Modeling Shock Waves Using Exponential Interpolation Functions with the Least-Squares Finite Element Method

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1 Old Dominion University ODU Digital Commons Mechanical & Aerospace Engineering Theses & Dissertations Mechanical & Aerospace Engineering Spring 26 Modeling Shock Waves Using Exponential Interpolation Functions with the Least-Squares Finite Element Method Bradford Scott Smith Jr. Old Dominion University Follow this and additional works at: Part of the Aerospace Engineering Commons, and the Applied Mathematics Commons Recommended Citation Smith, Bradford S.. "Modeling Shock Waves Using Exponential Interpolation Functions with the Least-Squares Finite Element Method" (26). Master of Science (MS), thesis, Mechanical & Aerospace Engineering, Old Dominion University, DOI:.25777/ ye3-av52 This Thesis is brought to you for free and open access by the Mechanical & Aerospace Engineering at ODU Digital Commons. It has been accepted for inclusion in Mechanical & Aerospace Engineering Theses & Dissertations by an authorized administrator of ODU Digital Commons. For more information, please contact

2 MODELING SHOCK WAVES USING EXPONENTIAL INTERPOLATION FUNCTIONS WITH THE LEAST-SQUARES FINITE ELEMENT METHOD by Bradford Scott Smith Jr. B.S. May 27, Virginia Polytechnic Institute and State University A Thesis Submitted to the Faculty of Old Dominion University in Partial Fulfillment of the Requirements for the Degree of MASTER OF SCIENCE AEROSPACE ENGINEERING OLD DOMINION UNIVERSITY May 26 Approved by: Gene J. Hou (Director) Miltiadis D. Kotinis (Member) Duc T. Nguyen (Member)

3 ABSTRACT MODELING SHOCK WAVES USING EXPONENTIAL INTERPOLATION FUNCTIONS WITH THE LEAST-SQUARES FINITE ELEMENT METHOD Bradford Scott Smith Jr. Old Dominion University, 26 Director: Dr. Gene J. Hou The hypothesis of this research is that exponential interpolation functions will approximate fluid properties at shock waves with less error than polynomial interpolation functions. Exponential interpolation functions are derived for the purpose of modeling sharp gradients. General equations for conservation of mass, momentum, and energy for an inviscid flow of a perfect gas are converted to finite element equations using the least-squares method. Boundary conditions and a mesh adaptation scheme are also presented. An oblique shock reflection problem is used as a benchmark to determine whether or not exponential interpolation provides any advantages over Lagrange polynomial interpolation. Using exponential interpolation in elements downstream of a shock and having edges coincident with the shock showed a slight reduction in the solution error. However there was very little qualitative difference between solutions using polynomial and exponential interpolation. Regardless of the type of interpolation used, the shocks were smeared and oscillations were present both upstream and downstream of the shock waves. When a mesh adaptation scheme was implemented, exponential elements adjacent to the shock waves became much smaller and the numerical solution diverged. Changing the exponential elements to polynomial elements yielded a convergent solution. There appears to be no significant advantage to using exponential interpolation in comparison to Lagrange polynomial interpolation.

4 Copyright, 26, by Bradford Scott Smith Jr., All Rights Reserved. iii

5 iv NOMENCLATURE H h β t x y l η η k γ Γ wall λ [J] [K] [Q] {κ} {R} {f} H P R R Ω ω Ω e φ ψ ρ ρ σ θ ε ϑ n Component of the Hessian reconstructed with positive eigenvalues Specific enthalpy Acute angle between a shock wave and the upstream velocity vector Time step Step in the x-direction Step in the y-direction Lagrange polynomial Local element coordinate in one dimension Location of node k in one-dimensional local coordinates Ratio of specific heat capacities Solid wall boundary Eigenvalue Jacobian Finite element coefficient matrix Finite element coefficient matrix for a solid wall boundary Eigenvector Vector of nonlinear residuals Finite element residual vector Component of the Hessian Spring potential Gas constant Nonlinear residual Domain of the finite element problem Relaxation parameter Domain of an element Potential field Two-dimensional interpolation function in global coordinates Density Free stream density Dummy variable that can represent density, velocity, or pressure Angle of flow deflection downstream of an oblique shock Exponential interpolation function in one dimension Penalty weight Outward pointing unit normal vector of an element

6 v v Velocity vector φ Two-dimensional shape function ψ Two-dimensional interpolation function in the local coordinate system ξ, η Local coordinate directions {U} Vector of dependent variables a Speed of sound C v E e f x f y f z g h I Specific heat capacity at constant volume Error Specific internal energy Body force in the x-direction Body force in the y-direction Body force in the z-direction Interpolation coefficients Exponential parameter Functional of residuals i, j, k, s Index variables I I e I GL k L M M n N GL P q R T t u V v v x v y v z Integral calculated using the quadgk function in MATLAB Functional of residuals on an element Integral calculated using Gauss-Legendre quadrature Spring stiffness Length Mach number Mach number of the velocity component normal to a shock wave Number of Gauss-Legendre quadrature points Pressure Heat transfer Linear residual Temperature Time Dummy dependent variable Magnitude of velocity Free stream velocity magnitude Component of velocity in the x-direction Component of velocity in the y-direction Component of velocity in the z-direction x, y, z Global coordinate directions

7 vi H Heaviside function

8 vii TABLE OF CONTENTS Page LIST OF TABLES ix LIST OF FIGURES x Chapter INTRODUCTION EXPONENTIAL INTERPOLATION FUNCTIONS Derivation Selection of the Exponent Parameters Gauss-Legendre Quadrature Extension to Higher Dimensions GOVERNING DIFFERENTIAL EQUATIONS Conservation of Mass Conservation of Momentum Conservation of Energy FINITE ELEMENT IMPLEMENTATION Linearization Least-Squares Finite Element Method Boundary Conditions Conversion to Local Coordinates Mesh Adaptation SHOCK REFLECTION EXAMPLE Analytical Solution Solution Using Uniform Grid and Polynomial Interpolation Comparison of Interpolation Functions on a Shock-Aligned Mesh Comparison of Interpolation Functions Using Mesh Adaptation CONCLUSION REFERENCES

9 viii Page APPENDIX VITA

10 ix LIST OF TABLES Table Page Factors and levels used to compare interpolation functions Factor combinations and error for density Factor combinations and error for magnitude of velocity Factor combinations and error for pressure Significant factors for magnitude of velocity

11 x LIST OF FIGURES Figure Page One-dimensional exponential interpolation functions for h = {,,, } One-dimensional exponential interpolation functions for h = {,,, } One-dimensional exponential interpolation functions for h = {,,, } Example of a function and a poor approximation to the function using exponential interpolation and h = {,, 7, } Example of a function and a good approximation to the function using exponential interpolation and h = {,, 7, } Number of Gauss-Legendre quadrature points needed to obtain E Integration error for h 9 = Interpolation functions defined in Eq. (66) Interpolation functions defined in Eq. (67) using the exponential parameters h = { 7, } Interpolation functions defined in Eq. (68) using the exponential parameters h = { 7, } Domain and shock wave locations Uniform bilinear mesh Pressure calculated using uniform mesh, polynomial interpolation, and t =.5 with dashed lines showing the analytical shock positions Pressure vs x at y = Shock-aligned bicubic mesh ρ ρ using t =.5 and polynomial interpolation ρ ρ using t =.5 and exponential interpolation downstream of the shock waves V V using t =.5 and polynomial interpolation V V using t =.5 and exponential interpolation downstream of the shock waves P P using t =.5 and polynomial interpolation P P using t =.5 and exponential interpolation downstream of the shock waves Density vs. x at y =.5, t = Density vs. x at y =.5, t = Velocity magnitude vs. x at y =.5, t = Velocity magnitude vs. x at y =.5, t =

12 xi Figure Page 26 Pressure vs. x at y =.5, t = Pressure vs. x at y =.5, t = Mesh containing exponential elements downstream of the shocks after the first adaptive cycle Mesh containing only polynomial elements after the first adaptive cycle Density vs. x at y =.5 after the first adaptive cycle Velocity magnitude vs. x at y =.5 after the first adaptive cycle Pressure vs. x at y =.5 after the first adaptive cycle Mesh containing exponential elements downstream of the shocks after the second adaptive cycle Mesh containing only polynomial elements after the second adaptive cycle Density vs. x at y =.5 after the second adaptive cycle Velocity magnitude vs. x at y =.5 after the second adaptive cycle Pressure vs. x at y =.5 after the second adaptive cycle Mesh containing exponential elements downstream of the shocks after the third adaptive cycle Mesh containing only polynomial elements after the third adaptive cycle Density vs. x at y =.5 after the third adaptive cycle Velocity magnitude vs. x at y =.5 after the third adaptive cycle Pressure vs. x at y =.5 after the third adaptive cycle Mesh containing exponential elements downstream of the shocks after the fourth adaptive cycle Mesh containing only polynomial elements after the fourth adaptive cycle Density vs. x at y =.5 after the fourth adaptive cycle Velocity magnitude vs. x at y =.5 after the fourth adaptive cycle Pressure vs. x at y =.5 after the fourth adaptive cycle Mesh containing exponential elements downstream of the shocks that were changed to polynomial elements during the first adaptive cycle Density vs. x at y =.5 after the exponential elements were changed to polynomial elements Velocity magnitude vs. x at y =.5 after the exponential elements were changed to polynomial elements Pressure vs. x at y =.5 after the exponential elements were changed to polynomial elements

13 CHAPTER INTRODUCTION Fluid properties in flows containing shock waves experience a sudden jump at the shock waves, similar to a Heaviside step function. Typical finite element models use polynomials as smooth approximations for the fluid properties which tend to either have a smearing effect on shock waves or create oscillations in the region of shock waves. Improvements to the finite element method for modeling shock waves fall into one or a mix of three general categories; mesh manipulation, discontinuous methods, and interpolation function refinement. Mesh manipulation typically involves subdividing elements into smaller elements, increasing the degrees of freedom available to approximate large gradients but also requiring more computational resources to generate the coefficient matrix. Another mesh manipulation approach is to move nodes closer to large gradients while keeping the total global degrees of freedom constant. Taghaddosi et al. developed a mesh adaptation method that also aligns the edges of elements with shock waves. The method estimates the error in the approximation of the dependent variables and moves nodes to distribute the error evenly over the mesh. Subdividing one element into smaller elements usually also requires the neighboring elements to be subdivided to maintain continuity. Discontinuous finite element methods circumvent that disadvantage by breaking the continuity between adjacent elements and imposing a constraint on the inter-element fluxes. Since inter-element continuity is not required, discontinuous Galerkin methods allow easy implementation of mesh refinement schemes. 2 The degree of interpolation polynomials can also be increased or decreased without affecting neighboring elements. 2 A discontinuous least-sqares method was published by Potanza and Reddy 3 but they did not address the subject of shock capturing. Another approach to shock capturing is to increase the degree of the interpolation polynomial. Like mesh refinement, polynomial refinement provides more degrees of freedom to approximate discontinuities and also requires refinement in neighboring elements unless discontinuous methods are used. If nodal interpolation functions are used, polynomial refinement requires adding more nodes to the elements. An alternative that does not require additional nodes to increase the polynomial degree is modal interpolation functions. When using modal functions, the finite element method solves for coefficients of the hierarchical modes of the functions. 4 The approach to shock modeling proposed here is to do away with polynomial interpolation functions in elements adjacent to shock waves in favor of exponential

14 2 interpolation functions. Since Heaviside functions can be approximated by continuous exponential functions, the hypothesis of this research is that exponential interpolation functions will approximate fluid properties at shock waves with less error than polynomial interpolation functions. The motivation to investigate the suitability of exponential functions is that the potential reduction of interpolation error near shock waves may also reduce or eliminate the need for mesh refinement in those regions. There are only a few published works in which Heaviside functions are used for finite element modeling. Meiring and Rosinger 5 used basis functions composed of products of Heaviside and continuous functions to solve nonlinear partial differential equations. Their results did not show an overall increase or decrease in error and they conclude that their basis functions lead to a system that is too loosely connected and that the possibility exists of improving the method by choosing basis functions which are not completely disjointed. Ichimura, Hori, and Wijerathne 6 applied Heaviside basis functions to finite element simulations of earthquake ground displacement. However, their objective was to obtain a diagonalized mass matrix without lumping, which can increase numerical error. The use of continuous approximations of Heaviside functions to model sharp gradients in the finite element method appears to have received very little attention. In Chapter 2, the exponential interpolation functions are derived and some properties of the functions are demonstrated. General expressions for conservation of mass, conservation of momentum, and conservation of energy are converted to the dimensionless Euler equations in Chapter 3. In Chapter 4, the Euler equations are linearized using Newton s method, the least-squares finite element method is used to generate the element equations, and the boundary conditions are explained. A shock reflection example is presented in Chapter 5. Numerical solutions of the shock reflection example are calculated using polynomial and exponential interpolation. The analytical solution is used to calculate the absolute error in the numerical solution and the performance of the exponential and polynomial interpolation functions is discussed.

15 3 CHAPTER 2 EXPONENTIAL INTERPOLATION FUNCTIONS This chapter introduces the exponential interpolation functions. Section 2. presents a detailed derivation of one-dimensional exponential interpolation functions along with a more general approach using Cramer s rule. In Section 2.2, exponential parameters are selected for interpolation of large gradients. Limits are presented to demonstrate the behavior of the exponential interpolation functions and several examples are used to show potential mistakes that could lead to large interpolation errors. An empirical equation is developed in Section 2.3 to determine the necessary number of quadrature points to integrate the exponential interpolation functions accurately. In Section 2.4, two-dimensional interpolation functions are derived using products of exponential interpolation functions and Lagrange polynomials. 2. Derivation The procedure to derive the exponential interpolation functions is demonstrated for a one-dimensional element with two nodes and C continuity. The nodes are located at η = ±. Start with a general approximation for some dependent variable u. u(η) g + g 2 e h η () The variables g and g 2 do not have any physical meaning. They are only used in the procedure to derive the interpolation equations. They are found by imposing the condition that u (η k ) = u k, where u k is the value of the dependent variable u at node k and η k is the location of node k, and solving the resulting set of equations. u ( ) = u = g + g 2 e h (2) Rearrange Eq. (3) and substitute it into Eq. (2). u () = u 2 = g + g 2 e h (3) g = u 2 g 2 e h (4)

16 4 u = u 2 g 2 e h + g 2 e h (5) Solve Eq. (5) for g 2. g 2 = u e h e h Equation (6) is substituted into Eq. (4), which is solved for g. u 2 e h e h (6) g = u 2 u e h e h e h + u 2e h e h e h (7) Equations (6) and (7) are substituted into Eq. () and the result is rearranged. u (η) u 2 u e h e h e h + u ( 2e h u u 2 e h e h + e h e h e h e h ( e h ) η e h = e h e h u + ( + eh ) e h η e h e h u 2 ( e h ) ( e h η e h ) η e h = u + u 2 2 sinh h 2 sinh h ) e h η Equation (8) is rewritten to resemble the typical representation of approximation equations used in the finite element method. (8) u(η) ε u + ε 2 u 2 = 2 ε k u k (9) k= ε = eh e h η 2 sinh h () ε 2 = eh η e h 2 sinh h () Hereafter ε k is used to represent the exponential interpolation functions. For higher degree functions, the preceding steps become tedious. A simpler way to generate high degree exponential interpolation functions is to use Cramer s rule. The first step is to write a general approximation function. u (η) g + g 2 e h η + g 3 e h 2η 2 + g 4 e h 3η 3 (2) Again, the condition that u (η k ) = u k is imposed. The resulting set of equations is arranged in matrix form.

17 5 e h η e h 2η 2 e h 3η 3 e h η 2 e h 2η2 2 e h 3η2 3 e h η 3 e h 2η3 2 e h 3η3 3 e h η 4 e h 2η4 2 e h 3η 4 3 g g 2 g 3 g 4 = The next step in Cramer s rule is to compute the determinant of the 4 4 matrix in Eq. (3). e h η e h 2η 2 e h 3η 3 e h η 2 e h 2η2 2 e h 3η 3 2 D = (4) e h η 3 e h 2η3 2 e h 3η3 3 e h η 4 e h 2η4 2 e h 3η4 3 The first column of the 4 4 matrix in Eq. (3) is replaced with {u u 2 u 3 u 4 } T and the determinant is calculated. u e h η e h 2η 2 e h 3η 3 u 2 e h η 2 e h 2η2 2 e h 3η 3 2 D = (5) u 3 e h η 3 e h 2η3 2 e h 3η3 3 u 4 e h η 4 e h 2η4 2 e h 3η4 3 u u 2 u 3 u 4 (3) The previous step is repeated, replacing each column with {u u 2 u 3 u 4 } T and calculating the determinant. u e h 2η 2 e h 3η 3 u 2 e h 2η2 2 e h 3η 3 2 D 2 = u 3 e h 2η3 2 e h 3η3 3 u 4 e h 2η4 2 e h 3η4 3 e h η u e h 3η 3 e h η 2 u 2 e h 3η 3 2 D 3 = e h η 3 u 3 e h 3η3 3 e h η 4 u 4 e h 3η4 3 (6) (7)

18 6 e h η e h 2η 2 u e h η 2 e h 2η2 2 u 2 D 4 = (8) e h η 3 e h 2η3 2 u 3 e h η 4 e h 2η4 2 u 4 The determinants D, D 2, D 3, and D 4 can be computed such that the result is a sum of coefficients multiplied by u, u 2, u 3, and u 4. The determinant D is constant for a given set of exponential parameters, h i, and node locations, η k. The coefficients of the interpolation equation are calculated using Eq. (9). g i = D i D The coefficients g i are then substituted into Eq. (2) and the result is rearranged to obtain Eq. (2). Cramer s rule can be implemented in software to generate interpolation functions of any degree. u (η) (9) 4 ε k u k (2) k= One-dimensional exponential interpolation functions for a five node element with nodes located at η = ±, ±, and are derived starting with Eq. (2). 2 u(η) g + g 2 e hη + g 3 e h 2η 2 + g 4 e h 3η 3 + g 5 e h 4η 4 (2) The resulting interpolation equations are shown without the remaining derivation steps. They will be used to demonstrate the effect of various choices of the exponential parameters.

19 ε (η) = ( e ηh ) sinh h ( sinh h 2 sinh h 3 sinh h sinh h 3 ( ) + e η2 h 2 ( (e sinh h h 3 2 ) ( ) e h ((e h 2 ) ( e h ) ( ) e η3 h 3 sinh h 2 2 ( sinh h 2 sinh h 3 sinh h sinh h 3 ( ) + e η4 h 4 ( (e sinh h h 2 2 ) ( ) e h ((e h 2 ) ( e h 4 6 ( e h 3 (e 8 h ) 4 )) + sinh h 3 8 ) ) ( e h ) ( e h 2 (e 4 h ) 3 )) + sinh h 3 8 ) ) ( e h 2 4 (( ) e h (e 2 h 4 ) ( e h ) ( e h 4 6 (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) ( (e h ) ( ) ( e h 2 4 e h 2 (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) )) ) (e h 2 )) (22) 7

20 ε 2 (η) = ( e ηh ) sinh h 3 2 ( sinh h 2 sinh h 3 sinh h sinh h 3 ( ) + e η2 h 2 (( ) sinh h e h 3 (e 8 h 4 ) ( e h 3 ) ( )) e h 4 6 ( (e h 2 2 ) ( ) ( ) e h 4 6 e h 2 4 ( ) + e η3 h 3 sinh h 2 ( ) sinh h 2 sinh h 3 sinh h sinh h 3 8 ( ) + e η4 h 4 sinh h (( e h 2 8 ) ) (e 4 h 3 ) ( e h 2 ) ( )) (( e h sinh h 3 ( ) ( ) 2 ((e h 2 ) e h 4 6 e h 2 4 ( (e h + sinh h 3 ) ( ) ( e h 4 6 e h (e 2 h ) (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) e h 4 )) ) (e 2 h 2 ) ( e h ) ( )) e h 2 4 (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) (23) ε 3 (η) = ( ) e η2 h 2 + ( ) e η4 h 4 (e h 2 ) ( e h 4 ( e h 4 ) ) (e h 4 6 ) ( ) 6 e h 2 4 (e h 4 ) ( ) e h 2 4 ( e h 2 ) ) (e h 2 ) ( e h 4 6 ( ) + e h 2 4 (e h 4 ) (24) 8

21 ε 4 (η) = ( e ηh ) sinh h 3 2 ( sinh h 2 sinh h 3 sinh h sinh h 3 ( ) + e η2 h 2 ( (e sinh h h 3 ) ( ) e h 4 6 ( ) 2 ((e h 2 ) e h 4 6 ( ) + e η3 h 3 sinh h 8 ) ( e h 3 8 ) (e h 2 ( sinh h 2 sinh h 3 sinh h sinh h 3 8 ) ( ) + e η4 h 4 ( (e sinh h h 2 ) ( ) e h ( e h 2 ((e h 2 ) ( e h 4 6 ( e h )) (( + sinh h 3 ) e h ) (e 2 h 4 ) ( e h ) ( )) e h 4 6 (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) ) (e 4 h 3 )) ( (e h + sinh h 3 ) ( ) e h 2 4 ) ( ) ) e h 2 (sinh ) 4 (e h 4 ) h 2 sinh h 3 sinh h sinh h 3 8 ( e h 2 ) (e h 2 )) (25) 9

22 ε 5 (η) = ( e ηh ) sinh h ( ) sinh h 2 sinh h 3 sinh h sinh h 3 ( ) 8 + e η2 h 2 (( ) sinh h 2 e h 3 (e 8 h 4 ) ( e h 3 ) ( )) e h 4 6 ( ) ( ) 2 ((e h 2 ) e h 4 6 e h 2 4 ( ) + e η3 h 3 sinh h 2 2 ( ) sinh h 2 sinh h 3 sinh h sinh h 3 ( ) 8 + e η4 h 4 (( ) sinh h 2 e h 2 (e 4 h 3 ) ( e h 2 ) ( )) e h 3 8 ( ) ( ) 2 ((e h 2 ) e h 4 6 e h 2 4 ( (e + sinh h 3 h 8 ) ( ) ( e h 4 6 e h 2 (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) ) (e h 4 )) (( ) + sinh h 3 8 e h (e 2 h 2 ) ( e h ) ( e h 2 4 (e h 4 ) ) (sinh h 2 sinh h 3 sinh h sinh h 3 8 ) )) (26)

23 2.2 Selection of the Exponent Parameters In order to completely define the exponential interpolation functions, the exponential parameters, h i, must be selected to suit the problem at hand. For inviscid compressible flows, the fluid properties experience a sudden jump at a shock wave, similar to a Heaviside step function. The dependent variable u defined in Eq. (27) is approximated by Eq. (28). The nodal values, u k, in Eq. (28) are the exact values of u at the nodes located at η = ±, ±, and. 2 u (η) = H (η η ) (27) 5 u (η) ε k u k (28) The fmincon function in MATLAB was used to minimize the norm of the difference between Eqs. (27) and (28) over η by changing the exponential parameters. The optimization algorithm was constrained to search for values of h i between ±7 to prevent MATLAB from evaluating ε k as ±. There is more than one combination of exponential parameters that yield satisfactory results and the exponential parameters returned by fmincon are sensitive to the initial guess. A simple set of parameters was found using trial and error to vary the inputs to the fmincon function. For η =, h = {,, 7, }. For η =, h = {,, 7, }. More generally, h j = 7 is used to approximate a discontinuity at the right edge of an element and h j = 7 is used to approximate a discontinuity at the left edge of an element where j is the highest odd numbered index of the exponential parameters. Suitable sets of parameters were not found for any case where η ±. A brief examination of Eqs. (22) to (26) shows that setting any exponential parameter equal to zero will result in division by zero. To circumvent that problem and to better illustrate the behavior of the exponential interpolation functions, the limits as h, h 2, and h 4 approach zero and h 3 approaches infinity were calculated using L Hopital s rule. The calculation steps are only shown for Eq. (22). The steps to calculate the limits of Eqs. (23) to (26) are the same. k=

24 First, the derivatives of the numerators are calculated. { 3 (e ηh ) ( sinh h 3 h h 2 h 4 8 = ( ηe ) ( ηh sinh h ) 3 (e h 4 ) ( ) 8 6 e h4 6 { 3 ( ) ( e η2 h 2 sinh h h h 2 h 4 2 ( ) ( + e η2 h 2 sinh h 3 8 ( ) ( = η 2 e η2 h 2 2 cosh h ) (e h 3 ) ( 2 ( ) ( + η 2 e η2 h 2 sinh h 3 8 { 3 ( ) ( e η3 h 3 sinh h h h 2 h 4 2 ( ) ( = e η3 h 3 2 cosh h 2 { 3 ( ) ( e η4 h 4 sinh h h h 2 h 4 2 ( ) ( + e η4 h 4 sinh h 3 8 ( ) ( = η 4 e η4 h 4 2 cosh h 2 ( η 4 e η4 h 4 ) ( sinh h 3 8 ) (e h 2 ) ( ) e h 4 6 ( e ηh ) ( sinh h 3 8 ( ηe ) ( ηh sinh h ) ( ) 3 (e 8 4 e h2 4 h 4 ) ) (e h 3 ) ( ) ( e h 4 6 e η2 h 2 ) ( ) e h (e 2 h 4 ) ) 6 e h4 6 ( ) ( e η2 h 2 sinh h 3 8 ( ) ( η 2 e η2 h 2 2 cosh h 2 ) ( ) h 2 e (e 2 h 4 ) ) (η ( 2 e η2 h 2 sinh h 3 8 ) (e h 2 ) ( ) ( e h 4 6 e η3 h 3 ) (e h 2 ) ( ) ( 6 e h4 6 e η3 h 3 ) ( 2 cosh h 2 ) (e h 2 ) ( ) ( e h 3 8 e η4 h 4 ) (e h ) ( e h 2 4 ) ) (e h 2 ) ( ) ( e h 3 8 ) (e h ) ( 4 e h2 4 ) Then the derivative of the denominator is calculated. ) ( sinh h 2 ) ( e h 2 4 ) (e h 4 )} ) ( e h 3 8 ) (e h 4 ) ) (e h ) ( e h 4 6 ) } ) ( e h 3 8 ) (e h 4 ) ) (e h ) ( 6 e h4 6 ) ) ( sinh h 2 ) ( ) (e 4 e h2 4 h 4 ) ) ( sinh h 2 ( ) ( e η4 h 4 sinh h 3 8 ) ( η 4 e η4 h 4 2 cosh h 2 ) ( ) h 2 e (e 2 h 2 ) ( η 4 e η4 h 4 ) ( sinh h 3 8 ) ( e h 2 4 ) (e h 4 )} ) ( e h 2 4 ) (e h 3 ) ) ( e h 2 ) (e h 2 )} ) ( 4 e h2 4 ) (e h 3 ) (29) (3) (3) (32) 2

25 3 h h 2 h 4 2 { 2 ( e h 2 ) ( ) ( e h 4 6 sinh h 2 ) ( ) e h 2 (e 4 h 4 ) ( sinh h 2 ) = 2 ( e ) ( h 2 6 e h4 6 2 ) ( 2 cosh h 2 ( ) (e 4 e h2 4 h 4 ) ( 2 cosh h 2 ) (sinh h 3 ) 2 ( e h 2 ) ( ) ( e h 4 6 (sinh h ) sinh h ) 3 8 (sinh h 3 ) + 2 ( e h 2 4 ) ( e h 4 ) ( (sinh h ) sinh h )} 3 8 (sinh h 3 ) 2 ( e ) ( ) ( h 2 6 e h4 6 (cosh h ) sinh h ) 3 8 ) ( ) ( (e (sinh h 3 ) e h2 4 h 4 ) (cosh h ) sinh h ) 3 8 (33) 3

26 4 Next, h = h 2 = h 4 = is subtituted into the numerators and the denominator. lim ε = h,h 2,h 4 2 sinh h 3 8 sinh h ( η 4 η 2) ( e h 3 8 ) + 6 [ ( ) e η3 h 3 + ( η 3η 2 + 4η 4) ( sinh h ) ( η 2 4η 4) ( (34) e h 3 )] Equation (34) is rearranged into a more convenient form. lim ε = h,h 2,h 4 ( 2 [ ( ) 3 ) e h 3 8 e h 3 8 (e h 3 e h 3) ( ) 3 e η3 h ( 4η 4 3η 2 η ) ( ) e h ( 4η 4 7η 2 + 3η ) ( ) e h ( η 2 4η 4) ( e h 3 )] (35) To find the limit as h 3 approaches infinity, a change of variables is used. As H 3 approaches infinity, h 3 also approaches infinity. h 3 = 8 ln H 3 (36) lim h 3 e h 3 8 ( ) = 2 e h 3 8 e h 3 8 (e h 3 e h 3) [ lim H 3 H3 9 H3 8 H H3 9 2H3 7 + ] = (37) lim h 3 e h 3 8 ( ) = 2 e h 3 8 e h 3 8 (e h 3 e h 3) [ lim H 3 H3 8 + H3 7 H H3 9 2H3 7 + ] = (38) lim h 3 e h 3 [ ( ) = lim 2 e h 3 8 e h 3 8 (e h 3 e h 3) H 3 H 6 ] H3 6 H3 8 = (39) 3 + 2H3 9 2H3 7 +

27 5 lim e η3 h 3 ( ) = h 3 2 e h 3 8 e h 3 8 (e h 3 e h 3) [ ] H 6η3 3 H 8 η [, ) 3 lim = H 3 H H3 9 2H3 7 + η = (4) Equation (4) behaves like a Heaviside step function, which is represented by H (η). Throughout this thesis, the convention that H () = is used. Equations (37) to (4) are 2 substituted into (35). ( ) lim lim ε h 3 h,h 2,h 4 = 2H ( η + ) 2 3 η η4 (4) The limits of the other one-dimensional exponential interpolation functions were calculated using the same procedure. Plots of the functions for h = {,, 7, }, which is a continuous approximation for h = {,,, }, are shown in Figure. ( ) lim lim ε 2 h 3 h,h 2,h 4 = 4H ( η + ) + 4 η + 3η 2 4η 4 (42) ( ) lim lim ε 4 h 3 h,h 2,h 4 lim ε 3 = 5η 2 + 4η 4 (43) h 2,h 4 ( ) lim lim ε 5 h 3 h,h 2,h 4 = 4H ( η + ) 4 + η η2 4 3 η4 (44) = 2H ( η + ) + 2 (45) Again, using a similar procedure the limits as h 3 approaches are found. Plots of the functions for h = {,, 7, }, which is a continuous approximation for h = {,,, }, are shown in Figure 2. ( ) lim lim ε h 3 h,h 2,h 4 ( ) lim lim ε 2 h 3 h,h 2,h 4 = 2H (η + ) + 2 (46) = 4H (η + ) 4 η η2 4 3 η4 (47) lim ε 3 = 5η 2 + 4η 4 (48) h 2,h 4

28 6 2 Function Value, ε ε ε 2 ε 3 ε 4 ε Local Coordinate, η Figure. One-dimensional exponential interpolation functions for h = {,,, }. ( ) lim lim ε 4 h 3 h,h 2,h 4 ( ) lim lim ε 5 h 3 h,h 2,h 4 = 4H (η + ) η + 3η 2 4η 4 (49) = 2H (η + ) 2 3 η η4 (5) The functions shown here can be generated without taking a limit by starting with Eq. (5) or Eq. (52) for the cases where h approaches {,,, } or {,,, }, respectively. u (η) g + g 2 η + g 3 η 2 + g 4 2H (η ) + g 5 η 4 (5) u (η) g + g 2 η + g 3 η 2 + g 4 2H ( η ) + g 5 η 4 (52)

29 7 2 Function Value, ε ε ε 2 ε 3 ε 4 ε Local Coordinate, η Figure 2. One-dimensional exponential interpolation functions for h = {,,, }. The condition that u (η k ) = u k is imposed. The resulting set of equations is arranged in matrix form as shown in Eqs. (53) and (54), Cramer s rule is used to find g k, and the remaining steps in Section 2. are applied. η η 2 η 4 η 2 η2 2 η 4 2 η 3 η3 2 η3 4 η 4 η4 2 η4 4 η 5 η5 2 η 5 4 g g 2 g 3 g 4 g 5 = u u 2 u 3 u 4 u 5 (53)

30 8 η η 2 η 4 η 2 η2 2 η 4 2 η 3 η3 2 η3 4 η 4 η4 2 η4 4 η 5 η5 2 η 5 4 g g 2 g 3 g 4 g 5 = u u 2 u 3 u 4 u 5 (54) One other noteworthy limit is that in which all of the exponential parameters approach zero. The exponential interpolation functions become Lagrange polynomials. Plots of the functions are shown in Figure 3. lim ε = h,h 2,h 3,h 4 6 η 6 η2 2 3 η η4 (55) lim ε 2 = 4 h,h 2,h 3,h 4 3 η η η3 8 3 η4 (56) lim ε 3 = 5η 2 + 4η 4 (57) h 2,h 4 lim ε 4 = 4 h,h 2,h 3,h 4 3 η η2 4 3 η3 8 3 η4 (58) lim ε 5 = h,h 2,h 3,h 4 6 η 6 η η η4 (59) Equations (4) to (5) show that the third degree of the interpolation functions is traded for a step at either the left or right edge of an element and that the resulting interpolation functions are not complete polynomials. To ensure that the polynomials are complete, elements should contain an even number of nodes. In that case, the highest polynomial degree is traded for a step at either η = or η = and the interpolation functions still contain all of the lower polynomial degrees. Figures 4 and 5 each show a function and an approximation to the function using exponential interpolation with h = {,, 7, }. The approximations in each figure use incomplete polynomials, which may work well in some cases, as shown in Figure 5 but they can lead to large approximation errors as shown in Figure 4.

31 9 Function Value, ε ε ε 2 ε 3 ε 4 ε Local Coordinate, η Figure 3. One-dimensional exponential interpolation functions for h = {,,, }.

32 2.5 Function Value Local Coordinate, η f = η 3 + H (η ) f ε f ( ) + ε 2 f (.5) + ε 3 f () + ε 4 f (.5) + ε 5 f () Figure 4. Example of a function and a poor approximation to the function using exponential interpolation and h = {,, 7, }.

33 Function Value Local Coordinate, η f = η 2 + H (η ) f ε f ( ) + ε 2 f (.5) + ε 3 f () + ε 4 f (.5) + ε 5 f () Figure 5. Example of a function and a good approximation to the function using exponential interpolation and h = {,, 7, }.

34 Gauss-Legendre Quadrature For the choice of exponential parameters described in the previous section, the resulting interpolation functions are the sum of a polynomial and a smooth approximation of a Heaviside function. The number of quadrature points required to integrate the exponential interpolation functions is driven by the term with a nonzero exponential parameter, which is the term that approximates the Heaviside function. A general expression for the term that approximates a step function is shown in Eq. (6). If i is even and h, there is a step at η = ±. If i is odd and h, there is a step at η =. If i is odd and h, there is a step at η =. H eh iη i e h i Gauss-Legendre quadrature is used for all of the numerical integration for the examples presented in Chapter 5. To determine the necessary number of quadrature points, h i and i in Eq. (6) were varied and the resulting functions were integrated with increasing numbers of quadrature points until the integration error, Eq. (6), fell below 3. Then least squares regression was used to fit a simple function to the data. For most choices of i, the exponential interpolation functions cannot be integrated analytically. The quadgk function in MATLAB was used to find a close approximation to the exact integral. The integral calculated using Gauss-Legendre quadrature is I GL and the integral calculated using quadgk is I. (6) E = I GL I I (6) Equation (62) is a least squares regression fit for the quadrature data. To ensure that enough quadrature points are used in every case, the slope of the regression equation was increased. The result is Eq. (63). Figure 6 is a plot of Eq. (62), Eq. (63), and the numerical integration data points. The one data point that lies above the lines in Figure 6 is an outlier. It corresponds to h 9 = 3. The integration error for this case is plotted in Figure 7. The first point where E 3 is N GL = 756, which is much greater than the number of quadrature points needed to obtain an accurate integral. The numbers of points predicted by Eqs. (62) and (63) are also plotted in Figure 7. N GL = h i i (62)

35 23 Number of Gauss-Legendre Points, NGL Quadrature Data N GL = h i i N GL = h i i hi i Figure Number of Gauss-Legendre quadrature points needed to obtain E N GL = h i i (63) In general terms, the larger h i and i become, the sharper the gradient at η = ± becomes and therefore more quadrature points are needed for integration. To integrate an exponential interpolation function, the number of quadrature points predicted by Eq. (63) is added to the number of points that would normally be used to integrate a polynomial. For example, 3 quadrature points are required to integrate a fourth degree polynomial using Gauss-Legendre quadrature. Numerical integration of Eq. (22) with h = {,, 7, } requires = 22 Gauss-Legendre quadrature points. The number of quadrature points needed to integrate the derivative of an exponential interpolation function is the same as the number needed for the original function, which is apparent if the derivative of Eq. (6) is calculated. No matter how many times the exponential interpolation functions are differentiated, h i η i is always in the exponent. In the finite element equations that will follow, most of the matrix terms will involve products of four interpolation functions. Raising the Heaviside approximation to the fourth power, as shown in Eq. (64), has the

36 24 3 Integration Error, E e3( η 9 ) dη (666, ) (74, ) (756, ) Number of Gauss-Legendre Points, N GL Figure 7. Integration error for h 9 = 3. same effect as multiplying the exponential parameter by four. H 4 ( ) 4 e h iη i = e4h iη i e h i e 4h i (64) Products of the exponential interpolation functions require more quadrature points than a single exponential interpolation function. Stated another way, numerically integrating ε 4 defined using h = {,, } requires the same number of quadrature points as the numerical integral of ε defined using h = {,, 4}. 2.4 Extension to Higher Dimensions In the formulas that follow ξ and η are used to represent orthogonal coordinate directions in the local coordinates of a master element. Lagrange polynomials in one dimension are represented by l i and two-dimensional interpolation functions in the local coordinate system are represented by ψ i. The typical method of defining interpolation functions in higher dimensions is to use products of one-dimensional interpolation functions. Equation (65) shows the formulation of interpolation functions for a two-dimensional element with three nodes along each coordinate direction and C continuity.

37 25 ψ ψ2 ψ3 l (η) { } ψ 4 ψ5 ψ6 = l 2 (η) l (ξ) l 2 (ξ) l 3 (ξ) ψ 7 ψ8 ψ9 l 3 (η) (65) The one-dimensional exponential interpolation functions are incorporated into the formulation of two-dimensional functions by replacing l i with ε i in one or both of the vectors on the right side of Eq. (65). A superscript is added to ψ i to distinguish the different types of interpolation functions. A superscript is used for two-dimensional C interpolation functions composed of one-dimensional Lagrange polynomials in each coordinate direction. The nine two-dimensional interpolation functions defined in Eq. (66) are plotted in Figure 8. ψ ψ 4 ψ 7 ψ 2 ψ 5 ψ 8 ψ 3 ψ 6 ψ 9 l (η) { } = l 2 (η) l (ξ) l 2 (ξ) l 3 (ξ) l 3 (η) (66)

38 26 ψ ψ 2 ψ 3 η ξ η ξ η ξ ψ 4 ψ 5 ψ 6 η ξ η ξ η ξ ψ 7 ψ 8 ψ 9 η ξ η ξ η ξ Figure 8. Interpolation functions defined in Eq. (66). A superscript is used for two-dimensional C interpolation functions composed of one-dimensional exponential interpolation functions in the ξ-direction and Lagrange polynomials in the η-direction. The nine two-dimensional interpolation functions defined in Eq. (67) are plotted in Figure 9. ψ ψ 4 ψ 7 ψ 2 ψ 5 ψ 8 ψ 3 ψ 6 ψ 9 l (η) { } = l 2 (η) ε (ξ) ε 2 (ξ) ε 3 (ξ) l 3 (η) (67)

39 27 ψ ψ 2 ψ 3 η ξ η ξ η ξ ψ 4 ψ 5 ψ 6 η ξ η ξ η ξ ψ 7 ψ 8 ψ 9 η ξ η ξ η ξ Figure 9. Interpolation functions defined in Eq. (67) using the exponential parameters h = { 7, }. A superscript 2 is used for two-dimensional C interpolation functions composed of one-dimensional Lagrange polynomials in the ξ-direction and exponential interpolation functions in the η-direction. The nine two-dimensional interpolation functions defined in Eq. (68) are plotted in Figure. ψ 2 ψ 2 4 ψ 2 7 ψ 2 2 ψ 2 5 ψ 2 8 ψ 2 3 ψ 2 6 ψ 2 9 ε (η) { } = ε 2 (η) l (ξ) l 2 (ξ) l 3 (ξ) ε 3 (η) (68)

40 28 ψ 2 ψ 2 2 ψ 2 3 η ξ η ξ η ξ ψ 2 4 ψ 2 5 ψ 2 6 η ξ η ξ η ξ ψ 2 7 ψ 2 8 ψ 2 9 η ξ η ξ η ξ Figure. Interpolation functions defined in Eq. (68) using the exponential parameters h = { 7, }.

41 29 CHAPTER 3 GOVERNING DIFFERENTIAL EQUATIONS General expressions for conservation of mass, momentum, and energy are converted to the form of the Euler equations used in Reference 4. For convenience, the equations are converted to dimensionless form. The fluid is assumed to be a perfect gas and body forces and heat transfer are assumed to be negligible. Viscosity is assumed to be zero. 3. Conservation of Mass Equation (69) is a general form of conservation of mass applied to a stationary point in a fluid. ρ t + x (ρv x) + (ρv y) + z (ρv z) = (69) The chain rule of differentiation is applied. ρ t + ρ v x x + v ρ x x + ρ v y + v ρ y + ρ v z z + v ρ z z = (7) Free stream quantities are used to convert the properties of the flow to dimensionless form. A subscript indicates a free stream quantity and bold font indicates a dimensionless quantity. v x = v x v v y = v y v v z = v z v (7) v x = v x v v y = v y v v z = v z v (72) ρ = ρ ρ (73) ρ = ρρ (74) To convert gradients to dimensionless form a constant reference length, L is used. x = x L y = y L z = z L (75)

42 3 x = xl y = yl z = zl (76) x = ( ) x L = ( ) y L z = ( ) (77) z L Since L is a constant, it can be moved outside of the partial derivatives in Eq. (77). x = L x = L z = L z (78) x = L x = L z = (79) L z Time and derivatives with respect to time are converted to dimensionless form using the free stream velocity, v and the reference length, L. t = tv L (8) t = t = tl v (8) ( tv L ) (82) Since v and L are constants, they can be moved outside of the partal derivative. t = L v t (83) t = v L t Equations (72), (74), (79), and (84) are substituted into Eq. (7) and the constant quantites are factored out of each term. ρ v L ( ρ t + ρ v x x + v x (84) ) ρ x + ρ v y + v ρ y + ρ v z z + v ρ z = (85) z The term outside of the parentheses in Eq. (85) is never zero unless the flow is static. Therefore the terms inside the parentheses must add to zero. Equation (86) is the dimensionless form of conservation of mass under the assumptions previously stated.

43 3 ρ t + ρ v x x + v ρ x x + ρ v y + v ρ y + ρ v z z + v ρ z z = (86) 3.2 Conservation of Momentum Equation (87) is a general form of conservation of momentum applied to a stationary point in a fluid with no viscosity. Momentum is conserved in each coordinate direction. Cartesian coordinates are used here. x-directon: y-directon: z-directon: t (ρv x) + x (ρv xv x ) + (ρv xv y ) + z (ρv xv z ) = P x + ρf x t (ρv y) + x (ρv yv x ) + (ρv yv y ) + z (ρv yv z ) = P + ρf y t (ρv z) + x (ρv zv x ) + (ρv zv y ) + z (ρv zv z ) = P z + ρf z (87) Body forces are assumed to be negligible. x-directon: y-directon: z-directon: t (ρv x) + x (ρv xv x ) + (ρv xv y ) + z (ρv xv z ) = P x t (ρv y) + x (ρv yv x ) + (ρv yv y ) + z (ρv yv z ) = P t (ρv z) + x (ρv zv x ) + (ρv zv y ) + z (ρv zv z ) = P z The chain rule of differentiation is applied to the terms to the left of the equal sign. (88)

44 x-directon: ρ v x t + v ρ x t + ρv v x x x + ρv v x x x + v ρ xv x x + ρv v y x + ρv v x y + v ρ xv y + ρv v z x z + ρv v x z z + v ρ xv z z = P x y-directon: ρ v y t + v ρ y t + ρv v x y x + ρv v y x x + v ρ yv x x + ρv v y y + ρv v y y + v ρ yv y + ρv v z y z + ρv v y z z + v ρ yv z z = P z-directon: ρ v z t + v ρ z t + ρv v x z x + ρv v z x x + v ρ zv x x + ρv v y z + ρv v z y + v ρ zv y + ρv v z z z + ρv v z z z + v ρ zv z z = P z (89) The terms to the left of the equal sign are rearranged. x-directon: y-directon: z-directon: [ ] ρ v x t + v ρ x x + ρ v x x + v ρ y + ρ v y + v ρ z z + ρ v z z [ ] ρ v y t + v ρ x x + ρ v x x + v ρ y + ρ v y + v ρ z z + ρ v z z [ ] ρ v z t + v ρ x x + ρ v x x + v ρ y + ρ v y + v ρ z z + ρ v z z + ρ v x t + ρv v x x x + ρv v x y + ρv v x z z = P x v y + ρ v y t + ρv v y x x + ρv v y y + ρv z + ρ v z t + ρv v z x x + ρv v z y + ρv z z = P v z z = P z (9) Due to the conservation mass, Eq. (7), the terms in brackets add to zero. The remaining terms are rearranged. 32

45 33 form. x-directon: y-directon: z-directon: v x t + v v x x x + v v x y + v v x z z + P ρ x = v y t + v v y x x + v v y y + v v y z z + P ρ = v z t + v v z x x + v v z y + v v z z z + P ρ z = Free stream quantities are used to convert the properties of the flow to dimensionless (9) P = P ρ v 2 (92) P = P ρ v 2 (93) Equations (72), (74), (79), (84), and (93) are substituted into Eq. (9) and the constant quantites are factored out of each term. x-directon: y-directon: z-directon: v 2 L v 2 L v 2 L [ vx t + v v x x x + v v x y + v v x z z + ] P = ρ x [ vy t + v v y x x + v v y y + v v y z z + ] P = ρ [ vz t + v v z x x + v v z y + v v z z z + ] P = ρ z The terms outside of the brackets in Eq. (94) are never zero unless the flow is static. Therefore, the terms inside the brackets must add to zero. Equation (95) is the dimensionless form of conservation of momenutm under the assumptions previously stated. (94) x-directon: y-directon: z-directon: v x t + v v x x x + v v x y + v v x z z + P ρ x = v y t + v v y x x + v v y y + v v y z z + P ρ = v z t + v v z x x + v v z y + v v z z z + P ρ z = (95) 3.3 Conservation of Energy Equation (96) is a general form of conservation of energy applied to a stationary point in a fluid with no viscosity.

46 34 [ ( ρ e + t 2 v2 x + 2 v2 y + )] 2 v2 z + φ + x ( [ρv y h + 2 v2 x + 2 v2 y + )] 2 v2 z + φ + z ( [ρv x h + 2 v2 x + 2 v2 y + )] 2 v2 z + φ [ρv z ( h + 2 v2 x + 2 v2 y + 2 v2 z + φ + )] = q x x q y q z z (96) It is assumed that there is no heat transfer between the fluid and its surroundings. [ ( ρ e + t 2 v2 x + 2 v2 y + )] 2 v2 z + φ + x ( [ρv y h + 2 v2 x + 2 v2 y + )] 2 v2 z + φ + z ( [ρv x h + 2 v2 x + 2 v2 y + )] 2 v2 z + φ )] [ρv z ( h + 2 v2 x + 2 v2 y + 2 v2 z + φ + = (97) The equation for specific enthalpy, Eq. (98), is substituted into Eq. (97). h = e + P ρ (98) [ ( ρ e + t 2 v2 x + 2 v2 y + )] 2 v2 z + φ + ( [ρv x e + P x ρ + 2 v2 x + 2 v2 y + )] 2 v2 z + φ + ( [ρv y e + P ρ + 2 v2 x + 2 v2 y + )] 2 v2 z + φ + ( [ρv z e + P z ρ + 2 v2 x + 2 v2 y + )] 2 v2 z + φ = The terms inside the brackets are rearranged. [ ( ρ e + t 2 v2 x + 2 v2 y + )] 2 v2 z + φ + ( [ρv x e + x 2 v2 x + 2 v2 y + ) ] 2 v2 z + φ + P v x + ( [ρv y e + 2 v2 x + 2 v2 y + ) ] 2 v2 z + φ + P v y + ( [ρv z e + z 2 v2 x + 2 v2 y + ) ] 2 v2 z + φ + P v z = (99) ()

47 35 The chain rule of differentiation is applied. ( ρ e + t 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ρ ( e + t 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( ρv x e + x 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( x (ρv x) e + 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( ρv y e + 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( (ρv y) e + 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( ρv z e + z 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( z (ρv z) e + 2 v2 x + 2 v2 y + ) 2 v2 z + φ + x (P v x) + (P v y) + z (P v z) = () The equation is rearranged. ( e + 2 v2 x + 2 v2 y + ) [ ρ 2 v2 z + φ t + x (ρv x) + (ρv y) + ] z (ρv z) + ρ ( e + t 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( ρv x e + x 2 v2 x + 2 v2 y + ) 2 v2 z + φ + x (P v x) + ( ρv y e + 2 v2 x + 2 v2 y + ) 2 v2 z + φ + (P v y) + ( ρv z e + z 2 v2 x + 2 v2 y + ) 2 v2 z + φ + z (P v z) = Due to conservation of mass, the terms in brackets add to zero. ρ ( e + t 2 v2 x + 2 v2 y + ) 2 v2 z + φ + ( ρv x e + x 2 v2 x + 2 v2 y + ) 2 v2 z + φ + x (P v x) + ( ρv y e + 2 v2 x + 2 v2 y + ) 2 v2 z + φ + (P v y) + ( ρv z e + z 2 v2 x + 2 v2 y + ) 2 v2 z + φ + z (P v z) = The partial derivatives are distributed to the terms in parentheses. (2) (3)

48 36 ( e ρ t + v v x x t + v v y y t + v v z z t + φ ) t ( e ρv x x + v v x x x + v v y y x + v v z z x + φ ) x ( e ρv y + v v x x + v v y y + v v z z + φ ) ρv z ( e z + v x v x z + v y v y z + v z v z z + φ z + + x (P v x) + + (P v y) + ) + z (P v z) = The variable φ represents a potential field. The negative gradient of a potential field ( φ) is a body force. (4) φ t = f t φ x = f x φ = f y φ z = f z (5) Body forces are assumed to be negligible. ( e ρ t + v v x x t + v v y y t + v z ρv x ( e x + v x ρv y ( e + v x ρv z ( e z + v x ) v z + t ) v x x + v v y y x + v v z z + x x (P v x) + ) v x + v v y y + v v z z + (P v y) + ) v x z + v v y y z + v v z z + z z (P v z) = The products of density and velocity are distributed to the terms in parentheses and the chain rule of differentiation is applied to the products of pressure and velocity. (6) ρ e t + ρv v x x t + ρv v y y t + ρv v z z t + ρv x e x + ρv xv x v x x + ρv xv y v y x + ρv xv z v z x + v x ρv y e + ρv yv x v x + ρv yv y v y + ρv yv z v z + v y ρv z e z + ρv zv x v x z + ρv zv y v y z + ρv zv z v z z + v z The terms are rearranged. P x + P v x x + P + P v y + P z + P v z z = (7)

49 37 ρ e t + ρv e x x + ρv e y + ρv e z z + P v x x + P v y + P v z z + ( vx ρv x t + v v x x x + v v x y + v v x z z + ) P + ρ x ( vy ρv y t + v v y x x + v v y y + v v y z z + ) P + ρ ( vz ρv z t + v v z x x + v v z y + v v z z z + ) P = ρ z (8) Due to conservation of momentum for a flow with no viscosity and negligible body force acting on it, the terms in parentheses add to zero. ρ e t + ρv e x x + ρv e y + ρv e z z + P v x x + P v y + P v z z = (9) For a calorically perfect gas, e = C v T, C v is constant, and C v = R/ (γ ). ρ R T γ t + ρv R T x γ x + ρv R T y γ + ρv R z γ T z + P v x x + P v y + P v z z = () For a perfect gas, T = P/ (ρr). ρ γ t ( ) P + ρv x ρ ρ γ ρv y γ γ x ( ) P + ρv y ρ ( P ρ t P ) ρ ρ 2 t ( P ρ P ) ρ ρ 2 γ + ρv x γ + ρv z γ ( ) P + ρv z ρ γ z ( ) P + ρ P v x x + P v y + P v z z = ( P ρ x P ) ρ + ρ 2 x ( P ρ z P ) ρ + ρ 2 z P v x x + P v y + P v z z = () (2) Density is canceled in the numerators and denominators and the entire equation is multiplied by (γ ).

50 38 ( P t P ρ ) ( ρ P + v x t x P ρ ) ( ρ P + v y x P ρ ) ( ρ P + v z z P ρ ) ρ + z (γ ) P v x x + (γ ) P v y + (γ ) P v z z = (3) The equation is expanded and all terms are multiplied by density. ρ P t P ρ t + ρv P x x v xp ρ x + ρv P y v yp ρ + ρv P z z v zp ρ z + γρp v x x ρp v x x + γρp v y ρp v y + γρp v z z ρp v (4) z z = The equation is rearranged. ρ P t + ρv P x x + ρv P y + ρv P z z + γρp v x x + γρp v y + γρp v z z ( ) ρ P t + ρ v x x + v ρ x x + ρ v y + v ρ y + ρ v z z + v ρ z = z (5) Due to conservation of mass, the terms in parentheses add to zero. Density is factored out of the remaining terms. ( P ρ t + v P x x + v P y + v P z z + γp v x x + γp v y + γp v ) z = (6) z The density of a fluid is always a positive number, therefore the terms in parentheses must add to zero. P v x x + v P y + v P z z + γp v x x + γp v y + γp v z z = (7) Equation (7) represents conservation of energy for an adiabatic flow of a perfect gas with no viscosity and no body force acting on it. It is converted to dimensionless form by substituting Eqs. (72), (74), (79), (84), and (93) and factoring out constant terms. ρ v 3 L ( P t + v P x x + v P y + v P z z + γp v x x + γp v y + γp v ) z = (8) z