Chapter 5: Isothermal Reactor Design: Conversion
Chapter 5 Example
Batch Reactor Optimization
The following irreversible liquid phase reaction follows an elementary rate law
A+B → C
Determine the minimum number of batch reactors 1.0 cubic meter in size to produce 10,000 moles of C in a 300 day period. The processing time to fill, empty and clean the reactor between batches is 4.5 hours. The feed is equal molar in A and B and the initial concentration of A is 1.0 moles per cubic meter. In a trial run it was found that 50% conversion was achieved in 2 hours.
Solution
Part 1. Find X as a function of t and then find k.
Batch
A+B → C, Gas Phase, Equil Molar Feed
Mole Balance:
\(\frac{dN_A}{dt} = r_A V\)
Rate Law:
\(-r_A = k C_A C_B\)
Stoichiometry: Gas, but \(V = V_0, \Theta_B = 1\)
\(C_A = \frac{N_A}{V_0} = \frac{N_{A0}(1 - X)}{V_0} = C_{A0}(1 - X)\)
The number of moles C formed in a batch is:
\(N_C = N_{A0} X = C_{A0} V X\)
Combine:
\(-r_A = kC_{A0}^2(1 - X)^2\)
\(\frac{dX}{dt} = \frac{-r_A V_0}{N_{A0}} = \frac{-r_A}{C_{A0}} = \frac{kC_A C_B}{C_{A0}}\)
\(\frac{dX}{dt} = kC_{A0}(1 - X)^2\)
Integrating:
\(\frac{X}{1 - X} = kC_{A0}t\)
Evaluate k from information given.
Evaluate:
\(\text{At } t = 2 \, \text{hr, then } X = 0.5\)
\(\frac{0.5}{1 - 0.5} = k \left( 1.0 \, \frac{\text{mol}}{\text{dm}^3} \right) 2 \, \text{hr}\)
\(k = 0.5 \, \frac{\text{dm}^3}{\text{mol} \cdot \text{hr}}\)
The conversion for a reaction time \(t_R\)
\(X = \frac{kC_{A0}t_R}{1 + kC_{A0}t_R}\)
Part 2. Find the minimum number of reactors
The total time to carry out one batch \( t \) is the reaction time \( t_R \) plus the processing time \( t_P \), which is the sum of the times to empty (\( t_e \)), to clean (\( t_{\text{clean}} \)), and to fill (\( t_f \)):
\( t = t_R + \overbrace{t_f + t_e + t_{\text{clean}}}^{t_P} \)
The total processing time, tP, between reactions is tP, = 4.5 hrs. How many 1 m3 reactors do you need to make 10,000 moles of C in 300 days? Consider a single reactor. The number of moles of C formed in one batch
\( N_C = N_{A0}X = C_{A0}VX \)
Total number of C moles made in one reactor for a 24 hour work day over 300 days is
\( N_C = C_{A0}VnX \)
n = number of batches in 300 days
n = 300 days x no. of batch per day
\( n = \frac{300 \cdot 24}{t_R + t_P} \)
\( N_{C_T} = C_{A0} V \cdot \frac{300 \cdot 24}{t_R + t_P} \cdot \frac{k C_{A0} t_R}{1 + k C_{A0} t_R} \)
Rearranging:
\( \frac{N_{C_T}}{[30 \cdot 24 \cdot k C_{A0}^2 V]} = \frac{1}{t_R + t_P} \cdot \frac{t_R}{1 + k C_{A0} t_R} \)
For max production rate
\( \frac{\frac{dN_{C_T}}{[30 \cdot 24 \cdot k C_{A0}^2 V]}}{dt_R} = 0 = \frac{t_R + k C_{A0} t_R^2 + t_P + k C_{A0} t_R t_P - t_R - 2k C_{A0} t_R^2 - 0 - k C_{A0} t_R t_P}{[(t_P + t_R)(1 + k C_{A0} t_R)]^2} \)
\( 0 = \frac{-k C_{A0} t_R^2 + t_P}{[(t_P + t_R)(1 + k C_{A0} t_R)]^2} \)
Solving for tR
\( t_R = \left(\frac{t_P}{kC_{A0}}\right)^{1/2} = \left[\frac{4.5}{(0.5)(1)}\right]^{1/2} = 3 \)
\( X = \frac{\sqrt{kC_{A0}t_P}}{1 + \sqrt{kC_{A0}t_P}} = \frac{\sqrt{kC_{A0}t_P}}{1 + \sqrt{kC_{A0}t_P}} \)
\( N_{C_T} = C_A V \frac{300 \cdot 24}{t_R + t_P} \cdot \frac{\sqrt{kC_{A0}t_P}}{1 + \sqrt{kC_{A0}t_P}} \)
\( X = \frac{\sqrt{kC_{A0}t_P}}{1 + \sqrt{kC_{A0}t_P}} = \frac{\sqrt{(0.5)(1)(4.5)}}{1 + \sqrt{(0.5)(1)(4.5)}} = 0.61 \)
\( N_{C_T} = \left(\frac{1 \, \text{mol}}{\text{m}^3}\right)\left(1 \, \text{m}^3\right) \left(\frac{300 \, \text{days} \left(\frac{24 \, \text{h}}{1 \, \text{day}}\right)}{4.5 \, \text{h} + 3 \, \text{h}}\right) (0.61) = 585.6 \, \text{moles} \)
The number of moles 1 reactor can make in 300 days is
\( N_C = 585.6 \, \text{moles} \)
The number of reactors, m, to make 10,000 moles of C in 300 days is
\( mN_{C_T} = 10{,}000, \quad m = \frac{10{,}000}{N_{C_T}} \)
\( m = \frac{10{,}000}{585.6} = 17 \, \text{reactors} \)