Energy Equation In the previous lecture, we derived the Navier-Stokes equation and towards the end, we made a conclusion, we arrived at a conclusion that there are situations when you require to solve the Navier-Stokes equation in conjunction with the energy equation. One example that we sighted to establish that fact is that when density is a function of temperature and pressure but there are several other considerations when you may not relate density as a function of pressure and temperature through equation of state. But the sensitivity of density with temperature is just given in terms of a volumetric expansion coefficient which comes into the picture by creating a density gradient due to temperature gradient that drives the flow that is called as natural convection or free convection. So there are various possibilities where in which the energy and the Navier-Stokes equation, they need to be dealt with together. So there are situations when they are coupled. There are situations when they are not coupled. For example in the heat transfer problem in which rho is a constant. Then you can solve the various Navier-Stokes equations separately, use that velocity field in the energy equation to solve for the temperature field but if rho is a variable, then you have to consider them coupled. So these are various possibilities. Now we will move on to the solution of the, the derivation of the energy equation. So to do that we will use the Reynolds transport theorem. In the Reynolds transport theorem now capital N is the total energy which is the summation of internal energy, kinetic energy and potential energy, that is what we consider in thermodynamics. In classical thermodynamics, we consider all forms of energy other than kinetic and potential energy. We club that together and give it a name internal energy. So just in thermodynamics books, here we just use a little bit of different notation as compared to the thermodynamics books. In thermodynamics books, the internal energy is given by u but we are already using u for velocity. So we are considering i as the internal energy per unit mass, just a notational difference but otherwise, these are very familiar forms of energy that you know from thermodynamic considerations. Small n is capital N per unit mass. So small n is what? Small n is i+V square/2+gz, okay. So now the Reynolds transport theorem, if we use n as capital E, the total energy of the system, then this is dEdt of the system. So dEdt of the system is what? dEdt of the system is nothing but the change within the control volume. So the del del t integral of rho edV+the outflow-inflow. Rho eV.dA. V.dA is actually V.etadA as scalar, where eta is the unit normal vector. So you have by this time very familiar with these types of terms and we are just going to use those here. So now under conditions of non-deformable control volume you can take this del del t out of the derivative and for stationary control volume the relative velocity is same as the absolute velocity but with the velocity that you have used here and you can convert this area integral into volume integral by using the divergence theorem. So you will get this particular form, right. So in place of eV.dA, it is divergence of eV. Sorry rho is there. So rho eV.dA is divergence of rho eV. So divergence of rho eV is del del xj of rho euj, okay. So that is what is written here. Now you can see that, so dEdt of the system can be written as the right-hand side can be simplified further. So this is what, what form? Conservative or non-conservative? This is a conservative form. Remember we can always convert the conservative form to the non-conservative form by using the continuity equation. So that is what we are doing in the next step. So what we are doing? You have rhodel e del t. So del del t of rho e is what? Rho del e del t+e del rho del t, these 2 terms. Then del del xj of rho euj, rho ujdel e del xj+edel del xj of rho uj, just like what we did for the Navier-Stokes equation, very similar. And then you can see that these 2 terms, they combine together, become 0 because that is nothing but the continuity equation. So you will get rhodel e del t+uj del e del xj. What is this? This is the total derivative of e, right. So rho DeDt. The total derivative of v is acceleration. The total derivative of D is just total derivative of e, you do not have any special name associated with that. So the right-hand side is the rhototal derivative of e dV and the left-hand side is dEdt of the system. So in the next slide, we will see how do we relate the dEdt of the system with head transfer and work done and for that, we relate that with the first law of thermodynamics. So let us go to the board a little bit to understand the background. So let us say you have a control volume. The first law for a system, so first let us consider this as a system. Why we have written in terms of the system because first law is traditionally defined for a system. In fact, the first law is traditionally defined for a system undergoing a cyclic process, that is the cyclic integral of heat is proportional to the cyclic integral of what? But that can be converted to a law that governs for a system undergoing any process, need not be a cyclic process. So for that we have the positive sign convention that we commonly follow is heat transfer to the system, that is considered to be positive, and work done by the system that is considered to be positive, okay. So energy entering the system in the form of heat is positive. Energy leaving the system in the form of work is positive work. So what we can write here is, so this is just an energy balance basically. Let us say you transfer some heat to the system, so what will the system do with that heat. It will do some work. If it does not do some work, the remaining part will change the, simply increase the energy of the system, right. So that is the basic energy balance, give some heat to the system, it can use the part of the heat to do work, the remaining part will increase the energy of the system itself. So that is what it is and this d, if you consider transport, this is actually meaning total derivative, this dEdt because this is about the system. So the total change because of transport and everything changed with respect to time, everything is considered with it. So you can see that dEdt of the system, you have to remember that it is not a must to take this sign convention. You could have taken a different sign convention for work and then simply this will become plus instead of minus, nothing more than that, okay. So the next step is that, remember that while deriving the Reynolds transport theorem, we took a limit as delta t tends to 0 so that the system is tending to the control volume. Therefore this for the system is as good as this for the control volume, okay. So these are thermodynamic parameters with which we are relating to the dEdt of the system and these parameters we will write in terms of the transport. So let us try to do that now. So Q.cv, what is the rate of heat transfer? Now just like in continuum mechanics, in force, you have surface force and body force. Similarly heat transfer, volumetric heat transfer and surface heat transfer. So volumetric heat transfer, let us say that Q triple prime is the rate of heat generation within a volume per unit volume. So for the volumetric term, this is this, very simple. This is Q triple prime is per unit volume. You multiply it with dV and integrate it over the control volume. What about the surface term? The surface term is given by what? Surface term is given by heat flux. In heat transfer, whatever is the, so let us say this is the control volume. Let us say there is a heat flux, Q, typically q double prime is considered as heat flux and triple prime as volumetric heat generation, that is the standard notation that we follow. So Q double prime, let us say this is a small area dA, so Q double prime, which is called as the heat flux vector, dot eta dA is the rate of heat transfer through dA. What is this eta? This eta is a unit normal vector. So we can integrate this but we have to keep in mind one thing. Here positive heat flux is in this direction but by the sign convention that we are using in first law of thermodynamics, positive heat transfer is heat transfer to the system or to the control volume. to Adjust for that, we use a negative sign here. So -integral ofů So what is this? -integral of divergence of, this is control surface, using the divergence theorem. So this in a short-hand notation, this is what? Index notation del qj del xj, okay. Now this is about the heat transfer. What about the work done? W.cv. For work done, what are the forces that we are considering? Surface force and body force. So first let us write the body force. I am not commenting about the algebraic sign. The algebraic sign we will discuss later on but first let us write the contributions. Let us first write for the body force, it is more trivial. So body force is what? BidV is if you take a volume element dV, then the force acting in the ith direction is bidV and work down is what? Dot product of the rate of work done. This dot product of force and velocity, right. So biui, this is basically b.V, right. So biuidV. For surface force, surface force is given by the traction vector. So +TiuidA. In one of our previous lecturers, we have discussed that this is nothing but the vector tau.eta but what is this tau? Tau i1i+tau i2j+tau i3k, right. So in terms of the divergence theorem, what it is? Integral of divergence of what? This is control surface, this is control volume. Control surface divergence of, it is not visible I think. Computer is blocking it. So write it separately. Write it here. So tau iů that is del del xj of tau ijui, okay. dV integral and all those things. So we have represented all terms in heat transfer and work in terms of volume integral. Only thing which we have not yet done appropriate is the algebraic sign of this Wcv. So why it is so? Because here we have considered work done by a force as positive work but in thermodynamics, positive work is done by a system when it does work against a force, that is it overcomes a resistance. A thermodynamic work is what? That you have a resistance, you overcome the resistance to do some work. So that means that is what is positive work, that is why energy leaves the system because you have to overcome the resistance to do the work and work done by the force is just having an opposite thing. So to make up for that, actually you have to put a minus sign here. I mean either in the left-hand side or in the right-hand side. So if you take care of all this, so let us see that, let us get back to the slide to summarise what we have learnt by now. First law of thermodynamics, then the expression for heat transfer and work done and of course with a positive, like the positive sign convention as per the standard consideration that we have in thermodynamic textbooks. These expressions we have just now derived. Now these expressions you can see left-hand side and right-hand side you have, left-hand side you have, in one side, right or left is up to you. So rhodEdtdV integral of that=the heat transfer and work done dV=0. So that means because your choice of the control volume is arbitrary, this is what you get, straightforward from the previous expression. So this is the rate of change of energy term, this is the heat transfer term, this is the work done, okay. So we will take it up from here. Now this is what? This is conservation of what? This is conservation of the total energy of the system. But in heat transfer, we are mainly interested in the conservation of the energy in its thermal form. That means we are basically talking about the internal energy. So we must sort of subtract the kinetic energy conservation from here to get the energy conservation in terms of internal energy. So that we will do. Let us go to the board to do that. So to get the kinetic energy equation, we start with the Navier's equation. See at this stage, we do not give a specific emphasis to whether the fluid is Newtonian, Stokesian, all those things. We start with a general equation of motion. So we have rho DuiDt=del tau ij del xj+bi, okay. So what we do is, we multiply this with ui. So you have rho DDt of ui square/2, right and then this is ui del tau ij del xj+uibi. What was our total energy equation? Let us write at the top, rho De/Dt=Q triple prime-del qj del xj+biui+del del xj of tau ijui. Now we subtract this equation from this equation. If we do that, you have rho DDt, see what we are claiming is that if we subtract u square/2 from e, we will get the internal energy. Question is where has the potential energy gone? This is a very important point and we must discuss it very carefully. See we have already considered the work done by the body force and the potential energy is nothing but the manifestation of the work done by the body force for a special case where the body force is the gravity force, okay. So we should not take one effect twice. Because we have already considered work done because of the body force, we should not duplicated it by considering the potential energy term again in the energy term. Either we will reflect it in the work term or we will reflect it in the energy term but not in both. So because we have already reflected in the work term, we will not reflect it in the energy term and simply if you subtract, we get the internal energy. Then this is =Q triple prime-del qj del xj. Now this term is a collection of 2 terms. Del del xj of tau ijui+tau ij del ui del xj. We can easily see that this term and this term, they get cancelled, right. ui del tau ij del xj. So what you are left with is tau ijdel ui del xj, okay. You may try to represent, this is the non-conservative form. You may as well get a conservative form by using the continuity equation. So for example this term rhoDiDt, so you can write rho del I del t+rho uj del i del xj, right. That is the definition, capital DDt is del del t+uj del del xj, okay. Now we just add +i del rho del t+del del xj of rho uj. This is just adding 0 because that is the continuity equation. So now what we can do is, you can combine this term with this term. So what is that? del del t of rho i and then this term with this term, del del xj of rho uji, right. The product rule. So you can see that you may very easily switch over from conservative to non-conservative and non-conservative to conservative just by using the continuity equation. Again let us stop for a moment and try to understand the physics of this equation before we move further. So this is what? This is the total rate of a change of internal energy and it is a combined consequence of heat transfer and work done. So the heat transfer is the volumetric heat transfer and the surface heat transfer and the work done, one important thing that you can understand is that work done due to body force cannot give rise to heating or cooling. It is not giving rise to a change in internal energy because it is just like work done by body force is just like creating a rigid body type of motion. So that cannot give rise to heating or cooling. So heating or cooling must be associated with what? The work done by the surface forces. So this surface forces should be responsible for heating or cooling and the next question is that we are saying heating or cooling but is it trivially heating or is it trivially cooling. Now we will show that it is trivially heating and not cooling and we will see that if we consider that that is cooling, then that will violate the second law of thermodynamics. So it is trivially heating and the reason is straightforward. So like viscous effect, you can see that this tau ij is eventually related to the viscous effect. So viscous effect gives a sort of shearing between layers of fluid molecules. So instead of fluid molecules, if you consider your palms. If you rub your palms, your palms will not become cooler but will become hotter because it is, what it is? It is an irreversible conversion of the work done due to viscous interaction to intermolecular form of energy that will increase the temperature. So basically this is like a heat source that will increase the temperature. So it is something like this that there is a work done to overcome the viscous resistance between various fluid layers. But this work is irreversibly converted into intermolecular form of energy and that will increase the temperature of the fluid. So this entire work is converted into the thermal form of energy. So let us go to the slides to a sort of summarise this discussion in the board. So we had the total energy equation, we subtracted the mechanical energy equation from there and then we got the equation in terms of the internal energy, okay. Now see let us get back to the slide, previous slide. See one important consideration is, this tau ij del ui del xj, what is this term? So we claim that this term should trivially give rise to a heat generation and not cooling. So it is possible to generalise this but to show this, it is more convenient if we consider a special type of fluid, that is the homogenous, isotropic, Newtonian and Stokesian fluid to demonstrate that this term which is called as viscous dissipation term in the language of heat transfer, that is always a positive term. So we will try to do that. Let us come to the board to show that. So tau ij del ui del xj, right. Let us assume that the fluid is homogeneous, isotropic and Newtonian to begin with. So tau ij is -p delta ij+lambda del uk del xk delta ij+mu del ui del xj+del uj del xi, for Stokesian fluid when we do the algebraic simplification, we will use lambda=-2/3mu, in some in between steps. This del ui del xj. So first term, -p delta ijdel ui del xj, what is this? Delta ij=1 only when j=i. So this becomes -p del ui del xi. So this is nothing but what? Divergence of the velocity vector. So this is like, see most of you have studied basic thermodynamics. So see one deficiency of a learning is we sort of try to in our mild interpret each course in the way in which we learn and do not relate that with the other. So we learn fluid mechanics separately. We learn thermodynamics separately. We learn heat transfer separately and it is not very common that we try to relate this but just try to observe this. Can you find any similarity of this with pdV, right, the thermodynamic, the work done in a thermodynamic process, pdV. So that, this is because this is what? This is like dV, the change in volume per unit volume. So this is interpreted like that pdV type of term, okay but this is not related to viscous effect. So this effect is there even if the fluid is inviscid. So then +lambda del uk del xk, delta ij del ui del xj is what? dell ui del xj, del ui del xi. Can you write del ui del xi as del uk del xk? We can do. There is no problem with that because it is basically repeated index, okay. +mu del ui del xj+del uj del xidel ui del xj=so -p del uk del xk+lambdaů this into this, del uk del xkdel uk del xk. So this into this is del this square. Then to simplify this term, we will do it in a little bit systematic way. So what are the values, i=1, j=1, I am just writing the values of i and j here. Again it is not visible I think. So let us write it somewhere else. May be at the top of the board. So i=1, j=1. This is 1 set. i=2, j=2, i=3, j=3, this is one set. i=1, j=2 and i=2, j=1, this is another set. i=2, j=3, i=3, j=2, this is one set. i=1, j=3, i=3, j=1, this is another set. Just see whether I have covered all, okay. So if you systematically club all this, it is possible to simplify the terms in a systematic way. So i=1, j=1, mu is common. i=1, j=1, del u1 del x1+del u1 del x1, so 2 del u1 del x1, then that multiplied by del u1 del x1. So 2 del u1 del x1 square. So this we have taken care of. Then if i=2, j=2, similar term. del u2 del x2 square. i=3, j=3, del u3 del x3 square. Then i=1, j=2, del u1 del x2+del u2 del x1, right and i=2, j=1, that is also del u1 del x2+del u2 del x1. So del u1 del x2+del u2 del x1del u1 del x2del u2 del x1, right. So +ů If I make any mistake, please let me know. You people are much younger people who do algebra much better than me. So if there is any mistake, please let me know. So this then, so this we have covered. i=2, j=3 and i=3, j=2 essentially the same term, so +del u2 del x3+del u3 del x2 square+this term del u1 del x3+del u3 del x1 square, okay. So see we have got already 3 positive terms and so remaining terms, we will manipulate with these 2. Now let us substitute lambda=-2/3mu. So this is -p del uk del xk+, let us take a term 2mu/3 common. So then this becomes 3 del u1 del x1 square+3 del u2 del x2 square+3 del u3 del x3 square-, this is A+B+C whole square. So A square+B square+C square+2AB+2BC+2CA. So -, and there is a 2, sorry, there is a 2mu/3 already there. So -ů this one, then -2 del u1 del x1del u2 del x2-2del u2 del x2del u3 del x3-2 del u1 del x1del u3 del x3, okay. Then this is 2mu/3. Then we have a + term, +mudel u1 del x2+del u2 del x1 whole square+del u2 del x3+del u3 del x2 whole square+del u1 del x3+del u3 del x1 whole square. I am afraid we are running out of space in the board but somehow we have to manage. So then let us simplify this further. So tau ij del ui del xj=-p del uk del xk+mu2/31/2ů, 1/2 is not required, yes, because del u1 del x1 square and del u1 del x1 square makes it 2 del u1 del x square, yes, so 1/2 is not required. Then the last term that is +ů So this entire term in the square bracket, this entire term in the square bracket, you can see is positive because it is a collection of some squares. This term this is called as a viscous dissipation function or phi. Viscous dissipation function. So this phi is greater than equal to 0. So this is -p del uk del xk+mu phi, okay. So let us get back to the slides to see, to summarise up to this stage. So we have derived the energy equation in terms of the internal energy where you have the tau ij del ui del xj term. And then the tau ij del ui del xj term, we have written in terms of the velocity gradients and the viscosity for the Newtonian and Stokesian fluid, of course with homogeneity and isotropy in addition in consideration. So you have basically substituted the viscous dissipation function here so the tau ij del ui del xj=-pdivergence of velocity+mu phi, okay. Now let us simplify this further. So what was our governing equation? Rho, del del t of rho i+del del xj ofů=-del qj del xj+Q triple prime+tau ij del ui del xj, right. So that is now -pdel uk del xk+mu phi, okay. Now our strategy will be that we have derived the equation in terms of the internal energy but normally we as an engineer or even a scientist, we do not measure internal energy. Internal energy is not a measurable quantity. Like we have thermometer to measure temperature, we do not have internal energy meter to measure internal energy. So what we will do is, we will try to convert this expression in terms of internal energy to an expression in terms of temperature and that we will do via the definition of enthalpy. So the enthalpy h=i+p/rho, like the common thermodynamic notation is h=u+pV, right. The specific volume is 1/density. So you can write rho i=rho h-p. So left-hand sideů So rho i is rho h-p. Here also we have substituted rho i as rho h-p, okay. So then you can write this as del del t of rho h+del del xj of rho ujh-del p del t-p del uj del xj-uj del p del xj, okay. This term we have written in terms of a product rule. Then these 2 terms together, how it can be simplified if you take help of the continuity equation. Now it should come to your practice, like immediately you can say. Rho capital DDt of h. Then - of, now del p del t+uj del p del xj, what is this? del p del t+uj del p del xj, what is this? This is the total derivative of p. So -DpDt-p, del uj del xj, you can write as good as del uk del xk, okay. Now you see the left-hand side and the right-hand side, the del uk del xk here gets cancelled with the, p del uk del xk gets cancelled with p del uk del xk. So this type of this term will not give rise to any effect in the enthalpy change because on both sides, it is cancelled. So you are left with rho DhDt-DpDt=del qj del xj+Q double prime+mu phi, sorry Q triple prime+mu phi, okay. So we have achieved one step. We have reduced the energy equation in terms of internal energy to energy equation in terms of enthalpy and then in our next class, we will reduce this in terms of temperature. So that will give rise to the well-known form of the energy equation. So to summarise, let us go to the next slide. I think or this slide itself which is there in this, in the view graph. So if you use h=i+p/rho, so you have rho DhDt-DpDt=Q tripel prime-del qj del x+mu phi. So this is the generalised thermal energy conservation equation. In our next lecture, we will take up from this equation and then write h as a function of t and p. We will discuss in detail that when we can write h as a function of t and p and when we cannot write h as a function of t and p. We will assume that we can write h as a function of t and p under those conditions when it is valid and then we will proceed to the derivation of the energy equation expressed in terms of temperature. So let us stop here today. Thank you very much.