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Reactor Conditions

  • 1.  Reactor Conditions

    Posted 9 days ago
    How does a change in the reactor operating pressure affect the residence time of the reactants in a packed bed catalytic reactor? 


  • 2.  RE: Reactor Conditions

    SENIOR MEMBER
    Posted 9 days ago
    The answer to this question is not straightforward.  What's the reaction, what's the packing?  Residence time in the pores of the catalyst?  Or the residence time determined by the flow rate of the reactants?

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    Svetlana Mitrovski PhDChemis,MSEchemEn
    Teaching Associate Professor
    University of Illinois
    Urbana IL
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  • 3.  RE: Reactor Conditions

    SENIOR MEMBER
    Posted 9 days ago

    Dear Apurva Bafana

    These packed bed catalytic reactors are generally tubular and are filled with solid catalyst particles, most often used to catalyze gas reactions. The chemical reaction takes place on the surface of the catalyst. The reaction conditions (feed stock concentrations, pressure, temperature profile, and residence time) can be found that lead to optimum yields and determine these parameters according to optimum yield at low input, Residence time also the reactor volume upon feed volume. In the case of selectivity-sensitive multi step reactions, any deviation from the optimum values inevitably leads to a decrease in yield. These must therefore be constructed so that the fixed bed or tube bundle is uniformly traversed and the gas residence time m each tube or each flow filament of the fixed bed is the same. The catalyst loading in such a way that placed on the catalyst is a low flow pressure loss. Mass transfer and heat transfer are strongly correlated with the pressure loss, because the pressure loss increases with the square of the flow velocity, a uniform distribution of the flow then occurs automatically. The pressure drop can be calculated from Hagen - Poiseuille equation. The pressure drop effect the residence time and pressure is the directly proportional to the reaction. The pressure drop depends very strongly on the void fraction of the packing. However weighting the performance function is not easy since for example. Small pressure losses, Uniform flow through the reactor and good mass- and heat-transfer properties generally represent opposing requirements.

    Regards,

    Prem Baboo



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    [Prem ] [Baboo] [DGM]
    [Mr.]
    [Dangote Fertilizer Projects]
    [Lekki] [Lagos]
    [Nigeria]
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  • 4.  RE: Reactor Conditions

    SENIOR MEMBER
    Posted 8 days ago
    The change in reactor residence time as a function of pressure depends on the compressibility factor (z) of the reacting fluid mixture.  Compressibility can be thought of as the change in fluid density as a function of pressure.

    For an ideal gas z = PV / nRT    (and z approaches 1).   
    P = pressure, V = volume, T = temperature, n = number of moles, R = ideal gas constant.

    Rearranging, molar density, (V / n) = RT / P , therefore density changes inversely with pressure.

    Other equations of state (EoS) can be used to more accurately calculate z for non-ideal fluids, particularly at very high pressures.

    If the fluid mixture is incompressible or has a very low compressibility such as a liquid, then increasing pressure does not increase the fluid density significantly so the residence time does not change significantly.  If the fluid mixture is a vapor which has high compressibility then for the same feed rate the vapor becomes denser (more compressed) so the residence time increases.

    If:
    t = reactor residence time, sec
    RV = reactor volume, cm3
    m = molar feed rate, mol/sec
    (V / n) = molar density, cm3/mol

    Then: t = RV / [m (V/n)]

    Hope that helps.  I would refer to an undergraduate kinetics course textbook for a refresher or to get more detail on various calculation methods.











  • 5.  RE: Reactor Conditions

    SENIOR MEMBER
    Posted 8 days ago
    Correction:  (V / n) is specific volume (the reciprocal of molar density), therefore density increases in proportion with increasing pressure.  Sorry about the mis-statement and any confusion.

    On Tue, Nov 5, 2019 at 11:25 AM John Waycuilis <jjwaycuilis@gmail.com> wrote:
    The change in reactor residence time as a function of pressure depends on the compressibility factor (z) of the reacting fluid mixture.  Compressibility can be thought of as the change in fluid density as a function of pressure.

    For an ideal gas z = PV / nRT    (and z approaches 1).   
    P = pressure, V = volume, T = temperature, n = number of moles, R = ideal gas constant.

    Rearranging, molar density, (V / n) = RT / P , therefore density changes inversely with pressure.

    Other equations of state (EoS) can be used to more accurately calculate z for non-ideal fluids, particularly at very high pressures.

    If the fluid mixture is incompressible or has a very low compressibility such as a liquid, then increasing pressure does not increase the fluid density significantly so the residence time does not change significantly.  If the fluid mixture is a vapor which has high compressibility then for the same feed rate the vapor becomes denser (more compressed) so the residence time increases.

    If:
    t = reactor residence time, sec
    RV = reactor volume, cm3
    m = molar feed rate, mol/sec
    (V / n) = molar density, cm3/mol

    Then: t = RV / [m (V/n)]

    Hope that helps.  I would refer to an undergraduate kinetics course textbook for a refresher or to get more detail on various calculation methods.