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1、CUMCM 2018 Problem A高温作业专用服装设计在高温环境下工作时，人们需要穿着专用服装以避免灼伤。专用服装通常由三层织物材料构成，记为I、II、III层，其中I层与外界环境接触，III层与皮肤之间还存在空隙，将此空隙记为IV层。为设计专用服装，将体内温度控制在37C的假人放置在实验室的高温环境中，测量假人皮肤外侧的温度。为了降低研发成本、缩短研发周期，请你们利用数学模型来确定假人皮肤外侧的温度变化情况，并解决以下问题：(1) 专用服装材料的某些参数值由附件1给出，对环境温度为75C、II层厚度为6 mm、IV层厚度为5 mm、工作时间为90分钟的情形开展实验，测量得到假人皮肤外侧
2、的温度（见附件2）。建立数学模型，计算温度分布，并生成温度分布的Excel文件（文件名为problem1.xlsx）。(2) 当环境温度为65C、IV层的厚度为5.5 mm时，确定II层的最优厚度，确保工作60分钟时，假人皮肤外侧温度不超过47C，且超过44C的时间不超过5分钟。(3) 当环境温度为80 时，确定II层和IV层的最优厚度，确保工作30分钟时，假人皮肤外侧温度不超过47C，且超过44C的时间不超过5分钟。Special clothing design for high temperature operationWhen working in a high temperature
3、environment, people need to wear special clothing to avoid burns. Special clothing usually consists of three layers of fabric material, I, II, III layer, in which layer I and the external environment contact, and between III layer and skin there is still a gap which is recorded as layer IV.In order
4、to design special clothes, place the body inside which the temperature is controlled at 37 in the high temperature environment of the laboratory, and measure the temperature outside the dummy skin. In order to reduce research and development costs and shorten the research and development cycle, plea
5、se use mathematical models to determine the temperature changes outside the dummy skin, and solve the following problems:(1) Some parameters of special clothing materials are given by Appendix 1. Experiments are carried out when the ambient temperature is 75, the thickness of layer II is 6 mm, the t
6、hickness of layer IV is 5 mm, and the working time is 90 minutes. The temperature outside the skin of the dummy is measured (Appendix 2). Establish a mathematical model, calculate the temperature distribution, and generate the temperature distribution Excel file (file named as problem1.xlsx).(2) Whe
7、n the ambient temperature is 65 and the thickness of layer IV is 5.5 mm, determine the optimal thickness of layer II to ensure the outer skin temperature of the dummy does not exceed 47 after 60 minutes and when temperature exceeds 44, the time does not exceed 5 minutes.(3) When the ambient temperat
8、ure is 80, determine the optimum thickness of layer II and layer IV to ensure that the outer skin temperature of the dummy does not exceed 47 after 30 minutes, and the when temperature exceeds 44, the time does not exceed 5 minutes.Answer:(1) We can get the relation of time and temperature on the du
9、mmys skin from Annex 2 (measurement temperature of the outer side of the dummys skin) . And convection on surface of layer I to the surrounding is negligible. So we can simplify this problem to the one-dimensional transient conduction from layer I to layer IV .Program as follow:k1=0.082;k2=0.37;k3=0
10、.045;k4=0.028; %四段材料的导热系数p1=300;p2=862;p3=74.2;p4=1.18; %四段材料的密度c1=1377;c2=2100;c3=1726;c4=1005; %四段材料的比热容t=1;x=0.0001; %时间，位置步长取值T=zeros(5401,153); %按时间，位置步长定义温度的二维矩阵f1=k1*t/(p1*c1*x2);f2=k2*t/(p2*c2*x2);f3=k3*t/(p3*c3*x2);f4=k4*t/(p4*c4*x2); %四段材料的傅里叶数T(1,:)=37; %时间为1的点温度简化为37T(:,1)=75; %位置为1的点温度简
11、化为75for i=1:5400 T(i+1,:)=T(i,:); %前一时间点的温度作为初始值赋给当前温度 while true s=T(i+1,:); %保留初始值 for j=2:6 T(i+1,j)=(f1*(T(i+1,j-1)+T(i+1,j+1)+T(i,j)/(2*f1+1); end T(i+1,7)=(f1*T(i+1,6)+f2*T(i+1,8)+T(i,7)/(f1+f2+1); for j=8:66 T(i+1,j)=(f2*(T(i+1,j-1)+T(i+1,j+1)+T(i,j)/(2*f2+1); end T(i+1,67)=(f2*T(i+1,66)+f3*T
12、(i+1,68)+T(i,67)/(f2+f3+1); for j=68:102 T(i+1,j)=(f3*(T(i+1,j-1)+T(i+1,j+1)+T(i,j)/(2*f3+1); end T(i+1,103)=(f3*T(i+1,102)+f4*T(i+1,104)+T(i,103)/(f3+f4+1); for j=104:152 T(i+1,j)=(f4*(T(i+1,j-1)+T(i+1,j+1)+T(i,j)/(2*f4+1); end %四段材料中温度离散化 T(i+1,153)=CUMCM2018ProblemAChineseAppendixS2(i); %导入位置为153
13、的点实测数据 if %控制收敛精度 break; end endendxlswrite( G:problem1,T,A1); %导出数据到excel表格pcolor(T),shading interp,colorbar %绘制二维伪彩色图Get temperature distribution (See ).Get isothermal diagram (See Annex4_problem1_ isothermal diagram.jpg).(2) Program as follow:k1=0.082;k2=0.37;k3=0.045;k4=0.028;p1=300;p2=862;p3=74
14、.2;p4=1.18;c1=1377;c2=2100;c3=1726;c4=1005;t=1;x=0.0001;h2=0.25;f1=k1*t/(p1*c1*x2);f2=k2*t/(p2*c2*x2);f3=k3*t/(p3*c3*x2);f4=k4*t/(p4*c4*x2);for r2=1:1000 T=zeros(3601,98+r2+101); T(1,:)=37; T(:,1)=65; for i=1:3600 T(i+1,:)=T(i,:); while true s=T(i+1,:); for j=2:6 T(i+1,j)=(f1*(T(i+1,j-1)+T(i+1,j+1)+T(i,j)/(2*f1+1); end T(i+1,7)=(f1*T(i+1,6)+f2*T(i+1,8)+T(i,7)/(f1+f2+1); for j=8:(6+r2) T(i+1,j)=(f2*(T(i+1,j-1)+T(i+1,j+1)+T(i,j)/(2*f2+1); end T(i+1,7+r2)=(f2*T(i+1,6+r2)+f3*T(i+1,8+r2)+T(i,7+r2)/(f2+f3+1); for j=(8+r2):(42+r2) T(i+1,j)=(f3*(T(i+1,j-
