Flow over an Obstruction MECH 523 Applied Computational Fluid Dynamics Presented by Srinivasan C...
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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