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+// Example 10_8
+clc;funcprot(0);
+// Given data
+m=2.80;// lbm/s
+// Station 1
+p_1=100;// psia
+T_1=500;// °F
+h_1=1279.1;// Btu/lbm
+s_1=1.7087;// Btu/lbm.R
+// Station 2
+p_2=10.0;// psia
+p_2s=p_2;// psia
+s_2f=0.2836;// Btu/lbm.R
+s_2fg=1.5043;// Btu/lbm.R
+s_2=s_1;// Btu/lbm.R
+h_2f=161.4;// Btu/lbm
+h_2fg=982.1;// Btu/lbm
+h_2s=1091.6;// Btu/lbm
+// Ground state
+x_0=0;// The quality of steam
+T_0=70.0;// °F
+s_0=0.0746;// Btu/lbm.R
+h_0=38.1;// Btu/lbm
+g_c=32.174;// lbm.ft/lbf.s^2
+g=32.174;// ft/s^2
+
+// Calculation
+// (a)
+V_1=0;// ft/s
+Z_2=0;// ft
+Z_1=Z_2;// ft
+V_2s=[2*g_c*(h_1-h_2s)*778.16]^(1/2);// ft/s
+V_2=(95/100)*V_2s;// ft/s
+h_2=h_1-((V_2^2)/(2*g_c*778.16));// Btu/lbm
+x_2=(h_2-h_2f)/h_2fg;// The quality of steam
+s_2=s_2f+(x_2*s_2fg);// Btu/lbm.R
+a_f1=(h_1-h_0)-((T_0+459.67)*(s_1-s_0))+(V_1^2/(2*g_c*778.16))+((g*Z_1)/g_c);// Btu/lbm
+// (b)
+a_f2=(h_2-h_0)-((T_0+459.67)*(s_2-s_0))+(V_2^2/(2*g_c*778.16))+((g*Z_2)/g_c);// Btu/lbm
+// (c)
+I=m*(a_f1-a_f2);// Btu/s
+printf("\n(a)The inlet specific flow availability,a_f1=%3.0f Btu/lbm \n(b)The exit specific flow availability,a_f2=%3.0f Btu/lbm  \n(c)The irreversibility rate inside the nozzle,I=%2.1f Btu/s",a_f1,a_f2,I);
+// The answer vary due to round off error