 In last two lectures, we dealt with similarity method for the velocity boundary layer. In lecture 6, we developed the similarity equation and in lecture 7, we obtained solutions to the similarity equation. Today, we will deal with similarity method for the temperature boundary layer or the energy equation. Like before, we shall first establish the condition for existence of similarity solutions to the temperature boundary layer equation and then derive the boundary conditions. As you will recall, the boundary layer form of the energy equation has convective terms on the left hand side, the diffusion term in the transverse direction, the viscous dissipation term, the pressure work term and two heat generation terms one due to chemical reactions, the other due to radiation. For our purposes presently, we shall ignore q dot chem and q dot rad. As we said earlier, U dp infinity by dx is important only in high speed compressible gas flows and therefore, as presently neglected. The last three terms are neglected because we are largely dealing with uniform property in compressible flows. The boundary conditions to this equation would be at the wall y equal to 0, t will equal t wall and that may or may not be a function of x. At y equal to infinity, t will be equal to t infinity and is assumed to be constant in the free stream. It does not vary with x. If you want to develop similarity equation, then first of all, we define tw x minus t infinity that is the reference between wall and infinity states as some function g x. We define theta eta where eta is the similarity variable introduced in the velocity boundary layer equation equal to t at any point in the boundary layer minus t infinity divided by t wall x minus t infinity and eta as you will recall is equals y into under root U infinity by nu x. If I make the substitutions of these two quantities in the differential equation, I am going to replace t in these three terms by theta. Then you will notice that d t d x will equal d by d x of t infinity plus theta into tw minus t infinity. Since t infinity is constant, it does not contribute to derivative and therefore, I would get tw minus t infinity d theta by dx plus theta into d by dx of tw because tw is a function of x. d t dy likewise will be d by dy of t infinity plus theta into tw minus t infinity and that will simply reduce to tw minus t infinity d theta by dy and likewise d 2 t dy square will simply reduce to tw minus t infinity d 2 theta by dy square. Therefore, upon substitution, you will see the equation would now become dividing through by rho C p. I will get U d t by dx plus V d t by dy equal to alpha d 2 t dy square plus mu by rho C p d U by dy square. If I substitute these three derivatives here, I will get tw minus t infinity into U d theta by dx plus V d theta by dy plus theta into U d tw by dx equal to alpha into tw minus t infinity into d 2 theta by dy square and plus mu by C p d U by dy whole square. That is what I will get. If I divide through now d t w by dx is nothing but d g by dx and tw minus t infinity is nothing but g. Therefore, if I divide through by g, I will get U d theta by dx plus V d theta by dy plus U theta by g d g by dx equal to alpha times d 2 theta by dy square plus mu by C p d w minus t infinity d U by dy whole square. This then is the last equation that you see here. We now wish to represent each term in terms of similarity variables. For the velocity boundary layer, the similarity variables are U infinity equal to C x raise to m. Eta as we saw before is y into square root of U infinity divided by nu x psi is f eta into n x a function of x. n x itself is under root nu U infinity x and we define U over U infinity equal to f dash. Therefore, U will become equal to U infinity f dash V which is equal to minus d psi by dx will become minus into f dash into n x into d eta by dx plus f into d n by dx follows from this definition of psi. d theta by dx would be simply theta dash into d eta by dx where theta dash is d theta by d eta. d theta by dy would be equal to theta dash d eta by dy and d eta by dy would be simply under root nu U infinity x. d 2 theta by dy square would be d theta double prime U infinity nu x into d U by dy whole square equal to f double prime square nu x. So, you will see that if I now substitute for U d theta by dx and so on and so forth in this equation U infinity into f dash into d theta by dx which is d theta I mean which is theta dash d eta by dx minus f dash n d eta by dx plus f d n by dx into d theta by dy which is theta dash into under root U infinity by nu x U infinity f dash theta divided by g d g by dx equal to alpha times U infinity by nu x into theta double prime plus nu by c p t wall minus t infinity into U infinity q by nu x into f double prime whole square. Now, notice that d n by dx since n is equal to under root nu U infinity x and U infinity is equal to c x raise to m and eta is equal to under root U infinity by nu x. So, I can show that n is equal to n divided by nu x which will be under root U infinity nu x. So, remember if I were to replace this n by nu x U infinity by nu x then you will see that I get U infinity f dash into theta dash d eta by dx minus f dash f dash n is equal to under root nu U infinity x d eta by dx plus f d n by dx into theta prime under root U infinity by nu x. Therefore, you will see this becomes U infinity f dash into theta dash d eta by dx minus f dash into theta dash U infinity d eta by dx plus f theta prime under root U infinity by nu x d n by dx. Therefore, you will see this term gets cancelled with this term and I essentially have now minus f theta dash this is the left hand side f theta dash into under root U infinity by nu x d n by dx plus f dash U infinity theta by g d g by dx equal to alpha U infinity by nu x theta double prime plus nu by C p T w minus T infinity U infinity q by nu x into f double prime whole square. This would be the equation you will see these two terms now get cancelled U infinity by nu x as you can see is simply n over nu x I divide through by U infinity by x then you will see you will get minus f theta prime into under root nu x by U infinity d n by dx plus f dash but it is quite correct n and n will get cancelled and this term gets cancelled with that one and minus theta dash n f d n by dx is equal to all this. So, my purpose is to then yes correct f dash theta and this will become U infinity square because I have divided through by this. So, this will go when this will go and I will get U infinity by nu x is really n over nu x and that is how this and this term gets cancelled. So, essentially you then have this term f dash theta d divided by g x and multiplying through by x I get f dash theta x d g by dx minus f theta dash x over n d n by dx equal to that is this term theta double prime divided by Prandtl plus 2 e c x f double prime square where a cut number e c x is defined as U infinity square divided by 2 divided by C p into T w minus T infinity. Now, to show this x over n d n by dx equal to m plus 1 by 2 well that is a straight forward thing n is equal to under root nu U infinity into x but U infinity is equal to C x raise to m and therefore, this becomes nothing but C times nu x raise to m plus 1 and that is equal to under root C nu x raise to m plus 1 by 2 that is n d n by dx therefore, will be under root C nu into m plus 1 by 2 into x raise to m minus 1 by 2 and therefore, x divided by n d n by dx will be equal to x divided by under root C nu x raise to m plus 1 by 2 into under root C nu into m plus 1 by 2 into x raise to m minus 1 by 2. So, you get this it is cancelled with that this is x raise to 1. So, this becomes x raise to minus 1 by 2. So, this this and this gets cancelled and you simply get m plus 1 by 2 that is what is shown here. This is the pressure gradient parameter m and therefore, our equation would become like this. I have transferred the terms all terms on the right hand side and multiplied through by Prandtl number then you will see our equation becomes theta double prime plus Prandtl into m plus 1 by 2 f theta dash minus f dash theta x d by g d g by dx which is the wall temperature variation function plus 2 e c x f double prime square equal to 0. Now, as we said before this equation will be a perfectly ordinary differential equation. If this quantity x divided by g d g by dx is a constant let us say it is gamma then the solution to g will be simply t w minus t is equal to some delta t ref yes x by g d g by dx equal to gamma gives me t g by g is equal to gamma times dx by x and the solution therefore, is g times some constant delta t ref into x raise to gamma. So, unless the wall temperature variation is of this form gamma is an arbitrary constant of course, then similarity solutions will be possible. Likewise, e c x should also be a constant, but what is e c x? e c x is u infinity square by 2 divided by c p into t w minus t infinity and this is equal to c square x raise to 2 m divided by 2 times c p and this is delta t ref x raise to gamma g is t w minus t f which is g, g is that that and therefore, e c x will be constant if this is constant then this and this must cancel which means gamma must equal to m if e c x is not equal to 0. So, remember similarity solutions to the temperature boundary layer are possible when e c is ignored that is viscous dissipation is ignored then gamma can take absolutely arbitrary values viscous dissipation is included gamma can take only restrictive values gamma equal to 2 m. This is because there is a connection between the viscous dissipation term and the velocity solution such odd constraints are put in similarity method. So, these are then the conditions for existence of similarity solution is that the wall temperature variation must follow this law must be proportional to x raise to gamma and e c x if e c x is included then gamma must equal to m. So, then our final energy equation is of this time theta double prime plus Prandtl m plus 1 by 2 f theta dash minus gamma f dash theta this is the presence of wall temperature variation this is the viscous dissipation and this is the pressure gradient and this is the Prandtl number with a reminder that where e c is equal to this and if e c is not equal to 0 then gamma is equal to 2 m otherwise gamma can take absolutely arbitrary values what would be the boundary conditions that is straight forward to see if you recall the temperature. So, at the wall eta equal to 0 theta 0 will be 1 and at infinity state the numerator will be 0. So, theta infinity will be 0 so that is what you see here. So, the boundary conditions are this and what would the solutions look like the theta eta will be functions first of all m and b f because f f dash and f double prime are functions of m and b f and in addition you have Prandtl number the wall temperature variation parameter and the viscous dissipation parameter e c this is what we expect how do we solve this second order differential equation well like we did in the case of velocity boundary layer where we had a third order equation. So, we split this up into two first order differential equations the first equation is simply d theta by d eta equal to theta dash with theta 0 equal to 1 which is known wall temperature is known and then d theta dash by d eta equal to theta prime which is equal to minus Prandtl all this, but without knowing theta dash 0 we do not know the theta dash 0. So, like we did in the velocity boundary layer equations so first of all we obtain for a given m and b f gives f f dash f double prime we then guess theta prime 0 and solve these two equations by Rangaketa method from eta equal to 0 to eta equal to eta max at each iteration we check the boundary condition at the outer edge whether theta eta max has tended to 0 or not if not we revise theta dash 0 and continue the solutions again when the solution is obtained we define thicknesses of the temperature boundary layer the physical thickness delta is remember our solution will look like this theta versus eta at wall it will be 1 and we expect the solution to go something like this with the 0 boundary condition at eta equal to some eta max. So, how do we choose the value of the thermal boundary layer thickness so just as we choose velocity boundary layer thickness and say that when f dash eta tends to about 0.99 we say that is velocity boundary layer thickness likewise in temperature boundary layer we say when theta tends to about 0.01 that is somewhere here then theta then that would represent, but there is a notional thickness there is no exactness we can impart exactness by defining another thickness called the enthalpy thickness and it is simply represents 0 to infinity rho c p u t minus t infinity which is the actual enthalpy within the boundary layer when integrated divide and this is the enthalpy that would be carried in the layer if there was no boundary layer for uniform property flow where rho c p gets cancelled with this u over u infinity will be simply f dash and t minus t infinity over t wall minus t is our theta and therefore, this relationship if I change y to eta then delta 2 star delta 2 by x into Reynolds x to the minus 0.5 equal to 0 to eta max f dash theta d eta this is the dimensionless form for a uniform property enthalpy thickness. Remember we obtained solutions by guessing the value of theta prime 0 how is it related to the heat transfer coefficient h x the local heat transfer coefficient will be q wall divided by t wall minus t infinity. Remember at any x if this is q wall t w and t infinity and this is q wall. Then h x is simply q w divided by t wall minus t infinity this is the definition you will recall and that would equal if this was the temperature profile the gradient of temperature therefore, this will be minus k d t by d y at y equal to 0 divided by t w minus t infinity but, notice I can since theta is equal to t minus t infinity over t w minus t infinity I can straight away say this is equal to minus k d theta by d y at y equal to 0 this is d t divided by t w minus t infinity and if I now change eta is equal to y times u infinity by nu x then you will see this can also be written as minus k d eta by d eta into d eta by d y at y equal to 0 and that is equal to minus k minus k theta dash 0 into under root u infinity by nu x under root nu infinity by nu x which I can develop further as h x into x by k which is what how I define the Nusit number n u x then you will see this becomes equal to minus x by k into k theta prime 0 into u infinity by nu x and you will see k and k gets cancelled and you will get this as minus theta prime 0 and under root u infinity x by nu and that is nothing but minus theta prime 0 R e x to the half and therefore I can say that N u x R e x to the half minus half is equal to minus theta prime 0 our R k solution essentially gives us the value of theta prime 0 which in turn gives us really the value of Nusit number that is what I have derived here N u x is equal to that sometimes in boundary layer theory we define Stanton x S t and say N u is Nusit named after the scientist Nusit. Similarly, S t is Stanton named after a scientist Stanton and the that is defined as h x divided by rho c p u infinity remember there is no length dimension in the definition of Stanton number and this is found sometimes quite useful in defining a dimensionless heat transfer coefficient it is simply Nusit x divided by Reynolds x into Prandtl it is a simple algebra to show that if I divide this quantity by Reynolds x and Prandtl you would get Stanton. So, then the solution is essentially Nusit x or Stanton x is a function of m v f Prandtl gamma e c and with a reminder if e c is not equal to 0 gamma is equal to m or knowing local value of h x you can always calculate average value of heat transfer coefficient and average value of Nusit number. So, here is the first solution obtained to the thermal boundary layer equation taking the case of a flat plate flat plate means m equal to 0 no suction blowing means this means b f is equal to 0 constant wall temperature this means gamma is equal to 0 no viscous dissipation no viscous dissipation this means e c is equal to 0. So, we have a very very simple equation then theta double prime plus Prandtl into f theta dash divided by 2 equal to 0 and the velocity equation would look like f triple prime plus half f triple prime plus half f double prime equal to 0. So, if I now in addition say besides these that Prandtl number is equal to 1 then theta double prime will equal 1 by 2 f theta prime equal to 0. The boundary conditions here are f 0 equal to f dash 0 equal to 0 and f dash infinity equal to 1 where the boundary conditions here are theta 0 equal to 1 and theta infinity equal to 0. You will see therefore, that there is a remarkable similarity between these two equations and if I were to say that theta eta equals 1 minus f times eta then the boundary conditions I do not have to solve the temperature equation simply the value of velocity boundary equation itself will show me that. So, you will see here earlier solution for f dash this is a velocity boundary equation for Prandtl equal to 1 and boundary layer thickness for the velocity was about 5 likewise the temperature profile theta is going sorry this should be 1 minus f dash eta. So, theta is nothing but 1 minus f dash eta and this is the f double prime 0 which is going so and here is theta prime at x equal to 0. So, this shows that perfect analogy between heat and momentum transfer exists for Prandtl equal to 1 remember Prandtl is just simply new by alpha. So, if this is equal to 1 the rate of diffusion of momentum and heat are equal and we say in such a case there is a perfect analogy between heat and momentum transfer and all thicknesses would be equal. So, thermal boundary layer thickness delta star would equal velocity boundary layer thickness delta star enthalpy thickness delta 2 star would equal momentum thickness delta 2 star minus theta prime 0 would simply equal f double prime 0 as seen in the figure here with the negative sign you see this is theta prime 0 I mean theta dash 0 and this is f double prime 0 that is what you see here. This reference case is of great value to us because we can now go on comparing the effects of m b f gamma and e c and Prandtl number on solutions. So, here are the solutions generated by Ranga-Kutta computer program and what I have now done is allowed for the variation of Prandtl number by keeping m b f gamma e c all equal to 0. So, we have case of flow over a flat plate at constant wall temperature no suction and blowing and viscous dissipation is ignored then you see value of minus theta prime 0 for I have computed from 0.7 to 25 as the Prandtl number. So, you see at Prandtl number 1 this is the well known velocity solution really where perfect analogy between momentum and heat transfer exist 0.33, 4.92 and but when the Prandtl number is reduced to 0.7 which is for the gas you get theta prime 0 is lowered to 0.291 the thermal boundary thickness increases and so does the enthalpy thickness increase. On the other hand at 5 which is more or less water you get higher heat transfer right 0.572, 2.73 and 231 still higher 0.721, 2.5 and these are really the organic liquid range in which this still increases. So, you have minus theta prime 0 increases with the Prandtl number. So, minus theta prime 0 is n u x r e x to the half. Now if I were to correlate these values theta prime 0 as a function of Prandtl number I would see that it will become 0.33 into Prandtl raise to one third and therefore you can develop from the numerical solutions or the similarity solution and expression from the set number as n u x is equal to 0.332 Reynolds x to the half Prandtl to the third. This result looks remarkably like an experimental correlation indeed this result has been found to be an excellent agreement with the experimental data. One point to note however is that delta star the thermal boundary layer thickness increases with decrease in Prandtl number or another way of saying thermal boundary layer thickness decreases with increase in Prandtl number. And therefore as we move towards oils and other things where the Prandtl number is very very large you would see that the thermal boundary layer thickness would be much much smaller than the velocity boundary layer thickness. So, in other words as we move towards oils which is Prandtl number very much greater than one then you will get situation like so this will be the velocity boundary layer and the temperature boundary layer would just develop like that. On the other hand for liquid metals when Prandtl number is very much less than one then you will get this with the temperature boundary layer and the velocity boundary layer will be this. So, here delta is much much smaller than delta where here delta would be much much greater than delta. This important deduction we will make use of it in several further developments of both the similarity method as well as the integral method that we will be discussing much later. So, with this I have given you a simple sample solution for a very special case of m equal to 0, B f equal to 0, gamma equal to 0, E c equal to 0. In the next lecture I will explore the influences precisely of these parameters on heat transfer rate.