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+// Example 12_6
+clc;clear;funcprot(0);
+//Given values
+T_1=400; // K
+P_1=100; // kPa
+Ma_1=0.3;// Mach number
+
+// Calculation
+//From table A-13.At the initial Mach number of Ma=0.3, we read
+// a_1=A1/A*; t_1=T1/T0; p_1=P1/P0;t_2=T1/T0;p_2=P2/P0;
+a_1=2.031;
+t_1=0.9823;
+p_1=0.9395;
+// A2=0.8*A1;
+//a_2=(A2/A*)=(A2/A1)*(A1/A*);
+a_2=0.8*a_1;
+//From table A-13,for the value of a_2
+t_2=0.9703;
+p_2=0.9000;
+Ma_2=0.391;
+T_2=T_1*(t_2/t_1);// K
+P_2=P_1*(p_2/p_1);// kPa
+printf('Mach number,Ma_2=%0.3f\n',Ma_2);
+printf('Temperature,T_2=%0.0f K\n',T_2);
+printf('Pressure,P_2=%0.1f kPa\n',P_2);