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// Example 5_6
clc;funcprot(0);
// Given data
// State 1
m=0.100;// lbm
p_1=100;// psia
T_1=180;// °F
// State 2
p_2=30.0;// psia
T_2=120;// °F
// State 3
p_3=p_2;// psia
// 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
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',W_12,Q_23,T_3);
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