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+// Example 8_10
+clc;funcprot(0);
+// Given data
+// State 1
+m=0.100;// lbm
+p_1=100;// psia
+T_1=180;// °F
+v_1=0.6210;// ft^3/lbm
+h_1=137.49;// Btu/lbm
+s_1=0.2595;// Btu/(lbm.R)
+// State 2
+p_2=30.0;// psia
+T_2=120;// °F
+v_2=1.9662;// ft^3/lbm
+h_2=126.39;// Btu/lbm
+s_2=0.2635;// Btu/(lbm.R)
+// State 3
+p_3=p_2;// psia
+v_3=v_1/2;// ft^3/lbm
+x_3=0.1952;// The quality of steam
+s_3=0.07241;// Btu/(lbm.R)
+K=5.00;// Btu/R
+
+// Calculation
+// (a)
+// From Table C.7e of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that at p1 = 100 psia and T1 = 180°F,
+v_1=0.6210;// ft^3/lbm
+u_1=125.99;// Btu/lbm
+// At p2= 30 psia and T2 = 120°F,
+v_2=1.966;// ft^3/lbm
+u_2=115.47;// Btu/lbm
+W_12=-m*(u_2-u_1);// Btu
+// (b)
+v_3=v_1/2;// ft^3/lbm
+// At p2= 30 psia 
+v_f3=0.01209;// ft^3/lbm
+v_g3=1.5408;// ft^3/lbm
+u_f3=16.24;// Btu/lbm
+u_g3=95.40;// Btu/lbm
+x_3=(v_3-v_f3)/(v_g3-v_f3);// The quality of steam
+x_3p=x_3*100;// %
+u_3=u_f3+(x_3*(u_g3-u_f3));// Btu/lbm
+Q_23=(m*(u_3-u_2))+(m*(p_3*144)*((v_3-v_2)*(1/778.17)));// Btu
+// (c)
+// From Table C.7b
+T_3=15.38;// °F
+dQ=0;// Btu
+S_p12=m*(s_1-s_2)-0;// Btu/R
+s_f3=0.0364;// Btu/(lbm.R)
+s_fg3=0.2209;// Btu/(lbm.R)
+s_3=s_f3+(x_3*(s_fg3-s_f3));// Btu/(lbm.R)
+S_p23=(m*[s_3-s_2])-(K*log((T_3+459.67)/(T_2+459.67)));// Btu/R
+S_p13=S_p12+S_p23;// Btu/R
+printf('\n(a)The work transport of energy during the adiabatic expansion,W_12=%1.2f Btu \n(b)The heat transport of energy during the isobaric compression,Q_23=%1.2f Btu \n(c)Since state 3 is saturated (a mixture of liquid and vapor), T3 must be equal to the saturation temperature at 30.0 psia,which, from Table C.7b, is T_3 =%2.2f°F \n(d)The total entropy production for both processes,1(S_p)3=%0.3f Btu/R',W_12,Q_23,T_3,S_p13);