4. The Scalar-Transport Equation - University of Manchester · 2021. 1. 4. · Example A thin rod...

4. The Scalar-Transport Equation

Structured Grid; Cell-Centred Storage

E

W N

S

B

T

P e

w

t

b

ns

j

i

k

Cell Indexing

GlobalLocal

i+1,ji-1,j i,j

i,j-1

i,j+1

i+2,ji-2,j

i,j-2

i,j+2

EW

N

S

P

n

e

s

w EEWW

NN

SS

Generic Scalar-Transport Equation

ϕ = concentration (amount per unit mass of fluid)

d

d𝑡(mass × ϕ) +

faces

( mass flux × ϕ −Γ𝜕ϕ

𝜕𝑛𝐴 ) = 𝑆

time derivative advection diffusion source

V

A

u

un

TIME DERIVATIVEof amount in 𝑉

+ADVECTION + DIFFUSIONthrough boundary of 𝑉

=SOURCEinside 𝑉

1-d Advection-Diffusion Equation

• Simple analysis

• Hand solution

• Straightforward generalisation to 2D/3D

• Coordinate-wise discretisation in practice

• Many important theoretical problems

Steady 1-d Advection-Diffusion Equation

flux𝑒 − flux𝑤 = sourceConservation:

Flux:

d

d𝑥(flux) = source per unit length

d

d𝑥(ρ𝑢𝐴ϕ − Γ𝐴

dϕ

d𝑥) = 𝑠

Conservative differential equation:

flux efluxw

x

area A

source

Source: 𝑠Δ𝑥

flux = ρ𝑢𝐴 ϕadvection

−Γ𝐴dϕ

d𝑥diffusion

Discretising Diffusion and Source

Example

A thin rod has length 1 m and cross-section1 cm 1 cm. The left-hand end is kept at 100 C, whilst the right-hand end is insulated.

The heat flux across any section of area 𝐴 isgiven by

−𝑘𝐴d𝑇

d𝑥

where the conductivity 𝑘 = 1000 Wm−1 K−1. The rod loses heat along its length at a rate proportional to the temperature excess over ambient (Newton’s law of cooling); i.e. the heat source per unit length is:

𝑠 = −𝑐(𝑇 − 𝑇∞)

where the ambient temperature 𝑇∞ = 20 °C and the coefficient 𝑐 = 2.5 Wm−1 K−1.

(a) Write down and solve the differential equation for temperature along the rod.

(b) Divide the rod into 5 control sections, with nodes at the centre of each section, and carry out a finite-volume analysis to find the temperature along the rod.

x = 0 1

T=100 Co

RoddTdx

=0

T = 20 Co

s = -c (T-T )

8

8

−𝑘𝐴d𝑇

d𝑥𝑒

− −𝑘𝐴d𝑇

d𝑥𝑤

= −𝑐(𝑇 − 𝑇∞)Δ𝑥

d

d𝑥( −𝑘𝐴

d𝑇

d𝑥) = −𝑐(𝑇 − 𝑇∞)

flux𝑒 − flux𝑤 = source

d2𝑇

d𝑥2−

𝑐

𝑘𝐴𝑇 = −

𝑐

𝑘𝐴𝑇∞

d2𝑇

d𝑥2− 25𝑇 = −500

𝑇 = 𝐴e5𝑥 + 𝐵e−5𝑥

complementaryfunction

+ ด20particularintegral

𝑇(0) = 100, 𝑇′(1) = 0

𝐴 = 0.0036 , 𝐵 = 79.9964

Differential Equation and Exact Solution

flux efluxw source

−𝑘𝐴d𝑇d𝑥 𝑒

− −𝑘𝐴d𝑇d𝑥 𝑤

Δ𝑥= −𝑐(𝑇 − 𝑇∞)

Finite-Volume Decomposition

x = 0.2

flux = 0T = 100L

A = 10-4

1 2 3 4 5R

Discretisation of Fluxes

T TTw eP EW

flux𝑒 = − ቤ𝑘𝐴d𝑇

d𝑥𝑒

−𝑘𝐴𝑇𝐸 − 𝑇𝑃Δ𝑥

→

x

T

TP

TE

x

e

T

Finite-Volume Discretisation

Fluxes:

source = −𝑐(𝑇𝑃 − 𝑇∞)Δ𝑥

𝑏𝑃 = 𝑐𝑇∞Δ𝑥 = 10 , 𝑠𝑃 = −𝑐Δ𝑥 = −0.5

Source:


flux𝑒 = − ቤ𝑘𝐴d𝑇

d𝑥𝑒

interior faces:

flux𝐿 = − ቤ𝑘𝐴d𝑇

d𝑥𝐿

left boundary:

flux𝑅 = 0right boundary:

𝐷 =𝑘𝐴

Δ𝑥= 0.5

flux efluxw source

T TTw eP EW

TPTL

TPflux = 0R

→ −𝑘𝐴𝑇𝐸 − 𝑇𝑃Δ𝑥

→ −𝐷(𝑇𝐸 − 𝑇𝑃)

→ −𝑘𝐴𝑇𝑃 − 𝑇𝐿12Δ𝑥

→ −2𝐷(𝑇𝑃 − 𝑇𝐿)

→ 𝑏𝑃 + 𝑠𝑃𝑇𝑃


−0.5(𝑇𝐸 − 𝑇𝑃) + 0.5(𝑇𝑃 − 𝑇𝑊) = 10 − 0.5𝑇𝑃Interior cells:

−0.5𝑇𝑊 + 1.5𝑇𝑃 − 0.5𝑇𝐸 = 10

Left boundary: −0.5(𝑇𝐸 − 𝑇𝑃) + 1.0(𝑇𝑃 − 100) = 10 − 0.5𝑇𝑃

2.0𝑇𝑃 − 0.5𝑇𝐸 = 110

Right boundary: 0 + 0.5(𝑇𝑃 − 𝑇𝑊) = 10 − 0.5𝑇𝑃

−0.5𝑇𝑊 + 1.0𝑇𝑃 = 10

flux𝑒 − flux𝑤 = source flux efluxw source

T TTw eP EW

Assembled Equations

2 −0.5 0 0 0−0.5 1.5 −0.5 0 00 −0.5 1.5 −0.5 00 0 −0.5 1.5 −0.50 0 0 −0.5 1

𝑇1𝑇2𝑇3𝑇4𝑇5

=

11010101010

x = 0.2

flux = 0T = 100L

A = 10-4

1 2 3 4 5R

2 −0.5 0 0 0−0.5 1.5 −0.5 0 00 −0.5 1.5 −0.5 00 0 −0.5 1.5 −0.50 0 0 −0.5 1


=

11010101010

All 𝑅 × 2:

4 −1 0 0 0 ∥ 220−1 3 −1 0 0 ∥ 200 −1 3 −1 0 ∥ 200 0 −1 3 −1 ∥ 200 0 0 −1 2 ∥ 20

𝑅2 → 4𝑅2 + 𝑅1:

𝑅3 → 11𝑅3 + 𝑅2:

4 −1 0 0 0 ∥ 2200 11 −4 0 0 ∥ 3000 −1 3 −1 0 ∥ 200 0 −1 3 −1 ∥ 200 0 0 −1 2 ∥ 20

4 −1 0 0 0 ∥ 2200 11 −4 0 0 ∥ 3000 0 29 −11 0 ∥ 5200 0 −1 3 −1 ∥ 200 0 0 −1 2 ∥ 20

𝑅4 → 29𝑅4 + 𝑅3:

4 −1 0 0 0 ∥ 2200 11 −4 0 0 ∥ 3000 0 29 −11 0 ∥ 5200 0 0 76 −29 ∥ 11000 0 0 −1 2 ∥ 20

𝑅5 → 76𝑅5 + 𝑅4:

4 −1 0 0 00 11 −4 0 00 0 29 −11 00 0 0 76 −290 0 0 0 123


=

22030052011002620

𝑇5 =2620

123= 21.30

𝑇4 =1100 + 29𝑇5

76= 22.60

𝑇3 =520 + 11𝑇4

29= 26.50

𝑇2 =300 + 4𝑇3

11= 36.91

𝑇1 =220 + 𝑇2

4= 64.23

Points to Note

• Each control volume gives one equation

• Each equation is a linear equation connecting the central node and nodes on either side:

−𝑎𝑤ϕ𝑊 + 𝑎𝑃ϕ𝑃 − 𝑎𝐸ϕ𝐸 = 𝑏𝑃

• Boundary values are incorporated into the source term

• The resulting matrix is tri-diagonal = b

Discretising Diffusion

−Γ𝐴 ቤdϕ

d𝑥𝑒

→ −(Γ𝐴)𝑒ϕ𝐸 −ϕ𝑃

Δ𝑥

x

P

E

x

e

Discretising the Source Term

source = 𝑏𝑃 + 𝑠𝑃ϕ𝑃 𝑠𝑃 ≤ 0

Examples:

source = 3 − 2ϕ 𝑏𝑝 = 3, 𝑠𝑝 = −2

source = 10 − 7ϕ3 𝑏𝑝 = 10, 𝑠𝑝 = −7ϕ∗2

Solve iteratively

Example

A 2-d finite-volume calculation is to be undertaken for fully-developed,laminar flow between stationary, plane, parallel walls. A streamwisepressure gradient d𝑝/d𝑥 = −𝐺 is imposed and the fluid viscosity is μ. Thedepth of the channel, 𝐻, is divided into 𝑁 equally-sized cells of dimension𝑥 × 𝑦 × 1 as shown, with the velocity 𝑢 stored at cell centres.

(a) What are the boundary conditions on velocity?

(b) What is the net pressure force on a single cell?

(c) Using a finite-difference approximation for velocity gradient, find expressions for the viscous forces on upper and lower faces of the jth cell in terms of the nodal velocities {𝑢𝑗}. (Deal separately with interior

cells and the boundary cells 𝑗 = 1 and 𝑗 = 𝑁.)

(d) By balancing pressure and viscous forces set up the finite-volume equations for velocity.

(e) Solve for the nodal velocities in the case 𝑁 = 6, leaving your answers as multiples of 𝑈0 = 𝐺𝐻2/μ. (You

are advised to note the symmetry of the problem.)

(f) Using your numerical solution, find the volume flow rate (per unit span) 𝑞 in terms of 𝑈0 and 𝐻.

(g) Find the wall shear stress, τ𝑤

(h) Compare your answers to (e), (f), (g) with the exact solution for plane Poiseuille flow.

u1

j-1u

ju

j+1u

Nu

y

x

H

u1

j-1u

ju

j+1u

Nu

y

x

H

A 2-d finite-volume calculation is to be undertaken for fully-developed, laminar flow between stationary, plane, parallel walls. A streamwise pressure gradient d𝑝/d𝑥 = −𝐺 is imposed and the fluid viscosity is μ. The depth of the channel, 𝐻, is divided into 𝑁 equally-sized cells of dimension 𝑥 × 𝑦 × 1 as shown, with the velocity 𝑢 stored at cell centres.

(a) What are the boundary conditions on velocity?

(b) What is the net pressure force on a single cell?

𝑢 = 0 on upper and lower walls

Net pressure force = 𝑝𝐿 × Δ𝑦 × 1 − 𝑝𝑅 × Δ𝑦 × 1

= (𝑝𝐿 − 𝑝𝑅) × Δ𝑦

= −d𝑝

d𝑥Δ𝑥 Δy = 𝐺Δ𝑥Δ𝑦

Interior faces: ≈ μ𝑢𝑗+1 − 𝑢𝑗

Δ𝑦Δ𝑥viscous force on upper face = μ

d𝑢

d𝑦× (Δ𝑥 × 1)

viscous force on lower face ≈ −μ𝑢𝑗 − 𝑢𝑗−1

Δ𝑦Δ𝑥

viscous force on upper face of top cell ≈ μ0 − 𝑢𝑗12Δ𝑦

Δ𝑥

≈ −μ𝑢𝑗 − 012Δ𝑦

Δ𝑥

Boundary faces:

viscous force on lower face of bottom cell

(c) Using a finite-difference approximation for velocity gradient, find expressions for the viscous forces on upper and lower faces of the jth cell in terms of the nodal velocities {𝑢𝑗}.

u1

j-1u

ju

j+1u

Nu

y

x

H


(d) By balancing pressure and viscous forces set up the finite-volume equations for velocity.

Interior cell: 𝐺Δ𝑥Δ𝑦 + μ𝑢𝑗+1 − 𝑢𝑗

Δ𝑦Δ𝑥 − μ

𝑢𝑗 − 𝑢𝑗−1

Δ𝑦Δ𝑥 = 0

𝐺Δ𝑦2

μ= −𝑢𝑗−1 + 2𝑢𝑗 − 𝑢𝑗+1

Top cell (𝑗 = 𝑁): 𝐺Δ𝑥Δ𝑦 + μ0 − 𝑢𝑗12Δ𝑦

Δ𝑥 − μ𝑢𝑗 − 𝑢𝑗−1

Δ𝑦Δ𝑥 = 0

𝐺Δ𝑦2

μ= −𝑢𝑗−1 + 3𝑢𝑗

Bottom cell (𝑗 = 1): 𝐺Δ𝑥Δ𝑦 + μ𝑢𝑗+1 − 𝑢𝑗

Δ𝑦Δ𝑥 − μ

𝑢𝑗 − 012Δ𝑦

Δ𝑥 = 0

𝐺Δ𝑦2

μ= 3𝑢𝑗 − 𝑢𝑗+1

(e) Solve for the nodal velocities in the case 𝑁 = 6, leaving your answers as multiples of 𝑈0 = 𝐺𝐻2/μ.

Symmetry: 𝑢1 = 𝑢6, 𝑢2 = 𝑢5, 𝑢3 = 𝑢4 Consider just 𝑗 = 1,2,3

𝐺Δ𝑦2

μ=𝐺𝐻2

μ

Δ𝑦

𝐻

2

=1

36𝑈0

u1

j-1u

ju

j+1u

Nu

y

x

H


(e) Solve for the nodal velocities in the case 𝑁 = 6, leaving your answers as multiples of 𝑈0 = 𝐺𝐻2/μ.

𝑗 = 1: 3𝑢1 − 𝑢2 =1

36𝑈0

𝑗 = 2:

𝑗 = 3:

−𝑢1 + 2𝑢2 − 𝑢3 =1

36𝑈0

−𝑢2 + 2𝑢3 − 𝑢4 =1

36𝑈0 −𝑢2 + 𝑢3 =

1

36𝑈0

𝑢1 =1

3(𝑢2 +

1

36𝑈0)

𝑢3 = 𝑢2 +1

36𝑈0

2

3𝑢2 =

7

3×

1

36𝑈0

𝑢2 =7

72𝑈0

𝑢1 = 𝑢6 =3

72𝑈0 𝑢2 = 𝑢5 =

7

72𝑈0 𝑢3 = 𝑢4 =

9

72𝑈0

−1

3(𝑢2 +

1

36𝑈0) + 2𝑢2 − (𝑢2 +

1

36𝑈0) =

1

36𝑈0

u1

j-1u

ju

j+1u

Nu

y

x

H


(g) Find the wall shear stress, τ𝑤

(f) Using your numerical solution, find the volume flow rate (per unit span) 𝑞 in terms of 𝑈0 and 𝐻.

𝑢1 = 𝑢6 =3

72𝑈0 𝑢2 = 𝑢5 =

7

72𝑈0 𝑢3 = 𝑢4 =

9

72𝑈0

𝑞 =

𝑗=1

6

𝑢𝑗Δ𝑦 = 2

𝑗=1

3

𝑢𝑗𝐻

6=𝐻

3𝑈0(

3

72+

7

72+

9

72) =

19

216𝑈0𝐻

2 × τ𝑤Δ𝑥 = (𝑝𝐿 − 𝑝𝑅)𝐻

τ𝑤 =1

2(𝑝𝐿 − 𝑝𝑅Δ𝑥

)𝐻 =1

2𝐺𝐻

u1

j-1u

ju

j+1u

Nu

y

x

H


(h) Compare your answers to (e), (f), (g) with the exact solution for plane Poiseuille flow.

Exact solution:𝑢

𝑈0=1

2

𝑦

𝐻1 −

𝑦

𝐻, 𝑈0 =

𝐺𝐻2

μ

𝑢1 = 𝑢6 =3

72𝑈0 𝑢2 = 𝑢5 =

7

72𝑈0 𝑢3 = 𝑢4 =

9

72𝑈0

Nodal values:

(numerical)

𝑢1 = 𝑢6 =2.75

72𝑈0 𝑢2 = 𝑢5 =

6.75

72𝑈0 𝑢3 = 𝑢4 =

8.75

72𝑈0(exact)

𝑞 = න0

𝐻

𝑢 d𝑦 =18

216𝑈0𝐻

(numerical) 𝑞 =19

216𝑈0𝐻

Flow rate:

(exact)

(numerical)

Wall stress:

(exact)

𝜏𝑤 =1

2𝐺𝐻

τ𝑤 = μ ቤ𝜕𝑢

𝜕𝑦𝑦=0

𝜏𝑤 =1

2𝐺𝐻

Discretising Advection (Part 1)

Example

A pipe of cross-section 𝐴 = 0.01 m2 and length 𝐿 = 1 m carries water (density ρ =1000 kg m−3) at velocity 𝑢 = 0.1 m s−1.

A faulty valve introduces a reactive chemical into the pipe half-way along its length at a rate of 0.01 kg s–1. The diffusivity of the chemical in water is Γ = 0.1 kg m−1s−1. The chemical is subsequently broken down at a rate proportional to its concentration ϕ (mass of chemical per unit mass of water), this rate amounting to – γϕ per metre, where γ =0.5 kg m−1 s−1 .

Approximating the downstream boundary condition by dϕ/d𝑥 = 0, set up a finite-volume calculation with 7 cells to estimate the concentration along the pipe using:(a) Central(b) Upwinddifferencing schemes for advection.

x = 0 L

pointsource

u= 0

= 0d

dx

Fluxes and Sources

Concentration ϕ (= mass of chemical per mass of water)

Source

‒ point source 𝑆𝑝𝑡 at 𝑥 = 0.5 m

‒ loss by chemical breakdown −γϕ per unit length

Flux (= rate of transport across a surface)

‒ diffusion: −Γ𝐴dϕ

d𝑥

‒ advection: 𝐶ϕ 𝐶 = ρ𝑢𝐴 (mass flux)

x = 0 L

pointsource

u= 0

= 0d

dx

Finite-Volume Decomposition

loss - per metre

S = 0.01 kg/spt

1 2 3 4 5 6 7= 0Ld

dx

= 0


source = ด𝑆ptcell 4

− 𝛾ϕ𝑃Δ𝑥 source = 𝑏𝑃 + 𝑠𝑃ϕ𝑃 𝑏𝑃 = 0.01 (cell 4)

𝑠𝑃 = −0.071

𝐶ϕ𝑒 − 𝐶ϕ𝑤advection

−𝐷ϕ𝑊 + 2𝐷ϕ𝑃 − 𝐷ϕ𝐸diffusion

= 𝑏𝑃 + 𝑠𝑃ϕ𝑃source

An approximation for cell-face values ϕe and ϕw is called an advection scheme or advection-differencing scheme

flux𝑒 − flux𝑤 = source

flux𝑒 = 𝐶ϕ𝑒 − 𝐷(ϕ𝐸 − ϕ𝑃)𝐶 = ρ𝑢𝐴 = 1.0

𝐷 =Γ𝐴

Δ𝑥= 0.007

flux = 𝐶ϕ − Γ𝐴dϕ

d𝑥

flux𝑤 = 𝐶ϕ𝑤 − 𝐷(ϕ𝑃 − ϕ𝑊)

w eP EW

Central Differencing

ϕ𝑒 =1

2(ϕ𝑃 + ϕ𝐸)

P Ee

PE

e

Finite-Volume Discretisation:Central Differencing For Advection




Central differencing:

−(𝐶

2+ 𝐷)ϕ𝑊 + (2𝐷 − 𝑠𝑃)ϕ𝑃 − (−

𝐶

2+ 𝐷)ϕ𝐸 = 𝑏𝑃

−0.507ϕ𝑊 + 0.085ϕ𝑃 + 0.493ϕ𝐸 = ถ0.01cell 4 only

𝐶ϕ𝑒 − 𝐶ϕ𝑤 = 𝐶1

2(ϕ𝑃 + ϕ𝐸) − 𝐶

1

2(ϕ𝑊 + ϕ𝑃)

w eP EW

=𝐶

2(ϕ𝐸 − ϕ𝑊)

Assembled EquationsCentral Advection Scheme

0.592 0.493 0 0 0 0 0−0.507 0.085 0.493 0 0 0 0

0 −0.507 0.085 0.493 0 0 00 0 −0.507 0.085 0.493 0 00 0 0 −0.507 0.085 0.493 00 0 0 0 −0.507 0.085 0.4930 0 0 0 0 −0.507 0.578

ϕ1ϕ2ϕ3ϕ4ϕ5ϕ6ϕ7

=

000

0.01000

Directional Bias

u

diffusion only

advection +diffusion

Upwind Differencing

ϕ𝑒 = ϕ𝑃 (if 𝑢 > 0) (if 𝑢 < 0)ϕ𝑒 = ϕ𝐸

P E

e

PE

e

P E

e

PEe

Upwind Differencing

Upwind differencing: 𝐶ϕ𝑒 − 𝐶ϕ𝑤 = 𝐶(ϕ𝑃 − ϕ𝑊)

−(𝐶 + 𝐷)ϕ𝑊 + (𝐶 + 2𝐷 − 𝑠𝑃)ϕ𝑃 − 𝐷ϕ𝐸 = 𝑏𝑃

−1.007ϕ𝑊 + 1.085ϕ𝑃 − 0.007ϕ𝐸 = ถ0.01cell 4 only




w eP EW

1.092 −0.007 0 0 0 0 0−1.007 1.085 −0.007 0 0 0 0

0 −1.007 1.085 −0.007 0 0 00 0 −1.007 1.085 −0.007 0 00 0 0 −1.007 1.085 −0.007 00 0 0 0 −1.007 1.085 −0.0070 0 0 0 0 −1.007 1.078

ϕ1ϕ2ϕ3ϕ4ϕ5ϕ6ϕ7

=

000

0.01000

Assembled EquationsUpwind Advection Scheme

Simple Advection Schemes Compared

−𝑎𝑊ϕ𝑊 + 𝑎𝑃ϕ𝑃 − 𝑎𝐸ϕ𝐸 = 𝑏𝑃

(if 𝐶 > 0)

No “wiggles” Pe ≤ 2

Pe =𝐶

𝐷=ρ𝑢Δ𝑥

ΓPeclet number

Theoretically, high accuracy ...... but unphysical “wiggles”

Lower accuracy ...... but no “wiggles”

Central differencing Upwind differencing

w eP EW

𝑎𝑊 =12𝐶 + 𝐷

𝑎𝐸 = −12𝐶 + 𝐷

𝑎𝑃 = 2𝐷 − 𝑠𝑃

𝑎𝑊 = 𝐶 + 𝐷

𝑎𝐸 = 𝐷

𝑎𝑃 = 𝐶 + 2𝐷 − 𝑠𝑃

Finite-Volume Analysis: 1-d



d𝑥

−𝑎𝑊ϕ𝑊 + 𝑎𝑃ϕ𝑃 − 𝑎𝐸ϕ𝐸 = 𝑏𝑃Discretisation:

Assembled matrix equation:

AΦ = b

flux efluxw source

W EP

= b

Finite-Volume Analysis: 2-d

flux𝑒 − flux𝑤 + flux𝑛 − flux𝑠 = sourceConservation:

flux𝑒,𝑤 = (ρ𝑢𝐴ϕ − Γ𝐴𝜕ϕ

𝜕𝑥)𝑒,𝑤

flux𝑛,𝑠 = (ρ𝑣𝐴ϕ − Γ𝐴𝜕ϕ

𝜕𝑦)𝑛,𝑠

−𝑎𝑆ϕ𝑆 − 𝑎𝑊ϕ𝑊 + 𝑎𝑃ϕ𝑃 − 𝑎𝐸ϕ𝐸 − 𝑎𝑁ϕ𝑁 = 𝑏𝑃Discretisation:

𝑎𝑃ϕ𝑃 −𝑎𝐹ϕ𝐹 = 𝑏𝑃

Assembled matrix equation:

AΦ = b

fluxefluxw source

fluxn

sflux

PW E

N

S

= b

General Discretisation Properties

Discretisation Properties

• Consistency

• Conservativeness

• Transportiveness

• Boundedness

• Stability

• Accuracy (“order”)

Consistency

Definition: an algebraic approximation is consistent if it tends to the exact governing equation as the mesh size tends to zero

ϕ𝐸 −ϕ𝑃Δ𝑥

→dϕ

d𝑥e.g.

Conservativeness

Definition: flux out of one cell = flux into adjacent cell

• Fluxes are worked out by face, not by cell

• Automatically built into the finite-volume method

flux

Transportiveness

Definition: upstream-biased

P E

e

PE

e

Boundedness

Definition. For advection-diffusion without sources:

(1) ϕP must lie between minimum and maximum values at surrounding nodes

(2) ϕ = constant must be a possible solution (if the boundary conditions permit)

Requirements for boundedness:

(1) Positive coefficients:

(2) Sum of neighbouring coefficients:

𝑎𝑃, 𝑎𝐹 ≥ 0

𝑎𝑃ϕ𝑃 −𝑎𝐹ϕ𝐹 = 0

PW E

N

S

𝑎𝑃 =𝑎𝐹

If ϕ𝐹 is the only non-zero adjacent value, ϕ𝑃 =𝑎𝐹𝑎𝑃

ϕ𝐹

If ϕ is constant. (𝑎𝑃−𝑎𝐹)ϕ = 0

Stability

Definition:

• Small errors do not grow in the course of the calculation

source = 𝑏𝑃 + 𝑠𝑃ϕ𝑃, 𝑠𝑃 ≤ 0

Requirement:

• Negative-slope linearisation of the source term (“negative feedback”)

Summary of Conditions on Matrix Coefficients

Negative-slope linearisation of the source term:

𝑠𝑃 ≤ 0

Positive coefficients:

𝑎𝐹 ≥ 0 for all 𝐹

Sum of neighbouring coefficients:

𝑎𝑃 =𝑎𝐹 − 𝑠𝑃

𝑎𝑃ϕ𝑃 −𝑎𝐹ϕ𝐹 = 𝑏𝑃

source = 𝑏𝑃 + 𝑠𝑃ϕ𝑃 PW E

N

S

Accuracy (“Order”)

• General definition:

𝐨𝐫𝐝𝐞𝐫 𝒏 ⇔ error ∝ Δ𝑥 𝑛 as Δ𝑥 → 0

• Determined by Taylor-series expansion (about a cell face)

• Common advection schemes:

‒ Upwind: 1st order

‒ Central: 2nd order

‒ LUD: 2nd order

‒ QUICK: 3rd order

P Ee

x

Order of AccuracyP Ee

x

ϕ𝐸 = ϕ𝑒 +dϕ

d𝑥𝑒

Δ𝑥

2+

1

2!

d2ϕ

d𝑥2𝑒

Δ𝑥

2

2

+1

3!

d3ϕ

d𝑥3𝑒

Δ𝑥

2

3

+1

4!

d4ϕ

d𝑥4𝑒

Δ𝑥

2

4

+ ⋯

ϕ𝐸 − ϕ𝑃 = 0 +dϕ

d𝑥𝑒

Δ𝑥 + 0 +2

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)3 + ⋯

Subtract:ϕ𝐸 − ϕ𝑃

Δ𝑥=

dϕ

d𝑥𝑒

+ O(Δ𝑥2)

Add:

ϕ𝑃 + ϕ𝐸 = 2ϕ𝑒 + 0 +d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 + ⋯ 1

2(ϕ𝑃 + ϕ𝐸) = ϕ𝑒 + O(Δ𝑥2)

2nd-order central differencing for diffusion

2nd-order central differencing for advection

ϕ𝑃 = ϕ𝑒 −dϕ

d𝑥𝑒

Δ𝑥

2+

1

2!

d2ϕ

d𝑥2𝑒

Δ𝑥

2

2

−1

3!

d3ϕ

d𝑥3𝑒

Δ𝑥

2

3

+1

4!

d4ϕ

d𝑥4𝑒

Δ𝑥

2

4

+ ⋯

First term:


d𝑥𝑒

Δ𝑥

2+ ⋯ ϕ𝑃 = ϕ𝑒 + O(Δ𝑥)

1st-order upwind differencing for advection

Example

Consider the uniform, one-dimensional arrangement of nodes shown. Face e lies half way between P and E nodes.

(a) Show that the central-differencing schemes

are second-order approximations for ϕ𝑒 and dϕ/d𝑥 𝑒respectively.

(b) Making use of the W and EE nodes also, find symmetric fourth-order approximations for ϕ𝑒 and dϕ/d𝑥 𝑒.

ϕ𝑒 ≈1

2(ϕ𝑃 + ϕ𝐸) ቤ

dϕ

d𝑥𝑒

≈ϕ𝐸 −ϕ𝑃

Δ𝑥

W EP EEe

x

3 x


W EP EEe

x

3 x


d𝑥𝑒

Δ𝑥

2+

1

2!

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 +

1

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)3 +

1

4!

d4ϕ

d𝑥4𝑒

(Δ𝑥

2)4 + ⋯


d𝑥𝑒

Δ𝑥

2+

1

2!

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 −

1

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)3 +

1

4!

d4ϕ

d𝑥4𝑒

(Δ𝑥

2)4 + ⋯

Add:

ϕ𝑃 + ϕ𝐸 = 2ϕ𝑒 + 0 +2

2!

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 + 0 +

d4ϕ

d𝑥4𝑒

(Δ𝑥

2)4 +⋯

ϕ𝑃 + ϕ𝐸2

= ϕ𝑒 +1

2!

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 + O(Δ𝑥)4

ϕ𝑊 + ϕ𝐸𝐸2

= ϕ𝑒 +1

2!

d2ϕ

d𝑥2𝑒

(3Δ𝑥

2)2 + O(Δ𝑥)4

αϕ𝑃 + ϕ𝐸

2+ β


= (α + β)ϕ𝑒 + (α + 9β)1

2

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 + O(Δ𝑥)4

Require α + β = 1

α + 9β = 0

8β = −1

β = −1

8α =

9

8

9

8

ϕ𝑃 + ϕ𝐸2

−1

8


= ϕ𝑒 + O(Δ𝑥)4


W EP EEe

x

3 x


d𝑥𝑒

Δ𝑥

2+

1

2!

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 +

1

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)3 +

1

4!

d4ϕ

d𝑥4𝑒

(Δ𝑥

2)4 + ⋯


d𝑥𝑒

Δ𝑥

2+

1

2!

d2ϕ

d𝑥2𝑒

(Δ𝑥

2)2 −

1

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)3 +

1

4!

d4ϕ

d𝑥4𝑒

(Δ𝑥

2)4 + ⋯

Subtract:

ϕ𝐸 − ϕ𝑃Δ𝑥

=dϕ

d𝑥𝑒

+1

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)2 + O(Δ𝑥)4

αϕ𝐸 − ϕ𝑃

Δ𝑥+ β

ϕ𝐸𝐸 − ϕ𝑊3Δ𝑥

= (α + β)dϕ

d𝑥𝑒

+ (α + 9β)1

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)2 + O(Δ𝑥)4

Require α + β = 1

α + 9β = 0

8β = −1

β = −1

8α =

9

8

9

8

ϕ𝐸 − ϕ𝑃Δ𝑥

−1

8


=dϕ

d𝑥𝑒

+ O(Δ𝑥)4

ϕ𝐸 − ϕ𝑃 = 0 +dϕ

d𝑥𝑒

Δ𝑥 + 0 +2

3!

d3ϕ

d𝑥3𝑒

(Δ𝑥

2)3 + 0 +

2

5!

d5ϕ

d𝑥5𝑒

(Δ𝑥

2)5 + ⋯


=dϕ

d𝑥𝑒

+1

3!

d3ϕ

d𝑥3𝑒

(3Δ𝑥

2)2 + O(Δ𝑥)4

Discretising Advection (Part 2)

Exponential Scheme

Exact solution of the 1-d advection-diffusion equation


d𝑥= constant

flux𝑒 = 𝐶ePeϕ𝑃 − ϕ𝐸

ePe − 1𝐶 = ρ𝑢𝐴, 𝐷 =

Γ𝐴

Δ𝑥, Pe =

𝐶

𝐷

flux𝑒 − flux𝑤 = 𝑎𝑃ϕ𝑃 −𝑎𝐹ϕ𝐹

𝑎𝑊 =𝐶ePe

ePe − 1, 𝑎𝐸 =

𝐶

ePe − 1

𝑎𝑃 = 𝑎𝐸 + 𝑎𝑊

(with constant coefficients and without sources)

P Ee

LUD (Linear Upwind Differencing)

• Extrapolates from the two upwind nodes

• 2nd-order accurate

• Transportive, but not bounded

• Generalises to unstructured meshes:

UU U D

faceW P E

P E EE

e

W

P

e

E

P

e

E

EE

u > 0

u < 0

e

ϕface = ϕ𝑈 +1

2(ϕ𝑈 − ϕ𝑈𝑈)

ϕface =3

2ϕ𝑈 −

1

2ϕ𝑈𝑈

ϕface = ϕ𝑈 + (xface − x𝑈) • (∇ϕ)𝑈

• 3-point, upwind-biased scheme

• Fits a quadratic through 3 nodes

• 3rd-order accurate

• Transportive, but not bounded

QUICK

W P E

P E EE

e

W

P

e

E

P

e

E

EE

u > 0

u < 0

e

UU U D

face

ϕface = −18ϕ𝑈𝑈 +

34ϕ𝑈 +

38ϕ𝐷

● 3-point, upwind-biased schemes; solution-dependent limiters

● General form:

● Non-linear ( iteration necessary)

● Total-Variation-Diminishing

Flux-Limited Schemes

ϕface = function(ϕ𝑈𝑈, ϕ𝑈 , ϕ𝐷)

ϕface = ቐϕ𝑈 +

1

2𝜓(𝑟)(ϕ𝐷 − ϕ𝑈) if monotonic

ϕ𝑈 otherwise

𝑟 =ϕ𝑈 − ϕ𝑈𝑈ϕ𝐷 − ϕ𝑈

UU U D

face

UU U D

monotonic

non-monotonic

UU U D

Scheme 𝛙(𝒓) (for 𝒓 > 𝟎)

UMISTUpstream Monotonic Interpolationfor Scalar Transport (Lien andLeschziner, 1993)

Harmonic Van Leer (1974)

Min-mod Roe (1985)

Van Albada et al. Van Albada et al., (1982)

min{ 2, 2𝑟, 141 + 3𝑟 ,

1

4(3 + 𝑟)}

2𝑟

1 + 𝑟

min{ 𝑟, 1}

𝑟 + 𝑟2

1 + 𝑟2

In all cases ψ 𝑟 is taken as 0 if non-monotonic (𝑟 < 0)

Example

The figure shows a single line of cells in a structured mesh. The values of a transported scalar ϕ at cell-centre nodes are given below the figure.

The Van Leer harmonic advection scheme is defined (for the immediate upstream and downstream nodes, U and D respectively) by:

ϕface = ϕ𝑈 +1

2ψ(𝑟)(ϕ𝐷 − ϕ𝑈)

ψ(𝑟) = ቐ2𝑟

1 + 𝑟if 𝑟 > 0

0 otherwise

Calculate the values of ϕ on the cell faces marked e and w in the figure using the Van Leer advection scheme if the velocity component 𝑢 is:(i) positive;(ii) negative.

P E EEWWW

ew

x


ϕ𝑊𝑊 = 1, ϕ𝑊= 4, ϕ𝑃= 5, ϕ𝐸= 7, ϕ𝐸𝐸= 3

Calculate the values of ϕ on the cell faces marked e and w in the figure using the Van Leer advection scheme if the velocity component 𝑢 is (i) positive; (ii) negative.

P E EEWWW

ew

x


ϕ𝑊𝑊 = 1, ϕ𝑊= 4, ϕ𝑃= 5, ϕ𝐸= 7, ϕ𝐸𝐸= 3

ϕface = ϕ𝑈 +1

2ψ(𝑟)(ϕ𝐷 − ϕ𝑈)

ψ(𝑟) = ቐ2𝑟

1 + 𝑟if 𝑟 > 0

0 otherwise

𝑢 positive

w: ϕ𝑈𝑈 = 1, ϕ𝑈 = 4, ϕ𝐷 = 5

𝑟 =4 − 1

5 − 4= 3

ψ =2 × 3

1 + 3=3

2

ϕ𝑓 = 4 +1

2×3

2(5 − 4) = 4.75

e: ϕ𝑈𝑈 = 4, ϕ𝑈 = 5, ϕ𝐷 = 7

𝑟 =5 − 4

7 − 5=1

2

ψ =2 × (1/2)

1 + (1/2)=2

3

ϕ𝑓 = 5 +1

2×2

3(7 − 5) = 5.667

𝑢 negative

w: ϕ𝑈𝑈 = 7, ϕ𝑈 = 5, ϕ𝐷 = 4

𝑟 =5 − 7

4 − 5= 2

ψ =2 × 2

1 + 2=4

3

ϕ𝑓 = 5 +1

2×4

3(4 − 5) = 4.333

e: ϕ𝑈𝑈 = 3, ϕ𝑈 = 7, ϕ𝐷 = 5

ϕ𝑓 = ϕ𝑈 = 7

Non-monotonic

Implementation of Higher-Order Schemes

The deferred correction is transferred to the source term and treated explicitly(i.e. not updated until the next iteration):

Write the cell-face value as upwind + correction: ϕ𝑓 = ถϕ𝑈upwind

+ (ϕ𝑓 − ϕ𝑈)

deferred correction

faces

( mass flux × ϕ −Γ𝜕ϕ

𝜕𝑛𝐴 ) = 𝑆

advection diffusion source

faces

[𝐶𝑓ϕ𝑓 + 𝐷𝑓(ϕ𝑃 − ϕ𝐹)] = 𝑏𝑃 + 𝑠𝑃ϕ𝑃

faces

[𝐶𝑓(ϕ𝑓 − ϕ𝑃) + 𝐷𝑓(ϕ𝑃 − ϕ𝐹)] = 𝑏𝑃 + 𝑠𝑃ϕ𝑃

𝐶𝑓(ϕ𝑓 − ϕ𝑃) = 𝐶𝑓(ϕ𝑈 − ϕ𝑃)

upwind

+𝐶𝑓(ϕ𝑓 − ϕ𝑈)

correction

= max( − 𝐶𝑓 , 0)(ϕ𝑃 − ϕ𝐹) +𝐶𝑓(ϕ𝑓 − ϕ𝑈)

𝑎𝐹 = max( − 𝐶𝑓 , 0) + 𝐷𝑓

𝐹

𝑎𝐹(ϕ𝑃 − ϕ𝐹) = 𝑏𝑃 + 𝑠𝑃ϕ𝑃−

faces

𝐶𝑓(ϕ𝑓 − ϕ𝑈)

deferred correction

The upwind part always gives positive coefficients and diagonal dominance

faces

𝐶𝑓 = 0

P Ff

Implementation of Boundary Conditions

Boundary fluxes are transferred to the source term:

notboundary

flux + fluxboundary = source

𝑎𝑃ϕ𝑃 −

notboundary

𝑎𝐹ϕ𝐹 = 𝑏𝑃 − fluxboundary

2 3 4 NI-1NI-2NI-3

boundary

boundary

P B

boundary

Matrix Solution Algorithms

Form of the Matrix Equations

𝑎𝑃ϕ𝑃 −

𝐹

𝑎𝐹ϕ𝐹 = 𝑏𝑃

1-d

−𝑎𝑊ϕ𝑊 + 𝑎𝑃ϕ𝑃 − 𝑎𝐸ϕ𝐸 = 𝑏𝑃

2-d

−𝑎𝑆ϕ𝑆 − 𝑎𝑊ϕ𝑊 + 𝑎𝑃ϕ𝑃 − 𝑎𝐸ϕ𝐸 − 𝑎𝑁ϕ𝑁 = 𝑏𝑃

W EP PW E

N

S

= b = b

Characteristics of the Equations

• Fluid-flow equations:

‒ non-linear

‒ coupled

• Matrix equations:

‒ sparse

‒ banded (for structured meshes)

• Solution methods:

‒ iterative

Common Solution Methods

• Gaussian elimination

• Gauss-Seidel

• Line Gauss-Seidel

Gaussian Elimination

• Direct, not iterative

• Sequence of row operations aiming to produce an upper-triangular matrix

• Tends to “fill in” sparse matrices

• Important exception: tri-diagonal systems(basis of the tri-diagonal matrix algorithm) = b

Gauss-Seidel

𝛟𝑷 =1

𝑎𝑃(𝑏𝑃 + 𝑎𝑆ϕ𝑆

∗ + 𝑎𝑊ϕ𝑊∗ + 𝑎𝐸ϕ𝐸

∗ + 𝑎𝑁ϕ𝑁∗ )

−𝑎𝑆ϕ𝑆 − 𝑎𝑊ϕ𝑊 + 𝒂𝑷𝛟𝑷 − 𝑎𝐸ϕ𝐸 − 𝑎𝑁ϕ𝑁 = 𝑏𝑃

P

N

S

W E

direction of sweep

being updated

already updated

not yet updated

Gauss-Seidel Example (i)

4 −1 0 0−1 4 −1 00 −1 4 −10 0 −1 4

𝐴𝐵𝐶𝐷

=

24613

4 −1 0 0−1 4 −1 00 −1 4 −10 0 −1 4

𝐴𝐵𝐶𝐷

=

24613

4𝐴 − 𝐵 = 2

−𝐴 + 4𝐵 − 𝐶 = 4

−𝐵 + 4𝐶 − 𝐷 = 6

−𝐶 + 4𝐷 = 13

𝐴 =2 + 𝐵

4

𝐵 =4 + 𝐴 + 𝐶

4

𝐶 =6 + 𝐵 + 𝐷

4

𝐷 =13 + 𝐶

4

A B C D

0 0 0 0

0.500 1.125 1.781 3.695

0.781 1.641 2.834 3.959

0.910 1.936 2.974 3.994

--- --- --- ---

1.000 2.000 3.000 4.000

Gauss-Seidel Example (ii)

1 −4 0 0−4 1 −4 00 −4 1 −40 0 −4 1

𝐴𝐵𝐶𝐷

=

−7−14−21−8

DIVERGES!

Gauss-Seidel Overview

• Iterative

• Simple to code

• Minimal storage requirements

• Can use for unstructured meshes

• Convergence slow for large matrices

• Constraints on matrix elements (e.g. diagonal dominance)

𝑎𝑖𝑖 ≥

𝑗≠𝑖

𝑎𝑖𝑗

Line Gauss-Seidel(Line-Iterative Procedures, LIP)

−𝒂𝑾𝛟𝑾 + 𝒂𝑷𝛟𝑷 − 𝒂𝑬𝛟𝑬 = 𝑏𝑃 + 𝑎𝑆ϕ𝑆∗ + 𝑎𝑁ϕ𝑁

∗

−𝑎𝑆ϕ𝑆 − 𝒂𝑾𝛟𝑾 + 𝒂𝑷𝛟𝑷 − 𝒂𝑬𝛟𝑬 − 𝑎𝑁ϕ𝑁 = 𝑏𝑃

= b - *

PW E

S

N

direct

ion

of sw

eep

Line Gauss-Seidel - Overview

• Iterative

• Repeated use of tri-diagonal solver:

‒ whole line updated in one go

‒ rapid propagation across domain

• Applicable to structured meshes only

• Basis of many research codes

Convergence Criteria

res = 𝑎𝑃ϕ𝑃 −

𝐹

𝑎𝐹ϕ𝐹 − 𝑏𝑃

Single-cell error:

residual:

cells

res

1

𝑁

cells

(res)2

sum of absolute residuals:

root-mean-square error:

Global error:

Convergence criteria:

error < toleranceor

error < fraction of initial error

𝑎𝑃ϕ𝑃 −

𝐹


Under-Relaxation

• The governing equations are coupled and non-linear

• The iterative solution method may be numerically unstable

• Under-relaxation tries to stabilise an iterative procedure by reducing the change in variables at each iteration

Under-Relaxation in CFD

𝑎𝑃ϕ𝑃 −𝑎𝐹ϕ𝐹 = 𝑏𝑃Standard linearised equation:

ϕ𝑃 =σ𝑎𝐹ϕ𝐹 + 𝑏𝑃

𝑎𝑃Rearrange as the change in ϕ:

ϕ𝑃 = ϕ𝑃prev

+σ𝑎𝐹ϕ𝐹 + 𝑏𝑃

𝑎𝑃− ϕ𝑃

prev

ϕ𝑃 = ϕ𝑃prev

+ ασ𝑎𝐹ϕ𝐹 + 𝑏𝑃

𝑎𝑃− ϕ𝑃

prevUnder-relax the change:

Rearrange back: 𝑎𝑃ϕ𝑃 −α𝑎𝐹ϕ𝐹 = α𝑏𝑃 + (1 − α)𝑎𝑃ϕ𝑃prev

Incorporate by modifying coefficients: 𝑎𝑃 → 𝑎𝑃′ =

𝑎𝑃α

𝑏𝑃 → 𝑏𝑃′ = 𝑏𝑃 + (1 − α)𝑎𝑃

′ ϕ𝑃prev

𝑎𝑃𝛼ϕ𝑃 −𝑎𝐹ϕ𝐹 = 𝑏𝑃 + (1 − α)

𝑎𝑃𝛼ϕ𝑃prev

• The generic scalar-transport equation for a control volume has the formrate of change + net outward flux = source

• Flux ≡ rate of transport through a surface and consists of:

‒ advection: transport with the flow (other authors prefer convection)

‒ diffusion: net transport by random molecular or turbulent fluctuations

• Discretisation of the (steady) scalar-transport equation yields a linear equation

𝑎𝑃ϕ𝑃 −

𝐹


for each control volume.

• The collection of these simultaneous equations on a structured mesh yields a matrix equation with limited bandwidth (i.e. few non-zero diagonals), typically solved by iterative methods such as Gauss-Seidel or line-Gauss-Seidel

Summary (1)

Summary (2)

• Source terms are linearised as

𝑏𝑃 + 𝑠𝑃ϕ𝑃 , 𝑠𝑃 ≤ 0

• Diffusive fluxes are usually discretised by central differencing; e.g.

−Γ ቤ𝜕ϕ

𝜕𝑥𝑒

𝐴 → −Γ𝐴

Δ𝑥(ϕ𝐸 − ϕ𝑃)

• Advection schemes are means of approximating ϕ on cell faces in order to compute advective fluxes. They include Upwind, Central, Exponential, LUD, QUICK, and various flux-limited schemes

• General desirable properties for a numerical scheme in CFD:

‒ consistency

‒ conservativeness

‒ transportiveness

‒ boundedness

‒ stability

‒ accuracy / order

● Boundedness and stability impose certain constraints:

𝑎𝐹 ≥ 0 (“positive coefficients”)

𝑠𝑃 ≤ 0 (“negative feedback in the source term”)

𝑎𝑃 = σ𝑎𝐹 − 𝑠𝑃 (“sum of the neighbouring coefficients”)

● To ensure positive coefficients (and implement non-linear schemes), advectivefluxes are often decomposed into

upwind + deferred correction

The latter is transferred to the source and treated explicitly

● Boundary conditions can be implemented by transferring boundary fluxes to the source term

● Under-relaxation is usually required to solve coupled and/or non-linear equations

Summary (3)

4. The Scalar-Transport Equation - University of Manchester · 2021. 1. 4. · Example A thin rod...

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