VOLUME 2 Post-Treatment, Reuse, and Disposal
4.5 Wastewater Flow Variation
4.5.1 Dry Weather Flow
Like water demand, wastewaterflows also vary with respect to time of the day, day of the week, weather condition, and season of the year. Under dry weather conditions, the daily wastewaterflow shows a diurnal pattern exhibiting a peak and a minimum in 24-h period. The diurnal wastewaterflow parallels that of water demand with a lag of few hours (Figure 4.1).
Hourly peak and minimum dry weather flows are generally estimated from several equations and graphical relationships developed from case studies. The ratio of hourly peak to daily average, and hourly minimum to daily averageflows depend upon the contributing population. Commonly used equations are given by (Equations 4.4 through 4.8).1,4,5,11
Mmax=1+ 14
4+
√P (4.4)
Mmax= 5
P0.167 (4.5)
Q′max=3.2(Q′avg)5/6 (4.6)
log(Mmax) = −0.19 log(P′) +0.74 (4.7)
log(Mmin) =0.142 log(P′) −0.682 (4.8)
where
Mmax=peaking factor, or ratio of hourly maximum to daily averageflows P =population in thousands
Q′max =hourly maximum dry weatherflow, MGD Q′avg =daily average dry weatherflow, MGD
Mmin=ratio of hourly minimum to daily averageflows P′ =population in thousands, range 1,P′,10,000
EXAMPLE 4.8: PEAK DRY WEATHER WASTEWATER FLOW CALCULATION FROM DIFFERENT EQUATIONS
A city has a population of 20,000 residents. The average wastewaterflow is 380 Lpcd. Calculate (1) daily average and hourly peak dry weatherflows in m3/d and MGD, (2) hourly minimumflow in m3/d, and (3) compare the results.
Solution
1. Calculate the daily average wastewaterflow.
Daily average wastewater flow = 380 L
person·d×20,000 persons× m3
1000 L=7600 m3/d or 2.0 MGD 2. Calculate the hourly peak dry weatherflows from Equations 4.4 through 4.7.
Peaking factor from Equation 4.4,Mmax=1+ 14
4+
√ =P 1+ 14
4+
√ =20 2.65 Hourly peak dry weather flow=2.65×7600 m3/d=20,140 m3/d or 5.30 MGD
4.5.2 Infiltration and Inflow
Infiltration is the groundwater that enters the sewers through service connections, cracked pipes, defective joints, and cracked manhole walls. Inflow is the directflow due to surface runoffthat may enter through a manhole cover, roof area drains, and cross connection from storm sewers and combined sewers. Flow from cellar and foundation drains, cooling water discharges, and drainage from springs and other steady flows are included with the infiltration. The conditions that apply to I/I are listed below:
1. The amount of I/I reaching a sewer system depends upon a. The length and age of the sewer
b. The construction material, methods, and workmanship c. Number of roof or drain connections
d. The groundwater table relative to sewer position e. Type of soil, ground cover, and topographic conditions
Peaking factor from Equation 4.5,Mmax= 5
P0.167= 5
(20)0.167=3.03
Hourly peak dry weather flow=3.03×7600 m3/d=23,028 m3/d or 6.06 MGD Hourly peak dry weatherflow Equation 4.6,
Q′max=3.2(Q′avg)5/6=3.2(2.0)5/6=5.70 MGD or 21,575 m3/d
Peaking factor from Equation 4.7,log(Mmax) = −0.19 log(P′) +0.74= −0.19 log(20) +0.74 Mmax=3.11
Hourly peak dry weather flow=3.11×7600 m3/d=23,636 m3/d or 6.22 MGD 3. Calculate the hourly minimumflow.
Ratio of hourly minimum to daily averageflows from Equation 4.8, log(Mmin) =0.142 log(P′) −0.682=0.142×log(20) −0.682= −0.497
Mmin=0.32
Hourly minimum dry weather flow=0.32×7600 m3/d=2432 m3/d or 0.64 MGD 4. Compare the results.
Flow Condition m3=d MGD Ratio to Average
Flow Hourly peak dry weatherflow from Equation 4.4 20,140 5.30 2.65 Hourly peak dry weatherflow from Equation 4.5 23,028 6.06 3.03 Hourly peak dry weatherflow from Equation 4.6 21,575 5.70 2.85 Hourly peak dry weatherflow from Equation 4.7 23,636 6.22 3.11 Hourly minimum dry weatherflow from Equation 4.8 2432 0.64 0.32
The results of hourly peak dry weatherflow from four equations are comparable. The results indi- cate that the ratio of hourly peak and minimum dry weatherflows is approximately 10:1.
2. Permissible I/I allowance is 780 Lpcd (205 gpcd) as long as there is no operational problems such as surcharges, bypasses, or poor treatment performance resulting from hydraulic overloading of pub- lically owned treatment work (POTW) during storm events.
3. In older sewers, infiltration is high because of deteriorated joints and masonry mortar. Newer sew- ers use joints sealed with rubber gaskets or synthetic material, and use precast manholes. Also, the I/I rate is significantly smaller.
4. When designing sewers, allowance must be made for old and new sewers. Average values of I/I allowance are 94–9400 L/cm (diameter)·km (linear length)·d (100–10,000 gpd/in·mile) or 200– 28,000 L/ha·d (20–3000 gpd/acre).
5. In the absence offlow records, many states specify 1500 Lpcd (400 gpcd) for design of laterals and submains, and 1300 Lpcd (350 gpcd) for mains and trunk sewers. These designflows include nor- mal I/I allowance.
6. For new sewers or recently constructed sewer systems having precast manholes and pipe joints made with gaskets of rubber or rubber like material, the average infiltration allowance may be found fromFigure 4.2.
4.5.3 Common Terms Used to Express Flow Variations
Several terms are commonly used to express theflow variations in the overall context of planning, design, and operation of wastewater treatment facilities. These terms and their significance are briefly presented below.
Daily Average Flow: The daily averageflow (also called the annual average dayflow) is based on annual flow rate data. It is the averageflow occurring over 24-h period under dry weather condition. This flow is very important and used to (1) evaluate treatment plant capacity, (2) developflow rate ratios, (3) size many treatment units, (4) calculate organic loadings, (5) estimate sludge solids, (6) estimate chemical needs, and (7) calculate pumping and treatment unit costs.
1 10 100
10 100 1000 10000
Peak infiltration rate, m3/ha·d
Service area, ha 10 ha, 48.5 m3/ha·d
Curve A for old sewers
5000 ha, 10 m3/ha·d
Curve B for new sewers
5000 ha, 3.3 m3/ha·d New sewers include existing sewers having pipe joints sealed
with compression gaskets of elastomeric materials material.
40.5 ha, 14 m3/ha·d
Note: ha × 2.4711 = acre; m3/ ha·d × 106.9 = gal/acre·d.
FIGURE 4.2 Average allowance for infiltration rate in new sewers.
Source:Adapted in part from Reference 4.
fl fl curve. Thisflow is important as it brings about dailyflushing action in sewers, conduits, and chan- nels. It is used to design and check the retention time of several critical components of a treatment facility.
Peak Wet Weather Flow: Peak wet weatherflow occurs after or during precipitation, and includes a substantial amount of I/I. Some regulatory agencies define peak wet weatherflow as the highest 2-hflow encountered under any operational condition, including times of highest rainfall (generally the 2-year, 24-h storm is assumed), and prolonged period of wet weather. The highest 2-hflow is very important in the design of wastewater treatment facilities. The ratio of wet weatherflow with respect to average dailyflow may reach 3–5:1. Peak hourlyflow (or highest 2-hflow) is needed to design (1) the interceptor sewers and collection system, (2) conduits and connecting pipes and channels in a treatment plant, (3) pumping stations, (4)flow meters, (5) bar screen and grit chan- nels, (6) the hydraulics of the influent and effluent structures of most treatment units, and (7) to check the retention period of many critical treatment units.
Minimum Hourly Flow: The minimum hourlyflow is the lowestflow on a typical dry weather diurnal flow curve (Figure 4.1). Lowflows may cause settling of solids in pipes and channels. Theseflow rates are needed for sizing of (1)flow meters, (2) lower range of chemical feeder, and (3) pumping equipment. Often recirculation is needed during lowflows to maintain hydraulic loadings in some treatment units.
Sustained Flows: Sustainedflows are theflows that persist over several days. Sustainedflows could be on the low or high sides. As an example, extraordinarily dry or hot weather may cause sustained low extremes. Likewise, prolong wet weathers and special events in a community such as fairs, conven- tions, games, exhibitions may be a source of population surges and could all cause high sustained flows. Data on sustainedflow rates may be needed in sizing equalization basins and other plant hydraulic components, and to check the design of many units under extreme conditions. The sus- tained-flow envelope should be developed from the longest available period of record. The typical sustainedflow envelope at a treatment plant is shown inFigure 4.3.
Emergency Flow Diversion: Many POTWs are equipped with aflow diversion facility to hold the excess or entireflow under emergency situations. Such conditions may develop if (1) power failure occurs, (2) the combined sewer overflows (CSOs) are directed, and (3) excessive I/I is diverted from the
5 10 15 20 25 30 35
1 2 3 4
0 0
Number of consecutive days during period of return that flow was sustained
Ratio of averaged sustained peak and low flows
Typical values Range of
values
FIGURE 4.3 The typical sustainedflow envelope at a wastewater treatment plant.
Source:Adapted in part from References 1, and 3 through 5.
sanitary sewers. The divertedflow is temporarily stored in a basin, and rerouted through the plant after restoration of power, or when the plant is able to handle the storedflows. Various types of flow-regulating devices are used to achieve proper flow diversion. Common diversion devices are side weir, transverse weir, leaping weir, orifice, relief siphon, tripping plate, and hydrobrake reg- ulators. References 1, 3, 4, 11, and 12 should be consulted for design information on these diversion devices.
Design Flows: The designflows are the ultimateflows when the facility is expected to reach its full design capacity. Designflows occur at the end of the design period, which may be 10–20 years after initial construction is completed. The designflows are peak wet weather, average day, and minimumflows.
EXAMPLE 4.9: PEAK I/////I
A city subdivision has a population density of 3500 residents per km2. During wet weather condition, the expected average I/I from this subdivision is 3800 L/ha·d and the peaking factor is 1.5. Estimate average and peak I/I in Lpcd.
Solution
1. Estimate the population per ha.
Population density=3500 persons km2 × km2
106m2×10,000 m2
ha =35 persons/ha 2. Estimate the average I/I.
Average I/I=3800 L ha·d
Average I/I per person=3800 L
ha·d × 1
35 person/ha=109 Lpcd 3. Estimate the peak I/I.
Peak I/I=1.5×109 Lpcd=164 Lpcd
EXAMPLE 4.10: PERMISSIBLE I/////I FOR NEW DEVELOPMENT
A new housing subdivision is planned. The total land area that will be developed is 2470 acres. Using Figure 4.2, estimate the average I/I allowance for design of intercepting sewer.
Solution
1. Determine the area served in ha.
Land area= ha
2.4711 acre×2470 acres=1000 ha
2. Estimate fromFigure 4.2the I/I allowance for new sewer system.
I/I allowance=5.7 m3/ha·d 3. Determine the average I/I.
Average I/I=5.7 m3/ha·d×1000 ha=5700 m3/d
EXAMPLE 4.11: PERMISSIBLE I/////I ALLOWANCE IN A SEWER AND PEAK WET WEATHER FLOW
The population of a city is 40,000. Average wastewaterflow is 400 Lpcd, and permissible average I/I allow- ance is 890 L/cm·km·d (961 gpd/in·mile). The average length of sanitary sewer is 8 m per capita. The sewer distribution by pipe size is as follows: 15 cm dia.=4%, 20 cm dia.=79%, 31 cm dia.=10%, 46 cm dia.=5%, and 61 cm dia.=2%. Estimate I/I in Lpcd and peak wet weatherflow in m3/d. Assume the peaking factor for I/I is 1.6.
Solution
1. Calculate the daily average and hourly peak dry weatherflows.
Daily average flow= 400 L
person·d×40,000 persons× m3
1000 L=16,000 m3/d Peaking factor from Equation 4.4,Mmax=1+ 14
4+
√ =P 1+ 14
4+
√ =40 2.36 Hourly peak dry weather flow=2.36×16,000 m3/d=37,760m3/d
Hourly peak dry weather flow in Lpcd=2.36×400 Lpcd=944 Lpcd 2. Calculate the length and equivalent sewer diameter in the collection system.
Total length of sanitary sewer= 8 m
person×40,000 persons× km
1000 m=320 km Equivalent diameter(weighted average)
=0.04×15 cm+0.79×20 cm+0.1×31 cm+0.05×46 cm+0.02×61 cm 0.04+0.79+0.1+0.05+0.02
=23.02 cm
1 =23 cm
3. Calculate the peak I/I.
Daily average I/I allowance= 890 L
cm·km·d× (23 cm×320 km) × m3
1000 L=6550 m3/d Daily average I/I allowance in Lpcd=6550 m3/d×1000 L
m3 × 1
40,000 persons=164 Lpcd Hourly peak I/I allowance=1.6×164 Lpcd=262 Lpcd
Hourly peak I/I allowance=40,000 persons× 262 L
person·d× m3
1000 L=10,480 m3/d 4. Calculate the hourly peak wet weatherflow.
Total hourly peak wet weatherflow to the plant
=hourly peak dry weatherflow+hourly peak I/I allowance
=(37,760+10,480) m3/d=48,240 m3/d
Total hourly peak wet weatherflow in Lpcd=(944+262) Lpcd=1206 Lpcd
EXAMPLE 4.12: I/////I AND DIURNAL FLOW CURVE
A sewer evaluation study of a service area was conducted to determine the infiltration and inflow reaching the collection system and to estimate the peak wet weatherflow. Aflow measurement device was installed in the interceptor that carried the entireflow from the service area. The hourlyflow measurements on a typical dry day gave the diurnal weatherflow pattern. The diurnalflow pattern of a typical dry day is shown inFigure 4.4. After a long dry spell, thefirst high-intensity storm caused heavy downpour in the service area. Since the groundwater table was low, the infiltration was negligible, and the excess flow over the dry weatherflow was the inflow. The measured hourlyflow under this condition is illus- trated inFigure 4.4. After this downpour, high-intensity storms and intermittent showers continued to occur over a 10-day period. Under this wet weather condition, the ground was completely saturated.
The recordedflow data are a measure of combined infiltration and diurnal dry weatherflow in the inter- ceptor. It is also shown inFigure 4.4. Determine (a) daily average dry weatherflow, (b) hourly peak dry weatherflow, (c) daily average infiltration, (d) 2-h peak inflow, (e) 2-h peak wet weatherflow, and (f) 2-h peaking factor, which is the ratio of 2-h peak wet weatherflow and daily average dry weatherflow.
Solution
1. Calculate the daily average and hourly peak dry weatherflows.
The daily average dry weatherflow is calculated from the diurnal dry weatherflow curve given in Figure 4.4.
Apply the trapezoidal rule (Equation 4.1).
Daily average dry weatherflow
=1
2×2 h× [40 m3/h+2× (24+20+26+40+75+110+105+51+54+77 +60)m3/h+40 m3/h]
=1364 m3total in 24 h or 56.8 m3/h as daily average
Hourly peak dry weatherflow as read between 12:00 and 1:00 p.m. inFigure 4.4=111 m3/h 0
20 40 60 80 100 120 140
0 2 4 6 8 10 12 14 16 18 20 22 24
Time of day Flow, m3/h
Duration of high-intensity
precipitation Inflow
Inflow
Peak dry weather flow Diurnal dry weather flow Infiltration + dry weather flow
FIGURE 4.4 Diurnal dry weatherflow, infiltration and inflow recorded in an interceptor (Example 4.12).
2. Calculate the daily average infiltration when soil is saturated completely.
The combined infiltration and dry weatherflow curve is also shown inFigure 4.4.
The average infiltration is also calculated from the curve using the trapezoidal rule (Equation 4.1).
Area under infiltrationflow curve (includes dry weatherflow)
=1
2×2 h× [59 m3/h+2× (42+39+45+58+94+128+123+70+73 +97+79)m3/h+59 m3/h]
=1814 m3total in 24 h
Daily average infiltration plus dry weather flow=1814 m3
24 h =75.6 m3/h Daily average infiltrationflow=(75.6−56.8) m3/h=18.8 m3/h 3. Estimate the 2-h peak inflow.
The inflow is estimated from the inflow curve between 3:30 and 5:30 a.m. inFigure 4.4.
2-h peak inflow plus base dry weatherflow=54 m3/h 2-h base dry weatherflow=21 m3/h
2-h peak inflow=(54−21) m3/h=33 m3/h 4. Estimate the 2-h peak wet weatherflow.
In the worst case scenario, assume peak inflow occurs between 12:00 and 2:00 p.m. when hourly peak dry weatherflow is reached. To be conservative, hourly peak dry weatherflow is used in the estimation.
2-h peak wet weatherflow¼hourly peak dry weatherflowþdaily average infiltration þ2-h peak inflow
¼(111þ18.8þ33) m3=h¼162.8 m3=h 5. Calculate the 2-h peaking factor.
2-h peaking factor=162.8 m3/h 56.8 m3/h =2.9
EXAMPLE 4.13: PEAKING FACTOR IN A CITY DEVELOPMENT PROJECT The future land use plan of a city development project has residential, commercial, and industrial areas, and a day school. Data on the expected saturation population densities, daily average wastewaterflows, and peaking factors for various types of land use plan of the development project are given inTable 4.5.
The day school has an enrollment of 2000 students. Estimate the daily average and peak wastewaterflow rates and the overall peaking factor.
Solution
1. Calculate the population andflows.
Set up a computational table for estimating the population andflows. The summary of computa- tions is provided inTable 4.6.
2. Estimate the daily averageflows.
Daily average dry weatherflow ¼19,029 m3=d Daily average I=Iflow ¼3700 m3=d Daily averageflow including average I=I ¼22,729 m3=d 3. Estimate the peakflows.
Peak dry weatherflow ¼46,092 m3=d
Peak I=Iflow ¼22,200 m3=d
Peak wet weatherflow ¼68,292 m3=d 4. Estimate the overall peaking factor.
Overall peaking factor=68,292 m3/d 22,729 m3/d=3.0
TABLE 4.6 Computation Table for Population and Flow Estimation (Example 4.13) Land Use Area, ha Expected
Population Density per ha
Population Unit Unit Loading
Daily Average
Flow, m3=d
Peaking Factor
Peak Flow, m3=d Single family
dwellings
170 35 5950 Lpcd 350 2083 3.2a 6666
Mixed residential dwellings
120 45 5400 Lpcd 260 1404 3.2a 4493
Apartments 150 130 19,500 Lpcd 220 4290 2.7a 11,583
School 40 2000 Lpcd 76 152 4.8 730
Commercial 150 m3=ha·d 30 4500 1.8 8100
Industrial 110 m3=ha·d 60 6600 2.2 14,520
Subtotal 740 32,850 19,029 2.4b 46,092
I=I 740 m3=ha·d 5 3700 6.0 22,200
Total=overall 740 32,850 22,729 3.0c 68,292
aValues calculated from Equation 4.4.
b46,092m3=d 19,029m3=d¼2:4.
cSee calculation in Step 4.
TABLE 4.5 Expected Saturation Population Densities, Wastewater Flows, and Peaking Factors for Various Land Use Plan of the City Development Project (Example 4.13)
Land Use Area, ha Expected Population
Density per ha Unit Unit
Loading
Peaking Factor
Single family dwellings 170 35 Lpcd 350 Equation 4.4
Mixed residential dwellings 120 45 Lpcd 260 Equation 4.4
Apartments 150 130 Lpcd 220 Equation 4.4
School 40 Lpcd 76 4.8
Commercial 150 m3=ha·d 30 1.8
Industrial 110 m3=ha·d 60 2.2
Infiltration=inflow Total area m3=ha·d 5 6.0
EXAMPLE 4.14: SUSTAINED AVERAGE FLOW RATIO (SAFR) ENVELOPE
Describe the procedure for developing sustained averageflow ratio (SAFR) envelope. Daily averageflow (DAF) data during 30 consecutive days at a wastewater treatment plant is summarized inTable 4.7. The annual average dayflow (AADF) at the plant is 0.58 m3/s. Develop the maximum and minimumSAFR envelope up to a 10-day period.
Solution
1. Describe the procedure for developing theSAFRenvelope.
a. Select the longest available period of consecutive days (m) withDAFdata.
b. Determine theAADF.
c. Calculate theSAFRifor eachiconsecutive days,i=1, 2, 3,…,n, andn,m.
d. Identify the maximum and minimum values ofSAFRi(i=1, 2, 3,…,n).
e. Summarize the maximum and minimumSAFRdata.
f. Plot the maximum and minimumSAFRdata as well as theAADFratio with respect toicon- secutive days theflow rate is sustained. TheAADFline will be horizontal and pass from the ratio of 1. The maximum and the minimum SAFR curves will be above and below the AADFline, respectively.
2. Develop the maximum and minimumSAFRenvelope for a 10-day period (n=10).
Calculation results are tabulated inTable 4.8. A brief description is provided below for each step of the calculations.
a. In this problem, a 30-day period (m=30) is selected in Column 1 and the consecutiveDAF data is summarized in Column 2.
b. TheAADFis 0.58 m3/s at the plant.
c. SAFRiis calculated for eachiconsecutive days (i=1, 2, 3,…, 10). The calculation results are summarized in Columns 4–12.
d. The maximum and minimum values ofSAFRiare identified in bold numbers in Columns 3–12 for eachiconsecutive days over the 30-day period.
TABLE 4.7 Average Flow (AF) Data during 30 Consecutive Days (Example 4.14)
Day DAF, m3=s Day DAF, m3=s Day DAF, m3=s
1 0.25 11 0.65 21 0.83
2 0.16 12 0.66 22 0.80
3 0.15 13 0.67 23 0.72
4 0.13 14 0.68 24 0.66
5 0.14 15 0.69 25 0.66
6 0.25 16 0.70 26 0.65
7 0.35 17 0.75 27 0.50
8 0.48 18 1.12 28 0.32
9 0.52 19 1.20 29 0.20
10 0.60 20 1.15 30 0.18
TABLE 4.8 Summary of Calculations of Maximum and Minimum of SAFRs (Example 4.14)
Day DAF, m3=s SAFRi
i¼1 i¼2 i¼3 i¼4 i¼5 i¼6 i¼7 i¼8 i¼9 i¼10
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)
1 0.25 0.43a
2 0.16 0.28a 0.35b
3 0.15 0.26 0.27b 0.32c
4 0.13 0.22c 0.24 0.25d 0.30
5 0.14 0.24 0.23e 0.24f 0.25 0.29
6 0.25 0.43 0.34 0.30 0.29 0.29 0.31
7 0.35 0.60 0.52 0.43 0.38 0.35 0.34 0.35
8 0.48 0.83 0.72 0.62 0.53 0.47 0.43 0.41 0.41
9 0.52 0.90 0.86 0.78 0.69 0.60 0.54 0.50 0.47 0.47
10 0.60 1.03 0.97 0.92 0.84 0.76 0.67 0.61 0.56 0.53 0.52
11 0.65 1.12 1.08 1.02 0.97 0.90 0.82 0.74 0.67 0.63 0.59
12 0.66 1.14 1.13 1.10 1.05 1.00 0.94 0.86 0.79 0.72 0.68
13 0.67 1.16 1.15 1.14 1.11 1.07 1.03 0.97 0.90 0.83 0.77
14 0.68 1.17 1.16 1.16 1.15 1.12 1.09 1.05 0.99 0.93 0.86
15 0.69 1.19 1.18 1.17 1.16 1.16 1.14 1.10 1.07 1.02 0.96
16 0.70 1.21 1.20 1.19 1.18 1.17 1.16 1.15 1.11 1.08 1.03
17 0.75 1.29 1.25 1.23 1.22 1.20 1.19 1.18 1.16 1.13 1.10
18 1.12 1.93 1.61 1.48 1.41 1.36 1.32 1.30 1.28 1.25 1.21
19 1.20 2.07c 2.00 1.76 1.63 1.54 1.48 1.43 1.39 1.36 1.33
20 1.15 1.98 2.03e 1.99f 1.82 1.70 1.61 1.55 1.50 1.46 1.43
21 0.83 1.43 1.71 1.83 1.85 1.74 1.65 1.59 1.53 1.49 1.46
22 0.80 1.38 1.41 1.60 1.72 1.76 1.68 1.61 1.56 1.52 1.48
23 0.72 1.24 1.31 1.35 1.51 1.62 1.67 1.62 1.57 1.52 1.49
24 0.66 1.14 1.19 1.25 1.30 1.43 1.54 1.60 1.56 1.52 1.49
25 0.66 1.14 1.14 1.17 1.22 1.27 1.39 1.48 1.54 1.51 1.48
26 0.65 1.12 1.13 1.13 1.16 1.20 1.24 1.35 1.44 1.49 1.47
27 0.50 0.86 0.99 1.04 1.06 1.10 1.15 1.19 1.29 1.37 1.43
28 0.32 0.55 0.71 0.84 0.92 0.96 1.01 1.06 1.11 1.20 1.29
29 0.20 0.34 0.45 0.59 0.72 0.80 0.86 0.91 0.97 1.02 1.12
30 0.18 0.31 0.33 0.40 0.52 0.64 0.72 0.78 0.84 0.90 0.95
aThe value ofSAFR1is separately calculated for each day in Column 3, Columnð3Þ ¼Columnð2Þ=ð0:58m3=sÞ.
For example,SAFR1¼AADFDAF ¼0:25 m0:58 m33=s=s¼0:43 for day 1, andSAFR1¼0:16 m0:58 m33=s=s¼0.28 for day 2.
bThe value ofSAFR2is separately calculated for each day in Column 4, Column (4)¼(1=2)(Sum of two consecutive values in Column (3)).
For example,SAFR2¼(1=2)(0.43þ0.28)¼0.35 during thefirst two days, andSAFR2¼(1=2)(0.28þ0.26)¼0.27 during the second and third days.
cThe maximum and minimum values of 2.07 and 0.22 are identified forSAFR1and highlighted in Column 3.
dThe value ofSAFR3is separately calculated for each day in Column 5, Column (5)¼(1=3)(Sum of three consecutive values in Column (3)).
For example,SAFR3¼(1=3)(0.43þ0.28þ0.26)¼0.32 during thefirst three days, andSAFR3¼(1=3)(0.28þ0.26þ 0.22)¼0.25 between the second and fourth days. Similarly, values ofSAFRi(i¼4, 5,…, 10) are calculated and summarized in Columns 6–10.
eThe maximum and minimum values of 2.03 and 0.23 are identified forSAFR2and highlighted in Column 4.
fThe maximum and minimum values of 1.99 and 0.24 are identified forSAFR2and highlighted in Column 5. Similarly, the maximum and minimum values are identified for allSAFRi(i¼4, 5,…, 10) and highlighted in Columns 6–10.
3. Summarize the maximum and minimumSAFRdata.
The maximum and minimumSAFRdata is summarized below.
Parameter
SAFRi
i¼1 i¼2 i¼3 i¼4 i¼5 i¼6 i¼7 i¼8 i¼9 i¼10
Maximum 2.07 2.03 1.99 1.85 1.76 1.68 1.62 1.57 1.52 1.49
Minimum 0.22 0.23 0.24 0.25 0.29 0.31 0.35 0.41 0.47 0.52
4. Plot the maximum and minimum values ofSAFRwith respect to the number of consecutive days dur- ing the 10-day period.
These plotted lines form an envelope. As the number of consecutive days is increased, the envelope will tend to converge to theAADFratio passing through 1. TheSAFRenvelope for 10 consecutive days is shown inFigure 4.5.
EXAMPLE 4.15: FORECASTING DESIGN FLOW RATES
A city has a current total population of 40,000 residents. The estimated design population is 55,000 at the end of a design period of 20 years. Determine the design average, peak, and minimumflow rates. Use the following data.
Current daily averageflow to the wastewater treatment plant ¼18,400 m3=d
Daily average I=Iflow ¼80 Lpcd
Hourly peak I=Iflow ¼180 Lpcd
A new junior college will attract 1500 nonresident students per day. The daily average wastewaterflow is 50 Lpcd. A new industry will add a daily averageflow of 900 m3/d for a 24 h/d operation. An hourly peakflow of 1500 m3/d will occur during the day shift. The plant will be shut down one day a week. It is expected that 10% per capita reduction in wastewaterflow will occur with the future population growth.
The minimum to averageflow ratio is 0.4 for the current dry weather wastewaterflow.
Solution
1. Calculate the current wastewaterflow in Lpcd.
Daily average I=Iflow ¼ 80 L
persond40,000 people m3
1000 L¼3200 m3=d Maximum SAFR
Minimum SAFR
AADF ratio = 1.0
15 10
5 0
Number of consecutive days the flow rates are sustained (d)
Maximum Minimum
0 1 2 3
SAFR
FIGURE 4.5 Sustained averageflow ratio (SAFR) envelop (Example 4.14).