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+// Example 13_13
+clc;funcprot(0);
+// Given data
+V_inlet=0;// ft/s
+V_exh=1560;// ft/s
+m_exh=270;// lbm/s
+g_c=32.174;// lbm.ft/(lbf.s^2)
+p_1=190;// psia
+T_1=2060;// R
+p_2s=28.0;// psia
+T_2=1350;// R
+p_3=14.7;// psia
+T_3=520;// R
+p_4s=200;// psia
+T_4=1175;// R
+k=1.40;// The specific heat ratio
+
+// Calculation
+// 1.The engine’s static thrust is given directly by Eq. (13.29) as
+T=m_exh*(V_exh-V_inlet)/g_c;// lbf
+// 2a.
+T_4s=T_3*((p_4s/p_3)^((k-1)/k));// °F
+n_s=((T_4s-T_3)/(T_4-T_3))*100;// The compressor’s isentropic efficiency in %
+T_2s=T_1*(p_2s/p_1)^((k-1)/k);// R
+n_s_pm=((T_1-T_2)/(T_1-T_2s))*100;// %
+// 3a.
+n_T_Bc=((T_1-T_2s-(T_4s-T_3))/(T_1-T_4s))*100;// The Brayton cold ASC thermal efficiency in %
+n_T_B=((T_1-T_2-(T_4-T_3))/(T_1-T_4))*100;// The actual thermal efficiency of the engine in %
+// 2b.
+// By using Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics
+p_r4=1.2147*(p_4s/p_3);
+T_4s=1084;// R
+T_4sF=624;// °F
+n_s_c=((T_4s-T_3)/(T_4-T_3))*100;// %
+p_r1=196.16;
+p_r2=p_r1*(p_2s/p_1);
+// By interpolation in Table C.16a,
+T_2s=1261-460;// °F 
+T_2s=1261;// R
+n_s_pm2=((T_1-T_2)/(T_1-T_2s))*100;// %
+// 3b.
+// From Table C.16a,
+h_3=124;// Btu/lbm 
+h_4s=262;// Btu/lbm
+h_1=521;// Btu/lbm
+h_2s=307;// Btu/lbm
+h_4=284.09;// Btu/lbm
+h_2=329.9;// Btu/lbm
+n_T_Bh=((h_1-h_2s-(h_4s-h_3))/(h_1-h_4s))*100;// %
+n_T_B2=((h_1-h_2-(h_4-h_3))/(h_1-h_4))*100;// %
+// 3c.
+n_T_max=(1-sqrt(T_3/T_1))*100;// The maximum work Brayton cold ASC thermal efficiency in %
+printf("\n(1)The engine’s static thrust is given directly,T=%5.0f lbf \n(2)(a)The compressor and turbine isentropic efficiencies for the Brayton cold air standard cycle,(n_s)_compressor=%2.1f percentage & (n_s)_pm=%2.1f percentage \n   (b)The compressor and turbine isentropic efficiencies for the Brayton hot air standard cycle using the gas tables for air,(n_T)_Brayton hot ASC=%2.1f percentage & (n_T)_Brayton actual=%2.1f percentage \n(3)(a)The ASC and actual thermal efficiencies for the Brayton cold air standard cycle,(n_T)_Brayton cold ASC =%2.1f percentage & (n_T)__Brayton actual=%1.1f percentage \n   (b)The Brayton hot air standard cycle using the gas tables for air,(n_T)_Brayton hot ASC =%2.1f percentage & (n_T)__Brayton actual=%1.1f percentage \n   (c)The maximum work Brayton cold ASC thermal efficiency,(n_T)_max work=%2.1f percentage",T,n_s,n_s_pm,n_s_c,n_s_pm2,n_T_Bc,n_T_B,n_T_Bh,n_T_B2,n_T_max);
+// The answer provided in the text book is wrong