A stream of argon flowing at the rate of 19842 lbmhr and a s

A stream of argon flowing at the rate of 19842 lbm/hr and a stream of helium flowing at the rate of 7937lbm/hr mix adiabatically in a steady flow process. If the gases are assumed ideal, what is the rate of entropy increase as a result of the process. The rate of entropy increase should be reported in kJK -1 s -1 .

Solution

The mass flow rates in kg/s are as follows:

m1_dot=19842 lbm/h=2.5 kg/s

m2_dot=7937 lbm/h=1 kg/s

cp_1=0.520 kJ/kg.K

cp_2=5.19 kJ/kg.K

Change in entropy = DeltaS1+DeltaS2

                              =m1_dot*c_p1*ln(Tf/T1)+m2_dot*c_p2*ln(Tf/T2)

                             =2.5*0.52*ln(Tf/T1)+1*5.19*ln(Tf/T2)

                             =1.3*ln(Tf/T1)+5.19*ln(Tf/T2)   kJ/(K.s)