11 files changed, 585 insertions, 0 deletions
diff --git a/1910/CH2/EX2.1/Chapter21.sce b/1910/CH2/EX2.1/Chapter21.sce
new file mode 100755
index 000000000..b6ef7a2ae
--- /dev/null
+++ b/1910/CH2/EX2.1/Chapter21.sce
@@ -0,0 +1,83 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 1")
+//The length of steel sheet (Ls)=1.5 mm and thermal conductivity (ks)=25 W/(mK) at the outer surface.
+Ls=1.5;
+ks=25;
+//The length of plywood (Lp)=10 mm and thermal conductivity (kp)= .05 W/(mK) at the inner surface.
+Lp=10;
+kp=.05;
+//The length of glass wool (Lg)=20 mm and thermal conductivity (kg)= .01W/(mK) in between steel sheet and plywood.
+Lg=20;
+kg=.01;
+//The temprature of van inside cold Enviroment is (Ti)= -15°C  while the outside surface is exposed to a surrounding ambient temprature (To)=24°C 
+To=24;
+Ti=-15;
+//The average value of heat transfer coefficients at the inner and outside surfaces of the wall are hi=12 W/(m^2*K) and ho= 20 W/(m^2*K) 
+hi=12;
+ho=20;
+//The surface area of wall (A)= .75 m^2 
+A=.75;
+//The convective resistance is Ro= 1/(ho*A) at the outer surface
+disp("The convective resistance Ro= 1/(ho*A) at the outer surface in KW^-1 is")
+Ro=1/(ho*A)
+//The conduction resistance is Rs= Ls/(ks*A) of steel sheet
+disp("The conduction resistance Rs= Ls/(ks*A) of steel sheet in KW^-1 is")
+Rs=Ls*10^-3/(ks*A)
+//The conduction resistance is Rg= Lg/(kg*A) of glass wool
+disp("The conduction resistance Rg= Lg/(kg*A) of glass wool in KW^-1 is")
+Rg= Lg*10^-3/(kg*A)
+//The conduction resistance is Rp= Lp/(kp*A) of plywood
+disp("The conduction resistance Rp= Lp/(kp*A) of plywood in KW^-1 is")
+Rp= Lp*10^-3/(kp*A)
+//The convective resistance is Ri= 1/(hi*A) at the outer surface
+disp("The convective resistance Ri= 1/(hi*A) at the outer surface in KW^-1 is")
+Ri= 1/(hi*A)
+//The rate of heat flow is Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)
+disp("The rate of heat flow Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri) in W is")
+Q=(To-Ti)/(Ro+Rs+Rg+Rp+Ri)
+//The tempraure at the outer surface of wall is T1.
+//The temprature at the interface b/w steel sheet and glass wool is T2.
+//The temprature at the interface b/w glass wool and plywood is T3.
+//The tempraure at the inner surface of wall is T4.
+disp("The temprature at the outer surface of wall is T1=To-(Q*Ro) in °C")
+T1=To-(Q*Ro)
+disp("The temprature at the interface b/w steel sheet and glass wool is T2=T1-(Q*Rs) in°C")
+T2=T1-(Q*Rs)
+disp("The temprature at the interface b/w glass wool and plywood is T3=T2-(Q*Rg) in°C")
+T3=T2-(Q*Rg)
+disp("The temprature at the inner surface of wall is T4=T3-(Q*Rp)in °C")
+T4=T3-(Q*Rp)
+//Check for Ti(Temprature inside the van)
+disp("Check for Ti(in °C)")
+Ti=T4-(Q*Ri)
+disp("The value is same as given in the problem")
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ 
+
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+
diff --git a/1910/CH2/EX2.10/Chapter210.sce b/1910/CH2/EX2.10/Chapter210.sce
new file mode 100755
index 000000000..75af849f3
--- /dev/null
+++ b/1910/CH2/EX2.10/Chapter210.sce
@@ -0,0 +1,48 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 10")
+//A copper pipe having 35mm outer diameter(Do) and 30mm inner diameter(Di) carries liquid oxygen to the storage site of a space shuttle at temprature,T1=-182°C
+//mass flow rate is ,mdot=0.06m^3/min. 
+Di=0.03;//in metre
+Do=0.035;// in metre
+T1=-182;
+mdot=0.06;
+//The ambient air is at temprature(Ta)=20°C and has a dew point(T3)=10°C.
+Ta=20;
+T3=10;
+//The thermal conductivity(k) of insulating material is 0.02W/(m*k)
+k=0.02;
+//The convective heat transfer coefficient on the outside is h=17W/(m^2*K)
+h=17;
+//The thermal conductivity of copper kcu=400W/(m*K)
+kcu=400;
+//We can write Q=((Ta-T1)/(R1+R2+R3))=((Ta-T3)/(R3)),Rearranging we get ((R1+R2+R3)/(R3))=((Ta-T1)/(Ta-T3))--------eq.1
+//The conduction Resistance of copper pipe(R1)=ln(0.035/0.03)/(2*%pi*L*kcu)=3.85*10^-4/(2*%pi*L)K/W
+//The conduction resistance of insulating material (R2)=ln(r3/0.035)/(2*%pi*L*k)=(1/(2*%pi*L))((50*ln(r3/0.035)))K/W where r3 is the outer radius of insulation in metres.
+//The convective resistance at the outer surface(R3)=1/(2*%pi*L*h*r3)=(1/2*%pi*L)*(mdot/r3)K/W
+//Substituting the values in eq.1 we have 1+((50*ln(r3/0.035)+(3.85*10^-4))/(mdot/r3))=20-(-182)/(20-10)
+//A rearrangement of the above equation gives r3*ln(r3)+3.35*r3=0.023
+//The equation is solved by trial and error method which finally gives r3=0.054m
+r3=0.054;//outer radius of insulation
+//Therefore the thickness of insulation is given by t=r3-Do
+disp("the thickness of insulation in metre is")
+t=r3-Do
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1910/CH2/EX2.11/Chapter211.sce b/1910/CH2/EX2.11/Chapter211.sce
new file mode 100755
index 000000000..15c5f01e0
--- /dev/null
+++ b/1910/CH2/EX2.11/Chapter211.sce
@@ -0,0 +1,37 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 11")
+//An electrical resistance wire 2.5mm or 2.5*10^-3m in diameter(D) and L=0.5m long has a measured voltage drop of E=25V for a current flow of I=40A.
+D=2.5*10^-3;
+I=40;
+E=25;
+L=0.5;
+ro=D/2;//ro is radius of wire
+//The thermal conductivity(k) of wire material is 24W/(m*K) 
+k=24;
+//The rate of generation of thermal energy per unit volume is given by qG=(E*I)/(L*%pi*D^2/4)
+disp("The rate of heat generation of thermal energy in W/m^3 is")
+qG=(E*I)/(L*%pi*D^2/4)
+//The temprature at the centre is given by To=Tw+((qG*ro^2)/(4*k)) where Tw=650K is surface temprature 
+Tw=650;
+disp("The temprature of wire at the centre in K is ")
+To=Tw+((qG*ro^2)/(4*k))//Note:The answer in the book is incorrect(value of D has been put instead of ro)
+
+
+
+
+
+
+
+
+
+
+
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+
+
+
diff --git a/1910/CH2/EX2.12/Chapter212.sce b/1910/CH2/EX2.12/Chapter212.sce
new file mode 100755
index 000000000..bcd831977
--- /dev/null
+++ b/1910/CH2/EX2.12/Chapter212.sce
@@ -0,0 +1,70 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 12")
+//A spherical tank of internal diameter, Di=5m ,made of t=25mm thick stainless steel(thermal conductivity,k=15W/(m*K)),is used to store water at temprature(Ti)=0°C.Do is outer diameter
+Di=5;
+t=25;
+Do=5+2*(t/1000);//in metre
+k=15;
+Ti=0;
+//The tank is located in a room whose temprature is (To)=20°C.
+To=20;
+//Emmisivity is 1.
+//The convection heat transfer coefficients at the inner and outer surfaces of the tank are hi=80W/(m^2*K) and ho=10W/(m^2*K)
+hi=80;
+ho=10;
+//The heat of fusion of ice at atmospheric pressure is deltahf=334kJ/kg.The stefan-boltzman constant(sigma) is 5.67*10^-8W/m^2.
+sigma=5.67*10^-8;
+deltahf=334;
+//The inner surface area is (A1) and outer surface area is (A2)of the tank
+disp("The inner(A1) and outer surfaces(A2) areas of the tank in m^2 are") 
+A1=%pi*Di^2
+A2=%pi*Do^2
+//The individual thermal resistances can be determined as
+//The convective resistance is (Ri)
+disp("The convective resistance(Ri) at the inner surface in K/W is ")
+Ri=1/(hi*A1)
+//The conduction resistance is(Rs)
+disp("The conduction resistance(Rs)of the tank in K/W is")
+Rs=(Do-Di)/(2*k*%pi*Di*Do)
+//The convective resistance is(Roc)
+disp("The convective resistance(Roc) at the outer surface in K/W is")
+Roc=1/(ho*A2)
+//The radiative resistance(Ror) at the outer surface is Ror=1/(A2*hr) 
+//The radiative heat transfer coefficient hr is determined by hr=sigma*(T2^2+293.15^2)*(T2+293.15)
+//But we do not know the outer surface temprature T2 of the tank and hence we cannot determine the value of hr.
+//Therfore we adopt an iterative procedure.We assume T2=4°C=277.15K.Putting the value in hr=sigma*(T2^2+293.15^2)*(T2+293.15) we get
+T2=277.15;
+disp("The radiative heat transfer coefficient hr in W/(m^2*K) is")
+hr=sigma*(T2^2+293.15^2)*(T2+293.15)
+disp("Therefore the radiative resistance(Ror) at the outer surface in K/W is")
+Ror=1/(A2*hr)
+//The two parallel resistances Roc and Ror can be replaced by an equivalent resistance Ro as X=(1/Ro)=(1/Roc)+(1/Ror)
+disp("The equivalent resistance in K/W is")
+X=(1/Roc)+(1/Ror);
+Ro=1/X
+//Now the resistance Ri,Rs and Ro are in series an hence the total resistance becomes Rtotal=Ri+Rs+Ro 
+disp("The total resistance in K/W is")
+Rtotal=Ri+Rs+Ro 
+//The rate of heat transfer is given by Q=(To-Ti)/Rtotal
+disp("The rate of heat transfer,Q in W is")
+Q=(To-Ti)/Rtotal
+//The outer surface(T2) is calculated as T2=To-Q*Ro
+disp("The outer surface temprature in °C is")
+T2=To-Q*Ro
+disp("which is sufficiently close to the assumption.So there is no need of further iteration")
+//The total heat transfer is (Qt),during a 24-hour period
+disp("The total heat transfer(Qt) during a 24-hour period in KJ is")
+Qt=Q*24*3600/1000
+//the amount of ice in kG which melts during a 24 hour period is (mice)
+disp("Therefore,the amount of ice(mice)in kG which melts during a 24 hour period is")
+mice=Qt/deltahf
+
+
+
+
+
diff --git a/1910/CH2/EX2.13/Chapter213.sce b/1910/CH2/EX2.13/Chapter213.sce
new file mode 100755
index 000000000..4524355eb
--- /dev/null
+++ b/1910/CH2/EX2.13/Chapter213.sce
@@ -0,0 +1,64 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 13")
+//A very long,10mm diameter(D) copper rod(thermal conductivity,k=370W/(m*K))is exposed to an enviroment at temprature,Tinf=20°C.
+D=0.01;
+k=370;
+Tinf=20;
+//The base temprature of the radius maintained at Tb=120°C.
+Tb=120;
+//The heat transfer coefficient between the rod and the surrounding air is h=10W/(m*K^2)
+h=10;
+//The rate of heat transfer for all finite lengths will be given by P/A=(4*pi*D)/(pi*D^2)
+//Let P/A=X
+disp("P/A in m^-1 is")
+X=(4*%pi*D)/(%pi*D^2)
+//m is defined as [(h*p)/(k*A)]^0.5 
+disp("m in m^-1 is")
+m=(h*X/k)^0.5
+//Let Y=h/(m*k)
+Y=h/(m*k)
+//Let M=(h*P*k*A)^0.5
+P=(%pi*D);//perimeter of the rod
+A=(%pi*D^2)/4;//Area of the rod
+disp("M in W/K is")
+M=(h*P*k*A)^0.5
+//thetab is the parameter that defines the base temprature
+disp("thetab in °C is ")
+thetab=Tb-Tinf
+//Heat loss from the rod is defined as Q=(h*P*k*A)*thetab*{[(h/m*k)+tanh(m*L)]/[1+(h/m*k)*tanh(m*L)]}
+disp("Heat loss from rod in Watt, for different value of length(in m) is ")
+L=0.02//Length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=0.04//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=0.08//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=0.20//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=0.40//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=0.80//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=1.00//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+L=10.00//length of rod
+Q=M*thetab*{[(Y)+tanh(m*L)]/[1+(Y)*tanh(m*L)]}
+//For an infinitely long rod we use heat loss as ,Qinf=(h*P*k*A)^0.5*thetab
+disp("For an infintely long rod heat loss in W is")
+Qinf=(h*P*k*A)^0.5*thetab
+disp("We see that since k is large there is significant difference  between the finite length and the infinte length cases")
+disp("However when the length of the rod approaches 1m,the result become almost same." )
+
+
+
+
+
+
+
+
+
diff --git a/1910/CH2/EX2.14/Chapter214.sce b/1910/CH2/EX2.14/Chapter214.sce
new file mode 100755
index 000000000..b7b3aeeca
--- /dev/null
+++ b/1910/CH2/EX2.14/Chapter214.sce
@@ -0,0 +1,55 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 14")
+//Considering two very long slender rods of the same diameter but of different materials.
+//The base of each rod is maintained at 100°C while the surfaces of the rod are exposed to 20°C
+//By traversing length of each rod with a thermocouple it was observed that tempratures of rod were equal at the position xA=0.15m and xB=0.075 from base.
+xA=0.15;
+xB=0.075;
+//Thermal conductivity of rod A is known to be kA=72 W/(m*K)
+kA=72;
+//In case of a very long slender rod we use the tip boundary condition thetaL=0 as L--->infinity
+//Therfore we can write for the locations where the tempratures are equal thetab*e^(-mA*xA)=thetab*e^(-mB*xB) or xA/xB=mB/mA,Again mB/mA=(kA/kB)^0.5
+//So kB=kA*(xB/xA)^2
+//The thermal conductivity of Rod B iskB
+disp("The thermal conductivity of Rod B kB in W/(m*K) is ")
+kB=kA*(xB/xA)^2
+
+
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diff --git a/1910/CH2/EX2.15/Chapter215.sce b/1910/CH2/EX2.15/Chapter215.sce
new file mode 100755
index 000000000..afc52051c
--- /dev/null
+++ b/1910/CH2/EX2.15/Chapter215.sce
@@ -0,0 +1,50 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 15")
+//A stack that is b=300mm wide and l=100mm deep contains N=60 fins each of length L=12mm.
+L=0.012;//in metre
+b=0.3;//in metre
+l=0.1;//in metre
+N=60;
+//The entire stack is made of aluminum which is everywhere t=1.0 mm thick.
+t=0.001;//in metre
+//The temprature limitations associated with electrical components are  Tb=400K and TL=350K.
+//Tb is base temprature and TL is end temprature
+Tb=400;
+TL=350; 
+//Given convection heat transfer coefficient(h=150W/(m^2*K)),Surrounding Temprature(Tinf=300K),thermal conductivity of aluminium(kaluminium=230W/(m*K))
+h=150;
+Tinf=300;
+kal=230;
+//Here both the ends of the fins are at fixed tempratures .Therefore we use M=(h*P*k*A)^0.5 and m=((h*P)/(k*A))^0.5,thetab=Tb-Tinf,thetaL=TL-Tinf
+//from the given data perimeter of each fin is given by P= 2*(l+t)in m and area of each fin is A=t*l
+disp(" perimeter of each fin in m is")
+P= 2*(l+t)
+disp("Cross sectional area of fin in m^2 is")
+A=t*l
+//M is defined as (h*P*kal*A)^0.5 and m is defined as ((h*P)/(kal*A))^0.5
+M=(h*P*kal*A)^0.5
+m=((h*P)/(kal*A))^0.5
+//thetab and thetaL are the parameters that define the fin tempratures at base and tip respectively.
+disp("Temprature parameter at fin base in K is")
+thetab=Tb-Tinf
+disp("Fixed temprature at fin tip in K is")
+thetaL=TL-Tinf
+//Heat loss from the plate is Qb
+disp("Heat loss from the plate at 400K in W is")
+Qb=[N*(h*P*kal*A)^0.5*thetab*((cosh(m*L)-(thetaL/thetab))/(sinh(m*L)))]+(((l*b)-(N*A))*h*thetab)+(l*b*h*thetab)
+
+
+
+
+
+
+
+
+
+
+
diff --git a/1910/CH2/EX2.2/Chapter22.sce b/1910/CH2/EX2.2/Chapter22.sce
new file mode 100755
index 000000000..e8e4f9c61
--- /dev/null
+++ b/1910/CH2/EX2.2/Chapter22.sce
@@ -0,0 +1,23 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 2")
+//The Thickness of fire clay bricks (Lb)=.2 m
+Lb=0.2;
+//The thermal conductivity of fire clay bricks(kb)=1.0 W/(m*K)
+kb=1;
+//the Thicknes of insulating material is L
+//The thermal conductivity of insulating material(ki)=.07 W/(m*K)
+ki=.07;
+//The furnace inner brick surface is at temprature Ti=1250 K
+Ti=1250;
+//The furnace outer brick surface is at temprature To=310 K
+To=310;
+//The maximum allowable heat transfer rate(Q) from wall = 900 W/m^2
+Q=900;
+//Q=(Ti-To)/((Lb/kb)+(L/ki)) so L=ki*(((Ti-To)/Q)-(Lb/kb))
+disp("The thickness of insulating material L=ki*(((Ti-To)/Q)-(Lb/kb)) in m")
+L=ki*(((Ti-To)/Q)-(Lb/kb))
diff --git a/1910/CH2/EX2.4/Chapter24.sce b/1910/CH2/EX2.4/Chapter24.sce
new file mode 100755
index 000000000..b44525438
--- /dev/null
+++ b/1910/CH2/EX2.4/Chapter24.sce
@@ -0,0 +1,53 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 4")
+//To cure the bond at a temprature(To),a radiant source used to provide a heat flux qo W/m^2 
+//The back of the substrate is maintained at a temprature T1
+//The film is exposed to air at a temprature (Tinf)
+//the convective heat transfer coefficient is h.
+//Given To=60°C,Tinf=20°C,h=25 W/(m^K)and T1=30°C
+To=60;
+Tinf=20;
+h=25;
+T1=30;
+//Ls is the thickness of substrate and Lf is thickness of film in mm.
+Ls=1.5;
+Lf=.25;
+//kf and ks are thermal conductivity of film and substrate respectively in W/(m*K)
+kf=.025;
+ks=.05;
+//qo is Heat flux.
+//qo=qf+qs where qf and qs are rate of heat transfer per unit surface area through the film and the substrate respectively.
+disp("Heat transfer per unit surface area through film qf=(To-Tinf)/((1/h)+(Lf/kf))in W/(m^2)")
+qf=(To-Tinf)/((1/h)+(Lf*10^-3/kf))
+disp("Heat transfer per unit surface area through substrate qs=(To-T1/(Ls/ks)in W/(m^2)")
+qs=(To-T1)/(Ls*10^-3/ks)
+disp("Heat flux qo=qs+qf in W/(m^2)")
+qo=qs+qf
+//If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then qo=q1+q2
+//q1 is rate of heat conduction through the film and substrate
+//q2 is rate of convective heat transfer from the upper surface of film to air
+disp("If the film is not transparent and all of the radiant heat flux is absorbed at its upper surface then")
+disp("Rate of heat conduction through the film and substrate q1=(To-T1)/(Ls/ks) in W/m^2")
+q1=(To-T1)/(Ls*10^-3/ks)
+//To determine q2 we need to find the temprature(T2) of the top surface 
+disp("The temprature of the top surface of the film T2=(q1*(Lf*10^-3/kf))+To in °C")
+T2=(q1*(Lf*10^-3/kf))+To
+disp("Rate of convective heat transfer from the upper surface of film to air q2=h*(T2-Tinf)in W/m^2")
+q2=h*(T2-Tinf)
+disp("Heat flux qo=q1+q2 in W/m^2 ")
+qo=q1+q2
+
+
+
+
+
+
+
+
+
+
diff --git a/1910/CH2/EX2.6/Chapter26.sce b/1910/CH2/EX2.6/Chapter26.sce
new file mode 100755
index 000000000..7bba1bcc1
--- /dev/null
+++ b/1910/CH2/EX2.6/Chapter26.sce
@@ -0,0 +1,46 @@
+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 6")
+//The thickness of plate 20mm or .02m with uniform heat generation qg=80MW/m^3
+//The thermal conductivity of wall(k)is 200W/m*K.
+k=200;
+//The left and right faces are kept at tempratures T=160°C and T=120°C respectively
+//We start with the equation (d^2T/dx^2)+(qg/k)=0 or T=-(qg/2k)*x^2 +(c1*x)+c2 
+//With qg=80*10^6W/m^3 and k=200W/(m*K)
+//Applying boundary condition at x=0,T=160°C and x=0.02,T=120°C we get constant c2=160°C and c2=2000m^-1
+//Hence T=160+2*10^3*(x-100*x^2)----->eq.1
+disp("(a)The expression for the temprature distribution in the plate T=160+2*10^3*(x-100*x^2)" )
+//For maximum temprature differentiating eq.1 with respect to x and equating it to zero...we get dT/dx=(2*10^3)-(4*10^5*x)=0,which gives x=0.005m=5mm
+disp("(b)The maximum temprature occurs at x=5mm or 0.005m away from the left surface towards the right")
+x=0.005;
+//The maximum temprature is Tmax
+disp("The maximum temprature(Tmax) in °C is")
+Tmax=160+2*10^3*(x-100*x^2)
+//The rate of heat transfer(q/A) is given by -k*(dT/dx)
+//Let dT/dx=X and (q/A)=Q
+disp("(c(i))The rate of heat transfer at the left face in MW/m^2 is")
+//For left face x=0
+x=0;
+X=(2*10^3)-(4*10^5*x);
+Q=-k*X/10^6
+disp("The minus sign indicates that the heat flow in the negative direction")
+disp("(c(ii))The rate of heat transfer at the right face in MW/m^2 is")
+//For right face x=0.02
+x=0.02;
+X=(2*10^3)-(4*10^5*x);
+Q=-k*X/10^6
+//(q/A)@x=0 implies rate of heat transfer at the position where x=0. 
+disp("The minus sign for (q/A)@x=0 and the plus sign for (q/A)@x=0.02 indicates that heat is lost from both the surfaces to surroundings")
+disp("(c(iii))The rate of heat transfer at the centre in MW/m^2 is")
+//For centre x=0.01
+x=0.01;
+X=(2*10^3)-(4*10^5*x);
+Q=-k*X/10^6
+//A check for the above results can be made from an energy balance of the plate as |(q/A)|@x=0+|(q/A)|@x=0.02=qG*0.02
+
+
+
diff --git a/1910/CH2/EX2.9/Chapter29.sce b/1910/CH2/EX2.9/Chapter29.sce
new file mode 100755
index 000000000..1fdf90849
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+++ b/1910/CH2/EX2.9/Chapter29.sce
@@ -0,0 +1,56 @@
+//Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Introduction to heat transfer by S.K.Som, Chapter 2, Example 09")
+//A thin walled copper tube of outside metal radius r=0.01m carries steam at temprature, T1=400K.It is inside a room where the surrounding air temprature is Tinf=300K.
+T1=400;
+Tinf=300;
+r=0.01;
+//The tube is insulated with magnesia insulation of an approximate thermal conductivity of k=0.07W/(m*K)
+k=0.07;
+//External convective Coefficient h=4W/(m^2*K)
+h=4;
+//Critical thickness(rc) is given by k/h
+disp("The critical thickness of insulation in metre is")
+rc=k/h
+//We use the rate of heat transfer per metre of tube length as Q=(Ti-Tinf)/((ln(r2/r1)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L))) where length,L=1m
+L=1;
+//When 0.002m thick layer of insulation r1=0.01m,r2=0.01+0.002=0.012m
+r1=0.01;//inner radius
+r2=0.012;//outer radius
+//Let ln(r2/r1)=X
+X=log(r2/r1)/log(2.718);
+//The heat transfer rate per metre of tube length is Q
+disp("The heat transfer rate Q per metre of tube length in W/m is ")
+Q=(T1-Tinf)/(((X)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L)))
+//When critical thickness of insulation r1=0.01m,r2=0.0175m
+r2=0.0175;//outer radius
+r1=0.01;//inner radius
+//Let ln(r2/r1)=X
+X=log(r2/r1)/log(2.718);
+//The heat transfer rate per metre of tube length is Q 
+disp("The heat transfer rate per metre of tube length Q in W/m is ")
+Q=(T1-Tinf)/(((X)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L)))
+//When there is a 0.05 m thick layer of insulation r1=0.01m,r2=.01+0.05=0.06m
+r1=0.01;//inner radius
+r2=0.06;//outer radius
+//Let ln(r2/r1)=X
+X=log(r2/r1)/log(2.718);
+//The heat transfer rate per metre of tube length is Q 
+disp("The heat transfer rate per metre of tube length Q in W/m is ")
+Q=(T1-Tinf)/(((X)/(2*%pi*L*k))+(1/(h*2*%pi*r2*L)))
+//It is important to note that Q increases by 5.2% when the insulation thickness increases from 0.002m to critical thickness. 
+//Addition of insulation beyond the critical thickness decreases the value of Q (The heat loss).
+
+
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+
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