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+// Example 3_4
+clc;funcprot(0);
+// Given data
+T=212;// °F
+V=3.00;// The total volume in ft^3
+m=0.200;// lbm
+p=14.696;// psia
+v_f=0.01672;// ft^3/lbm
+v_g=26.80;// ft^3/lbm
+u_f=180.1;// Btu/lbm
+u_g=1077.6;// Btu/lbm
+h_f=180.1;// Btu/lbm
+h_g=1150.5;// Btu/lbm
+
+// Solution
+// (a)
+v=V/m;// The specific volume in ft^3/lbm
+// (b)
+v_fg=v_g-v_f;// ft^3/lbm
+x=(v-v_f)/v_fg;// The quality
+x_m=1-x;// The amount of moisture present
+// (c)
+u_fg=u_g-u_f;// Btu/lbm
+u=u_f+(x*u_fg);// The specific internal energy in Btu/lbm
+// (d)
+h_fg=h_g-h_f;// Btu/lbm
+h=h_f+(x*h_fg);// The specific enthalpy in Btu/lbm
+// (e)
+m_g=x*m;// The mass of water in the vapor phase in lbm
+m_f=m-m_g;// The mass of water in the liquid phase in lbm
+printf('\n(a)The specific volume,v=%2.0f ft^3/lbm \n(b)The quality,x=%0.3f (or) %2.1f percentage \n The amount of moisture present,1-x=%0.3f (or) %2.1f percentage \n(c)The specific internal energy,u=%3.0f Btu/lbm \n(d)The specific enthalpy,h=%3.0f Btu/lbm \n(e)The mass of water in the liquid and vapor phases,m_f=%0.3f lbm & m_g=%0.3f lbm',v,x,x*100,x_m,x_m*100,u,h,m_f,m_g);