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關於用1stopt解方程的問題?

函數F=myfun(X)

t 1 = X(1);T2 = X(2);T3 = X(3);T4 = X(4);TF = X(5);q = X(6);

c 1 = 0.003;

c2 = 0.003

h = 0.002

w 1 = 0.002;

w2 = 0.002

v = 5;

T0 = 3287

%氣體對流

Pr hc1處的syms sigma

dt = 0.013;d = 0.03pc=3.5*10^6;c = 2000c0=9.18*10^5;CP = 1500;u=1.56*10^(-5);

m = 0.022k = 1.26;ma = 3;

sigma=(t1*(1+(k-1)*ma*ma/2)/(2*t0)+0.5)^(-0.68)*(1+(k-1)*ma*ma/2)^(-0.12);

at = pi * dt * dt/4;a = pi * d * d/4;

pr = u * Cp/m;

hc1=(0.026/dt^0.2)*(u^0.2*cp/pr^0.6)*(pc/c)^0.8*(at/a)^0.9*sigma;

f(1)= HC 1 *(T0-t 1)+C0-q;

%內壁熱傳導

syms k2

m 1 = 387;

k2 = m 1/c 1;

f(2)= k2 *(t 1-T2)-q;

%冷卻劑對流

劉釔彤

CPL = 4.1 * Tf+1892.2;

ml =-0.0001 * Tf+0.1383;lf=10^(-8)*(-5.3*tf+683.7);lt=10^(-8)*(-5.3*t2+683.7);

pf =-0.8102 * Tf+831.4661;pt =-0.8102 * T2+831.4661;uf = pf * lfut = pt * lt

de = 2 * h * w 1/(h+w 1);ref = v * w 1/lf;PRF = uf * Cpl/ml;

hc20=0.023*ref^0.8*prf^0.34*(uf/ut)*ml/de;

m2 = 202.4m=(2*hc20/(m2*w2))^0.5;劉釔彤= tanh(m * h)/(m * h);

hc2 = hc20 *劉釔彤;

f(3)= hc2 *(T2-Tf)-q;

f(4)= T2+T3-2 * Tf;

%外墻熱傳導

m3 = 387

f(5)=(T3-T4)/(C2/m3)-q;

空氣對流百分比

syms av Gra hc3

g = 9.8Ta = 293av = 1/((T4+Ta)/2);d = 0.2pa = 1.06;Pra = 0.7ma = 0.04la=4*10^(-5);

gra=g*av*(t4-ta)*d^3/la^2;hc3=0.48*(gra*pra)^0.25*ma/d;

f(6)= hc3 *(T4-Ta)-q;

結束