The majority of gas-phase reactions are catalyzed by passing the reactant through a packed bed of catalyst particles.
The equation used most often to calculate pressure drop in a packed porous bed is the Ergun equation4,5:
Term 1 is dominant for laminar flow, and Term 2 is dominant for turbulent flow, where
In calculating the pressure drop using the Ergun equation, the only parameter that varies with pressure on the right-hand side of Equation (5-22) is the gas density, ρ. We are now going to calculate the pressure drop through a packed bed reactor.
Because the PBR is operated at steady state, the mass flow rate at any point down the reactor, (kg/s), is equal to the entering mass flow rate, (i.e., equation of continuity),
Recalling Equation (4-16), we have
Combining Equations (5-22) and (5-23) gives
Simplifying yields
where β0 is a constant that depends only on the properties of the packed bed (φ, DP) and the fluid properties at the entrance conditions (i.e., μ, G, ρ0, T0, P0).
For tubular packed-bed reactors, we are more interested in catalyst weight, rather than the distance z down the reactor. The catalyst weight up to a distance of z down the reactor is
where Ac is the cross-sectional area. The bulk density of the catalyst, ρb (mass of catalyst per volume of reactor bed), is just the product of the density of the solid catalyst particles, ρc, and the fraction of solids, (1 – φ):
ρb = ρc(1 – φ)
Using the relationship between z and W [Equation (5-26)] we can change our variables to express the Ergun equation in terms of catalyst weight:
Further simplification yields
where
We will use Equation (5-28) when multiple reactions are occurring or when there is pressure drop in a membrane reactor. However, for single reactions in packed-bed reactors, it is more convenient to express the Ergun equation in terms of the conversion X. Recalling Equation (4-20) for FT,
where, as before,
[Nomenclature note: y with the subscript A0—i.e., yA0—is the inlet mole fraction of species A, while y without any subscript is the pressure ratio—i.e., y = (P/P0)].
Substituting for the ratio (FT/FT0), Equation (5-28) can now be written as
We note that when ε is negative, the pressure drop ΔP will be less (i.e., higher pressure) than that for ε = 0. When ε is positive, the pressure drop ΔP will be greater than when ε = 0.
For isothermal operation, Equation (5-30) is only a function of conversion and pressure:
Recalling Equation (5-21), for the combined mole balance, rate law, and stoichiometry,
we see that we have two coupled first-order differential equations, (5-31) and (5-21), that must be solved simultaneously. A variety of software packages (e.g., Polymath) and numerical integration schemes are available for this purpose.
Analytical Solution. If ε = 0, or if we can neglect (εX) with respect to 1.0 (i.e., ), we can obtain an analytical solution to Equation (5-30) for isothermal operation (i.e., T = T0). For isothermal operation with ε = 0, Equation (5-30) becomes
Rearranging gives us
Taking y inside the derivative, we have
Integrating with y = 1 (P = P0) at W = 0 yields
(y)2 = 1 – αW
Taking the square root of both sides gives
Be sure not to use this equation if ε ≠ 0 or if the reaction is not carried out isothermally, where again
Equation (5-33) can be used to substitute for the pressure in the rate law, in which case the mole balance can then be written solely as a function of conversion and catalyst weight. The resulting equation can readily be solved either analytically or numerically.
If we wish to express the pressure in terms of reactor length z, we can use Equation (5-26) to substitute for W in Equation (5-33). Then
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