initial commit / add all books

9 files changed, 159 insertions, 0 deletions

diff --git a/1304/CH1/EX1.1/1_1.sce b/1304/CH1/EX1.1/1_1.sce
new file mode 100644
index 000000000..3e6f12441
--- /dev/null
+++ b/1304/CH1/EX1.1/1_1.sce
@@ -0,0 +1,15 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.1\n\n\n");

+// heat transfer through brick wall 

+// illustration1.1

+// solution

+

+L = 0.25; // [m]Thickness of brick wall 

+dT = 20-10;//[degree celsius] temperature difference

+

+A= 5;//[m^2] surface area of the wall

+//calculating rate of heat flow across the brick wall

+k=0.69;//[W/m degree celcius] thermal conductivity

+q = (k*A*dT)/L;//[W] 

+printf("rate of heat flow across the brick wall is %f W",q);

diff --git a/1304/CH1/EX1.2/1_2.sce b/1304/CH1/EX1.2/1_2.sce
new file mode 100644
index 000000000..505a170c6
--- /dev/null
+++ b/1304/CH1/EX1.2/1_2.sce
@@ -0,0 +1,17 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.2\n\n\n");

+// heat loss through a glass window 

+// illustration1.2

+// solution

+

+L = 0.015; // [m]Thickness of glass window

+dT = 20-(-5);//[degree celsius] temperature difference

+

+A= 0.5;//[m^2] surface area of the glass window

+//calculating rate of heat flow across the brick wall

+k=0.78;//[W/m degree celcius] thermal conductivity of the glass window

+q = (k*A*dT)/(1000*L);//[kW] rate of heat loss

+t=2;//[h] time in hours

+H=q*t;//[kW.h] total heat loss in time t 

+printf("Total heat loss across the glass window in 2 hours is %f kW.H",H);

diff --git a/1304/CH1/EX1.3/1_3.sce b/1304/CH1/EX1.3/1_3.sce
new file mode 100644
index 000000000..5d4b86773
--- /dev/null
+++ b/1304/CH1/EX1.3/1_3.sce
@@ -0,0 +1,16 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.3\n\n\n");

+// limiting the heat loss through a fierglass insulating board

+// illustration1.

+// solution

+

+dT = 150;//[degree celsius] temperature drop

+k=0.05;//[W/m degree celcius] thermal conductivity of the board

+q = 100;//[W/m^2] rate of heat loss per unit area

+// heat flux is directly proportional to temperature difference and inversely to thickness

+L = (k*dT)/q; // [m]Thickness of board

+

+

+printf("The thickness of the fibre glass board to limit heat losses to 100W/m^2 should be %f m",L);

+

diff --git a/1304/CH1/EX1.4/1_4.sce b/1304/CH1/EX1.4/1_4.sce
new file mode 100644
index 000000000..9c150f08e
--- /dev/null
+++ b/1304/CH1/EX1.4/1_4.sce
@@ -0,0 +1,15 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.4\n\n\n");

+// calculating heat flow from plate to the air flowing aboe it 

+// solution

+

+Tw =40 ;//[degree celsius] temperature of the plate

+

+Tf =10 ;//[degree celsius] temperature of the air

+hm=30;//[W/m^2 degree celcius] mean heat transfer coefficient

+A = 2;//[m^2]area of the plate

+qm=hm*(Tw-Tf);//[W/m^2] heat flux

+Q=(qm*A)/1000;//[kW] Heat flow

+printf("The heat flow through the area is %f kW",Q);

+

diff --git a/1304/CH1/EX1.5/1_5.sce b/1304/CH1/EX1.5/1_5.sce
new file mode 100644
index 000000000..b4b1ca379
--- /dev/null
+++ b/1304/CH1/EX1.5/1_5.sce
@@ -0,0 +1,18 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.5\n\n\n");

+// calculating heat flow from tube to the air flowing across it 

+// solution

+

+Tw =110 ;//[degree celsius] temperature of the tube

+

+Tf =10 ;//[degree celsius] temperature of the air

+v=5;//[m/s] velocity of air flow

+hm=85;//[W/m^2 degree celcius] mean heat transfer coefficient between tube surface and air for air velocity of 5 m/s across a cylinder

+d=0.01;//[m] outer diameter of the tube

+L=5;

+A= %pi*d*L;//[m^2]total area of the cylinder 

+qm=hm*(Tw-Tf);//[W/m^2] heat flux, by Newton's law of cooling

+Q=(qm*A)/1000;//[kW] Heat flow

+printf("The heat flow from the tube to the flowing air  is %f kW",Q);

+

diff --git a/1304/CH1/EX1.6/1_6.sce b/1304/CH1/EX1.6/1_6.sce
new file mode 100644
index 000000000..a5959a51a
--- /dev/null
+++ b/1304/CH1/EX1.6/1_6.sce
@@ -0,0 +1,18 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.6\n\n\n");

+// calculating heat flow from tube to the air flowing across it 

+// solution

+

+Q=100;//[W] heat transfer rate from sphere to the surrounding

+d=0.1; //[m] diameter of sphere

+dT=50; //[degree celcius] temeprature difference between sphere and surrounding ambient air 

+//heat transfer rate Q is given by Q=hm*A*dT (by Newton's law of cooling)

+//where hm= mean convection heat transfer coefficient

+//A=Surface area of sphere = %pi*d^2

+//Thus

+hm=Q/(dT*%pi*d^(2));//[W/m^2 degree celcius]

+

+

+printf("The convection heat transfer coefficient between the sphere and the ambient air is %f W/m^2 degree celcius",hm);

+

diff --git a/1304/CH1/EX1.7/1_7.sce b/1304/CH1/EX1.7/1_7.sce
new file mode 100644
index 000000000..e403b103b
--- /dev/null
+++ b/1304/CH1/EX1.7/1_7.sce
@@ -0,0 +1,18 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.7\n\n\n");

+// calculating rate of heat loss from cylinder per unit length

+// solution

+

+Tw =50 ;//[degree celsius] temperature of the cylinder

+

+Tf =25 ;//[degree celsius] temperature of the air

+D=0.02;//[m] diameter of the cylinder

+hm=8;//[W/m^2 degree celcius] mean heat transfer coefficient

+qm=hm*(Tw-Tf);//[W/m^2] heat flux, by Newton's law of cooling

+// Total heat flow rate is qm*A

+//hence heat flow rate per unit length is Q=qm*A/L= qm*%piDL/L = qm*%pi*D

+Q=qm*%pi*D; //[W/m]heat flow rate per unit length

+

+

+printf("The heat flow rate per unit length is %f W/m",Q);

diff --git a/1304/CH1/EX1.8/1_8.sce b/1304/CH1/EX1.8/1_8.sce
new file mode 100644
index 000000000..87c27f42c
--- /dev/null
+++ b/1304/CH1/EX1.8/1_8.sce
@@ -0,0 +1,27 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.8\n\n\n");

+// calculating minimum heat coefficient at the outer surface to maintain outer surface temperature at 100 degree celcius

+// solution

+

+

+T1 =200 ;//[degree celsius] temperature of the inner surface of the insulation layer

+

+

+Tf =20 ;//[degree celsius] temperature of the air

+k=1.5; //[W/m degree celcius] heat conductivity of insulator

+L=0.05;//[m] insulator thickness

+// Let T2 be the temperature of outer layer of the insulator

+// qconv = heat transfer rate from outer layer to the air

+//qcond = heat transfer rate across the insulator

+// if qconv > qcond, then T2 will decrease

+// if qcond > qconv, then T2 will increase

+// for T2 to not go above 100 degree celcius, qcond should be less than equal to qconv , and equality will be when T2 is equal to 100

+//thus qcond = qconv at T2=100

+// qcond = k*(T1-T2)/L = h*(T2-Tf)  = qconv

+//thus

+T2 =100 ;//[degree celsius]maximum permissible temperature of the outer surface 

+

+

+h = k*(T1-T2)/(L*(T2-Tf));// [W/m^2 degree celcius] convection heat trasnfer coefficient between insulator's outer surface and air 

+printf("The minimum convection heat transfer coefficient required to maintain outer surface tempereature below 100 is %f W/m^2 degreee celcius",h);

diff --git a/1304/CH1/EX1.9/1_9.sce b/1304/CH1/EX1.9/1_9.sce
new file mode 100644
index 000000000..c744e12fd
--- /dev/null
+++ b/1304/CH1/EX1.9/1_9.sce
@@ -0,0 +1,15 @@
+clear;

+clc;

+printf("\t\t\tExample Number 1.9\n\n\n");

+// calculating increasing in emmisive power

+// solution

+

+sigma= 5.6697*10^(-8);// [W/m^2 K^4] Stefan Boltzmann constant

+T1=20+273.15;//[K] initial temperature

+

+T2=100+273.15;//[K] final temperature

+//emmissive power E is given as sigma*T^4

+//hence 

+dE = sigma*((T2)^(4)-(T1)^(4));//[W/m^2] difference in emmissive power

+

+printf("The increase in emmissive power of the blackbody after heating is %f W/m^2",dE);