initial commit / add all books

11 files changed, 275 insertions, 0 deletions

diff --git a/2969/CH3/EX3.1/Ex3_1.sce b/2969/CH3/EX3.1/Ex3_1.sce
new file mode 100755
index 000000000..274269311
--- /dev/null
+++ b/2969/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,11 @@
+clc

+clear

+//DATA GIVEN

+Q=-50;                          //heat rejected to cooling water in kJ/kg

+W=-100;                         //work input in kJ/kg

+

+//using First Law of Thermodynamics, Q=(u2-u1)+W

+Du=Q-W;                         //(u2-u1) change in internal energy in kJ/kg

+//since Du is +ve, there is gain in internal energy

+

+printf('The GAIN in internal energy is: %2.0f kJ/kg. \n',Du);

diff --git a/2969/CH3/EX3.10/Ex3_10.sce b/2969/CH3/EX3.10/Ex3_10.sce
new file mode 100755
index 000000000..e10c09b95
--- /dev/null
+++ b/2969/CH3/EX3.10/Ex3_10.sce
@@ -0,0 +1,37 @@
+clc

+clear

+//DATA GIVEN

+//initial state

+p1=10^5;                        //initial pressure of gas in Pa

+V1=0.45;                        //initial volume of gas in m^3

+T1=80+273;                      //initial temperature of gas in K

+//final state

+p2=5*10^5;                      //final pressure of gas in Pa

+V2=0.13;                        //final volume of gas in m^3

+

+//gamma for air, g

+g=1.4;

+R=294.2                         //J/kgK

+

+m=p1*V1/R/T1;                   //mass in kg

+

+//p1*(V1^n)=p2*(V2^n)

+n=log(p1/p2)/log(V2/V1);        //index n

+

+//In a polytropic process

+//(T2/T1)=(V1/V2)^(n-1);

+T2=T1*(V1/V2)^(n-1);            //temp. T2 in K

+

+Cv=R/(g-1);

+Du=m*Cv*(T2-T1)/1000;           //increase in internal energy in kJ

+

+//using First Law of Thermodynamics, Q=(u2-u1)+W

+//W12=(p1*V1-p2*V2)/(n-1)=mR(T2-T1)/(n-1)

+W12=m*R*(T1-T2)/(n-1)/1000;

+Q=Du+W12;

+//since Q is -ve, there is rejection of heat from system to surroundings

+

+printf(' (i) The Mass of the gas is: %1.3f kg. \n',(m));

+printf(' (ii) The index n is: %1.3f. \n',(n));

+printf('(iii) The change in internal energy is: %2.1f kJ. \n',(Du));

+printf(' (iv) The Heat REJECTED is: %2.2f kJ. \n',(-Q));

diff --git a/2969/CH3/EX3.11/Ex3_11.sce b/2969/CH3/EX3.11/Ex3_11.sce
new file mode 100755
index 000000000..62b63315e
--- /dev/null
+++ b/2969/CH3/EX3.11/Ex3_11.sce
@@ -0,0 +1,31 @@
+clc

+clear

+//DATA GIVEN

+//initial state

+p1=1.02;                        //initial pressure of air in bar

+V1=0.015;                       //initial volume of air in m^3

+T1=22+273;                      //initial temperature of air in K

+//final state

+p2=6.8;                         //final pressure of air in bar

+//Law of adiabatic compression, pV^g=C

+

+//gamma for air, g

+g=1.4

+R=0.287;

+

+//In a adiabatic process

+//(T2/T1)=(p2/p1)^((g-1)/g);

+T2=T1*(p2/p1)^((g-1)/g);;       //final temp. T2 in K

+

+//p1*(V1^g)=p2*(V2^g)

+V2=V1*(p1/p2)^(1/g);            //final volume in m^3

+

+m=p1*10^5*V1/10^3/R/T1;         //mass in kg

+

+//W=(p1*V1-p2*V2)/(g-1)=mR(T2-T1)/(g-1)

+W=m*R*(T1-T2)/(g-1);

+//since W is -ve, the work is done on the air

+

+printf(' (i) The Final temperature is: %3.2f deg. celsius. \n',(T2-273));

+printf(' (ii) The Final Volume is: %1.5f m^3. \n',V2);

+printf('(iii) The Work done on the air is: %1.3f kJ. \n',(-W));

diff --git a/2969/CH3/EX3.12/Ex3_12.sce b/2969/CH3/EX3.12/Ex3_12.sce
new file mode 100755
index 000000000..2ccbea2f8
--- /dev/null
+++ b/2969/CH3/EX3.12/Ex3_12.sce
@@ -0,0 +1,28 @@
+clc

+clear

+//DATA GIVEN

+m=0.44;                         //mass of air in kg

+T1=180+273;                     //initial temperature of air in K

+T2=15+273;                      //final temperature of air in K

+W12=52.5;                       //work done during the process in kJ

+//V2/V1=3

+Vr=3;                           //volume ratio, Vr=V2/V1

+

+//Law of adiabatic expansion, pV^g=C

+

+//In an adiabatic process

+//(T2/T1)=(V1/V2)^(g-1);

+g=1+[(log(T2/T1)/log(1/Vr))];              //gamma for air, g=Cp/Cv

+

+//W12=(p1*V1-p2*V2)/(n-1)=mR(T2-T1)/(g-1)

+R=W12/m/(T1-T2)*(g-1);

+//R=Cp-Cv

+

+Cv=R/(g-1);

+Cp=g*Cv;

+

+printf(' (i) The value of Cv is: %1.3f kJ/kgK. \n',Cv);

+printf(' (ii) The value of Cp is: %1.3f kJ/kgK. \n',Cp);

+

+//NOTE:

+//there is slight variation in answers of the book due to rounding off of the values 

diff --git a/2969/CH3/EX3.13/Ex3_13.sce b/2969/CH3/EX3.13/Ex3_13.sce
new file mode 100755
index 000000000..73e7aea89
--- /dev/null
+++ b/2969/CH3/EX3.13/Ex3_13.sce
@@ -0,0 +1,31 @@
+clc

+clear

+//DATA GIVEN

+m=1;                            //mass of etahne gas in kg

+M=30;                           //molecular weight of ethane

+p1=1.1;                         //initial pressure in bar

+T1=27+273;                      //initial temperature in K

+p2=6.6;                         //final pressure in bar

+Cp=1.75;                        //in kJ/kgK

+

+//Law of compression, pV^1.3=C

+n=1.3;

+

+//Characteristic gas constant, R = Universal gas constant (Ro)/Molecular weight(M)

+Ro=8314;

+R=Ro/M/1000;                    //kJ/kgK

+

+//R=Cp-Cv

+Cv=Cp-R;

+g=Cp/Cv;                        //gamma g

+

+//In a polytropic process

+//(T2/T1)=(p2/p1)^((n-1)/n);

+T2=T1*(p2/p1)^((n-1)/n);;       //final temp. T2 in K

+

+//W=(p1*V1-p2*V2)/(n-1)=mR(T2-T1)/(g-1)

+W=m*R*(T1-T2)/(n-1);

+

+Q=[(g-n)/(g-1)]*W;              //heat flow in kJ/kg

+

+printf(' The Heat SUPPLIED is: %2.1f kJ/kg. \n',(Q));

diff --git a/2969/CH3/EX3.2/Ex3_2.sce b/2969/CH3/EX3.2/Ex3_2.sce
new file mode 100755
index 000000000..114ed7333
--- /dev/null
+++ b/2969/CH3/EX3.2/Ex3_2.sce
@@ -0,0 +1,12 @@
+clc

+clear

+//DATA GIVEN

+u1=450;                         //internal energy at beginning of the expansion in kJ/kg

+u2=220;                         //internal energy after expansion in kJ/kg

+W=120;                          //work done by the air during expansion in kJ/kg

+

+//using First Law of Thermodynamics, Q=(u2-u1)+W

+Q=(u2-u1)+W;                    //heat flow in kJ/kg

+//since Q is -ve, there is rejection of heat

+

+printf('The heat REJECTED by air is: %3.0f kJ/kg. \n',(-Q));

diff --git a/2969/CH3/EX3.3/Ex3_3.sce b/2969/CH3/EX3.3/Ex3_3.sce
new file mode 100755
index 000000000..2620c936c
--- /dev/null
+++ b/2969/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,18 @@
+clc

+clear

+//DATA GIVEN

+m=0.3;                          //mass of nitrogen in kg

+p1=0.1;                         //pressure in MPa

+T1=40+273;                      //temperature before compression in K

+p2=1;                           //pressure in MPa

+T2=160+273;                     //temperature after compression in K

+W=-30;                          //work done during the compression in kJ/kg

+Cv=0.75                         //in kJ/kgK

+

+//using First Law of Thermodynamics, Q=(u2-u1)+W

+//(u2-u1)=m*Cv*(T2-T1)

+Du=m*Cv*(T2-T1);

+Q=Du+W;                         //heat flow in kJ/kg

+//since Q is -ve, there is rejection of heat

+

+printf('The heat REJECTED by air is: %1.0f kJ. \n',(-Q));

diff --git a/2969/CH3/EX3.4/Ex3_4.sce b/2969/CH3/EX3.4/Ex3_4.sce
new file mode 100755
index 000000000..59fcab224
--- /dev/null
+++ b/2969/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,20 @@
+clc

+clear

+//DATA GIVEN

+//initial state

+p1=0.105;                       //pressure of gas in MPa

+V1=0.4;                         //volume of gas in m^3

+//final state

+p2=0.105;                       //pressure of gas in MPa

+V2=0.20;                        //volume of gas in m^3

+

+Q=-42.5;                        //heat transferred in kJ

+p=p1;

+

+//process used- ISOBARIC (Constant pressure)

+W12=p*(V2-V1)*1000;             //work in kJ

+//using First Law of Thermodynamics, Q=(u2-u1)+W

+Du=Q-W12;                       //(u2-u1) change in internal energy in kJ

+//since Du is -ve, there is decrease in internal energy

+

+printf('The DECREASE in internal energy is: %2.1f kJ. \n',(-Du));

diff --git a/2969/CH3/EX3.5/Ex3_5.sce b/2969/CH3/EX3.5/Ex3_5.sce
new file mode 100755
index 000000000..f34d24f82
--- /dev/null
+++ b/2969/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,18 @@
+clc

+clear

+//DATA GIVEN

+//part-1

+//pressure=p1,temperature=T1

+//part-2

+//pressure=p2,temperature=T2

+

+//Acc. First Law of Thermodynamics, Q=(u2-u1)+W

+//when partition moved

+DQ=0;

+DW=0;

+DU=DQ-DW;

+//DU=0

+

+printf('     CONCLUSION: \n');

+printf('      Acc. to First Law of Thermodynamics, \n');

+printf('      When partion moved, there is conservation of internal energy. \n');

diff --git a/2969/CH3/EX3.6/Ex3_6.sce b/2969/CH3/EX3.6/Ex3_6.sce
new file mode 100755
index 000000000..6e2efbc99
--- /dev/null
+++ b/2969/CH3/EX3.6/Ex3_6.sce
@@ -0,0 +1,25 @@
+clc

+clear

+//DATA GIVEN

+//initial state

+p1=10^5;                        //initial pressure of air in Pa

+v1=1.8;                         //volume of air in m^3/kg

+T1=25+273;                      //initial temperature of air in K

+//final state

+p2=5*10^5;                      //final pressure of air in Pa

+T2=25+273;                      //final temperature of air in K

+

+//process used- ISOTHERMAL (Constant temperature)

+W12=[p1*v1*log(p1/p2)]/1000;    //work in kJ/kg

+//since W is -ve, work is supplied to the air

+

+//since temperature is constant

+Du=0;                           //(u2-u1) change in internal energy in kJ/kg

+

+//using First Law of Thermodynamics, Q=(u2-u1)+W

+Q=Du+W12;

+//since Q is -ve, there is rejection of heat from system to surroundings

+

+printf(' (i) The Work done on the air is: %3.1f kJ/kg. \n',(-W12));

+printf(' (ii) The change in internal energy is: %1.0f kJ/kg. \n',(Du));

+printf('(iii) The Heat REJECTED is: %3.1f kJ/kg. \n',(-Q));

diff --git a/2969/CH3/EX3.8/Ex3_8.sce b/2969/CH3/EX3.8/Ex3_8.sce
new file mode 100755
index 000000000..da26d5ab7
--- /dev/null
+++ b/2969/CH3/EX3.8/Ex3_8.sce
@@ -0,0 +1,44 @@
+clc

+clear

+//DATA GIVEN

+p1=4*10^5;                      //initial pressure in N/m^2

+V1=0.2;                         //initial volume in m^3

+T1=130+273;                     //initial temperature in K

+p2=1.02*10^5;                   //final pressure after adiabatic expansion in N/m^2

+Q23=72.5;                       //increase in enthalpy during constant pressure process in kJ

+Cp=1;                           //in kJ/kgK

+Cv=0.714;                       //in kJ/khK

+

+//gamma for air, g

+g=Cp/Cv;

+R=(Cp-Cv)*1000;

+

+//for reversible adiabatic process 1-2

+//p1*(V1^g)=p2*(V2^g)

+V2=V1*(p1/p2)^(1/g);            //final volume in m^3

+//(T2/T1)=(p2/p1)^((g-1)/g);

+T2=T1*(p2/p1)^((g-1)/g);;       //final temp. T2 in K

+

+m=p1*V1/R/T1;                   //mass in kg

+

+//for constant pressure process 2-3

+//Q23=m*Cp*(T3-T2);

+T3=Q23/m/Cp+T2;

+//V2/T2=V3/T3

+V3=V2/T2*T3;

+

+//Work done by the path 1-2-3, W123=W12+W23

+W12=(p1*V1-p2*V2)/(g-1);

+W23=p2*(V3-V2);

+W123=W12+W23;

+

+//if the above processes are replaced by a single reversible polytropic process giving the same work between initial and final states,

+//W13=W123=(p1V1-p3V3)/(n-1)

+p3=p2;

+n=1+(p1*V1-p3*V3)/W123;         //index of expansion, n

+

+printf(' (i) The Total Work done is: %5.0f Nm or J. \n',W123);

+printf(' (ii) The value of index of expansion, n is: %1.3f. \n',n);

+

+//NOTE:

+//there is slight variation in answers of the book due to rounding off of the values