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+// Example 9_4
+clc;funcprot(0);
+// Given data
+m=0.100;// lbm/s
+// Station 1
+x_1=0.00;// The quality of steam at inlet
+T_1=100;// °F
+// Station 2
+x_2=0.530;// The quality of steam at exit
+T_2=20;// °F
+T_b=60.0;// °F
+
+// Calculation
+// (a)
+// From Table C.7a for R-134a, we find
+h_f1=44.23;// Btu/lbm
+h_1=h_f1;// Btu/lbm
+s_f1=0.0898;// Btu/(lbm.R)
+s_1=s_f1;// Btu/(lbm.R)
+h_f2=17.74;// Btu/lbm
+h_fg2=86.87;// Btu/lbm
+s_f2=0.0393;// Btu/(lbm.R)
+s_fg2=0.2206-s_f2;// Btu/(lbm.R)
+h_2=h_f2+(x_2*h_fg2);// Btu/lbm
+s_2=s_f2+(x_2*s_fg2);// Btu/(lbm.R)
+Q=m*(h_2-h_1);// Btu/s
+S_pa=((m*(s_2-s_1))-(Q/(T_b+459.67)));// The entropy production rate inside the valve in Btu/(s.R)
+S_p=S_pa*778.17;// ft.lbf/(s.R)
+printf("\n(a)The entropy production rate inside the valve if the valve is not insulated and has an isothermal external surface temperature of 60.0°F,S_p=%0.4f ft.lbf/(s.R)",S_p);
+// (b)
+h_2=h_1;// Btu/lbm
+x_2=(h_2-h_f2)/h_fg2;// The quality of steam
+x_2p=x_2*100;// % (in x_2p,p refers the quality of steam in percentage)
+s_2=s_f2+(x_2*s_fg2);// Btu/(lbm.R)
+Q=0;// W
+S_pb=m*(s_2-s_1)-(Q/T_b);// Btu/(s.R) 
+S_p=S_pb*778.17;// lbf/(s.R)
+printf("\n(b)The entropy production rate inside the valve if it is insulated and assuming it has the same inlet conditions and exit temperature,S_p=%0.3f ft.lbf/(s.R)",S_p);
+//(c)
+S_p_pd=((S_pa-S_pb)/S_pa)*100;// The percentage decrease in S_p brought about by adding the insulation in %
+printf("\n(c)The percentage decrease in S_p brought about by adding the insulation is %2.1f percentage.",S_p_pd);