CFD Online Logo CFD Online URL
Home > Forums > Main CFD Forum

Computational cost - Re ^3

Register Blogs Members List Search Today's Posts Mark Forums Read

Like Tree6Likes
  • 5 Post By michujo
  • 1 Post By sbaffini

LinkBack Thread Tools Display Modes
Old   May 27, 2013, 05:18
Default Computational cost - Re ^3
New Member
Join Date: May 2013
Posts: 1
Rep Power: 0 is on a distinguished road
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 is offline   Reply With Quote

Old   May 27, 2013, 06:38
Senior Member
Join Date: Dec 2011
Location: Madrid, Spain
Posts: 134
Rep Power: 8
michujo is on a distinguished road
Hi. Imagine you want to compute the turbulent flow over a body of characteristic length L.

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 t_L 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.

michujo is offline   Reply With Quote

Old   May 27, 2013, 07:27
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 3,504
Rep Power: 40
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
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)
FMDenaro is offline   Reply With Quote

Old   May 28, 2013, 04:28
Senior Member
sbaffini's Avatar
Paolo Lampitella
Join Date: Mar 2009
Location: Italy
Posts: 768
Blog Entries: 17
Rep Power: 21
sbaffini will become famous soon enough
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.
sbaffini is online now   Reply With Quote



Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On

Similar Threads
Thread Thread Starter Forum Replies Last Post
Coupled solver, computational cost Bollonga FLUENT 46 April 19, 2014 07:02
computational cost of ILU Shenren_CN Main CFD Forum 1 September 20, 2012 18:10
URANS & SAS-SST: Computational Cost siw CFX 3 May 31, 2012 18:58
Computational cost of transient turbulence model costinruja CFX 1 March 8, 2012 18:10
Short Course: Computational Thermal Analysis Dean S. Schrage Main CFD Forum 11 September 27, 2000 17:46

All times are GMT -4. The time now is 06:46.