 How to derive similarity equations and solve these equations to obtain friction factor and a set number for two dimensional laminar boundary layers? We now move to the next method that is the integral method for solving boundary layer equations. The task for this lecture therefore, is to derive the integral equations of a boundary layer. The development of firstly the integral equation for a velocity boundary layer and then the integral equation for a temperature boundary layer. Equations and solve these equations to obtain friction factor and a set number for two dimensional laminar boundary layers. We now move to the next method that is the integral method for solving boundary layer equations. The task for this lecture therefore, is to derive the integral equations of a boundary layer. We shall begin with the development of firstly the equation for a temperature boundary layer. The integral method represents a class of approximate methods capable of handling arbitrary variations of u infinity, suction or blowing velocity v w and wall temperature variation T w. In this respect the method removes you will recall that similarity method permitted only certain types of variations of the free stream velocity, the wall velocity and the wall temperature variations. The method is called approximate method not because its equations are approximate. So, the method actually derives exact boundary layer equations, but in integral form rather than differential. It is the solution methodology which is approximate and not the equations. A point very important to remember in all further development. Unlike the similarity method, this method is attractive because at least in some simple cases closed form solutions can be obtained with little algebraic effort. In this sense at one time people used to call the integral because you could do all the calculations on a by simple algebra without requiring having to go to a computer. The second advantage of the method is of course, that it can now deal with any arbitrary variations of u infinity v w and T w. Let us first of all consider the boundary layer d u by d x plus d v by d y equal to 0 and the convection terms or the momentum transfer terms and the pressure gradient term and the viscous term. These equations are integrated term by term with respect to y of the boundary layer from y equal to 0 where u is equal to 0 and v is equal to v w to y equal to l where u equals u infinity and v equals some fictitious velocity v l l is greater than the delta x in the region 0 to x 0 to l as you will see on the next slide. This figure shows a surface on which a boundary layer grows. Of course, a boundary layer development can be uneven. The boundary layer can grow then shrink and then grow depending on the pressure gradient that is applied and when u infinity varies arbitrarily with x of course, the boundary layer thickness too will vary arbitrarily with x. But we always choose to analyze a boundary layer over a given length l and in this length l we make sure that we choose a dimension del l which is greater than delta max in this region as shown here. Continuity equation then term by term integration will give 0 to l d u by d x d y. This term will simply reduce to v l minus v w and this term is taken to the right hand side as minus d 0 to l d u by d x d y and I can take the differential out and write it as minus d by d x of 0 to l u d y. You will see then that v l minus v w simply represents the change in the flow rate in between two successive stations. v l is a fictitious velocity at l and v w x is the familiar suction or blowing velocity. Likewise, if I were to integrate momentum equation term by term then you will see 0 to l d by d x of u u 0 to l d by d y of u v equal to 0 to l u infinity d u infinity by d x d y plus nu times d u by d y square d y. Then this term taking the differential out will simply d by d x of 0 to l u u d y plus this term will become u infinity v l minus u w v w equal to remember this is not a function of y u infinity d u infinity by d x is not a function of y and therefore, can be taken out and integral 0 to l d y would simply yield a l whereas, this in term would integrate to nu times d u by d y at y equal to l minus d u by d y at y equal to 0. But notice this term will vanish because u w the pose that u w is equal to 0. So, that term goes to 0 likewise d u by d y at l is also equal to 0 and the term that survives is mu by rho d u by d y at y equal to 0. But mu into d u by d y at y equal to 0 is nothing but the shear stress divided by rho and that is what you see on the right hand side here. This term is represented here I have replaced v l into u infinity as v l from the previous result as shown here as v w minus d by d x of u d y and then of course, this term survives. So, we move further remember u infinity d u infinity by d x l is nothing but u infinity d u infinity by d x 0 to l d y and since u infinity not a function of y I can absorb it inside the integral and write it as d u infinity by d x 0 to l u infinity d y. Then read as the first term here u infinity into v w minus d by d x u d y as before. This term is now replaced by that d u by the infinity is minus tau valve x by rho. Now, consider the identity d by d x of a product of n t by d x integral 0 to l u d y plus u infinity d by d x integral 0 to l u d y and this is precisely the term you see here. So, I am going to replace here if I do that then you will notice that I can write this equation in this manner d by d x 0 to l u into u minus u infinity d y which is this term plus d u infinity by d x into 0 to l u d y which is this term and then this term which is minus u infinity into d y and then minus tau valve rho x by d x I mean minus tau valve x divided by rho and minus u infinity into v w. So, if I divide and multiply each term by the same quantity meaning if I multiply this equation by u infinity square and divide by u infinity square if I multiply this equation by u infinity and divide by u infinity then you will notice I can write the equation in this manner d by d x of u infinity square u over u infinity into u over u infinity minus 1 d y plus u infinity d u infinity by d x equal to 0 to l u over u infinity minus 1 d y and all this. But, recall that this integral is nothing but our minus 2 or the momentum thickness and this term is nothing but minus delta 1 or the displacement thickness and therefore this term will become minus d by d x u infinity square delta 2 this term will become minus u infinity d u infinity by d x into delta 1 and two terms. So, cancelling the minus sign which appears in each term we would have the equation which looks like this d by d x u infinity square delta 2 plus u infinity d u by d u infinity by d x delta 1 equal to the shear stress term and the suction and blowing term. I will further manipulate this equation as shown on the next slide. So, if I divide through by delta u infinity square and open up this differential of a product as u infinity square d delta 2 by d x into delta 2 into d u infinity square by d x then you will notice that the equation will read like this d delta 2 by d x plus 1 over u infinity d u infinity by d x into d u delta 2 plus delta 1 equal to c f x by 2 plus v w by u infinity. So, each term now dimension so is x u infinity and u infinity has the velocity dimension delta 2 and delta 1 have length dimension so is does x. So, the term is dimensionless and so are the terms on the right hand side completely dimensionless. It is this equation which is known as the integral momentum equation it is an exact equation we have because we have not introduced any assumptions in its derivation. We have simply integrated the partial differential equation from 0 to L because all quantities vary only with x. So, the partial differential equation of the boundary layer is converted to an ODE for an integral parameter delta 2. This is very similar to what we did in similarity method where the PDE is very converted to third order ordinary differential equation. Here the equations are in converted to a first order ordinary differential equation for an integral parameter delta 2. C f x is the coefficient of friction and that is simply defined as tau all x over rho infinity square divided by 2. So, this is well known to you. So, this then is the integral momentum equation. There is another variant of the equation which is called the integral kinetic energy equation it can be derived from the earlier momentum equation, but I will derive it from first principles. So, recall our momentum equation reads like that. These are the terms. If I multiply each term by u divided by rho throughout, then you will see and define e as u square by 2. Then you will see this term simply can be written as d u e by dx. This term can be written as dv e by dy equal to u u infinity d u infinity by dx plus nu u d 2 u by dy square. I repeat e is the energy of the axial velocity u square by 2. Again, if we integrate this equation from y equal to 0 to y equal to l or delta where does not really matter what we do, then this term will simply become u into u square by 2. So, that becomes u cube by 2 d by dx integral 0 to delta dy. This term will be v e at delta minus v e at y equal to 0, but at y equal to 0 u is equal to 0. Therefore, e is equal to 0 and that term will go out. So, the v l into e at l or infinity is simply v w minus d by dx of 0 to delta u dy u infinity square by 2. This is the value of e infinity and that would equal d u by d into d by dx. Now, u u infinity is absorbed inside the integral dy plus nu integral 0 to delta u d 2 u by dy square dy. It is this last term because this is just an result on the next slide. So, integration by parts will give nu 0 to delta u d 2 u by dy square dy equal to minus nu d 0 to delta d u by dy whole square. Now, in the internal kinetic energy equation, we introduce another thickness called the kinetic energy thickness and it is defined very much like the momentum thickness, except that we now have 1 minus u over u infinity square, which represents the kinetic energy deficit caused by interpolation of the previous equation, which is shown here. You will see that this can be manipulated to read as d by dx of u infinity cube delta 3 equal to v w u infinity square plus 2 nu d u by dy whole square, which is this term. Integral kinetic energy equation is sometimes used in boundary layers, analysis of boundary layers with suction and blowing. We now turn to integral energy equation. Again, consider the boundary layer, the surface on which a boundary layer is developing and this solid line represents the velocity boundary layer. Now, the thermal boundary layer will have a thickness either greater than the velocity boundary layer thickness or smaller than velocity boundary layer thickness. You already know from your similarity solutions, then when Prandtl number is greater than 1, the thermal boundary layer thickness is smaller than the velocity boundary layer thickness. Whereas, if Prandtl number was less, the thermal layer thickness is greater than the velocity boundary layer thickness delta. While integrating this energy equation, which includes this viscous dissipation term, we will choose L big enough, so that it is either greater than delta for case of Prandtl greater than 1 or it is greater than thermal boundary layer thickness capital delta for Prandtl number less than 1. We define for convenience theta equal to t, where t infinity is constant with x, but t wall is some function of x and I will introduce that function shortly. So, then this term will become du theta by dx. Additional term arising out of the fact that, t w now is a function of x and therefore, you will have a term called by t w minus t infinity into d by dx of t w minus t infinity. This term will become alpha delta 2 theta by dy square and this term will become nu c p t w minus t infinity du by dy whole square. It is this equation that we shall integrate from 0 to L, where L is greater than delta max for Prandtl equal to 1, Prandtl greater than 1 and L is greater than capital delta max for Prandtl less than 1. Let us do the integration on the next line. You will notice that integration of this will simply become d by dx of integral u theta dy and if I divide and multiply by u infinity, it will read as d by dx of u infinity 0 to L u infinity u by u infinity theta dy. The integration of this term will simply give v L theta infinity minus v w theta w and theta w as you know is 1, then you will see v L theta L minus v w theta w plus this term will be simply 1 over t w minus t infinity by dx of t w minus integral u theta dy and that is what I have written and u theta dy I have divided by u infinity and multiplied by u infinity. The diffusion term will simply give alpha d theta by dy at L minus alpha d theta by dy at 0 plus nu square dy. Now, if you recall that our enthalpy thickness delta 2 was defined as 0 to infinity u over u infinity t minus t infinity over t wall minus t infinity, then it is nothing but 0 to L u over u infinity theta dy. In other words, this quantity is nothing but delta 2 capital delta 2 theta L because of its definition is 0 and therefore, that term vanishes. Whereas, theta w is equal to 1 equal to L is 0 because the temperature gradient at the infinity at the age of the boundary layer is 0, but see what this term means d theta by dy at y equal to 0 multiplied by alpha. Well, the d theta by dy will simply be equal to d t dy divided by t w minus t infinity and therefore, q w into rho c p with a negative sign and q w divided by t w minus t infinity is nothing but local heat transfer coefficient h x divided by rho into c p. So, this term will be replaced by minus h x rho c p and therefore, you will see the equation would now read as u infinity d by dx of u infinity delta 2 by dx u infinity delta 2 into t w minus t infinity d by dx t w minus t infinity and h x divided by rho c p plus v w plus nu by c p 0 to L du by dy whole square into dy. If I divide through by u infinity after opening up this differential, then I would get what is called as the integral energy equation d delta 2 by dx plus delta 2 into the wall variation term into the free stream variation or the pressure gradient term stanton x which is h x divided by rho c p u infinity as I have shown here, which is nothing but Nusselt number divided by Reynolds number into Prandtl number. We have seen this before, then the wall velocity variation term v w by u infinity and then the viscous dissipation term as this one. Each term here is dimensionless as you can see, this is dimensionless, this is dimensionless, so is delta 2, this is dimensionless, this is dimensionless and so is this dimension, it can be shown quite easily. So, each term now is dimensionless and therefore, we have an ordinary differential equation for capital delta 2 analogous to an ODE for small delta 2 in the velocity boundary equation. So, let me summarize all the equations that we have derived. So, integral continuity equation simply gives v l minus v w equal to minus d by dx of 0 to L u dy. The integral momentum equation gives variation of d delta 2 by dx as that and accounts for the wall velocity and the integral energy equation gives the rate of change of enthalpy thickness delta 2 and accounts for the wall temperature variation, the free stream variation and wall velocity variation and the viscous dissipation. Our interest always would be to determine stanton x and c f x. With this, I will take up in the next lecture the solution of the velocity boundary layer equations in their integral form. Thank you.