initial commit / add all books

13 files changed, 447 insertions, 0 deletions

diff --git a/1752/CH4/EX4.1/exa_4_1.sce b/1752/CH4/EX4.1/exa_4_1.sce
new file mode 100755
index 000000000..14298e329
--- /dev/null
+++ b/1752/CH4/EX4.1/exa_4_1.sce
@@ -0,0 +1,39 @@
+//Exa 4.1

+clc;

+clear;

+close;

+//given data

+format('v',9)

+L=1;// in m

+rho=1600;// in kg/m^3

+k=40;// in w/mK

+Cp=4*10^3;// in J/kgK

+a=900;// in  degree C

+b=-300;// in  degree C/m

+c=-50;// in  degree C/m^2

+Qg=1*10^3;// in kW/m^2

+A=10;// area in m^2

+//t=a+b*x+c*x^2 at any instant, so

+// dtBYdx= b+2*c*x

+// d2tBYdx2 = 2*c, then

+

+// Part(a)

+//q1= -k*A*dtBYdx , at

+x=0;

+q1= -k*A*(b+2*c*x);// in w

+//q2= -k*A*dtBYdx , at

+x=L;

+q2= -k*A*(b+2*c*x);// in w

+E_stored= (q1-q2)+Qg*A*L;// in watt

+disp(E_stored,"The rate of change of energy storage in watt")

+

+// Part(b)

+alpha= k/(rho*Cp);// in m^2s

+d2tBYdx2 = 2*c;

+dtBYdtoh= alpha*(d2tBYdx2+Qg/k );// in degree C/sec

+disp(dtBYdtoh,"Rate of change of temperature in degree C/sec");

+disp("Since dt by dx is independent of x. Hence time rate of charge of temperature throughout wall will remain same.")

+

+

+

+

diff --git a/1752/CH4/EX4.10/exa_4_10.sce b/1752/CH4/EX4.10/exa_4_10.sce
new file mode 100755
index 000000000..0a9dc1450
--- /dev/null
+++ b/1752/CH4/EX4.10/exa_4_10.sce
@@ -0,0 +1,45 @@
+//Exa 4.10

+clc;

+clear;

+close;

+//given data

+L=40*10^-2;// in m

+k=1.5;// in W/mK

+A=4;// in square meter

+alpha=1.65*10^-3;// in m^2/h

+//T = 50-40*x+10*x^2+20*x^3-15*x^4 , so

+// dtBYdx= -40+20*x+60*x^2-60*x^3

+// d2tBYdx2 = 20+120*x-180*x^2

+

+// Part (a) Heat entering the slab

+//q1= -k*A*dtBYdx , at

+x=0;

+qi= -k*A*(-40+20*x+60*x^2-60*x^3);// in w

+disp(qi,"Heat entering the slab in watt")

+// Heat leaving the slab

+//ql= -k*A*dtBYdx , at

+x=L;

+ql= -k*A*(-40+20*x+60*x^2-60*x^3);// in w

+disp(ql,"Heat leaving the slab in watt")

+

+// Part (b) Rate of heat storage

+RateOfHeatStorage = qi-ql;// in watt

+disp(RateOfHeatStorage,"Rate of heat storage in watt");

+

+// Part (c) Rate of temperature change

+// d2tBYdx2 = 1/alpha*dtBYdtoh

+// dtBYdtoh= alpha*d2tBYdx2, at

+x=0;

+dtBYdtoh = alpha*(20+120*x-180*x^2);// in degree C/h

+disp(dtBYdtoh,"The rate of temperature change at entering the slab in degree C/h")

+// dtBYdtoh= alpha*d2tBYdx2, at

+x=L

+dtBYdtoh = alpha*(20+120*x-180*x^2);// in degree C/h

+disp(dtBYdtoh,"The rate of temperature change at leaving the slab in degree C/h")

+

+// Part (d) for the rate of heating or cooling to be maximum

+// dBYdx of dtBYdtoh = 0

+// dBYdx of (alpha*d2tBYdx2) =0

+// d3tBYdx3 = 0

+x=120/360;// in meter

+disp(x,"The point where rate of heating or cooling is maximum in meter")

diff --git a/1752/CH4/EX4.11/exa_4_11.sce b/1752/CH4/EX4.11/exa_4_11.sce
new file mode 100755
index 000000000..a25e248ea
--- /dev/null
+++ b/1752/CH4/EX4.11/exa_4_11.sce
@@ -0,0 +1,21 @@
+//Exa 4.11

+clc;

+clear;

+close;

+//given data

+k=40;// in W/m degree C

+d =12*10^-3;// in meter

+t=127;// in degree C

+t_i=877;// in degree C

+t_infinite=52;// in degree C

+h= 20;// in W/m^2 degree C

+rho=7800;// in W/m^2K

+C=600;// in J/kg K

+r=d/2;// in meter

+//l_s = V/A = r/3

+l_s =  r/3;

+Bi= h*l_s/k;

+// since Bi < 0.1 , hence lumped heat capacity analysis can be applied

+// (t-t_infinite)/(t_i-t_infinite)  =  %e^(-h*A*toh /(rho*V*C)) = %e^(-h*toh/(rho*l_s*C)) = %e^(-h*toh/(rho*C*l_s))

+toh = -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h;// in sec

+disp("Time required for cooling process : "+string(toh)+" seconds or "+string(toh/60)+" minutes")
\ No newline at end of file
diff --git a/1752/CH4/EX4.12/exa_4_12.sce b/1752/CH4/EX4.12/exa_4_12.sce
new file mode 100755
index 000000000..c98316b32
--- /dev/null
+++ b/1752/CH4/EX4.12/exa_4_12.sce
@@ -0,0 +1,45 @@
+//Exa 4.12

+clc;

+clear;

+close;

+//given data

+D=10*10^-2;// in m

+b=D/2;

+h= 100;// in W/m^2 degree C

+T_o=418;// in degree C

+T_i=30;// in degree C

+T_infinite=1000;// in degree C

+

+disp(" (A) For copper cylinder ");

+k=350;// in W/mK

+alpha=114*10^-7;// in m^2/s

+Bi= h*b/k;

+theta_0_t = (T_o-T_infinite)/(T_i-T_infinite);

+Fo=18.8;

+// Formula Fo= alpha*t/b^2

+t=Fo*b^2/alpha;

+disp("Time required to reach for the cylinder centreline temperature 418 degree C : "+string(t)+" seconds or "+string(t/3600)+" hours")

+

+// (2) Temperature at the radius of 4 cm

+theta_0_t = 0.985;

+// Formula theta_0_t = (T-T_infinite)/(T_o-T_infinite)

+T= theta_0_t*(T_o-T_infinite)+T_infinite;// in degree C

+disp(T,"Temperature at the radius of 4 cm ") 

+disp("It has very less temperature gradients over 4 cm radius")

+

+disp(" (B) For asbestos cylinder ");

+k=0.11;// in W/mK

+alpha=0.28*10^-7;// in m^2/s

+Bi= h*b/k;

+theta_0_t = (T_o-T_infinite)/(T_i-T_infinite);

+Fo=0.21;

+// Formula Fo= alpha*t/b^2

+t=Fo*b^2/alpha;

+disp("Time required to reach for the cylinder centreline temperature 418 degree C : "+string(t)+" seconds or "+string(t/3600)+" hours")

+

+// (2) Temperature at the radius of 4 cm

+theta_x_t = 0.286;

+// Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)

+T= theta_x_t*(T_o-T_infinite)+T_infinite;// in degree C

+disp(T,"Temperature at the radius of 4 cm ") 

+disp("It has large temperature gradients")

diff --git a/1752/CH4/EX4.13/exa_4_13.sce b/1752/CH4/EX4.13/exa_4_13.sce
new file mode 100755
index 000000000..2b561fc54
--- /dev/null
+++ b/1752/CH4/EX4.13/exa_4_13.sce
@@ -0,0 +1,60 @@
+//Exa 4.13

+clc;

+clear;

+close;

+//given data

+D=5*10^-2;// in m

+b=D/2;

+h= 500;// in W/m^2 degree C

+k=60;// in W/m^2K

+rho=7850;// in kg/m^3

+C=460;// in J/kg

+alpha=1.6*10^-5;// in m^2/s

+T_i=225;// in degree C

+T_infinite=25;// in degree C

+t=2;// in minute

+

+// Part(i)

+Bi= h*b/k;

+Fo= alpha*t/b^2;

+theta_0_t = 0.18;

+// Formula theta_0_t = (T_o-T_infinite)/(T_i-T_infinite)

+T_o= theta_0_t*(T_i-T_infinite)+T_infinite;// in degree C

+disp(T_o,"Centreline Temperature of the sphere after 2 minutes of exposure in degree C ") ;

+

+// Part(2)

+depth= 10*10^-3;// in meter

+r=b-depth;// in meter

+rBYb=r/b;

+theta_x_t = 0.95;

+// Formula theta_x_t = (T-T_infinite)/(T_o-T_infinite)

+T= theta_x_t*(T_o-T_infinite)+T_infinite;// in degree C

+disp(T,"The Temperature at the depth of 1 cm from the surface after 2 minutes in degree C ") ;

+

+// Part (3)

+BiSquareFo= Bi^2*Fo;

+QbyQo= 0.8;// in kJ

+A=4/3*%pi*b^3;

+Qo= rho*A*C*(T_i-T_infinite);// in J

+Qo=Qo*10^-3;// in kJ

+// The heat transffered during 2 minute, 

+Q= Qo*QbyQo;// in kJ

+disp(Q,"The heat transffered during 2 minutes in kJ")

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

diff --git a/1752/CH4/EX4.2/exa_4_2.sce b/1752/CH4/EX4.2/exa_4_2.sce
new file mode 100755
index 000000000..094d15c3c
--- /dev/null
+++ b/1752/CH4/EX4.2/exa_4_2.sce
@@ -0,0 +1,41 @@
+//Exa 4.2

+clc;

+clear;

+close;

+//given data

+k=40;// in W/mK

+rho=7800;// in kg/m^3

+C=450;// in J/kgK

+d=20*10^-3;// in m

+r=d/2;

+t_i=400;// in degree C

+t=85;// in degree C

+t_infinite=25;// in degree C

+h=80;// in W/m^2K

+//l_s=V/A = (4/3*%pi*r^3)/(4*%pi*r^2) = r/3

+l_s=r/3;// in m

+Bi= h*l_s/k;

+// since Biot number is less than 0.1, hence lumped heat capacity system analysis can be applied

+

+// Part(a)

+// Formula (t-t_infinite)/(t_i-t_infinite)= %e^(-h*A*toh/(rho*V*C)) = %e^(-h*toh/(rho*l_s*C))

+toh= -log((t-t_infinite)/(t_i-t_infinite))*(rho*l_s*C)/h;// in sec

+disp(toh,"The time require to cool the sphere in sec");

+

+// Part(b)

+// dtBYdtoh = h*A*(t_i-t_infinite)/(rho*V*C) = h*(t_i-t_infinite)/(rho*l_s*C)

+dtBYdtoh = h*(t_i-t_infinite)/(rho*l_s*C);// in degree C/sec

+disp(dtBYdtoh,"Initial rate of cooling in degree C/sec");

+

+// Part(c)

+A=4*%pi*r^2;

+toh=60;

+q_in= h*A*(t_i-t_infinite)*%e^(-h*toh/(rho*l_s*C));// in watt

+disp(q_in,"Instantaneous heat transfer rate in watt");

+

+// Part(d) Total energy transferred during first one minute

+V=4/3*%pi*r^3;

+TotalEnergy = rho*C*V*(t_i-t_infinite)*(1-%e^(-h*toh/(rho*C*l_s)));

+disp(TotalEnergy,"Total energy transferred during first one minute in watt")

+

+// Note: Answer of first and last part in the book is wrong

diff --git a/1752/CH4/EX4.3/exa_4_3.sce b/1752/CH4/EX4.3/exa_4_3.sce
new file mode 100755
index 000000000..705f00c2e
--- /dev/null
+++ b/1752/CH4/EX4.3/exa_4_3.sce
@@ -0,0 +1,42 @@
+//Exa 4.3

+clc;

+clear;

+close;

+//given data

+k=40;// in W/mK

+rho=8200;// in kg/m^3

+C=400;// in J/kgK

+D=6*10^-3;// in m

+R=D/2;

+t_i=30;// in degree C

+t_infinite1=400;// for 10 sec in degree C

+t_infinite2=20;// for 10 sec in degree C

+h=50;// in W/m^2K

+

+// Part(a)

+//l_s= V/A = R/3

+l_s= R/3;// in m

+//toh= rho*V*C/(h*A) = rho*C*l_s/h

+toh= rho*C*l_s/h;// in sec

+disp(toh,"Time constance in sec")

+

+// Part (b)

+Bi= h*l_s/k;

+// since Bi < 0.1 , hence lumped heat capacity analysis is valid. Now , temperature attained by junction in 10 seconds when exposed to hot air at 400 degree C

+toh=10;// in sec

+// (t-t_infinite1)/(t_i-t_infinite1)= %e^(-h*A*toh/(rho*V*C)) = %e^(-h*toh/(rho*l_s*C))

+t= %e^(-h*toh/(rho*l_s*C))*(t_i-t_infinite1)+t_infinite1;// in degree C

+

+disp("The junction is taken out from hot air stream and placed in stream of still air 20 degree C. The initial temperature in this case will be "+string(t)+" .")

+t_i=t;

+toh=20;// in sec

+t= %e^(-h*toh/(rho*l_s*C))*(t_i-t_infinite2)+t_infinite2;// in degree C

+disp(t,"The temperature attained by junction in degree C");

+

+// Note: In the last, calculation to find the value of t is wrong so Answer in the book is wrong

+

+

+

+

+

+

diff --git a/1752/CH4/EX4.4/exa_4_4.sce b/1752/CH4/EX4.4/exa_4_4.sce
new file mode 100755
index 000000000..2a921d740
--- /dev/null
+++ b/1752/CH4/EX4.4/exa_4_4.sce
@@ -0,0 +1,23 @@
+//Exa 4.4

+clc;

+clear;

+close;

+//given data

+k=8;// in W/mK

+alpha=4*10^-6;// in m^2/s

+h=50;// in W/m^2K

+D=6*10^-3;// in m

+R=D/2;

+T=0.5;// where T = (t-t_infinite)/(t_i-t_infinite)

+//l_s= V/A = R/3

+l_s= R/2;// in m

+Bi= h*l_s/k;

+// since Bi < 0.1 , hence lumped heat capacity analysis can be applied

+// toh= rho*V*C/(h*A) = rho*C*l_s/h = k*l_s/(h*alpha)

+toh=  k*l_s/(h*alpha);// in seconds

+disp(toh,"time constant in seconds");

+// It is given that (t-t_infinite)/(t_i-t_infinite) = 0.5 =  %e^(-h*A*c /(rho*V*C)) = %e^(-h*c/(rho*l_s*C)) = %e^(-h*alpha*c/(l_s))

+// or (t-t_infinite)/(t_i-t_infinite) = %e^(-h*alpha*c/(l_s);

+c= -log(T)*l_s/(h*alpha);// in sec

+disp(c,"The time required to temperature change to reach half of its initial value in seconds");

+

diff --git a/1752/CH4/EX4.5/exa_4_5.sce b/1752/CH4/EX4.5/exa_4_5.sce
new file mode 100755
index 000000000..b8f768bbb
--- /dev/null
+++ b/1752/CH4/EX4.5/exa_4_5.sce
@@ -0,0 +1,19 @@
+//Exa 4.5

+clc;

+clear;

+close;

+//given data

+//t=450-500*x+100*x^2+150*x^3 at any instant, so

+// dtBYdx= -500+200*x+450*x^2

+

+L=0.5;// thickness of the wall in meter

+k=10;// in W/mK

+// Rate of heating entering in the wall per unit area, at

+x=0;

+//q1= -k*dtBYdx

+q1= -k*(-500+200*x+450*x^2);// in W/m^2

+// Rate of heat going out of the wall per unit area , at

+x=L;

+q2= -k*(-500+200*x+450*x^2);// in W/m^2

+E=q1-q2;// in W/m^2

+disp(E,"Heat energy stored per unit area in W/m^2")

diff --git a/1752/CH4/EX4.6/exa_4_6.sce b/1752/CH4/EX4.6/exa_4_6.sce
new file mode 100755
index 000000000..aabca8fc0
--- /dev/null
+++ b/1752/CH4/EX4.6/exa_4_6.sce
@@ -0,0 +1,29 @@
+//Exa 4.6

+clc;

+clear;

+close;

+//given data

+k=385;// in W/mK

+h=100;// in W/m^2K

+delta =2*10^-3;// thickness of plate in meter

+A=25*25;// area of plate in square meter

+rho=8800;// kg/m^3

+C=400;// J/kg-K

+// l_s= V/A= L*B*delta/(2*L*B) = delta/2

+l_s= delta/2;// in meter

+Bi= h*l_s/k;

+// since Bi < 0.1 , hence lumped heat capacity analysis can be applied

+

+// Part(i)

+// toh= rho*V*C/(h*A) = rho*C*l_s/h

+toh= rho*C*l_s/h;// in second

+disp(toh,"Time constant in seconds");

+

+// Part(ii)

+t_i=400;// in degree C

+t=40;// in degree C

+t_infinite=25;// in degree C

+// (t-t_infinite)/(t_i-t_infinite) =  %e^(-h*A*toh /(rho*V*C)) = %e^(-h*toh/(rho*l_s*C)) 

+toh= -log((t-t_infinite)/(t_i-t_infinite))*rho*C*l_s/h;// in sec

+disp(toh,"The time required for the plate to reach the temperature of 40 degree C in seconds");

+

diff --git a/1752/CH4/EX4.7/exa_4_7.sce b/1752/CH4/EX4.7/exa_4_7.sce
new file mode 100755
index 000000000..b60250c65
--- /dev/null
+++ b/1752/CH4/EX4.7/exa_4_7.sce
@@ -0,0 +1,29 @@
+//Exa 4.7

+clc;

+clear;

+close;

+//given data

+k=380;// in W/mK

+delta =6*10^-2;// thickness of plate in meter

+rho=8800;// kg/m^3

+C=400;// J/kg-K

+// l_s= V/A = delta/2

+l_s= delta/2;// in meter

+t=80;// in degree C

+t_i=200;// in degree C

+t_inf=30;// in degree C

+hw= 75;// in W/m^2K

+ha= 10;// in W/m^2K

+

+// Part(i)

+// ha*A*(t-t_inf_a)+ hw*A*(t-t_inf_w) = -rho*V*C*dtBYdtho, since t_ini_a = t_inf_w = t_inf = 30 degree C

+// (ha+hw)*A*(t-t_inf)= -rho*V*C*dtBYdtho

+// (ha+hw)/(rho*C*V)*A*dtoh = -dt/(t-t_inf)

+// integrate('(ha+hw)/(rho*V*C)*A','toh',0,toh) = integrate('1/(t-t_inf)','t',t_i,t)

+toh= -rho*l_s*C/(ha+hw)*log((t-t_inf)/(t_i-t_inf));

+disp("Time required to cool plate to 80 degree C is : "+string(toh)+" seconds = "+string(toh/60)+" minutes");

+

+// Part (ii)

+t= -rho*l_s*C/(2*ha)*log((t-t_inf)/(t_i-t_inf));

+disp("Time required to cool plate in only air is : "+string(t)+" seconds = "+string(t/60)+" minutes");

+

diff --git a/1752/CH4/EX4.8/exa_4_8.sce b/1752/CH4/EX4.8/exa_4_8.sce
new file mode 100755
index 000000000..039f0899a
--- /dev/null
+++ b/1752/CH4/EX4.8/exa_4_8.sce
@@ -0,0 +1,28 @@
+//Exa 4.8

+clc;

+clear;

+close;

+//given data

+k=45;// in W/m degree C

+d =0.1;// in meter

+l =0.30;// in meter

+t=800;// in degree C

+t_i=100;// in degree C

+t_infinite=1200;// in degree C

+h= 120;// in W/m^2 degree C

+alpha=1.2*10^-5;// in meter

+rhoC= k/alpha;

+V=%pi/4*d^2*l;// in m^3

+A= %pi*d*l + 2*%pi/4*d^2;// in m^2

+// l_s= V/A = (%pi/4*d^2*l)/(%pi*d*l + 2*%pi/4*d^2) = d*l/(4*l+2*d^2)

+l_s = d*l/(4*l+2*d^2);

+Bi= h*l_s/k;

+// since Bi < 0.1 , hence lumped heat capacity analysis can be applied

+// (t-t_infinite)/(t_i-t_infinite)  =  %e^(-h*A*toh /(rho*V*C)) = %e^(-h*toh/(rho*l_s*C)) = %e^(-h*toh/(rhoC*l_s))

+toh = -log((t-t_infinite)/(t_i-t_infinite))*rhoC*l_s/h;// in sec

+

+// So, the velocity of ingot passing through the furnace

+FurnaceLength = 8*100;// in cm

+time = toh;

+Velocity = FurnaceLength/time;// in cm/sec

+disp(Velocity,"Maximum speed in cm/sec")

diff --git a/1752/CH4/EX4.9/exa_4_9.sce b/1752/CH4/EX4.9/exa_4_9.sce
new file mode 100755
index 000000000..516bbffcc
--- /dev/null
+++ b/1752/CH4/EX4.9/exa_4_9.sce
@@ -0,0 +1,26 @@
+//Exa 4.9

+clc;

+clear;

+close;

+//given data

+rho=8500;// in kg/m^3

+C=400;// J/kgK

+toh=1;// in sec

+h= 400;// in W/m^2 degree C

+t=198;// in degree C

+t_i=25;// in degree C

+t_infinite=200;// in degree C

+

+// Part (1)

+// toh =rho*V*C/(h*A) = rho*C*l_s/h

+l_s= toh*h/(rho*C);

+// l_s = V/A = r/3 

+r=3*l_s;// in m

+r=r*10^3;// in mm

+d=2*r;// in m

+disp(d,"Junction diameter needed for the thermocouple in mili miter");

+

+// Part(ii)

+// toh= -rho*V*C/(h*A)*log((t-t_infinite)/(t_i-t_infinite))  

+toh = -toh*log((t-t_infinite)/(t_i-t_infinite));

+disp(toh,"Time required for the thermocouple junction to reach 198 degree C in seconds");