 I am going to Hyderabad now 1032 Mufakkam Jha College Hyderabad over to you. I have two questions so you need to answer both the questions graphically as well as experimentally sir my first question is for an ideal gas dou H by dou P at constant temperature into dou P is equal to 0 and my second question is what is the difference between partial differential equation and exact differential equations you need to answer both the questions graphically as well as experimentally I don't understand what do you mean by graphically as well as experimentally I can't do an experiment here and explaining what is the first question pertaining to ideal gas the first question is dou H by dou P at constant temperature into dp is equal to 0 for an ideal gas no but how can something into dt be equal to 0 and partial of what with respect to partial of change in enthalpy with respect to change in pressure at a constant temperature into dp is equal to your one equation you have draw it is already for an ideal gas you want to show that partial of enthalpy with respect to pressure at constant temperature is 0 okay that can be derived in fact we can derive an expression for partial of enthalpy with respect to pressure at constant temperature I am not sure whether that is that is a derivation here but that's one of those standard relations which we have and then substitute the PVT relationship of an ideal gas in that and you will be able to show that it is 0 what's the problem sir how about second question sir partial differential equation and exact differential equation okay this is actually a mathematics problem but as I understand it a partial differential equation turns up when you have the function to be solved for is a function of two variables for example in heat transfer say conduction in a two-dimensional situation the temperature could be a function of location in the x direction as well as location in the y direction so temperature first is a function of x and y and the differential equation contains the derivatives of T with respect to x as well as the derivative of T with respect to y if the order of differentiation is more than one they may even include partial derivative for example our conduction equation says that if the conductivity is uniform and if you have a steady state then the temperature is governed by the differential equation second partial derivative of T with respect to x plus second partial derivative of T with respect to y is 0 such an equation where the solution is expected to be a function of more than one variable and the in the differential equation derivatives with respect to more than one independent variables occur is known as a partial differential equation now an exact differential equation is a term used for an ordinary differential equation so suppose you have a differential equation which finally can be written down as say if it's a first-order ordinary differential equation suppose it is to be solved for y equal to as y as a function of x then if you can write it down as simply d by dx of some function of some non-differential function of y and x to be equal to 0 then it turns out to be a exact differential equation if you want a mathematically neat and proper and complete definition you will have to look up a book on calculus I think you will have something about this in either Thomas and Finney or Kresig these are the two standard books in engineering mathematics which we use here and these are the two very popular books on engineering Solapur 1131 Baichand Institute of Technology over to you how to solve problem number 5 in which part SL 0.5 SL 0.5 SL second law okay that is about the insulated cylinder insulated chamber of volume 2 V0 we have an insulated chamber thin separating partition separating it into two parts one volume V0 another volume V0 one chamber contains an ideal gas at a pressure P0 and temperature T0 the other chamber is evacuated nothing vacuum it's insulated so the whole thing is adiabatic the partition is suddenly removed show that when equilibrium is reestablished the temperature is T0 determine the change in entropy which is reversible process has taken place let us select our system as the gas in the non-evacuated partition let that be our system the process is the partition is removed and the gas expands into the free space the final volume of the gas will be 2 V0 so let's see what we can sketch on the process diagram initially pressure is P0 volume is V0 this is the initial state let me call it state 1 final state I only know at that at this time that the final volume will be 2 V0 so state 1 is V0 P0 P0 final state 2 let me call it 2 V0 some P2 some T2 now with the reading we can't do anything more than this now let's start thinking we have to apply first law since it is given to be insulated at least one term in the first law is known to us Q equals delta E plus W which is first law and Q equals 0 because it is insulated now let's see what is about W there is no mention of a stirrer there is no mention of an electric connection so W if at all will be the expansion work but is the gas expanding the gas is expanding freely is there another surrounding system or is there a piston against which it is doing work no it is a free expansion so the gas cannot do any work although its volume increases from V0 to 2 V0 and that means W equals W expansion plus W other W other is 0 because there is no hint otherwise and W expansion also is 0 since it is free expansion remember that for work to be done there must be a donor system and a recipient system here there is no other system involved it's just expanding freely into vacuum hence it's free expansion no expansion work is done as a consequence our W is 0 and as a consequence our delta E is 0 and delta E can be written down as delta U plus delta E other and since it's a gas and since it is we assume it to be stationary so we assume that delta E other is 0 so this leaves us with delta U equal to 0 and since it's an ideal gas U is a function only of T so if delta U is 0 delta T is 0 and that means final temperature is the internal temperature and that means now we can proceed with the process diagram the initial temperature will be T0 if this is the isotherm pertaining to T0 the final state will be this so the final state is T0 temperature to V0 as volume. And then use PV equals RT or P1 V1 by T1 or P0 V0 by T0 is P2 V2 by T2 temperature is the same volume doubles so the pressure must become half so that also means that final pressure is initial pressure by 2 going back the pressure in the final state will have to be P0 by 2 the next part of the problem we have shown when equilibrium is reestablished that equilibrium is reestablished means you have to wait long enough till it reaches equilibrium that means it reaches uniform values of pressure temperature of course volume soon will become to V0 but it will take some time for the pressure to be uniformly P0 by 2 and temperature to be uniformly settling again to T0. Remember that this is totally a non quasi static process so you should not show it along an isotherm you should just show it this is this dotted line is actually the isotherm the actual process we can show only as a dotted line joining 1 by 2 1 to 2 there is no meaning no significance at all in the way this line is drawn just see to it that you do not mechanically you are by default draw the isothermal line and that too as a continuous line that is not the process which is executed it is not a quasi static process and since now we know the initial state and the final state the entropy difference changes depends only on that state on the 2 states so calculate the entropy difference and since well that is what you have to determine that entropy difference will turn out to be I think P0 V0 into logarithm of or P0 V0 by T0 into logarithm of 2 I think I remember the solution this will be a positive number and the last question is which irreversible process has taken place the answer is free expansion is an irreversible process and for this free expansion to take place you do not really have to make the partition valid remove the partition suddenly even if you make a small hole in the partition that is sufficient maybe it will take more time to move from the left chamber to right chamber but eventually equilibrium will be established and the same end state will be reached over 1 2 3 RBS engineering Bichpuri Uttar Pradesh over to you if we have two fluid one is water at 100 degree centigrade and another is steam at 100 degree centigrade then what are the difference between entropy of both the fluid I am assuming you say water and 100 degree centigrade you mean saturated liquid water at 100 degree centigrade and when you say steam at 100 degree centigrade I assume you are meaning you mean the dry saturated steam at 100 degree centigrade come to your steam tables come to your steam tables this data is found either on page 4 or in page 6 I am looking at page 6 look at this 100 degree centigrade line below that I have blanked things out so the last line on this page as seen on this 100 degree centigrade saturation temperature so the system pressure will have to be 1.01325 bar come to the entropy column the SF column says entropy of the saturated liquid is 1.307 kilo joule per kilogram Kelvin that of the dry saturated vapor is in the last column 7.355 kilo joule per kilogram Kelvin this is the answer you wanted for liquid it is 1.307 for vapor it is 7.355 over to you. So there is one more question if we are having two fluids one is steam dry saturated steam and one is saturated water at the same temperature which one will have the higher internal energy what we have steam at you are comparing dry saturated steam and saturated liquid water right at the same temperature yeah at the same temperature you have the data here for example you look up the two columns under internal energy I am again taking the example of 100 degree C because I already have the data in front of me and you will notice that the saturated liquid water has a internal energy of 418.9 kilo joule per kg and dry saturated vapor has an internal energy of 2506.5 kilo joule per kg. Yes go on. There is another question sir. Yes madam. Sir I want to know sir there is a any practical example of conversion of vapour into solid. Practical example of conversion of vapor into solid difficult to find out because you will have to really go to really cold places but since dry ice at atmospheric pressure evaporates or sublimates directly into its vapor that means if you take carbon dioxide and start cooling it at atmospheric pressure you will find that at sufficiently low temperature it will simply start getting flakes of solid which will start falling down rather than droplets of liquids which will start falling down or sticking to the surface of it. If you want you can do that experiment it is not very uncommon. There is a problem in sir numerical SL12, which exercise SL12 you have water in a cylinder piston assembly at shown in the figure there are two stops a lower one at which the volume enclosed is 1 meter cube and the upper one at which the volume enclosed is 3 meter cube. The piston mass and the atmospheric pressure are such that the piston floats at 500 mega Pascal the above described system sorry 500 kilo Pascal half a mega Pascal the above described system has water initially at 1 mega Pascal and 500 degree C. This is allowed to cool to 100 degree C by rejecting heat to the atmosphere at 30 degree C. Calculate the total entropy generated in the process first we have to appreciate the process itself. We are told that if the pressure of the system is 500 kilo Pascal the weight of the piston and the atmospheric pressure is such that the piston will float on the fluid and that means if the pressure is higher than 500 kilo Pascal the pressure of the fluid will be large enough to start moving the piston upwards and then the piston will go to the upper stop and the volume enclosed will be 3 meter cube. If the system pressure falls below 500 kilo Pascal then the piston will start moving down and it will be stopped at the lower stops where the enclosed volume is 1 meter cube. So the idea is if the pressure is higher than 500 kilo Pascal the system will have a volume of 300 3 meter cube and any process will then be a constant volume process so long as the pressure remains above 500 kilo Pascal. At 500 kilo Pascal the piston will start floating and the volume will start reducing but the volume cannot reduce below 1 kilo Pascal. If you try to volume cannot reduce below 1 meter cube because that is the limit and if you still try to reduce the volume the volume will not reduce but the pressure will reduce below 500 kilo Pascal. So the possible process is like this. So let us say this is P V and let us say this is 500 kilo Pascal that is 0.5 mega Pascal and this is the volume 1 meter cube and this is the volume 3 meter cube. The volume of the system can never go below 1 meter cube the volume of the system can never go above 5 meter cube. The initial state is 1 mega Pascal and 500 degree C since the pressure is higher than 500 kilo Pascal the piston will be at the upper stop and the volume will be 3 meter cube. Let us say this is 1 mega Pascal. The initial thing would be initial state 1 will be 1 mega Pascal 3 meter cube and T 1 is 500 degree C. So the state will be 500 degree C, V 1 is 3 meter cube and P is 1 mega Pascal. Now with this P V T we can determine all properties from the steam tables because 500 degree C and 1 mega Pascal will tell you 1 mega Pascal is 10 bar. So 10 bar 500 degree C is definitely superheated. Look up the superheated steam tables will get specific volume so you will get mass and you will get all other properties. Now what happens is it is allowed to cool to 100 degree C by rejecting heat to atmosphere at 30 degree C. So our system which is this cylinder piston rejects heat to atmosphere which is another system like a reservoir at 30 degree C and as it reduces now trace the process. At some stage it will reach a pressure of 500 kilo Pascal hopefully the process is not over till then. So the first part of the process will be from state 1 to state A where the volume will be 3 meter cube but the pressure would reduce to 500 kilo Pascal. So 3 meter cube 500 kilo Pascal and we know the mass of the system now we can determine what its temperature will be. If the process continues it will now continue at constant pressure and the next change will be at B. At B it would have reached the lower stops the volume would be 1 meter cube and further reduction in temperature would bring you to the final state 2. So this is the estimated process you can determine assuming these 3 processes to be quasi static processes and determine the net heat transfer out of the system when it is cooled and when it is cooled the net heat which comes out of the system will be absorbed by the reservoir and you can determine the change in entropy of the system and change in entropy of the reservoir. Hence the total entropy generated in the process I hope this is understood over to you. Sir there is one more question you told a free expansion is a reversible process why it is so? Free expansion is an irreversible process because you cannot do anything by which the gas in one half after filled it will automatically go back and assume its old position of just half the chamber nothing you can do once the gas expands and fills up the whole chamber it will continue to occupy the whole chamber what can you do without now remember there is no heat interaction there is no work interaction. So to reverse this if at all it is reversible that means you should not have any heat interaction you should not have any work interaction but even then you should do something I do not know what by which the gas in the half of this chamber I think I having here gas which occupies this right half of the chamber decides not to occupy it and go back to the left hand side and none of the molecules there to come back on the right hand side again that is just not possible and that is why it is called reverse irreversible and this is demonstrated because if you calculate the change in entropy of this adiabatic system actually it is an isolated system more than adiabatic. So for this isolated system the entropy change turns out to be positive so even from the principles of thermodynamics it is a it is an irreversible process over. One more question sir is there any natural process which can be considered as absolute reversible process is there any natural process which is considered as an absolute reversible process no I do not think there is any process in nature which is an absolutely reversible process and that is why I have said in the beginning today that a reversible process is a thought process it can be thought about we can do lots of thinking about it but we cannot execute it in practice because no process in nature is a reversible process. Anything that happens in nature leaves its stamp somewhere. Sir in problem SL 5 the partition is suddenly removed why the word suddenly is not needed you can simply say the partition is removed or a hole is made in the partition the problem does not change when the partition is suddenly removed the process should not be adiabatic process why adiabatic process is what you do at the boundaries if I insulate the boundaries perfectly I can have an adiabatic process what is wrong in it. But the boundaries are fixed and not heat is removed from the boundary. You can say that I just punched a hole in the boundary small hole in the boundary it did not take much of an effort so I agree that suddenly is not a good word there the partition is removed hole is made in the partition over last visit today 1098 Marathwada Institute of Technology Aurangabad over to you sir what is the significance of Helmholtz function and Gibbs function in thermodynamics okay this Helmholtz function and Gibbs functions are derived for they are mainly used in physical chemistry we do not have much use of it in mechanical engineering unless you do either combustion calculations or detailed calculations of the equations of state and these are derived because they have to be found they are found useful as I said in that panel I have already put up all my today's morning slides on the net and on the moodle so you will notice that these are potential so something like they indicate the capacity to extract work and we have shown that the Helmholtz function is a potential the decrease in which represent the maximum work that can be extracted under isothermal conditions and Gibbs function is a potential the reduction in which represents the maximum work that you can do in an isothermal come isobaric process which typically occurs during chemical reactions apart from this the Helmholtz function has the specific pick property because for that the natural variables are T and V unlike the other variables where you have Gibbs function which has T and P if you specify Helmholtz function as a function of T and V throughout that state space then that has information from which you can extract critical point you can extract all properties of state all over the state space that is the advantage of the Helmholtz function over to you hello one more question sir yes the transitional form of energy we said only heat and work for the modern energy yes energy transfer can take place only by two modes one is the heat mode and one is the work mode electrical mode electrical transitional form of energy no those are various types of work interactions they are not different modes electrical work magnetic work surface tension work these are all work modes they are not any distinct type of interactions the interactions can be distinguished only as work type in which all these expansions stirring twisting bending electrical magnetic is included and heat type heat is defined as the non-work type of interaction that is it over to you one more question yes heat is equals to change in enthalpy enthalpy is property it is H1 minus H2 whereas Q is a path function this enthalpy is a point function your first statement that heat is the change in enthalpy itself is not right heat is an interaction the only link between that and the rest of the world is it equals delta E plus W now depending on the process it may turn out that delta E plus W is delta H it will happen I know under the following conditions it will happen if the system is a closed system the process is a quasi static isobaric process and there is no work done other than expansion work then you will have Q equals delta H but that is a derivation and finally depending on the process the interactions will be related to properties of the initial state properties of the final state and the details of the process so that is what is happening finally you will have Q equals something which will finally written in terms of properties of the initial state and final state there is no contradiction here that we even isothermal work or any other type of work when you integrate that expression integral pdv to determine the expansion work you will get in terms of the initial state and final state so what is the problem the problem is the fact is given the same initial state and final state I can have different processes so different expressions which represent work the very first exercise in your exercise cheat exactly does that I have one question yes sir as we have a specific heat at constant pressure specific heat at constant volume dp and pdv so is there any specific heat for a different process like no specific heat is defined as a change in some property like u or h with respect to temperature at some other property being kept constant so we define specific heats only two ways cv and cp there is no need or usually it is not defined it in any other way for example in principle if you want you can define du with partial of u with respect to t at constant pressure at some other type of specific heat but that is not needed and that is usually not done but you can evaluate du by dp at du by dt at constant p that can always be written down in terms of cp cv and pvt data the exercise cheat pr is all about that and you can create any number of exercises in the property relationship thank you very much I am stopping here today.