initial commit / add all books

174 files changed, 6923 insertions, 0 deletions

diff --git a/2705/CH1/EX1.1/Ex1_1.sce b/2705/CH1/EX1.1/Ex1_1.sce
new file mode 100755
index 000000000..e41b309e6
--- /dev/null
+++ b/2705/CH1/EX1.1/Ex1_1.sce
@@ -0,0 +1,17 @@
+clear ;
+clc;
+disp('Example 1.1');
+
+
+
+// Given values
+P = 700;   //pressure,[kN/m^2]
+V1 = .28;   //initial volume,[m^3]
+V2 = 1.68;   //final volume,[m^3]
+
+//solution
+
+W = P*(V2-V1);// // Formula for work done at constant pressure is, [kJ]
+mprintf('\n The Work done is = %f MJ\n',W*10^-3);
+
+//End
diff --git a/2705/CH1/EX1.10/Ex1_10.sce b/2705/CH1/EX1.10/Ex1_10.sce
new file mode 100755
index 000000000..801ac3a6f
--- /dev/null
+++ b/2705/CH1/EX1.10/Ex1_10.sce
@@ -0,0 +1,21 @@
+clear;
+clc;
+disp('Example 1.10');
+
+
+
+//  Given values
+m_dot = 3.045; //  use of coal, [tonne/h]
+c = 28; // calorific value of the coal, [MJ/kg]
+P_out = 4.1; //  output of turbine, [MW]
+
+//  solution
+m_dot = m_dot*10^3/3600; // [kg/s]
+
+P_in = m_dot*c; //  power input by coal, [MW]
+
+n = P_out/P_in; //  thermal efficiency formula
+
+mprintf('\n Thermal efficiency of the plant is  =  %f \n',n);
+
+//End
diff --git a/2705/CH1/EX1.11/Ex1_11.sce b/2705/CH1/EX1.11/Ex1_11.sce
new file mode 100755
index 000000000..de23f795c
--- /dev/null
+++ b/2705/CH1/EX1.11/Ex1_11.sce
@@ -0,0 +1,17 @@
+clear;
+clc;
+disp('Example 1.11');
+
+
+//  Given values
+v = 50; //  speed, [km/h]
+F = 900; //  Resistance to the motion of a car
+
+//  solution
+v = v*10^3/3600; //  [m/s]
+ Power = F*v; //  Power formula, [W]
+
+mprintf('\n The power output of the engine is  =  %f  kW\n',Power*10^-3);
+ 
+ //  End
+ 
diff --git a/2705/CH1/EX1.12/Ex1_12.sce b/2705/CH1/EX1.12/Ex1_12.sce
new file mode 100755
index 000000000..426f6c26d
--- /dev/null
+++ b/2705/CH1/EX1.12/Ex1_12.sce
@@ -0,0 +1,25 @@
+
+clear;
+clc;
+disp('Example 1.12');
+
+
+
+//  Given values
+V = 230; //  volatage, [volts]
+I = 60; //  current, [amps]
+n_gen = .95; //  efficiency of generator
+n_eng = .92; //  efficiency of engine
+
+//  solution
+
+P_gen = V*I; //  Power delivered by generator, [W]
+P_gen=P_gen*10^-3; //  [kW]
+
+P_in_eng=P_gen/n_gen;//Power input from engine,[kW]
+
+P_out_eng=P_in_eng/n_eng;//Power output from engine,[kW]
+
+mprintf('\n The power output from the engine is  =  %f  kW\n',P_out_eng);
+
+//  End
diff --git a/2705/CH1/EX1.13/Ex1_13.sce b/2705/CH1/EX1.13/Ex1_13.sce
new file mode 100755
index 000000000..c91fa71a0
--- /dev/null
+++ b/2705/CH1/EX1.13/Ex1_13.sce
@@ -0,0 +1,17 @@
+clear;
+clc;
+disp('Example 1.13');
+
+
+
+//  Given values
+V = 230; // Voltage, [volts]
+W = 4; //  Power of heater, [kW]
+
+//  solution
+
+//  using equation P=VI
+I = W/V; //  current, [K amps]
+mprintf('\n The current taken by heater is  =  %f  amps \n',I*10^3);
+
+//  End
diff --git a/2705/CH1/EX1.14/Ex1_14.sce b/2705/CH1/EX1.14/Ex1_14.sce
new file mode 100755
index 000000000..857faf8fc
--- /dev/null
+++ b/2705/CH1/EX1.14/Ex1_14.sce
@@ -0,0 +1,24 @@
+clear;
+clc;
+disp('Example 1.14');
+
+
+
+//  Given values
+P_out = 500; //  output of power station, [MW]
+c = 29.5; //  calorific value of coal, [MJ/kg]
+r=.28; 
+
+//  solution
+
+//  since P represents only 28 percent of energy available from coal
+P_coal = P_out/r; //  [MW]
+ 
+m_coal = P_coal/c; //  Mass of coal used, [kg/s]
+m_coal = m_coal*3600; //  [kg/h]
+
+//After one hour
+m_coal = m_coal*1*10^-3; //  [tonne]
+mprintf('\n Mass of coal burnt by the power station in 1 hour is  =  %f  tonne \n',m_coal);
+
+//  End
diff --git a/2705/CH1/EX1.2/Ex1_2.sce b/2705/CH1/EX1.2/Ex1_2.sce
new file mode 100755
index 000000000..dab25d43a
--- /dev/null
+++ b/2705/CH1/EX1.2/Ex1_2.sce
@@ -0,0 +1,21 @@
+clear;
+clc;
+disp('Example 1.2');
+
+
+
+//Given values
+P1 = 138;  // initial pressure,[kN/m^2]
+V1 = .112;  //initial volume,[m^3]
+P2 = 690; //  final pressure,[kN/m^2]
+Gama=1.4; //  heat capacity ratio
+
+//  solution
+
+//  since gas is following, PV^1.4=constant,hence
+
+V2 =V1*(P1/P2)^(1/Gama); //   final volume, [m^3] 
+
+mprintf('\n The new volume of the gas is = %f m^3\n',V2)
+
+//End
diff --git a/2705/CH1/EX1.3/Ex1_3.sce b/2705/CH1/EX1.3/Ex1_3.sce
new file mode 100755
index 000000000..bddd96cd3
--- /dev/null
+++ b/2705/CH1/EX1.3/Ex1_3.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+disp('Example 1.3');
+
+
+
+//  Given values
+P1 = 2070;  //  initial pressure,  [kN/m^2]
+V1 = .014;  //  initial volume,  [m^3]
+P2 = 207;  //  final pressure, [kN/m^2]
+n=1.35;  //  polytropic index
+
+//  solution
+
+//  since gas is following PV^n=constant
+//  hence 
+
+V2 = V1*(P1/P2)^(1/n);  // final volume,  [m^3]
+
+//  calculation of workdone
+
+W=(P1*V1-P2*V2)/(1.35-1);  // using work done formula for polytropic process, [kJ]
+
+mprintf('\n The Work done by gas during expansion is  =  %f kJ\n',W);
+
+//End
diff --git a/2705/CH1/EX1.4/Ex1_4.sce b/2705/CH1/EX1.4/Ex1_4.sce
new file mode 100755
index 000000000..6ae65d27c
--- /dev/null
+++ b/2705/CH1/EX1.4/Ex1_4.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+disp('Example 1.4');
+
+
+
+//  Given values
+P1 = 100;  //  initial pressure, [kN/m^2]
+V1 = .056;  //  initial volume,  [m^3]
+V2 = .007;  //  final volume,  [m^3]
+
+//  To know  P2
+//  since process is hyperbolic so, PV=constant
+//  hence
+
+P2 = P1*V1/V2;  //  final pressure, [kN/m^2]
+
+mprintf('\n  The final pressure is  =  %f  kN/m^2\n',P2);
+
+// calculation of workdone
+
+W = P1*V1*log(V2/V1); //  formula for work done in this process, [kJ]
+
+mprintf('\n  Work done on the gas is  =  %f  kJ\n',W);
+
+//End
diff --git a/2705/CH1/EX1.5/Ex1_5.sce b/2705/CH1/EX1.5/Ex1_5.sce
new file mode 100755
index 000000000..2e651d7e1
--- /dev/null
+++ b/2705/CH1/EX1.5/Ex1_5.sce
@@ -0,0 +1,19 @@
+clear;
+clc;
+disp('Example 1.5');
+
+
+
+//  Given values
+m = 5; //  mass,  [kg]
+t1 = 15; //  inital temperature, [C]
+t2 = 100; //  final temperature, [C]
+c = 450; //  specific heat capacity, [J/kg K]
+
+//  solution
+
+//  using heat transfer equation,[1]
+Q = m*c*(t2-t1); //  [J]
+mprintf('\n The heat required is  =  %f  kJ\n',Q*10^-3);
+
+//End
diff --git a/2705/CH1/EX1.6/Ex1_6.sce b/2705/CH1/EX1.6/Ex1_6.sce
new file mode 100755
index 000000000..fe3ceb552
--- /dev/null
+++ b/2705/CH1/EX1.6/Ex1_6.sce
@@ -0,0 +1,28 @@
+clear;
+clc;
+disp('Example 1.6');
+
+
+//  Given values
+m_cop = 2; //  mass of copper vessel, [kg]
+m_wat = 6; //  mass of water, [kg]
+c_wat = 4.19; //  specific heat capacity of water,  [kJ/kg K]
+
+t1 = 20; //  initial temperature, [C]
+t2 = 90; //  final temperature, [C]
+
+// From the table of average specific heat capacities
+c_cop = .390;  // specific heat capacity of copper,[kJ/kg k]
+
+// solution
+Q_cop = m_cop*c_cop*(t2-t1); //  heat required by copper vessel, [kJ]
+
+Q_wat = m_wat*c_wat*(t2-t1); //  heat required by water, [kJ]
+
+// since there is no heat loss,so total heat transfer is sum of both
+Q_total = Q_cop+Q_wat ; //  [kJ]
+
+mprintf(' \n Required heat transfer to accomplish the change  =  %f  kJ\n',Q_total);
+
+//End
+
diff --git a/2705/CH1/EX1.7/Ex1_7.sce b/2705/CH1/EX1.7/Ex1_7.sce
new file mode 100755
index 000000000..5fb3d347d
--- /dev/null
+++ b/2705/CH1/EX1.7/Ex1_7.sce
@@ -0,0 +1,23 @@
+
+clear;
+clc;
+disp('Example 1.7');
+
+
+//  Given values
+m = 10; //  mass of iron casting, [kg]
+t1 = 200; //  initial temperature, [C]
+Q = -715.5; //  [kJ], since heat is lost in this process
+
+// From the table of average specific heat capacities
+c = .50; //  specific heat capacity of casting iron, [kJ/kg K]
+
+//  solution
+//  using heat equation
+//  Q = m*c*(t2-t1)
+
+t2 = t1+Q/(m*c); // [C]
+
+mprintf('\n The final temperature is  t2 =  %f C\n',t2);
+
+// End
diff --git a/2705/CH1/EX1.8/Ex1_8.sce b/2705/CH1/EX1.8/Ex1_8.sce
new file mode 100755
index 000000000..6febbea57
--- /dev/null
+++ b/2705/CH1/EX1.8/Ex1_8.sce
@@ -0,0 +1,22 @@
+clear;
+clc;
+disp('Example 1.8');
+
+ 
+
+// Given values
+m = 4; //  mass of the liquid, [kg]
+t1 = 15; //  initial temperature, [C]
+t2 = 100; //  final temperature, [C]
+Q = 714; //  [kJ],required heat to accomplish this change
+
+//  solution
+//  using heat equation
+//  Q=m*c*(t2-t1)
+
+//  calculation of c
+c=Q/(m*(t2-t1)); //  heat capacity, [kJ/kg K] 
+
+mprintf('\n The specific heat capacity of the liquid is  c =  %f  kJ/kg K\n',c);
+
+//End
diff --git a/2705/CH1/EX1.9/Ex1_9.sce b/2705/CH1/EX1.9/Ex1_9.sce
new file mode 100755
index 000000000..a5cc15bad
--- /dev/null
+++ b/2705/CH1/EX1.9/Ex1_9.sce
@@ -0,0 +1,27 @@
+clear;
+clc;
+disp('Example 1.9');
+
+
+//  Given values
+m_dot = 20.4; //  mass flowrate of petrol, [kg/h]
+c = 43; //  calorific  value of petrol, [MJ/kg]
+n = .2; //  Thermal efficiency of engine
+
+//  solution
+m_dot = 20.4/3600; // [kg/s]
+c = 43*10^6; //  [J/kg]
+
+//  power output
+P_out = n*m_dot*c; //  [W]
+
+mprintf('\n The power output of the engine is  =  %f  kJ\n',P_out*10^-3);
+ 
+//  power rejected
+
+P_rej = m_dot*c*(1-n); //  [W]
+P_rej = P_rej*60*10^-6; //  [MJ/min]
+
+mprintf('\n The energy rejected by the engine is  =   %f  MJ/min \n',P_rej);
+
+//End
diff --git a/2705/CH10/EX10.1/Ex10_1.sce b/2705/CH10/EX10.1/Ex10_1.sce
new file mode 100755
index 000000000..563d4d450
--- /dev/null
+++ b/2705/CH10/EX10.1/Ex10_1.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+disp('Example 10.1');

+

+//  aim : To determine

+//   the equivalent evaporation

+

+//  Given

+P = 1.4;//  [MN/m^2]

+m = 8;//  mass of water,[kg]

+T1 = 39;//  entering temperature,[C]

+T2 = 100;//  [C]

+x = .95;//dryness fraction 

+

+//  solution

+hf = 830.1;//  [kJ/kg]

+hfg = 1957.7;//  [kJ/kg]

+//  steam is wet so specific enthalpy of steam is

+h = hf+x*hfg;//  [kJ/kg]

+

+//  at 39 C

+h1 = 163.4;//  [kJ/kg]

+//  hence

+q = h-h1;//  [kJ/kg]

+Q = m*q;//  [kJ]

+

+evap = Q/2256.9;//  equivalent evaporation[kg steam/(kg coal)]

+

+mprintf('\n The equivalent evaporation, from and at 100 C is  =  %f  kg steam/kg coal\n ',evap);

+

+// End

diff --git a/2705/CH10/EX10.10/Ex10_10.sce b/2705/CH10/EX10.10/Ex10_10.sce
new file mode 100755
index 000000000..c0d346f7f
--- /dev/null
+++ b/2705/CH10/EX10.10/Ex10_10.sce
@@ -0,0 +1,64 @@
+clear;

+clc;

+disp('Example 10.10');

+

+// aim : To determine

+// (a) the mass of steam bled to each feed heater in kg/kg of supply steam

+// (b) the thermal efficiency of the arrangement

+

+// given values

+P1 = 7;// steam initial pressure, [MN/m^2]

+T1 = 273+500;// steam initil temperature, [K]

+P2 = 2;// pressure at stage 1, [MN/m^2]

+P3 = .5;// pressure at stage 2, [MN/m^2]

+P4 = .05;// condenser pressure,[MN/m^2]

+SE = .82;// stage efficiency of turbine

+

+// solution

+// from the enthalpy-entropy chart(Fig10.23) values of specific enthalpies are

+h1 = 3410;// [kJ/kg]

+h2_prim = 3045;// [kJ/kg]

+// h1-h2=SE*(h1-h2_prim), so

+h2 = h1-SE*(h1-h2_prim);// [kJ/kg]

+

+h3_prim = 2790;// [kJ/kg]

+// h2-h3=SE*(h2-h3_prim), so

+h3 = h2-SE*(h2-h3_prim);// [kJ/kg]

+

+h4_prim = 2450;// [kJ/kg]

+// h3-h4 = SE*(h3-h4_prim), so

+h4 = h3-SE*(h3-h4_prim);// [kJ/kg]

+

+// from steam table

+// @ 2 MN/m^2

+hf2 = 908.6;// [kJ/kg]

+// @ .5 MN/m^2

+hf3 = 640.1;// [kJ/kg] 

+// @ .05 MN/m^2

+hf4 = 340.6;// [kJ/kg]

+

+// (a) 

+// for feed heater1

+m1 = (hf2-hf3)/(h2-hf3);// mass of bled steam, [kg/kg supplied steam]

+// for feed heater2

+m2 = (1-m1)*(hf3-hf4)/(h3-hf4);// 

+mprintf('\n (a) The mass of steam bled in feed heater 1 is  =  %f  kg/kg supply steam\n',m1);

+mprintf('\n      The mass of steam bled in feed heater 2 is  =  %f kg/kg supply steam\n',m2);

+

+// (b)

+W = (h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);// theoretical work done, [kJ/kg]

+Eb = h1-hf2;// energy input in the boiler, [kJ/kg]

+TE1 = W/Eb;// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the arrangement is  =  %f percent\n',TE1*100);

+

+// If there is no feed heating

+hf5 = hf4;

+h5_prim = 2370;// [kJ/kg]

+// h1-h5 = SE*(h1-h5_prim), so

+h5 = h1-SE*(h1-h5_prim);// [kJ/kg]

+Ei = h1-hf5;//energy input, [kJ/kg]

+W = h1-h5;//  theoretical work, [kJ/kg]

+TE2 = W/Ei;// thermal efficiency

+mprintf('\n      The thermal efficiency if there is no feed heating is  =  %f percent\n',TE2*100);

+

+//  End 

diff --git a/2705/CH10/EX10.2/Ex10_2.sce b/2705/CH10/EX10.2/Ex10_2.sce
new file mode 100755
index 000000000..7993e77ac
--- /dev/null
+++ b/2705/CH10/EX10.2/Ex10_2.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+disp('Example 10.2');

+

+// aim : To determine 

+//  the mass of oil used per hour and the fraction of enthalpy drop through the turbine

+//  heat transfer available per kilogram of exhaust steam

+

+//  Given values

+ms_dot = 5000;// generation of steam, [kg/h]

+P1 = 1.8;// generated steam pressure, [MN/m^2]

+T1 = 273+325;// generated steam temperature, [K]

+Tf = 273+49.4;// feed temperature, [K]

+neta = .8;// efficiency of boiler plant 

+c = 45500;// calorific value, [kJ/kg]

+P = 500;// turbine generated power, [kW]

+Pt = .18;// turbine exhaust pressure, [MN/m^2]

+x = .98;// dryness farction of steam

+

+//  solution

+//  using steam table at 1.8 MN/m^2

+hf1 = 3106;// [kJ/kg]

+hg1 = 3080;// [kJ/kg]

+//  so

+h1 = hf1-neta*(hf1-hg1);// [kJ/kg]

+//  again using steam table specific enthalpy of feed water is

+hwf = 206.9;// [kJ/kg]

+h_rais = ms_dot*(h1-hwf);// energy to raise steam, [kJ]

+

+h_fue = h_rais/neta;// energy from fuel per hour, [kJ]

+m_oil = h_fue/c;// mass of fuel per hour, [kg]

+

+//  from steam table at exhaust

+hf = 490.7;// [kJ/kg]

+hfg = 2210.8;//  [kJ/kg]

+//  hence

+h = hf+x*hfg;// [kJ/kg]

+//  now

+h_drop = (h1-h)*ms_dot/3600;// specific enthalpy drop in turbine [kJ]

+f = P/h_drop;// fraction ofenthalpy drop converted into work

+//  heat transfer available in exhaust is

+Q = h-hwf;// [kJ/kg]

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

+mprintf('\n The fraction of the enthalpy drop through the turbine that is converted into useful work is  =  %f\n',f);

+mprintf('\n The heat transfer available in exhaust steam above 49.4 C is  =  %f  kJ/kg\n',Q);

+

+//  End

diff --git a/2705/CH10/EX10.3/Ex10_3.sce b/2705/CH10/EX10.3/Ex10_3.sce
new file mode 100755
index 000000000..5ea7c08db
--- /dev/null
+++ b/2705/CH10/EX10.3/Ex10_3.sce
@@ -0,0 +1,44 @@
+clear;

+clc;

+disp('Example 10.3');

+

+//  aim : To determine

+//  (a) the thermal efficiency of the boiler

+//  (b) the equivalent evaporation of the boiler

+//  (c) the new coal consumption 

+

+//  given values

+ms_dot = 5400;// steam feed rate, [kg/h]

+P = 750;// steam pressure, [kN/m^2]

+x = .98;// steam dryness fraction

+Tf1 = 41.5;// feed water temperature, [C]

+CV = 31000;// calorific value of coal used in the boiler, [kJ/kg]

+mc1 = 670;// rate of burning of coal/h, [kg]

+Tf2 = 100;// increased water temperature, [C]

+

+// solution

+//   (a)

+SRC = ms_dot/mc1;// steam raised/kg coal, [kg]

+hf = 709.3;// [kJ/kg]

+hfg = 2055.5;// [kJ/kg]

+h1 = hf+x*hfg;// specific enthalpy of steam raised, [kJ/kg]

+//  from steam table 

+hfw = 173.9;// specific enthalpy of feed water, [kJ/kg]

+EOB = SRC*(h1-hfw)/CV;// efficiency of boiler

+mprintf('\n (a) The thermal efficiency of the boiler is  =  %f percent\n',EOB*100);

+

+// (b)

+he = 2256.9;// specific enthalpy of evaporation, [kJ/kg]

+Ee = SRC*(h1-hfw)/he;// equivalent evaporation[kg/kg coal]

+mprintf('\n (b) The equivalent evaporation of boiler is  =  %f  kg/kg coal\n',Ee);

+

+// (c)

+hw = 419.1;// specific enthalpy of feed water at 100 C, [kJ/kg]

+Eos = ms_dot*(h1-hw);// energy of steam under new condition, [kJ/h]

+neb = EOB+.05;// given condition new efficiency of boiler if 5%more than previous

+Ec = Eos/neb;// energy from coal, [kJ/h]

+mc2 = Ec/CV;// mass of coal used per hour in new condition, [kg]

+mprintf('\n (c) Mass of coal used in new condition is  =  %f kg\n',mc2);

+mprintf('\n      The saving in coal per hour is  =  %f  kg\n',mc1-mc2);

+

+//  End

diff --git a/2705/CH10/EX10.4/Ex10_4.sce b/2705/CH10/EX10.4/Ex10_4.sce
new file mode 100755
index 000000000..c7d9ea745
--- /dev/null
+++ b/2705/CH10/EX10.4/Ex10_4.sce
@@ -0,0 +1,46 @@
+clear;

+clc;

+disp('Example 10.4');

+

+//  aim : To determine the

+//  (a) Heat transfer in the boiler

+//   (b) Heat transfer in the superheater

+//  (c) Gas used

+

+//  given values

+P = 100;// boiler operating pressure, [bar]

+Tf = 256;// feed water temperature, [C]

+x = .9;// steam dryness fraction.

+Th = 450;// superheater exit temperature, [C]

+m = 1200;// steam generation/h, [tonne]

+TE = .92;// thermal efficiency

+CV = 38;// calorific value of fuel, [MJ/m^3]

+

+// solution

+// (a)

+// from steam table

+hw = 1115.4;// specific enthalpy of feed water, [kJ/kg]

+// for wet steam

+hf = 1408;// specific enthalpy, [kJ/kg]

+hg = 2727.7;// specific enthalpy, [kJ/kg]

+//  so

+h = hf+x*(hg-hf);// total specific enthalpy of wet steam, [kJ/kg]

+//  hence

+Qb = m*(h-hw);// heat transfer/h for wet steam, [MJ]

+mprintf('\n (a) The heat transfer/h in producing wet steam in the boiler is  =  %f  MJ\n',Qb);

+

+// (b)

+// again from steam table

+// specific enthalpy of superheated stem at given condition is,

+hs = 3244;// [kJ/kg]

+

+Qs = m*(hs-h);// heat transfer/h in superheater, [MJ]

+mprintf('\n (b) The heat transfer/h in superheater is  =  %f  MJ\n',Qs);

+

+// (c)

+V = (Qb+Qs)/(TE*CV);// volume of gs used/h, [m^3]

+mprintf('\n (c) The volume of gas used/h is  =  %f m^3\n',V);

+

+//  There is calculation mistake in the book so our answer is not matching

+

+//  End

diff --git a/2705/CH10/EX10.5/Ex10_5.sce b/2705/CH10/EX10.5/Ex10_5.sce
new file mode 100755
index 000000000..e7086049a
--- /dev/null
+++ b/2705/CH10/EX10.5/Ex10_5.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+disp('Example 10.5');

+

+//aim : To determine 

+// the flow rate of cooling water

+

+//Given values

+P=24;//pressure, [kN/m^2]

+ms_dot=1.8;//steam condense rate,[tonne/h]

+x=.98;//dryness fraction

+T1=21;//entrance temperature of cooling water,[C]

+T2=57;//outlet temperature of cooling water,[C]

+

+//solution

+//at 24 kN/m^2, for steam

+hfg=2616.8;//[kJ/kg]

+hf1=268.2;//[kJ/kg]

+//hence

+h1=hf1+x*(hfg-hf1);//[kJ/kg]

+

+//for cooling water

+hf3=238.6;//[kJ/kg]

+hf2=88.1;//[kJ/kg]

+

+//using equation [3]

+//ms_dot*(hf3-hf2)=mw_dot*(h1-hf1),so

+mw_dot=ms_dot*(h1-hf1)/(hf3-hf2);//[tonne/h]

+disp('tonne/h',mw_dot,'The flow rate of the cooling water is =')

+

+//End

diff --git a/2705/CH10/EX10.6/Ex10_6.sce b/2705/CH10/EX10.6/Ex10_6.sce
new file mode 100755
index 000000000..b4d0e04fe
--- /dev/null
+++ b/2705/CH10/EX10.6/Ex10_6.sce
@@ -0,0 +1,49 @@
+clear;

+clc;

+disp('Example 10.6');

+

+//  aim : To determine

+//  (a) the energy supplied in the boiler

+//  (b) the dryness fraction of the steam entering the condenser

+//  (c) the rankine efficiency

+

+//  given values

+P1 = 3.5;// steam entering pressure, [MN/m^2]

+T1 = 273+350;// entering temperature, [K]

+P2 = 10;//steam exhaust pressure, [kN/m^2]

+

+// solution

+//  (a)

+//  from steam table, at P1 is,

+hf1 = 3139;// [kJ/kg]

+hg1 = 3095;// [kJ/kg]

+h1 = hf1-1.5/2*(hf1-hg1);

+// at Point 3

+h3 = 191.8;// [kJ/kg]

+Es = h1-h3;// energy supplied, [kJ/kg]

+mprintf('\n (a) The energy supplied in boiler/kg steam is  =  %f kJ/kg\n',Es);

+

+// (b)

+// at P1

+sf1 = 6.960;// [kJ/kg K]

+sg1 = 6.587;// [kJ/kg K]

+s1 = sf1-1.5/2*(sf1-sg1);// [kJ/kg K]

+// at P2

+sf2 = .649;// [kJ/kg K] 

+ sg2 = 8.151;// [kJ/kg K]

+ // s2=sf2+x2(sg2-sf2)

+ // theoretically expansion through turbine is isentropic so s1=s2

+ // hence

+ s2 = s1;

+ x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+ mprintf('\n (b) The dryness fraction of steam entering the condenser is =  %f \n',x2);

+ 

+ // (c)

+ // at point 2

+ hf2 = 191.8;// [kJ/kg]

+ hfg2 = 2392.9;// [kJ/kg]

+ h2 = hf2+x2*hfg2;// [kJ/kg]

+ Re = (h1-h2)/(h1-h3);// rankine efficiency

+ mprintf('\n (c) The Rankine efficiency is  =  %f percent\n',Re*100);

+ 

+ //  End

diff --git a/2705/CH10/EX10.7/Ex10_7.sce b/2705/CH10/EX10.7/Ex10_7.sce
new file mode 100755
index 000000000..b3e5be740
--- /dev/null
+++ b/2705/CH10/EX10.7/Ex10_7.sce
@@ -0,0 +1,61 @@
+clear;

+clc;

+disp('Example 10.7');

+

+// aim : To determine

+//  the specific work done and compare this with that obtained when determining the rankine effficiency

+

+// given values

+P1 = 1000;// steam entering pressure, [kN/m^2]

+x1 = .97;// steam entering dryness fraction

+P2 = 15;//steam exhaust pressure, [kN/m^2]

+n = 1.135;// polytropic index

+

+// solution

+// (a)

+// from steam table, at P1 is

+hf1 = 762.6;// [kJ/kg]

+hfg1 = 2013.6;// [kJ/kg]

+h1 = hf1+hfg1; // [kJ/kg]

+

+sf1 = 2.138;// [kJ/kg K]

+sg1 = 6.583;// [kJ/kg K]

+s1 = sf1+x1*(sg1-sf1);// [kJ/kg K]

+

+// at P2

+sf2 = .755;// [kJ/kg K] 

+ sg2 = 8.009;// [kJ/kg K]

+// s2 = sf2+x2(sg2-sf2)

+// since expansion through turbine is isentropic so s1=s2

+ // hence

+ s2 = s1;

+ x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+ 

+  // at point 2

+ hf2 = 226.0;// [kJ/kg]

+ hfg2 = 2373.2;// [kJ/kg]

+ h2 = hf2+x2*hfg2;// [kJ/kg]

+ 

+// at Point 3

+h3 = 226.0;// [kJ/kg]

+

+// (a)

+ Re = (h1-h2)/(h1-h3);// rankine efficiency

+ mprintf('\n (a) The Rankine efficiency is  =  %f percent\n',Re*100);

+ 

+// (b)

+vg1 = .1943;// specific volume at P1, [m^3/kg]

+vg2 = 10.02;// specific volume at P2, [m^3/kg]

+V1 = x1*vg1;// [m^3/kg]

+V2 = x2*vg2;// [m^3/kg]

+

+W1 = n/(n-1)*(P1*V1-P2*V2);// specific work done, [kJ/kg]

+

+//  from rankine cycle

+W2 = h1-h2;// [kJ/kg]

+mprintf('\n (b) The specific work done is  =  %f kJ/kg\n',W1);

+mprintf('\n     The specific work done (from rankine) is  =  %f kJ/kg\n',W2);

+

+// there is calculation mistake in the book so our answer is not matching

+

+//  End

diff --git a/2705/CH10/EX10.8/Ex10_8.sce b/2705/CH10/EX10.8/Ex10_8.sce
new file mode 100755
index 000000000..f65703173
--- /dev/null
+++ b/2705/CH10/EX10.8/Ex10_8.sce
@@ -0,0 +1,63 @@
+clear;

+clc;

+disp('Example 10.8');

+

+//  aim : To determine

+//  (a) the rankine fficiency

+//  (b) the specific steam consumption

+//  (c) the carnot efficiency of the cycle

+

+// given values

+P1 = 1100;// steam entering pressure, [kN/m^2]

+T1 = 273+250;// steam entering temperature, [K]

+P2 = 280;// pressure at point 2, [kN/m^2]

+P3 = 35;// pressure at point 3, [kN/m^2]

+

+// solution

+// (a)

+// from steam table, at P1  and T1 is

+hf1 = 2943;// [kJ/kg]

+hg1 = 2902;// [kJ/kg]

+h1 = hf1-.1*(hf1-hg1); // [kJ/kg]

+

+sf1 = 6.926;// [kJ/kg K]

+sg1 = 6.545;// [kJ/kg K]

+s1 = sf1-.1*(sf1-sg1);// [kJ/kg K]

+

+// at P2

+sf2 = 1.647;// [kJ/kg K] 

+ sg2 = 7.014;// [kJ/kg K]

+// s2=sf2+x2(sg2-sf2)

+// since expansion through turbine is isentropic so s1=s2

+ // hence

+ s2  = s1;

+ x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+ 

+ // at point 2

+ hf2 = 551.4;// [kJ/kg]

+ hfg2 = 2170.1;// [kJ/kg]

+ h2 = hf2+x2*hfg2;// [kJ/kg]

+ vg2 = .646;// [m^3/kg]

+ v2 = x2*vg2;// [m^3/kg]

+ 

+ // by Fig10.20.

+ A6125 = h1-h2;// area of 6125, [kJ/kg]

+ A5234 = v2*(P2-P3);// area 5234, [kJ/kg]

+ W = A6125+A5234;// work done 

+ hf = 304.3;// specific enthalpy of water at condenser pressuer, [kJ/kg]

+ ER = h1-hf;// energy received, [kJ/kg]

+ Re = W/ER;// rankine efficiency

+ mprintf('\n (a) The rankine efficiency is  =  %f percent\n',Re*100);

+ 

+ // (b)

+ kWh = 3600;// [kJ]

+ SSC = kWh/W;// specific steam consumption, [kJ/kWh]

+ mprintf('\n (b) The specific steam consumption is = %f kJ/kWh\n',SSC);

+ 

+ // (c)

+ // from steam table 

+T3 = 273+72.7;// temperature at point 3

+CE = (T1-T3)/T1;// carnot efficiency

+mprintf('\n (c) The carnot efficiency of the cycle is  =  %f percent\n',CE*100);

+

+//  End

diff --git a/2705/CH10/EX10.9/Ex10_9.sce b/2705/CH10/EX10.9/Ex10_9.sce
new file mode 100755
index 000000000..cd4a60ca0
--- /dev/null
+++ b/2705/CH10/EX10.9/Ex10_9.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 10.9');

+

+// aim : To determine

+// (a) the theoretical power of steam passing through the turbine

+// (b) the thermal efficiency of the cycle

+// (c) the thermal efficiency of the cycle assuming there is no reheat

+

+// given values

+P1 = 6;// initial pressure, [MN/m^2]

+T1 = 450;// initial temperature, [C]

+P2 = 1;// pressure at stage 1, [MN/m^2]

+P3 = 1;// pressure at stage 2, [MN/m^2]

+T3 = 370;// temperature, [C]

+P4 = .02;// pressure at stage 3, [MN/m^2]

+P5 = .02;// pressure at stage 4, [MN/m^2]

+T5 = 320;// temperature, [C]

+P6 = .02;// pressure at stage 5, [MN/m^2]

+P7 = .02;// final pressure , [MN/m^2]

+

+// solution

+// (a) 

+// using Fig 10.21

+h1 = 3305;// specific enthalpy, [kJ/kg]

+h2 = 2850;// specific enthalpy, [kJ/kg]

+h3 = 3202;// specific enthalpy, [kJ/kg]

+h4 = 2810;// specific enthalpy, [kJ/kg]

+h5 = 3115;// specific enthalpy, [kJ/kg]

+h6 = 2630;// specific enthalpy, [kJ/kg]

+h7 = 2215;// specific enthalpy, [kJ/kg]

+W = (h1-h2)+(h3-h4)+(h5-h6);// specific work through the turbine, [kJ/kg]

+mprintf('\n (a) The theoretical power/kg steam/s is  =  %f kW\n',W);

+

+// (b)

+// from steam table

+hf6 = 251.5;// [kJ/kg]

+

+TE1 = ((h1-h2)+(h3-h4)+(h5-h6))/((h1-hf6)+(h3-h2)+(h5-h4));// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',TE1*100);

+

+// (c)

+// if there is no heat

+hf7 = hf6;

+TE2 = (h1-h7)/(h1-hf7);// thermal efficiency

+mprintf('\n (c) The thermal efficiency of the cycle if there is no heat is  =  %f percent\n',TE2*100);

+

+//  End

diff --git a/2705/CH11/EX11.1/Ex11_1.sce b/2705/CH11/EX11.1/Ex11_1.sce
new file mode 100755
index 000000000..973fd235e
--- /dev/null
+++ b/2705/CH11/EX11.1/Ex11_1.sce
@@ -0,0 +1,45 @@
+clear;

+clc;

+disp('Example 11.1')

+

+//  aim : To determine the 

+// (a) bore of the cylinder

+// (b) piston stroke

+// (c) speed of the engine

+

+//  Given values

+P_req = 60;// power required to develop, [kW]

+P = 1.25;// boiler pressure, [MN/m^2]

+Pb = .13;// back pressure, [MN/m^2]

+cut_off = .3;// [stroke]

+k = .82;// diagram factor

+n = .78;// mechanical efficiency

+LN = 3;// mean piston speed, [m/s]

+

+// solution

+// (a)

+r = 1/cut_off;// expansion ratio

+Pm = P/r*(1+log(r))-Pb;// mean effective pressure, [MN/m^2]

+P_ind = P_req/n;// Actual indicated power developed, [kW]

+P_the = P_ind/k;// Theoretical indicated power developed, [kW]

+

+//  using indicated_power=Pm*LN*A

+//  Hence

+A = P_the/(Pm*LN)*10^-3;// piston area,[m^2]

+d = sqrt(4*A/%pi)*10^3;// bore ,[mm]

+mprintf('\n (a) The bore of the cylinder is  =  %f mm\n',d);

+

+// (b)

+// given that stroke is 1.25 times bore

+L = 1.25*d;// [mm]

+mprintf('\n (b) The piston stroke is  =  %f mm\n',L);

+

+// (c)

+// LN=mean piston speed, where L is stroke in meter and N is 2*rev/s,(since engine is double_acting)

+//  hence

+rev_per_sec = LN/(2*L*10^-3);// [rev/s]

+

+rev_per_min = rev_per_sec*60;// [rev/min]

+mprintf('\n (c) The speed of the engine is  =  %f rev/min\n',rev_per_min);

+

+//  End

diff --git a/2705/CH11/EX11.2/Ex11_2.sce b/2705/CH11/EX11.2/Ex11_2.sce
new file mode 100755
index 000000000..e7f862fc2
--- /dev/null
+++ b/2705/CH11/EX11.2/Ex11_2.sce
@@ -0,0 +1,51 @@
+clear;

+clc;

+disp('Example 11.2')

+

+//  aim : To determine the 

+//  (a) the diameter of the cylinder

+//  (b) piston stroke

+//  (c) actual steam consumption and indicated thermal efficiency

+

+//  Given values

+P = 900;// inlet pressure, [kN/m^2]

+Pb = 140;//  exhaust pressure, [kN/m^2]

+cut_off =.4;// [stroke]

+k = .8;// diagram factor

+rs = 1.2;// stroke to bore ratio

+N = 4;// engine speed, [rev/s]

+ip = 22.5;// power output from the engine, [kW]

+

+// solution

+// (a)

+r = 1/cut_off;// expansion ratio

+Pm = P/r*(1+log(r))-Pb;// mean effective pressure, [kN/m^2]

+Pm = Pm*k;// actual mean effective pressure, [kN/m^2]

+

+// using ip=Pm*L*A*N

+// and L=r*d; where L is stroke and d is bore

+d = (ip/(Pm*rs*%pi/4*2*N))^(1/3);// diameter of the cylinder, [m]

+

+mprintf('\n (a) The diameter of the cylinder is  =  %f  mm\n',d*1000);

+

+// (b)

+L = rs*d;// stroke, [m]

+mprintf('\n (b) The piston stroke is  =  %f  mm\n',L*1000);

+

+// (c)

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

+V = SV*cut_off*2*240*60;// volume of steam consumed per  hour, [m^3]

+v = .2148;// specific volume at 900 kN/m^2, [m^3/kg]

+SC = V/v;// steam consumed/h, [kg]

+ASC = 1.5*SC;// actual steam consumption/h, [kg]

+mprintf('\n (c) The actual steam consumption/h is  =  %f kg\n',ASC);

+

+m_dot = ASC/3600;// steam consumption,[kg/s] 

+// from steam table

+hg = 2772.1;// specific enthalpy of inlet steam, [kJ/kg]

+hfe = 458.4;// specific liquid enthalpy at exhaust pressure, [kJ/kg]

+

+ITE = ip/(m_dot*(hg-hfe));// indicated thermal efficiency

+mprintf('\n     The indicated thermal efficiency is  =  %f  percent\n',ITE*100);

+

+//  End

diff --git a/2705/CH11/EX11.3/Ex11_3.sce b/2705/CH11/EX11.3/Ex11_3.sce
new file mode 100755
index 000000000..14b2cf23b
--- /dev/null
+++ b/2705/CH11/EX11.3/Ex11_3.sce
@@ -0,0 +1,41 @@
+clear;

+clc;

+disp('Example 11.3');

+

+//  aim : To determine

+//  (a) the diagram factor

+//  (b) the indicated thermal efficiency of the engine

+

+// given values

+d = 250*10^-3;// cylinder diameter, [m]

+L = 375*10^-3;// length of stroke, [m]

+P = 1000;// steam pressure , [kPa]

+x = .96;// dryness fraction of steam

+Pb = 55;// exhaust pressure, [kPa]

+r = 6;// expansion ratio

+ip = 45;// indicated power developed, [kW]

+N = 3.5;// speed of engine, [rev/s]

+m = 460;// steam consumption, [kg/h]

+

+// solution

+// (a)

+Pm = P/r*(1+log(r))-Pb;// [kN/m^3]

+A = %pi*(d)^2/4;// area, [m^2]

+tip = Pm*L*A*N*2;// theoretical indicated power, [kW]

+k = ip/tip;// diagram factor

+mprintf('\n (a) The diagram factor is  =  %f\n',k);

+

+// (b)

+// from steam table at 1 MN/m^2

+hf = 762.6;// [kJ/kg]

+hfg = 2013.6;// [kJ/kg]

+// so 

+h1 = hf+x*hfg;// specific enthalpy of steam at 1MN/m^2, [kJ/kg]

+// minimum specific enthalpy in engine at 55 kPa 

+hf = 350.6;// [kJ/kg]

+// maximum energy available in engine is

+h = h1-hf;// [kJ/kg]

+ITE = ip/(m*h/3600)*100;// indicated thermal efficiency

+mprintf('\n (b) The indicated thermal efficiency is  =  %f  percent\n ',ITE);

+

+//  End

diff --git a/2705/CH11/EX11.4/Ex11_4.sce b/2705/CH11/EX11.4/Ex11_4.sce
new file mode 100755
index 000000000..4e480bf97
--- /dev/null
+++ b/2705/CH11/EX11.4/Ex11_4.sce
@@ -0,0 +1,20 @@
+clear;

+clc;

+disp('Example 11.4');

+

+// aim : To determine

+// steam consumption

+

+// given values

+P1 = 11;// power, [kW]

+m1 = 276;// steam use/h when developing power P1,[kW]

+ip = 8;// indicated power output, [kW]

+B = 45;// steam used/h at no load, [kg]

+

+// solution

+// using graph of Fig.11.9 

+A = (m1-B)/P1;// slop of line, [kg/kWh]

+W = A*ip+B;// output, [kg/h]

+mprintf('\n The steam consumption is  =  %f  kg/h\n ',W);

+

+//  End

diff --git a/2705/CH11/EX11.5/Ex11_5.sce b/2705/CH11/EX11.5/Ex11_5.sce
new file mode 100755
index 000000000..b4106e20d
--- /dev/null
+++ b/2705/CH11/EX11.5/Ex11_5.sce
@@ -0,0 +1,52 @@
+clear;

+clc;

+disp('Example 11.5');

+

+//  aim : To determine

+// (a) the intermediate pressure

+// (b) the indicated power output

+// (c) the steam consumption of the engine

+

+// given values

+P1 = 1400;// initial pressure, [kN/m^2]

+x = .9;// dryness fraction

+P5 = 35;// exhaust pressure

+k = .8;// diagram factor of low-pressure cylindaer

+L = 350*10^-3;// stroke of both the cylinder, [m]

+dhp = 200*10^-3;// diameter of high pressure cylinder, [m]

+dlp = 300*10^-3;// diameter of low-pressure cylinder, [m]

+N = 300;// engine speed, [rev/min]

+

+// solution

+// taking reference Fig.11.13

+Ahp = %pi/4*dhp^2;// area of high-pressure cylinder, [m^2]

+Alp = %pi/4*dlp^2;// area of low-pressure cylinder, [m^2]

+// for equal initial piston loads

+// (P1-P7)Ahp=(P7-P5)Alp

+deff('[x]=f(P7)','x=(P1-P7)*Ahp-(P7-P5)*Alp');

+P7 = fsolve(0,f);// intermediate pressure, [kN/m^2]

+mprintf('\n (a) The intermediate pressure is  =  %f kN/m^2\n ',P7);

+

+// (b)

+V6 = Ahp*L;// volume of high-pressure cylinder, [m^3]

+P2 = P1;

+P6 = P7;

+// using P2*V2=P6*V6

+V2 = P6*V6/P2; // [m^3]

+V1 = Alp*L;// volume of low-pressure cylinder, [m^3]

+R = V1/V2;// expansion ratio

+Pm = P1/R*(1+log(R))-P5;// effective pressure of low-pressure cylinder, [kn/m^2]

+Pm = k*Pm;// actual effective pressure, [kN/m^2]

+ip = Pm*L*Alp*N*2/60;// indicated power, [kW]

+mprintf('\n (b) The indicated power is  =  %f kW\n',ip);

+

+// (c) 

+COV = V1/ R;// cut-off  volume in high-pressure cylinder, [m^3]

+V = COV*N*2*60;// volume of steam admitted/h

+// from steam table

+vg = .1407;// [m^3/kg]

+AV = x*vg;// specific volume of admission steam, [m^3/kg]

+m = V/AV;// steam consumption, [kg/h]

+mprintf('\n (c) The steam consumption of the engine is  =  %f  kg/h\n',m);

+

+//  End 

diff --git a/2705/CH11/EX11.6/Ex11_6.sce b/2705/CH11/EX11.6/Ex11_6.sce
new file mode 100755
index 000000000..5cbaf958a
--- /dev/null
+++ b/2705/CH11/EX11.6/Ex11_6.sce
@@ -0,0 +1,45 @@
+clear;

+clc;

+disp('Example 11.6');

+

+// aim : To determine

+// (a) the indicated power output

+// (b) the diameter of high-pressure cylinder of the engine

+// (c) the intermediate pressure

+

+// given values

+P = 1100;// initial pressure, [kN/m^2]

+Pb = 28;// exhaust pressure

+k = .82;// diagram factor of low-pressure cylindaer

+L = 600*10^-3;// stroke of both the cylinder, [m]

+dlp = 600*10^-3;// diameter of low-pressure cylinder, [m]

+N = 4;// engine speed, [rev/s]

+R = 8;// expansion ratio

+

+// solution

+// taking reference Fig.11.13

+// (a)

+Pm = P/R*(1+log(R))-Pb;// effective pressure of low-pressure cylinder, [kn/m^2]

+Pm = k*Pm;// actual effective pressure, [kN/m^2]

+Alp = %pi/4*dlp^2;// area of low-pressure cylinder, [m^2]

+ip = Pm*L*Alp*N*2;// indicated power, [kW]

+mprintf('\n (a) The indicated power is  =  %f kW\n',ip);

+

+// (b)

+// work done by both cylinder is same as area of diagram

+w = Pm*Alp*L;// [kJ]

+W = w/2;// work done/cylinder, [kJ]

+V2 = Alp*L/8;// volume, [m63]

+P2 = P;// [kN/m^2]

+//  using area A1267=P2*V2*log(V6/V2)=W

+V6 = V2*exp(W/(P2*V2));// intermediate volume, [m^3]

+// using Ahp*L=%pi/4*dhp^2*L=V6

+dhp = sqrt(V6*4/L/%pi);// diameter of high-pressure cylinder, [m]

+mprintf('\n (b) The diameter of high-pressure cylinder is  =  %f  mm\n',dhp*1000);

+

+// (c)

+// using P2*V2=P6*V6

+P6 = P2*V2/V6; // intermediate pressure, [kN/m^2]

+mprintf('\n (c) The intermediate opressure is =   %f  kN/m^2\n',P6);

+

+//  End 

diff --git a/2705/CH11/EX11.7/Ex11_7.sce b/2705/CH11/EX11.7/Ex11_7.sce
new file mode 100755
index 000000000..da5f5dd4e
--- /dev/null
+++ b/2705/CH11/EX11.7/Ex11_7.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+disp('Example 11.7');

+

+// aim : To determine

+// (a) The speed of the engine

+// (b) the diameter of the high pressure cylinder

+

+// given values

+ip = 230;// indicated power, [kW]

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

+Pb = 35;// exhaust pressure, [kN/m^2]

+R = 12.5;// expansion ratio

+d1 = 400*10^-3;// diameter of low pressure cylinder, [m]

+L = 500*10^-3;// stroke of both the cylinder, [m]

+k = .78;// diagram factor

+rv = 2.5;// expansion ratio of high pressure cylinder

+

+// solution

+// (a)

+Pm = P/R*(1+log(R))-Pb;// mean effective pressure in low pressure cylinder, [kN/m^2]

+ipt = ip/k;// theoretical indicated power, [kw]

+// using ip=Pm*L*A*N

+A = %pi/4*d1^2;// area , [m^2]

+N = ipt/(Pm*L*A*2);// speed, [rev/s]

+mprintf('\n (a) The engine speed is  =  %f  rev/s\n',N);

+

+// (b)

+Vl = A*L;// volume of low pressure cylinder, [m^3]

+COV = Vl/R;// cutt off volume of hp cylinder, [m^3]

+V = COV*rv;// total volume, [m^3]

+

+//  V = %pi/4*d^2*L, so

+d = sqrt(4*V/%pi/L);// diameter of high pressure cylinder, [m]

+mprintf('\n (b) The diameter of the high pressure cylinder is  =  %f  mm\n',d*1000);

+

+//  End

diff --git a/2705/CH11/EX11.8/Ex11_8.sce b/2705/CH11/EX11.8/Ex11_8.sce
new file mode 100755
index 000000000..4189dbf58
--- /dev/null
+++ b/2705/CH11/EX11.8/Ex11_8.sce
@@ -0,0 +1,54 @@
+clear;

+clc;

+disp('Example 11.8');

+

+// aim : To determine

+// (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder

+// (b) the overall diagram factor

+// (c) the indicated power 

+

+// given values

+P = 1100;// steam supply pressure, [kN/m^2]

+Pb = 32;// back pressure, [kN/m^2]

+d1 = 300*10^-3;// cylinder1 diameter, [m]

+d2 = 600*10^-3;// cylinder2 diameter, [m]

+L = 400*10^-3;// common stroke of both cylinder, [m]

+

+A1 = 12.5;//  average area of indicated diagram for HP, [cm^2]

+A2 = 11.4;//  average area of indicated diagram for LP, [cm^2]

+

+P1 = 270;// indicator calibration, [kN/m^2/ cm]

+P2 = 80;// spring calibration, [kN/m^2/ cm]

+N = 2.7;// engine speed, [rev/s]

+l = .75;// length of both diagram, [m]

+

+// solution

+// (a)

+// for HP cylinder

+Pmh = P1*A1/7.5;// [kN/m^2]

+F = Pmh*%pi/4*d1^2;// force on HP, [kN]

+PmH = Pmh*(d1/d2)^2;// pressure referred to LP cylinder, [kN/m^2]

+PmL = P2*A2/7.5;// pressure for LP cylinder, [kN/m^2]

+PmA = PmH+PmL;// actual effective pressure referred to LP cylinder, [kN/m^2]

+

+Ah = %pi/4*d1^2;// area of HP cylinder, [m^2]

+Vh = Ah*L;// volume of HP cylinder, [m^3]

+CVh = Vh/3;// cut-off volume of HP cylinder, [m^3]

+Al = %pi/4*d2^2;// area of LP cylinder, [m^2]

+Vl = Al*L;// volume of LP cylinder, [m^3]

+

+R = Vl/CVh;// expansion ratio

+Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2]

+

+mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is  =  %f kN/m^2\n',PmA);

+mprintf('\n     The hypothetical mean effective pressure referred to LP cylinder is  =  %f kN/m^2\n',Pm);

+

+// (a)

+ko = PmA/Pm;// overall diagram factor

+mprintf('\n (b) The overall diagram factor is  =  %f\n',ko);

+

+// (c) 

+ip = PmA*L*Al*N*2;// indicated power, [kW]

+mprintf('\n (c) The indicated power is  =  %f kW\n',ip);

+

+//   End

diff --git a/2705/CH11/EX11.9/Ex11_9.sce b/2705/CH11/EX11.9/Ex11_9.sce
new file mode 100755
index 000000000..e020aaa3e
--- /dev/null
+++ b/2705/CH11/EX11.9/Ex11_9.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 11.9');

+

+// aim : To determine

+// (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder

+// (b) the overall diagram factor

+// (c) the pecentage of the total indicated power developed in each cylinder

+

+// given values

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

+Pb = 20;// back pressure, [kN/m^2]

+Chp = .6;// cut-off in HP cylinder, [stroke]

+dh = 300*10^-3;// HP diameter, [m]

+di = 500*10^-3;// IP diameter, [m]

+dl = 900*10^-3;// LP diameter, [m]

+

+Pm1 = 590;// actual pressure of HP cylinder, [kN/m^2]

+Pm2 = 214;// actual pressure of IP cylinder, [kN/m^2]

+Pm3 = 88;// actual pressure of LP cylinder, [kN/m^2]

+

+// solution

+// (a)

+// for HP cylinder

+PmH = Pm1*(dh/dl)^2;// PmH referred to LP cylinder, [kN/m^2]

+// for IP cylinder

+PmI = Pm2*(di/dl)^2;// PmI referred to LP cylinder, [kN/m^2]

+PmA = PmH+PmI+Pm3;// actual mean effective pressure referred to LP cylinder, [kN/m^2]

+

+R = dl^2/(dh^2*Chp);// expansion ratio

+Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2]

+

+mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is  =  %f  kN/m^2\n',PmA);

+mprintf('\n      The hypothetical mean effective pressure referred to LP cylinder is  =  %f  kN/m^2\n',Pm);

+

+// (b)

+ko = PmA/Pm;// overall diagram factor

+mprintf('\n (b) The overall diagram factor is  =  %f\n',ko);

+

+// (c)

+HP = PmH/PmA*100;// %age of indicated power developed in HP

+IP = PmI/PmA*100; // %age of indicated power developed in IP

+LP = Pm3/PmA*100; // %age of indicated power developed in LP

+mprintf('\n (c) The pecentage of the total indicated power developed in HP cylinder is  =  %f percent\n',HP);

+mprintf('\n     The pecentage of the total indicated power developed in IP cylinder is  =  %f percent\n',IP);

+mprintf('\n     The pecentage of the total indicated power developed in LP cylinder is  =  %f  percent\n',LP);

+

+//   End

diff --git a/2705/CH12/EX12.1/Ex12_1.sce b/2705/CH12/EX12.1/Ex12_1.sce
new file mode 100755
index 000000000..6acd59e30
--- /dev/null
+++ b/2705/CH12/EX12.1/Ex12_1.sce
@@ -0,0 +1,43 @@
+clear;

+clc;

+disp('Example 12.1');

+

+//  aim : To determine the

+//  (a) throat area

+//   (b) exit area

+//   (c) Mach number at exit

+

+// Given values

+P1 = 3.5;// inlet pressure of air, [MN/m^2]

+T1 = 273+500;// inlet temperature of air, [MN/m^2]

+P2 = .7;// exit pressure, [MN/m^2]

+m_dot = 1.3;// flow rate of air, [kg/s]

+Gamma = 1.4;// heat capacity ratio

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

+

+// solution

+// given expansion may be considered to be adiabatic and to follow the law PV^Gamma=constant

+// using ideal gas law

+v1 = R*T1/P1*10^-3;// [m^3/kg]

+Pt = P1*(2/(Gamma+1))^(Gamma/(Gamma-1));// critical pressure, [MN/m^2]

+

+//  velocity at throat is

+Ct = sqrt(2*Gamma/(Gamma-1)*P1*10^6*v1*(1-(Pt/P1)^(((Gamma-1)/Gamma))));// [m/s]

+vt = v1*(P1/Pt)^(1/Gamma);// [m^3/kg]

+// using m_dot/At=Ct/vt

+At = m_dot*vt/Ct*10^6;// throat area, [mm^2]

+mprintf('\n (a) The throat area is  =  %f mm^2\n',At);

+

+// (b)

+//  at exit

+C2 = sqrt(2*Gamma/(Gamma-1)*P1*10^6*v1*(1-(P2/P1)^(((Gamma-1)/Gamma))));// [m/s]

+v2 = v1*(P1/P2)^(1/Gamma);// [m^3/kg]

+A2 = m_dot*v2/C2*10^6;// exit area, [mm^2]

+

+mprintf('\n (b) The exit area is  =  %f  mm^2\n',A2);

+

+//  (c)

+M = C2/Ct;

+mprintf('\n (c) The Mach number at exit is  =  %f\n',M);

+

+//  End

diff --git a/2705/CH12/EX12.2/Ex12_2.sce b/2705/CH12/EX12.2/Ex12_2.sce
new file mode 100755
index 000000000..fd3cd191b
--- /dev/null
+++ b/2705/CH12/EX12.2/Ex12_2.sce
@@ -0,0 +1,33 @@
+clear;

+clc;

+disp('Example 12.2');

+

+//  aim : To determine the increases in pressure, temperature and internal energy per kg of air

+

+//  Given values

+T1 = 273;// [K]

+P1 = 140;// [kN/m^2]

+C1 = 900;// [m/s]

+C2 = 300;// [m/s]

+cp = 1.006;// [kJ/kg K]

+cv =.717;// [kJ/kg K]

+

+// solution

+R = cp-cv;// [kJ/kg K]

+Gamma = cp/cv;// heat capacity ratio

+//  for frictionless adiabatic flow, (C2^2-C1^2)/2=Gamma/(Gamma-1)*R*(T1-T2)

+

+T2 =T1-((C2^2-C1^2)*(Gamma-1)/(2*Gamma*R))*10^-3; //  [K]

+T_inc = T2-T1;// increase in temperature [K]

+

+P2 = P1*(T2/T1)^(Gamma/(Gamma-1));// [MN/m^2]

+P_inc = (P2-P1)*10^-3;// increase in pressure,[MN/m^2]

+

+U_inc = cv*(T2-T1);//  Increase in internal energy per kg,[kJ/kg]

+mprintf('\n The increase in pressure is  =  %f  MN/m^2\n',P_inc);

+mprintf('\n Increase in temperature is  =  %f  K\n',T_inc);

+mprintf('\n Increase in internal energy is  =   %f  kJ/kg\n',U_inc);

+

+// there is minor variation in result

+

+//  End

diff --git a/2705/CH12/EX12.3/Ex12_3.sce b/2705/CH12/EX12.3/Ex12_3.sce
new file mode 100755
index 000000000..37084b47a
--- /dev/null
+++ b/2705/CH12/EX12.3/Ex12_3.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+disp('Example 12.3');

+

+// aim : To determine the 

+//  (a) throat and exit areas

+//   (b) degree of undercooling at exit

+// Given values

+P1 = 2;// inlet pressure of air, [MN/m^2]

+T1 = 273+325;// inlet temperature of air, [MN/m^2]

+P2 = .36;// exit pressure, [MN/m^2]

+m_dot = 7.5;// flow rate of air, [kg/s]

+n = 1.3;// polytropic index

+

+// solution

+// (a)

+//  using steam table

+v1 = .132;// [m^3/kg]

+// given expansion following law PV^n=constant

+

+Pt = P1*(2/(n+1))^(n/(n-1));// critical pressure, [MN/m^2]

+

+//velocity at throat is

+Ct = sqrt(2*n/(n-1)*P1*10^6*v1*(1-(Pt/P1)^(((n-1)/n))));// [m/s]

+vt = v1*(P1/Pt)^(1/n);// [m^3/kg]

+// using m_dot/At=Ct/vt

+At = m_dot*vt/Ct*10^6;// throat area, [mm^2]

+mprintf('\n (a) The throat area is =  %f  mm^2\n',At);

+

+// at exit

+C2 = sqrt(2*n/(n-1)*P1*10^6*v1*(1-(P2/P1)^(((n-1)/n))));// [m/s]

+v2 = v1*(P1/P2)^(1/n);// [m^3/kg]

+A2 = m_dot*v2/C2*10^6;// exit area, [mm^2]

+

+mprintf('\n      The exit area is  =  %f  mm^2\n',A2);

+

+//  (b)

+T2 = T1*(P2/P1)^((n-1)/n);//outlet temperature, [K]

+t2 = T2-273;//[C]

+//  at exit pressure saturation temperature is

+ts = 139.9;// saturation temperature,[C]

+Doc = ts-t2;// Degree of undercooling,[C]

+mprintf('\n (b) The Degree of undercooling at exit is  =  %f C\n',Doc);

+

+// There is some calculation mistake in the book so answer is not matching

+

+//  End

diff --git a/2705/CH12/EX12.4/Ex12_4.sce b/2705/CH12/EX12.4/Ex12_4.sce
new file mode 100755
index 000000000..0bc59c65e
--- /dev/null
+++ b/2705/CH12/EX12.4/Ex12_4.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 12.4');

+

+// aim : To determine the 

+//  (a) throat and exit velocities

+//  (b) throat and exit areas

+

+// Given values

+P1 = 2.2;// inlet pressure, [MN/m^2]

+T1 = 273+260;// inlet temperature, [K]

+P2 = .4;// exit pressure,[MN/m^2]

+eff = .85;// efficiency of the nozzle after throat

+m_dot = 11;// steam flow rate in the nozzle, [kg/s]

+

+// solution

+// (a)

+// assuming steam is following same law as previous question 12.3

+Pt = .546*P1;// critical pressure,[MN/m^2]

+// from Fig. 12.6

+h1 = 2940;// [kJ/kg]

+ht = 2790;// [kJ/kg]

+

+Ct = sqrt(2*(h1-ht)*10^3);// [m/s]

+

+//  again from Fig. 12.6

+h2_prime = 2590;// [kJ/kg]

+// using eff = (ht-h2)/(ht-h2_prime)

+

+h2 = ht-eff*(ht-h2_prime); // [kJ/kg]

+

+C2 = sqrt(2*(h1-h2)*10^3);// [m/s]

+

+// (b)

+// from chart

+vt = .16;// [m^3/kg]

+v2 = .44;// [m^3/kg]

+//  using m_dot*v=A*C

+At = m_dot*vt/Ct*10^6;// throat area, [mm^2]

+

+A2 = m_dot*v2/C2*10^6;// throat area, [mm^2]

+

+mprintf('\n (a) The throat velocity is  =  %f  m/s\n',Ct);

+mprintf('\n      The exit velocity is  =  %f  m/s\n',C2);

+mprintf('\n (b) The throat area is  =  %f  mm^2\n',At);

+mprintf('\n      The throat area is  =  %f  mm^2\n',A2);

+

+//  End

diff --git a/2705/CH13/EX13.1/Ex13_1.sce b/2705/CH13/EX13.1/Ex13_1.sce
new file mode 100755
index 000000000..55203b728
--- /dev/null
+++ b/2705/CH13/EX13.1/Ex13_1.sce
@@ -0,0 +1,24 @@
+clear;

+clc;

+disp('Example 13.1');

+

+//  aim : To determine 

+//  the power developed for a steam flow of 1 kg/s at the blades and the kinetic energy of the steam finally leaving the wheel

+

+//  Given values

+alfa = 20;//  blade angle, [degree]

+Cai = 375;// steam exit velocity in the nozzle,[m/s]

+U = 165;// blade speed, [m/s]

+loss = .15;//  loss of velocity due to friction

+

+//  solution

+//  using Fig13.12,

+Cvw = 320;// change in velocity of whirl, [m/s]

+cae = 132.5;// absolute velocity at exit, [m/s]

+Pds = U*Cvw*10^-3;// Power developed for steam flow of 1 kg/s, [kW]

+Kes = cae^2/2*10^-3;// Kinetic energy change of steam, [kW/kg] 

+

+mprintf('\n The power developed for a steam flow of 1 kg/s is  =  %f kW\n',Pds)

+mprintf('\n The energy of steam finally leaving the wheel is  =  %f  kW/kg\n',Kes);

+

+//  End

diff --git a/2705/CH13/EX13.2/Ex13_2.sce b/2705/CH13/EX13.2/Ex13_2.sce
new file mode 100755
index 000000000..601a28f65
--- /dev/null
+++ b/2705/CH13/EX13.2/Ex13_2.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+disp('Example 13.2');

+

+// aim : To determine

+// (a) the entry angle of  the blades

+// (b) the work done per kilogram of steam per second

+// (c) the diagram efficiency

+// (d) the end-thrust per kilogram of steam per second

+

+// given values

+Cai = 600;// steam velocity, [m/s]

+sia = 25;// steam inlet angle with blade, [degree]

+U = 255;// mean blade speed, [m/s]

+sea = 30;// steam exit angle with blade,[degree] 

+

+// solution

+// (a)

+// using Fig.13.13(diagram for example 13.2)

+eab = 41.5;// entry angle of blades, [degree]

+mprintf('\n (a) The angle of blades is  =  %f  degree\n',eab);

+

+// (b)

+Cwi_plus_Cwe = 590;// velocity of whirl, [m/s]

+W = U*(Cwi_plus_Cwe);// work done on the blade,[W/kg]

+mprintf('\n (b) The work done on the blade is  =  %f kW/kg\n',W*10^-3);

+

+// (c)

+De = 2*U*(Cwi_plus_Cwe)/Cai^2;// diagram efficiency 

+mprintf('\n (c) The diagram efficiency is  =  %f percent\n',De*100);

+

+// (d)

+// again from the diagram

+Cfe_minus_Cfi = -90;// change invelocity of flow, [m/s]

+Eth = Cfe_minus_Cfi;// end-thrust, [N/kg s]

+mprintf('\n (d) The End-thrust is  =  %f  N/kg',Eth);

+

+//  End

diff --git a/2705/CH13/EX13.3/Ex13_3.sce b/2705/CH13/EX13.3/Ex13_3.sce
new file mode 100755
index 000000000..511e4e183
--- /dev/null
+++ b/2705/CH13/EX13.3/Ex13_3.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+disp('Example 13.3');

+

+// aim : To determine

+// (a) the power output of the turbine

+// (b) the diagram efficiency

+

+// given values

+U = 150;// mean blade speed, [m/s]

+Cai1 = 675;// nozzle speed, [m/s]

+na = 20;// nozzle angle, [degree]

+m_dot = 4.5;// steam flow rate, [kg/s]

+

+// solution

+// from Fig. 13.15(diagram 13.3)

+Cw1 = 915;// [m/s]

+Cw2 = 280;// [m/s]

+

+// (a)

+P = m_dot*U*(Cw1+Cw2);// power of turbine,[W]

+mprintf('\n (a) The power of turbine is  =  %f kW\n',P*10^-3);

+

+// (b)

+De = 2*U*(Cw1+Cw2)/Cai1^2;// diagram efficiency

+mprintf('\n (b) The diagram efficiency is  =  %f percent\n',De*100);

+

+//  End

diff --git a/2705/CH13/EX13.4/Ex13_4.sce b/2705/CH13/EX13.4/Ex13_4.sce
new file mode 100755
index 000000000..2f9f9c42a
--- /dev/null
+++ b/2705/CH13/EX13.4/Ex13_4.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+disp('Example 13.4');

+

+// aim : To determine

+// (a) the power output of the stage

+// (b) the specific enthalpy drop in the stage

+// (c) the percentage increase in relative velocity in the moving blades due to expansion in the bladse

+

+// given values

+N = 50;// speed, [m/s]

+d = 1;// blade ring diameter, [m]

+nai = 50;// nozzle inlet angle, [degree]

+nae = 30;// nozzle exit angle, [degree]

+m_dot = 600000;// steam flow rate, [kg/h]

+se = .85;// stage efficiency

+

+// solution

+// (a)

+U = %pi*d*N;// mean blade speed, [m/s]

+// from Fig. 13.17(diagram 13.4)

+Cwi_plus_Cwe = 444;// change in whirl speed, [m/s]

+P = m_dot*U*Cwi_plus_Cwe/3600;// power output of the stage, [W]

+mprintf('\n (a) The power output of the stage is  =  %f MW\n',P*10^-6);

+

+// (b)

+h = U*Cwi_plus_Cwe/se;// specific enthalpy,[J/kg]

+mprintf('\n (b) The specific enthalpy drop in the stage is  =  %f  kJ/kg\n ',h*10^-3);

+

+// (c)

+// again from diagram

+Cri = 224;// [m/s]

+Cre = 341;// [m/s]

+Iir = (Cre-Cri)/Cri;// increase in relative velocity

+mprintf('\n (c) The increase in relative velocity is  =  %f  percent\n',Iir*100);

+

+//  End

diff --git a/2705/CH13/EX13.5/Ex13_5.sce b/2705/CH13/EX13.5/Ex13_5.sce
new file mode 100755
index 000000000..b2c3c6672
--- /dev/null
+++ b/2705/CH13/EX13.5/Ex13_5.sce
@@ -0,0 +1,36 @@
+clear;

+clc;

+disp('Example 13.5');

+

+// aim : To determine

+// (a) the blade height of the stage

+// (b) the power developed in the stage

+// (c) the specific enthalpy drop at the stage

+

+// given values

+U = 60;// mean blade speed, [m/s]

+P = 350;// steam pressure, [kN/m^2]

+T = 175;// steam temperature, [C]

+nai = 30;// stage inlet angle, [degree]

+nae = 20;// stage exit angle, [degree] 

+

+// solution

+// (a)

+m_dot = 13.5;// steam flow rate, [kg/s]

+// at given T and P

+v = .589;// specific volume, [m^3/kg]

+// given H=d/10, so

+H = sqrt(m_dot*v/(%pi*10*60));// blade height, [m]

+mprintf('\n (a) The blade height at this stage is  =  %f  mm\n',H*10^3);

+

+// (b)

+Cwi_plus_Cwe = 270;// change in whirl speed, [m/s]

+P = m_dot*U*(Cwi_plus_Cwe);// power developed, [W]

+mprintf('\n (b) The power developed is  =  %f  kW\n',P*10^-3);

+

+// (c)

+s = .85;// stage efficiency

+h = U*Cwi_plus_Cwe/s;// specific enthalpy,[J/kg]

+mprintf('\n (a) The specific enthalpy drop in the stage is  =  %f  kJ/kg',h*10^-3);

+

+//  End

diff --git a/2705/CH14/EX14.1/Ex14_1.sce b/2705/CH14/EX14.1/Ex14_1.sce
new file mode 100755
index 000000000..404a8d8d4
--- /dev/null
+++ b/2705/CH14/EX14.1/Ex14_1.sce
@@ -0,0 +1,55 @@
+clear;

+clc;

+disp(' Example 14.1');

+

+// aim : To determine 

+// (a) the free air delivered

+// (b) the volumetric efficiency

+// (c) the air delivery temperature

+// (d) the cycle power

+// (e) the isothermal efficiency

+

+// given values

+d = 200*10^-3;// bore, [m]

+L = 300*10^-3;// stroke, [m]

+N = 500;// speed, [rev/min]

+n = 1.3;// polytropic index

+P1 = 97;// intake pressure, [kN/m^2]

+T1 = 273+20;// intake temperature, [K]

+P3 = 550;// compression pressure, [kN/m^2]

+

+// solution

+// (a)

+P4 = P1;

+P2 = P3;

+Pf = 101.325;// free air pressure, [kN/m^2]

+Tf = 273+15;// free air temperature, [K]

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

+V3 = .05*SV;// [m^3]

+V1 = SV+V3;// [m^3]

+V4 = V3*(P3/P4)^(1/n);// [m^3]

+ESV = (V1-V4)*N;// effective swept volume/min, [m^3]

+// using PV/T=constant

+Vf = P1*ESV*Tf/(Pf*T1);// free air delivered, [m^3/min]

+mprintf('\n (a) The free air delivered is  =  %f  m^3/min\n',Vf);

+

+// (b)

+VE = Vf/(N*(V1-V3));// volumetric efficiency

+mprintf('\n (b) The volumetric efficiency is  =  %f  percent\n',VE*100);

+

+// (c)

+T2 = T1*(P2/P1)^((n-1)/n);//  free air temperature, [K]

+mprintf('\n (c) The air delivery temperature is  =  %f  C\n',T2-273);

+

+//  (d)

+CP = n/(n-1)*P1*(V1-V4)*((P2/P1)^((n-1)/n)-1)*N/60;// cycle power, [kW]

+ mprintf('\n (d) The cycle power is  =  %f  kW\n',CP);

+

+// (e)

+// neglecting clearence

+W = n/(n-1)*P1*V1*((P2/P1)^((n-1)/n)-1)

+Wi = P1*V1*log(P2/P1);// isothermal efficiency

+IE = Wi/W;// isothermal efficiency

+mprintf('\n (e) The isothermal efficiency neglecting clearence  is  =  %f percent\n',IE*100);

+

+// End

diff --git a/2705/CH14/EX14.2/Ex14_2.sce b/2705/CH14/EX14.2/Ex14_2.sce
new file mode 100755
index 000000000..ec512aa18
--- /dev/null
+++ b/2705/CH14/EX14.2/Ex14_2.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+disp(' Example 14.2');

+

+// aim : To determine 

+// (a) the intermediate pressure

+// (b) the total volume of each cylinder

+// (c) the cycle power

+

+// given values

+v1 = .2;// air intake, [m^3/s]

+P1 = .1;// intake pressure, [MN/m^2]

+T1 = 273+16;// intake temperature, [K]

+P3 = .7;// final pressure, [MN/m^2]

+n = 1.25;// compression index

+N = 10;// speed, [rev/s]

+

+// solution

+// (a)

+P2 = sqrt(P1*P3);// intermediate pressure, [MN/m^2]

+mprintf('\n (a) The intermediate pressure is  =  %f  MN/m^2\n',P2);

+

+// (b)

+V1  = v1/N;// total volume,[m^3]

+// since intercooling is perfect so 2 lie on the isothermal through1, P1*V1=P2*V2

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

+mprintf('\n (b) The total volume of the HP cylinder is  =  %f  litres\n',V2*10^3);

+

+ // (c)

+ CP = 2*n/(n-1)*P1*v1*((P2/P1)^((n-1)/n)-1);// cycle power, [MW]

+  mprintf('\n (c) The cycle power is  =  %f  MW\n',CP*10^3);

+ 

+ // there is calculation mistake in the book so answer is not matching

+ 

+ //  End

diff --git a/2705/CH14/EX14.3/Ex14_3.sce b/2705/CH14/EX14.3/Ex14_3.sce
new file mode 100755
index 000000000..7b899feb5
--- /dev/null
+++ b/2705/CH14/EX14.3/Ex14_3.sce
@@ -0,0 +1,59 @@
+clear;

+clc;

+disp(' Example 14.3');

+

+// aim : To determine 

+// (a) the intermediate pressures

+// (b) the effective swept volume of the LP cylinder

+// (c) the temperature and the volume of air delivered per stroke at 15 bar

+// (d) the work done per kilogram of air

+

+// given values

+d = 450*10^-3;//  bore , [m]

+L = 300*10^-3;//  stroke, [m]

+cl = .05;//  clearence

+P1 = 1; // intake pressure, [bar]

+T1 = 273+18;// intake temperature, [K]

+P4 = 15;// final delivery pressure, [bar]

+n = 1.3;//  compression and expansion index

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

+

+// solution

+// (a)

+k=(P4/P1)^(1/3); 

+// hence

+P2 = k*P1;//  intermediare pressure, [bar]

+P3 = k*P2;//  intermediate pressure, [bar]

+

+mprintf('\n (a) The intermediate pressure is P2 =  %f  bar\n',P2);

+mprintf('\n     The intermediate pressure is  P3=  %f  bar\n',P3);

+

+// (b)

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

+// hence

+V7 = cl*SV;// volume, [m^3]

+V1 = SV+V7;// volume, [m^3]

+// also

+P7 = P2;

+P8 = P1;

+V8 = V7*(P7/P8)^(1/n);//  volume, [m^3]

+ESV = V1-V8;// effective swept volume of LP cylinder, [m^3]

+

+mprintf('\n (b) The effective swept volume of the LP cylinder is  =  %f litres\n',ESV*10^3);

+

+// (c)

+T9 = T1;

+P9 = P3;

+T4 = T9*(P4/P9)^((n-1)/n);// delivery temperature, [K]

+// now using P4*(V4-V5)/T4=P1*(V1-V8)/T1

+V4_minus_V5 = P1*T4*(V1-V8)/(P4*T1);// delivery volume, [m^3]

+ 

+mprintf('\n (c) The delivery temperature is =  %f  C\n',T4-273);

+mprintf('\n      The delivery volume is =  %f  litres\n',V4_minus_V5*10^3);

+

+//  (d)

+

+W = 3*n*R*T1*((P2/P1)^((n-1)/n)-1)/(n-1);//  work done/kg ,[kJ]

+mprintf('\n (d) The work done per kilogram of air is =  %f  kJ\n',W);

+ 

+// End

diff --git a/2705/CH14/EX14.4/Ex14_4.sce b/2705/CH14/EX14.4/Ex14_4.sce
new file mode 100755
index 000000000..8cfa8e686
--- /dev/null
+++ b/2705/CH14/EX14.4/Ex14_4.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+disp(' Example 14.4');

+

+// aim : To determine 

+// (a) the final pressure and temperature

+// (b) the energy required to drive the compressor

+

+// given values

+rv = 5;// pressure compression ratio

+m_dot = 10;// air flow rate, [kg/s]

+P1 = 100;// initial pressure, [kN/m^2]

+T1 = 273+20;// initial temperature, [K]

+n_com = .85;// isentropic efficiency of compressor

+Gama = 1.4;// heat capacity ratio

+cp = 1.005;// specific heat capacity, [kJ/kg K]

+

+// solution

+// (a)

+T2_prim = T1*(rv)^((Gama-1)/Gama);// temperature after compression, [K]

+// using isentropic efficiency=(T2_prim-T1)/(T2-T1)

+T2 = T1+(T2_prim-T1)/n_com;//  final temperature, [K]

+P2 = rv*P1;// final pressure, [kN/m^2]

+mprintf('\n (a) The final temperature is  =  %f  C\n',T2-273);

+mprintf('\n (b) The final pressure is  =  %f  kN/m^2\n',P2);

+

+// (b)

+E = m_dot*cp*(T1-T2);// energy required, [kW]

+mprintf('\n (b) The energy required to drive the compressor is  =  %f  kW',E);

+if(E<0)

+    disp('The negative sign indicates energy input');

+else

+    disp('The positive sign indicates energy output');

+end

+

+ //  End

+

+

diff --git a/2705/CH14/EX14.5/Ex14_5.sce b/2705/CH14/EX14.5/Ex14_5.sce
new file mode 100755
index 000000000..c9d23cd92
--- /dev/null
+++ b/2705/CH14/EX14.5/Ex14_5.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+disp(' Example 14.5');

+

+// aim : To determine 

+//  the power absorbed in driving the compressor

+

+// given values

+FC = .68;// fuel consumption rate, [kg/min]

+P1 = 93;// initial pressure, [kN/m^2]

+P2 = 200;// final pressure, [kN/m^2]

+T1 = 273+15;// initial temperature, [K]

+d = 1.3;// density of mixture, [kg/m^3]

+n_com = .82;// isentropic efficiency of compressor

+Gama = 1.38;// heat capacity ratio

+

+// solution

+R = P1/(d*T1);// gas constant, [kJ/kg K]

+// for mixture

+cp = Gama*R/(Gama-1);// heat capacity, [kJ/kg K]

+T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// temperature after compression, [K]

+// using isentropic efficiency=(T2_prim-T1)/(T2-T1)

+T2 = T1+(T2_prim-T1)/n_com;//  final temperature, [K]

+m_dot = FC*15/60;// given condition, [kg/s]

+P = m_dot*cp*(T2-T1);// power absorbed by compressor, [kW]

+mprintf('\n The power absorbed by compressor is  =  %f kW\n',P);

+

+// End

diff --git a/2705/CH14/EX14.6/Ex14_6.sce b/2705/CH14/EX14.6/Ex14_6.sce
new file mode 100755
index 000000000..7e971e4b6
--- /dev/null
+++ b/2705/CH14/EX14.6/Ex14_6.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+disp(' Example 14.6');

+

+// aim : To determine 

+//  the power required to drive the blower

+

+// given values

+m_dot = 1;// air capacity, [kg/s]

+rp = 2;// pressure ratio

+P1 = 1*10^5;// intake pressure, [N/m^2]

+T1 = 273+70;// intake temperature, [K]

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

+

+// solution

+V1_dot = m_dot*R*T1/P1*10^3;// [m^3/s]

+P2 = rp*P1;// final pressure, [n/m^2]

+P = V1_dot*(P2-P1);// power required, [W]

+mprintf('\n The power required to drive the blower is  =  %f kW\n',P*10^-3);

+

+// End

diff --git a/2705/CH14/EX14.7/Ex14_7.sce b/2705/CH14/EX14.7/Ex14_7.sce
new file mode 100755
index 000000000..a367f2e9e
--- /dev/null
+++ b/2705/CH14/EX14.7/Ex14_7.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp(' Example 14.7');

+

+// aim : To determine 

+//  the power required to drive the vane pump

+

+// given values

+m_dot = 1;// air capacity, [kg/s]

+rp = 2;// pressure ratio

+P1 = 1*10^5;// intake pressure, [N/m^2]

+T1 = 273+70;// intake temperature, [K]

+Gama = 1.4;// heat capacity ratio

+rv = .7;// volume ratio

+

+// solution

+V1 = .995;// intake pressure(as given previous question),[m^3/s] 

+// using P1*V1^Gama=P2*V2^Gama, so

+P2 = P1*(1/rv)^Gama;// pressure, [N/m^2]

+V2 = rv*V1;// volume,[m^3/s]

+P3 = rp*P1;// final pressure, [N/m^2]

+P = Gama/(Gama-1)*P1*V1*((P2/P1)^((Gama-1)/Gama)-1)+V2*(P3-P2);// power required,[W]

+mprintf('\n The power required to drive the vane pump is  =  %f  kW\n',P*10^-3);

+

+//  End

diff --git a/2705/CH14/EX14.8/Ex14_8.sce b/2705/CH14/EX14.8/Ex14_8.sce
new file mode 100755
index 000000000..e851d6a3b
--- /dev/null
+++ b/2705/CH14/EX14.8/Ex14_8.sce
@@ -0,0 +1,36 @@
+clear;

+clc;

+disp(' Example 14.8');

+

+// aim : To determine 

+//  the total temperature and pressure of the mixture

+

+// given values

+rp = 2.5;// static pressure ratio

+FC = .04;// fuel consumption rate, [kg/min]

+P1 = 60;// inilet pressure, [kN/m^2]

+T1 = 273+5;// inilet temperature, [K]

+n_com = .84;// isentropic efficiency of compressor

+Gama = 1.39;// heat capacity ratio

+C2 = 120;//exit velocity from compressor, [m/s]

+rm = 13;// air-fuel ratio

+cp = 1.005;// heat capacity ratio

+

+// solution

+P2 = rp*P1;// given condition, [kN/m^2]

+T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// temperature after compression, [K]

+// using isentropic efficiency=(T2_prim-T1)/(T2-T1)

+T2 = T1+(T2_prim-T1)/n_com;//  final temperature, [K]

+m_dot = FC*(rm+1);// mass of air-fuel mixture, [kg/s]

+P = m_dot*cp*(T2-T1);// power to drive compressor, [kW]

+mprintf('\n The power required to drive compressor is  =  %f  kW\n',P);

+

+Tt2 = T2+C2^2/(2*cp*10^3);// total temperature,[K]

+Pt2 = P2*(Tt2/T2)^(Gama/(Gama-1));// total pressure, [kN/m^2]

+mprintf('\n The temperature in the engine is  =  %f  C\n',Tt2-273);

+mprintf('\n The pressure in the engine cylinder is  =  %f  kN/m^2\n',Pt2);

+

+//  There is calculation mistake in the book

+

+

+//  End

diff --git a/2705/CH15/EX15.1/Ex15_1.sce b/2705/CH15/EX15.1/Ex15_1.sce
new file mode 100755
index 000000000..153a3eea6
--- /dev/null
+++ b/2705/CH15/EX15.1/Ex15_1.sce
@@ -0,0 +1,17 @@
+clear;

+clc;

+disp('Example 15.1');

+

+// aim : To determine 

+// the thermal efficiency of the cycle

+

+// given values

+T1 = 273+400;//  temperature limit, [K]

+T3 = 273+70;// temperature limit, [K]

+

+//  solution

+//  using equation [15] of section 15.3

+n_the = (T1-T3)/T1*100;// thermal efficiency 

+mprintf('\n The thermal efficiency of the cycle is  =  %f percent\n',n_the);

+

+//  End

diff --git a/2705/CH15/EX15.10/Ex15_10.sce b/2705/CH15/EX15.10/Ex15_10.sce
new file mode 100755
index 000000000..5748bff66
--- /dev/null
+++ b/2705/CH15/EX15.10/Ex15_10.sce
@@ -0,0 +1,32 @@
+clear;

+clc;

+disp('Example 10');

+

+// aim : To determine

+// (a) the maximum temperature attained during the cycle

+// (b) the thermal efficiency of the cycle

+

+// given value

+rva  =7.5;// volume ratio of adiabatic expansion

+rvc  =15;// volume ratio of compression

+P1 = 98;// initial pressure, [kn/m^2]

+T1 = 273+44;// initial temperature, [K]

+P4 = 258;// pressure at the end of the adiabatic expansion, [kN/m^2]

+Gama = 1.4;//  heat capacity  ratio

+

+// solution

+// by seeing diagram

+// for process 4-1, P4/T4=P1/T1

+T4 = T1*(P4/P1);// [K]

+// for process 3-4

+T3 = T4*(rva)^(Gama-1);

+mprintf('\n (a) The maximum temperature during the cycle is  =  %f  C\n',T3-273);

+

+// (b)

+

+// for process 1-2,

+T2 = T1*(rvc)^(Gama-1);// [K]

+n_the = 1-(T4-T1)/((Gama)*(T3-T2));// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',n_the*100);

+

+//  End

diff --git a/2705/CH15/EX15.11/Ex15_11.sce b/2705/CH15/EX15.11/Ex15_11.sce
new file mode 100755
index 000000000..9045f1cd3
--- /dev/null
+++ b/2705/CH15/EX15.11/Ex15_11.sce
@@ -0,0 +1,37 @@
+clear;

+clc;

+disp('Example 15.11');

+

+// aim : To determine

+// (a) the thermal efficiency of the cycle

+// (b) the indicared power of the cycle

+

+// given values

+// taking basis one second

+rv = 11;// volume ratio

+P1 = 96;// initial pressure , [kN/m^2]

+T1 = 273+18;// initial temperature, [K]

+Gama = 1.4;// heat capacity ratio

+

+// solution

+// taking reference  Fig. 15.24

+// (a)

+Beta = 2;// ratio of V3 and V2

+TE = 1-(Beta^(Gama)-1)/((rv^(Gama-1))*Gama*(Beta-1));// thermal efficiency

+mprintf('\n (a) the thermal efficiency of the cycle is  =  %f percent\n ',TE*100);

+

+// (b) 

+// let V1-V2=.05, so

+V2 = .05*.1;// [m^3]

+// from this

+V1 = rv*V2;// [m^3]

+V3 = Beta*V2;// [m^3]

+V4 = V1;// [m^3]

+P2 = P1*(V1/V2)^(Gama);// [kN/m^2]

+P3 = P2;// [kn/m^2]

+P4=P3*(V3/V4)^(Gama);// [kN/m^2]

+// indicated power

+W = P2*(V3-V2)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// indicated power, [kW]

+mprintf('\n (c) The indicated power of the cycle is  =  %f kW\n',W);

+

+//  End

diff --git a/2705/CH15/EX15.12/Ex15_12.sce b/2705/CH15/EX15.12/Ex15_12.sce
new file mode 100755
index 000000000..43bddcdf8
--- /dev/null
+++ b/2705/CH15/EX15.12/Ex15_12.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 15.12');

+

+// aim : To determine

+// (a) the pressure and temperature at the end of compression

+// (b) the pressure and temperature at the end of the constant volume process

+// (c) the temperature at the end of constant pressure process

+

+// given values

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

+T1 = 273+22;// initial temperature, [K]

+rv = 16;// volume ratio of the compression

+Q = 244;//heat added, [kJ/kg]

+Gama = 1.4;// heat capacity ratio

+cv = .717;// heat capacity, [kJ/kg k]

+

+// solution

+//  taking reference as Fig.15.26

+// (a)

+// for compression

+// rv = V1/V2

+P2 = P1*(rv)^Gama;// pressure at end of compression, [kN/m^2]

+T2 = T1*(rv)^(Gama-1);// temperature at end of compression, [K]

+mprintf('\n (a) The pressure at the end of compression is  =  %f  MN/m^2\n',P2*10^-3);

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

+

+// (b)

+// for constant volume process, 

+// Q = cv*(T3-T2), so

+T3 = T2+Q/cv;// temperature at the end of constant volume, [K]

+

+// so for constant volume, P/T=constant, hence

+P3 = P2*(T3/T2);// pressure at the end of constant volume process, [kN/m^2]

+mprintf('\n (b) The pressure at the end of constant volume process is  =  %f  MN/m^2\n ',P3*10^-3);

+mprintf('\n     The temperature at the end of constant volume process is  =  %f C\n',T3-273);

+

+// (c)

+S = rv-1;// stroke

+// assuming 

+V3 = 1;// [volume]

+//so

+V4 = V3+S*.03;// [volume]

+// also for constant process V/T=constant, hence

+T4 = T3*(V4/V3);// temperature at the end of constant presure process, [k] 

+mprintf('\n (c) The temperature at the end of constant pressure process is  =  %f C\n',T4-273);

+

+//  End

diff --git a/2705/CH15/EX15.13/Ex15_13.sce b/2705/CH15/EX15.13/Ex15_13.sce
new file mode 100755
index 000000000..50608fe00
--- /dev/null
+++ b/2705/CH15/EX15.13/Ex15_13.sce
@@ -0,0 +1,79 @@
+clear;

+clc;

+disp('Example 15.13');

+

+// aim : To determine

+// (a) the pressure, volume and temperature at cycle process change points

+// (b) the net work done 

+// (c)  the thermal efficiency

+// (d) the heat received

+// (e)  the work ratio

+// (f)  the mean effective pressure

+// (g)  the carnot efficiency

+

+

+// given values

+rv = 15;// volume ratio

+P1 = 97*10^-3;// initial pressure , [MN/m^2]

+V1 = .084;// initial volume, [m^3]

+T1 = 273+28;// initial temperature, [K]

+T4 = 273+1320;// maximum temperature, [K]

+P3 = 6.2;// maximum pressure, [MN/m^2]

+cp = 1.005;// specific heat capacity at constant pressure, [kJ/kg K]

+cv = .717;// specific heat capacity at constant volume, [kJ/kg K]

+

+// solution

+// taking reference  Fig. 15.27

+// (a)

+R = cp-cv;// gas constant, [kJ/kg K]

+Gama = cp/cv;// heat capacity ratio

+// for process 1-2

+V2 = V1/rv;// volume at stage2, [m^3] 

+// using PV^(Gama)=constant for process 1-2

+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2]

+T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K]

+

+// for process 2-3

+// since volumee is constant in process 2-3 , so using P/T=constant, so

+T3 = T2*(P3/P2);// volume at stage 3, [K]

+V3 = V2;// volume at stage 3, [MN/m^2]

+

+// for process 3-4

+P4 = P3;// pressure at stage 4, [m^3]

+// since in stage 3-4 P is constant, so V/T=constant, 

+V4 = V3*(T4/T3);// temperature at stage  4,[K]

+

+// for process 4-5

+V5 = V1;// volume at stage 5, [m^3]

+P5 = P4*(V4/V5)^(Gama);// pressure at stage5,. [MN/m^2]

+T5 = T4*(V4/V5)^(Gama-1);// temperature at stage 5, [K]

+

+mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f MN/m^2,    V2  =  %f m^3,        t2  =  %f C,\n      P3  =  %f MN/m^2,    V3  =  %f m^3,         t3  =  %f C,\n      P4  =  %f MN/m^2,      V4  =  %f m^3,       t4  =  %f C,\n      P5  =   %fkN/m^2,    V5  =  %fm^3,        t5  =  %fC\n',P1*10^3,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273,P5*10^3,V5,T5-273);

+

+

+// (b)

+W = (P3*(V4-V3)+((P4*V4-P5*V5)-(P2*V2-P1*V1))/(Gama-1))*10^3;// work done, [kJ]

+mprintf('\n (b) The net work done is  =  %f kJ\n',W);

+

+// (c) 

+TE = 1-(T5-T1)/((T3-T2)+Gama*(T4-T3));// thermal efficiency

+mprintf('\n (c) The thermal efficiency is  =  %f percent\n',TE*100);

+

+// (d)

+Q = W/TE;// heat received, [kJ]

+mprintf('\n (d) The heat received is  =  %f kJ\n',Q);

+

+// (e)

+PW = P3*(V4-V3)+(P4*V4-P5*V5)/(Gama-1)

+WR = W*10^-3/PW;//  work ratio

+mprintf('\n (f) The work ratio is  =  %f\n',WR);

+

+// (e)

+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]

+mprintf('\n (e) The mean effective pressure is  =  %f kN/m^2\n',Pm);

+

+// (f)

+CE = (T4-T1)/T4;// carnot efficiency

+mprintf('\n (f) The carnot efficiency is  =  %f percent\n',CE*100);

+

+//  End

diff --git a/2705/CH15/EX15.14/Ex15_14.sce b/2705/CH15/EX15.14/Ex15_14.sce
new file mode 100755
index 000000000..ebf067b5b
--- /dev/null
+++ b/2705/CH15/EX15.14/Ex15_14.sce
@@ -0,0 +1,77 @@
+clear;

+clc;

+disp('Example 15.14');

+

+// aim : To determine

+// (a)  the thermal efficiency

+// (b)  the heat received

+// (c) the heat rejected

+// (d) the net work 

+// (e)  the work ratio

+// (f)  the mean effective pressure

+// (g)  the carnot efficiency

+

+

+// given values

+P1 = 101;// initial pressure , [kN/m^2]

+V1 = 14*10^-3;// initial volume, [m^3]

+T1 = 273+15;// initial temperature, [K]

+P3 = 1850;// maximum pressure, [kN/m^2]

+V2 = 2.8*10^-3;// compressed volume, [m^3]

+Gama = 1.4;// heat capacity

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

+

+// solution

+// taking reference  Fig. 15.29

+// (a)

+// for process 1-2

+// using PV^(Gama)=constant for process 1-2

+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2]

+T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K]

+

+// for process 2-3

+// since volumee is constant in process 2-3 , so using P/T=constant, so

+T3 = T2*(P3/P2);// volume at stage 3, [K]

+

+// for process 3-4

+P4 = P1;

+T4 = T3*(P4/P3)^((Gama-1)/Gama);// temperature

+

+TE = 1-Gama*(T4-T1)/(T3-T2);// thermal efficiency

+mprintf('\n (a) The thermal efficiency is  =  %f percent\n',TE*100);

+

+// (b)

+cv = R/(Gama-1);// heat capacity at copnstant volume, [kJ/kg k]

+m = P1*V1/(R*T1);// mass of gas, [kg]

+Q1 = m*cv*(T3-T2);// heat received, [kJ/cycle]

+mprintf('\n (b) The heat received is  =  %f kJ/cycle\n',Q1);

+

+// (c)

+cp = Gama*cv;// heat capacity at constant at constant pressure, [kJ/kg K]

+Q2 = m*cp*(T4-T1);// heat rejected, [kJ/cycle]

+mprintf('\n (c) The heat rejected is  =  %f kJ/cycle\n',Q2);

+

+// (d)

+W = Q1-Q2;// net work , [kJ/cycle]

+mprintf('\n (d) The net work is  =  %f  kJ/cycle\n',W);

+

+// (e)

+// pressure is constant for process 1-4, so V/T=constant

+V4 = V1*(T4/T1);// volume, [m^3]

+V3 = V2;// for process 2-3

+P4 = P1;// for process 1-4

+PW = (P3*V3-P1*V1)/(Gama-1);// positive work done, [kJ/cycle]

+WR = W/PW;//  work ratio

+mprintf('\n (e) The work ratio is  =  %f\n',WR);

+

+// (f)

+Pm = W/(V4-V2);// mean effective pressure, [kN/m^2]

+mprintf('\n (f) The mean effefctive pressure is  =  %f  kN/m^2\n',Pm);

+

+// (g)

+CE = (T3-T1)/T3;// carnot efficiency

+mprintf('\n (g) The carnot efficiency is  =  %f percent\n',CE*100);

+

+// there is minor variation in answer reported in the book

+

+//  End

diff --git a/2705/CH15/EX15.15/Ex15_15.sce b/2705/CH15/EX15.15/Ex15_15.sce
new file mode 100755
index 000000000..d6b24e902
--- /dev/null
+++ b/2705/CH15/EX15.15/Ex15_15.sce
@@ -0,0 +1,41 @@
+clear;

+clc;

+disp('Example 15.15');

+

+// aim : To determine

+// (a) the net work done

+// (b) the ideal thermal efficiency

+// (c) the thermal efficiency if the process of generation is not included

+

+// given values

+P1 = 110;// initial pressure, [kN/m^2)

+T1 = 273+30;// initial temperature, [K]

+V1 = .05;// initial volume, [m^3]

+V2 = .005;// volume, [m^3]

+T3 = 273+700;// temperature, [m^3]

+R = .289;// gas constant, [kJ/kg K]

+cv = .718;// heat capacity, [kJ/kg K]

+

+// solution

+// (a)

+m = P1*V1/(R*T1);// mass , [kg]

+W = m*R*(T3-T1)*log(V1/V2);// work done, [kJ]

+mprintf('\n (a) The net work done is  =  %f kJ\n',W);

+

+// (b)

+n_the = (T3-T1)/T3;// ideal thermal efficiency

+mprintf('\n (b) The ideal thermal efficiency is  =  %f percent\n',n_the*100);

+

+// (c)

+V4 = V1;

+V3 = V2;

+T4 = T3;

+T2 = T1;

+

+Q_rej = m*cv*(T4-T1)+m*R*T1*log(V1/V2);// heat rejected

+Q_rec = m*cv*(T3-T2)+m*R*T3*log(V4/V3);// heat received

+

+n_th = (1-Q_rej/Q_rec);// thermal efficiency

+mprintf('\n (c) the thermal efficiency if the process of regeneration is not included is  =  %f percent\n',n_th*100);

+

+//  End

diff --git a/2705/CH15/EX15.16/Ex15_16.sce b/2705/CH15/EX15.16/Ex15_16.sce
new file mode 100755
index 000000000..5417fa898
--- /dev/null
+++ b/2705/CH15/EX15.16/Ex15_16.sce
@@ -0,0 +1,48 @@
+clear;

+clc;

+disp('Example 15.16');

+

+// aim : To determine

+// (a) the maximum temperature

+// (b) the net work done

+// (c) the ideal thermal efficiency

+// (d) the thermal efficiency if the process of regeneration is not included

+

+// given values

+P1 = 100;// initial pressure, [kN/m^2)

+T1 = 273+20;// initial temperature, [K]

+V1 = .08;// initial volume, [m^3]

+rv = 5;// volume ratio

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

+cp = 1.006;// heat capacity, [kJ/kg K]

+V3_by_V2 = 2;

+

+// solution

+// (a)

+// using Fig.15.33

+// process 1-2 is isothermal

+T2 = T1;

+// since process 2-3 isisobaric, so V/T=constant

+T3 = T2*(V3_by_V2);// maximumtemperature, [K]

+mprintf('\n (a) The maximum temperature is  =  %f C\n',T3-273);

+

+// (b)

+m = P1*V1/(R*T1);// mass , [kg]

+W = m*R*(T3-T1)*log(rv);// work done, [kJ]

+mprintf('\n (b) The net work done is  =  %f  kJ\n',W);

+

+// (c)

+TE = (T3-T1)/T3;// ideal thermal efficiency

+mprintf('\n (c) The ideal thermal efficiency is  =  %f  percent\n',TE*100);

+

+// (d)

+T4 = T3;

+T2 = T1;

+

+Q_rej = m*cp*(T4-T1)+m*R*T1*log(rv);// heat rejected

+Q_rec = m*cp*(T3-T2)+m*R*T3*log(rv);// heat received

+

+n_th = (1-Q_rej/Q_rec);// thermal efficiency

+mprintf('\n (d) the thermal efficiency if the process of regeneration is not included is  =  %f  percent\n',n_th*100);

+

+//  End

diff --git a/2705/CH15/EX15.17/Ex15_17.sce b/2705/CH15/EX15.17/Ex15_17.sce
new file mode 100755
index 000000000..5182839fe
--- /dev/null
+++ b/2705/CH15/EX15.17/Ex15_17.sce
@@ -0,0 +1,36 @@
+clear;

+clc;

+disp('Example 15.17');

+

+// aim : To determine 

+// (a) the net work done

+// (b) thethermal efficiency

+

+// given values

+m = 1;// mass of air, [kg]

+T1 = 273+230;// initial temperature, [K]

+P1 = 3450;// initial pressure, [kN/m^2]

+P2 = 2000;// pressure, [kN/m^2]

+P3 = 140;// pressure, [kN/m^2]

+P4 = P3;

+Gama = 1.4; // heat capacity ratio

+cp = 1.006;// heat capacity, [kJ/kg k]

+

+// solution

+T2 =T1;// isothermal process 1-2

+// process 2-3 and 1-4 are adiabatic so

+T3 = T2*(P3/P2)^((Gama-1)/Gama);// temperature, [K] 

+T4 = T1*(P4/P1)^((Gama-1)/Gama);// [K]

+R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K]

+Q1 = m*R*T1*log(P1/P2);// heat received, [kJ]

+Q2 = m*cp*(T3-T4);// heat rejected

+

+//hence

+W = Q1-Q2;// work done

+mprintf('\n (a) The net work done is  =  %f kJ\n',W);

+

+// (b)

+TE = 1-Q2/Q1;// thermal efficiency

+mprintf('\n (b) The thermal efficiency is  =  %f  percent\n',TE*100);

+

+//  End

diff --git a/2705/CH15/EX15.18/Ex15_18.sce b/2705/CH15/EX15.18/Ex15_18.sce
new file mode 100755
index 000000000..2932ac9d0
--- /dev/null
+++ b/2705/CH15/EX15.18/Ex15_18.sce
@@ -0,0 +1,22 @@
+clear;

+clc;

+disp('Example 15.18');

+

+// aim : To determine 

+// thermal eficiency

+// carnot efficiency

+

+// given values

+rv = 5;// volume ratio

+Gama = 1.4;// heat capacity ratio

+

+// solution

+// under given condition

+

+TE = 1-(1/Gama*(2-1/rv^(Gama-1)))/(1+2*((Gama-1)/Gama)*log(rv/2));// thermal efficiency

+mprintf('\n The thermal efficiency is  =  %f percent\n',TE*100);

+

+CE = 1-1/(2*rv^(Gama-1));// carnot efficiency

+mprintf('\n The carnot efficiency is  =  %f \n',CE*100);

+

+//  End

diff --git a/2705/CH15/EX15.2/Ex15_2.sce b/2705/CH15/EX15.2/Ex15_2.sce
new file mode 100755
index 000000000..2d6440345
--- /dev/null
+++ b/2705/CH15/EX15.2/Ex15_2.sce
@@ -0,0 +1,29 @@
+clear;

+clc;

+disp('Example 15.2');

+

+// aim : To determine

+// (a) the volume ratios of the isothermal and adiabatic processes

+// (b) the thermal efficiency of the cycle

+

+// given values

+T1 = 273+260;// temperature, [K]

+T3 = 273+21;// temperature, [K]

+er = 15;// expansion ratio

+Gama = 1.4;// heat capacity ratio

+

+// solution

+// (a)

+T2 = T1;

+T4 = T3;

+// for adiabatic process

+rva = (T1/T4)^(1/(Gama-1));// volume ratio of adiabatic

+rvi = er/rva;// volume ratio of isothermal

+mprintf('\n (a) The volume ratio of the adiabatic process is  =  %f\n',rva);

+mprintf('\n      The volume ratio of the isothermal process is  =  %f\n',rvi);

+

+// (b)

+n_the = (T1-T4)/T1*100;// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',n_the);

+

+//  End

diff --git a/2705/CH15/EX15.3/Ex15_3.sce b/2705/CH15/EX15.3/Ex15_3.sce
new file mode 100755
index 000000000..8cab979fa
--- /dev/null
+++ b/2705/CH15/EX15.3/Ex15_3.sce
@@ -0,0 +1,69 @@
+clear;

+clc;

+disp('Example 15.3');

+

+// aim : To determine

+// (a) the pressure, volume and temperature at each corner of the cycle

+// (b) the thermal efficiency of the cycle

+// (c) the work done per cycle

+// (d) the work ratio

+

+// given values

+m = 1;// mass of air, [kg]

+P1 = 1730;// initial pressure of carnot engine, [kN/m^2]

+T1 = 273+300;// initial temperature, [K]

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

+Gama = 1.4;// heat capacity ratio

+

+// solution

+// taking reference  Fig. 15.15

+// (a)

+// for the isothermal process 1-2

+// using ideal gas law

+V1 = m*R*T1/P1;// initial volume, [m^3]

+T2 = T1;

+V2 = 3*V1;// given condition

+// for isothermal process, P1*V1=P2*V2, so

+P2 = P1*(V1/V2);// [MN/m^2]

+// for the adiabatic process 2-3

+V3 = 6*V1;// given condition

+T3 = T2*(V2/V3)^(Gama-1);

+// also for adiabatic process, P2*V2^Gama=P3*V3^Gama, so

+P3 = P2*(V2/V3)^Gama;

+// for the isothermal process 3-4

+T4 = T3;

+// for both adiabatic processes, the temperataure ratio is same, 

+// T1/T4 = T2/T3=(V4/V1)^(Gama-1)=(V3/V2)^(Gama-1), so

+V4 = 2*V1;

+// for isothermal process, 3-4, P3*V3=P4*V4, so

+P4 = P3*(V3/V4);

+disp('(a) At line 1');

+mprintf('\n V1  =  %f  m^3, t1  =  %f  C,  P1  =  %f  kN/m^2\n',V1,T1-273,P1);

+

+disp('At line 2');

+mprintf('\n V2  =  %f  m^3, t2  =  %f  C,  P2  =  %f  kN/m^2\n',V2,T2-273,P2);

+

+disp('At line 3');

+mprintf('\n V3  =  %f  m^3, t3  =  %f  C,  P3  =  %f  kN/m^2\n',V3,T3-273,P3);

+

+

+disp('At line 4');

+mprintf('\n V4  =  %f  m^3, t4  =  %f  C,  P4  =  %f  kN/m^2\n',V4,T4-273,P4);

+

+

+// (b)

+n_the = (T1-T3)/T1;// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',n_the*100);

+

+// (c)

+W = m*R*T1*log(V2/V1)*n_the;//  work done, [J]

+mprintf('\n (c) The work done per cycle is  =  %f  kJ\n',W);

+

+//  (d)

+wr = (T1-T3)*log(V2/V1)/(T1*log(V2/V1)+(T1-T3)/(Gama-1));// work ratio

+mprintf('\n (d) The work ratio is  =  %f\n',wr);

+

+// there is calculation mistake in the book so answer is not matching

+

+//   End

+

diff --git a/2705/CH15/EX15.4/Ex15_4.sce b/2705/CH15/EX15.4/Ex15_4.sce
new file mode 100755
index 000000000..e13c274f1
--- /dev/null
+++ b/2705/CH15/EX15.4/Ex15_4.sce
@@ -0,0 +1,72 @@
+clear;

+clc;

+disp('Example 15.4');

+

+// aim : To determine

+// (a) the pressure, volume and temperature at cycle state points

+// (b) the heat received

+// (c) the work done 

+// (d) the thermal efficiency

+// (e) the carnot efficiency

+// (f) the work ration

+// (g) the mean effective pressure

+

+// given values

+ro = 8;// overall volume ratio;

+rv = 6;// volume ratio of adiabatic compression

+P1 = 100;// initial pressure , [kN/m^2]

+V1 = .084;// initial volume, [m^3]

+T1 = 273+28;// initial temperature, [K]

+Gama = 1.4;// heat capacity ratio

+cp = 1.006;// specific heat capacity, [kJ/kg K]

+

+// solution

+// taking reference  Fig. 15.18

+// (a)

+V2 = V1/rv;// volume at stage2, [m^3] 

+V4 = ro*V2;// volume at stage 4;[m^3]

+// using PV^(Gama)=constant for process 1-2

+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2]

+T2 = T1*(V1/V2)^(Gama-1);// [K]

+

+P3 = P2;// pressure at stage 3, [kN/m^2]

+V3 = V4/rv;// volume at stage 3, [m^3]

+// since pressure is constant in process 2-3 , so using V/T=constant, so

+T3 = T2*(V3/V2);// temperature at stage 3, [K]

+

+// for process 1-4

+T4 = T1*(V4/V1);// temperature at stage4, [K

+P4 = P1;// pressure at stage4, [kN/m^2]

+

+mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =    %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4  =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);

+

+// (b)

+R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K]

+m = P1*V1/(R*T1);// mass of gas, [kg]

+Q = m*cp*(T3-T2);// heat received, [kJ]

+mprintf('\n (b) The heat received is  =  %f kJ\n',Q);

+

+// (c) 

+W = P2*(V3-V2)-P1*(V4-V1)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// work done, [kJ]

+mprintf('\n (c) The work done is  =  %f  kJ\n',W);

+

+//  (d)

+TE = 1-T1/T2;// thermal efficiency

+mprintf('\n (d) The thermal efficiency is  =  %f  percent\n',TE*100);

+

+// (e)

+CE = (T3-T1)/T3;// carnot efficiency

+mprintf('\n (e) The carnot efficiency is  =  %f  percent\n',CE*100);

+

+// (f)

+PW = P2*(V3-V2)+(P3*V3-P4*V4)/(Gama-1);// positive work done, [kj]

+WR = W/PW;//  work ratio

+mprintf('\n (f) The work ratio is  =  %f\n',WR);

+

+// (g)

+Pm = W/(V4-V2);// mean effective pressure, [kN/m^2]

+mprintf('\n (g) The mean effective pressure is  =  %f  kN/m^2\n',Pm);

+

+//  there is minor variation in answer reported in the book

+

+//  End

diff --git a/2705/CH15/EX15.5/Ex15_5.sce b/2705/CH15/EX15.5/Ex15_5.sce
new file mode 100755
index 000000000..7de39d424
--- /dev/null
+++ b/2705/CH15/EX15.5/Ex15_5.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+disp('Example 15.5');

+

+// aim : To determine

+// (a) the actual thermal efficiency of the turbine

+// (b) the specific fuel consumption of the turbine in kg/kWh

+

+// given values

+P2_by_P1 = 8;

+n_tur = .6;// ideal turbine thermal efficiency

+c = 43*10^3;// calorific value of fuel, [kJ/kg]

+Gama = 1.4;// heat capacity ratio

+

+// solution

+// (a)

+rv = P2_by_P1;

+n_tur_ide = 1-1/(P2_by_P1)^((Gama-1)/Gama);// ideal thermal efficiency

+ate = n_tur_ide*n_tur;// actual thermal efficiency

+mprintf('\n (a) The actual thermal efficiency of the turbine is  =  %f percent\n',ate*100);

+

+// (b)

+ewf = c*ate;// energy to work fuel, [kJ/kg]

+kWh = 3600;// energy equivalent ,[kJ]

+sfc = kWh/ewf;// specific fuel consumption, [kg/kWh]

+mprintf('\n (b) The specific fuel consumption of the turbine is  =  %f  kg/kWh',sfc);

+

+//  End

diff --git a/2705/CH15/EX15.6/Ex15_6.sce b/2705/CH15/EX15.6/Ex15_6.sce
new file mode 100755
index 000000000..d0bad01c2
--- /dev/null
+++ b/2705/CH15/EX15.6/Ex15_6.sce
@@ -0,0 +1,23 @@
+clear;

+clc;

+disp('Example 15.6');

+

+// aim : To determine

+// the relative efficiency of the engine

+

+// given values

+d = 80;// bore, [mm]

+l = 85;// stroke, [mm]

+V1 = .06*10^6;// clearence volume, [mm^3]

+ate = .22;// actual thermal efficiency of the engine

+Gama = 1.4;// heat capacity ratio

+

+// solution

+sv = %pi*d^2/4*l;// stroke volume, [mm^3]

+V2 = sv+V1;// [mm^3]

+rv = V2/V1;

+ite = 1-(1/rv)^(Gama-1);// ideal thermal efficiency

+re = ate/ite;// relative thermal efficiency

+mprintf('\n The relative efficiency of the engine is  =  %f percent\n',re*100);

+

+//  End

diff --git a/2705/CH15/EX15.7/Ex15_7.sce b/2705/CH15/EX15.7/Ex15_7.sce
new file mode 100755
index 000000000..0f950939f
--- /dev/null
+++ b/2705/CH15/EX15.7/Ex15_7.sce
@@ -0,0 +1,68 @@
+clear;

+clc;

+disp('Example 15.7');

+

+// aim : To determine

+// (a) the pressure, volume and temperature at each cycle process change points

+// (b) the heat transferred to air

+// (c) the heat rejected by the air

+// (d) the ideal thermal efficiency

+// (e) the work done 

+// (f) the mean effective pressure

+

+// given values

+m = 1;// mass of air, [kg]

+rv = 6;// volume ratio of adiabatic compression

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

+T1 = 273+100;// initial temperature, [K]

+P3 = 3450;// maximum pressure, [kN/m^2]

+Gama = 1.4;// heat capacity ratio

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

+

+// solution

+// taking reference  Fig. 15.20

+// (a)

+// for point 1

+V1 = m*R*T1/P1;// initial volume, [m^3]

+

+// for point 2

+V2 = V1/rv;// volume at point 2, [m^3] 

+// using PV^(Gama)=constant for process 1-2

+P2 = P1*(V1/V2)^(Gama);// pressure at point 2,. [kN/m^2]

+T2 = T1*(V1/V2)^(Gama-1);// temperature at point 2,[K]

+

+// for point 3

+V3 = V2;// volume at point 3, [m^3]

+// since volume is constant in process 2-3 , so using P/T=constant, so

+T3 = T2*(P3/P2);// temperature at stage 3, [K]

+

+// for point 4

+V4 = V1;//  volume at point 4, [m^3]

+P4 = P3*(V3/V4)^Gama;// pressure at point 4, [kN/m^2] 

+// again  since volume is constant in process 4-1 , so using P/T=constant, so

+T4 = T1*(P4/P1);// temperature at point 4, [K]

+

+mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4  =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);

+

+//  (b)

+cv = R/(Gama-1);//  specific heat capacity, [kJ/kg K]

+Q23 = m*cv*(T3-T2);// heat transferred, [kJ]

+mprintf('\n (b) The heat transferred to the air is  =  %f  kJ\n',Q23);

+

+// (c) 

+Q34 = m*cv*(T4-T1);// heat rejected by air, [kJ]

+mprintf('\n (c) The heat rejected by the air is  =  %f  kJ\n',Q34);

+

+// (d)

+TE = 1-Q34/Q23;// ideal thermal efficiency

+mprintf('\n (d) The ideal thermal efficiency is  =  %f percent\n',TE*100);

+

+// (e)

+W = Q23-Q34;// work done ,[kJ]

+mprintf('\n (e) The work done  is  =  %f  kJ\n',W);

+

+// (f)

+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]

+mprintf('\n (f)  The mean effefctive pressure is  =  %f kN/m^2\n',Pm);

+

+//  End

diff --git a/2705/CH15/EX15.8/Ex15_8.sce b/2705/CH15/EX15.8/Ex15_8.sce
new file mode 100755
index 000000000..a37208dcc
--- /dev/null
+++ b/2705/CH15/EX15.8/Ex15_8.sce
@@ -0,0 +1,65 @@
+clear;

+clc;

+disp('Example 15.8');

+

+// aim : To determine

+// (a) the pressure, volume and temperature at cycle state points

+// (b) the thermal efficiency

+// (c) the theoretical output

+// (d) the mean effective pressure

+// (e) the carnot efficiency

+

+// given values

+rv = 9;// volume ratio

+P1 = 101;// initial pressure , [kN/m^2]

+V1 = .003;// initial volume, [m^3]

+T1 = 273+18;// initial temperature, [K]

+P3 = 4500;// maximum pressure, [kN/m^2]

+N = 3000;

+cp = 1.006;// specific heat capacity at constant pressure, [kJ/kg K]

+cv = .716;// specific heat capacity at constant volume, [kJ/kg K]

+

+// solution

+// taking reference  Fig. 15.20

+// (a)

+// for process 1-2

+Gama = cp/cv;// heat capacity ratio

+R = cp-cv;// gas constant, [kJ/kg K]

+V2 = V1/rv;// volume at stage2, [m^3] 

+// using PV^(Gama)=constant for process 1-2

+P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2]

+T2 = T1*(V1/V2)^(Gama-1);// [K]

+

+// for process 2-3

+V3 = V2;// volume at stage 3, [m^3]

+// since volume is constant in process 2-3 , so using P/T=constant, so

+T3 = T2*(P3/P2);// temperature at stage 3, [K]

+

+// for process 3-4

+V4 = V1;// volume at stage 4

+// using PV^(Gama)=constant for process 3-4

+P4 = P3*(V3/V4)^(Gama);// pressure at stage2,. [kN/m^2]

+T4 = T3*(V3/V4)^(Gama-1);// temperature at stage  4,[K]

+

+mprintf('\n (a)  P1  =  %f kN/m^2,    V1  =  %f m^3,       t1  =  %f C,\n      P2  =  %f kN/m^2,    V2  =  %f m^3,      t2  =  %f C,\n      P3  =  %f kN/m^2,    V3  =  %f m^3,       t3  =  %f C,\n      P4 =  %f kN/m^2,      V4  =  %f m^3,       t4  =  %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273);

+

+// (b)

+TE = 1-(T4-T1)/(T3-T2);// thermal efficiency

+mprintf('\n (b) The thermal efficiency is  =  %f percent\n',TE*100);

+

+// (c)

+m = P1*V1/(R*T1);// mass os gas, [kg] 

+W = m*cv*((T3-T2)-(T4-T1));// work done, [kJ]

+Wt = W*N/60;// workdone per minute, [kW]

+mprintf('\n (c) The theoretical output  is  =  %f  kW\n',Wt);

+

+// (d)

+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]

+mprintf('\n (g) The mean effefctive pressure is  =  %f  kN/m^2\n',Pm);

+

+// (e)

+CE = (T3-T1)/T3;// carnot efficiency

+mprintf('\n (e) The carnot efficiency is  =  %f percent\n',CE*100);

+

+

+//  End

diff --git a/2705/CH15/EX15.9/Ex15_9.sce b/2705/CH15/EX15.9/Ex15_9.sce
new file mode 100755
index 000000000..7482cbe06
--- /dev/null
+++ b/2705/CH15/EX15.9/Ex15_9.sce
@@ -0,0 +1,72 @@
+clear;

+clc;

+disp('Example 15.9');

+

+// aim : To determine

+// (a) the pressure and temperature at cycle process change points

+// (b) the work done 

+// (c)  the thermal efficiency

+// (d)  the work ratio

+// (e)  the mean effective pressure

+// (f)  the carnot efficiency

+

+

+// given values

+rv = 16;// volume ratio of compression

+P1 = 90;// initial pressure , [kN/m^2]

+T1 = 273+40;// initial temperature, [K]

+T3 = 273+1400;// maximum temperature, [K]

+cp = 1.004;// specific heat capacity at constant pressure, [kJ/kg K]

+Gama = 1.4;// heat capacoty ratio

+

+// solution

+cv = cp/Gama;// specific heat capacity at constant volume, [kJ/kg K]

+R = cp-cv;// gas constant, [kJ/kg K]

+// for one kg of gas

+V1 = R*T1/P1;// initial volume, [m^3]

+// taking reference  Fig. 15.22

+// (a)

+// for process 1-2

+// using PV^(Gama)=constant for process 1-2

+// also rv = V1/V2

+P2 = P1*(rv)^(Gama);// pressure at stage2,. [kN/m^2]

+T2 = T1*(rv)^(Gama-1);// temperature at stage 2, [K]

+

+// for process 2-3

+P3 = P2;// pressure at stage 3, [kN/m^2]

+V2 = V1/rv;//[m^3]

+// since pressure is constant in process 2-3 , so using V/T=constant, so

+V3 = V2*(T3/T2);// volume at stage 3, [m^3]

+

+// for process 1-4

+V4 = V1;// [m^3]

+P4 = P3*(V3/V4)^(Gama)

+// since in stage 1-4 volume is constant, so P/T=constant, 

+T4 = T1*(P4/P1);// temperature at stage  4,[K]

+

+mprintf('\n (a)  P1  =  %f kN/m^2,          t1  =  %f C,\n      P2  =  %f kN/m^2,        t2  =  %f C,\n      P3  =  %f kN/m^2,         t3  =  %f C,\n      P4  =  %f kN/m^2,          t4  =  %f C\n',P1,T1-273,P2,T2-273,P3,T3-273,P4,T4-273);

+

+// (b)

+W = cp*(T3-T2)-cv*(T4-T1);// work done, [kJ]

+mprintf('\n (b) The work done is  =  %f kJ\n',W);

+

+// (c) 

+TE = 1-(T4-T1)/((T3-T2)*Gama);// thermal efficiency

+mprintf('\n (c) The thermal efficiency is  =  %f percent\n',TE*100);

+

+// (d)

+PW = cp*(T3-T2)+R*(T3-T4)/(Gama-1);// positive work done

+WR = W/PW;//  work ratio

+mprintf('\n (d) The work ratio is  =  %f\n',WR);

+

+// (e)

+Pm = W/(V1-V2);// mean effective pressure, [kN/m^2]

+mprintf('\n (e) The mean effefctive pressure is  =  %f kN/m^2\n',Pm);

+

+// (f)

+CE = (T3-T1)/T3;// carnot efficiency

+mprintf('\n (f) The carnot efficiency is  =  %f percent\n',CE*100);

+

+// value of t2 printed in the book is incorrect

+

+// End

diff --git a/2705/CH16/EX16.1/Ex16_1.sce b/2705/CH16/EX16.1/Ex16_1.sce
new file mode 100755
index 000000000..b3251fa8a
--- /dev/null
+++ b/2705/CH16/EX16.1/Ex16_1.sce
@@ -0,0 +1,45 @@
+clear;

+clc;

+disp('Example 16.1');

+

+// aim : To determine 

+// (a) the net power output of the turbine plant if the turbine is coupled to the compresser

+// (b) the thermal efficiency of the plant

+// (c) the work ratio

+

+// Given values

+P1 = 100;// inlet pressure of compressor, [kN/m^2]

+T1 = 273+18;// inlet temperature, [K]

+P2 = 8*P1;// outlet pressure of compressor, [kN/m^2]

+n_com = .85;// isentropic efficiency of compressor

+T3 = 273+1000;//inlet temperature of turbine, [K]

+P3 = P2;// inlet pressure of turbine, [kN/m^2]

+P4 = 100;// outlet pressure of turbine, [kN/m^2]

+n_tur = .88;// isentropic efficiency of turbine

+m_dot = 4.5;// air mass flow rate, [kg/s]

+cp = 1.006;// [kJ/kg K]

+Gamma = 1.4;// heat capacity ratio

+

+// (a)

+// For the compressor

+T2_prime = T1*(P2/P1)^((Gamma-1)/Gamma);// [K]

+T2 = T1+(T2_prime-T1)/n_com;//  exit pressure of compressor, [K]

+

+// for turbine

+T4_prime = T3*(P4/P3)^((Gamma-1)/Gamma);// [K]

+T4 = T3-(T3-T4_prime)*n_tur;// exit temperature of turbine, [K]

+

+P_output = m_dot*cp*((T3-T4)-(T2-T1));// [kW]

+mprintf('\n (a) The net power output is  =  %f  kW\n',P_output);

+

+// (b)

+n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;// thermal efficiency

+mprintf('\n (b) The thermal efficiency of the plant is  =  %f percent\n',n_the);

+

+// (c)

+P_pos = m_dot*cp*(T3-T4);// Positive cycle work, [kW]

+

+W_ratio = P_output/P_pos;// work ratio

+mprintf('\n (c) The work ratio is  =  %f\n',W_ratio)

+

+//  End

diff --git a/2705/CH16/EX16.2/Ex16_2.sce b/2705/CH16/EX16.2/Ex16_2.sce
new file mode 100755
index 000000000..e76ab804c
--- /dev/null
+++ b/2705/CH16/EX16.2/Ex16_2.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+disp('Example 16.2');

+

+// aim : To determine

+// (a) the pressure ratiowhich will give the maximum net work output

+// (b) the maximum net specific work output

+// (c) the thermal efficiency at maximum work output

+// (d) the work ratio at maximum work output

+// (e) the carnot efficiency within the cycle temperature limits

+

+// Given values

+// taking the refrence as Fig.16.35

+T3 = 273+1080;// [K]

+T1 = 273+10;// [K]

+cp = 1.007;// [kJ/kg K]

+Gamma = 1.41;//  heat capacity ratio

+

+// (a)

+r_pmax = (T3/T1)^((Gamma)/(Gamma-1));// maximum pressure ratio

+// for maximum net work output

+r_p = sqrt(r_pmax);

+mprintf('\n (a) The pressure ratio which give the maximum network output is  =  %f\n',r_p);

+

+// (b)

+T2 = T1*(r_p)^((Gamma-1)/Gamma);// [K]

+// From equation [23]

+T4 = T2;

+W_max = cp*((T3-T4)-(T2-T1));// Maximum net specific work output, [kJ/kg]

+

+mprintf('\n (b) The maximum net specific work output is  =  %f kJ/kg\n',W_max);

+

+// (c)

+W = cp*(T3-T2);

+n_the = W_max/W;// thermal efficiency

+mprintf('\n (c) The thermal efficiency at maximum work output is  = %f percent\n ',n_the*100);

+

+// (d)

+// From the equation [26]

+W_ratio = n_the;// Work ratio

+mprintf('\n (d) The work ratio at maximum work output is  =  %f\n',W_ratio);

+

+// (e)

+n_carnot = (T3-T1)/T3*100;// carnot efficiency

+mprintf('\n (e) The carnot efficiency within the cycle temperature limits is  =  %f  percent\n',n_carnot);

+

+// End

diff --git a/2705/CH16/EX16.3/Ex16_3.sce b/2705/CH16/EX16.3/Ex16_3.sce
new file mode 100755
index 000000000..de93273ac
--- /dev/null
+++ b/2705/CH16/EX16.3/Ex16_3.sce
@@ -0,0 +1,62 @@
+clc;

+disp('Example 16.3');

+

+// aim : To determine

+// (a) the net power output of the plant

+// (b) the exhaust temperature from the heat exchanger

+// (c) the thermal efficiency of the plant

+// (d) the thermal efficiency of the plant if there were no heat exchanger

+// (e) the work ratio

+

+// Given values

+T1 = 273+15;// temperature, [K]

+P1 = 101;// pressure, [kN/m^2]

+P2 = 6*P1; // [kN/m^2]

+eff = .65;// effectiveness of the heat exchanger, 

+T3 = 273+870;// temperature, [K]

+P4 = 101;// [kN/m^2]

+n_com = .85;// efficiency of compressor, 

+n_tur = .80;// efficiency of turbine

+m_dot = 4;// mass flow rate, [kg/s]

+Gama = 1.4;// heat capacity ratio

+cp = 1.005;// [kJ/kg K]

+

+// solution

+// (a)

+// For compressor

+T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K]

+

+// using n_com = (T2_prim-T1)/(T2-T1)')

+

+T2 = T1+(T2_prim-T1)/n_com

+// For turbine

+P3 = P2;

+T4_prim = T3*(P4/P3)^((Gama-1)/Gama);// [K]

+

+T4=T3-n_tur*(T3-T4_prim); //  [K]

+P_out = m_dot*cp*((T3-T4)-(T2-T1));//  net power output, [kW]

+mprintf('\n (a) The net power output of the plant is  =  %f kW\n',P_out);

+

+// (b)

+mtd = T4-T2;// maximum temperature drop for heat transfer, [K]

+atd = eff*mtd;// actual temperature, [K]

+et = T4-atd;// Exhaust temperature from heat exchanger, [K]

+t6 = et-273;// [C]

+mprintf('\n (b) The exhaust temperature from the heat exchanger is  =  %f C\n',t6);

+

+// (c)

+T5 = T2+atd;// [K]

+n_the = ((T3-T4)-(T2-T1))/(T3-T5)*100;// thermal effficiency 

+mprintf('\n (c) The thermal efficiency of the plant is  =  %f percent\n',n_the);

+

+// (d)

+// with no heat exchanger

+n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;// thermal efficiency without heat exchanger

+mprintf('\n (d) The thermal efficiency of the plant if there wereno heat exchanger is  =  %f  percent\n',n_the);

+

+// (e)

+P_pos = m_dot*cp*(T3-T4);// positive cycle work;// [kW]

+w_rat = P_out/P_pos;// work ratio

+mprintf('\n (e) The work ratio is  =  %f\n',w_rat)

+

+//  End

diff --git a/2705/CH16/EX16.4/Ex16_4.sce b/2705/CH16/EX16.4/Ex16_4.sce
new file mode 100755
index 000000000..01cd7211b
--- /dev/null
+++ b/2705/CH16/EX16.4/Ex16_4.sce
@@ -0,0 +1,68 @@
+clear;

+clc;

+disp('Example 16.4');

+

+// aim : To determine

+// (a) the pressure and temperature as the air leaves the compressor turbine

+// (b) the power output from the free power turbine

+// (c) the thermal efficiency of the plant

+// (d) the work ratio

+// (e) the carnot efficiency within the cycle temperature limits

+

+// Given values

+T1 = 273+19;// temperature, [K]

+P1 = 100;// pressure, [kN/m^2]

+P2 = 8*P1; // [kN/m^2]

+P3 = P2;// [kN/m^2]

+T3 = 273+980;// temperature, [K]

+n_com = .85;// efficiency of rotary compressor

+P5 = 100;// [kN/m^2]

+n_cum = .88;// isentropic efficiency of combustion chamber compressor, 

+n_tur = .86;// isentropic efficiency of turbine

+m_dot = 7;// mass flow rate of air, [kg/s]

+Gama = 1.4;// heat capacity ratio

+cp = 1.006;// [kJ/kg K]

+

+// solution

+// (a)

+// For compressor

+T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K]

+

+T2 = T1+(T2_prim-T1)/n_com;// temperature, [K]

+

+// for compressor turbine

+// T3-T4 = T2-T1,because compressor turbine power=compressor power so

+T4 = T3-(T2-T1);//turbine exit temperature, [K]

+T4_prim = T3-(T3-T4)/n_cum;// [K]

+

+// For turbine

+// T4_prim = T3*(P4/P3)^((Gama-1)/Gama)

+P4 = P3*(T4_prim/T3)^(Gama/(Gama-1));// exit air pressure of air, [kN/m^2]

+

+mprintf('\n (a) The temperature as the air leaves the compressor turbine is  =  %f  C\n',T4-273);

+mprintf('\n      The pressure as the air leaves the compressor turbine is  =  %f  kN/m^2\n',P4);

+

+// (b)

+T5_prim = T4*(P5/P4)^((Gama-1)/Gama);// [K]

+

+

+T5 = T4-n_tur*(T4-T5_prim);// temperature, [K]

+

+PO = m_dot*cp*(T4-T5);// power output

+mprintf('\n (b) The power output from the free power turbine is  =  %f kW\n',PO);

+

+// (c)

+

+n_the = (T4-T5)/(T3-T2)*100;// thermal effficiency 

+mprintf('\n (c) The thermal efficiency of the plant is  =  %f percent\n',n_the);

+

+// (d)

+

+WR = (T4-T5)/(T3-T5);// work ratio

+mprintf('\n (d) The work ratio is  =  %f\n',WR);

+

+// (e)

+CE = (T3-T1)/T3;// carnot efficiency

+mprintf('\n (e) The carnot efficiency is  =  %f percent\n',CE*100);

+

+//  End

diff --git a/2705/CH16/EX16.5/Ex16_5.sce b/2705/CH16/EX16.5/Ex16_5.sce
new file mode 100755
index 000000000..3b2cd40b4
--- /dev/null
+++ b/2705/CH16/EX16.5/Ex16_5.sce
@@ -0,0 +1,51 @@
+clear;

+clc;

+disp('Example 16.5');

+

+// aim : To determine

+// (a) the pressure and temperature of the air compression 

+// (b) the power developed by the gas turbine

+// (c) the temperature and pressure of the airentering the exhaust jet as it leaves the gas turbine 

+

+// Given values

+T1 = 273-22.4;// temperature, [K]

+P1 = 470;// pressure, [bar]

+P2 = 30*P1; // [kN/m^2]

+P3 = P2;// [kN/m^2]

+T3 = 273+960;// temperature, [K]

+r = 1.25;// ratio of turbine power to compressor power

+n_tur = .86;// isentropic efficiency of turbine

+m_dot = 80;// mass flow rate of air, [kg/s]

+Gama = 1.41;// heat capacity ratio

+cp = 1.05;// [kJ/kg K]

+

+// solution

+// (a)

+// For compressor

+T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K]

+// using n_tur=(T2_prim-T1)/(T2-T1)

+T2 = T1+(T2_prim-T1)/n_tur;// temperature, [K]

+

+mprintf('\n (a) The pressure of the air after compression is  =  %f  bar\n',P2);

+

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

+

+// (b)

+Td  = r*(T2-T1);// temperature drop in turbine, [K]

+PO = m_dot*cp*Td;// power output, [kW]

+mprintf('\n (b) The power developed by the gas turbine is  =  %f  MW\n',PO*10^-3);

+

+// (c)

+t3 = T3-273;// [C]

+t4 = t3-Td;// temeprerature of air leaving turbine,[K]

+Tdi = Td/n_tur;// isentropic temperature drop, [K]

+T4_prim = t3-Tdi+273;// temperature, [K]

+// using T4_prim=T3*(P4/P3)^((Gama-1)/Gama)

+P4 = P3*(T4_prim/T3)^(Gama/(Gama-1));// exit air pressure of air, [kN/m^2]

+

+mprintf('\n (c) The air pressure as it leaves the gas turbine is  =  %f  bar\n',P4);

+

+// Result in the book is not matching because they have taken pressure in mbar  but in in question it is given in bar 

+

+//   End

+

diff --git a/2705/CH16/EX16.6/Ex16_6.sce b/2705/CH16/EX16.6/Ex16_6.sce
new file mode 100755
index 000000000..5a9318ec8
--- /dev/null
+++ b/2705/CH16/EX16.6/Ex16_6.sce
@@ -0,0 +1,65 @@
+clc;

+disp('Example 16.6');

+

+// aim : To determine

+// (a) the mass of fuel oil used by the gas turbine

+// (b) the mass flow of steam from the boiler 

+// (c) the theoretical output from the steam turbine

+// (d) the overall theoretical thermal efficiency of the plant

+

+// given values

+Po = 150;// generating plant output, [MW]

+n_the1 = .35;// thermal efficiency

+CV = 43;// calorific value of fuel, [MJ]

+me = 400;// flow rate of exhaust gas, [kg/s]

+T = 90;// boiler exit temperature, [C]

+T1 = 550;// exhaust gas temperature, [C]

+P2 = 10;// steam generation pressure, [MN/m^2]

+T2 = 450;// boiler exit temperature, [C]

+Tf = 140;// feed water temperature, [C]

+n_tur = .86;// turbine efficiency

+P3 = .5;// exhaust temperature, [MN/m^2]

+n_boi = .92;// boiler thermal efficiency

+cp = 1.1;// heat capacity, [kJ/kg]

+

+

+//  solution

+// (a)

+ER = Po*3600/n_the1;// energy requirement from the fuel, [MJ/h]

+mf = ER/CV*10^-3;// fuel required, [tonne/h]

+mprintf('\n (a) The mass of fuel oil used by the gas is  =  %f  tonne/h\n',mf);

+

+// (b) 

+

+ET = me*cp*(T1-T)*3600*n_boi;// energy transferred to steam,[kJ/h]

+// from steam table

+h1 = 3244;// specific enthalpy, [kJ/kg]

+hf = 588.5;// specific enthalpy, [kJ/kg]

+ERR = h1-hf;// energy required to raise steam, [kJ/kg]

+ms = ET/ERR*10^-3;// mass flow of steam, [tonne/h]

+mprintf('\n (b) The mass flow  rate of steam from the boiler is  =  %f  tonne/h\n',ms);

+

+// again from steam table

+s1 = 6.424;// specific entropy, [kJ/kg K]

+sf2 = 1.86;// specific entropy, [kJ/kg K

+sg2 = 6.819;// specific entropy, [kJ/kg K]

+

+hf2 = 640.1;// specific enthalpy,[kJ/kg]

+hg2 = 2747.5;// specific enthalpy, [kJ/kg]

+// for ths process s1=s2=sf2+x2*(sg2-sf2)

+s2 = s1;

+// hence

+x2 = (s2-sf2)/(sg2-sf2);// dryness fraction

+

+h2_prim = hf2+x2*(hg2-hf2);// specific enthalpy of steam, [kJ/kg]

+

+TO = n_tur*(h1-h2_prim);//theoretical steam turbine output, [kJ/kg]

+TOt = TO*ms/3600;// total theoretical steam turbine output, [MW]

+

+mprintf('\n (c) The theoretical output from the steam turbine is  =  %f  MW\n',TOt);

+

+// (d)

+n_tho = (Po+TOt)*n_the1/Po;// overall theoretical thermal efficiency

+mprintf('\n (d) The overall thermal efficiency is  =  %f  percent\n',n_tho*100);

+

+//  End

diff --git a/2705/CH17/EX17.1/Ex17_1.sce b/2705/CH17/EX17.1/Ex17_1.sce
new file mode 100755
index 000000000..1222d2ae4
--- /dev/null
+++ b/2705/CH17/EX17.1/Ex17_1.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+disp('Example 17.1');

+

+// aim : To determine

+// the indicated and brake output and the mechanicl efficiency

+// draw up an overall energy balance and as % age

+

+// given values

+h = 21;// height of indicator diagram, [mm]

+ic = 27;// indicator calibration, [kN/m^2 per mm]

+sv = 14*10^-3;// swept volume of the cylinder;,[m^3]

+N = 6.6;// speed of engine, [rev/s]

+ebl = 77;// effective brake load, [kg]

+ebr = .7;// effective brake radious, [m]

+fc = .002;// fuel consumption, [kg/s]

+CV = 44000;// calorific value of fuel, [kJ/kg]

+cwc = .15;// cooling water circulation, [kg/s]

+Ti = 38;// cooling water inlet temperature, [C]

+To = 71;// cooling water outlet temperature, [C]

+c = 4.18;// specific heat capacity of water, [kJ/kg]

+eeg = 33.6;// energy to exhaust gases, [kJ/s]

+g = 9.81;// gravitational acceleration, [m/s^2]

+

+// solution

+PM = ic*h;// mean effective pressure, [kN/m^2]

+LA = sv;// swept volume of the cylinder, [m^3]

+ip = PM*LA*N/2;// indicated power,[kW]

+T = ebl*g*ebr;// torque, [N*m]

+bp = 2*%pi*N*T;// brake power, [W]

+n_mech = bp/ip*10^-3;// mechanical efficiency

+mprintf('\n The Indicated power is  =  %f kW\n',ip);

+mprintf('\n The Brake power is  =  %f  kW\n',bp*10^-3);

+mprintf('\n The mechanical efficiency is  =  %f  percent\n',n_mech);

+

+ef = CV*fc;// energy from fuel, [kJ/s]

+eb = bp*10^-3;// energy to brake power,[kJ/s]

+ec = cwc*c*(To-Ti);// energy to coolant,[kJ/s]

+es = ef-(eb+ec+eeg);// energy to surrounding,[kJ/s]

+

+disp('Energy can be tabulated as :-');

+disp('----------------------------------------------------------------------------------------------------');

+disp('                                                  kJ/s                           Percentage   ')

+disp('----------------------------------------------------------------------------------------------------');

+mprintf('\n Energy from fuel                         %f                   %f\n Energy to brake power                %f                    %f\n Energy to coolant                       %f                    %f\n Energy to exhaust                      %f                    %f\n Energy to suroundings,etc.         %f                    %f\n',ef,ef/ef*100,eb,eb/ef*100,ec,ec/ef*100,eeg,eeg/ef*100,es,es/ef*100);

+

+//  End

diff --git a/2705/CH17/EX17.2/Ex17_2.sce b/2705/CH17/EX17.2/Ex17_2.sce
new file mode 100755
index 000000000..16f568583
--- /dev/null
+++ b/2705/CH17/EX17.2/Ex17_2.sce
@@ -0,0 +1,87 @@
+clear;

+clc;

+disp('Example 17.2');

+

+// aim : To determine

+// (a) bp

+// (b) ip

+// (c) mechanical efficiency

+// (d) indicated thermal efficiency

+// (e) brake specific steam consumption

+// (f) draw up complete energy account for the test one-minute basis taking 0 C as datum

+

+// given values

+d = 200*10^-3;// cylinder diameter, [mm]

+L = 250*10^-3;// stroke, [mm]

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

+r = .75/2;// effective radious of brake wheel, [m]

+Ps = 800;// stop valve pressure, [kN/m^2]

+x = .97;// dryness fraction of steam

+BL = 136;// brake load, [kg]

+SL = 90;// spring balance load, [N]

+PM = 232;// mean effective pressure, [kN/m^2]

+Pc = 10;// condenser pressure, [kN/m^2]

+m_dot = 3.36;// steam consumption, [kg/min]

+CC = 113;// condenser cooling water, [kg/min]

+Tr = 11;// temperature rise of condenser cooling water, [K]

+Tc = 38;// condensate temperature, [C]

+C = 4.18;// heat capacity of water, [kJ/kg K]

+g = 9.81;// gravitational acceleration, [m/s^2]

+

+// solution

+// from steam table

+// at 800 kN/m^2

+tf1 = 170.4;// saturation temperature, [C]

+hf1 = 720.9;// [kJ/kg]

+hfg1 = 2046.5;// [kJ/kg]

+hg1 = 2767.5;// [kJ/kg]

+vg1 = .2403;// [m^3/kg]

+

+// at 10 kN/m^2

+tf2 = 45.8;// saturation temperature, [C]

+hf2 = 191.8;// [kJ/kg]

+hfg2 = 2392.9;// [kJ/kg]

+hg2 = 2584.8;// [kJ/kg]

+vg2 = 14.67;// [m^3/kg]

+

+// (a)

+T = (BL*g-SL)*r;// torque, [Nm]

+bp = 2*%pi*N*T*10^-3;// brake power,[W]

+mprintf('\n (a) The brake power is  =   %f  kW\n',bp);

+

+// (b)

+A = %pi*d^2/4;// area, [m^2]

+ip = PM*L*A*N*2;// double-acting so*2, [kW]

+mprintf('\n (b) The indicated power is  =  %f kW\n',ip);

+

+// (c)

+n_mec = bp/ip;// mechanical efficiency

+mprintf('\n (c) The mechanical efficiency is  =  %f  percent\n',n_mec*100);

+

+// (d)

+h = hf1+x*hfg1;// [kJ/kg]

+hf = hf2;

+ITE = ip/((m_dot/60)*(h-hf));// indicated thermal efficiency

+mprintf('\n (d) The indicated thermal efficiency is  =  %f percent\n',ITE*100);

+// (e)

+Bsc=m_dot*60/bp;// brake specific steam consumption, [kg/kWh]

+mprintf('\n (e) The brake steam consumption is  =  %f  kg/kWh\n',Bsc);

+

+// (f)

+// energy balanvce reckoned from 0 C

+Es = m_dot*h;// energy supplied, [kJ]

+Eb = bp*60;// energy to bp, [kJ]

+Ecc = CC*C*Tr;// energy to condensate cooling water, [kJ]

+Ec = m_dot*C*Tc;// energy to condensate, [kJ]

+Ese = Es-Eb-Ecc-Ec;// energy to surrounding,etc, [kJ]

+

+mprintf('\n (f) Energy supplied/min is  =  %f  kJ\n',Es);

+

+mprintf('\n    Energy to bp/min is  =  %f  kJ\n',Eb);

+mprintf('\n    Energy to condenser cooling water/min is  =  %f kJ\n',Ecc);

+mprintf('\n    Energy to condensate/min is  =  %f kJ\n',Ec);

+mprintf('\n    Energy to surrounding, etc/min is  =  %f  kJ\n',Ese);

+

+// answer in the book is misprinted

+

+//  End

diff --git a/2705/CH17/EX17.3/Ex17_3.sce b/2705/CH17/EX17.3/Ex17_3.sce
new file mode 100755
index 000000000..baddc9cc9
--- /dev/null
+++ b/2705/CH17/EX17.3/Ex17_3.sce
@@ -0,0 +1,55 @@
+clear;

+clc;

+disp('Example 17.3');

+

+// aim : To determine

+// (a) the brake power

+// (b) the brake specific fuel consumption

+// (c) the indicated thermal efficiency

+// (d) the energy balance, expressing the various items

+

+// given values

+t = 30;// duration of trial, [min]

+N = 1750;// speed of engine, [rev/min]

+T = 330;// brake torque, [Nm]

+mf = 9.35;// fuel consumption, [kg]

+CV = 42300;// calorific value of fuel, [kJ/kg]

+cwc = 483;// jacket cooling water circulation, [kg]

+Ti = 17;// inlet temperature, [C]

+To = 77;// outlet  temperature, [C]

+ma = 182;// air consumption, [kg]

+Te = 486;// exhaust temperature, [C]

+Ta = 17;// atmospheric temperature, [C]

+n_mec = .83;// mechanical efficiency

+c = 1.25;// mean specific heat capacity of exhaust gas, [kJ/kg K]

+C = 4.18;// specific heat capacity, [kJ/kg K]

+

+// solution

+// (a)

+bp = 2*%pi*N*T/60*10^-3;// brake power, [kW]

+mprintf('\n (a) The Brake power is  =  %f  kW\n',bp);

+

+// (b)

+bsf = mf*2/bp;//brake specific fuel consumption, [kg/kWh]

+mprintf('\n (b) The brake specific fuel consumption is  =  %f kg/kWh\n',bsf);

+

+// (c)

+ip = bp/n_mec;// indicated power, [kW]

+ITE = ip/(2*mf*CV/3600);// indicated thermal efficiency

+mprintf('\n (c) The indicated thermal efficiency is  =  %f percent\n',ITE*100);

+

+// (d)

+// taking  basis one minute 

+ef = CV*mf/30;// energy from fuel, [kJ]

+eb = bp*60;// energy to brake power,[kJ]

+ec = cwc/30*C*(To-Ti);// energy to cooling water,[kJ]

+ee = (ma+mf)/30*c*(Te-Ta);// energy to exhaust, [kJ]

+es = ef-(eb+ec+ee);// energy to surrounding,etc,[kJ]

+

+mprintf('\n (d) Energy from fuel is  =  %f kJ\n',ef);

+mprintf('\n      Energy to brake power is  =  %f kJ\n',eb);

+mprintf('\n      Energy to cooling water is  =  %f  kJ\n',ec);

+mprintf('\n      Energy to exhaust is  =  %f  kJ\n',ee);

+mprintf('\n      Energy to surrounding, etc is  =  %f kJ\n',es);

+ 

+//  End

diff --git a/2705/CH17/EX17.4/Ex17_4.sce b/2705/CH17/EX17.4/Ex17_4.sce
new file mode 100755
index 000000000..e50619a8c
--- /dev/null
+++ b/2705/CH17/EX17.4/Ex17_4.sce
@@ -0,0 +1,33 @@
+clear;

+clc;

+disp('Example 17.4');

+

+// aim : To determine

+// (a) the indicated power of the engine

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

+

+// given values

+bp = 52;// brake power output, [kW]

+bp1 = 40.5;// brake power of cylinder cut1, [kW]

+bp2 = 40.2;// brake power of cylinder cut2, [kW]

+bp3 = 40.1;// brake power of cylinder cut3, [kW]

+bp4 = 40.6;// brake power of cylinder cut4, [kW]

+bp5 = 40.7;// brake power of cylinder cut5, [kW]

+bp6 = 40.0;// brake power of cylinder cut6, [kW]

+

+// sollution

+ip1 = bp-bp1;// indicated power of cylinder cut1, [kW]

+ip2 = bp-bp2;// indicated power of cylinder cut2, [kW]

+ip3 = bp-bp3;// indicated power of cylinder cut3, [kW]

+ip4 = bp-bp4;// indicated power of cylinder cut4, [kW]

+ip5 = bp-bp5;// indicated power of cylinder cut5, [kW]

+ip6 = bp-bp6;// indicated power of cylinder cut6, [kW]

+

+ip = ip1+ip2+ip3+ip4+ip5+ip6;// indicated power of engine,[kW]

+mprintf('\n (a) The indicated power of the engine is  =  %f  kW\n',ip);

+

+// (b)

+n_mec = bp/ip;// mechanical efficiency

+mprintf('\n (b) The mechanical  efficiency of the engine is  =  %f percent\n',n_mec*100);

+

+// End

diff --git a/2705/CH17/EX17.5/Ex17_5.sce b/2705/CH17/EX17.5/Ex17_5.sce
new file mode 100755
index 000000000..23733b131
--- /dev/null
+++ b/2705/CH17/EX17.5/Ex17_5.sce
@@ -0,0 +1,61 @@
+clear;

+clc;

+disp('Example 17.5');

+

+// aim : To determine

+// the brake power,indicated power and mechanicl efficiency

+// draw up an energy balance and as % age of the energy supplied

+

+// given values

+N = 50;// speed, [rev/s]

+BL = 267;// break load.,[N]

+BL1 = 178;// break load of cylinder cut1, [N]

+BL2 = 187;// break load of cylinder cut2, [N]

+BL3 = 182;// break load of cylinder cut3, [N]

+BL4 = 182;// break load of cylinder cut4, [N]

+

+FC = .568/130;// fuel consumption, [L/s]

+s = .72;// specific gravity of fuel

+CV = 43000;// calorific value of fuel, [kJ/kg]

+

+Te = 760;// exhaust temperature, [C]

+c = 1.015;// specific heat capacity of exhaust gas, [kJ/kg K]

+Ti = 18;// cooling water inlet temperature, [C]

+To = 56;// cooling water outlet temperature, [C]

+mw = .28;// cooling water flow rate, [kg/s]

+Ta = 21;// ambient tempearture, [C]

+C = 4.18;// specific heat capacity of cooling water, [kJ/kg K]

+

+// solution

+bp = BL*N/455;// brake power of engine, [kW]

+bp1 = BL1*N/455;// brake power of cylinder cut1, [kW]

+i1 = bp-bp1;// indicated power of cylinder cut1, [kW]

+bp2 = BL2*N/455;// brake power of cylinder cut2, [kW]

+i2 = bp-bp2;// indicated power of cylinder cut2, [kW]

+bp3 = BL3*N/455;// brake power of cylinder cut3, [kW]

+i3 = bp-bp3;// indicated power of cylinder cut3, [kW]

+bp4 = BL4*N/455;// brake power of cylinder cut4, [kW]

+i4 = bp-bp4;// indicated power of cylinder cut4, [kW]

+

+ip = i1+i2+i3+i4;// indicated power of engine, [kW]

+n_mec = bp/ip;// mechanical efficiency

+

+mprintf('\n The Brake power is  =  %f kW\n',bp);

+mprintf('\n The Indicated power is  =  %f kW\n',ip);

+mprintf('\n The mechanical efficiency is  =  %f percent\n',n_mec*100);

+

+mf = FC*s;// mass of fuel/s, [kg]

+ef = CV*mf;// energy from fuel/s, [kJ]

+me = 15*mf;// mass of exhaust/s,[kg],(given in condition)

+ee = me*c*(Te-Ta);// energy to exhaust/s,[kJ]

+ec = mw*C*(To-Ti);// energy to cooling water/s,[kJ]

+es = ef-(ee+ec+bp);// energy to surrounding,etc/s,[kJ]

+

+disp('Energy can be tabulated as :-');

+disp('----------------------------------------------------------------------------------------------------');

+disp('                                                  kJ/s                           Percentage   ')

+disp('----------------------------------------------------------------------------------------------------');

+mprintf('\n Energy from fuel                         %f                 %f\n Energy to brake power                 %f                   %f\n Energy to exhaust                       %f                   %f\n Energy to coolant                        %f                   %f\n Energy to suroundings,etc.           %f                  %f\n',ef,ef/ef*100,bp,bp/ef*100,ee,ee/ef*100,ec,ec/ef*100,es,es/ef*100);

+

+// there is minor variation in the result reported in the book

+//  End

diff --git a/2705/CH17/EX17.6/Ex17_6.sce b/2705/CH17/EX17.6/Ex17_6.sce
new file mode 100755
index 000000000..7f34d5492
--- /dev/null
+++ b/2705/CH17/EX17.6/Ex17_6.sce
@@ -0,0 +1,33 @@
+clear;

+clc;

+disp('Example 17.6');

+

+// aim : To determine 

+// (a) the break power of engine

+// (b) the fuel consumption of the engine

+// (c) the brake thermal efficiency of the engine

+

+// given values

+d = 850*10^-3;// bore , [m]

+L = 2200*10^-3;// stroke, [m]

+PMb = 15;// BMEP of cylinder, [bar]

+N = 95/60;// speed of engine, [rev/s]

+sfc = .2;// specific fuel oil consumption, [kg/kWh]

+CV = 43000;// calorific value of the fuel oil, [kJ/kg]

+

+// solution

+// (a)

+A = %pi*d^2/4;// area, [m^2]

+bp = PMb*L*A*N*8/10;// brake power,[MW]

+mprintf('\n (a) The brake power is  =  %f  MW\n',bp);

+

+ // (b)

+ FC = bp*sfc;// fuel consumption, [kg/h]

+ mprintf('\n (b) The fuel consumption is  =  %f  tonne/h\n',FC);

+ 

+ // (c)

+ mf = FC/3600;// fuel used, [kg/s]

+ n_the = bp/(mf*CV);// brake thermal efficiency

+ mprintf('\n (c) The brake thermal efficiency is  =  %f  percent\n',n_the*100);

+ 

+ // End

diff --git a/2705/CH18/EX18.1/Ex18_1.sce b/2705/CH18/EX18.1/Ex18_1.sce
new file mode 100755
index 000000000..85da4efdc
--- /dev/null
+++ b/2705/CH18/EX18.1/Ex18_1.sce
@@ -0,0 +1,61 @@
+clear;

+clc;

+disp('Example 18.1');

+

+// aim : To determine

+// (a) the coefficient of performance

+// (b) the mass flow of the refrigerant

+// (c) the cooling water required by the condenser

+

+// given values

+P1 = 462.47;// pressure limit, [kN/m^2]

+P3 = 1785.90;// pressure limit, [kN/m^2]

+T2 = 273+59;// entering saturation temperature, [K]

+T5 = 273+32;// exit temperature of condenser, [K]

+d = 75*10^-3;// bore, [m]

+L = d;// stroke, [m]

+N = 8;// engine speed, [rev/s]

+VE = .8;// olumetric efficiency

+cpL = 1.32;// heat capacity of liquid, [kJ/kg K]

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

+

+// solution

+// from given table

+// at P1

+h1 = 231.4;// specific enthalpy, [kJ/kg]

+s1 = .8614;// specific entropy,[ kJ/kg K

+v1 = .04573;// specific volume, [m^3/kg]

+

+// at P3

+h3 = 246.4;// specific enthalpy, [kJ/kg]

+s3 = .8093;// specific entropy,[ kJ/kg K

+v3 = .04573;// specific volume, [m^3/kg]

+T3= 273+40;// saturation temperature, [K]

+h4 = 99.27;// specific enthalpy, [kJ/kg]

+// (a)

+s2 = s1;// specific entropy, [kJ/kg k]

+// using s2=s3+cpv*log(T2/T3)

+cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k]

+

+// from Fig.18.8

+T4 = T3;

+h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg]

+h5 = h4-cpL*(T4-T5);// specific enthalpy, [kJ/kg]

+h6 = h5;

+COP = (h1-h6)/(h2-h1);// coefficient of performance

+mprintf('\n (a) The coefficient of performance of the refrigerator is  =  %f\n',COP);

+

+// (b)

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

+ESV = SV*VE*N*3600;// effective swept volume/h, [m^3]

+m = ESV/v1;// mass flow of refrigerant/h,[kg]

+mprintf('\n (b) The mass flow of refrigerant/h is  =  %f  kg\n',m);

+

+// (c)

+dT = 12;// temperature limit, [C]

+Q = m*(h2-h5);// heat transfer in condenser/h, [kJ]

+// using Q=m_dot*c*dT, so

+m_dot = Q/(c*dT);// mass flow of water required, [kg/h]

+mprintf('\n (c) The mass flow of water required is  =  %f kg/h\n',m_dot);

+

+//  End

diff --git a/2705/CH18/EX18.2/Ex18_2.sce b/2705/CH18/EX18.2/Ex18_2.sce
new file mode 100755
index 000000000..7d917dfd9
--- /dev/null
+++ b/2705/CH18/EX18.2/Ex18_2.sce
@@ -0,0 +1,54 @@
+clear;

+clc;

+disp('Example 18.2');

+

+// aim : To determine

+// (a) the mass flow of R401

+// (b) the dryness fraction of  R401 at the entry to the evaporator

+// (c) the power of driving motor

+// (d) the ratio of heat transferred from condenser to the power required to the motor

+

+// given values

+P1 = 411.2;// pressure limit, [kN/m^2]

+P3 = 1118.9;// pressure limit, [kN/m^2]

+Q = 100*10^3;// heat transfer from the condenser,[kJ/h]

+T2 = 273+60;// entering saturation temperature, [K]

+

+// given

+// from given table

+// at P1

+h1 = 409.3;// specific enthalpy, [kJ/kg]

+s1 = 1.7431;// specific entropy,[ kJ/kg K

+

+// at P3

+h3 = 426.4;// specific enthalpy, [kJ/kg]

+s3 = 1.7192;// specific entropy,[ kJ/kg K

+T3 = 273+50;// saturation temperature, [K]

+h4 = 265.5;// specific enthalpy, [kJ/kg]

+// (a)

+s2 = s1;// specific entropy, [kJ/kg k]

+// using s2=s3+cpv*log(T2/T3)

+cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k]

+

+// from Fig.18.8

+h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg]

+Qc = h2-h4;// heat transfer from condenser, [kJ/kg]

+mR401 = Q/Qc;// mass flow of R401, [kg]

+ mprintf('\n (a) The mass flow of R401 is  =  %f kg/h\n',mR401);

+

+// (b)

+hf1 = 219;// specific enthalpy, [kJ/kg]

+h5 = h4;

+// using h5=hf1+s5*(h1-hf1),so

+x5 = (h5-hf1)/(h1-hf1);// dryness fraction

+mprintf('\n (b) The dryness fraction of R401 at the entry to the evaporator is  =  %f\n',x5);

+

+// (c)

+P = mR401*(h2-h1)/3600/.7;// power to driving motor, [kW]

+ mprintf('\n (c) The power to driving motor is  =  %f kW\n',P);

+

+// (d)

+r = Q/3600/P;// ratio

+mprintf('\n (d) The ratio of heat transferred from condenser to the power required to the motor is  =  %f : 1\n',r);

+

+//  End

diff --git a/2705/CH19/EX19.1/Ex19_1.sce b/2705/CH19/EX19.1/Ex19_1.sce
new file mode 100755
index 000000000..feed53150
--- /dev/null
+++ b/2705/CH19/EX19.1/Ex19_1.sce
@@ -0,0 +1,46 @@
+clear;

+clc;

+disp('Example 19.1');

+

+// aim : To compare the moisture content and the true specific volumes of  atmosphere air 

+// (a) temperature is 12 C and the air is saturaded

+// (b) temperature is 31 C and air is .75 saturated

+

+// Given values

+P_atm = 101.4;// atmospheric pressure, [kN/m^2]

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

+

+// solution

+// (a)

+T = 273+12;// air temperature, [K]

+// From steam table at 12 C

+p = 1.4;// [kN/m^2]

+vg = 93.9;// [m^3/kg]

+pa = P_atm-p;// partial pressure of the dry air, [kN/m^2]

+va = R*T/pa;// [m^3/kg]

+

+mw = va/vg;// mass of water vapor in the air,[kg]

+v = va/(1+mw);// specific volume of humid air, [m^3/kg]

+

+mprintf('\n (a) The mass of water vapor in the humid air is  =  %f  kg\n',mw);

+mprintf('\n      The specific volume of humid air is  =  %f m^3/kg\n',v);

+

+// (b)

+x = .75;// dryness fraction

+T = 273+31;// air temperature, [K]

+// From steam table

+p = 4.5;// [kN/m^2]

+vg = 31.1;// [m^3/kg]

+pa = P_atm-p;// [kN/m^2]

+va = R*T/pa;// [m^3/kg]

+mw1= va/vg;// mass of water vapor in the air, [kg]

+mw_actual = mw1*x;// actual mass of vapor, [kg]

+v = va/(1+mw_actual);// true specific volume of humid air,[m^3/kg] 

+

+mprintf('\n (b) The mass of water vapor in the humid air is  =  %f  kg\n',mw1);

+mprintf('\n     The specific volume of humid air is  =  %f  m^3/kg\n',v);

+

+ewv = mw_actual/mw ;

+mprintf('\n On the warm day the air conteains %f  times the mass of water vapor as on the cool day \n',ewv);

+

+//  End

diff --git a/2705/CH19/EX19.2/Ex19_2.sce b/2705/CH19/EX19.2/Ex19_2.sce
new file mode 100755
index 000000000..99dc20724
--- /dev/null
+++ b/2705/CH19/EX19.2/Ex19_2.sce
@@ -0,0 +1,34 @@
+clear;

+clc;

+disp('Example 19.2');

+

+// aim : To determine

+// (a) the partial pressures of the vapor and the dry air

+// (b) the specific humidity of the mixture

+// (c) the composition of the mixture

+

+//  Given values

+phi = .65;// Relative humidity

+T = 2733+20;// temperature, [K]

+p = 100;// barometric pressure, [kN/m^2]

+

+// solution

+// (a)

+//  From the steam table at 20 C

+pg = 2.34;// [kN/m^2]

+ps = phi*pg;// partial pressure of vapor, [kN/m^2]

+pa = p-ps;// partial pressure of dry air, [kN/m^2]

+mprintf('\n (a) The partial pressure of vapor is  =  %f  kN/m^2\n',ps);

+mprintf('\n     The partial pressure of dry air is  =  %f kN/m^2\n',pa);

+

+// (b)

+// from equation [15]

+omega = .622*ps/(p-ps);// specific humidity of the mixture

+mprintf('\n (b) The specific humidity of the mixture is  =  %f kg/kg dry air\n',omega);

+

+// (c)

+// using eqn [1] from section 19.2

+y = 1/(1+omega);// composition of the mixture

+mprintf('\n (c) The composition of the mixture is  =  %f\n',y);

+

+// End

diff --git a/2705/CH19/EX19.3/Ex19_3.sce b/2705/CH19/EX19.3/Ex19_3.sce
new file mode 100755
index 000000000..54f4d0111
--- /dev/null
+++ b/2705/CH19/EX19.3/Ex19_3.sce
@@ -0,0 +1,52 @@
+clear;

+clc;

+disp('Example 19.3');

+

+// aim : To determine

+// (a) the specific humidity

+// (b) the dew point

+// (c) the degree of superheat of the superheated vapor

+// (d) the mass of condensate formed per kg of dry air if the moist air is cooled to 12 C

+

+// Given values

+t = 25;// C

+T = 273+25;// moist air temperature, [K]

+phi = .6;// relative humidity

+p = 101.3;// barometric pressure, [kN/m^2]

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

+

+// solution

+// (a)

+// From steam table at 25 C

+pg = 3.17;// [kN/m^2]

+ps = phi*pg;// partial pressure of the vapor, [kN/m^2]

+omega = .622*ps/(p-ps);// the specific humidity of air

+

+mprintf('\n (a) The specific humidity is  =  %f  kg/kg air\n',omega);

+

+// (b)

+// Dew point is saturated temperature at ps is,

+t_dew = 16+2*(1.092-1.817)/(2.062-1.817);// [C]

+mprintf('\n (b) The dew point is  =  %f C\n',t_dew);

+

+// (c)

+Dos = t-t_dew;// degree of superheat, [C]

+mprintf('\n (c) The degree of superheat is  =  %f C\n',Dos);

+

+// (d)

+// at 25 C

+pa = p-ps;// [kN/m^2]

+va = R*T/pa;// [m^3/kg]

+// at 16.69 C

+vg = 73.4-(73.4-65.1)*.69/2;// [m^3/kg]

+ms1= va/vg; 

+// at 12 C

+vg = 93.8;// [m^3/kg]

+ms2 = va/vg;

+

+m = ms1-ms2;// mas of condensate

+mprintf('\n (d) The mass of condensate is  =  %f  kg/kg dry air\n',m);

+

+//  there is calculation mistake in the book so answer is no matching

+

+//  End

diff --git a/2705/CH19/EX19.4/Ex19_4.sce b/2705/CH19/EX19.4/Ex19_4.sce
new file mode 100755
index 000000000..0cec4f91c
--- /dev/null
+++ b/2705/CH19/EX19.4/Ex19_4.sce
@@ -0,0 +1,73 @@
+clear;

+clc;

+disp(' Example 19.4');

+

+// aim : To determine

+// (a) the volume of external saturated air

+// (b) the mass of air

+// (c) the heat transfer

+// (d) the heat transfer required by the combind water vapour

+

+// given values

+Vb = 56000;// volume of building, [m^3]

+T2 = 273+20;// temperature of air in thebuilding, [K]

+phi = .6;// relative humidity

+T1 = 8+273;// external air saturated temperature, [K]

+p0 = 101.3;// atmospheric pressure, [kN/m^2]

+cp = 2.093;// heat capacity of saturated steam, [kJ/kg K]

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

+

+// solution

+// from steam table at 20 C saturation pressure of steam is,

+pg = 2.34;// [kN/m^2]

+

+// (a)

+pvap = phi*pg;// partial pressure of vapor, [kN/m^2] 

+P = p0-pvap;// partial pressure of air, [kN/m^2]

+V = 2*Vb;// air required, [m^3]

+// at 8 C saturation pressure ia

+pvap = 1.072;// [kN/m^2]

+P2 = p0-pvap;// partial pressure of entry at 8 C, [kN/m^2]

+

+// using P1*V1/T1=P2*V2/T2;

+V2 = P*V*T1/(T2*P2);// air required at 8 C, [m^3/h]

+mprintf('\n (a) The volume of air required is  =  %f  m^3/h\n',V2);

+

+// (b)

+//  assuming

+pg = 1.401;// pressure, [kN/m^2]

+Tg = 273+12;// [K]

+vg = 93.8;// [m^3/kg]

+// at constant pressure

+v = vg*T2/Tg;// volume[m^3/kg]

+mv = V/v;// mass of vapor in building at 20 C, [kg/h]

+// from steam table at 8 C

+vg2 = 121;// [m^3/kg]

+mve = V2/vg2;// mass of vapor supplied with saturated entry air, [kg/h]

+mw = mv-mve;// mass of water added, [kg/h]

+mprintf('\n (b) The mass of water added is  =  %f  kg/h\n ',mw);

+

+// (c)

+// for perfect gas

+m = P2*V2/(R*T1);// [kg/h]

+Cp = .287;// heat capacity, [kJ/kg K]

+Q = m*Cp*(T2-T1);// heat transfer by dry air,[kJ/h]

+mprintf('\n (c) The heat transfer required by dry air is  =  %f MJ/h\n',Q*10^-3);

+

+// (d)

+// from steam table

+h1 = 2516.2;// specific enthalpy of saturated vapor at 8 C,[kJ/kg]

+hs = 2523.6;// specific enthalpy of saturated vapor at 20 C, [kJ/kg]

+h2 = hs+cp*(T2-T1);// specific enthalpy of vapor at 20 c, [kJ/kg]

+Q1 = mve*(h2-h1);// heat transfer required for vapor, [kJ]

+

+// again from steam table

+hf1 = 33.6;// [kJ/kg]

+hg3 = 2538.2;// [kJ/kg]

+Q2 = mw*(hg3-hf1);// heat transfer required for water, [kJ/h]

+Qt = Q1+Q2;// total heat transfer, [kJ/h]

+mprintf('\n (d) The heat transferred required for vapor+supply water is  =  %f  MJ/h\n',Qt*10^-3);

+

+//  there is minor variation in the answer reported in the book

+

+//  End

diff --git a/2705/CH2/EX2.1/Ex2_1.sce b/2705/CH2/EX2.1/Ex2_1.sce
new file mode 100755
index 000000000..1d5f07b56
--- /dev/null
+++ b/2705/CH2/EX2.1/Ex2_1.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+disp('Example 2.1');
+
+
+
+//  Given values
+Q = 2500; // Heat transferred into the system, [kJ]
+W = 1400; //  Work transferred from the system,  [kJ]
+ 
+//   solution
+
+//  since process carried out on a closed system, so using equation [4]
+del_E = Q-W; //  Change in total energy, [kJ]
+
+mprintf('\n The Change in total energy is, del_E =  %f  kJ\n',del_E);
+
+if(del_E&gt;0)
+    disp('Since del_E is positive, so there is an increase in total enery')
+else
+    disp('Since del_E is negative, so there is an decrease in total enery')
+end
+
+// There is mistake in the book's results unit
+
+//  End
diff --git a/2705/CH2/EX2.2/Ex2_2.sce b/2705/CH2/EX2.2/Ex2_2.sce
new file mode 100755
index 000000000..095f933f2
--- /dev/null
+++ b/2705/CH2/EX2.2/Ex2_2.sce
@@ -0,0 +1,22 @@
+clear;
+clc;
+disp('Example 2.2');
+
+
+//  Given values
+del_E = 3500; // Increase in total energy of the system, [kJ]
+W = -4200; // Work transfer into the system, [kJ]
+
+//  solution
+//  since process carried out on a closed system, so using equation [3]
+Q = del_E+W;// [kJ]
+
+mprintf('\n The Heat transfer is, Q =  %f kJ \n',Q);
+
+if(Q&gt;0)
+    disp('Since Q&gt;0, so heat is transferred into the system')
+else
+    disp('Since Q&lt;0, so heat is transferred from the system')
+end
+
+//  End
diff --git a/2705/CH2/EX2.3/Ex2_3.sce b/2705/CH2/EX2.3/Ex2_3.sce
new file mode 100755
index 000000000..0f6f799d4
--- /dev/null
+++ b/2705/CH2/EX2.3/Ex2_3.sce
@@ -0,0 +1,23 @@
+clear;
+clc;
+disp('Example 2.3');
+
+
+
+//  Given values
+Q = -150; // Heat transferred out of the system, [kJ/kg]
+del_u = -400; // Internal energy decreased ,[kJ/kg]
+
+//  solution
+//  using equation [3],the non flow energy equation
+//  Q=del_u+W
+W = Q-del_u; //  [kJ/kg]
+mprintf('\n The Work done is, W =  %f kJ/kg \n',W);
+
+if(W&gt;0)
+      disp('Since W&gt;0, so Work done by the engine per kilogram of working substance')
+else
+   disp('Since &lt;0, so Work done on the engine per kilogram of working substance')
+end
+
+//  End
diff --git a/2705/CH2/EX2.4/Ex2_4.sce b/2705/CH2/EX2.4/Ex2_4.sce
new file mode 100755
index 000000000..9102fc8b3
--- /dev/null
+++ b/2705/CH2/EX2.4/Ex2_4.sce
@@ -0,0 +1,43 @@
+clear;
+clc;
+disp('Example 2.4');
+
+
+
+//  Given values
+m_dot = 4; //  fluid flow rate, [kg/s]
+Q = -40;  //  Heat loss to the surrounding, [kJ/kg]
+
+//  At inlet 
+P1 = 600;  // pressure  ,[kn/m^2]
+C1 = 220;  // velocity  ,[m/s]
+u1 = 2200;  // internal energy,  [kJ/kg]
+v1 = .42; // specific volume, [m^3/kg]
+
+//  At outlet
+P2 = 150; // pressure, [kN/m^2]
+C2 = 145;  // velocity,  [m/s]
+u2 = 1650;  // internal energy,  [kJ/kg]
+v2 = 1.5;  // specific volume,  [m^3/kg]
+
+//  solution
+//  for steady flow energy equation for the open system is given by
+//  u1+P1*v1+C1^2/2+Q=u2+P2*v2+C2^2/2+W
+//  hence
+
+W = (u1-u2)+(P1*v1-P2*v2)+(C1^2/2-C2^2/2)*10^-3+Q; // [kJ/kg]
+
+mprintf('\n workdone is, W =  %f kJ/kg ',W);
+
+if(W&gt;0)
+      disp('Since W&gt;0, so Power is output from the system')
+else
+   disp('Since &lt;0, so Power is input to the system')
+end
+ 
+//  Hence
+
+P_out = W*m_dot; // power out put from the system, [kW]
+mprintf('\n The power output from the system is  =  %f  kW \n',P_out);
+
+//  End
diff --git a/2705/CH2/EX2.5/Ex2_5.sce b/2705/CH2/EX2.5/Ex2_5.sce
new file mode 100755
index 000000000..f43f37c08
--- /dev/null
+++ b/2705/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,27 @@
+clear;
+clc;
+disp('Example 2.5');
+
+
+
+//  Given values
+del_P = 154.45; // pressure difference across the die, [MN/m^2]
+rho = 11360; // Density of the lead, [kg/m^3]
+c = 130; // specific heat capacity of the lead, [J/kg*K]
+
+//  solution
+//  since there is no cooling and no externel work is done, so energy balane becomes
+//  P1*V1+U1=P2*V2+U2 ,so
+//  del_U=U2-U1=P1*V1-P2*V2
+
+//  also, for temperature rise, del_U=m*c*t, where, m is mass; c is specific heat capacity; and t is temperature rise
+
+//  Also given that lead is incompressible, so V1=V2=V and assuming one m^3 of lead
+
+//  using above equations
+t = del_P/(rho*c)*10^6 ;// temperature rise [C]
+
+mprintf('\n The temperature rise of the lead is  =  %f  C\n',t);
+
+//  End
+
diff --git a/2705/CH2/EX2.6/Ex2_6.sce b/2705/CH2/EX2.6/Ex2_6.sce
new file mode 100755
index 000000000..186ca15e2
--- /dev/null
+++ b/2705/CH2/EX2.6/Ex2_6.sce
@@ -0,0 +1,38 @@
+clear;
+clc;
+disp('Example 2.6');
+
+
+//  Given values
+m_dot = 4.5; // mass flow rate of air, [kg/s]
+Q = -40; // Heat transfer loss, [kJ/kg]
+del_h = -200; // specific enthalpy reduce, [kJ/kg]
+
+C1 = 90; // inlet velocity, [m/s]
+v1 = .85; // inlet specific volume, [m^3/kg]
+
+v2 = 1.45; //  exit specific volume, [m^3/kg]
+A2 = .038;  //  exit area of turbine, [m^2]
+
+//  solution
+
+//  part (a)
+//  At inlet, by equation[4], m_dot=A1*C1/v1
+A1 = m_dot*v1/C1;//inlet area, [m^2]
+mprintf('\n (a) The inlet area is, A1 =  %f  m^2 \n',A1);
+
+//  part (b), 
+//  At outlet, since mass flow rate is same, so m_dot=A2*C2/v2, hence
+C2 = m_dot*v2/A2; // Exit velocity,[m/s]
+mprintf('\n (b) The exit velocity is, C2 =  %f  m/s \n',C2);
+
+//  part (c)
+//  using steady flow equation, h1+C1^2/2+Q=h2+C2^2/2+W
+W = -del_h+(C1^2/2-C2^2/2)*10^-3+Q; //  [kJ/kg]
+
+//  Hence power developed is
+P = W*m_dot;//  [kW]
+mprintf('\n (c) The power developed by the turbine system is  =  %f kW \n',P);
+
+//  End
+
diff --git a/2705/CH4/EX4.1/Ex4_1.sce b/2705/CH4/EX4.1/Ex4_1.sce
new file mode 100755
index 000000000..afa29662b
--- /dev/null
+++ b/2705/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,23 @@
+clear;

+clc;

+disp('Example 4.1');

+

+//  aim : To determine

+//  the enthalpy

+

+//  Given values

+P = .50;// Pressure, [MN/m^2]

+

+//  solution

+

+//  From steam tables, at given pressure

+hf = 640.1;//  specific liquid enthalpy ,[kJ/kg]

+hfg = 2107.4;//  specific enthalpy of evaporation ,[kJ/kg]

+hg = 2747.5; //  specific enthalpy of dry saturated steam ,[kJ/kg]

+tf = 151.8; //  saturation temperature,[C]

+

+mprintf('\n The specific liquid enthalpy is  =  %f kJ/kg \n',hf);

+mprintf('\n The specific enthalpy of evaporation is  =  %f  kJ/kg \n',hfg);

+mprintf('\n The specific enthalpy of dry saturated steam is  =  %f kJ/kg \n',hg);

+

+//  End

diff --git a/2705/CH4/EX4.10/Ex4_10.sce b/2705/CH4/EX4.10/Ex4_10.sce
new file mode 100755
index 000000000..972a1eed3
--- /dev/null
+++ b/2705/CH4/EX4.10/Ex4_10.sce
@@ -0,0 +1,38 @@
+clear;

+clc;

+disp('Example .10');

+

+// aim : To determine

+// (a) the mass of steam entering the heater

+// (b) the mass of water entering the heater

+

+//  Given values

+x = .95;//  Dryness fraction

+P = .7;//  pressure,[MN/m^2]

+d = 25;//  internal diameter of heater,[mm]

+C = 12; //  steam velocity in the pipe,[m/s]

+

+//  solution

+//  from steam table at .7 MN/m^2 pressure

+hf = 697.1;//  [kJ/kg]

+hfg = 2064.9;//  [kJ/kg]

+hg = 2762.0; //  [kJ/kg]

+vg = .273; //  [m^3/kg]

+

+//  (a)

+v = x*vg; //  [m^3/kg]

+ms_dot = %pi*(d*10^-3)^2*C*3600/(4*v);//  mass of steam entering, [kg/h]

+mprintf('\n (a) The mass of steam entering the heater is  =  %f kg/h \n',ms_dot);

+

+//  (b)

+h = hf+x*hfg;//  specific enthalpy of steam entering heater,[kJ/kg]

+//  again from steam tables

+hf1 = 376.8;//  [kJ/kg] at 90 C

+hf2 = 79.8;//  [kJ/kg] at 19 C

+

+//  using energy balance,mw_dot*(hf1-hf2)=ms_dot*(h-hf1)

+mw_dot = ms_dot*(h-hf1)/(hf1-hf2);//  mass of water entering to heater,[kg/h]

+

+mprintf('\n (b) The mass of water entering the heater is  =   %f  kg/h \n',mw_dot);

+

+//  End

diff --git a/2705/CH4/EX4.11/Ex4_11.sce b/2705/CH4/EX4.11/Ex4_11.sce
new file mode 100755
index 000000000..89819556c
--- /dev/null
+++ b/2705/CH4/EX4.11/Ex4_11.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+disp('Example 4.11');

+

+// aim: To determine

+//  the change of internal energy

+

+//  Given values

+m = 1.5;//  mass of steam,[kg]

+P1 = 1;//  initial pressure, [MN/m^2]

+t = 225;//  temperature, [C]

+P2 = .28;//  final pressure, [MN/m^2]

+x = .9;//  dryness fraction of steam at P2

+

+//  solution

+

+// from steam table at P1

+h1 = 2886;//  [kJ/kg]

+v1 = .2198; //  [m^3/kg]

+//  hence

+u1 = h1-P1*v1*10^3;// internal energy [kJ/kg]

+

+// at P2

+hf2 = 551.4;// [kJ/kg]

+hfg2 = 2170.1;// [kJ/kg]

+vg2 = .646; //  [m^3/kg]

+//  so

+h2 = hf2+x*hfg2;//  [kj/kg]

+v2 = x*vg2;//   [m^3/kg]

+

+// now

+u2 = h2-P2*v2*10^3;//  [kJ/kg]

+

+//  hence change in specific internal energy is

+del_u = u2-u1;//  [kJ/kg]

+

+del_u = m*del_u;//  [kJ];

+mprintf('\n The change in internal energy is  =  %f  kJ \n',del_u);

+

+//  End

diff --git a/2705/CH4/EX4.12/Ex4_12.sce b/2705/CH4/EX4.12/Ex4_12.sce
new file mode 100755
index 000000000..2ef1bbed3
--- /dev/null
+++ b/2705/CH4/EX4.12/Ex4_12.sce
@@ -0,0 +1,29 @@
+clear;

+clc;

+disp('Example 4.12');

+

+//  aim : To determine 

+//  the dryness fraction of steam after throttling

+

+//  given values

+P1 = 1.4;//  pressure before throttling, [MN/m^2]

+x1 = .7;//  dryness fraction before throttling

+P2 = .11;//  pressure after throttling, [MN/m^2]

+

+//  solution

+//  from steam table

+hf1 = 830.1;//  [kJ/kg]

+hfg1 = 1957.7;//  [kJ/kg]

+h1 = hf1 + x1*hfg1; //  [kJ/kg]

+

+hf2 = 428.8;//  [kJ/kg]

+hfg2 = 2250.8;//  [kJ/kg]

+

+//  now for throttling,

+//  hf1+x1*hfg1=hf2+x2*hfg2; where x2 is dryness fraction after throttling

+

+x2=(h1-hf2)/hfg2; // final dryness fraction

+

+mprintf('\n Dryness fraction of steam after throttling is  =  %f \n',x2);

+

+//  End

diff --git a/2705/CH4/EX4.13/Ex4_13.sce b/2705/CH4/EX4.13/Ex4_13.sce
new file mode 100755
index 000000000..108099539
--- /dev/null
+++ b/2705/CH4/EX4.13/Ex4_13.sce
@@ -0,0 +1,65 @@
+clear;

+clc;

+disp('Example 4.13');

+

+//  aim : To determine 

+//  the dryness fraction of steam 

+//  and the internal diameter of the pipe

+

+//  Given values

+

+//  steam1

+P1 = 2;// pressure before throttling, [MN/m^2]

+t = 300;//  temperature,[C]

+ms1_dot = 2;// steam flow rate, [kg/s]

+P2 = 800;//  pressure after throttling, [kN/m^2]

+

+//  steam2

+P = 800;// pressure, [N/m^2]

+x2 = .9;//  dryness fraction

+ms2_dot = 5; //  [kg/s]

+

+//  solution

+//  (a)

+//  from steam table specific enthalpy of steam1 before throttling is

+hf1 = 3025;//  [kJ/kg]

+//  for throttling process specific enthalpy will same so final specific enthalpy of steam1 is

+hf2 = hf1;

+// hence

+h1 = ms1_dot*hf2;// [kJ/s]

+

+//  calculation of specific enthalpy of steam2

+hf2 = 720.9;//  [kJ/kg]

+hfg2 = 2046.5;//  [kJ/kg]

+//  hence

+h2 = hf2+x2*hfg2;//  specific enthalpy, [kJ/kg]

+h2 = ms2_dot*h2;//  total enthalpy, [kJ/s]

+

+//  after mixing

+m_dot = ms1_dot+ms2_dot;//  total mass of mixture,[kg/s]

+h = h1+h2;//  Total enthalpy of the mixture,[kJ/s]

+h = h/7;//  [kJ/kg]

+

+//  At pressure 800 N/m^2 

+hf = 720.9;//  [kJ/kg]

+hfg = 2046.5;//  [kJ/kg]

+//  so total enthalpy is,hf+x*hfg, where x is dryness fraction of mixture and which is equal to h

+//  hence

+x = (h-hf)/hfg;// dryness fraction after mixing

+mprintf('\n (a) The condition of the resulting mixture is dry with dryness fraction  =  %f \n',x);

+

+//  (b)

+// Given

+C = 15;//  velocity, [m/s]

+//  from steam table

+v = .1255;//  [m^/kg]

+A = ms1_dot*v/C;//  area, [m^2]

+//  using ms1_dot = A*C/v, where A is cross section area in m^2 and

+//  A = %pi*d^2/4, where d is diameter of the pipe 

+

+//  calculation of d

+d = sqrt(4*A/%pi); // diameter, [m]

+

+mprintf('\n (b) The internal diameter of the pipe is  =  %f mm \n',d*1000);

+

+//  End

diff --git a/2705/CH4/EX4.14/Ex4_14.sce b/2705/CH4/EX4.14/Ex4_14.sce
new file mode 100755
index 000000000..caec80859
--- /dev/null
+++ b/2705/CH4/EX4.14/Ex4_14.sce
@@ -0,0 +1,17 @@
+clear;

+clc;

+disp('Example 4.14');

+

+//  aim : To estimate 

+//  the dryness fraction

+

+//  Given values

+M = 1.8;//  mass of condensate, [kg]

+m = .2;//  water collected, [kg]

+

+//  solution

+x = M/(M+m);//  formula for calculation of dryness fraction using seprating calorimeter

+

+mprintf(' \n The dryness fraction of the steam entering seprating calorimeter is  =  %f \n',x);

+

+//  End

diff --git a/2705/CH4/EX4.15/Ex4_15.sce b/2705/CH4/EX4.15/Ex4_15.sce
new file mode 100755
index 000000000..e28726c11
--- /dev/null
+++ b/2705/CH4/EX4.15/Ex4_15.sce
@@ -0,0 +1,51 @@
+clear;

+clc;

+disp('Example 4.15');

+

+//  aim : To determine

+//  the dryness fraction of the steam at 2.2 MN/m^2

+

+//  Given values

+P1 = 2.2;//  [MN/m^2]

+P2 = .13;//  [MN/m^2]

+t2 = 112;//  [C]

+tf2 = 150;//  temperature, [C]

+

+// solution

+// from steam table, at 2.2 MN/m^2

+//  saturated steam at 2 MN/m^2 Pressure

+hf1 = 931;//  [kJ/kg]

+hfg1 = 1870;//  [kJ/kg]

+hg1 = 2801;//  [kJ/kg]

+

+// for superheated steam

+//  at .1 MN/m^2

+hg2 = 2675;//  [kJ/kg]

+hg2_150 = 2777;// specific enthalpy at 150 C, [kJ/kg]

+tf2 = 99.6;//  saturation temperature, [C]

+

+// at .5 MN/m^2

+hg3 = 2693;//  [kJ/kg]

+hg3_150 = 2773;// specific enthalpy at 150 C, [kJ/kg]

+tf3 = 111.4;//  saturation temperature, [C]

+

+Table_P_h1 = [[.1,.5];[hg2,hg3]];// where, P in MN/m^2 and h in [kJ/kg]

+hg = interpln(Table_P_h1,.13);//  specific entahlpy at .13 MN/m^2, [kJ/kg]

+

+Table_P_h2 = [[.1,.5];[hg2_150,hg3_150]];//  where, P in MN/m^2 and h in [kJ/kg]

+hg_150 = interpln(Table_P_h2,.13);//  specific entahlpy at .13 MN/m^2 and 150 C, [kJ/kg]

+

+Table_P_tf = [[.1,.5];[tf2,tf3]];// where, P in MN/m^2 and h in [kJ/kg]

+tf = interpln(Table_P_tf,.13);//  saturation temperature, [C]

+

+//  hence

+h2 = hg+(hg_150-hg)/(t2-tf)/(tf2-tf);//  specific enthalpy at .13 MN/m^2 and 112 C, [kJ/kg]

+

+// now since process is throttling so h2=h1

+// and h1 = hf1+x1*hfg1, so

+x1 = (h2-hf1)/hfg1;// dryness fraction

+mprintf(' \n The dryness fraction of steam is  =  %f \n',x1);

+

+// There is a calculation mistake in book so answer is not matching

+

+//  End

diff --git a/2705/CH4/EX4.16/Ex4_16.sce b/2705/CH4/EX4.16/Ex4_16.sce
new file mode 100755
index 000000000..895917d3a
--- /dev/null
+++ b/2705/CH4/EX4.16/Ex4_16.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+disp('Example 4.16');

+

+//  aim : To determine 

+//  the minimum dryness fraction of steam

+

+//  Given values

+P1 = 1.8;//  testing pressure,[MN/m^2]

+P2 = .11;//  pressure after throttling,[MN/m^2]

+

+//  solution

+//  from steam table

+//  at .11 MN/m^2 steam is completely dry and specific enthalpy is

+hg = 2680;//  [kJ/kg]

+

+//  before throttling steam is wet, so specific enthalpy is=hf+x*hfg, where x is dryness fraction

+//  from steam table

+hf = 885;//  [kJ/kg]

+hfg = 1912;//  [kJ/kg]

+

+//  now for throttling process,specific enthalpy will same before and after

+//  hence

+x = (hg-hf)/hfg;

+mprintf('\n The minimum dryness fraction of steam is x  =  %f \n',x);

+

+//  End

+

diff --git a/2705/CH4/EX4.17/Ex4_17.sce b/2705/CH4/EX4.17/Ex4_17.sce
new file mode 100755
index 000000000..b85b38646
--- /dev/null
+++ b/2705/CH4/EX4.17/Ex4_17.sce
@@ -0,0 +1,52 @@
+clear;

+clc;

+disp('Example 4.17');

+

+//  aim : To determine the

+//  (a) mass of steam in the vessel

+//  (b) final dryness of the steam

+//  (c) amount of heat transferrred during the cooling process

+

+//  Given values

+V1 = .8;//  [m^3]

+P1 = 360;//  [kN/m^2]

+P2 = 200;//  [kN/m^2]

+

+//  solution

+

+//  (a)

+// at 360 kN/m^2

+vg1 = .510;//  [m^3]

+m = V1/vg1;//  mass of steam,[kg]

+mprintf('\n (a) The mass of steam in the vessel is =  %f kg\n',m);

+

+//  (b)

+//  at 200 kN/m^2

+vg2 = .885;// [m^3/kg]

+//  the volume remains constant so

+x = vg1/vg2;// final dryness fraction

+mprintf('\n (b) The final dryness fraction of the steam is  =  %f \n',x);

+

+// (c)

+//  at 360 kN/m^2

+h1 = 2732.9;// [kJ/kg]

+//  hence

+u1 = h1-P1*vg1;//  [kJ/kg]

+

+//  at 200 kN/m^2

+hf = 504.7;// [kJ/kg]

+hfg=2201.6;//[kJ/kg]

+//  hence

+h2 = hf+x*hfg;// [kJ/kg]

+//  now

+u2 = h2-P2*vg1;//  [kJ/kg]

+//  so

+del_u = u2-u1;//  [kJ/kg]

+//  from the first law of thermodynamics del_U+W=Q, 

+W = 0;//  because volume is constant

+del_U = m*del_u;//  [kJ]

+//  hence

+Q = del_U;//  [kJ]

+mprintf('\n (c) The amount of heat transferred during cooling process is  =  %f kJ \n',Q);

+

+//  End

diff --git a/2705/CH4/EX4.18/Ex4_18.sce b/2705/CH4/EX4.18/Ex4_18.sce
new file mode 100755
index 000000000..61a3ef637
--- /dev/null
+++ b/2705/CH4/EX4.18/Ex4_18.sce
@@ -0,0 +1,31 @@
+clear;

+clc;

+disp('Example 4.18');

+

+//  aim : To determine

+//  the heat received by the steam per kilogram

+

+// Given values

+// initial

+P1 = 4;// pressure, [MN/m^2]

+x1 = .95; //  dryness fraction

+

+//  final

+t2 = 350;//  temperature,[C]

+

+//  solution

+

+// from steam table, at 4 MN/m^2 and x1=.95

+hf = 1087.4;//  [kJ/kg]

+hfg = 1712.9;//  [kJ/kg]

+//  hence

+h1 = hf+x1*hfg;//  [kJ/kg]

+

+//  since pressure is kept constant ant temperature is raised so at this condition

+h2 = 3095;//  [kJ/kg]

+

+//  so by energy balance

+Q = h2-h1;//  Heat received,[kJ/kg]

+mprintf('\n The heat received by the steam is  =  %f  kJ/kg \n',Q);

+

+//  End

diff --git a/2705/CH4/EX4.19/Ex4_19.sce b/2705/CH4/EX4.19/Ex4_19.sce
new file mode 100755
index 000000000..edc6f9d44
--- /dev/null
+++ b/2705/CH4/EX4.19/Ex4_19.sce
@@ -0,0 +1,50 @@
+clear;

+clc;

+disp('Example 4.19');

+

+//  aim : To determine the condition of the steam after 

+//  (a) isothermal compression to half its initial volume,heat rejected

+//  (b) hyperbolic compression to half its initial volume

+

+//  Given values

+V1 = .3951;//  initial volume,[m^3]

+P1 = 1.5;//  initial pressure,[MN/m^2]

+

+//  solution

+

+//  (a)

+//  from steam table, at 1.5 MN/m^2 

+hf1 = 844.7;//  [kJ/kg]

+hfg1 = 1945.2;//  [kJ/kg]

+hg1 = 2789.9;// [kJ/kg]

+vg1 = .1317;//  [m^3/kg]

+

+//  calculation

+m = V1/vg1;//  mass of steam,[kg]

+vg2b = vg1/2;// given,[m^3/kg](vg2b is actual specific volume before compression)

+x1 = vg2b/vg1;//  dryness fraction

+h1 = m*(hf1+x1*hfg1);//  [kJ]

+Q = m*x1*hfg1;//  heat loss,[kJ]

+mprintf('\n (a) The Quantity of steam present is  =  %f kg \n',m);

+mprintf('\n      Dryness fraction is  =  %f \n',x1);

+mprintf('\n      The enthalpy is  =  %f kJ \n',h1);

+mprintf('\n      The heat loss is  =  %f kJ \n',Q);

+

+//  (b)

+V2 = V1/2;

+//  Given compression is according to the law PV=Constant,so

+P2 = P1*V1/V2;//  [MN/m^2]

+//  from steam table at P2

+hf2 = 1008.4;// [kJ/kg]

+hfg2 = 1793.9;//  [kJ/kg]

+hg2 = 2802.3;//  [kJ/kg]

+vg2 = .0666;//  [m^3/kg]

+

+//  calculation

+x2 = vg2b/vg2;//  dryness fraction

+h2 = m*(hf2+x2*hfg2);//  [kJ]

+

+mprintf('\n (b) The dryness fraction is  =  %f \n',x2);

+mprintf('\n     The enthalpy is  =  %f kJ\n',h2);

+

+//  End

diff --git a/2705/CH4/EX4.2/Ex4_2.sce b/2705/CH4/EX4.2/Ex4_2.sce
new file mode 100755
index 000000000..d3c8e64dd
--- /dev/null
+++ b/2705/CH4/EX4.2/Ex4_2.sce
@@ -0,0 +1,42 @@
+clear;

+clc;

+disp('Example 4.2');

+

+//  aim : To determine 

+//  saturation temperature and enthalpy

+

+//  Given values

+P = 2.04;// pressure, [MN/m^2]

+

+//  solution

+//  since in the steam table values of enthalpy and saturation temperature at 2 and 2.1 MN?m^2 are   given, so for knowing required values at given pressure,there is need to do interpolation

+

+//  calculation of saturation temperature

+//  from steam table

+Table_P_tf = [[2.1,2.0];[214.9,212.4]]; //  P in [MN/m^2] and tf in [C]

+// using interpolation

+tf = interpln(Table_P_tf,2.04);//  saturation temperature at given condition

+mprintf('\n The Saturation temperature is  =  %f C \n',tf);

+

+// calculation of specific liquid enthalpy

+//  from steam table

+Table_P_hf = [[2.1,2.0];[920.0,908.6]];//  P in [MN/m^2] and hf in [kJ/kg]

+// using interpolation

+hf = interpln(Table_P_hf,2.04); //  enthalpy at given condition, [kJ/kg]

+mprintf('\n The Specific liquid enthalpy is  =  %f kJ/kg \n',hf);

+

+// calculation of specific enthalpy of evaporation

+//  from steam table

+Table_P_hfg = [[2.1,2.0];[1878.2,1888.6]];//  P in [MN/m^2] and hfg in [kJ/kg]

+// using interpolation

+hfg = interpln(Table_P_hfg,2.04); //  enthalpy at given condition, [kJ/kg]

+mprintf('\n The Specific enthalpy of evaporation is  =  %f kJ/kg \n',hfg);

+

+//  calculation of specific enthalpy of dry saturated steam

+//  from steam table

+Table_P_hg = [[2.1,2.0];[2798.2,2797.2]];//P in [MN/m^2] and hg in [kJ/kg]

+// using interpolation

+hg = interpln(Table_P_hg,2.04); //  enthalpy at given condition, [kJ/kg]

+mprintf('\n The Specific enthalpy of dry saturated steam is  =  %f kJ/kg \n',hg);

+

+//  End

diff --git a/2705/CH4/EX4.20/Ex4_20.sce b/2705/CH4/EX4.20/Ex4_20.sce
new file mode 100755
index 000000000..47eddb9ac
--- /dev/null
+++ b/2705/CH4/EX4.20/Ex4_20.sce
@@ -0,0 +1,76 @@
+clear;

+clc;

+disp('Example 4.20');

+

+//  aim : To determine the

+//  (a) mass of steam 

+//  (b) work transfer

+//  (c) change of internal energy

+//  (d) heat exchange b/w the steam and surroundings

+

+//  Given values

+P1 = 2.1;//  initial pressure,[MN/m^2]

+x1 = .9;//  dryness fraction

+V1 = .427;//  initial volume,[m^3]

+P2 = .7;//  final pressure,[MN/m^2]

+//  Given process is polytropic with

+n = 1.25; // polytropic index

+

+//  solution

+//  from steam table

+

+//  at 2.1 MN/m^2

+hf1 = 920.0;//  [kJ/kg]

+hfg1=1878.2;//  [kJ/kg]

+hg1=2798.2;//  [kJ/kg]

+vg1 = .0949;//  [m^3/kg]

+

+//  and at .7 MN/m^2

+hf2 = 697.1;//  [kJ/kg]

+hfg2 = 2064.9;//  [kJ/kg]

+hg2 = 2762.0;// [kJ/kg]

+vg2 = .273;//  [m^3/kg]

+

+//  (a)

+v1 = x1*vg1;//  [m^3/kg]

+m = V1/v1;//  [kg]

+mprintf('\n (a) The mass of steam present is  =  %f kg\n',m);

+

+//  (b)

+//  for polytropic process

+v2 = v1*(P1/P2)^(1/n);//  [m^3/kg]

+

+x2 = v2/vg2;//  final dryness fraction

+//  work transfer

+W = m*(P1*v1-P2*v2)*10^3/(n-1);//  [kJ]

+mprintf('\n (b) The work transfer is  =  %f  kJ\n',W);

+

+//  (c)

+//  initial

+h1 = hf1+x1*hfg1;//  [kJ/kg]

+u1 = h1-P1*v1*10^3;//  [kJ/kg]

+

+//  final

+h2 = hf2+x2*hfg2;//  [kJ/kg]

+u2 = h2-P2*v2*10^3;//  [kJ/kg]

+

+del_U = m*(u2-u1);//  [kJ]

+mprintf('\n (c) The change in internal energy is  =  %f kJ',del_U);

+if(del_U<0)

+    disp('since del_U<0,so this is loss of internal energy')

+else

+    disp('since del_U>0,so this is gain in internal energy')

+end

+

+//  (d)

+Q = del_U+W;//  [kJ]

+mprintf('\n (d) The heat exchange between the steam and surrounding is  =  %f  kJ',Q);

+if(Q<0)

+    disp('since Q<0,so this is loss of heat energy to surrounding')

+else

+    disp('since Q>0,so this is gain in heat energy to the steam')

+end

+

+// there are minor vairations in the values reported in the book

+

+//  End

diff --git a/2705/CH4/EX4.21/Ex4_21.sce b/2705/CH4/EX4.21/Ex4_21.sce
new file mode 100755
index 000000000..6aafe47b0
--- /dev/null
+++ b/2705/CH4/EX4.21/Ex4_21.sce
@@ -0,0 +1,42 @@
+clear;

+clc;

+disp('Example 4.21');

+

+//  aim : To determine the 

+//  (a) volume occupied by steam

+//  (b)(1) final dryness fraction of steam

+//       (2) Change of internal energy during expansion

+

+//  (a)

+//  Given values

+P1 = .85;//  [mN/m^2]

+x1 = .97;

+

+//  solution

+//  from steam table, at .85 MN/m^2,

+vg1 = .2268;//  [m^3/kg]

+//  hence

+v1 = x1*vg1;//  [m^3/kg]

+mprintf('\n (a) The volume occupied by 1 kg of steam is  =  %f m^3/kg\n',v1);

+

+// (b)(1)

+P2 = .17;//  [MN/m^2]

+// since process is polytropic process with

+n = 1.13; //  polytropic index

+// hence

+v2 = v1*(P1/P2)^(1/n);// [m^3/kg]

+

+// from steam table at .17 MN/m^2

+vg2 = 1.031;// [m^3/kg]

+// steam is wet so

+x2 = v2/vg2;//  final dryness fraction

+mprintf('\n (b)(1) The final dryness fraction of the steam is  =  %f \n',x2);

+

+//  (2)

+W = (P1*v1-P2*v2)*10^3/(n-1);// [kJ/kg]

+//  since process is adiabatic, so

+del_u = -W;// [kJ/kg]

+mprintf('\n     (2) The change in internal energy of the steam during expansion is  =  %f kJ/kg  (This is a loss of internal energy)\n',del_u);

+// There are minor variation in the answer

+

+//  End

diff --git a/2705/CH4/EX4.3/Ex4_3.sce b/2705/CH4/EX4.3/Ex4_3.sce
new file mode 100755
index 000000000..7b6b2dd19
--- /dev/null
+++ b/2705/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+disp('Example 4.3');

+

+//  aim : To determine

+//  the specific enthalpy

+

+//  given values

+P = 2; //  pressure ,[MN/m^2]

+t = 250; //  Temperature, [C]

+cp = 2.0934; //  average value of specific heat capacity, [kJ/kg K]

+

+//  solution

+

+//  looking up steam table it shows that at given pressure saturation temperature is 212.4 C,so

+tf = 212.4; //  [C]

+//  hence,

+Degree_of_superheat = t-tf;//  [C]

+//  from table at given temperature 250 C

+h = 2902; //  specific enthalpy of steam at 250 C ,[kJ/kg]

+mprintf('\nThe specific enthalpy of steam at 2 MN/m^2 with temperature 250 C is  =  %f  kJ/kg \n',h);

+

+//  Also from steam table enthalpy at saturation temperature is

+hf = 2797.2 ;//  [kJ/kg]

+//  so enthalpy at given temperature is

+h = hf+cp*(t-tf);//  [kJ/kg]

+

+mprintf('\n The specific enthalpy at given T and P by alternative path is  =  %f kJ/kg \n',h);

+

+//  End

diff --git a/2705/CH4/EX4.4/Ex4_4.sce b/2705/CH4/EX4.4/Ex4_4.sce
new file mode 100755
index 000000000..2af9a5594
--- /dev/null
+++ b/2705/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,43 @@
+clear;

+clc;

+disp('Example 4.4');

+

+//  aim : To determine

+//  the specific enthalpy of steam

+

+//  Given values

+P = 2.5;//  pressure, [MN/m^2]

+t = 320; //  temperature, [C]

+

+//  solution

+//  from steam table at given condition the saturation temperature of steam is 223.9 C, therefore steam is superheated

+tf = 223.9;//  [C]

+

+//  first let's calculate estimated enthalpy

+//  again from steam table 

+

+hg = 2800.9;//  enthalpy at saturation temp, [kJ/kg]

+cp =2.0934;//  specific heat capacity of steam,[kJ/kg K]

+

+//  so enthalpy at given condition is

+h = hg+cp*(t-tf);//  [kJ/kg]

+mprintf('\n The estimated specific enthalpy is  =  %f kJ/kg \n',h);

+

+//  calculation of accurate specific enthalpy

+//  we need double interpolation for this

+

+//  first interpolation w.r.t. to temperature

+//  At 2 MN/m^2

+Table_t_h = [[325,300];[3083,3025]];// where, t in [C] and h in [kJ/kg]

+h1 = interpln(Table_t_h,320); //  [kJ/kg]

+

+//  at 4 MN/m^2

+Table_t_h = [[325,300];[3031,2962]]; //  t in [C] and h in [kJ/kg]

+h2 = interpln(Table_t_h,320); //  [kJ/kg]

+

+//  now interpolation w.r.t. pressure

+Table_P_h = [[4,2];[h2,h1]]; //  where P in NM/m^2 and h1,h2 in kJ/kg

+h = interpln(Table_P_h,2.5);//  [kJ/kg]

+mprintf('\n The accurate specific enthalpy of steam at pressure of 2.5 MN/m^2 and with a temperature 320 C is  =   %f kJ/kg \n',h);

+

+//  End

diff --git a/2705/CH4/EX4.5/Ex4_5.sce b/2705/CH4/EX4.5/Ex4_5.sce
new file mode 100755
index 000000000..c2c030046
--- /dev/null
+++ b/2705/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp('Example 4.5');

+

+// aim : To determine 

+//  the specific enthalpy 

+

+//  Given values

+P = 70; //  pressure, [kn/m^2]

+x = .85; //  Dryness fraction

+

+// solution

+

+//  from steam table, at given pressure 

+hf = 376.8;//  [kJ/kg]

+hfg = 2283.3;//  [kJ/kg]

+

+// now using equation [2]

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

+

+mprintf('\n The specific enthalpy of wet steam is  =  %f kJ/kg \n',h);

+

+// There is minor variation in the book's answer

+

+//  End

diff --git a/2705/CH4/EX4.8/Ex4_8.sce b/2705/CH4/EX4.8/Ex4_8.sce
new file mode 100755
index 000000000..7cfe99980
--- /dev/null
+++ b/2705/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,20 @@
+clear;

+clc;

+disp('Example 4.8');

+

+//  aim : To determine 

+//  the specific volume of wet steam

+

+//  Given values

+P = 1.25; //  pressure, [MN/m^2]

+x = .9; //  dry fraction

+

+//  solution

+//  from steam table at given pressure

+vg = .1569;//  [m^3/kg]

+//  hence

+v = x*vg; //  [m^3/kg]

+

+mprintf('\nThe specific volume of wet steam is  =  %f  m^3/kg \n',v);

+

+//  End

diff --git a/2705/CH4/EX4.9/Ex4_9.sce b/2705/CH4/EX4.9/Ex4_9.sce
new file mode 100755
index 000000000..d9417bfea
--- /dev/null
+++ b/2705/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+disp('Example 4.9');

+

+//  aim : To determine

+//  the specific volume 

+

+//  Given values

+t = 325; //  temperature, [C]

+P = 2; //  pressure, [MN/m^2]

+

+//  solution

+// from steam table at given t and P

+vf = .1321; //  [m^3/kg]

+tf = 212.4; //  saturation temperature, [C]

+

+mprintf('\n The specific volume of steam at pressure of 2 MN/m^2 and with temperature 325 C is  =  %f  m^3/kg \n',vf);

+doh= t-tf; //  degree of superheat, [C]

+mprintf('\n The degree of superheat is  =  %f C\n',doh);

+

+// End 

diff --git a/2705/CH5/EX5.1/Ex5_1.sce b/2705/CH5/EX5.1/Ex5_1.sce
new file mode 100755
index 000000000..dd8a53487
--- /dev/null
+++ b/2705/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,23 @@
+clear;

+clc;

+disp('Example 5.1');

+

+//  aim : To determine 

+//  new pressure exerted on the air and the difference in two mercury column level

+

+//  Given values

+P1 = 765;//  atmospheric pressure, [mmHg]

+V1 = 20000;//  [mm^3]

+V2 = 17000;//  [mm^3]

+

+//  solution

+

+//  using boyle's law P*V=constant

+//  hence

+P2 = P1*V1/V2;//  [mmHg]

+mprintf('\n The new pressure exerted on the air is  =  %f  mmHg \n',P2);

+

+del_h = P2-P1;//  difference in Height of mercury column level

+mprintf('\n The difference in the two mercury column level is  =  %f  mm\n',del_h);

+

+//  End

diff --git a/2705/CH5/EX5.10/Ex5_10.sce b/2705/CH5/EX5.10/Ex5_10.sce
new file mode 100755
index 000000000..b5d45cd32
--- /dev/null
+++ b/2705/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,27 @@
+clear;

+clc;

+disp('Example 5.10');

+

+//  aim : To determine

+//  the new temperature of the gas

+

+//  Given values

+V1 = .015;//  original volume,[m^3]

+T1 = 273+285;//  original temperature,[K]

+V2 = .09;//  final volume,[m^3]

+

+//  solution 

+//  Given gas is following the law,P*V^1.35=constant

+//  so process is polytropic with

+n = 1.35; // polytropic index

+

+// hence

+T2 = T1*(V1/V2)^(n-1);//  final temperature, [K]

+

+t2 = T2-273;//  [C]

+

+mprintf('\n The new temperature of the gas is  =  %f C \n',t2);

+

+// there is minor error in book's answer

+

+//  End

diff --git a/2705/CH5/EX5.11/Ex5_11.sce b/2705/CH5/EX5.11/Ex5_11.sce
new file mode 100755
index 000000000..42acb405c
--- /dev/null
+++ b/2705/CH5/EX5.11/Ex5_11.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+disp('Example 5.11');

+

+//  aim : To determine the

+//  (a) original and final volume of the gas

+//  (b) final pressure of the gas

+//  (c) final temperature of the gas

+

+//  Given values

+m = .675;//  mass of the gas,[kg]

+P1 = 1.4;//  original pressure,[MN/m^2]

+T1 = 273+280;//  original temperature,[K]

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

+

+//  solution

+

+//  (a)

+//  using characteristic equation, P1*V1=m*R*T1

+V1 = m*R*T1*10^-3/P1;//  [m^3]

+//  also Given 

+V2 =  4*V1;// [m^3]

+mprintf('\n (a) The original volume of the gas is  =  %f  m^3\n',V1);

+mprintf('\n      and The final volume of the gas is  =  %f  m^3\n',V2);

+

+//  (b)

+//  Given that gas is following the law P*V^1.3=constant

+//  hence process is polytropic with 

+n = 1.3; //  polytropic index

+P2 = P1*(V1/V2)^n;//  formula for polytropic process,[MN/m^2]

+mprintf('\n (b) The final pressure of the gas is  =  %f  kN/m^2\n',P2*10^3);

+

+//  (c)

+//  since mass is constant so,using P*V/T=constant

+//  hence

+T2 = P2*V2*T1/(P1*V1);//  [K]

+t2 = T2-273;//  [C]

+mprintf('\n (c) The final temperature of the gas is  =  %f C\n',t2);

+

+//  End

diff --git a/2705/CH5/EX5.12/Ex5_12.sce b/2705/CH5/EX5.12/Ex5_12.sce
new file mode 100755
index 000000000..e522db8a2
--- /dev/null
+++ b/2705/CH5/EX5.12/Ex5_12.sce
@@ -0,0 +1,57 @@
+clear;

+clc;

+disp('Example 5.12');

+

+//  aim : T0 determine 

+//  (a) change in internal nergy of the air

+//  (b) work done

+//  (c) heat transfer

+

+//  Given values

+m = .25;//  mass, [kg]

+P1 = 140;//  initial pressure, [kN/m^2]

+V1 = .15;//  initial volume, [m^3]

+P2 = 1400;//  final volume, [m^3]

+cp = 1.005;//  [kJ/kg K]

+cv = .718;//  [kJ/kg K]

+

+//  solution

+

+//  (a)

+//  assuming ideal gas

+R = cp-cv;//  [kJ/kg K]

+//  also, P1*V1=m*R*T1,hence

+T1 = P1*V1/(m*R);//  [K]

+

+//  given that process is polytropic with 

+n = 1.25; //  polytropic index

+T2 = T1*(P2/P1)^((n-1)/n);//  [K]

+

+//  Hence, change in internal energy is,

+del_U = m*cv*(T2-T1);//  [kJ]

+mprintf('\n (a) The change in internal energy of the air is del_U  =  %f  kJ',del_U);

+if(del_U>0)

+    disp('since del_U>0, so it is gain of internal energy to the air')

+else

+    disp('since del_U<0, so it is gain of internal energy to the surrounding')

+end

+

+//  (b)

+W = m*R*(T1-T2)/(n-1);//  formula of work done for polytropic process,[kJ]

+mprintf('\n (b) The work done is W =  %f  kJ',W);

+if(W>0)

+    disp('since W>0, so the work is done by  the air')

+else

+    disp('since W<0, so the work is done on the air')

+end

+

+//  (c)

+Q = del_U+W;//  using 1st law of thermodynamics,[kJ]

+mprintf('\n (c) The heat transfer is Q  =  %f  kJ',Q);

+if(Q>0)

+    disp('since Q>0, so the heat is received by  the air')

+else

+    disp('since Q<0, so the heat is rejected by the air')

+end

+

+//  End

diff --git a/2705/CH5/EX5.13/Ex5_13.sce b/2705/CH5/EX5.13/Ex5_13.sce
new file mode 100755
index 000000000..e2fc9c1b2
--- /dev/null
+++ b/2705/CH5/EX5.13/Ex5_13.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+disp('Example 5.13');

+

+//  aim : To determine the

+//  final volume, work done and the change in internal energy

+

+//  Given values

+P1 = 700;//  initial pressure,[kN/m^2]

+V1 = .015;// initial volume, [m^3]

+P2 = 140;//  final pressure, [kN/m^2]

+cp = 1.046;//  [kJ/kg K]

+cv = .752; //  [kJ/kg K]

+

+//  solution

+

+Gamma = cp/cv;

+//  for adiabatic expansion, P*V^gamma=constant, so

+V2 = V1*(P1/P2)^(1/Gamma);//  final volume, [m^3]

+mprintf('\n The final volume of the gas is V2  =  %f  m^3\n',V2);

+

+//  work done

+W = (P1*V1-P2*V2)/(Gamma-1);//  [kJ]

+mprintf('\n The work done by the gas is  =  %f  kJ\n',W);

+

+//  for adiabatic process

+del_U = -W;//  [kJ]

+mprintf('\n The change of internal energy is  =  %f kJ',del_U);

+if(del_U>0)

+    disp('since del_U>0, so the the gain in internal energy of the gas ')

+else

+    disp('since del_U<0, so this is a loss of internal energy from the gas')

+end

+

+//  End

diff --git a/2705/CH5/EX5.14/Ex5_14.sce b/2705/CH5/EX5.14/Ex5_14.sce
new file mode 100755
index 000000000..25771ce9f
--- /dev/null
+++ b/2705/CH5/EX5.14/Ex5_14.sce
@@ -0,0 +1,54 @@
+clear;

+clc;

+disp('Example 5.14');

+

+//  aim : To determine the

+//  (a)heat transfer

+//  (b)change of internal energy

+//  (c)mass of gas

+

+//   Given values

+V1 = .4;//  initial volume, [m^3]

+P1 = 100;//  initial pressure, [kN/m^2]

+T1 = 273+20;//  temperature, [K]

+P2 = 450;//  final pressure,[kN/m^2]

+cp = 1.0;//  [kJ/kg K]

+Gamma = 1.4; //  heat capacity ratio

+

+//  solution

+

+//  (a)

+//  for the isothermal compression,P*V=constant,so

+V2 = V1*P1/P2;//  [m^3]

+W = P1*V1*log(P1/P2);//  formula of workdone for isothermal process,[kJ]

+

+//  for isothermal process, del_U=0;so

+Q = W;

+mprintf('\n (a) The heat transferred during compression is Q  =  %f kJ\n',Q);

+

+

+//  (b)

+V3 = V1;

+//  for adiabatic expansion

+//  also

+

+P3 = P2*(V2/V3)^Gamma;//  [kN/m^2]

+W = -(P3*V3-P2*V2)/(Gamma-1);//  work done formula for adiabatic process,[kJ]

+//  also, Q=0,so using Q=del_U+W

+del_U = -W;//  [kJ]

+mprintf('\n (b) The change of the internal energy during the expansion is,del_U  =  %f kJ\n',del_U);

+

+//  (c)

+//  for ideal gas

+//  cp-cv=R, and cp/cv=gamma, hence

+R = cp*(1-1/Gamma);//  [kj/kg K]

+

+//  now using ideal gas equation

+m = P1*V1/(R*T1);//  mass of the gas,[kg]

+mprintf('\n (c) The mass of the gas is,m  =  %f  kg\n',m);

+

+// There is calculation mistake in the book

+

+

+//  End

+

diff --git a/2705/CH5/EX5.15/Ex5_15.sce b/2705/CH5/EX5.15/Ex5_15.sce
new file mode 100755
index 000000000..e5ae9aa6a
--- /dev/null
+++ b/2705/CH5/EX5.15/Ex5_15.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+disp('Example 5.15');

+

+//  aim : To determine 

+//  the heat transferred and polytropic specific heat capacity

+

+//  Given values

+P1 = 1;//  initial pressure, [MN/m^2]

+V1 = .003;//  initial volume, [m^3]

+P2 = .1;//  final pressure,[MN/m^2]

+cv = .718;//  [kJ/kg*K]

+Gamma=1.4;//  heat capacity ratio

+

+//  solution

+//  Given process is polytropic with

+n = 1.3;//  polytropic index

+//  hence

+V2 = V1*(P1/P2)^(1/n);//  final volume,[m^3]

+W = (P1*V1-P2*V2)*10^3/(n-1);//  work done,[kJ]

+//  so

+Q = (Gamma-n)*W/(Gamma-1);//  heat transferred,[kJ]

+

+mprintf('\n The heat received or rejected by the gas during this process is Q  =  %f kJ',Q);

+if(Q>0)

+    disp('since Q>0, so heat is received by the gas')

+else

+    disp('since Q<0, so heat is rejected by the gas')

+end

+

+//  now

+cn = cv*(Gamma-n)/(n-1);//  polytropic specific heat capacity,[kJ/kg K]

+mprintf('\n The polytropic specific heat capacity is cn  =  %f kJ/kg K\n',cn);

+

+//  End

diff --git a/2705/CH5/EX5.16/Ex5_16.sce b/2705/CH5/EX5.16/Ex5_16.sce
new file mode 100755
index 000000000..37712f1ac
--- /dev/null
+++ b/2705/CH5/EX5.16/Ex5_16.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+disp('Example 5.16');

+

+//  aim : To determine the 

+//  (a) initial partial pressure of the steam and air

+//  (b) final partial pressure of the steam and air

+//  (c) total pressure in the container after heating

+

+//  Given values

+T1 = 273+39;//  initial temperature,[K]

+P1 = 100;//  pressure, [MN/m^2]

+T2 = 273+120.2;//  final temperature,[K]

+

+//  solution

+

+//  (a)

+//  from the steam tables, the pressure of wet steam at 39 C is

+Pw1 = 7;//  partial pressure of wet steam,[kN/m^2]

+//  and by Dalton's law

+Pa1 = P1-Pw1;//  initial pressure of air, [kN/m^2]

+

+mprintf('\n (a) The initial partial pressure of the steam is  =  %f  kN/m^2',Pw1);

+mprintf('\n      The initial partial pressure of the air is  =  %f kN/m^2\n',Pa1);

+

+//  (b)

+//  again from steam table, at 120.2 C the pressure of wet steam is

+Pw2 = 200;//  [kN/m^2]

+

+//  now since volume is constant so assuming air to be ideal gas so for air  P/T=contant, hence

+Pa2 = Pa1*T2/T1 ;//  [kN/m^2]

+

+mprintf(' \n(b) The final partial pressure of the steam is  =  %f kN/m^2',Pw2);

+mprintf('\n     The final partial pressure of the air is  =  %f  kN/m^2\n',Pa2);

+

+//  (c)

+Pt = Pa2+Pw2;//  using dalton's law, total pressure,[kN/m^2]

+mprintf('\n (c) The total pressure after heating is  =  %f  kN/m^2\n',Pt);

+

+//  End

diff --git a/2705/CH5/EX5.17/Ex5_17.sce b/2705/CH5/EX5.17/Ex5_17.sce
new file mode 100755
index 000000000..187f6250a
--- /dev/null
+++ b/2705/CH5/EX5.17/Ex5_17.sce
@@ -0,0 +1,40 @@
+clear;

+clc;

+disp('Example 5.17');

+

+//  aim : To determine 

+//  the partial pressure of the air and steam, and the mass of the air

+

+//  Given values

+P1 = 660;//  vaccum gauge pressure on condenser [mmHg]

+P = 765;//  atmospheric pressure, [mmHg]

+x = .8;//  dryness fraction 

+T = 273+41.5;//  temperature,[K]

+ms_dot = 1500;//  condense rate of steam,[kg/h]

+R = .29;//  [kJ/kg]

+

+//  solution

+Pa = (P-P1)*.1334;// absolute pressure,[kN/m^2]

+// from steam table, at 41.5 C partial pressure of steam is

+Ps = 8;//  [kN/m^2]

+//  by dalton's law, partial pressure of air is

+Pg = Pa-Ps;//  [kN/m^2]

+

+mprintf('\n The partial pressure of the air in the condenser is  =  %f  kN/m^2\n',Pg);

+mprintf('\n The partial pressure of the steam in the condenser is  =  %f kN/m^2\n',Ps);

+

+//  also

+vg = 18.1;//  [m^3/kg]

+//  so

+V = x*vg;//  [m^3/kg]

+//  The air associated with 1 kg of the steam will occupiy this same volume

+//  for air, Pg*V=m*R*T,so

+m = Pg*V/(R*T);//  [kg/kg steam]

+//  hence

+ma = m*ms_dot;//  [kg/h]

+

+mprintf('\n The mass of air which will associated with this steam is  =  %f  kg\n',ma);

+

+// There is misprint in book

+

+//  End

diff --git a/2705/CH5/EX5.18/Ex5_18.sce b/2705/CH5/EX5.18/Ex5_18.sce
new file mode 100755
index 000000000..3266c108e
--- /dev/null
+++ b/2705/CH5/EX5.18/Ex5_18.sce
@@ -0,0 +1,41 @@
+clear;

+clc;

+disp('Example 5.18');

+

+//  aim : To determine the

+//  (a) final pressure

+//   (b) final dryness fraction of the steam

+

+//  Given values

+P1 = 130;//  initial pressure, [kN/m^2]

+T1 = 273+75.9;//  initial temperature, [K]

+x1 = .92;//  initial dryness fraction

+T2 = 273+120.2;//  final temperature, [K]

+

+//  solution

+

+//  (a)

+//  from steam table, at 75.9 C

+Pws = 40;//  partial pressure of wet steam[kN/m^2]

+Pa = P1-Pws;//  partial pressure of air, [kN/m^2]

+vg = 3.99//  specific volume of the wet steam, [m^3/kg]

+// hence

+V1 = x1*vg;//  [m^3/kg]

+V2 = V1/5;//  [m^3/kg]

+//  for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so

+P2 = Pa*V1*T2/(V2*T1);//  final pressure,[kN/m^2]

+

+//  now for steam at 120.2 C

+Ps = 200;//  final partial pressure of steam,[kN/m^2]

+//  so by dalton's law total pressure in cylindert is

+Pt = P2+Ps;//  [kN/m^2]

+mprintf('\n (a) The final pressure in the cylinder is  =  %f  kN/m^2\n',Pt);

+

+//  (b)

+//  from steam table at 200 kN/m^2 

+vg = .885;//  [m^3/kg]

+//  hence

+x2 = V2/vg;//  final dryness fraction of the steam

+mprintf('\n (b) The final dryness fraction of the steam is  =  %f\n ',x2);

+

+//  End

diff --git a/2705/CH5/EX5.19/Ex5_19.sce b/2705/CH5/EX5.19/Ex5_19.sce
new file mode 100755
index 000000000..830fd787f
--- /dev/null
+++ b/2705/CH5/EX5.19/Ex5_19.sce
@@ -0,0 +1,39 @@
+clear;

+clc;

+disp('Example 5.19')

+

+//  aim : To determine the 

+//  (a) Gamma,

+//  (b) del_U

+

+//  Given Values

+P1 = 1400;//  [kN/m^2]

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

+P3 = 220;//  [kN/m^2]

+T1 = 273+360;//  [K]

+m = .23;// [kg]

+cp = 1.005;//  [kJ/kg*K]

+

+//  Solution

+T3 = T1;//  since process 1-3 is isothermal

+

+//  (a)

+//  for process 1-3, P1*V1=P3*V3,so

+V3_by_V1 = P1/P3;

+//  also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence

+//  and process process 2-3 is iso-choric so,V3=V2 and

+V2_by_V1 = V3_by_V1;

+//  hence,

+Gamma = log(P1/P2)/log(P1/P3); //  heat capacity ratio

+

+mprintf('\n (a) The value of adiabatic index Gamma is  =  %f\n',Gamma);

+

+//  (b)

+cv = cp/Gamma;//  [kJ/kg K]

+//  for process 2-3,P3/T3=P2/T2,so

+T2 = P2*T3/P3;//  [K]

+

+//   now

+del_U = m*cv*(T2-T1);//  [kJ]

+mprintf('\n (b) The change in internal energy during the adiabatic expansion is U2-U1 =  %f kJ (This is loss of internal energy)\n',del_U);

+//   End

diff --git a/2705/CH5/EX5.2/Ex5_2.sce b/2705/CH5/EX5.2/Ex5_2.sce
new file mode 100755
index 000000000..3e80ae408
--- /dev/null
+++ b/2705/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,20 @@
+clear;

+clc;

+disp('Example 5.2');

+

+//  aim : To determine 

+//  the new volume

+

+//  Given values

+P1 = 300;//  original pressure,[kN/m^2]

+V1 = .14;//  original volume,[m^3]

+

+P2 = 60;//  new pressure after expansion,[kn/m^2]

+

+//  solution

+//  since temperature is constant so using boyle's law P*V=constant

+V2 = V1*P1/P2;//  [m^3]

+

+mprintf('\n The new volume after expansion is  =  %f  m^3\n',V2);

+

+//  End

diff --git a/2705/CH5/EX5.20/Ex5_20.sce b/2705/CH5/EX5.20/Ex5_20.sce
new file mode 100755
index 000000000..daf3343a7
--- /dev/null
+++ b/2705/CH5/EX5.20/Ex5_20.sce
@@ -0,0 +1,43 @@
+clear;

+clc;

+disp('Example 5.20');

+

+//  aim : To determine 

+//  the mass of oxygen and heat transferred

+

+//  Given values

+V1 = 300;//  [L]

+P1 = 3.1;//  [MN/m^2]

+T1 = 273+18;//  [K]

+P2 = 1.7;//  [MN/m^2]

+T2 = 273+15;//  [K]

+Gamma = 1.4; //  heat capacity ratio

+//  density condition

+P = .101325;// [MN/m^2]

+T = 273;// [K]

+V = 1;//  [m^3]

+m = 1.429;// [kg]

+

+//  hence

+R = P*V*10^3/(m*T);//  [kJ/kg*K]

+//  since volume is constant

+V2 = V1;// [L]

+//  for the initial conditions in the cylinder,P1*V1=m1*R*T1

+m1 = P1*V1/(R*T1);//  [kg]

+

+// after some of the gas is used

+m2 = P2*V2/(R*T2);//  [kg]

+//  The mass of oxygen remaining in cylinder is m2 kg,so

+//  Mass of oxygen used is

+m_used = m1-m2;//  [kg]

+mprintf('\n The mass of oxygen used  =  %f kg\n',m_used);

+

+//  for non-flow process,Q=del_U+W

+//  volume is constant so no external work is done so,Q=del_U

+cv = R/(Gamma-1);//  [kJ/kg*K]

+

+//  heat transfer is

+Q = m2*cv*(T1-T2);//  (kJ)

+mprintf('\n The amount of heat transferred through the cylinder wall is  =  %f kJ\n',Q);

+

+//  End

diff --git a/2705/CH5/EX5.21/Ex5_21.sce b/2705/CH5/EX5.21/Ex5_21.sce
new file mode 100755
index 000000000..a5cdcb432
--- /dev/null
+++ b/2705/CH5/EX5.21/Ex5_21.sce
@@ -0,0 +1,43 @@
+clear;

+clc;

+disp('Example 5.21');

+

+//  aim : To determine the

+//  (a) work transferred during the compression

+//  (b) change in internal energy

+//  (c) heat transferred during the compression

+

+//  Given values

+V1 = .1;//  initial volume, [m^3]

+P1 = 120;//  initial pressure, [kN/m^2]

+P2 = 1200; // final pressure, [kN/m^2]

+T1 = 273+25;//  initial temperature, [K]

+cv = .72;//  [kJ/kg*K]

+R = .285;//  [kJ/kg*K]

+

+//  solution

+

+//  (a)

+//  given process is polytropic with

+n = 1.2;  // polytropic index

+//  hence

+V2 = V1*(P1/P2)^(1/n);//  [m^3]

+W = (P1*V1-P2*V2)/(n-1);//  workdone formula, [kJ]

+mprintf('\n (a) The work transferred during the compression is  =  %f  kJ\n',W);

+

+//  (b)

+//  now mass is constant so,

+T2 = P2*V2*T1/(P1*V1);//  [K]

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

+m = P1*V1/(R*T1);//  [kg]

+

+//  change in internal energy is

+del_U = m*cv*(T2-T1);//  [kJ]

+mprintf('\n (b) The change in internal energy is  =  %f  kJ\n',del_U);

+

+//  (c)

+Q = del_U+W;//  [kJ]

+mprintf('\n (c) The heat transferred during the compression is  =  %f  kJ\n',Q);

+

+//  End

+

diff --git a/2705/CH5/EX5.22/Ex5_22.sce b/2705/CH5/EX5.22/Ex5_22.sce
new file mode 100755
index 000000000..756233f57
--- /dev/null
+++ b/2705/CH5/EX5.22/Ex5_22.sce
@@ -0,0 +1,36 @@
+clear;

+clc;

+disp('Example 5.22');

+

+//  aim : To determine the

+//  (a) new pressure of the air in the receiver

+//  (b) specific enthalpy of air at 15 C

+

+//  Given values

+V1 = .85;//  [m^3]

+T1 = 15+273;//  [K]

+P1 = 275;// pressure,[kN/m^2]

+m = 1.7;//  [kg]

+cp = 1.005;//  [kJ/kg*K]

+cv = .715;//  [kJ/kg*K]

+

+//  solution

+

+//  (a)

+

+R = cp-cv;//  [kJ/kg*K]

+//  assuming m1 is original mass of the air, using P*V=m*R*T

+m1 = P1*V1/(R*T1);//  [kg]

+m2 = m1+m;//  [kg]

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

+//  P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so

+P2 = P1*m2/m1;//  [kN/m^2]

+mprintf('\n (a) The new pressure of the air in the receiver is = %f kN/m^2\n',P2);

+

+//  (b)

+//  for 1 kg of air, h2-h1=cp*(T1-T0)

+//  and if 0 is chosen as the zero enthalpy, then

+h = cp*(T1-273);//  [kJ/kg]

+mprintf('\n (b) The specific  enthalpy of the air at 15 C is =  %f kJ/kg\n',h);

+

+//  End

diff --git a/2705/CH5/EX5.23/Ex5_23.sce b/2705/CH5/EX5.23/Ex5_23.sce
new file mode 100755
index 000000000..69c2910b0
--- /dev/null
+++ b/2705/CH5/EX5.23/Ex5_23.sce
@@ -0,0 +1,47 @@
+clear;

+clc;

+disp('Example 5.23');

+

+//  aim : T determine the

+//  (a) characteristic gas constant of the gas

+//  (b) cp,

+//  (c) cv,

+//  (d) del_u 

+//  (e) work transfer

+

+//  Given values

+P = 1;//  [bar] 

+T1 = 273+15;//  [K]

+m = .9;//  [kg]

+T2 = 273+250;//  [K]

+Q = 175;//  heat transfer,[kJ]

+

+//  solution

+

+//  (a)

+//  using, P*V=m*R*T, given,

+m_by_V = 1.875;

+//  hence

+R = P*100/(T1*m_by_V);//  [kJ/kg*K]

+mprintf('\n (a) The characteristic gas constant of the gas is R  =  %f kJ/kg K\n',R);

+

+//  (b)

+//  using, Q=m*cp*(T2-T1)

+cp = Q/(m*(T2-T1));//  [kJ/kg K]

+mprintf('\n (b) The specific heat capacity of the gas at constant pressure cp  =  %f kJ/kg K\n',cp);

+

+// (c)

+//  we have, cp-cv=R,so

+cv = cp-R;// [kJ/kg*K]

+mprintf('\n (c) The specific heat capacity of the gas at constant volume cv  =  %f  kJ/kg K\n',cv);

+

+//  (d)

+del_U = m*cv*(T2-T1);//  [kJ]

+mprintf('\n (d) The change in internal energy is  =  %f kJ\n',del_U);

+

+//  (e)

+// using, Q=del_U+W

+W = Q-del_U;//  [kJ]

+mprintf('\n (e) The work transfer is W  =  %f kJ\n',W);

+

+//  End

diff --git a/2705/CH5/EX5.24/Ex5_24.sce b/2705/CH5/EX5.24/Ex5_24.sce
new file mode 100755
index 000000000..70324999b
--- /dev/null
+++ b/2705/CH5/EX5.24/Ex5_24.sce
@@ -0,0 +1,42 @@
+clear;

+clc;

+disp('Example 5.24');

+

+//  aim : To determine the

+//  (a) work transfer,

+//  (b)del_U  and,

+//  (c)heat transfer

+

+//  Given values

+V1 = .15;//  [m^3]

+P1 = 1200;//  [kN/m^2]

+T1 = 273+120;// [K]

+P2 = 200;//  [kN/m^2]

+cp = 1.006;//[kJ/kg K]

+cv = .717;//  [kJ/kg K]

+

+//  solution

+

+//  (a)

+//  Given, PV^1.32=constant, so it is polytropic process with

+n = 1.32;//  polytropic index

+// hence

+V2 = V1*(P1/P2)^(1/n);//  [m^3]

+//  now, W

+W = (P1*V1-P2*V2)/(n-1);// [kJ]

+mprintf('\n (a) The work transfer is W  =  %f kJ\n',W);

+

+//  (b)

+R = cp-cv;//  [kJ/kg K]

+m = P1*V1/(R*T1);//  gas law,[kg]

+//  also for polytropic process

+T2 = T1*(P2/P1)^((n-1)/n);//  [K]

+//  now for gas,

+del_U = m*cv*(T2-T1);//  [kJ]

+mprintf('\n (b) The change of internal energy is del_U =  %f kJ\n',del_U);

+

+//  (c)

+Q = del_U+W;//  first law of thermodynamics,[kJ]

+mprintf('\n (c) The heat transfer Q  =  %f kJ\n',Q);

+

+//  End

diff --git a/2705/CH5/EX5.26/Ex5_26.sce b/2705/CH5/EX5.26/Ex5_26.sce
new file mode 100755
index 000000000..83f0cc59c
--- /dev/null
+++ b/2705/CH5/EX5.26/Ex5_26.sce
@@ -0,0 +1,45 @@
+clear;

+clc;

+disp('Example 5.26');

+

+//  aim : To determine

+//  the volume of the pressure vessel and the volume of the gas before transfer

+

+//  Given values

+

+P1 = 1400;//  initial pressure,[kN/m^2]

+T1 = 273+85;//  initial temperature,[K]

+

+P2 = 700;//  final pressure,[kN/m^2]

+T2 = 273+60;//  final temperature,[K]

+

+m = 2.7;// mass of the gas passes,[kg]

+cp = .88;//  [kJ/kg]

+cv = .67;//  [kJ/kg]

+

+//  solution

+

+//  steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W  [1], 

+//  given, there is no kinetic energy change and neglecting potential energy term

+W = 0;// no external work done

+//  so final equation is,u1+P1*v1+Q=u2   [2]

+// also u2-u1=cv*(T2-T1)

+// hence Q=cv*(T2-T1)-P1*v1    [3]

+//  and for unit mass P1*v1=R*T1=(cp-cv)*T1  [4]

+//  so finally

+Q = cv*(T2-T1)-(cp-cv)*T1;//  [kJ/kg]

+//  so total heat transferred is

+Q = m*Q;//  [kJ] 

+

+//  using eqn [4]

+v1 = (cp-cv)*T1/P1;//  [m^3/kg]

+//  Total volume is

+V1 = m*v1;//  [m^3]

+

+//  using ideal gas equation P1*V1/T1=P2*V2/T2

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

+

+mprintf('\n The volume of gas before transfer is  =  %f  m^3\n',V1);

+mprintf('\n The volume of pressure vessel is  =  %f  m^3\n',V2);

+ 

+//  End

diff --git a/2705/CH5/EX5.3/Ex5_3.sce b/2705/CH5/EX5.3/Ex5_3.sce
new file mode 100755
index 000000000..71188b43b
--- /dev/null
+++ b/2705/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,20 @@
+clear;

+clc;

+disp('Example 5.3');

+

+//  aim : To determine 

+//  the new volume of the gas

+

+//  Given values

+V1 = 10000;//  [mm^3]

+T1 = 273+18;//  [K]

+T2 = 273+85;//  [K]

+

+//  solution

+//  since pressure exerted on the apparatus is constant so using charle's law V/T=constant

+//  hence

+V2 = V1*T2/T1;//  [mm^3]

+

+mprintf('\n The new volume of the gas trapped in the apparatus is  =  %f  mm^3\n',V2);

+

+//  End

diff --git a/2705/CH5/EX5.4/Ex5_4.sce b/2705/CH5/EX5.4/Ex5_4.sce
new file mode 100755
index 000000000..954ced5f4
--- /dev/null
+++ b/2705/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,20 @@
+clear;

+clc;

+disp('Example 5.4');

+

+//  aim : To determine 

+//  the final temperature

+

+//  Given values

+V1 = .2;//  original volume,[m^3]

+T1 = 273+303;// original temperature, [K]

+V2 = .1;// final volume, [m^3]

+

+//  solution

+//  since pressure is constant, so using charle's law V/T=constant

+//  hence

+T2 = T1*V2/V1;//  [K]

+t2 = T2-273;//  [C]

+mprintf('\n The final temperature of the gas is  =  %f  C\n',t2);

+

+//  End

diff --git a/2705/CH5/EX5.5/Ex5_5.sce b/2705/CH5/EX5.5/Ex5_5.sce
new file mode 100755
index 000000000..894496057
--- /dev/null
+++ b/2705/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp('Example 5.5');

+

+//  aim : To determine 

+//  the new volume of the gas

+

+//  Given values

+

+//  initial codition

+P1 = 140;//  [kN/m^2]

+V1 = .1;//   [m^3]

+T1 = 273+25;// [K]

+

+//  final condition

+P2 = 700;//  [kN/m^2]

+T2 = 273+60;//  [K]

+

+//  by charasteristic equation, P1*V1/T1=P2*V2/T2

+

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

+

+mprintf('\nThe new volume of the gas is  =  %f  m^3\n',V2);

+

+//  End

diff --git a/2705/CH5/EX5.6/Ex5_6.sce b/2705/CH5/EX5.6/Ex5_6.sce
new file mode 100755
index 000000000..40f5fc559
--- /dev/null
+++ b/2705/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,29 @@
+clear;

+clc;

+disp('Example 5.6');

+

+//  aim : To determine

+//  the mas of the gas and new temperature

+

+//  Given values

+P1 = 350;//  [kN/m^2]

+V1 = .03;//  [m^3]

+T1 = 273+35;//  [K]

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

+

+//  solution

+//  using charasteristic equation, P*V=m*R*T

+m = P1*V1/(R*T1);//  [Kg]

+mprintf('\n The mass of the gas present is  =  %f kg\n',m);

+

+//  Now the gas is compressed

+P2 = 1050;//  [kN/m^2]

+V2 = V1;

+// since mass of the gas is constant so using, P*V/T=constant

+//  hence

+T2 = T1*P2/P1//  [K]

+t2 = T2-273;//  [C]

+

+mprintf('\n The new temperature of the gas is  =  %f  C\n',t2);

+

+//  End

diff --git a/2705/CH5/EX5.7/Ex5_7.sce b/2705/CH5/EX5.7/Ex5_7.sce
new file mode 100755
index 000000000..85cf1c2f9
--- /dev/null
+++ b/2705/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,28 @@
+clear;

+clc;

+disp('Example 5.7');

+

+//  aim : To determine 

+//  the heat transferred to the gas and its final pressure

+

+//  Given values

+m = 2;//  masss of the gas, [kg]

+V1 = .7;//  volume,[m^3]

+T1 = 273+15;//  original temperature,[K]

+T2 = 273+135;//  final temperature,[K]

+cv = .72;//  specific heat capacity at constant volume,[kJ/kg K]

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

+

+//  solution

+Q = m*cv*(T2-T1);//  Heat transferred at constant volume,[kJ]

+mprintf('\n The heat transferred to the gas is  =  %f  kJ\n',Q);

+

+//  Now,using P1*V1=m*R*T1

+P1 = m*R*T1/V1;//  [kN/m^2]

+

+//  since volume of the system is constant, so P1/T1=P2/T2

+//  hence

+P2 = P1*T2/T1;//  final pressure,[kN/m^2]

+mprintf('\n The final pressure of the gas is  =  %f  kN/m^2 \n',P2);

+

+//  End

diff --git a/2705/CH5/EX5.8/Ex5_8.sce b/2705/CH5/EX5.8/Ex5_8.sce
new file mode 100755
index 000000000..b99a3fdae
--- /dev/null
+++ b/2705/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,33 @@
+clear;

+clc;

+disp('Example 5.8');

+

+//  aim : To determine 

+//  the heat transferred from the gas and the work done on the gas

+

+//  Given values

+P1 = 275;//  pressure, [kN/m^2]

+V1 = .09;//  volume,[m^3]

+T1 = 273+185;//  initial temperature,[K]

+T2 = 273+15;//  final temperature,[K]

+cp = 1.005;//  specific heat capacity at constant pressure,[kJ/kg K]

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

+

+//  solution

+//  using P1*V1=m*R*T1

+m = P1*V1/(R*T1);//  mass of the gas

+

+//  calculation of heat transfer

+Q = m*cp*(T2-T1);//  Heat transferred at constant pressure,[kJ]

+mprintf('\n The heat transferred to the gas is  =  %f  kJ\n',Q);

+

+//  calculation of work done

+//  Now,since pressure is constant so, V/T=constant

+// hence

+V2 = V1*T2/T1;//  [m^3]

+

+W = P1*(V2-V1);//  formula for work done at constant pressure,[kJ]

+mprintf('\n Work done on the gas during the process is  =  %f  kJ\n',W);

+

+//  End

+

diff --git a/2705/CH5/EX5.9/Ex5_9.sce b/2705/CH5/EX5.9/Ex5_9.sce
new file mode 100755
index 000000000..016e3d2ee
--- /dev/null
+++ b/2705/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,22 @@
+clear;

+clc;

+disp('Example 5.9');

+

+//  aim : To determine

+//  the new pressure of the gas

+

+//  Given values

+P1 = 300;//  original pressure,[kN/m^2]

+T1 = 273+25;//  original temperature,[K]

+T2 = 273+180;//  final temperature,[K]

+

+//  solution

+//  since gas compressing according to the law,P*V^1.4=constant

+//  so,for polytropic process,T1/T2=(P1/P2)^((n-1)/n),here n=1.4

+

+//  hence

+P2 = P1*(T2/T1)^((1.4)/(1.4-1));//  [kN/m^2]

+

+mprintf('\n The new pressure of the gas is  =  %f  kN/m^2\n',P2);

+

+//  End

diff --git a/2705/CH7/EX7.1/Ex7_1.sce b/2705/CH7/EX7.1/Ex7_1.sce
new file mode 100755
index 000000000..3a5463218
--- /dev/null
+++ b/2705/CH7/EX7.1/Ex7_1.sce
@@ -0,0 +1,25 @@
+clear;

+clc;

+disp('Example 7.1');

+

+//  aim : To determine

+//  the specific enthalpy of water

+

+//  Given values

+Tf = 273+100;//  Temperature,[K]

+

+//  solution

+//  from steam table

+cpl = 4.187;// [kJ/kg K]

+//  using equation [8]

+sf = cpl*log(Tf/273.16);//  [kJ/kg*K]

+mprintf('\n The specific entropy of water is  =  %f kJ/kg K\n',sf);

+

+//  using steam table

+sf = 1.307;//  [kJ/kg K]

+mprintf('\n From table The accurate value of sf in this case is  =  %f  kJ/kg K\n',sf);

+

+// There is small error in book's final value of sf

+

+

+//  End

diff --git a/2705/CH7/EX7.2/Ex7_2.sce b/2705/CH7/EX7.2/Ex7_2.sce
new file mode 100755
index 000000000..38d974b14
--- /dev/null
+++ b/2705/CH7/EX7.2/Ex7_2.sce
@@ -0,0 +1,33 @@
+

+clear;

+clc;

+disp('Example 7.2');

+

+//  aim : To determine

+//  the specific entropy

+

+//  Given values

+P = 2;//  pressure,[MN/m^2]

+x = .8;//  dryness fraction

+

+//  solution

+//  from steam table at given pressure

+Tf = 485.4;//  [K]

+cpl = 4.187;//  [kJ/kg K]

+hfg = 1888.6;//  [kJ/kg]

+

+//  (a) finding entropy by calculation

+s = cpl*log(Tf/273.16)+x*hfg/Tf;//  formula for entropy calculation

+

+mprintf('\n (a) The specific entropy of wet steam is  =  %f kJ/kg K\n',s);

+

+//  (b) calculation of entropy using steam table

+//  from steam table at given pressure

+sf = 2.447;//  [kJ/kg K]

+sfg = 3.89;//  [kJ/kg K]

+//  hence

+s = sf+x*sfg;//  [kJ/kg K]

+

+mprintf('\n (b) The specific entropy using steam table is  =  %f kJ/kg K\n',s);

+

+//  End

diff --git a/2705/CH7/EX7.3/Ex7_3.sce b/2705/CH7/EX7.3/Ex7_3.sce
new file mode 100755
index 000000000..21e0a7087
--- /dev/null
+++ b/2705/CH7/EX7.3/Ex7_3.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+disp('Example 7.3');

+

+//  aim : To determine

+//  the specific entropy of steam

+

+//  Given values

+P = 1.5;//pressure,[MN/m^2]

+T = 273+300;//temperature,[K]

+

+//  solution

+

+//  (a)

+//  from steam table

+cpl = 4.187;//  [kJ/kg K]

+Tf = 471.3;//  [K]

+hfg = 1946;//  [kJ/kg]

+cpv = 2.093;//  [kJ/kg K]

+

+//  usung equation [2]

+s = cpl*log(Tf/273.15)+hfg/Tf+cpv*log(T/Tf);//  [kJ/kg K]

+mprintf('\n (a) The specific entropy of steam is  =  %f kJ/kg K\n',s);

+

+//  (b)

+//  from steam tables

+s = 6.919;//  [kJ/kg K]

+mprintf('\n (b) The accurate value of specific entropy from steam table is  =  %f kJ/kg K\n',s);

+

+//  End

diff --git a/2705/CH7/EX7.4/Ex7_4.sce b/2705/CH7/EX7.4/Ex7_4.sce
new file mode 100755
index 000000000..c908c859e
--- /dev/null
+++ b/2705/CH7/EX7.4/Ex7_4.sce
@@ -0,0 +1,30 @@
+clear;

+clc;

+disp('Example 7.4');

+

+//  aim : To determine

+//  the dryness fraction of steam

+

+//  Given values

+P1 = 2;//  initial pressure, [MN/m^2]

+t = 350;//  temperature, [C]

+P2 = .28;//  final pressure, [MN/m^2]

+

+//  solution

+//  at 2 MN/m^2 and 350 C,steam is superheated because the saturation temperature is 212.4 C

+//  From steam table

+s1 = 6.957;//  [kJ/kg K]

+

+//  for isentropic process

+s2 = s1;

+//  also

+sf2 = 1.647;//  [kJ/kg K]

+sfg2 = 5.368;//  [kJ/kg K]

+

+//  using

+//  s2 = sf2+x2*sfg2, where x2 is dryness fraction of steam

+//  hence

+x2 = (s2-sf2)/sfg2;

+mprintf('\n The final dryness fraction of steam is x2  =  %f\n',x2);

+

+//  End

diff --git a/2705/CH7/EX7.5/Ex7_5.sce b/2705/CH7/EX7.5/Ex7_5.sce
new file mode 100755
index 000000000..53cec3d7a
--- /dev/null
+++ b/2705/CH7/EX7.5/Ex7_5.sce
@@ -0,0 +1,53 @@
+clear;

+clc;

+disp('Example 7.5');

+

+//  aim : To determine

+//  the  final condition of steam...

+//  the change in specific entropy during hyperbolic process

+

+//  Given values

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

+t = 250;//  temperature, [C]

+P2 = .36;//  pressure, [MN/m^2]

+P3 = .06;//  pressure, [MN/m^2]

+

+//  solution

+

+//  (a)

+//  from steam table

+s1 = 6.545;//  [kJ/kg K]

+//  at .36 MN/m^2

+sg = 6.930;//  [kJ/kg*K]

+

+sf2 = 1.738;//  [kJ/kg K]

+sfg2 = 5.192;//  [kJ/kg K]

+vg2 = .510;//  [m^3]

+

+//  so after isentropic expansion, steam is wet

+//  hence, s2=sf2+x2*sfg2, where x2 is dryness fraction

+//  also

+s2 = s1;

+//  so

+x2 = (s2-sf2)/sfg2;

+//  and

+v2 = x2*vg2;//  [m^3]

+

+//  for  hyperbolic process

+//  P2*v2=P3*v3

+//  hence

+v3 = P2*v2/P3;//  [m^3]

+	

+mprintf('\n (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is  =  %f  m^3/kg\n',v3);

+

+//  (b)

+//  at this condition

+s3 = 7.609;//  [kJ/kg*K]

+//  hence

+change_s23 = s3-sg;//  change in specific entropy during the hyperblic process[kJ/kg*K]

+mprintf('\n (b) The change in specific entropy during the hyperbolic process is  =  %f kJ/kg K\n',change_s23);

+

+// In the book they have taken sg instead of s2 for part (b), so answer is not matching

+

+//  End

+

diff --git a/2705/CH7/EX7.6/Ex7_6.sce b/2705/CH7/EX7.6/Ex7_6.sce
new file mode 100755
index 000000000..b9e9a3beb
--- /dev/null
+++ b/2705/CH7/EX7.6/Ex7_6.sce
@@ -0,0 +1,67 @@
+

+clear;

+clc;

+disp('Example 7.6');

+

+//  aim : To determine the

+//  (a) heat transfer during the expansion and

+//  (b) work done durind the expansion

+

+//  given values

+m = 4.5;  //  mass of steam,[kg]

+P1 = 3;  //  initial pressure,[MN/m^2]

+T1 = 300+273;  //  initial temperature,[K]

+

+P2 = .1;  //  final pressure,[MN/m^2]

+x2 = .96;  //  dryness fraction at final stage

+

+//  solution

+//  for state point 1,using steam table

+s1 = 6.541;//  [kJ/kg/K]

+u1 = 2751;// [kJ/kg]

+ 

+ //  for state point 2

+ sf2 = 1.303;//  [kJ/kg/K]

+ sfg2 = 6.056;//  [kJ/kg/k]

+ T2 = 273+99.6;//  [K]

+ hf2 = 417;//  [kJ/kg]

+ hfg2 = 2258;//  [kJ/kg]

+ vg2 = 1.694;//   [m^3/kg]

+ 

+ //  hence

+ s2 = sf2+x2*sfg2;// [kJ/kg/k]

+ h2 = hf2+x2*hfg2;// [kJ/kg]

+ u2 = h2-P2*x2*vg2*10^3;// [kJ/kg]

+ 

+ //  Diagram of example 7.6

+ x = [s1 s2];

+ y = [T1 T2];

+plot2d(x,y);

+ xtitle('Diagram for example 7.6(T vs s)');

+ xlabel('Entropy (kJ/kg K)');

+ ylabel('Temperature (K)');

+

+x = [s1,s1];

+y = [0,T1];

+plot2d(x,y);

+

+x = [s2,s2];

+y = [0,T2];

+plot2d(x,y);

+

+ //  (a)

+ //  Q_rev is area of T-s diagram

+ Q_rev = (T1+T2)/2*(s2-s1);// [kJ/kg]

+ //  so total heat transfer is

+ Q_rev = m*Q_rev;// [kJ]

+ 

+ //  (b)

+ del_u = u2-u1;//  change in internal energy, [kJ/kg]

+ //  using 1st law of thermodynamics

+ W = Q_rev-m*del_u;//  [kJ]

+ 

+mprintf('\n (a) The heat transfer during the expansion is  =  %f kJ  (received)\n',Q_rev);

+

+ mprintf('\n (b) The work done during the expansion is  =  %f kJ\n',W);

+ 

+ //  End

diff --git a/2705/CH7/EX7.7/Ex7_7.sce b/2705/CH7/EX7.7/Ex7_7.sce
new file mode 100755
index 000000000..9a68eddbd
--- /dev/null
+++ b/2705/CH7/EX7.7/Ex7_7.sce
@@ -0,0 +1,45 @@
+

+clear;

+clc;

+disp('Example 7.7');

+

+//  aim : To determine the  

+//  (a) change of entropy

+//  (b)  The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression

+

+//  Given values

+P1 = 140;// initial pressure,[kN/m^2]

+V1 = .14;// initial volume, [m^3]

+T1 = 273+25;// initial temperature,[K]

+ P2 = 1400;// final pressure [kN/m^2]

+ n = 1.25; // polytropic index

+ cp = 1.041;// [kJ/kg K]

+ cv = .743;// [kJ/kg K]

+ 

+ //  solution

+ // (a)

+ R = cp-cv;// [kJ/kg/K]

+ //  using ideal gas equation 

+ m = P1*V1/(R*T1);// mass of gas,[kg]

+ //  since gas is following law P*V^n=constant ,so 

+ V2 = V1*(P1/P2)^(1/n);//  [m^3]

+ 

+ //  using eqn [9]

+ del_s = m*(cp*log(V2/V1)+cv*log(P2/P1));//  [kJ/K]

+ mprintf('\n (a) The change of entropy is  =  %f kJ/K\n',del_s);

+ 

+ //  (b)

+ W = (P1*V1-P2*V2)/(n-1);// polytropic work,[kJ]

+ Gamma = cp/cv;// heat capacity ratio

+ Q = (Gamma-n)/(Gamma-1)*W;// heat transferred,[kJ]

+ 

+ // Again using polytropic law

+ T2 = T1*(V1/V2)^(n-1);// final temperature, [K]

+ T_avg = (T1+T2)/2;// mean absolute temperature, [K]

+ 

+ // so approximate change in entropy is

+ del_s = Q/T_avg;// [kJ/K]

+ 

+ mprintf('\n (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression  =  %f  kJ/K\n',del_s);

+ 

+ //  End

diff --git a/2705/CH7/EX7.8/Ex7_8.sce b/2705/CH7/EX7.8/Ex7_8.sce
new file mode 100755
index 000000000..711505b0c
--- /dev/null
+++ b/2705/CH7/EX7.8/Ex7_8.sce
@@ -0,0 +1,39 @@
+clear;

+clc;

+disp('Example 7.8');

+

+//  aim : To determine

+//  the change of entropy

+

+//  Given values

+m = .3;//  [kg]

+P1 = 350;//  [kN/m^2]

+T1 = 273+35;//  [K]

+P2 = 700;//  [kN/m^2]

+V3 = .2289;//  [m^3]

+cp = 1.006;//  [kJ/kg K]

+cv = .717;//  [kJ/kg K]

+

+//  solution

+//  for constant volume process

+R = cp-cv;//  [kJ/kg K]

+//  using PV=mRT

+V1 = m*R*T1/P1;//  [m^3]

+

+//  for constant volume process P/T=constant,so

+T2 = T1*P2/P1;//  [K]

+s21 = m*cv*log(P2/P1);//  formula for entropy change for constant volume process

+mprintf('\n The change of entropy in constant volume process is  =  %f  kJ/kg K\n',s21);

+

+// 'For the above part result given in the book is wrong

+

+V2 = V1;

+// for constant pressure process

+T3 = T2*V3/V2;//  [K]

+s32 = m*cp*log(V3/V2);//  [kJ/kg K]

+

+mprintf('\n The change of entropy in constant pressure process is  =  %f  kJ/kg K\n',s32);

+

+// there is misprint in the book's result

+

+// End

diff --git a/2705/CH7/EX7.9/Ex7_9.sce b/2705/CH7/EX7.9/Ex7_9.sce
new file mode 100755
index 000000000..05f58353b
--- /dev/null
+++ b/2705/CH7/EX7.9/Ex7_9.sce
@@ -0,0 +1,21 @@
+clear;

+clc;

+disp('Example 7.9');

+

+//  aim : To determine

+//  the  change of entropy

+

+// Given values

+P1 = 700;// initial pressure, [kN/m^2]

+T1 = 273+150;//  Temperature ,[K]

+V1 = .014;// initial volume, [m^3]

+V2 = .084;//  final volume, [m^3]

+

+//  solution

+//  since process is isothermal so

+T2 = T1;

+// and using fig.7.10

+del_s = P1*V1*log(V2/V1)/T1 ;//  [kJ/K]

+mprintf('\n The change of entropy is  =  %f  kJ/kg K\n',del_s);

+

+//  End

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

diff --git a/2705/CH9/EX9.1/Ex9_1.sce b/2705/CH9/EX9.1/Ex9_1.sce
new file mode 100755
index 000000000..4a04d7b87
--- /dev/null
+++ b/2705/CH9/EX9.1/Ex9_1.sce
@@ -0,0 +1,32 @@
+clear;

+clc;

+disp('Example 9.1');

+

+//  aim : To determine 

+//  the heat loss per hour through the wall and interface temperature

+

+//  Given values

+x1 = .25;// thickness of brick,[m]

+x2 = .05;// thickness of concrete,[m]

+t1 = 30;// brick face temperature,[C]

+t3 = 5;// concrete face temperature,[C]

+l = 10;// length of the wall, [m]

+h = 5;// height of the wall, [m]

+k1 = .69;// thermal conductivity of brick,[W/m/K]

+k2 = .93;// thermal conductivity of concrete,[W/m/K]

+

+//  solution

+A = l*h;// area of heat transfer,[m^2]

+Q_dot = A*(t1-t3)/(x1/k1+x2/k2);// heat transferred, [J/s]

+

+//  so heat loss per hour is

+Q = Q_dot*3600*10^-3;// [kJ]

+mprintf('\n The heat lost per hour is  =  %f  kJ\n',Q);

+

+//  interface temperature calculation

+//   for  the brick wall, Q_dot=k1*A*(t1-t2)/x1;

+//  hence

+t2 = t1-Q_dot*x1/k1/A;// [C]

+mprintf('\n The interface temperature is  =  %f C\n',t2);

+

+//  End

diff --git a/2705/CH9/EX9.2/Ex9_2.sce b/2705/CH9/EX9.2/Ex9_2.sce
new file mode 100755
index 000000000..1a60b1763
--- /dev/null
+++ b/2705/CH9/EX9.2/Ex9_2.sce
@@ -0,0 +1,35 @@
+clear;

+clc;

+disp('Example 9.2');

+

+//  aim : To determine

+//  the minimum 

+//  thickness of the lagging required

+

+//  Given values

+r1 = 75/2;// external radious of the pipe,[mm]

+L = 80;// length of the pipe,[m]

+m_dot = 1000;// flow of steam, [kg/h]

+P = 2;// pressure, [MN/m^2]

+x1 = .98;// inlet dryness fraction

+x2 = .96;// outlet dryness fraction

+k = .08;// thermal conductivity of of pipe, [W/m/K]

+t2 = 27;// outside temperature,[C]

+

+//  solution

+//  using steam table  at 2 MN/m^2 the enthalpy of evaporation of steam is,

+hfg = 1888.6;// [kJ/kg]

+//  so heat loss through the pipe is

+Q_dot = m_dot*(x1-x2)*hfg/3600;// [kJ]

+

+// also from steam table saturation temperature of steam at 2 MN/m^2 is,

+t1 = 212.4;// [C]

+// and for thick pipe, Q_dot=k*2*%pi*L*(t1-t2)/log(r2/r1)

+// hence

+r2 = r1*exp(k*2*%pi*L*(t1-t2)*10^-3/Q_dot);// [mm]

+

+t = r2-r1;// thickness, [mm]

+

+mprintf('\n The minimum thickness of the lagging required is  =  %f  mm\n',t);

+

+//  End

diff --git a/2705/CH9/EX9.3/Ex9_3.sce b/2705/CH9/EX9.3/Ex9_3.sce
new file mode 100755
index 000000000..0901fd861
--- /dev/null
+++ b/2705/CH9/EX9.3/Ex9_3.sce
@@ -0,0 +1,41 @@
+clear;

+clc;

+disp('Example 9.3');

+

+//  aim : To determine the

+//  (a) heat loss per hour

+//   (b) interface temperature og lagging

+

+// Given values

+r1 = 50; // radious of steam main,[mm]

+r2 = 90;// radious with first lagging,[mm]

+r3 = 115;// outside radious os steam main with lagging,[mm]

+k1 = .07;// thermal conductivity of 1st lagging,[W/m/K]

+k2 = .1;// thermal conductivity of 2nd lagging, [W/m/K]

+P = 1.7;// steam pressure,[MN/m^2]

+t_superheat = 30;// superheat of steam, [K]

+t3 = 24;// outside temperature of the lagging,[C]

+L = 20;// length of the steam main,[m]

+

+//  solution

+//  (a)

+//  using steam table saturation temperature of steam at 1.7 MN/m^2 is

+t_sat = 204.3;// [C]

+// hence

+t1 = t_sat+t_superheat;// temperature of steam,[C]

+

+Q_dot = 2*%pi*L*(t1-t3)/(log(r2/r1)/k1+log(r3/r2)/k2);// heat loss,[W]

+//  heat loss in hour is

+Q = Q_dot*3600*10^-3;// [kJ]

+

+mprintf('\n (a) The heat lost per hour is  =  %f kJ\n',Q);

+

+// (b)

+//  using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1) 

+t2 = t1-Q_dot*log(r2/r1)/(2*%pi*k1*L);// interface temperature of lagging,[C]

+

+mprintf('\n (b) The interface temperature of the lagging is  =  %f C\n',t2);

+

+//  There is some calculation mistake in the book so answer is not matching

+

+//  End

diff --git a/2705/CH9/EX9.4/Ex9_4.sce b/2705/CH9/EX9.4/Ex9_4.sce
new file mode 100755
index 000000000..b6f5f4106
--- /dev/null
+++ b/2705/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,22 @@
+clear;

+clc;

+disp('Example 9.4');

+

+// aim : To determine 

+//  the energy emetted from the surface

+

+//  Given values

+h = 3;// height of surface, [m]

+b = 4;// width of surface, [m]

+epsilon_s = .9;// emissivity of the surface

+T = 273+600;// surface temperature ,[K]

+sigma = 5.67*10^-8;// [W/m^2/K^4]

+

+//  solution

+As = h*b;// area of the surface, [m^2]

+

+Q_dot = epsilon_s*sigma*As*T^4*10^-3;// energy emitted, [kW]

+

+mprintf('\n The energy emitted from the surface is  =  %f  kW\n',Q_dot);

+

+//  End

diff --git a/2705/CH9/EX9.5/Ex9_5.sce b/2705/CH9/EX9.5/Ex9_5.sce
new file mode 100755
index 000000000..e916c7e40
--- /dev/null
+++ b/2705/CH9/EX9.5/Ex9_5.sce
@@ -0,0 +1,34 @@
+clear;

+clc;

+disp('Example 9.5');

+

+//  aim : To determine 

+//  the rate of energy transfer between furnace and the sphere and its direction

+

+//  Given values

+l = 1.25;// internal side of cubical furnace, [m]

+ti = 800+273;// internal surface temperature of the furnace,[K]

+r = .2;// sphere radious, [m]

+epsilon = .6;// emissivity of sphere

+ts = 300+273;// surface temperature of sphere, [K]

+sigma = 5.67*10^-8;// [W/m^2/K^4]

+

+//  Solution

+Af = 6*l^2;// internal surface area of furnace, [m^2]

+As =4 *%pi*r^2;// surface area of sphere, [m^2]

+

+//  considering internal furnace to be black

+Qf = sigma*Af*ti^4*10^-3;// [kW]

+

+//  radiation emitted by sphere is

+Qs = epsilon*sigma*As*ts^4*10^-3; // [kW]

+

+//  Hence transfer of energy is

+Q = Qf-Qs;// [kW]

+

+mprintf('\n The transfer of energy will be from furnace to sphere and transfer rate is  =  %f kW\n',Q);

+

+//   There is some calculation mistake in the book so answer is not matching

+

+//  End

+

diff --git a/2705/CH9/EX9.6/Ex9_6.sce b/2705/CH9/EX9.6/Ex9_6.sce
new file mode 100755
index 000000000..07386e7e2
--- /dev/null
+++ b/2705/CH9/EX9.6/Ex9_6.sce
@@ -0,0 +1,34 @@
+clear;

+clc;

+disp('Example 9.6');

+

+//  aim : To determine

+//  the overall transfer coefficient and the heat loss per hour

+

+//  Given values

+x1 = 25*10^-3;// Thickness of insulating board, [m]

+x2 = 75*10^-3;// Thickness of fibreglass, [m]

+x3 = 110*10^-3;// Thickness of brickwork, [m]

+k1 = .06;// Thermal conductivity of insulating board, [W/m K]

+k2 = .04;// Thermal conductivity of fibreglass, [W/m K]

+k3 = .6;// Thermal conductivity of brickwork, [W/m K]

+Us1 = 2.5;//  surface heat transfer coefficient of the inside wall,[W/m^2 K]

+Us2 = 3.1;//  surface heat transfer coefficient of the outside wall,[W/m^2 K]

+ta1 = 27;// internal ambient temperature, [C]

+ta2 = 10;// external ambient temperature, [C]

+h = 6;// height of the wall, [m]

+l = 10;// length of the wall, [m]

+

+//  solution

+U = 1/(1/Us1+x1/k1+x2/k2+x3/k3+1/Us2);// overall heta transfer coefficient,[W/m^2 K]

+

+A = l*h;// area ,[m^2]

+

+Q_dot = U*A*(ta1-ta2);// heat loss [W]

+

+//  so heat loss per hour is

+Q = Q_dot*3600*10^-3;// [kJ]

+mprintf('\n The overall heat transfer coefficient for the wall is   =  %f  W/m^2 K\n',U);

+mprintf('\n The heat loss per hour through the wall is  =  %f kJ\n',Q);

+

+//  End

diff --git a/2705/CH9/EX9.7/Ex9_7.sce b/2705/CH9/EX9.7/Ex9_7.sce
new file mode 100755
index 000000000..4af2d8450
--- /dev/null
+++ b/2705/CH9/EX9.7/Ex9_7.sce
@@ -0,0 +1,44 @@
+clear;

+clc;

+disp('Example 9.7');

+

+//  aim : To determine  

+//  the heat loss per hour and the surface temperature of the lagging

+

+//  Given values

+r1 = 75*10^-3;// External radiou of the pipe, [m]

+t_l1 = 40*10^-3;// Thickness of lagging1, [m]

+t_l2 = t_l1;

+k1 = .07;// thermal conductivity of lagging1, [W/m K]

+k2 = .1;// thermal conductivity of lagging2, [W/m K]

+Us = 7;// surface transfer coefficient for outer surface, [W/m^2 K]

+L = 50;// length of the pipe, [m]

+ta = 27;// ambient temperature, [C]

+P = 3.6;// wet steam pressure, [MN/m^2]

+

+//  solution

+//  from steam table saturation temperature of the steam at given pressure is,

+t1 =  244.2;// [C]

+r2 = r1+t_l1;// radious of pipe with lagging1,[m]

+r3 = r2+t_l2;// radious of pipe with both the lagging, [m]

+

+R1 = log(r2/r1)/(2*%pi*L*k1);// resistance due to lagging1,[C/W]

+R2 = log(r3/r2)/(2*%pi*L*k2);// resistance due to lagging2,[C/W]

+R3 = 1/(Us*2*%pi*r3*L);// ambient resistance, [C/W]

+

+//  hence overall resistance is,

+Req = R1+R2+R3;// [C/W]

+tdf = t1-ta;// temperature driving force, [C]

+Q_dot = tdf/Req;// rate of heat loss, [W]

+//  so heat loss per hour is,

+Q = Q_dot*3600*10^-3;// heat loss per hour, [kJ]

+

+//  using eqn [3]

+t3 = ta+Q_dot*R3;// surface temperature of the lagging, [C]

+

+mprintf('\n The heat loss per hour is  =  %f  kJ\n',Q);

+mprintf('\n The surface temperature of the lagging is  =  %f C\n',t3);

+

+//  there is minor variation in the answer

+

+//  End