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  • 大小: 3KB
    文件類型: .m
    金幣: 1
    下載: 0 次
    發布日期: 2021-05-05
  • 語言: Matlab
  • 標簽: 噴管流??

資源簡介

亞聲速—超聲速等熵噴管流動的CFD解法,基于Matlab采用麥考馬克方法為一維噴管流動建立一種時間推進的有限差分解法。一維噴管流動控制體

資源截圖

代碼片段和文件信息

clc
clear

L=3;???????????????%?length?of?nozzle
nx=31;?????????????%?number?of?grid?points
dx=L/(nx-1);???????%?mesh?size??
C=0.5;?????????????%?courant?number
shr=1.4;???????????%?specific?heat?ratio
N=1000;????????????%?number?of?time?steps
x=0:dx:L;??????????%?mesh
A=zeros(1nx);?????%?section?area
rho=zeros(Nnx);???%?density
T=zeros(Nnx);?????%?nondimensional?temperature
V=zeros(Nnx);?????%?nondimensional?velocity
P=zeros(Nnx);?????%?nondimensional?pressure
Ma=zeros(Nnx);????%?Mach?number

%?initial?conditions
for?j=1:nx
????A(j)=1+2.2*(x(j)-1.5)^2;
????rho(1j)=1-0.3146*x(j);
????T(1j)=1-0.2314*x(j);
????V(1j)=(0.1+1.09*x(j))*T(1j)^0.5;
????P(1j)=rho(1j)*T(1j);
????Ma(1j)=V(1j)./(T(1j).^0.5);
end


for?i=1:N
????%?time?step
????temp_dt=zeros(1nx);
????for?ii=1:nx
????????temp_dt(ii)=C*dx/(V(iii)+T(iii)^0.5);
????end
????dt=min(temp_dt);???????
????
????%?prediction?step
????d_rho=zeros(2nx-1);
????d_V=zeros(2nx-1);
????d_T=zeros(2nx-1);
????pre_rho=zeros(nx-1);
????pre_V=zeros(nx-1);
????pre_T=zeros(nx-1);
????for?m=1:nx-1
????????d_rho(1m)=-V(im)*(rho(im+1)-rho(im))/dx-rho(im)*(V(im+1)-V(im))/dx-rho(im)*V(im)*(log(A(m+1))-log(A(m)))/dx;
????????d_V(1m)=-V(im)*(V(im+1)-V(im))/dx-(1/shr)*((T(im+1)-T(im))/dx+(T(im)/rho(im))*(rho(im+1)-rho(im))/dx);
????????d_T(1m)=-V(im)*(T(im+1)-T(im))/dx-(shr-1)*T(im)*((V(im+1)-V(im))/dx+V(im)*(log(A(m+1))-log(A(m)))/dx);
????????pre_rho(m)=rho(im)+d_rho(1m)*dt;????????
????????pre_V(m)=V(im)+d_V(1m)*dt;
????????pre_T(m)=T(im)+d_T(1m)*dt;
????end
????
????%?correction?step
????d_rho_av=zeros(1nx-1);
????d_V_av=zeros(1nx-1);
????d_T_av=zeros(1nx-1);
????for?m=2:nx-1
????????

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