VOLUME 2 Post-Treatment, Reuse, and Disposal
3.5 Plug Flow Reactors with Dispersion and Conversion
Plugflow with dispersion is an intermediateflow regime between the ideal plugflow and ideal completely mixedflow. It is also referred to asintermediate-mixedflow,arbitraryflow, nonidealflow, orflow with axial dispersion. All dispersion problems are three dimensional, and dispersion coefficient varies with direction and degree of turbulence. To simplify the analysis, one-dimensional dispersion with longitudinal mixing (axial mixing) is usually assumed.
3.5.1 Flow Regime and Dispersion of Tracer
If a slug of conservative tracer is released into a rector near the inlet, a front is formed. The front is not straight because of longitudinal transport of material due to turbulence and molecular diffusion. This initial effect on front formation is shown inFigure 3.29a. As the tracer moves through the reactor, the mixing and dispersion lengthen the zone of tracer (Figure 3.29b). Finally, the tracer exits from the effluent zone (Figure 3.29c).
3.5.2 Performance Evaluation of Sedimentation Basin
Tracer Exit Profile: The tracer exit profile (or observed recovery of tracer) from a sedimentation basin with dispersion is shown inFigure 3.30. Several terms and relationships are used to define theflow char- acteristics of the basin.8These are:
1. Flow through time or standard detention time. It is the time to reach the peak tracer concentration.
It is also calledmodaltimeθSD
2. Average detention time. This time corresponds to thecentroidof the observed recovery curve of the tracer; it is also calledmediantime (θAD).
3. Nominal, theoretical, or mean detention time. It is the hydraulic retention time expressed by V/Qorθ.
The following observations can be made from the tracer profile:
• In absence of short-circuiting, the standard, average, and nominal detention times must coincide.
• The ratios of standard and average detention time to mean detention time are,1.
• The ratios of difference between nominal and standard, (θ−θSD)/θand nominal and average (θ−θAD)/θto nominal detention time increase with dispersion.
• In basins with low short circuiting, the relationship between the detention times may be expressed by Equation 3.32.9,10
(a)
(b)
(c)
FIGURE 3.29 Progress of slug tracer input in a plugflow reactor with dispersion: (a) tracer front formation, (b) tracer front movement, and (c) tracer exit.
θSD=θAD−3(θ−θAD), (3.32) where
θSD=standard detention time, min θAD=average detention time, min θ =nominal detention time, min
The actual shape of tracer exit profile can only be determined by tracer studies because the shape is a function of multiple parameters that may include (1) geometry and relative dimensions of the reactor, (2) mixing intensity, (3) dispersion coefficient, (4) dead volume, (5) short circuiting, and (6) density stratification.
Most reactors used in water and wastewater applications exhibitflow regimes that are plugflow with dispersion. Engineers prefer to design them as a plugflow system and apply a correction factor to simulate the dispersed plugflow condition.5The correction factor may vary from 0.1 for unbaffled low length-to- width ratio, to 1.0 for very high length-to-width ratio (pipelineflow) basins. The standard detention time for a well-designed tank is expected to be larger than 30% of nominal detention time.
Mostly tracer tests are utilized to estimate the actual efficiency factor. The slug dye tracer profiles for several types of sedimentation basins are shown inFigure 3.31.11
Center of gravity
1.0 t/θ
C/C0 θSD
θAD θ
FIGURE 3.30 Slug tracer exit curve for plugflow with dispersion.
0 0.25 0.75 1.00
0 0.25 0.50 0.75 1 1.25 1.50 1.75 2.00
A B C
D
E F
0.50 C/C0
t/q FIGURE 3.31 Typical slug dye tracer curves for several tanks.
Note:Curve Ais for ideal CFSTR (complete dispersion);Curve Bis for radialflow circular tank (large dispersion);Curve Cis for wide rectangular tank with relatively shallow depth (medium to large dispersion);Curve Dis for long narrow rectangular tank (medium or intermediate dispersion);Curve Eis for around-the-end baffled tank (small dispersion); andCurve Fis for ideal PF tank (no dispersion).
Performance Evaluation fi
tation basin. Fielder and Fitch developed empirical relationships between the dye tracer test data and the sedimentation efficiency.8These relationships are expressed by Equations 3.33 through 3.39.
(Y-function)=
C−1 2ΔC
Δt, mg·min
L (3.33)
(Z-function)=(Y-function) t−1
2Δt
, mg
L (3.34)
(W-function)=
(Y-function) t−1 2Δt
, mg·min2
L (3.35)
θSD=Σ(Y-function)
Σ(Z-function) (3.36)
θAD=Σ(W-function)
Σ(Y-function) (3.37)
SDE=θSD
θ ×100% (3.38)
PDV=θ−θAD
θ ×100% (3.39)
where
C =dye concentration, mg/L
ΔC =change in dye concentration, mg/L Δt =time increment, min
SDE =standard detention efficiency, % PDV=percent dead volume, %
EXAMPLE 3.43: PERFORMANCE OF A SEDIMENTATION BASIN WITH CONSERVATIVE DYE TRACER
A relatively long sedimentation basin has a volume of 100 m3, and it receives aflow of 2 m3/min. A 320 gram slug of a conservative tracer is applied. The effluent samples were collected at suitable time intervals and concentration of dye was measured in each sample. The results are summarized inTable 3.7.
a. CalculateC0andθ.
b. Draw the dye tracer profile curve as mg/L versus time.
c. Draw the dye tracer profile curve asC/C0versust/θ.
d. Determine the standard detention efficiency, and percent dead volume in the basin.
Solution
1. Calculate the initial concentrationC0of a slug input from Equation 3.9b.
C0is the average theoretical initial concentration of dye if it is completely mixed in the basin.
C0=wtracer
V = 320 g
100 m3=3.2 g/m3or mg/L
2. Calculate the nominal detention timeθfrom Equation 3.10c.
θ=V
Q= 100 m3
2 m3/min=50 min
3. Plot the tracer profile ofC, mg/L versus timet, min.
The tracer concentration in the effluent with respect to time of samplingtand other parameters are calculated inTable 3.7. These values are plotted inFigure 3.32a.
TABLE 3.7 Tracer Profile Data and Calculated Values of Parameters (Example 3.43) t,
min C, mg=L t
θ C C0
Δt, min
mg=LΔC, C1=2ΔC,
mg=L Y-function,
mg·min=L t1=2Δt, min
Z-function,
mg=L W-function, mg·min2=L
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11)
0 0 0 0 – – – – – – –
4 0 0.08 0 4 – – – – – –
7 0.10 0.14a 0.03b 3c 0.10d 0.05e 0.15f 5.50g 0.027h 0.83i
10 0.25 0.20 0.08 3 0.15 0.18 0.53 8.50 0.062 4.46
13 0.83 0.26 0.26 3 0.58 0.54 1.62 11.5 0.141 18.6
16 1.48 0.32 0.46 3 0.65 1.16 3.47 14.5 0.239 50.2
20 2.45 0.40 0.77 4 0.97 1.97 7.86 18.0 0.437 141
25 2.85 0.50 0.89 5 0.40 2.65 13.3 22.5 0.589 298
30 2.32 0.60 0.73 5 0.53 2.59 12.9 27.5 0.470 355
35 1.75 0.70 0.55 5 0.57 2.04 10.2 32.5 0.313 331
40 1.40 0.80 0.44 5 0.35 1.58 7.88 37.5 0.210 295
50 0.92 1.00 0.29 10 0.48 1.16 11.6 45.0 0.258 522
60 0.62 1.20 0.19 10 0.30 0.77 7.70 55.0 0.140 424
70 0.41 1.40 0.13 10 0.21 0.52 5.15 65.0 0.079 335
80 0.28 1.60 0.09 10 0.13 0.35 3.45 75.0 0.046 259
90 0.18 1.80 0.06 10 0.10 0.23 2.30 85.0 0.027 196
100 0.12 2.00 0.04 10 0.06 0.15 1.50 95.0 0.016 143
110 0.03 2.20 0.01 10 0.09 0.08 0.75 105 0.007 78.8
118 0 2.36 0.00 8 0.03 0.02 0.12 114 0.001 13.7
ΣY-function
¼90.4
ΣZ-function
¼3.06
ΣW-function
¼3465
a 7 min 50 min¼0:14
b0:10 mg=L 3 mg=L ¼0:13
c7 min–4 min¼3 min
d0.10 mg=L–0 mg=L¼0.10 mg=L
e0:10 mg=L1
20:1 mg=L¼0:05 mg=L
f0.05 mg=L3 min¼0.15 mg·min=L
g7 min1
23 min¼5:5 min
h0:15 mgmin=L
5:50 min ¼0:027 mg=L
i0.15 mg·min=L5.5 min¼0.83 mg·min2=L
4. Plot the tracer profileC/C0versust/θ.
The value ofC/C0andt/θat different time intervals are calculated in Columns (3) and (4) of Table 3.7. The results are plotted inFigure 3.32b.
5. Calculate the standard detention efficiency and fraction dead volume.
The procedure for calculatingY-function,Z-function, andW-function from Equations 3.33 through 3.35 is tabulated in Columns (5) to (11) ofTable 3.7.
Standard detention time from Equation 3.36, θSD=
(Y-function)
(Z-function)=90.4 mg·min/L
3.06 mg/L =29.5 min Average detention time from Equation 3.37,
θAD=
(W-function)
(Y-function) =3465 mg·min2/L
90.4 mg·min/L =38.3 min Standard detention efficiency Equation 3.38, SDE=θSD
θ ×100%=29.5 min
50.0 min×100%=59% Percent dead volume, PDV=θ−θAD
θ ×100%=50−38.5 min
50 min ×100%=23%
C, mg/L
(b) (a)
C/C0
t, min t/θ
CC0=1 t=1
θ = 50 min θ
0 1 2 3 4
0 25 50 75 100 125 0.00
0.25 0.50 0.75 1.00 1.25
0.0 0.5 1.0 1.5 2.0 2.5
C0 = 3.2 mg/L
FIGURE 3.32 Tracer profile in the effluent from a plugflow reactor with dispersion during a slug tracer test:
(a)Cversust, and (b)C/C0versust/θ(Example 3.34).
EXAMPLE 3.44: DEAD VOLUME IN A CFSTR
A tracer study was conducted in a CFSTR. The tracer concentrations in the effluent at different time inter- vals are given below. The volume of the basin, the averageflow, and influent concentrationC0to the basin are 10 m3, 0.04 m3/s, and 2.2 mg/L, respectively. Determine the dead volume.
t, s 50 100 200 300 400 500
C, mg=L 1.78 1.42 0.88 0.48 0.20 0.10
Solution
1. Calculate the ratio of tracer concentrationC/C0in an ideal CFSTR from the experimental data.
The ratio of tracer concentration in an ideal CFSTR is expressed by Equation 3.10b.
C C0=e−t/θ
C/C0is calculated from actual tracer concentration and tabulated in Column (3) ofTable 3.8.
2. Determine the actual average experimental detention timeθfrom experimental tracer data.
Rearrange Equation 3.10b to obtain the relationship.
θ′= −t ln(C/C0)
SubstituteC/C0andtfor each data point to determineθ′. These values are summarized in Column (4) ofTable 3.8.
3. Calculate the nominal detention timeθfrom Equation 3.10c.
θ=V
Q= 10 m3
0.04 m3/s=250 s
4. Calculate the arithmetic average detention time from experimental data in Column (4).
θAD=Σθ′
n =1208 s 6 =201 s
5. Calculate the theoretical tracer concentrationCTfrom nominal detention timeθusing Equation 3.10a.
CT=C0e−t/θ=2.2 mg/L×e−t/(250 s)
CalculateCTfor differenttvalues. These values are tabulated in Column (5) ofTable 3.8.
6. Plot the actual and theoretical tracer concentrations in the effluent from the CFSTR.
The measured concentrations in Column (2) and theoretical concentrations in Column (5) are plot- ted againsttinFigure 3.33.
7. Compare the results.
The plotted values show that dead volume exists because the tracer concentrations in the experi- mental curve are lower than that in the theoretical curve. This is due to tracer washout.
8. Determine the dead volume.
The nominal detention time is 250 s. The average detention time is 201 s. Clearly, there is tracer washout.
Percent dead volume,PDV=θ−θAD
θ ×100%=250 s−201 s
250 s ×100%=20%
TABLE 3.8 Experimental and Theoretical Concentrations of Tracer in Effluent from a CFSTR (Example 3.44)
t, s C, mg=L C=C0 θ0, s CT, mg=L
(1) (2) (3) (4) (5)
50 1.78 0.81 236a 1.80
100 1.42 0.65 228 1.47
200 0.88 0.40 218 0.99
300 0.48 0.22 197 0.66
400 0.20 0.09 167 0.44
500 0.10 0.05 162 0.30
Pθ0¼1208
a 50 min ln 1:78 mg=L
2:2 mg=L
¼236 min
3.5.3 Dispersion with Conversion
The fundamental approach to understanding nonidealflow in reactors was proposed by Donckwerts.12 Wehner and Wilhelm developed steady-state solution for thefirst-order reaction (r=kC). The solution is expressed by Equation 3.40, which is independent of inlet and outlet conditions, and depends upon dispersion number (Equation 3.41).13
0.0 0.5 1.0 1.5 2.0
50 100 150 200 250 300 350 400 450 500
C, mg/L
t, s
Experimental Theoretical Theoretical
Experimental
FIGURE 3.33 Comparison of theoretical and experimental tracer concentrations in the CFSTR (Example 3.44).
Value of kθ
8
7
6
5
4
3
2
1
0.02 0.04 0.06 0.08 0.10 0.20 0.40 0.60
Fraction remaining, C/C0
FIGURE 3.34 Relationship betweenkθand fraction of concentration remaining for different dispersion numbers.
(From Reference 14 used with permission of American Society of Civil Engineers).
C
C0= 4ae1/2d
(1+a)2ea/2d−(1−a)2e−a/2d (3.40)
d= D
vL or d=Dθ
L2 (3.41)
where
a =coefficient, dimensionless
a=
1+4kθd
√
d =dispersion number, dimensionless
D=longitudinal axial dispersion coefficient, m2/h (ft2/h) v =fluid axial velocity, m/h (ft/h)
L=reactor length, m (ft) θ =hydraulic retention time, h
Thirumurthi developed Figure 3.34to facilitate the solution of Equation 3.40.14 In this figure, the dimensionless termkθ is plotted against percent C/C0 (remaining) for dispersion number dvarying from 0 for ideal PFR to infinity (∞) for an ideal CFSTR. The dispersion numberdvaries for different reactors. Typical values of dispersion numberdfor some wastewater treatment facilities are summarized inTable 3.9.
EXAMPLE 3.45: EFFLUENT QUALITY FROM A REACTOR WITH DISPERSION A stabilization pond is operating at 90% BOD5removal efficiency. The dispersion numberdof the pond is 0.1. The averageflow andfirst-order reaction rate constant are 1000 m3/d and 0.4 d−1, respectively. Cal- culate the volume of the pond.
Solution
1. Determine theC/C0ratio.
TABLE 3.9 Typical Dispersion Numbers for Various Treatment Facilities
Treatment Facility Dispersion Number
Waste stabilization pond
Single pond 1–4
Multiple ponds in series 0.1–1
Aerated lagoon
Long rectangular shape 1–4
Square shaped 3–4
Rectangular sedimentation basin 0.2–2
Aeration basin
Long plugflow 0.1–1
Complete mix 3–4
Oxidation ditch 3–4
Chlorine contact basin 0.02–0.08
Source: Adapted in part from Reference 3.
Removal efficiency=90%
(1−C/C0)=0.9 C/C0=(1−0.9)=0.1 2. Determinekθ.
Read the value ofkθforC/C0=0.1 andd=0.1 fromFigure 3.34.
kθ=2.75
3. Calculate the volume of reactor.
θ=2.75 k = 2.75
0.4 d−1=6.9 d
V=Qθ=1000 m3/d×6.9 d=6900 m3
EXAMPLE 3.46: DISPERSION COEFFICIENT
A UV disinfection facility is designed for reduction of coliform count in the secondary effluent of a wastewater treatment plant. Total length of UV exposure is 300 cm. The velocity through the channel is 22.5 cm/s. Performance of UV disinfection facility is high under plugflow condition. At a dispersion coefficientD=200 cm2/s low to moderate dispersion exists. Calculate the dispersion numberd(Equation 3.41). Also, calculate (a) N/No from Equation 3.40 and compare it with the value obtained from Figure 3.34, (b) coliform number remaining and percent reduction from UV radiation. Because of high rate of kill of coliform organism by UV radiation thefirst-order reaction ratek is assumed 0.29 s−1.N0=104organisms/100 mL.
Solution
1. Calculate the dispersion numberdforD=200 cm2/s from Equation 3.41.
d=D
vL= 200 cm2/s
22.5 cm/s×300 cm=0.03 2. Calculate theθvalue.
θ=L(reactor length)
v(velocity) = 300 cm
22.5 cm/s=13.3 s 3. Calculate the dimensionless factorkθ.
kθ=0.29 s−1×13.3 s=3.86 4. Calculate the dimensionless factora.
a=
1+4kθd
√ =
1+4×3.86×0.03
√ =1.21
5. Calculate the dimensionless ratioN/N0from Equation 3.40.
N
N0= 4ae1/2d
(1+a)2ea/2d−(1−a)2e−a/2d= 4×1.21×e2×0.031
(1+1.21)2e2×0.031.21 −(1−1.21)2e−2×0.031.21
= 4.84e16.7
4.88×e20.2−0.044×e−20.2= 8.4×107
2.8×109−7.7×10−11=0.03 6. Estimate the value ofN/N0fromFigure 3.34.
Estimate the value ofN/N0forkθ=3.86 andd=0.03.
N/N0≈0.03
7. Compare the calculated and estimated values.
The calculated and estimated values are the same.
8. Calculate the coliform number remaining and percent reduction.
N=0.03 ×104organisms/100 mL=300 organisms per 100 mL Percent reduction=(104−300) organisms per 100 mL
104organisms per 100 mL ×100%=97%
EXAMPLE 3.47: OBSERVEDE.COLIDIE-OFF IN A SERIES OF STABILIZATION PONDS
Two stabilization ponds are operating in series. The operational data on both ponds are summarized below. Estimate theE. colinumber in the effluent from the second pond. The averageflow and influent E. colicounts are 4000 m3/d and 107organisms/100 mL, respectively.
Operational Data First Pond Second Pond
Volume, ha·m 2 5
Dispersion number,d 0.25 0.1
Die-offcoefficient (k), d1 1.3 0.4
Solution
1. Calculate theθvalues for both ponds using Equation 3.10c.
In the first pond, θ1=V1
Q=2 ha·m×10,000 m3/ha 4000 m3/d =5 d In the second pond, θ2=V2
Q=5 ha·m×10,000 m3/ha
4000 m3/d =12.5 d 2. Calculatekθfor both ponds.
In thefirst pond,k1θ1=1.3 d−1×5 d=6.5
In the second pond,k2θ2=0.4 d−1×12.5 d=5 3. DetermineN/N0for both ponds usingFigure 3.34.
Ford1=0.25 in thefirst pond, (N/N0)1=0.023 andk1θ1=6.5 Ford2=0.1 in the second pond, (N/N0)2=0.022 andk2θ2=5 4. EstimateE.colinumber in the effluent from the second pond.
(N/N0)overall=(N/N0)1×(N/N0)1=0.023×0.022=0.00051=5.1×10−4 Nin the effluent from the second pond
=N0×(N/N0)overall=107organisms/100 mL×5.1×10−4=5100 organisms/100 mL
EXAMPLE 3.48: DISPERSION FACTOR AND PERFORMANCE OF A CHLORINE CONTACT BASIN
A chlorine contact basin is 3 m wide and has three-pass baffle arrangement as shown inFigure 3.35. At a flow of 5000 m3/d, the depth in the basin is 1 m. The dispersion numberdand reaction rate constant are 0.1 and 2.9 h−1. Determine (a) contact timeθ, (b) dispersion coefficientD, (c)N/N0from Equation 3.40, and (d)N/N0, fromFigure 3.34. The baffle wall thickness is 0.3 m.
Solution
1. Calculate the contact timeθ.
Volume of basin,V ¼(lengthwidthdepth)(volume of baffle walls)
¼(11.5 m9.6 m1 m)2(11.53) m0.3 m1 m¼105 m3 Calculateθfrom Equation 3.10c.
θ=V
Q= 105 m3
5000 m3/d=0.021 d orθ=0.021 d×24 h/d=0.5 h 5000 m3/d
11.5 m
3 m 3 m 3.3 m
9.6 m 5000 m3/d 0.3 m
FIGURE 3.35 Chlorine contact basin layout (Example 3.48).
EXAMPLE 3.49: COMPARISON OF RESIDENCE TIMES FOR IDEAL AND DISPERSED PFRs TREATING A NONCONSERVATIVE SUBSTANCE
An idealfirst-order PFR receives a nonconservative substance that undergoes 90%first-order conversion.
The residence timeθis 6 h. Determine the residence time of a dispersed PFR that has dispersion num- berd=1.0 and achieves the same removal.
Solution
1. Determine theC/C0ratio for 90% removal.
C/C0=(1−0.90)=0.1
2. Determine thekvalue of an ideal PFR withθ=6 h.
In an ideal PFR, theC/C0ratio is expressed by Equation 3.16d.
C C0=e−kθ
2. Calculate the dispersion coefficientD.
Velocityv=Q
A=5000 m3/d
3 m×1 m=1667 m/d orv=1667 m/d
24 h/d =69.5 m/h
Total approximate length offlow pass=2×(11.5−1.5) m+2×(3+0.3) m+(11.5−3) m
=35.1 m
Rearrange Equation 3.41 and calculate dispersion coefficientD.
D=d v L=0.1×69.5 m/h×35.1 m=224 m2/h 3. Calculate the ratio ofN/N0from Equation 3.40.
kθ=2.9 h−1×0.5 h=1.45
a=
1+4kθd
√ =
1+4×1.45×0.1
√ =1.26
N
N0= 4a e1/2d
(1+a)2ea/2d−(1−a)2e−a/2d= 4×1.26×e2×0.11
(1+1.26)2e2×0.11.26 −(1−1.26)2e−2×0.11.26
= 5.04e5
5.11×e6.3−0.068×e−6.3= 748
2783−1.25×10−4=0.27 4. Determine the ratio ofN/N0fromFigure 3.34.
Read the value ofN/N0forkθ=1.45 andd=0.1.
N/N0=0.27
The results obtained from Equation 3.40 andFigure 3.34are compatible.
SubstituteC/C0ratio=0.1 in the equation and solve forkθ. 0.1=e−kθ
kθ= −ln (0.1)=2.3 k=2.3
θ =2.3
6 h=0.38 h−1
3. Determine theθvalue of a dispersed PFR withk=0.38 h−1.
Read the value ofkθforC/C0=0.1 andd=1.0 in a dispersed PFR fromFigure 3.34.
kθ=5 θ=5
k= 5
0.38 h−1=13.2 h
4. Comparison of residence times of ideal and dispersed PFRs.
For achieving a given conversion rate of 90%, the required residence time in a dispersed PFR is more than double of that in an ideal PFR.
EXAMPLE 3.50: COMPARISON OF FRACTION REMAINING OF A NONCONSERVATIVE SUBSTANCE IN IDEAL AND DISPERSED PFRs
An ideal PFR hask=0.35 h−1andθ=8 h. Compare the fraction of a substance remaining in an ideal PFR and in a dispersed PFR havingd=1.0.
Solution
1. Determine theC/C0ratio for an ideal PFR from Figure 3.34.
kθ=0.35 h−1×8 h=2.8
ReadC/C0=0.06 forkθ=2.8 andd=0 in an ideal PFR.
2. Determine theC/C0ratio for a dispersed PFR withd=1.0 fromFigure 3.34.
ReadC/C0=0.2 forkθ=2.8 andd=1.0.
3. Compare the fraction remaining.
Fraction remaining in an ideal PFR is 0.06, and fraction remaining in a dispersed PFR is 0.2. The fraction remaining in a dispersed PFR is more than three times that of an ideal PFR.
EXAMPLE 3.51: REACTORS IN SERIES TO ACHIEVE A GIVEN REMOVAL OF A NONCONSERVATIVE SUBSTANCE
A reactor has total volume of 50 m3. It receives aflow of 150 m3/d. Thefirst-order reaction rate constant for removal of a species A is 0.5 h−1. Determine the following:
a. The fraction of A remaining if the reactor is an ideal PFR, and
b. Number of equal volume reactors in series with dispersion numberdof 0.1 each. The total volume of all reactors is 50 m3and overall efficiency being the same as that for the ideal PFR in (a).
Solution
1. Determine the fraction A remaining in the effluent from the ideal PFR.
Calculateθfrom Equation 3.10c.
θPFR=VPFR
Q = 50 m3
150 m3/d×24 h/d=8 h
C/C0ratio is expressed by Equation 3.16d for an ideal PFR.
C
C0=e−kθPFR =e−0.5 h−1×8 h=0.018
2. Determine the number of reactors in series with dispersion numberdof 0.1 that giveC/C0=0.018.
Assume the number of reactorsn=3.
Volume of each reactorV′PFR=VPFR
n =50 m3
3 =16.7 m3 Hydraulic residence time in each reactorθ′PFR= 16.7 m3
150 m3/d×24 h/d=2.7 h kθ′PFR=0.5 h−1×2.7 h=1.4
ObtainC/C0ratio of 0.26 for each reactor fromFigure 3.34atkθ′PFR=1.4 andd=0.1.
OverallC/C0ratios for three reactors in series=(0.26)3=0.018
Three reactors each of volume 16.7 m3and dispersion number of 0.1 in series will produce an effluent quality approximately equal to that of a single ideal PFR with a total volume of 50 m3. The ideal PFR hasd=0.