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9 files changed, 158 insertions, 0 deletions

diff --git a/617/CH3/EX3.1/Example3_1.sci b/617/CH3/EX3.1/Example3_1.sci
new file mode 100755
index 000000000..6a40fa31a
--- /dev/null
+++ b/617/CH3/EX3.1/Example3_1.sci
@@ -0,0 +1,12 @@
+clear;
+clc();
+
+// To find heat loss per square feet of wall surface per hour
+
+deltax=9/12;                 // thickness of wall in ft
+k=0.18;                // thermal conductivity of wall in B/hr-ft-degF
+t1=1500;                     // inside temperature of oven wall in degF
+t2=400;                      // outside temperature of oven wall in degF
+
+q=k*(t1-t2)/deltax;          // heat loss in Btu/hr
+printf("\n The heat loss for each square foot of wall surface is %d Btu/hr-ft^2",q);
\ No newline at end of file
diff --git a/617/CH3/EX3.10/Example3_10.sci b/617/CH3/EX3.10/Example3_10.sci
new file mode 100755
index 000000000..bdd0540cd
--- /dev/null
+++ b/617/CH3/EX3.10/Example3_10.sci
@@ -0,0 +1,19 @@
+clear;

+clc();

+

+// To find the total heat flow per foot of depth through the sction and the shape factor

+

+k=0.9;                        // thermal conductivity of section material in Btu/hr-ft-degF

+

+// Heat is considered to flow through fictitious rods and only half of the heat flows through symmetry axes    

+

+Ta=300;Tb=441;Tc=600;Td=300;Te=432;Tf=600;Tg=600;Th=600;Ti=300;Tj=384;Tk=461;Tl=485;Tm=490;Tn=300;To=340;Tp=372;Tq=387;Tr=391;Ts=300;Tt=300;Tu=300; Tv=300;Tw=300;

+// Above grid point temperatures are given in the question for the quarter section considered in degF(a,b,c...w are grid points)

+

+q1=4*k*((Tc-Tb)/2+(Tf-Te)+(Tf-Tk)+(Tg-Tl)+(Th-Tm)/2);                           // Amount of heat coming from inside in Btu/hr

+q2=4*k*((Tb-Ta)/2+(Te-Td)+(Tj-Ti)+(To-Tn)+(To-Tt)+(Tp-Tu)+(Tq-Tu)+(Tr-Tw)/2);   // Amount of heat going outside in Btu/hr

+q=(q1+q2)/2;                  // average of heat going in and heat coming out

+printf("\n Total heat flow per unit depth is %.1fBtu/hr",q);

+

+S=q/(k*(Tc-Ta));              // shape factor in ft

+printf("\n Shape factor is %.2fft",S)

diff --git a/617/CH3/EX3.2/Example3_2.sci b/617/CH3/EX3.2/Example3_2.sci
new file mode 100755
index 000000000..3673dcca4
--- /dev/null
+++ b/617/CH3/EX3.2/Example3_2.sci
@@ -0,0 +1,21 @@
+clear;

+clc();

+

+// To compute tempertures at the contact surfaces inside the furnaces

+

+x1=9/12;             // thickness of firebrick in ft

+k1=0.72;             // thermal conductivity of firebrick in Btu/hr-ft-degF

+x2=5/12;             // thickness of insulating brick in ft

+k2=0.08;             // thermal conductivity of insulating brick in Btu/hr-ft-degF

+x3=7.5/12;           // thickness of redbrick in ft

+k3=0.5;              // thermal conductivity of firebrick in Btu/hr-ft-degF

+t1=1500;             // inner temperature of wall in degF

+t2=150;              // outer temperature of wall in degF

+

+// resistances of mortar joints are neglected

+q=(t1-t2)/(x1/k1+x2/k2+x3/k3);   // heat flow per square ft in Btu/hr

+t2=t1-(q*x1/k1);                 // first contact temperature in degF

+printf("\n The temperature at the contact of firebrick and insulating brick is %d degF",t2);

+

+t3=t2-(q*x2/k2);                 // second contact temperature in degF

+printf("\n The temperature at the contact of insulating brick and red brick is %d degF",t3);
\ No newline at end of file
diff --git a/617/CH3/EX3.3/Example3_3.sci b/617/CH3/EX3.3/Example3_3.sci
new file mode 100755
index 000000000..6c40083d0
--- /dev/null
+++ b/617/CH3/EX3.3/Example3_3.sci
@@ -0,0 +1,19 @@
+clear;

+clc();

+

+ // to calculate the heat loss from pipe

+ 

+ d1=2.375/12;                   // internal diameter of pipe in ft

+ t=1/12;                        // thickness of insulating material in ft

+ d2=d1+2*t;                     // external (insulation)diameter of pipe in ft

+ k=0.0375;                      // thermal conductivity of insulating material in Btu/hr-ft-F

+ l=30;                          // length of pipe in ft

+ t1=380;                        // inner surface temperature of insulation

+ t2=80;                         // outer surface temperature of insulation

+ 

+ q=2*%pi*k*(t1-t2)/log(d2/d1);  // heat loss per unit length

+ printf("\n Heat loss per linear foot is %.d Btu/hr",q)

+ 

+ qtot=round(q)*l;               // heat loss for 30 ft pipe

+ printf("\n Total heat loss through 30 ft of pipe is %d Btu/hr",qtot)

+ 
\ No newline at end of file
diff --git a/617/CH3/EX3.4/Example3_4.sci b/617/CH3/EX3.4/Example3_4.sci
new file mode 100755
index 000000000..4577fdada
--- /dev/null
+++ b/617/CH3/EX3.4/Example3_4.sci
@@ -0,0 +1,17 @@
+clear;

+clc();

+

+// To calculate heat loss from pipe

+

+d1=10.75/12;            // outer diameter of pipe in ft

+x1=1.5/12;              // thickness of insulation 1 in ft

+x2=2/12;                // thickness of insulation 2 in ft

+d2= d1+2*x1;            // diameter of insulation 1 in ft

+d3=d2+2*x2;             // diameter of insulation 1 in ft

+t1=700;                 // inner surface temperature of composite insulation in degF

+t2=110;                 // outer surface temperature of composite insulation in degF

+k1=0.05;                //thermal conductivity of material 1 in Btu/hr-ft-degF

+k2=0.039;               // thermal conductivity of material 2 in Btu/hr-ft-degF

+

+q=2*%pi*(t1-t2)/(log(d2/d1)/k1+log(d3/d2)/k2);                                     // heat loss per linear foot in Btu/hr

+printf("\n The heat loss is found to be %d Btu/hr-ft", q);
\ No newline at end of file
diff --git a/617/CH3/EX3.5/Example3_5.sci b/617/CH3/EX3.5/Example3_5.sci
new file mode 100755
index 000000000..0805ef320
--- /dev/null
+++ b/617/CH3/EX3.5/Example3_5.sci
@@ -0,0 +1,13 @@
+clear;

+clc();

+

+// To find out heat loss through 1 sq. ft of flat slab of 85%magnesia and 15% asbestos 

+

+km=0.0377;               // Mean thermal conductivity at 220degF

+t1=260;                  // Inner surface temperature of slab in degF 

+t2=180;                  // Outer surface temperature of slab in degF

+A=1;                     // Area of slab in ft

+x=2/12;                  // Thickness of insulation in ft

+

+q=km*A*(t1-t2)/x;        // Heat loss through slab in Btu/hr

+printf("\n Heat loss through flat slab is %.1f Btu/hr",q);

diff --git a/617/CH3/EX3.6/Example3_6.sci b/617/CH3/EX3.6/Example3_6.sci
new file mode 100755
index 000000000..f225ce8c2
--- /dev/null
+++ b/617/CH3/EX3.6/Example3_6.sci
@@ -0,0 +1,23 @@
+clear all

+clc()

+

+// To find out heat loss through conduction through a furnace

+k=0.8                     // Avg. thermal conductivity in Btu/hr-ft-degF

+T1=400                    // Inner surface temperature of furnace in degF 

+T2=100                    // Outer surface temperature of furnace in degF

+a=3                       // Length of furnace in ft

+b=4                       // Breadth of furnace in ft

+c=2.5                     // Height of furnace in ft

+Aa=2*a*b                  // Area of surface A in ft^2

+Ab=2*b*c                  // Area of surface A in ft^2

+Ac=2*a*c                  // Area of surface A in ft^2

+x=4.5/12                  // Thickness of insulation in ft

+t=24                      // Time elapsed in hr

+M=4                       // Number of edges

+N=8                       // Number of corners

+

+S=Aa/x+Ab/x+Ac/x+0.54*(a+b+c)*M+0.15*x*N       // Shape factor

+qo=S*k*(T1-T2)                                 // Heat flow per hour

+q=qo*t                                         // Heat loss in 24 hr

+

+printf("The heat loss in 24 hr is %d Btu",q)
\ No newline at end of file
diff --git a/617/CH3/EX3.7/Example3_7.sci b/617/CH3/EX3.7/Example3_7.sci
new file mode 100755
index 000000000..851fd9565
--- /dev/null
+++ b/617/CH3/EX3.7/Example3_7.sci
@@ -0,0 +1,13 @@
+clear;

+clc();

+

+// To compute shape factor  for the special section in figure

+

+// Ratio of diameter of circle to the side of square is 0.5. Hence required lines have been estabilished by trial and error method.

+

+M=8*9;                  // number of flow channels for the entire section

+N=8.37;                 // number of equal channel intervals

+// the fractional part arises due to the fractional part of temperature close to border EG

+

+k = M/N;               // Ratio of shape factor to wall length

+printf("\n Shape factor for the special section (where the ratio of radius of circle to half side length is 0.5),S is %.2fL", k );

diff --git a/617/CH3/EX3.8/Example3_8.sci b/617/CH3/EX3.8/Example3_8.sci
new file mode 100755
index 000000000..931c74cf1
--- /dev/null
+++ b/617/CH3/EX3.8/Example3_8.sci
@@ -0,0 +1,21 @@
+clear;

+clc();

+

+// To find the temperature of planes indicated by grid points using relaxation method

+t1=800;            // inner surface temperature of wall in degF

+t4=200;           //  outer surface temperature of wall in degF

+

+//Grids are square in shape so delx =dely where delx,y sre dimensions of square grid 

+

+t2=[700 550 550 587.5 587.5 596.9 596.9 599.3 599.3 599.8];             // Assumed temperature of grid point 1

+t3=[300 300 375 375 393.8 393.8 398.5 398.5 399.6 399.6];               // Assumed temperature of grid point 2

+ 

+for i=1:9

+     th2(i)=t1+t3(i)-2*t2(i);; // th1= q/kz at grid pt1

+     th3(i)=t2(i)+t4-2*t3(i);// th2= q/kz at grid pt2

+     printf("\n Assuming t2=%.1f  degF and t2=%.1f degF \n th1[%d]=%.1f degF and th2[%d]=%.1f degF \n",t2(i),t3(i),i,th2(i),i,th3(i));     

+     printf(" Since th2[%d] is not equal to th3[%d], hence other values of t2 and t3 are to be assumed\n",i,i);

+end

+

+printf("\nAssuming t2=600 degF and t3=400 degF, th2=th3.");

+printf("\nHence Steady state condition is satisfied at grid temperatures of 400 degF and 600 degF");