10 files changed, 245 insertions, 0 deletions
diff --git a/1910/CH1/EX1.1/Chapter11.sce b/1910/CH1/EX1.1/Chapter11.sce
new file mode 100755
index 000000000..85e657cd2
--- /dev/null
+++ b/1910/CH1/EX1.1/Chapter11.sce
@@ -0,0 +1,17 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 1")
+//The temprature of two faces of the slabs are T1=40°C & T2=20°C 
+//The thickness of the slab(L) is 80mm or .08m
+//The thermal conductivity(k)of the material is .20 W/(m*K)
+T1=40;
+T2=20;
+L=.08;
+k=.20;
+//The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L
+disp ("The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m^2 ")
+q=k*(T1-T2)/L
diff --git a/1910/CH1/EX1.10/Chapter110.sce b/1910/CH1/EX1.10/Chapter110.sce
new file mode 100755
index 000000000..ff7a2511e
--- /dev/null
+++ b/1910/CH1/EX1.10/Chapter110.sce
@@ -0,0 +1,49 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 10")
+//The spacecraft panel has thickness(L)=.01 m
+//The spacecraft has inner temprature (Ti)=298 K
+//The spacecraft has outer temprature(T2)
+//The panel is exposed to deep space where temprature(To)= 0K
+//The material has Thermal conductivity(k)= 5.0 W/(m*K)
+//The emissivity(emi)=0.8
+//The inner surface of the panel is exposed to airflow resulting in an average heat transfer coefficient(hbri)=70 W/(m^2*K)
+L=.01;
+Ti=298;
+To=0;
+k=5;
+emi=0.8;
+hbri=70;
+//The stefan Boltzman constant(sigma)= 5.67*10^-8 W/(m^2/K^4)
+sigma=5.67*10^-8;
+//Heat transfer from the outer surface takes place only by radiation is given by  Q/A=emi*sigma*(T2^4-T0^4)in W/m^2=F1
+//heat transfer from the outer surface can also be written as Q/A=(Ti-To)/((1/hbri)+(L/k)+(1/hr))=F2
+//Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)
+//so hr=4.536*10^-8*T2^3
+disp("Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2^4-T0^4)in W/m^2 for different values of tempratures in K")
+disp("heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m^2 at different tempratures in K" )
+disp("The values of temprature that are considered are <298 K")
+for (i=285:292)
+    T2=i,hr=4.536*10^-8*T2^3; F1=emi*sigma*(T2^4-To^4),F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))
+end
+if F1==F2 then T2=i
+    else T2=292.5,hr=4.536*10^-8*T2^3; F1=emi*sigma*(T2^4-To^4),F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))
+end
+disp("Satisfactory solutions for Temprature in K is")
+disp(Temprature = T2)
+disp("Approximate Rate of Heat Transfer in W/m^2 is")
+disp(332)
+
+
+
+
+
+
+
+
+
+
diff --git a/1910/CH1/EX1.11/Chapter111.sce b/1910/CH1/EX1.11/Chapter111.sce
new file mode 100755
index 000000000..f26bc4e83
--- /dev/null
+++ b/1910/CH1/EX1.11/Chapter111.sce
@@ -0,0 +1,27 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 11")
+//The horizontal steel pipe has outer diameter(D)=80 mm or.08 m
+//The pipe is maintained at a temprature(T1)=60°C where the air and wall temprature(T2)=20 °C 
+//The average free convective heat transfer coefficient(hbr)=6.5 W/(m^2/K) b/w the outer surface of the pipe and air
+D=.08;
+T1=60;
+T2=20;
+hbr=6.5;
+//Length(L=1) since per unit length is considered
+L=1;
+//The surface area of pipe is given by A=(%pi*D*L)
+A=(%pi*D*L);
+//The surface emissivity(emi) of steel = 0.8
+//The stefan -Boltzman constant(sigma)= 5.7*10^-8 W/(m^2*K^4)
+sigma=5.67*10^-8;
+emi=.8;
+//The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1^4-T2^4)
+disp("The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1^4-T2^4) in W/m")
+//Let Q/L=F
+F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)^4-(T2+273.15)^4)
+
diff --git a/1910/CH1/EX1.2/Chapter12.sce b/1910/CH1/EX1.2/Chapter12.sce
new file mode 100755
index 000000000..6f7eef26e
--- /dev/null
+++ b/1910/CH1/EX1.2/Chapter12.sce
@@ -0,0 +1,20 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 2")
+//The thermal conductivity(km)of masonry wall is .8 W/(mK)
+//The thermal conductivity(kc)of composite wall is .2 W/(mK)
+//The thickness of composite wall(Lc) is 100 mm or .1 m
+km=.8;
+kc=.2;
+Lc=.1;
+//The thickness of masonry wall(Lm) is to be found. 
+//The steady state heat flow(qm)through masonry wall is km(T1-T2)/L
+// The steady state heat flow(qc)through composite wall is kc(T1-T2)/L
+//As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)
+//The thickness of masonry wall is Lm.
+disp ("The thickness of masonry wall is Lm in m")
+Lm=(km/kc)*(Lc/(0.8))
diff --git a/1910/CH1/EX1.4/Chapter14.sce b/1910/CH1/EX1.4/Chapter14.sce
new file mode 100755
index 000000000..ababfe72d
--- /dev/null
+++ b/1910/CH1/EX1.4/Chapter14.sce
@@ -0,0 +1,16 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 4")
+//The average forced convective heat transfer coefficient(hbr) is 200 W/( m^2 °C)
+//The fluid temprature(Tinf) upstream of the cold surface is 100°C
+//The surface temprature(Ts) is 20°C
+hbr=200;
+Tinf=100;
+Ts=20;
+//The rate of heat transfer per unit area is q
+disp ("The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m^2")
+q=hbr*(Tinf-Ts)
diff --git a/1910/CH1/EX1.5/Chapter15.sce b/1910/CH1/EX1.5/Chapter15.sce
new file mode 100755
index 000000000..9e4910a43
--- /dev/null
+++ b/1910/CH1/EX1.5/Chapter15.sce
@@ -0,0 +1,16 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 5")
+//The average heat transfer coefficient(hbr) is 800 W/(m^2°C)
+//The surface temprature of heat exchanger is 75°C and air temprature is 25°C so deltaT=(75-25)
+//The amount of heat exchanged(Q) is 20 MJ/h
+//The heat exchanger surface area(A) is given by A=Q/(hbr*∆T)
+hbr=800;
+deltaT=(75-25);
+Q=20;
+disp("The heat exchanger surface area(A)in m^2 required for 20 MJ/h of heating is ")
+A = (Q*10^6)/(3600*hbr*deltaT)
diff --git a/1910/CH1/EX1.6/Chapter16.sce b/1910/CH1/EX1.6/Chapter16.sce
new file mode 100755
index 000000000..953c5daaf
--- /dev/null
+++ b/1910/CH1/EX1.6/Chapter16.sce
@@ -0,0 +1,39 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 6")
+//The temprature of the plate(Ts) is 225°C
+//The ambient temprature (Tinf) is 25°C
+//The change in plate temprature with time is dT/dt=-.02K/s
+//The plate area (A)=.1m^2 , mass(m)= 4Kg and specific heat(cp)=2.8KJ/(Kg*K)
+//The average free convective heat coefficient(hbr) is to be found
+Ts=225;
+Tinf=25;
+//|dT/dt|=0.2,because it is modulus function and it converts negative values to positive value.
+//Let |dT/dt|=X
+X=0.02;
+A=.1;
+m=4;
+cp=2.8;
+disp("The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)")
+disp("The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.")
+disp("Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m^2°C)")
+hbr=(m*cp*10^3*X)/(A*(Ts-Tinf))
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1910/CH1/EX1.7/Chapter17.sce b/1910/CH1/EX1.7/Chapter17.sce
new file mode 100755
index 000000000..9c5862730
--- /dev/null
+++ b/1910/CH1/EX1.7/Chapter17.sce
@@ -0,0 +1,17 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 7")
+//The temprature(T) of brick wall after sunset is 50°C
+//The emissity value(emi)=0.9
+//The radiant heat flux per square meter =E/A Where E is radiant heat energy and A is area of brick wall.
+//The stefan-Boltzman constant(sigma)=5.6697*10^-8 W/(m^2*K^4).
+T=50;
+emi=.9;
+sigma=5.6697*10^-8;
+disp("The heat flux per square meter is given by E/A=emi*sigma*T^4 in W/m^2")
+//Let E/A=F
+F=emi*sigma*(T+273.15)^4
diff --git a/1910/CH1/EX1.8/Chapter18.sce b/1910/CH1/EX1.8/Chapter18.sce
new file mode 100755
index 000000000..5c4c7db64
--- /dev/null
+++ b/1910/CH1/EX1.8/Chapter18.sce
@@ -0,0 +1,15 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 8")
+//The temprature(T) of asphalt pavement = 50°C
+//The stefan-Boltzman constant(sigma)=5.6697*10^-8 W/(m^2*K^4).
+T=50;
+sigma=5.6697*10^-8;
+//The emitted radiant energy per unit surface area is given by (Eb/A)=sigma*T^4
+disp ("The emitted radiant energy per unit surface area is given by Eb/A=sigma*T^4 in W/m^2")
+//Let Eb/A=F
+F=sigma*(50+273.15)^4
diff --git a/1910/CH1/EX1.9/Chapter19.sce b/1910/CH1/EX1.9/Chapter19.sce
new file mode 100755
index 000000000..761767c11
--- /dev/null
+++ b/1910/CH1/EX1.9/Chapter19.sce
@@ -0,0 +1,29 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 1, Example 9")
+//The Thickness(L) of wall= 150 mm or 0.15 m.
+//The wall on one side is exposed to air at temprature(Ta)= 60°C and on the other side to air at temprature(Tb) = 20°C
+//The average convective heat transfer coefficients are hbr1=40 W/(m^2°C) on the 60°C and hbr2= 10 W/(m^2°C) on 20°C side.
+//The thermal conductivity(k)=.8 W/(m°C)
+L=0.15;
+Ta=60;
+Tb=20;
+hbr1=40;
+hbr2=10;
+k=0.8;
+//Area(A=1 m^2 )since unit surface area is required.
+A=1;
+//The rate of heat transfer per unit surface area of wall is given by (Q/A)=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))
+disp("The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m^2")
+//Let Q/A=F
+F=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))
+//The surface tempratures of wall on 60°C side is T1 and on 20°C side is T2
+disp("The surface tempratures of wall on 60°C side is T1 =Ta-(Q/(A*hbr1)) in °C")
+T1 =Ta-(F/hbr1)
+disp("The surface tempratures of wall on 20°C side is T2 =Tb+(Q/(A*hbr2)) in °C")
+T2 =Tb+(F/hbr2)
+