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

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Applications Chip cooling Heat sinks Use of fire extinguishers at obstructions Though there are no significant applications for flow over obstructions, this model is the bench work for researchers to compare their work and findings.

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Case 1 10 cm uTuT 4 cm TwTw 2 cm 0.5 cm

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Case 2 TwTw 0.5 cm 10 cm uTuT 4 cm 0.5 cm

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Navier-Stokes Equations Continuity No mass source has been assumed.

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Momentum is the molecular viscosity of the fluid.

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Energy Turbulent thermal conductivity k eff = k + k t S h – Volumetric heat source Brinkman Number where U e is the velocity of undisturbed free stream Viscous heating will be important when Br approaches or exceeds unity.

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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.

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In solid regions, Energy equation is

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Standard k-є Turbulence Model k - Turbulent Kinetic energy є - rate of dissipation of turbulent kinetic energy

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k and є equations k and є are obtained from the following transport equations:

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where G k represents the generation of turbulent kinetic energy due to mean velocity gradients G b is the generation of turbulent kinetic energy due to buoyancy Y M represents the contribution of the fluctuating dilatation in compressible turbulence to the overall dissipation rate C 1є, C 2є, C 3є are constants k and є are the turbulent Prandtl numbers for k and є respectively

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Eddy or Turbulent viscosity The model constants C 1 = 1.44, C 2 = 1.92, C = 0.09, k = 1.0, = 1.3 (Typical experimental values for these constants)

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Turbulence Intensity

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Discretization

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Discretization …continued

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Ideal gas model for Density calculations and Sutherland model for Viscosity calculations Density is calculated based on the Ideal gas equation. Viscosity calculations C 1 and C 2 are constants for a given gas. For air at moderate temperatures (about 300 – 500 K), C 1 = 1.458 x 10-6 kg/(m s K 0.5 ) C 2 = 110.4 K

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Reynolds Number calculation For flow over an obstruction, is the density of the fluid V is the average velocity (inlet velocity for internal flows) D is the hydraulic diameter is the Dynamic viscosity of the fluid

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Re for V = 0.5 m/sec For this problem, V = 0.5 m/sec, air = 1.225 kg/m 3, air = 1.7894 e–5 kg/m-sec

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Solver and Boundary conditions Solver – 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

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Wall boundary conditions Heat sources No heat flux at top and bottom walls Stationary top and bottom walls Volumetric heat source for the (solid) obstruction – 100,000 W/m 3

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Under relaxation factors Pressure0.3 Momentum0.7 Energy1 k0.8 Viscosity1 Density1 Body forces1

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Convergence criteria Continuity0.001 x – velocity0.001 y – velocity0.001 Energy1e-6 k0.001

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Case 1 – Grids Number of nodes - 4200 Number of nodes - 162938 Number of nodes - 208372

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Case 1 – Velocity contours

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Case 1 – Temperature contours

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Case 1 - Velocity Vectors

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Case 1 –Contours of Stream function

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Case 1 – Plot of Velocity Vs X-location

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Case 1 – Plot of Temperature Vs X-location

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

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Case 2 - Grids 4220 nodes 42515 nodes 79984 nodes

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Case 2 – Contours of Velocity

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Case 2 – Contours of Temperature

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Case 2 – Contours of Stream function

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Case 2 - Velocity vectors

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Case 2 – Plot of Velocity Vs X -location

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Case 2 – Plot of temperature Vs X-location

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

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