VOLUME 2 Post-Treatment, Reuse, and Disposal
3.6 Equalization of Flow and Mass Loadings
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.
ff tion basins.15,16
3.6.2 Design Considerations
The design of equalization basin should utilize mixing and aeration to prevent solids deposition and odor problems. Location below grit removal or primary sedimentation basin is desirable. Rectangular, square, and circular shapes with 4–6 m liquid depth have been adapted by the designers. Controlledflow pumping is necessary from the equalization basin.
EXAMPLE 3.52: PROCESS TRAIN WITH EQUALIZATION BASINS
Equalization basin is provided between grit and primary sedimentation basins. Draw the process trains to show in-line and off-line arrangements of the equalization basins.
Solution
The process trains for in-line and off-lineflow equalization basins, andflow patterns before and after equalization are shown inFigure 3.36.
Constant flow pumping
Flow measurement
t Q
Flow pattern at A Flow pattern at B
t Q
Controlled flow pumping Diversion structure Bar screen Grit
removal In-line flow equalization
Primary sedimentation
Bar screen Grit removal
Flow
measurement Primary sedimentation Off-line flow
equalization (a)
(b)
F
A B F
A B
(c)
FIGURE 3.36 Process trains andflow pattern with equalization basins: (a) in-line, (b) off-line, and (c)flow patterns (Example 3.52).
3.6.3 Design Volume of Equalization Basin
The volume of an equalization basin is established for (a)flow equalization and (b) mass loading equal- ization. Basin sizing for both conditions are given below:
Flow Equalization: The volume or storage capacity of an equalization basin is determined by either graphical or analytical technique. The graphical technique requires preparation of a mass diagram in which cumulative inflow volume and cumulative average day volume are plotted on the same plot against the time of the day.17,18The analytical technique is based on cumulative change in volume deficiency between inflow and outflow as expressed by Equations 3.42a and 3.42b.
ΔV=Vin−Vout (3.42a)
ΔV=(Qin−Qout)Δt (3.42b)
where
ΔV =change in storage volume during a specified time interval, volume Vin =total inflow volume during the specified time interval, volume Vout=total outflow volume during the specified time interval, volume Qin =total inflow, volume/time
Qout=total outflow, volume/time Δt =time interval, time
Mass Loading Equalization: In an in-line equalization basin, the variable mass loading rate is also equalized along with theflow rate. The degree of equalization depends upon the size of the basin. The basic expression used to calculate the effluent concentration is usually expressed by the general mass balance relationship of a completely mixed reactor without conversion. This relationship for an equalization basin is expressed by Equation 3.43.19
VdC
(accumulation)
=Caqadt
(input)
−Cqadt
(output) (3.43)
where
V =volume of the tank, m3
dC=change in concentration in the tank, mg/L or g/m3 qa =averageflow rate over small time interval, m3/s dt =small time increment of time, s
Ca=average concentration of material in the influent, mg/L or g/m3 C =concentration of material in the effluent (variable), mg/L or g/m3 VdC=qa(Ca−C)dt
dt=V qa
dC (Ca−C)
Using the integration limitsC1att1andC2att2, integrate the equation to obtain Equation 3.44.
t2 t1
dt=V qa
C2 C1
dC (Ca−C) (t2−t1)=V
qa
ln Ca−C1
Ca−C2
(3.44)
t1 =time at the start of the time incrementdt, s t2 =time at the end of the time incrementdt, s
C1=concentration in the effluent at the timet1, mg/L or g/m3 C2=concentration in the effluent at the timet2, mg/L or g/m3
EXAMPLE 3.53: STORAGE VOLUME BASED ON FLOW DIAGRAM AND CHANGE IN VOLUME
The hourlyflow pattern of an industrial process is given below. The treatment facility is designed for a constantflow rate of 1500 m3/h. Determine the volume of equalization basin byflow diagram and change in volume calculations. Also, draw the line representing the volume remaining in the basin.
Time Period Midnight–
1:00 a.m. 1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9 9–10 10–11 11:00 a.m.
noon Flow, m3=h 1050 960 930 930 975 1080 1170 1350 1570 1800 2100 2235
Time Period Noon–1:00
p.m. 1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9 9–10 10–11 11:00 p.m.–
midnight Flow, m3=h 2250 2235 2100 1920 1680 1380 1230 1350 1574 1574 1383 1170
Solution
1. Develop the calculation table forflow diagram and change in volume.
The calculations are provided inTable 3.10.
2. Develop theflow diagram.
a. Plot cumulative inflow and outflow with respect to time of the day. This plot is shown in Figure 3.37.
b. Draw two tangents at points A and B on cumulative inflow line. Points A and B are at the valley and hump of the curvatures. The tangents are parallel to cumulative outflow or pumping line.
3. Determine the required volume of the equalization basin using change in volume calculations.
The change in volume calculations are summarized inTable 3.11. The following information may be drawn:
a. The basin is empty at 8:00 a.m.
b. The basin startsfilling after 8:00 a.m. by the incomingflow to the plant.
c. It is completely full to its maximum volume of 4390 m3at 5:00 p.m.
d. The basin starts emptying after 5:00 p.m.
e. The basin is totally empty at 8:00 a.m on the next day.
f. This cycle repeats again.
4. Determine the required volume of equalization basin fromflow diagram.
The vertical distance between the two tangents is the required theoretical volume of the equalization basin. This volume is 4390 m3.
5. Draw the line representing the basin volume remaining based on cumulative change in volume (Figure 3.37).
a. From 8:00 a.m. to 5:00 p.m., is thefilling cycle. The basin isfilled in 9 h. From 5:00 p.m. to 8:00 a.
m., is the emptying cycle. The basin is emptied in 15 h.
b. During thefilling and emptying cycles, a constantflow of 1500 m3/h is withdrawn from the basin.
c. The maximum volume of 4390 m3on the curve represents the required theoretical volume of the equalization basin.
TABLE 3.10 Calculations for Determination of Volume of Equalization Basin (Example 3.53) Time Δt, h Qin, m3=h Qout, m3=h P
ΔVin;m3 P
ΔVout, m3 ΔV, m3 P ΔV, m3
(1) (2) (3) (4) (5) (6) (7) (8)
Midnight – – – – – – –
1:00 a.m. 1 1050 1500 1050 1500 450 3105
2 1 960 1500 2010 3000 540 2565
3 1 930 1500 2940 4500 570 1995
4 1 930 1500 3870 6000 570 1425
5 1 975 1500 4845 7500 525 900
6 1 1080 1500 5925 9000 420 480
7 1 1170 1500 7095 10,500 330 150
8 1 1350 1500 8445 12,000 150 0a
9 1 1570 1500 10,015 13,500 70b 70
10 1 1800 1500 11,815 15,000 300 370
11 1 2100 1500 13,915 16,500 600 970
Noon 1 2235 1500 16,150 18,000 735 1705
1:00 p.m. 1 2250 1500 18,400 19,500 750 2455
2 1 2235 1500 20,635 21,000 735 3190
3 1 2100 1500 22,735 22,500 600 3790
4 1 1920 1500 24,655 24,000 420 4210
5 1 1680 1500 26,335 25,500 180 4390c
6 1 1380 1500 27,715 27,000 120d 4270
7 1 1230 1500 28,945 28,500 270 4000
8 1 1350 1500 30,295 30,000 150 3850
9 1 1574 1500 31,869 31,500 74 3924
10 1 1574 1500 33,443 33,000 74 3998
11 1 1383 1500 34,826 34,500 117 3881
Midnight 1 1174 1500 36,000 36,000 326 3555
aBasin is empty.
bBasin startsfilling.
cBasin full. Required theoretical capacity of the basin is reached.
dBasin starts emptying.
0 10,000 20,000 30,000 40,000
Required equalization volume 4390 m3
Volume, 1000 m3
A
B
Midnight 2 4 6 8 10 Noon 2 4 6 8 10 Midnight
Time Cumulative
outflow or pumping
Tangent at B
Tangent at A Cumulative
change in volume Cumulative
inflow Basin empty
Basin full
FIGURE 3.37 Mass diagram for determination of the capacity of the equalization basin and volume remain- ing (Example 3.53).
EXAMPLE 3.54: VOLUME OF AN IN-LINE EQUALIZATION BASIN
An in-line equalization basin receives variableflow. Determine the volume of an equalization basin to withdraw a constantflow of 652 m3/min into a biological wastewater treatment plant. Use volume change calculations andflow diagram to determine the volume. Indicate what time of the day the basin will be full and empty.
Time Period Midnight–
1:00 a.m. 1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9 9–10 10–11 11:00 a.m.–
noon Averageflow
rate during time period, m3=min
582 468 348 276 222 210 252 432 750 870 900 912
Time Period Noon–1:00
p.m. 1–2 2–3 3–4 4–5 5–6 6–7 7–8 8–9 9–10 10–11 11:00 p.m.–
midnight Averageflow
rate during time period, m3=min
900 858 817 744 690 690 696 774 864 846 804 743
Solution
1. Determine the constant withdrawal rate from the equalization basin during each hour of pumping.
Constantflow rate=652 m3/min×60 min/h=39.1×103m3/h 2. Determine the average variableflow during each hour time period.
Change in volume due toflow into the basin during midnight and 1:00 a.m.
¼582 m3=min60 min=h1 h (duration)
¼34.9103m3 Repeat calculations for each hour time period.
3. Prepare a calculation table to determine the volume of the basin.
The cumulative inflow and cumulative outflow, and change in volume are summarized in Table 3.11.
4. Determine the basin volume based on volume change.
The theoretical volume of the equalization basin is 146×103m3. The basin will be empty at 8:00 a.m. and it will be full at midnight.
5. Determine the basin volume from theflow diagram.
The basinflow diagram and volume remaining at any time of the day are plotted inFigure 3.38. The basin volume when full is 146×103m3.
6. Draw the volume change curve.
The equalization basin startsfilling at 8:00 a.m. and is full at midnight in a period of 16 h. The basin starts emptying at midnight and becomes empty at 8:00 a.m. in a period of 8 h. The volume change curve is also shown inFigure 3.38.
TABLE 3.11 Calculations to Determine Equalization Basin Volume (Example 3.54) Time of
Day
Qin,
m3=min Vin, 103m3
PΔVin, 103 m3
Vout, 103m3
PΔVout, 103m3
ΔV, 103m3
PΔV, 103m3
Midnight
1:00 a.m. 582 34.9 34.9 39.1 39.1 4.20a 141
2 468 28.1 63.0 39.1 78.2 11.0 130
3 348 20.9 83.9 39.1 117 18.2 112
4 276 16.6 100 39.1 156 22.6 89.5
5 222 13.3 114 39.1 196 25.8 63.7
6 210 12.6 126 39.1 235 26.5 37.2
7 252 15.1 141 39.1 274 24.0 13.2
8 432 25.9 167 39.1 313 13.2 0.00b
9 750 45.0 212 39.1 352 5.88c 5.88
10 870 52.2 265 39.1 391 13.1 19.0
11 900 54.0 319 39.1 430 14.9 33.8
Noon 912 54.7 373 39.1 469 15.6 49.4
1:00 p.m. 900 54.0 427 39.1 509 14.9 64.3
2 858 51.5 479 39.1 548 12.4 76.7
3 817 49.0 528 39.1 587 9.90 86.5
4 744 44.6 572 39.1 626 5.52 92.1
5 690 41.4 614 39.1 665 2.28 94.3
6 690 41.4 655 39.1 704 2.28 96.6
7 696 41.8 697 39.1 743 2.64 99.2
8 774 46.4 743 39.1 782 7.32 107
9 864 51.8 795 39.1 822 12.7 119
10 846 50.8 846 39.1 861 11.6 131
11 804 48.2 894 39.1 900 9.12 140
Midnight 745 44.7 939 39.1 939 5.58 146d
aBasin starts emptying cycle.
bBasin is empty.
cBasin startsfilling.
dBasin is full. Required theoretical capacity of the basin is reached.
0 200 400 600 800 1000
Midnight 2 4 6 8 10 Noon 2 4 6 8 10 Midnight
Time A
B
Volume, 1000 m3
Basin empty
Basin volume remaining
Basin full Cumulative inflow
Tangent at A
Basin volume
= 146 × 103 m3 Cumulative
outflow and tangent at B
FIGURE 3.38 Flow diagram for determination of the capacity of equalization basin and volume remaining (Example 3.54).
EXAMPLE 3.55: EQUALIZATION OF MASS LOADING
An equalization basin has a volume of 1 million liters (ML). The influentflow and BOD5data for several time periods are given below. Calculate the BOD5concentration in the basin effluent. It is given that the BOD5concentration in the effluent at 8:00 a.m. is 160 mg/L.
Time Influent Flow (MLDa) Influent BOD5(mg=L)
8:00 a.m. 0.2 150
9:00 a.m. 0.3 250
10:00 a.m. 0.4 180
11:00 a.m. 0.5 350
12:00 a.m. (Noon) 0.3 200
aMLD: million liters per day.
Solution
1. CalculateqaandCaatt1=8:00 a.m., andt2=9:00 a.m.
qa=(0.2+0.3) MLD
2 =0.25 MLD orqa=0.25 MLD× d
24 h=0.0104 ML/h Ca=(150+250) mg/L
2 =200 mg/L
2. Calculate the effluent BOD5concentration (C9) at 9:00 a.m.
Use Equation 3.44 to calculate effluent concentration. It is given that the effluent BOD5concentra- tion at 8:00 a.m. is 160 mg/L.
(t2−t1)=V qa
ln Ca−C8
Ca−C9
(9−8) h= 1 ML
0.0104 ML/h×ln (200−160) mg/L (200−C9) mg/L
ln 40
200−C9
=0.0104 40
200−C9=e0.0104=1.01 SolvingC9=160 mg/L
3. Calculate the effluent BOD5concentrationC10at 10:00 a.m.
qa=(0.3+0.4) MLD
2 =0.35 MLD orqa=0.35 MLD× d
24 h=0.0146 ML/h Ca=(250+180) mg/L
2 =215 mg/L (10−9) h= 1 ML
0.0146 ML/h×ln (215−160) mg/L (215−C10) mg/L
ln 55
215−C10
=0.0146 55
215−C10=e0.0146=1.01 SolvingC10=161 mg/L
4. Calculate the effluent BOD5concentrationC11at 11:00 a.m.
qa=(0.4+0.5) MLD
2 =0.45 MLD orqa=0.45 MLD× d
24 h=0.0188 ML/h Ca=(180+350) mg/L
2 =265 mg/L (11−10) h= 1 ML
0.0188 ML/h×ln (265−161) mg/L (265−C11) mg/L
ln 104
265−C11
=0.0188 104
265−C11=e0.0188=1.02 SolvingC11=163 mg/L
5. Calculate the effluent BOD5concentrationC12at noon.
qa=(0.5+0.3) MLD
2 =0.4 MLD orqa=0.4 MLD× d
24 h=0.0167 ML/h Ca=(350+200) mg/L
2 =275 mg/L
(12−11) h= 1 ML
0.0167 ML/h×ln (275−163) mg/L (275−C12) mg/L
ln 112
275−C12
=0.0167 112
275−C12=e0.0167=1.02 SolvingC12=165 mg/L 6. Summarize the results.
Influentflows and BOD5concentrations, and the effluent BOD5concentrations are summarized inTable 3.12. It may be noted that the effluent concentrations at different times are based on the given or assumed value of effluent BOD5of 160 mg/L at 8:00 a.m. If the calculations are continued over 24-h period, the concentration of BOD5at 8:00 a.m. after 24-h cycle should match 160 mg/L. If it does not match, that means the given or assumed value of 160 mg/L of BOD5at 8:00 a.m. is not reliable enough.
The procedure should be carried out over 24 h for several iterations until a stable value is reached.
TABLE 3.12 Effluent BOD5from Equalization Basin (Example 3.55) Time Influent Flow,
MLD
Influent
BOD5, mg=L qa, MLD Ca, mg=L Effluent BOD5, mg=L
8:00 a.m. 0.2 150 – – 160 (given)
9:00 a.m. 0.3 250 0.25 200 160
10:00 a.m. 0.4 180 0.35 215 161
11:00 a.m. 0.5 350 0.45 265 163
Noon 0.3 200 0.40 275 165
EXAMPLE 3.56: EQUALIZATION OF MASS LOADING OVER 24-h CYCLE An equalization basin receivesflow from an industrial plant. The hourlyflow and COD concentration data of the influent to the equalization basin are given below. Determine: (a) averageflow rate over 24-h period, (b) the theoretical volume of the equalization basin in ML that will provide constantflow to the treatment plant, and (c) hourly concentration of COD in the effluent from the equalization basin that has a volume equal to the theoretical volume, 0.35 and 0.7 ML. Draw the influent and effluentflow profiles, and influent and effluent COD concentration profiles at all three volumes of the equalization basin.
Time 7:00 a.m. 8 9 10 11 Noon 1:00 p.m. 2 3 4 5 6
Influentflow, MLD
0.40 0.50 0.80 1.20 1.30 1.40 1.30 0.80 0.70 0.60 0.50 0.50 Influent COD,
mg=L 100 115 125 150 200 280 300 300 150 125 100 100
Time 7:00 p.m. 8 9 10 11 Midnight 1:00 a.m. 2 3 4 5 6
Influentflow, MLD
0.60 0.80 1.00 0.90 0.60 0.40 0.30 0.25 0.20 0.20 0.25 0.30 Influent COD,
mg=L 150 175 225 225 175 150 100 90 80 70 80 90
Solution
1. Determine the theoretical volume of the equalization basin by change in volume method.
a. Prepare the calculation table to determine constantflow and the theoretical volume of the basin (Table 3.13).
b. Determine the constant withdrawalflow from the basin.
There are 24 hourly observations (7:00 a.m. to 6:00 a.m.). The average withdrawalflow based on 24 observations is 0.658 MLD.
Mean constant withdrawal flow based on 24 observations=15.8 MLD
24 =0.658 MLD Average flow per hour=0.658 MLD× d
24 h×106L
ML =27.4×103L/h 2. Determine the theoretical volume of the equalization basin.
The theoretical volume of the equalization basin is obtained from the maximum value of the changes in volume (Column [7],Table 3.13). The procedure to calculate the required theoretical vol- ume of aflow equalization basin is shown in Examples 3.53 and 3.54.
Theoretical volume=133×103L or 0.133×106L (0.133 ML) 3. Plot the influent and effluentflow profiles from the equalization basins.
The influentflow and constant effluentflow profiles from an equalization basin with the theoretical volume of 0.133 ML are shown inFigure 3.39. If there is noflow accumulation in the basin, the influent and effluentflow profiles are independent of the equalization basin volume. Therefore, the influent and
TABLE3.13CalculationTableforDeterminationofTheoreticalVolumeoftheBasin,andCODConcentrationintheEffluentfromBasinVolumesof0.133,0.35,and 0.7MG(Example3.56) tQin,MLDCin,mg=LVin,103 LVout,103 LΔV,103 LP ΔV,103 Lqa,103 L=hCa,mg=LCoutforDifferentBainVolume,mg=L V¼0.133MLV¼0.35MLV¼0.7ML (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) 7a.m.0.4010016.7a146b169176 80.5011520.827.46.6018.8c108d141e166174 90.8012533.327.45.95.9027.1120137163172 101.2015050.027.422.628.541.7138137160170 111.3020054.227.426.755.252.1175149162171 Noon1.4028058.327.430.986.156.3240181174176 1p.m.1.3030054.227.426.711356.3290218191185 20.8030033.327.45.911943.8300241204192 30.7015029.227.41.712031.3225238205193 40.6012525.027.42.411827.1138219200191 50.5010020.827.46.611122.9113202195189 60.5010020.827.46.610520.8100188189186 70.6015025.027.42.410222.9125178185184 80.8017533.327.45.910829.2163175183183 91.0022541.727.414.212337.5200181185184 100.9022537.527.410.1133f39.6225192189186 110.6017525.027.42.413031.3200194190187 Midnight0.4015016.727.410.811920.8163189189186 1a.m.0.3010012.527.414.910514.6125183186185 20.259010.427.417.087.511.595175183183 30.20808.327.419.168.49.485169181182 40.20708.327.419.149.38.375164178181 (Continued)
TABLE3.13(Continued)CalculationTableforDeterminationofTheoreticalVolumeoftheBasin,andCODConcentrationintheEffluentfromBasinVolumesof 0.133,0.35,and0.7MG(Example3.56) tQin,MLDCin,mg=LVin,103 LVout,103 LΔV,103 LP ΔV,103 Lqa,103 L=hCa,mg=LCoutforDifferentBainVolume,mg=L V¼0.133MLV¼0.35MLV¼0.7ML 50.258010.427.417.032.39.475158175179 60.309012.527.414.917.411.585152172178 70.4010016.727.410.87.1214.695146g 169176 Total658658 Average0.65815227.427.427.4152 a0:4MLD106L MLd 24h1h¼16:7103L bAssumeaneffluentconcentrationC1¼146mg=Ltostartthecalculation. cqa¼(0:4þ0:5)MLD 2¼0:45MLDorqa¼0:45MLD106L MLd 24h¼18:8103ML=h dCa¼(100þ115)mg=L 2¼108mg=L eRearrangeEquation3.44tocalculateC2. C2¼Ca(CaC1)eqa V(t2t1) ¼108mg=L(108mg=L146mg=L)e18:8103ML=h 133ML(87)h¼141mg=L fTheoreticalvolumeofthebasin. gThecalculatedeffluentconcentrationat7:00a.m.issameasthatassumedinthefirstrow.
effluentflow profiles at basin volumes of 0.35 and 0.7 ML will also be the same as those of 0.133 ML (Figure 3.39).
4. Determine the effluent concentration of COD from the equalization basin.
An iterative procedure is used to determine the COD concentration in the effluent.
a. Calculate averageflow rateqabetween the consecutive time intervals.
b. Calculate average concentrationCain the effluent between the consecutive time intervals.
c. Assume that the effluent COD concentrationC1at the start of time (7:00 a.m.) for a given basin volumeV. For example,C1=146 mg/L forV=0.133 ML.
d. Calculate effluent COD concentrationC2at the next time period for the given basin volume. In this case, the next time period is 8:00 a.m. Use the procedure shown in Example 3.55 to calculate C2.
e. Calculate effluent COD concentration for different time periodsC3,C4,…C24. After a 24-h cycle, the calculated effluent concentration at 7:00 a.m. in the last row should match the assumed value in thefirst row. If not, repeat the procedure until the values fall within an accept- able limit. The entire procedure for effluent COD concentration from 0.133 ML basin is sum- marized inTable 3.13.
f. Apply the similar procedure to determine the effluent COD concentrations from the basin with a volume of 0.35 and 0.7 ML, respectively. The results are summarized inTable 3.13.
5. Draw the influent and effluent COD concentration profiles.
The influent and effluent COD concentration profiles from the equalization basin with a total vol- ume of 0.133, 0.35, and 0.7 ML are respectively shown inFigure 3.40a through c. It may be noted that the effluent concentration is more uniform from the basin with a larger volume.
0 10 20 30 40 50 60 70
7:00 a.m.
9:00 a.m.
11:00 a.m.
1:00 p.m.
3:00 p.m.
5:00 p.m.
7:00 p.m.
9:00 p.m.
11:00 p.m.
1:00 a.m.
3:00 a.m.
5:00 a.m.
7:00 a.m.
Flow, 103 L/h
Time
Influent flow Effluent flow
Influent flow
Effluent flow
FIGURE 3.39 Influent and effluentflow profiles at theoretical volume of 0.133 ML (Example 3.56).
Note: The profiles for basin volumes of 0.35 and 0.7 ML will be the same as that for the theoretical volume of 0.133 ML.
0 100 200 300 (a) 400
(b)
(c) 7:00 a.m.
9:00 a.m.
11:00 a.m.
1:00 p.m.
3:00 p.m.
5:00 p.m.
7:00 p.m.
9:00 p.m.
11:00 p.m.
1:00 a.m.
3:00 a.m.
5:00 a.m.
7:00 a.m.
COD concentration, mg/L
Time
Influent COD concentration Effluent COD concentration
0 100 200 300 400
7:00 a.m.
9:00 a.m.
11:00 a.m.
1:00 p.m.
3:00 p.m.
5:00 p.m.
7:00 p.m.
9:00 p.m.
11:00 p.m.
1:00 a.m.
3:00 a.m.
5:00 a.m.
7:00 a.m.
COD concentration, mg/L
Time
Influent COD concentration Effluent COD concentration
0 100 200 300 400
7:00 a.m.
9:00 a.m.
11:00 a.m.
1:00 p.m.
3:00 p.m.
5:00 p.m.
7:00 p.m.
9:00 p.m.
11:00 p.m.
1:00 a.m.
3:00 a.m.
5:00 a.m.
7:00 a.m.
COD concentration, mg/L
Time
Influent COD concentration Effluent COD concentration Influent COD concentration
Effluent COD concentration
Influent COD concentration
Effluent COD concentration
Influent COD concentration
Effluent COD concentration
FIGURE 3.40 Influent and effluent COD concentration profiles: (a) basin volume of 0.133 ML, (b) basin vol- ume of 0.35 MG, and (c) basin volume of 0.7 MG (Example 3.56).
Discussion Topics and Review Problems
3.1 Wastewaterflows from manhole 1–2. Manhole 1 is connected to three sewer lines that bring average flow of 4.5 L/s, 80 gal/min, and 38 L/min. Manhole 2 receives additionalflow of 35 L/s. Determine theflow in the outgoing sewer.
3.2 An industrial sewer survey was conducted to establish the concentrations of heavy metals in the intercepting sewer. Averageflow and concentrations of heavy metals in different streams are given below. Determine the concentrations of various heavy metals in theflow from the intercepting sewer. Manhole (1) Q=3 L/s, Cu=4 mg/L, Zn=10 mg/L; Q=15 L/s, Cd=15 mg/L;Q=8 L/s, Cd=8 mg/L, Ni=8 mg/L, Manhole (2)Q=10 L/s, Cu=8 mg/L;Q=5 L/s, Ni=28 mg/L, Manhole (3)Q=8 L/s, Cd=8 mg/L, Ni=8 mg/L;Q=10 L/s, Cu=8 mg/L, Co=2 mg/L;Q= 5 L/s, Zn=6 mg/L. Manhole (4) receivesflows from three manholes.
3.3 A domestic water softener produces water with hardness of 5 mg/L as CaCO3. The raw water has hardness of 225 mg/L. The averagefinished water demand is 1600 gpd with hardness of 80 mg/L as CaCO3. Determine theflow through water softener.
3.4 In an ideal process, a chemical reaction with two liquidsF1andF2produces one liquid productP and one gaseous productG. The gas is produced at a mass ratio of 1 unit of gas for average 1000 units of liquid product. The recycle streamR=0.5P. Determine the weight and volume of liquid and gaseous products generated, and recycle stream. Use the following data: theflow rates of streams F1andF2are 1000 L/min and 6000 L/min; the densities of streamsF1andF2are 1.5 and 1.2 kg/L;
and the densities of the productsPandGare 1.3 kg/L and 1.0 kg/m3.
3.5 A primary sedimentation basin receives an average wastewaterflow of 1500 m3/d. The suspended solids in the raw wastewater is 240 mg/L. The solids removal efficiency of the basin is 60 %. The sludge has 4% solids and specific gravity is 1.03. Determine (a) TSS in the primary settled wastewa- ter, and (b) average quantity of sludge withdrawn from the basin in kg/d and m3/d.
3.6 A CFSTR is receiving an averageflow of 240 m3/d. The volume of the reactor is 48 m3. A slug of 25- L stock solution containing 6.0 g/L dye tracer is released at the influent channel of the reactor. Cal- culate the dye concentration in the effluent at time intervals of 0.5, 1.0, 2.0, 4.0, 5.0, 20.0, and 0.0 h from the time of release of the slug. Draw the tracer profiles as a function of (a) tracer concentration versus time, and (b) dimensionless functionsC/C0versust/θ.
3.7 The volume of a CFSTR is 45 m3and average constantflow in the reactor is 300 m3/d. The reactor flow regime was established by feeding continuously 6 g/L stock tracer solution in the influent line at a rate of 125 L per day. Determine the theoretical tracer concentration in the effluent at 3 and 10 h after the tracer injection. Plot the tracer profile as a function of (a) concentration versust, and (b) C/C0versust/θ.
3.8 A CFSTR receives 150 g/m3reactive material that undergoesfirst-order conversion reaction with k=0.20 h−1. The volume of the reactor is 400 m3, and the influentflow rate is 150 m3/d. Determine the exit concentration of the material and percent stabilization if the process is operating under steady state condition. What is the value oftin days to achieve 95% conversion?
3.9 A CFSTR receives industrial material for product conversion. The operational data of the reactor are:V=450 m3,Q=35 m3/d,C0=100 g/m3, andk=0.25 d−1. The reactor is operating under nonsteady-state conditions. Calculate (a) the concentration of the material at 0.5 d, 1 d, 2 d, 4 d, 6 d, 8 d, 10 d, 15 d, and 20 d, and (b) plot the graph of concentration remaining versus time.
3.10 A constantflow is maintained through a CFSTR. Att=0 a reactive substance is added into the influent stream at a constant rate. The conversion reaction is of thefirst order. Calculate the output concentration as g/m3for the following time intervals: 0.5 h, 1.0 h, 1.5 h, 3 h, 6 h, 8 h, and 10 h. The reactor data are given below:V=30 m3,Q=250 m3/d,C0=100 g/m3, andk=0.3 h−1. 3.11 A reactive substance with influent concentration of 75 g/m3is treated in a plugflow reactor. The
first-order reaction constant is 0.15/h. Calculate the effluent concentration. The volume of the reac- tor is 50 m3and average constantflow through the reactor is 150 m3/d.
3.12 fi
consists of two identical CFSTRs in series. The hydraulic retention time of each reactor is 60 h.
Oxygen is transferred by means of aerators at a rate which depends on the oxygen deficit. The DO transfer rate is expressed by the equationraer=k2(CDOs−CDOt). The temperature is 25◦C.
where
raer =oxygen transfer rate mg/L·d k2 =reaction rate constant=70 d−1
CDOs=DO saturation concentration=8.24 mg/L at 25◦C
The influent NH3-N concentration is 40 mg/L. The NH3-N conversion rate is given by Equa- tion 3.45.
CNH3-N= −k(CNH3-N)
Ks+(CNH3-N) (3.45)
where
k =0.12 g/m3·h Ks=0.74 g/m3
The DO concentration in the influent is zero. Calculate the dissolved oxygen and ammonia nitro- gen concentrations in (a) reactor 1 and (b) reactor 2. Assume steady-state conditions. Assume that there is no carbonaceous oxygen demand and oxygen consumption is due to only nitrification.
3.13 One CFSTR is arranged in between two PFRs in series (Figure 3.41). The arrangement is shown below. The influent BOD5is 200 g/m3. The wastewaterflow is 200 m3/d. Calculate thefinal effluent BOD5. Thefirst-order reaction rate constant is 0.08 h−1. Assume that the system is at steady state.
3.14 Four identical CFSTRs are operating in series. The influent concentration of the reactive substance is 250 mg/L. Thefirst-order reaction rate constant for each reactor is 0.74 h−1. Volumes of reac- tors 1, 2, 3, and 4 are 20 m3, 25 m3, 30 m3, and 35 m3respectively. Theflow through the assembly is 50 m3/h. Calculate the effluent concentration from all four reactors by using analytical and graphical method.
3.15 Calculate thefirst-order reaction rate constant in days for the given data. The number of CFSTRs in series are 5. The influent BOD5is 160 g/m3andfinal effluent BOD5is 10 g/m3. The hydraulic res- idence time of each reactor is 1 h. Determine the detention time of a PFR that will produce same effluent quality as that from the series of reactors.
3.16 Derive the expression for PFR treating a reactive substance under steady state. The second-order reaction kinetics apply. The reaction rates are: (a) second-order reaction rate r=–kC12
and (b) r= −kC11.5
.
3.17 A sedimentation basin receives an averageflow of 4.8 m3/min. The volume of the basin is 205 m3. A 5-L slug of dye tracer solution containing 50,000 mg/L of dye was introduced at the influent zone of the basin. The effluent samples were collected at different time intervals and dye concentrations were measured. The sampling time and dye concentrations are given below.
BOD5 = 200 g/m3
PFR I CFSTR PFR II
V = 100 m3 V = 150 m3 V = 100 m3 Q = 200 m3/d
FIGURE 3.41 Definition sketch of one CFSTR between two PFRs (Problem 3.13).