clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 3.3 Page 109 \n'); //Example 3.3 // Find Thermal conductivity of Carbon Nanotube D = 14 * 10^-9; // [m]Dia of Nanotube s = 5*10^-6; // [m]Distance between the islands Ts = 308.4; //[K] Temp of sensing island Tsurr = 300; //[K] Temp of surrounding q = 11.3*10^-6; //[W] Total Rate of Heat flow //Dimension of platinum line wpt = 10^-6; //[m] tpt = 0.2*10^-6; //[m] Lpt = 250*10^-6; //[m] //Dimension of Silicon nitride line wsn = 3*10^-6; //[m] tsn = 0.5*10^-6; //[m] Lsn = 250*10^-6; //[m] //From Table A.1 Platinum Temp Assumed = 325K kpt = 71.6; //[W/m.K] //From Table A.2, Silicon Nitride Temp Assumed = 325K ksn = 15.5; //[W/m.K] Apt = wpt*tpt; //Cross sectional area of platinum support beam Asn = wsn*tsn-Apt; //Cross sectional area of Silicon Nitride support beam Acn = %pi*D^2/4; //Cross sectional Area of Carbon nanotube Rtsupp = [kpt*Apt/Lpt + ksn*Asn/Lsn]^-1; //[K/W] Thermal Resistance of each support qs = 2*(Ts-Tsurr)/Rtsupp; //[W] Heat loss through sensing island support qh = q - qs; //[W] Heat loss through heating island support Th = Tsurr + qh*Rtsupp/2; //[K] Temp of Heating island //For portion Through Carbon Nanotube //qs = (Th-Ts)/(s/(kcn*Acn)); kcn = qs*s/(Acn*(Th-Ts)); printf("\n\n Thermal Conductivity of Carbon nanotube = %.2f W/m.K",kcn); //END