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

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.

Case 1 10 cm uTuT 4 cm TwTw 2 cm 0.5 cm

Case 2 TwTw 0.5 cm 10 cm uTuT 4 cm 0.5 cm

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

Momentum  is the molecular viscosity of the fluid.

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.

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

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

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

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

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)

Turbulence Intensity

Discretization

Discretization …continued

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 = x 10-6 kg/(m s K 0.5 ) C 2 = K

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

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

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

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

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

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

Case 1 – Grids Number of nodes Number of nodes Number of nodes

Case 1 – Velocity contours

Case 1 – Temperature contours

Case 1 - Velocity Vectors

Case 1 –Contours of Stream function

Case 1 – Plot of Velocity Vs X-location

Case 1 – Plot of Temperature Vs X-location

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

Case 2 - Grids 4220 nodes nodes nodes

Case 2 – Contours of Velocity

Case 2 – Contours of Temperature

Case 2 – Contours of Stream function

Case 2 - Velocity vectors

Case 2 – Plot of Velocity Vs X -location

Case 2 – Plot of temperature Vs X-location

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