cmcc1.rar_18数学建模A题代码

clc; clear; %分为五层，假人为一层，各项参数近似与真人相同 Lambda = [0.082 0.37 0.045 0.028 0.432];%热导率 Ci = [1377 2100 1726 1005 4.2];%比热容 Rho = [300 862 74.2 1.18 1060];%密度 T = 3600;%时间分为T个单位 deltat = 3600 / T;%每个单位时间 deltax = 0.1 * 1e-3;%取衣物及人体厚度0.01mm为单位长度求解 Tmax = 65.00;%最外层温度 Tmin = 37.00;%最内层温度 n = 61; %第一问算出假人有效吸热厚度为61个单位长度 Hb = 0; Ub = zeros(T); for hi = 25 : -0.1 : 0.6 H = [0.6 hi 3.6 5.5 n * deltax * 1e+3] * 1e-3; %各层厚度 Hsum = zeros(1,6); for i = 1 : 5 Hsum(i + 1) = Hsum(i) + H(i); end L = Hsum(6); M = int32(L / deltax);%总单位长度数 m = M - n; x = linspace(0,L,M);%将厚度分为M块 Lambdax = zeros(1,M);%得到各个单位长度lambda Cix = zeros(1,M);%得到各个单位长度Ci Rhox = zeros(1,M);%得到各个单位长度Rho for i = 1 : M for j = 1 : 5 if x(i) <= Hsum(j + 1) Lambdax(i) = Lambda(j); Cix(i) = Ci(j); Rhox(i) = Rho(j); break; end end end alphax = [0 Lambdax * deltat ./ Cix ./ Rhox / 2 / deltax^2 0]; U = zeros(T,M + 2);%温度分布 U(1,:) = zeros(1,M + 2) + Tmin;%初始化温度分布 U(1,1) = Tmax; for i = 2 : T%微分方程求解 MatrixU = zeros(M + 2); D_last = zeros(M + 2,1); MatrixU(1,1) = 1 + 2 * alphax(1);MatrixU(1,2) = -alphax(1); D_last(1) = Tmax; for row = 2 : M + 1 MatrixU(row,row - 1) = -alphax(row); MatrixU(row,row) = 2 * alphax(row); MatrixU(row,row + 1) = -alphax(row); D_last(row) = alphax(row) * U(i - 1,row + 1) + alphax(row) * U(i - 1,row - 1) + (1 - 2 * alphax(row)) * U(i - 1,row); end MatrixU(M + 2,M + 1) = -alphax(M + 2); MatrixU(M + 2,M + 2) = 1 + 2 * alphax(M + 2); D_last(M + 2) = Tmin; U(i,:) = reshape(MatrixU \ D_last,1,M + 2); for row = 1 : M + 2 if U(i,row) > 65.00 U(i,row) = 65.00; end end end temp = 0; tcount = 0; for i = 1 : T if U(i,m + 1) > 47.00 || tcount > 5 * 60 temp = 1; break; end if U(i,m + 1) > 44.00 tcount = tcount + 1; end end if temp == 0 Hb = hi; Ub = U(:,m + 1); end end Hb %8.9 figure(1); axis([0 5500 35 55]); hold on; plot(1:T,Ub);%最优解为Ub(m+1)