 any questions from you, over to you. Sir, good afternoon to you. Myself Parag Mule, attending your course from KJ Somaya. Sir, in problem number three there are two unknowns which are coming, one is V2 and another is T2. So, if we can use the simultaneous equation or the deriving one expression as Q is equal to mcp delta t plus p delta v that is the first law, in that there are two unknowns, one is T2 and another is V2 and for then establishing a relation between T2 and V2, if we substitute either of the variable and then some. Yes, yes it is not necessary that you get the answers one after another. And sir, question number six, in that we have used an approach of the first law it is Q is equal to delta i plus w, you can take it either as dQ is equal to du plus pdv or you can take it as dQ is equal to dh plus pdv and solve for proving that pv raise to k equal to constant that. I think there is some confusion in what you said, you can say dQ equals du plus pdv or dQ equals dh plus pdv. What is this equation dQ equals dh plus pdv? There is no such equation like this. Sir, we are taking it into the differential form. Yes, but in a differential form dQ will always be equal to du minus pdv. Yes, there is no such thing as small dQ, it is dQ is dQ is du minus pdv. Yes, plus pdv is small dv is small v. Okay, but then how? So, with that if we proceed by taking the values we are arriving at an expression p raise to k, v raise to k minus 1 is equal to constant, which is not matching with what is required. We are asked pv raise to k equal to constant and we are getting p raise to k, v raise to k minus 1 equal to constant. Is it acceptable, sir, if we arrive at that expression? See, pv raise to k is constant is the form we want. Now, if you instead of this form get say p raise to m, v raise to m minus 1 is constant, then just do the algebra and convert it into the form pv raise to k is constant because I know the answer when it is obtained for k in this form. Over to you, Truba. Hello, sir, my question is that as per you said, specific heat is unfortunately not related with the heat. It is an intrinsic property, but specific heat is defined as the heat required to raise the unit mass of an object by 1 degree centigrade. So, there is a relationship of heat with a specific heat. So, can we elaborate peace? Over to you, sir. That is what I said today morning that historically that is the way it was defined, but what were our definitions? Our definitions and henceforth in this course the definition of cv will be partial derivative of u with respect to t at constant volume. The definition of cp will be partial derivative of h with respect to temperature at constant pressure. So, even in our derivation, we have not brought in the term heat. This definition is not consistent directly with our school bookish definition that the heat required to raise the temperature of a unit mass of something by 1 degree C or by a unit temperature. That is an incomplete definition because that does not say what are the other conditions needed. The other conditions needed if you really look at it would be for cv maintaining at constant volume and c to it that no work interaction is involved. And for cp, I think the requirements would be the process should be at constant pressure and no work interaction except the expansion work is involved. All these details are not written down in school textbooks. But for some reason we continue with those definitions, but forget those definitions now as good teachers of mechanical engineering thermodynamics. Our definitions would be cv is partial of u with respect to t at constant v and cp is partial of h with respect to t at constant p that is it over. And I hope all of you are in the process of solving these four exercises. I will take up the questions now. I have received a question from Calicut through chat. The question was or they wanted more discussion on w cube as energy in transit. See, I have already given the rain analogy that is rain is water in transit. We do not say that when rain fills up a lake, we do not say that the lake contains so many meter cube of rain or so many liters of rain. We say it contains so many meter cube of water. So, similarly the clouds do not contain rain. They contain so many kg of water either in vapor form or droplet form. So, just there is something in transit. So, r w and q are energies in transit. So, the similarity is to the word rain or to the word flow or to the word even to some extent traffic. So, when you say flow, you have water in one tank and you have water in another tank. So, when you connect the two tanks, there is a flow of water you get a stream of water. So, this stream or flow is something similar to our q and w. The water here is like energy. So, the water here goes down equivalent to delta E being negative. Water amount of water here goes up equivalent to delta E being positive. But you do not say there is so much flow of water here is there is so much stream of water here. Water here is just bulk water. Water here is also bulk water. Only when it is flowing you say you have a flow of water or you have a stream of water. C O E P over to you. I am asking question on behalf of one of the participants. The question is like this for constant volume process q equal to m C v delta T and q equal for constant pressure process q is equal to m C p delta T. Will you please explain again the concept of C p C v and how they are related to heat transfer q? I think the whole thing starts with the final form of the first law. The whole thing starts with the final form of first law q equals delta E plus w. And at that time I made a statement may be many of us were not attentive or maybe because it came at the end of I think day one it was not noticed. I said that q is related to the rest of the world only by this relation and by nothing else. So, that means any other relation which you just mentioned our I call them school book is relation that q equals m C v delta T for a constant volume process and q equals m C p delta T the constant pressure process that must also come out of this. And again at someone asked before the tea break you know we have defined in school that C v and C p are the amount of heat required to raise the temperature of 1 kg of some system by a unit temperature either at constant volume or constant pressure. But we have defined C v as partial of u with respect to T at constant volume definition. And similarly C p is partial of h with respect to T at constant pressure along with this there is a hidden definition h is nothing but u plus p v. Now, let us analyze this when I say that this is the relation between q equals q and delta E and W let us look at a constant volume process. So, let me continue let us consider a constant v process it is this equation which is applicable q equals delta E plus W. Now constant v process a school kid will assume W is p dv but no W is p dv plus W other constant v process only means that if I expand W as W expansion plus W other which could be a stirrer electric or anything this will be 0 because it is constant volume. But the other part will not be 0 or may not be 0 similarly we have to expand delta E as delta u plus delta E other this could be changes in kinetic energy potential energy or what have you plus W other. And then if you have a fluid system I will recreate the page quickly I said that q must be equal to delta E plus W and then we said that we have a constant volume process. So, adding this basic equation final form of first law but which essentially is the definition of q and only relationship between q and the rest of the world will be this will be delta u plus delta E other plus W expansion plus W other. Since it is a constant volume process that means W expansion is 0 because if the volume cannot change then the there is no p dv type of work. And now if we have a fluid system and no change of phase then delta u is m integral c v d t which becomes m c v delta t if c v is constant. And then finally, with that you end up with q equals m c v delta t plus delta E other plus W other. So, if we want our school book is relation that in a constant volume process q should be m c v delta t look at the conditions which have to be satisfied delta E other has to be 0 W other has to be 0 apart from it being a constant volume process. It has to be a fluid system where internal energy is a function of temperature and volume there has to be no change of phase instead of a fluid system if it is a solid system then also it will work, but again condition no change of phase. So, that our definition that c v is partial of u with respect to t at constant v is applicable. And in a similar way now I am told the tablet is operational. So, let me try that otherwise I will quickly come back here. Now, let us in a similar fashion analyze the constant temperature process again we start with q equals delta E plus W and then expand this as earlier to delta u plus delta E other plus W expansion plus W other. And we say it is a constant p process hence W expansion will be integral p dv because it is constant p it will be p delta v and again because it is constant p that will be delta of p v. Now, combine this into this and you will get q equals delta u I am transposing some terms or no let me not do that. So, that things are clear plus delta p v plus W other and then I will combine delta u plus delta p v those two will become delta h plus we have delta E other plus W other. And then if you have a fluid or a fluid like system and again if you have no change of phase then you will have delta h equal to m c p delta t that is because under those conditions we can use the definition that c p is partial of delta h sorry partial of h with respect to t at constant p. And I am writing this that means I am also assuming c p is constant otherwise we will have to write it as an integral. But remember that there are two other terms delta E other plus W other. So, as in the case of the earlier case where you had the constant volume process our school bookish expression that q equals m c p delta t is valid only when minor assumption c p is constant. But major assumptions delta E other other than u is 0 and W other that means other than p delta v p d v expansion work is also 0. I hope that explains the issue. Raisoni-Nagpur any questions over to you. Sir, in exercise number 1.4 in exercise 1.4 yes there is a slimmer work and net work. So, what is the difference between the question pertains to f 1.4 and the question is there is a mention of sterrer work and there is a mention of net work. Net work is our W term which comes out of first law. So, if you write the first law and I recommend that you always write the first law as q equals delta E plus W or delta E equals q minus W that means it is transpose form. Do not write a shortened version like q equals delta u plus W or delta u plus p delta v or something like that let it come out of the formulation. Now, here we have a perfectly insulated system that means q is 0 perfectly insulated adiabatic system. It contains 0.1 meter cube of hydrogen that means some volume of a gas at 30 degree C 5 bar. It is stirred at constant pressure till the temperature resets some other value and you have to determine heat transfer change in internal energies sterrer work and net work. Now, the W because it is a constant pressure process and hence there will be expansion because the pressure remains constant. So, if the temperature changes volume has to change and that means there will be an expansion work and since there is a sterrer work, sterrer mention there will be a sterrer work. So, this W is net work and this W s t is sterrer work. Does that answer your question? Hello sir, yes sir in the same question if the sterrer work is supplied then there is heat transfer, heat addition will be there? It is specified in exercise F 1.4 that it is a perfectly insulated system. So, there is no question of any heat transfer there. Our equation for exercise 1.4 this equation reduces to 0 on the left hand side that you should never change delta u plus delta e other plus W expansion plus W sterrer. Now, there is no mention of any change in energy other than the internal energy. So, we have assumed or we will have to assume this to be 0. That means you are assuming that the system containing the hydrogen remains at one place. It does not change in velocity nor does it go up and down. And as in the previous illustration because it is a constant pressure process, this will turn out to be first integral p dv then because it is a constant pressure process that becomes p delta v and because it is a constant pressure process and initial pressure and final pressure this will become delta pv under this assumption. And delta u plus delta pv will then be combined and you will get delta h plus W sterrer with 0 delta h will then be m c p delta t plus W sterrer. So, here you see we have a constant situation where delta t is positive. So, delta h is positive m c p delta t is positive, but there is no heat transfer. The first law is balanced because this is a positive number and the sterrer work is a negative number. In fact, it is this equation which will be used to determine the sterrer work which is required in the solution. Sir, in the morning session we have discussed about Kelvin scale as well as ideal gas scales where under discussion of Kelvin scale of temperature we have taken 273.16 Kelvin as triple point of water. So, will you give some justification for this point sir? Remember the definition of the Kelvin scale is T by T ref is pv by pv at the reference state. It has been found that the reference state an excellent reference state is the triple point of water not very difficult to create all that you need with some amount of absolutely pure water and an excellent skilled glass blower and a very good vacuum pump. That is all that you need and you can create a triple point of water cell. So, creating this reference state is not a problem at all that is why it is taken. Now, the question is which T ref and the way the Kelvin scale was attempted to be designed whenever we design a new thing we expected to be compatible and reasonably nicely related to some old established thing something which is already known and already in use. So, we wanted a determination or a selection of T ref in such a way that the Kelvin scale and the Celsius scale get aligned to each other very well. Now, a simple relation that there are two possible simple relation either one should be fixed factor of the other. For example, Kelvin could have been 3.75 into Celsius that would also be easily acceptable. Another thing Kelvin could be Celsius plus something or Celsius minus something. It turns out that if you select T ref of a particular value then you can obtain T ice point and T steam point. T ice point will turn out to be this P v at ice point divided by P v ref into T ref. This will turn out to be P v steam point divided by P v ref into T ref. We have not yet decide what T ref is, but even if we do not decide now we can get on the Kelvin scale T steam point minus T ice point to be T ref into P v at steam point minus P v at ice point divided by P v ref. This is experimentally determined and then you notice that the value of T ref will determine what would be the Kelvin temperature difference between steam point and ice point. At this stage it was realized that the temperature difference on the Celsius scale between steam point and ice point is 100 degree C. So, why not keep the temperature difference on the Kelvin scale between steam point and ice point also to be 100 Kelvin. That way the temperature differences would be aligned on the Celsius scale as well as on the Kelvin scale. Hence, the team of scientists which developed this said let us have this 100 and let us calculate what T ref is and that T ref turned out to be so near 1 6 Kelvin. Sir, this is regarding the basic question in the first class regarding the properties of the system can be classified in the two groups. As you said it is extensive property and extensive property in which case neither extensive property nor extensive property can you elaborate further by means of examples. The question is we said that a property can be classified as extensive or intensive and our definition was that if we have a system with a property phi and we partition it in two parts and if the two parts end up with the same values of properties of say phi a and phi b then in case phi is an extensive property then phi should turn out to be phi a plus phi b and for an intensive property phi should turn out to be phi a equal to phi b. That means this just depends on the intent that is what the system is whereas in this case it depends on the extent you make a bigger system phi will be bigger. I had asked a question I think on moodle or somewhere that I suppose we have enough illustrations of intensive properties. The two inherently intensive properties are temperature and pressure extensive properties are mass, volume, energy, enthalpy, entropy, etcetera. But if you divide any one of these and create specific properties then for example density specific volume, specific energy, specific enthalpy, specific entropy are all intensive properties. But remember properties are something which we can create. So for example if you create just for the sake of it the square of mass, mass is a property so the square of mass is also a property but you will notice that the square of mass will neither be an extensive property nor an intensive property. Because if you take a system with a mass of 10 k g square of 10 k g is 100 partition into 2 of say 6 k g and 4 k g 36 plus 16 that is 6 square plus 4 square is not 100. So this is a situation where property is neither intensive nor extensive. If you think mass square or square of mass is an esoteric thing just consider the property surface area tell me whether it is intensive or extensive. It is not intensive because if I take a part of the system naturally the area will be different. So the area of a part of a system is not the area of the system. Is it extensive? You will notice that the area of the whole system is the external surface but the area of a subsystem is that part of external surface plus some inner surface. So consequently surface area is a property which is neither extensive nor intensive. So that is another illustration of a property which is neither intensive nor extensive. But such properties are rare most of our properties will be either intensive or extensive over to you. Let us not discuss about the first test anymore. Tomorrow afternoon I am going to have another test and hope things will improve and remember these tests are tests as much as for you as much as for us. So whether somebody has scored 0 marks or 32 out of 32 marks that just does not matter. We are not going to give any prices or any certificates based on the scores in these tests. These tests are feedback for you, feedback for us. Do not look at the marks. If you know and if you are confident that you have answered the questions correctly be happy. Do not look at that 0 marks. See the vibration of the interface means a movement of the interface. So if you are really looking at vibration that interaction will be something like a PDV interaction or an FDX interaction depending on the actual detail of the surface movement. I have not really thought over this in any detail because a vibratory mode is usually not we do not generally come across in thermal systems. But now that you have mentioned it wheels will start churning and maybe I will set up some exercise problems or test problems based on that. But give me some time but I assure you and others do not get scared. I am not going to set up any vibration based test or exercise problems during this course. But the idea is nice but the only of the curve I can tell you is a vibration would mean movement of a surface and that would mean that it would be something akin to a PDV work or an FDX work depending on the local detail of the movement. Somaya, over to you. Hello sir. Go ahead. Go ahead. How do you solve problem 1.6? Somaya wants to look at the detail solution of 1.6. Let us come to that. Before that let me look at Thruba. Hello sir. That sir, question number F 1.6. Sir, can you give me some guideline regarding this question? How to solve this question? Yeah, your problem is the same thing as Somaya. Let me come to 1.6 in a few minutes. Over and out. Question is that, can I know answer of F 1.4 work done by a steward and network? I do not have the numerical answers but I think I discussed sometime ago F 1.4. So, for F 1.4 the formulation is here on page, I think they are seeing this slide. So, here is the complete formulation for 1.4 but of course, if you write it like this I would not give you many marks because I have not sketched here the system diagram, the process diagram. But finally, after doing all that and writing all this thing you reach here and if you notice in 1.4 you have the volume of hydrogen is given, its initial temperature is given, its initial pressure is given, the molecular weight of hydrogen is given and the gamma for hydrogen is given. Maybe what I should do is I will just use another sheet and tell you the detail of the process. I will do some detailing of 1.4 and then we will go to 1.6. We have a perfectly insulated system initially 0.1 meter cube of hydrogen 30 degree C 5 bar, it is expanding at constant pressure. So, the there is a stirrer work. So, let me say this is a piston, there is going to be a W expansion, there is going to be a W stirrer because there is a stirrer, our system is this. Let me write m k g hydrogen and it is a perfectly insulated system. So, Q is 0 given, the data we have is initial state 1, volume 1 is 0.1 meter cube, T 1 is 30 degree C converted to Kelvin as needed, P 1 is 5 bar, state 2 it is stirred at constant pressure till the temperature reaches 60 degree C. So, P 2 is P 1 is 5 bar, T 2 is 60 degree C. The other data given for hydrogen is molecular weight of 2 k g per k mole and gamma which is C p by C v is 1.4. Now, first you calculate R as universal gas constant divided by molecular weight, substitute the values and get the value of R. Then remember that we have C p minus C v equals R and it is a constant pressure process. So, in all probability we will need C p as we go. So, you write this as C p into 1 minus C v by C p into R. Now, this C v by C p is reciprocal of gamma. So, you know this ratio, you know R. So, this gives you sorry this is equal to R. So, this gives you C p. So, now, C p is known. Now, come to our formulation and that formulation began with Q equals delta E plus W and we ended up with I am not doing it again it is in the previous slide. I will just go back here it is in slide 39 we ended up with M C p delta t plus W stirrer. So, from here we came to 0 equals M C p T 2 minus T 1 plus W stirrer. Out of this T 2 is known, T 1 is known 60 degree 30 degree C, C p has been calculated and M. M calculated using this data and the value of R. Remember the equation of state for the first initial state is P 1 V 1 is M R T 1. Do not make the mistake of substituting T 1 in Celsius convert 30 degree C into Kelvin. You have T 1 R has been calculated earlier P 1 is given V 1 is given. So, you can determine M the moment you determine M in this equation the only unknown is W stirrer that is how you calculated. Now, change in internal energy is delta U delta U is M C v delta T and in the right hand side we have determined M we have determined C v once you get C p C p minus R will be C v and delta T is already known T 2 minus T 1. So, you can determine delta U and the network will be W stirrer plus W p dv or you can calculate the network from our first law because we already know delta E is delta U. So, from here go to a part of this was delta U plus W total this is W expansion plus W stirrer W expansion was absorbed here. So, use this delta U has been calculated. So, you know what W is I think that should answer your question good question question is what type of property is thermal conductivity. First thing is remember that thermal conductivity is not a thermodynamic property because remember that our properties are characteristic of a thermodynamic system which is in equilibrium and thermal conductivity is not the property of a thermodynamic system in equilibrium thermal conductivity is the property which is of a system which is not in equilibrium in which there are some temperature differences. So, it is properly known as a transport property and not a basic thermodynamic property after having said that if you assume that during the process of conduction the departure or disturbance from equilibrium is not very significant. Then with some minor assumptions one is able to show that the thermal conductivity is something akin to a an intensive property. I think that should satisfy you over. I want to know the term absolute temperature and why we are adding only 273 in degree Celsius to get the absolute value over to use. See the term absolute temperature is a misnomer it is only of historical interest. Now we have we have the ideal gas temperature on the Kelvin scale and later on we will have the thermodynamic temperature again on the Kelvin scale. This we will define later just now we have defined this and we will stick to that earlier when they started looking at gases there was some idea that there is some temperature which directly can be substituted in the ideal gas equation of state. For some reason it was known as the absolute temperature, but we do not we do not use this term absolute anymore about your second part of the question why 273.16 that I think I have answered a few minutes ago when I explained how that came about let me go back a few slides and see whether I can show it to you. It is here Kelvin scale of ideal gas temperature that T reference here was determined to be 273.16 from the consideration that the temperature on the Kelvin scale the difference of temperature on the Kelvin scale between the steam point and the ice point should turn out to be 100 units on the Kelvin scale and that was decided to keep it aligned with the Celsius scale. And when that happens your T ref turns out to be 273.15 and the relation between Kelvin and Celsius turns out to be a simple relation that temperature on the Kelvin scale is temperature on the Celsius scale plus 273.15 that difference of 0.01 turns out to be because the ice point is at 0 Celsius 0 degree Celsius whereas the triple point is 0.01 degree Celsius over there is a technical question from S G S I T S S G S I T S wanted to know why should the process in exercise 1.4 stop when the system reaches 60 degrees C. The there are two answers to this first thing is it may stop because we have stopped stirring and stirring and we have frozen the movement of the piston or you can say that the process continues but we have computed it only up to 60 degrees C whichever way you look at it that does not change the technicality of the problem. So, we have a system containing an ideal gas with constant specific heats. Let us say M K G ideal gas we are given C P C V are constants there is a W expansion and that is W only expansion work is done and the heat transferred is represented in such a way that D Q is C D T or D Q is M C D T where C is a specified constant show that the equation for this process can be represented as P V raise to K is constant and determine K in terms of C P C V and C here remember it is an ideal gas and C P C V's are given. So, the equations which we have is the first law the equation of state and the equation relating D U to D T through C V or D H to D T through C P. So, let us write first law first law for differential process will be D Q is D E plus D W let us assume although it is not said. So, we cannot proceed let us assume that D E is D W that means there are no changes in any other component of energy other than the thermal or internal energy. So, that gives us this to be D E plus D U plus D W and since only expansion work is done that gives me this D U plus P D V D Q is specified to be M C D T I will divide throughout by M and I will end up with C D T is D U plus P D V and since it is a gas ideal gas with constant specific heats this turns out to be C V D T plus P D V this is one equation, but here we have a relationship between D T and D V there is a P there, but finally we want the relationship between P and V. So, we will have to use the equation of state the equation of state is P V equals R T. So, for any small process we are going to have P D V plus V D P plus sorry equal to R D T. Now, let me call this earlier equation let me call this equation 1 the equation on this page as equation 2 I will just tell you that eliminate between 1 and 2 and there is no T existing anywhere everything is D T if T were to occur you will eliminate T writing in terms of P and V. So, if you eliminate D T between 1 and 2 you will get a relation only between D V and D P integrate that and you should get that relation as far as I know and I remember K turns out to be C P minus C divided by C V minus C, but I would like you to check this and again there are special cases here. You will notice that if C happens to be 0 then C happens to be 0 means no heat is transferred an adiabatic process and here then we have an adiabatic process in which only expansion work is done quasi static adiabatic process of an ideal gas constant specific heat only expansion work is done and D E equals D U you will end up with K being equal to C P minus C V which is by definition gamma. So, the relationship turns to be the classical P V raise to gamma is constant for an adiabatic quasi static process of an ideal gas then you put C equal to C V then this K turns out to be very large that means the process is now P V raise to infinity is constant or that essentially means P is constant sorry V is constant because you can write it as P raise to 1 by K into V is constant and 1 by K becomes 0. So, you have a constant volume process whenever you have C equals 0 similarly for a constant pressure process you will notice that C has to be equal to C P if you make C equal to C P then K turns out to be 0 and the process turns out to be P equals constant and I will leave it to you to answer the question what happens if K turns out to be 1 when will K turn out to be 1 K will turn out to be 1 when the value of C is very large and value of C being very large means hardly any temperature difference d t being pretty small, but C is large. So, enough amount of heat is being added without any change in temperature. So, that means you are ending up with an isothermal process and check that for a very large C K equals 1 and for an isothermal process of an ideal gas P V equals constant is the result. This is regarding problem number 1 point it is about a closed system undergoing an isothermal process for which Q minus W will be 0 work which is equal to the sum of stator work and the expansion work. When you calculate the heat transfer whether do we have to add the expansion work and the stator work or stator work is to be deducted from that as the stator work is the work done on the system. Look at 1.5 we have a closed system 2 kg of air at 3 bar some temperature stirred and expands till pressure reduces the to 1 bar. So, you have expansion and you have stirring sketch of the system 2 kg air W expansion W stirrer let there be some Q there is nothing mentioned about Q we have to determine that. Then initial state 3 bar 150 degree C final state 1 bar final temperature 150 degree C process stirrer plus expansion isothermal that means by some trick we have maintained the temperature uniform constant. I think I have noted down the stirrer does 120 kilo joules of work that means W stirrer equals minus 120 kilo joule the way it is work done on the system not work done by the system. We have to determine expansion work done and heat transfer. The tools with us are first law we will assume delta E equals delta U, but then we will know this will be 0 why because T 1 equals T 2 and we have an ideal gas. So, this now becomes 0 plus W which is W expansion plus W stirrer. Now, on the right hand side W expansion is simply m integral p d V from the initial state to final state or p d capital V from the initial state to final state plus W stirrer. W stirrer is given expansion work you can simply compute out that is one of the answer and you sum the two up and you will get the second answer Q is that clear. Sir there the stirrer work is having a negative side. Yes stirrer work see our convention is work done by the system is positive. So, when you insert a stirrer in the system and stir it will only be negative work because work will be done on the system. So, that is why we write minus 120 kilo joules. I hope that was the only confusion. Thank you.