Astream of air flowing at 20 liters/min with p = 0.20 mpa and t = 400 k is mixed with a stream ofmethane flowing at 5 liters/min with p = 0.20 mpa and t = 300 k. the combined gas stream exitingthe mixer is at p = 0.10 mpa and t = 370 k. what is the volumetric flow rate and composition of theexiting gas stream?

a) the mole fraction of air in the exiting stream is Xa=0.25 (25%) and for methane  Xm=0.75 (75%)

b) the volumetric flow rate is 49.33 L/s

Explanation:

Assuming ideal gas behaviour, then

for air

Pa*Va=Na*R*Ta

for methane

Pm*Vm=Nm*R*Tm

dividing both equations

(Pa/Pm)*(Va/Vm)= (Na/Nm)*(Ta/Tm)

Na/Nm = (Pa/Pm)*(Va/Vm) * (Tm/Ta) = (0.2/0.2)*(20/5)*(300/400) =  1*4*3/4 = 3

Na=3*Nm

therefore the moles of gas of the outflowing stream are (assuming that the methane does not react with the air):

Ng= Na+Nm = 4*Na

the mole fraction of A is

Xa= Na/Ng= Na/(4*Na) = 1/4 (25%)

and

Xm= 1-Xa = 3/4 (75%)

also for the exiting gas

Pg*Vg=Ng*R*Tg  = Na*R*Tg + Nm*R*Tg = Pa*Va * (Tg/Ta) + Pm*Vm * (Tg/Tm)

Vg = Va * (Pa/Pg)*(Tg/Ta) + Vm *(Pm/Pg)* (Tg/Tm)

Vg = 20 L/min * (0.2/0.1)*(370/400) + 5 L/min * (0.2/0.1)*(370/300) = 49.33 L/s

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this