# Hegedus Aerodynamics

Governing Equations for Aero Troll CFD

Note: the equations below require MathML to be displayed.

Solution Vector

Note: the total energy variable, ${e}_{0}$, is clarified in the previous Non-Dimensionalization section.

The following was adapted from reference (1).

$Q = ρ ρ u ρ v ρ w ρ e 0$

Governing Equations

The following was adapted from reference (1).

$1 J ∂ Q ∂ t + ∂ F ^ - F ^ v ∂ ξ + ∂ G ^ - G ^ v ∂ η + ∂ H ^ - H ^ v ∂ ζ - D ξ - D η - D ζ = 0$ $F ^ = 1 J ρ ξ → · U → ρ U → ξ → · U → + p ξ → ρ e 0 + p ξ → · U →$ $G ^ = 1 J ρ η → · U → ρ U → η → · U → + p η → ρ e 0 + p η → · U →$ $H ^ = 1 J ρ ζ → · U → ρ U → ζ → · U → + p ζ → ρ e 0 + p ζ → · U →$
The variables ${\stackrel{^}{F}}_{v}$, ${\stackrel{^}{G}}_{v}$, and ${\stackrel{^}{H}}_{v}$ are described in the Viscous Terms section.

Since central differencing is used, artificial dissipation $\left({D}_{\xi },{D}_{\eta },{D}_{\zeta }\right)$ is added.  Artificial dissipation is described in the Artificial Dissipation section.

Time Discretization

The following was adapted from reference (1), (2), and (3).

$1 + ϕ J Δ Q n + 1 Δ t + ∂ A ∂ ξ + ∂ B ∂ η + ∂ C ∂ ζ - ∂ D ξ ∂ Q - ∂ D η ∂ Q - ∂ D ζ ∂ Q n Δ Q n + 1 = ϕ J Δ Q n Δ t - ∂ F ^ - F ^ v ∂ ξ + ∂ G ^ - G ^ v ∂ η + ∂ H ^ - H ^ v ∂ ζ - D ξ - D η - D ζ n$ Where $\varphi =0$ for first order time accurate and $\varphi =0.5$ for second order time accurate. $Δ Q n + 1 = Q n + 1 - Q n$ $A = ∂ F ^ - F ^ v ∂ Q$ $B = ∂ G ^ - G ^ v ∂ Q$ $C = ∂ H ^ - H ^ v ∂ Q$ The flux Jacobians are defined as follows:

$∂ F ^ ∂ Q = 1 J 0 ξ x ξ y ξ z 0 ξ x ψ - u θ ξ ξ x u - ξ x γ - 1 u + θ ξ ξ y u - ξ x γ - 1 v ξ z u - ξ x γ - 1 w ξ x γ - 1 ξ y ψ - v θ ξ ξ x v - ξ y γ - 1 u ξ y v - ξ y γ - 1 v + θ ξ ξ z v - ξ y γ - 1 w ξ y γ - 1 ξ z ψ - w θ ξ ξ x w - ξ z γ - 1 u ξ y w - ξ z γ - 1 v ξ z w - ξ z γ - 1 w + θ ξ ξ z γ - 1 θ ξ ψ - h θ ξ ξ x h - θ ξ γ - 1 u ξ y h - θ ξ γ - 1 v ξ z h - θ ξ γ - 1 w γ θ ξ$
$ψ = γ - 1 2 U → · U →$ $h = γ ρ e 0 ρ - ψ$ $θ ξ = ξ → · U →$
$∂ G ^ ∂ Q = 1 J 0 η x η y η z 0 η x ψ - u θ η η x u - η x γ - 1 u + θ η η y u - η x γ - 1 v η z u - η x γ - 1 w η x γ - 1 η y ψ - v θ η η x v - η y γ - 1 u η y v - η y γ - 1 v + θ η η z v - η y γ - 1 w η y γ - 1 η z ψ - w θ η η x w - η z γ - 1 u η y w - η z γ - 1 v η z w - η z γ - 1 w + θ η η z γ - 1 θ η ψ - h θ η η x h - θ η γ - 1 u η y h - θ η γ - 1 v η z h - θ η γ - 1 w γ θ η$
$θ η = η → · U →$
$∂ H ^ ∂ Q = 1 J 0 ζ x ζ y ζ z 0 ζ x ψ - u θ ζ ζ x u - ζ x γ - 1 u + θ ζ ζ y u - ζ x γ - 1 v ζ z u - ζ x γ - 1 w ζ x γ - 1 ζ y ψ - v θ ζ ζ x v - ζ y γ - 1 u ζ y v - ζ y γ - 1 v + θ ζ ζ z v - ζ y γ - 1 w ζ y γ - 1 ζ z ψ - w θ ζ ζ x w - ζ z γ - 1 u ζ y w - ζ z γ - 1 v ζ z w - ζ z γ - 1 w + θ ζ ζ z γ - 1 θ ζ ψ - h θ ζ ζ x h - θ ζ γ - 1 u ζ y h - θ ζ γ - 1 v ζ z h - θ ζ γ - 1 w γ θ ζ$
$θ ζ = ζ → · U →$

The viscous flux Jacobians are described in the Viscous Terms section.

The artificial dissipation Jacobians are described in the Artificial Dissipation section.

Spatial Discretization

Assuming the coordinates in the computational plane are separated by one unit, the following are true:

$J = 1 vol cell$ $k → J = S → k$ Where $k=\xi \text{,}\phantom{\rule{0.5em}{0ex}}\eta \text{,}\phantom{\rule{0.5em}{0ex}}\zeta$, ${\mathrm{vol}}_{\mathrm{cell}}$ is the volume of the cell, and ${\stackrel{\to }{S}}^{k}$ is the cell face area in the k direction.

One method for inverting the left hand side (LHS) implicit matrix to solve for $\Delta {Q}^{n+1}$ is the Alternating Direction Implicit (ADI) method (4).

Starting with:

$A ^ = ∂ F ^ ∂ Q , B ^ = ∂ G ^ ∂ Q , C ^ = ∂ H ^ ∂ Q$ $I + Δ t 1 + ϕ J ∂ A ^ ∂ ξ + ∂ B ^ ∂ η + ∂ C ^ ∂ ζ n Δ Q n + 1 = ϕ 1 + ϕ Δ Q n - Δ t 1 + ϕ J ∂ F ^ - F ^ v ∂ ξ + ∂ G ^ - G ^ v ∂ η + ∂ H ^ - H ^ v ∂ ζ n$ $R ^ n = ϕ 1 + ϕ Δ Q n - Δ t 1 + ϕ J ∂ F ^ - F ^ v ∂ ξ + ∂ G ^ - G ^ v ∂ η + ∂ H ^ - H ^ v ∂ ζ n$ $I + Δ t 1 + ϕ J ∂ A ^ ∂ ξ I + Δ t 1 + ϕ J ∂ B ^ ∂ η I + Δ t 1 + ϕ J ∂ C ^ ∂ ζ n Δ Q n + 1 = R ^ n$ $h = Δ t 1 + ϕ J$ Results in:

$I + h ∂ A ^ ∂ ξ I + h ∂ B ^ ∂ η I + h ∂ C ^ ∂ ζ n Δ Q n + 1 = R ^ n$

Diagonalized Alternating Direction Implicit (DADI) Method

Another method for inverting the left hand side (LHS) implicit matrix to solve for $\Delta {Q}^{n+1}$ is the Diagonalized Alternating Direction Implicit (DADI) method (5).

Starting with:

$I + h ∂ A ^ ∂ ξ I + h ∂ B ^ ∂ η I + h ∂ C ^ ∂ ζ n Δ Q n + 1 = R ^ n$ $A ^ = T ξ Λ ^ ξ T ξ - 1 , B ^ η = T η Λ ^ η T η - 1 , C ^ ζ = T ζ Λ ^ ζ T ζ - 1$ $Λ ^ k = k → · V → 0 0 0 0 0 k → · V → 0 0 0 0 0 k → · V → 0 0 0 0 0 k → · V → + a k → · k → 1 2 0 0 0 0 0 k → · V → - a k → · k → 1 2$ $k = ξ , η , ζ$ $T k = k ~ x k ~ y k ~ z α α k ~ x u k ~ y u - k ~ z ρ k ~ z u + k ~ y ρ α u + k ~ x a α u - k ~ x a k ~ x v + k ~ z ρ k ~ y v k ~ z v - k ~ x ρ α v + k ~ y a α v - k ~ y a k ~ x w - k ~ y ρ k ~ y w + k ~ x ρ k ~ z w α w + k ~ z a α w - k ~ z a k ~ x ϕ 2 γ - 1 + ρ k ~ z v - k ~ y w k ~ y ϕ 2 γ - 1 + ρ k ~ x w - k ~ z u k ~ z ϕ 2 γ - 1 + ρ k ~ y u - k ~ x v α ϕ 2 + a 2 γ - 1 + a θ ~ α ϕ 2 + a 2 γ - 1 - a θ ~$ $T k - 1 = k ~ x 1 - ϕ 2 a 2 - ρ - 1 k ~ z v - k ~ y w k ~ x γ - 1 u a - 2 k ~ z ρ - 1 + k ~ x γ - 1 v a - 2 - k ~ y ρ - 1 + k ~ x γ - 1 w a - 2 - k ~ x γ - 1 a - 2 k ~ y 1 - ϕ 2 a 2 - ρ - 1 k ~ x w - k ~ z u - k ~ z ρ - 1 + k ~ y γ - 1 u a - 2 k ~ y γ - 1 v a - 2 k ~ x ρ - 1 + k ~ y γ - 1 w a - 2 - k ~ y γ - 1 a - 2 k ~ z 1 - ϕ 2 a 2 - ρ - 1 k ~ y u - k ~ x v k ~ y ρ - 1 + k ~ z γ - 1 u a - 2 - k ~ x ρ - 1 + k ~ z γ - 1 v a - 2 k ~ z γ - 1 w a - 2 - k ~ z γ - 1 a - 2 β ϕ 2 - a θ ~ β k ~ x a - γ - 1 u β k ~ y a - γ - 1 v β k ~ z a - γ - 1 w β γ - 1 β ϕ 2 + a θ ~ - β k ~ x a + γ - 1 u - β k ~ y a + γ - 1 v - β k ~ z a + γ - 1 w β γ - 1$ $ϕ 2 = 1 2 γ - 1 u 2 + v 2 + w 2$ $θ ~ = k ~ x u + k ~ y v + k ~ z w$ $α = ρ 2 a$ $β = 1 2 ρ a$ $k ~ x = k x k x 2 + k y 2 + k z 2 , etc.$ $μ = 1 2$ $I + h ∂ T ξ Λ ^ ξ T ξ - 1 ∂ ξ I + h ∂ T η Λ ^ η T η - 1 ∂ η I + h ∂ T ζ Λ ^ ζ T ζ - 1 ∂ ζ n Δ Q n + 1 = R ^ n$ Results in:

$T ξ I + h ∂ Λ ^ ξ ∂ ξ N ^ I + h ∂ Λ ^ η ∂ η P ^ I + h ∂ Λ ^ ζ ∂ ζ T ζ - 1 n Δ Q n + 1 = R ^ n$ Where

$N ^ = T ξ - 1 T η , N ^ - 1 = T η - 1 T ξ , P ^ = T η - 1 T ζ , P ^ - 1 = T ζ - 1 T η$ $T k - 1 T l = m 1 m 2 m 3 - μ m 4 μ m 4 - m 2 m 1 m 4 μ m 3 - μ m 3 - m 3 - m 4 m 1 - μ m 2 μ m 2 μ m 4 - μ m 3 μ m 2 μ 2 1 + m 1 μ 2 1 - m 1 - μ m 4 μ m 3 - μ m 2 μ 2 1 - m 1 μ 2 1 + m 1$ $m 1 = k ^ → · l ^ → , m 2 = k ^ x l ^ y - k ^ y l ^ x , m 3 = k ^ x l ^ z - k ^ z l ^ x , m 4 = k ^ y l ^ z - k ^ z l ^ y$ and ${\Lambda }_{k}$, ${T}_{k}$, ${T}_{k}^{-1}$, and $h$ are given above.

References

1) Krist, S. L., Biedron, R. T., and Rumsey, C. L., "CFL3D User's Manual (Version 5.0)," NASA TM 1998-208444, June 1998.
2) Nichols, R. H., and Buning P. G., "User's Manual for OVERFLOW 2.1, Version 2.1t Aug 2008," Aug 2008.
3) Pulliam, T. H., "Solution Methods In Computational Fluid Dynamics," Jan 1986.
4) Beam, R. and Warming, R. F., "An Implicit Finite-Difference Algorithm for Hyperbolic Systems in Conservation-Law From," J. Comput. Phys. 22, pp. 87-110 (1976).
5) Pulliam, T. H., and Chaussee, D. S., "A Diagonal Form of an Implicit Approximate-Factorization Algorithm," J. Comput. Phys. 39, pp. 347-363 (1981).