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+// Example 10_3
+clc;funcprot(0);
+// Given data
+T_1=50.0;// °F
+V_1=500;// mph
+Z=30.0*10^3;// ft
+T_0=70.0;// °F
+p_0=136.12;// psia
+m=5.00;// lbm
+g=32.174;// ft/s^2
+g_c=32.174;// lbm.ft/lbf.s^2
+
+// Calculation
+// State 1 (flying)
+x_1=0.00;// The quality of steam
+T_1=50.0;// °F
+v_1=0.0128;// ft^3/lbm
+u_1=24.04;// Btu/lbm
+s_1=0.0519;// Btu/lbm.R
+V_1=500;// mph
+Z_1=30000;// ft
+// State 2 (landed)
+p_2=100;// psia
+T_2=400;// °F
+v_2=1.046;// ft^3/lbm
+u_2=154.77;// Btu/lbm
+s_2=0.31464;// Btu/lbm.R
+V_2=0;// mph
+Z_2=0;// ft
+// Ground state
+x_0=0.00;// The quality of steam
+T_0=70.0;// °F
+v_0=0.01325;// ft^3/lbm
+u_0=29.78;// Btu/lbm
+s_0=0.06296;// Btu/lbm.R
+p_0=136.12;// psia
+A_1=(m*[(u_1-u_0)+((p_0)*(144/778.16)*(v_1-v_0))])-(m*(T_0+459.67)*(s_1-s_0))+(m*[(([V_1*5280*(1/3600)]^2)/(2*g_c*778.16))+((g*Z_1)/(g_c*778.16))]);// Btu
+A_2=(m*[(u_2-u_0)+((p_0)*(144/778.16)*(v_2-v_0))])-(m*(T_0+459.67)*(s_2-s_0))+(m*[(([V_2*5280*(1/3600)]^2)/(2*g_c*778.16))+((g*Z_2)/(g_c*778.16))]);// Btu
+// (b)
+dA=A_2-A_1;// Btu
+printf("\n(a)The total availability of the refrigerant before and after the aircraft lands,A_1=%3.0f Btu & A_2=%2.1f Btu. \n(b)The change in total availability during the landing,A_2-A_1=%3.0f Btu",A_1,A_2,dA);