 So, let us get started with the energy equation same as Newton's second law we will invoke thermodynamics what does first law of thermodynamics say. So, increment change in the internal energy of the system is equal to heat transfer to the system plus work done on the system. So, if I take a total derivative this internal energy per unit mass of the system consists of the kinetic component kinetic energy component for those of you who cannot see this it is u squared plus v squared by 2. I am taking two dimensional I am not taking three dimensional because it becomes too lengthy and thermal internal energy is given by C v into t. So, if I write this internal energy change and the heat transfer to the system and to the work done on the system in terms of moving system that is if I apply the control volume approach that is if I apply the substantial derivative I get dE by dt equal to dq by dt dw by dt note that this is substantial derivative note that this is total derivative. So, now if I do that remember what was there on the left hand side dE by dt so dE by dt that is if I have to write rate of increase of energy E of fluid element equal to rate of heat transfer to fluid element and rate of work done on the fluid element by surface and body forces. See where is this derivation I have taken from we have taken this derivation from if anyone wants to see the original where this derivation has taken from this is boundary layer flows by Sebasi and Bradshaw it is a wonderful book if anyone wants to read and understand much about convective heat transfer this is the book I do not I forget exactly the author's name Sebasi and Bradshaw but sorry the title I think if I am right it is boundary layer boundary layers boundary layer flows I am not too sure about this if someone is really interested if you put the question in model I will put it across in the model. So, this is rate of increase of energy E of fluid element equal to rate of heat transfer to the fluid element plus rate of work done on the fluid element by surface and body forces but then I have to apply this for control volume. So, applying Reynolds transport theorem rate of increase of E in CV it is like you remember always this business of E dot in minus E dot out plus E dot G equal to E dot S T essentially I am doing that only nothing new you should not get lost at all this is rate of increase of E in CV and rate at which E is entering through the surface of CV and rate at which E is leaving leaves through surface of CV and now who is doing the heat transfer heat is transferred to the fluid element by conduction and who is doing the work who can do the work only the forces which are acting and we had realized in the momentum equation derivation the forces acting are surface forces and body forces. So, those are the rate at which the surface and body forces are doing their work on my control volume is this rate of surface of surface this is that now let us take only the left hand side only the left hand side. So, understand this left hand side what are we up to that is the rate of increase of internal energy in the control volume rate at which the internal energy enters and rate at which the internal energy leaves the best way to do is to take the control volume that is what I have done in this next transfer. So, if you do this this is the control volume I think we are all well familiar with this control volume approach now delta x delta y delta z. So, the internal energy entering is E but how much it can enter mass flow rate in the x direction into the internal energy per unit mass that is rho u d y d z that is rho area is d y d z into velocity rho. So, this is the internal energy entering getting out is by Taylor's expansion series by now you are well familiar with this the same term rho u e d y d z plus rho of rho e u upon del x d x d y d z is that now similarly in the y direction I have rho e v d x d z and rho e v d x d z plus del by del y of rho e v d y d x d z and there is a mistake here del by rho what is the rate of increase of e in c v this plus is missing here and remove this equal to sign and put plus here. So, that is del rho e by del t what is this term what is this term rate of increase of e in c v. So, this is del of rho e by del t plus why I have only one term because this is this term is plus minus of this term that means this term and this term get cancelled out similarly this term and this term get cancelled out I am left out with del rho e by del t plus del rho e u by del x plus del rho e v by del y d x d y d z is that what is this term this is the left hand side of this my energy equation which I have formulated let us keep it like that let us keep it like that now let us take that the same thing and expand that what am I doing here. I am expanding keeping rho constant differentiating e keeping e constant differentiating rho keeping rho u keeping e constant and differentiating rho u keeping rho u constant and differentiating Similarly, keeping E constant differentiating rho v and keeping rho v constant differentiating E. So what do I if I group this that is I take this term then this term and then this term and take out E what is this term this is nothing but my continuity equation. So this whole of this vanishes. So I am left out with rho del E del t plus rho u del E del x plus rho v del E del y. I am not going to keep telling all the time d x d y d z because it is there in the back of the minds all the time. So I have rho del E del t plus rho u del E del x plus rho v rho E del rho del E by del y. So what is this term what is this term E is changing not only with time but also changing with space x and y by virtue of velocity u and v. So I have this is the total derivative. So rho d by dt of E what is E I told in the first transparency E is sum of internal energy and thermal internal energy I am using for both the terms capital E I am saying internal energy but small e is thermal internal energy because virtue of by virtue of temperature that exists. So E is c v t for those guys who have forgotten that plus u squared plus v squared by 2 because I have taken only 2 d. So this is my what is this this is my left hand side that is the rate of what is the change of E with respect to space and time we have accounted for but now what is that I need to account what is that I need to account if you remember what was that we said rate of heat transfer into c v by conduction that is the first term on the right hand side of my energy equation simply nothing we are doing you do not have to get carried away or lost with these equations. So for those guys who are forgotten why are we deriving this equation we will why are we deriving this energy equation for getting temperature distribution from momentum equation we intend to get velocity profile or velocity distribution from energy equation we are supposed to get from energy equation we are supposed to get temperature distribution why do I need temperature distribution H remember H equal to minus k del T by del y at y equal to 0 upon T s minus T infinity to get this temperature gradient we are setting up an energy equation so that we get temperature distribution let us not forget this let us not forget this okay. So now we get back to rate of heat conduction into control volume by conduction so this is I guess pretty straightforward why I say straightforward because essentially this is what we did in heat diffusion equation. So q x dot is the heat transfer in the x direction that is entering q x dot into area dy dz and q x dot dy dz plus del q x dot by del x dx dy dz similarly q dot y dx dz and q dot y dx dz plus del q dot y del y dx dy dz now what is q dot x what is q dot x actually we should be using q x double dash but I am replacing the using these notations please bear with me for these notations so that is minus of y this minus where did this minus come from in minus out so this minus this becomes these two terms get cancelled out and I have minus del q x dot by del x now what is q x dot by Fourier's law of conduction minus k del T by del x plus del by del y minus actually minus sign is there for the whole bracket remember that del by del y minus k del T by del y dx dy dz negative sign arises because heat transfer is counted as positive in the positive coordinate direction okay so this is just a reminder so this is what I get from the for the rate of heat transfer into control volume by conduction now let me take up rate of work done by pressure forces so the term is if you remember rate of surface of surface and body forces doing work on control so first let me take surface forces means what are those pressure force then I have normal stress then I have shear stress then I have to take body force similarly I have so I will take first pressure force work done by pressure force so I have the control volume okay so what is the pressure force what is work done by a force what is work done by a force force into velocity that is what I have written rate of doing work is force into velocity force is what in the x direction because of pressure p into dy dz into u okay so actually there is a important concept here we need to emphasize students people ask always sir pressure does not have direction very right if we study Pascal's law if you remember in fluid mechanics we study that in Pascal's law px equal to py equal to pz equal to p pressure is independent of direction that is what Pascal derives but what are we saying what are we saying here is it pressure in x direction no it is the work done because of pressure force and by virtue of velocity it is taking this direction so please remember this when you are telling this it is not the pressure which is having the direction it is the velocity which is giving this direction so work is being done okay so fine rate of work doing work is p dy dz into u so this is what is acting in this direction but again which is acting in the inward direction someone on the other day when we were signing off on Friday we were asking why pressure is both on the inward direction for temperature and heat flux and all you are writing one inward and one outward pressure can act only inward or outward we take by notation then it cannot become pressure no if pressure is inward here and pressure is outward here then nothing is acting so pressure can occur either when it is inward or when it is outward so it has to be either inward or outward this is true for normal stresses also so even for shear stresses it can shear only when it is there in opposite direction shearing is essentially because of either tensile or compression but acting in different direction so it has to be in opposing direction so please note this please note this do not get confused we are doing whatever we feel like we are not doing that okay so p into dy dz into u and here p u dy dz plus del up by del x dx dy dz similarly for y direction p dx dz into v pv dx dz plus del vp by del y dx dy dz if I do this budgeting this is this is positive so this and this get cancelled out I am ending up with minus del up by del x remember why this is taking negative sign because it is in the negative x direction so similarly I end up getting minus del up by del x plus del vp by what is this this is the work done by pressure force in x and y direction now now what is the work done by the normal stresses so we have normal stresses and shear stresses sigma xx sigma xy sigma xx sigma yy and sigma xx are the normal stresses sigma xy sigma yx and sigma xz sigma yz are shear stresses so the way we did for pressure again we are doing for energy sorry for normal stresses rate of work done by normal stresses only thing here is notation is there is again a mistake here this should be this arrow mark should be outward this should be outward this is again this also should be outward that is this should be downward and this should be left side this arrow mark this is a mistake because of this is simple copy paste mistake copy paste is a major problem okay so because I just pasted the pressure thing and that is what I did not bother to change this arrow sorry for that so this is sigma xx so same thing if I put sigma xx dy dz into u sigma xx u dy dz plus del u sigma xx upon del x all this volume similarly you have in the y direction so what do I end up this term this term this term this term get cancelled out with each other and I end up with del u sigma xx by del x plus del v sigma yy by del y note that these two terms are now positive unlike pressure because the notation for both are opposite that is all okay so that is why this has become positive now what is the work done by shear stress so you have the shear stress sigma yx minus sigma yx velocity here is velocity here is v velocity here is v area is dy dz and if I expand this by Taylor's expansion I get sigma yx plus del yx del x into dx into v plus del v by del x into dx dy dz so when I multiply this I am going to get sigma yx v del y dz that gets cancelled out with this and I get multiplication of these two terms that I am neglecting and that is it that is it and then I have what are the other terms that is it so the other terms get neglected because they are small so similarly I have minus sigma xy into u dx dz I get four terms sigma yx v dy dz that is what is this that gets cancelled out and the second term is v sigma yx del x dx plus sigma yx del v del x dx these two are coupled and written as del v sigma yx by del x plus del sigma yx by del x into del v by del x into dx because these two are pretty small gradients multiplication of two gradients we are neglecting so essentially this is two term one term get cancelled out and the other term we are neglecting this is reducing to two terms that is what professor wants me to emphasize okay so similarly here I get del u sigma xy by del y remember this is work done by work done by shear stresses then so next is we have to take stock of everything so this is what we got for for the rate of increase of internal energy this is the heat transfer and these three terms are work done by this is work done by pressure force work done by normal stresses work done by shear stresses and this is these two terms are work done by body forces so remember here body force is per unit volume so I have to multiply by rho u to make it work done okay so rho u fx is the work done by body force in x direction rho v fy is the work done by body force in the y direction now so this is the whole term you do not have to lose heart because of so many terms so now we have the full equation so now it is just the algebra there is no thinking business now it is just the algebra I will tend to go little fast now so now let me take just for the heck of it let us keep this is a equation a keep this in the back of your mind now let me take the momentum equation rho du dt equal to minus del p del x plus del sigma xx by del x plus del sigma yx plus del y plus del sigma zx plus del z plus rho fx now let me multiply this by u and divide it by rho okay so minus u by rho del p del x plus u by rho del u sigma xx by del x plus u by rho del sigma yx by del y plus u fx this is what I get now what do I do with this so u del u by del t the same thing I can write this is du squared by 2 dt squared 2 u 2 2 get cancelled I will be getting u du by dt so I can write this is du squared by 2 you might be seeing why am I doing this I am doing this circus essentially to write this equation because I have a term here u squared by 2 I want to do some circus there so that is the reason I am writing this as du squared by 2 so I get whole of this so similarly I can write v dv by dt as dv squared by 2 dt and this is written from y momentum equation if I add these two equations I get d by dt of u squared by 2 plus v squared by 2 additional term that is 1 minus 1 by rho u del p by del x plus v del p by del y plus u by rho this term this term and then this term and this term so now what am I doing I am deducting this equation from equation a that means this is going to be if I deduct that I get on the left hand side u squared plus v squared by 2 get cancelled so I get d e by dt and what do I get plus the other terms are anyway there that is these terms are anyway there these are not going to go anywhere these two terms are there that is what I have written here these two terms but the only thing is that I have made a simplification here what is that I have taken k is independent of direction and pulled the k out so k by rho del square t by del x squared plus del square t by del y squared what is the other thing I have done can you notice that what is that other thing I have done that is u del p by del x plus del v p by del y and I have here on the right hand side u del p by del x plus v del p by del y if I expand that I will be getting two terms will get cancelled out I get p by rho del u by del x plus del v by del y so I do not think I think I will give up doing the algebra on the transparency I do not think I can do that so I would request you to please please I cannot do that a minus b you do and see it for yourself indeed these are the terms you are going to get so that is what I am going to get that is this is these are the terms you are going to get so what is sigma xx we have not yet substituted sigma xx sigma yx and sigma yy we already know sigma xx sigma yy if I substitute that if I substitute all those terms it reduces to this essentially what you can see here is that they are nothing but the velocity gradients I think I have written the same thing little bolder no this is these are the velocity gradients so this is nothing but this is what is called as phi this is what is called as phi viscous dissipation because these are all velocity gradients and that is there with viscosity multiplied if I substitute that if I substitute that I get do I have reached the goal where I am supposed to reach have I reached the goal where I am supposed to reach do I see everywhere temperature no where I do not see temperature in e d e by d t I do not see temperature so I need to write this d e by d t in terms of temperature so that is what I am going to do so that is e that is h equal to e plus p by rho I think we all know h equal to e plus p v I am not writing it as specific volume I am putting specific volume as density so if I take total derivative d h by d t equal to d e by d t plus d p by rho by d t if I expand this d p d of p by rho by d t I get 1 by rho d p by d t minus p by rho square d rho by d t so 1 by rho d p by d t is the plus what is this p by d rho plus p by rho square d rho by d t is what d rho by d t continuity equation says d rho by d t plus rho del dot v equal to 0 remember whatever equation I am deriving I have never made any simplification till now incompressible this equation is still valid for compressible so for this is the compressible this is the general equation continuity for d rho by d t I can substitute as more minus rho del dot v so that is minus into minus becomes plus p by rho square rho into del dot v is del u by del x plus del v by del y equal to 1 by rho d p by d t plus rho rho gets cancelled p by rho del u by del x plus del v by del y so now what am I going to do I will substitute this for d h by d t d h by d t equal to I will substitute this complete thing for d e by d t I can substitute as d h by d t minus d by d t of p by rho in my energy equation this energy equation is what I have borrowed from previous transparency so that is d h by d t equal to this is all d e by d t that is this term and this is this term plus I am not writing all of this I am just writing that as phi by rho because this is my phi plus where did this come from this is nothing but my d p by d of p by rho by d t which is given by these two terms that is what I have written so this term and this term gets cancelled out and I get d h by d t equal to k by rho del square t by del x square plus del square t by del y square plus 1 by rho d p by d t plus phi by rho so yes d last step we are there we do not have to be really impatient ok so that is d h by d t equal to k by rho I think at least this step we shall write we shall write that is I want all of you to write along with me so that we do not lose each other d h by d t d h by d t equal to k by rho k by rho del square by del square by del square t by del x square plus del square t by del y square plus plus 1 by rho d p by d t plus 1 by rho d p by d t plus phi by rho do not forget phi is nothing but viscous this patient term it just contains viscosity and velocity gradients that is both normal and also the shearing ok so now what is that I need to substitute I substitute for h equal to C p into p ok so if I substitute that what do I get rho C p of course I am taking this rho C p as constant property I get d t by d t equal to do I do that let me check it out yes for constant specific heat and for constant density I get this because these are I am taking it out now so I get k by rho only k right k into because I am multiplying by rho sorry see what am I putting h equal to C p into t so that means when I am pulling that C p outside all that I am assuming is C p constant not rho as constant I am just multiplying by rho why I said that very carefully because I have still now not made any assumption that my flow is incompressible that is very important that was the wrong statement what I did 2 minutes back so I have just multiplying by rho that does not make rho is constant ok so rho C p k del square t by del x square plus del square t by del y square plus d p by d t plus phi by plus phi ok so this I think at least we can realize the way how we see now let us expand this d t d t term also rho C p what is this d t I think I had taken the last time example you should be able to write this very easily del t by del t plus u del t by del x del square t by del x square sorry plus del v del t by del y equal to k del square t by del x square plus del square t by del y square plus d p by d t plus phi so what is this term this is the unsteady term what is this term this is the unsteady term what is this term this is the convection term remember I had taken an example where in which I move myself from an outside hot air room to AC room my velocity is u and del t by del x is the velocity gradient which I experienced because of my movement into an AC room that is this convection term so that is why these are called as convection term why they are convection terms because temperature gradient is getting convicted by virtue of velocity by virtue of velocity and what is this term this is conduction and this is pressure term and this is viscous dissipation term viscous dissipation term so let me come back here this is written it nicely here convection term conduction term pressure work and convex viscous dissipation term and now I am handing over no do not worry about these two terms you can very easily see these two terms you will understand the importance of these two terms now professor Arun is going to non dimensionalize these equations and explain as the significance of these terms so do not worry keep your question for these two terms for a while professor Arun is going to teach us principle of similarity. Good morning just take up little bit from where professor Prabhu had left see energy equation was derived with the idea again just to reemphasize in couple of minutes that we want to find the temperature field as I had told in conduction also if I get the temperature distribution I can get the gradient so I should not have equations in terms of derivatives of enthalpy derivatives of internal energy so on and so forth. So, the dependent variable the primary dependent variable should become temperature that is the first thing second thing is I just remembered fleetingly when professor Prabhu was doing this why is this not an acceptable form of the energy equation this part I convert this to temperature what about these things my primary dependent variables in all continuity momentum equation are velocity u v w if you are talking of three dimensions and pressure. So, I do not like all this sigma x x sigma y y tau x x tau y y I do not like all those things because my primary variables are u and v and w. So, these also necessarily have to be written in terms of the velocity gradients or velocity where derivatives of the velocity with respect to either x y or z that is why we go into so much details to convert all this to velocities and temperatures where energy these will be converted to temperature these will be converted to the velocity gradients. So, now what do I have as a result of all this what do you call algebra and you know one hour of discussion I have one equation which essentially tells me how energy how heat is going to get transported from one part of the fluid to another and we said that this involves convection which is convection plus the conduction part there is a pressure work we will see what this is there is what is called as a viscous dissipation. Discipation as we all know is nothing, but when we use the word dissipated means it is lost it is going as a waste. So, this viscous dissipation term essentially is where does it come from let us just go back to the physics where does this phi come from this phi comes from this sigma etcetera which is here. So, in other words look at this phi this phi is gradients of velocity either in the direction of the velocity or in the direction perpendicular to the direction of velocity. So, I have d u by d x d v by d y d u by d v by d x so on and so forth. So, these are essentially the shear normal stresses and so on and so forth what are they caused because of they are caused because of viscosity. So, friction as we know from high school we have been studying friction is bad because it involves loss of energy it involves loss of power and our simple example will be from riding a car or a vehicle or whatever pushing a block on a table you will say friction larger the friction greater is the power that I have to supply it is going to go as a waste. So, here when I say this is energy this is the shear and normal stresses the work which is being lost because of the shear and the normal stresses can this be prevented no this cannot be prevented because one layer of fluid is going to slide over another layer of fluid one there are bound to be rotational aspect there are bound to be angular deformations of the fluid whenever the fluid is moving. So, maybe some of the terms here can become 0, but this term can never become equal to 0. So, what we are trying to say this phi is an we can we can we eliminate friction we cannot eliminate friction. In fact, friction needs to be there in many applications because otherwise things will just slide. So, in order that friction is there we have to pay a penalty something useful is coming out of friction. So, we have to pay a penalty that penalty is this loss of energy which we call as the viscous dissipation the word viscous implies it is coming because of viscosity dissipation implies it is a loss it is going away it is useless you cannot retrieve it back it is irreversible. If I have to go to thermodynamics and say it is an irreversible irrevocable loss of energy it is like you know to maintain a credit card many times you have to pay an annual fee that annual fee gives you the privileges of using that credit card wherever you want. So, that annual fee cannot be got back same kind of thing to have this flow to have this energy transport there has to be a penalty there has to be a loss which we call as the viscous dissipation when will this be small this will be small when the velocity gradients are small that means when the velocities are reasonably small the gradients are expected to be small. So, probably in those cases these terms can be small they will never go to 0 they can be neglected, but they will never be equal to 0. Then other situations where the viscosity is much smaller very very small flows in our fluid mechanics we have dealt with potential flow solutions or where viscosity is neglected in viscid flows that we have seen in that case with where mu is equal to 0 or very very small I can put this equal to 0. So, what am I trying to say all this physics of this is very very important because in most UG textbooks undergraduate universities colleges I do not think this equation is derived I may be wrong, but we do not derive this, but we do not even present this probably, but I think we have to at least show this present this and probably spend some time understanding what this is. So, this is the motion energy transport because the fluid is moving this is the energy transport because of conduction even when velocity equal to 0 even when the fluid is brought to rest 0 velocities this thing will go to 0 this thing will go to 0 this material derivative will go to 0 because this involves d p by d t so on and so forth pressure is there only when there is a flow you can look at it like that or you can look at it from the fact of expanding this bracket d p by d t plus U d p by d x plus V d p by d y. However, way you look at this this term is going to be 0 when velocities are 0 and the fluid is stationary phi again remember this phi was all viscous dissipation grotten due to velocity gradients. So, when the fluid is stationary this whole thing is going to go to 0 what do I get back I get back my nice beautiful conduction equation two dimensional k d square t by d x square plus d square t by d y square is equal to rho c p d t by d t bring this k down here I will get d square t by d x square plus d square t by d y square is equal to 1 by alpha d rho d tau d t by d time this was what we derived in the first or second lecture heat diffusion equation that was not from did not jump from anywhere it is essentially this energy equation where all the velocities all the flows were brought to 0. So, this general energy equation we cannot say it is only conduction base or convection related or it is the most general equation and again where did this come from this came from your first law of thermodynamics. So, what you have seen in thermodynamics is this now I am putting first law of thermodynamics was q minus d e by d t is equal to q dot minus w dot plus so on and so forth that q dot and w dot were all black boxes you are supplied q's or w's you had to calculate other quantities. So, you did not care where this q was lost by conduction convection radiation now we are saying I know little bit more than what I knew when I knew thermodynamics. So, I will write expressions for q and w what are these w's these w's are cause because of the shear and the normal stresses. So, with this so much of talk on this energy equation I hope the physics of this thing is clear that in our sleep after doing this heat transfer course I think we have to be able to write this one at least this one 3 dimensional if you cannot write it is 2 dimensional we should be able to write and a word of caution all participants have to have to the slides are there with you it is about few pages of algebra we urge all of you to please go and derive this equation ok what we will where we will ask you during the final test on Friday you will definitely see a question related to energy equation in some form ok. So, we urge all of you even though this is given to you a priory please derive every step put in the maths here fill in the blanks those of you who are not participants in sitting here also should derive this equation step by step you come up with the final form substitute for these sigma's x sigma y put it here and indeed show that viscous dissipation is like this and then come up with this final form after all the algebra we want you to do this we urge you to do this we will test you on this also ok. So, now we will go to the ok I have got temperature distribution a solution of the what kind of a differential equation is this this is the real messy equation meaning it has independent variables x y and t dependent variable is temperature also let me go back to viscous dissipation phi is here this one, but what is phi that phi is this part correct. So, if I want to rewrite this equation I will write this as rho C p d t by d t plus u d t by d x plus v d t by d y equal to k d square t by d x square plus d square t by d y square plus d p by d t plus viscous dissipation I do not remember directly I will write it here from document and you will see that mu d square u by d x square 2 mu d u by d x square plus 2 mu d v by d y plus mu d u by d y plus d v by d x whole square minus 2 by 3 mu this one minus 2 by 3 mu d u by d x plus d v by d y whole square ok why did I write this just a minute we will see why did I write this this one is written because we want to try to understand what are the dependent and independent variables. So, my continuity equation this is my energy equation continuity equation was d u by d x plus d v by d y is equal to 0 if I call it unsteady or if I want to write it in this form I will write d rho by d t plus I will write it in this form. So, my variables are if I look at this temperature u and v. So, I wanted to derive an energy equation where temperature was the independent was the dependent variable I wanted to get what temperature distribution to get what heat transfer coefficient can I do it directly if I have a differential equation d square y by d x square plus d y by d x plus y equal to 0 where I have only independent variable x and dependent variable y I can solve to get y. Right now if you see here this is an x equation which involves three variables dependent variable z t which we have not seen before, but u and v which we have carried forward from continuity and momentum equation. So, I cannot solve this independently what does that mean I have to be able to get u's and v's from the momentum continuity equation to substitute that here substitute for u's and v's here then I will get differential equation which has only temperature as the dependent variable. So, very very important thing from fundamentals point of view is that this continuity equation gives me u v can I solve this and get both no one equation two unknowns not possible to get both very good then momentum equation x momentum y momentum it gives me u p and p. So, this I can probably solve three equations x y momentum continuity equation three unknowns what are the three unknowns u v and p. So, it is the closed system of equation when I have number of equations is equal to number of unknowns I can solve it yeah solution techniques might be difficult might be a lengthy, but we can solve it on paper at least energy equation now gives me what is it a vector or a scalar it is a scalar. So, I have only one equation it looks from that point of view energy is good it has only one equation momentum is a pain because I have x y z three equation vector quantity here I have t u v what else pressure. So, I have all these things. So, typically what I have to do is these have to be solved together simultaneously either I should get the flow distribution flow field substitute for the flow distribution and then get t as the function of x y z and t. So, we cannot just get t independent what I am trying to say conduction nothing was there. So, I could get t directly here I need to be able to understand the physics of the flow from these two equations continuity and momentum get the velocity distribution substitute that here and then be able to get the temperature distribution. Once I get the temperature distribution I can do the heat transfer coefficient etcetera. So, that is what we will do what are you trying to do now first we will simplify this and try to throw away terms as we have done because we do not like to carry such a big equation with so many terms. So, first assumption we make is steady flow. So, steady flow means this whole thing does not become zero please bear in mind for those of you who are unable to see this professor has already expanded this on the white board let us go back here. This is the expansion of d t by d t this expansion when I say it is for steady flow this term will go to 0 these two will remain. Similarly, d p by d t when expanded will have partial derivative of pressure with respect to time that will go to 0 these terms will remain. So, I will have this form of the equation. Now, when I non why do I need to non dimensional what is the reason for this non dimensionalization because normally what we try to do is in any system of equation when we have to solve we have to take into account various non dimensional quantities which might affect the distribution. So, in fluid mechanics we knew Reynolds number affected the distribution that Reynolds number etcetera how did it come. So, that is what we are trying to see if x star this is again just algebra. So, we will go quickly x star is x by L y star is y by L what is this L you will leave it for now this L is a characteristic length scale this question was asked in the Moodle forum why do we take L what is the characteristic dimension for Nusselt number and bio number are they the same one was v by a surface area another we use L or d e depending on what we want it is not d by 2 d by 3 etcetera why the question is answered by this see this is a representative scale of in the x direction this characteristic dimension it has nothing to do with the volume surface area etcetera that bio number definition of characteristic length was a result of writing energy transfer by convection on this outside to energy transfer by conduction through the solid you came up with v by a s and that v by a s represents a length because it is of the units meter we call it as a characteristic length and we give it different things for plane wall L by 2 for a mean plane wall of thickness L it will come as L by 2 so on and so forth. So, that has nothing to do with this this is a length scale. So, what could what could be the length scales for a flat plate of length L of 1 meter that 1 meter is a representative length scale meaning the physical dimension which is important that is the length scale I cannot have you know size of this room when I am doing an experiment you know of the where the plate is of this dimension because it has to be related to the problem. Now why in a pipe flow diameter is the characteristic dimension and length is not the characteristic dimension that we will see when we go to internal flow do not break your head on that our request to you actually most participants is put questions on Moodle for things that have been covered till that day because most questions I have answered I have been please wait we will see that later because many times writing an answer is basically reinventing the wheel since we are going to study it here anyway. So, u star let us scale the velocity by u star is equal to u by v now what is this v there is a question again on that what velocity is this flat plate we have understood we know we are acclimatized with what we call as free stream velocity. So, that could be the free stream velocity in the internal flows is there a free stream velocity no there is no free stream velocity fluid will come in. So, and it will become some kind of a velocity profile will be there and this u I mean this v could be the average velocity. So, we have used all these unknowingly we have used root force we have used now we are going to see why we are using what we are using then pressure p star is non-dimensionalized by half rho v square what is this half rho v square we could have used one half rho v square, but we are just using rho v square this rho v square represents what all of us know this all static dynamic all these things right. So, this represents the dynamic pressure and p star p star is a non-dimensional temperature scale what is that how many temperatures are involved in the heat transfer problem that is very important in fluid mechanics we had u local u free stream we did not have u surface because that was 0 we because of no slip condition even the if the surface had a velocity by relative velocity concept that could be brought to 0 here is therefore, we say you had a local velocity u you had a free stream velocity u infinity only two variables whereas, in heat transfer unfortunately I have three variables local temperature surface temperature free stream temperature t infinity. So, these have to be taken together and why do we take this who has given me this bright idea that it has to be like that this bright idea essentially comes from this fact that p star somebody has told me this is t s minus t local divided by t s minus t infinity why t s minus t why not t minus t infinity. So, the answer lies in this fact. So, if I have a flat plate which is a temperature t s fluid is coming in at a well at free stream temperature t infinity cold air at 30 degree centigrade is flowing over a hot plate which is at 100 degree centigrade classic simple problem there will be some kind of a boundary layer formed we have understood all that there will be a temperature distribution which will go like this is also we have understood what are we saying this is the surface temperature this at any point is the local fluid temperature t infinity is something which is on the outside unaffected temperature beyond the boundary layer. So, this t minus t s is representing what a local temperature difference between what this is a local temperature everybody agrees or between what is what is necessary for heat transfer for heat transfer to occur there has to be a temperature difference this denominator represents the maximum possible temperature difference that the fluid can experience this represents the local temperature difference that the fluid experiences between when this fluid at say 45 degree centigrade and this is at 30 this is at 100 what we are saying is this fluid cannot carry fluid particle at this temperature 45 cannot carry as much heat away with it as a particle at 30 degree centigrade can carry that is the heat transfer characteristic because of a higher temperature it has lost some of it. So, this temperature ratio is going to always be between 0 and 1 that is the way we look at non-dimensional quantities we non-dimensional is typically such that the non-dimensional parameter lies between 0 and 1 that is a general rule such that the non-dimensional parameter lies between 0 and 1 at the surface what are we going to have at the surface we are going to have this become equal to t s minus t s 0 at the free stream t infinity minus t s divided by t infinity minus t s non-dimensional temperature becomes equal to 1 take the case for velocity free stream u same as v it will become 1 that is the way now what I do is I substitute for we have done non-dimension before why I am not going to redo this whole thing as we did because it is essentially algebra I will just show you one term you can do the other things from point of view of homework. So, first thing that we will see is this rho C p u d t by d x I will show and then the remaining you can do rho C p u d t by d x this is what we are going to non-dimensionalize we have u star is equal to u by v. So, therefore, I will write u is equal what is my aim I have I want to throw away all the variables u v t x etcetera and write it in terms of a non-dimensional quantity. So, this I will substitute what else is there x by l is x star. So, I can write d x is equal to I will write it a little bit d x is equal to l times d x star this is one thing third thing is T s minus t divided or t minus T s is equal to t divided by t infinity minus T s is t star why cannot it be the other way it can be the other way T s minus t divided by T s minus t infinity I keep it this way because I want t easily they cannot without doing algebra for minus sign. So, this will tell me T infinity minus T s is constant. So, I can write this like this T infinity minus T s times t star differentiate this d t is nothing but derivative of T s is 0. So, this goes to 0 derivative of this quantity is 0 T infinity minus T s is a constant. So, that stays as it is I get d t star. So, now I am going to substitute all this here. So, rho C p u instead of u I am going to use v times u star d by d x is replaced by 1 by l d by d x star is that correct d x is been replaced by l d x star from here what am I left d t d t therefore, is going to be d t star times t infinity minus t s I will just unfortunately keep it like this this I can write therefore, as rho C p v t infinity minus t s divided by l u star d t star by d x star is that correct next term I will get second term of the equation I will get rho C p without even seeing I am going to write let me see if I do it correctly rho C p v t infinity minus t s divided by l v star d t star divided by d y star this is the left hand side of my energy equation is that correct let us see rho C p v this one please look here rho C p v t infinity minus t s by l u star d t star by d x star plus v star please note here please note this term second term of the first bracket left hand side the stars are missing stars are missing this is v star d t star by d y star and that missing gets carried over all through this derivation. So, please make a note this term does not have a star by mistake please write it and the right hand side if I non-dimensionalize I will get similar kind of term what am I trying to do we will just see in a minute viscous dissipation I have expanded I have written those 6 terms associated with viscous dissipation that also needs to be non-dimensionalized. So, I will substitute all the v's and u's and so on and so forth and try to non-dimensionalize this d u by d x therefore, would become v d u star by l d x star that is what is there square of that you will get v square by l square. So, this whole thing is a non-dimensionalized viscous dissipation this is the non-dimensionalized conduction term this is the non-dimensionalized pressure work term how did you get v square because pressure is non-dimensionalized by a factor which has v squared already and then there is a u which will have a v when it is non-dimensionalized. So, I have rho v cubed coming in here and this is phi star mu v square by l square times a bunch of terms associated with that what is that bunch of terms associated with that what will happen to that we will see. So, this is the non-dimensional form what is the u let us just do the algebra I do not like all this even after non-dimensionalization I would want the equation to look in a similar form. So, what am I doing next. So, this is a rho v c p l t infinity minus t s I will take to the other side. So, my left hand side of this non-dimensionalized equation I would want like this only that is u star d t star by d x star plus v star d t star by d y star this is the left hand side I want why do I want this we will see in just a minute equal to I will take all those things several terms are there I will divide by this rho v c p t infinity minus t s several terms will be there when I do all that and do a bunch of things which I have to do for algebra I will come up with this very nice form that I will rewrite here after lot of algebra I will get so u star d t star by d x star plus v star d t star by d y star this is the left hand side of the equation what was the left hand side of the dimensional equation the dimensional equation was this rho c p let me not worry about it for now I will I will write it I will write it the dimensional equation was rho c p u d t by d x plus v d t by d y something like that so I am having now the non-dimensional part essentially the same thing u star d star x star like that I am having and the right hand side comes out to be 1 over r e p r did we non-dimensionalize the momentum equation you have not when you non-dimensionalize the momentum equation in fact you will see we will do it next step you will see r e coming in there one by r e p r d square t star by d x star square plus d square t star by d y star square plus v square by c p t infinity minus t s u star plus v square by c p t infinity u star d p star by d x plus v star d p star by d y star see I missed the star here plus phi star mu v rho c p l t infinity minus so what is the great deal that we have done we have tried to keep the equation in a similar form except bring some coefficients in front of them right hand side I had conduction term d square t by d x square plus v square t by d y square instead of that I have the star terms but in front of it I have something which I like Reynolds and Prandtl number Reynolds number rings a bell fluid mechanics somewhere we are getting connected now this was the pressure work term it had what material derivative of pressure so steady flow we drop the time derivative and then you would have had u d p by d x plus v d p by d y I have a similar form with some coefficient in front of it last term was viscous dissipation phi I have a similar term with some coefficient in front of it and the left hand side was essentially that rho c p u d t by d x term I have brought rho c p down and I have got a similar term on the left hand side what are the dimensions of this dimensions of the original equation was energy right that is what is energy it was in it was actually in terms of watts u d t by d x rho c p etcetera right conduction term all these things would give us in terms of watts whereas this is a non dimensional equation why is non dimensionality so important very important because if I have any situation if I have any situation any flow situation any heat transfer situation the non dimensionalization will give me a solution which if I cast the problem in terms of a non dimensional quantity if I know the solution of a non dimensional equation and if I can cast the new problem in terms of similar non dimensional numbers I already have a solution in place I can use it directly. So, this is the non dimensional energy equation now we will call this we will call this v squared by c p t infinity minus t s we will call this as the Eckert number e c what is this it essentially again comes from interconversion of kinetic energy and enthalpy if you realize your first law of thermodynamics for a nozzle when you wrote nozzle is a device where interconversion of kinetic energy and enthalpy occurs right h plus v squared by 2 is equal to h e plus v squared by 2 for an ideal gas delta h would be c p delta t. So, c p delta t is equal to v squared exit minus v square in let by 2 similar thing this shows the effectiveness or how good is this conversion from sensible kinetic energy to sensible heat is there and that is what is given here in the non dimensional equation Eckert number is the measure of dissipation effects in the flow that is since this grows in proportion to the square of the velocity how what is happening in a nozzle velocity is increasing enthalpy is getting converted decreasing right. So, if we just go back to nozzle that probably people can connect quite easily let us just go here I would have first law I am just writing h i plus v i squared by 2 is equal to h e plus v e squared by 2. So, this i v i squared minus v e squared by 2 or let me write it the other way because nozzle exit velocity is more v e squared minus v i squared by 2 is equal to h i minus h e which is nothing but c p t i minus t e correct. So, v e squared minus v i squared is equal to 2 c p t i minus t e what does this tell me exit velocity is more greater than 0 that means this is greater than 0 t i is higher than t e temperature has gone down as the velocity has increased. So, inter conversion of this to this has happened Eckert number is trying to tell me that only how much how is this dissipation of velocity is getting converted to heat kinetic energy is getting converted to heat velocity has no meaning here kinetic energy is getting converted to heat first law only nothing else here. So, since this quantity has v squared we are going to have this quantity very important in high speed flows. In the second term that viscous dissipation term comes out to be 5 star Eckert by Reynolds number. So, when is Eckert number important Eckert number is important in high speed flows it can be neglected for small velocities. So, incompressible flow situations that we are going to study in this part of heat transfer course we will not take this term and this term into account that means what these are negligible compared to these terms that is what we are saying. It is not equal to 0 it will be there there will be a finite value there will be an effect of velocity and on temperature inter conversion will happen Eckert number will be there, but compared order of magnitude compared to the convection term and the heat conduction term. So, viscous dissipation term and the pressure work term can be neglected. So, Eckert number simple example given velocity is 10 meter per second the temperature difference reference rate is 10 Kelvin Eckert number is 0.01 when the temperature difference is larger Eckert number still goes smaller in typical speeds now that we see water or air or whatever you will see 10 or such velocities for water smaller velocities for incompressible fluids this number would still be small therefore we can throw away these two terms and when I throw away these two terms now you will appreciate that this equation which is left with only the non dimensional left hand side what am I left with non dimensional u star d t star by d x star plus v star d t star by d y star is equal to 1 by r e p r d square t star by d x star square plus d square t star by d y star this is a much more tractable equation in terms that this is only a number in front of me as opposed to bunch of terms which were there why did we do this non dimensionalization if we had not done this non dimensionalization never in my sleep I would have known that viscous dissipation could be negligible for normal regular low speed situations nor would I have known how to carry this pressure work term now because of non dimensionalization I have understood that I can throw away these two terms at least for low speed flow situations ok. So, with this background let us just quickly go to the combined now we have to see everything in full perspective the continuity equation is here two dimensional continuity equation viscous dissipation neglected pressure work term neglected we have thought of steady flow situation incompressible laminar flow with constant properties this is my continuity equation this is my what momentum x momentum equation we will see what happens to y momentum equation rho u d u by d x plus v d u by d y is equal to mu d square u by d x square minus d p by d x and this is the energy equation just chew on this during the t break these two equations look identical if rho and rho c p you do not worry about or take the rows on that side left hand side is identical except that the form the form of the equation is identical except that you have instead of u here you have a t here you have a u here you have a t here u square by y square t square by y square t square by x square and that is the d p by d x something else is there when d p by d x goes to 0 the nature or the appearance at least of this dimensional form of the equation looks the same.