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Reactor design chemical engineering question ?

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2 Answers

  • 4 weeks ago

    A=ammonia, B=oxygen, C=nitrogen, D=water vapor

    Basis; 100 moles/s of dry air (FD0=0 moles/s)is fed to the CSTR having FB0=21moles/s of B and FC0=79moles/s of C 

    Reaction is A+0.75B->0.5C+1.5D 

    Delta= 0.5+1.5-0.75-1=0.25

    This means 21moles/s of B will stoichiometrically need 21x4/3 =28moles/s of A=FA0 moles/s

    So total molar flow rate of gas fed to the CSTR is 128moles/s

    Mole fraction of A in feed gas is 28/128=0.21875=yA0

    Mole fraction of B in feed gas is 21/128=0.16406=yB0

    Mole fraction of C in feed gas is 79/128=0.61719=yC0

    Mole fraction of D in feed gas is 1-yA0-yB0-yC0=0=yD0

    epsilon=yA0 x delta=0.0546875=0.055

    The feed gas enters the CSTR at 1200K and 3atm absolute pressure

    Assuming ideal gas law at the above temperature and pressure conditions and for gases A, B and C, Volumetric flow rate of the feed gas is v0=(128 x 8.314 x 1200)/(3x101325)=4201103.4cm3/s

    Molar concentration of A in feed gas is CA0=28/4201103.4=6.664916e-6moles/cm3

    X=conversion of reactant A, x=multiplication

    v=v0(1+epsilonxX)=cm3/s of gas leaving the reactor




    At T=1200K, k=19830268.4141 (moles/cm3)^1.95/s

    Mole Balance on reactant A using well-mixed steady state model gives FA0=-rAxV+FA0x(1-X)


    reactor volume V cm3= FA0xX/-rA, V=v0xCA0xX/-rA

    residence time tau seconds = V/v0=CA0xX/-rA

    Moles/s of A out from the reactor =FA0x(1-X)=FA=28x(1-X)

    Moles/s of B out from the reactor=FB0-0.75xFA0xX=FB=0.75xFA0x(1-X)=21x(1-X)

    Moles/s of C out from the reactor=FC0+0.5xFA0xX=FC=79+14xX

    Moles/s of D out from the reactor=FD0+1.5xFA0xX=FD=0+42xX



    using the value of k =19830268.411and above two expressions for CA and CB find -rA

    tau= CA0xXx(1/k)x[v0x(1+epsilonxX)/FA0/(1-X)]^1.95x(1/0.75)^1.2



    For X=0.95, tau = 1.83s

    (FA0xCpA+FB0xCpB+FC0xCpC)x(1200-300)+316600xFA0xX-(FAxCpA+FBxCpB+FCXCpC+FDxCpD)x(1200-300)=Heat removal rate.

    CpA=37.2J/mole/K, CpB=29.4J/mole/K, CpC=29.1J/mole/K, CpD=43.7J/mole/K


    For X=0.95 heat of reaction is 316600x28x0.95=8421560J/s






    Heat removal rate=7922411J/s =282943.25 J/mole of A fed

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  • Anonymous
    1 month ago

    It's not a question though. 

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