 We will now look at solutions to the similarity temperature boundary layer equation. By way of remainder, the equation is theta double prime equal to Prandtl number into m plus 1 by 2 f theta dash minus gamma f dash theta plus 2 e c f double prime 0 square equal to 0 with the boundary conditions that at the wall theta 0 will be 1 and in the infinite state theta n pretty will be 0. We are interested in developing solution so that we look at effect of Prandtl number, we look at effect of pressure gradient and of course, also the suction of blowing parameter which is embedded in the solutions for f f dash and f double prime. We look at effect of wall temperature variation gamma and we look at effect of viscous dissipation on the nature of the solutions of things. Remember, I obtain already solution for Prandtl equal to 1, m equal to 0, gamma equal to 0 and e c equal to 0. That is the solution I presented in the last lecture. Now, I want to look at effect of very high and very low Prandtl numbers and this is what you see. For liquid metals, the value of eta, remember at 0.005 which is very low Prandtl number, the value of theta max or the value of delta star is nearly 53.3 whereas, for Prandtl number equal to 1, it was 4.92 as you remember. So, as the Prandtl number is reduced, the thermal boundary layer thickness goes on lower and lower temperature gradient. If I were to correlate this theta prime 0, it would be correlated reasonably well by this correlation n u x is equal to 0.564 Reynolds x Prandtl to the half. On the other hand, on the oil side, you will see Prandtl equal to 1 gives you this which is the reference solution about 4.92 as the temperature boundary layer thickness. But for 500, 500 and 1000, you see the temperature boundary layer thickness goes on reducing and I have given some values here. For 1000, it is as low as 0.495 which is 10 times smaller than the velocity boundary layer thickness which was about 5 and this is 0.6 and at Prandtl number of 100, it is just one-fifth of the boundary layer thickness for Prandtl equal to 1 and minus theta preserve goes on increasing with Prandtl number and a good curve fit which is 0.339 Reynolds x to the power 5 Prandtl to the third. But we will discuss both these solutions a little later. We want to study effect of m first of all and for which we will assume that there is no suction or blowing. We will assume that the wall temperature is constant. We will assume that there is no viscous dissipation included. We are only including the effect of m along with that of the Prandtl number. So, the governing equation as you will see for gamma equal to 0 and E c equal to 0 would simply be that theta double prime Prandtl m plus 1 by 2 f theta dash equal to 0 which I can write as d theta dash over d theta dash because theta double prime is simply d theta dash and that would equal minus Prandtl m plus 1 by 2 f. If I integrate this equation once, I will get ln theta prime from 0 to eta equal to minus Prandtl m plus 1 by 2 which is a constant into 0 to eta f d eta. Another way of saying it is theta prime at any eta would be equal to theta prime at 0 exponential of minus Prandtl into m plus 1 by 2 0 to eta f d eta. If I integrate this equation once again, then I get theta equal to minus theta prime 0 0 to eta exponential of this quantity into d eta plus a constant of integration C 1 which I discover firstly from the boundary condition theta 0 is equal to 1. So, if I set theta 0 which means 0 to 0, therefore, this contributes nothing and theta 0 being 1, I get C 1 equal to 1. Now, I impose the boundary condition at infinity. So, theta infinity is 0. This integration will become 0 to infinity and therefore, I will get theta prime 0 which is of interest to me because that represents the Nusselt number equal to 0 to infinity exponential of all this raise to minus 1. Now, we will see some special cases of f. Since velocity solution f d eta is known, we can evaluate this integral. The solutions that I show here have been evaluated for moderate Prandtl numbers 0.75, 10 and 25 allowing for M negative. Remember, I cannot go below M equal to minus 0.091 because that is where the separation occurs. So, I begin with M equal to minus 0.085, minus 0.065, minus 0.04, 0.331 and 4. These are the axial rating flows. Now, remember H x will be function of x M minus 1 by 2, the heat transfer coefficient. Remember, N u x is H x x by k is equal to some constant into Reynolds x to the half and this constant is a function of Prandtl number. So, what do we get? H x will be equal to k by x into u infinity x by nu raise to half into or proportional to that. What is u infinity? It is equal to C x. So, I get this as C nu by half into x into M plus 1 by 2, which is nothing but proportional to k C by nu raise to half x raise to M minus 1 by 2. Remember, heat transfer coefficient, the pressure gradient has this effect on heat transfer coefficient. If M is equal to 0, H x will simply decrease with x. So, for M less than 1, H x will decrease with x, but for M greater than 1, you could also get H x relatively increasing with x. This is something very, very important. This is for M greater than 1 H x versus x and for M equal to 1, H x will be constant M equal to 1. Now, this is a very, very important point to remember that for stagnation point flow, heat transfer coefficient is essentially constant. You will recall that we had also seen that delta 2 star and delta 1 star etcetera, all the thicknesses are also constant for stagnation point flow. So, let us look at the solutions for showing the effect of M at any one Prandtl number. Let us take Prandtl number 0.7 and then you will see that when M is equal to 0 for flat plate, the thickness is 0.5.6, but when you have decelerating flow, the thickness goes on increasing like the velocity boundary layer thickness. When you have accelerating flow, the temperature boundary layer thickness also decreases as shown here. This state of affairs prevails at all Prandtl numbers that I have listed here. Now, let us look at the effect of Prandtl number at any pressure gradient. Let us look at stagnation itself. You will see that for Prandtl number 7, minus theta prime 0 is 0.49. It increases to 1.03 at Prandtl 5, 1.32 and 1.81. What have we seen? As the value of M increases, the heat transfer coefficient increases. As the Prandtl number increases, the heat transfer coefficient increases. We have captured both these effects in the solution. Of course, it would be possible to develop a correlation of the form N u x is equal to C times N u x. R e x to the half, where C is now a function of Prandtl number and M for Prandtl number between 0.7 and 25 and minus 0.085 M 4. You will see that this will take the form of an experimental correlation, which can be used for further design work at any time we wish and one need not generate new solutions every time by means of a computer. I now want to look at the effect of very high and very low Prandtl numbers in the presence of pressure gradient. Remember, for Prandtl number very much greater than 1, delta would be much less than 0. Therefore, you will have a situation of a velocity boundary layer like that and a temperature boundary layer just like that. In other words, delta is much, much smaller than delta. For all practical purposes, I can say that the velocity profile is really linear in the temperature boundary layer equation and that we have seen many times in our earlier solution. In other words, f dash eta within the temperature boundary layer will simply be equal to f dash 0 for that value of M into eta. Integration will give me f eta is equal to f double prime 0 M, which is a constant eta square by 2 plus a constant. But f 0 is equal to 0 and therefore, c is equal to 0. So, essentially I get f dash eta equal to f double prime 0 into eta square by 2. That is what I have shown here. f eta would be approximately equal to f double prime 0 M eta square by 2. If I now include this solution in this here, which is an approximate solution to the velocity in this, I can readily evaluate minus theta prime 0 and that is what I have done. If I substitute f double prime 0 M eta square by 2 for f, then I get that. This would simply transform to minus Prandtl M plus 1 f double prime 0 M divided by 12 into eta cube because this is equal to eta square is eta cube by 3. 2 into 2 is 4 into 3 is 12 into d eta minus 1. Now, if I use the definition of this and that is what I will show you how to evaluate that integral. It is not a very difficult thing to do, but we will nonetheless do that. So, for the moment, let us say let Prandtl M plus 1 f double prime 0 M divided by 12. Let us say it is equal to a and define a eta cube equal to x, then 3 into eta square 3 a eta square d eta will equal d x. Therefore, d eta will equal d x divided by 3 a eta square. And eta square will be eta cube raised to 2 by 3 and that will equal x by a raised to 2 by 3. I get d x divided by 3 a into x by a raised to 2 by 3. 1 over 3 a into 1 by 3 into d x over x raised to 2 by 3. Therefore, our integral is simply 0 to infinity and exponential of minus a eta cube d eta can be written as 0 to infinity exponential of minus x into 3 a raised to 2 by 3 1 by 3 sorry into x raised to minus 2 by 3 into d x or this is equal to 1 raised to 3 a raised to 1 by 3 integral 0 to infinity x raised to minus 2 by 3 exponential of minus x d x. Now, if you recall the definition of gamma function, it is given as 0 to infinity x raised to n minus 1 e raised to minus x d x. So, in our case n is nothing but n minus 1 is equal to minus 2 by 3 and therefore, n will be equal 1 minus 2 by 3 equal to 1 by 3 equal to 1 by 3. So, essentially we have integration integral itself is equal to 1 over 3 a raised to 1 by 3 is equal into gamma times 1 by 3, but usually gamma functions are plotted for values of gamma greater than or equal to 1 and therefore, we make use of the relationship that gamma n plus 1 is also equal to n times gamma n. So, in other words gamma 1 by 3 can be written as 1 over 1 by 3 into gamma 4 by 3 into that is equal to 3 times gamma 4 by 3. This is gamma 4 by 3 and therefore, you will see our integral can be written as gamma 4 by 3 divided by a raised to 1 by 3. The integral itself will become minus theta prime 0 a raised to 1 by 3 divided by gamma 4 by 3 because this is raised to minus 1. So, this goes into the denominator, the a goes into the numerator and this is the solution. The value of gamma 4 by 3 you can look up the tables of course, it is about 0.893. So, if you look at m equal to 0, if I put m equal to 0 and recall that for m equal to 0 f double prime 0 is 0.33 and use this equal to 0.893 then you will see a correlation is possible for very very high Prandtl numbers n u x 0.339 r e x to the 0.5 Prandtl third. Now, I had curve fitted the solutions exactly in the same manner here for oils as you will see here 0.339 r e x to the Prandtl third. These are numerical solutions and this is obtained from our assumption that the velocity profile is linear. So, this kind of an assumption helps you to evaluate closed form solutions which we can call as correlations, but you can do this for any value of m for which f double prime 0 m has been calculated from the velocity boundary layer solution. Likewise, let us look at the case of very low Prandtl number in which case as I said the velocity boundary layer thickness will be solved and the temperature boundary layer thickness will be solved T b l and this will be the v b l. So, you will see for greater part of the thickness of the boundary layer u is in fact equal to u infinity and therefore, f dash eta in the term is about 1 throughout the boundary layer which gives me f eta equal to eta itself. In this equation for theta prime 0 if I substitute f equal to eta then I will get the relationship as that this will simply become eta square by 2. So, this becomes Prandtl m plus 1 by 4 eta square whole thing is integrated raise to minus 1. If you put this as some y or something like that you can do this carry out this integration quite easily and use the error function definition e r f x is equal to 2 by root pi 0 to x e raise to minus eta square d eta then you will see and note that r f infinity is equal to 1. Then you will it is very easy to show that n u x r e x to the minus 5 which is minus theta prime 0 will simply be equal to Prandtl m plus 1 by pi and that is what you see for again for m equal to 0 you will see this becomes simply Prandtl divided by pi and as a result 0.564 r e x in Prandtl to the half. The great thing about this solution is the following at liquid metals for liquid metals you see both the exponent of Reynolds number and Prandtl number is identical. Now what does that mean r e x and Prandtl that is u infinity x by nu into nu by alpha the thermal. So, nu and nu gets cancelled that means in liquid metals viscosity has no influence and that matches with the idea that viscosity is really if its effect is confined to such a narrow region of the thermal boundary layer thickness that viscosity simply has no effect on the rate of heat. This result is very typical of liquid metal heat transfer whereas, for all other higher Prandtl numbers you will always get a separate effect of Reynolds and Prandtl number. But here incidentally this product this is Reynolds number this is Prandtl number, but the product Reynolds Prandtl is called Peclet number Peclet was a French scientist and Peclet number is of importance in liquid metal heat transfer. Liquid metals as you know I used in very high heat flux heat transfer cooling such as breeder reactors and others. So, these kinds of solutions are great value and the constant is 0.564 if you recall I will go back to the numerical solution that we had obtained. Yes indeed the solutions at very low Prandtl numbers do match this correlation here and that is precisely where it comes from it comes from this assumption that F dash eta is about 1. So, having studied the effect of M the pressure gradient and Prandtl number high and low Prandtl under essentially constant wall temperatures and no suction and blowing and no viscous dissipation. We now move to the case of looking at effect of wall temperature variation again I am going to consider the flat plate boundary layer no suction or blowing and viscous dissipation is neglected. Then the governing equation read something like this now there is no closed form solution possible for this because of this additional extra term and I need to now look at the numerical solution which I have presented here from very moderate Prandtl numbers wall temperature variation gamma positive indicates that the temperature of the wall is increasing with x whereas, this indicates that the temperature is decreasing with x negative value of gamma indicates the temperature is decreasing with x. So, what do we find let us look at results at any one Prandtl number and you will see compared to gamma equal to 0 where wall temperature is constant there is as gamma increases there is a wall temperature increases in the x direction then so does the Nusselt number or the solution which is minus theta prime 0 increase when the temperature decreases with x you in fact the heat transfer decreases and in fact you hit the situation at gamma equal to minus 0.5 where there is absolutely no heat transfer this is called the adiabatic case. If you reduce gamma still further then you will actually get negative heat transfer although the wall temperature is bigger than the free stream temperature you will actually get a negative heat transfer. Now, I have allowed gamma to take any arbitrary value simply because E c is equal to 0 and we did say that gamma is restricted to 2 m only when E c is included we have not included E c and therefore, this is a remarkable result that when the temperature wall temperature decreases with x you could get a situation of adiabatic heat transfer or in fact even negative heat transfer and we did alert right in the beginning that we are interested in looking at situations like this for gamma equal to minus 0.5 temperature gradient at the wall is 0 is an adiabatic case although T w is greater than T infinity. Now, we can explain this why this happens. So, let us take our flat plate and the wall temperature is now decreasing with x this is the wall temperature. So, you can imagine at any x the particles close to the surface would arrive from a region of higher temperature from a region of higher temperature. So, much so that the convicted heat into the layer the temperature here could well become absolutely equal to the temperature of the wall at that point. So, in effect I get a situation where I have a profile which will look like that which will look like that. So, basically although T w is bigger than T infinity T w is bigger than T infinity I could get a situation where it is like that if the slope is still negative this is corresponding to gamma equal to minus 0.5 I could even get a situation which is like that this is gamma greater less than minus 0.5 this is let us say minus 0.6 as I have shown. So, you could get situation where the heat transfer will be heat will going in although T w is greater than T infinity. The situation arises because the fluid particles coming from the upstream region would be a hotter than the value of the wall temperature itself. So, these order fluid may thus inhibit heat transfer from the surface to the cooler free stream by the same reasoning for gamma equal to minus 0.5 theta would increase beyond 1 at some distance close to the surface and heat will flow into the surface even if T w is greater than hence T w heat transfer coefficient will be negative. We verify these in the next slide where I have shown effect of value of gamma on the predicted temperature profile. The curve here shows the constant wall temperature case the frontal number is chosen to be 0.7 therefore, the boundary layer thickness is slightly bigger than 5. Let us look at the acceleration side this is gamma equal to 1, this is 2, this is 4 and you get steeper gradients and near the wall and thinner and thinner temperature boundary layer thickness as wall temperature increasing with x. But on the negative side that is wall temperature decreasing with x you will see you get a shallower gradient near the wall and the boundary layer thickness itself goes on increasing at gamma equal to minus 0.5 you can see very clearly the 0 gradients and for gamma equal to minus 0.6 value of theta here at this point has exceeded the value at the wall which was 1. So, our temperature solution does indeed confirm what we anticipated now we look at the effect of viscous dissipation and viscous dissipation typically is accounted when you have very high velocity gradients d u by d y whole square or when you have very high viscosity which is mu. So, I am taking here the case of pranthole equal to 0.7 and assuming that the velocity gradients near the wall are very very high then the solution would be and the wall temperature is constant there is no suction or blowing and the pressure gradient is 0. Then the governing equation will be this and because the wall temperature is constant E c would be simply this and theta is this. So, E c less than 0 implies that T w is less than T infinity E c greater than 0 implies T w is greater than T infinity. So, here are the solutions to so this is 0.292 for pranthole number 0.7 is the value of theta dash 0 with 5.26 as the temperature boundary level of thickness. When E c is positive that is when T w is greater than T infinity you in fact do find an almost adiabatic situation at E c equal to 1.2 and negative heat transfer when E c becomes 2.4 and 4.8. On the other hand when E c is negative that is if T wall is less than T infinity that is cooling case then you find that the heat transfer rate increases in the presence of effect of it. So, E c equal to 1.2 is a special adiabatic case is a very very special adiabatic case and here are the solutions for different values of E c the a cut number. So, here the solution is for a cut number equals 0 this is theta dash eta and the wall value is 1 you do see that at 1.2 the gradient is 0 and therefore it turns out to be an adiabatic case and for 2.4 the temperature is exceeded value of 1 and it has exceeded value of 1. So, although T w is greater than T infinity you will get negative heat transfer for these two values of a cut numbers. On the negative side you see the temperature gradient becomes sharper and sharper. Because of the viscous dissipation see at some point the temperature actually goes below the free stream temperature because it is less than 0. So, just as wall temperature is exceeded for positive E c free stream temperature is exceeded. So, E c less than 1.2 theta is less than 0 at some positions indicating that T eta E can be greater than T infinity within the boundary layer and E c greater than 1.2 minus theta h x is less than 0 even when T w is given. Both these are effects of viscous dissipation due to mu du device square. This kind of viscous heating is of great concern viscous heating effect in reentry vehicles of into the space from the space into the upper atmosphere. When sudden increase in viscosity of the atmosphere would generate and the reentry vehicle is hurtling down at very high velocities of the order of 5 to 6000 meters per second and you get enormous gradients very close to the wall as a result the viscous dissipation term becomes very dominant and you would generate large amounts of heat and if the cut number turns out to be bigger than 1.2 then you could well get heat transfer into the reentry vehicle surface a dangerous and that is why the reentry vehicles are coated with ablating materials. So, that when the heat transfer takes place inside and the surface temperature goes up the material evaporates and carries away the heat of which has been ingress from the outside into the wall of the reentry vehicle. So, study of such effects is of great consequence in practice. Now, we look at finally, the effect of B F the suction and blowing parameter and I am going to consider again the constant wall temperature and ignore viscous dissipation but I will allow for effect of M, I will allow for effect of Prandtl number and B F. So, we are looking at three parameters here B F, Prandtl number and M and I am restricting this to the case of gases because these B F cases are of interest in gas turbine cooling problems. So, you will see in suction case when B F is negative compared to B F equal to 0 which is no suction and I am looking at flat plate solution then when there is suction there is increase in heat transfer but when there is a blowing there is a decrease in heat transfer and you will recall that B F equal to 0.612 was the case where separation occurs. So, we are not interested in B F equal to 1 but we are interested for M equal to 1 in which you will see that again the same thing prevails that the heat transfer increases on the suction side whereas, it decreases on the blowing side and this essentially occurs because the thickening and thinning of the boundary layers associated with suction or blowing. Suction thins the boundary layer and therefore, increases the rate of heat transfer blowing thickens the boundary layer and therefore, lowers the heat transfer coefficient. For M equal to 0, B F equal to 0.612 is a very special case and that is where the separation occurs. You will see on the next slide the graphs these are the flat plate solutions and these are the stagnation point solution. This is B F equal to 0.5, 0.3, 0, minus 0.5, 1 and 2. As I said B F equal to 0.612 is an adiabatic case. You will see on the suction side we get very large temperature gradients and on the blowing side we get shallower temperature gradients compared to M equal to 0. Now, if I were to compute this for potential number of 0.5 or 1, the results are very similar and therefore, not shown here. So, with this I conclude my discussion on the similarity method as a whole and therefore, it is time that we look at the summary of the equations and the boundary conditions. For the velocity boundary layer similarity equation is this F triple prime plus M plus 1 F F double prime plus M 1 minus F times square equal to 0, where M is the pressure gradient parameter. The suction and blowing parameter comes in through the boundary condition F 0. The similarity conditions or the variables are U infinity is equal to C x raise to M. B F V W x U infinity x R e x to the half must be constant and the variable eta must be y under root U infinity by nu x and the similarity solutions are C F x M B F which is a function of M and B F equal to 2 times F double prime 0 R e x to the minus 0.5. Likewise, the temperature equation is theta double prime into Prandtl M plus 1 by 2 F theta prime minus Prandtl into the wall temperature gradient parameter and the viscous dissipation parameter. The boundary conditions are simply this theta 0 is equal to 1 theta infinity is 0. The similarity condition is that the wall temperature can vary only as x to the power of gamma. If a cut number or the viscous dissipation is present however, gamma must remain equal to 2 M and the similarity solutions as a whole are multi-parameter solutions involving pressure gradient M, Prandtl number, suction and blowing parameter, wall temperature and E c that is what I have shown you. We have seen a wide variety of solutions of this type. It is not very difficult to write computer programs both for solving the velocity boundary layer equation as well as the temperature boundary layer and I have already given you the method in the last two lectures. So, I suppose with that I conclude my discussion of similarity method. In the lectures to follow, we will take up integral methods of solving boundary layer equations.