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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example 12_16
+clc;funcprot(0);
+// Given data
+X_N_2=0.7809;// The mole fraction for nitrogen
+X_O_2=0.2095;// The mole fraction for oxygen
+X_Ar=0.00930;// The mole fraction for Argon
+X_CO_2=0.0003;// The mole fraction for Carbondioxide
+
+// Calculation
+// Using Eqs. (12.39) and (12.40), the composition data given in Example 12.2 and the critical point data given in Table C.12b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics give
+p_c_N_2=3.39;// MPa
+p_c_O_2=5.08;// MPa
+p_c_Ar=4.86;// MPa
+p_c_CO_2=7.39;// MPa
+p_c_air=(X_N_2*p_c_N_2)+(X_O_2*p_c_O_2)+(X_CO_2*p_c_CO_2);// MPa
+T_c_N_2=126.2;// K
+T_c_O_2=154.8;// K
+T_c_Ar=151;// K
+T_c_CO_2=304.2;// K
+T_c_air=(X_N_2*T_c_N_2)+(X_O_2*T_c_O_2)+(X_CO_2*T_c_CO_2);// K
+printf("\nThe critical pressure for air,(p_c)_air=%1.2f MPa \nThe critical temperature for air,(T_c)_air=%3.0f K",p_c_air,T_c_air);
+// The answer vary due to round off error