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+// Example 11_12
+clc;funcprot(0);
+// Given data
+T=60.0;// °F
+p_1=14.7;// psia
+r_c=19.2:1;// Compression ratio
+
+// Solution
+// From Table C.16a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics,we find that, at 60.0°F = 520.R,
+u_1=88.62;// Btu/lbm
+p_r1=1.2147;
+v_r1=158.58;
+v_2byv_1=1/19.2;
+v_r2=v_r1*v_2byv_1;
+// Scanning down the v_r column in Table C.16a, we find that vr = 8.26 at about
+T_2=1600-459.67;// °F
+u_2=286.06;// Btu/lbm
+p_r2=71.73;
+p_2=p_1*(p_r2/p_r1);// psia
+W_12bym=u_1-u_2;// Btu/lbm
+printf("\nThe final temperature and pressure of the air at the end of the compression stroke,T_2=%4.0f°F and p_2=%3.1f psia. \nThe work required per lbm of air present,1W2/m=%3.2f Btu/lbm",T_2,p_2,W_12bym);