The answer would be 398.40 K and 4.71 * 10^-6 kW/K.
Temp. of Argon stream : 80 F = 299.817 K
Temp of CO2 stream : 550 F = 533.15 K
CO2 mass, flow rate : 0.5 lbm/h = 6.299 * 10^-5 kg/s
Ar,mass, flow rate : 1 lbm / h = 0.000125 kg /s
(a) Particular heat capacity of Ar : 0.523 kJ/Kg.K
Particular heat capacity of CO2 : 0.756 kJ/kg.K
Mass balance expression = mo3 = mo1 + m02 ; mo3 is mass mixture.
Formula to find heat capacity of mixed particle : cpm mo1*cpa + mo2cpc/mo1 + mo2
= 0.6010 kJ/ kg.K
Formula for energy balance law mo3*cpm*T03 = mo1*cpa*To1 + mo2*cpc*To2
Put the values , and the temperature of the mixture is 398.40 K
(b) Formula for rate of entropy geneartion for process is
del So= mo1*cpa ln(T03 / T01) + mo2*cpc in (T03 / T02)
Put the values and we get del So = 4.71 * 10^-6 kW/K
The given question is incomplete : (a) Final Temperature ?
(b) Rate of entropy generation of the process ?
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