okay thank you so much take care so let me start one more time let me uh okay so what I was saying is this diagram you may have seen already I have already I think shown you earlier now here total energy is basically two types of stored energy and Transit energy what is stored energy stored energy is basically if you have some Mass any mass then that mass will store energy in the form of kinetic energy potential energy internal energy chemical energy magnetic energy and other kind of energies okay now this I am not going to discuss I think you are all aware what is kinetic energy MV Square we all have been doing this from 11th and 12th okay and what is potential energy m g z what I am going to discuss I will discuss a bit about internal energy heat and enthalpy these are the three things I am going to discuss and then we will move to open system okay so you see work transfer we have discussed in detail heat transfer internal energy we are going to discuss and anyone can tell me what is this quickly tell me what is this quickly what is this box that I have marked highlighted anyone first law for what for what process and for a closed system yes for a closed system okay so be very clear about what we are talking about do not get confused about okay I mean you use some equation you have some equation you apply everywhere you know you can't do that okay so it is basically for a closed system and for a process I very clearly defined this and for a cycle basically Del U becomes 0 because it is a property and Q 1 2 is equals to W 1 2 this is first law of thermodynamics for a cycle okay let's move ahead now what is heat transfer basically heat transfer is whenever energy interaction takes place that is responsible for either temperature change or phase change okay that can cause so if energy any energy is coming inside okay enthalpy comes yes we will discuss that don't worry don't um go ahead I mean no we will discuss that okay and that we will discuss today so energy interaction whenever energy is consumed or it is I mean it is coming in or going out and it is causing either temperature change or phase change you can say that heat transfer is happening now heat transfer that is responsible for temperature change Isco is called as sensible heat okay here heat heat transfer that is responsible for temperature change is called as sensible heat now if heat transfer is causing phase change then it is called as latent heat now let's see you have this you have this uh bucket and in this you have some water this water is at let's say 20 degree Celsius okay now I am giving some heat maybe I have put on an electric heater or something I mean there can be so many ways or I am burning some coal here there are so many ways I can give heat so I am giving some heat and then its temperature rises to let us say 50. so this was its initial temperature after some time the final temperature is 50 degree so which heat is getting what is this heat what is this heat that is getting transferred quickly quickly answer in the chat that is sensible heat yes sensible heat very nice okay now I keep giving I keep giving this heat I keep giving this heat and let us say this temperature reaches this temperature reaches 100 degree Celsius and I still keep giving heat I still keep giving heat then tell me what will happen then tell me quickly what will happen any idea yes so this water will start evaporating there are so many things that you can say so it will convert into steam it will start vaporizing it will start having phase change it will start app operating so there are so many ways of saying the same thing but basically it will convert from liquid to Vapor in a in very simple way and when that is happening basically one is converting into gaseous phase liquid phase which is water is getting converted into gaseous phase which is Steam so phase change is happening and the heat but this phase change of course will happen at 100 degree Celsius okay it will not change it will remain at 100 okay and that heat that I am transferring is called as latent heat is called as latent heat okay so basically the heat that is involved during phase change is called as latent heat the heat which is responsible for temperature change is called as sensible heat okay so sensible heat you can calculate by this because it is proportional to mass specific heat and temperature difference and latent heat you can calculate by this you cannot use this formula to calculate latent heat because delta T is zero you see when phase change is happening delta T will be zero so if you use this formula Q will be 0 but that is not the case you are transferring heat okay so you cannot use this you cannot use this you have to use this formula okay now sign convention I hope you are all aware of sine convention now when let's say you have this is your system and outside it is surrounding okay now I am transferring heat to the system Q in it is getting transferred so this heat transfer Q in is basically positive it is always considered as possible Heat supplied to the system is taken as positive and it is just a random assumption okay you can take it negative as well but whenever you are solving a question you have to always assume a particular sign okay you cannot in up in the same question you cannot take Q in somewhere negative and Q out somewhere negative okay so you have to follow this once follow only one convention now if heat is going out if this system is rejecting Heat then heat supplied we say heat supplied by the system by the system and it is negative so it is negative very simple okay now what are units Q is equals to m c d t what is mass this is kg specific heat kilo Joule per kg Kelvin and temperature difference is Kelvin so this will get canceled and basically units are kilo Joule Joule Mega Joule or these things okay so these are basically the mostly used unit is kilojoule okay now what is specific heat so basically you see your main equation is Q Z equals to MC DT okay how did it come we saw that Q is proportional to Mass experimentally and Q is also proportional to temperature difference if mass is more we need more heat for the same temperature difference or if temperature difference is more then for the same mass you are transferring more heat so experimentally we have seen and then we added a proportionality constant C here okay so that's it that is how we got this equation now this is the defining equation for specific heat so specific heat is basically represents the heat absorbing capacity now I don't know if you okay heat uh heat transfer if you have done you may have already aware of the thermal conductivity so like we Define it there similarly uh we Define it here that what do we do we say specific heat very clear you know very uh I mean what I am saying just pay a little more attention here specific heat is the quantity of heat required to raise the temperature of one gram of substance so let's say you have some ma mass of one gram okay and you are raising one unit temperature difference so delta T is equals to one so if you put in this equation so C is equals to Q upon 1 into 1. so basically if you have one gram of mass and you are changing its temperature to 1 degree Celsius okay or one unit whatever because delta T Kelvin degree celsius will remain same it will not matter so whatever amount of heat that you require whatever amount of heat that is required is called as sensible heat is called as sensible heat okay so it is not I mean I'm just telling you this but basically to understand its significance you have to understand that it gives you the heat absorption capacity you see look Q is equals to m c d t so let us say mass is constant okay so when mass is constant now uh if you are giving for example you are giving just one second let me explain it by Numbers if Q you are giving 100 units okay so see let's say it's fifth it's 60 okay so what will be delta T what will be delta T delta T will be let me write it 10 actually not 60 so that you can directly divide it if I say it is 10. then what is delta T delta T will be 10. okay now if I say Q is same C is let us say 20 so what is delta T delta T is 5. what does this mean let me explain what does this mean it means if a body has higher specific heat capacity if our value if our body has high C it means it you keep giving it heat it will keep absorbing heat it will not allow the temperature difference I will repeat it again okay if a body has higher specific heat then if you keep giving heat it will keep absorbing heat and it will not allow a temper higher temperature difference okay and if it has low specific heat then the temperature difference will be higher okay so if I ask you one first of all tell me is this clear this much is clear if a body has high specific heat then for the same heat transfer temperature difference will be less okay now and can anyone tell me can anyone tell me that when phase change is happening when phase change is happening phase change then what is the specific heat of the fluid what is the specific heat of the fluid any guess no not zero zero means very high temperature difference by you see Delta Q is equals to C into delta T I am not saying zero but let us say tends to zero so if it tends to 0 that means delta T will be tends to infinite okay so basically when phase change is happening C is in finite so basically you see if it is water you keep giving heat you are giving yeah yes infinite okay so you keep giving heat you keep giving heat its C is infinite very high C it is it keep absorbing heat and it is not letting any temperature say change so it is equal to zero okay so again if C is high delta T is low Force same for same heat transfer C is low delta T is high I hope now it is making sense okay now specific heat is basically two types constant pressure and constant volume okay let's see them how about constant volume so I hope you remember this formula first law of thermodynamics for a process quora process now I am talking about constant volume I am talking about constant volume so tell me constant volume DW is 0 right because DV is 0 yes quickly you have to be more because I don't want to wait for your answers you have to say yes no it feels it's clear constant volume process DW db0 so this term will be 0 yes if this is 0 then that means DQ is equals to d u so I can write heat transferred at constant volume increases the internal energy of the system it is only if you are giving heat it is only changing the internal energy because work no it is not doing any work work is zero okay so basically C is equals to DQ upon basically I can write here d u is equals to MC DT okay so DQ is equals to M Delta t or d u C is equals to DQ yeah so d u is equals to m c v this is basically if I put here CV this will be this constant volume sorry constant volume okay CV delta T okay T1 to T2 so this is basically heat transfer for constant volume I have already proved this for you in the LA I mean I have already given you this equation in the last class as well that for constant volume Q is equals to Delta U but I am just proving this for you okay now what about constant pressure so constant pressure is again this is also I have done it for you okay d u plus I know it is constant pressure so I can take P inside in this last class I have done the same thing DQ is equals to d u plus PV DQ is equals to BH so can I write here that heat transferred at constant pressure increases the enthalpy of the system now what is d h d h is equals to m c p DT so I can write T1 to T2 and this is basically heat transfer at constant fashion this much is clear this thing we have already seen in the last class actually this I have defined it for you in the last class so I'm just elaborating on the same concept yes or no is it clear yes no anyone okay now so these are the formulas basically d u is equals to now why I have written it like it because CV and CP they are they are a function of temperature so if you go any you please repeat last one which one last one this one it's very simple no look DH what what have we done we have assumed here here we have assumed that pressure is constant so we have taken pressure inside okay I know pressure spelling is wrong but I am not repeating it again pressure is constant so so just one second okay so I have taken P inside I have simplified it and I have gotten this but this equation is only valid for constant pressure because this is what I have assumed in the beginning and then I have just written because I know what is DQ DQ is equals to m c c but I have it in c p y p because I am using this for this pressure equation so I have specified for constant Dasher only okay and it is equals to heat transfer okay so basically what you will be using is q d q for constant pressure is equals to d h is equals to m c p DT this is what you will be using okay what first slide this one okay shivendra Ras can we move ahead or do you are you writing something okay so now this is the equation that will be using ah and this is equals to heat transfer at constant volume this is equals to heat transfer constant pressure now you say Sir why I why are you integrating c p and C with temperature because basically they are a function of techno uh books what you study in your b-tech so on the back of those books you will see CP and CV tables and you see that those basically will be changing the temperature so you will be getting some you know like uh some something like this some equation that CP is equals to and some constant will be given to you N1 plus 80 plus b t square plus C by T Cube something like this okay but and similarly for CV as well some relationship so basically it is saying that CP is a function of temperature and how it how it is a function of temperature through this equation okay it can be linear it can be polynomial it can be hyperbolic whatever relationship it could be now but most of our applications most of our problems in any competitive exam we assume that CP and CV are constant they are not changing much with respect to temperature so we take their average values we take their average value for example for ideal gas we can just take heat transfer constant volume is equals to d u is equals to MCV average at T2 minus T1 simple we will talk about internal energy in some time then I'll talk about this as well but yes that is because of the uh vibrate that rotation vibrational motion translation motion of the molecules so internal energy is basically what by this is coming from internal energy right U is equals to m c v DT so this is internal energy is what it is because of the rotational translational and vibrational energy of the molecules okay so and if temperature is higher that vibration is higher okay so you see if the CV will change and it is identified by you okay so anyway so but you can take average values for your problem solving okay similarly p is equals to C is equals to d h is equals to m c p average A2 minus T1 so you can basically whatever values you will get in your exam for CP and CV those are average values for example for a we take c p is equals to 1.013 I think if I am correct okay kilo Joule per kg Kelvin CV is equals to 0.718 kilo Joule per kg Kelvin but are these no the temperature that we deal with Okay so 1.005 yeah I think that is that is more correct yes but he can let me I mean I think it's approximate value so what I am trying to say is that whatever temperatures that we deal with for example we usually deal around uh deal with around you know uh let us say minus 10 2 let us say minus 400 plus 400 degree or 500 degree celsius okay so in that temperature range the values of CP and CV is not changing much I mean they will be changing for sure but not much so we can assume that it is the average value that we can take okay so and we use this continuously if you do not use the relationship we just directly put CV and CP is that okay can we move ahead quickly okay so basically CP and CV values are for it is applicable for ideal gases and for fluids and solids fluids liquids and solids okay now Yeah so basically in solids and liquid what happens is here though you have to use both of them so for air we always Define CV and CV but for for solids and Lucas basically what happens is let's say h is equals to U plus PV okay so d h is equals to d u plus p d v plus v d p now you see for solids for a given pressure for a given pressure DV is almost a zero like for solids and liquid they're incompressible okay you cannot compress them gases are compressible so for a given pressure for if you are applying some pressure DV is equals to zero so these terms will be equal to 0 and also change in pressure for a given volume change in pressure will also be zero so basically d h is equals to d u so m c p d t is equals to m c v DT everything will get canceled if mass and DT are constant they are same I mean we are assuming they are same so c p is equals to CV in what case in solids and liquids but in gases again you see when we talk about water so I'll just give you one value see which you can use a cpcv but it is 4.18 kilo Joule per kg Kelvin similarly if I talk about any solid now I don't remember any values for example ice you can take I don't remember the values but basically I will give you a single value okay but when I talk about air or I talk about Vapor or I talk about uh some other gas okay then I will give you CP I'll give you CV got it because a lot of people remain confused sir water they have given water should I apply CV or CV CP it does not matter DH is equals to Du you apply whatever you want doesn't matter okay now let's compare the values of CP and CP okay so we have two systems so in this also we are giving some heat and in this also we are giving some heat so we have one is constant volume process one is constant pressure so here our piston is fixed so volume will remain constant volume will remain constant now I am giving some heat and you know DQ is equals to d u plus DW so we have bu we have DW okay now I know for constant volume d w is 0 this is what I know also I want same internal energy change for example I want 50 or 50. okay so now you see here if I am for this 50 internal energy change I only need to give 50 heat so this will be 50 because everything is converted into d u by for constant volume DQ is equals to d u so for this 50 change in internal energy I just need to give 50 teeth units of heat but in order to get this 50 if I give here 50 then that means work will be 0 but in constant pressure work is non-zero so let us say your work is your work is let us say 20 whatever some some value some non-zero value okay so basically in this time in constant pressure you would be needing you would be needing 70 T transfer when you put 7t constant pressure then change in internal energy will be 50 and the Piston has moved you see piston has moved from here to here so the work transfer that is involved is what p v two minus V1 is basically 20. these are some random values I am giving but I just want to tell you that all for same change in internal energy you have to give more heat transfer for constant pressure than constant volume so mCP DT will be greater than m c v DT and if for same mass transfer and delta T CP is always greater than CV and the ratio between them is called as gamma is called as gamma so I always know gamma is always greater than one because CP is greater than CB always no doubt about it no doubt about it okay is it clear should we move ahead it it changes with other properties but okay just wait then there we will discuss this okay so basically just briefly I'll tell you for ideal gas ideal gas internal energy is a function of temperature only okay this is also called as joules law it is also called as joules law only for temperature because internal energy is basically again I am devoting from the topic I will explain this later okay don't worry about it I'll explain it in a while today only I'll explain yes yes yes yes okay so let's not deviate because if I debate I will have to explain again and again okay so so this much is clear or not first of all you tell me foreign so what is the relationship between c p c v and C gamma okay now you see H is equals to U plus VV DH is equals to d u plus d V now what is ideal gas equation ideal gas equation e v is equals to MRT so I can write d h is equals to d u Plus d m r t so now I can expand this CP DT MCV DT plus m r DT I can cancel this so c p minus CV is equals to R C D minus CV so this is the relationship between CP and CV now c p minus CV okay CP minus CV is equals to R and we have we also know that CP by c v is equals to gamma so can I say c p is equals to gamma CV and let me substitute it here so gamma CV minus c v is equals to r c v gamma minus 1 is equals to r c v is equals to R upon gamma minus 1. okay and now let me substitute it here so c p is equals to gamma R upon gamma minus 1. okay so very rarely we use these two equations the most we use is basically these two equations this is very frequently used very very frequent use this I think let us say maybe five times out of 100 we use this very rarely and anyway it came from here itself so basically if you are using this you are fine okay clear yes no quickly just be with me answer quickly so that we can cover enough okay now so we have seen what I if you remember that uh thing so energy is basically stored and Transit heat we have discussed work we have discussed in this we have macroscopic and microscopic but we have kinetic energy potential energy internal energy and chemical energy so this is what chemical energy we are not going to discuss and also I have already told you kinetic and potential energy is basically we all know kinetic energy is what because of the velocity half MV Square what is potential energy mgz okay so our focus is internal energy now just I have written it here the stored energy this portion can be classified as macroscopic form so macroscopic form of energy are defined as the energy with respect to some outside reference okay so you see potential energy is basically always potential energy is defined with respect to a reference so if you take some reference you see this is z one this is Z2 or whatever and also kinetic energy that means velocity is always defined with respect to something okay there is no absolute velocity okay so you define key with respect to this body this body has some velocity okay so this basically depends on outside references that's why they are called as macroscopic form microscopic is basically that depends only on the system only on the system so basically ah related to the molecular structure and degree of molecular activities and are independent of outside reference so these are base internal energy is a microscopic and total energy is basically inter total stored energy is internal energy kinetic energy and potential energy okay now what is internal energy so energy is stored in the molecules of the substance due to vibration and collision between the atoms and molecules that's why I was not explaining it there because it was only anywhere coming okay so energy stored in the molecules due to the vibration and collision now why this fibration molecules happens it happens because of the temperatures so basically you see molecules have some molecules have some translation okay some they will move from one place to so that is called as translation motion they can also have some rotation with respect to each other okay so they can rotate so rotation or rotation or they also have vibration with respect to each other so because of this higher the temperature higher these motions okay higher the temperature higher the translation higher the rotation higher the vibration so the energy that is because of this is called as internal energy so basically the energy stored by the okay is it just for you or for everyone else can everyone else hear me like am I clear to everyone okay so whatever energy is because the what whatever okay thank you thank you thank you everyone okay so whatever energy is possessed by the molecules so if some mass is there Mars will have what molecules molecules will have some translational energy rotation energy vibrational energy if you give more heat more temperature that means higher energy higher translation higher rotation higher vibration but the energy processed or stored by this mass is called as and okay let me write it one more time energy possessed by this Mass due to temperature due to temperature because this mass will have some energy due to Velocity as well that is kinetic energy due to position as well that is potential energy but due to temperature the energy that is possessed is called as internal energy is it clear to everyone quickly okay so it is internal energy is an extensive property I have already explained this and specific internal energy is basically internal energy divided by mass so it's a intensive property all these things we have discussed in the first class I am just repeating it again so that it will you will get some revision as well okay and I have explained you here that there is a joules law this basically tells you that internal energy for a ideal gas is a function of temperature only so this can be objective objective question as well internal energy for ideal gas is a function of temperature only okay now what is enthalpy the sum of internal energy and the product of the pressure and volume of thermodynamic system is called so enthalpy is basically more of a convenient term so we have this term we use very frequently and this is basically determined by the energy processed by open system now right now I am not explaining this I will today come to open system and again I will again explain this okay because if x minute here it will take some time and I have to anyway explain later on so I'm not explaining right now but just remember U plus TV is the energy possessed I know this spelling is wrong okay possessed by open system flow work at open look Sanji wait that's what I am saying wait we will come to that whatever I am saying just try to because we are on closed system you see so don't worry about this we will come to open system we will discuss this okay now but what I'm trying to say is this is one term that we use very frequently and just for convenience we have given it a name H and H is called as enthalpy now enthalpy is an extensive property if you divide it by mass then it will be specific enthalpy and it is intensive property okay so again this I will explain in some time do not worry about it okay I hope internal energy is clear enthalpy will be clear in some time but right now just understand H is equals to U plus PV okay can we move ahead quickly okay let's go now perpetual motion machine of kind so it says it is impossible to develop a device that works on a thermodynamic cycle and produces work continuously without consuming any form of energy so basically it is saying you have a device let us say you have a device and it works on a thermodynamic cycle and it produces work continuous that means it is giving you work continuously work output without consuming any form of energy that means it is not interacting with anything is coming in but work is going out but if you see according to first law of thermodynamics for a cycle q12 will be equal to W12 right that means I am saying I am not giving it anything that means Q 1 2 is 0. and I am saying work output is not zero it is giving me work output but first law is saying if this is 0 then this must be zero so that means this device whatever device this is it is not possible because it is violating first law of thermodynamics and if this device is possible somehow it is called as pmm1 perpetual motion machine I hope you all understand what is Perpetual for future can anyone tell me what is the meaning of permission like in physical what is the Perpetual meaning like in general continuous yeah continuous or always or whatever you want to call it so basically a machine that can always be in motion it does not matter you are giving some heat or not but it's always in motion okay so this is not possible this is because it is violating the first law this machine is saying that work output you are getting but you are not giving anything but first law is saying you have to give work then only you have to give heat then only you will get work okay clear yes no okay now very some important points so heat and work okay so heat and work both are path function they both are boundary phenomena they are not a property inexact differential and they are energy in transit okay and there is nothing like heat and work it's always heat transfer or work transfer we I mean he there is no absolute value of work or heat okay it's always because it's an energy in transit so we always say heat transfer work transfer always okay now so I have summarized everything for you this is a ideal gas index okay we discussed about K value right so this is that value isobaric so we have everything in terms of you see uh if you remember uh just one second we defined everything in terms of p v k or M is equals to constant so n value is this n is equals to 0 for constant volume it's infinite isothermal it's one adiabatic it's gamma polytropic it's n heat transfer isobaric we have just discussed this today is equals to DH isokoric du isothermal we have discussed by d q is equals to d u plus d w a bit is isothermal so this is what m c d t isothermal means this is zero that means this is 0 so DQ is equals to DW already discussed in the last class okay so here he transfer is equals to work transfer adiabatic heat transfer is zero and polytropic is zero we very rarely use this so I have not defined this okay word transfer isobaric PV to minus V1 this is zero this is worth transfer this everything we have defined ideal gas equation also we have defined everything we have discussed I have just summarized you because I know I have covered so many formulas so at some point of time you will get confused so you can refer to this table and maybe it will clarify a few things for you got it nothing new I am teaching here it's just the same thing we have discussed in now yes no okay very nice let's move ahead let's move ahead so till now what we were discussing is this portion of the first law of thermodynamic closed system now we will move ahead to open system now in open system there are two classifications steady flow and an STD flow I think I already explained what is TD flow and what is an STD flow okay we will discuss again but anyway I think I hope I have discussed this so STD flow is where Mass flow rate in is equals to mass Vera out energy flow rate in is equals to energy flow it out and there is nothing stored into the system but in an STD flow so HTT flow this is zero but in an STD flow this is not non-zero okay so we will discuss this now let's start with steady flow so as I said STD flow means a rate of work rate of flow of mass and energy across the control volume are constant it is simply telling you that whatever mass is going in or what is the mass flow rate in is equals to mass flow rate out okay so let us say I will rub this I am just explaining this and then I'll drop this if inside mass of this system is let us say uh uh 100 kg okay now I am sending in m dot in is equals to 10 kg per second and with the same rate I am taking it out 10 kg per second so if the initial is 100 kg what will be final Mass inside the system quickly yes so basically finally there will not be any change because whatever is coming in is same thing is going out so basically the change of mass inside the system or storage is equals to zero okay because m dot in is equals to m dot out okay in fact let me write this here because I will drop this figure so m dot in this is the condition for CD flow okay I am not making it up myself and therefore change of mass inside the system storage is equals to zero same thing happens with energy so energy B whatever is in is equals to energy are out and therefore change in energy inside the system will be zero okay now what is happening here you see look mass is Flowing inside so mass is coming from outside some mass is coming from here and this system has already some Mass you see this system is already filled some with some Mass okay now when this green mass is coming from outside it is going to push see I have mass still here okay this is completely filled I have already filled this mask and it is also filled here now when this new mass will come in so it will try to push this Mass a little bit inside yes it will try to come in right so it will push the boundary of that blue Mass which is already inside okay now when it is pushing the boundary that means when boundary is changing that means what when boundary yes work is happening so when this mass is coming in it is pushing the boundary inside so it is a compression work it is type of a compression work it is compressing the mass inside so that is called as negative if you are pushing the boundary inside it is negative so the work that is it is doing to come inside is minus P1 V1 so this is the work done by Mass entering at the boundary okay now now what is happening now there is some Mass here you see let me okay look it is here we have a boundary right here we have a boundary okay now this blue mass is trying to go out it will try to flow out when it is try trying to flow out it will try to push the boundary outside it will try to push the boundary outside so if it is pushing the boundary outside volume will increase and it is kind of an expansion work expansion work is what positive okay it is work done by the system on the boundary or on the surrounding by if this is my boundary so outside anything is just one second if this is my system inside then outside anything I am just you know running out of colors so outside anything is what surrounding so this when this mass is going out it is pushing the boundary outside towards the surrounding so it is work done by the system on the surrounding and what was this this was work done by the surrounding on the system or work done by the mass on the system okay so this Mass when it is going out it is going to push the boundary outside it is kind of expansion work and this work is given positive and it is p2v2 so it is basically work done by Mars exiting outside the system okay so these are two extra one now there could be one other type of work for example what do you have let's say you have you have a turbine okay so turbine is something like this okay so when this turbine when mass is coming in then this will happen when mass is going out this will happen Okay because you see it is pushing the boundary this is the boundary but one shaft is also attached with it so this mass that is coming in is trying to rotate this shaft and this is called a shaft work so it is called as shaft work or the work obtained by the system and this is called as control volume or the extra work that the system is from the system that you are getting is called as control system so the total work during this process when the mass is coming in and mass is going out and also system is giving you some work then it is called as the total work is basically this now this work you not you will not always get for example if you know like you have this nozzle you have this nozzle so only mass is coming in and going out there is no control volume but so this work will depend upon what type of system you have for example if you have heat exchanger we will talk about everything but only mass is coming in and going out coming in and going out no one so sometimes it could be zero as well but basically total work is this is it clear till now is it clear then we will move ahead okay now so let us Define our steady flow energy equation let us Define our CD flow energy equation okay so as I said always start with the first law of thermodynamics now I am not saying for closed system for open system for this for steady flow work okay look you see we have a boundary here okay so when this Mass from outside will come in it will try to push this boundary inside so it is basically the work done by the mass on the system so it is basically work done on the system and work done on the system is negative so this mass that is entering it is doing some work on the system and what is this work it is minus P1 V1 similarly when this mass that is going out and this is your boundary let us say so it is trying to push the boundary outside so it is kind of expanding the volume and volume is expanded it is kind of an expansion work okay so it is basically work done by the mass or work done by the system on the surrounding so it's positive work so this is PV P2 V2 so this is that now there could be a different kind of work depending on your system so for example you have a turbine so turbine this is minus P1 V1 this is plus p2v2 but something else is also happening what is happening this fluid is rotating your shaft the shaft work you are also getting so control volume work it is called as which is happening due to the system inside to the mass inside the system okay and it is crossing the boundary of course that is called as control volume if it is compressor then it will be minus DW if it is nozzle then DW control volume will be 0 because only mass is coming in mass is going out that's all heat exchanger mass is coming in mass is going out the W control volume will be equal to zero got it okay now so you always start with first law of thermodynamics and first law of thermodynamics everyone knows it is the energy conservation equation I am not saying for open system closed system whatever it is just first law of thermodynamics it is energy conservation and that means energy in is equals to energy out Plus change in energy of the system or the storage now I have already told you for STD state mass flow rate in is equals to mass flow rate out and therefore change of mass of storage will be equal to zero similarly energy okay just wait let me write it here because I need this space okay mass flow rate in is equals to mass flow rate out and therefore change of mass inside the system or the storage mass is basically zero similarly energy dot in whatever energy is coming in same energy is going out therefore there will not be any change of energy of the system and it is equals to zero so by that logic now if I say first law of thermodynamics for steady flow so STD flow means open system okay steady flow so now I can say e in is equals to e out y because Delta e storage is equals to zero is it clear to everyone first of all till tell me till now is it clear okay now so as I said so what do we have energy in is equals to energy out now how this energy is coming in energy is coming in two ways what some heat is going in and some mass is going in and that mass is carrying some energy you see Mass can store energy in how how can it is stored energy tell me very quickly how a mass can store energy very quickly tell me no one is telling me really today I only explain this yes internal energy potential energy kinetic energy very nice okay so some mass is going in and it is carrying some energy with it also what is what energy Q in is going okay what is going out some mass is going out and it is carrying some energy with it plus some work output if you may be getting out from it okay so I know that this mask can store this energy in the form of internal energy kinetic energy and potential energy Q in this mass that is going out it is storing energy in the form of exit plus v exit Square by 2 plus g z exit Plus work out okay now I know what is this work output I have already defined it for you so let me write it here plus what is this work output control volume work so whatever this volume is doing one then what work is coming in and what work is going out okay so here it is work is coming in minus P1 V1 here it is going out plus P2 V2 okay now I can write m dot i u i let me take this P1 V1 here so I can write plus P1 Capital V1 plus velocity now this is volumes you see this is volume this velocity let me write it like this can anyone tell me am I doing anything wrong here just in this equation have I done anything wrong quickly tell me I will be very happy if you can find out what mistake I am doing you multiplied yes very good very nice yeah that is what I was expecting from you so you see I cannot write capital V here why it should be small V okay because here it is capital V here it is capital V but here mass is outside okay so I have to write a small V here that is specific volume okay similarly here also I have to write a small V because mass is outside okay clear so I have given you a bit big equation now I can write m i now this is what enthalpy h i plus v i Square by 2 plus g z i plus Q in is equals to m dot e u e sorry sorry sorry sorry h e plus v e Square by two plus g z e Plus work of control volume anyone need any explanation but rate should be balanced dimensionally yeah yeah I mean okay yeah you can do this that's fine I mean you can do this okay if you can't see I'll do it with red you can do this this okay is it clear I mean yeah that was my mistake actually you have to do this because you are taking mass flow rate here so this you have to do dot dot U that means per unit time okay so basically this is our STD flow this equation is a steady flow energy equation okay very frequently you will be using that when we solve our questions okay so you will be using this equation and it is also called as first law of thermodynamic for open system for CD flows okay so this is my first law of thermal let me uh okay so this is basically uh STD flow energy equation or you can say first law of thermodynamics for open system but for what steady flow honestly not an HD flow STD flow okay now someone was saying that enthalpy and all and what is open system let me explain this that now I think now we are on the stage that we can discuss that so now you see when I take this next okay when I take this system if I have let us say if I have this and this is constant volume okay so whatever heat I am giving whatever heat I am giving that heat is basically stored in terms of internal energy only so internal energy is basically energy possessed by Mass in our constant volume or a you know like closed system or a closed system okay where there is no extra mass that is coming in how much energy du energy it is possessment means the mass that has that energy whatever mass is there it will only store d u but here here what will happen whatever mass is there that will store energy in the form of U and then you see here P1 V1 is coming here P2 V2 is coming so basically U plus KV so basically d m change in mass will be d u plus P 1 V 1 plus P2 V2 so basically or change I'm sorry not change in mass I'm really sorry it is basically the energy that is possessed by the mass this is d e that is possessed by the mass so here you have this d u here you have d u plus P1 V minus P 1 V 1 plus P two V two so you can say energy possessed by a open system is U plus PV so U plus PV which is also DH it is the energy by that energy that is coming in in the form of work is observed but by what it is absorbed by this Mass only right and this energy that is going out it is mass that is sticking out that in terms of work where is it going from it is going from Mass only okay so it is energy possessed by open system okay yes again it is energy it is energy okay but energy of what system okay hello now let's come to open system work done so open system is so most of your you know like whatever uh applications that you have in power plant most of them are isentropic so is this PV internally stored energy with us yes it is stored internally yes it is the energy possessed by the molecules I mean again it is a it is work or it is heat but it is energy only no so PV is basically the energy that is given to the Mass okay that mass is where is that mass going mass is going inside only right okay so you see most of the systems are you see wherever you solve your questions you will see ice and Tropic turbine or isentropic compressor so most of the processes are isentropic now this is what this first law of thermodynamic for open system right and how will you get this how will you get this you you neglect neglect if you neglect kinetic and potential energy changes so m dot i h i plus Q dot in is equals to m dot e h e plus control volume so Q dot in is equals to m dot e h e minus h i y because m dot e is equals to m dot I okay Plus work of control volume so yes you can write Q dot in is equals to d h plus DW so this is for what open system okay so this is your first law of thermodynamics for open system and STD flow and there is one more gifts equation that anyway I am not going to cover but it's come comes in very end okay so uh click its gives equation which relates temperature and drop maybe I'll cover in entropy probably I'll cover okay and VDP but it's not as important because you don't have to derive it just you have to know that how enthalpy and entropy are related how and enthalpy and entropy are related now most of our systems turbine compressors they are all isentropic now what is isentropic isentropic is reversible idea batik so adiabatic means Q is equals to zero isentropic means DS is equals to zero okay so in this equation if d s is equals to 0 and in this equation if Q is equals to 0 and if I replace a d h by VDP so open work is equals to from this equation I can write control volume work is equals to minus DH and then by this equation I can write is equals to minus V dtp minus VDP but you can only apply this in case of turbine and compressors because turbine and compression 99 they are isentropic the question they will give you asymptote okay so if you remember closed system work was what can can you tell me what was closed system work General equation General equation quickly quickly pdv was it not one two two now you remember yes so basically at a given pressure at an intermediate pressure change in volume so what did we took we took a small pressure and we defined it like this and then we did it in finite number of times and we said that this area when extended on volume axis gives you closed system work now same thing we are going to do with open system one so here it is VDP it is VDP so what we are going to do we are going to take a very small volume intermediate volume and see what is the change in pressure here so this is what we will take it is p e Dash sorry V Dash V Dash DP then we will do it in finite number of times very very small V Double Dash DP then it will do here then we will do here and basically we will cover this whole area and this is basically area extended on pressure axis on PV curve is what is open system work open system work again if you don't remember let me write it area under curve when extended on volume axis is closed system work okay T V where I am going to extend it volume axis so this is your work closed okay then area under curve when extended on pressure axis is open system work pressure volume where are you going to extend it pressure axis so this is work open am I clear quickly okay now let us see the applications let us see the applications office 3D flow all the all the devices that you have in power plant they are all STD flow okay they're all open system and STD flow so this is my main equation I am going to apply all my assumptions based on this okay so turbine we all know what is turbine what are the turbine into turbine is a work absorbing device producing so mass is entering here mass is going out work output we are getting okay and always it is isentropic it is isentropic so isentropic means what how much is DQ how much is DQ is 0 yes so it is well insulated that means Q is equals to zero and I'm also going to neglect kinetic and potential energy changes so if I neglect kinetic energy potential energy kinetic energy potential energy also well insulated means isentropic is basically so this is zero so I have m dot i h i is equals to m dot e h e plus control volume one so control volume work is equals to m h i minus h e so this is how you can calculate work transfer for a turbine you can directly apply this and kilojoule okay I don't think there is much to explain here it's just the application of steady flow in a turbine okay if kinetic and potential energy are given then you have to put it the values I am assuming that we are taking it at zero but in question they might give you kinetic and velocity and Z they will give you so you have to take it that's all do not blindly assume that this is the formula that you are going to use clear I don't think there is much to explain here then we have compressor quickly I'll you know go up through everyone so here mass in Mass out their work is in it will be negative okay again it is a work consuming device compressor is a work consuming device so I will neglect kinetic and personalities changes well insulated so this is zero this will go away this will go away this will go away mass in h i is equals to mass exit H exit now you see this is basically we have taken if you remember we have taken it on the energy out I hope you remember this okay we defined energy out where is it okay next page we defined energy out and we took it as work output but here in this case it is work input in the term in compressor it is work input so I will write minus work of control volume okay so work of control volume is equals to m h exit minus h i compressor so you can calculate it like this every equation is applicable for this particular scenario okay next is what I am just quickly going through it because there's not much text in here okay now if you have heat exchangers yes of course okay you want to explain me how that is basically the uh you know like more of uh applications part but you see what is what is and what is enthalpy can you tell me what is enthalpy enthalpy is the energy U plus PV so it is basically the energy that the mass has now when this mass is coming in it is at very high temperature and very high pressure okay and yes and then you have rotors here okay you have rotors here now when this coming at a very high pressure this pressure is applied to the rotors and these root rotors this energy is utilized to rotate the rotors okay and this rotors are connected to the shaft so you will rotate the shaft you will connect the shaft and you can you know it is the work output basically it is the mechanical work output but the energy there this Mass has has been given to the rotors this pressure is given to the rotor step pressure will reduce temperature will reduce so if temperature will reduce you will reduce this will reduce so basically and capable also reduce okay simple in simple words I would tell you that the energy that the mass has is basically given to the rotors and that energy is reduced when the mass comes out and that's the enthalpy reduction enthalpy energy reduction okay clear yes I mean you can say that yes because temperature is also you know varying in the inside the inside the turbine are you relating it with the heat transfer do not say you want me to consider heat transfer no really I mean it's not heat transfer heat transfers what heat transfer basically should happen from the boundary this is the more this is the energy contained by the molecules you see the thermal energy that you are talking about is basically energy contained by the molecules so it is an internal phenomena it is not outer it is not crossing any boundary or something okay hello now let's go to boiler or evaporators okay basically a heat exchanger let us say now what is happening hot fluid is flowing through here cold fluid is flowing through here and heat transfer is happening between them Q heat transfer is happening between them but the outer area of the heat exchanger is basically insulated so no heat exchange is happening between system and surrounding so Q will be zero Q will be zero or okay this is q in it is taking just one second okay okay here what we can do is we can take one system so you cannot take the whole system then actually you can do this two ways uh okay let me let me explain you both the ways so so that just one second okay let me explain two ways first let me take the whole boundary let me take the whole boundary as a system so what is happening here pot fluid is flowing in cold fluid is flowing in heat transfer is happening internally but there is no heat transfer between the system and surrounding so Q is equals to zero okay so in that case what will happen is in that case what will happen is that this kinetic and potential energy we are neglecting anyway okay and in heat transfer we all know that there is no work done so work is equals to zero so this is also zero this is negligible this is negligible this is negligible no work transfer no heat transfer so basically what can I say in this particular scenario m dot in h i is equals to m dot e h e okay now what mass is coming in mass is coming in M mass of hot fluid h of hot fluid in okay mass of hot fluid out sorry mass of cold fluid in cold fluid in enthalpy of cold fluid in mass of hot fluid exit enthalpy of hot fluid exit plus mass of cold fluid exit enthalpy of cold fluid exit so this equation you can use to calculate your heat basically it's you have you are equating this is if I say m dot h h I minus hhe is equals to m dot c h c e minus HTI so this is basically the Heat rejected by hot fluid and this is the heat absorbed or heat again by fold fluid okay this is something that you will use in heat exchanger as well okay so this is the scenario when we are assuming that the how I mean we are assuming that this is my boundary okay now now what can I do is let me take one more page and if I say only this is my boundary only the hot fluid is my boundary okay so it in it out is happening here from hot fruit to cold fluid okay and hot fluid is coming m dot h cold fluid so now what will happen now which one the last one this one okay it's it's actually simple it's not I mean okay let me explain it look let me assume that this is the boundary of the system okay now when the hot fluid is Flowing here cold fluid is Flowing here heat transfer is happening internally it is not crossing the boundary so if no it is not crossing the boundary that means Q is equals to zero now work is any rate zero for a heat exchanger I am neglecting kinetic and potential energies so what is left in the above equation it is simple m i h i is equals to m e h e okay now what is entering you see this is entering and this is entering so hot fluid is entering m dot h i and enthalpy of hot fluid in plus what is entering here cold fluid is entering so m dot C entering and H dot C entering now what is going out what fluid is going out m dot h exit into H dot h exit when you will do questions then it will be more clear okay plus what is going out cold fluid is going out okay so it is a general equation now if you want to simplify it more what you can do is you can bring all the hot fluid one side so h i minus you will take the other side h h e cold fluid will take the other side okay m dot c h c e minus HCI okay again by whatever is entering m dot h is constant right so m dot h entering and leaving will remain same similarly m dot C entering and leaving will remain same now what is this this is basically energy this queue you see this hot fluid is losing some energy that energy is going to the cold fluid so it is the Heat let me write in short uh okay what did I rub so this is basically heat rejected by hot fluid and this is Q absorbed by both through it okay I hope it is clear this time when you do more questions it will be clear okay now let me take one example now this is this time this is my system okay so in this scenario I will have m dot i h i Plus Q in is equals to m dot e h e because work is zero kinetic and potential energy are neglected okay so in this case m dot i h i minus Q is equals to m dot e h e okay now here Q is equals to m dot i h i minus m dot e h e okay now this is for what this is for if you want to calculate how much heat is gained by the cold fluid so Q absorbed is equals to m dot C h i minus h e so you can calculate it using this formula okay now same thing applies for a condenser it's it's again a heat exchanger okay but condenser what happens it is rejected okay sorry heat is absorbed actually uh hot vapor condenses to liquid by rejecting heat to cold fluid Okay so here cold fluid will be rejecting Heat listen heat to cold fluid at constant pressure okay so condenser okay what happens in condenser anyone what is the purpose of contain what is condensing I think we discussed this as well what is condensing condensing is basically [Music] condensing is basically conversion of vapor who looked so that means if I have a fluid and I need to condense this then it need to reject some heat okay it need to yes hot to no condensed you know condensation is the process of converting Vapor tool condensed vapor to liquid okay evaporation is the process of converting liquid to Vapor or boil or boiling okay so same thing you are going to do here you can m dot i h i Q in is equals to m dot e h e so Q in U of condenser is equals to m dot h i minus h a simple okay so basically you have to apply this steady flow energy equation based upon what is given to you in question there are other applications as well for example you have nozzle so what is the purpose of nozzle the sole purpose of nozzle is to con increase the kinetic energy of the fluid okay the device that increase the kinetic energy of the fluid okay so all other change now it it does not do any work it is insulated so Q is equals to zero also the purpose is to increase the kinetic energy means exit velocity which is V2 it is much much greater than V1 yes to accelerate the fluid so if I apply this uh in our equation what will I have m i h i Plus I can write v i Square by 2000 here but I am not writing it y because V1 is negligible so I am not writing this because it is neglected okay Plus this is also 0 is equals to m dot e plus v e Square by 2000 plus work is 0. so exit velocity will be equal to under root two thousand h i minus h e now I know that you will have some doubt that where this this 2000 came from do you have this doubt or someone can explain me m i n m e is same right whatever is entering same is going out that is the basic nature of steady flow right so you can cancel that you can you can cancel that okay now they're also basically here also you will have me sorry here also you will have m e and then it can get canceled okay now so does anyone had this doubt where this this 2000 came from 1000 came from because our equation was this this was I my question I write it 2 here and I write 1 here I write 2 here and one here yes very nice very good okay now the problem here is you see if you are using this equation if you are putting this enthalpy in joules then you will use V1 Square by 2 and g z i plus Q in this should also be in Joule okay m e this should also be in June plus v e Square by 2 plus g z e by 1 and this should also be in Joule okay so if this equation is in Joule you can use this equation but you see sometimes enthalpy values given in kilo Joule what heat transfer is given in kilo Joule again this is given in closure or this is given in closure but you see if you multiply mass with velocity mass and velocity okay so basically this will be given in kilo Joule per kg because it is small H okay now mass and velocity is basically what kg meter per second okay so if we if we you know Explore More I I can do that here as well so basically Joule is Newton meter Newton is what Newton is forced into acceleration so force is what uh no one second yeah Force into acceleration no one second one second Joule is Newton meter and Newton is what Newton is force and force is mass into acceleration here uh just one second again do not get confused with this just one second I just want to show you something Joule is what Joule is Newton meter okay now Newton is what force we explain Force by we express Force by Newton now force is equals to mass into acceleration what is mass mass is kg acceleration it was meter per second Square now we I take this meter as well here so kg meter Square second Square so this is equal to basically Joule okay so here also we have v i Square so this is equal to joule now to convert because this is given in kilo Joule to convert this value also in kilo Joule I have to divide by 1000 so basically instead of 2 you will take it two thousand similarly here instead of one you will take it 1000 because this this is also in joules this is also in joules so if enthalpy is in kilo Joule you will use this equation is it clear so this is kilojoule equation okay and this is Joule equation is it clear okay very nice next is diffuser diffuser the I mean diffuser what does diffuser do it decreases the velocity and increase the pressure okay it increases the pressure but velocity means exit velocity is much less so now I can neglect exit velocity again work is zero it is insulated uh potential energy and kinetic energy you cannot neglect okay so basically what do you have m dot i h i plus v i Square v i Square by 2000 is equals to m e h e okay here also you will get here also you will have plus m i now mi you can take common or okay just one second plus v i okay so you can cancel m i m i v i e is equals to under root 2000 h e minus h i meter per second okay so you can calculate what is the inlet velocity okay similarly there is adiabatic mixing process so it is same thing it is what we have discussed earlier so it's like fluids three fluids are coming in one fluid is coming in here one fluid is coming in here and this fluid is going out okay so let us say m one is here M2 H2 H1 at temperature T1 at temperature T2 and here what is going out M3 is going out so what do we know I know that M1 plus M2 will be equal to M3 because Mass conservation but if I apply this here so what is I can neglect kinetic and potential energies no work done and again it is insulated so no heat interaction we are just applying the energy equation here nothing else okay so M1 H1 this is entering what else is entering M2 H2 is entering what is going out M3 S3 I can write M1 H1 plus M2 H2 is equals to M1 plus M2 S3 so we can use this equation to calculate enthalpy of the Final Exit fluid okay now again I know look I do not need to do so many application and I don't think anyone does so many applications in the class they usually what you do is they will teach you energy question they will give you two three applications maybe a heat exchanger nozzle turbine compressor and they will be done with it then once the question comes then you will struggle with you know what uh concept I can apply here but I am giving you questions in the form of application so these are all questions that they have asked earlier but I am just giving you everything in terms of applications okay just writing the application and giving you that thing so now you can directly whenever you see these type of questions you can directly apply okay then next is throttling process the throttling process I hope you all know throttling process is basically when you know like when you let the fluid pass to a very small pore so like here it is filled with fluid and you open this wall a little bit and the fluid drop by drop drop by drop it will come here okay so this is called as throttling so flow through a constricted passage like an open wall or if I is called as throttling and of course you see here the friction will be highest because whatever is coming in it is in contact with this wall with a solid body so here you cannot neglect friction okay because you see if this is pipe here and the fluid is Flowing so only friction this fluid is going to experience friction here at the boundary not in the middle okay so here maybe you can neglect uh friction and you can assume it is a reversible process but in throttling process whatever is going to come in it is going to cross this little space it is going to rub in this wall and therefore you cannot neglect the irreversibility here so it is a irreversible process okay now we are neglecting kinetic and personal energy no heat interaction no work interaction so this is zero this is zero everything is zero this is zero so basically m i h i is equals to m e h e okay and we also know that Mi me is same so enthalpy same so basically I can say that throatling is a ISO enthalpic process enthalpy remains constant ISO enthalpic ISO and tell pick plus it is again objective question okay it is again an objective question okay so I hope we are good till here are we fine again we will solve so many questions that it everything will be clear okay I know that in this class I am little fast because I want to cover the concepts so now on a CD flow is left we will cover next 15-20 minutes but we will compress an expansion process in open system analysis what do you mean compression expansion uh in open system turbine or compressor are you asking mass in and mass out I have I think I have explained in general I mean what look it's a very wide question that you have asked I am re I am really finding it hard to narrow it down but what should I explain you okay ask something specific and I I can explain that but if you just ask you sir explain mass in a mass out it's really difficult for me to narrow it down I think what you are trying to ask is um no I I think I got what you are trying to ask uh you see this is open system STD flow unsteady flow now steady flow Mass was entering the boundary compression phenomena okay look if what I have explained you shubham I really request you to because I have explained that two times okay now what I want you to do is you go whatever I have explained you I know that you want to ex uh you're asking me to explain that in detail but that is absolutely not required whatever I have explained you I have explained in very detail and that is more than enough okay but what you can do is you will get this recording uh in the night or tomorrow morning you can go back and try to you know go through it once if you have any doubt we have next class on Friday you can ask any doubt because if I go through that again I think I already explained that two times in class today itself okay so if I go that again then I think you know like uh we will have time issues please just go through the recording try to I have explained it more than enough that whatever is required of course we can go in much detail with each and every topic and I can go believe me I can go okay I have done mtech from IIT and I have done projection thermal engineering I have worked on a convection I have worked on open systems so it's not that I don't want to go I really want to go but it's just a waste of time for I mean waste of your time as well as mine because we are here for a particular objective okay so uh I think then we can move ahead let's finish an STD State systems and then tomorrow we'll focus on I mean nectar we will focus on questions and we will see each and every type of questions so whatever doubt we have today it will be clear okay and I would request to revise also before the last next class because we will be using each and every formula that we have discussed today so now our focus is this an STD state our focus is on STD State now here steady state this was Zero y because energy in was equal to energy out and mass in was equal to mass out so this was Zero but here it is non-zero here it is non-zero so what is an STD State honest TD State flow on STD flow is basically rate of flow of mass and energy across the control volume variably respect to time simple meaning is mass entering let us say is 10 units or 10 kg per second and Mars that is coming out is 5 units so that means mass of the system is bound to change it will increase right because more mass is coming in less is going out so mass of system the mass of storage rate of change of mass of storage is not equal to 0 yes similarly energy that is coming in is not equal to by mass is bringing energy right so if mass is not equal that means energy is also not equal so energy out and therefore energy of storage or energy of system is also not equal to 0 in this particular case of an STD true okay now I would request each and every one to be very focused whatever you are doing because I am going to give you a very big equation and you I will try to explain each and every word out of it but if you do not focus then you will ask me three times to explain that again and again but I will explain no problem but I would ask you to for 10 15 minutes be very very focused and I will explain each and every point and you will understand also but if you don't understand now I think you will not understand it again okay because I cannot explain it again with that efficiency with which I am going to explain it in the first time okay so again like we started with the steady flow energy equation we will start with the first law of thermodynamic what is that what is first law of thermodynamics quickly energy conservation that means energy in is equals to energy out plus change in energy of stored into the system okay STD flow we took it at zero but this time we cannot take because it is not zero it is not zero okay now let's see what is going in what is going in so you see again I will explain you two three things here let us assume that the initial mass of the system is MI so mi very be very focused okay so m i is initial Mass inside control volume every point is important inside control volume what is M final you see some mass is coming in some mass is going out but it is not same that means M final will be something else and that is final Mass inside control volume what is M1 M1 is mass entering the control volume and drink the control volume what is M2 M2 is Mass leaving the control volume mass leaving the control volume so whatever Mass okay okay okay okay little little little little I have just one second just excuse me I have messed up with the conventions am I we are always taking as in okay let me m i is basically we are always taking Mass entering in the control volume M exit we are taking Mass leaving the control volume this one M1 is basically the initial Mass inside control volume and M2 is the final Mass M2 is the final mask this much is clear or not this much is clear just convention just convention and it is very important I am telling you because you will get confused okay everyone is it clear convention I am already warning you will get confused that's why I want to make sure it's very clear okay so I can say that M what is entering minus what is exiting is equals to change of mass of control volume which is m 2 minus M1 M2 minus number okay same thing we are going to do with energy same thing we are going to do with energy so you see e i e i is equals to m i e i is equals to energy entering in the system okay M exit m e e is equals to energy leaving or exiting the system E1 is equals to M1 E1 is equals to energy or initial energy let me write initial energy inside the control volume whatever the mass here is due to which it will have some energy even okay now this is because M2 E2 it is the final energy final energy every point is important be very aware of the convention so this is I this is e this is 1 and this is 2 initial and final is it clear I want answer from everyone is it clear okay now as I said okay as I said okay I have exhausted the pages okay so as I already explained you that when I apply first law I will get energy in is equals to energy out plus change of energy of the system okay now what is entering m i e i is entering plus Q in is entering now what is going out M exit e exit plus workout this much we have seen in the STD flow yes but there is a extra term here energy of storage energy of system sir e e what ee yes sorry e s e e okay okay work output so this must we have seen but their Mass was say mass in and mass out was same so this term was getting 0 but here it is not same so some mass will be changing some energy of the system will be changing and what is this so it will be equal to Plus by final energy inside the system minus an initial energy okay so what is final energy what is final energy well final energy is M2 E2 minus M1 even okay now let me explain I have already explained you how does the mass store energy m i UI [Music] plus v i Square by 2 plus g z i plus Q in is equals to m e u x z plus v x z Square by 2 plus g z x z Plus work output okay now what is this work output already explained you it is Plus work of control volume plus minus P 1 V 1 plus P2 V2 okay uh uh okay okay one second it is okay I mean I just want to be write it in a way that is much clearer let me write it again one more time let me write it here so that okay so m i u i plus v i Square by 2 plus g z i plus Q in is equals to m e u exit plus v exit Square by 2 G exit plus control volume minus P 1 V 1 plus p 2 V 2. okay so this is basically this which is mass and energy entering this is heat transfer this is this and this is one transfer now let us move to further part so this one plus M2 now energy u 2 plus v 2 square by 2 plus g z 2 minus M1 U1 plus v 1 square by 2 plus g z okay so this is the very now this is what this is this this is what this is disk are you able to make sense out of this equation Z yeah this one is Z1 yes I'm really sorry I because I just you know uh right right very fast to the make minute mistakes but I think are we making sense here are you able to understand this equation you have to check okay what we are looking at now see what is this this is let me explain here okay so this is my equation everything I have written here now what is this q in it is the heat entering in the system what is this it is the energy given into the control volume by the mass entering so this mass that is going in it is taking this energy with it what is this this mass that is going out it is taking this energy out with it okay work done by the system is this one done now energy due to final Mass inside the system by initial Mass was M1 final mass is M2 so because of this final Mass what is the energy is this and this is energy to initial Mass so I hope this equation you have to write it at least 10 times so that it's absolutely clear to you okay don't make mistake in i e 1 and to be very clear what is I I is in e is exit one is initial inside two is final inside Elliot clear okay quickly let us I'll take five five seven minutes so what is the application of this charging of a tank you see what is happening charging for example when you fill your bucket right in your bathroom or anywhere so what is happening you just keep it keep it feeling like you know like it's just going inside that thing is coming out or sometimes you even take it out but you see what is going in is not equal to what you are taking out so sometimes the water level in your bucket increases or sometimes decreases depending upon what is happening here and what I mean what how much is going in how much is going out similarly charging of a battery so when you charge when the power is on you charge your battery so you do not use your battery only you give power and you charge your battery right so energy is going in only nothing is coming out then when the power goes off then nothing is going in only Power only power from the battery is coming out so that is discharging okay so these are all example of on a CD flow energy equation the first is charging of tank now charging what is mass in mass and ring mass in initial mass is M1 final mass is M2 now in this case I am assuming that nothing is going out exit is 0. exit is 0 okay so what are the assumptions it's a adiabatic tank so that means Q is equals to zero of course only filling or emptying you are doing so work will also be zero I am assuming M exit is equals to zero initially that means M1 is equals to zero of course these assumptions will always not be valid you have to see what they are asking in question always start with the general equation this is your most General equation and it's difficult energy equation then see what assumptions are valid so assumptions you can apply first assumption kinetic and potential energy are neglected second assumption no exit nothing is going out that means m e is equals to zero so everything this is zero next initial Mass initial initially it is empty that means M1 is equals to that means this is also zero so everything here is zero okay also no work done so this is also zero adiabatic yes shivendras this is zero so what are we left with now it is very simple m i h i is equals to M2 U2 and now see if if you logically think if the initial Mass was 0 and nothing is going out that means whatever came in will be equal to M2 yes so I can m i is equals to M2 so basically h i is equals to U2 so c p initial temperature is equals to CP Final temperature this is in temperature this is final temperature inside the control volume so it is c p by c v t i or you can say T2 is equals to gamma t i again make sure what are the assumptions that are applicable in the question do not apply there blindly all the assumptions but this is the most sound case actually this question has been asked in Gate that's why I have SIM I have applied all the assumptions and given you directly what and they have asked this temperature so you have to and they have given this temperature in the question and it's a very big question if you see that question it's like 10 15 lines but basically everything you need to cancel and T2 you can calculate by this formula that's all nothing else okay quickly what is discharging emptying of tank so now nothing is coming in okay so M out is basically what is going out mass x m e is mass exiting M1 is initial M2 is final okay what are the assumptions nothing is coming in Mass n is equals to zero and full tank initially so it is M2 M1 sorry M1 is full okay now so you see adiabatic nothing is coming in that means this everything is zero no work transfer okay and what else finally empty tank okay I can assume one more thing here if you want I mean if if we can't we can neglect it as well so now what is left 0 is equals to m e h e because kinetic and potential energy are also neglected okay plus M2 U2 minus M1 U1 so either you can use this equation M1 U1 minus M2 U2 is equals to m x z h e okay so either you can use this equation or if in question yes yes thank you or if in question they have given that finally empty tank so finally empty tank a camera what does it mean yes M2 is equals to zero so this time M2 will also be 0 so M1 U1 is equals to m e h e and you can apply the same logic we did earlier and you can find out your final temperatures as well okay so basically this is uh you know the applications for energy equation I know that I have gone a little fast because I need to cover so many things but I think I was at descent speed uh and uh I think we covered it decently okay now in the next class I will come up with all the questions on first law grow system open system so we solve so many questions that every doubt if you have even now it will be crystal clear okay and I am always here you can ask whatever you want anytime okay so let's close the class here and I will see you in on Friday I think got it okay thank you thank you so much