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-rw-r--r--534/CH5/EX5.1/5_1_Thermocouple_junction.sce27
-rw-r--r--534/CH5/EX5.10/5_10_Finite_Difference2_slab.sce108
-rw-r--r--534/CH5/EX5.2/5_2_Thermocouple_junction2.sce49
-rw-r--r--534/CH5/EX5.3/5_3_Two_step_process.sce75
-rw-r--r--534/CH5/EX5.4/5_4_Radial_Two_Step.sce40
-rw-r--r--534/CH5/EX5.5/5_5_Sphere_Two_Step.sce34
-rw-r--r--534/CH5/EX5.6/5_6_Burial_Depth.sce22
-rw-r--r--534/CH5/EX5.7/5_7_Spherical_Tumor.sce38
-rw-r--r--534/CH5/EX5.8/5_8_Nanostructured_Material.sce33
-rw-r--r--534/CH5/EX5.9/5_9_Finite_Difference1.sce56
10 files changed, 482 insertions, 0 deletions
diff --git a/534/CH5/EX5.1/5_1_Thermocouple_junction.sce b/534/CH5/EX5.1/5_1_Thermocouple_junction.sce
new file mode 100644
index 000000000..f41c10e04
--- /dev/null
+++ b/534/CH5/EX5.1/5_1_Thermocouple_junction.sce
@@ -0,0 +1,27 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.1 Page 261 \n'); //Example 5.1
+// Junction Diameter and Time Calculation to attain certain temp
+
+//Operating Conditions
+
+h = 400; //[W/m^2.K] Heat Convection coefficient
+k = 20; //[W/m.K] Thermal Conductivity of Blade
+c = 400; //[J/kg.K] Specific Heat
+rho = 8500; //[kg/m^3] Density
+Ti = 25+273; //[K] Temp of Air
+Tsurr = 200+273; //[K] Temp of Gas Stream
+TimeConstt = 1; //[sec]
+
+//From Eqn 5.7
+D = 6*h*TimeConstt/(rho*c);
+Lc = D/6;
+Bi = h*Lc/k;
+
+//From eqn 5.5 for time to reach
+T = 199+273; //[K] Required temperature
+
+t = rho*D*c*2.30*log10((Ti-Tsurr)/(T-Tsurr))/(h*6);
+
+printf("\n\n Junction Diameter needed for a time constant of 1 s = %.2e m \n\n Time Required to reach 199degC in a gas stream = %.1f sec ", D, t);
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.10/5_10_Finite_Difference2_slab.sce b/534/CH5/EX5.10/5_10_Finite_Difference2_slab.sce
new file mode 100644
index 000000000..5db5cf98b
--- /dev/null
+++ b/534/CH5/EX5.10/5_10_Finite_Difference2_slab.sce
@@ -0,0 +1,108 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.10 Page 311 \n'); //Example 5.10
+// Using Explicit Finite Difference method, determine temperatures at the surface and 150 mm from the surface after an elapsed time of 2 min
+// Repeat the calculations using the Implicit Finite Difference Method
+// Determine the same temperatures analytically
+
+//Operating Conditions
+
+delx = .075; //[m] Metre
+T = 20+273; //[K] Temperature
+q = 3*10^5; //[W/m^3] Volumetric Rate
+
+//From Table A.1 copper 300 K
+k = 401; //[W/m.K] Conductivity
+a = 117*10^-6; //[m^2/s]
+
+//By using stability criterion reducing further Fourier Number
+Fo = (2)^-1;
+//By definition
+delt = Fo*delx^2/a;
+format('v',5);
+
+//System of Equation for Explicit Finite difference Fo = 1/2
+Tv1(1,:) = [20 20 20 20 20]; //At p=0 Initial Temperature t - 20 degC
+for i = 2:6
+ Tv1(i,1) = 56.1 + Tv1(i-1,2);
+ Tv1(i,2) = (Tv1(i-1,3) + Tv1(i-1,1))/2;
+ Tv1(i,3) = (Tv1(i-1,4) + Tv1(i-1,2))/2;
+ Tv1(i,4) = (Tv1(i-1,5) + Tv1(i-1,3))/2;
+ Tv1(i,5) = Tv1(i-1,5);
+end
+for j=1:6
+ T1(j,:)=[j-1 delt*(j-1) Tv1(j,:)];
+end
+printf("\n\n EXPLICIT FINITE-DIFFERENCE SOLUTION FOR Fo = 1/2\n p t(s) T0 T1 T2 T3 T4\n");
+disp(T1);
+printf('\n Hence after 2 min, the surface and the desirde interior temperature T0 = %.2f degC and T2 = %.1f degC',T1(6,3),T1(6,5));
+
+//By using stability criterion reducing further Fourier Number
+Fo = (4)^-1;
+//By definition
+delt = Fo*delx^2/a;
+//System of Equation for Explicit Finite difference for Fo = 1/4
+Tv2(1,:) = [20 20 20 20 20 20 20 20 20]; //At p=0 Initial Temperature t - 20 degC
+for i=2:11
+ Tv2(i,1)=1/2*(q*delx/k + Tv2(i-1,2)) +Tv2(i-1,1)/2;
+ Tv2(i,2)=(Tv2(i-1,1)+Tv2(i-1,3))/4 + Tv2(i-1,2)/2;
+ Tv2(i,3)=(Tv2(i-1,2)+Tv2(i-1,4))/4 + Tv2(i-1,3)/2;
+ Tv2(i,4)=(Tv2(i-1,3)+Tv2(i-1,5))/4 + Tv2(i-1,4)/2;
+ Tv2(i,5)=(Tv2(i-1,4)+Tv2(i-1,6))/4 + Tv2(i-1,5)/2;
+ Tv2(i,6)=(Tv2(i-1,5)+Tv2(i-1,7))/4 + Tv2(i-1,6)/2;
+ Tv2(i,7)=(Tv2(i-1,6)+Tv2(i-1,8))/4 + Tv2(i-1,7)/2;
+ Tv2(i,8)=(Tv2(i-1,7)+Tv2(i-1,9))/4 + Tv2(i-1,8)/2;
+ Tv2(i,9)= Tv2(i-1,9);
+end
+for j=1:11
+ T2(j,:)=[j-1 delt*(j-1) Tv2(j,:)];
+end
+printf("\n\n EXPLICIT FINITE-DIFFERENCE SOLUTION FOR Fo = 1/4\n p t(s) T0 T1 T2 T3 T4 T5 T6 T7 T8\n")
+disp(T2)
+printf('\n Hence after 2 min, the surface and the desirde interior temperature T0 = %.2f degC and T2 = %.1f degC',T2(11,3),T2(11,5))
+
+
+//(b)Implicit Finite Difference solution
+Fo = (4)^-1;
+//By definition
+delt = Fo*delx^2/a;
+
+T3 = rand(6,11); //Random Initital Distribution
+function[Tm]=Tvalue(i)
+function[f]=F(x)
+ f(1)= 2*x(1) - x(2) - q*delx/k - T3(i,3);
+ f(2)= -x(1)+4*x(2)-x(3)-2*T3(i,4);
+ f(3)= -x(2)+4*x(3)-x(4)-2*T3(i,5);
+ f(4)= -x(3)+4*x(4)-x(5)-2*T3(i,6);
+ f(5)= -x(4)+4*x(5)-x(6)-2*T3(i,7);
+ f(6)= -x(5)+4*x(6)-x(7)-2*T3(i,8);
+ f(7)= -x(6)+4*x(7)-x(8)-2*T3(i,9);
+ f(8)= -x(7)+4*x(8)-x(9)-2*T3(i,10);
+ f(9)= -x(9)+T3(i,11);
+ funcprot(0);
+endfunction
+x = [30 30 30 30 30 30 30 30 30];
+Tm = fsolve(x,F);
+ funcprot(0)
+endfunction
+
+//At p=0 Initial Temperature t - 20 degC
+T3(1,:) = [0 delt*0 20 20 20 20 20 20 20 20 20];
+for j=1:5
+ T3(j+1,:)=[j delt*j Tvalue(j)];
+end
+printf("\n\n IMPLICIT FINITE-DIFFERENCE SOLUTION FOR Fo = 1/4\n p t(s) T0 T1 T2 T3 T4 T5 T6 T7 T8\n");
+disp(T3);
+printf('\n Hence after 2 min, the surface and the desirde interior temperature T0 = %.2f degC and T2 = %.1f degC',T3(6,3),T3(6,5));
+
+t = 120; //[seconds]
+//(c) Approximating slab as semi-infinte medium
+Tc = T -273 + 2*q*(a*t/%pi)^.5/k;
+
+//At interior point x=0.15 m
+x =.15; //[metre]
+//Analytical Expression
+Tc2 = T -273 + 2*q*(a*t/%pi)^.5/k*exp(-x^2/(4*a*t))-q*x/k*[1-erf(.15/(2*sqrt(a*t)))];
+
+printf(' \n\n (c) Approximating slab as a semi infinte medium, Analytical epression yields \n At surface after 120 seconds = %.1f degC \n At x=.15 m after 120 seconds = %.1f degC',Tc,Tc2);
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.2/5_2_Thermocouple_junction2.sce b/534/CH5/EX5.2/5_2_Thermocouple_junction2.sce
new file mode 100644
index 000000000..cedc27399
--- /dev/null
+++ b/534/CH5/EX5.2/5_2_Thermocouple_junction2.sce
@@ -0,0 +1,49 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.2 Page 265 \n'); //Example 5.2
+// Steady State Temperature of junction
+// Time Required for thermocouple to reach a temp that is within 1 degc of its steady-state value
+
+//Operating Conditions
+
+h = 400; //[W/m^2.K] Heat Convection coefficient
+k = 20; //[W/m.K] Thermal Conductivity of Blade
+c = 400; //[J/kg.K] Specific Heat
+e = .9; //Absorptivity
+rho = 8500; //[kg/m^3] Density
+Ti = 25+273; //[K] Temp of Air
+Tsurr = 400+273; //[K] Temp of duct wall
+Tg = 200+273; //[K] Temp of Gas Stream
+TimeConstt = 1; //[sec]
+stfncnstt=5.67*10^(-8); // [W/m^2.K^4] - Stefan Boltzmann Constant
+
+//From Eqn 5.7
+D = 6*h*TimeConstt/(rho*c);
+As = %pi*D^2;
+V = %pi*D^3/6;
+
+//Balancing Energy on thermocouple Junction
+//Newton Raphson method for 4th order eqn
+T=500;
+while(1>0)
+f=(e*stfncnstt*(Tsurr^4-T^4)-(h*(T-Tg)));
+fd=(-3*e*stfncnstt*T^3)-h;
+Tn=T-f/fd;
+if((e*stfncnstt*(Tsurr^4-Tn^4)-(h*(Tn-Tg)))<=.01)
+ break;
+end;
+T=Tn;
+end
+printf("\n (a) Steady State Temperature of junction = %.2f degC\n",T-273);
+
+//Using Eqn 5.15 and Integrating the ODE
+// Integration of the differential equation
+// dT/dt=-A*[h*(T-Tg)+e*stefncnstt*(T^4-Tsurr^4)]/(rho*V*c) , T(0)=25+273, and finds the minimum time t such that T(t)=217.7+273.15
+deff("[Tdot]=f(t,T)","Tdot=-As*[h*(T-Tg)+e*stfncnstt*(T^4-Tsurr^4)]/(rho*V*c)");
+deff("[z]=g(t,T)","z=T-217.7-273");
+
+T0=25+273;ng=1;
+[T,rd]=ode("roots",T0,0,217.7+273,f,ng,g);
+printf("\n (b) Time Required for thermocouple to reach a temp that is within 1 degc of its steady-state value = %.2f s\n",rd(1));
+
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.3/5_3_Two_step_process.sce b/534/CH5/EX5.3/5_3_Two_step_process.sce
new file mode 100644
index 000000000..b481ac7dc
--- /dev/null
+++ b/534/CH5/EX5.3/5_3_Two_step_process.sce
@@ -0,0 +1,75 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.2 Page 267 \n'); //Example 5.3
+// Total Time t required for two step process
+
+//Operating Conditions
+
+ho = 40; //[W/m^2.K] Heat Convection coefficient
+hc = 10; //[W/m^2.K] Heat Convection coefficient
+k = 177; //[W/m.K] Thermal Conductivity
+e = .8; //Absorptivity
+L = 3*10^-3/2; //[m] Metre
+Ti = 25+273; //[K] Temp of Aluminium
+Tsurro = 175+273; //[K] Temp of duct wall heating
+Tsurrc = 25+273; //[K] Temp of duct wall
+Tit = 37+273; //[K] Temp at cooling
+Tc = 150+273; //[K] Temp critical
+
+stfncnstt=5.67*10^(-8); // [W/m^2.K^4] - Stefan Boltzmann Constant
+p = 2770; //[kg/m^3] density of aluminium
+c = 875; //[J/kg.K] Specific Heat
+
+//To assess the validity of the lumped capacitance approximation
+Bih = ho*L/k;
+Bic = hc*L/k;
+printf("\n Lumped capacitance approximation is valid as Bih = %f and Bic = %f", Bih, Bic);
+
+//Eqn 1.9
+hro = e*stfncnstt*(Tc+Tsurro)*(Tc^2+Tsurro^2);
+hrc = e*stfncnstt*(Tc+Tsurrc)*(Tc^2+Tsurrc^2);
+printf("\n Since The values of hro = %.1f and hrc = %.1f are comparable to those of ho and hc, respectively radiation effects must be considered", hro,hrc);
+
+// Integration of the differential equation
+// dy/dt=-1/(p*c*L)*[ho*(y-Tsurro)+e*stfncnstt*(y^4 - Tsurro^4)] , y(0)=Ti, and finds the minimum time t such that y(t)=150 degC
+deff("[ydot]=f1(t,y)","ydot=-1/(p*c*L)*[ho*(y-Tsurro)+e*stfncnstt*(y^4 - Tsurro^4)]");
+deff("[z]=g1(t,y)","z=y-150-273");
+y0=Ti;
+[y,tc]=ode("root",y0,0,150+273,f1,1,g1);
+te = tc(1) + 300;
+
+//From equation 5.15 and solving the two step process using integration
+function Tydot=f(t,T)
+ Tydot=-1/(p*c*L)*[ho*(T-Tsurro)+e*stfncnstt*(T^4 - Tsurro^4)];
+ funcprot(0)
+endfunction
+Ty0=Ti;
+t0=0;
+t=0:10:te;
+Ty=ode("rk",Ty0,t0,t,f);
+
+// solution of integration of the differential equation
+// dy/dt=-1/(p*c*L)*[hc*(y-Tsurrc)+e*stfncnstt*(y^4 - Tsurrc^4)] , y(rd(1))=Ty(43), and finds the minimum time t such that y(t)=37 degC=Tit
+deff("[Tdot]=f2(t,T)","Tdot=-1/(p*c*L)*[hc*(T-Tsurrc)+e*stfncnstt*(T^4 - Tsurrc^4)]");
+for(tt=0:1:900)
+ tq=ode(Ty(43),0,tt,f2);
+ if(tq-Tit<=10^-2)
+ break;
+ end
+end
+
+function Ty2dot=f2(t,T)
+ Ty2dot=-1/(p*c*L)*[hc*(T-Tsurrc)+e*stfncnstt*(T^4 - Tsurrc^4)];
+ funcprot(0)
+endfunction
+Ty20=Ty(43);
+t20=te;
+t2=te:10:1200;
+Ty2=ode("rk",Ty20,t20,t2,f2);
+clf();
+plot(t,Ty-273,t2,Ty2-273,[tc(1) tc(1)],[0 Tc-273],[te te],[0 Ty(43)-273],[tt+te tt+te],[0 tq-273]);
+xtitle('Plot of the Two-Step Process','t (s)','T (degC)');
+legend('Heating','Cooling','tc','te','tt');
+
+printf('\n\n Total time for the two-step process is t = %i s with intermediate times of tc = %i s and te = %i s.',tt+te,tc(1),te);
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.4/5_4_Radial_Two_Step.sce b/534/CH5/EX5.4/5_4_Radial_Two_Step.sce
new file mode 100644
index 000000000..64b3b4b7b
--- /dev/null
+++ b/534/CH5/EX5.4/5_4_Radial_Two_Step.sce
@@ -0,0 +1,40 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.4 Page 278 \n'); //Example 5.4
+// Radial System with Convection
+
+//Operating Conditions
+
+h = 500; //[W/m^2.K] Heat Convection coefficientat inner surface
+k = 63.9; //[W/m.K] Thermal Conductivity
+rho = 7832; //[kg/m^3] Density
+c = 434; //[J/kg.K] Specific Heat
+alpha = 18.8*10^-6; //[m^2/s]
+L = 40*10^-3; //[m] Metre
+Ti = -20+273; //[K] Initial Temp
+Tsurr = 60+273; //[K] Temp of oil
+t = 8*60 ; //[sec] time
+D = 1 ; //[m] Diameter of pipe
+
+//Using eqn 5.10 and 5.12
+Bi = h*L/k;
+Fo = alpha*t/L^2;
+
+//From Table 5.1 at this Bi
+C1 = 1.047;
+eta = 0.531;
+theta0=C1*exp(-eta^2*Fo);
+T = Tsurr+theta0*(Ti-Tsurr);
+
+//Using eqn 5.40b
+x=1;
+theta = theta0*cos(eta);
+Tl = Tsurr + (Ti-Tsurr)*theta;
+q = h*[Tl - Tsurr];
+
+//Using Eqn 5.44, 5.46 and Vol per unit length V = pi*D*L
+Q = [1-(sin(eta)/eta)*theta0]*rho*c*%pi*D*L*(Ti-Tsurr);
+
+printf("\n (a) After 8 min Biot number = %.2f and Fourier Numer = %.2f \n\n (b) Temperature of exterior pipe surface after 8 min = %i degC \n\n (c) Heat Flux to the wall at 8 min = %i W/m^2 \n\n (d) Energy transferred to pipe per unit length after 8 min = %.2e J/m",Bi,Fo, T-273,q,Q);
+
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.5/5_5_Sphere_Two_Step.sce b/534/CH5/EX5.5/5_5_Sphere_Two_Step.sce
new file mode 100644
index 000000000..f0fb0366d
--- /dev/null
+++ b/534/CH5/EX5.5/5_5_Sphere_Two_Step.sce
@@ -0,0 +1,34 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.5 Page 280 \n'); //Example 5.5
+// Two step cooling process of Sphere
+
+//Operating Conditions
+
+ha = 10; //[W/m^2.K] Heat Convection coefficientat air
+hw = 6000; //[W/m^2.K] Heat Convection coefficientat water
+k = 20; //[W/m.K] Thermal Conductivity
+rho = 3000; //[kg/m^3] Density
+c = 1000; //[J/kg.K] Specific Heat
+alpha = 6.66*10^-6; //[m^2/s]
+Tiw = 335+273; //[K] Initial Temp
+Tia = 400+273; //[K] Initial Temp
+Tsurr = 20+273; //[K] Temp of surrounding
+T = 50+273; //[K] Temp of center
+ro = .005; //[m] radius of sphere
+
+//Using eqn 5.10 and
+Lc = ro/3;
+Bi = ha*Lc/k;
+ta = rho*ro*c*2.30*(log10((Tia-Tsurr)/(Tiw-Tsurr)))/(3*ha);
+
+//From Table 5.1 at this Bi
+C1 = 1.367;
+eta = 1.8;
+Fo = -1*2.30*log10((T-Tsurr)/((Tiw-Tsurr)*C1))/eta^2;
+
+tw = Fo*ro^2/alpha;
+
+printf("\n (a) Time required to accomplish desired cooling in air ta = %.1f s\n\n (b) Time required to accomplish desired cooling in water bath tw = %.2f s",ta,tw);
+
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.6/5_6_Burial_Depth.sce b/534/CH5/EX5.6/5_6_Burial_Depth.sce
new file mode 100644
index 000000000..7cc458dd7
--- /dev/null
+++ b/534/CH5/EX5.6/5_6_Burial_Depth.sce
@@ -0,0 +1,22 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.6 Page 288 \n'); //Example 5.6
+// Burial Depth
+
+//Operating Conditions
+
+k = .52; //[W/m.K] Thermal Conductivity
+rho = 2050; //[kg/m^3] Density
+c = 1840; //[J/kg.K] Specific Heat
+Ti = 20+273; //[K] Initial Temp
+Ts = -15+273; //[K] Temp of surrounding
+T = 0+273; //[K] Temp at depth xm after 60 days
+t = 60*24*3600; //[sec] time perod
+
+alpha = k/(rho*c); //[m^2/s]
+//Using eqn 5.57
+xm = erfinv((T-Ts)/(Ti-Ts))*2*(alpha*t)^.5;
+
+printf("\n Depth at which after 60 days soil freeze = %.2f m",xm);
+
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.7/5_7_Spherical_Tumor.sce b/534/CH5/EX5.7/5_7_Spherical_Tumor.sce
new file mode 100644
index 000000000..3104f4932
--- /dev/null
+++ b/534/CH5/EX5.7/5_7_Spherical_Tumor.sce
@@ -0,0 +1,38 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.7 Page 293 \n'); //Example 5.7
+// Spherical Tumor
+
+//Operating Conditions
+
+k = .5; //[W/m.K] Thermal Conductivity Healthy Tissue
+kappa = .02*10^3; //[m] extinction coefficient
+p = .05; // reflectivity of skin
+D = .005; //[m] Laser beam Dia
+rho = 989.1 ; //[kg/m^3] Density
+c = 4180 ; //[J/kg.K] Specific Heat
+Tb = 37+273; //[K] Temp of healthy tissue
+Dt = .003 ; //[m] Dia of tissue
+d = .02 ; //[m] depth beneath the skin
+Ttss = 55+273 ; //[K] Steady State Temperature
+Tb = 37+273 ; //[K] Body Temperature
+Tt = 52+273 ; //[K] Tissue Temperature
+q = .170 ; //[W]
+
+//Case 12 of Table 4.1
+q = 2*%pi*k*Dt*(Ttss-Tb);
+
+//Energy Balancing
+P = q*(D^2)*exp(kappa*d)/((1-p)*Dt^2);
+
+//Using Eqn 5.14
+t = rho*(%pi*Dt^3/6)*c*(Tt-Tb)/q;
+
+alpha=k/(rho*c);
+Fo = 10.3;
+//Using Eqn 5.68
+t2 = Fo*Dt^2/(4*alpha);
+
+printf("\n (a) Heat transferred from the tumor to maintain its surface temperature at Ttss = 55 degC is %.2f W \n\n (b) Laser power needed to sustain the tumor surface temperautre at Ttss = 55 degC is %.2f W \n\n (c) Time for tumor to reach Tt = 52 degC when heat transfer to the surrounding tissue is neglected is %.2f sec \n\n (d) Time for tumor to reach Tt = 52 degC when Heat transfer to thesurrounding tissue is considered and teh thermal mass of tumor is neglected is %.2f sec" ,q,P,t,t2);
+
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.8/5_8_Nanostructured_Material.sce b/534/CH5/EX5.8/5_8_Nanostructured_Material.sce
new file mode 100644
index 000000000..71fc149ad
--- /dev/null
+++ b/534/CH5/EX5.8/5_8_Nanostructured_Material.sce
@@ -0,0 +1,33 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.8 Page 300 \n'); //Example 5.8
+// Thermal Conductivity of Nanostructured material
+
+//Operating Conditions
+
+k = 1.11 ; //[W/m.K] Thermal Conductivity
+rho = 3100; //[kg/m^3] Density
+c = 820 ; //[J/kg.K] Specific Heat
+//Dimensions of Strip
+w = 100*10^-6; //[m] Width
+L = .0035 ; //[m] Long
+d = 3000*10^-10; //[m] Thickness
+delq = 3.5*10^-3; //[W] heating Rate
+delT1 =1.37 ; //[K] Temperature 1
+f1 = 2*%pi ; //[rad/s] Frequency 1
+delT2 =.71 ; //[K] Temperature 2
+f2 = 200*%pi; //[rad/s] Frequency 2
+
+A = [delT1 -delq/(L*%pi);
+ delT2 -delq/(L*%pi)] ;
+
+C= [delq*-2.30*log10(f1/2)/(2*L*%pi);
+ delq*-2.30*log10(f2/2)/(2*L*%pi)] ;
+
+B = inv(A)*C;
+
+alpha = k/(rho*c);
+delp = [(alpha/f1)^.5 (alpha/f2)^.5];
+printf("\n C2 = %.2f k = %.2f W/m.K \n\n Thermal Penetration depths are %.2e m and %.2e m at frequency 2*pi rad/s and 200*pi rad/s" ,B(2),B(1), delp);
+
+//END \ No newline at end of file
diff --git a/534/CH5/EX5.9/5_9_Finite_Difference1.sce b/534/CH5/EX5.9/5_9_Finite_Difference1.sce
new file mode 100644
index 000000000..b527b3273
--- /dev/null
+++ b/534/CH5/EX5.9/5_9_Finite_Difference1.sce
@@ -0,0 +1,56 @@
+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 5.9 Page 305 \n'); //Example 5.9
+// Temperature distribution 1.5s after a change in operating power
+
+//Operating Conditions
+
+L = .01; //[m] Metre
+Tsurr = 250+273; //[K] Temperature
+h = 1100; //[W/m^2.K] Heat Convective Coefficient
+q1 = 10^7; //[W/m^3] Volumetric Rate
+q2 = 2*10^7; //[W/m^3] Volumetric Rate
+k = 30; //[W/m.K] Conductivity
+a = 5*10^-6; //[m^2/s]
+
+delx = L/5; //Space increment for numerical solution
+Bi = h*delx/k; //Biot Number
+//By using stability criterion for Fourier Number
+Fo = (2*(1+Bi))^-1;
+//By definition
+t = Fo*delx^2/a;
+printf('\n As per stability criterion delt = %.3f s, hence setting stability limit as .3 s.',t)
+// Using Finite time increment of .3s
+delt = 1*.3;
+Fo1 = a*delt/delx^2;
+x = [0 delx delx*2 delx*3 delx*4 delx*5];
+
+//At p=0 Using equation 3.46
+for i = 1: length(x)
+T(1,i) = q1*L^2/(2*k)*(1-x(i)^2/L^2)+Tsurr + q1*L/h -273 ;
+end
+//System of Equation in Finite Difference method
+for j = 2:6
+ T(j,1)=Fo1*(2*T(j-1,2)+q2*delx^2/k) + (1 -2*Fo1)*T(j-1,1);
+ T(j,2)=Fo1*(T(j-1,1)+T(j-1,3)+q2*delx^2/k) + (1 -2*Fo1)*T(j-1,2);
+ T(j,3)=Fo1*(T(j-1,2)+T(j-1,4)+q2*delx^2/k) + (1 -2*Fo1)*T(j-1,3);
+ T(j,4)=Fo1*(T(j-1,3)+T(j-1,5)+q2*delx^2/k) + (1 -2*Fo1)*T(j-1,4);
+ T(j,5)=Fo1*(T(j-1,4)+T(j-1,6)+q2*delx^2/k) + (1 -2*Fo1)*T(j-1,5);
+ T(j,6)=2*Fo1*(T(j-1,5)+Bi*(Tsurr-273)+q2*delx^2/(2*k)) + (1 -2*Fo1-2*Bi*Fo1)*T(j-1,6);
+end
+//At p=infinity Using equation 3.46
+x = [0 delx delx*2 delx*3 delx*4 delx*5];
+for i = 1:length(x)
+T(7,i) = q2*L^2/(2*k)*(1-x(i)^2/L^2)+Tsurr+q2*L/h-273;
+end
+
+for j= 1:6
+Tans(j,:) = [j-1 delt*(j-1) T(j,:)];
+end
+
+printf("\n\n Tabulated Nodal Temperatures \n\n p t(s) T0 T1 T2 T3 T4 T5\n");
+format('v',6);
+disp(Tans);
+printf(" inf inf %.1f %.1f %.1f %.1f %.1f %.1f",T(7,1),T(7,2),T(7,3),T(7,4),T(7,5),T(7,6));
+
+//END \ No newline at end of file
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