1d colocated simple solution

1

METU Mechanical Engineering Department ME 485 Computational Fluid Dynamics using Finite Volume Method

Spring 2014 (Dr. Sert)

Demonstration of How SIMPLE Algorithm Works on 1D Co-located Meshes

Problem Definition

Simplify the incompressible flow in the following converging nozzle to be 1D and inviscid. Obtain the finite volume

solution using 5 cells of equal length. Density of the fluid is 1 kg/m3. As boundary conditions inlet velocity is given

1 m/s and exit pressure is specified as 0 Pa.

Cross sectional area of the nozzle decreases linearly

𝐴(𝑥) = 0.5 − 0.2𝑥

Exit pressure is set to zero for simplicity. Actually the value of this reference pressure is not important for an

incompressible flow, because in INS only the pressure gradient (𝑑𝑝/𝑑𝑥) is important, not the actual pressure

values. If the actual exit pressure is 𝑝𝑎𝑡𝑚, pressure values that we will obtain by setting exit pressure to zero will

be gage pressures. To obtain absolute values we can add 𝑝𝑎𝑡𝑚 to all the pressure values.

Analytical Solution

Analytical solution of the problem is governed by the Bernoulli equation.

𝑝 +𝜌𝑢2

2= constant

Mass flow rate inside the nozzle is constant and its value is known due to the given inlet speed

�̇� = 𝜌 𝑢𝑖𝑛 𝐴𝑖𝑛 = 0.5 kg s⁄

Using this value we can calculate the speed at any point inside the nozzle. Exit speed is

𝑢𝑒𝑥𝑖𝑡 =�̇�

𝜌𝐴𝑒𝑥𝑖𝑡= 5m s⁄

𝑥

𝐴𝑖𝑛 = 0.5 m2

𝑢𝑖𝑛 = 1 m/s

𝐴𝑒𝑥𝑖𝑡 = 0.1 m2

𝑝𝑒𝑥𝑖𝑡 = 0 Pa

𝐿 = 2 m

𝜌 = 1 kg/m3

2

Using the known speed and pressure at the exit the constant of the Bernoulli equation can be calculated as

𝑝𝑒𝑥𝑖𝑡 +𝜌𝑢𝑒𝑥𝑖𝑡

2

2= 0 +

(1)(5)2

2= 12.5 Pa

With the �̇� = 0.5 equation and the above equation speed and pressure at any point inside the nozzle can be

determined.

Nodes and faces of the 5 cell mesh are shown below

Analytical solution at the faces and nodes is as follows

Face Node 𝑥 [m] 𝐴 [m2] 𝑢𝑒𝑥𝑎𝑐𝑡 [m/s] 𝑝𝑒𝑥𝑎𝑐𝑡 [Pa]

𝑓1 0.0 0.50 1.0000 12.0000

1 0.2 0.46 1.0870 11.9093

𝑓2 0.4 0.42 1.1905 11.7914

2 0.6 0.38 1.3158 11.6343

𝑓3 0.8 0.34 1.4706 11.4187

3 1.0 0.30 1.6667 11.1111

𝑓4 1.2 0.26 1.9231 10.6509

4 1.4 0.22 2.2727 9.9174

𝑓5 1.6 0.18 2.7778 8.6420

5 1.8 0.14 3.5714 6.1224

𝑓6 2.0 0.10 5.0000 0.0000

Discretization of the x-Momentum Equation

Consider the following cell P with W and E neighbors

For cell P, the discretized x-momentum equation without the viscous and source terms is

𝐹𝑒𝐴𝑒𝑢𝑒 − 𝐹𝑤𝐴𝑤𝑢𝑤 = −𝑑𝑝

𝑑𝑥|𝑃∆∀𝑃 (1)

where 𝐹𝑒 = (𝜌𝑢)𝑒 , 𝐹𝑤 = (𝜌𝑢)𝑤 are known, calculated using initial guesses or previous iteration values. 𝑢𝑒 and

𝑢𝑤 of Eqn (1) can be expressed in terms of speeds at nodes using various schemes. Here upwind scheme is used

as follows

𝑓1 𝑓2 𝑓3 𝑓4 𝑓5 𝑓6

1 2 3 4 5

𝑥

∆𝑥 = 0.4 m

𝑤 𝑒

W P E

3

𝑢𝑒 = {

𝑢𝑃 𝑢𝐸 if if 𝐹𝑒 > 0

𝐹𝑒 < 0 → 𝑢𝑒 = 𝑢𝑃

max(𝐹𝑒 , 0)

𝐹𝑒+ 𝑢𝐸

max(−𝐹𝑒 , 0)

−𝐹𝑒

𝑢𝑤 = {

𝑢𝑊 𝑢𝑃 if if 𝐹𝑤 > 0

𝐹𝑤 < 0 → 𝑢𝑤 = 𝑢𝑊

max(𝐹𝑤 , 0)

𝐹𝑤+ 𝑢𝑃

max(−𝐹𝑤, 0)

−𝐹𝑤

Pressure derivative of Eqn (1) is discretized as

𝑑𝑝

𝑑𝑥|𝑃=𝑝𝐸 − 𝑝𝑊2∆𝑥

Volume of cell P is

∆∀𝑃= 𝐴𝑃∆𝑥

where 𝐴𝑃 is the cross sectional area at node P.

Substituting these details into Eqn (1) we get

𝐹𝑒𝐴𝑒 [𝑢𝑃max(𝐹𝑒 , 0)

𝐹𝑒+ 𝑢𝐸

max(−𝐹𝑒, 0)

−𝐹𝑒] − 𝐹𝑤𝐴𝑤 [𝑢𝑊

max(𝐹𝑤, 0)

𝐹𝑤+ 𝑢𝑃

max(−𝐹𝑤, 0)

−𝐹𝑤] = 𝑆𝑃

𝑢 − 𝐴𝑃𝑝𝐸 − 𝑝𝑊

2 (2)

which can be arranged as

𝑎𝑊𝑢 𝑢𝑊 + 𝑎𝑃

𝑢𝑢𝑃 + 𝑎𝐸𝑢𝑢𝐸 = 𝑆𝑃

𝑢 − 𝐴𝑃𝑝𝐸 − 𝑝𝑊

2 (3)

where

𝑎𝑊𝑢 = −𝐴𝑤max (𝐹𝑤 , 0)

𝑎𝐸𝑢 = −𝐴𝑒max(−𝐹𝑒 , 0)

𝑎𝑃𝑢 = 𝐴𝑤max(−𝐹𝑤 , 0) + 𝐴𝑒max(𝐹𝑒 , 0)

𝑆𝑃𝑢 = 0

Eqn (3) can also be written as

𝑢𝑃 = �̂�𝑃 − 𝑑𝑃𝑢𝑝𝐸 − 𝑝𝑊

2 (4)

where

�̂�𝑃 =1

𝑎𝑃𝑢 (𝑆𝑃

𝑢 −∑𝑎𝑛𝑏𝑢 𝑢𝑛𝑏

𝑛𝑏

) and 𝑑𝑃𝑢 =

𝐴𝑃𝑎𝑃𝑢 (5)

Although the source term is zero, it is kept in the equations because for boundary cells there may be nonzero

contributions to it.

4

Modification of the x-Momentum Equation for Boundary Cells

Cell 1:

At the west face inlet velocity is known, i.e. in the x-momentum equation flux at the west face is known

𝐹𝑤𝐴𝑤𝑢𝑤 = 𝜌𝑢𝑖𝑛2 𝐴𝑖𝑛 = known

This can be taken to the right hand side of the equation to act as a source term.

For the pressure derivative one-sided difference can be used instead of central differencing

𝑑𝑝

𝑑𝑥|𝑃=𝑝2 − 𝑝1∆𝑥

With these, Eqn (3) for cell 1 becomes (modified terms are shown in red)


𝑢𝑢𝑃 + 𝑎𝐸𝑢𝑢𝐸 = 𝑆𝑃

𝑢 − 𝐴𝑃(𝑝𝐸 − 𝑝𝑃)

where

𝑎𝑊𝑢 = 0

𝑎𝐸𝑢 = −𝐴𝑒max(−𝐹𝑒 , 0)

𝑎𝑃𝑢 = 𝐴𝑒max(𝐹𝑒 , 0)

𝑆𝑃𝑢 = 𝜌𝑢𝑖𝑛

2 𝐴𝑖𝑛

Therefore at an inlet boundary where velocity is given, following changes occur in the x-momentum equation

Coefficient of the ghost neighbor is set to zero.

𝑎𝑃𝑢 coefficient changes because there is no contribution from the ghost neighbor.

Momentum flux created by the known inlet velocity acts as a source term.

Pressure discretization becomes one-sided (not central).

Cell 5:

Considering that the right end of the domain is an exit boundary we can use

𝑢𝑒 = 𝑢𝑃

Pressure gradient term can be discretized to make use of the given 𝑝𝑒𝑥𝑖𝑡 value.

𝑑𝑝

𝑑𝑥|𝑃=𝑝𝑒 − 𝑝𝑤∆𝑥

=𝑝𝑒𝑥𝑖𝑡 −

𝑝𝑊 + 𝑝𝑃2

∆𝑥=2𝑝𝑒𝑥𝑖𝑡 − 𝑝𝑊 − 𝑝𝑃

2∆𝑥

𝑤 𝑒

1 2

𝑢𝑤 = 𝑢𝑖𝑛

P E

𝑤 𝑒

4 5

𝑝𝑒 = 𝑝𝑒𝑥𝑖𝑡

W P

5

With these, Eqn (3) for cell 1 becomes (modified terms are shown in red)


𝑢𝑢𝑃 + 𝑎𝐸𝑢𝑢𝐸 = −𝐴𝑃

2𝑝𝑒𝑥𝑖𝑡 − 𝑝𝑊 − 𝑝𝑃2∆𝑥

where

𝑎𝑊𝑢 = −𝐴𝑤max(𝐹𝑤 , 0)

𝑎𝐸𝑢 = 0

𝑎𝑃𝑢 = 𝐴𝑤max(−𝐹𝑤 , 0) + 𝐴𝑒max(𝐹𝑒 , 0)

Therefore at an exit boundary where pressure is given, following changes occur in the x-momentum equation

Coefficient of the ghost neighbor is set to zero.

Pressure discretization uses the known exit pressure.

Face Velocity Calculation using Rhie-Chow (Momentum) Interpolation and Relaxation

In total, there are 6 faces in the mesh. Consider the following face 𝑓 with neighboring cells L (left) and R (right).

Using Rhie-Chow interpolation and velocity under-relaxation, velocity at face 𝑓 is calculated as

𝑢𝑓 = 𝛼𝑢[�̂�𝑓 − 𝑑𝑓𝑢(𝑝𝑅 − 𝑝𝐿)] + (1 − 𝛼𝑢)𝑢𝑓

𝑂𝐿𝐷 (6)

where

�̂�𝑓 =�̂�𝐿 + �̂�𝑅

2 , 𝑑𝑓

𝑢 =𝑑𝐿𝑢 + 𝑑𝑅

𝑢

2

𝛼𝑢 is the velocity under-relaxation factor and 𝑢𝑓𝑂𝐿𝐷 is the face velocity of the previous iteration.

Eqn (6) and the above expressions for �̂�𝑓 and 𝑑𝑓𝑢 need to be modified at the boundary faces.

Modification at face 1:

Face 1 at the left boundary has a specified inlet velocity, so we do not use Eqn (6) at face 1, i.e. we do not need

�̂�𝑓1 or 𝑑𝑓1𝑢 .

𝑓

L R

𝑓1

𝑢𝑓1 = 𝑢𝑖𝑛

6

Modification at face 6:

There is no cell on the right of face 6. Instead of central interpolation, one-sided interpolation can be used to

calculate �̂�𝑓6 and 𝑑𝑓6𝑢 . With the assumption of constant cell size

�̂�𝑓6 = �̂�5 +�̂�5 − �̂�42

, 𝑑𝑓6𝑢 = 𝑑5

𝑢 +𝑑5𝑢 − 𝑑4

𝑢

2

Also (𝑝𝑅 − 𝑝𝐿) term of Eqn (6) can be expressed in terms of the known exit pressure as

(𝑝𝑒𝑥𝑖𝑡 − 𝑝5)

1/2

which comes from 𝑑𝑝

𝑑𝑥|𝑓6≈

𝑝𝑒𝑥𝑖𝑡−𝑝5

∆𝑥/2

Pressure Correction (PC) Equation

PC equation is

𝐹𝑒′𝐴𝑒 − 𝐹𝑤

′𝐴𝑤 = −𝐴𝑒𝐹𝑒∗ + 𝐴𝑤𝐹𝑤

∗

Relating velocity corrections to pressure corrections as follows

𝐹𝑒′ = 𝜌𝑢𝑒

′ = −𝜌𝑑𝑒𝑢(𝑝𝐸

′ − 𝑝𝑃′ ) and 𝐹𝑤

′ = 𝜌𝑢𝑤′ = −𝜌𝑑𝑤

𝑢 (𝑝𝑃′ − 𝑝𝑊

′ )

PC equation becomes

𝑎𝑊𝑃𝐶𝑝𝑊

′ + 𝑎𝑃𝑃𝐶𝑝𝑃

′ + 𝑎𝐸𝑃𝐶𝑝𝐸

′ = −𝐴𝑒𝐹𝑒∗ + 𝐴𝑤𝐹𝑤

∗ (7)

𝑎𝑊𝑃𝐶 = −𝜌𝑑𝑤𝐴𝑤

𝑎𝑃𝑃𝐶 = 𝜌𝑑𝑤𝐴𝑤 + 𝜌𝑑𝑒𝐴𝑒

𝑎𝐸𝑃𝐶 = −𝜌𝑑𝑒𝐴𝑒

Eqn (7) is modified as follows for boundary cells.

Modification for cell 1:

At west face inlet velocity is specified and 𝑢𝑤′ = 0. Due to this following changes happen

𝑎𝑊𝑃𝐶 = 0 , 𝑎𝑃

𝑃𝐶 = 𝜌𝑑𝑒𝐴𝑒

𝑓6


4 5

𝑤 𝑒

1 2

𝑢𝑤 = 𝑢𝑖𝑛

P E

7

Modification for cell 5:

𝑢𝑒′ expression need to be modified because there is no 𝑝𝐸

′ . Instead of using (𝑝𝐸′ − 𝑝𝑃

′ ) we can use half cell as

𝑢𝑒′ =

𝑑𝑒𝑢(𝑝𝑒𝑥𝑖𝑡

′ − 𝑝𝑃′ )

1/2

where 𝑝𝑒𝑥𝑖𝑡′ = 0 because exit pressure is fixed. Due to this coefficients of the PC change as follows

𝑎𝐸𝑃𝐶 = 0 , 𝑎𝑃

𝑃𝐶 = 2𝜌𝑑𝑒𝐴𝑒 + 𝜌𝑑𝑤𝐴𝑤

Face Velocity Corrections

For the above face 𝑓, velocity correction is done as follows

𝑢𝑓 = 𝑢𝑓∗ − 𝑑𝑓

𝑢(𝑝𝑅′ − 𝑝𝐿

′ ) (8)

where 𝑢𝑓∗ is the velocity calculated previously using Rhie-Chow interpolation. For the boundary faces Eqn (8) is

used as follows

Modification for face 1: Inlet velocity is given and no correction is done.

Modification for face 6: Similar to the previous step we use

𝑢𝑓6 = 𝑢𝑓6∗ − 𝑑𝑓6

𝑢(𝑝𝑒𝑥𝑖𝑡′ − 𝑝5

′ )

1/2 where 𝑝𝑒𝑥𝑖𝑡

′ = 0

Correct Cell Center Velocities

𝑢𝑃 = 𝑢𝑃∗ − 𝑑𝑃

𝑢(𝑝𝐸′ − 𝑝𝑊

′ )

2 (9)

Eqn (9) will be modified for the boundary cells.

𝑤 𝑒

4 5


W P

𝑓

L R

𝑤 𝑒

W P E

8

Modification for cell 1: West cell does not exist. Pressure correction difference can be done in a one-sided way.

𝑢1 = 𝑢1∗ − 𝑑1

𝑢(𝑝2′ − 𝑝1

′ )

1

Modification for cell 5: East cell does not exist. Pressure correction difference can be done in a one-sided way

using the fact that exit pressure is fixed.

𝑢5 = 𝑢5∗ − 𝑑5

𝑢(𝑝𝑒𝑥𝑖𝑡′ − 𝑝5

′ )

1/2 where 𝑝𝑒𝑥𝑖𝑡

′ = 0

Correct Pressures

𝑝𝑃 = 𝑝𝑃∗ + 𝛼𝑃𝑝1

′ (10)

where 𝑝𝑃∗ is the pressure of the previous iteration (or the initial guess) and 𝛼𝑃 is the pressure relaxation factor.

SIMPLE Iterations

STEP 1:

As initial guess we can use the inlet velocity and exit pressure at all nodes and faces.

𝑢1 = 𝑢2 = 𝑢3 = 𝑢4 = 𝑢5 = 1.0

𝑢𝑓1 = 𝑢𝑓2 = 𝑢𝑓3 = 𝑢𝑓4 = 𝑢𝑓5 = 𝑢𝑓6 = 1.0

𝑝1 = 𝑝2 = 𝑝3 = 𝑝4 = 𝑝5 = 0.0

ITERATION 1:

STEP 2: Setup x-momentum equation set to solve for 𝑢∗.

Cell 1:

Knowns: 𝐹𝑒 = (𝜌𝑢)𝑒 = 1.0 , 𝐹𝑤 = (𝜌𝑢)𝑤 = 1.0 , 𝐴𝑤 = 0.5 , 𝐴𝑒 = 0.42 , 𝑝𝑃 = 𝑝𝐸 = 0.0

𝑎𝑊𝑢 = 0.0 , 𝑎𝑃

𝑢 = 0.42 , 𝑎𝐸𝑢 = 0.0 , 𝑆𝑃

𝑢 = 0.5

�̂�1 =0.5 − 0.0

0.42= 1.1905 , 𝑑1

𝑢 =0.46

0.42= 1.0952

Cell 1 eqn is : 0.42𝑢1∗ = 0.5 − 0.46(0.0 − 0.0)

9

Cell 2:

Knowns: 𝐹𝑒 = (𝜌𝑢)𝑒 = 1.0 , 𝐹𝑤 = (𝜌𝑢)𝑤 = 1.0 , 𝐴𝑤 = 0.42 , 𝐴𝑒 = 0.34 , 𝑝𝑊 = 𝑝𝐸 = 0.0

𝑎𝑊𝑢 = −0.42 , 𝑎𝑃

𝑢 = 0.34 , 𝑎𝐸𝑢 = 0.0 , 𝑆𝑃

𝑢 = 0.0

�̂�2 =0.0 − ((−0.42)(1.0))

0.34= 1.1905 , 𝑑2

𝑢 =0.38

0.34= 1.0952

Cell 2 eqn is : −0.42𝑢1∗ + 0.34𝑢2

∗ = −0.38(0.0 − 0.0)/2

Cell 3:


𝑎𝑊𝑢 = −0.34 , 𝑎𝑃

𝑢 = 0.26 , 𝑎𝐸𝑢 = 0.0 , 𝑆𝑃

𝑢 = 0.0

�̂�3 =0.0 − ((−0.34)(1.0))

0.26= 1.3077 , 𝑑3

𝑢 =0.30

0.26= 1.1538

Cell 3 eqn is : −0.34𝑢2∗ + 0.26𝑢3

∗ = −0.30(0.0 − 0.0)/2

Cell 4:


𝑎𝑊𝑢 = −0.26 , 𝑎𝑃

𝑢 = 0.18 , 𝑎𝐸𝑢 = 0.0 , 𝑆𝑃

𝑢 = 0.0

�̂�4 =0.0 − ((−0.26)(1.0))

0.18= 1.4444 , 𝑑4

𝑢 =0.22

0.18= 1.2222

Cell 4 eqn is : −0.26𝑢3∗ + 0.18𝑢4

∗ = −0.22(0.0 − 0.0)/2

Cell 5:

Knowns: 𝐹𝑒 = (𝜌𝑢)𝑒 = 1.0 , 𝐹𝑤 = (𝜌𝑢)𝑤 = 1.0 , 𝐴𝑤 = 0.18 , 𝐴𝑒 = 0.1 , 𝑝𝑊 = 𝑝𝑃 = 𝑝𝑒𝑥𝑖𝑡 = 0.0

𝑎𝑊𝑢 = −0.18 , 𝑎𝑃

𝑢 = 0.1 , 𝑎𝐸𝑢 = 0.0 , 𝑆𝑃

𝑢 = 0.0

�̂�5 =0.0 − ((−0.18)(1.0))

0.1= 1.8000 , 𝑑5

𝑢 =0.14

0.1= 1.4000

Cell 5 eqn is : −0.18𝑢4∗ + 0.1𝑢5

∗ = −0.14(2(0.0) − 0.0 − 0.0)/2

Discretized x-momentum equation system and the solution for 𝑢∗ is

[ 0.42 0 0 0 0−0.42 0.34 0 0 00 −0.34 0.26 0 00 0 −0.26 0.18 00 0 0 −0.18 0.1]

{

𝑢1∗

𝑢2∗

𝑢3∗

𝑢4∗

𝑢5∗}

=

{

0.50000 }

→

{

𝑢1∗

𝑢2∗

𝑢3∗

𝑢4∗

𝑢5∗}

=

{

1.19051.47061.92312.77780.5000}

10

STEP 3: Calculate face velocities using Rhie-Chow interpolation and 𝛼𝑢 = 0.6. These velocities will be used as

“star” velocities in Step 5.

Face 1: 𝑢𝑓1 = 1.0

Face 2: �̂�𝑓2 =�̂�1+�̂�2

2= 1.2129 , 𝑑𝑓2

𝑢 =𝑑1𝑢+𝑑2

𝑢

2= 1.1064 →

𝑢𝑓2 = (0.6)[1.2129 − 1.1064(0.0 − 0.0)] + (1 − 0.6)(1.0) = 1.1277

Face 3: �̂�𝑓3 =�̂�2+�̂�3

2= 1.2715 , 𝑑𝑓3


𝑢

2= 1.1357 →

𝑢𝑓3 = (0.6)[1.2715 − 1.1357(0.0 − 0.0)] + (1 − 0.6)(1.0) = 1.1629

Face 4: �̂�𝑓4 =�̂�3+�̂�4

2= 1.3761 , 𝑑𝑓4


𝑢

2= 1.1880 →

𝑢𝑓4 = (0.6)[1.3761 − 1.1880(0.0 − 0.0)] + (1 − 0.6)(1.0) = 1.2256

Face 5: �̂�𝑓5 =�̂�4+�̂�5

2= 1.6222 , 𝑑𝑓5


𝑢

2= 1.3111 →

𝑢𝑓5 = (0.6)[1.6222 − 1.3111(0.0 − 0.0)] + (1 − 0.6)(1.0) = 1.3733

Face 6: �̂�𝑓6 = �̂�5 +�̂�5−�̂�4

2= 1.9778 , 𝑑𝑓6

𝑢 = 𝑑5𝑢 +

𝑑5𝑢−𝑑4

𝑢

2= 1.4889 →

𝑢𝑓5 = (0.6)[1.9778 − 1.4889(0.0 − 0.0)] + (1 − 0.6)(1.0) = 1.5867

STEP 4: Calculate the coefficients of the pressure correction equation system solve for 𝑝′ values.

Cell 1: 𝑎𝑊𝑃𝐶 = 0.0

𝑎𝐸𝑃𝐶 = −𝜌𝑑𝑒

𝑢𝐴𝑒 = −(1)(1.1064)(0.42) = −0.4647

𝑎𝑃𝑃𝐶 = 𝜌𝑑𝑒

𝑢𝐴𝑒 = 0.4647

RHS value: −𝐴𝑒𝐹𝑒∗ + 𝐴𝑤𝐹𝑤

∗ = −(0.42)(1.1277) + (0.5)(1.0) = 0.0264

Cell 2: 𝑎𝑊𝑃𝐶 = −𝜌𝑑𝑤

𝑢𝐴𝑤 = −(1)(1.1064)(0.42) = −0.4647


𝑢𝐴𝑒 = −(1)(1.1357)(0.34) = −0.3862

𝑎𝑃𝑃𝐶 = 𝜌𝑑𝑤

𝑢𝐴𝑤 + 𝜌𝑑𝑒𝑢𝐴𝑒 = 0.8509


∗ = −(0.34)(1.1629) + (0.42)(1.1277) = 0.0783


𝑢𝐴𝑤 = −(1)(1.1357)(0.34) = −0.3862


𝑢𝐴𝑒 = −(1)(1.1880)(0.26) = −0.3089




∗ = −(0.26)(1.2256) + (0.34)(1.1629) = 0.0767


𝑢𝐴𝑤 = −(1)(1.1880)(0.26) = −0.3089


𝑢𝐴𝑒 = −(1)(1.3111)(0.18) = −0.2360




∗ = −(0.18)(1.3733) + (0.26)(1.2256) = 0.0715

11


𝑢𝐴𝑤 = −(1)(1.3111)(0.18) = −0.2360

𝑎𝐸𝑃𝐶 = 0.0


𝑢𝐴𝑤 + 2𝜌𝑑𝑒𝑢𝐴𝑒 = 0.5338


∗ = −(0.1)(1.5867) + (0.18)(1.3733) = 0.0885

Discretized PC equation system and the solution for 𝑝′ is

[ 0.4647 −0.4647 0 0 0−0.0467 0.8509 −0.3862 0 0

0 −0.3862 0.6950 −0.3089 00 0 −0.3089 0.5449 −0.23600 0 0 −0.2360 0.5338 ]

{

𝑝1′

𝑝2′

𝑝3′

𝑝4′

𝑝5′}

=

{

0.02640.07830.07670.07150.0885}

→

{

𝑝1′

𝑝2′

𝑝3′

𝑝4′

𝑝5′}

=

{

3.13213.07542.80452.21751.1463}

STEP 5: Correct face velocities using Eqn (8).

Face 1: 𝑢𝑓1 = 1.0 (No correction for the inlet velocity)

Face 2: 𝑢𝑓2 = 𝑢𝑓2∗ + 𝑑𝑓2

𝑢 (𝑝2′ − 𝑝1

′ ) = 1.1277 − 1.1064(3.0754 − 3.1321) = 1.1905


𝑢 (𝑝3′ − 𝑝2

′ ) = 1.1629 − 1.1357(2.8045 − 3.0754) = 1.4706


𝑢 (𝑝4′ − 𝑝3

′ ) = 1.2256 − 1.1880(2.2175 − 2.8045) = 1.9231


𝑢 (𝑝5′ − 𝑝4

′ ) = 1.3733 − 1.3111(1.1463 − 2.2175) = 2.7778


𝑢 (𝑝𝑒𝑥𝑖𝑡′ −𝑝5

′)

0.5= 1.5867 − 1.4889

(0.0−1.1463)

0.5= 5.0000

Important note: As seen, corrected face velocities satisfy the continuity equation exactly, i.e. mass is

conserved exactly in each cell with the corrected face velocities. For example consider cell 3.

Mass balance for cell 3: 𝜌𝑢𝑤𝐴𝑤 − 𝜌𝑢𝑒𝐴𝑒 = (1)(0.26)(1.9231) − (1)(0.34)(1.4706) = 0.0000

This is true for all cells. For this 1D problem with specified inlet velocity, these corrected face velocities are

the same as the analytical values. For a 2D or 3D problem mass balance in each cell will again be exact, but

the face velocities cannot be equal to the analytical values, which are probably not known anyway. This

“exact mass balance” at each iteration is an important power of the SIMPLE algorithm. Although in this

problem face velocities are exact, cell center velocities and pressures will require a number of iterations to

converge and the converged values will not be equal to the analytical values. Accuracy will depend on the

used mesh and the selected convergence tolerance.

5 plus/minus typos. Results are correct.

12

STEP 6: Correct cell center velocities using Eqn (9).

Cell 1: 𝑢1 = 𝑢1∗ + 𝑑1

𝑢 (𝑝2′ − 𝑝1

′ ) (1)⁄ = 1.1905 + 1.0952(3.0754 − 3.1321) = 1.2526

Cell 2: 𝑢2 = 𝑢2∗ + 𝑑2

𝑢 (𝑝3′ − 𝑝1

′ ) (2)⁄ = 1.4706 + 1.1176(2.8045 − 3.0754) = 1.6537

Cell 3: 𝑢3 = 𝑢3∗ + 𝑑3

𝑢 (𝑝4′ − 𝑝2

′ ) (2)⁄ = 1.9231 + 1.1538(2.2175 − 2.8045) = 2.4181

Cell 4: 𝑢4 = 𝑢4∗ + 𝑑4

𝑢 (𝑝5′ − 𝑝3

′ ) (2)⁄ = 2.7778 + 1.2222(1.1463 − 2.2175) = 3.7911

Cell 5: 𝑢5 = 𝑢5∗ + 𝑑5

𝑢 (𝑝𝑒𝑥𝑖𝑡′ − 𝑝5

′ ) (1)⁄ = 5.0000 + 1.2222(1.1463 − 2.2175) = 8.2096

STEP 7: Correct pressures using Eqn (10). Use a pressure relaxation value of 𝛼𝑝 = 0.4.

Cell 1: 𝑝1 = 𝑝1∗ + 𝛼𝑝𝑝1

′ = 0.0 + 0.6 ∗ 3.1321 = 1.2529

Cell 2: 𝑝2 = 𝑝2∗ + 𝛼𝑝𝑝2

′ = 0.0 + 0.6 ∗ 3.0754 = 1.2302

Cell 3: 𝑝3 = 𝑝3∗ + 𝛼𝑝𝑝3

′ = 0.0 + 0.6 ∗ 2.8045 = 1.1218

Cell 4: 𝑝4 = 𝑝4∗ + 𝛼𝑝𝑝4

′ = 0.0 + 0.6 ∗ 2.2175 = 0.8870

Cell 5: 𝑝5 = 𝑝5∗ + 𝛼𝑝𝑝5

′ = 0.0 + 0.6 ∗ 1.1463 = 0.4585

STEP 8: Check for convergence. There are many possibilities here. One simple way is to compare velocity and

pressure differences of two consecutive iterations. Convergence is declared if the maximum of such

differences is less than a certain user specified tolerance value. It is preferred to use not just differences but

normalize them in a logical way, e.g. normalize the velocity differences with respect to the given inlet velocity.

It is always a good idea to select couple of critical monitoring points in the flow field and watch how variables

change at those points before declaring convergence.

If the convergence check fails go to Step 2 and perform one more iteration. Face and cell center velocities and

cell center pressures calculated in this iteration will be used in the next iteration.

The solution may also divergence, depending on how far the initial guess is from the exact solution, mesh

density and relaxation factors.

ITERATION 2:

You can use the MATLAB code NS_1D_Colocated.m available at the course web site to perform a full solution.

You can try different meshes, initial guesses, boundary condition implementations, relaxation factors, etc.

Results of the second iteration are given below for you to check the correctness of your hand calculations.

STEP 2:

𝑑1𝑢 = 0.9200 �̂�1 = 1.0000 𝑢1

∗ = 1.0209

𝑑2𝑢 = 0.7600 �̂�2 = 1.2526 𝑢2

∗ = 1.0707

𝑑3𝑢 = 0.6000 �̂�3 = 1.6537 𝑢3

∗ = 1.1736

𝑑4𝑢 = 0.4400 �̂�4 = 2.4181 𝑢4

∗ = 1.3195

𝑑5𝑢 = 0.2800 �̂�5 = 3.7911 𝑢5

∗ = 1.5079

13

STEP 3:

𝑑𝑓1𝑢 = 1.0000 �̂�𝑓1 = 0.8737 𝑢𝑓1

∗ = 1.0000

𝑑𝑓2𝑢 = 0.8400 �̂�𝑓2 = 1.1263 𝑢𝑓2

∗ = 1.1634

𝑑𝑓3𝑢 = 0.6800 �̂�𝑓3 = 1.5043 𝑢𝑓3

∗ = 1.5043

𝑑𝑓4𝑢 = 0.5200 �̂�𝑓4 = 2.0640 𝑢𝑓4

∗ = 2.0640

𝑑𝑓5𝑢 = 0.3600 �̂�𝑓5 = 3.0664 𝑢𝑓5

∗ = 3.0664

𝑑𝑓6𝑢 = 0.2000 �̂�𝑓6 = 4.7967 𝑢𝑓6

∗ = 4.7957

STEP 4: STEP 5: STEP 6: STEP 7:

𝑝1′ = −0.5818 𝑢𝑓1 = 1.0000 𝑢1 = 1.0505 𝑝1 = 1.0201

𝑝2′ = −0.6141 𝑢𝑓2 = 1.1905 𝑢2 = 1.0641 𝑝2 = 0.9845

𝑝3′ = −0.5645 𝑢𝑓3 = 1.4706 𝑢3 = 1.0774 𝑝3 = 0.8960

𝑝4′ = −0.2934 𝑢𝑓4 = 1.9231 𝑢4 = 1.0835 𝑝4 = 0.7996

𝑝5′ = −0.5084 𝑢𝑓5 = 2.7778 𝑢5 = 1.7926 𝑝5 = 0.6619

𝑢𝑓6 = 5.0000

Converged Solution with 5 cells

14

Converged Solution with 100 cells

1d colocated simple solution

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