Isothermal Reactor Design – Part 2 Pressure Drop In Reactors

Design a PBR In case of 2nd order rxn, gas phase, isothermal mole balance rate laws stoichiometry combination Need to relate pressure drop to catalyst weight (in order to determine conversion)

Design a PBR Ergun equation = pressure (kPa) = inlet pressure (kPa) = temperature (K) = inlet temperature (K) = porosity = volume of void = void fraction = volume of solid total bed volume total bed volume = cross sectional area (m2) =diameter of particle in the bed, ft (m) =viscosity of gas passing through the bed, (kg/m.s) =length down the packed bed of pipe, ft (m) =superficial velocity = volumetric flow ÷ cross-sectional area of pipe (m/s) =gas density (kg/m3) = solid density (kg/m3) = inlet gas density = = superficial mass velocity, (kg/m2.s) = 32.174lbmft/s2lbf (conversion factor)

Design a PBR For isothermal operation, we have two sets of equation with two unknowns, X & P (1) (2) Special case: if ε=0, an analytical solution to second equation is obtained as follows Used only when ε=0

Design a PBR In case of 2nd order rxn, gas phase, isothermal mole balance rate laws stoichiometry Combination Solve By integration; When ε=0

Design a PBR Solving for conversion gives: Solving for catalyst weight,

For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop. ↑ W, ↓P, ↑ΔP ↑ ΔP, ↓P, ↓CA, ↓-rA ↑ W, ↓P, ↓CA ↑W, ↑X

Effect of pressure drop on the conversion profile Consider a packed bed column with a second order reaction is taking place in 20 meters of a 1 ½ schedule 40 pipe packed with catalyst. 2A  B + C The following data are given: Inlet pressure, P0 = 10 atm=1013 kPa Entering flowrate, v0 = 7.15 m3/h Catalyst pellet size, Dp = 0.006 m Solid catalyst density: ρc = 1923 kg/m3 Cross sectional area of 1 ½ -in schedule 40 pipe: AC =0.0013 m2 Pressure drop parameter, β0 = 25.8 kPa/m Reactor length, L = 20 m Void fraction = 45% Calculate the conversion in the absence of pressure drop. Calculate the conversion accounting for pressure drop. What is conversion in part (b) if the catalyst particle diameter were doubled. The entering concentration of A is 0.1 kmol/m3 and the specific reaction rate is .

Effect of pressure drop on the conversion profile (a) Conversion for ΔP = 0 α = 0 thus, [Volume of catalyst] x [catalyst density]

Effect of pressure drop on the conversion profile (b) Conversion with pressure drop

Effect of pressure drop on the conversion profile (c) Conversion when catalyst diameter were doubled. (increase by a factor of 2, ) From , thus dominant

Effect of pressure drop on the conversion profile (c) Conversion when catalyst diameter were doubled. Thus, Conversion increases from 0.693 to 0.774 by increasing catalyst diameter by a factor of 2. Increasing particle size decrease the pressure drop parameter, increase conversion & reaction rate