Problem Description: A gaseous fuel (A) is reacted over a catalyst packed-bed reactor with a gaseous oxidant (B) and is converted into gases C and D by the following reaction: + 2B 2 + The reaction is exothermic and carried out over a catalyst bed; there is a pressure drop along the length of the reactor that needs to be taken care of in the solution approach. The process is represented schematically below

Exothermic Gas Phase Catalytic Reactor

Define the parameters for the problem:

The list of the parameters that we will be used in the problem are: Cpa = 35.0 J/g-mol K,

R = 8.314 J/g-mol K

Fao = 5.0 g-mol/min

To = 450 K

HR = -22000 J/g-mol

Ea = 41,800 J/g-mol K

K = k1*Exp(Ea/R*((1/To)-(1/T)))

k1= 0.5 (appropriate units @ 450 K)

= 0.015 kg-1

Po = 10 atm

Cao = 0.271 g-mol/dm3

Cbo = 0.55555 g-mol/dm3

Cco = Cdo =0

Equations and Additional Information: Plug flow (no radial velocity, concentration, or temperature gradients) may be assumed inside the reactor. Based on the information provided, the following equation will be resulted. The reactor equations will be in the form of conversion of A denoted by X and temperature T and pressure P, all functions of axial location within the catalyst bed specified by catalyst weight W. General Reactor Design Equation based on mole balance for A: = F. = Simple Catalytic Reaction Rate Expression: = k. C 0.5 1.5 Rate Constant as a Function of A and T (Arrhenius expression): k = k1.exp{Ea/R*[(1/450) (1/T)]} (initial value of k at 450 k) Use the reaction stoichiometry and define the concentrations as functions of conversion (remember this is a gas phase reaction and the number of moles is changing, for every one mole of A reacted, two moles of B, Two moles of C and one mole of C change. Temperature is changed by the heat of reaction; pressure drop is caused by the flow resistance through the packed bed of catalyst. Here are the different relationships which you should be able to derive (for stoichiometry principals see Chapter4 of H Scot Fogler Book, 6th Edition): Ca = Cao*(1 X)*(P/Po)*(To/T) = 0.55555 0.271 Cb = Cao*[( X) *(P/Po)*(To/T) y = P/Po Cc = (2Cao*X)*y*(To/T) Cc = (Cao*X) *y*(To/T) Pressure Drop Expression: d(P/Po)/dW = -/2*(1-X)*(Po/P)*(T/To) or you can use dy/dW = -/2*(1-X)*y*(T/To) Energy Balance: dT/dW = ra*HR]/[Fao*Cpa] dQ/dW = ra*Hr These equations will result from the rate law, energy balance and the stoichiometry and should be used for examining the reactor conversion, pressure, temperature, and reaction rate and heat generated.

Questions: Plot the heat generated (Q), reduced pressure (y), temperature (T) and conversion along the reactor for catalyst weight from W = 0 kg to W = 20 kg. Around W = 8-12 kg (or may be other valus), you will observe a knee in the conversion profile. Explain why this occurs and what parameters affect the knee. Plot the concentration profiles for A, B, C and D for catalyst weight from W = 0 kg to 20 kg. To solve this problem a spreadsheet (e.g. Excel) or MATLAB or a computer programme (which you have learnt in your previous semesters) need to be used. The original spreadsheet, MATLAB or Programming files which were used for solving the assignment problem need to be submitted