 We were on this slide where we looked at the, compared the equation of first law for the closed system and open system and before that we have looked at the much simpler equation of conservation of mass and if you have looked at the comparison you will notice that the interaction terms remain the same or more or less the same and terms pertaining to mass inflow and mass outflow they get involved on the right hand side. So if we do this in an analogous manner what shall we get for the second law? The second law quite often known as the entropy principle for a closed system in terms of I have written in terms of the entropy generation term S dot P that means I have just taken the equation delta S equals dQ by T or integral dQ by T plus S P differentiated that in differential form and we get dS of the system by dt or d by dt of the entropy of the system is summation of Q by T. I have written summation of Q by T just for simplicity this could be summation if there are different Q dots at different T's. If as Q dot takes place the temperature changes you will have to have a dQ dot by T integration and in a most general case is summation of an integration I think you will be able to take care of that plus S dot P. Actually this the equation on the left hand side here is the definition of S dot P the second law actually dictates that S dot P is greater than or equal to 0. Now I think I forgot to mention yesterday for some reason it is my habit to write greater than or equal to or less than or equal to by two symbols stacked over each other and not the mathematical symbol where greater than or equal to is a greater than sign with the lower arm doubled up and less than or equal to is a less than sign with the lower arm doubled up that is okay but I prefer this nomenclature for the simple reason that it tells us that it can be greater than 0 or it can be equal to 0 and the equal to 0 has a spatial connotation that means it is the reversible limit and that is why I would like at least I would like to continue using this nomenclature but I do not insist that others this nomenclature although I would prefer this nomenclature to be used while doing word processing in latech I have to create such a symbol because if you write greater than or equal to it will use the standard mathematical greater than or equal to okay. Going back to our second law now if you convert this from closed system to open system what is going to happen I think you would have guessed that the system here will be replaced by control volume these two terms will not change but you will have a term which would be m.i si minus m.e si added on to the right hand side so notice that as in the earlier two equations conservation of mass and conservation of energy we have two appropriate terms added on the right hand side on the left hand side the subscript system has been replaced by control volume otherwise nothing else had changed I recommend that you spend time on this so that you appreciate the similarity and the small specific differences which exist between the two this is necessary because later on when we consider more complex systems where along with a fluid various components flow in okay different chemical species as for example in case of psychrometry along with air moisture flows in and moisture flows out we can write a an equation for conservation of moisture or tomorrow you have some other property suppose hydrogen and oxygen go in you can write an equation for conservation of hydrogen and you can write an equation for conservation of oxygen and there everywhere on the left hand side you will have a rate of change term on the right hand side you will have the inflow and outflow terms and on the right hand side you will have some other terms which in general are known as the source and sink terms so that is why I referred to show this to you by analogy if you do not believe analogy is proper convince yourself that this is the true derivation by going back to our control volume and deriving this expression from first principle now so far we have considered our open system to behave in a more or less general way we will say let the state of the system change with time on the left hand side for the mass conservation equation we have dm Cv by dt we have d e Cv by dt on the left hand side of the energy equation and we have d s Cv by dt for the second law that is the entropy equation quite often we find that our equipment works in so called steady state a steady state is one in which things do not generally vary with time that means the interaction u dot w dot s m dot i and m dot e are constant constant means invariant with time and not only that the state of the system does not change with time so the mass of control volume does not change with time energy of the control volume does not change with time and even the entropy of the control volume does not change with time thus such a situation is known as a steady state it is a common fault or common not uncommon trap to assume that a steady state only means these three steady state does not mean just these three derivatives to be 0 it also means that q dot w dot s m dot i and m dot e are constant and invariable in time at least over a given period of time may not be indefinitely now in a steady state what happens let substitute what we have decided to define the steady state that these are invariable with time but we do not have the derivatives of this so the expressions do not change but these are put to 0 so we put our left hand side to 0 and for convenience we will transpose some terms from the right hand side to the left hand side or flip the equation so the conservation of mass becomes m dot i equals m dot e and if you have one inlet and one exit both are equal and we use a common symbol m dot for them and we call this not as mass inflow and mass outflow but the mass flow rate through the system the flow of mass through the system enters as well as leaves the first law becomes q dot minus w dot s this term is brought to the left hand side on the right hand side we have m dot into h e plus v square by 2 plus g z e that is the exit velocity exit term the inflow term is m dot into h i plus v i square by 2 plus g z i and since now m dot e equals m dot i and is replaced by m dot it is common to combine terms like this we write it as m dot into h e minus h i plus v square by 2 plus v square by 2 minus v i square by 2 plus g z e minus z i and again it is common to write this h e minus h i as delta h the increase in enthalpy from inlet to exit delta e k this is the change in enthalpy from inlet to exit so delta e k is the change in kinetic energy from inlet to exit and delta e p is the change in gravitational potential energy from inlet to exit the second law becomes m dot into h e minus h i is sigma of that q by t term plus s dot p with the second law dictating that s dot p is greater than or equal to 0 and here you should notice that there are three term and you can play with these three terms the way we have played with delta s dq by t or ds dq by t and dsp so if q dot by t is 0 we will say that our control volume is adiabatic if s dot p is 0 we will say that the processes of flow and thermodynamics involved with the control volume are reversible and m dot into h e minus s i means the inlet if this is 0 the inlet and exit states are isentropic we do not know about the process inside we have not looked at it but we will say that the inlet and exit states are isentropic and that means that you can have the inlet and exit isentropic but you can have this positive this negative so you need not have an adiabatic system you need not have reversible situation similarly you can have inlet and exit states isentropic but you can have adiabatic control volume but that need not lead to isentropic inlet and exit states because you could have s e greater than s i and s dot p greater than 0 that also means that if you have an adiabatic open system with one inlet and one exit the exit entropy has to be higher than or at most equal to the inlet entropy and in the similar fashion a an open system in which all reversible processes are taking place need not be a situation where inlet and exit states are isentropic you could have s e greater than or less than s i depending on the appropriate values of q dot as dictated by q dot by t so this is a small exercise of thinking and discussion which I will leave to yourself and since most of our situations in practice or a large number are steady state situations we will use these equations very often. Now as teachers we make a mistake that we consider that the steady state situations are very very common and tend to neglect unsteady state situations that is not true unsteady state situations are also common in fact most of the processing is unsteady the steady state situation is only an approximation which is true in a few cases. Now we will use simplified forms for special cases why do we need those simplified forms because not all terms in the equations which we have derived are often significant and we also need to make suitable assumptions quite often of course whenever we make these assumptions we better be realistic and we check whether these assumptions are valid and we will now look at some typical classes of devices and look at the default assumptions and the simplified forms that these lead us to simplified forms of the first law of thermodynamics and the second law of thermodynamics for open systems. First we look at heat transfer devices the so called heat transfer devices such as boilers, condensers, heaters, coolers, heat exchangers. The purpose of this is to absorb or reject heat and one consequence is the any fluid stream which goes through such heat transfer device will have a significant delta H either positive or negative. In case of say boilers, heaters delta H is likely to be significantly positive for coolers and condensers delta H is generally significantly negative. For heat exchangers there are usually two streams sometimes more and when there are two streams one stream will have a significantly positive delta H another stream will have a significantly negative delta H. A typical thing is for such things there is no W dot S or hardly any W dot S. A boiler may have a small pump or circulator somewhere inside but the interaction is negligible compared to other term. Although I may say no W dot S well you may have a boiler with a W dot S either take care of it or assume and demonstrate that the value of W dot S is negligible compared to other terms. So our symbolism reduces a heat transfer device in steady state to something like this. M dot is the mass flow through it inlet and exit are the states I and E q dot is absorbed and W dot S is 0. Not only that we find that q dot and delta H are significant and quite often usually delta E k and delta E p are pretty small negligible. So if no information is provided or no assumptions or no information is provided or no assumptions or no information is provided or we cannot compute these it is quite often convenient to assume these to be 0 delta E k and delta E p. You should also note that for many of these devices delta p from inlet to exit is also pretty small. The first law in this case reduces to this simple equation q dot is M dot into H E minus H I. There is no special simplification for the second law. We now come to work transfer devices. These are of more interest because of the work transfer interaction and the second law coming into play very significantly. Work transfer devices are turbines, compressors, pumps, fans, blowers, etcetera. Very common devices. The purpose in these devices is to have significant power transfer. All these are consumers or producers of significant power and significant delta H. Turbines as we know are producers of power, turbines including wind turbines and hydro turbines, compressors, pumps, fans, blowers are consumers of power. And for all of these the delta H of the stream is significant. For turbines the delta H of the fluid flowing is significantly negative number. For compressors, pumps, fans and blowers it is a positive number. These things are not heat transfer devices. In fact heat transfer quite often creates problems or reduces efficiency of these devices and hence if they work at very high or very low temperature they are well insulated. So the heat transfer rate between the system and the surroundings is negligible and the adiabatic of assumption is often valid and is one of the default assumptions made. For these devices we have some standard symbols which are often used. For a turbine we have a channel with increasing area from left to right. For a compressors we have a channel with a decreasing area from left to right and for a pump we often have a circle with an arrow showing the rate of the direction of increase of pressure from inlet to exit. Of course for pumps there are other symbols used. For example sometimes the exit will be shown as a tangent to the circle with inlet going right up to the center of the circle. That symbol is quite often used for a centrifugal pump. Similarly a symbol with a rectangle with a piston at the bottom showing inlet and exit both at the top is sometimes used for a reciprocating pump. But those are technical details used in appropriate industry and there may be some industry standards for these. But for our course in thermodynamics something like a rectangle for a heat transfer device and these three symbols for work transfer devices are good enough. Now such devices as I have said are usually adiabatic. W dot S and delta H are the most significant terms. Again quite often the changes in kinetic energy and potential energy are small can be assumed negligible. Once you remember that for these devices most of these devices except perhaps for fans the delta P is significant from inlet to exit. Significant rise in pressure or significant drop in pressure. So the first law is reduced to with these assumptions W dot S is m dot into H i minus H e. Similar to the work transfer law but here on the right hand side H i and H e come in different order. That is because we have two different signs for Q and W on the right hand side of the first law. The second law becomes m dot into H e minus H i which represents the rate at which entropy is produced S dot P. This should be greater than or equal to 0. And that automatically means that under default assumptions these pieces of equipment will have their entropy exit entropy higher than or in the limiting case equal to the inlet entropy. And then it is very important for us to take care of this whenever we consider work transfer devices as open systems. And it will be very useful for us to look at the enthalpy entropy diagrams for such systems. Now that brings us to the importance of enthalpy entropy diagrams. Why are they important because notice go back to the previous slide. Enthalpy difference dictates the work transfer or power output. The entropy difference tells us how reversible or how near to the reversible limit it is. And it is for this difference with these reasons that the significance of the entropy enthalpy entropy or the molier diagram was realized decades ago may be a century ago. In fact that diagram was developed even before a complete and full understanding of the principles of thermodynamics. So that is another illustration which tells us that industry goes ahead whether science explains all the phenomena properly or not. So let us look at the H s diagram of adiabatic work transfer devices. Since we have devices in which enthalpy can write significantly or enthalpy can drop significantly we have to look at these two diagrams in two different ways. On the left hand side here I have a enthalpy entropy diagram for a typical compressor. And on the right hand side I have an enthalpy entropy diagram for a typical turbine. And in all these cases it is assumed that we have a steady state delta E k delta E p are negligible and the equipment is adiabatic. Now here you will notice that the inlet state i and the exit state e are such that S e is greater than S i. The limiting case will be e star where the exit pressure will be maintained but the exit entropy would be equal to the inlet entropy. It is traditional to show the ideal line by means of a continuous line this is the isentropic line. But the actual process being irreversible it is shown by a dotted line. Actually there is no reason for us to even show this continuously because we do not know what the intermediate states are but we could simply join i and e by means of another dotted line. But tradition says join this by a continuous line join this by a dotted line. Similarly for a turbine this is the inlet the actual exit will be generally at a entropy higher than the inlet entropy. But there is an ideal exit state indicated by e star which will be at the same entropy as inlet. And again by tradition and convention we for e and e star the exit pressure maintained both for the compressor as well as the turbine. And now one thing you should remember that whether it is compressor or whether it is turbine the state e lies at higher entropy compared to state e star or state i whether it is a compressor or a turbine entropy of e is higher than entropy of e star which equals the entropy of i. And the enthalpy of e is always higher than the enthalpy of e star whether it is a compressor or a turbine enthalpy of e is higher than enthalpy of e star. Now based on these diagrams for adiabatic turbine and adiabatic compressor we have some definition which are very common in theory of thermodynamic and its application. So first let us look at an adiabatic turbine and its H s diagram. A turbine is one in which the enthalpy of the fluid is reduced significantly from H i to H e. There is not no significance in delta e k and delta e p and this reduction in enthalpy is used for producing power which is delivered to the outside world. The exit pressure is below the inlet pressure significantly. Now notice this entropy of a star is entropy at i but entropy at s e is higher than entropy at s e star. Also enthalpy at h e is greater than the enthalpy at h e star. Now subtract this inequality from h i since we are subtracting the inequality will change direction h i minus h e h i minus h e this one the real enthalpy drop will be less than or equal to the ideal enthalpy drop h i minus h e star. And w dot s which is m dot into h i minus h e would be less than w dot s star which is the power output under the conditions of isentropic inlet and exit states. Both will be positive numbers and we define the isentropic efficiency of the turbine to be the actual power output of the turbine divided by the ideal power output of the turbine. The ideal power output of the turbine is defined as the one in which the entropy of the exit state would equal the entropy of the inlet state at unchanged exit pressure. And one should remember that this isentropic efficiency of the turbine is defined and is define able only for an adiabatic turbine. If it is not adiabatic then this relation is not valid this relation is not valid and hence the further derivations are on weak foundation. In a similar fashion let us look at an adiabatic compressor and its H s diagram. All that has happened is p i and p e have shifted their locations. Now p i is lower than p e, p e is higher than p i, otherwise the other relations are still applicable. S e star is still equal to S i and S e is still higher than S e star and hence h e is still higher than h e star. Now what we do is subtract from this inequality h e greater than or equal to h e star h i. So the actual enthalpy rise is higher than the ideal enthalpy rise. Now remember that w dot s is m dot into h i minus h e. So we will get w dot s to be less than or equal to w dot s star but both are equal to negative and since both are negative in magnitude w dot s star will be lower than w dot s. It is like w dot s being something like minus 15 kilowatt whereas w dot s star being minus 10 kilowatt. So both are negative but minus 15 is algebraically less than minus 10. So in magnitude w dot s star will be lower than w dot s and hence we define the isentropic efficiency of the compressor in a slightly different way compared to the definition of the isentropic efficiency of a turbine. For a turbine it is w dot s divided by w dot s star. Both in case of turbine both w dot s and w dot s star are positive numbers. In case of turbine in case of a compressor both w dot s and w dot s star are negative numbers and the isentropic efficiency of the compressor is defined as w dot s star divided by w dot s and this difference should be noted and it is defined like this for no particular reason except that psychologically we are tuned to the fact or tuned to the belief that efficiency should be a number always less than or equal to 1 but that does not mean that we force all efficiencies to be less than 1. All of us are mechanical engineers and we know that welding is a very popular and very useful metal joining process and the efficiency of weld is defined in such a way that quite often the efficiency of a weldment turns out to be greater than 1 and when it turns out to be greater than 1 the user of that equipment is happy because there is no deterioration in its performance. There are no additional limits on its performance. Let us go further. There are some other adiabatic devices and one of the most important adiabatic device that we will come across is the nozzle. The nozzle is not a work transfer device. The nozzle is not a heat transfer device. Nozzles and diffusers are unique in their own way. For a nozzle which is a very common component in steam turbines, the purpose is to reduce enthalpy but increase velocity. A typical symbol for the nozzle is a converging long duct sometimes shown as converging diverging with just inlet and exit and the exit velocity is shown because that is the important outlet parameter that we need. Typically for a nozzle the exit pressure is significantly below the inlet pressure. The exit enthalpy is significantly below the inlet enthalpy. So these two characteristics the nozzle shares with a typical turbine. Exit pressure lower than inlet pressure exit enthalpy lower than inlet enthalpy. But the difference is we are producing no power. Our job is to see to it that our exit velocity is large and is definitely significantly higher than the inlet velocity. There is no attempt to extract power so W dot S is 0 and we try to maintain no heat transfer. So a nozzle, a proper nozzle is a good adiabatic device and because it is adiabatic and because pressure reduces the H S diagram, the enthalpy, entropy or molier diagram for a nozzle is similar to that of the turbine. So let us look at the H S diagram of a typical adiabatic nozzle. Notice that if we apply first law in steady state all terms will vanish and if we neglect the delta E p only delta H and delta E k terms will remain and first law will reduce to H E plus V E squared by 2 is H i plus V i squared by 2. Because it is adiabatic the ideal situation would be an exit state of E star at a lower enthalpy and lower entropy than the actual exit state E. Suppose I were to have an ideal nozzle which is adiabatic and with reversible and reversible and inlet entropy equal to exit entropy then the first law will reduce to H E star plus V E star squared by 2 is H i plus V i squared by 2. Now since H E star is less than H E, V E star for the ideal nozzle would be higher than the V E which is the exit velocity of the actual nozzle and hence we define the isentropic efficiency of nozzle as the actual specific kinetic energy at the exit of the nozzle. Divided by the ideal the kinetic energy at the exit under the ideal situation isentropic situation and which generally will be less than 1 in the limit it will be equal to 1. Notice that the isentropic efficiency of the nozzle is defined not in terms of velocity but in terms of kinetic energy because when it comes to efficiencies of turbines and nozzles turbine compressors and nozzles the efficiency is expected to be a ratio of entities like work or energy or power that is why we do not use the ratio of velocity. However in actual turbine theory the ratio of velocity is also used as a parameter and that is quite often known as the velocity coefficient of the nozzle. In thermodynamics we do not have to talk about velocity coefficient. Finally let us look at a very special case that of an adiabatic duct. Let us say that I have a control volume shown by dotted line just for the sake of it and a long pipe or duct flows through it and this is adiabatic so as the fluid flows from the inlet I to the exit E there is no heat transfer and since it is a simple duct I have made no attempt to extract or provide any power W dot S is also 0. There may be some flow work which will be there if something is flowing we do not worry about it. In this case the first law will reduce to in short symbols delta H plus delta E K plus delta E P is 0 or separating inlet keeping inlet terms on one side and exit terms on other side we will get H i plus V i square by 2 plus G z i equals H e plus V square by 2 plus G z e and if the state at inlet and exit such that the inlet specific energy specific internal energy equals the exit specific internal energy then this reduces to P i V i which I write as P i divided by rho i plus V i square by 2 plus G z i is P e by rho e plus V square by 2 plus G z e which all of us will realize is the Bernoulli equation. Now a question here traditionally the Bernoulli derived in fluid dynamics in fact thermodynamics does not claim to own the Bernoulli equation it is the science of fluid dynamics which claims to own the Bernoulli equation and in fluid dynamic the standard way of deriving the Bernoulli equation is to consider a stream tube or a stream line in the limit up to apply the conservation of momentum to a slice in the stream tube is known as the Euler equation and then integrate the Euler equation along conservation of momentum not looked at the conservation of momentum. We have looked at only conservation of energy and this is now a representation of conservation of energy. If you have studied fluid dynamics ask your this question or if you consider not an expert in fluid dynamics then get hold of a colleague or a friend who is an expert in fluid dynamics explain this situation to him and you yours or you two together should discuss this matter that whether Bernoulli equation really represents conservation of energy or does it represent conservation of momentum or does it represent two different things and only the end result seems to be similar if so why we will not discuss this particular aspect any further because that would mean encroaching on the domain of fluid dynamics bias which we will not do that encroachment is allowed only during a study of compressible fluid flow which you will do with Professor Balchandra Puranic on Monday and Tuesday. So at the end of this I have latched on to 1, 2, 5, 7, HCTM Technical Campus, Kaital, Haryana over to you. Yesterday you were talking about the entropy as the entropy is always increasing so what was the limit of increasing entropy in the universe? This is a typical question entropy increases but entropy increases only in an adiabatic system and the other issue which we generally come across is the confusion over the term universe. Universe in thermodynamics means any suitable adiabatic system the universe in thermodynamics has nothing to do with our actual physical universe and because of that although thermodynamics tells us that the entropy of an adiabatic system always increases at least today in these days we do not really have to worry about the rise in entropy of the universe because the physical universe is not really model able today as a thermodynamic universe we are not sure whether it is adiabatic system at all over to you. In the thermodynamic nations there is a availability, will it touch it or not in thermodynamics? Yes, we will be discussing terms like availability and exergy beginning Venice day in topic 12 which is called combined first and second laws. I have devoted one full day typically for three fourths of a day to that discussion. What is the difference between a classical thermodynamics and statistical thermodynamics? Can you throw some light on it sir? See by classical thermodynamics we mean the following if you see the history of thermodynamics thermodynamics was developed in two or three different streams. One was the thermodynamics which was developed by engineers for the thermodynamics the developed was technology development by engineers for water wheel steam turbines wind turbines boilers and all that without even understanding what they are doing that was one development of thermodynamics where people handled fluids like air water various gases. Then physicist and chemist started working on materials in their lab and they developed thermodynamics in their own ways. They realized that you know Dalton's law of partial pressures and various laws of physical chemistry made us realize that material is made up of atoms and molecules and everything is a collection of these. When thermodynamics was formalized it was realized that there are some very general principles of thermodynamics for which we need consider our material to be only a continuum and based on observations on the behavior of continuum we will be able to derive our laws of thermodynamics. That today is known as classical thermodynamics whereas the modern thermodynamics considers the fact that each and everything is made up of atoms and molecules and these atoms and molecules will behave like either point particles or extended particles or minute rigid bodies they will interact by collision, electron, magnetic forces. But then to keep track of these we need to handle a very large number of pieces of information. For example you know that even if you take just one mole of a fluid it will have the Avogadro number of molecules N A. So each molecule even if you consider it to be a particle it will have three positions point particle. So that means 3 into Avogadro number of positions all changing with respect to time. So you will have to write ordinary differential equations, a set of ordinary differential equations the order of that set will be 3 into Avogadro number. Now imagine solving these in time when we know that on an average their average momentum is going to be the same, average energy is going to be the same. So we do averaging and we end up with kinetic theory and statistical thermodynamics. When we are in engineering particularly in mechanical engineering we find that a vast majority of our applications do not need us to look at all these details. Hence we work with classical thermodynamics and naturally if we work with classical thermodynamics our aim is to remain in the classical domain to the extent possible. So even for situations where we have reactions or we have pictures we will try to extend our classical domain to the extent possible. And today we have a situation in our understanding it is so neat that even reacting things boilers, turbines, turbines do not react but we have mixture there. All these things can be handled by minor modifications to the classical theory of thermodynamics. I hope this explains it. Over to you. Sir I have a doubt regarding the refrigeration system. In the refrigeration system how about increasing the number of blowers in the freezer? What would be the impact if you know increase the number of blowers in the freezer? Will it affect the efficiency of the freezer sir? It is a question which is on the borderline of thermodynamics and heat transfer. We know that the evaporator which is at the heart of the freezer absorbs heat from the whatever is the material surrounding it. We want good heat transfer between the freezing chest or the frozen material and the evaporator so that the material is kept cool. It does not get heated up because of heat gain from outside. So we want good heat transfer and for this we provide a blower or a fan. Now the fan provides increases the heat transfer coefficient. So reduces the temperature drop or temperature difference between the cold space and the evaporator. Now this increases the evaporator temperature and hence it increases the coefficient of performance of the refrigeration cycle. But the blower requires a power to be consumed. So now the question that arises is that if I do not have a blower I do not consume blower power but my evaporator temperature goes down and hence my COP decreases and hence the refrigerator power goes up. If I use a blower refrigerator power goes down but the blower power goes up. The answer to this is not unique. It depends on the situation at hand and you will have to do the calculations in detail of the two situations considering the characteristic of your refrigeration system as well as the characteristic of the blower and then decide whether to have a blower or not and if you have a blower what should be its capacity, power consumption, flow rate and all that. Over to you.