I have to write a code with using these formulas. (Gama=1.4 M1=3.4) Then I have to reach p02/p01~= 1. So beta in the code should be changable. At the end M2 will also found

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Eda Deniz Soydinc 2022-6-8

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评论： Eda Deniz Soydinc 2022-6-13

采纳的回答： Torsten

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Answer 1

Torsten 2022-6-10

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编辑：Torsten 2022-6-10

在 MATLAB Online 中打开

fun = @(beta,p02divp01,M1,gamma)p02divp01-((gamma+1)*M1.^2.*sin(beta).^2./...
       (2+(gamma-1)*M1.^2.*sin(beta).^2)).^(gamma/(gamma-1)).*...
       ((gamma+1)./(2*gamma*M1.^2.*sin(beta).^2-(gamma-1))).^1/(gamma-1);
p02divp01 = 1.0;
M1 = 3.4;
gamma = 1.4;
beta0 = pi/4;
beta = fzero(@(beta)fun(beta,p02divp01,M1,gamma),beta0);  
beta_degree = beta*180/pi
beta_degree = 13.1745
theta = atan(2*cot(beta).*M1.^2.*sin(beta).^2./(M1.^2.*(gamma+cos(2*beta))+2.0));
theta_degree = theta*180/pi
theta_degree = 10.1904
M2 = sqrt((2+(gamma-1)*M1.^2.*sin(beta).^2)./...
    ((2.0*gamma*M1.^2.*sin(beta).^2-(gamma-1)).*sin(beta-theta).^2))
M2 = 25.3984

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Torsten 2022-6-13

I assumed p02/p01 and M1 as given. From these inputs, I calculated beta from the second equation, theta from the first equation and at last M2 from the third equation.

So with the above code, you could vary p02/p01 and M1 as inputs and get different beta, theta and M2 as output.

Eda Deniz Soydinc 2022-6-13