Computational cost - Re ^3
I just started reading a textbook on cfd and found this statement- The computational cost for a DNS increases as the cube of the Reynold number.
Can someone please explain how this order of magnitude estimate was arrived at.
Hi. Imagine you want to compute the turbulent flow over a body of characteristic length .
Suppose your mesh is uniform. Turbulence is unsteady so you will need to perform a computation over one characteristic time of the largest turbulence scale at least. Also, turbulence is 3D so the dimensions of your domain are, at least, .
The minimum number of operations are the result of multiplying the number of cells by the number of time steps: .
The Kolmogorov scale characteristic length and turnover time scale with those of the largest scale as:
In a DNS you have to resolve all the turbulence scales, both in space and time. Therefore, the cell size and time step in your simulation must be and (length scale and characteristic turnover time of the smallest, i.e. Kolmogorov, scale).
The total number of cells is therefore (notice that your domain is 3D) . Likewise, the number of time steps is .
The total number of operations scales as .
Notice that these are minimum requirements, so I guess you can easily reach the scaling with the third power of the Reynolds number. A detailed explanation can be found in the book of Pope.
a very quick (and brutal) estimation can be done assuming that you have to work with a cell Reynolds number = O(1). Therefore, for each direction, you have
Reh = u h /ni = ReL h/L = ReL / N = O(1)
From the michujo estimate, the exact Re^3 scaling is then reached when considering an unitary Courant number (as most DNS methods rely on explicit convection schemes). As a result, the time step also scales like the grid step and the final estimate is reached.
|All times are GMT -4. The time now is 10:17.|