a. Use the experimental data obtained at day 3 and 6.
Time, d C, mole/L
Ct112Ct21
2 <dCt
dt
0 250
1 70
2 42
3 30 (23 2 42)/2 29.5
4 23
5 18
6 16 (13 2 18)/2 22.5
7 13
8 12
b. Substitute and solve for n.
n5 log (9.5)2log (2.5)
log (30.0)2log (16.0)52.07 use n5 2 c. The reaction is second order.
d. The reaction rate constant is 1
C2 1 Co5kt 1
422 1
250 5k(2)
k 5 0.0103/d, use k 5 0.010/d
In the applications described above, the initial concentration of a constituent is generally known. However, in the conventional BOD test, described in Chap. 2, both UBOD (ulti- mate biological oxygen demand) and k1 are unknown. To determine these values, the usual procedure is to run a series of BOD measurements with time. Using these measurements, the UBOD and k1 values can be determined using a number of methods including the method of least-squares, the method of moments, the daily-difference method, the rapid- ratio method, the Thomas method, and the Fujimoto method, as discussed in Sec. 2–6 in Chap. 2.
1–11 Introduction to Treatment Process Modeling
43
Batch Reactor with Reaction
The derivation of the materials mass balance equation for a batch reactor [see Fig. 1–7 (a)]
for a reactive constituent is written as follows:
Accumulation 5 inflow 2 outflow 1 generation dC
dtV5QCo2QC1rCV (1–57) Because Q 5 0 the resulting equation for a batch reactor is
dC
dt 5rC (1–58) Before proceeding further, it will be instructive to explore the difference between the rate of change term that appears as part of the accumulation term and the rate of generation or utilization or decay term. In general, these terms are not equal, except in the special case of a batch reactor in which there is no inflow or outflow from the control volume. The key point to remember is that when flow is not occurring, the concentration per unit volume is changing according to the applicable rate expression. On the other hand, when flow is occurring, the concentration in the reactor is also being modified by the inflow and outflow from the reactor.
If the rate of reaction is defined as first order (i.e., rc52kC ), integrating between the limits C 5 Co and C 5 C and t 5 0 and t 5 t yields
#
CC55CCodC
C 5 2k
#
t5t5t0dt5kt (1–59) The resulting expression is
C Co
5e2kt (1–60) Equation (1–60) is the same as the BOD equation Eq. (2–59) considered subsequently in Chap. 2.
Complete-Mix Reactor with Reaction
The general form of the mass-balance equation for a complete-mix reactor as shown on Figs. 1–4(b) and 1–5(a-1), in which the liquid in the reactor is mixed completely, follows:
Accumulation 5 inflow 2 outflow 1 generation dC
dtV5QCo2QC1rcV (1–61) Assuming first order removal kinetics (rc52kC), Eq. (1–61) can be rearranged and writ- ten as follows
C9 1bC5Q
VCo (1–62) where C9 5dC/dt
b5k1Q/V
To solve Eq. (1–62) both sides of the expression are multiplied by the integrating factor ebt: ebt(C9 1bC)5 Q
VCoebt (1–63)
The left side of the above expression can be written as a differential as follows:
(Cebt)9 5Q
VCoebt (1–64) The differential sign is removed by integrating the above expression
Cebt5 Q
VCo
#
ebt (1–65)Integration of Eq. (1–65) yields Cebt5 Q
V Co
bebt1K (1–66) Dividing by ebt yields
C5Q V
Co
b 1Ke2bt (1–67) But when t 5 0, C 5 Co and K is equal to
K5Co2Q V
Co
b (1–68)
Substituting for K in Eq. (1–68) and simplifying yields the following expression, which is the non-steady state solution of Eq. (1–61):
C5Q V
Co
b(12e2bt)1Coe2bt (1–69) The solution to Eq. (1–61) under steady-state conditions (i.e., the rate accumulation term is equal to zero [dC/dt 5 0)] is given below.
C5 Co
[11k(V/Q)] 5 Co
(11kt) (1–70) It should also be noted that when t S ', Eq. (1–69) becomes the same as Eq. (1–67).
Complete-Mix Reactors in Series with Reaction
When complete-mix reactors are used in series, the steady-state solution is of concern as it is used for design. Two approaches are presented for the analysis of reactors in series:
(1) analytical and (2) graphical. The graphical approach also applies to cascades of reac- tors, as discussed for mass transfer equilibria.
Analytical Solution. The steady-state form of the mass balance for the second reac- tor of the three reactor system (see Fig. 1–10), is given by
Accumulation 5 inflow 2 outflow 1 generation dC2
dt V
2505QC12QC21rc
V
2 (1–71) Assuming first order removal kinetics (rc52kC2), Eq. (1–71) can be rearranged and solved for C2 yielding
C25 C1
[11(kV/2Q)] (1–72)
But from Eq. (1–70), the value of C1 is equal to C25 Co
[11(kV/2Q)] (1–73)
Combining the above two expressions yields C25 Co
[11(kV/2Q)]2 (1–74)
For n reactors in series the corresponding expression is Cn5 Co
[11(kV/nQ)]n5 Co
[11(k/2t)]n (1–75) For example, consider three 1000 m3 complete-mix reactors in series with a flowrate of 100 m3/d and first order kinetics with a reaction rate coefficient value of k 5 0.1/d. Using Eq. (1–75), the effluent concentration from the third reactor, assuming the starting concen- tration was 100 mg/L is
C35 Co
[11(kV/3Q)]35 100
e11 c(0.1/d)(3000 m3) (33100 m3/d)d f
3512.5 mg/L
Solving Eq. (1–75) for the detention time yields t5 V
Q5 c 1
(Cn/Co)1/n21d an
kb or t5 c aCo
Cn
b
1 n
21d an
kb (1–76) Graphical Solution. The graphical solution for 3 (or for n) reactors in series is obtained as follows. For a single reactor, Eq (1–71) can be written as follows:
Accumulation 5 inflow 2 outflow 1 generation
05QCo2QC12 rcV (1–77) The first step in developing a graphical solution is to draw a graph of rc versus C (see Fig. 1–10). To plot rc versus C, Eq. (1–77) is now rewritten as follows:
rc5 2Q
V(C12Co)5 21
t(C12Co) (1–78) The above equation can be represented graphically by a straight line drawn from the point rc5 0 and C 5 100 mg/L with a slope of 21/t. The line drawn will intersect the graph of rc versus C at rc5 5.0 and C15 50 mg/L as shown on Fig. 1–11. The value C15 50 mg/L is the solution of Eq. (1–78) for a single rector for the stated conditions used to derive the
Inflow Outflow
Q, Co Q, C1 Q, C2 Q, Cn
Mixer Mixer Mixer
Vi Vi Vi
Figure 1–10
Definition sketch for the analysis of complete-reactors in series.
1–11 Introduction to Treatment Process Modeling
45
Figure 1–11
Graphical analysis used to determine the effluent concentration from a series of complete-mix reactors.
C3
C2 C1
–r c
= kC
0 2
0 4 6 8 10
20 40 60 80 C0 = 100
Concentration, C, mg/L
Slope = – 1/t = 0.1 d –1 –rc = kC = (0.1/d)(100 mg/L)
analytical solution, presented previously. If the procedure is repeated for a second and a third reactor, the final effluent concentration from the third reactor is found to be 12.5 mg/L, which is the same as the analytical solution determined above. The graphical approach is especially useful in solving phase separation processes as described previously in Sec. 1–9. To use the graphical procedure, the reaction rate coefficient must be a function of a single variable (e.g., C). Use of both the analytical and graphical methods of analysis are illustrated in Example 1–3. Additional details on the graphical solution of design equa- tions may be found in Eldridge and Piret (1950) and Smith (1981).
Analysis of Reactors in Series Using Both an Analytical and Graphical Approach Two 1000 m3 complete-mix reactors are to be used in series with a flow- rate of 500 m3/d and second order kinetics with a k value of 0.01/d. Determine the effluent concentration from the second reactor assuming the starting concentration is 100 mg/L.
1. Determine the effluent concentration from the series of two complete-mix reactors analytically.
a. At steady-state, the mass balance for the first complete-mix reactor is:
05QCo2QC12kcC21V
Substituting the given values and solving for C1 yields 05 (500 m3/d)
1000 m3 (100 mg/L)2 (500 m3/d)
1000 m3 C12(0.01/d)C21 C15 50 mg/L
b. At steady-state, the mass balance for the second complete-mix reactor is 05QC12QC22kC22V
Substituting the given values and solving for C2 yields 05 (500 m3/d)
1000 m3 (50 mg/L)2 (500 m3/d)
1000 m3 C22(0.01/d)C22
C25 30 mg/L EXAMPLE 1–3
Solution
2. Determine the effluent concentration from the series of two complete-mix reactors graphically.
a. Prepare a plot of rc versus C as shown below:
0 0.2 0.4 0.6 0.8 1
0 20 40 60 80 100
Concentration C, mg/L rC = kC2
rC = 0.01C2
b. Linearize the mass balance equation from Step 1a rc52Q
V(C12Co)
c. On the plot prepared above draw a straight line from the point rc5 0 and C 5 100 mg/L with a slope of 2Q / V equal to 20.5/d [2(500 m3/d)/1000 m3]. The line drawn intersects the graph of rc versus C at rc5 0.25 and C15 50 mg/L.
Repeating the above procedure the final effluent concentration from the second reactor is found to be 30 mg/L, which is the same as the analytical solution determined in Step 1.
Ideal Plug-Flow Reactor with Reaction
The derivation of the materials balance equation for an ideal plug-flow reactor, in which the concentration, C, of the constituent is uniformly distributed across the cross-sectional area of the control volume, and there is no longitudinal dispersion, can be illustrated by considering the differential volume element shown on Fig. 1–6. For the differential volume element ¢V shown on Fig. 1–6, the materials balance on a reactive constituent C is written as follows:
Accumulation 5 inflow 2 outflow 1 generation 0C
0t≤V5QCkx2QCkx1≤x1rc≤V (1–79) Accumulation 5 inflow 2 outflow 1 generation
where 0C/0t 5 change in constituent concentration with time, ML23T21 (g/m3?s) C 5 constituent concentration, ML23 (g/m3)
¢V 5 differential volume element, L3 (m3) Q 5 volumetric flowrate, L3T21 (m3/s)
rc5 reaction rate for constituent C, ML23T21, (g/m3?s)
Substituting the differential form for the term QC Z x1¢x in Eq. (1–79) results in 0C
0t≤V5QC2QaC1≤C
≤x≤xb 1rc≤V (1–80) 1–11 Introduction to Treatment Process Modeling
47
Substituting A¢x for ¢V and dividing by A and ¢x yields 0C
0t 5 2Q A
≤C
≤x 1rc (1–81) Taking the limit as ¢x approaches zero yields
0C 0t 5 2Q
A 0C
0x 1rc (1–82) If steady-state conditions are assumed (,C/ ,t 5 0) and the rate of reaction is defined as rc52kC n, integrating between the limits C 5 Co and C 5 C and x 5 0 and x 5 L yields
#
CCodC
Cn5 2kA Q
#
0Ldx5 2k AL
Q 5 2kV
Q5 2kt (1–83) Equation (1–83) is the steady-state solution to the materials balance equation for a plug- flow reactor without dispersion. If it is assumed that n is equal to one, Eq. (1–83) becomes
C Co
5e2kt (1–84) which is equivalent to Eq. 1–60, derived previously for the batch reactor.
Comparison of Complete-Mix and Plug-Flow Reactors with Reaction
The combined effect of reactor type (e.g., complete-mix versus plug-flow) and kinetics is also of interest. The total volume required for various removal efficiencies for first order kinetics, using 1, 2, 4, 6, 8, or 10 reactors in series is reported in Table 1–13 and shown graphically on Fig. 1–12. The corresponding volume required for a plug-flow reactor is also reported in Table 1–13 As shown in Table 1–13, as the number of reactors in series is increased, the total reactor volume required approaches that of a plug-flow reactor. A com- parison of reactor types for second-order kinetics is examined in Example 1–4.
It should be noted, however, that for zero order kinetics the volume of the two reactors will be the same. It is also important to note that biological processes do not obey the results presented in Table 1–13 (i.e., plug-flow is more efficient than complete-mix) because biological process are modeled using BOD and COD which includes microbial products in addition to any residual substrate. As a result, the volumes required for the two reactors will be the same. The use of a plug-flow reactor, or mixed cells in series, is often favored to help control the growth of filamentous organisms (see discussion in Chap. 7).
Table 1–13