4 files changed, 171 insertions, 0 deletions
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