 Now, we come back to where we left and we will now look at exercises 3 which are essentially based on steam except a few in between which are based on equation of state other than Van der Waals gas that is other than steam that is on Van der Waals gas. If you just go through you will find that the exercises 3.1, 3.2 and 3.3 these are essentially warm up exercises for the students to get used to steam debuts except that there are a few exceptions which I will mention we come to exercises based steam that is set 3.3.1 and 3.2 ask you to determine the state given two properties and it is good idea to not for example in 3.2 it is as classify the following states of 1 kg of water substance as wet dry saturated super heated steam subcooled liquid etc etc you should also extend this by saying and determine all other relevant property so that they do not just stop by saying subcooled liquid but they also calculate other properties. Now, some of these for example the first two maybe I should go here and show it here in 3.2 the first two pressure temperature pressure temperature it is directly based on the phase plot so just look at the saturation line and you can determine whether it is subcooled liquid super heated vapor or saturated liquid then you have a few others where it is pressure and some property other than temperature similarly pressure some property other than temperature pressure some property other than temperature temperature some property other than pressure and of course this is a another one. So here one property is purity and the other property is either V or U or H or S then one should realize by looking at a study of the steam tables that if say this property is phi let me use the symbol phi for any one of these property then it turns out that phi of subcooled liquid is less than phi of saturated liquid is less than phi of wet steam is less than phi of dry set vapor is less than phi of super heated steam when the pressure or temperature is below critical above the critical temperature it is only a single phase so it does not matter and based on this for example pressure 2 bar internal energy 6.2 kilo joule per Kelvin but since it is given 1 kilogram specific entropy will be 6.2 kilo joule per kilogram Kelvin you go to the steam tables and you will notice that at 2 bar since the pressure is provided let us go to the pressure base table and at 2 bar you will find that the saturated liquid enthalpy is 1.530 dry saturated vapor sorry entropy is 7.127 the value needed is 6.2 kilo joule per kilogram Kelvin which lies in between these two so it is wet vapor if the value were less than 1.530 it would be subcooled liquid if the value were to be higher than 7.127 kilo joule per kilogram Kelvin it would be super heated steam if it were exactly this then saturated liquid exactly then dry saturated vapor since it lies in between you can determine the dryness fraction first and from the dryness fraction using these other values all other properties. So these cases C similarly D also H and also G can be handled this way because one of the two properties given is pressure or temperature the second property is one of these four and mind you since this density is the reciprocal of specific volume if you use densities it would be the other way round but fortunately for us densities are not tabulated so it is a good idea to tell the students that as far as possible in thermodynamics work with specific volumes because the trends of specific volume are similar to trend of U H and S if you have to work with density remember that it is a reciprocal of specific volume so the trends would just be the other way round and another reason is most of the steam tables will tabulate only specific volumes because the basic formulation is in terms of PVT not P rho T there is one case here where temperature is 100 degree C and dryness fraction is 0.8 temperature is 100 degree C dryness fraction is 0.8 here it is obviously wet steam because dryness fraction is directly given okay. Again sometimes the students have a mental set that 0 is the dryness fraction of liquid saturated liquid 1 is the dryness fraction of dry saturated vapor so if you substitute in that formula they will say superheated vapor is a dryness fraction greater than 1 subcooled liquid is a dryness fraction less than 1 you should tell them that this is meaning less dryness fraction is not defined for subcooled liquid and not defined for superheated vapor okay. So a dryness fraction a formula will give you a value it is a formula that is not thermodynamic formula will give you a value for superheated steam dryness fraction of more than 1 subcooled water dryness fraction less than 0 but that is meaningless that is not meaningful at all so that is what you should impress on. Now here there are two cases E and F one is specific enthalpy of 2900 kilojoule per kilogram Kelvin and specific entropy of 6.2 kilojoule sorry enthalpy of 2900 kilojoule per kilogram entropy of 6.2 kilojoule per kilogram Kelvin 1 kg so these are also the specific values a similar value here 2500 kilojoule per kg 8.8 kilojoule per kg K since this is enthalpy and entropy it is not easy to find out the state directly from the steam tables okay. I have kept these two just to get used to some general characteristic of steam for example if you look at E let us come to the pressure table and it is good to point out to them that if you look at the specific enthalpy of dry saturated steam Hg you will notice that the values start from 2501.4 at triple point they go on increasing and you will find something funny they increase up to 7 bar 2763.5 2799.5 but on this page you will notice 2800.5 going up to 2804.2 and again decreasing okay so the enthalpy increases and then decreases for dry saturated steam and the highest value found is at 30 bar 2804.2 kilojoule per kilogram since our value is 2900 kilojoule per kilogram there is no way it can be sub cooled or wet vapor it cannot even be dry saturated vapor it will be some pressure some temperature which is in the superheated domain so that is what here superheated steam but when it comes to determining the properties you will have to hunt it out in the superheated table and do a two dimensional interpolation may be at this stage it is good to introduce them to the enthalpy entropy diagram but there are a few difficulties here I am putting it here but it would not be only a small part will be seen so we open it out there are two problems here first thing is a minor problem is that enthalpy entropy are non-standard coordinates not even one of them is priority entropy is a property which is not known to them so what we tell them is this if you plot against pvuhs any one of them either p or t here or vuuhs here the general characteristic is like this with the critical point always at the top this is a general characteristic that means you have an inverted dome with the critical point at the top the relative slope of this side and this side may be different it is possible that this side may even go back like this as in case of enthalpy okay in case of for example pH diagram it is very clear and we have seen it enthalpy increases and then decreases to the critical enthalpy so this could be here this could be here but it is an inverted dome a crooked inverted dome in sometimes this will be sharper this will be very shallow for example if you take p and v this is very sharp whereas this is very shallow but when it comes to enthalpy entropy first thing why do we plot enthalpy and entropy that will be clear when you complete the study of second law the importance of entropy will be clear and when you study open systems the importance of enthalpy will be clear and hence for open systems some like turbines compressors boilers the enthalpy entropy diagram is of great significance and that is why it is plotted so on the enthalpy entropy diagram if you plot the full entropy entropy diagram entropy enthalpy for example if this is triple point triple point pressure or temperature and if you say this is our triple point the saturated line that is the liquid vapor line goes something like this this line from here to here gets mapped on to this and this dome gets mapped in a very crooked way like this this shape may not be correct but all this zone critical point and all that gets mapped in a very small zone here so small that it becomes very difficult to read at any scale and that is why what is plotted on the enthalpy entropy diagram at the end of your steam tables is essentially a sub part which is something like this you will see a diagonal line going this line going across like this that is although it is written saturation line it is the dry saturated vapor line the critical point will be somewhere here on the enthalpy entropy diagram it is not on the top it is somewhere on the left it is out of this and again you will notice that at around 30 bar we have the maximum enthalpy dry saturated vapor enthalpy and then it decreases on either side of that point and our value of 2900 kilo joule per kilogram kelvin and what was the entropy 6.2 6.2 kilo joule per kilogram kelvin this is the point in the state space that we were looking at that gives you roughly the idea where it lies so if you look at it these are the constant temperature lines so this point is somewhere between 280 and 290 degree C and these are the constant pressure lines so this pertains to 50 bar this pertains to 40 bar so we are very near the 45 bar line so that allows us to you know reduce our trial and error zone if you really want to determine the properties by interpolation and if you look at the next thing enthalpy of 2500 kilo joule per kg entropy of 8.8 kilo joule per kg kelvin you will notice 2500 is here 8.8 is here so you are out here which is wet zone again in wet zone you notice that we have it in terms of enthalpy and entropy so again you will have to interpolate between pressure lines or temperature lines to get the required values 3.3 is similar except that we are talking of if u is mentioned unfortunately you will have to work only with tables in h and s at least we have a diagrammatic representation in the Moldier diagram but if you have to suppose you are analyzing a situation where quite a number of times u and s would be the variables then you will sit with a steam table and a plotter and plot your us diagram it will be similar to the hs diagram you will plot the us diagram so you will get a feel for it and then start working with it because you know that is what we tell our students that a figure is worth a thousand words the figure simplifies a lot so if you have to work with u and s you will have to do 1000 searches of the steam table so plot it once and your searches from 1000 will reduce to 2 or 3. Now coming to 3.3 the only thing is here there are two states provided you are given an initial state a final state and the link between the two states so you have to first find out the for example in a initial state is given as p1 is 5 bar v1 is 0.3 meter cube per kg and the process is constant volume so v2 is also equal to v1 but p2 is 3 bar so instead of 1 you have to determine two points find out two properties and find out differences in them because this is also something that they will be doing quite often for the time being we will skip 3.4, 3.5, 3.6, 7, 8 and 9 and some parts of later problem which pertain to van der Waals gas time permitting we will come to that otherwise we will solve them in an offline mode when the module session is on. Now let us start solving some problems and first we will look at 3.11 then we will look at 3.14 then we will look at 3.12 we quickly see what is the characteristic of each of these problems are. So we will first look at 3.11 since we have solved a number of problems using first law yesterday the basic discussion of first law we will not do we will assume that we are comfortable with it we will quickly see what differences we have to come across we will face when we look at steam first thing remember steam is nowhere near an ideal gas it can condense it can vaporize. So du equals cv dt, dh equals cp dt is an absolute no-no similarly pv equal to RT is an absolute no-no I am so cranky about this that for a steam if a student like anyone of this I draw a red mark there and I will not look further in that answer at all that means you have not really notice that it is a problem with water and steam and not with some ideal gas will you have the tables yes. So you have to use there is absolutely no reason to use pv equals RT for any part of steam absolutely no reason particularly in a course in thermodynamics of course you can use that if you ask a student that look people say that at low pressures and high temperatures steam like any gas is expected to behave like an ideal gas. So at a pressure of 0.5 bar and 600 degree c find out the volume of steam as predicted assuming it to be an ideal gas in fact something like that is there in the van der Waals gas problems instead of ideal gas we are asking it to be compared to a van der Waals gas but such comparison if you want to do then you can use ideal gas equation for steam to get the volume as predicted by the ideal gas equation and compare it with the volume as given by steam. Yes but in that the vapor pressures are so small they are very small fraction of the atmosphere the temperature region is small so that becomes a suitable approximation but in principle in psychrometry you can use the ideal gas equation or relations for the vapor part steam table sorry for the gas part steam table for the vapor part and you need not ever open the psychrometric chart with these two you can do all your psychrometric calculation they may be a bit cumbersome but they are from a thermodynamic point of view they are more robust and they are more neat they may not be more accurate because we know in that range it is a good approximation but remember there we do not calculate Cp as Cp minus Cv is R and Cp by Cv is gamma we calculate Cp by fitting a curve and say let the mean Cp be something like 1.88 kilo joule per kilogram kilo so there is a data fit used there so a full ideal gas relation is not used. So coming to 3.11 you have two kg of steam at critical conditions sealed in a rigid container rigid container immediately means no Pd viewer constant volume process it is then placed inside an oven at 349.9 degree C and allowed to reach thermal equilibrium so that means it is given that initial condition is critical state final condition is the same value as the value of volume as the initial state rigid container final temperature is 349.9 degree C so initial state is given final state is given determine the masses of water and steam in the final state and the heat transfer. So we start with the system diagram rigid container so I just show a box let us say this is my system it contains 2 kg initially at the critical point and what do I do I put it in an oven which is at 349.9 degree C and we definitely know that the temperature is higher than 349.9 degree C something like 374 the temperature will reduce we expect some heat transfer to take place but we expect W expansion to be 0 there is no stirrer so we expect even W other to be 0 this is an assumption this comes out of the fact that it is a rigid container and we say let there be some heat transfer Q notice that the value of Q may turn out to be positive or may be negative but the arrow of Q we will always show into the system because that is the default positive direction. Now since it is a fluid and in particular since a constant volume process is specified we will sketch the process on the PV diagram the initial state 1 is given to be critical point the final state is given to be a temperature of 349.9 degree C so let me plot an isotherm of 349.9 degree C it will go something like this this is T 349.9 degree C it is a constant volume process so the final state will be on this line and at 349.9 degree C so this is the final state this is the final state state 2 and in terms of specific volume V2 equals V1 is V critical point so we know the initial state fully properties at the critical point we know the final state fully in terms of specific volume and temperature so all other properties as needed can be determined so we have to determine the masses of water and steam in the final state and the heat transfer simple enough. We know total mass is 2 kg we know V2 is V1 is Vcp read it out from the steam tables critical point so end of table 1 or end of table 2 your choice if I go to end of table 2 you will find Vf and Vg are both listed but they have equal value so this is 0.003155 meter cube per kg then you notice that T2 is 349.9 degree C I think this is given because although it is not a this is I think historic because earlier the saturation table in this set of steam tables stopped at 100 but now you have a 350 degree C line there but here you look up the table 2 you have a temperature of 349.9 degree C saturation temperature pressure of 165 bar okay but that is not the issue we now notice that our state is now provided in terms of V2 and T2 so we use this formulation for specific volume and we notice immediately that at 349.9 degree C what is Vf and what is Vg at 349.9 degree C we will notice Vf is 001740 meter cube per kilogram Vg is 0.00883 okay where does V2 lie V2 lies in between you said Vf at 2 is less than V2 is less than Vg at 2 hence 2 is wet steam force the students to write this once it is wet steam then you can calculate x2 is V2 minus Vf2 divided by Vg 2 minus Vf2 I have used commas here so I should also have commas here and that means you calculate x2 once you calculate x2 well from here it is straight forward mass of water mass of water in the final state 2 is 1 minus x2 into the total mass mass of steam in the final state 2 is x2 into the total mass so one part of the answer we can calculate the second part of the answer is the heat transferred naturally for heat transferred we have to use first law do not think of anything else the moment you have to determine heat transfer means it has to be delta E plus W start with this what about W it is given that it is a rigid container so W expansion is 0 there is no hint of any other work mode so we have assumed W other to be 0 that brings us to the conclusion that W is 0 the whole thing is sitting still in a furnace not moving up and down or accelerating decelerating so we will assume delta E to be equal to delta U we will assume other components to be 0 let the students at least for the initial fuse problems go through this and write it down and delta U is nothing going to be M into U2 minus U1 U1 can be read off from steam tables it is already there tabulated 2029.6 kJ per kg at critical point and U2 is obtained from x2 Uf2 and Ug2 Uf2 Ug2 read off from the steam table x2 already completed mass is given if you know this you know Q okay now let us look at 3.14 we have water at some temperature pressure it executes a constant pressure process absorbing heat the amount of heat absorbed is given determine the final state of the system in terms of enthalpy dryness fraction temperature the system is a closed one that is a tautology we are only looking at closed system okay 3.14 since it is expanding we should show it by means of a cylinder piston arrangement the initial state is 100 degree C and 6 bar and since it is going to remain at constant pressure we can show P equals 6 bar that will always remain and since everything is per kg let us assume the mass to be M kg everything will be in terms of M but properties will not depend on it a Q is provided and Q is given to be M into 1500 kilo joule per kg the state space we can draw PV does not matter expansion work so PV is a good idea let us say that this is the saturated liquid line this is the dry saturated vapor line and we are given that it is 6 bar so the whole process will take place on this line of 6 bar the initial state is given at 6 bar 100 degree C when that happens you should immediately check what is the saturation temperature at 6 bar at 6 bar you will notice that the saturation temperature is 158.9 degree C our temperature is below that 100 so we have subcooled liquid or you can say at 100 degree C what is the saturation pressure our pressure will be higher than that either way you will come to the conclusion that it is subcooled liquid so if you plot 100 degree C line it will go something like this and so this will be our initial state state 1 it is subcooled liquid so using the subcooled liquid model I mean incompressible liquid model we know how to calculate the properties HUVS as appropriate now it absorbs heat and expands at constant pressure so the process will go something like this it will end up somewhere we do not know let it end up in some 2 at this stage we do not know whether 2 is in the subcooled liquid zone or wet zone or exact dry saturated vapor or superheated vapor but it is somewhere on that line and now we have to determine the final state of the system and of course it is not asked but we have to we can determine even the amount of work done because once we start this we know what is or what all can be done we know initial state fully we do not know the final state except that at the final state the pressure is going to be 6 part so one part of the final state is known so we have to determine some other property now instead of some other property here an interaction is provided Q is 1000 kilo joule per kilogram so the other property remember heat is related, heat interaction is related to work done and change in energy that is the only relation between heat interaction and change in a property so to make use of that information we will have to use first law so we start with first delta U we will keep as delta U but W expansion it is a constant pressure process so integral PDV assuming it to be quasi-static and since it is constant pressure process a quasi-static assumption is in order if it is not in order we cannot proceed so it becomes P delta V and since it is a constant pressure process you can write this as delta PV because P is constant I can take it inside delta and then I can write it as again because of this and this is delta H which why definition of H. So notice what we noticed yesterday yesterday we said that in a constant pressure process Q equals delta H only when the work done is only expansion work no other work and delta E is delta U we are finding it here if you are knowledgeable experience you could straight away say Q equals delta H because delta E must be delta U and there is no other work other than expansion work but only when you understand. So now Q is provided Q is 1500 kilo joule sorry kilo joule per kilogram multiplied by M this is delta H which we write as mass into H2 minus H1 okay mass cancels out this and this so we have H2 equals H1 plus 1500 kilo joule per kg at this time I will leave the rest to you. So state 2 now gets defined in terms of P2 and H2 then determine whether it is wet steam I think it turns out to be wet steam in this case determine dryness fraction if it is wet steam determine other properties if you want to determine work done P into delta V will be the work done okay. So with this we conclude this problem the calculation I left due to you. Now let us look at the 3.12 now in 3.12 we are told that some saturated liquid at a given pressure is mixed with 1 kg of superheated steam at 12 bar same pressure notice and 300 degree C the mixing is adiabatic and at constant pressure notice it is very clearly given that it is adiabatic and constant pressure you should not object to it because this is possible determine the changes in volume and internal energy the dryness fraction in the final state and work done exercise 3.12 we have 2 kg saturated liquid at 12 bar mixed with 1 kg of superheated steam at what temperature 300 degree C 12 bar you have a process adiabatic plus isobaric we have no difficulty with isobaric because the initial pressure of both components is 12 bar so the initial pressure of the system you can also call is 12 bar so that means if this is the state 1A and this is the state 1B the state 2 will be such that P2 will be 12 bar that conclusion we already have come to. Now how do you sketch the system diagram and the process diagram this is where some explanation is ordered to students let us say that we have a cylinder piston arrangement and we have a thin partition here separating it into 2 parts this is our piston and the whole thing is adiabatic for the whole system Q is 0 let us say this is the part 1A and this is the part 1B initially nothing happens this is a thin insulated partition just to keep the 2 separate it can be thin because pressures on either side are the same but it better be a thermal resistor not a conductor so that 2 kg saturated liquid at 12 bar the temperature at 12 bar is 188.0 C whereas at other side it is 300 degree C so let us assume that there is such a partition which keeps them separate and of course there is a piston to take care of any changes in volume during mixing and we say that the process is begun by removing this partition I can do it like this oh no this should not be and now instead of 1A and 1B finally it ends up at 2 now how do you depict the process on a PV diagram notice that I am showing P specific volume V so that the masses do not matter and since the action takes place at 12 bar we have 1A initially here saturated liquid at 12 bar and if this is the dry saturated vapor line 300 degree C will be here 1B because we know that this temperature is how much 188 such as so if you want you can show an isotherm going through this that will be 188 degree C so this part between 1A and this point will be at 188 degree C the state 2 will lie somewhere here I am showing it approximately in the wet zone but it could be anywhere in the wet zone or even in the superheated zone depending on the relative masses of 1A and 1B here 1A as a large mass 1B as a small mass so maybe it will be wet you agree that if instead of 2 kg and 1 kg if it were 2 kg and 0.1 kg we would have said almost wet where it will 2 kg and 200 kg we would have said well it is most likely superheated but it is 2 and 1 so we do not know but let us assume it to be 2 just for now it is possible that the process takes place like this but that would indicate that 1B comes to 2 1A comes to 2 and then they sort of come in equilibrium but the process does not really take place like that when the partition is removed the pressure is the same so there will be no immediate mixing but the temperatures are different so there will be some amount of heat transfer taking place maybe because of the resulting thing the volumes will change what we will assume is that throughout the process the pressure remains at 12 bar isobarie and we will assume that the process is quasi-static this is very slippery this is an assumption and in this particular case it is a significant assumption because although we can say that the pressure is being maintained constant quasi-static means that there are not going to be sudden and jerky jumps in volume that is what it means so although these two arrows are not a real depiction something like that will happen changes in volume would be such that the piston from here will go to the final volume in a reasonably smooth way so that at any time we photograph it we see no significant velocity no significant acceleration now how do you relate state 2 to state 1 one specification is isobarie another specification is adiabatic adiabatic means heat transfer is 0 take it as heat transfer specified to be 0 so just as in the earlier problem we wrote q equals delta e plus w here also we wrote write q equals delta e plus w then we say because it is adiabatic q is 0 we make an assumption that no other component other than delta u and we make an assumption that there will be volume changes so there will be pressure changes sorry work done by expansion but we will assume w to be w expansion okay and again we will say that since w expansion takes place at constant pressure this will be represented by p into v2 minus v1 and by our earlier thing this will finally come to delta h the same step as in the previous problem so the whole thing finally reduces to delta h equals 0 and this we expand as equal to h2 minus h1 and write it as h2 equal to m2 h2 equal to h1 which is m1a h1a plus m1b h1b where 1a and 1b are the two parts so m1a is 2 kg m1b is 1 kg and obviously we always have conservation of mass so along with this comes m2 is m1a plus m1b now at this stage we know m2 because m1a and m1b are known we know everything about states 1a and 1b so we know h1a h1b so we know h2 because m2 has already been calculated that means state 2 is available to us in terms of p2 and h2 and that means we know everything about state 2 the moment we know everything about state 2 we can determine v2 minus v1 work done change in internal energy change in enthalpy has already been given as you know but anything else that you want to calculate you can calculate remember that what you have to do here is that major assumption that it is a quasi static process and the quasi static process here means since pressure is already given we do not expect during a quasi static mixing process jerky or sudden changes in volume so the expansion work can be written down as p delta v if there are certain changes in volume that may not be true then it is a non quasi static we cannot proceed with solving the problem yes sir if the mixing is carried out in the rigid container then the mixing is carried out in rigid container instant of the piston cylinder arrangement then there is a no question of work done expansion work etc. Yes we have a problem like that so in this case it is not directly written that mixing is carried out in piston cylinder arrangement I can take into assumption that it should be carried out in the rigid container. When you carry it out in the rigid container you cannot expect it to be a constant pressure pressure is going to remain constant sir how in a rigid container what will you do to maintain the pressure constant remember that it is given to be adiabatic so you cannot say that I am hitting it to maintain the pressure constant but how will it evaporate to maintain pressure constant you are giving a rigid container so you are saying no change in volume okay wait wait wait let me come to this if you are saying a rigid container that means you are excluding expansion work you are excluding other type of work also so there is no work interaction you are saying it is adiabatic so there is no heat interaction so you are saying there is no going to be there is no change in energy so there is going to be no change in energy there is going to be no change in volume based on this your final pressure will be dictated the problem will reduce to v2 is v1 because of rigid container and u2 is u1 because adiabatic and rigid container means no expansion work other work anyway is assumed to be 0 and that is exactly the problem 3.15 slightly different but you can have a sub problem of 3.15 a rigid metallic container it separated into two equal parts by volume by a thin partition one part contains 1 kg of saturated liquid at 100 degree C the other part is evacuated for simplicity the other part is m equals 0 a very special case the partition is broken and equilibrium reestablished after some time now here is the difference during the process the temperature of the system is maintained by immersing it in an oil bath maintained at 100 degree C like that critical pressure problem which we first looked at 3.11 but I can have a different problem saying during the process the temperature during the process the rigid container is adiabatic so here because it is in a oil bath maintained at 100 degree C I know the final temperature but if I say adiabatic I would not even know the final temperature determine what the state is we can do that I think there is a problem of yes 3.17 is exactly you need trial and error previous problem is which number 3.14 the one which we are solving we have solved just now okay no how do you decide that for a quasi static process entropy change is 0 that means I can execute any process quasi statically and all quasi static processes will become isentropic processes that is not true no we have not even defined what is a reversible quasi static only means that I can trace it nicely through the state space when we come to reversible we will realize that for a process to be reversible it definitely has to be quasi static but the other way is not true for example a process later on we will see a process of stirring so long as you stir it very slowly it is a quasi static process but it is nowhere near a reversible process it is just not reversible. First of all this is a quasi static in the book it is later it is mentioned as a reversible. Then there is something wrong there do not ever assume that everything in a book is true challenge that as one of my professors Professor Achyutan says that do not accept any statement without challenging it including this statement of mine okay so we meet after lunch.