blob: 20d19fdc086ef7530539d1a752864a6d654c6ecd (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
|
//developed in windows 8 operating system 64bit
//platform Scilab 5.4.1
//example 28_5w
clc;clear;
//Given Data
area=100*10^-4; //Area of the cross section (Unit:m^-2)
thick_a=0.04; //Thickness of part A (Unit: m)
thick_b=0.025; //Thickness of part B (Unit:m)
k_a=200; //Thermal conductivity of A (Unit: W/m-degree centigrade)
k_b=400; //Thermal conductivity of B (Unit: W/m-degree centigrade)
theta_a=100; //Temperature at A (Unit:degree centigrade)
theta_b=0; //Temperature at B (Unit:degree centigrade)
//calculation
rate_heat_flow=area*(theta_a-theta_b)/(thick_a/k_a+thick_b/k_b); //Calculation of The rate of heat flow through any cross section (Unit:Watt)
temp=rate_heat_flow*thick_b/(area*k_b); //Calculation of The temperature at the interface (Unit: degree centigrade)
equivalent_k=(thick_a+thick_b)/(thick_a/k_a+thick_b/k_b); //Calculation of The equivalent thermal conductivity of the compound plate (Unit: W/m-degree centigrade)
disp(rate_heat_flow,"The rate of heat flow through any cross section is (Unit:Watt)");
disp(temp,"The temperature at the interface is (Unit: degree centigrade)");
disp(equivalent_k,"The equivalent thermal conductivity of the compound plate is (Unit: W/m-degree centigrade)");
|
