时间t是一个0到40范围的变量,pl,pg,pwe取值都是t=t2时的值
时间: 2024-04-03 10:36:45 浏览: 125
好的,那么根据您的要求,以下是修改后的程序:
```matlab
% 清空工作区变量
clear
% 定义参数
TJ = 13.8;
KG = 100/6.7;
TG = 5;
HW = 10.38;
KW1 = 20;
KW2 = 50;
R = 35;
beta = 0;
V = 10;
% 定义变量
t = sdpvar(1);
w = sdpvar(1);
Pun = sdpvar(1);
Pl = sdpvar(1);
% 定义表达式
A = (TG + KW2 + KW1*TG)/(-2*TG*(TG + KW2));
W = sqrt((KG + KW1)/(TG*(TJ + KW2))) - A^2;
M = PL/(W*(TJ + KW2))*sqrt(KG/(KG + KW1));
Thet = acos(W*(TG + KW2)/sqrt(KG*(KG + KW1)));
C = 1 - PL/(KG + KW1);
f = M*exp(t*A)*cos(t*W + Thet) + C;
df_dt = jacobian(f, t);
df_dw = jacobian(f, w);
t1 = (pi/2 - Thet - atan(W/A))/W;
A1 = (TG*TG)/(-2*TG*(TG));
W1 = sqrt((KG)/(TG*TJ)) - A^2;
Theta1 = acos(W*(TG)/sqrt(KG^2));
% 定义约束条件
Pwe = 97.81*w^3;
s = tf('s');
Pg = (1 - f)*(-14.92)/(1 + s*5);
Pun = Pl - Pg - Pwe;
C1 = f(t) - Pun/KG;
fsecmax = 1 - M*exp(A*(t - t1))*cos(W*(t - t1) + Thet) - C1;
t2 = (pi/2 - Thet - atan(W/A))/W;
t3 = (pi/2 - Theta1 - atan(W1/A1))/W1 - t2;
constraints = [t > t1, t < t3, w > 1.47, w < 2.52];
dw_dt = (0.248*w - 1.16*w - 50*df_dt - 20*(1 - f))/(2.1*0.875);
constraints = [constraints, dw_dt == (0.248*w - 1.16*w - 50*df_dt - 20*(1 - f))/(2.1*0.875)];
% 定义目标函数
objective = fsecmax;
% 求解优化问题
options = sdpsettings('verbose', 0);
sol = optimize(constraints, objective, options);
% 输出结果
if sol.problem == 0
fprintf('最优的 t2 = %.2f\n', value(t));
fprintf('fsecmax 的最小值 = %.4f\n', value(objective));
else
disp('优化问题求解失败');
disp(sol.info);
end
```
需要注意的是,由于您没有给出 `PL` 的值,所以我在程序中将其定义为 `Pl`,并假设其为一个变量。如果您有具体的 `PL` 值,请在程序中修改。另外,由于 `Pun` 的值是根据 `Pl`、`f` 和 `Pwe` 计算得出的,所以在程序中将 `Pun` 定义为一个变量,并在约束条件中添加 `Pun = Pl - Pg - Pwe` 的限制。
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