saturated because this thermometer is actually measuring only the temperature of this wet bulb. We are assuming that whatever vapor has gone into the air is diffusing so fast that whatever quantity of air from which you picked up the heat from that much quantity is immediately going to absorb that entire moisture and come out with W star. That is what we mean by saying Lewis number is 1 that whatever quantity of air from which you removed the heat if that diffusion of the mass of water vapor in the same quantity is equally fast then that entire quantity of water then leaves with saturated condition at T star that will only happen in this case whereas here unless the Lewis number is 1 that thing so fast diffusion cannot happen but roughly they are the same as far as water vapor is concerned and that is why there is only a slight difference between the thermodynamic wet bulb temperature and the empirical wet bulb temperature thank you. Sir I have gone through some books and some authors use the externally reversible and internally reversible instead of reversible so what does it mean sir? Ok so in this case I am not very sure but what I would assume is that if you have an adiabatic system and you are not interacting with anyone else then you can undergo a process and I can say that is reversible or not so everything is internal to the system and that would be internally reversible whereas I can always have a heat transfer process which is also reversible in nature so I can make it as close to reversible as possible by ensuring that the heat transfer takes place with a small temperature difference always and that is where I can ensure external reversibility but that is whenever we are having an external interaction. Ok so those are processes where the system is interacting with surrounding and I can make it reversible that is where probably the external reversibility comes from whereas if it is an adiabatic process entirely the system is on its own it is isolated and probably that is where the internal reversibility is talked of. I am not so familiar with such separateness I can just say that whether a process is reversible or not is a very standard way of looking at it. My question regarding dew point temperature and dry ball of temperature at what point they are thin? So as I said you know if your saturation pressure is the same so I have drawn this TS diagram so often now ok so on the TS diagram if I draw this pressure line here ok your temperature is here and this is the pressure line however if your temperature was exactly here on the dome then your dew point temperature and dry ball temperature would have been the same. So if you are 100% saturated then definitely your actual temperature is the same as the dew point temperature you cannot cool it any further because it will immediately start condensing out thank you. I want to ask that earlier today we derived relation for speed of sound and the process turned out to be isentropic but when we derived formula for normal shock wave we said that its entropy is greater than or equal to 0. So why is the difference? Why isn't it an isentropic process like sound wave? Yeah so the question is between the difference between a sound wave process and a shock wave process and in particular for sound wave we said that the process is isentropic whereas for shock wave it is not. We haven't completely talked about the non-isentropic nature of the shock wave process hopefully we can do that tomorrow little more but as far as you can see if you just go to the analysis of both these situations if you look at the control volume what we see in case of a sound wave process is that the conditions from one end of the control volume to the other end change infinitesimally right so the changes in the density pressure temperature etc in case of a sound wave process are extremely small whereas those in the case of shock are not. So just because of the fact that the changes in the shock wave process are finite in amount in terms of the pressure change the temperature change etc we cannot really call that situation a reversible or isentropic type situation. In fact what we are going to show tomorrow is that there is a there is a change in the stagnation pressure across the shock and that comes directly as the change in the entropy. In fact what ends up happening is that within the shock process there are very high gradients in all quantities and these high gradients will make sure that the process is not reversible. So fundamentally if you want to look at it from a purely physical point of view the very small gradients that occur in a sound wave process make it reversible and hence isentropic whereas finite gradients in case of a shock wave process make it a non-isentropic process. Thank you. 1064. My question was regarding the psychometric chart basically and if the quality of air for example it is not a standard air and the composition suppose the pollution is there so is the psychometric chart needs to be some correction factor or by using molecular weight of air or some gases can be incorporated and the same chart can be used for these applications? Yes I would think so see for example if you think that there is polluted and you are having probably a more than a percent of some other thing like CO2 etcetera those will also behave like an ideal gas and what you can do is suitably modify the molecular weight of air so hence your partial pressure calculations and specific humidity calculations is slightly deeper. So I mean there is no harm as long as you know exactly what the composition is I do not think there is any harm if you stick to the basic formula that we have used today. Thank you. By psychometric chart are as good as a steam table for one bar pressure? Well it depends on what you want isn't it? You get many other quantities from the steam table and at one atmospheric pressure if you want to use the psychometric chart it is very good because it gives you other quantities too like air and specific volume then and there it is very fast calculation so if you are a regular air conditioner or refrigerant person then I think you should use the psychometric chart at one atmosphere otherwise there is really no comparison between the two because the information available is much different in both of these. So it is up to you if you are sure your pressure is one atmosphere and you are doing only AC calculation you know you are feel free to go ahead and use the psychometric chart. Thank you. Sir these psychometric charts are empirically or purely experimental? So I mean you can say that in some ways all of these things especially the steam table is some kind of a fit to empirical data and it has been checked out that most of these calculations will give you what is regularly you know obtained empirically. In fact only the wet bulb temperature is the empirical part out of there otherwise everything else if you realize Cp, Ca we have all calculated this using some way or the other you know comparison between empirical and some kind of theoretical basis. So I do not see any difference if you say it is theoretical and empirical so it is the same thing. 1056 How to find the flow properties within the shock layer? Okay so the question is how to find the flow properties within the shock region that is a good question actually. See within the shock region it is not that easy to obtain the property variation. One way it can be obtained is by doing a very detailed fluid mechanic type calculation but that calculation will not be really using any continuum type models because as we have noted in the morning earlier the extent of the shock region is roughly 10 to the power minus 7 meters or so and it is comparable to the mean free paths that you can normally experience. So the region within the shock as such is not really a continuum region and it is not that obvious to get the property variation within the shock region. What we end up doing as mostly engineering type calculation is that we are interested in finding out the change in the properties from ahead of the shock to behind the shock. So we do not really look at what is happening through the shock. We simply say that we come up with a control volume and that control volume encloses the entire shock region and what we are interested is relating the properties ahead of the shock with those behind the shock through our standard governing equations of fluid mechanics. If you want to perform calculations and find out how the properties are varying through the shock then you have to resort to fairly sophisticated type calculations which will involve molecular methods to find out how the variation is occurring through the shock. Thank you. One more question, how to find the distance of shock? At which location it is going to be obtained? The floor at one body or in a shock tube or in a divergence section of nozzle. So you are going to obtain the shock at different? Yes, right. So it depends upon which application you are looking at. So the first application that you mentioned, shock over a blunt body, it is theoretically impossible to locate the shock for which you have to perform CFD type simulations or you have to actually perform laboratory experiments and you have to find out where the shock location is. So theoretically where the shock is going to occur can be done only for fairly simplistic situations like convergent-divergent nozzles which are very, very close to being one-dimensional. If you deviate from one-dimensionality again within the nozzle also it is not that easy to figure out where exactly the shock is going to occur. So my answer is that depending on which problem you are looking at where the shock is going to occur is going to be dependent. Many times analytically it is not possible to find out. You mentioned also a shock tube. So shock tube is a situation which is an unsteady flow situation and in a shock tube fortunately there is an analytical solution available. However, the shock tube analysis is completely out of the scope of this particular course for which you will have to refer to a compressible flow type course where you will have the details of the shock tube type flow covered and there you will know how to calculate how the shock forms and how it moves within the shock tube. Thank you. One more question. Why total pressure is decreasing across a shock and total temperature is constant across a shock? Yeah, so the question is why total pressure decreases across the shock and total temperature is constant across the shock. However, if you look at the energy equation across the shock what we have is h1 plus v1 squared over 2 equal to h2 plus v2 squared over 2 and this h plus v squared over 2 is nothing but the stagnation enthalpy which you can express in terms of a stagnation temperature for an ideal gas using h equal to Cp times T and you can see that the stagnation enthalpy remains constant across a shock from which we can immediately infer that the stagnation temperature remains across the constant across the shock. On the other hand, because of the irreversibility associated within the shock what we will see tomorrow is that you can show that the stagnation pressure actually ends up decreasing across the shock. So, I will request you to wait till tomorrow when we complete our discussion of the nozzle. We will come back to the change in the stagnation properties across the shock and we will show that the stagnation pressure there is going to reduce but the physical reason why the stagnation pressure reduces across the shock is because of the irreversibility associated with very large gradients within the shock region that is the primary reason. Thank you. Yeah, 1028. Basically, we know that air having the water holding capacity at a given temperature as temperature is going to be increases there water holding capacity is going to be increases there. We know that. It is maximum matter, even air is a saturated one and we can calculate that using the saturation temperature corresponding to driable temperatures. So, my question is that why air cannot hold the more amount of the water vapor corresponding to the driable temperatures? So, what is physically what happens in the air? It is not what physically happens in the air but what is happening to the water vapor. So, you can imagine it as you are boiling water. So, as you start boiling or heating the temperature of the water slowly the vapor pressure above the water it gets completely made up of steam and there is an equilibrium there. Of course, if you start giving more and more heat you will go ahead that is you will evaporate all the water but if there is an equilibrium between the vapor and the liquid then what happens is you know particles go from both the liquid to the gas and vice versa. So, if I start trying to put in more vapor it just starts condensing out because that is it you have actually tried to put see this is what will happen. Let us say that I am having let us say water boiling at a particular temperature 100 degree C. Now, if I start putting more water vapor on top of this there are two things will happen the water vapor will condense or I can just allow the net pressure to go up that is if I start holding the volume constant. So, only these two things can happen. So, if you try to hold the volume constant and start increasing the water vapor pressure then you can go higher and higher up but your temperature also should start increasing. So, it is just a relationship between the boiling point. So, water is actually boiling at that point and there is just an equilibrium between you know something with a just boil and something which is just in the liquid state that is all that is happening there. Thank you. In some cases, actual biometric reading is given more than say, 160 mm of Hg. What are the thermodynamic fundamentals behind this? Yeah, so it is entirely possible I mean you are not you are not claiming here that the highest pressure on earth is 1.01325 into 10. It is just what is called as a mean sea level what is the pressure there. There are definitely some places where the pressure is more than the so called one atmospheric pressure which is more or less a mean pressure for mean sea level. So, I mean I do not see why at all the pressure can't be more. You can go to a lower lying region lower than the mean sea level your net atmosphere on top of you is higher you will have higher atmospheric pressure. Okay, so you should not confuse this with you know it is not the lowest possible temperature on earth. So, I can only say that thank you. The question regarding the second chart sir sir the dry bulb temperature is maximum 50 degree centigrade sir. Can't it go beyond that this is my question. No, the dry bulb temperature can be anything you know it just depends on what is available on earth. I mean maybe in some Sahara desert you can get 55 degrees I mean we are not contesting that. Okay, for our conditions we are just drawing it between 0 and 50. Okay, if you are living in the desert probably you should have a psychrometric chart extending there. So, that is what I am saying. Don't just use the psychrometric chart as it is because those are made region specific. Okay, hence it is always better to have a thermometer have a barometer and your steam tables in hand with these three things you can perform all your calculations and not rely on any kind of psychrometric chart. Thank you. Thank you. Thank you. So, 9-2. Sir, what happens in the winter when you are scratching the skin somewhat lines are there because due to dryness and in the I think in warm or in the summer season or in the summer season that things are different one. This is related to the this atmospheric condition so that this I want somewhat answer. Correct, correct. This part is for sure, I mean, you know, when your skin has some moisture and if the surrounding air is very dry, you start losing moisture and that is why it becomes so dry that, you know, even if you do not scratch it, you can get lines out there. So, that is the only reason. So, if the air is moist enough, then your skin will not start losing so much moisture. In a way, the principle is the same as in an air cooler. If you use it in a dry region, you can pump in as much water as possible. The same thing here, in a dry atmosphere, you can remove as much or a reasonable amount of water vapor from your skin, whereas you cannot do it so easily in a moist atmosphere. Thank you. How the vapor pressure is required or in the air-conditioning system, how the vapor pressure will be playing an important role so that the vapor pressure should be maintained properly in that particular room or any system? Now, see, what normally people have found out empirically is some kind of a comfort situation, which they claim is, you know, some temperature lying between 24 to 26 degrees and roughly 50 percent humidity. So, for example, if the humidity falls below 50 percent, you start feeling dry, as I said, okay, and itchy. And if the humidity starts increasing to 80, 90 degrees, you start feeling very sweaty and again uncomfortable. So, overall people have, you know, looking at many humans try to come up with some kind of a relative humidity, where the human being seems comfortable, and that seems to be around 50 percent relative humidity and the temperature should be around 25 degrees, let us say. So, what people would normally do is, if you have very moist air, then you will ensure that you condense out enough so that whatever is going out is kept at 50 percent relative humidity. Whereas, you should realize that, you know, especially in a place like Mumbai or any coastal place, it is pretty humid and hot. So, to do the air-conditioning is probably a slightly straightforward job because all you need is coils which are at a temperature lower than the dew point, which will ensure that moisture just drains off and you ensure a reasonable bypass coil ratio such that you get whatever temperature you want. The bigger difficulty comes in drier climates when, you know, the thing is so dry, let us say you are very cold and very dry, okay, and this is what happens in northern countries. And at this place, if you just start on the heater and increase the temperature to roughly 30 degrees, already it is pretty dry, your skin starts losing more, you know, vapor and you start feeling even more drier. In fact, it is pretty bad. And there, you really have to ensure that there is some system which will pump in steam into the air to make it more moist, okay. So, that is actually a far worse situation than what we have here. As long as you have a reasonably cool coil, you can get a reasonably good comfort condition and it is not so easy to do air-conditioning in really cold and dry climates. Thank you. 207. In the industry, where the humidification process is required, so can we clearly be able to model these humidification plants that how much load it is taking and based on the air-conditioning criteria including this humidification process which require a big fan blower rather? Okay. See, I will try to answer that question as much as I know. I am not very clear about how the humidification plants work, but what you really have in most industrial processes where you require high humidity and temperature is that you have air coming in at a particular temperature and specific humidity and you want to increase its temperature and specific humidity that is the moisture content and normally you would pass in a lot of steam into this. So, you would require some kind of a temperature, sorry, some kind of a volume in which you would mix these two. Now, I am sure you can easily model this as some kind of control volume where incoming air is at a particular temperature and specific humidity and you are pumping in steam at a particular rate or you would require to find out how much steam you should put in so that you get out a particular temperature as well as specific humidity. Apart from this, you may also need to provide some kind of heating surface because it is just providing steam may not necessarily give you the temperature you want. So, you may want to say there is some Q input, some mass input due to steam and net if there is a good mixing then you will come out with a particular temperature and specific humidity that you want. In fact, those based on those simple models, we have the last three or four questions in our sheet. So, you can just assume a control volume, assume incoming conditions, assume steam, assume some kind of heat input and then try to see whether you get the required output you want. Thank you. Sir, I would like to know that whether the energy consumption will change according to the weather or not. Yes, so if you really want high temperature and your air is already hot that is surrounding air is hot, maybe you are in a good situation and you really do not require so much of energy into the plant. So, definitely it will depend a lot. Similarly, if you have something like a boiler where you want to prevent heat loss to the atmosphere, your insulation will matter a lot and if you are really having cool surrounding your efficiency will drop down. So, definitely energy consumption is a function of outside or ambient condition that is for sure. Sir, what is stagnation property and may I silence and accent zone in make cone sir, please sir explain. Yes, actually we have not discussed the stagnation properties completely today. In fact, in tomorrow's class from 11.30 to 1 we are going to look at the stagnation properties. So, perhaps I will request you to wait till tomorrow's lecture where we will complete the discussion of the stagnation properties. The Mach cone concept that you talked about is really a two dimensional situation in compressible flow and physically what it is supposed to mean is that if you have a small object like a small point source as it is said which is moving with a supersonic velocity in otherwise stagnant air. What will happen is that there will be a certain portion of the entire fluid domain which will be affected by the presence of that moving object and that domain which is affected by the presence of the moving object is in the form of a cone. That cone is what is called as the Mach cone. Physically what is supposed to signify is that whatever acoustic field that this moving object is generating is going to be completely encompassed inside that Mach cone. So, that only the region inside that Mach cone has information that such and such object is moving. However, we are not going to talk about these Mach cone ideas here in this class. We are going to restrict ourselves only to a one dimensional situation. So, by definition or by design this Mach cone idea is a two dimensional situation and hopefully I have explained it a little bit. You can look at it in any of the standard compressible flows, but that is where I will limit my answer to right now. Sir, I would like to know regarding the solar dryers. Is it so that in some particular region it is quite difficult to get drying with solar radiations because the humidity has got particular range in that region. Is it so like? Yeah, I would think so. I mean you yourself know that if you want to dry clothes in the Mach cone it is very difficult and if you want to dry it in dry weather it is pretty easy. Even if the temperature is very low you can easily dry your clothes if it is a pretty dry weather. So, if you can manage to increase a dry bulb temperature and you know increase the water holding capacity of air then probably you can do a better job of drying, but the increased temperature may lead to other consequences. But that is definitely true that if you have moist situation it is difficult to dry. Thank you. One double to eight. Go ahead. What is the significance of wet bulb temperature? And is there any connection between wet bulb temperature and temperature of water? Okay, so I think I have mentioned this earlier that as far as the wet bulb temperature is concerned the physical significance is that we will use this to get the properties of air at this moment, because that is what we require for calculating our properties. For example, I need to know what is the specific humidity at the particular temperature. I just can measure the temperature, but I cannot measure the specific humidity. So, the only way to do it is either to get the dew point temperature or the wet bulb temperature and I usually use one of these to get to what my specific humidity is. So, that is the practical need to get the wet bulb temperature otherwise really there is no extra need to go for this. And if you want temperature of water and temperature of wet bulb, they would be different. I mean normally in equilibrium your temperature of water would be the same as your room temperature whereas the wet bulb temperature is definitely going to be lower than your dry bulb temperature. And I think I explained this using the TS diagram. So, you know we drew this pressure line here and said that you know this is the dry bulb temperature, this is the dew point temperature and this wick of cloth that you have wrapped around the thermometer initially would be at this temperature which is the dry bulb temperature. But once you start evaporating water from it from somewhere or the other the energy has to come and it will come from both the air initially as well as the water. So, this is the same as when you are trying to let us say cool boiled water. It is not purely convective cooling that is happening but also the fact that some of the water is actually evaporating into the air and the energy it is picking up is from the water also apart from some energy that it is picking up from water and hence the water will cool faster. And this is what you are doing here. You are trying to ensure that some air passes over the cloth and evaporate some amount of water vapor from this cloth and hence some energy will be drawn from the air, some energy will be drawn from the cloth and the cloth temperature will steadily decrease till you reach a steady state and the wet bulb temperature will lie below the dry bulb temperature but above the dew point temperature. So, we have learnt that shock wave. Normal shock waves are formed when a flow dissolves from a supersonic state to a subsonic state. Why is it that it has to go through a shock wave? Why does it happen smoothly without the shock wave being formed? Yeah. So, the answer is let us say you are looking at a solid body and that solid body is inserted in a supersonic stream. So, what you will know is that ahead of the solid body there will be a normal shock or some sort of a shock form across which the flow will decelerate from supersonic to subsonic. So, the real answer to this is actually related to that coalescence of sound waves forming a shock is what I discussed about if you remember. So, what you can imagine is that if you have a body which is inserted in a flow, you can imagine that the body starts accelerating from a lower value of velocity to a higher value. What ends up happening is that as this body is trying to move through the fluid, it continuously generates these acoustic waves because it is continuously generating these disturbances at locations wherever it is going to. What ends up happening is that beyond a certain limit of the flow or beyond a certain limit of the speed, these acoustic waves are not able to go beyond a certain distance and what ends up happening is that they will actually coalesce or merge together to form a shock in front of the body. So, anywhere the shock is forming, the fundamental physical mechanism that is causing the shock to form is collection or merging or coalescence as we say of these acoustic waves into a shock. So, that is really the basic physical reason why a shock forms and this happens only when a certain speed is crossed and that happens to be these subsonic to supersonic transition where these shock waves are supposed to occur. So, my answer to your question is that the reason lies in the fact that a shock occurs when we get into these coalescence or merging of these acoustic waves and that will happen when the flow is becoming supersonic. Thank you. Yes, 1166. Go ahead. What are the effects of supersonic flow on gas turbine efficiency and mass flow rate? In case you get into supersonic flow, typically there is a very high probability of shock waves forming and when you get into a situation where shock waves are forming, what ends up happening is because of the irreversibility associated with the shock wave, you get into a very high non-isentropic processes because of which the efficiency, the isentropic efficiency if you want of the turbine will deteriorate or go down significantly. So, that is all that really I can say based on my understanding. Professor Bandarkar wants to add to this. In both steam turbines and gas turbines, normally there are stationary nozzles followed by moving glades and people do try to get in supersonic velocities, but again the point remains that if you have supersonic velocities, even if you go slightly off design and which is very easily done in all these turbines, you will get into shock which will cause entropy increase and hence you will not be efficient. The other reason is that if you try to go to very high velocity, even your frictional pressure drops start becoming higher and what people do normally is try to restrict the velocity is roughly to around sonic or just below it and go for reaction turbines where you know you can have a compact turbine without accelerating the flow too much in the nozzles. You accelerate it only partly and then in the moving glades you further accelerate it, but you try to ensure that the velocity is nowhere cross the sonic or just about cross the sonic velocities and hence you can get a far more efficient turbine in a compact format. So, that is roughly what would happen and the people would not normally go for having very high velocity in this turbine. Thank you. What is the Mach number for oblique shock wave? So, whenever you are dealing with a shock, the upstream Mach number or the upstream flow will always be supersonic. So, it can always go from Mach number of 1 to as high as you want. In that sense, there is no difference between oblique shock and normal shock. The only difference between oblique shock and normal shock is that in the case of normal shock, the downstream Mach number is definitely subsonic whereas in the case of oblique shock, the downstream Mach number can also be supersonic. So, that is the only difference, but the upstream Mach number has to be always supersonic. Thank you. 1 to 1, 9. Sir, my question is that sir, if you want to develop the segmentation from temperature 5 degree to the 60 degree, so how can it be developed? I think we did talk about this in the morning. I mean, nowhere did we mention that we are going to have temperatures necessarily between 0 and 50. I said that most psychrometric charts are made between 0 and 50, but there is no one going to stop you if you want to make it from 10 degrees to 90 degrees. I mean, you can go ahead. All your formulas are known. I mean, you do not really require a psychrometric chart, but you can easily make whatever you want. So, I do not see any harm in making a psychrometric chart if you think your weather is between these two temperatures. Sir, it is psychrometry that is for air, and if you want to develop the psychrometry for another gases, then is it maybe can be developed? Yeah, why not? I mean, but are you considering that you are going to live in an atmosphere made of other gases? Yeah, why not? I mean, you can have surrounding as anything you want in which you are putting in water vapor. Just get the correct molecular weight for that gas and see if it behaves like an ideal gas because all we did was you just ideal gas equation and the steam table. So, you can use any other gas if you want and make a psychrometric chart, but right now it will only be an intellectual exercise. I mean, it will not have a practical value at all. Thank you. 1241, please go ahead if you have a question. Sir, for analysis of the thermal power plant, there is a different component of use, just like a pump, turbine, boiler, and condenser. How to evaluate different types of parameters enthalpy, entropy, and pressure, and temperature, etc. Because of there is a difference between the molier chart and the steam table. So, how to evaluate, how to match and how to verify this component and this parameter? Over, sir. I see if you use the steam table, you will get as accurate answer as you can. If you use the molier chart, it will be more approximate and you will only get rough values. So, I would say that if you want to do exact calculations, you better go to using the steam table. And normally we use the molier chart only when we are setting problems in exams just to ensure that we are getting roughly the values in the ballpark that we wanted. Otherwise, I would suggest that we should use nowadays there are even standard software which have hard coded the steam table and you can use those. But it is better to use the steam table because they would tend to be far more accurate. Thank you. Sir, we are using the steam table and molier chart simultaneously and we are using the superheater and regenerative cycle in the rank and cycle. So, there is a difference between the some data of molier chart and steam table enthalpy and entropy and pressure also. So, how to evaluate different types of work done and etc. in the turbines and condenser boiler and pump? I think I did just answer that question. I said please use the steam table, it is better. Of course, even in the steam table in the superheated region you may have to interpolate. But as long as they are consistent and if your steam tables are accurate, I do not see it just depends on the accuracy of both. So, I would rather use the steam table because I think I can interpolate to as much as I want and get as accurate values as I want. Whereas, if I use the molier chart, okay, things are still very easy. I just have to put rulers exactly and get the values which I want and, you know, points lie between two values which are very close by. I would rather use the steam table than the molier chart. Sir, we have just evolved the different types of work done by the using of the steady flow energy equations. That is the application of the first law of thermodynamics. So, how to just justify the work done by this evolution of the enthalpy by the enthalpy H1 minus H2. And because of in the Rankine cycle, the fluid has transferred to some phases also. The fluid phase changing cycle means. So, there is some problem of the condenser also. So, how to evolve the enthalpy and entropy in condenser? Yeah, so see this is a regular open system analysis. It does not matter in the control volume whether the fluid is changing phase or not because as long as you get the enthalpy values correctly recorded, okay, it is just a control volume. You are saying fluid enters with some enthalpy, leaves with some enthalpy. Whether it changed its phase, you have already accounted for that in the steam table by putting HFG. So, you know, there is absolutely no problem if the phase has changed because that has already been accounted for. Sir, is there any measuring device for the calculation of entropy, enthalpy, etc.? Yeah, definitely there is nothing direct. There is no such quantification of entropy available from a meter or something. We always use other properties to get our estimate of what the entropy should be, okay. Whereas, enthalpy is more or less straightforward. You get your CCCV values and you choose a reference value for your enthalpy and measure the difference using your CT value and hence you get your enthalpy. So, that is more straightforward. I mean, you just have to ensure that your system is exactly as you wanted. For example, let us say that, you know, you have built in a system where you are trying to put in some kind of heat energy and if you want to get enthalpy, if you want to have a very straightforward way to do it, then you just have an adiabatic piston-cylinder arrangement and try to see if you can measure the heat input and try to get a very good estimate of the enthalpy, okay. Otherwise, I do not see what other methods there could exist out there. Thank you. These are reversible adiabatic processes, always isentropic. This is number one. That is true. Number two is an isentropic process. We are having doubt with the second clause. Is an isentropic process always reversible adiabatic? Okay. So, let me put it the other way. The first question is, if you have an adiabatic and reversible process, yes, it has to be isentropic. But if you have an isentropic process, it need not be adiabatic and reversible. It can be just an isentropic process, but it can be an irreversible process. So, that is for sure. But if you had an isentropic process, and somehow you ensured that it was reversible, then it has to be an adiabatic process also, okay. But by itself, a process could have been purely isentropic, and it need not be adiabatic also, and it need not be reversible also. You can just have at the end of the process, the entropy is same, but the process could have been irreversible. So, there are analysis of the second law of Tharunanamit, the Clausius Statement and their Kelvin-Plank process. What is the basic difference between the heat pump and refrigerator? So, only the end effect is important. That is the only thing. In the refrigerator system, you are interested in having something which is cool, and that is what you call the end effect. If what you want is the hot thing, you will call the same system as a heat pump. So, absolutely no difference between the two systems, okay. They are exactly the same, okay. So, that is about it. There is some discussion, there is some few days, yesterday's. So, there is some difference of the, what is the significance of the entropy, and some parameters, how to evaluate when you are using the Moliachard and the steam tables also. There is some problems about the vapours and steam. So, there is a big confusion of vapours and steam, and how to evaluate the enthalpy and entropy. Because of their use in the evolution of breakdown and etc., for the thermal machines, thermodynamics machines, which is the steam turbine and gas turbine also. Okay. See, if you are solving for the steam turbine, then you just purely use the steam tables. Okay. You use the steam tables. If you know your pressure and temperature, all the properties are listed, that the enthalpy, the entropy, etc. So, I do not see what the problem is. Whereas, if you are using a gas turbine, you will purely go with ideal gas equation as far as possible, even though you are operating at a reasonably high pressure. But, we will still use an ideal gas equation and go ahead, and it gives reasonable answer. But, I think you already repeated this question thrice, and I am not sure whether I am getting it correctly. I mean, I am not very sure that I am answering your question correctly. So, you know, it is better if you can just write this question and send it to us and we will see what it is that you are trying to say, because I have already answered this and then, again you are asking the same question. So, probably I am not able to get what you really want to ask. So, you seem to be, there seems to be some problem with the steam table calculations that you seem to have. Is there some problem with how you are using the steam tables? 1195. Please go ahead. In which the pressure and temperature is given, we have to find out the, how many G is acceleration and air particles is subjected. Actually, I could not pose it for this. Please explain it. Yeah, so the question is on Cf, problem number Cf5. So, I will just explain to you what is expected here. With the data that is provided in the problem, you can find out the velocity of the air flow ahead of the shock and you can also find out the velocity of the air flow after the shock. So, as you can imagine that the velocity of the flow before the shock is going to be supersonic and the velocity of the flow after the shock is going to be subsonic. So, that is v1 and v2 is both something that you can find out. Is that fine? And once you find out both v1 and v2, the acceleration that an air particle that is going to experience is going to be something like v2 squared minus v1 squared divided by 2 multiplied by the extent of the shock. This is coming directly from mechanics if you recall the expression for acceleration, which we many times write as v times dv ds or equivalently you can write as dds of v squared over 2. So, the dds of v squared over 2 is simply designated as v2 squared over 2 minus v1 squared over 2 divided by the extent of the shock which is given as 10 to the power minus 7 meter. So, this is a rough estimation and the idea for the problem is to just to give you an idea of how large a deceleration an air particle is going to experience when it moves across a shock. Hopefully you could understand what I was talking about. Mach number so far we have discussed it is the function of velocity, fluid velocity and speed of sound. How it behave with the temperature when we are considering this constant velocity. See if you want to look at how the Mach number is going to behave with temperature, you can just look at its expression that it is the velocity of the fluid divided by the velocity of the sound and velocity of the sound is given in terms of the square root of temperature. So, in that sense Mach number is going to behave as 1 over square root of temperature as long as the fluid velocity is going to remain constant. So, in that sense m is inversely proportional to the square root of temperature. 1, 2, 6, 1. Sir, if I have heard one of the statement of second law that heat can't flow from lower temperature to higher temperature, okay. So, in case of earthen pot or in Marathi you can say it as a Mach or Madhaka in that how cooling occurs and will you please explain that process over to you sir. Okay, so you have made a very general statement of the second law which is not as it is, it is you are actually trying to look at the clausier statement okay that it is in given in terms of temperature and obviously without any external aid that this temperature the energy flow will be from we will not occur between from a lower temperature to a higher temperature. In case of a Mach as you are calling it okay. So, what really a Mach is of course the same it will behave the same as a water cooler. It is very effective in dry temperature okay where if you arrange enough surface area for the water to evaporate then what you are doing in essence is the energy for evaporation you are removing some from the water and some from the surrounding air and hence you are cooling the air. So, this is just like how water cooler work and that is how the Mach is also working. So, it is not that you have not used any external aid there is some external aid happening water does go through the pores and it comes in contact with the air and it can evaporate there because there is a chance for it to evaporate and hence some energy will be removed from the water itself. Thank you. In some substances like melting point and boiling point there is one also one of the point occurs called as fire point okay. So, why it occurs and as previously that Gaitondeser has drawn a diagram for water that is that phase diagram. So, is it possible to draw for possible to show that fire point on that phase diagram? Okay, I mean these would be normally given for fuel you know at what point see the fire point and flash point are normally given for fuel which will mix with an oxidant and give fire okay. So, they are not normally given for every kind of fluid and you cannot put them on a regular TS diagram like this okay. What you really require is combustion principles we will cover combustion later but what you really require is for so many grams of fuel if you can actually match the so many grams of air and give the initial energy to ensure that combustion starts that is when the fire will start okay. So, it has to it is to do with how much energy you are providing and whether you are using the fire fuel mixture okay. So, the fire point is totally a different concept you would not normally put it on something like a TS diagram or some phase diagram thank you. What is the significance of Mach number from fluid mechanics point of view and from the thermodynamics point of view? Actually from thermodynamics point of view is something that we would like to talk about here and from thermodynamics point of view set of light today you can show that the square of the Mach number if you see so m square if you form which will be basically velocity of the fluid square divided by the square of the speed of sound you can show that the square of the Mach number is a measure of the bulk kinetic energy the ratio of the bulk kinetic energy to the internal energy which is because of the random molecular motion. So, the square of the Mach number gives you how much bulk kinetic energy is contained within the flow with respect to the random thermodynamic energy. This is what from a fluid mechanics point of sorry from a thermodynamics point of view from a fluid mechanics point of view you can talk about different types of forces that can exist and in term in case of this Mach number you can talk about the forces that are coming because of the compressibility of the material however in this class I would prefer to restrict myself to the thermodynamic interpretation which is that ratio of bulk kinetic energy to that of internal energy but that is given by the square of the Mach number. So, please keep that in mind. Thank you. My question is that can we say that entropy is a property is evolved from a reversible process if it is so then it is okay but can we apply the property entropy to the irreversible cycle or process over to you sir. Yeah I mean entropy is just state property so it does not depend on what kind of process you have used at all okay. So, you have one state there is an entropy associated with it you have another state you have an entropy associated with it if you go from one state to other state depending on you know the difference in entropy either your entropy can increase you can decrease anything can happen now it is up to you whether you want to use it or not okay. So, as engineers we would normally use it in heat engines in turbines in compressor we make use of entropy to ensure that our engine is as good as possible okay. Otherwise from a physical point of view you will give entropy just a number that you assign to a state and do not do anything with it okay. So, it is something that we engineers really use to get our machines as good as within the constraint that is about it thank you. One zero eight line please go ahead sir you have told regarding the Mach number like if M is less than one it is subsonic if M is greater than one it is supersonic and M is equal to one it is a sonic flow. So, my question is Mach number will not only depend on the V. So, when do you say that M is infinite or M is equal to zero. Yeah. So, the question is about how do you want to interpret the Mach number as being very large and very small that is a good point actually. So, you can really talk about Mach number being very large if the speed of sound is very small compared to the fluid velocity itself. So, it is as you say that as you can see it is a relative measure of the fluid velocity to the speed of sound in that fluid. So, to get very high Mach number either the fluid velocity has to be very large as compared to the speed of sound or vice versa you can say that the speed of sound has to be much smaller as compared to the fluid velocity. So, this is as far as large value of the Mach number is concerned. Likewise you can say similar statement for a small value of the Mach number. So, typically for incompressible type flows what we have is that the flow velocity is very very small in comparison with the speed of sound in incompressible situation and that is the reason for incompressible situation you will see that the Mach number is very small. But your point is well taken that it can be either the flow velocity or the speed of sound both can decide whether the Mach number is going to be large or small. Thank you. If we increase the speed of sound that time it means that m is infinite. Well as I said it is the ratio of the speed of the flow to the speed of sound that is going to decide whether you are going to call Mach number very large or very small. So, in case you are in a situation for example as an example if you have a very high speed flow really high speed flow and the temperature within the flow is very very low ok. So, what is going to happen is that the speed of sound which is given by your square root of gamma r times temperature is going to be very small. So, here the ratio of the speed of the flow to the speed of sound because the flow speed is very high and because the speed of sound is very low is going to be very large and therefore you will say that the Mach number is going to be large. So, this is the way you can explain anywhere by the ratio of the speed of the flow itself and the speed of the sound. Center 1118 you have had a question let me just read the question and then hopefully I will try to answer it. So, the question is what is the difference between normal shock wave and oblique shock wave. So, the difference in the normal shock wave and the oblique shock wave is in the fact that when we are dealing with a normal shock wave the discontinuity or the shock itself as we say is standing perpendicular to the flow streamline whereas in the case of an oblique shock wave the discontinuity which is the shock is standing typically at some angle which is not equal to 90 degrees with respect to the flow stream lines that is the difference between normal shock and oblique shock. The second part is which one is more effective and this is slightly subjective question in the sense that what is it that you want the shock to do. So, we have seen that today the shock will increase the pressure for example from stream to downstream. So, in case your objective is to increase pressure you will see that the normal shock is much more effective than the oblique shock and the final part of the question was what kind of shock wave what kind of a wave or a shock wave do we expect in a helicopter. This probably is the most difficult question to answer for me the in case of helicopter what you have is a spinning the fan type situation. But the fan is essentially if you see clearly it has whatever number of blades but that blade has finally an aerodynamic airfoil type cross section. So, if you want to look at the flow as a relative flow between the or past the airfoil. Then I would expect that in case we are getting into situations where that relative flow is going to be supersonic with the blade. We will get into something like an oblique shock type situation. In fact, it is going to be more than oblique it is going to be something called a blunt body shock which will be detached shock from the body. So, hopefully that answers your three questions. 1047 SBM college Darwad please go ahead. Sir, in the case of back pressure turbines and condensing turbines how can we really compare these back pressure turbines and condensing turbines on T s diagram and H s diagram and how the state end points can be represented. I mean I am not very sure why you know what is the problem if you draw it on a T s diagram. So, I will just draw the T s diagram and I will just draw let us say a Rankine cycle here. Now, a back pressure turbine normally is something where you will say that the exit pressure is more than the atmospheric pressure and I would not normally use this if I am in a purely power generation situation. Whereas if I am using an industrial process that is when I really use a back pressure turbine because the exit of the turbine I need some pressure for some industrial process. Otherwise here you have a condensing line and I would rather have the whole thing as a condenser because I will get maximum work output. So, this is my pressure line here where I will be in a condensing turbine. But of course I there is no nothing stopping me if I want to draw you know I want to have a higher pressure at the exit here. So, here let us say I am at a back pressure turbine it does not matter on the T s and H s diagram I just see where the line will end up on that pressure line. It is just that you would not use it normally if it is purely power generation you will use it only if you are doing some kind of you know co-generation that is that exiting steam you will use for some other process. That is when you will really require the steam for a higher pressure otherwise normally I would not go for a back pressure turbine. Based on the cooling power actually in cooling power analysis we are using psychometric charge. But psychometric charge are prepared at a constant pressure but in a natural rock cooling tower that is a natural rock cooling tower inside the cooling tower the pressure changes from one section to other section is it viable or some other methods are there to see if the total pressure actually changes and if it changes reasonably let us say by 5% or so or within 5% it does not matter if you use psychometric charge but I think if the pressure starts changing far more than that you know you can use the steam table nothing is stopping you from using the steam table it is as accurate as it gets and as I said you know at least when we are teaching thermodynamics we do not want people to start using the psychometric charge you better be comfortable using only the steam tables because there is no assumption anywhere that the total pressure is more or less than one atmosphere you just are going to measure the barometric pressure and do all your calculations using the steam table. So using the steam tables your calculations will be as accurate as possible but of course you will need to know the pressures at different points in the cooling tower but that is something you will have to get via measurement. Thank you and see you tomorrow.