initial commit / add all books

10 files changed, 245 insertions, 0 deletions

diff --git a/479/CH4/EX4.1/Example_4_1.sce b/479/CH4/EX4.1/Example_4_1.sce
new file mode 100755
index 000000000..a33d16f7d
--- /dev/null
+++ b/479/CH4/EX4.1/Example_4_1.sce
@@ -0,0 +1,32 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.1

+clear;

+clc;

+

+//Given

+Q1 = 250;//Heat absorbed in Kcal

+T1 = (260+273);//Temperature at which engine absorbs heat

+T0 = (40+273);//Temperature at which engine discards heat

+//To Calculate work output, heat rejected, entropy change of system,surronding & total change in entropy and the efficiency of the heat engine

+

+//(i)Calculation of work output

+W = (Q1*((T1-T0)/T1));//Work done using equations 4.7 & 4.9 given on page no 98

+mprintf('(i)The work output of the heat engine is %f Kcal',W);

+

+//(ii)Calculation of heat rejected

+Q2 = (Q1*T0)/T1;

+mprintf('\n (ii)The heat rejected is %f Kcal',Q2);

+

+//(iii)Calculation of entropy

+del_S1 = -(Q1/T1);//Change in the entropy of source in Kcal/Kg K

+del_S2 = Q2/T0;//Change in the entropy of sink in Kcal/Kg K

+del_St = del_S1+del_S2;//Total change in entropy in Kcal/Kg K

+mprintf('\n (iii)Total change in entropy is %d confirming that the process is reversible',del_St);

+

+//(iv)Calculation of efficiency

+n = (W/Q1)*100;

+mprintf('\n (iv)The efficiency of the heat engine is %f percent',n);

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.10/Example_4_10.sce b/479/CH4/EX4.10/Example_4_10.sce
new file mode 100755
index 000000000..65432306e
--- /dev/null
+++ b/479/CH4/EX4.10/Example_4_10.sce
@@ -0,0 +1,11 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.10

+clear;

+clc;

+

+//Given

+//The given example is a theoretical problem and it does not involve any numerical computation

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.2/Example_4_2.sce b/479/CH4/EX4.2/Example_4_2.sce
new file mode 100755
index 000000000..013071aa1
--- /dev/null
+++ b/479/CH4/EX4.2/Example_4_2.sce
@@ -0,0 +1,31 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.2

+clear;

+clc;

+

+//Given

+T1 = 373;//Temperature of the saturated steam in K

+T2 = 298;//Temperature of the saturated water in K

+//To calculate the total change in entropy and hence determine the reversibility of the process

+

+//del_H = del_Q+(V*del_P)

+//del_H =del_Q; since it is a constant pressure process

+

+//From steam table,

+//enthalpy of saturated steam at 373K is

+H1 = 6348.5;// in Kcal/Kg

+//enthalpy of saturated liquid water at 373K is

+H2 = 99.15;//in Kcal/Kg

+Q = H2-H1;//heat rejected in Kcal/Kg

+del_S1 = Q/T1;//change in entropy of the system in Kcal/Kg K

+del_S2 = Q/T2;//change in entropy of the surronding in Kcal/Kg K

+del_St = del_S1+del_S2;//total change in the entropy in Kcal/Kg K

+if(del_St == 0)

+    mprintf('Process is reversible');

+else

+    mprintf('Process is irreversible');

+end

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.3/Example_4_3.sce b/479/CH4/EX4.3/Example_4_3.sce
new file mode 100755
index 000000000..f72a688e8
--- /dev/null
+++ b/479/CH4/EX4.3/Example_4_3.sce
@@ -0,0 +1,34 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.3

+clear;

+clc;

+

+//Given

+Cp = 0.09;//specific heat of metal block in Kcal/Kg K

+m = 10;//mass of metal block in Kg

+T1 = 323;//initial temperature of the block in K

+T2 = 298;//final temperature of the block in K

+//constant pressure process

+//To find out entropy change of block,air and total entropy change

+

+//(i)To calculate the entropy change of block

+del_S1 = m*Cp*log(T2/T1);

+mprintf('(i)Entropy change of block is %f Kcal/Kg K',del_S1);

+

+//(ii)To calculate the entropy change of air

+Q = m*Cp*(T1-T2);//heat absorbed by air = heat rejected by block in Kcal

+del_S2 = (Q/T2);

+mprintf('\n (ii)Entropy change of air is %f Kcal/Kg K',del_S2);

+

+//(iii)To calculate the total entropy change

+del_St = del_S1+del_S2;

+mprintf('\n (iii)Total entropy change is %f Kcal/Kg K',del_St);

+if(del_St == 0)

+    mprintf('\n Process is reversible');

+else

+    mprintf('\n Process is irreversible');

+end

+//end 
\ No newline at end of file
diff --git a/479/CH4/EX4.4/Example__4_4.sce b/479/CH4/EX4.4/Example__4_4.sce
new file mode 100755
index 000000000..8ad2ad919
--- /dev/null
+++ b/479/CH4/EX4.4/Example__4_4.sce
@@ -0,0 +1,25 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.4

+clear;

+clc;

+

+//Given

+m1 = 10;//mass of metal block in Kg

+m2 = 50;//mass of water in Kg

+Cp1 = 0.09;//Specific heat of metal block in Kcal/Kg K

+Cp2 = 1;//Specific heat of water in Kcal/Kg K

+T1 = 50;//Initial temperature of block in deg celsius

+T2 = 25;//Final temperature of block in deg celsius

+

+//To calculate the total change in entropy

+//Heat lost by block = Heat gained by water

+Tf = ((m1*Cp1*T1)+(m2*Cp2*T2))/((m1*Cp1)+(m2*Cp2));//final temperature of water in deg celsius

+Tf1 = Tf+273.16;//final temperature in K

+del_S1 = m1*Cp1*log(Tf1/(T1+273));//change in entropy of the block in Kcal/K

+del_S2 = m2*Cp2*log(Tf1/(T2+273));//change in entropy of the block in Kcal/K

+del_St = del_S1+del_S2;

+mprintf('The total change entropy is %f Kcal/K',del_St);

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.5/Example_4_5.sce b/479/CH4/EX4.5/Example_4_5.sce
new file mode 100755
index 000000000..c5cd1b51b
--- /dev/null
+++ b/479/CH4/EX4.5/Example_4_5.sce
@@ -0,0 +1,21 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.5

+clear;

+clc;

+

+//Given

+//Air at 20 deg celsius

+//P1 = 250;initial pressure in atm

+//P2 = 10;final pressure after throttling in atm

+

+//To calculate the entropy change

+//According to the given conditions from figure4.5(page no 103)

+S1 = -0.38;//initial entropy in Kcal/Kg K

+S2 = -0.15;//final entroy in Kcal/Kg K

+del_S = S2-S1;

+mprintf('Change in entropy for the throttling process is %f Kcal/Kg K',del_S);

+//From figure 4.6(page no 104), the final temperature is -10 deg celsius

+//end 
\ No newline at end of file
diff --git a/479/CH4/EX4.6/Example_4_6.sce b/479/CH4/EX4.6/Example_4_6.sce
new file mode 100755
index 000000000..38c7e19c9
--- /dev/null
+++ b/479/CH4/EX4.6/Example_4_6.sce
@@ -0,0 +1,11 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.6

+clear;

+clc;

+

+//Given

+//The given problem does not involve any numerical comptation

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.7/Example_4_7.sce b/479/CH4/EX4.7/Example_4_7.sce
new file mode 100755
index 000000000..c3fe7aebb
--- /dev/null
+++ b/479/CH4/EX4.7/Example_4_7.sce
@@ -0,0 +1,23 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.7

+clear;

+clc;

+

+//Given

+//Basis: 1 hour

+m = 10;//mass of air in Kg

+T = 293;//Constant temperature throughout the process in K

+//P1 = 1;//Initial pressure in atm

+//P2 = 30;//Final pressure in atm

+//According to the given data and using the graph  or figure A.2.7 given in page no 105

+S1 = 0.02;//Initial entropy in Kcal/Kg

+S2 = -0.23;//Final entropy in Kcal/Kg

+H1 = 5;//Initial enthalpy in Kcal/Kg

+H2 = 3;//Final enthalpy in Kcal/Kg

+

+W = -((H2-H1)+T*(S2-S1))*m*(427/(3600*75));

+mprintf('The horse power of the compressor is %f hp',W);

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.8/Example_4_8.sce b/479/CH4/EX4.8/Example_4_8.sce
new file mode 100755
index 000000000..8164a0694
--- /dev/null
+++ b/479/CH4/EX4.8/Example_4_8.sce
@@ -0,0 +1,28 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.1

+clear;

+clc;

+

+//Given

+//Basis: 1 Kg of steam

+//P1 = 30;Intial pressure in Kgf/cm^2

+//P2 = 3;Final pressure in Kgf/cm^2

+//T = 300;//Operating temperature

+//From figure A.2.8, 

+H1 = 715;//Initial enthalpy of steam in Kcal/Kg

+H2 = 625;//Final enthalpy of steam in Kcal/Kg

+S1 = 1.56;//Initial entropy of steam in Kcal/Kg K

+S2 = 1.61;//Final entropy of steam in Kcal/Kg K

+Q = -1;//heat loss in Kcal/Kg

+To = 298;//The lowest surronding temperature in K

+

+//To calculate the effectiveness of the process

+W = (-(H2-H1)+Q);//Actual work output by the turbine in Kcal

+//The maximum or available work can be calculated from equation 4.14

+del_B = -((H2-H1)-(To*(S2-S1)));// Maximum work that can be obtained in Kcal

+E = (W/del_B)*100;

+mprintf('The effectiveness of the process is %f percent',E);

+//end
\ No newline at end of file
diff --git a/479/CH4/EX4.9/Example_4_9.sce b/479/CH4/EX4.9/Example_4_9.sce
new file mode 100755
index 000000000..9f32b8e39
--- /dev/null
+++ b/479/CH4/EX4.9/Example_4_9.sce
@@ -0,0 +1,29 @@
+//Chemical Engineering Thermodynamics

+//Chapter 4

+//Second Law of Thermodynamics

+

+//Example 4.9

+clear;

+clc;

+

+//Given

+m = 1;//mass of liquid water in Kg

+T1 = 1350;//initial temperature in deg celsius

+T2 = 400;//final temperature in deg celsius

+Cp = 1;//Specific heat of water in Kcal/Kg K

+Cpg = 0.2;//Specific heat of combustion gases in Kcal/Kg K

+Hv = 468.35;//Heat of vapourisation at 14 Kgf/cm^2 and 194.16 deg celsius in Kcak/Kg

+To = 298;//Surronding temperature

+Tb = 194.16+273;//Boiling point of liquid water

+

+//To Calculate the maximum work obtained and the entropy change

+//(i)Calculation of maximum work

+//Q = del_H = m*Cp*(T2-T1); gas can be assumed to cool at constant pressure

+//From equation 4.14 (page no 110)

+del_B = -((m*Cpg*(T2-T1))-(To*m*Cp*log((T2+273)/(T1+273))));

+mprintf('(i)The maximum work that can be obtained is %f Kcal/Kg of gas',del_B);

+

+//(ii)To Calculate the change in entropy

+del_S =(m*Cp*log(Tb/To))+((m*Hv)/Tb);

+mprintf('\n(ii)The entropy change per Kg of water is %f',del_S);

+//end
\ No newline at end of file