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diff --git a/2705/CH13/EX13.1/Ex13_1.sce b/2705/CH13/EX13.1/Ex13_1.sce
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+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
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+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
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+++ 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
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+++ 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
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+++ 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