Computational cost - Re ^3

karthikg.ae · May 27, 2013, 06:18

Hi,
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.

Thanks

michujo · May 27, 2013, 07:38

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, $L \times L \times L$ .

The minimum number of operations are the result of multiplying the number of cells by the number of time steps: $N_{ops}= N_{cells} \cdot N_{timesteps}$ .

The Kolmogorov scale characteristic length and turnover time scale with those of the largest scale as:
$\eta/L\sim Re_L^{-3/4}$ .
$t_\eta/t_L\sim Re_L^{-1/2}$ .

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 $\Delta x \sim \eta$ and $\Delta t \sim t_\eta$ (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) $N_{cells} \sim (L/\Delta x)^3 \sim (Re_L^{3/4})^3$ . Likewise, the number of time steps is $N_{timesteps} \sim t_L/\Delta t \sim Re_L^{1/2}$ .

The total number of operations scales as $N_{ops}= N_{cells} \cdot N_{timesteps} \sim (Re_L^{3/4})^3 \cdot Re_L^{1/2} = Re_L^{11/4} = Re_L^{2.75}$ .

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.

Cheers,
Michujo.

FMDenaro · May 27, 2013, 08:27

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)

sbaffini · May 28, 2013, 05:28

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.

May 27, 2013, 06:18	Computational cost - Re ^3	#1
karthikg.ae New Member Karthik Join Date: May 2013 Posts: 1 Rep Power: 0	Hi, 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. Thanks

May 27, 2013, 07:38		#2
michujo Senior Member Join Date: Dec 2011 Location: Madrid, Spain Posts: 134 Rep Power: 16	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, $L \times L \times L$ . The minimum number of operations are the result of multiplying the number of cells by the number of time steps: $N_{ops}= N_{cells} \cdot N_{timesteps}$ . The Kolmogorov scale characteristic length and turnover time scale with those of the largest scale as: $\eta/L\sim Re_L^{-3/4}$ . $t_\eta/t_L\sim Re_L^{-1/2}$ . 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 $\Delta x \sim \eta$ and $\Delta t \sim t_\eta$ (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) $N_{cells} \sim (L/\Delta x)^3 \sim (Re_L^{3/4})^3$ . Likewise, the number of time steps is $N_{timesteps} \sim t_L/\Delta t \sim Re_L^{1/2}$ . The total number of operations scales as $N_{ops}= N_{cells} \cdot N_{timesteps} \sim (Re_L^{3/4})^3 \cdot Re_L^{1/2} = Re_L^{11/4} = Re_L^{2.75}$ . 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. Cheers, Michujo. Far, JimKnopf, RodriguezFatz and 2 others like this.

May 28, 2013, 05:28		#4
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	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. michujo likes this.

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May 27, 2013, 08:27		#3
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,896 Rep Power: 73	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)