Conversion and Reactor Sizing Lec 5 week 6

Definition of Conversion for the following reaction The reaction can be arranged as follows: Now we ask such questions as "How can we quantify how far the above reaction proceeds to the right?" or “How many moles of C are formed for every mole A consumed? A convenient way to answer these questions is to define a parameter called conversion. The conversion XA is the number of moles of A that have reacted per mole of A fed to the system.

Batch Reactor Design Equations in terms of conversion In most batch reactors. the longer a reactant stays in the reactor, the more the reactant is converted to product until either equilibrium is reached or the reactant is exhausted, Consequently. in batch systems the conversion X is a function of the time the reactants spend in the reactor. If NAO is the number of moles of A initially in the reactor then the total number of moles of A that have reacted after a time t is [NA0 *X]

Batch Reactor Design Equations in terms of conversion the mole balance on species A for a batch system is given by the following equation: reactant A is disappearing: therefore, we multiply both sides of Equation by -1 then

Batch Reactor Design Equations in terms of conversion For batch reactors. we are interested in determining how long to leave the reactants in the reactor to achieve a certain conversion X. To determine this length of time, we write the mole balance. Equation in terms of conversion. NA=NA0(1-XA) by differentiating the above equation with respect to time, remembering that NAo is the number of moles of A initially present and is therefore a constant with respect to time.

Batch Reactor Design Equations in terms of conversion To determine the time to achieve a specified conversion X This equation is now integrated with the limits that the reaction begins at time equal zero where there is no conversion initially (i.e., t = 0, X = 0).

Design Equations for Flow Reactors For a batch reactor. we saw that conversion increases with time spent in the reactor. For continuous-flow systems, this time usually increases with reactor volume. E.g. the bigger /longer the reactor, the more time it will take the reactants to flow completely through the reactor and thus, the more time to react. The conversion X is a Function of reactor volume V. If FA0 is the molar flow rate of species A fed to a system operated at steady state. The molar rate at which species A is reacting within the entire system will be FA0X.

Design Equations for Flow Reactors

Design Equations for Flow Reactors For liquid systems, CA0, is commonly given in terms of molarities, for example, CAO = 2 moll/dm3. For gas systems, CAo can be calculated from the entering temperature and pressure using the Ideal gas law.

Design Equations for Flow Reactors

Example(1 ) A gas of pure A at 830 kPa (8.2 atm) enters a reactor with a volumetric flow rate,v0 of 2 dm3/s. at 500 K. Calculate the entering concentration of A, CA0, and the entering molar flow rate. FAo.

solution For flow reactors (CSTR) For gas phase reactor.

Tubular Flow Reactor (PFR) For a flow system, FA has previously been given in terms of the entering molar flow rare FA0 and the conversion X By differentiate Substitute in the 1st equation to give the differential form of the design equation for a plug-flow reactor (PFR): We now separate the variables and integrate with the limits V = 0 when X = 0 to obtain the plug-flow reactor volume necessary to achieve a specified conversion X:

Packed-Bed Reactor

Example Consider the liquid phase reaction which we will write symbolically as A B The first order (-rA = k CA) reaction is carried out in a tubular reactor in which the volumetric flow rate, v, Is constant i.e. v =v0. (a) Derive an equation relating the reactor volume to the, entering and exiting concentrations of A the rate constant k, and the volumetric flow rate v. (b) Determine the reactor volume necessary to reduce the exiting concentration to 10% of the entering concentration when the volumetric flow rate is I0(dm3/min) and the specific reaction rate, k. is 0.23 min-1 .

Solution