5. Pressure and Velocity

Part 1. Pressure-Velocity CouplingPart 2. Pressure-Correction Methods

Scalar-Transport Equation For Momentum

• concentration, ϕ velocity component (𝑢, 𝑣, 𝑤)

• diffusivity, viscosity,

• source, 𝑆 non-viscous forces

Each component of momentum satisfies its own scalar-transport equation:

d

d𝑡(𝑚𝑎𝑠𝑠 × ϕ) +

𝑓𝑎𝑐𝑒𝑠

( 𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × ϕ −Γ𝜕

𝜕𝑛𝐴 ) = 𝑆

𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑎𝑑𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑠𝑜𝑢𝑟𝑐𝑒

Special Features of the Momentum Equations

• The momentum equations are:

‒ non-linear

‒ coupled

‒ required also to be mass-consistent

• As a result they must be solved:

‒ iteratively

‒ together

‒ in conjunction with the continuity equation

• And we also need to specify pressure …

(mass flux) velocity → (ρ𝑢𝐴)𝑢

(ρ𝑢𝐴)𝑣

uv

mass flux

uA

Solving Mass and Momentum Equations

DO WHILE (not_converged)

CALL SCALAR_TRANSPORT( u )

CALL SCALAR_TRANSPORT( v )

CALL SCALAR_TRANSPORT( w )

CALL MASS_CONSERVATION

END DO

Mass Conservation

General Scalar-Transport Equation

d

d𝑡(𝑚𝑎𝑠𝑠) +

𝑓𝑎𝑐𝑒𝑠

(𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥) = 0

d

d𝑡(𝑚𝑎𝑠𝑠 × ϕ) +

𝑓𝑎𝑐𝑒𝑠

(𝑚𝑎𝑠𝑠 𝑓𝑙𝑢𝑥 × ϕ − Γ𝜕ϕ

𝜕𝑛𝐴) = 𝑆

For incompressible flow, the mass equation could be regarded as a special case of the scalar-transport equation ... but only if:

ϕ 1 !!!

Γ 0

𝑆 0

For compressible flow, the mass equation is a transport equation for density

V

A

u

un

Status of the Mass-Conservation Equation?

How is Pressure Determined?

• Compressible flow:

‒ mass conservation → transport equation for density,

‒ transport equation for energy→ temperature, 𝑇

‒ equation of state (e.g. ideal gas law 𝑝 = 𝑅𝑇) → pressure, 𝑝

• Incompressible flow:

‒ the momentum equations link velocity and pressure

‒ substitution in the mass equation yields an equation for pressure

A pressure equation arises from the requirement thatsolutions of the momentum equation be mass-consistent.

Solving Mass and Momentum Equations

DO WHILE (not_converged)

CALL SCALAR_TRANSPORT( u )

CALL SCALAR_TRANSPORT( v )

CALL SCALAR_TRANSPORT( w )

CALL MASS_CONSERVATION( p )

END DO

Pressure-Velocity Coupling

Q1. How are velocity and pressure linked?

Q2. How does a pressure equation arise?

Q3. Should velocity and pressure be stored at the same locations?

Q1. How are Pressure and Velocity Linked?

Momentum equation:

𝑎𝑃𝑢𝑃 −𝑎𝐹𝑢𝐹

𝑛𝑒𝑡 𝑓𝑙𝑢𝑥

= 𝐴(𝑝𝑤 − 𝑝𝑒)𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑜𝑟𝑐𝑒

+ 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑐𝑒𝑠

Velocity-pressure linkage:

𝑢𝑃 = 𝑑𝑃(𝑝𝑤 − 𝑝𝑒) + ⋯ 𝑑𝑝 =𝐴

𝑎𝑃

A1. (a) The force terms in the momentum equation provide a link between velocity and pressure.

(b) Velocity depends on the pressure gradient or, when discretised, on the differencebetween pressure values ½ cell either side.

𝑢 = −𝑑Δ𝑝 +⋯

uPp p

w e

area A

Q2. How Does a Pressure Equation Arise?

𝑢 = −𝑑Δ𝑝 +⋯

The momentum equation links velocity and pressure:

Substituting in the mass equation gives an equation for pressure:

0 = (ρ𝑢𝐴)𝑒 − (ρ𝑢𝐴)𝑤 +⋯

A2. A pressure equation arises from the requirement that solutions of the momentum equation be mass-consistent.

EW

N

S

P

n

e

s

w

= (ρ𝐴𝑑)𝑒(𝑝𝑃 − 𝑝𝐸) − (ρ𝐴𝑑)𝑤(𝑝𝑊 − 𝑝𝑃) + ⋯

= −𝑎𝑊𝑝𝑊 + 𝑎𝑝𝑝𝑝 − 𝑎𝐸𝑝𝐸 +⋯

Q3. Should velocity and pressure be stored at the same locations?

Co-located Pressure and Velocity(i) Effect in the Momentum Equation

𝐴(𝑝𝑤 − 𝑝𝑒) =𝐴

2(𝑝𝑖−1 − 𝑝𝑖+1)Net pressure force: No 𝑝𝑖 !!!

𝑝𝑤 =1

2(𝑝𝑖−1 + 𝑝𝑖) 𝑝𝑒 =

1

2(𝑝𝑖 + 𝑝𝑖+1)

w eii-1 i+1

Co-located Pressure and Velocity(ii) Effect in the Continuity Equation

Continuity:

=1

2ρ𝐴[𝑢𝑖+1 − 𝑢𝑖−1] + ⋯

𝑢𝑤 =1

2(𝑢𝑖−1 + 𝑢𝑖) 𝑢𝑒 =

1

2(𝑢𝑖 + 𝑢𝑖+1)

Involves alternate 𝑝 values only

w eii-1 i+1i-2 i+2

0 = (ρ𝐴𝑢)𝑒 − (ρ𝐴𝑢)𝑤 + …

= ρ𝐴1

2(𝑢𝑖 + 𝑢𝑖+1) −

1

2(𝑢𝑖−1 + 𝑢𝑖) + ⋯

=1

2ρ𝐴

1

2𝑑𝑖+1(𝑝𝑖 − 𝑝𝑖+2) −

12𝑑𝑖−1(𝑝𝑖−2 − 𝑝𝑖) + ⋯

Pressure/Velocity Coupling

Momentum & continuity → pressure equation

Remedies:

(1) staggered grid, or

(2) Rhie-Chow interpolation for advective velocities

• co-located 𝑢, 𝑝 and

• linear interpolation for advective velocities

but the combination of

odd-even decoupling(“checkerboard” effect)

}

x

p

1 2 3 4 5 6 7 8 9 10

Staggered Grid(Harlow and Welch, 1965)

u p

v

u p

v

p p

p

pressure/scalarcontrol volume

u control volume

v control volume

Staggered Grid: Advantages

On a cartesian mesh:

• Pressure is stored at the points required to compute pressure forces

𝑢𝑖 = 𝑑𝑖(𝑝𝑖−1 − 𝑝𝑖) + ⋯

• Velocity is stored at the points required to compute mass fluxes

up pi-1 i i

pp pi-1 i i+1ui ui+1

0 = 𝑢𝑖+1 − 𝑢𝑖 +⋯

= 𝑑𝑖+1 𝑝𝑖 − 𝑝𝑖+1 − 𝑑𝑖 𝑝𝑖−1 − 𝑝𝑖 +⋯

= −𝑑𝑖𝑝𝑖−1 + (𝑑𝑖 + 𝑑𝑖+1)𝑝𝑖 − 𝑑𝑖+1𝑝𝑖+1 +⋯

Staggered Grid: Disadvantages

• Added geometric complexity

• Rationale fails on non-cartesian grids

• Impossible to implement in unstructured meshes

Non-Staggered/Co-located Grid(Rhie and Chow, 1983)

Momentum equation:

𝑢 = ො𝑢 − 𝑑Δ𝑝

Rhie-Chow interpolation: apply this symbolic relation at both nodes and faces

𝑢𝑓𝑎𝑐𝑒 = ො𝑢𝑓𝑎𝑐𝑒 − 𝑑𝑓𝑎𝑐𝑒Δ𝑝

𝑎𝑃𝑢𝑃 −𝑎𝐹𝑢𝐹 = −𝐴(𝑝𝑒 − 𝑝𝑤) + ⋯

ො𝑢 = 𝑢 + 𝑑Δ𝑝 Work out pseudovelocity at nodes

Then interpolate to faces

𝑢𝑒 = (𝑢 + 𝑑Δ𝑝)𝑒 − 𝑑𝑒(𝑝𝐸 − 𝑝𝑃)

𝑢𝑃 =σ𝑎𝐹𝑢𝐹𝑎𝑃

− 𝑑𝑃(𝑝𝑒 − 𝑝𝑤) + ⋯

w eW P E

Symbolically:

Mass Conservation → Pressure Equation

𝑢 = ො𝑢 − 𝑑Δ𝑝

In practice, one solves for a pressure correction:

0 = (ρ𝐴𝑢∗)𝑒 − (ρ𝐴𝑢∗)𝑤 − 𝑎𝑊𝑝𝑊′ + 𝑎𝑃𝑝𝑃

′ − 𝑎𝐸𝑝𝐸′ +⋯

−𝑎𝑊𝑝𝑊′ + 𝑎𝑃𝑝𝑃

′ − 𝑎𝐸𝑝𝐸′ +⋯ = − 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑚𝑎𝑠𝑠 𝑜𝑢𝑡𝑓𝑙𝑜𝑤

w eW P E

0 = (ρ𝐴𝑢)𝑒 − (ρ𝐴𝑢)𝑤 +⋯

= (ρ𝐴ො𝑢)𝑒 − (ρ𝐴ො𝑢)𝑤 + (ρ𝐴𝑑)𝑒(𝑝𝑃 − 𝑝𝐸) − (ρ𝐴𝑑)𝑤(𝑝𝑊 − 𝑝𝑃) + ⋯

= (ρ𝐴ො𝑢)𝑒 − (ρ𝐴ො𝑢)𝑤 − 𝑎𝑊𝑝𝑊 + 𝑎𝑃𝑝𝑃 − 𝑎𝐸𝑝𝐸 +⋯

Example

For the uniform cartesian mesh shown below the momentum equation gives a velocity/pressure relationship

𝑢 = −4Δ𝑝 +⋯

For the 𝑢 and 𝑝 values given, calculate the advective velocity on the cell face f :

(a) by linear interpolation;(b) by Rhie-Chow interpolation if 𝑝𝐴 = 0.6;(c) by Rhie-Chow interpolation if 𝑝𝐴 = 0.9.

u =

p =

f

A

1 2 3 3

p 0.8 0.7 0.6

Calculate the advective velocity on the cell face f :(a) by linear interpolation;

u =

p =

f

A

1 2 3 3

p 0.8 0.7 0.6

𝑢 = −4Δ𝑝 +⋯

𝑢𝑓 =1

2(2 + 3) = 2.5

(b) by Rhie-Chow interpolation if 𝑝𝐴 = 0.6;

𝑢 = ො𝑢 − 4Δ𝑝

ො𝑢 = 𝑢 + 4Δ𝑝

... needs ො𝑢 at face ... interpolated from ො𝑢 at nodes ... needs 𝑝 at faces

𝑢 (𝑛𝑜𝑑𝑒)

𝑝 (𝑛𝑜𝑑𝑒)

𝑝 (𝑓𝑎𝑐𝑒)interpolate:

ො𝑢 (𝑛𝑜𝑑𝑒)ො𝑢 = 𝑢 + 4Δ𝑝

ො𝑢 (𝑓𝑎𝑐𝑒)

𝑢 (𝑓𝑎𝑐𝑒)

𝑢 = ො𝑢 − 4Δ𝑝

1 2 3 3

0.6 0.8 0.7 0.6

0.7 0.75 0.65

2.2 2.6

2.4

2.8

interpolate:

Calculate the advective velocity on the cell face f :(c) by Rhie-Chow interpolation if 𝑝𝐴 = 0.9.

u =

p =

f

A

1 2 3 3

p 0.8 0.7 0.6

𝑢 = −4Δ𝑝 +⋯

𝑢 = ො𝑢 − 4Δ𝑝

ො𝑢 = 𝑢 + 4Δ𝑝

𝑢 (𝑛𝑜𝑑𝑒)

𝑝 (𝑛𝑜𝑑𝑒)

𝑝 (𝑓𝑎𝑐𝑒)interpolate:

ො𝑢 (𝑛𝑜𝑑𝑒)ො𝑢 = 𝑢 + 4Δ𝑝

ො𝑢 (𝑓𝑎𝑐𝑒)

𝑢 (𝑓𝑎𝑐𝑒)

𝑢 = ො𝑢 − 4Δ𝑝

1 2 3 3

0.9 0.8 0.7 0.6

0.85 0.75 0.65

1.6 2.6

2.1

2.5

interpolate:

Looking Ahead ...

• Pressure/velocity coupling is the dominant feature in incompressible flow

• Mass and momentum equations must be satisfied simultaneously

• The most popular type of solution algorithm is called a pressure-correction method

Part 1. Pressure-Velocity Coupling

Part 2. Pressure-Correction Methods

Correcting Pressure to Enforce Mass Conservation

Net mass flux in;increase cell pressure

Net mass flux out;decrease cell pressure

Pressure-Correction Methods

• Iterative numerical schemes for pressure-linked equations

• Seek to produce velocity and pressure fields satisfying both mass and momentum equations

• Consist of alternating updates of velocity and pressure:

‒ solve momentum equation with current pressure

‒ use velocity-pressure link to rephrase continuity as a pressure-correction equation

‒ solve for pressure corrections to “nudge” velocity towards mass conservation

• Popular methods:

‒ SIMPLE

‒ PISO

Velocity and Pressure Corrections

• The momentum equation links velocity and pressure: 𝑢 = 𝑑(𝑝−1/2 − 𝑝+1/2) + ⋯

• Must simultaneously correct pressure to retain a solution of the momentum equation: 𝑢′ = 𝑑(𝑝−1/2

′ − 𝑝+1/2′ ) + ⋯

• Must correct velocity to satisfy continuity: 𝑢 → 𝑢∗ + 𝑢′

𝑢 → 𝑢∗ + 𝑑(𝑝−1/2′ − 𝑝+1/2

′ )Velocity correction formula:

Example

In the steady, one-dimensional, constant-density situation shown, the pressure pis stored at locations 1, 2 and 3, whilst velocity u is stored at the staggered locations A, B and C. The velocity-correction formula is

𝑢 = 𝑢∗ + 𝑢′ where 𝑢′ = 𝑑(𝑝𝑖−1′ − 𝑝𝑖

′)

where pressure nodes 𝑖– 1 and 𝑖 lie on either side of the location for 𝑢. The value of 𝑑 is 2 everywhere. The boundary condition is 𝑢𝐴 = 10. If, at a given stage in the iteration process, the momentum equations give 𝑢𝐵

∗ = 8 and 𝑢𝐶∗ = 11, calculate

the values of 𝑝1, 𝑝2, 𝑝3 and the resulting corrected velocities.

A 1 B 2 C 3

In the steady, one-dimensional, constant-density situation shown, the pressure p is stored at locations 1, 2 and 3, whilst velocity u is stored at the staggered locations A, B and C. The velocity-correction formula is

𝑢 = 𝑢∗ + 𝑢′ where 𝑢′ = 𝑑(𝑝𝑖−1′ − 𝑝𝑖

′)

where pressure nodes 𝑖– 1 and 𝑖 lie on either side of the location for 𝑢. The value of 𝑑 is 2 everywhere. The boundary condition is 𝑢𝐴 = 10. If, at a given stage in the iteration process, the momentum equations give𝑢𝐵∗ = 8 and 𝑢𝐶

∗ = 11, calculate the values of 𝑝1, 𝑝2, 𝑝3 and the resulting corrected velocities.

A 1 B 2 C 3

𝑢𝐴 = 10 𝑢𝐵 = 8 + 2(𝑝′1− 𝑝′

2)

Cell 1 Cell 2

𝑢𝐵 − 𝑢𝐴 = 0

ρ𝑢𝐵𝐴 − ρ𝑢𝐴𝐴 = 0

)8 + 2(𝑝′1 − 𝑝′2 − 10 = 0

2𝑝′1 − 2𝑝′2= −(8 − 10)

𝑝′1 − 𝑝′2 = 1

𝑢𝐶 − 𝑢𝐵 = 0

11 + 2 𝑝′2− 𝑝′

3− 8 − 2(𝑝′

1− 𝑝′

2) = 0

−2𝑝′1 + 4𝑝′2− 2𝑝′3 = −(11 − 8)

−𝑝′1+ 2𝑝′

2− 𝑝′

3= −3/2

𝑢𝐶 = 11 + 2(𝑝′2 − 𝑝′3)

WLOG 𝑝′3 = 0 𝑝′1 − 𝑝′2 = 1

−𝑝′1+ 2𝑝′

2= −3/2 𝑝′

2= −1/2

𝑝′1= +1/2 𝑢𝐵 = 10

𝑢𝐶 = 10

1 2 3A B C

Comments

• The continuity equation doesn’t explicitly contain pressure ... but constraints imposed by the momentum equation lead to a pressure equation

• Matrix equations for pressure are similar to those for the scalar-transport equation

• The source term is minus the current mass outflow

• (In incompressible flow) pressure is fixed only up to a constant

ExampleIn the 2-d staggered-grid arrangement shown below, 𝑢 and 𝑣 (the 𝑥 and 𝑦 components of velocity), are stored at nodes indicated by arrows, whilst pressure 𝑝 is stored at the intermediate nodes A-D. The grid spacing is uniform and the same in both directions. Velocity is fixed on the boundaries as shown. The velocity components at the interior nodes (𝑢𝐵, 𝑢𝐷, 𝑣𝐶 and 𝑣𝐷) are to be found.

At an intermediate stage of calculation the internal velocity values are found to be

𝑢𝐵 = 11, 𝑢𝐷 = 14, 𝑣𝐶 = 8, 𝑣𝐷 = 5

whilst correction formulae derived from the momentum equation are

𝑢′ = 2(𝑝𝑤′ − 𝑝𝑒

′ ) , 𝑣′ = 3(𝑝𝑠′ − 𝑝𝑛

′ )

with geographical (w,e,s,n) notation indicating the relative location of pressure nodes.

Show that applying mass conservation to control volumes centred on pressure nodes leads to simultaneous equations for the pressure corrections. Solve for the pressure corrections and use them to generate a mass-consistent flow field.

10 20

5 5

5 10

15 10

A B

D

vC vD

uD

uB

x

y

C

At an intermediate stage of calculation the internal velocity values are found to be

𝑢𝐵 = 11, 𝑢𝐷 = 14, 𝑣𝐶 = 8, 𝑣𝐷 = 5

whilst correction formulae derived from the momentum equation are

𝑢′ = 2(𝑝𝑤′ − 𝑝𝑒

′ ) , 𝑣′ = 3(𝑝𝑠′ − 𝑝𝑛

′ )

10 20

5 5

5 10

15 10

A B

D

vC vD

uD

uB

x

y

C

𝑢𝐵 = 11 + 2(𝑝′𝐴 − 𝑝′𝐵)

𝑢𝐷 = 14 + 2(𝑝′𝐶 − 𝑝′𝐷)

𝑣𝐶 = 8 + 3(𝑝′𝐴 − 𝑝′𝐶)

𝑣𝐷 = 5 + 3(𝑝′𝐵 − 𝑝′𝐷)

𝑢𝐵 − 5 + 𝑣𝐶 − 15 = 0

Cell A:

11 + 2(𝑝′𝐴 − 𝑝′𝐵) − 5 + 8 + 3(𝑝′𝐴 − 𝑝′𝐶) − 15 = 0

5𝑝′𝐴 − 2𝑝′𝐵 − 3𝑝′𝐶 = 1

Cell B:

5 − 𝑢𝐵 + 𝑣𝐷 − 10 = 0

5 − 11 − 2(𝑝′𝐴 − 𝑝′𝐵) + 5 + 3(𝑝′𝐵 − 𝑝′𝐷) − 10 = 0

−2𝑝′𝐴 + 5𝑝′𝐵 − 3𝑝′𝐷 = 11

Cell C:

Cell D:

−3𝑝′𝐴 + 5𝑝′𝐶 − 2𝑝′𝐷 = −1

−3𝑝′𝐵 − 2𝑝′𝐶 + 5𝑝′𝐷 = −11

5 −2 −3 0−2 5 0 −3−3 0 5 −20 −3 −2 5

𝑝′𝐴𝑝′𝐵𝑝′𝐶𝑝′𝐷

=

111−1−11

10 20

5 5

5 10

15 10

A B

D

vC vD

uD

uB

x

y

C

5 −2 −3 0 ∥ 1−2 5 0 −3 ∥ 11−3 0 5 −2 ∥ −10 −3 −2 5 ∥ −11

5 −2 −3 0 ∥ 10 21 −6 −15 ∥ 570 −6 16 −10 ∥ −20 −3 −2 5 ∥ −11

𝑅2 → 5𝑅2 + 2𝑅1𝑅3 → 5𝑅3 + 3𝑅1

5 −2 −3 0 ∥ 10 7 −2 −5 ∥ 190 −3 8 −5 ∥ −10 −3 −2 5 ∥ −11

𝑅2 → 𝑅2/3𝑅3 → 𝑅3/2

5 −2 −3 0 ∥ 10 7 −2 −5 ∥ 190 0 50 −50 ∥ 500 0 −20 20 ∥ −20

𝑅3 → 7𝑅3 + 3𝑅2𝑅4 → 7𝑅4 + 3𝑅2

5 −2 −3 0 ∥ 10 7 −2 −5 ∥ 190 0 1 −1 ∥ 10 0 −1 1 ∥ −1

𝑅3 → 𝑅3/50𝑅4 → 𝑅4/20

5 −2 −3 00 7 −2 −50 0 1 −10 0 0 0

𝑝′𝐴𝑝′𝐵𝑝′𝐶𝑝′𝐷

=

11910𝑅4 → 𝑅4 + 𝑅3

𝑝′𝐶 − 𝑝′𝐷 = 1 𝑝′𝐶 = 1 + 𝑝′𝐷

7𝑝′𝐵 − 2𝑝′𝐶 − 5𝑝′𝐷 = 19

𝑝′𝐵 =19 + 2𝑝′𝐶 + 5𝑝′𝐷

7

5𝑝′𝐴 − 2𝑝′𝐵 − 3𝑝′𝐶 = 1

𝑝′𝐴 =1 + 2𝑝′𝐵 + 3𝑝′𝐶

5

𝑝′𝐵 = 3 + 𝑝′𝐷

𝑝′𝐴 = 2 + 𝑝′𝐷

10 20

5 5

5 10

15 10

A B

D

vC vD

uD

uB

x

y

C

𝑝′𝐶 = 1 + 𝑝′𝐷

𝑝′𝐵 = 3 + 𝑝′𝐷

𝑝′𝐴 = 2 + 𝑝′𝐷

𝑢𝐵 = 11 + 2(𝑝′𝐴 − 𝑝′𝐵)

𝑢𝐷 = 14 + 2(𝑝′𝐶 − 𝑝′𝐷)

𝑣𝐶 = 8 + 3(𝑝′𝐴 − 𝑝′𝐶)

𝑣𝐷 = 5 + 3(𝑝′𝐵 − 𝑝′𝐷)

= 9

= 16

= 11

= 14

SIMPLESemi-Implicit Method for Pressure-Linked Equations

solve pressure-correction equation

correct velocityand pressure

converged?

START

END

yes

no

solve momentumequations

SIMPLE

1. Solve momentum equation with the current pressure.

2. Formulate pressure-correction equation:

(i) relate changes in 𝑢 and 𝑝; 𝑢𝑃′ =

σ𝑎𝐹𝑢𝐹′

𝑎𝑃+ 𝑑𝑃(𝑝𝑤

′ − 𝑝𝑒′ )

(iv) rewrite in terms of 𝑝. (ρ𝐴𝑑)𝑒(𝑝𝑃′ − 𝑝𝐸

′ ) − (ρ𝐴𝑑)𝑤(𝑝𝑊′ − 𝑝𝑃

′ )] + ⋯ = − ሶ𝑚∗

3. Solve pressure-correction equation: 𝑎𝑃𝑝𝑃′ −𝑎𝐹𝑝𝐹

′ = − ሶ𝑚∗

4. Correct velocity and pressure: 𝑝𝑃 → 𝑝𝑃∗ + 𝑝𝑃

′

𝑢𝑃 → 𝑢𝑃∗ + 𝑑𝑃(𝑝𝑤

′ − 𝑝𝑒′ )

(iii) apply mass conservation; [ρ𝐴(𝑢∗ + 𝑢′)]𝑒 − [ρ𝐴(𝑢∗ + 𝑢′)]𝑤 +⋯ = 0

(ii) make SIMPLE approximation;

𝑢𝑝 =Σ𝑎𝐹𝑢𝐹𝑎𝑃

+ 𝑑𝑃(𝑝𝑤∗ − 𝑝𝑒

∗) + ⋯

(ρ𝐴𝑢′)𝑒 − (ρ𝐴𝑢′)𝑤 +⋯ = − ሶ𝑚∗

SIMPLE (continued)

• The source of the pressure-correction equation is minus the current mass imbalance

• (Substantial) pressure under-relaxation is usually required:𝑝 → 𝑝∗ + α𝑝𝑝

′

• Iterative process – no need to solve equations exactly at each stage. Either:‒ do 𝑚 iterations of velocity, 𝑛 iterations of pressure; or‒ do enough iterations to reduce the error by a set amount

Variants of SIMPLE

SIMPLE 𝑢𝑃′ =

σ𝑎𝐹𝑢𝐹′

𝑎𝑃+ 𝑑𝑃(𝑝𝑤

′ − 𝑝𝑒′ )

SIMPLEC: alternative correction formula:

𝑢𝑃′ ≈

𝑑𝑃1 − σ𝑎𝐹/𝑎𝑃

(𝑝𝑤′ − 𝑝𝑒

′ )

SIMPLER: precede momentum update with exact pressure equation:

SIMPLEX: solve equations for correction coefficients δ𝑃:

𝑢𝑃′ ≈ δ𝑃(𝑝𝑤

′ − 𝑝𝑒′ )

𝑎𝑃𝑝𝑃 −𝑎𝐹𝑝𝐹 = −(𝑛𝑒𝑡 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝑜𝑓 ො𝑢)

PISO(Pressure-Implicit with Splitting of Operators)

• Time-dependent pressure-correction method

• Each timestep 𝑡𝑜𝑙𝑑 → 𝑡𝑛𝑒𝑤 is a non-iterative sequence:

1. solve time-dependent momentum eqns with 𝑡𝑜𝑙𝑑 pressure

2. solve a pressure-correction equation and update pressure and velocity

3. second mass-corrector step with time-advanced pressure

• More efficient than SIMPLE for time-dependent problems(allegedly)

Summary (1)

• Each momentum component satisfies its own scalar-transport equation

• The momentum equations require special treatment because they are:

‒ non-linear

‒ coupled

‒ required also to be mass-consistent

• In incompressible flow, continuity (mass conservation) leads to a pressure equation

• Odd-even decoupling of pressure can be addressed by either:

‒ staggered velocity grid

‒ non-staggered grid, but Rhie-Chow interpolation for advective velocities

Summary (2)

• Pressure-correction methods are iterative schemes for solving mass and momentum equations simultaneously

• They consist of alternating solutions of:

‒ momentum equation (with pressure fixed)

‒ pressure-correction equation, nudging velocity toward mass conservation

• Widely used pressure-correction methods are:

‒ SIMPLE

‒ PISO