Local Reynolds-Number?

Kappa · January 4, 2017, 06:25

Hey,
I need to compute Reynolds-Numbers in a turbulent flow. Assuming I got a 2D-flow and my domain is divided into squares, my question is how to compute the local Reynold-numbers.

Imagine my domain is 1m x 1m and it is divided into 100x100 squares (each having a length of 1cm). Computing the global Reynolds-number (like in the OpenFOAM-Driven Cavity example) looks as follows: Re_global = rho*1m*U_inlet / nu.
If I want to compute the local Reynolds-number for each square I consider 1cm instead of 1m but the remaining variables (rho,U,nu) stay the same, at least for incompressible flow.

As a result, my local Re is only 1% of the global Re. But actually, it should be the same since the velocity, kinematic viscosity and density is the same in the whole 1m x 1m region.
Can you tell me how to compute the local Reynolds-Number in such a flow so that it is an sufficiently accurate representation of the flow in the square?

Thanks to all

FMDenaro · January 4, 2017, 06:35

The local Re number is actually somehow a pointwise function ... it represents a Re value for that position. Think about the flow over a flat plate starting at x=0. The local Re number is a function of x. Small Rex says the flow is laminar, then transitional then turbulent.

In you case, using the cell size you are considering the cell Re number, not the local Re number

January 4, 2017, 06:25	Local Reynolds-Number?	#1
Kappa New Member Join Date: Oct 2016 Posts: 10 Rep Power: 10	Hey, I need to compute Reynolds-Numbers in a turbulent flow. Assuming I got a 2D-flow and my domain is divided into squares, my question is how to compute the local Reynold-numbers. Imagine my domain is 1m x 1m and it is divided into 100x100 squares (each having a length of 1cm). Computing the global Reynolds-number (like in the OpenFOAM-Driven Cavity example) looks as follows: Re_global = rho1mU_inlet / nu. If I want to compute the local Reynolds-number for each square I consider 1cm instead of 1m but the remaining variables (rho,U,nu) stay the same, at least for incompressible flow. As a result, my local Re is only 1% of the global Re. But actually, it should be the same since the velocity, kinematic viscosity and density is the same in the whole 1m x 1m region. Can you tell me how to compute the local Reynolds-Number in such a flow so that it is an sufficiently accurate representation of the flow in the square? Thanks to all

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January 4, 2017, 06:35		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	The local Re number is actually somehow a pointwise function ... it represents a Re value for that position. Think about the flow over a flat plate starting at x=0. The local Re number is a function of x. Small Rex says the flow is laminar, then transitional then turbulent. In you case, using the cell size you are considering the cell Re number, not the local Re number