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28 May, 18:11

A stream 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?

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  1. 28 May, 22:03
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    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
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