Getting started: CFD notation. PDE of p-th order f (u,x, t,. ∂u. ∂x1. ,...,. ∂u. ∂xn. ,
. ∂u. ∂t. ,. ∂2u. ∂x1∂x2. ,...,. ∂pu. ∂tp ) = 0 scalar unknowns u = u(x,t), ...
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 = Ω\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).
