# How to iterate with fsolve to find Mach number

조회 수: 16(최근 30일)
Hailey 10 Sep 2019
편집: Torsten 10 Sep 2019
I'm trying to interate static pressure ratio to find Mach number using fsolve, but I can't get the syntax correct. The Mathworks explaination didn't really help me. Could anyone out there help?
I have a set range for the pressure ratio, but the Mach number is unknown. Here is the equation:
This is what I have so far for the code:
pt2pinf = [1.9, 2.7, 3.5, 4.3, 5.1, 5.9, 6.7, 7.5, 27.5, 47.5, 67.5 ,87.5 ,107.5,127.5, 147.5, 200, 252.5, 305, 357.5, 410, 462.5 ,515 ,567.5, 620, 672.5, 725, 777.5, 830 ,882.5, 935, 987.5, 1000];
Me0=ones(1,32);
g=1.2;
pt2pinf_test = fsolve(@(Me)(((1+((g-1)/2)*Me.^2)^(g/(g-1)))/((((2*g)/(g+1))*Me.^2)-((g-1)/(g+1)))),Me0);

#### 댓글 수: 8

표시 이전 댓글 수: 5
Hailey 10 Sep 2019
But I need to have the Mach number values. I need an array of Mach numbers that match up with it's respective pressure ratio value. This did not do that.
Torsten 10 Sep 2019
They are stored in the fsolve return parameter (pt2pinf_test in your case).
Hailey 10 Sep 2019
The values are too small to be a Mach number after a shock wave.
Not sure what I did wrong.

댓글을 달려면 로그인하십시오.

### 채택된 답변

Torsten 10 Sep 2019
편집: Torsten 10 Sep 2019
function main
pt2pinf = [1.9, 2.7, 3.5, 4.3, 5.1, 5.9, 6.7, 7.5, 27.5, 47.5, 67.5 ,87.5 ,107.5,127.5, 147.5, 200, 252.5, 305, 357.5, 410, 462.5 ,515 ,567.5, 620, 672.5, 725, 777.5, 830 ,882.5, 935, 987.5, 1000];
Me0 = 5.0;
g = 1.2;
fun = @(Me,pquot)((((1+((g-1)/2)*Me.^2).^(g/(g-1)))./((((2*g)/(g+1))*Me.^2)-((g-1)/(g+1))))-pquot).^2
for i = 1:numel(pt2pinf)
pquot = pt2pinf(i);
machnumber(i) = fminsearch(@(Me)fun(Me,pquot),Me0,optimset('TolFun',1e-10));
end
machnumber
fun(machnumber,pt2pinf)
end

댓글을 달려면 로그인하십시오.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by