initial commit / add all books

7 files changed, 281 insertions, 0 deletions

diff --git a/1910/CH11/EX11.10/Chapter1110.sce b/1910/CH11/EX11.10/Chapter1110.sce
new file mode 100755
index 000000000..08e28ea04
--- /dev/null
+++ b/1910/CH11/EX11.10/Chapter1110.sce
@@ -0,0 +1,60 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 10")

+//Given a furnace which can be approximated as an equuilateral triangle duct

+//The hot wall is maintained at temprature (T1)=1000K and has emmisivity(emi1)=0.75

+//The cold wall is at temprature(T2)=350K and has emmisivity(emi2)=0.7

+T1=1000;

+T2=350;

+emi1=0.75;

+emi2=0.7;

+//Stefan-Boltzman constant(sigma)=5.67*10^-8 W/(m^2*K^4)

+sigma=5.67*10^-8;

+//The third wall is reradiating zone having Q3=0

+//The radiation flux leaving the hot wall is Q/A=[sigma*(T1^4-T2^4)]/(A*R)

+//By summation rule F33+F31+F32=1

+//F33=0(in consideration of surface to be plane)

+//From symmetry F31=F32

+F31=0.5;//View factors

+F32=F31;//View factors

+F33=0;//View factors

+//From reciprocity theorem F13=F31 and F23=F32=0.5 (since A1=A2=A3=A)

+F13=F31;//View factors

+F23=F32;//View factors

+//Again F11+F12+F13=1 from summation rule

+F11=0;//View factors

+F12=1-F13-F11;//View factors

+//R1,R2,R12,R13,R23 are the resistances

+//R is equivalent resistance of thermal network is given by R1+[(1/R12)+(1/(R13+R23))]^-1+R2

+R1=(1-emi1)/(emi1);

+R2=(1-emi2)/(emi2);

+R12=1/(F12);

+R13=1/(F13);

+R23=1/(F23);

+//R is equivalent resistance of thermal network 

+disp("Equivalent resistance of thermal network (R) is given by R1+[(1/R12)+(1/(R13+R23))]^-1+R2")

+R=R1+[(1/R12)+(1/(R13+R23))]^-1+R2

+//The radiation flux leaving the hot wall is Q/A.

+disp("The radiation flux leaving the hot wall is Q/A=[sigma*(T1^4-T2^4)]/(A*R) in W/m^2")

+//Since A gets cancelled in the factor (A*R)

+//So Q/A=[sigma*(T1^4-T2^4)]/(R)

+//Let Q/A=H

+H=[sigma*(T1^4-T2^4)]/(R)

+

+

+

+

+

+

+

+

+

+

+

+

+

+

diff --git a/1910/CH11/EX11.3/Chapter113.sce b/1910/CH11/EX11.3/Chapter113.sce
new file mode 100755
index 000000000..ad2c4a0be
--- /dev/null
+++ b/1910/CH11/EX11.3/Chapter113.sce
@@ -0,0 +1,21 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 3")

+disp("The view factors F13 and F31 between the surfaces 1 and 3 are ")

+//Determine the view factors F13 and F31 between the surfaces 1 and 3.

+//F1-2,3=F12+F13

+//So F13=F1-2,3-F12

+//Let F1-2,3=F123

+//From Radiation Shape factor b/w two perpendicular rectangles with a commom edge table we get F12=.027,F1-2,3=0.31

+F123=0.31;//View factor

+F12=.27;//View factor

+F13=F123-F12//View factor

+//A1,A2 and A3 are the emitting surface areas

+//From reciprocity relation F31=(A1/A3)/F13

+A1=2;

+A3=2.5;

+F31=(A1/A3)*F13

diff --git a/1910/CH11/EX11.4/Chapter114.sce b/1910/CH11/EX11.4/Chapter114.sce
new file mode 100755
index 000000000..2cb3b56b0
--- /dev/null
+++ b/1910/CH11/EX11.4/Chapter114.sce
@@ -0,0 +1,31 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 4")

+//Determine the view factors F14 for the composite surface .

+//From the table of radiation shape factor b/w two perpendicular surfaces F1,2-3,4=0.14 and F1,2-3=0.1

+//By subdivision of the recieving surfaces we get F1,2-4=F1,2-3,4-F1,2-3

+//Let F1,2-4=F124 , F1,2-3,4=F1234 , F1,2-3=F123

+F1234=0.14;//View factor

+F123=0.1;//View factor

+F124=F1234-F123;//View factor

+//Again from the table of radiation shape factor b/w two perpendicular surfaces F2-3,4=0.24 , F23=0.18

+//Let F2-3,4=F234

+F234=0.24;//View factor

+F23=0.18;//View factor

+//By subdivision of the recieving surfaces we get F24=F2-3,4-F23

+F24=F234-F23;//view factor

+//A1 and A2 are the emitting surface areas.

+A1=12;

+A2=12;

+//Now by subdivision of emitting surfaces F1,2-4=(1/(A1+A2))*(A1*F14+A2*F24)

+//This implies F14=((F1,2-4*(A1+A2)))-A2*F24)/A2

+disp("The view factor F14=((F1,2-4*(A1+A2)))-A2*F24)/A2")

+F14=((F124*(A1+A2))-(A2*F24))/A2

+

+

+

+ 

diff --git a/1910/CH11/EX11.5/Chapter115.sce b/1910/CH11/EX11.5/Chapter115.sce
new file mode 100755
index 000000000..e4a80e3ce
--- /dev/null
+++ b/1910/CH11/EX11.5/Chapter115.sce
@@ -0,0 +1,21 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 5")

+//Consider a cylinder having length,L=2r determine the view factor of cylindrical surface with respect to the base.

+//From the graph of radiation shape factor b/w parallel coaxial disks of equal diameter F12=0.16

+F12=0.16;//View factor

+//By the summation rule of an enclosure F11+F12+F13=1

+//But F11=0(since the base surface is flat)

+F11=0;//View factor

+disp ("The view factors of cylindrical surface with respect to the base are")

+F13=1-F12-F11//view factor

+//By making use of reciprocity theorem we have F31=(A1/A2)*F13

+//A1 and A2 are emitting surface areas

+//A1/A2=(pi*r^2)/(2*pi*r*2*r)=1/4

+//Let A1/A2=A

+A=1/4;

+F31=(A)*F13

diff --git a/1910/CH11/EX11.6/Chapter116.sce b/1910/CH11/EX11.6/Chapter116.sce
new file mode 100755
index 000000000..7e337f9cf
--- /dev/null
+++ b/1910/CH11/EX11.6/Chapter116.sce
@@ -0,0 +1,63 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 6")

+//Two rectangles length,L=1.5m by breadth,B=3.0m are parallel and directly opposed.

+L=1.5;

+B=3;

+//They are 3m apart

+//Temprature(T1) of surface 1 is 127°C or 400K and temprature(T2) of surface 2 is 327°C or 600K 

+T1=400;

+T2=600;

+//Area (A) is the product of L and B

+A1=L*B;

+//Stefan -Boltzman constant(sigma)=5.67*10^-8 W/(m^2*K^4)

+sigma=5.67*10^-8;

+//From the graph of radiation shape factor b/w parallel rectangles F12=0.11

+F12=0.11;//View factor

+//The rate of heat transfer is given by Q=A1*F12*sigma*(T1^4-T2^4)

+disp("The rate of heat transfer is given by Q=A1*F12*sigma*(T1^4-T2^4) in W")

+Q=A1*F12*sigma*(T1^4-T2^4)

+disp("Here minus sign indicates that the net heat transfer is from surface2 to surface1")

+//Surface1 recieves energy only from surface 2,since the surrounding is at 0K.

+//Therefore Q1=A1*Eb1-A2*F21*Eb2

+//This implies Q1 can also be written as A1*sigma*(T1^4-F12*T2^4)

+//From reciprocity theorem F21=F12 (since A1=A2)

+F21=F12;//view factor

+disp("The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 0K in W" )

+Q1=A1*sigma*(T1^4-F12*T2^4)

+//In the case when surrounding is at temprature, Ts=300K ,the energy recieved from the surrounding by the surface 1 has to be considered.

+Ts=300;

+//Applying summation rule of view factors F11+F12+F1s=1

+F11=0;//view factor

+disp("The view factor of surface 1 with respect to surrounding is")

+F1s=1-F11-F12

+//subscript s denotes the surroundings

+//Q1=A1*Eb1-A2*F21*Eb2-As*Fs1*Ebs

+//With the help of reciprocity theorem A2*F21=A1*F12 , As*Fs1=A1*F1s

+//Therefore we can write Q1=A1*sigma*(T1^4-F12*T2^4-F1s*Ts^4)

+disp("The net rate of energy loss from the surface at 127°C if the surrounding other than the two surfaces act as black body at 300K in W ")

+Q1=A1*sigma*(T1^4-F12*T2^4-F1s*Ts^4)

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

diff --git a/1910/CH11/EX11.7/Chapter117.sce b/1910/CH11/EX11.7/Chapter117.sce
new file mode 100755
index 000000000..4af4f58cf
--- /dev/null
+++ b/1910/CH11/EX11.7/Chapter117.sce
@@ -0,0 +1,34 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 7")

+//Two parallel infinite surafces are maintained at tempratures T2=200°C or 473.15K and T1=300°C or 573.15K

+T1=573.15;

+T2=473.15;

+//The emissivity(emi) is 0.7 for both the surfaces which are gray.

+emi1=0.7;

+emi2=0.7;

+//stefan=boltzman constant(sigma)=5.67*10^-8W/(m^2*K^4)

+sigma=5.67*10^-8;

+//The net rate of heat transfer per unit area is given Q/A=(sigma*(T1^4-T2^4))/[(1/Ɛ1)+(1/Ɛ2)-1]

+//Let Q/A=H

+disp("The net rate of heat transfer per unit area is given Q/A=(sigma*(T1^4-T2^4))/[(1/Ɛ1)+(1/Ɛ2)-1] in W")

+H=(sigma*(T1^4-T2^4))/[(1/emi1)+(1/emi2)-1]

+//When the two surfaces are black

+//This implies emiisivity(emi)=1 for both surfaces

+//So,The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1^4-T2^4)

+disp("The net rate of heat transfer when the two surfaces are black is Q/A=sigma*(T1^4-T2^4) in W")

+H=sigma*(T1^4-T2^4)

+

+

+

+

+

+

+

+

+

+

diff --git a/1910/CH11/EX11.8/Chapter118.sce b/1910/CH11/EX11.8/Chapter118.sce
new file mode 100755
index 000000000..8f83095c5
--- /dev/null
+++ b/1910/CH11/EX11.8/Chapter118.sce
@@ -0,0 +1,51 @@
+// Display mode

+mode(0);

+// Display warning for floating point exception

+ieee(1);

+clear;

+clc;

+disp("Introduction to heat transfer by S.K.Som, Chapter 11, Example 8")

+//Two concentric spheres of diameters D1=0.5m and D2=1m are separated by an air space.

+//The surface tempratures are T1=400K and T2=300K

+T1=400;

+T2=300;

+D1=0.5;

+D2=1;

+//A1 and A2 are the areas in m^2 of surface 1 and surface 2 respectively

+A1=(%pi*D1^2);

+A2=(%pi*D2^2);

+//Stefan-Boltzman constant(sigma)=5.67*10^-8 W/(m^2*K^4)

+sigma=5.67*10^-8;

+//The emissivity is represented by emi 

+//The radiation heat exchange in case of two concentric sphere is given by Q=[A1*sigma*(T1^4-T2^4)]/[(1/emi1)+(A1/A2)*(1/emi2-1)] 

+//When the spheres are black emi1=emi2=1

+emi1=1;

+emi2=1;

+//Hence Q=A1*sigma*(T1^4-T2^4)

+disp("The net rate of heat exchange between the spheres when the surfaces are black is Q=A1*sigma*(T1^4-T2^4) in W ")

+Q=A1*sigma*(T1^4-T2^4)

+//The net rate of radiation exchange when one surface is gray and other is diffuse having emi1=0.5 and emi2=0.5  

+emi1=0.5;

+emi2=0.5;

+disp("The net rate of radiation exchange when one surface is gray and other is diffuse is given by Q1=[A1*sigma*(T1^4-T2^4)]/[(1/emi1)+(A1/A2)*(1/emi2-1)] in W")  

+Q1=[A1*sigma*(T1^4-T2^4)]/[(1/emi1)+(A1/A2)*(1/emi2-1)]

+//The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1)

+emi2=1;//emissivity of outer surface

+disp("The net rate of radiation exchange when outer surface is assumed to be black body i;e(emi2=1) in W")

+Q2=[A1*sigma*(T1^4-T2^4)]/[(1/emi1)+(A1/A2)*(1/emi2-1)]

+disp("Error(E) is given By [(Q2-Q1)/Q1]*100 in percentage")

+E=[(Q2-Q1)/Q1]*100

+

+

+

+

+

+

+

+

+

+

+

+

+

+