 Today, we will be looking at chemical reactor energy balances. Well, energy balance is something that is so central to chemical reaction engineering, because we have to supply heat or we have to remove heat and then we must design our systems. So, that you know the amount of area that is required for the heat addition or heat removal have to be appropriately correctly done. So, that you know the reaction proceeds as we expected to proceed. So, this is a very important part of the course in chemical reaction engineering. So, this is something that we get started today and then see how we can understand what happens. Now, we write our energy balance for a vessel of our interest. I have deliberately taken a vessel of this very crooked shape just this not to indicate that the balances that we write take into account any shape in principle any shape, but of course we construct shapes that are of you know useful to us etcetera. So, we have here a stirred equipment by stirred we mean that the intensive properties of the system at different positions in the equipment is the same. So, that temperature compositions are uniform everywhere that is the assumption. So, that stirring means that every point inside the equipment has the same intensive property like temperature and composition. We have here f j 0 moles of component j entry and because of chemical reaction f j moles of component j comes out. So, f j and f j 0 if they are different to the extent that they are different that is the extent to which that particular reaction is taking place if there is one reaction for many reactions also we look at. Now, as per the first law convention we have taken heat addition into the system as positive and then heat I mean work that is produced by the system we have taken as positive. Therefore, our statement of conservation says input minus of output plus generation equal to accumulation something that we have been writing for a long time our statement of conservation of mass. We wrote input minus of output plus generation equal to accumulation and we wrote this about conservation with respect to component j. We always wrote with respect to material balance we wrote generation of that component j. So, since the rate of chemical reaction is known we are able to write the rate of generation of component j. Therefore, input output generation it must be equal to accumulation. Now, when it comes to energy balance since energy cannot be generated we say that generation term in the energy balance should not be there that is why I have removed the generation term. So, input of energy minus of output of energy must be equal to accumulation of energy. So, our energy balance is different from material balance because of the fact that generation term is not there in the energy balance but generation term is there in the mole balance not I am not saying material balance mole balance. Now, even if you are talking about material balance in terms of kilograms whatever the generation of a particular term may be there because it is getting converted to something else. So, the overall balances there is no change but individual component balances there will be a change because of chemical reaction. So, what are we saying now if you are talking about energy balance in a stirred vessel there is input of energy because of component j is bringing some energy. There is an output of energy because component j is taking away some energy then there is input of heat which is some energy that is coming in and there is some output of work which is some energy that is coming out. Therefore, you have input minus of output of generation equal to accumulation is a statement of consummation of energy. So, let us now translate all this in terms of some symbols just to understand what might be going on in terms of the symbols easier to work with. So, what is means mentioned here is d by dt of E system S Y S refers to the energy inside the system E is energy S Y S refers to system E subscript system is the energy of the system. So, left hand side says rate at which energy of the system changes with time. So, that is the accumulation term as you can see here the accumulation term here this is the term that is written E system is the energy inside the system S S V C I will put the S here d by dt of E system is the rate of change of energy inside the system. So, that is the left hand side now the right hand side it is summation I equal to 1 to n E I F I E I E I is the energy of component I and F I is the molar flow of component I therefore, E I F I represents the energy that is associated with component I at the input that means here here at the input here. So, summation over all components tells you the total amount of energy that this components are bringing in to the system. Similarly, is sigma I equal to 1 to n E I F I out represents the energy that is going out the energy that is going out with all these material. So, E I F I summation means you are looking at the total amount of energy that is associated with all the components that is going out. Notice here we have not taken into account what is called as the heat of mixing or free mean the enthalpy change of mixing. So, as assumption here is that E I F I this total energy of E I F I and total energy of E I F I out and E I F I in there is no what is called as the loss gain of energy because of mixing. So, that is the so that is what is the assumption here now we have plus Q minus W S. So, Q is the amount of heat that you put into the system and W dot is the amount of work that you take out of the system. So, accumulation equal to input of energy minus output of energy there is no generation of energy, but there is amount of heat that is put in this input of this another this is energy that is coming in with the materials this is the energy that you putting in through your heat transfer pipes and so on and W dot is the amount of work that you might be getting into your shaft that might be made to spin which is turning a motor and so on. So, this is work this is heat this is the energy that is coming in because of the fluids that is coming into the system. So, you have here E I as the energy that is associated with component I which consists of three components one is internal energy is kinetic energy and potential energy. So, we take total energy of component I consisting of internal energy potential energy potential energy and kinetic energy. So, E I represents the total energy of the system of the of component I and E I F I therefore, represents total amount of energy that component I brings in and some amount over all species gives you the total amount of energy that is coming in with the species similarly, for the output this one term which all of us may be familiar, but to put it in the context. Suppose, we have some material that is going into the system and coming out of the system I have got here F I P T omega I out minus F I P T omega I in what is this word F I P T omega I if you look at the units of F I P T omega I omega I is specific volume P T is pressure F I is moles per. So, this essentially this is a work term we can understand it here itself. So, let us say it is moles per time and this is Newton's per square meter this is specific volume that is cubic meter per mole. So, cancels off. So, it is meter square meter cube. So, this is meter. So, this is Newton meter per second which is joule. So, this is per second. So, this is joule per second is what. So, essentially it is energy per unit time is what is the. So, F I P T omega I in is kind of it is coming in. So, much of energy is coming in the form of F I P T omega I and similarly, so much of energy is coming in. So, much of energy is going out. So, this difference what meaning can be attached to this difference. Now, if you look carefully at this suppose there is a chemical reaction in which F I this term sigma F I P T omega I and this term sigma F P T omega in and this is out they are not the same. Now, here is an instance which says that even in the absence of friction this term F I P T omega I out is different from F I P T omega I in. Now, this is greater than this it means this is positive. If it is this term is less than this whole term is negative. So, what is that means W F W F which we call as flow work. The flow work into and out of a system that means in this out of a system and into the system this difference if it is positive it means that this particular system it gives you some flow work. If this is negative which means that you have to put in work to make this fluid enter the system and come out even in the absence of friction. So, what we are trying to say is that there can be instances in which you have to put in work because even in the absence of friction this term is this minus this is negative. This is something that is to be understood that you may need to put in work to be able to get system materials in and out of the system even in the absence of friction. So, this is the point that I am trying to get across to you that flow work represents the difference between F I P T omega I out minus F I P T omega I in and this term in a reacting system can be different. Therefore, this minus this can be positive or negative and if it is negative you have to put in work to make fluids go in and come out of the system. So, flow work is an instance where you may need to put in work to get system fluids in and out of the system. So, what are we saying then if you are looking at our system here and we are saying that W dot is the amount of energy in the form of work that comes out of our system. We recognize that this W dot actually consists of two types of work one is W s which is shaft work and one is W f which is flow work. Now, in an instance where W f is negative that means you are have to put some energy to get fluids in and out of the system W s is negative and therefore, the total amount of work this W s is different from W dot that is the point we are trying to put across to you. So, W s represents the shaft work that we will get out of the system and therefore, it is that work which is going to appear to us it is going to turn our turbines and it is going to turn our motors and so on that is the amount of energy that you and I will be able to make use of. So, what are we saying now in our first law where W dot represents the work that comes out of the system now can be broken up into two parts part one is shaft work and part two which is what is called as flow work which is F i p t omega i out minus F i p t omega i in. So, the total amount of work that comes out of our system which is W dot we have broken up into two parts part one is shaft work shaft two is flow work and we have put the flow work separately. Let us see what advantages we get out of doing this what I am saying is that this flow work let me just sort of go through this once again W s is the measurable work since it is available on the shaft. That means we can actually put in our measuring systems and actually measure W s. So, that is why it is a measurable quantity and this W which is the true work can be measured only by measuring W f and W s. This can be measured because we know p t omega i and F i therefore, this is a known quantity this is an experimentally measured quantity. And therefore, W becomes a quantity that you and I can determine because W f is a quantity that we know because F i p t omega i for all the components are known. You can sum over all species at inlet and outlet and W s is the work that we we can measure at the motor because it is turning and we can measure the work output in various ways. Therefore, W is a quantity that is measured by measuring separately both W s and W f. Let us try to illustrate this because some of these things may become clearer if we look at a small example. Let us look at a sulphur dioxide plant reacting with oxygen in air give you sulphur trioxide. It is a well known reaction it is been going on in the process industry for last many many years and it is a catalytic reaction and so on. Generally, air is used for oxidation therefore, you get air oxygen contained in air is used for the reaction and so on. So, I put some numbers here just taken some data from a factory which processing about 3000 tons per day of sulphur dioxide. So, which is reacting to form sulphur trioxide. So, the numbers look something like this for the case of this kind of flow you can calculate all this I made some estimates the flow V naught turns out to be about 250 cubic meters per second and this flow is measured at some standards at the reactor conditions and what is the pressure typically the pressure at which you will pump the fluids near atmosphere or slightly greater than atmosphere. So, let us say it is 10 to the power of 5 Newton per square meter. So, our situation here is that we have a flow of 250 meters per cube per second and at pressure of 10 to the power of 5 Newton per meter square is coming. So, what is to V naught F i times omega i is the volumetric flow. So, if sum over all species it becomes the total volumetric flow as in this particular term can be seen as volumetric flow multiplied by the molar flow of components. So, this one way we can understand this similarly, you can understand this as flow multiplied by and the number of in the sense F i omega i is flow and then P t is the pressure. So, pressure multiplied by flow gives the units of energy. So, this difference is what is the flow work. Now, if you put all the numbers P t is given V naught is given. So, this P t omega i summation experience says that about this difference is typically about 0.5 percent of that means this minus this turns out to be about 0.5 percent of what is at the outlet. So, we have written this as in the form of P t times V naught multiplied by 0.5 percent which is 0.5 by 100 is 0.05. So, what we try to do we are try to make an estimate of flow work for the case of a 3000 tons per day sulphur dioxide plant and when you make these estimates for the numbers that we have taken this turns out to be about 125 kilowatt with a minus sign showing that this amount of work will have to be put in for the fluids to be put in put into the system and taken out. In the absence even in the absence of friction you need so much of work to be put in. So, that the fluids can go in and come out of the system why is this. So, it is. So, because F i P t omega i at the inlet and F i P t omega i at the outlet are not the same and this. So, that energy will have to be put in appropriately. So, just to cut the long story short what we have said here is that flow work is a term which is the energy that is associated with getting material in out of the system in the absence of friction. And for a plant of 3000 tons per day sulphur dioxide we have made an estimate that it turns out to be 125 kilowatt with a minus sign showing that we have to put work into the system. I mean we have shown at an earlier date that the pressure drop associated with sulphur dioxide sulphur trioxide. There we estimated that the energy that is required to push the material in out of the system in the is something like 2.5 megawatt taking pressure drops into account. So, taking pressure drops into account the energy required is 2.5 megawatts if you do not take pressure drops into account it is only 125 kilowatt. What are we saying pressure drop is a huge activity in the process industry it requires to be properly managed. Now, why are we saying this the context of saying this is that when the pressure drops in a process is something we cannot avoid. So, improving catalyst design of course, will reduce pressure drops understood, but the least pressure drop would be that of flow work. So, flow work is something that is the minimum at which you can work that means that amount of energy you have to put in any way otherwise fluids will not go in and come out of the system. So, flow work is a measure of the minimum amount of energy that is required to put the materials in and out of the system. So, with this let us look at some examples to understand energy in terms of our interest. I have taken the example of a steam turbine where what do we have? We have a boiler we have a boiler to which fluids are coming in and it becomes steam that steam goes into the turbine and here it expands and this steam finally condenses and then comes out into the condenser where it condenses and then that fluid that liquid is again pumped back into the process. So, what is the boiler of what is the power plant power plant is an instance of a process in which steam is generated in a boiler it is put into the it is put into a turbine where it turns the wheels of the turbine in which in turns turns the motor and so on for a generator. Now, you have so much of energy in the form of heat q 1 is going into the boiler and so much of energy q 2 which is thrown into the environment. So, q 1 minus of q 2 is really the energy that has been lost by the working fluid which is in your boiler minus condenser. So, this q 1 minus of q 2 must appear as work. So, we have got here w t is the work for the turbine w p is the work of the in the pump therefore, w t minus of w p is what you will you will be able to produce in this system giving where you are putting q 1 of heat and then q 2 of heat output q 1 of heat input. So, q 1 minus of q 2 and ideal conditions q 1 minus of q 2 equal to w t minus of 4 w t minus of w p sorry is that clear. What are we saying q 1 minus of q 2 is the total amount of energy that is absorbed by the system and therefore, that must appear as the shaft work which is w t minus plus w p. So, so much of shaft work is done therefore, that difference must be equal to shaft work that is what we have said. So, what we are saying here we are essentially this is also an instance of energy balance where the energy of steam is used to run a turbine. And therefore, q 1 minus of q 2 which is the energy net energy that is absorbed by steam must appear as turbine work which is w t and w p. Now, having said all these things let us now just look at what we are saying in terms of the equations that generate that describe the energy balance. So, we have said that f i e i in f i e i out this term w dot can be replaced as w s plus w flow work that is what has been done here the word w dot has been replaced as w s plus flow work that is what has been done. So, essentially the same energy balance we still work with the left hand side is the state of accumulation of energy in the system. This is the amount of energy that is coming in with the fluids amount of energy that is going out of the fluids by this is the amount of heat that is transferred into the fluid. And this term minus is the term that is associated flow work or the work associated with getting material in and out of the system in the absence of friction. So, with this kind of formulation now we can take this f i p t omega i along with the input terms and output terms and so on. So, that we can combine this and make it it might look far easier to understand now. So, d by d t of energy of the system now we have is sigma i equal to 1 to n f i is the molar flow of e i and p t omega i. You notice here f i e i this f i p t omega i can be combined. So, that this term now looks like e i which is the internal energy of the system and p t omega i it is it is an external phenomena because of pressure and so on. So, we have put e i and p t omega i together because of the advantages that it will give us. So, just like we have done for the inlet term you can do the same thing for the outlet term where f i is multiplied by e i by p t omega i. So, what is it that we are saying? What we are saying is that that if you have an energy balance if you take the flow work into account and then the flow work term appears in the energy balance in this form you have f i is the molar flow and e i plus p t omega you can notice here in term here it is a negative correct negative negative positive. Therefore, e i plus p t omega i is the energy of the system. Similarly, f i p t energy of this out. So, you have internal energy plus p t omega i is the energy in and that is the energy out. So, since we shall we say that input energy output energy this is the amount of heat that you put in amount of work that you put in. Therefore, this represents the statement of energy balance after we have made necessary substitutions for our terms. So, our statement of energy balance as we have said is d by d t. Now, you can see here on your left hand side the system on the left hand side also I have replaced the e system by our energy for each component multiplied by the number of moles of component i inside the system. The number of moles of component i inside the system we have taken as n i see what I am saying here is that the number of moles here the we are assuming that the n i moles of components present here each of this n i moles has specific energy sorry e i and then e i consists of u i g i z i and then small u i. So, all that is taken into account and that is how our system now looks like this. So, you have on the left hand side e system this is e system which consists of n i multiplied by this is internal energy this is kinetic energy this is potential energy this is the term associated with the flow work which is now we have brought in appropriately it goes from i equal to 1 to n number of species all this i equal to 1 to n which is number of species. So, notice here that this term this whole term u i plus p u i b s this is the total energy. So, this is something like what is called as internal energy and this is something like p v term which is. So, this looks like our conventional nomenclature what is called enthalpy. So, you have inputs and you have the outputs plus q minus w s notice here this w s previously we call it as w dot which we said was equal to w s plus flow work and that is how we this flow work we put it in terms of p t omega i and so on and then rearrange this equation. So, that this equation now looks slightly different from what we started with and the form in which this equation now looks is something like this with all the simplifications what we have is the this term you notice here this term u i u i plus p t omega i it is termed as h i that is what we have done here h i and then you have this kinetic energy this is the potential energy notice here this term is not there. So, notice here that on the left hand side left hand side the term p t omega i does not appear, but on the right hand side it appears is it clear because this term flow work is associated with flow and therefore, in the left hand side is only accumulation therefore, flow work does not appear that term does not appear. Therefore, on the left hand side I have subtracted my by putting p t omega i. So, that this term h i minus p t omega i by definition h i minus p t omega i by definition is u i you see this that is how it is written on the left hand side. So, this is the statement of energy balance for a system which is well stirred which means every position inside the equipment has same composition same temperature or in other words all intensive properties are the same at different positions of the system. Now, we can simplify this further which I have done. So, that we have some more advantages which I will do right now earlier. So, what we have done now is that the left hand side p t omega i minus. So, this term if this term p t omega i is not very large particularly now a reacting systems that we are talking about this is not a term which is going to be very important. Therefore, it is possible to look at the whole equation in a slightly simpler form by saying the following that the left hand side this term and this term and these two terms are not very important. In the sense that the real energy that is associated with our chemical reaction is in h i and therefore, this term u i square by 2 g z i and p t omega i are terms which may not be very important. Now, having said this. So, of course, this is something that we must verify for our every case we apply these equations we must convince ourselves that our assumptions are correct and if it is not we must come back and use the general form which we have written now. So, that we do not make mistakes. So, but in situations of most of the situations of chemical reaction engineering you will find that these terms can be deleted these terms can be deleted. So, that the simplified form of the energy balance might look something like this d by d t of n i h i i equal to 1 to n i equal to 1 to n. So, what is it that our equations look like our equations look like that left hand side is n i h i summation over all species right hand side is f i h i summation over all species and then again out term is f i h i summation over all species plus q minus w s. So, what we have done is that we have started with the general statement of conservation of energy and we made certain simplifications and then we have written our equations now in terms of enthalpy which is a term which is well tabulated etcetera and then n i is the number of moles inside the system h i is the number of moles coming in this is the this term h i f i is the amount of energy that is coming in and this h i f i out is the amount of energy that is going out all measured in terms of h i enthalpy. Since enthalpy is a well tabulated in this representation form there are certain advantages because this can be read out of tables and therefore, our effort in trying to put the numbers becomes much easier. So, just put it in the context what we have done is that we have set up the general balance equation and then we have written our equations in a form. So, that we can simplify some of the terms can be deleted and so on and got the final form in which we have on the left hand side n i h i summed over all species the right hand side f i h i summed over all species in flow stream as well as out flow stream q and my w s are heat and work that goes into an order system. Now, having said this so we would like to see how this representation that we have done how does it sort of explain or shall we say how it is able to describe what happens in different situations of our interest. So, we will apply our equations to a variety of reacting systems. So, every system that we apply our equations should hold correct. So, we want to see how our equations sort of help us to understand what goes on in different systems etcetera. So, what are we said we said that our energy balance equation looks like this our energy balance equation is f n i h i summed over all species f i h i summed over all species f i h i summed over all species plus q minus w. So, let us take the example of a power station let us say we have a hydro power station. Now, how does our equations help us to tell what happens in this hydro power station. So, what is the hydro power station hydro power station is an instance where you have a reservoir containing a large quantity of water huge many of our irrigation dams are often used to generate energy because the water is available at a very high level and can be made to run through a turbine. So, you have a reservoir which holds huge quantity of water and then it is at a level which is much higher than our hydro turbine which is at lower level. Therefore, you have a difference of height that is a difference of height. So, many meters so many some height difference is there and therefore, that height difference can be exploited to derive energy from that water. So, what does it do what is it doing this water of the reservoir is going through a pipeline into this turbine and this turns the turbine and this turbine turns a generator and that is how we get electricity and so on. So, I have taken this as position 0 and this is position 2. So, we are writing our material energy balance between 2 position 0 and position 2 how does it look. So, we said on the left hand side our equations as you can see here there is d by d t of n i h i is equal to f i equal to 1 to n f i h i in i equal to 1 to n f i h i out plus q minus w s. Now, what is q what is w s q is the amount of heat that is put in or amount of heat that is lost. Now, if you assume that this q is not very significant or in other words for the application of our interest we may assume that this q is not very high. It is not such a bad assumption as far as a hydropower stations are concerned, but the fact is that it is an assumption and every assumption will have to be appropriately validated checked to make sure that you know our assumptions are consistent with what happens in the field. So, let us form and assume that q is not very important that means heat losses are not very important. Therefore, this statement of energy balance says that left hand side accumulation of energy equal to input of energy minus output of energy plus flow energy in the form of q that is going into the system energy in the form of w going out of the system. So, you notice here what I have done is I have written the energy f i h i and f i n i h i these two terms I have written it as a pressure energy p by rho plus c v times t naught minus of t r plus u naught squared by 2 g z naught that means energy that is associated with our reservoir at this point 0. What are the energies that is containing it is there is an atmospheric pressure. So, this is p 0 by rho and what is this t naught minus of t r t naught is the temperature at which this fluid is there is taken as t naught and our reference is at t r we always take a reference whatever that may be. So, t r therefore, the energies associated with this fluid present in the reservoir at temperature t is given by c v t naught minus of t r. So, you have so much of energy in the reservoir. Now, you also have situations where u naught is the velocity of the fluid and z naught is the potential energy of the fluid. So, in this particular case if you assume that the energy or velocity of the fluid at the surface of this reservoir is not very high or not very it is not very important. Therefore, this term u naught squared can be neglected why can we neglect this because on the top of the reservoir where you have fluid essentially at rest or may be moving at small velocities and so on. So, for our purpose we can take it as 0 or in other words u naught is 0 is g z naught it is the potential energy at which water is available g z naught is the amount of potential energy. So, what is it that we have the right hand side you have pressure energy this is sensible heat energy that is associated with the fluid at temperature t naught this is kinetic energy this potential energy you multiply this by f f is what is the molar flow of the fluid that is going through the hydro turbine. So, you have the input what is the output output is pressure at p 2 and then it is c v which is the energy associated with the sensible heat and so on. So, that we written as c v times t 2 minus t r and this is kinetic energy this is potential energy. So, what we have done is that we have we have used this equation we have used this equation and then based on our understanding of the of the of the situation we have converted this h i in and h out in terms of c v d t. So, that we can take care of the change in the change in the enthalpy of the fluids. So, that is taking care of c v t naught minus of t or this is for the reagents this is for the this one is it all right. So, you have pressure you have enthalpies you have kinetic energy you have potential energy this minus this how does it look now when you look carefully when you look carefully at this hydro reservoir what is what is the pressure at this point it is the atmospheric pressure. Similarly, at point 2 which is emerging into the atmosphere therefore, pressure at position 0 and position 2 are atmosphere. So, what I have done here I put p naught equal to p 2 second thing I have done here is that the rate of change of the reservoir capacity or reservoir fluid accumulation is not very large. Therefore, we can we can delete the accumulation term in our energy balance what you are saying is that we can delete this term because this term is not very large. So, putting these two simplifications left hand side is 0 and the right hand side we knock out terms which you think is not important therefore, we are left with a slightly simpler form of the energy balance which says 0 which is on the left hand side equal to g times z naught minus z 2 plus c v t naught minus t. So, we have now said that the energy that is associated with the height or in the with the fluid at a height now it is able to exchange it and converted it into this term c v t naught minus t 2 or enthalpy. So, what is it that our hydro turbine was doing it is using the energy of energy of potential energy and converting it into energy that you and I can use. So, this is the statement of conservation. So, let us do a small example to illustrate how far we can go this is an example this is an example where what we have done is we have done is that W s by f we have written our our energy balance equation this energy balance equation in this energy balance equation we have deleted all terms accepting the term corresponding to g z naught and z 2 and t naught and t 2 all other terms are disappearing accepting of course, the flow work from the flow. So, when we do all this when we do all this what is it that we get we get that f s W s by f is given by this relationship it is fairly straight forward because simply by replacing this terms simply by replacing this terms in terms of terms that we know we will find that W s by f which is the energy that is produced divided by the throughput of fluids divided by throughput of fluids is given by gravity plus this term. So, what we are saying now is that the energy in gravity term can be converted to energy in fluids in the form of c v and then that relationship is what is given here. We can do this number of small calculations to understand these numbers let us let us look at one such number I have taken here t naught is t 2 and f is 4000 kilograms 4000 tons per second very large very large and z naught minus z 2 is 5. So, these are the things that are some numbers we assume to get to feel for the for the size of the plant that you are going to be building. So, I have done here you have 10 what is 10 50 and 4000 4000 is the flow this is the flow and what is 50 what is 50 this is m this is g this is m g h. So, this is m sorry this is m. So, this is g this is h. So, what we are trying to do is that we have a fluid the hydro reservoir in which you have fluid available at a height of 50 meters. Now, if it flows through a hydraulic turbine with the g value of 10 what will we get that is what is the answer that is mentioned here. Then if you multiply all this you get something like 2 megawatts what we are saying is that if you have a hydro power station which is able to supply at a head of 50 meters. That means, you are able to lift the water to that height g is the gravitational acceleration constant for 4000 is the flow. So, m g h that is the term associated with the shaft work that turns out to be 2 megawatts. So, what we are trying to say here is that if you have a hydro power station in which water is available at 50 meters height and then if kg per second second. And if so many kg per second water is flowing then you will be able to generate 2 megawatt of power. So, this is the point we mentioned here that if you have a water flow of 4000 kg per second and then if you have a water flow such that the gravitational acceleration is still 10 meters square per second squared and then this 50 refers to the height at which the water is available. So, now what we have done is that we have shown that our energy balance can explain situation that happen in a hydro power station. So, we go on to see how best we can make use of what we know for running our daily activities. Now, let us say instead of doing hydraulic turbine we do a steam turbine. What are we doing now? Instead of doing a hydraulic turbine we are doing a steam turbine. How does it help us or what difficulties does it bring us into? Let us just quickly understand this. Once again when we have a steam turbine instead of a hydraulic turbine we want to apply the energy balance and the basic procedure is identical. This is no change. You have a steam turbine where d by dt of left hand side equal to in minus of out f i h i in f i h i out out plus q minus w s. So, this q and this is w s and this is in and this is out. This is 0. Why has we taken the left hand side as 0? Because accumulation terms are quite small in relation to the flow of energy that is through the system. Therefore, we find taking the left hand side equal to 0 is not a serious compromise on the quality of the descriptions that we can provide. Now, what might happen is that this q is the amount of energy that we put in. Now, we may be thinking that we are putting 100 units of energy, but what actually goes in only 30. This could happen to us. How do you overcome this? We overcome this by recognizing that this q may not be a complete description of what is going on. Therefore, we must know all the heat pathways to understand where the problems are. So, what this equation explains to you is that how we can interpret that information to understand our system appropriately. Now, having said this, I want you to go through this exercise fully so that we appreciate what we are saying. Our hydro turbine water was coming in and going out and we have water coming at 50 meters. Now, if you put all our numbers, let us take the example of water is coming in at T 2 4000 kg per second of water is coming in and let us say Z naught which is minus Z 2 is 50 meters. That means, water is flowing from 50 meter height down. Now, if this term if T 2 minus of T 0, if T 2 equal to T 0, what we are saying is that the temperature here T 2 and temperature here is T 0. If they are the same, then we would produce for this 4000 kg per second of flow 2 megawatt of power. If the flow is 4000, if Z naught minus of Z 2 is 50, then we will produce in the shaft. That means, your generator because shaft work is so much 2 megawatts of work. Now, if it so happens that if T 2 minus of T naught, there is a slight temperature difference of 0.05 degree centigrade. Suppose, which means what this term is not 0, this term is T naught minus this is actually T 2 minus 0.05. Therefore, T naught minus T 2 is minus 0.05. So, I have just put those things here. That means, our shaft work now it is not 10 multiplied by 50 multiplied by 4000, but it is subtract you have to subtract this term. Now, we notice here that this term is something like 290. This whole term turns out to be 290 units. So, that 500 minus of 290 is 210 multiplied by 4000. So, now, you are producing 0.81 megawatt instead of 2 megawatt. Now, what we are trying to put across here is that in the energy balance, because you have taken into account the effect of change in the temperature of the fluids that is going in and coming out going in and coming out that effect in a hydro turbine as per this calculation says can be very significant. In fact, we would expect after all energy in water is form a mechanical energy. Therefore, we would have thought mechanical energies can be transmitted without any loss of efficiency. Here is an instance where you are transmitting mechanical energy with an efficiency of only 50 percent or less. This is something that you should bear in mind. Therefore, we have to take into account the fact how do we design our pipelines taking the water into our turbine without allowing it to get heated up even by a very very small margin. This is an important part of design which we should take care. I will stop there when we come next time and we will take these ideas further. Thank you very much.