# General CFD FAQ

### From CFD-Wiki

This FAQ is empty. This is just a suggestion on how to structure it. Please feel free to add questions and answers here!

## Contents |

## Physics

### What is the Reynolds decomposition?

The Reynolds decomposition is used to separate the scales in a turbulent flow and resolve the velocity field (or any other scalar field) as the sum of an average component and a fluctuating component. The time average of the fluctuating field is identically zero.

### What is the difference between the Reynolds decomposition and LES filtering?

The Reynolds decomposition separates the velocity field into an average component and a fluctuating component. **The time average of the fluctuating component is identically zero**. In LES, a filtering operation is performed to decompose the velocity field into a *filtered* velocity field and a *filtered residual* field. The crucial difference between both methods is

- The filtered velocity field is a
**random variable**while the average field (Reynolds) is not - The time average of the residual field is generally
**nonzero**while it is identically zero for fluctuating component in the Reynolds decomposition.

## Numerics

### What is the difference between FEM, FVM and FDM?

Describe differences between the finite volume method, the finite element method and the finite difference method...

### How do i count the number of floating point operations in an algorithm?

First of all, write down your algorithm as pseudocode. Now simply count the number of operations that the algorithm is performing (per iteration of course). all operations: addition, subtraction, multiplication, and division count the same, i.e. **1 flop**.

This is not exact since multiplication includes several additions, and division includes several multiplications... But it seems reasonable to assume that on average, all operations are counted in the same manner.

Here is an example (just for illustration):

for i = 0 to P

for n = 1 to N (number of elements in array)

B(n) = a(n)*a(n-1) - 2*c(n) + 3

next n

next i

For each **n** there are 2 multiplications, one subtraction, and one addition resulting in **four** operations. For all **N** there are **4N** operations. This is the the order of the algorithm. In this example, its is O(4N) or simply O(N) (constants do not count).

For all iterations, there are 4N(P+1) operations. (remember, when starting the loop from zero to p, there are p+1 steps).