VOLUME 2 Post-Treatment, Reuse, and Disposal
3.4 Types of Reactors
Solution
1. Draw the definition sketch.
The definition sketch of batch reactor is shown inFigure 3.14.
2. Write the generalized mass balance equation.
The generalized mass balance equation is expressed by Equation 3.1.
[Accumulation rate]=[Input rate]−[Output rate]+[Utilization or conversion rate]
Since the input and output rates are zero, the equation is written as Equation 3.7.
[Accumulation]=[Decrease due to conversion]
rA=dCA
dt = −kCA (3.7)
3. Rearrange and integrate Equation 3.7 within the limits to obtain Equations 3.8a through 3.8c.
C
C0
dCA
CA = −k
t
0
dt
ln C
C0 = −kt
t= −1 kln C
C0
(3.8a)
C=C0e−kt (3.8b)
C
C0=e−kt (3.8c)
where
C0=initial concentration of a substance att=0, mg/L C =concentration of a substance at timet, mg/L
V, C
Reaction vessel where the composition changes with time Mixing
FIGURE 3.14 Definition of batch reactor (Example 3.14).
EXAMPLE 3.15: MASS CONVERSION IN A BATCH REACTOR
A batch reactor is designed for removal or conversion of a substance. If thefirst-order reaction rate constant at 20◦C is 0.21 h−1and required removal at 10◦C is 90%, determine the reaction time. Assume θT=1.047. Also, calculate the concentration ratioC/C0at timet=4, 10, and 20 h.
Solution
1. Calculate the reaction rate from Equation 2.10 at 10◦C.
k10=k20(1.047)T−20=0.21 h−1×(1.047)10−20=0.21 h−1×(1.047)−10=0.13 h−1 2. Calculate the time required to achieve 90% conversion from Equation 3.8a.
For 90%, removedC/C0remaining ratio=(1−0.9)=0.1.
Substitute the remaining ratio in Equation 3.8a to determinet.
C
C0=0.1, t= − 1
0.13 h−1ln(0.1)=17.7 h
3. Calculate theCtoC0ratio at different time intervals from Equation 3.8c.
t=4 h, C
C0=e−0.13 h−1×4 h=0.6 t=10 h, C
C0=e−0.13 h−1×10 h=0.27 t=20 h, C
C0=e−0.13 h−1×20 h=0.07
3.4.2 Continuous-Flow Stirred-Tank Reactor
In a CFSTR, material entering is dispersed instantly throughout the reactor. As a result, the concentration of material leaving the reactor is same as that at any point in the reactor. Mixing in a CFSTR is extremely important. A round, square, or slightly rectangular reactor may be used. Baffles may be necessary to control vortexing.
Theflow schematic of a CFSTR is shown inFigure 3.15.
Conservative Tracer Response in a CFSTR: Tracer studies are conducted in a CFSTR to determine the reactor response andflow regime. A nonreactive conservative tracer is normally injected at the influent zone, and its concentration is measured with respect to time. The tracer may be applied as a slug input or as a step (continuous) feed. In a slug input, a known volume of stock solution containing known
Q, C Q, C0
V, C
Mixing FIGURE 3.15 Flow schematic of a CFSTR.
quantity of tracer is released into the influent zone. A constantflow is maintained through the reactor. The tracer concentration in the effluent is measured with respect to time. A step-feed tracer study utilizes con- tinuous feed of knownflow and concentration of tracer at the influent, and its concentration is monitored in the effluent.4,6
Conversion of Nonconservative (Reactive) Substance in a CFSTR: A nonconservative substance undergoes conversion reaction in a CFSTR. The reaction may be zero order,first order, or second order.
The concentration equation at any timetcan be developed from mass balance relationship. The solution of mass balance equations has been presented earlier. The concentration equation withfirst-order decay under steady-state and nonsteady-state conditions are expressed by Equation 3.3.
EXAMPLE 3.16: C/////C0RATIO OF A CONSERVATIVE TRACER IN A CFSTR A conservative tracer is injected into a CFSTR. Develop the generalized concentration equations for the effluent from a CFSTR receiving (a) slug tracer input and (b) step (continuous) tracer feed.
Solution
1. Write the generalized mass balance equations (Equations 3.1 and 3.3).
Accumulation rate
= Input rate
− Output rate
+ Utilization or conversion rate
VdC
dt =QC0−QC+V(−kC)
For a conservative substance, the conversion term −kC=0. Equation 3.3 is simplified by Equation 3.9a.
VdC
dt =QC0−QC (3.9a)
where
Q =flow rate through the reactor, m3/s V =volume of reactor, m3
C0=tracer input concentration, mg/L C =tracer output concentration, mg/L
For slug input,C0, slug=initial concentration of tracer in the reactor volume right after the tracer is fed att=0, mg/L. It is determined by Equation 3.9b.
C0,slug=Vtracer
V Ctracer or C0,slug=wtracer
V (3.9b)
where
Vtracer =volume of slug tracer, m3
Ctracer =concentration of trace in the slug feed, mg/L (g/m3) wreactor=weight of reactor, g
For step feed,C0,step=feed concentration of tracer in the influent into the reactor att.0. It is expressed by Equation 3.9c.
C0,step= Qtracer
Q+Qtracer
Ctracer (3.9c)
WhenQ≫Qtracer, (Q+Qtracer)≈Q. Equation 3.9c is simplified to Equation 3.9d.
C0,step=Qtracer
Q Ctracer (3.9d)
where
Qtracer=tracerflow rate fed into influent, m3/s Ctracer=concentration of trace in the tracerflow, mg/L 2. Develop equations for slug input.
Apply the condition of slug input to Equation 3.9a. For a slug input,C0,slug=0 att.0 after slug is added. Rearrange Equation 3.9a atQC0,slug=0.
dC C = −Q
Vdt
Use the integration limits:C0,slugatt=0, andCatt, integrate the equation to obtain a remaining concentration equation (Equation 3.10a) and a ratio equation (Equation 3.10b).
C
C0,slug
dC C = −Q
V
t
0
dt
ln C
C0,slug
= −Q Vt= −t
V Q
= −t θ
C=C0,sluge−t/θ (3.10a)
C
C0,slug=e−t/θ (3.10b)
whereθ=theoretical detention time, h. It is expressed by Equation 3.10c.
θ=V
Q (3.10c)
3. Develop the equations for step feed.
Apply the condition of step feed to Equation 3.9a. For a step feed,C0,step≠0 att.0 after tracer injection begins. Rearrange Equation 3.9a.
dC C0,step−C=Q
Vdt
Use the integration limits:C0,step=0 att=0, andCatt, integrate the equation to obtain a remain- ing concentration equation (Equation 3.10d) and a ratio equation (Equation 3.10e).
C
0
dC
(C0,step−C)=Q V
t
0
dt
ln C0,step−C C0,step
= −t Q
V = −t/θ C0,step−C
C0,step =e−t/θ or 1− C
C0,step=e−t/θ
C=C0,step(1−e−t/θ) (3.10d)
C
C0,step=1−e−t/θ (3.10e)
EXAMPLE 3.17: CONSERVATIVE TRACER PROFILE FOR SLUG AND STEP FEED IN A CFSTR
A CFSTR has a volume of 35 m3. The constantflow in the reactor is 200 m3/d. The reactorflow regime was established by (a) releasing in the influent zones a slug of 20 L stock solution containing 3.5 g/L tracer and (b) feeding continuously 3.5 g/L stock tracer solution in the influent line at a rate of 100 L per day.
Determine the following for both test conditions (a) expected tracer concentration in the effluent at 2 and 6 h after the tracer injection, and (b) tracer profile as a function ofC/C0versust/θ.
Solution
1. Determine the theoretical detention timeθof the CFSTR from Equation 3.10c.
θ=V
Q= 35 m3
200 m3/d×24 h/d=4.2 h
2. Determine initial tracer concentrationC0and concentrationCat time 2 h (C2) and 6 h (C6) since tracer injection. Also, calculate dimensionless ratiosC/C0andt/θ.
a. Slug input.
Calculate the initial concentrationC0,slugfrom Equation 3.9b.
C0,slug=3.5 g/L×20 L
35 m3 =2 g/m3
Calculate the concentrationC2att2=2 h from Equation 3.10a.
C2=C0,sluge−t2/θ=2 g/m3×e−2 h/4.2 h=1.24 g/m3
Calculate the concentration ratioC2/C0,slugfrom Equation 3.10b.
C2
C0,slug=e−t2/θ=e−2 h/4.2 h=0.62 t2
θ = 2 h 4.2 h=0.48
Repeat the calculations fort6=6 h.
C6=C0,sluge−t6/θ=2 g/m3×e−6 h/4.2 h=0.48 g/m3 C6
C0,slug=e−t6/θ=e−6 h/4.2 h=0.24 t6
θ = 6 h 4.2 h=1.43 b. Step feed.
Calculate the initial concentrationC0,stepfrom Equation 3.9d.
C0,step= 100 L/d
200 m3/d×3.5 g/L=1.75 g/m3
Calculate the concentrationC2att2=2 h from Equation 3.10d.
C2=C0,step(1−e−t2/θ)=1.75 g/m3×(1−e−2 h/4.2 h)=0.66 g/m3 Calculate the concentration ratio (C2/C0,step) from Equation 3.10e.
C2
C0,slug=1−e−t2/θ=1−e−2 h/4.2 h=0.38 Repeat the calculations fort6=6 h.
C6=C0,step(1−e−t6/θ)=1.75 g/m3×(1−e−6 h/4.2 h)=1.33 g/m3 C6
C0,slug=1−e−t6/θ=1−e−6h/4.2h=0.76 3. Draw the tracer profiles.
Assume a series of time intervals (t) after the slug is added or since the tracer injection began and calculatet/θ, (C/C0,slug), and (C/C0,step) ratios. The tracer profile data for the slug input and step feed are summarized inTable 3.1.
The profiles are shown for (a) (C/C0,slug) versust/θ, and (b) (C/C0,step) versust/θinFigure 3.16.
TABLE 3.1 Tracer Profile Data for Slug and Step Feed (Example 3.17)
Time Step (t), h t=θ Slug Input Step Feed
C, g=m3 C=C0, slug C, g=m3 C=C0, step
0 0 2.00 1.00 0.00 0.00
0.5 0.12 1.77 0.89 0.20 0.11
1 0.23 1.59 0.80 0.36 0.21
2 0.48 1.24 0.62 0.66 0.38
4 0.95 0.77 0.39 1.07 0.61
6 1.43 0.49 0.25 1.33 0.76
10 2.38 0.19 0.10 1.59 0.91
20 4.76 0.02 0.01 1.74 0.99
EXAMPLE 3.18: STEADY-STATE OPERATION OF A NONCONSERVATIVE SUBSTANCE WITH FIRST-ORDER DECAY
A reactor receives industrial material for product conversion. The reactor volume is 500 m3, and influ- ent and effluent flow rates are 50 m3/d. The concentration of feed material is 650 mg/L, and it is consumed according to first-order kinetics with k=0.28 day−1. Develop the kinetic equation, and determine the exit concentration of the material. The process is operating under steady-state condition.
Solution
Use Equation 3.3, and apply steady-state condition (V(dC/dt)=0).
The solution is expressed by Equations 3.11a through 3.11c.
0=QC0−QC−VkC C= C0
1+kV Q
(3.11a)
C= C0
1+kθ (3.11b)
C C0= 1
1+kθ (3.11c)
Substitute the data in Equation 3.11a to determine the value ofC.
C= 650 g/m3 1+0.28 d−1× 500 m3
50 m3/d
=650 g/m3
1+2.8 =171 g/m3
EXAMPLE 3.19: TIME-DEPENDENT CONCENTRATION OF A NONCONSERVATIVE SUBSTANCE
A nonconservative substance has an influent concentration ofC0. It undergoes conversion reaction A→B which is known to befirst order (r= −kC). Conduct mass balance analysis, and (a) develop
(a) (b)
0.0 0.2 0.4 0.6 0.8 1.0 1.2
0 2 4 6
C/C0, slug
t/θ 0 2 4 6
t/θ 0.0
0.2 0.4 0.6 0.8 1.0 1.2
C/C0, step
C0, slug C0, step
C = e−t/θ C = 1–e−t/θ
FIGURE 3.16 Tracer profile in CFSTR: (a) slug input, and (b) step feed (Example 3.17).
time-dependent concentration equation, (b) determine the steady-state concentration equation when t=∞ and compare the concentration with Equation 3.11b, (c) calculate the concentra- tion of the substance at time t=1 and 3 day, and∞, and (d) calculate the steady-state concentra- tion from Equation 3.11b and compare it with time-dependent concentration from time- dependent equation when t=∞. Use the following data: V=480 m3, Q=30 m3/d, C0=40 g/m3, and k=0.18 d−1.
Solution
1. Develop the time-dependent concentration equation.
a. Develop the mass balance equation from Equation 3.3.
VdC
dt =QC0−QC+V(−kC) b. Simplify and rearrange the above equation.
dC dt =1
V[−C(Q+kV)+QC0]
= −C Q V+k
+C0
Q V dC
dt +C Q V+k
=C0
Q V c. Integrate the above equation.
i. Substitute (Q/V)+kby the following equation (Equation 3.12).
β=Q
V+k or β=1
θ+k=1+kθ
θ (3.12)
dC
dt+βC=C0
Q V
ii. Multiply both sides by the integrating factoreβt. dC
dt+βC
eβt= C0
Q V
eβt
iii. The left side of the equation is written as a differential.
d
dt(Ceβt)=C0
Q Veβt d(Ceβt) =C0
Q Veβtdt Ceβt =Q
VC0 eβtdt
iv. Integration yields the following (Equation 3.13a).
Ceβt=Q V
C0
βeβt+k (3.13a)
when t=0, C=C0
k=C0−Q V
C0
β
v. Substitute the value ofkin Equation 3.13a to obtain the nonsteady-state solution of time- dependent concentration expressed by Equation 3.13b.
C=Q V
C0
β (1−e−βt)+C0e−βt (3.13b)
2. Determine the effluent concentrationCwhent=∞. a. Substitutet=∞and solve.
e−βt=e−β1= 1 e10 C=Q
V C0
β (1−e−β1)+C0e−β1 C=Q
V C0
β C=Q
V C0
Q V+k =Q
V C0
Q
V 1+kV Q
C= C0
1+kV Q
= C0
1+kθ
b. Compare the above equation with the steady-state equation (Equation 3.11b).
The steady-state equation Equations 3.11b and 3.13b are identical whent=∞.
3. Calculate the concentration of nonconservative waste in the effluent whent=1 day, 3 days, and∞.
a. Calculateβfrom Equation 3.12.
β=Q
V+k=30 m3/d
480 m3 +0.18 d−1=0.243 d−1
b. CalculateCfort=1 day, 3 days, and∞from Equation 3.13b.
t=1 day C=Q
V C0
β (1−e−βt)+C0e−βt
=30 m3/d
480 m3 × 40 g/m3
0.243 d−1×(1−e−0.243d−1×1d)+40 g/m3×e−0.243d−1×1d
=10.29 g/m3×(1−0.784)+40 g/m3×0.784
=(2.22+31.36) g/m3=33.6 g/m3
t=3 day
C=10.29 g/m3×(1−e−0.243d−1×3d)+40 g/m3×e−0.243d−1×3d
=10.29 g/m3×(1−0.482)+40 g/m3×0.482
=(5.33+19.28) g/m3=24.6 g/m3 t=1
C=10.29 g/m3(1−0)+40 g/m3×0=10.3 g/m3
4. Calculate the steady-state concentrationCfrom Equation 3.11b and compare it with the concentration obtained from Equation 3.13b whent=∞.
Steady-state concentrationC= C0
1+kθ= 40 g/m3 1+0.18 d−1× 480 m3
30 m3/d
=10.3 g/m3
ConcentrationC=10.3 g/m3whent=∞(see Step 3b above)
This comparison clearly shows that the time-dependent nonsteady-state equation yields steady- state result whentis very large.
EXAMPLE 3.20: STEADY-STATE CONVERSION OF A REACTIVE SUBSTANCE IN A CFSTR
A CFSTR receives a reactive substance. Thefirst-order reaction rate constantk=0.30 h−1. What is the residence timeθto achieve 90% conversion under steady-state condition? Plot the fraction remaining (C/ C0), and fraction removed (1−C/C0) with respect to corresponding residence timeθ. Also, derive linear relationships that can be used to determine the reaction rate constantkbased on the experimental data.
Solution
1. Plot fraction remaining and removal curves.
a. Select the conversion equation.
The first-order conversion relationship under steady-state condition is expressed by Equation 3.11b.
C= C0
1+kθ
b. Rearrange Equation 3.11b to solve for the residence timeθ. θ=C0−C
kC
c. Equation 3.11c gives a relationship betweenC/C0andθ. C
C0= 1 1+kθ
d. Rearrange Equation 3.11b to obtain the relationship between (1−C/C0) andθ. 1−C
C0= kθ 1+kθ
e. Develop the data for plotting the above two relationships.
Use 90% conversion as an example.
C=C0−0.9×C0=0.1C0(fraction remaining) C/C0=0.1
1−C/C0=0.9 (fraction removed) θ=C0−C
kC =C0−0.1C0
k×(0.1C0)= 0.9
0.30 h−1×0.1=30 h
Assume different values of conversions and repeat the calculations. The results are tabulated in Table 3.2.
Note: If the experimental data are utilized, the relationships between the fraction remaining (C/C0), and fraction removed (1−C/C0) with respect to corresponding residence timeθare developed and summarized as inTable 3.2.
f. Plot the curves.
The relationships between the fraction remaining (C/C0) and fraction removed (1−C/C0) with respect to corresponding residence timeθare plotted inFigure 3.17a.
2. Derive linear relationships for determiningk.
a. Equation 3.11b is rearranged as follows.
C0
C =1+kθ
A linear relationship betweenC0/C(a reverse of fraction remaining) andθis developed. The first-order reaction rate constantkcan be determined from the slope of the linear relationship from experimental data (Figure 3.17b).
b. Equation 3.11b can be rearranged into the following relationship.
C0
C0−C=1+1 k×1
θ
A linear relationship betweenC0/(C0−C) (a reverse of fraction removed) and 1/θ(a reverse of residence time) is developed. Thefirst-order reaction rate constantkcan be determined by revers- ing the slope of the linear relationship obtained from experimental data (Figure 3.17c).
c. Develop the data for plotting above two linear relationships.
Use 90% conversion as an example.
C0/C=1 ÷ 0.1=10
TABLE 3.2 Fraction Remaining and Removed of a Reactive Substance in a CFSTR with Respect to Residence Time (Example 3.20)
Parameter Conversion, %
100 90 80 60 40 20 10 0
C=C0, fraction remaining 0 0.1 0.2 0.4 0.6 0.8 0.9 1
1C=C0, fraction removed 1 0.9 0.8 0.6 0.4 0.2 0.1 0
θ, h ∞ 30 13.3 5 2.22 0.83 0.37 0
Relationship betweenC0=Candθ
θ, h ∞ 30 13.3 5 2.22 0.83 0.37 0
C0=C ∞ 10 5 2.5 1.67 1.25 1.11 1
Relationship betweenC0=(C0–C) and 1=θ
1=θ, h1 0 0.033 0.075 0.2 0.45 1.2 2.7 ∞
C0=(C0C) 1 1.11 1.25 1.67 2.5 5 10 ∞
C0/(C0−C)=1 ÷ 0.9=1.1 1/θ=1 ÷ 30 h=0.033 h−1
Assume different values of conversions and repeat the calculations. The results are also pro- vided inTable 3.2.
d. Determinekfrom linear plots.
The linear relationships are also plotted inFigure 3.17bandc.
e. ValidatekfromFigure 3.17bandc.
i. FromFigure 3.17b.
k=(10−1)
30 h =0.3 h−1 ii. FromFigure 3.17c.
1
k=(10−1)
2.7 h−1 =3.33 h k= 1
3.33 h=0.3 h−1
iii. Compare the results with the givenk.
Thekvalue determined from eitherFigure 3.17borcis exactly same of that given in the example statement.
0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6 0.8 1.0 (a)
(b) (c)
0 10 20 30 40
Fraction removed, 1−C/C0
Fraction remaining, C/C0
θ (h)
Fraction remaining Fraction removed Fraction removed
Fraction remaining
0 2 4 6 8 10 12
0 1 2 3 4
Reverse of fraction removed, C0/(C0 − C)
1/θ (h−1) 1 1×
= 1 + k C0 − C
C0
θ 0
2 4 6 8 10 12
0 10 20 30 40
Reverse of fraction remaining, C0/C
θ, h C C0
= 1 + kθ
FIGURE 3.17 Concentration profiles and linear relationships of reactive substance in the effluent under steady-state conditions: (a)C/C0and (1−C/C0) versusθ, (b)C0/Cversusθ, and (c)C0/(C0−C) versus 1/θ(Example 3.20).
EXAMPLE 3.21: NONSTEADY-STATE CONVERSION OF A REACTIVE SUBSTANCE IN A CFSTR
A reactor receives industrial waste for destruction of hazardous substance. The influentflow and concen- tration vary considerably. The reactor data are summarized below:
Reactor volume,V¼480 m3 Average waste concentration,C0 ¼40 mg=L Averageflow,Q ¼30 m3=d First-order reaction kinetic coefficient,k¼0.18 d1
The reactor is operating under nonsteady-state condition. Using Equation 3.13b develop the following plots: (a) concentration profileCversus reaction timet; and (b) Dimensionless parameters C/C0versust/θ.
Also, indicate steady-state concentration line on these plots.
Solution
1. Calculateθfrom Equation 3.10c.
θ=V
Q= 480 m3 30 m3/d=16 d
2. Calculateβfrom Equation 3.12, and steady-state concentration of hazardous substance.
β=Q
V+k=30 m3/d
480 m3 +0.18 d−1=0.243 d−1 Q
V×C0
β = 1
16 d× 40 g/m3
0.243 d−1=10.29 g/m3 (steady-state concentration) 3. Substitute the above values in Equation 3.13b and simplify to obtain Equation 3.14.
C=Q V×C0
β (1−e−βt)+C0e−βt
C=10.29 g/m3×(1−e−0.243 d−1×t)+40 g/m3×e−0.243 d−1×t (3.14) 4. Tabulate the data for plotting the concentration profile curves (Table 3.3).
TABLE 3.3 Nonsteady-State Conversion of a Reactive Substance in a CFSTR (Example 3.21)
t, d e0:243 d1t (1e0:243 d1t) 10:29(1e0:243 d1t) 40e0:243 d1t C, g=m3 t=θ C=C0
0 1 0 0 40 40 0 1
0.5 0.886 0.114 1.18 35.4 36.6 0.0313 0.92
1 0.784 0.216 2.22 31.4 33.6 0.0625 0.84
1.5 0.695 0.305 3.14 27.8 30.9 0.0938 0.77
2 0.615 0.385 3.96 24.6 28.6 0.125 0.71
3 0.482 0.518 5.33 19.3 24.6 0.188 0.62
6 0.233 0.767 7.89 9.31 17.2 0.375 0.43
10 0.0880 0.912 9.38 3.52 12.9 0.625 0.32
15 0.0261 0.974 10.0 1.04 11.1 0.938 0.28
20 0.00775 0.992 10.2 0.310 10.5 1.25 0.26
5. Plot the concentration profileCversustfrom the data tabulated above.
The values are plotted inFigure 3.18a. The steady-state concentration of 10.29 g/m3obtained from Equation 3.11b is also shown inFigure 3.18a.
6. Plot the dimensionless valuesC/C0versust/θ(Figure 3.18b).
The steady-state concentration ratioC/C0=(10.29 g/m3)/(40 g/m3)=0.26 (Figure 3.18b).
0 10 20 30 40
(a) 50 (b)
0 5 10 15 20 25
C, mg/L C/C0
t, d
0.0 0.2 0.4 0.6 0.8 1.0 1.2
0.0 0.5 1.0 1.5
t/θ Vβ
QC0
C = (1 – e−βt) + C0e−βt Vβ(1 – e−βt) + C0e−βt C0
C
C0 C
=Q
C/C0 = 0.26
Steady-state condition
= 1
C0
C =1 + kθ 1 + kθ
Steady-state condition Nonsteady-state condition
C0 = 10.29 mg/L
Nonsteady-state condition
FIGURE 3.18 Concentration profile in the effluent under steady-state and nonsteady-state conditions: (a)C versust, and (b) dimensionless parametersC/C0versust/θ(Example 3.21).
EXAMPLE 3.22: STEP FEED OF A REACTIVE SUBSTANCE IN A CFSTR
A constantflow is maintained through a CFSTR. Att=0, a reactive substance is added in the influent stream at a constant rate. The conversion reaction is of thefirst order. Determine (a) the output con- centration equation as a function of time, (b) output concentration whent=∞, (c) output concentration as g/m3att=1 and 4 h, and∞from the start of step feed, and (d) output concentration profile as g/m3 and as dimensionless parameters C/C0 and t/θ for both the reactive and conservative substances.
The operational conditions of the reactor are:V=20 m3,Q=200 m3/d,C0in step feed=120 g/m3, andk=0.2 h−1.
Solution
1. Development of the output concentration equation of a reactive (nonconservative) substance.
Write the mass balance equation for the reactor (Equation 3.3).
VdC
dt =QC0−QC+V(−kC)
Rearrange the equation to obtain Equation 3.15a.
dC dt =Q
VC0−Q VC−kC dC
dt =Q VC0−Q
VC 1+V Qk
=1
θ[C0−C(1+θk)] (3.15a)
Rearrange and integrate Equation 3.15a to obtain Equations 3.15b and 3.15c.
C
C0
dC
C0−C(1+kθ)=1 θ
t
0
dt− 1 1+kθ
ln C0−(1+kθ)C C0
=t θ
C=C0(1−e−(1+kθ)t/θ)
1+kθ (3.15b)
C
C0=(1−e−(1+kθ)t/θ)
1+kθ (3.15c)
2. Develop steady-state equation whent=∞.
Substitute t=∞in Equation 3.15b and solve. Steady-state equationC=C0/(1+kθ) (Equation 3.11b) is reached.
3. Determine output concentration in g/m3. Calculateθfrom Equation 3.10c.
θ=V
Q= 20 m3
200 m3/d×24 h/d=2.4 h
CalculateCvalues from Equation 3.15b whent=1, 4 h and∞.
whent=1h, C=120 g/m3×(1−e−(1+0.2 h−1×2.4 h)×(1 h/2.4 h)) (1+0.2h−1×2.4h)
=120 g/m3×(1−e−0.61) 1.48
=37.3 g/m3
whent=4h, C=120 g/m3×(1−e−(1+0.2 h−1×2.4 h)×(4 h/2.4 h))
(1+0.2 h−1×2.4 h) =74.2 g/m3
whent=1, C=120 g/m3×(1−e−1)
1.48 =120 g/m3
1.48 =81.1 g/m3 4. Develop the output time response calculation table.
The calculations are developed from Equation 3.15b for reactive substance. The data for conserva- tive substances utilize Equation 3.10d. These calculations are summarized inTable 3.4.
TABLE 3.4 Output Concentrations of Reactive and Conservative Substance in a CFSTR (Example 3.22) t, h t=θ
Reactive Substance Conservative Substance
Cfrom Equation 3.15b,
g=m3
C=C0from Equation 3.15c
Cfrom Equation 3.10d,
g=m3
C=C0from Equation 3.10e
0 0 0 0 0 0
0.5 0.21 21.5 0.18 22.6 0.19
1 0.42 37.3 0.31 40.9 0.34
2 0.83 57.5 0.48 67.8 0.57
4 1.67 74.2 0.62 97.3 0.81
10 4.17 80.9 0.67 118 0.98
15 6.25 81.1 0.68 120 1.00
20 8.33 81.1 0.68 120 1.00
5. Plot concentration profiles.
The output time response curve of reactive (nonconservative) substance in g/m3versus reaction timetis shown inFigure 3.19a. The output time response curves for reactive and conservative sub- stances as dimensionless parametersC/C0versust/θare shown inFigure 3.19b.
0 20 40 60 80
(a) 100 (b)
0 5 10 15 20 25
C, g/m3
t, h t/θ
0.0 0.2 0.4 0.6 0.8 1.0 1.2
0 2 4 6 8 10
C/C0
Reactive C0 (1–e−(1+kθ)t/θ)
1 + kθ C =
(1 + kθ) (1 – e–(1+kθ)t/θ ) C0
C =
C0
C = (1 – e−t/θ ) Conservative
FIGURE 3.19 Concentration profiles: (a) output concentration of reactive (nonconservative) substance, and (b) comparison of concentration profiles of reactive (nonconservative) and nonreactive (conservative) substances in dimensionless parameters (Example 3.22).
EXAMPLE 3.23: NONSTEADY STATE REACTIVE PURGING OF A REACTOR WITH FIRST-ORDER DECAY
A manufacturing plant stabilizes its waste in a reactor before discharging the effluent into a sanitary sewer. The raw waste concentration is 250 mg/L. The reactor volume is 750 m3, and volumetricflow rate is 30 m3/d. The waste is stabilized under steady-state condition exhibitingfirst-order reaction kinetic (k=0.37 d−1). For annual maintenance, the waste input into the reactor is terminated att=0, while theflow of clean liquid is continued for reactive purging of the reactor.
a. Determine the steady-state concentration of waste before purging started.
b. Develop the generalized equation to express the reactive purging of the reactor. Check the boundary conditions (t=0 andt=∞).
c. Determine the concentration of waste in the reactor after 2 days since waste input is terminated and purging started, att=0.
d. Draw the concentration profile in the reactor for steady-state condition until the reactor purging is complete.
Solution
1. Determine the steady-state concentration of waste in the reactor before purging started.
Calculateθfrom Equation 3.10c.
θ=V
Q= 750 m3 30 m3/d=25 d
Substitute the data in Equation 3.11b to determine the steady-state concentration in the reactor dur- ing continuous feed.
C= C0
1+kθ= 250 mg/L
1+0.37 d−1×25 d=250 mg/L
10.25 =24.4 mg/L
2. Develop the generalized equation to express the reactive purging of the reactor before purging started.
Simplify Equation 3.3 to determineCas a function of time.
VdC
dt =QC0−QC−VkC
Since waste input is terminated,C0=0.
VdC
dt = −C(Q+Vk)
Integrate the differential equation to solve forC.
C
C0
dC C =
t
0
−(Q/V+k)dt
where
t =0, exponential terme0=1
C=C0, steady-state concentration in the reactor is reached.
t =∞, exponential terme−∞=0 C=0, the reactor purging is complete.
ln(C)−ln(C0)= −(Q/V+k)t ln C
C0 = −(Q/V+k) C=C0e−(Q/V+k)t C=C0e−(1+kθ)t/θ C
C0=e−(1+kθ)t/θ
3. Determine the concentration of waste in the effluent after 2 days (t=2) since purging began.
C=C0e−(1+kθ)t/θ
C=24.4 mg/L×e−(1+0.37 d−1×25 d)×(2 d/25 d)
C=24.4 mg/L×e−0.82=24.4 mg/L×0.44=10.7 mg/L
4. Draw the concentration profile in the effluent before and after reactive purging began.
a. Under steady-state condition,C0=24.4 mg/L.
b. Assume different values oftand repeat the calculations to determine values ofC. The results are provided inTable 3.5.
5. Data plot is shown inFigure 3.20.
3.4.3 Plug Flow Reactor
In an ideal plugflow reactor, the elements offluid pass through the reactor and are discharged in the same sequence in which they enter the reactor. Theflow regimes are characterized by the following (1) eachfluid particle remains in the reactor for a time period equal to the theoretical detention time, and (2) there is no longitudinal dispersion or mixing of the fluid elements as they move through the system. Plug flow regime is approached in systems that have large length-to-width ratio for rectangular basin or large length-to-diameter ratios for circular pipes. At a length-to-diameter ratio of 50:1, theflow approaches plugflow regime if the velocity is not excessive.4,5
Conservative Tracer Response with Slug Input: If a slug of dye tracer is released in an ideal plugflow reactor, it will move and form a band. As theflow continues, the band will move through the reactor. It will emerge as a band in the effluent at the theoretical detention timeθ(Equation 3.10c). The generalized tracer band movement is shown inFigure 3.21a.
0 5 10 15 20 25 30
0 3 6 9 12 15
C, mg/L
t, d
C0, 24.4 mg/L (Steady-state condition)
C = C0e−(1+kθ)t/θ
FIGURE 3.20 Concentration profile of industrial waste undergoing reactive purging from steady-state oper- ation (Example 3.23).
TABLE 3.5 Nonsteady State Reactive Purging Data of a Reactor (Example 3.23) t, d
Reactive Substance
(1þkθ)θt e(1þkθ)θt C, mg=L
0 0 1 24.4
0.5 0.205 0.815 19.9
1 0.410 0.664 16.2
1.5 0.615 0.541 13.2
2 0.820 0.440 10.7
2.5 1.03 0.359 8.8
3.5 1.44 0.238 5.8
5 2.05 0.129 3.1
7 2.87 0.057 1.4
9 3.69 0.025 0.6
12 4.92 0.00730 0.2
Conservative Tracer Response with Step Feed: In a plugflow reactor, the continuous dye tracer feed results in a continuous discharge of concentrationC0after the theoretical detention timeθ. Prior toθ, no dye will be detected in the effluent. The step-feed tracer profile is shownFigure 3.21b.
Stabilization of Nonconservative (Reactive) Substance in a PFR: A nonconservative substance undergoes zero-,first-, or second-order decay in a PFR. The concentration of reactive substance in the reactor varies from point to point along theflow path. The concentration equations for steady-state and nonsteady-state conditions can be developed from mass balance procedure.4,5
Saturation-Type Reaction: The residence time equations of CFSTR and PFR for saturation-type reac- tion under steady-state condition (r=–kC/[Ks+C]) were presented in Section 2.3.2. The saturation-type reaction requires values ofC0,Ks, andk. Thefinal results are quite sensitive to these values. Readers may refer to Equation 2.7a, and Examples 2.10, 3.32, and 3.33 for more information.3,4,7
Concentration, C
Band travels
t1 t2 t3
t0
Reactor length or time, t Reactor length or time, t C0 Tracer
input
(a) (b)
0 θ
FIGURE 3.21 Movement of tracer in an ideal plugflow reactor: (a) slug band and (b) step-feed profile.
EXAMPLE 3.24: SLUG TRACER INPUT IN A PLUG FLOW REACTOR
An ideal plugflow reactor is a long channel. The channel volume is 38 m3andflow is 500 m3/d. The tracer solution contains 4 g dye/L. A slug of 19 L dye solution is released into the influent zone. The tracer band has a concentration of 8.4 timesC0. Determine (a) average expected tracer concentrationC0if it is completely mixed in the basin, (b) theoretical detention timeθ, (c) tracer concentration in the band, and (d) time for emergence of the band in the effluent. Draw the tracer profile.
Solution
1. CalculateC0from Equation 3.9b.
C0,slug=Vtracer
V Ctracer= 19 L
38 m3×4 g/L=2 g/m3or 2 mg/L 2. Calculate the theoretical detention timeθfrom Equation 3.10c.
θ=V
Q= 38 m3
500 m3/d=0.076 d orθ=0.076 d×24 h/d=1.8 h
3. Calculate the tracer concentration in the band, and time for the band to exit in the effluent.
Tracer concentration in the band or piston=2 g/m3×8.4=16.8 g/m3 Time for dye concentration band to reach effluent isθor 1.8 h.