Flow over an ObstructionFlow over an Obstruction

MECH 523 Applied Computational Fluid

Dynamics

Presented bySrinivasan C Rasipuram

ApplicationsApplications

Chip coolingHeat sinksUse of fire extinguishers at

obstructionsThough there are no significant

applications for flow over obstructions, this model is the bench work for researchers to compare their work and findings.

Case 1Case 1

10 cm

u

T

4

cm

Tw2

cm

0.5 cm

Case 2Case 2

Tw

0.5 cm 10 cm

u

T

4

cm

0.5 cm

0.5 cm

0.5 cm

0.5 cm

Navier-Stokes Navier-Stokes EquationsEquations

Continuity

No mass source has been assumed.

0

x u

i

i

Momentum

is the molecular viscosity of the fluid.

1,2 kj, i,: x u

32 -

x

u

x u

tensor Stress the is where

x

x p

- u u x

ji

k

k

i

j

j

iji

ji

j

ji

i

ji

j

Energy

Turbulent thermal conductivity keff = k + kt

Sh – Volumetric heat source

Brinkman Number

where Ue is the velocity of undisturbed free stream

Viscous heating will be important when Br approaches or exceeds unity.

hi

effi

ii

S x T

k x

p e u x

T kU

Br 2

e

Typically, Br ≥ 1 for compressible flows.

But viscous heating has been neglected in the simulations as Segregated solver assumes negligible viscous dissipation as its default setting.

Viscous dissipation – thermal energy created by viscous shear in the flow.

In solid regions, Energy equation is

sourceheat Volumetric - q

re Temperatu- T

tyconductivi - k

dT C henthalpy sensible

material the ofdensity -

q x T

k x

h u x

T

Tp

conduction to due FluxHeat

iii

i

K 298.15

ref

StandardStandard k-є k-є Turbulence Turbulence ModelModel

k - Turbulent Kinetic energyє - rate of dissipation of turbulent

kinetic energy

kk and and є є equationsequationsk and є are obtained from the

following transport equations:

kε ρ C - G C G

kε

C ix ε

εσtμ

μ ix

t Dε D

ρ

Y - ε ρ - G G ix

k

kσtμ

μ ix

t Dk D

ρ

2

2εb3εk1ε

Mbk

where Gk represents the generation of turbulent

kinetic energy due to mean velocity gradients Gb is the generation of turbulent kinetic energy

due to buoyancy YM represents the contribution of the

fluctuating dilatation in compressible turbulence to the overall dissipation rate

C1є, C2є, C3є are constantsk and є are the turbulent Prandtl numbers for

k and є respectively

Eddy or Turbulent viscosity

The model constants C1 = 1.44, C2 = 1.92, C = 0.09,

k = 1.0, = 1.3

(Typical experimental values for these constants)

constant.a isC where

k C

2

t

Turbulence IntensityTurbulence Intensity

Turbulence Intensity

avgu ity,flow veloc meannsfluctuatio velocity the of rms

I

DiscretizationDiscretization

0

0

100000

0

0

0

P ,

with termsy

k with terms x y T

k - vp e

- v

- p u v

v

G

with terms

k with termsx T

k -u p e

- u v

- p u

u

F

P yG

x F

eff

yy2

xy

eff

yx

xx2

Discretization …Discretization …continuedcontinued

2

S S

2

S S V

P S.F S.F S.F S.FV1

21j2

1j21i2

1iij

21j1-ji,

21jji,

21ij1,-i

21iji,

ij

Ideal gas model for Density Ideal gas model for Density calculations and Sutherland calculations and Sutherland

model for Viscosity calculationsmodel for Viscosity calculations Density is calculated based on the Ideal gas equation.

Viscosity calculations

C1 and C2 are constants for a given gas. For air at moderate temperatures (about 300 – 500 K),

C1 = 1.458 x 10-6 kg/(m s K0.5)

C2 = 110.4 K

2

23

1 C TT

C

μ

R T p

Reynolds Number Reynolds Number calculationcalculation

For flow over an obstruction,

is the density of the fluidV is the average velocity (inlet velocity

for internal flows)D is the hydraulic diameter is the Dynamic viscosity of the fluid

μDV ρ

Re

Re for V = 0.5 m/secRe for V = 0.5 m/sec

For this problem, V = 0.5 m/sec, air = 1.225 kg/m3, air = 1.7894 e–5

kg/m-sec

Laminar 2300 1198 Re

cases) these(for cm .53 D

Solver and Boundary Solver and Boundary conditionsconditions

Solver – SegregatedSolver – Segregated

Inlet Boundary– Velocity at inlet

0.5 m/sec.

– Temperature at inlet 300 K

– Turbulence intensity 10%

– Hydraulic diameter

3.5 cm

Outlet boundary– Gage Pressure at

outlet 0 Pa– Backflow total

temperature – 300 K– Turbulence intensity 10%– Hydraulic diameter 3.5 cm

Wall boundary conditionsWall boundary conditionsHeat sourcesHeat sources

No heat flux at top and bottom walls

Stationary top and bottom walls

Volumetric heat source for the (solid) obstruction – 100,000 W/m3

Under relaxation factorsUnder relaxation factors

Pressure 0.3

Momentum

0.7

Energy 1

k 0.8

0.8

Viscosity 1

Density 1

Body forces

1

Convergence criteriaConvergence criteria

Continuity 0.001

x – velocity 0.001

y – velocity 0.001

Energy 1e-6

k 0.001

0.001

Case 1 – GridsCase 1 – Grids

Number of nodes - 4200

Number of nodes - 162938

Number of nodes - 208372

Case 1 – Velocity Case 1 – Velocity contourscontours

Case 1 – Temperature contoursCase 1 – Temperature contours

Case 1 - Velocity VectorsCase 1 - Velocity Vectors

Case 1 –Contours of Stream Case 1 –Contours of Stream functionfunction

Case 1 – Plot of Velocity Vs X-Case 1 – Plot of Velocity Vs X-locationlocation

Case 1 – Plot of Temperature Vs X-Case 1 – Plot of Temperature Vs X-locationlocation

Case 1 – Plot of Surface Nusselt number Vs X-Case 1 – Plot of Surface Nusselt number Vs X-locationlocation

Case 2 - GridsCase 2 - Grids

4220 nodes

42515 nodes

79984 nodes

Case 2 – Contours of Case 2 – Contours of VelocityVelocity

Case 2 – Contours of Case 2 – Contours of TemperatureTemperature

Case 2 – Contours of Stream Case 2 – Contours of Stream functionfunction

Case 2 - Velocity vectorsCase 2 - Velocity vectors

Case 2 – Plot of Velocity Vs X -Case 2 – Plot of Velocity Vs X -locationlocation

Case 2 – Plot of temperature Vs X-Case 2 – Plot of temperature Vs X-locationlocation

Case 2 – Plot of Surface Nusselt number Case 2 – Plot of Surface Nusselt number Vs x-locationVs x-location