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//Engineering and Chemical Thermodynamics
// Example 3.16
//Page no:144
clear ; clc ;
//Given
P_1 = 120 * 10^3 ; //[N]
P_2 = 900 * 10^3 ; //[N]
h_4 = 25.486 ; //[kJ/mol], From table
h_1 = h_4 ;
h_2 = 39.295 ; //[kJ/mol], From table
S_2 = 177.89 ; //[kJ/molK], From table
S_3 = S_2 ; //[kJ/mol]
h_3 = 43.578 ; //[kJ/mol] , Enthalpy corresponding to S3 value which equales to S2
Q_dot_c_des = 10 ; //[kW]
q_c = h_2 - h_1 ;
Q_dot_c = h_2 - h_1 ;
W_dot_c = h_3 - h_2 ;
COP = Q_dot_c / W_dot_c ;
n_dot = Q_dot_c_des / q_c ;
disp(" Example: 3.16 Page no : 144") ;
printf("\n COP of the refrigerator is = %.2f \n\n Mass flow rate needed = %.3f mol/s",COP,n_dot)
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