// 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);