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Output - Derived Variables • Derived Variables are quantities evaluated from the primitive (or solved) variables by PHOENICS. • It means, PHOENICS first solve U1, V1, W1, P1, TEM1 etc. After achieving convergence it evaluates the Derived Variables. • Usually they are: Heat Transfer Coefficients Friction Factors Shear Stresses at walls Wall Distances, etc

Output - Derived Variables Derived Variables are quantities evaluated from the primitive (or solved) variables by PHOENICS. It means, PHOENICS first solve

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Output - Derived Variables

• Derived Variables are quantities evaluated from the primitive (or solved) variables by PHOENICS.• It means, PHOENICS first solve U1, V1, W1, P1, TEM1 etc. After achieving convergence it evaluates the Derived Variables.• Usually they are:

Heat Transfer CoefficientsFriction Factors Shear Stresses at walls Wall Distances, etc

• Placing the Skin friction coefficient, Stanton Number, Shear stress (actually friction velocity squared, equivalent to shear stress divided by density), Yplus (non-dimensional distance to the wall) and heat transfer coefficient (in W/m2/K) into 3-D storage allows them to be plotted in the Viewer or PHOTON, as well as appearing in the RESULT file.

2

w

+

wf 21

2 ref

L

p ref

''w ref

Wall Shear Stress u *

Yplus=distance from the wall y yu *

Skin Friction Coef CU

Nu hStanton Number St

Re Pr C U

Local Heat Transf COeff q h T T kdT / dn

•The major difficult to establish these variables in a general purpose code lays on the determination of a reference velocity and temperature.

Output - Derived Variables – Usual Definitions

Output - Derived VariablesPHOENICS DEFINITIONS

• The phoenics uses as a reference velocity the wall velocity, i.e. , the velocity evaluated at the volume adjacent to the wall, here defined as Uw

w2W

2 wW

w

w 1

w

p w w p w

Skin Friction Coef = SKIN=U

Stress = STRS = SKIN U =

q kHeat Transf COeff = HTCO =

T T

q HTCOStanton Number = STAN=

C U T C U

wall

Output - Derived Variables

• Note that the friction and heat-transfer coefficients are only calculated for turbulent flow. To make them appear for laminar cases, the turbulent viscosity should be set to a very small value – say 1.0E-10.• •The Stanton Number must be stored for the heat-transfer coefficients to be calculated.

Output - Derived Variables

• The friction force components SHRX, SHRY and SHRZ are used in the force-integration routines to add the friction force to the pressure force acting on each object. If they are not stored, the integrated force will only contain the pressure component.

•The Total or Stagnation Pressure (PTOT) is only calculated if the Mach Number is stored. If the Reference Pressure (Main Menu – Properties) is set to zero, the total pressure may go below zero, leading to an error-stop.

Workshop #1 Friction factor and heat transfer coefficients for a laminar flow inside a circular section pipe.

• Pipe radius = 0.005 m, Pipe length = 1 m, = 1rad.• Inlet velocity and temperature 1.0m/s and 100oC (Re = 1000)• Wall velocity and temperature = 0 m/s and 0oC• Outlet at atmospheric pressure

• Grid size: ny = 20 & nz = 60 (Z power 2.0)• Model : activate energy (static temp)

turbulent with constant T of 1E-10

• Set Properties Manually to: RHO = 1; ENUL = 10-5; CP = 1000 & k = 0.02 ReD = 1000 e Pr = 0.5 (dH < dT)

Workshop #1• Numerics: ITER = 200;

• automatic relax OFF, • use: P1 = 1; V1=0.01; W1=0.01 & TEM1 = 1E+09

• OUTPUT – activate storage for: SKIN, SHEAR, HTCO, STAN

• If every thing went wrong: Upload Q1

Check W profiles

• Parabolic velocity profile, max velocity is twice the mean velocity! (IZ = 58)

• Axial centerline velocity achieves 2.0 m/s indicating the fully developed flow.

Check T profiles

• Axial temperature distribution at the pipe centerline (Y=1) and at the pipe wall (Y=20)

Skin friction

• Shear stress evaluation at the fully developed region:

• Please, compare to the analytical solution, Cf = 16/ReD

2 3W WSKIN W 8 10

HTCO & Heat Transferred

• Temperature diff. at the wall: Tw – T1

• Tw = 0 & T1 = TEM1TEM

1

w

w 1

q kHTCO = 160

T T

TEM1

(1)

(2)(3)

(4)(5)

TOTAL HEAT TRANSFER [W]

Y 20 W

1

0

q HTCO T T

dQ q dA q R dz

dQ 160 T 1 0.005 dz

Q 0.80 T dz

•The area under the curve is approximated by three triangles and rectangles. Summing the areas one will find Q ~ 1,12 W.• Compare this result with the phoenics evaluation of the total heat transfer for object plate (1.19 W).

Forces & Momentum in Objetcs

• One of the output options is to calculate the forces and moments on blockage objects, and print them to the result file.

• To activate this feature, open the Main Menu, then click on 'Output'. Turn 'Output of forces and moments' to ON.

• By default, only the pressure forces are integrated over the surface of each body. To add in the friction components, we must store the Friction Forces. Click on 'Derived variables' and turn ON the 3D storage of the X, Y and Z friction force components.

Test case: Drag on a 2D cylinder with square cross section

Available q1

i. Fluid: air (0)

ii. Flow model: Laminar & Steady State

i. Relax: insert the reference vel and length

ii. Z length = 0.1 m

Activating the object force evaluator functiongo to OUTPUT box

• The force on the selected objects are assessed from the RESULT.

• Else one can go to VR Viewer, select an object, say B3, right click and select 'Show results'. The following dialog will appear:

• Unfortunately in my 2010 64 bits this function did not work. The forces were assessed through the result file only.

• The results are not accurate, the grid is very poor.

Streamlines for a Re = 6 regime; steady state flow

Transient Case• Case: vortex shedding of a cylinder in cross flow. • The upstream steady flow becomes a periodic flow downstream due

to the flow instabilities developed at the cylinder boundary layer.• The vortices detach from the cylinder surface periodically inducing

longitudinal and transversal forces. • This case uses the storage of surface forces for transient case.

Transient Case: problem set up• Properties: = 1 kg/m3, = 0.006 m2/s• Grid: Time = 25s, time steps 1600, LX = 10m, LY = 6m, LZ = 1m,

Cylinder = 0.6m, NX = 67, NY = 50, NZ = 1• Numerics: Iter = 20, Relax (manual) P1 = 1.0, U1 = V1 = 5,

difference scheme QUICK.

U = 3 m/sAvailable q1 for download

Note about q1: a. it uses a coarse grid, better results

come with 165x100 volumes grid.b. it is included inform lines to

evaluate the flow vorticity.

Transient Case: force history

• Look at d_priv1 phoenics folder for ‘forces-t.csv’:

Reference: cylinder drag coefficient

for Re = 300, Cd 1,3

Cylinder: drag and liftExtracts from ‘forces-t.csv’

• Fx force in N parallel to the main flow direction

• Fy force in N transversal to the main flow direction

• Fx is a drag force and has a frequency half of the Fy

• Fy is periodic, the net Fy is null.

• CD and CL are the drag and lift coefficients

• CD is close to the experimental data.

• Only the absolute value of CL is shown.