initial commit / add all books

22 files changed, 894 insertions, 0 deletions

diff --git a/2705/CH8/EX8.1/Ex8_1.sce b/2705/CH8/EX8.1/Ex8_1.sce
new file mode 100755
index 000000000..dc5881db5
--- /dev/null
+++ b/2705/CH8/EX8.1/Ex8_1.sce
@@ -0,0 +1,24 @@
+

+clear;

+clc;

+disp('Example 8.1');

+

+//  aim : To determine

+//   the stoichiometric mass of air required to burn 1 kg the fuel 

+

+//  Given values

+C = .72;//  mass fraction of C; [kg/kg]

+H2 = .20;//  mass fraction of H2;, [kg/kg]

+O2 = .08;//  mass fraction of O2, [kg/kg]

+aO2=.232;//  composition of oxygen in air

+

+//  solution

+//  for 1kg of fuel

+mO2  = 8/3*C+8*H2-O2;//  mass of O2, [kg]

+

+//  hence stoichiometric mass of O2 required is

+msO2  = mO2/aO2;// [kg]

+

+mprintf('\n The stoichiometric mass of air required to burn 1 kg the fuel should be =  %f  kg\n',msO2);

+

+//  End

diff --git a/2705/CH8/EX8.10/Ex8_10.sce b/2705/CH8/EX8.10/Ex8_10.sce
new file mode 100755
index 000000000..079878869
--- /dev/null
+++ b/2705/CH8/EX8.10/Ex8_10.sce
@@ -0,0 +1,44 @@
+clear;

+clc;

+disp('Example 8.10');

+

+// aim : To determine

+// volumetric composition of the products of combustion

+

+// given values

+C = .86;//  mass composition of carbon

+H = .14;// mass composition of hydrogen

+Ea = .20;//  excess air for combustion

+O2 = .23;// mass composition of O2 in air 

+

+MCO2 = 44;// moleculer mass of CO2

+MH2O = 18;// moleculer mass of H2O

+MO2 = 32;// moleculer mass of O2

+MN2 = 28;// moleculer mass of N2,

+

+

+// solution

+sO2 = (8/3*C+8*H);// stoichiometric O2 required, [kg/kg petrol]

+sair = sO2/O2;// stoichiometric air required, [kg/kg petrol]

+// for one kg petrol

+mCO2 = 11/3*C;// mass of CO2,[kg]

+mH2O = 9*H;// mass of H2O, [kg]

+mO2 = Ea*sO2;// mass of O2, [kg]

+mN2 = 14.84*(1+Ea)*(1-O2);// mass of N2, [kg]

+

+mt = mCO2+mH2O+mO2+mN2;// total mass, [kg]

+// percentage mass composition

+x1 = mCO2/mt*100;// mass composition of CO2

+x2 = mH2O/mt*100;// mass composition of H2O

+x3 = mO2/mt*100;// mass composition of O2

+x4 = mN2/mt*100;// mass composition of N2

+

+vt = x1/MCO2+x2/MH2O+x3/MO2+x4/MN2;// total volume of petrol

+v1 = x1/MCO2/vt*100;// %age composition of CO2 by volume

+v2 = x2/MH2O/vt*100;// %age composition  of H2O by volume

+v3 = x3/MO2/vt*100;// %age composition of O2 by volume

+v4 = x4/MN2/vt*100;// %age composition of N2 by volume

+ 

+mprintf('\nThe percentage composition of CO2 by volume is  =  %f\n,\nThe percentage composition of H2O by volume is  =  %f\n,\nThe percentage composition of O2 by volume is  =  %f\n,\nThe percentage composition of N2 by volume is  =  %f\n',v1,v2,v3,v4);

+

+//  End

diff --git a/2705/CH8/EX8.11/Ex8_11.sce b/2705/CH8/EX8.11/Ex8_11.sce
new file mode 100755
index 000000000..f9ee92977
--- /dev/null
+++ b/2705/CH8/EX8.11/Ex8_11.sce
@@ -0,0 +1,33 @@
+clear;

+clc;

+disp('Example 8.11');

+

+// aim : To determine

+//  the energy carried away by the dry flue gas/kg of fuel burned

+

+// given values

+C = .78;//  mass composition of carbon

+H2 = .06;// mass composition of hydrogen

+O2 = .09;// mass composition of oxygen

+Ash = .07;// mass composition of ash

+Ea = .50;//  excess air for combustion

+aO2 = .23;// mass composition of O2 in air 

+Tb = 273+20;// boiler house temperature, [K]

+Tf = 273+320;// flue gas temperature, [K]

+c = 1.006;// specific heat capacity of dry flue gas, [kJ/kg K]

+

+// solution

+// for one kg of fuel

+sO2 = (8/3*C+8*H2);// stoichiometric O2 required, [kg/kg fuel]

+sO2a = sO2-O2;// stoichiometric O2 required from air, [kg/kg fuel]

+sair = sO2a/aO2;// stoichiometric air required, [kg/kg fuel]

+ma = sair*(1+Ea);// actual air supplied/kg of fuel, [kg]

+// total mass of flue gas/kg fuel is

+mf = ma+1;// [kg]

+mH2 = 9*H2;//H2 produced, [kg] 

+// hence, mass of dry flue gas/kg coall is

+m = mf-mH2;// [kg]

+Q = m*c*(Tf-Tb);// energy carried away by flue gas, [kJ]

+mprintf('\n The energy carried away by the dry flue gas/kg is  =  %f kg\n',Q);

+

+//  End

diff --git a/2705/CH8/EX8.12/Ex8_12.sce b/2705/CH8/EX8.12/Ex8_12.sce
new file mode 100755
index 000000000..ec6dd7d76
--- /dev/null
+++ b/2705/CH8/EX8.12/Ex8_12.sce
@@ -0,0 +1,49 @@
+clear;

+clc;

+disp('Example 8.12');

+

+// aim : To determine

+// (a) the stoichiometric volume of air for the complete combustion of 1 m^3

+// (b) the percentage volumetric analysis of the products of combustion

+

+// given values

+N2 = .018;// volumetric composition of N2

+CH4 = .94;// volumetric composition of CH4

+C2H6 = .035;// volumetric composition of C2H6

+C3H8 = .007;// volumetric composition of C3H8

+aO2 = .21;// O2 composition in air

+

+// solution

+// (a)

+// for CH4

+// CH4 +2 O2= CO2 + 2 H2O

+sva1 = 2/aO2;// stoichiometric volume of air, [m^3/m^3 CH4]

+svn1 = sva1*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 CH4]

+

+// for C2H6

+// 2 C2H6 +7 O2= 4 CO2 + 6 H2O

+sva2 = 7/2/aO2;// stoichiometric volume of air, [m^3/m^3 C2H6]

+svn2 = sva2*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 C2H6]

+

+// for C3H8

+// C3H8 +5 O2=3 CO2 + 4 H2O

+sva3 = 5/aO2;// stoichiometric volume of air, [m^3/m^3 C3H8]

+svn3 = sva3*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 C3H8]

+

+Sva = CH4*sva1+C2H6*sva2+C3H8*sva3;// stoichiometric volume of air required, [m^3/m^3 gas]

+mprintf('\n (a) The stoichiometric volume of air for the complete combustion  =  %f m^3m^3 gas\n',Sva);

+

+// (b)

+// for one m^3 of natural gas

+vCO2 = CH4*1+C2H6*2+C3H8*3;// volume of CO2 produced, [m^3]

+vH2O = CH4*2+C2H6*3+C3H8*4;// volume of H2O produced, [m^3]

+vN2 = CH4*svn1+C2H6*svn2+C3H8*svn3+N2;// volume of N2 produced, [m^3]

+

+vg = vCO2+vH2O+vN2;// total volume of gas, [m^3]

+x1 = vCO2/vg*100;// volume percentage of CO2 produced

+x2 = vH2O/vg*100;// volume percentage of H2O produced

+x3 = vN2/vg*100;// volume percentage of N2 produced

+

+mprintf('\n (b) The percentage volumetric composition of CO2 in produced is  =  %f\n,\n     The percentage volumetric composition of H2O in produced is  =  %f\n,\n     The percentage volumetric composition of N2 in produced is  =  %f\n',x1,x2,x3);

+

+// End

diff --git a/2705/CH8/EX8.13/Ex8_13.sce b/2705/CH8/EX8.13/Ex8_13.sce
new file mode 100755
index 000000000..87f8573ec
--- /dev/null
+++ b/2705/CH8/EX8.13/Ex8_13.sce
@@ -0,0 +1,46 @@
+clear;

+clc;

+disp('Example 8.13');

+

+// aim : To determine

+// (a) the volume of air taken by the fan

+// (b) the percentage composition of dry flue gas

+

+// gien values

+C = .82;// mass composition of carbon

+H = .08;// mass composition of hydrogen

+ O = .03;// mass composition of oxygen

+ A = .07;// mass composition of ash

+mc = .19;// coal uses, [kg/s] 

+ ea = .3;// percentage excess air of oxygen in the air required for combustion

+Oa = .23;// percentage of oxygen by mass in the air

+  

+ // solution

+ // (a)

+ P = 100;// air pressure, [kN/m^2]

+ T = 18+273;// air temperature, [K]

+ R = .287;// [kJ/kg K]

+ // basis one kg coal

+ sO2 = 8/3*C+8*H;// stoichiometric O2 required, [kg]

+ aO2 = sO2-.03;// actual O2 required, [kg]

+tO2 = aO2/Oa;// theoretical O2 required, [kg]

+Aa = tO2*(1+ea);// actual air supplied, [kg]

+m = Aa*mc;// Air supplied, [kg/s]

+

+// now using P*V=m*R*T

+V = m*R*T/P;// volume of air taken ,[m^3/s]

+mprintf('\n (a) Volume of air taken by fan is  =  %f  m^3/s\n',V);

+

+// (b)

+mCO2 = 11/3*C;// mass of CO2 produced, [kg]

+mO2 = aO2*.3;// mass of O2 produces, [kg]

+mN2 = Aa*.77;// mass of N2 produced, [[kg]

+mt = mCO2+mO2+mN2;// total mass, [kg]

+

+mprintf('\n (b) Percentage mass composition of CO2 is  =  %f percent \n',mCO2/mt*100);

+mprintf('\n     Percentage mass composition of O2 is  =  %f  percent\n',mO2/mt*100)

+mprintf('\n     Percentage mass composition of N2 is  =  %f  percent \n',mN2/mt*100)

+

+

+

+//  End

diff --git a/2705/CH8/EX8.14/Ex8_14.sce b/2705/CH8/EX8.14/Ex8_14.sce
new file mode 100755
index 000000000..a93839efd
--- /dev/null
+++ b/2705/CH8/EX8.14/Ex8_14.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+disp('Example 8.14');

+

+// aim : To determine 

+// (a) the mass of fuel used per cycle

+// (b) the actual mass of air taken in per cycle

+// (c) the volume of air taken in per cycle

+

+// given values

+W = 15;// work done, [kJ/s]

+N = 5;// speed, [rev/s]

+C = .84;// mass composition of carbon

+H = .16;// mass composition of hydrogen

+ea = 1;// percentage excess air supplied 

+CV = 45000;// calorificvalue of fuel, [kJ/kg]

+n_the = .3;// thermal efficiency

+P = 100;// pressuer, [kN/m^2]

+T = 273+15;// temperature, [K]

+R = .29;// gas constant, [kJ/kg K]

+

+// solution

+// (a)

+E = W*2/N/n_the;// energy supplied, [kJ/cycle]

+mf = E/CV;// mass of fuell used, [kg]

+mprintf('\n (a) Mass of fuel used per cycle is  =  %f g\n',mf*10^3);

+

+// (b)

+// basis 1 kg fuel

+mO2 = C*8/3+8*H;// mass of O2 requirea, [kg]

+smO2 = mO2/.23;// stoichiometric mass of air, [kg]

+ma = smO2*(1+ea);// actual mass of air supplied, [kg]

+m = ma*mf;// mass of air supplied, [kg/cycle]

+mprintf('\n (b) The mass of air supplied per cycle is  =  %f  kg\n',m);

+

+// (c)

+V = m*R*T/P;// volume of air, [m^3]

+mprintf('\n (c) The volume of air taken in per cycle is  =  %f  m^3\n',V);

+

+// End

diff --git a/2705/CH8/EX8.15/Ex8_15.sce b/2705/CH8/EX8.15/Ex8_15.sce
new file mode 100755
index 000000000..a4adab52d
--- /dev/null
+++ b/2705/CH8/EX8.15/Ex8_15.sce
@@ -0,0 +1,70 @@
+clear;

+clc;

+disp('Example 8.15');

+

+//  aim : To determine

+//  (a) the mass of coal used per hour

+//  (b) the mass of air used per hour

+//  (c) the percentage analysis of the flue gases by mass

+

+//  given values

+m = 900;// mass of steam boiler generate/h, [kg]

+x = .96;// steam dryness fraction

+P = 1400;// steam pressure, [kN/m^2]

+Tf = 52;// feed water temperature, [C]

+BE = .71;// boiler efficiency

+CV = 33000;// calorific value  of coal, [kJkg[

+ea = .22;// excess air supply

+aO2 = .23;// oxygen composition in air

+c = 4.187;// specific heat capacity of water, [kJ/kg K]

+

+//  coal composition

+C = .83;// mass composition of carbon

+H2 = .05;// mass composition of hydrogen

+O2 = .03;// mass composition of oxygen

+ash = .09;// mass composition of ash

+

+// solution

+// from steam table at pressure P

+hf = 830.1;// specific enthalpy, [kJ/kg]

+hfg = 1957.1;// specific enthalpy, [kJ/kg]

+hg = 2728.8;// specific enthalpy, [kJ/kg]

+

+// (a)

+h = hf+x*hfg;// specific enthalpy of steam generated by boiler, [kJ/kg]

+hfw = c*Tf;// specific enthalpy of feed water, [kJ/kg]

+Q = m*(h-hfw);// energy to steam/h, [kJ]

+Qf = Q/BE;// energy required from fuel/h, [kJ]

+mc = Qf/CV;// mass of coal/h,[kg]

+mprintf('\n (a) The mass of coal used per hour is  =  %f kg\n',mc);

+

+// (b)

+// for one kg coal

+mO2 = 8/3*C+8*H2+-O2;// actual mass of O2 required, [kg]

+mta = mO2/aO2;// theoretical mass of air, [kg]

+ma = mta*(1+ea);// mass of air supplied, [kg]

+mas = ma*mc;// mass of air supplied/h, [kg]

+mprintf('\n (b) The mass of air supplied per hour is  =  %f kg\n',mas);

+

+ 

+// (c)

+// for one kg coal

+mCO2 = 11/3*C;// mass of CO2 produced, [kg]

+mH2O = 9*H2;// mass of H2O produced, [kg]

+mO2 = mO2*ea;// mass of excess O2 in flue gas, [kg]

+mN2 = ma*(1-aO2);// mass of N2 in flue gas, [kg]

+

+mt = mCO2+mH2O+mO2+mN2;// total mass of gas

+x1 = mCO2/mt*100;// mass percentage composition of CO2

+x2 = mH2O/mt*100;// mass percentage composition of H2O

+x3 = mO2/mt*100;// mass percentage composition of O2

+x4 = mN2/mt*100;// mass percentage composition of N2

+

+mprintf('\n (c) The mass percentage composition of CO2  =  %f,\n      The mass percentage composition of H2O  =  %f,\n      The mass percentage composition of O2  =  %f,\n      The mass percentage composition of N2  =  %f',x1,x2,x3,x4);

+

+//  mass of coal taken in part (b) is wrong so answer is not matching

+

+//  End

+

+

+

diff --git a/2705/CH8/EX8.16/Ex8_16.sce b/2705/CH8/EX8.16/Ex8_16.sce
new file mode 100755
index 000000000..5e1808b5d
--- /dev/null
+++ b/2705/CH8/EX8.16/Ex8_16.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+disp('Example 8.16');

+

+// aim : To determine

+// (a) volume of gas

+// (b) (1) the average molecular mass of air

+//       (2) the value of R

+//       (3) the mass of 1 m^3 of air at STP

+

+// given values

+n = 1;// moles of gas, [kmol]

+P = 101.32;// standard pressure, [kN/m^2]

+T = 273;// gas tempearture, [K]

+

+O2 = 21;// percentage volume composition of oxygen in air

+N2 = 79;// percentage volume composition of nitrogen in air

+R = 8.3143;// molar gas constant, [kJ/kg K]

+mO2 = 32;// moleculer mass of O2

+mN2 = 28;// moleculer mass of N2

+

+// solution

+// (a)

+V = n*R*T/P;// volume of gas, [m^3]

+mprintf('\n (a) The volume of the gas is  =  %f m^3\n',V);

+

+// (b)

+//(1)

+Mav = (O2*mO2+N2*mN2)/(O2+N2);// average moleculer mass of air

+mprintf('\n (b)(1) The average moleculer mass of air is  =  %f g/mol\n',Mav);

+

+// (2)

+Rav = R/Mav;// characteristic gas constant, [kJ/kg k]

+mprintf('\n     (2) The value of R is  =  %f  kJ/kg K\n',Rav);

+

+// (3)

+rho = Mav/V;// density of air, [kg/m^3]

+mprintf('\n     (3) The mass of one cubic metre of air at STP is  =  %f  kg/m^3\n',rho);

+

+//  End

diff --git a/2705/CH8/EX8.17/Ex8_17.sce b/2705/CH8/EX8.17/Ex8_17.sce
new file mode 100755
index 000000000..d1099ee95
--- /dev/null
+++ b/2705/CH8/EX8.17/Ex8_17.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 8.17');

+

+// aim : To determine

+// (a) the partial pressure of each gas in the vessel

+// (b) the volume of the vessel

+// (c)  the total pressure in the gas when temperature is raised to228 C

+

+// given values

+MO2 = 8;//  mass of O2, [kg]

+MN2 = 7;//  mass of N2, [kg]

+MCO2 = 22;//  mass of CO2, [kg]

+

+P = 416;// total pressure in the vessel, [kN/m^2]

+T = 273+60;// vessel temperature, [K]

+R = 8.3143;// gas constant, [kJ/kmol K]

+

+mO2 = 32;// molculer mass of O2 

+mN2 = 28;// molculer mass of N2

+mCO2 = 44;// molculer mass of CO2

+

+// solution

+// (a)

+n1 = MO2/mO2;// moles of O2, [kmol]

+n2 = MN2/mN2;// moles of N2, [kmol]

+n3 = MCO2/mCO2;// moles of CO2, [kmol]

+

+n = n1+n2+n3;// total moles in the vessel, [kmol]

+// since,Partial pressure is proportinal, so

+P1 = n1*P/n;// partial pressure of O2, [kN/m^2]

+P2 = n2*P/n;// partial pressure of N2, [kN/m^2]

+P3 = n3*P/n;// partial pressure of CO2, [kN/m^2]

+

+mprintf('\n (a)The partial pressure of O2 is  =  %f  kN/m^2,\n,   The partial pressure of N2 is  =  %f  kN/m^2,\n    The partial pressure of CO2 is  =  %f  kN/m^2,\n',P1,P2,P3);

+

+// (b)

+// assuming ideal gas 

+V = n*R*T/P;// volume of the container, [m^3]

+mprintf('\n (b) The volume of the container is  =  %f  m^3\n',V);

+

+// (c)

+T2 = 273+228;// raised vessel temperature, [K]

+// so volume of vessel  will constant , P/T=constant

+P2 = P*T2/T;// new pressure in the vessel , [kn/m62]

+mprintf('\n (c) The new total pressure in the vessel is  =  %f  kN/m^2\n',P2);

+

+//  End

diff --git a/2705/CH8/EX8.18/Ex8_18.sce b/2705/CH8/EX8.18/Ex8_18.sce
new file mode 100755
index 000000000..c0961970a
--- /dev/null
+++ b/2705/CH8/EX8.18/Ex8_18.sce
@@ -0,0 +1,65 @@
+clear;

+clc;

+disp('Example 8.18');

+

+// aim : To determine

+// the actual mass of air supplied/kg coal

+// the velocity of flue gas

+

+// given values

+mc = 635;// mass of coal burn/h, [kg]

+ea = .25;// excess air required

+C = .84;// mass composition of carbon

+H2 = .04;// mass composition of hydrogen

+O2 = .05;// mass composition of oxygen

+ash = 1-(C+H2+O2);// mass composition of ash

+

+P1 = 101.3;// pressure, [kJn/m^2]

+T1 = 273;// temperature, [K]

+V1 = 22.4;// volume, [m^3]

+

+T2 = 273+344;// gas temperature, [K]

+P2 = 100;// gas pressure, [kN/m^2]

+A = 1.1;// cross section area, [m^2]

+aO2 = .23;// composition of O2 in air

+

+mCO2 = 44;// moleculer mass of carbon

+mH2O = 18;// molecular mass of hydrogen

+mO2 = 32;// moleculer mas of oxygen

+mN2 = 28;// moleculer mass of nitrogen

+

+// solution

+mtO2 = 8/3*C+8*H2-O2;// theoretical O2 required/kg coal, [kg]

+msa= mtO2/aO2;// stoichiometric mass of  air supplied/kg coal, [kg]

+mas = msa*(1+ea);// actual mass of air supplied/kg coal, [kg]

+

+m1 = 11/3*C;// mass of CO2/kg coal produced, [kg]

+m2 = 9*H2;// mass of H2/kg coal produced, [kg]

+m3 = mtO2*ea;// mass of O2/kg coal produced, [kg]

+m4 = mas*(1-aO2);// mass of N2/kg coal produced, [kg]

+

+mt = m1+m2+m3+m4;// total mass, [kg]

+x1 = m1/mt*100;// %age mass composition of CO2 produced

+x2 = m2/mt*100;// %age mass composition of H2O produced

+x3 = m3/mt*100;// %age mass composition of O2 produced

+x4 = m4/mt*100;// %age mass composition of N2 produced

+

+vt = x1/mCO2+x2/mH2O+x3/mO2+x4/mN2;// total volume

+v1 = x1/mCO2/vt*100;// %age volume composition of CO2

+v2 = x2/mH2O/vt*100;// %age volume composition of H2O

+v3 = x3/mO2/vt*100;// %age volume composition of O2

+v4 = x4/mN2/vt*100;// %age volume composition of N2

+

+Mav = (v1*mCO2+v2*mH2O+v3*mO2+v4*mN2)/(v1+v2+v3+v4);// average moleculer mass, [kg/kmol]

+// since no of moles is constant so PV/T=constant

+V2 = P1*V1*T2/(P2*T1);//volume, [m^3]

+

+mp = mt*mc/3600;// mass of product of combustion/s, [kg]

+

+V = V2*mp/Mav;// volume of flowing gas /s,[m^3]

+

+v = V/A;// velocity of flue gas, [m/s]

+mprintf('\n The actual mass of air supplied is  =  %f  kg/kg coal\n',mas);

+mprintf('\n The velocity of flue gas is  =  %f m/s\n',v);

+

+//  End

diff --git a/2705/CH8/EX8.19/Ex8_19.sce b/2705/CH8/EX8.19/Ex8_19.sce
new file mode 100755
index 000000000..30bfc9211
--- /dev/null
+++ b/2705/CH8/EX8.19/Ex8_19.sce
@@ -0,0 +1,51 @@
+clear;

+clc;

+disp('Example 8.19');

+

+// aim : To determine

+// (a) the temperature of the gas after compression

+// (b) the density of the air-gas mixture

+

+// given values

+CO = 26;// %age volume composition of CO 

+H2 = 16;// %age volume composition of H2

+CH4 = 7;// %age volume composition of CH4 

+N2 = 51;// %age volume composition of N2

+

+P1 = 103;// gas pressure, [kN/m^2]

+T1 = 273+21;// gas temperature, [K]

+rv = 7;// volume ratio

+

+aO2 = 21;// %age volume composition of O2 in the air

+c = 21;// specific heat capacity of diatomic gas, [kJ/kg K]

+cCH4 = 36;// specific heat capacity of CH4, [kJ/kg K]

+R = 8.3143;// gas constant, [kJ/kg K]

+

+mCO = 28;// moleculer mass of carbon

+mH2 = 2;// molecular mass of hydrogen

+mCH4 = 16;// moleculer mas of methane

+mN2 = 28;// moleculer mass of nitrogen

+mO2 = 32;// moleculer mass of oxygen

+

+// solution

+// (a)

+Cav = (CO*c+H2*c+CH4*cCH4+N2*c+100*2*c)/(100+200);// heat capacity, [kJ/kg K]

+

+Gama = (Cav+R)/Cav;// heat capacity ratio

+// rv = V1/V2

+// process is polytropic, so

+T2 = T1*(rv)^(Gama-1);// final tempearture, [K]

+mprintf('\n (a) The temperature of the gas after compression is  =  %f C\n',T2-273);

+

+// (b)

+

+Mav = (CO*mCO+H2*mH2+CH4*mCH4+N2*mN2+42*mO2+158*mN2)/(100+200)

+

+// for 1 kmol of gas

+V = R*T1/P1;// volume of one kmol of gas, [m^3]

+// hence

+rho = Mav/V;// density of gas, [kg/m^3]

+

+mprintf('\n (b) The density of air-gas mixture is  =  %f  kg/m^3\n',rho);

+

+//  End

diff --git a/2705/CH8/EX8.20/Ex8_20.sce b/2705/CH8/EX8.20/Ex8_20.sce
new file mode 100755
index 000000000..51e9a6f72
--- /dev/null
+++ b/2705/CH8/EX8.20/Ex8_20.sce
@@ -0,0 +1,19 @@
+clear;

+clc;

+disp('Example 8.20');

+

+// aim : to determine 

+// stoichiometric equation for combustion of hydrogen

+

+// solution

+// equation with algebric coefficient is

+// H2+aO2+79/21*aN2=bH2O+79/21*aN2

+// by equating coefficients

+b = 1;

+a = b/2;

+// so equation becomes

+// 2 H2+ O2+3.76 N2=2 H2O+3.76 N2

+disp('The required stoichiometric equation is  =  ');

+disp('2 H2+ O2+3.76 N2 = 2 H2O+3.76 N2');

+

+// End

diff --git a/2705/CH8/EX8.22/Ex8_22.sce b/2705/CH8/EX8.22/Ex8_22.sce
new file mode 100755
index 000000000..9e5bddbd5
--- /dev/null
+++ b/2705/CH8/EX8.22/Ex8_22.sce
@@ -0,0 +1,53 @@
+clear;

+clc;

+disp('Example 8.22');

+

+// aim : To determine

+// the percentage gravimetric analysis of the total products of combustion

+

+// given values

+CO = 12;//  %age volume composition of CO

+H2 = 41;//  %age volume composition of H2

+CH4 = 27;//  %age volume composition of CH4

+O2 = 2;//  %age volume composition of O2

+CO2 = 3;//  %age volume composition of CO2

+N2 = 15;//  %age volume composition of N2

+

+mCO2 = 44;// moleculer mass of CO2,[kg/kmol]

+mH2O = 18;// moleculer mass of H2O, [kg/kmol]

+mO2 = 32;// moleculer mass of O2, [kg/kmol]

+mN2 = 28;// moleculer mass of N2, [kg/kmol]

+ 

+ea = 15;// %age excess air required

+aO2 = 21;// %age air composition in the air

+

+// solution

+// combustion equation by no. of moles

+// 12CO + 41H2 + 27CH4 + 2O2 + 3CO2 + 15N2 + aO2+79/21*aN2 = bCO2 + dH2O + eO2 + 15N2 +79/21*aN2

+// equating C coefficient

+b = 12+27+3;// [mol]

+// equatimg H2 coefficient

+d = 41+2*27;// [mol]

+// O2 required is 15 % extra,so

+// e/(e-a)=.15 so e=.13a

+// equating O2 coefficient

+// 2+3+a=b+d/2 +e

+

+a = (b+d/2-5)/(1-.13);

+e = .13*a;// [mol]

+

+// gravimetric analysis of product

+v1 =  b*mCO2;// gravimetric volume of CO2 

+v2 =  d*mH2O ;// gravimetric volume of H2O 

+v3 = e*mO2;// gravimetric volume of O2

+v4 = 15*mN2 +79/21*a*mN2;// gravimetric volume of N2 

+

+vt = v1+v2+v3+v4;// total

+x1 = v1/vt*100;// percentage gravimetric of CO2

+x2 = v2/vt*100;// percentage gravimetric of H2O

+x3 = v3/vt*100;// percentage gravimetric of O2

+x4 = v4/vt*100;// percentage gravimetric of N2

+

+mprintf('\n Percentage gravimetric composition of CO2  =  %f\n ,\n Percentage gravimetric composition of H2O  =  %f\n\n Percentage gravimetric composition of O2  =  %f\n\n Percentage gravimetric composition of N2  =  %f\n',x1,x2,x3,x4);

+

+//  End 

diff --git a/2705/CH8/EX8.23/Ex8_23.sce b/2705/CH8/EX8.23/Ex8_23.sce
new file mode 100755
index 000000000..6a5a20ce2
--- /dev/null
+++ b/2705/CH8/EX8.23/Ex8_23.sce
@@ -0,0 +1,54 @@
+clear;

+clc;

+disp('Example 8.23');

+

+//  aim : To determine

+// (a) the actual quantity of air supplied/kg of fuel

+// (b) the volumetric efficiency of the engine

+

+// given values

+d = 300*10^-3;// bore,[m]

+L = 460*10^-3;// stroke,[m]

+N = 200;// engine speed, [rev/min]

+

+C = 87;//  %age mass composition of Carbon in the fuel

+H2 = 13;//  %age mass composition of H2 in the fuel

+

+mc = 6.75;// fuel consumption, [kg/h]

+

+CO2 = 7;// %age composition of CO2 by volume

+O2 = 10.5;// %age composition of O2 by volume

+N2 = 7;// %age composition of N2 by volume

+

+mC = 12;// moleculer mass of CO2,[kg/kmol]

+mH2 = 2;// moleculer mass of H2, [kg/kmol]

+mO2 = 32;// moleculer mass of O2, [kg/kmol]

+mN2 = 28;// moleculer mass of N2, [kg/kmol]

+

+T = 273+17;// atmospheric temperature, [K]

+P = 100;// atmospheric pressure, [kn/m^2]

+R =.287;// gas constant, [kJ/kg k]

+

+// solution

+// (a)

+// combustion equation by no. of moles

+// 87/12 C + 13/2 H2 + a O2+79/21*a N2 = b CO2 + d H2O + eO2 + f N2

+// equating  coefficient

+b = 87/12;// [mol]

+a = 22.7;// [mol]

+e = 10.875;// [mol]

+f = 11.8*b;// [mol]

+// so fuel side combustion equation is

+// 87/12 C + 13/2 H2 +22.7 O2 +85.5 N2

+mair = ( 22.7*mO2 +85.5*mN2)/100;// mass of air/kg fuel, [kg]

+mprintf('\n (a) The mass of actual air supplied per kg of fuel is  =  %f  kg\n',mair);

+

+// (b)

+m = mair*mc/60;// mass of air/min, [kg]

+V = m*R*T/P;// volumetric flow of air/min, [m^3]

+SV = %pi/4*d^2*L*N/2;// swept volume/min, [m^3]

+

+VE = V/SV;// volumetric efficiency

+mprintf('\n (b) The volumetric efficiency of the engine is  =  %fpercent\n',VE*100);

+

+//  End

diff --git a/2705/CH8/EX8.24/Ex8_24.sce b/2705/CH8/EX8.24/Ex8_24.sce
new file mode 100755
index 000000000..42ac180f3
--- /dev/null
+++ b/2705/CH8/EX8.24/Ex8_24.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+disp('Example 8.24');

+

+// aim : To determine

+//  the mass of air supplied/kg of fuel burnt

+

+// given values

+// gas composition in the fuel

+C = 84;//  %age mass composition of Carbon in the fuel

+H2 = 14;//  %age mass composition of H2 in the fuel

+O2f = 2;// %age mass composition of O2 in the fuel

+

+// exhaust gas composition

+CO2 = 8.85;// %age composition of CO2 by volume

+CO = 1.2// %age composition of CO by volume

+O2 = 6.8;// %age composition of O2 by volume

+N2 = 83.15;// %age composition of N2 by volume

+

+mC = 12;// moleculer mass of CO2,[kg/kmol]

+mH2 = 2;// moleculer mass of H2, [kg/kmol]

+mO2 = 32;// moleculer mass of O2, [kg/kmol]

+mN2 = 28;// moleculer mass of N2, [kg/kmol]

+

+// solution

+// combustion equation by no. of moles

+// 84/12 C + 14/2 H2 +2/32 O2 + a O2+79.3/20.7*a N2 = b CO2 + d CO2+  eO2 + f N2 +g H2

+// equating  coefficient and given condition

+b = 6.16;// [mol]

+a = 15.14;// [mol]

+d = .836;// [mol]

+f = 69.3*d;// [mol]

+// so fuel side combustion equation is

+// 84/12 C + 14/2 H2 +2/32 O2 +  15.14 O2 +85.5 N2

+mair = ( a*mO2 +f*mN2)/100;// mass of air/kg fuel, [kg]

+mprintf('\n The mass of air supplied per kg of fuel is  =  %f  kg\n',mair);

+

+//  End

diff --git a/2705/CH8/EX8.3/Ex8_3.sce b/2705/CH8/EX8.3/Ex8_3.sce
new file mode 100755
index 000000000..6050727e8
--- /dev/null
+++ b/2705/CH8/EX8.3/Ex8_3.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+disp('Example 8.3');

+

+//  aim : To determine 

+//  the stoichiometric mass of air

+//  the products of combustion both by mass and as percentage 

+

+//  Given values

+C = .82;// mass composition C

+H2 = .12;// mass composition of H2

+O2 = .02;// mass composition of O2

+S = .01;// mass composition of S

+N2 = .03;// mass composition of N2

+

+ //  solution

+//  for 1kg fuel

+mo2 = 8/3*C+8*H2-O2+S*1;// total mass of  O2 required, [kg]

+sa = mo2/.232;// stoichimetric  air, [kg]

+mprintf('\n The stoichiometric mass of air is  =  %f kg/kg fuel\n',sa);

+

+// for one kg fuel

+mCO2 = C*11/3;// mass of CO2 produced, [kg]

+mH2O = H2*9;// mass of H2O produced, [kg]

+mSO2 = S*2;// mass of SO2 produce, [kg]

+mN2 = C*8.84+H2*26.5-O2*.768/.232+S*3.3+N2;// mass of N2 produced, [kg]

+

+mt = mCO2+mH2O+mSO2+mN2;// total mass of product, [kg]

+

+x1 = mCO2/mt*100;// %age mass composition of CO2 produced

+x2 = mH2O/mt*100;// %age mass composition of H2O produced

+x3 = mSO2/mt*100;// %age mass composition of SO2 produced

+x4 = mN2/mt*100;// %age mass composition of N2 produced

+

+mprintf('\n CO2 produced  =  %f  kg/kg fuel,  percentage composition  =  %f,\n H2O produced  =  %f  kg/kg fuel,  percentage composition  =  %f,\n SO2 produced  =  %f  kg/kg fuel,  percentage composition  =  %f,\n N2 produced  =  %f  kg/kg fuel,  percentage composition  =  %f',mCO2,x1,mH2O,x2,mSO2,x3,mN2,x4);

+

+//  End

diff --git a/2705/CH8/EX8.4/Ex8_4.sce b/2705/CH8/EX8.4/Ex8_4.sce
new file mode 100755
index 000000000..6938b2413
--- /dev/null
+++ b/2705/CH8/EX8.4/Ex8_4.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp('Example 8.4');

+

+//  aim : To determine 

+//  the stoichiometric volume of air required for complete combution of 1 m^3 of the gas

+

+//  Given values

+H2 = .14;// volume fraction of H2

+CH4 = .02;// volume fraction of CH4

+CO = .22;// volume fraction of CO

+CO2 = .05;// volume fraction of CO2

+O2 = .02;// volume fraction of O2

+N2 = .55;// volume fraction of N2

+

+// solution

+//  for 1 m^3 of fuel

+Va = .5*H2+2*CH4+.5*CO-O2;// [m^3]

+

+//  stoichiometric air required is

+Vsa = Va/.21;//  [m^3]

+

+mprintf('\n The  stoichiometric volume of air required for complete combustion is  =  %f  m^3/m^3  fuel\n',Vsa);

+

+// End

diff --git a/2705/CH8/EX8.5/Ex8_5.sce b/2705/CH8/EX8.5/Ex8_5.sce
new file mode 100755
index 000000000..a69209cb1
--- /dev/null
+++ b/2705/CH8/EX8.5/Ex8_5.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+disp('Example 8.5');

+

+//  aim : To determine

+//   the volume of the air required 

+

+//  Given values

+H2 = .45;// volume fraction of H2

+CO = .40;// volume fraction of CO

+CH4 = .15;// volume fraction of CH4

+

+//  solution

+V = 2.38*(H2+CO)+9.52*CH4;// stoichimetric volume of air, [m^3]

+

+mprintf('\n The volume of air required is  =  %f  m^3/m^3 fuel\n',V);

+

+// Result in the book is misprinted

+

+// End

+

diff --git a/2705/CH8/EX8.6/Ex8_6.sce b/2705/CH8/EX8.6/Ex8_6.sce
new file mode 100755
index 000000000..fe97bb2fb
--- /dev/null
+++ b/2705/CH8/EX8.6/Ex8_6.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+disp('Example 8.6');

+

+// aim : To determine

+// the stoichiometric volume of air for the complete combustion

+// the products of combustion

+

+// given values

+CH4 = .142;// volumetric composition of CH4

+CO2 = .059;// volumetric composition of CO2

+CO = .360;// volumetric composition of CO

+H2 = .405;// volumetric composition of H2

+O2 = .005;// volumetric composition of O2

+N2 = .029;// volumetric composition of N2

+

+aO2 = .21;// O2 composition into air by volume

+

+//  solution

+svO2 = CH4*2+CO*.5+H2*.5-O2;// stroichiometric volume of O2 required, [m^3/m^3 fuel]

+svair = svO2/aO2;// stroichiometric volume of air required, [m^3/m^3 fuel]

+mprintf('\n Stoichiometric volume of air required is  =  %f  m^3/m^3 fuel\n',svair);

+

+// for one m^3 fuel

+vN2 = CH4*7.52+CO*1.88+H2*1.88-O2*.79/.21+N2;// volume of N2 produced, [m^3]

+vCO2 = CH4*1+CO2+CO*1;// volume of CO2 produced, [m^3]

+vH2O = CH4*2+H2*1;// volume of H2O produced, [m^3]

+

+vt = vN2+vCO2+vH2O;// total volume of product, [m^3]

+

+x1 = vN2/vt*100;// %age composition of N2 in product,

+x2 = vCO2/vt*100;// %age composition of CO2 in product

+x3 = vH2O/vt*100;// %age composition of H2O in product

+

+mprintf('\n N2 in products  =  %fm^3/m^3  fuel,  percentage composition  =  %f,\n CO2 in products =  %f  m^3/m^3 fuel,  percentage composition  =  %f,\n H2O in products  =  %fm^3/m^3  fuel,  percentage composition  =  %f',vN2,x1,vCO2,x2,vH2O,x3);

+

+// End 

diff --git a/2705/CH8/EX8.7/Ex8_7.sce b/2705/CH8/EX8.7/Ex8_7.sce
new file mode 100755
index 000000000..534f4f01e
--- /dev/null
+++ b/2705/CH8/EX8.7/Ex8_7.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp('Example 8.7');

+

+//  aim : To determine

+//  the percentage analysis of the gas by mass

+

+//  Given values

+CO2 = 20;// percentage volumetric composition  of CO2

+N2 = 70;// percentage volumetric composition of N2

+O2 = 10;// percentage volumetric composition  of O2

+

+mCO2 = 44;//  moleculer mas of CO2

+mN2 = 28;//  moleculer mass of N2

+mO2 = 32;//  moleculer mass of O2

+

+//  solution

+mgas = CO2*mCO2+N2*mN2+O2*mO2;//  moleculer mass of gas 

+m1 = CO2*mCO2/mgas*100;// percentage composition of CO2 by mass 

+m2 = N2*mN2/mgas*100;// percentage composition of N2 by mass 

+m3 = O2*mO2/mgas*100;// percentage composition of O2 by mass 

+

+mprintf('\n Mass percentage of CO2 is  =  %f\n\n Mass percentage of N2 is  =  %f\n\n Mass percentage of O2 is  =  %f\n',m1,m2,m3 )

+

+// End

diff --git a/2705/CH8/EX8.8/Ex8_8.sce b/2705/CH8/EX8.8/Ex8_8.sce
new file mode 100755
index 000000000..347ec4c01
--- /dev/null
+++ b/2705/CH8/EX8.8/Ex8_8.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+disp('Example 8.8');

+

+//  aim : To determine

+//  the percentage composition of the gas by volume

+

+//  given values

+CO = 30;// %age mass composition of CO

+N2 = 20;// %age mass composition of N2

+CH4 = 15;// %age mass composition of CH4

+H2 = 25;// %age mass composition of H2

+O2 = 10;// %age mass composition of O2

+

+mCO = 28;// molculer mass of CO

+mN2 = 28;// molculer mass of N2

+mCH4 = 16;// molculer mass of CH4

+mH2 = 2;// molculer mass of H2

+mO2 = 32;// molculer mass of O2

+

+// solution

+vg = CO/mCO+N2/mN2+CH4/mCH4+H2/mH2+O2/mO2;

+v1 = CO/mCO/vg*100;// %age volume composition of CO

+v2 = N2/mN2/vg*100;// %age volume composition of N2

+v3 = CH4/mCH4/vg*100;// %age volume composition of CH4

+v4 = H2/mH2/vg*100;// %age volume composition of H2

+v5 = O2/mO2/vg*100;// %age volume composition of O2

+

+mprintf('\n The percentage composition of CO by volume is  =  %f\n,\nThe percentage composition of N2 by volume is  =  %f\n\nThe percentage composition of CH4 by volume is  =  %f\n\nThe percentage composition of H2 by volume is  =  %f\n\nThe percentage composition of O2by volume is=%f',v1,v2,v3,v4,v5);

+

+// End

diff --git a/2705/CH8/EX8.9/Ex8_9.sce b/2705/CH8/EX8.9/Ex8_9.sce
new file mode 100755
index 000000000..0528a2575
--- /dev/null
+++ b/2705/CH8/EX8.9/Ex8_9.sce
@@ -0,0 +1,44 @@
+clear;

+clc;

+disp('Example 8.9');

+

+// aim : To determine

+// the mass of air supplied per kilogram of fuel burnt

+

+// given values

+CO2 = 8.85;// volume composition of CO2

+CO = 1.2;//  volume composition of CO

+O2 = 6.8;//  volume composition of O2

+N2 = 83.15;//  volume composition of N2 

+

+// composition of gases in the fuel oil

+C = .84;// mass composition of carbon 

+H = .14;// mass composition of hydrogen

+o2 = .02;// mass composition of oxygen

+

+mC = 12;// moleculer mass of Carbon

+mCO2 = 44;// molculer mass of CO2

+mCO = 28;// molculer mass of CO

+mN2 = 28;// molculer mass of N2

+mO2 = 32;// molculer mass of O2

+aO2 = .23;// mass composition of O2 in air

+

+// solution

+ma = (8/3*C+8*H-o2)/aO2;// theoretical mass of air/kg fuel, [kg]

+

+mgas = CO2*mCO2+CO*mCO+N2*mN2+O2*mO2;//  total mass of gas/kg fuel, [kg]

+x1 = CO2*mCO2/mgas;//  composition of CO2 by mass 

+x2 = CO*mCO/mgas;// composition of CO by mass

+x3 = O2*mO2/mgas;//  composition of O2 by mass 

+x4 = N2*mN2/mgas;//  composition of N2 by mass 

+

+m1 = x1*mC/mCO2+x2*mC/mCO;// mass of C/kg of dry flue gas, [kg]

+m2 = C;// mass of C/kg fuel, [kg]

+mf = m2/m1;// mass of dry flue gas/kg fuel, [kg]

+mo2 = mf*x3;// mass of excess O2/kg fuel, [kg]

+mair = mo2/aO2;// mass of excess air/kg fuel, [kg]

+m = ma+mair;// mass of excess air supplied/kg fuel, [kg]

+

+mprintf('\n The mass of air supplied per/kg of fuel burnt is  =  %f  kg\n',m);

+ 

+// End