diff options
author | priyanka | 2015-06-24 15:03:17 +0530 |
---|---|---|
committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /617/CH3 | |
download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip |
initial commit / add all books
Diffstat (limited to '617/CH3')
-rwxr-xr-x | 617/CH3/EX3.1/Example3_1.sci | 12 | ||||
-rwxr-xr-x | 617/CH3/EX3.10/Example3_10.sci | 19 | ||||
-rwxr-xr-x | 617/CH3/EX3.2/Example3_2.sci | 21 | ||||
-rwxr-xr-x | 617/CH3/EX3.3/Example3_3.sci | 19 | ||||
-rwxr-xr-x | 617/CH3/EX3.4/Example3_4.sci | 17 | ||||
-rwxr-xr-x | 617/CH3/EX3.5/Example3_5.sci | 13 | ||||
-rwxr-xr-x | 617/CH3/EX3.6/Example3_6.sci | 23 | ||||
-rwxr-xr-x | 617/CH3/EX3.7/Example3_7.sci | 13 | ||||
-rwxr-xr-x | 617/CH3/EX3.8/Example3_8.sci | 21 |
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");
|