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Diffstat (limited to '3843/CH12/EX12.12')
-rw-r--r-- | 3843/CH12/EX12.12/Ex12_12.sce | 51 |
1 files changed, 51 insertions, 0 deletions
diff --git a/3843/CH12/EX12.12/Ex12_12.sce b/3843/CH12/EX12.12/Ex12_12.sce new file mode 100644 index 000000000..0063e49e3 --- /dev/null +++ b/3843/CH12/EX12.12/Ex12_12.sce @@ -0,0 +1,51 @@ +// Example 12_12
+clc;funcprot(0);
+// Given data
+A=2;// The surface area in m^2
+U=0.5;// The over all heat transfer coefficient in kW/m^2.K
+mdot_p=0.2;// The mass flow rate of propane in kg/s
+M_p=44;// The molecular weight of the propane in kg/kmol
+T_E=25+273;// K
+P=1;// atm
+// From example 12.11
+// The combustion equation C_3H_8+5(O_2+3.76N_2)--->3CO_2+4H_2O+18.8N_2
+N_CO2=3;// mol
+N_H2O=4;// mol
+N_N2=18.8;// mol
+hbar0_fp=-103850;// kJ/kmol (C_3H_8)
+hbar0_fCO2=-393520;// kJ/kmol
+hbar0_CO2=9360;// kJ/kmol
+hbar0_fH2O=-241810;// kJ/kmol
+hbar0_H2O=9900;// kJ/kmol
+hbar0_fN2=0;// kJ/kmol
+hbar0_N2=8670;// kJ/kmol
+
+// Calculation
+mdot_fuel=mdot_p/M_p;// The molar influx in kg/s
+M_CO2=N_CO2*mdot_fuel;// kmol/s
+M_H2O=N_H2O*mdot_fuel;// kmol/s
+M_N2=N_N2*mdot_fuel;// kmol/s
+// LHS=Q+H_R
+// RHS=H_P
+// For a first guess at T_P let us assume a some what lower temperature than that of Example 12.11,since energy leaving the combustion chamber.The guesses follow
+T_P1=1600;// K
+LHS_1=(-U*A*(T_P1-T_E))+(mdot_fuel*hbar0_fp);// kJ/kmol fuel
+hbar_CO2=76944;// kJ/kmol
+hbar_H2O=62748;// kJ/kmol
+hbar_N2=50571;// kJ/kmol
+RHS_1=(M_CO2*(hbar0_fCO2+hbar_CO2-hbar0_CO2))+(M_H2O*(hbar0_fH2O+hbar_H2O-hbar0_H2O))+(M_N2*(hbar0_fN2+hbar_N2-hbar0_N2));// The enthalpy of the products of combustion in kJ/kmol fuel
+T_P2=2000;// K
+LHS_2=(-U*A*(T_P2-T_E))+(mdot_fuel*hbar0_fp);// kJ/kmol fuel
+hbar_CO2=100804;// kJ/kmol
+hbar_H2O=82593;// kJ/kmol
+hbar_N2=64810;// kJ/kmol
+RHS_2=(M_CO2*(hbar0_fCO2+hbar_CO2-hbar0_CO2))+(M_H2O*(hbar0_fH2O+hbar_H2O-hbar0_H2O))+(M_N2*(hbar0_fN2+hbar_N2-hbar0_N2));// The enthalpy of the products of combustion in kJ/kmol fuel
+T_P3=1900;// K
+LHS_3=(-U*A*(T_P3-T_E))+(mdot_fuel*hbar0_fp);// kJ/kmol fuel
+hbar_CO2=94793;// kJ/kmol
+hbar_H2O=77517;// kJ/kmol
+hbar_N2=61220;// kJ/kmol
+RHS_3=(M_CO2*(hbar0_fCO2+hbar_CO2-hbar0_CO2))+(M_H2O*(hbar0_fH2O+hbar_H2O-hbar0_H2O))+(M_N2*(hbar0_fN2+hbar_N2-hbar0_N2));// The enthalpy of the products of combustion in kJ/kmol fuel
+// Interpolation between the last two entries gives
+T_P=1970;// K
+printf("\nThe temperature of products of combustion,T_P=%4.0f K",T_P);
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