© B. Subramaniam, C&PE Dept., University of Kansas Notes on Chemical Reaction Engineering

5.1

5. Semi-Batch Reactor 5.1 Mathematical Model

Consider the liquid-phase reaction A + B ---> C in which reactant A is continuously added to a stirred vessel containing reactant B (Figure 5.1). It is assumed that the reactor is operated isothermally at a temperature T. It is required to determine how the concentrations of the reactants (CA, CB) change in the reactor with time (see also example 4.10 of Fogler's text).

F , v ,C Ao Ao

V , C , C o Bi

o

Ai at t = 0

V, C , C BA at any t

Figure 5.1. Semi-Batch Reactor Schematic

The definitions of the variables in Fig. 5.1 are as follows:

vo volumetric flow rate of feed stream containing species A, vol/time CAo concentration of species A in feed stream, mol/vol CAi concentration of species A in reactor initially (t = 0), mol/vol CA concentration of species A in reactor at any time t, mol/vol CBi concentration of species B in reactor initially (t = 0), mol/vol CB concentration of species B in reactor at any time t, mol/vol Vo volume of reaction mixture initially (t = 0), vol V volume of reaction mixture at any time t, vol Nj Moles of species j in reactor at any time t, mol

© B. Subramaniam, C&PE Dept., University of Kansas Notes on Chemical Reaction Engineering

5.2

A material balance for species A around the reactor yields,

FAo − FA + rA V = dNAdt =

d (V CA)

dt(5.1)

vo CAo + rA V = V dCAdt + CA

dVdt

(5.2)

0

The reaction volume changes with time as reactant A is added. An expression for the

volume change with time is obtained by writing an overall mass balance as follows:

ρ V = ρo Vo + ∫t

0 ρf vo dt (5.3)

If ρ ≅ ρo ≅ ρf (constant−density reactions), eq. (5.3) becomes

V = Vo + vo t (5.4)

⇒dVdt = vo (5.5)

Substituting eqs. (5.4−5) in eq. (5.2) yields

vo CAo + rA (Vo + vo t) = (Vo + vo t) dCAdt + vo CA (5.6)

⇒dCAdt =

vo (CAo − CA)

(Vo + vo t) + rA = f (Cj' s, t) (5.7)

Initial Condition: t = 0, CA = CAi

For first-order rate expression (i.e., if B is in excess), - rA = k CA. In such a case, eq. (5.7) may be solved analytically to yield CA (t), as given by eq. (4.61) of Fogler's text.

If the reaction is second-order, eq. (5.7) must be solved numerically. If - rA = k CA CB, eq. (5.7) becomes

dCAdt =

vo (CAo − CA)

(Vo + vo t) − k CA CB = f1 (CA, CB, t) (5.8)

Eq. (5.8) involves two variables, CA and CB. Hence, the material balance for species

'B' is needed to solve for CA and CB. A material balance for species B yields:

© B. Subramaniam, C&PE Dept., University of Kansas Notes on Chemical Reaction Engineering

5.3

FBo − F B + rB V = dNB

dt = d (V CB)

dt(5.9)

+ rB V = V dCBdt + CB

dVdt

(5.10)

0 0

Recognizing that - rB = k CA CB, and substituting eqs. (5.4-5) into eq. (5.10) yields:

(− k CA CB) (Vo + vo t) = (Vo + vo t) dCBdt + vo CB (5.11)

⇒dCBdt = − k CA CB −

vo CB(Vo + vo t)

= f2 (CA, CB, t) (5.12)

Initial Condition: t = 0, CA = CAi, CB = CBi

Eqs. (5.11 and 5.12) may be solved simultaneously using a numerical scheme such as fourth-order Runge-Kutta or Mathcad software.

Alternate Approach: CB can be expressed in terms of CA as follows:

CB(t) = NBV =

Moles of B in reactor at time tVolume of reaction mixture at time t

= (Moles of B initially) − (Moles of B reacted)

V

= (Moles of B initially) − (Moles of A reacted)

V

= (Vo CBi) − {(Vo CAi + vo CAo t) − V CA}

V(5.13)

⇒ CB = CA − {Vo (CAi − CBi) + vo CAo t}

V(5.14)

Substituting eq. (5.14) in eq. (5.8) makes the RHS of eq. (5.8) a function of only CA. In other words, only one ODE needs to be solved. In general, it is possible to reduce the number of ODEs to one in the case of an isothermal semi-batch reactor in which a single reaction occurs. However, the resulting equation would still require a numerical solution.