# Difference between first order and second order

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 December 4, 2012, 05:59 Difference between first order and second order #1 Member   Marco Join Date: Aug 2012 Location: Spain Posts: 41 Rep Power: 7 Hi everyone, I've done a simulation of a fluid through some narrowings and turns in a pipe. The first time i did the iterations with the pressure (solution methods) in first order, getting a good solution. Then, i defined the pressure in second order. I did the simulation (it got me much more time), and the results were identical to those i got in the first simulation (with the pressure in first order). Is there any difference of defining the pressure in first or second order? What is the meaning of the order (first or second)? Thanks for your answers.

 December 4, 2012, 08:46 #2 Senior Member     Philipp Join Date: Jun 2011 Location: Germany Posts: 1,297 Rep Power: 20 The actual equations (Navier-Stokes) you would like to solve are coupled differential euqations. Once you decide to solve them on a computer you do it numerically, i.e. you solve the equations on discrete points/volume elements. Now you have to get from the original differntial equations (dp/dx ...) to a finite set of equations for the unknown grid values. This can be done, by switching from derivatives to some finite difference approximations, such as df/dx = [f(x+1) - f(x-1)] / 2dx. Unfortunately these approximations are just approximations and the "=" in the last equation isn't really true. The order of an approximation says, that the error that is introduced by it is proportional to the grid spacing to the power of your order. Thus, first order means that a certain derivative is approximated by the finite difference PLUS "something times (dx)^1". Second order means plus "somthing times (dx)^2" and so on. If a grid is small enough the error of a second order scheme will always be smaller than a first order error, because "dx^2" -> 0 faster than just "dx". I guess in your case you just don't see the difference because it is much smaller than the relevant values in your domain. But you can be sure that for small "dx", second order will be better. msatrustegui, Saideep, naviones and 1 others like this. __________________ The skeleton ran out of shampoo in the shower.

December 5, 2012, 06:54
#3
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Marco
Join Date: Aug 2012
Location: Spain
Posts: 41
Rep Power: 7
Quote:
 Originally Posted by RodriguezFatz The actual equations (Navier-Stokes) you would like to solve are coupled differential euqations. Once you decide to solve them on a computer you do it numerically, i.e. you solve the equations on discrete points/volume elements. Now you have to get from the original differntial equations (dp/dx ...) to a finite set of equations for the unknown grid values. This can be done, by switching from derivatives to some finite difference approximations, such as df/dx = [f(x+1) - f(x-1)] / 2dx. Unfortunately these approximations are just approximations and the "=" in the last equation isn't really true. The order of an approximation says, that the error that is introduced by it is proportional to the grid spacing to the power of your order. Thus, first order means that a certain derivative is approximated by the finite difference PLUS "something times (dx)^1". Second order means plus "somthing times (dx)^2" and so on. If a grid is small enough the error of a second order scheme will always be smaller than a first order error, because "dx^2" -> 0 faster than just "dx". I guess in your case you just don't see the difference because it is much smaller than the relevant values in your domain. But you can be sure that for small "dx", second order will be better.