Module-5: Hypersonic Boundary Layer theory. Lecture-20: Hypersonic boundary equation

Module-5: Hypersonic Boundary Layer theory Lecture-0: Hypersonic boundary equation 0.1 Governing Equations for Viscous Flows The Navier-Stokes (NS) equaadtions are the governing equations for the viscous compressible flow and hence are the governing equations for hypersonic flows. This section deals with the basics of NS equations and its non-dimensionalization. Continuity Equation: Considering steady state conditions we have; ( ρu) ( ρv) + = 0 i.e. (0.1) X Momentum Equation: D( ρu) ρdu udρ = +, now since D ρ = 0; Dt Dt Dt Dt D( ρu) ρdu = ; therefore the L.H.S. simplifies to give X momentum equation in Dt Dt steady state conditions as; u u p v u u u ρu + ρ v = + µ + + µ + (0.) Y Momentum Equation: D( ρv) ρdv vdρ = +, now since D ρ = 0; Dt Dt Dt Dt D( ρv) ρdv = ; therefore the L.H.S. simplifies to give Y momentum equation in Dt Dt steady State Conditions as; Joint initiative of IITs and IISc Funded by MHRD Page 1 of 39

v v p v u v v ρu + ρ v = + µ + + µ + (0.3) Energy Equation: ( u + v ) ( u + v ) ( u + v ) D ρ e+ ρd e e Dρ + + = +, Dt Dt Dt Since D ρ = 0; Dt ( u + v ) ( u + v ) D ρ e+ ρd e+ =, Dt Dt Therefore the L.H.S. simplifies to give Energy Equation in steady State Conditions as; ( ) ( ) u e+ = ρq+ κ + κ + + τ + τ + τ + τ V T T pu pv u u v v xx xy xy yy (0.4) The above equations are written for steady, compressible, viscous, two dimensional flows in Cartesian coordinates. Where u and v are velocities in x and y directions respectively; e represents internal energy per unit mass and q represents the volumetric heating that might occur. All other notations carry their usual meaning. We can simplify the above set of equations via appropriate assumptions, and obtain approximate viscous flow results. Joint initiative of IITs and IISc Funded by MHRD Page of 39

0. Non Dimensional Form of Governing Equations The non dimensional form of Navier-Stokes equations can be obtained as follows. Here we have considered a two dimensional steady flow and ignored the normal stresses τ xx and τ yy. Reference variables of the flow can be used for nondimensionalization. Non Dimensional Variables: u = u V v = v V x x = c y y c p p = ρv = e = e CT v µ µ = µ κ κ = κ ρ = ρ ρ Where V, T, ρ, µ, κ are free stream parameters and c reference length. Therefore the Non Dimensional Equations are given as: Non Dimensional Continuity Equation: ( ρu) ( ρv) + = 0 (0.5) Non Dimensional X Momentum Equation: ρ u u 1 p 1 u v v u x ρ y γ M x Re y µ + = + + (0.6) Non Dimensional Y Momentum Equation: ρ v v 1 p 1 u v v u x ρ y γ M y Re x µ + = + + (0.7) Joint initiative of IITs and IISc Funded by MHRD Page 3 of 39

Non Dimensional Energy Equation: ( ) ( ) e e γ u v ( 1) M u u v v u v T ρ ρ γγ ρ ρ κ κ T + = + + + + + Pr Re M v u v u + γ ( γ 1) µν + + µ u + (0.8) Re 0.3 Process of Non-dimensionlisation of Governing Equations:- Continuity Equation:- We Have, ( ρu) ( ρv) + = 0, With non-dimensional parameters ρ V ( ρu/ ρ V ) ( ρv/ ρ V ). 0 c + ( x/ c) ( y/ c) = ( ρu) ( ρv) + = 0 X & Y momentum Equations:- The process of non dimensionlisation is almost the same for both X & Y momentum equations, therefore we have for X momentum equation; u u p v u u u ρu + ρ v = + µ + + µ + Rewriting the above Equation with Non Dimensional parameters, we get; ρ V ρu ( u/ V ) ρu ( v/ V ) c + = ρ V ( x/ c) ρ V ( y/ c) p ( p/ p ) c ( x/ c) µ V µ ( v/ V ) ( u/ V ) µ ( v/ V ) ( u/ V ) + + + + c ( y/ c) µ ( x/ c) ( y/ c) ( x/ c) µ ( x/ c) ( y/ c) Joint initiative of IITs and IISc Funded by MHRD Page 4 of 39

Now dividing the above equation on both sides by the factor wise ρ V c ; we get term p c p First, the pressure term on Right Hand side (R.H.S); = c ρ V ρ V Since Mach Number can be written as M V = therefore; a p ρ V p = and furthermore, p ρ RT ρ M a = and a = γ RT ; the term reduces to p RT 1 = = ρ M a M γrt γm µ V c µ 1 Now the viscous term, = = c ρ V ρ V c Re Thus the X momentum Equation reduces to, ρ u u 1 p 1 u v v u x ρ y γ M x Re y µ + = + + & similarly Y momentum Equation reduces to, ρ v v 1 p 1 u v v u x ρ y γ M y Re x µ + = + + Normal stresses τ xx and τ yy are ignored for the sake of simplicity. Joint initiative of IITs and IISc Funded by MHRD Page 5 of 39

Lecture-1: Non-dimensionalisation of governing equations 1.1 Process of Non-dimensionlisation of Governing Equations Energy Equation: We know the energy equation as Eq. (0.4). Expanding the LHS of this equation, we get, ( u v ) ( u v ) ρu + e ρv + + + e+ = T T ( pu) ( pv) ( uτ xx) ( uτ xy) ( vτ xy) ( vτ yy) κ + κ + + + + + Now simplifying the above equation by neglecting normal stresses τ xx and τ yy, we get e e ( u v ) ( u v ) ρu ρv ρu + ρv + + = + T T ( pu) ( pv) v u v u κ + κ + + v µ + + u µ + Now non dimensionalising Left hand side (L.H.S.) & R.H.S. terms separately; L.H.S.= ρ V CT v ρu ( e/ CT v ) ρv ( e/ CT v ) ρ V CT v e e ρu ρv c + ρ V ( x/ c) ρ V ( y/ c) = + c Non-dimensional R.H.S., 1 st & nd term; 3 ρ V ρu ( u/ V ) + ( v/ V ) ρv ( u/ V ) + ( v/ V ) + c ρ V ( x/ c) ρ V ( y/ c) = ρ V ( u + v ) ( u + v ) ρu + ρv c 3 Joint initiative of IITs and IISc Funded by MHRD Page 6 of 39

Non-dimensional R.H.S., 3 rd & 4 th term; κ T κ ( T / T ) κ ( T / T ) κ T T T + = κ κ + c ( x/ c) κ ( x/ c) ( y/ c) κ ( y/ c) c Non-dimensional R.H.S., 6 th & 7 th term reduces to; µ V v µ ( v/ V ) ( u/ V ) µ V v u v µ + = + c ( x/ c) V µ ( x/ c) ( y/ c) c µ V u µ ( v/ V ) ( u/ V ) µ V v u u µ + = + c ( y/ c) V µ ( x/ c) ( y/ c) c Non-dimensional R.H.S., 5 th term reduces to; pv ( p/ pu / V ) pv ( pu) c ( x/ c) = c Now dividing the L.H.S. & R.H.S. by the factor as; c ρ V CT v, we obtain R.H.S. terms For term 1 st & nd ; V c V V ( 1) V ( 1) c ρ V CT CT RT a 3 ρ γ γγ = = = = M γγ v v ( 1) For term 3 rd & 4 th ; κ T c κ µ κγ γ = = = Re Pr c ρ V CvT cρ V Cv cρ V µ Cp For term 5 th ; p V c p RT ( γ 1) = = = ( γ 1) c ρ V CT v ρ CT v RT Joint initiative of IITs and IISc Funded by MHRD Page 7 of 39

For term 6 th & 7 th ; µ V c µ V 1 ( γ 1) V 1 γγ ( 1) V γγ ( 1) M = = = = c ρ V CT v cρ V CT v Re RT Re a Re Finally the Non dimensional Energy equation is given as; ( ) ( ) e e γ u v ( 1) M u u v v u v T ρ ρ γγ ρ ρ κ κ T + = + + + + + Pr Re M v u v u + γ ( γ 1) µν + + µ u + Re The non-dimensional parameters of the Navier-Stokes equation are, Ratio of specific heats: C γ = C p v Mach number: M V = a Reynolds number: Re ρ V µ = c Prandtl number: Pr = µ Cp k These four dimensionless parameters are called similarity parameters, and are very important in determining the nature of a given viscous flow problem. Thermodynamic properties, as reflected by γ, are important for any high speed flow problem. A combination of thermodynamics and flow kinetic energy can be found in M, and it is known that, Flow Kinetic Energy M ~ Flow Internal Energy For Reynolds number, we have Re ~ Inertia Force Viscous Force Joint initiative of IITs and IISc Funded by MHRD Page 8 of 39

Prandtl number, appearing in the energy equation, has the physical interpretation as, Pr ~ Frictional dissipation Thermal conduction Here γ and Pr are the properties of gas while Re and M involve flow properties as well. Joint initiative of IITs and IISc Funded by MHRD Page 9 of 39

Lecture-: Order of magnitude estimate.1 Boundary conditions An important difference between inviscid and viscous flows can be seen explicitly in the boundary conditions at the wall. The usual boundary condition for an inviscid flow is no mass transfer through the wall which mathematically gets expressed as the normal component of velocity to be zero at the wall. This boundary condition is termed as free slip along the wall. This boundary condition gets added with the cancellation of tangential velocity at the wall due to the existence of friction. This boundary condition is termed as no slip along the wall. Therefore both the components of velocity become zero for viscous wall boundary condition, that is, Wall boundary condition: u=v=0 If there is mass transfer at the wall, then we have to express the normal velocity at the wall as per the known mass flow rate to satisfy the mass conservation equation. However tangential component of velocity will still remain zero at the wall if wall has zero velocity while for conservation of momentum. There are two types of boundary conditions related with the energy equation. In one of them, wall is treated with isothermal wall temperature where the known temperature is assigned at the wall as, Constant wall temperature boundary condition: T=T w Here T w is the specified wall temperature. For non uniform temperature distribution along the surface we have, Variable wall temperature boundary condition: T=T w (s) here T w (s) is the specified wall temperature variation as a function of distance along the surface (s). This boundary condition is very much suitable for high conductivity wall materials so as to keep the wall at known constant temperature variation. Joint initiative of IITs and IISc Funded by MHRD Page 10 of 39

However in case of insulators, where thermal conductivity is very low, the wall temperature usually remains unknown. In such cases, wall heat flux is treated as zero or wall is treated as adiabatic wall. The mathematical representation of this boundary condition is, Adiabatic wall boundary condition: q = k w T n = 0 Here q w is the wall heat flux. Moreover in some cased wall heat flux distribution can be apriorily known. Therefore q w or the wall heat transfer rate should be specified as the boundary condition. This wall heat flux is dependent on temperature gradient normal to the wall in the gas immediately above the wall.. Application to boundary layer flow Consider the boundary layer along a flat plate of length c as shown in Fig..1. V δ c Fig..1 Hypersonic flow over flat plate. A thin layer of fluid is assumed to be decelerated in the presence of the wall. This assumption leads to the mathematical expressionδ < c. Here δ is the local boundary layer thickness. Apart from this, for hypersonic flow, we can also assume that, that are; v<< u and << y Now consider the Continuity Equation in Non Dimensional form; ( ρu) ( ρv) + = 0 Joint initiative of IITs and IISc Funded by MHRD Page 11 of 39

Here u varies from 0 at the wall to 1 at the edge of the boundary layer. Therefore we can consider that u has order of magnitude equal to 1. It is mathematically represented as O(1). On similar lines, we can as well mention for density as, ρ = O(1). Actually the x-coordinate of all the points in the fluid domain vary from 0 to c which is length of the plate. Therefore the non-dimensional x length scale can be represented as x =O(1). However the y co-ordinate of all the points at a particular x- location varies from 0 toδ where δ is the local boundary layer thickness. Hence the non-dimensional length scale y is smaller magnitude in comparison with other length scales. This can be represented as y =O( δ / c ). For unit flat plate length, we have y =O(δ ). Therefore from the continuity equation in terms of order of magnitude is, [ 0(1) ][ 0(1) ] [ 0(1) ] v + = 0 0(1) 0( δ ) From the above equation it is clear that v must be of an order of magnitude equal to the local boundary layer thickness, δ, i.e. v =O(δ ). We know the non-dimensional form of X-momentum Eq. (0.6). Consider the order of magnitude form of each term as, u ρu = 0(1) x u ρv = 0(1) y p = 0(1) v µ = 0(1) u 1 µ = 0 δ Thus the order of magnitude equation for X momentum can be written as, 1 1 1 0(1) + 0(1) = 0(1) + 0(1) γm Re + δ Joint initiative of IITs and IISc Funded by MHRD Page 1 of 39

Lets assume that the Reynolds number is large. Therefore the term with Reynolds number in the denominator is of small magnitude which can be mathematically mentioned as, 1 Re = 0( ) δ Therefore the above equation now becomes; 1 1 0(1) + 0(1) = 0(1) + 0( δ ) 0(1) γm + δ It is clear from this figure that product of 0( δ )[0(1)] 0( δ ) = has very low order of magnitude in comparison with the rest of the terms in the same equation. This term actually is 1 Re v µ in X momentum equation (Eq. 0.6). Since this term is very small in magnitude we can neglect it. Therefore the non-dimensional X- momentum equation can now be written as, u u 1 p 1 ρu + ρv = + M Re x y γ x y u µ y (.1) The same in the dimensional form is, u u p u ρu + ρ v = + µ This is X-momentum is valid for high Reynolds number flows having thin boundary layer at the wall. Consider the Y momentum equation in non dimensional form given by Eq. (0.7); ρ v v 1 p 1 u v v u x ρ y γ M y Re x µ + = + + Lets do the order of magnitude analysis of this equation. 1 p 1 + = + + γm y δ 0( δ ) 0( δ ) 0( δ ) 0(1) Joint initiative of IITs and IISc Funded by MHRD Page 13 of 39

Here all the terms are of very small magnitude 0( δ ) except the term of pressure gradient if γ M is of unity order of magnitude. Hence that term should also have p very small order of magnitude. It implies that = 0( δ ). Therefore, from Y- y momentum equation given by Eq. 0.7 gets transformed for the boundary layer theory as; p = 0 y (.) This equation clearly states that pressure at a particular X location does not change with Y- coordinate such that the gradient of pressure remains zero in the boundary layer. Therefore the pressure is only function of X co-ordinate, p= px ( ) = pe( x), where pe( x ) is the pressure distribution outside the boundary layer. However if freestream Mach number, M, is very large so as to have 1/ M 0( ) γ = δ. In such cases, the pressure gradient in normal direction cab be large and still satisfy the Y- momentum equation. Hence for large Mach numbers, p/ y might be large enough to be expressed as 0(1). Hence pressure is not constant in the direction normal to the wall for hypersonic flows. Joint initiative of IITs and IISc Funded by MHRD Page 14 of 39

Lecture-3: Boundary layer equations 3.1 Boundary layer equations Let s derive the energy equation under the assumption of very thin boundary layer for very high Reynolds number hypersonic flows. We know that the non-dimensional energy equation is given by Eq. 0.8. This energy equation is for total energy which means the summation of kinetic energy and internal energy. We have neglected the potential energy of the fluid particle. Therefore let us derive the energy equation for kinetic energy alone. Consider the X-momentum and Y-momentum equations given by Eq. (0.) and (0.3). in the dimensional form. Multiplying the X momentum equation by u velocity, we get; Du D u p τ τ ρu = ρ = u + u + u Dt Dt ( / ) xx yx Similarly multiply Y momentum equation by v velocity, we get; Dv D v p ρv = ρ = v + v + v Dt Dt ( / ) τ xy τ yy Adding the above both the equations we have; Du ( / + v /) DV ( /) p p xx yx xy yy u v u τ τ ρ ρ v τ τ = = + + + + Dt Dt (3.1) But the energy equation given by Eq. (0.4), ( u v ) ( u v ) T T ρu + e ρv + e κ + + + = + κ y ( pu) ( pv) ( uτ xx) ( uτ xy) ( vτ xy) ( vτ yy) + + + + + Joint initiative of IITs and IISc Funded by MHRD Page 15 of 39

Replacing the L.H.S. in terms of the substantial derivative form we get, V V De ( + V / ) ρu e+ + ρv e+ = ρ = RHS... y Dt (3.) Subtracting Eq. (3.1) from (3.), we get, the heat energy equation is; D() e T T ( pu) ( pv) ρ = κ + κ + Dt ( u τ xx) ( u τ xy) ( v τ xy) ( v τ yy) p p xx yx xy yy u v u τ τ v τ τ + + + + + + + + Let s simplify the L.H.S., De () e e LHS... = ρ = ρu + ρv Dt From the definition of enthalpy we have, e= h p/ ρ Therefore the heat energy equation is, ( h p/ ρ) ( h p/ ρ) h h p/ ρ p/ ρ ρu + ρv = ρu + ρv ρu ρv, Further simplification of this equation leads to, h h p / ρ p / ρ h h ( pu) ( pv) ( ρu) ( ρv) ρu + ρv ρu ρv = ρu + ρv + pρ + Cancellation of similar terms from both the sides we have, L.H.S.= h h ρu + ρv Now, R.H.S. = T T ( uτ xx) ( uτ xy) ( vτ xy) ( vτ yy) κ + κ + + + + Joint initiative of IITs and IISc Funded by MHRD Page 16 of 39

p p τ xx τ yx τ xy τ yy + u + v u + v + Simplifying the viscous terms, ( u τ xx) ( u τ xy) ( v τ xy) ( v τ yy) xx yx xy yy u v u v u τ τ v τ τ + + + + + = τ xx + τ yy + τ yx + u v u v u τ v xx τ yy τ yx λ µ u v u v + + + = + + + + + Therefore we have heat energy equation as, h h ρu + ρv T T = κ + κ p p + u + v u v λ µ u v u v + + + + + + p Now from Y momentum Boundary Layer Equation we have, = 0. Let us carry out the order of magnitude analysis for the above Energy Equation; 1 1 0(1) + 0(1) = 0(1) + + 0(1) + λ ( 0(1) + 0(1) ) + µ * 0(1) + * 0(1) + + δ 0( δ ) δ From the above equation it is very much clear that temperature gradient along Y axis has larger magnitude as compared to the temperature gradient along X axis. Moreover, we know from the momentum equation, that, v<< u & << for boundary layer assumption. Therefore gradients of lower magnitude, Energy Equation can be expressed as h h T dpe u ρu ρv κ u µ + = + + dx (3.3) Joint initiative of IITs and IISc Funded by MHRD Page 17 of 39

The mass, X-momentum, Y-momentum and energy equations for the boundary layer are respectively as follows, ρu ρv + = 0 u u p u ρu + ρ v = + µ p = 0 h h T dpe u ρu ρv κ u µ + = + + dx These equations are non-linear. However assumptions of boundary layer theory make solution procedure simpler. Apart from this the pressure is only function of X- coordinate hence can be represented using an ordinary differential equation rather than a partial differential equation. The variables u, v, p, ρ, T and h are the unknowns in these equations. However, p can be known from p=p e (x). Rest of the variables likeµ and κ are properties of fluid and are temperature dependant. The following perfect gas relations should also be used to complete the set of equations. p = ρrt h= CT p Boundary conditions to be considered to solve above equations are T At Wall: y=0, u=0, v=0, T=T w or = 0 n wall) w (adiabatic At boundary layer edge: y, u u e, T T e Here subscript e stands for the values measured or known at the edge of the boundary layer. The boundary layer equations are valid for compressible subsonic or supersonic flow. In case of application for hypersonic flows, it should again be noted that the Y- momentum equation p/ y = 0 should be changed for high Mach number cases. Joint initiative of IITs and IISc Funded by MHRD Page 18 of 39

Lecture-4: Similarity solution for boundary layer equation 4.1 Similarity solution of compressible boundary layer equation. Boundary layer theory makes it convenient to reduce the complexity of basic governing equation which has been derived using the order of magnitude estimate for the non-dimensional form of the governing equations. Hence following are the mass, momentum and energy equation which are to be solved for compressible boundary layer. ρu ρv + = 0 (4.1) u u p u ρu + ρ v = + µ (4.) p = 0 y (4.3) h h T dpe u ρu ρv κ u µ + = + + dx (4.4) These equations are derived for the X-Y coordinate system. Hence the variation of all the properties is assumed to be dependant on X and Y co-ordinates. However, lets transform the dependence of all the variables from X and Y to new dependant variables ( ξη, ). This transformation ensures the self similar solution for the velocity profile where u = u( η) and independent of ξ. This transformation has the following dependant variables as, x = eue edx (4.1) ξ ρ µ 0 ue v η = ρdy ξ (4.) 0 Joint initiative of IITs and IISc Funded by MHRD Page 19 of 39

Here ρ e, ue and µ e are the density, velocity and viscosity coefficients at the edge of the boundary layer and are functions of x only. Therefore ξ = ξ( x). These special variables chosen for transformation should be implemented for the governing equations derived especially for the boundary layer. This would lead to new form of the same governing equations. Following are the basic steps involved in this transformation. Step 1. Replacement of derivatives of independent variables. The new independent variables are expressed in terms of the old independent variables using Eq. (4.1) and (4.). Now we have to express the derivatives of them in terms of new independent variables. ξ η = + ξ η ξ η = + ξ η (4.3) (4.4) From the definition of ξ = ξ( x) given by Eq. (4.1) we can write, ξ = ρ eue µ e (4.5) ξ = 0 (4.6) From the definition of η given by Eq. (4.) we can write, η = ρ ξ η ue (4.7) Substituting Eqs. (4.5)-(4.7) into Eqs. (4.3) and (4.4), we can the derivatives as, η = + ρ eue µ e ξ x η (4.8) = ρ ξ η ue (4.9) Joint initiative of IITs and IISc Funded by MHRD Page 0 of 39

We know the definition of the stream function ψ defined as ψ = ρ u ψ = ρv (4.10) (4.11) The X momentum boundary layer equation given by Eq. (4.) in terms of ψ is, ψ u ψ u dpe u = + µ dx (4.1) Let s introduce the derivatives from eq. (4.8) and (4.9) in the eq. (4.1) we get, ueρ ψ u η u ψ η ψ ueρ u ρ eueµ e + ρ eueµ e + ξ η ξ x η ξ x η ξ η dpe ueρ ueρµ u = ρ eueµ e + d ξ ξ η ξ η (4.13) Multiplying Eq. (4.13) by ξ / ueρ, we get, ψ u η u ψ η ψ u ρ eueµ e ρ eueµ + e + η ξ x η ξ x η η ρ e dpe ueρµ u = ξ µ e + ρ dξ η ξ η (4.14) Step. Replacement of dependent variables. Let us define a function f of ξ and η, f ( ξη, ), such that u f = ue η f (4.15) Joint initiative of IITs and IISc Funded by MHRD Page 1 of 39

Here prime denotes the partial derivative of f with respect to η. We know that the velocity at the boundary layer edge is function of X alone. Hence it is function of ξ only ( ue= ue( ξ )). Therefore we can get the following u due f = f + ue ξ dξ ξ (4.16) u = η uf e (4.17) These two steps make it easy to express the governing equations of the thin boundary layer in terms of new variables so as to make it easier to solve. Joint initiative of IITs and IISc Funded by MHRD Page of 39

Lecture-5: Transformation of boundary layer equations 5.1 Transformation of boundary layer momentum equation. We have already seen the two steps involved in getting simpler boundary layer equation. The steps followed to those are seen here. Step 3. Expression of f in terms of ψ. The stream function ψ has been introduced to reduce the number of dependant variables of the governing equations. Therefore let us express the new dependent variable f ( ξη, ) in terms of stream function. From eq. (4.9), (4.10) and (4.15) we can write the following, ue ρ ψ = ρ u= ρ fu ξ η e ψ or = ξ f η (5.1) Integrating this equation with respect to η, we get, ψ = ξ f + F( ξ) (5.) Here F( ξ ) is any arbitrary function of ξ. However, from the definition of stream function, we know that difference in stream function casts mass flow rate. Therefore for the stream function should be anchored to zero at the non-blowing wall, ψξ (,0) = 0. Hence f = 0 and F( ξ ) = 0 ensure the zero value of ψ at the wall. This makes it clear that the any arbitrary function, represented by F( ξ ) must be zero, which leads to, ψ = ξ f (5.3) We will also have, ψ f 1 = ξ + f ξ ξ ξ (5.4) Joint initiative of IITs and IISc Funded by MHRD Page 3 of 39

Step 4. Derivation for final equation. We can derive the final expression using Eqs. (4.16)- (4.17) and (5.1) and substituting into Eq. (4.14) which is X-momentum boundary layer equation, we get, due f η ξ f ρ eueµ e f + ue + uf e dξ ξ f 1 η ρ eueµ e ξ + f + ξ f uf e ξ ξ ρ e dpe ue ρµ = ξ µ e + f (5.5) ρ dξ η ξ The Euler equation for the outer flow is, (5.6) Using Eq. (5.6) we can get the Eq. (5.5) as ue f ' η ξρ eueµ e( f ') ξ f ' ρ eueµ e ξµ eff + + ξ ξ = f ρ eueµ e η ξρ eue µ ef ff ξµ ef f ξ ξ Therefore, ue f ' e e e e e e ( ) ξρ u µ f ' + ξ f ' ρ u µ ξ ξ ( e) due ue ρ ρµ = ξ ue µ e + f ρ dξ η ξ (5.7) Joint initiative of IITs and IISc Funded by MHRD Page 4 of 39

It can be seen that the term involving η / does not appear in the Eq. (5.7). This the main reason for non evaluation of the same equation is, η / x earlier explicitly. Simplified form of 1 ( ) due f f ρ e 1 due f f f 1 ρµ + = + f ue dξ ξ ξ ρue dξ η ξρµ e e (5.8) Let s introduce a variable, C = ρµ / ρ eµ e and obtain the final form of the X- momentum equation in the transformed state. ξ ρ du f f Cf + ff = f + ξ f f u ρ dξ ξ ξ e e ( ) ( ) e (5.9) This transformed equation is valid for steady compressible flow in the thin boundary layer. The Y-momentum boundary layer y-momentum equation, gets transformed as, p = 0 η (5.10) 5. Transformation of boundary layer energy equation We can obtain the energy equation in the transformed variables using same strategy incorporated earlier. Let s substitute Eq. (4.8)- (4.11) in the L.H.S. of Eq. (4.4), we obtain the transformed R.H.S. term of energy equation as ψ h ψ h T dpe u κ u µ = + + dx (5.11) Using Eq.(4.8), we can express the first term on LHS as, ψ h ue ρ ψ h h ρ eu η = eµ e + y x ξ η ξ η (5.1) Joint initiative of IITs and IISc Funded by MHRD Page 5 of 39

Let s define the static enthalpy as the non dimensional variable, h g = g( ξη, ) = (5.13) he Hence we can write as, h g he = h e + g ξ ξ ξ (5.14) h = η hg e (5.15) Here g = g/ η, since he= he( x) hence its derivation with respect to η is zero. Therefore using Eq. (5.14), (5.15), we can re-express the first term of LHS of Eq. (5.1) as, ψ h ue ρ ψ h h ρ eu η = eµ e + y x ξ η ξ η ueρ g he η = ξ f ρ eueµ e he + g + hg e ξ ξ ξ g he η = ( ueρ f ) ρ eueµ ehe + ρ eueµ eg + heg ξ ξ ψ h g he η = ρρ e ue µ ehf e + ρρ e ue µ efg + ueρf hg e ξ ξ (5.16) Consider the second term of L.HS. of Eq.(5.1) using Eqs. (4.8) and (4.9) in the same way, we get, ψ h ue h eu ψ η ρ ψ ρ = eµ e + ξ η ξ η (5.17) Joint initiative of IITs and IISc Funded by MHRD Page 6 of 39

Now substituting Eq.(5.15),(5.1) and (5.4) in Eq. (5.17), we have ψ h ψ h f 1 η ueρ = ρ eueµ e ξ + f + ξ f hg e ξ ξ ξ f ρρ e ue µ ehefg η = ρρ e ueµ ehg e + + ρuhgf e e ξ ξ (5.18) Therefore complete L.H.S. of Eq.(5.11) using Eq. (5.16) and (5.18) is ψ h ψ h LHS... = g he η = ρρ e ue µ ehf e + ρρ e ue µ efg + ueρf hg e ξ ξ f ρρ e ue µ ehefg η ρρ e ueµ ehg e ρuhgf e e ξ ξ g he f hefg LHS... = ρρ e ue µ e hf e fg hg e + ξ ξ ξ ξ (5.19) Joint initiative of IITs and IISc Funded by MHRD Page 7 of 39

Lecture-6: Similarity solution of boundary layer equation 6.1 Transformation of boundary layer energy equation and similarity solution The simpler forms of X-momentum and Y-momentum have been obtained. The energy equation has also been obtained in the transformed form for LHS. Let s consider the RHS of Eq. (5.11) for the transformation. RHS of the Eq. (5.11) is as. T dpe u RHS... κ u µ = + + dx (6.1) Lets consider the first term of this RHS and express the temperature in terms of enthalpy as T = h/ Cp we get, T ueρ κueρ h κ = ξ η ξ η Cp (6.) If we consider the gas to be calorically perfect and variation of C p to be negligible then, we get T ueρ ρκue h ueρ ρκue κ = = hg e ξ η ξ η Cp ξ η ξcp T hu e eρ ρµ κ = g ξ η Pr (6.3) For the nd term of Eq.(6.1), we know, dpe dpe u = uf e ρ eueµ e dx d ξ dpe dpe due 3 due u = uf e ρ eueµ e ρ eueµ ef ρeue ρ eueµ ef dx dξ = = dξ dξ (6.4) Joint initiative of IITs and IISc Funded by MHRD Page 8 of 39

For the 3 rd term, from Eq.(6.1), we get 4 e e e u u ρ u u ρµ u ρµ µ = µ = = ξ ξ ξ ξ ( uf e ) ( f ) (6.5) Therefore the complete form of RHS of Eq. (6.1), 4 e e eρ ρµ 3 e eρ µ e e e T dp u h u du u RHS... = κ + u + µ = g ρ u µ f + f y dx y ξ η Pr dξ ξ (6.6) Therefore the complete form of Eq. (6.1) using Eq. (5.19) and (6.3)-(6.5) is ( ) Dividing both sides by ρρ u e e e µ, we get g he f hfg e he ρµ ρ euf e due ueρµ hf e + fg hg e = g + f ξ ξ ξ ξ ξρ eµ e η Pr ρ dξ ξρ eµ e ( ) Multiplying both sides by ξ / he, introducing the term C = ρµ / ρ eµ ethe above equation simplifies to, g ξ he f C ξρ euef due ue C ξ f + f g ξg fg = g + f ξ he ξ ξ Pr ρhe dξ he Finally rearranging the terms we get, ( ) C g f g he f ρ euef due ue C g + fg = ξ f g f Pr + + ξ he ξ ξ ρhe dξ he ( ) (6.7) Hence, the governing Eqs. (5.9), (5.10) and (6.7) form the transformed set of equations for compressible hypersonic boundary layer. The boundary conditions to be considered for solution are, η = 0, f = f ' = 0 and g = gw for isothermal wall boundary condition While, η = 0, f = f ' = 0 and g ' = 0 for adiabatic wall boundary condition Joint initiative of IITs and IISc Funded by MHRD Page 9 of 39

The boundary condition at the edge of the boundary layer is, η, f ' = 1 and g = 1 We can solve the transformed boundary layer equations which are partial differential equations. The necessary outcome for this is the velocity and enthalpy variation in the boundary layer. Since pressure is assumed to have same variation in the boundary layer as that at the outer flow, we can always evaluate the other thermodynamic properties in the boundary layer. Moreover the main parameters which we can evaluate are skin friction coefficient and Stanton Number. We know that the skin friction coefficient is defined as, C f = τw 1 ρeu e Where, the wall shear can be calculated as, u τ W = µ w Hence, C f u = µ ρ w eu e w Using the expression given by Eq. (4.9) we can write the equation for skin friction coefficient as, C f u ρ u = ρ e w µ w eu e ξ η w C f = ρ u e e u eρw µ w f ( ξ,0) ξ C f µρ w w = f ( ξ,0) (6.8) ρ ξ e Joint initiative of IITs and IISc Funded by MHRD Page 30 of 39

We can also calculate the Stanton number from the solution of boundary layer equations. Here Stanton number is defined as, S t qw = ρ u ( h h ) e e aw w Here h aw is the adiabatic wall enthalpy, which is the enthalpy at the wall when there is no heat transfer from the fluid to the wall. As well, h w is the wall static enthalpy corresponding to the temperature of the wall at isothermal wall temperature condition. Here, q W is the heat flux to the wall from the fluid which can be evaluated as, T = qw k y w or qw k h = cp w Assumption made here is the constancy of specific heat or calorically perfect gas. Thus using this expression for heatflux and from Eq. (4.9) and (5.13) we can calculate the Stanton number as, S t 1 k h = ρeue( haw hw ) cp w S S t t 1 k ueρhe h = ρeue( haw hw ) cp ξ η 1 k ueρhe = g'( ξ, 0) ρ u ( h h ) c ξ e e aw w p w (6.9) Joint initiative of IITs and IISc Funded by MHRD Page 31 of 39

Lecture-7: Hypersonic flow over flat plate 7.1 Hypersonic flow over flat plate. The equations (5.9)(5.10) and (6.7) form the boundary layer equations. Let s consider these equations for solution of hypersonic flow over flat plate. As we had seen earlier, there are two prominent boundary conditions considered for this flow over flat plate as, Isothermal wall, T w =const T Or an adiabatic wall, = 0 w All the freestream variables of the variables at the edge of the boundary layer are now assumed be constant. Hence The variables ue, he, ρ e are of constant values and independent of ξ and η. Therefore the governing equations reduce to, f f ( Cf ) + ff = ξ f f ξ ξ (7.1) C g f ue C g fg ξ f g + = f Pr ξ ξ he ( ) (7.) These equations are the partial differential equations. Let s assume that the functions f and g are functions of Eqs.(4.40) and (4.41) are still partial differential equations. Let us assume that f and g are the functions of η only. Hence f and g are independent of ξ. For these assumptions the same equations reduce to, ( Cf ) + ff = 0 (7.3) C ue C g + fg + ( f ) = 0 Pr he (7.4) These equations are single independent variable equations, hence are the nonlinear ordinary differential equations. These equations are valid for a compressible boundary layer over a flat plate with constant wall conditions. Here both the constants Joint initiative of IITs and IISc Funded by MHRD Page 3 of 39

C = ρµ / ρ eµ e and Pr = µ cp / κ are meant for the local values in the boundary layer. We can you shooting technique for solving these equations. Thus obtained velocity and thermal boundary layers can be used to obtain the wall heat flux and shear stress in turn the Stanton number and skin friction coefficient. Here the ratio of Stanton number and skin friction coefficient can be approximated as the function of Prandtl s number by Reynolds analogy. 7. Hypersonic flow around a stagnation point Hypersonic flow around a blunt body forms a stagnation point which can also be evaluated using boundary layer equations. Consider hypersonic flow around a blunt body which marks a stagnation region, as sketched in Fig. 7.1. Consider the flow to be D flow for simplicity; hence the span of the cylinder is infinity. Let X be the direction of freestream flow and R be the radius of curvature at the surface. Fig. 7.1 Hypersonic flow around the stagnation point region [1]. Joint initiative of IITs and IISc Funded by MHRD Page 33 of 39

Let us consider that f and g from Eq. (5.9) and (6.7) are functions of η alone, Therefore, = = = 0 This leads to the following changed in Eq. (5.9) and (6.7) as, (CCff ) + ffff = ξξ [(ff ) ρρ ee ] dduu ee uu ee ρρ dddd (7.5) And CC PPPP gg + ffff = ξξ[ ρρ eeuu ee ρρh ee ff dduu ee dddd ] CC uu ee h ee (ff ) (7.6) These equations are still ξξ dependent. Moreover, we can assume that the velocity at the edge of the boundary layer, uu ee is very small and static enthalpy at the edge of the boundary layer is h ee = h 0 (stagnation enthalpy). These facts lead to the assumption that, uu ee h ee 0 (7.7) We can as well assume that, the flow velocity is the boundary layer at edge of the boundary layer at the stagnation point behind the normal shock is low as to be considered in the incompressible flow regime. Hence, we can use the result of incompressible and inviscid flow at the stagnation point, which expresses the boundary layer velocity as, uu ee = dduu ee xx (7.8) dddd ss Here dduu ee is the velocity gradient at the stagnation point external to the boundary dddd ss layer. Using these assumptions we can re-express the ξξ as, xx ξξ = ρρ ee uu ee μμ ee dddd = ρρ ee μμ ee dduu ee xxxxxx 0 0 dddd ss Or xx ξξ = ρρ ee μμ ee dduu ee xx dddd ss (7.9) Joint initiative of IITs and IISc Funded by MHRD Page 34 of 39

The velocity gradient required for calculation here can be evaluated as, dduu ee dddd = dduu ee dddd dddd dddd = (dduu ee dddd ) (dddd dddd ) (7.10) But from the definition of ξξ, Eq. (4.1) we know that, dddd dddd = ρρ eeuu ee μμ ee (7.11) Substituting this Eq. (7.11) into (7.10), we get dduu ee = 1 dduu ee dddd ρρ ee uu ee μμ ee dddd (7.1) From Eq. (7.11) into (7.1), we write, dduu ee dddd ss = 1 ρρ ee μμ ee xx Or, dduu ee dddd = ss 1 μμ ee ρρ ee (dduu ee dddd) ss xx dduu ee dddd ss Let s consider the term (ξξ uu ee ) dduu ee dddd which appears in the Eq. 7.1. We can rewrite this term and can derive for the same as, ξξ uu ee dduu ee ) ss xx ] = [ρρ eeμμ ee (dduu ee dddd dddd (dduu ee dddd ) ss xx 1 ρρ ee μμ ee xx = 1 Similarly consider the term (ξξ uu ee )(ρρ ee uu ee ρρh ee ) dduu ee dddd appearing in Eq. (7.). We can re-arrange this term as well, ξξ ρρ eeuu ee dduu ee = ρρ ee ρρ ρρh ee dddd ρρh ee uu ee dduu ee xx ee dddd ss dduu ee xx 1 = ρρ ee dduu ee dddd ss ρρ ee μμ ee xx ρρh ee dddd ss xx However at the stagnation point, x=0. This fact leads to, ξξ ρρ eeuu ee ρρh ee dduu ee dddd = 0 Moreover, for a calorically perfect gas, we have, ρρ ee = pp ee TT ee = pp ee h = h gg ρρ pp TT pp h ee h ee Joint initiative of IITs and IISc Funded by MHRD Page 35 of 39

Using all these short expression for various terms involved in Eq.(7.1) and (7.) we can get these equations as, (CCCC ) + ffff = (ff ) gg (7.13) CC PPPP gg + ffgg = 0 (7.14) These equations are special equation for stagnation point flow. These equations are independent of ξξ. We can use numerical techniques shooting technique to solve these equations. Joint initiative of IITs and IISc Funded by MHRD Page 36 of 39

Lecture-8: Stagnation point flow field 8.1 Hypersonic flow around a stagnation point. The known correlation for stagnation point heat flux for cylinder is Cylinder: (8.1) This equation is valid for D configurations. Moreover for, sphere or axi-symmetric configuration the stagnation point heat flux can be predicted using, Sphere: 0.763PPPP 0.6 (ρρ ee μμ ee ) 1 dduu ee (h dddd aaaa h ww ) (8.) These equations are called as Fay and Riddle equations for stagnation point heat flux prediction. The stagnation point heat flux is more for sphere in comparison for that of the cylinder of same diameter. The main reason for this discrimination is the dimensionality of the flow. The hypersonic flow is two dimensional for flow over cylinder hence it has two possible direction for passing over the cylinder however the flow over the sphere is three dimensional. Due to the extra available dimension, flow passes around the object easily hence the shock stand off distance and the boundary layer thickness decrease for the sphere in comparison with the cylinder of same diameter. The decreased boundary layer thickness increases the gradient and hence the shear stress and heat flux at the wall for sphere in comparison with the cylinder. The closer observation to the Eq. (8.1) and (8.) suggests that the wall heat flux is propotional to the square root of the stream wise velocity gradient,( dduu ee ) along the dddd stagnation streamline. dp = ρ u du e e e e Hence, du 1 dp e e = (8.3) dx ρ u dx e e Joint initiative of IITs and IISc Funded by MHRD Page 37 of 39

We can evaluate the pressure gradient of this equation from the known pressure variation given by Newtonian method. C p = sin θ Here θθ is flow deflection angle, the angle between a tangent at any point on the surface and the freestream direction. Let s define as the angle between freestream velocity and the local surface normal. Hence the pressure distribution gets transformed to, C p = cos ϕ From definition of pressure coefficient we have, Or, pp ee pp qq = cccccc cos p = q ϕ + p e Differentiating this equation w.r.t X we get, dpe 4q = dϕ cosϕsin ϕ dx ρ u dx e e (8.4) Substituting the Eq. (8.4) in to the Eq. (8.3) we get, dduu ee dddd = 4qq cccccc ssssss dd uu ee ρρ ee dddd (8.5) All the terms involved in this equation can be evaluated using various approximations. uu ee = dduu ee xx (8.6) dddd ss Joint initiative of IITs and IISc Funded by MHRD Page 38 of 39

cccccc 1 ssssss xx RR dd dddd = 1 RR (8.7) qq = 1 (pp ee pp ) (8.8) Now, substituting Eqs. (8.6) (8.8) in (8.5), we get, Or dduu ee dddd = (pp ee pp ) xx ρρ ee xx RR 1 RR dduu ee dddd = 1 RR (pp ee pp ) ρρ (8.9) This is the approximate expression for velocity gradient encountered in the Eq. (8.1) or (8.). The expression clearly suggests that the wall heat flux is inversely proportional to the nose radius or radius at the stagnation point. This is the main reason for having hypersonic vehicles being blunt nosed to reduce the heat load at the compromise of the drag. Joint initiative of IITs and IISc Funded by MHRD Page 39 of 39