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Getting started: CFD notation ∂u ∂u ∂2u ∂pu ∂u , . . . , , , , . . . , ∂x1 ∂xn ∂t ∂x1 ∂x2 ∂tp PDE of p-th order f u, x, t, scalar unknowns u = u(x, t), vector unknowns v = v(x, t), Nabla operator ∂ ∂ ∂ ∇ = i ∂x + j ∂y + k ∂z ∇u = i ∂u ∂x ∇·v = + ∂vx ∂x j ∂u ∂y + ∇ × v = det + ∂vy ∂y i ∂ ∂x k ∂u ∂z + = h x ∈ Rn , t ∈ R, v ∈ Rm , ∂u ∂u ∂u ∂x , ∂y , ∂z ∂ ∂y k ∂ ∂z vx vy vz ∆u = ∇ · (∇u) = ∇2 u = m = 1, 2, . . . v = (vx , vy , vz ) gradient z ∂vz ∂z j =0 n = 1, 2, 3 x = (x, y, z), iT v divergence = ∂2u ∂x2 + ∂vz ∂y ∂vx ∂z ∂vy ∂x ∂2u ∂y 2 − − − + ∂vy ∂z ∂vz ∂x ∂vx ∂y ∂2u ∂z 2 k curl i x Laplacian j y Tensorial quantities in fluid dynamics Velocity gradient ∇v = [∇vx , ∇vy , ∇vz ] = ∂vy ∂x ∂vy ∂y ∂vy ∂z ∂vx ∂x ∂vx ∂y ∂vx ∂z ∂vz ∂x ∂vz ∂y ∂vz ∂z 1111111111111111 0000000000000000 0000000000000000 1111111111111111 v 1111111111111111 0000000000000000 0000000000000000 1111111111111111 Remark. The trace (sum of diagonal elements) of ∇v equals ∇ · v. Deformation rate tensor (symmetric part of ∇v) 1 T D(v) = (∇v + ∇v ) = 2 Spin tensor 1 2 1 2 S(v) = ∇v − D(v) ∂vx ∂x ∂vx ∂y + ∂vx ∂z + ∂vy ∂x + ∂vy ∂y 1 ∂vy 2 ∂z + 1 2 ∂vy ∂x ∂vz ∂x ∂vx ∂y ∂vz ∂y 1 2 1 2 (skew-symmetric part of ∇v) ∂vz ∂x + ∂vz ∂y + ∂vz ∂z ∂vx ∂z ∂vy ∂z Vector multiplication rules Scalar product of two vectors a, b ∈ R3 , a · b = aT b = [a1 a2 a3 ] b2 = a1 b1 + a2 b2 + a3 b3 ∈ R b3 v · ∇u = vx Example. b1 ∂u ∂u ∂u + vy + vz ∂x ∂y ∂z convective derivative Dyadic product of two vectors 3 a, b ∈ R , a1 a1 b1 a ⊗ b = abT = a2 [b1 b2 b3 ] = a2 b1 a3 a3 b1 a1 b2 a2 b2 a3 b2 a1 b3 3×3 a2 b3 ∈R a3 b3 Elementary tensor calculus T = {tij } ∈ R3×3 , α ∈ R 1. αT = {αtij }, 2. T 1 , T 2 ∈ R3×3 , a ∈ R3 T 1 + T 2 = {t1ij + t2ij }, t t t 3 11 12 13 P ai [ti1 , ti2 , ti3 ] a · T = [a1 , a2 , a3 ] t21 t22 t23 = | {z } i=1 t31 t32 t33 i-th row t a t t t 3 1j 11 12 13 1 P T · a = t21 t22 t23 a2 = t2j aj (j-th column) j=1 t3j a3 t31 t32 t33 2 2 2 1 1 1 3 t11 t12 t13 t11 t12 t13 P t1ik t2kj T 1 · T 2 = t121 t122 t123 t221 t222 t223 = k=1 t231 t232 t233 t131 t132 t133 3. 4. 5. 6. 1 2 1 2 T T : T = tr (T · (T ) ) = 3 3 P P i=1 k=1 t1ik t2ik Divergence theorem of Gauß Let Ω ∈ R3 and n be the outward unit normal to the boundary Γ = Ω̄\Ω. Then Z Ω ∇ · f dx = Z Γ f · n ds for any differentiable function f (x) Example. A sphere: Ω = {x ∈ R3 : ||x|| < 1}, where ||x|| = √ Consider f (x) = x volume integral: surface integral: x·x= p so that x2 + y 2 + z 2 Γ = {x ∈ R3 : ||x|| = 1} is the Euclidean norm of x ∇ · f ≡ 3 in Ω and n = x ||x|| on Γ 4 3 ∇ · f dx = 3 dx = 3|Ω| = 3 π1 = 4π 3 Ω Ω Z Z Z Z x·x f · n ds = ds = ||x|| ds = ds = 4π ||x|| Γ Γ Γ Γ Z Z Governing equations of fluid dynamics Physical principles Mathematical equations 1. Mass is conserved • continuity equation 2. Newton’s second law • momentum equations 3. Energy is conserved • energy equation It is important to understand the meaning and significance of each equation in order to develop a good numerical method and properly interpret the results Description of fluid motion z v Eulerian monitor the flow characteristics (x1 ; y1 ; z1 ) in a fixed control volume i Lagrangian track individual fluid particles as they move through the flow field (x0 ; y0 ; z0 ) k x j y Description of fluid motion Trajectory of a fluid particle z x = x(x0 , t) v (x1 ; y1 ; z1 ) (x0 ; y0 ; z0 ) k i j y x = x(x0 , y0 , z0 , t) y = y(x0 , y0 , z0 , t) z = z(x0 , y0 , z0 , t) dx = vx (x, y, z, t), dt dy = vy (x, y, z, t), dt dz = vz (x, y, z, t), dt x|t0 = x0 y|t0 = y0 z|t0 = z0 x Definition. A streamline is a curve which is tangent to the velocity vector v = (vx , vy , vz ) at every point. It is given by the relation y dx dy dz = = vx vy vz dy dx v y(x) = vy vx x Streamlines can be visualized by injecting tracer particles into the flow field. Flow models and reference frames Lagrangian S V fixed CV of a finite size dS dV fixed infinitesimal CV V integral S moving CV of a finite size dS dV moving infinitesimal CV Good news: all flow models lead to the same equations differential Eulerian Eulerian vs. Lagrangian viewpoint d is the rate of change for a moving Definition. Substantial time derivative dt ∂ fluid particle. Local time derivative ∂t is the rate of change at a fixed point. Let u = u(x, t), where x = x(x0 , t). The chain rule yields ∂u ∂u dx ∂u dy ∂u dz ∂u du = + + + = + v · ∇u dt ∂t ∂x dt ∂y dt ∂z dt ∂t substantial derivative = local derivative + convective derivative Reynolds transport theorem d dt Z u(x, t) dV = V ≡Vt Vt rate of change in a moving volume Z = ∂u(x, t) dV + ∂t rate of change in a fixed volume Z + S≡St u(x, t)v · n dS convective transfer through the surface Derivation of the governing equations Modeling philosophy 1. Choose a physical principle • conservation of mass • conservation of momentum • conservation of energy 2. Apply it to a suitable flow model • Eulerian/Lagrangian approach • for a finite/infinitesimal CV 3. Extract integral relations or PDEs which embody the physical principle Generic conservation law Z Z Z ∂ u dV + f · n dS = q dV ∂t V S V S f = vu − d∇u n V dS f flux function Divergence theorem yields Z Z Z ∂u dV + ∇ · f dV = q dV ∂t V V V Partial differential equation ∂u +∇·f =q ∂t in V Derivation of the continuity equation Physical principle: conservation of mass Z Z Z d ∂ρ dm = dV + ρ dV = ρv · n dS = 0 dt dt Vt ∂t S≡St V ≡Vt accumulation of mass inside CV = net influx through the surface Divergence theorem yields Z ∂ρ + ∇ · (ρv) dV = 0 ∂t V Continuity equation ∂ρ + ∇ · (ρv) = 0 ∂t ⇒ Lagrangian representation ∇ · (ρv) = v · ∇ρ + ρ∇ · v Incompressible flows: dρ dt =∇·v =0 ⇒ dρ + ρ∇ · v = 0 dt (constant density) Conservation of momentum Physical principle: dV dS n h g f = ma (Newton’s second law) total force f = ρg dV + h dS, body forces g gravitational, electromagnetic,. . . surface forces h pressure + viscous stress Stress tensor σ = −pI + τ where h=σ·n momentum flux For a newtonian fluid viscous stress is proportional to velocity gradients: τ = (λ∇ · v)I + 2µD(v), where τxx = τyy = τzz = 1 (∇v + ∇vT ), 2 2 λ≈− µ 3 Shear stress: deformation Normal stress: stretching x λ∇ · v + 2µ ∂v ∂x ∂v λ∇ · v + 2µ ∂yy z λ∇ · v + 2µ ∂v ∂z D(v) = τxy = τyx = µ y xx τxz = τzx τyz = τzy x “ ∂vy ∂x + ∂vx ∂y ” ` x ´ ∂vz = µ ∂v + ∂x ” “ ∂z ∂v z + ∂zy = µ ∂v ∂y y yx x Derivation of the momentum equations Newton’s law for a moving volume Z Z Z d ∂(ρv) ρv dV = dV + (ρv ⊗ v) · n dS dt Vt ∂t V ≡Vt S≡St Z Z = ρg dV + σ · n dS V ≡Vt S≡St Transformation of surface integrals Z Z ∂(ρv) + ∇ · (ρv ⊗ v) dV = [∇ · σ + ρg] dV, ∂t V V σ = −pI + τ ∂(ρv) + ∇ · (ρv ⊗ v) = −∇p + ∇ · τ + ρg ∂t ∂(ρv) ∂v ∂ρ dv + ∇ · (ρv ⊗ v) = ρ + v · ∇v + v + ∇ · (ρv) = ρ ∂t ∂t ∂t dt | | {z } {z } Momentum equations substantial derivative continuity equation Conservation of energy Physical principle: dV dS n h g δe = s + w (first law of thermodynamics) δe accumulation of internal energy s heat transmitted to the fluid particle w rate of work done by external forces Heating: s = ρq dV − fq dS q internal heat sources fq diffusive heat transfer T κ absolute temperature thermal conductivity Work done per unit time = Fourier’s law of heat conduction fq = −κ∇T the heat flux is proportional to the local temperature gradient total force × velocity w = f · v = ρg · v dV + v · (σ · n) dS, σ = −pI + τ Derivation of the energy equation Total energy per unit mass: |v|2 2 specific internal energy due to random molecular motion e |v| 2 E =e+ 2 specific kinetic energy due to translational motion Integral conservation law for a moving volume Z Z Z ∂(ρE) d ρE dV = ρE v · n dS dV + dt Vt ∂t V ≡Vt S≡St Z Z = ρq dV + κ∇T · n dS V ≡Vt S≡St Z Z v · (σ · n) dS ρg · v dV + + V ≡Vt accumulation heating work done S≡St Transformation of surface integrals Z Z ∂(ρE) + ∇ · (ρEv) dV = [∇ · (κ∇T ) + ρq + ∇ · (σ · v) + ρg · v] dV, ∂t V V where ∇ · (σ · v) = −∇ · (pv) + ∇ · (τ · v) = −∇ · (pv) + v · (∇ · τ ) + ∇v : τ Different forms of the energy equation Total energy equation ∂(ρE) + ∇ · (ρEv) = ∇ · (κ∇T ) + ρq − ∇ · (pv) + v · (∇ · τ ) + ∇v : τ + ρg · v ∂t ∂ρ dE ∂(ρE) ∂E + ∇ · (ρEv) = ρ + v · ∇E + E + ∇ · (ρv) = ρ ∂t ∂t ∂t dt {z } {z } | | substantial derivative Momentum equations ρ ρ dv = −∇p + ∇ · τ + ρg dt continuity equation (Lagrangian form) dE de dv ∂(ρe) =ρ +v·ρ = + ∇ · (ρev) + v · [−∇p + ∇ · τ + ρg] dt dt dt ∂t Internal energy equation ∂(ρe) + ∇ · (ρev) = ∇ · (κ∇T ) + ρq − p∇ · v + ∇v : τ ∂t Summary of the governing equations 1. Continuity equation / conservation of mass ∂ρ + ∇ · (ρv) = 0 ∂t 2. Momentum equations / Newton’s second law ∂(ρv) + ∇ · (ρv ⊗ v) = −∇p + ∇ · τ + ρg ∂t 3. Energy equation / first law of thermodynamics ∂(ρE) + ∇ · (ρEv) = ∇ · (κ∇T ) + ρq − ∇ · (pv) + v · (∇ · τ ) + ∇v : τ + ρg · v ∂t |v|2 E =e+ , 2 ∂(ρe) + ∇ · (ρev) = ∇ · (κ∇T ) + ρq − p∇ · v + ∇v : τ ∂t This PDE system is referred to as the compressible Navier-Stokes equations Conservation form of the governing equations Generic conservation law for a scalar quantity ∂u + ∇ · f = q, ∂t where f = f (u, x, t) Conservative variables, fluxes and sources ρv ρ U = ρv ⊗ v + pI − τ ρv , F = (ρE + p)v − κ∇T − τ · v ρE is the flux function , Q= 0 ρg ρ(q + g · v) Navier-Stokes equations in divergence form ∂U +∇·F=Q ∂t U ∈ R5 , F ∈ R3×5 , Q ∈ R5 • representing all equations in the same generic form simplifies the programming • it suffices to develop discretization techniques for the generic conservation law Constitutive relations Variables: ρ, v, e, p, τ , T Equations: continuity, momentum, energy The number of unknowns exceeds the number of equations. 1. Newtonian stress tensor τ = (λ∇ · v)I + 2µD(v), D(v) = 1 (∇v + ∇vT ), 2 2 λ≈− µ 3 2. Thermodynamic relations, e.g. p = ρRT ideal gas law R specific gas constant e = cv T caloric equation of state cv specific heat at constant volume Now the system is closed: it contains five PDEs for five independent variables ρ, v, e and algebraic formulae for the computation of p, τ and T . It remains to specify appropriate initial and boundary conditions. Initial and boundary conditions Initial conditions ρ|t=0 = ρ0 (x), v|t=0 = v0 (x), e|t=0 = e0 (x) Let Γ = Γin ∪ Γw ∪ Γout Boundary conditions w Γin = {x ∈ Γ : v · n < 0} Inlet ρ = ρin , v = vin , in Ω e = ein in out prescribed density, energy and velocity w Solid wall Γw = {x ∈ Γ : v · n = 0} Outlet Γout = {x ∈ Γ : v · n > 0} v · n = vn or v=0 no-slip condition T = Tw fq ∂T ∂n = − κ given temperature or v · s = vs prescribed heat flux prescribed velocity or −p + n · τ · n = 0 s·τ ·n=0 vanishing stress The problem is well-posed if the solution exists, is unique and depends continuously on IC and BC. Insufficient or incorrect IC/BC may lead to wrong results (if any).