Solutions End of Chapter Problems Chapter 7
1. A developer in Lexington, Virginia is proposing a new retirement community called “Keydet
Acres” which is expected to have a build-out population of 3000 people by the year 2025.
Assume a domestic wastewater with weak characteristics and use the values in Tables 7.2 and
7.4 for estimating the following:
a) average daily flow (MGD) in 2025
b) peak hour flow (MGD) in 2025
c) minimum hour flow (MGD) in 2025
d) average daily COD mass loading (lb/d) in 2025
e) average daily TN mass loading (lb/d) in 2025
f) average daily TP mass loading (lb/d) in 2025
Solution Part A
The average daily flow in 2025 is estimated by multiplying the build-out population of 3000
people by the per capita wastewater generation rate of 120 gallons per capita per day.
ADF 3000 people×
120gal 1MG
×
capita×d 106 gal
0.36 MGD
Solution Part B
From Table 7.2, assume a PHF: ADF peaking factor of 3:1. Calculate the peak hour flow for
2025 by multiplying ADF by 3:1 as follows:
PHF ADF× PF
PHF ADF×
PHF
ADF
0.36
MG 3
D 1
1.08MGD
Solution Part C
The minimum hour flow (MHF) is calculated similarly. From Table 7.2, assume a MHF: ADF
peaking factor of 0.33:1. Calculate the minimum hour flow for 2025 by multiplying ADF by
0.33:1 as follows:
MHF ADF× PF
1
MHF ADF×
MHF
ADF
0.36
MG 0.33
D
1
0.12 MGD
Solution Part D
Average daily mass loading rates are calculated using Equation (7.1). The average daily COD
mass loading rate is calculated by multiplying the ADF by the assumed COD concentration in
Table 7.4 for a weak wastewater as follows:
m
lb
d
Q (MGD)× C
mg
L
m
lb
d
0.36 MGD× 250
mg
L
8.34lb
MG mg L
8.34lb
MG mg L
751
lb COD
d
Solution Part E
Using Equation (7.1) and the assumed TN concentration in Table 7.4 for a weak wastewater as
follows:
m
m
lb
d
lb
d
Q (MGD)× C
0.36 MGD× 20
mg
L
mg
L
8.34lb
MG mg L
8.34lb
MG mg L
60.0
lb TN
d
Solution Part F
Using Equation (7.1) and the assumed TP concentration in Table 7.4 for a weak wastewater of
4.0 mg/L as follows:
m
m
lb
d
lb
d
Q (MGD)× C
0.36 MGD× 4.0
mg
L
mg
L
8.34lb
MG mg L
8.34lb
MG mg L
12.0
lb TP
d
2. Tabulated below are the average monthly influent flow and wastewater characteristics for a
wastewater treatment plant near Orlando, Florida. Estimate the following:
a) Annual average daily, maximum month daily, and minimum month daily flows in
MGD.
2
b) Peaking factors for maximum month and minimum month daily flows.
c) Average influent COD, maximum month COD, and minimum month COD
concentrations in mg/L.
d) Peaking factors for maximum month and minimum month COD concentrations.
e) Average influent TP, maximum month TP, and minimum month TP concentrations in
mg/L.
f) Peaking factors for maximum month and minimum month TP concentrations.
g) Average influent COD, maximum month COD, and minimum month COD mass
loadings in lb/d.
h) Peaking factors for maximum month and minimum month COD mass loadings.
i) Average influent TP, maximum month TP, and minimum month TP mass loadings in
lb/d.
j) Peaking factors for maximum month and minimum month TP mass loadings.
D
ate
Influent
flow,
Influent COD
Influent TP
concentration,
concentration,
mg/L
mg/L
(2)
(3)
MGD
(1)
Jan-09
3.72
362
7.00
Feb-09
4.48
416
6.35
Mar-09
4.73
386
7.29
Apr-09
4.65
419
7.09
May-09
4.63
427
6.70
Jun-09
5.41
394
7.15
Jul-09
5.69
365
7.15
Aug-09
5.59
353
7.16
Sep-09
4.87
292
7.00
Oct-09
5.70
325
6.52
3
Nov-09
5.95
333
8.11
Dec-09
5.28
283
9.05
Solution
For this type of problem it is best to use some type of spreadsheet to perform the calculations.
The following table shows the results of the spreadsheet analysis. At the bottom of the table the
peaking factors for flows, concentrations, and mass loadings are listed. Sample calculations are
provided beneath the table to identify how the values were determined.
Date
Influent
flow,
Influent COD
Influent TP
Influent COD
Influent TP
concentration,
concentration,
mass loading,
mass loading,
mg/L
mg/L
lb/d
lb/d
(2)
(3)
(4)
(5)
MGD
(1)
9-Jan
3.72
362
7.00
11231
217
9-Feb
4.48
416
6.35
15543
237
9-Mar
4.73
386
7.29
15227
288
9-Apr
4.65
419
7.09
16249
275
9-May
4.63
427
6.70
16488
259
Jun-09
5.41
394
7.15
17777
323
Jul-09
5.69
365
7.15
17321
339
Aug-09
5.59
353
7.16
16457
334
Sep-09
4.87
292
7.00
11860
284
Oct-09
5.70
325
6.52
15450
310
Nov-09
5.95
333
8.11
16524
402
Dec-09
5.28
283
9.05
12462
399
(A)
5.06
363
7.21
15216
306
Max.
month
(B)
5.95
427
9.05
17777
402
Average
4
Min.
month
(C)
3.72
283
6.35
11231
217
1.18
1.18
1.25
1.17
1.31
0.74
0.78
0.88
0.74
0.71
Max.
month PF
(D)
Min.
month PF
(E)
Solution Part A
The solution to Part A is determined as follows. The annual average daily flow is the arithmetic
average or mean of the 12 months of data shown in Column 1 (Column 1, Row A lists the
answer). For this data set, the annual average daily flow rate is 5.06 MGD.
ADF
3.72+4.48+ 4.73 4.65+ 4.63+5.41+5.69 +5.59 + 4.87 +5.70 + 5.95+5.28
12
ADF
5.06 MGD
The maximum month flow for this data set is the maximum average month flow value which is
5.95 MGD for November 2009, whereas, the minimum month flow is the minimum average
month flow which corresponds to 3.72 MGD for January 2009.
Solution Part B
The peaking factors for flow are calculated as follows. The maximum month flow to average
daily flow is determined by dividing the maximum month flow by the average daily flow shown
below (See Column 1, Row D).
PFMax month flow =
maximum month flow 5.95MGD
=
average daily flow
5.06 MGD
1.18
The WEF MOP #8 (1998) indicates this PF typically ranges from 0.9 to 1.2.
The minimum month flow to average daily flow is determined by dividing the minimum month
flow by the average daily flow shown below (See Column 1, Row E).
5
PFMin month flow =
minimum month flow 3.72 MGD
=
average daily flow
5.06 MGD
0.74
The WEF MOP #8 (1998) indicates this PF typically ranges from 0.9 to 1.1.
Solution Part C
The solution to Part C is determined as follows. The annual average COD concentration is the
arithmetic average or mean of the 12 months of data shown in Column 2 ( Column 2, Row A
gives the answer). For this data set, the annual average COD concentration is 363 mg/L.
Average COD
362 + 416 +386 + 419 + 427 +394 +365+353+ 292 +325 333 283
12
Average COD
363mg/L
The maximum month COD concentration for this data set is the maximum average month COD
value which is 427 mg/L for May 2009, whereas, the minimum average month COD
concentration is the minimum average month value which corresponds to 283 mg/L for
December 2009.
Solution Part D
The peaking factors for COD concentrations are calculated as follows. The maximum month
COD concentration to average COD concentration is determined by dividing the maximum
month COD concentration by the average COD concentration shown below (See Column 2, Row
D).
PFMax month conc =
maximum month COD concentration 427 mg L
=
average COD concentration
363mg L
1.18
The minimum month COD concentration to average COD concentration is determined by
dividing the minimum month COD value by the average COD value shown below (See Column
2, Row E).
PFMinimum month =
minimum month COD concentration 283mg/L
=
average daily COD concentration
363mg/L
Solution Part E
6
0.78
The solution to Part E is determined as follows. The annual average TP concentration is the
arithmetic average or mean of the 12 months of data shown in Column 3 ( Column 3, Row A
gives the answer). For this data set, the annual average TP concentration is 7.21 mg/L.
Average TP
Average TP
7.00 + 6.35+ 7.29 +7.09 + 6.70 +7.15+7.15+7.16 + 7.00 6.52 +8.11+9.05
12
7.21mg/L
The maximum month TP concentration for this data set is the maximum average month TP value
which is 9.05 mg/L for December 2009, whereas, the minimum average month TP concentration
is the minimum average month value which corresponds to 6.35 mg/L for February 2009.
Solution Part F
The peaking factors for TP concentrations are calculated as follows. The maximum month TP
concentration to average TP concentration is determined by dividing the maximum month TP
concentration by the average TP concentration shown below (See Column 3, Row D).
PFMax month conc =
maximum month TP concentration 9.05 mg L
=
average TP concentration
7.21mg L
1.25
The minimum month TP concentration to average TP concentration is determined by dividing
the minimum month TP value by the average TP value shown below (See Column 3, Row E).
PFMinimum month =
minimum month TP concentration 6.35mg/L
=
average daily TP concentration
7.21mg/L
0.88
Solution Part G
The solution to Part G is determined as follows. The annual average COD mass loading rate is
the arithmetic average or mean of the 12 months of data shown in Column 4 ( Column 4, Row A
gives the answer). For this data set, the annual average COD mass loading rate is 15,216 lb/d.
Average COD mass loading
11231+15543+15227 +16249 +16488+17777 17321 16457 + 11860 +15450 +16524 +12462
12
Average CODmass loading
15, 216 lb/d
The maximum month COD concentration for this data set is the maximum average month COD
value which is 17,777 lb/d for June 2009, whereas, the minimum average month COD
7
concentration is the minimum average month value which corresponds to 11,231 lb/d for
January 2009.
Solution Part H
The peaking factors for COD mass loadings are calculated as follows. The maximum month
COD mass loading rate to average COD mass loading rate is determined by dividing the
maximum month COD mass loading value by the average COD mass loading value shown
below (See Column 4, Row D).
PFMax month load =
maximum month COD loading 17, 777 lb d
=
average COD loading
15, 216lb d
1.17
The minimum month COD mass loading rate to average COD mass loading rate is determined
by dividing the minimum month COD mass loading value by the average COD mass loading
value shown below (See Column 4, Row E).
PFMinimum month load =
minimum month COD loading 11,231 lb/d
=
average COD loading
15, 216 lb/d
0.74
Solution Part I
The solution to Part I is determined as follows. The annual average TP mass loading rate is the
arithmetic average or mean of the 12 months of data shown in Column 5 ( Column 5, Row A
gives the answer). For this data set, the annual average TP mass loading rate is 306 lb/d.
Average TP mass loading
217 + 237 + 288 + 275 + 259 + 323+ 339 + 334 + 284 + 310 + 402 + 399
12
Average TP mass loading
306 lb/d
The maximum month TP mass loading rate for this data set is the maximum average month TP
mass loading value which is 402 lb/d for November 2009, whereas, the minimum average month
TP mass loading rate is the minimum average month TP mass loading value which corresponds
to 217 lb/d for January 2009.
Solution Part J
The peaking factors for TP mass loadings are calculated as follows. The maximum month TP
mass loading rate to average TP mass loading rate is determined by dividing the maximum
8
month TP mass loading value by the average TP mass loading value shown below (See Column
5, Row D).
PFMax month load =
maximum month TP loading 402lb d
=
average TP loading
306lb d
1.31
The maximum month TP mass loading rate to average TP mass loading rate in Table 7.5 is 1.3.
The minimum month TP mass loading rate to average TP mass loading rate is determined by
dividing the minimum month TP mass loading value by the average TP mass loading value
shown below (See Column 5, Row E).
PFMinimum month load =
minimum month TP loading 217 lb/d
=
average TP loading
306 lb/d
0.71
The minimum month TP mass loading rate to average TP mass loading rate in Table 7.5 is 0.88
which is slightly larger than our calculated value of 0.71.
3. A mechanical bar screen with 1/2 inch openings and 3/8 inch bars is installed in a rectangular
channel where the approach velocity should not exceed 2.0 ft per second. Estimate:
a) The velocity between the bars.
b) The headloss through the screen assuming it is clean.
Solution Part A
Assume that the width and depth of flow in the rectangular channel are W and D, respectively.
Estimate the net area of the openings in the bar screen by multiplying the cross-sectional area of
the rectangular channel by the ratio of the width of the bar screen openings divided by the width
of opening plus width of the bar as:
net area of openings WD
0.5
0.5 3 8
0.57 WD
From the continuity equation, Equation (7.26): Q = VAAA= VBSABS
Calculate the velocity of flow through the bar screen, VBS, as:
VBS
VA AA
ABS
2.0 fps WD
0.57 WD
3.5 fps
where:
9
AA = cross-sectional area of approach channel, WD, and
ABS = net cross-sectional area of openings in bar screen, 0.57WD.
Note that the velocity through the bar screen (3.5 fps) is significantly larger than the velocity of
approach (2.0 fps).
Solution Part B
Estimate the headloss through the bar screen using Equation (7.3).
hL
1 VBS2 VA2
C
2g
1
0.7
2
3.5fps
2.0 fps
2 32.2 ft s 2
2
0.18ft
This is not an appreciable loss of energy or pressure drop. The headloss through mechanically
cleaned coarse screens is typically 6 inches or 0.5 ft (150 mm).
4. Estimate the headloss through a coarse screen before and after the accumulation of solids
occurs. Assume the following conditions for solving the problem. The approach velocity and
velocity through the screen are 0.5 meters per second and 0.9 meters per second, respectively.
The open area through the clean bar screen is 0.20 m2.
a) Estimate the headloss through the clean bar screen assuming a discharge
coefficient of 0.7 for a clean screen.
b) Estimate the headloss through the clogged bar screen assuming a discharge
coefficient of 0.6 for a clogged screen and that 50% of the flow area has been
blocked by debris.
Solution Part A
Estimate the headloss through the bar screen using Equation (7.3).
hL
1 VBS2 VA2
C
2g
1
0.7
2
0.9 mps
0.5 mps
2 9.81m s 2
2
0.041m
This is not an appreciable loss of energy or pressure drop. The headloss through mechanically
cleaned coarse screens is typically 150 mm or 0.15 m.
Solution Part B
Determine the area through the bar screen assuming a 50% reduction in flow area. Use the
continuity equation (7.26) to calculate the volumetric flow rate and new velocity when the screen
becomes clogged.
10
Q
A V
V
Q
A
0.20 m 2 0.9
0.18 ms s
0.50 0.20 m 2
m
s
1.8
m3
s
0.18
m
s
Estimate the headloss through the bar screen using Equation (7.3).
hL
1 VBS2 VA2
C
2g
1
0.6
2
1.8 mps
0.5 mps
2 9.81m s 2
2
0.25 m
5. Two horizontal-flow type of grit chambers are designed to remove grit particles with a
diameter of 0.15 mm (100 mesh) and specific gravity of 2.65. A flow-through velocity of 1.0 ft/s
will be maintained by a proportioning weir. The average daily wastewater flow is 2.5 MGD. The
PHF:ADF ratio is 2.5:1.0. Determine the channel dimensions (in feet) for the PHF.
Solution
First, calculate the PHF as follows:
PHF 2.5 ADF 2.5MGD 2.5 6.25MGD
Assume that a rectangular cross section for the grit chamber will be used and that the depth of
the chamber is 1.5 width at maximum flow. The cross sectional area is determined using the
continuity equation, Equation (7.26).
Q
AV
where:
Q = volumetric flow rate,
ft 3
s
m3
,
s
A = cross-sectional area, ft2 (m2), and
V = velocity of flow, fps (m/s).
A
Q
V
6.25 106gal d 1d
1.0ft s
24 h
1h
3600s
1
2chambers
ft 3
7.48gal
Next, determine the width (W) and depth (D) of the channel as follows:
A W D
4.84 ft 2
Recall that D 1.5W
11
4.84ft 2
A W
1.5W
1.5W 2
4.84 ft 2
4.84ft 2
1.5
W
D 1.5W
4.84ft 2
1.80 ft
1.5 1.80ft
2.70ft
Estimate the settling velocity of a 0.15 mm diameter particle with a specific gravity of 2.65 using
Equation (7.25).
Vs
Vs
3.3 g S p 1 d
0.0895
0.5
m 3.281ft
s
m
1m
3.3 9.81m s 2.65 1 0.15 mm
1000 mm
0.5
2
0.29
ft
s
The detention time, , is calculated by dividing the depth by the particle settling velocity as
follows:
D
Vs
2.7 ft
0.29 ft s
9.3s
Length of the grit chamber is equal to the detention time multiplied by the horizontal flowthrough velocity as follows:
L
Vh
9.3s 1.0
ft
s
9.3ft
Metcalf and Eddy (2003) recommend that the theoretical length be increased by 50% to account
for influent and effluent turbulence. Therefore, the overall length should be equal to 1.5 9.3 ft
= 14.0 ft .
6. Design an aerated grit chamber for an average daily wastewater flow rate of 10 MGD.
Assume two grit chambers are operating in parallel. The peak hour flow rate is three times the
ADF. Use the design criteria in Table 7.8 to determine the following:
a)
b)
Solution Part A
The dimensions of each grit chamber.
The total air required (ft3/d).
The peak flow passing through the grit chamber must be calculated first.
12
PHF 3 ADF 3 10 MGD= 30 MGD
The volume of the grit chambers is calculated using the detention time equation, Equation (7.28):
V
Q
where:
= detention time, min
V = volume of the grit chambers, ft3, and
Q = volumetric flow rate, ft3/d.
Assuming a detention time of 3.0 min at PHF, the volume of the grit chambers is 4180 ft3.
V
Q 3.0 min
1h
60 min
V
grit chamber
1d
24 h
8360ft 3
2
30 106
gal
ft 3
d 7.48gal
8360ft 3
4180ft 3
Assuming a length:width ratio of 4:1 and a width: depth ratio of 1.5:1 from Table 7.8, calculate
the width as follows.
L
W
W
D
4
1
L
4W
1
1.5
1
D
1W
1.5
4W
Recall the definition of volume:
V
L W D
4180ft 3
W3
4W W
W
1.5
4180ft 3 1.5
4
13
W
3
4180ft 3 1.5
4
4×11.6ft = 46.4ft
L 4W
D
11.6 ft
1W
1.5
11.6ft
1.5
7.7 ft
Solution Part B
From Table 7.8 the quantity of air required per unit of length is assumed to be 5 ft3/(min ft). The
total quantity of air required is determined as follows:
air required 5
air required
ft 3
60 min
min ft
h
24 h
d
2 chambers 46.4ft
ft 3
6.68 10
d
5
7. A municipal WWTP receives an average daily wastewater flow of 10.0 MGD. Two,
rectangular, primary clarifiers operating in parallel will treat the flow. The peak hour flow
anticipated is 2.75 times the average daily flow. Use the design criteria in Tables 7.9 and 7.10,
and assume that the effluent weir in each clarifier is twelve times the clarifier width.
Determine:
a) The dimensions of each primary clarifier (ft).
b) The detention time (h) in each clarifier.
c) The weir loading rate (gpd/ft) for each clarifier at PHF.
d) The BOD and suspended solids removal efficiencies (%) at ADF.
14
Solution Part A
5 MGD
5 MGD
Assume overflow rates of 800 and 2500 gpd/ft2 at average and peak flows and then calculate the
surface areas of the clarifiers. At ADF, the surface area is:
AS
Q
Vo
10.0 106 gpd
12,500 ft 2
800gpd ft 2
At PHF, the surface area is:
AS
Q
Vo
2.75
10.0 106 gpd
2500 gpd ft
2
11, 000 ft 2
Use the larger of the two areas which is 12,500 ft2 or 6250 ft2 per clarifier. Using a length:width
ratio, calculate the length and width of each clarifier.
L
W
4
1
A
L W
W
39.5
L 4W
or
4W 2
L 4 W
6250 ft 2
40ft
4 40ft = 160 ft
From Table 7.9 select a side water depth (SWD) of 10 ft. Add two feet of freeboard to allow for
peak flows. Freeboard is additional length added to the depth to provide for variations in flow.
SWD
10 ft
Overall depth = 10 ft+2 ft
15
12 ft
Solution Part B
Calculate the detention time by dividing the clarifier volume by the flow rate. The detention
time at average flow in a primary clarifier at average flow should range from 45 minutes to 2.0
hours (Richards and Reynolds, 1996).
V
Q
160ft×40ft×10ft 7.48gal
5.0×106gpd
1ft 3
24 h
1d
2.3h
The detention time is slightly greater than what is required; however, this will provide a longer
detention time at peak flow too.
Solution Part C
The weir loading rate is the volumetric flow rate divided by the weir length; it is determined as
follows.
q
Q
weir length
2.75
10.0 106 gal d
28, 646
12 40 ft 2 basins
gpd
ft
30,000
gpd
OK
ft
m3
gpd
Peak weir loading rates generally should not exceed 30, 000
(Reynolds and
373
d m
ft
Richards, 1996).
Solution Part D
The efficiency of removal for BOD and suspended solids in a primary clarifier is estimated from
Figure 7.6. To use the diagram, the overflow rate at ADF must be calculated and converted to
m/d. Surface overflow rate at ADF is calculated as:
Vo
Q
AS
Vo
781
10.0 106 gpd
2
160 ft×40 ft
gpd 3.785L
ft 2
gal
781
gpd
ft 2
3.2812 ft 2
m2
1m3
1000 L
31.8
m
d
From Figure 7.6, the % removal for BOD and suspended solids is 31% and 60%, respectively.
8. An industrial WWTP receives an average daily wastewater flow of 38,000 m3/d. Two,
circular, primary clarifiers operating in parallel will treat the flow. The peak hour flow
anticipated is 1.5 times the average daily flow. Use the design criteria in Tables 7.9 and 7.10,
and assume that a peripheral effluent weir is used for each clarifier.
16
Determine:
a) The diameter (m) of each primary clarifier.
b) The detention time (h) in each clarifier at ADF.
c) The weir loading rate (m3/dm) for each clarifier at PHF.
d) The BOD and suspended solids removal efficiencies (%) at ADF.
19,000 m3/d
19,000 m3/d
Solution Part A
Assume overflow rates of 30 and 80 m3/(dm2) at average and peak flows and then calculate the
surface areas of the clarifiers. At ADF, the surface area is:
AS
Q
Vo
1.9 104 m3 /d
30 m3 d m2
633 m2
At PHF, the surface area is:
AS
Q
Vo
1.5
1.9 104 m3 /d
3
80 m d m
2
356 m 2
Use the larger of the two areas which is 633 m2 per clarifier.
A
D2
4
D
28.4 m
633m2
17
From Table 7.9 select a side water depth (SWD) of 3 m. Add 0.5 meters of freeboard to allow
for peak flows. Freeboard is additional length added to the depth to provide for variations in
flow.
SWD
Overall depth = 3.0 m+0.5 m
3.0 m
3.5 m
Solution Part B
Calculate the detention time by dividing the clarifier volume by the flow rate. The detention
time at average flow in a primary clarifier at average flow should range from 45 minutes to 2.0
hours (Richards and Reynolds, 1996).
2
V
Q
28.4 m
3.0 m
24 h
4
4 3
1.9×10 m /d
1d
2.4 h
The detention time is slightly longer than the typical range.
Solution Part C
The weir loading rate is the volumetric flow rate divided by the weir length; it is determined as
follows.
q
Q
weir length
1.5
1.9 104 m3 d
319
28.4 m
m3
d m
500
m3
OK
d m
Solution Part D
The efficiency of removal for BOD and suspended solids in a primary clarifier is estimated from
Figure 7.6. To use the diagram, the overflow rate at ADF must be calculated. Surface overflow
rate at ADF is calculated as:
Vo
Q
AS
1.9 104 m3 /d
28.4 m
4
2
30
m
d
From Figure 7.6, the % removal for BOD and suspended solids is 32% and 62%, respectively.
9. A conventional activated sludge process treats 3,785 m3/d of wastewater containing 250 g/m3
BOD5 and produces an effluent containing 20 g/ m3 BOD5. The nominal detention time in the
aeration basin excluding the return activated sludge flow is 6 hours and the MLSS concentration
is 3000 g/ m3. Determine the following:
a) The aeration basin volume (m3).
18
b) F/M ratio (d-1).
c) Specific substrate utilization rate (d-1).
d) Substrate removal efficiency (%).
Solution Part A
Calculate the volume using Equation (7.49).
V
Q 6h
3,785m3
d
d
24 h
946 m3
Solution Part B
F/M ratio is calculated using Equation (7.50).
Q Si
X V
F M
g BOD5
3,785m3 /d×250g m3
= 0.33
3
3
g SS d
3000g m ×946 m
Solution Part C
U is calculated using Equation (7.51) as follows:
U
3
3
3
Q Si Se 3, 785 m /d 250 g m - 20 g m
=
XV
3000 g m3 ×946 m3
0.31
g BOD5
g TSS d
Solution Part D
Treatment efficiency is calculated using Equation (7.52).
E
Ci
Ce
Ci
100
250 g m3 - 20 g m3
250 g m3
100
92%
10. A step-aeration activated sludge process treats 38,000 m3/d of wastewater containing 220
g/m3 BOD5. The F/M ratio based on VSS in the aeration basin is 0.30 d-1 and the MLVSS
concentration is 3000 g/ m3. Determine the following:
a) The aeration basin volume (m3).
b) The detention time (h).
c) The effluent substrate concentration if the specific substrate utilization rate is 0.28 d-1.
d) Substrate removal efficiency (%).
19
Solution Part A
Calculate the volume using Equation (7.50).
Q Si
X V
F M
V
38,000 m3 /d×220g m3
= 0.30d -1
3
3000g VSS m × V
9, 289 m3
Solution Part B
Detention time is calculated using Equation (7.49).
V
Q
9289 m3
24 h
3
38,000 m d d
5.9 h
Solution Part C
The effluent substrate concentration in terms of BOD5 is determined by rearranging Equation
(7.51) for U as follows:
U
3
3
Q Si Se 38, 000 m /d 220g m - Se
=
XV
3000g m3 ×9289m3
Se =14.7
0.28d -1
g
m3
Solution Part D
Treatment efficiency is calculated using Equation (7.52).
E
Ci
Ce
Ci
100
220 g m3 -14.7 g m3
220 g m3
100
93.3%
11. NPDES permits typically require that activated sludge WWTPs meet an annual average
effluent BOD5 concentration of 20 mg/L. If an oxidation ditch type of activated sludge process
produces an effluent containing 20 mg/L of TSS, estimate the soluble BOD5 necessary to meet
the effluent BOD5 standard. Assume that the volatile fraction of the effluent suspended solids is
65%.
Solution
Estimate the particulate BOD5 in the effluent using Equation (7.76).
20
VSSe
PBOD5
TSSe
VSSe
TSSe
1.42VSSe
20 mg TSS
0.65
L
2
3
1.42
13mg VSS
L
13mg
L
2
3
12.3mg
L
The effluent soluble BOD5 can be calculated from Equation (7.75).
TBOD5 = PBOD5 + SBOD5
Se
12.3mg
+ Se
L
20.0 mg
L
7.7 mg
L
12. A complete-mix activated sludge process (CMAS) is to be designed to treat 5.0 million
gallons per day (MGD) of primary effluent having a BOD5 of 180 mg/L. The NPDES permit
requires that the effluent BOD5 and TSS concentrations should be 20 mg/L or less on an annual
average basis. The following biokinetic coefficients obtained at 20ºC will be used in designing
the process: Y = 0.6 mg VSS/mg BOD5, k = 4 d-1, Ks = 70 mg/L BOD5, and kd = 0.05 d-1.
Assume that the MLVSS concentration in the aeration basin is maintained at 2500 mg/L and the
VSS:TSS ratio is 0.75. The temperature of the wastewater during the winter months is expected
to remain at 15ºC for extended periods. During the summer, the wastewater temperature may
reach 30ºC for several weeks. Determine the following:
a) Effluent soluble BOD5 (SBOD5) concentration in mg/L necessary to meet the effluent
total BOD5 (TBOD5) requirement of 20 mg/L.
b) Mean cell residence time (d) necessary to meet the NPDES permit during the winter
months.
c) Volume of the aeration basin in cubic feet.
d) Mean cell residence time (d) necessary to meet the NPDES permit during the summer
months.
e) Oxygen requirements (lb/d) assuming complete nitrification during the summer
months and a TKN0=35 mg/L.
f) The quantity of excess biomass produced (lb/d) in terms of TSS at the shortest MCRT
that the facility will be operated.
Solution Part A
First, the effluent particulate BOD5 (PBOD5) concentration is calculated from Equation 7.76).
21
VSSe
VSSe
TSSe
TSSe
PBOD5
1.42VSSe
PBOD5
1.42
20 mg
L
15mg
L
0.75
2
3
15mg
L
2
3
14.2 mg
L
The effluent soluble BOD5 (SBOD5) concentration is estimated from Equation (7.75).
TBOD5 = PBOD5 + SBOD5
SBOD5
TBOD5 - PBOD5
20 14.2
5.8
mg
L
Solution Part B
Before the MCRT can be determined, the biokinetic coefficients k, Ks, and kd must be corrected
from a temperature of 20ºC to 15ºC. It’s unnecessary to correct the yield coefficient for
temperature variations. Temperature correction coefficients ( ) of 1.07, 1.00, and 1.04 will be
used to correct k, Ks, and kd, respectively. Equation (7.101) is used to correct for temperature
variations.
K2
k
15 C
Ks
kd
o
15o C
15o C
K1
4 d -1 1.07
70
T2 T1
15 - 20o C
mg
1.00
L
0.05 d-1 1.04
2.85 d -1
15 - 20o C
15 - 20o C
70
mg
L
0.041 d -1
Calculate the MCRT using Equations (7.58) and (7.59).
dX
dt
Q
NG
Qw X e
X V
Qw X r
1
Y
c
22
dS dt
X
U
kd
dS
dt
k X Se
K S Se
U
1
dS dt
Y
X
c
1
c
U
kd
k Se
K S Se
Y
2.85d -1 5.8mg L
70 mg L 5.8mg L
0.6 mg VSS
mg BOD5
1
0.09 d -1
c
kd
0.041d -1
0.09d -1
11.1d
There is no need to use a safety factor since the calculated SRT is much greater than 5 days.
Solution Part C
The volume of the aeration basin is calculated from Equation (7.84).
Y Q Si
V
V
1 kd
0.60 mg VSS
gal
5.0 106
mg BOD5
d
1 0.041d
V
Se
-1
c
c
X
180 mg
L
5.8 mg
L
11.1d
1.59 106 gal
1ft 3
7.48gal
11.1d
1.59 106 gal
2500 mgVSS L
2.13 105 ft 3
Solution Part D
Calculate the MCRT necessary to meet effluent requirements during the summer months. First,
the biokinetic coefficients must be corrected from 20ºC to 30ºC. Use the same temperature
correction coefficients that were used in Part B.
K2
k
o
30 C
Ks
o
30 C
4 d -1 1.07
70
T2 T1
K1
30 - 20o C
mg
1.00
L
7.87 d -1
30 - 20o C
23
70
mg
L
kd
0.05 d -1 1.04
o
30 C
30 - 20o C
0.074 d -1
Use Equations (7.58) and (7.59) to determine MCRT.
1
Y
dS dt
c
kd
X
c
1
U
0.6 mg VSS
mg BOD5
k Se
K S Se
kd
7.87 d -1 5.8mg L
70 mg L 5.8mg L
1
0.287 d -1
c
Y
0.074d-1
0.287 d-1
3.5d
Solution Part E
To calculate the oxygen required during the summer months, we first determine X at the design
MCRT of 3.5 days. The actual biomass concentration X at a MCRT of 3.5 days is determined
from Equation (7.83). Recall, that the volume of the aeration basin is 1.59106 gal = 1.59 MG.
The detention time is determined below.
V
Q
X
X
1.59 MG
5.0 MGD
Y Si S e
1 kd c
0.32 d
c
0.60 mg VSS 180 mg 5.8mg
mg BOD5
L
L
3.5d
-1
0.32d
1 0.074d 3.5d
908mg VSS
L
Finally, the oxygen requirements are determined from Equation (7.104).
O2
Q Si
Se 1 1.42 Y
1.42 kd X V
NOD
The nitrogenous oxygen demand (NOD) is calculated as follows.
NOD = Q(TKN 0 )(4.57) 5 MGD 35
NOD 800
mg MG
L d
24
mg
L
4.57
O2
5.0 MG
mg
0.60 mg VSS
180 5.8
1 1.42
d
L
mg BOD5
O2
9020
1.42 0.074d -1
908mg
8.34lb
1.59 MG 800
mg
L
MG×
L
lb O 2
d
Solution Part F
The largest quantity of excess biomass produced occurs at the shortest MCRT, which in this case
is 3.5 days. Excess biomass is determined from Equation (7.92).
Px
Px
Yobs Q Si
1 kd
Yobs Q Si
1 kd
Se
c
Se
c
Px
3460
lb VSS
d
Px
3460
lb VSS
d
0.60 mg VSS
180 mg 5.8 mg
5.0 MGD
mg BOD5
L
L
-1
1 0.074 d 3.5d
1 lb TSS
0.75 lb VSS
4610
8.34 lb
mg
MG
L
lb TSS
d
13. A complete-mix activated sludge process (CMAS) is to be designed to treat 19,000 m3/d of
raw wastewater having a BOD5 of 200 mg/L. The NPDES permit requires that the effluent
BOD5 and TSS concentrations should be 20 mg/L or less on an annual average basis. The
following biokinetic coefficients obtained at 20ºC will be used in designing the process: Y = 0.6
mg VSS/mg BOD5, k = 4 d-1, Ks = 50 mg/L BOD5, and kd = 0.05 d-1. Assume that the MLVSS
concentration in the aeration basin is maintained at 2800 mg/L and the VSS:TSS ratio is 0.70.
The temperature of the wastewater during the winter months is expected to remain at 17ºC for
extended periods. During the summer, the wastewater temperature may reach 28ºC for several
weeks. Determine the following:
a) Effluent soluble BOD5 (SBOD5) concentration in mg/L necessary to meet the effluent
total BOD5 (TBOD5) requirement of 20 mg/L.
b) Mean cell residence time (d) necessary to meet the NPDES permit during the winter
months.
25
c) Volume of the aeration basin in cubic meters.
d) Mean cell residence time (d) necessary to meet the NPDES permit during the summer
months.
e) Oxygen requirements (kg/d) assuming complete nitrification during the summer
months and a TKN0 = 30 mg/L.
f) The quantity of excess biomass produced (kg/d) in terms of TSS at the shortest MCRT
that the facility will be operated.
Solution Part A
First, the effluent particulate BOD5 (PBOD5) concentration is calculated from Equation
(7.76). VSSe
VSSe
TSSe
TSSe
PBOD5
1.42VSSe
PBOD5
1.42
20 mg
L
0.70
14 mg
L
2
3
14 mg
L
2
3
13.3mg
L
The effluent soluble BOD5 (SBOD5) concentration is estimated from Equation.
TBOD5 = PBOD5 + SBOD5
SBOD5
TBOD5 - PBOD5
20 13.3
6.7
mg
L
Solution Part B
Before the MCRT can be determined, the biokinetic coefficients k, Ks, and kd must be corrected
from a temperature of 20ºC to 17ºC. It’s unnecessary to correct the yield coefficient for
temperature variations. Temperature correction coefficients ( ) of 1.07, 1.00, and 1.04 will be
used to correct k, Ks, and kd, respectively. Equation (7.101) is used to correct for temperature
variations.
K2
k
o
17 C
K1
4 d-1 1.07
T2 T1
17 - 20o C
26
3.3 d -1
Ks
kd
50
17o C
mg
1.00
L
0.05 d -1 1.04
17o C
17 - 20o C
50
17 - 20o C
mg
L
0.044 d -1
Calculate the MCRT using Equations (7.58) and (7.59).
dX
dt
dS
dt
1
Q Qw X e Qw X r
X V
NG
dS dt
U
X
c
1
kd
dS dt
c
c
Y
k Se
K S Se
U
X
kd
kd
3.3d -1 6.7 mg L
50 mg L 6.7 mg L
0.6 mg VSS
mg BOD5
c
Y
k X Se
K S Se
U
Y
1
1
0.19 d -1
0.044d -1
0.19d -1
5.3d
Let’s increase the MCRT by using a safety factor of 1.5.
c design
5.3d 1.5 = 7.95
8d
Solution Part C
The volume of the aeration basin is calculated from Equation (7.84).
V
Y Q Si
1 kd
Se
c
c
X
First, we need to calculate the effluent substrate concentration using Equation (7.72) at a MCRT
K s 1 kd c
Se
of 8 days.
kd 1
c Y k
Se
K s 1 kd
kd
c Y k
50 mg L 1 0.044 d -1 8.0 d
c
1
8.0 d
0.60 mg VSS
3.3d -1 0.044 d -1
mg BOD5
27
4.7
1
mg BOD5
L
V
V
0.60 mg VSS
m3
1.9 104
mg BOD5
d
1 0.044 d
-1
200 g
m3
4.7 g
m3
8.0 d
8.0 d
2800 gVSS m 3
4700 m3
Solution Part D
Calculate the MCRT necessary to meet effluent requirements during the summer months. First,
the biokinetic coefficients must be corrected from 20ºC to 30ºC using Equation (7.101). Use the
same temperature correction coefficients that were used in Part B.
K2
k
28 C
Ks
kd
o
28o C
o
28 C
28 - 20o C
4 d-1 1.07
50
T2 T1
K1
mg
1.00
L
0.05 d -1 1.04
6.9 d -1
28 - 20o C
28 - 20o C
50
mg
L
0.068 d -1
Use Equations (7.58) and (7.59) to determine MCRT.
1
c
1
c
c
Y
dS dt
U
X
0.6 mg VSS
mg BOD5
1
0.42d -1
kd
Y
k Se
K S Se
kd
6.9d -1 6.7 mg L
50 mg L 6.7 mg L
0.068d -1
0.42d -1
2.4d
Solution Part E
To calculate the oxygen required during the summer months, we first determine Se and X at the
design MCRT of 2.4 days. The actual soluble BOD5 in the effluent at 2.4 days is determined
from Equation.
28
Se
Se
K s 1 kd
kd
c Y k
K s 1 kd
kd
c Y k
c
1
50 mg L 1 0.068d -1 2.4 d
c
1
2.4 d
0.60 mg VSS
6.9 d -1 0.068d -1
mg BOD5
1
6.6 mg
L
The actual biomass concentration X at a MCRT of 2.4 days is determined from Equation (7.83).
Recall, that the volume of the aeration basin is 4700 m3. The detention time is determined
V
Q
below.
X
X
4700 m3
1.9 104 m3 /d
Y Si S e
1 kd c
0.25d
c
0.60 mg VSS 200 mg 6.6 mg
mg BOD5
L
L
2.4d
-1
0.25d
1 0.068d 2.4d
958mg VSS
L
Finally, the oxygen requirements are determined from Equation (7.104).
O2
Q Si
Se 1 1.42 Y
1.42 kd X V
NOD
The nitrogenous oxygen demand (NOD) is calculated as follows.
m3
g
NOD = Q(TKN 0 )(4.57) 1.9 10
30 3 (4.57)
d
m
g
NOD 2.60 106
d
4
O2
1.9 104 m3
g
0.60 mg VSS
200 6.6 3 1 1.42
d
m
mg BOD5
O2
3580
1.42 0.068d -1
958g
m3
4700 m3
2.60 106
g 1kg
d 1000 g
kg O 2
d
Solution Part F
The largest quantity of excess biomass produced occurs at the shortest MCRT, which in this case
is 2.4 days. Excess biomass is determined from Equation (7.92).
29
Px
Px
Yobs Q Si
1 kd
Se
c
Se
c
1895
kg VSS
d
Px 1895
lb VSS
d
Px
14.
Yobs Q Si
1 kd
0.60 mg VSS
200 mg 6.6 mg
1.9 104 m3 /d
mg BOD5
L
L
-1
1 0.068d 2.4 d
1 lb TSS
0.70 lb VSS
2710
1kg
1000g
kg TSS
d
A wastewater treatability study was performed on a municipal wastewater and the data in
the table below were collected in a bench-scale CMAS reactor. Determine the following
biokinetic coefficients: Y, k, kd, Ks, a, and b.
MCRT
MLVSS
Si
Se
OUR
(days)
(mg/L)
(d)
(mg/L)
(mg/L)
(mg/L h)
5
2,280
0.167
200
7.00
48
10
3,590
0.167
200
4.30
59
15
4,420
0.167
200
3.50
66
20
5,000
0.167
200
3.10
71
25
5,420
0.167
200
2.80
75
Solution
A spreadsheet is an excellent resource for solving these types of problems and for graphing. The
values in the table below are used to make the necessary plots to determine the biokinetic
coefficients.
30
1
c
c
1/Se
1/U
SOUR
(L/mg)
(d)
(d-1)
(d)
(d-1)
U
(d-1)
5
0.200
0.508
0.142
1.969
0.505
10
0.100
0.327
0.232
3.057
0.394
15
0.067
0.267
0.285
3.749
0.358
20
0.050
0.236
0.322
4.232
0.341
25
0.040
0.218
0.357
4.581
0.332
Sample Calculations
Calculations showing how each value in the above table for SRT=5 days will be presented.
1
c
1
5.0d
0.20d
1
The specific substrate utilization rate U is determined using Equation (7.51).
Q Si Se
U
XV
Recall the definition of detention time from Equation (7.49).
Substituting Equation (7.49) into Equation (7.51) results in the following equation for U.
U
Si Se
X
200 mg L 7 mg L
2280 mg L 0.167 d
1
1
1.96 d
U 0.51d -1
0.508 d -1
The specific oxygen utilization rate is determined using Equation (7.114).
OUR
X
48mg O2 L h
24 h
2280 mg MLVSS L d
SOUR
SOUR
31
0.505d -1
Plot Equation (7.106) to determine Y and kd.
1
YU
(7.106)
kd
c
0.25
U - 0.0806
Yield coefficient (Y ) = slope= 0.55 g VSS/ g BOD 1/SRT = 0.5525
R2 = 1
Y -intercept = k d = 0.08 d-1
1/SRT, 1/d
0.20
0.15
0.10
0.05
0.00
0.00
0.10
0.20
0.30
0.40
0.50
0.60
Specific Substrate Utilization Rate (U ), 1/d
Plot Equation (7.112) to determine k and Ks.
1
U
Ks 1
k Se
1
k
(7.112)
5.00
4.50
4.00
1/U = 12.36 1/Se + 0.2032
R2 = 0.9991
3.50
1/U, d
3.00
2.50
Slope = K s /k = 12.36
Y-intecept = 1/k = 0.20
k = 5 d-1
Ks= 12.36 5-1 = 61.8 mg/L
2.00
1.50
1.00
0.50
0.00
0.00
0.05
0.10
0.15
0.20
1/Se, L/mg
32
0.25
0.30
0.35
0.40
Plot Equation (7.115) to determine a and b.
SOUR a U
b
(7.115)
0.60
Specific Oxygen Utilization Rate, 1/d
SOUR = 0.6021 U + 0.1988
R2 = 0.9997
0.50
0.40
0.30
Slope = a = 0.60 mg O2/mg BOD
Y-intecept = b = 0.20 mg O2/mg MLVSS d
0.20
0.10
0.00
0.00
0.10
0.20
0.30
0.40
0.50
0.60
Specific Substrate Utilization Rate (U ), 1/d
15.
A wastewater treatability study was performed on a nitrogenous wastewater with an
influent ammonia concentration of 300 mg/L as N and the data in the table below were
collected in a bench-scale CMAS reactor. Determine the following biokinetic coefficients
for nitrification: Y, k, kd, and Ks.
MCRT
MLVSS
Si
Se
(days)
(mg/L)
(d)
(mg/L)
(mg/L)
6
875
0.167
300
0.36
12
1,322
0.167
300
0.21
16
1,515
0.167
300
0.17
20
1,661
0.167
300
0.16
24
1,774
0.167
300
0.14
Solution
A spreadsheet is an excellent resource for solving these types of problems and for graphing. The
values in the table below are used to make the necessary plots to determine the biokinetic
33
coefficients.
1
c
c
1/Se
1/U
(L/mg)
(d)
(d)
(d-1)
U
(d-1)
6
0.167
2.056
2.757
0.486
12
0.083
1.363
4.853
0.734
16
0.063
1.184
5.761
0.845
20
0.050
1.084
6.444
0.923
24
0.042
1.016
6.978
0.984
Sample Calculations
Calculations showing how each value in the above table for SRT=5 days will be presented.
1
c
1
6.0d
0.167 d
1
The specific substrate utilization rate U is determined using Equation (7.51).
Q Si Se
XV
U
Recall the definition of detention time from Equation (7.49).
Substituting Equation (7.49) into Equation (7.51) results in the following equation for U.
300 mg L 0.36 mg L
875 mg L 0.167 d
1
1
0.48d
U 2.05d -1
Plot Equation (7.106) to determine Y and kd.
U
Si S e
X
1
YU
kd
c
34
2.05 d-1
(7.106)
0.18
0.16
0.14
Yield coefficient (Y ) = slope= 0.12 g VSS/ g NH4+
Y -intercept = k d = 0.08 d-1
1/SRT, 1/d
0.12
0.10
0.08
1/SRT = 0.12 U - 0.0801
R2 = 1
0.06
0.04
0.02
0.00
0.00
0.50
1.00
1.50
2.00
2.50
Specific Substrate Utilization Rate, 1/d
Plot Equation (7.112) to determine k and Ks.
1
U
Ks 1
k Se
1
k
(7.112)
1.20
1/U = 0.1181(1/S e ) + 0.1612
R2 = 0.9999
1.00
1/U , days
0.80
0.60
Slope = K s /k = 0.12
Y -intecept = 1/k = 0.16
k = 6.25 d-1
-1
K s = 0.12 6.25d = 0.75 mg/L
0.40
0.20
0.00
0.00
1.00
2.00
3.00
4.00
1/Se, L/mg
35
5.00
6.00
7.00
8.00
16.
A wastewater treatability study was performed on a soluble synthetic wastewater and the
data in the table below were collected in a bench-scale CMAS reactor. The substrate
concentration was measured as COD. The reactor volume was 5 liters and the volumetric
flow rate averaged 20 liters per day. Determine the following biokinetic coefficients: Y,
k, kd, Ks, a, and b.
MCRT
MLVSS
Si
Se
OUR
(days)
(mg/L)
(mg/L)
(mg/L)
(mg/L h)
3.9
1,030
300
20.00
50
4.2
1,100
300
18.00
60
8.5
1,940
300
10.00
65
13.8
2,650
300
7.00
70
21.0
3,320
300
6.00
75
Solution
A spreadsheet is an excellent resource for solving these types of problems and for graphing. The
values in the table below are used to make the necessary plots to determine the biokinetic
coefficients.
c
1
c
1/Se
1/U
SOUR
(L/mg)
(d)
(d-1)
(d)
(d-1)
U
(d-1)
3.9
0.256
1.087
0.050
0.920
1.165
4.2
0.238
1.025
0.055
0.975
1.309
8.5
0.118
0.598
0.100
1.672
0.804
13.8
0.072
0.442
0.142
2.261
0.634
21.0
0.048
0.354
0.166
2.823
0.542
Sample Calculations
Calculations showing how each value in the above table for SRT=3.9 days will be presented.
36
1
c
1
3.9d
0.256d
1
The specific substrate utilization rate U is determined using Equation (7.51).
U
Q Si Se
XV
Recall the definition of detention time from Equation (7.49)
V
Q
5.0 L
0.25d
20.0 L
d
Substituting Equation (7.49) into Equation (7.51) results in the following equation for U.
U
Si S e
X
300 mg L 20 mg L
1030 mg L 0.25d
1
U
1
1.087 d -1
1.087 d -1
0.92 d
The specific oxygen utilization rate is determined using Equation (7.114).
OUR
SOUR
X
50 mg O2 L h
24 h
SOUR
1.165d-1
1030 mg MLVSS L d
Plot Equation (7.106) to determine Y and kd.
1
Y U kd
c
37
0.30
1/SRT= 0.2843 U - 0.053
R2 = 1
0.25
Yield coefficient (Y ) = slope= 0.28 g VSS/ g COD
Y -intercept = k d = 0.053 d-1
1/SRT, 1/d
0.20
0.15
0.10
0.05
0.00
0.00
0.20
0.40
0.60
0.80
1.00
1.20
Specific Substrate Utilization Rate, 1/d
Plot Equation (7.112) to determine k and Ks.
1
U
Ks 1
k Se
1
k
3.00
1/U = 15.843 1/S e + 0.0982
R2 = 0.9933
2.50
1/U, days
2.00
1.50
Slope = K s /k = 15.8
Y -intecept = 1/k = 0.098
k = 10.2 d-1
K s = 15.8 10.2d-1 = 161 mg/L
1.00
0.50
0.00
0.00
0.02
0.04
0.06
0.08
0.10
1/Se, L/mg
Plot Equation (7.115) to determine a and b.
38
0.12
0.14
0.16
0.18
SOUR a U
b
(7.115)
1.40
SOUR = 0.9681 U + 0.2118
R2 = 0.9524
1.20
SOUR, 1/d
1.00
0.80
0.60
Slope = a = 0.97mg O2/mg COD
Y-intecept = b = 0.21 mg O2/mg MLVSS d
0.40
0.20
0.00
0.00
0.20
0.40
0.60
0.80
1.00
1.20
Specific Substrate Utilization Rate, 1/d
17. Design a sequencing batch reactor WWTP to treat an average daily flow of 11,400 m3/d and
a peak flow of three times the ADF. Use three, circular SBRs, and 6 cycles per day operating
scheme, and a low-water-level (LWL) of 3 meters.
Determine:
a) The volume of each SBR.
b) Dimensions of each SBR.
Solution Part A
SBRs must be sized to treat the peak wastewater flow rate. The design of the aeration system
should be based on the average daily flow rate. A simplified procedure will be used in this
example.
First, calculate the volume necessary to treat the average daily flow using the Equation (7.153).
V
V
Q
CPD # of SBRs
11, 400 m3 d
6 cycles
3
d
633m3
where:
V = volume of SBR required to treat a specified flow, ft3 (m3),
39
CPD = cycles per day (4 to 6),
Q = volumetric flow rate, gpd (Lpd), and
# of SBRs = number of sequencing batch reactors in operation.
The volume of 633 m3 must be allocated between the LWL and the average-water-level (AWL)
as shown in the figure provided in the Part B solution.
Next, the volume necessary to accommodate the peak wastewater flow is calculated using
Equation (7.153).
V
3 11, 400 m3 d
1,900 m3
6cycles
3
d
Q
CPD # of SBRs
The volume of 1,900 m3 must be allocated between the AWL and the high-water-level (HWL).
Solution Part B
Determine the depth and diameter of the circular reactors. Assume that the low-water-level is 3
meters deep and each reactor is 30 meters in diameter. Typically, the LWL ranges from 2.7 m to
4.0 m (9 to 13 feet).
The surface area (As) of each reactor is:
D2
4
As
30 m
4
2
= 707 m2
The depth required to handle the average daily flow is calculated as follows.
D
V
As
where:
D = depth of wastewater, ft (m), and
As = surface area of SBRs, ft2 (m2).
D
V
As
633m3
707 m2
0.90 m
Therefore, the AWL is 3 m + 0.90 m = 3.90 m.
The depth required to handle the peak flow is calculated as follows.
40
V
As
D
1900 m3
707 m2
2.69 m
Therefore, the HWL is 3.90 m + 2.69 m = 6.59 m. Make the total depth of each SBR 7 meters;
this allows 0.41 meters of freeboard. Freeboard is additional depth to allow for unexpected
flows. The figure below shows the cross-sectional area of one of the SBRs.
18. Given the following information for an extended aeration activated sludge process: Q = 3.0
MGD, influent soluble BOD5 = 220 mg/L, Y = 0.7 g VSS/ g BOD5, kd = 0.05 d-1, KS = 60
mg/L BOD5, k = 5 d-1, MLVSS = 0.75 MLSS, TKN0 = 35 mg/L, MLSS = 3000 mg/L, and
DO = 2 mg/L. α = 0.80 = 0.98 . Assume the biokinetic coefficients have been corrected for
a wastewater temperature of 30ºC which is expected during the summer months.
Determine: a) Volume of the aeration basin (MG) at an SRT = 25 days.
b) Oxygen required (lb/d), SRT = 25 days.
c) Number and hp of mechanical aerators assuming SOTR/aerator = 3.0
O2/hp h.
Solution Part A
First, find the effluent soluble BOD5 concentration at a 10 day SRT using Equation (7.72)
Se
K s 1 kd
kd
c Y k
60 mg L 1 0.05d -1 25d
c
1
25d 0.7 g g 5d
-1
0.05d
-1
1
1.6
mg BOD5
L
Determine the volume of the aeration basin (MG) using Equation (7.84).
41
lb
V
V
Y Q S0
1 kd
Se
c
X
c
(0.7 g VSS g COD ) 3MGD 220 1.6 mg L 25d
1 0.05d -1 25d
3000 mgSS L 0.75VSS TSS
2.26 MG
Solution Part B
Determine the oxygen process requirements using Equations (7.104) and (7.105).
NOD Q TKN0 4.57
O2
O2
Q S0
3MGD 35
Se 1 1.42 Y
(3MGD) 220 1.6
(33
mg
1 1.42 0.7
L
lb O2
)
d
4.57 8.34
1.42 kd X V
1.42 0.05d -1 (3000
O2
mg
L
lb MG
mg L
lb O2
d
NOD
8.34 lb MG
mg L
mg TSS
g VSS
0.75
) 2.26 MG
L
g TSS
3010
4000
lb O2
d
4000
lb O2
d
8.34 lb MG
mg L
7040
4000
lb O 2
d
Solution Part C
Design the mechanical aerator system for summer conditions, the worst case scenario assuming
the temperature of the wastewater rises to 30°C. Assume the alpha and beta coefficients for the
mechanical aerators as supplied by the aeration manufacturer are 0.80 and 0.98, respectively and
use a temperature correction factor ( ) of 1.024. The dissolved oxygen (DO) saturation
concentration of water is 7.54 mg/L at sea level and 30°C. Typically, a DO concentration, Ct, of
2.0 mg/L is maintained in the aeration basin so that oxygen does not limit the biological process.
Calculate the actual oxygen transfer rate (AOTR) of the mechanical aerators using Equation
AOTR
(7.175).
AOTR
SOTR
Cs
Ct
9.17
T C
20 C
lb O2 0.80 0.98 7.54 2.0 mg L 1.024
3.0
hp h
9.17
30 C 20 C
1.8
lb O2
hp h
The total horsepower required for the mechanical aerators is estimated by dividing the total
oxygen requirements determined in Part B by the AOTR determined above.
42
7040 ppd O2
1.8pph O2 hp
total hp
1d
24 h
163
Assume there are two aeration basins and two aerators in the aeration basin; therefore, the hp per
aerator is determined by dividing the total hp by four:
hp
163 hp
=
= 41hp
aerator 2 basins 2 aerators
Therefore, size up to 45 hp per aerator.
Next, it is necessary to check the mixing requirements to ensure that the microorganisms remain
suspension. Recall that mixing requirements dictate that the hp per 1000 ft3 of basin should
range from 0.75 to 1.5.
Actual hp
1000 ft 3
4 45 hp 1 MG
2.26 MG 106 gal
1000 ft 3
1000 ft 3
7.48 gal
ft 3
0.60
There is sufficient power available to keep the biomass in suspension.
19. Given the following information for an oxidation ditch activated sludge process: Q = 11,400
m3/d, influent soluble BOD5 = 250 mg/L, Y = 0.6 g VSS/g BOD5, kd = 0.06d-1, KS = 70 mg/L
BOD5, k = 6 d-1, MLVSS = 0.70 MLSS, TKN0 = 40 mg/L, MLSS = 3500 mg/L, and DO = 2
mg/L. α = 0.88 = 0.95 . Assume the biokinetic coefficients have been corrected for a
wastewater temperature of 28ºC which is expected during the summer months and three
aeration basins with two aerators per basin.
Determine: a) Volume of the aeration basin (m3) at an SRT = 30 days.
b) Oxygen required (kg/d), SRT = 30 days.
c) Number and kw of mechanical aerators assuming SOTR/aerator = 1.0 kg
O2/kw h.
Solution Part A
First, find the effluent soluble BOD5 concentration at a 30 day SRT using Equation (7.72).
Se
K s 1 kd
kd
c Y k
60 mg L 1 0.06 d -1 30 d
c
1
30 d 0.6 g g 6 d
-1
0.06 d
-1
1
1.6
Determine the volume of the aeration basin (m3) using Equation (7.84).
43
mg BOD5
L
V
V
Y Q S0
1 kd
Se
c
c
X
(0.6 g VSS g COD ) 11, 400 m3 250 1.6 mg L 25d
1 0.06 d -1 30 d
3000 mg SS L 0.70 VSS TSS
7220 m3
Solution Part B
Determine the oxygen process requirements using Equations (7.104) and (7.105).
NOD Q TKN0 4.57
11, 400 m3
O2
Q S0
Se 1 1.42 Y
O2
(11, 400 m3 /d) 250 1.6
1.42 0.06 d -1 (3000
O2
(419
kg O2
)
d
40
g
m3
1.42 kd X V
g
m3
4.57
1kg
1000g
kg O2
d
1kg
1000 g
1 1.42 0.6
2080
kg O2
d
NOD
mg TSS
g VSS
0.70
) 7220 m3
L
g TSS
1290
2080
kg O2
d
3800
1kg
1000 g
2080
kg
d
kg O2
d
Solution Part C
Design the mechanical aerator system for summer conditions, the worst case scenario assuming
the temperature of the wastewater rises to 30°C. Assume the alpha and beta coefficients for the
mechanical aerators as supplied by the aeration manufacturer are 0.88 and 0.95, respectively and
use a temperature correction factor ( ) of 1.024. The dissolved oxygen (DO) saturation
concentration of water is 7.81 mg/L at sea level and 28°C. Typically, a DO concentration, Ct, of
2.0 mg/L is maintained in the aeration basin so that oxygen does not limit the biological process.
Calculate the actual oxygen transfer rate (AOTR) of the mechanical aerators using Equation
AOTR
(7.175).
AOTR
1.0
SOTR
Cs
Ct
9.17
T C
20 C
kg O2 0.88 0.98 7.81 2.0 mg L 1.024
kw h
9.17
28 C 20 C
0.66
kg O2
kw h
The total horsepower required for the mechanical aerators is estimated by dividing the total
oxygen requirements determined in Part B by the AOTR determined above.
44
total kw
3800 kg/d O2
0.66 kg O2 kw h
1d
24 h
240
Assume there are three aeration basins and two aerators each aeration basin; therefore, the kw
per aerator is determined by dividing the total kw by four:
kw
240 kw
=
= 40 kw
aerator 3 basins 2 aerators
Next, it is necessary to check the mixing requirements to ensure that the microorganisms remain
suspension. Recall that mixing requirements dictate that the kw per 1000 m3 of basin should
range from 20 to 40.
Actual kw
1000 m3
2 40 kw
7220 m3 3basins
1000 m3
1000 m3
33.2
There is sufficient power to meet the mixing requirements.
20. Design a coarse-bubble diffused aeration system to meet an oxygen demand of 10,000
pounds per day. The plant is located at an elevation of 1,000 feet. There are two aeration basins
with a total volume = 1.43 MG, assume the alpha and beta coefficients for the diffusers are 0.45
and 0.98, respectively and use a temperature correction factor ( ) of 1.024. The dissolved
oxygen (DO) saturation concentration of water is 7.54 mg/L at sea level and 30°C. A DO
concentration of 2.0 mg/L is maintained in the aeration basins. The oxygen transfer efficiency is
assumed to be 10% since coarse-bubble diffusers are being used. The side water depth of the
aeration basin is 15 ft with diffusers placed one foot from the bottom of the tank. The
atmospheric pressure at an elevation of 1,000 feet is 733 mm Hg. Assume that the specific
weight of water is 62.4 lb/ft3.
Determine:
a) Volume of the air (scfm) to meet oxygen requirement of 10,000 lb per day.
b) Horsepower required for centrifugal blowers.
Solution Part A
The dissolved oxygen saturation concentration at mid-depth must be calculated using Equation
(7.179). We first must estimate the oxygen concentration (%) in the off-gas. The concentration
of oxygen in the atmosphere is approximately 21% by volume and 23% by weight (Metcalf and
Eddy, 2003). The concentration of nitrogen in the atmosphere is approximately 79% by volume.
45
Assuming 100 moles of air to be compressed, there would be 21 moles of oxygen nO2 and 79
moles of nitrogen nN2 . The moles of oxygen in the off-gas are estimated as follows.
nO2
21moles O 2 1 OTE
nO2
21moles O 2 1
10%
100%
19 moles
The percentage of oxygen in the off-gas may now be calculated.
Oe
nO2
nO2
nN2
19
100
19 79
100
19%
The hydrostatic pressure due to water is calculated using Equation (7.181).
p
h
where:
p = hydrostatic pressure,
lb N
,
ft 2 m 2
= specific weight of water, 62.4
lb
ft 3
9800
N
m3
and,
h = depth of water, ft (m).
The pressure at the point of air release (Pr) at the bottom of the aeration basin is calculated
below and must be expressed in terms of absolute pressure. Absolute pressure (pabs) is gauge
pressure (pgauge) plus barometric pressure (pbar).
pabs
Pr
15ft 1ft
Pr
20.2 psia
62.4lb
ft 3
1ft 2
144in 2
pgauge
14.7 psia
pbar
733mm Hg
760 mm Hg
Now, calculate the dissolved oxygen saturation concentration at mid-depth.
CM
Cs
Pr
C
Oe
42
46
(7.182)
CM
0.98
7.54 mg
L
20.2 psia
29.4
19%
42
8.4 mg
L
Use Equation (7.178) to determine the quantity of oxygen required at standard conditions.
AOTR
SOTR
CM
Ct
9.17
T °C 20°C
0.45 8.4 mg L - 2.0 mg L 1.024
SOTR
9.17
lb
2.51 104
d
lb
10, 000
d
SOTR
30°C 20°C
Calculate the density of air at standard conditions (20°C and 1 atmosphere of pressure).
air
14.7 lb
P
RT
53.3
ft lb
lb air o R
in 2
68 460 R
144in 2
ft 2
0.075
lb
ft 3
The standard cubic feet of air required per minute is calculated by dividing the SOTR by the
specific weight of air
0.075 lb air
at standard conditions (20°C and 1 atmosphere of pressure)
ft 3 air
and by the concentration of oxygen in the atmosphere by weight
scfm = 2.51 104
lb O2
1 ft 3 air
d
0.075 lb air
1 lb air
0.23 lb O
2
1d
24 h
0.23 lb O2
.
lb air
1h
60 min
1
0.10
10,100
Check to see if enough air has been provided to ensure adequate mixing of the wastewater in the
aeration basin i.e. divide the standard cubic feet per minute of air required by the volume of the
aeration basin.
Mixing air =
10,100scfm 7.48gal
1.43 106 gal
ft 3
1000 ft 3
1000 ft 3
53 scfm
1000 ft 3
The amount of air required for mixing typically ranges from 20 to 30 scfm/1000 ft3. We have
more than enough to meet mixing requirements, but cannot reduce this amount since 10,100
scfm are required to meet process air requirements. As a design engineer, other types of diffuser
systems could be evaluated that have a higher transfer efficiency which would reduce the amount
of air required.
47
Solution Part B
Next calculate the power required by the centrifugal blower for compressing the air using
Equation (7.180). We will assume that the air temperature may get up to 100°F during the
summer months.
Air temp. = 100°F
460 560o R
Std. temp.= 68°F 460 528o R
Must use combined gas law to compute the air required at a different temperature and pressure.
PV
1 1
T1
V2
PV
2 2
T2
PV
1 1T2
P2T1
(7.179)
Recall that the atmospheric pressure is 733 mm Hg at an elevation of 1000 feet. Determine the
density of air at 1000 feet and at a temperature of 100 ºF.
10,100 scfm
14.7 psia
14.2 psia
560°R
528°R
11,100 cfm
Calculate the density of air at standard conditions (100°F and 733 mm Hg of pressure).
lb
in 2
733mm Hg
760 mm Hg
14.7
air
air
P
RT
53.3
ft lb
lb air o R
100 460 R
144in 2
ft 2
0.068
lb
ft 3
0.068lb ft 3 at1000ft and 100o F
w (11,100cfm) 0.068
lb
ft 3
1min
60s
12.6
lb
s
Assume that the inlet pressure (p ) = 14.2 psia (733 mm Hg) at 1000 ft and that the outlet
1
pressure (p ) = 20.2 psia = Pr, T = 100°F + 460 = 560°R. Also assume the efficiency of the
2
1
blower, e = 75%, efficiency of the electrical motor, em = 95%, R = 53.5 and n = 0.283 for air.
48
Pw
w RT1
C1 n e
12.6 lb s 53.3
Pw
ft lb
550
s hp
p2
p1
ft lb
lb air
0.283
1
528 R
0.283 0.75
20.2 psia
14.2 psia
0.283
1
319 hp
bhp = 319
mhp = motor horsepower
mhp =
bhp
em
mhp =
319 hp
0.95
336 hp
21. Design a fine-bubble diffused aeration system to meet an oxygen demand of 5,000 kg/d.
The plant is located at an elevation of 610 meters (2,000 ft). There are two aeration basins with a
total volume = 5,400 m3, assume the alpha and beta coefficients for the diffusers are 0.35 and
0.95, respectively and use a temperature correction factor ( ) of 1.024. The dissolved oxygen
(DO) saturation concentration of water is 7.81 mg/L at sea level and 28°C. A DO concentration
of 2.0 mg/L is maintained in the aeration basins. The oxygen transfer efficiency is assumed to be
30% since fine-bubble diffusers are being used. The side water depth of the aeration basin is 4.9
m with diffusers placed 0.5 m from the bottom of the tank. The atmospheric pressure at an
elevation of 610 meters is 706 mm Hg. Assume that the specific weight of water is 9810 N/m3.
Determine:
a) Volumetric flow rate of the air (m3/min) to meet oxygen requirement of 5,000
kg per day.
b) Power required (kw) for centrifugal blowers.
Solution Part A
The dissolved oxygen saturation concentration at mid-depth must be calculated using Equation
(7.179). We first must estimate the oxygen concentration (%) in the off-gas. The concentration
of oxygen in the atmosphere is approximately 21% by volume and 23% by weight (Metcalf and
Eddy, 2003). The concentration of nitrogen in the atmosphere is approximately 79% by volume.
49
Assuming 100 moles of air to be compressed, there would be 21 moles of oxygen nO2 and 79
moles of nitrogen nN2 . The moles of oxygen in the off-gas are estimated as follows.
nO2
21moles O 2 1 OTE
nO2
21moles O 2 1
30%
100%
15 moles
The percentage of oxygen in the off-gas may now be calculated.
Oe
nO2
nO2
nN2
100
15
100
15 79
16%
The hydrostatic pressure due to water is calculated using Equation (7.181).
p
h
where:
p = hydrostatic pressure,
lb N
,
ft 2 m 2
= specific weight of water, 62.4
lb
ft 3
9810
N
m3
and,
h = depth of water, ft (m).
The pressure at the point of air release (Pr) at the bottom of the aeration basin is calculated
below and must be expressed in terms of absolute pressure. Absolute pressure (pabs) is gauge
pressure (pgauge) plus barometric pressure (pbar).
pabs
pgauge
(7.182)
pbar
The specific gravity of mercury is 13.6 and is used to get pressure in terms of meters of water
pressure.
Pr
4.9 m 0.5 m
Pr 1.37 105
N
m2
706 mm Hg 13.6
137 kPa
1atm
101.37 kPa
1m
1000 mm
9810 N
m3
1.35atm
Now, calculate the dissolved oxygen saturation concentration at mid-depth.
50
CM
CM
0.95
Pr
C
Cs
7.81mg
L
Oe
42
137 kPa
203
16%
42
7.8mg
L
Use Equation (7.178) to determine the quantity of oxygen required at standard conditions.
AOTR
CM
SOTR
T °C 20°C
Ct
9.17
0.35 7.8 mg L - 2.0 mg L 1.024
SOTR
9.17
kg
SOTR 1.87 104
d
kg
5, 000
d
28°C 20°C
Calculate the density of air at standard conditions (20°C and 1 atmosphere of pressure).
air
PM
RT
1.01 105
8314
N
m2
28.97
N m
kg mole air o K
kg
kg mole
1.20
20 273K
kg
m3
The standard cubic meters of air required per minute is calculated by dividing the SOTR by the
specific weight of air
1.20 kg air
at standard conditions (20°C and 1 atmosphere of pressure)
m3 air
and by the concentration of oxygen in the atmosphere by weight
kg O2 1 m3 air
m3
= 1.87 104
min
d
1.20 kg air
1 kg air
0.23 kg O
2
1d
24 h
0.23 kg O2
.
kg air
1h
60 min
1
0.30
157
Check to see if enough air has been provided to ensure adequate mixing of the wastewater in the
aeration basin i.e. divide the standard cubic feet per minute of air required by the volume of the
aeration basin.
Mixing air =
157 m3 min 1000 m3
5, 400 m3 1000 m3
29 m 3 /min
1000 m 3
The amount of air required for mixing typically ranges from 20 to 30 m3/(1000 m3min).
51
Solution Part B
Next calculate the power required by the centrifugal blower for compressing the air using
Equation (7.180). We will assume that the air temperature may get up to 38°C during the
summer months.
Air temp. = 38°C
273 311 K
Std. temp.= 20°C 273 293 K
Must use combined gas law to compute the air required at a different temperature and pressure.
PV
1 1
T1
V2
PV
2 2
T2
(7.179)
PV
1 1T2
P2T1
Recall that the atmospheric pressure is 706 mm Hg at an elevation of 610 meters. Determine the
density of air at 610 m and at a temperature of 38 ºC.
157
m3
min
760 mm
706 mm
311 K
293 K
179
m3
min
Calculate the density of air at standard conditions (38°C and 706 mm Hg of pressure).
air
PM
RT
w (179
9.42 104
8314
N
m2
28.97
N m
kg mole air o K
m3
kg
) 1.06 3
min
m
1min
60s
kg
kg mole
38 273K
3.2
1.06
kg
m3
kg
s
Assume that the inlet pressure (p ) = 706 mm Hg at 610 m or 0.93 atm and that the outlet
1
pressure (p ) = 1.35 atm = Pr, and T = 38°F + 273 = 311 K. Also assume the efficiency of the
2
1
blower, e = 75%, efficiency of the electrical motor, em = 95%, R = 8.314 and n = 0.283 for air.
52
Pw
w RT1
C1 n e
p2
p1
3.2 kg s 8.314
Pw
0.283
1
N m
kg air K
m N
29.7
s kw
311K
0.283 0.75
1.35atm
0.93atm
0.283
1
146 kw
bp =146 kw
mp = motor power
mp =
bp
em
mp =
146 bp
0.95
154 kw
22. A single-stage trickling filter is to be designed to treat primary effluent containing a BOD5
concentration of 120 g/m3. The minimum wastewater temperature anticipated is 15ºC.
Using a recycle ratio of 1.0, determine the minimum allowable BOD5 loading rate for a
stone-media filter based on the NRC equation if an average effluent BOD5 concentration of
30 g/m3 is to be achieved.
120g m3 30g m3
E15 C
100% 75%
120g m3
Use Equation (7.184) to correct for a temperature of 15ºC.
E15 C
E20 C 1.035
75%
E20 C 1.035
E20 C
89%
15 C 20 C
(7.184)
15 C 20 C
Calculate the recirculation factor using Equation (7.182).
F
1 R
1 0.1 R
1 1
2
1 0.1 1
2
1.65
53
Calculate the BOD loading rate using Equation (7.181).
100
E
W1
1 0.444
V F
0.5
100
89
W1
1 0.443
V 1.65
W1
V
0.5
0.13kg
m3 d
23. A single-stage trickling filter is to be designed to treat primary effluent containing a BOD5
concentration of 170 mg/L. The minimum wastewater temperature anticipated is 16ºC.
Using a recycle ratio of 0.5, determine the minimum allowable BOD5 loading rate (lb/1,000
ft3 d) for a stone-media filter, based on the NRC equation, if an average effluent BOD5
concentration of 30 mg/L is to be achieved.
170g m3 30g m3
170g m3
E16 C
100% 82.35%
Use Equation (7.184) to correct for a temperature of 16ºC.
E16 C
E20 C 1.035
82.35%
E20 C
16 C 20 C
16 C 20 C
E20 C 1.035
94.5%
Calculate the recirculation factor using Equation (7.182).
F
1 R
1 0.1 R
E
1 0.5
2
1 0.1 0.5
2
1.36
100
W1
1 0.056
V F
0.5
Calculate the BOD loading rate using Equation (7.181).
54
100
94.5
W1
1 0.0561
V 1.36
W1
V
0.5
1.46 lb
1000 ft 3 d
24. A 3.0 MGD trickling filter (TF) plant consists of a primary clarifier, followed by a singlestage trickling filter containing stone media, which is followed by a secondary clarifier. The
influent to the WWTP contains 150 mg/L of BOD5 and the primary clarifier removes 35% of the
BOD5 and the temperature of the wastewater during the winter averages 14ºC. Determine the
effluent BOD5 concentration (mg/L) if the diameter of the TF is 80 ft and the depth is 8 ft. Use a
recycle ratio of 1.0 and assume there are two parallel treatment trains.
Calculate the recirculation factor using Equation (7.182).
F
1 R
1 0.1 R
1 1
2
1 0.1 1.0
2
1.65
Calculate the BOD loading as follows:
8.34 lb
MG mg L
W1
Q BOD5
W1
1.5 MGD 150
W1 1220
mg
1 0.35
L
8.34 lb
MG mg L
lb BOD5
d
Next, calculate the volume of media and express the BOD loading rate in terms of BOD/(1000
ft d).
3
V
W1
V
D2
Depth
4
80ft
4
1220lb d 1000ft 3
4.02×104 ft 3 1000ft 3
2
8ft 4.02 104 ft 4
30.3
lb BOD5
1000ft 3 d
Determine the efficiency of the trickling filter using Equation (7.181).
55
100
E
0.5
W1
1 0.056
V F
E20 C
100
30.3
1 0.056
1.65
0.5
80.6%
Correct the efficiency for a temperature of 14ºC using Equation (7.184).
14 C 20 C
E14 C
E20 C 1.035
E14 C
80.6% 1.035
E14 C
65.6%
Effluent BOD5 150
14 C 20 C
mg
1 0.35 (1 0.656)
L
33.5
mg
L
25.
A two-stage high-rate trickling filter with 200% recycle will be used to treat 5.0 million
gallons per day (MGD). The influent BOD to the primary clarifier is 225 mg/L and the primary
removes 30% of the incoming BOD. The effluent BOD should be 20 mg/L or less. Assume that
the design BOD loading to the trickling filter system is 40 lb BOD/(d·1,000 ft3) and a depth of 6
ft will be used. Use the NRC Equations for designing the trickling filter system and assume that
there are two treatment trains operating in parallel.
Determine:
a) The diameter (ft) of the high-rate trickling filter system.
c) The effluent BOD concentration (mg/L).
Solution Part A
First, calculate the actual BOD loading to the trickling filter system.
W1
2.5MGD 1 0.30 225mg BOD L
8.34lb MG
mg L
3280 ppd
The factor (1-0.30) accounts for the BOD that is removed in the primary clarifier.
Next, calculate the volume of the trickling filter media by dividing the actual BOD loading by
the design BOD loading.
V
3280 ppd
40 lb d 1000 ft 3
82, 000 ft 3
56
Volume per trickling filter is calculated by dividing the total volume by 2, since this is a twostage trickling filter.
82,000 ft 3
= 41,000ft 3
2
V
trickling filter
The area of each trickling filter is determined by dividing the volume by the depth of 6 ft.
A
41,000 ft 3
= 6833 6830ft 2
6ft
This area is set equal to the area of a circle ( r2) to determine the radius and diameter of the
trickling filter.
A
6830 ft 2
r2
r
6830 ft 2 /
46.6ft 47 ft
Diameter = 2r = 2(47 ft) = 94 ft, therefore use a diameter of 95 ft. Wastewater equipment
normally is manufactured in increments of 5 ft.
Next, calculate the new area based on a diameter of 95 ft.
A
D2
4
95ft
4
2
= 7090ft 2
Actual volume of each trickling filter is calculated by multiplying the area by the 6 ft depth.
V
A
Depth
7090 ft 2 6 ft
42,540 ft 3
Determine the recirculation factor (F) using Equation (7.182).
F
1 R
1 2
1 0.1 R
2
1 0.1 2
2
2.08
Solution Part B
Estimate the efficiency of the first-stage trickling filter using Equation (7.181).
E1
100
W1
1 0.0561
V F
100
0.5
3280 ppd
1 0.0561
(42,540 1000) 2.08
Next, calculate the BOD loading to the 2nd-stage trickling filter (W2).
57
0.5
74.5%
W2
1 0.745 3280 ppd
836 ppd
Now, it is possible to calculate the efficiency of the 2nd-stage using Equation (7.183).
100
E2
1
0.0561/ 1 E1
W2
V F
0.5
100
E2
0.5
1
0.0561/ 1 0.745
59.6%
836 ppd
42,540 1000 2.08
Estimate the final BOD in the effluent to see if it meets the standard.
Effluent BODe
225
mg
1 0.30 1 0.745 1 0.596
L
16.2
mg
L
20
mg
L
26. A two-stage high-rate trickling filter with 100% recycle will be used to treat 5,000 m3/d.
The influent BOD to the primary clarifier is 220 mg/L and the primary removes 33% of the
incoming BOD. The effluent BOD should be 30 mg/L or less. Assume that the design BOD
loading to the trickling filter system is 1.0 kg BOD/(d · m3) and a depth of 1.83 m will be used.
Use the NRC Equations for designing the trickling filter system and assume that the operating
temperature is 18ºC. The coefficient 0.0561 in Equation (7.181) can be replaced with 0.443 and
the organic loadings must be expressed in kg/d and the volume of media in cubic meters.
Determine:
a) The diameter (m) of the high-rate trickling filter system.
d) The effluent BOD concentration (mg/L).
Solution Part A
First, calculate the actual BOD loading to the trickling filter system.
W1
5000
m3
d
1 0.33 220g BOD m3
1kg
1000g
737
kg
d
The factor (1-0.33) accounts for the BOD that is removed in the primary clarifier.
Next, calculate the volume of the trickling filter media by dividing the actual BOD loading by
the design BOD loading.
V
737 kg / d
1.0 kg d m3
737 m3
58
Volume per trickling filter is calculated by dividing the total volume by 2, since this is a twostage trickling filter.
737 m3
= 369 m3
2
V
trickling filter
The area of each trickling filter is determined by dividing the volume by the depth of 1.83 m.
A
369 m3
= 202 m 2
1.83m
This area is set equal to the area of a circle ( r2) to determine the radius and diameter of the
trickling filter.
A
202 m 2
r2
r
202 m2 /
8.0 m
Diameter = 2r = 2(8 m) = 16 m, therefore use a diameter of 16 m.
Next, calculate the new area based on a diameter of 16 m.
A
D2
4
16 m
4
2
= 201m2
Actual volume of each trickling filter is calculated by multiplying the area by the 1.83 m depth.
V A Depth 201m 2 1.83m 368 m3
Determine the recirculation factor (F) using Equation (7.182).
F
1 R
1 0.1 R
1 1
2
1 0.1 1
2
1.65
Solution Part B
Estimate the efficiency of the first-stage trickling filter using Equation (7.181).
E1
100
100
W1
1 0.443
V F
0.5
0.5
1 0.443
67.2 %
737 kg/d
368 m3 1.65
The above efficiency must be corrected for a temperature of 18ºC using the following equation.
ET C
E20 C 1.035
T C 20 C
67.2% 1.035
59
18 C 20 C
62.7%
Next, calculate the BOD loading to the 2nd-stage trickling filter (W2).
W2
1 0.627 737 kg/d
275 kg/d
Now, it is possible to calculate the efficiency of the 2nd-stage using Equation (7.183).
100
E2
1
0.443 / 1 E1
100
E2
0.5
1
ET C
0.5
W2
V F
0.443 / 1 0.627
E20 C 1.035
T C 20 C
55.6%
275 kg/d
368 m3 2.08
55.6% 1.035
18 C 20 C
51.9%
Estimate the final BOD in the effluent to see if it meets the standard.
Effluent BODe
225
mg
1 0.33 1 0.627 1 0.556
L
25
mg
L
30
mg
L
27. Two biotowers operating in parallel are to be designed to treat a flow of 6.0 million gallons
per day (MGD). The BOD5 in the influent to the primary clarifier is 200 mg/L and the clarifier
removes 35% of the BOD5. The minimum wastewater temperature is expected to be 15°C. The
value of k is assumed to be 0.078 (gpm)0.5/ft2 at 20°C and an effluent BOD5 concentration of 20
mg/L is to be achieved. The biotower depth = 20 ft.
Determine:
a) The radius (ft) of each biotower.
b) The volume (ft3) of biotowers.
c) The recycle flow (gpm/ft2) around the biotowers if the minimum wetting rate is
0.75 gpm/ft2 and the recycle ratio, R.
Solution Part A
Normalize k for site specific depth and influent BOD5 concentration using Equation (7.187).
S0
1 0.35 200
mg
mg
130
L
L
60
D1
D2
0.5
S1
S2
k2
k1
k2
0.084 gpm
0.5
0.5
0.5
0.078 gpm
ft
2
0.5
20 ft
20 ft
150 mg L
130 mg L
0.5
ft 2
Correct k2 for a temperature of 15°C using Equation (7.185).
kT
C
k20 C
k
20 C
1.035
T C 20 C
0.5
0.071 gpm
0.084 gpm
0.5
/ ft 2 1.035
15 C 20 C
/ ft 2
The hydraulic loading rate, q, is determined but substituting the appropriate values into Equation
(7.186).
Se
So
e
k D qn
20 mg L
130 mg L
q
e
0.071 gpm
0.5
ln
0.5
0.071 gpm
0.5
/ft 2 20ft q0.5
/ ft 2 20ft
20 mg L
130 mg L
1
0.071 gpm
q
ln
0.5
0.5
/ ft 2 20ft
0.58
20 mg L
130 mg L
gpm
ft 2
The surface area of the biotowers can now be determined by dividing the volumetric flow rate by
the hydraulic loading rate as follows:
Across section
Q
q
6.0 106 gal d 1d
0.58gpm ft 2 24 h
biotower
7.18 103 ft 2
2
Across section
3.59 103 ft 2
Across section
D
4 3.59 103 ft 2
3.59 103 ft 2
D2
4
70 ft
67.6 ft
61
1h
60 min
7.18 103 ft 2
The volume of the biotower media is determined by multiplying the cross sectional area by the
depth or height.
D2
Depth
4
V
70ft
4
7.70 104 ft 3
VTotal
2
7.70 104 ft 3
20ft
2 biotowers = 1.54 105 ft 3
The minimum wetting rate has been specified as 0.75 gpm/ft2 so the sum of the hydraulic loading
rate (q) and recycle loading rate (qr) must be at least equal to or greater than the minimum
wetting rate.
q
qr
qr
0.75 gpm ft 2
0.75 gpm ft 2
0.17 gpm ft 2
0.58 gpm ft 2
The recycle ratio is calculated as follows:
R
qr
q
0.17 gpm ft 2
0.58gpm ft 2
0.29
28. Two biotowers operating in parallel are to be designed to treat a flow of 16,000 m3/d. The
BOD5 in the influent to the primary clarifier is 200 mg/L and the clarifier removes 35% of the
BOD5. The minimum wastewater temperature is expected to be 14°C. The value of k is
assumed to be 0.210 (Lps)0.5/m2 at 20°C and an effluent BOD5 concentration of 20 mg/L is to be
achieved. The biotower depth = 6.1 m.
Determine:
a) The radius (m) of each biotower.
b) The volume (m3) of biotowers.
c) The recycle flow (Lps/ft2) around the biotowers if the minimum wetting rate is
0.5 Lps/ft2 and the recycle ratio, R.
Solution Part A
Normalize k for site specific depth and influent BOD5 concentration using Equation (7.187).
S0
1 0.35 200
mg
mg
130
L
L
62
D1
D2
0.5
0.5
S1
S2
k2
k1
k2
0.226 Lps
0.210 Lps
0.5
0.5
m
2
6.1 m
6.1m
0.5
150 mg L
130 mg L
0.5
m2
Correct k2 for a temperature of 14°C using Equation (7.185).
kT
C
k20 C
k
20 C
T C 20 C
1.035
0.184 Lps
0.5
0.226 Lps
0.5
/ m 2 1.035
14 C 20 C
/ m2
The hydraulic loading rate, q, is determined but substituting the appropriate values into Equation
(7.186).
Se
So
e
k D qn
20 mg L
130 mg L
q
0.5
0.184 Lps
ln
e
0.5
0.184 Lps
0.5
/m2 6.1m q0.5
/ m 2 6.1m
20 mg L
130 mg L
1
q
0.184 Lps
ln
0.5
/ m 2 6.1m
0.5
0.36
20 mg L
130 mg L
Lps
m2
The surface area of the biotowers can now be determined by dividing the volumetric flow rate by
the hydraulic loading rate as follows:
63
Across section
Across section
biotower
Across section
D
Q
q
1.6 104 m 3 d 1d
0.36 Lps m 2 24 h
5.14 103 m 2
2
2
2.57 10 m
4 2.57 102 m 2
2
1h
60 min
1min
60s
1000 L
m3
5.14 102 m 2
2.57 102 m 2
D2
4
18.1 m
The volume of the biotower media is determined by multiplying the cross sectional area by the
depth or height.
D2
Depth
4
V
18.1m
4
1.57 103 m3
VTotal
2
6.1m 1.57 103 m3
2 biotowers = 3.14 103 m3
The minimum wetting rate has been specified as 0.5 Lps/m2 so the sum of the hydraulic loading
rate (q) and recycle loading rate (qr) must be at least equal to or greater than the minimum
wetting rate.
q qr
qr
0.50 Lps m2
0.50 Lps m2
0.14 Lps m2
0.36 Lps m2
The recycle ratio is calculated as follows:
R
qr
q
0.14 Lps m2
0.36 Lps m2
0.39
29. Design a multi-stage RBC WWTP to treat the following wastewater for BOD removal.
Influent flow = 6,000 m3/d, influent BOD5 = 260 g/m3, primary clarifier removes 35% of BOD5,
total effluent BOD5 and effluent TSS = 20 g/m3, and effluent soluble BOD5 (sBOD5 ) = 10 g/m3.
T = 10°C
Determine:
a) The number of stages and number of trains.
b) The soluble BOD5 in the effluent from each stage (g/m3).
c) The organic loading on the first stage (g/m2d).
d) The overall organic loading to the RBC WWTP (g/m2d).
64
e) The overall hydraulic loading rate to the RBC WWTP (m3/m2d).
Solution Part A
Determine the number of RBCs shafts for the first stage assuming that the primary effluent
soluble BOD5 is 50% of the primary effluent total BOD5 and that the total soluble BOD5 loading
rate is 15 g sBOD5/(m2 d).
First, calculate the total BOD5 in the primary effluent.
Total BOD5 in primary effluent = 1 - 0.35 ×260
g
g
= 169 3
3
m
m
Calculate the soluble BOD5 in the primary effluent.
Soluble BOD5 in primary effluent =169
g
g
0.5 84.5 3
3
m
m
Calculate the soluble BOD5 loading to the first stage.
Soluble BOD5 Loading =(6.00 103
m3
)
d
84.5
g
m3
5.07 105
g
d
Determine the number of shafts in the first stage.
5.07 105 g sBOD5 d
Media Surface Area =
15.0g sBOD5
m2 d
Number of Shafts
3.38 104 m2
9,300 m2 shaft
3.38 104 m 2
3.6 4.0shafts per stage
Assume there are three treatment trains and four stages with 4.0 shafts per stage. Next, calculate
the flow per train.
Solution Part B
Flow per train
6.00 103 m3 d
3trains
2.00 103 m3 d
train
Determine the ratio of surface area to the flow rate as follows:
As
Q
9300 m2
2.00 103 m3 d
4.65d
m
65
Calculate the soluble BOD5 from each stage using Equation (7.188).
1
S1
1
2 0.00974 As Q
1
1
S1
1
1
1
S2
S3
1
S4
g
m3
1
4 0.00974 As Q S2
2 0.00974 As Q
4 0.00974 4.65d m 18.3g m3
S4
1
18.3
4 0.00974 4.65d m 33.5g m3
2 0.00974 4.65d m
1
g
m3
4 0.00974 As Q S1
2 0.00974 4.65d m
S3
33.5
2 0.00974 As Q
1
1
4 0.00974 4.65d m 84.5g m3
2 0.00974 4.65d m
S2
1
4 0.00974 As Q S0
1
1
11.9
g
m3
4 0.00974 As Q S3
2 0.00974 As Q
4 0.00974 4.65d m 11.9 g m 3
2 0.00974 4.65d m
8.6
g
m3
10.0
g
Good!
m3
Solution Part C
The soluble organic loading on the first-stage is calculated below.
First-stage soluble BOD5 loading
Within range 12 15
6.00×103 m3 d 84.5g m3
2
4shafts 9300 m shaft
g sBOD5
m2 d
66
13.6
g sBOD5
m2 d
Solution Part D
The total organic loading on the RBC WWTP is calculated as follows.
6.00×103 m3 d 169 g m3
Total BOD5 loading
Within range 8 20
2
3 trains 4shafts stage 9300 m shaft
9.1
g BOD5
m2 d
g BOD5
m2 d
Solution Part E
The hydraulic loading on the RBC WWTP is calculated as follows.
Hydraulic loading
6.00×103 m3 d
0.054
3 trains 4shafts stage 9300 m 2 shaft
Slightly below range but still should work fine. 0.08 - 0.16
m3
m2 d
m3
m2 d
30. A modified Ludzack-Ettinger (MLE) biological nutrient removal process is to be designed to
meet a 20/20/10 mg/L effluent standard for BOD5, TSS, and TN, respectively. The influent
wastewater characteristics, mixed liquor parameters, and biokinetic coefficients to be used in
design are presented in the tables below. Assume that the influent ammonium concentration is
equal to the TKN concentration, RAS flow = 1Q and MLR flow = 3Q, DO in the RAS and MLR
recycle is 1.0 g/m3.
Wastewater characterization
Influent
Mixed liquor
Q=
38,000 m3/d
DO =
2.0 mg/L
pH =
7.0
MLSS =
3,500 mg/L
Minimum
temperature
13°C
MLVSS =
0.75 MLSS
TKN =
30 mg/L
KDO =
0.5 mg/L
Alkalinity =
200 mg/L as CaCO3
MLVSS =
2,625 mg/L
67
BOD5 =
200 mg/L
TSS =
167 mg/L
Inerts (XL) =
60 mg/L
Heterotrophic biokinetic coefficients at 20°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.06 d-1
k=
5 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 20°C
Coefficient
Coefficient
YNS =
0.15 g VSS/g NH+4 -N
(kd)NS =
0.05 d-1
kNS =
3 d-1
(Ks)NS =
0.74 mg/L NH+4 -N
Heterotrophic biokinetic coefficients at 13°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.046 d-1
k=
3.1 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 13°C
Coefficient
Coefficient
YNS =
0.15 g VSS/g NH+4 -N
(kd)NS =
0.038 d-1
kNS =
1.9 d-1
(Ks)NS =
0.52 mg/L NH+4 -N
A temperature correction coefficient ( ) of 1.04 should be used for correcting both the
heterotrophic and Nitrosomonas kd values, whereas a of 1.07 should be used for correcting both
the heterotrophic and Nitrosomonas k values. The heterotrophic Ks value is to be corrected with
a of 1.00 and the Nitrosomonas Ks is corrected with a value of 1.053. All temperature
correction coefficients were obtained from Metcalf and Eddy (2003). Use a safety factor of 1.25
and peaking factor of 1.2 and assume that Fn = 0.10.
68
Solution
First Oxic Zone Calculations
1. Calculate the maximum growth rate of Nitrosomonas for the ambient temperature of
DO
13°C.
(7.192)
1-0.833 7.2 - pH
MAX NS = YNS k NS
K DO + DO
MAX
NS
=
0.15gVSS
gNH +4 -N
1.9d-1
2 mg L
0.5mg L +2 mg L
1-0.833 7.2 - 7.0
0.19d
1
2. Correct the Nitrosomonas decay coefficient for temperature.
kd
NS T C
kd
kd
NS 13 C
1.04
NS 20 C
0.05d -1
NS 20 C
T 20 C
13
1.04
20 C
0.038d -1
3. Calculate the minimum MCRT required for nitrification.
( max )NS NH 4 N
1
0
=
- kd
K NS + NH 4 N
c min
NS
(7.193)
0
1
=
c
min
c
min
0.19d
-1
30 mg L
0.52 mg L + 30 mg L
- 0.038d -1
1
= 6.7 d
0.15d -1
4. Calculate the design MCRT based on a safety factor of 1.25 and peaking factor of 1.2 as
follows:
(7.194)
c design =
c min SF PF
= 6.7 d 1.25 1.2
c design
10.1d
5. Estimate the overall MCRT by multiplying the design MCRT by a multiplication factor.
The multiplication factor is estimated as follows by taking the reciprocal of 1- the
fraction of the assumed % of the anoxic volume. Assume that the anoxic zone is 35% of
the overall volume of the MLE process.
1
MF
1.54
(7.195)
1 0.35
c overall
c overall
=
c design
= 10.1d 1.54
69
MF
15.6d
6. For a given set of heterotrophic biokinetic coefficients, calculate the effluent soluble
substrate concentration (Se).
K s 1 + kd c overall
Se =
(7.197)
Yk - kd - 1
c overall
60
Se =
15.6 d
mg
1 + 0.046 d -1 15.6 d
L
g VSS
0.6
3.1d -1 - 0.046 d -1
g BOD5
3.8
-1
mg
L
7. For a given set of autotrophic biokinetic coefficients, calculate the effluent
ammonium nitrogen concentration using the following equation:
K NS 1 + kd
+
NH 4 - N e =
c design
NS
c design
YNS k NS - kd
NS
(7.198)
-1
mg
1 + 0.038d -1 10.1d
L
g VSS
0.15
1.9 d -1 - 0.038d -1
g NH +4 -N
0.52
+
NH 4 - N e =
10.1d
+
NH 4 - N e = 0.48
-1
mg
L
8. The nitrogen utilized in synthesis neglecting the small amount of nitrogen synthesized by
the nitrifiers is calculated using Equation and assuming FN = 0.10:
N syn =
0.60
N syn =
Y Si
Se FN
1+ k d
g VSS
mg
mg
200
3.8
0.10
g BOD5
L
L
mg TSS
+ 20
-1
L
1+ 0.046d
15.6d
N syn = 8.4
(7.199)
+ X e FN
c overall
0.75
mg VSS
0.10
mg TSS
mg N
L
9. Determine the amount of nitrogen to be oxidized using Equation (7.200).
NO = TKN 0 - NH 4
NO = 30
mg
L
0.48
N
e
- N syn
mg
mg
8.4
L
L
70
21.1
mg
L
10. Calculate the volume of the first oxic zone necessary to achieve nitrification for a given
temperature, pH, and DO as follows:
Q
Voxic1 =
Y Si - S e
c design
X
1 + kd
m3
10.1d
d
mg
3500
L
0.6
38, 000
Voxic1 =
+ XL
(7.201)
c design
mgVSS
g TSS
mg
mg
200
- 3.8
mg BOD5 0.75 gVSS
L
L
-1
1 + 0.046d 10.1d
+ 60
mg
L
Voxic1 = 18,300 m3
First-Stage Anoxic Zone Calculations
1. Estimate the nitrate nitrogen concentration in the mixed liquor recycle returned to the
first stage anoxic zone using the following equation:
NO(Q)
(7.202)
N=
(Qmlr +Qr +Q)
21.1
N=
mg
L
38, 000
3
38, 000
m
d
m3
d
3
3 38, 000
m
d
3
38, 000
m
d
4.2
mg
L
2. Estimate the nitrate equivalence of dissolved oxygen in the mixed liquor recycle as
follows:
g NO3 - N
1 kg
(7.203)
NO3 N = DO mlr 0.35
Qmlr
eq 1
g O2
1000 g
NO3
N
eq1
= 1.0
g
m3
NO3
0.35
N
eq1
g NO3 - N
g O2
= 40.0
38,000
m3
3
d
1 kg
1000 g
kg
d
3. Calculate the mass of nitrates to be removed in the first stage anoxic zone using the
following equation and assuming that the nitrate nitrogen concentration in the effluent is
the same as N =4.2 mg/L:
1 kg
(7.204)
NOR1 = Qr NO3 e + Qmlr (N)
1000 g
71
m3
g
m3
g
4.2 2 + 38,000
3 4.2 2
d
m
d
m
kg
NOR1 = 638
d
1 kg
1000 g
NOR1 = 38,000
4. The total mass of nitrates to be removed in the first stage anoxic zone is calculated as
follows:
(7.205)
TNOR1 = NOR1 + NO3 N
eq1
TNOR1 = 638
kg
kg
+ 40.0
d
d
678
kg
d
5. The volume of the 1st-stage anoxic zone can be determined using Equation (7.209) .
Vanoxic1
TNOR1 1000 g kg
0.03 Q Si 1.06
0.029 X 1.06
T 20
0.03 38, 000 m3 d
678 kg d 1000 g kg
Vanoxic1
mg
L
7800 m3
0.029 3500
Vanoxic1
T 20
1.06
200
mg
1.06
L
13 20 C
13 20 C
Check the Overall Design
1. First, calculate the total volume of the biological system using the following equation:
Vtotal = Vanoxic1 Voxic1
Vtotal = 7800 m3 18,300 m3
26,100 m3
2. Determine the total quantity of sludge produced as follows:
Px =
0.60
Px =
Y Si - S e
1 + kd
+XL
Q
c overall
g VSS
g TSS
g
g
200 3 - 3.8 3
g BOD 5 0.75g VSS
m
m
-1
1+ 0.046 d
15.6 d
Px = 5750
kg
d
72
1 kg
1000 g
+ 60
(7.217)
g
m3
38, 000
m3
d
1 kg
1000 g
3. Compute the new overall MCRT using Equation (7.218).
c overall
c overall
% Difference
=
=
Vtotal
X
Px (1000 g/kg)
26,100 m3 3500g m3
kg
5750
d
15.9d 15.6d
g
1000
kg
100%
15.9d
15.9d
1.9% 5.0%OK
Since the difference between the original and calculated value for the overall MCRT is within 5
percent, there it is not necessary to repeat the procedure
Oxygen Requirements
Oxygen is required to meet both the carbonaceous and nitrogenous demand. The total kilograms
of oxygen required can be estimated using the following equations:
O 2 = CBOD + NOD - DOC
CBOD =
Q ( 1-1.42 Y ) S i - S e
CBOD = 38,000
+ 1.42 k d
X Voxic1
m3
g VSS
g
g
(1 1.42 0.6
) 200 3 3.8 3
d
g BOD5
m
m
+1.42 0.046 d -1
2625
g
18,300 m3
3
m
m3
4.57
d
gO 2
21.1
mg
L
1 kg
1000 g
1 kg
1000 g
NOD = Q 4.57 NO
NOD = 38,000
1 kg
(7.220)
1000 g
1 kg
1000 g
1 kg
1000 g
(7.221)
3660
kg
gd
1 kg
1000 g
(7.222)
e
g NO3- - N
gO 2
m3
mg
mg
1 kg
kg
DOC = 38,000
2.86
21.1
4.2
1840
d
g NO3- - N
L
L
1000 g
d
kg
kg
kg
kg
O2 = CBOD + NOD - DOC = 4240
3660
1840
6060
d
d
d
d
Alkalinity Requirements
DOC = Q 2.86
NO - NO3
The effluent alkalinity can be calculated as follows from the 4-stage anoxic/oxic process:
73
ALK e = ALKo - 7.14 NO + 3.57 NO - NO3 - N
ALK e = 200
(7.223)
e
mg
mg
mg
mg
- 7.14 21.1
+ 3.57 21.1
- 4.2
L
L
L
L
110
mg
L
31. A modified Ludzack-Ettinger (MLE) biological nutrient removal process is to be designed to
meet a 20/20/10 mg/L effluent standard for BOD5, TSS, and TN, respectively. The influent
wastewater characteristics, mixed liquor parameters, and biokinetic coefficients to be used in the
design are presented in the tables below. Assume that the influent ammonium concentration is
equal to the TKN concentration, RAS flow = 1Q and MLR flow = 3Q, DO in the RAS and MLR
recycle is 1.0 g/m3.
Wastewater characterization
Influent
Mixed liquor
Q=
38,000 m3/d
DO =
2.0 mg/L
pH =
7.1
MLSS =
3,500 mg/L
Minimum
temperature
10°C
MLVSS =
0.75 MLSS
TKN =
30 mg/L
KDO =
0.5 mg/L
Alkalinity =
225 mg/L as CaCO3
MLVSS =
2,625 mg/L
BOD5 =
200 mg/L
TSS =
167 mg/L
Inerts (XL) =
60 mg/L
Heterotrophic biokinetic coefficients at 20°C
Coefficient
Coefficient
74
Y=
0.6 g VSS/g BOD5
kd =
0.06 d-1
k=
5 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 20°C
Coefficient
Coefficient
YNS =
0.15 g VSS/g NH+4 -N
(kd)NS =
0.05 d-1
kNS =
3 d-1
(Ks)NS =
0.74 mg/L NH+4 -N
Heterotrophic biokinetic coefficients at 10°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.041 d-1
k=
2.5 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 10°C
Coefficient
Coefficient
YNS =
0.15 g VSS/g NH+4 -N
(kd)NS =
0.034 d-1
kNS =
1.5 d-1
(Ks)NS =
0.44 mg/L NH+4 -N
A temperature correction coefficient ( ) of 1.04 should be used for correcting both the
heterotrophic and Nitrosomonas kd values, whereas a of 1.07 should be used for correcting both
the heterotrophic and Nitrosomonas k values. The heterotrophic Ks value is corrected with a of
1.00 and the Nitrosomonas Ks is corrected with a value of 1.053. All temperature correction
coefficients were obtained from Metcalf and Eddy (2003). Use a safety factor of 1.25 and
peaking factor of 1.2 and assume that Fn = 0.10.
Solution
First Oxic Zone Calculations
11. Calculate the maximum growth rate of Nitrosomonas for the ambient temperature of
DO
1-0.833 7.2 - pH
10°C. MAX NS = YNS k NS
(7.192)
K DO + DO
75
MAX
NS
=
0.15gVSS
gNH+4 -N
1.5d-1
2 mg L
0.5mg L +2 mg L
1-0.833 7.2 - 7.1
0.165d
1
12. Correct the Nitrosomonas decay coefficient for temperature.
kd
NS T C
kd
kd
NS 10 C
1.04
NS 20 C
0.05d -1
NS 20 C
T 20 C
10
1.04
20 C
0.034 d -1
13. Calculate the minimum MCRT required for nitrification.
(
1
c
1
=
c
min
c
min
=
min
)
max NS
NH 4
K NS + NH 4
0.165d
-1
N
N
30 mg L
0.44 mg L + 30 mg L
0
- kd
NS
(7.193)
0
- 0.034d -1
1
= 7.8d
0.129 d -1
14. Calculate the design MCRT based on a safety factor of 1.25 and peaking factor of 1.2 as
follows:
(7.194)
c design =
c min SF PF
= 7.8d 1.25 1.2
c design
11.7 d
15. Estimate the overall MCRT by multiplying the design MCRT by a multiplication factor.
The multiplication factor is estimated as follows by taking the reciprocal of 1- the
fraction of the assumed % of the anoxic volume. Assume that the anoxic zone is 40% of
the overall volume of the MLE process.
1
MF
1.67
(7.195)
1 0.40
c overall
c overall
=
c design
= 11.7 d 1.67
MF
19.5d
16. For a given set of heterotrophic biokinetic coefficients, calculate the effluent soluble
substrate concentration (Se).
76
Se =
K s 1 + kd
c overall
60
Se =
19.5d
c overall
Yk - kd
(7.197)
-1
mg
1 + 0.041d -1 19.5d
L
g VSS
0.6
2.5d -1 - 0.041d -1
g BOD5
3.9
-1
mg
L
17. For a given set of autotrophic biokinetic coefficients, calculate the effluent
ammonium nitrogen concentration using the following equation:
K NS 1 + kd
+
NH 4 - N e =
c design
NS
c design
YNS k NS - kd
NS
(7.198)
-1
mg
1 + 0.034 d -1 11.7 d
L
g VSS
0.15
1.5d -1 - 0.034 d -1
+
g NH 4 -N
0.44
+
NH 4 - N e =
11.7 d
+
NH 4 - N e = 0.50
-1
mg
L
18. The nitrogen utilized in synthesis neglecting the small amount of nitrogen synthesized by
the nitrifiers is calculated using Equation and assuming FN = 0.10:
Y Si Se FN
N syn =
+ X e FN
(7.199)
1+ kd c overall
0.60
N syn =
g VSS
mg
mg
200
3.9
0.10
g BOD5
L
L
mg TSS
+ 20
-1
L
1+ 0.041d 19.5d
N syn = 8.0
0.75
mg VSS
0.10
mg TSS
mg N
L
19. Determine the amount of nitrogen to be oxidized using Equation (7.200).
NO = TKN 0 - NH 4 N - N syn
e
mg
mg
mg
mg
0.50
8.0
21.5
L
L
L
L
20. Calculate the volume of the first oxic zone necessary to achieve nitrification for a given
temperature, pH, and DO as follows:
NO = 30
77
Q
Y Si - S e
c design
Voxic1 =
X
m3
38, 000
11.7 d
d
Voxic1 =
mg
3500
L
0.6
1 + kd
+ XL
(7.201)
c design
mgVSS
g TSS
mg
mg
200
3.9
mg BOD5 0.75 gVSS
L
L
-1
1 + 0.041d 11.7 d
+ 60
mg
L
Voxic1 = 21,100 m3
First-Stage Anoxic Zone Calculations
6. Estimate the nitrate nitrogen concentration in the mixed liquor recycle returned to the
first stage anoxic zone using the following equation:
NO(Q)
(7.202)
N=
(Qmlr +Qr +Q)
mg
38, 000 m3 d
L
N=
3
38, 000 m d 3 38, 000 m3 d 38, 000 m3 d
21.5
4.3
mg
L
7. Estimate the nitrate equivalence of dissolved oxygen in the mixed liquor recycle as
follows:
g NO3 - N
lb
NO3 N = DO mlr 0.35
Qmlr 8.34
eq 1
g O2
MG mg L
NO3
N
eq1
= 1.0
mg
L
NO3
0.35
N
eq1
g NO3
N
g O2
= 40.0
38,000 m3 d 3
kg
103 g
kg
d
8. Calculate the mass of nitrates to be removed in the first stage anoxic zone using the
following equation and assuming that the nitrate nitrogen concentration in the effluent is
the same as N = 4.34 mg/L:
kg
NOR1 = Qr NO3 e + Qmlr (N)
103 g
NOR1 = 38,000
m3
mg
m3
mg
4.3
+ 38,000
3 4.3
d
L
d
L
78
kg
103 g
(7.204)
NOR1 = 654
kg
d
9. The total mass of nitrates to be removed in the first stage anoxic zone is calculated as
follows:
(7.205)
TNOR1 = NOR1 + NO3 N
eq1
kg
kg
kg
+ 40.0
694
d
d
d
st
10. The volume of the 1 -stage anoxic zone can be determined using Equation (7.209).
TNOR1 = 654
Vanoxic1
694 kg d
Vanoxic1
kg
103 g
TNOR1 0.03 Q Si 1.06
0.029 X 1.06
0.03 38,000
0.029 3500
mg
L
m3
d
1.06
T 20
T 20
200
mg
L
1.06
10 20 C
10 20 C
10, 000 m3
Vanoxic1
Check the Overall Design
4. First, calculate the total volume of the biological system using the following equation:
Vtotal = Vanoxic1 Voxic1
Vtotal = 10,000 m3 21,100 m3
31,100 m3
5. Determine the total quantity of sludge produced as follows:
Px =
0.60
Px =
Y Si - S e
1 + kd
+XL
Q
(7.217)
c overall
g VSS
g TSS
mg
mg
200
3.9
g BOD 5 0.75g VSS
L
L
-1
1+ 0.041d 19.5d
Px = 5590
+ 60
mg
L
kg
d
6. Compute the new overall MCRT using Equation (7.218).
c overall
=
79
Vtotal
Px
X
38, 000
m3
d
kg
103 g
c overall
=
31,100 m3 3500 mg L
kg
5590
d
19.5d 19.5d
% Difference
19.5d
kg
103 g
100%
0.0% 5.0%OK
19.5d
Since the difference between the original and calculated value for the overall MCRT is within 5
percent, there it is not necessary to repeat the procedure
Oxygen Requirements
Oxygen is required to meet both the carbonaceous and nitrogenous demand. The total kilograms
of oxygen required can be estimated using the following equations:
O 2 = CBOD + NOD
CBOD =
Q ( 1-1.42 Y ) S i - S e
CBOD = 38, 000
1.42 0.041d -1
DOC
m3
d
1 1.42
(7.219)
+ 1.42 k d
0.6 g VSS
g BOD5
200 mg
L
X Voxic1
3.9 mg
L
kg
103 g
2625 mg
kg
21,100 m3
L
103 g
CBOD =4320 kg d
NOD = Q 4.57 NO
NOD =
38, 000 m3
4.57
d
DOC = Q 2.86
DOC =
O2 = CBOD + NOD
21.5mg
L
gO 2
NO
3
g NO - N
gO2
38,000 m3
2.86
d
g NO3- - N
8.34lb
MG mg L
kg
103 g
NO3
(7.221)
3730 kg
d
e
kg
103 g
(7.222)
mg
mg
kg
kg
4.3
1870
3
L
L
10 g
d
4320 kg 3730 kg 1870kg
kg
DOC =
9920
d
d
d
d
21.5
Alkalinity Requirements
The effluent alkalinity can be calculated as follows from the 4-stage anoxic/oxic process:
ALK e = ALKo - 7.14 NO + 3.57 NO - NO3 - N
80
e
(7.223)
ALK e = 225
mg
mg
mg
mg
- 7.14 21.5
+ 3.57 21.5
- 4.3
L
L
L
L
133
mg
L
32. A 4-stage Bardenpho process is to be designed to meet a 5/5/3/ mg/L effluent standard for
BOD5, TSS, and TN, respectively. The influent wastewater characteristics, mixed liquor
parameters, and biokinetic coefficients are to be used in design are presented in the tables below.
Assume that the influent ammonium concentration is equal to the TKN concentration, RAS flow
= 1Q and MLR flow = 3Q, DO in the RAS and MLR recycle is 0.5 mg/L and 1.0 mg/L entering
the second anoxic zone.
Wastewater characterization
Influent
Mixed liquor
Q=
38,000 m3/d
DO =
2.0 mg/L
pH =
7.1
MLSS =
3,500 mg/L
Minimum
temperature
15°C
MLVSS =
0.75 MLSS
TKN =
30 mg/L
KDO =
0.5 mg/L
Alkalinity =
200 mg/L as CaCO3
MLVSS =
2,625 mg/L
BOD5 =
200 mg/L
TSS =
167 mg/L
Inerts (XL) =
50 mg/L
Heterotrophic biokinetic coefficients at 20°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.06 d-1
k=
5 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 20°C
Coefficient
YNS =
Coefficient
0.15 g VSS/g NH+4 -N
(kd)NS =
81
0.05 d-1
3 d-1
kNS =
(Ks)NS =
0.74 mg/L NH+4 -N
Heterotrophic biokinetic coefficients at 15°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.05 d-1
k=
3.6 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 15°C
Coefficient
Coefficient
YNS =
0.15 g VSS/g NH+4 -N
(kd)NS =
0.04 d-1
kNS =
2.1 d-1
(Ks)NS =
0.57 mg/L NH+4 -N
A temperature correction coefficient ( ) of 1.04 should be used for correcting both the
heterotrophic and Nitrosomonas kd values, whereas a of 1.07 should be used for correcting both
the heterotrophic and Nitrosomonas k values. The heterotrophic Ks value is to be corrected with
a of 1.00 and the Nitrosomonas Ks is corrected with a value of 1.053. All temperature
correction coefficients were obtained from Metcalf and Eddy (2003). Use a safety factor of 1.5
and peaking factor of 1.2 and assume that Fn =0.12.
Solution
First Oxic Zone Calculations
21. Calculate the maximum growth rate of Nitrosomonas for the ambient temperature of
DO
1-0.833 7.2 - pH
15°C.
(7.192)
MAX NS = YNS k NS
K DO + DO
MAX
NS
=
0.15gVSS
gNH+4 -N
2.1d-1
2 mg L
0.5mg L +2 mg L
1-0.833 7.2 - 7.1
22. Correct the Nitrosomonas decay coefficient for temperature.
kd
NS T C
kd
NS 15 C
kd
NS 20 C
0.05d -1
1.04
NS 20 C
T 20 C
1.04
15
20 C
0.04 d -1
23. Calculate the minimum MCRT required for nitrification.
82
0.23d
1
(
1
c
1
=
c
min
c
min
=
)
max NS
NH 4
K NS + NH 4
min
N
N
0.23d -1 30 mg L
0.57 mg L + 30 mg L
0
- kd
NS
0
- 0.04d -1
1
= 5.3d
0.19 d -1
24. Calculate the design MCRT based on a safety factor of 1.5 and peaking factor of 1.2 as
follows:
c design =
c min SF PF
c design
= 5.3d 1.5 1.2
9.5d
25. Estimate the overall MCRT by multiplying the design MCRT by a factor ranging from
1.8 to 2.0, assume a value of 2.0.
c overall =
c design 1.8 to 2.0
c overall
= 9.5d 2.0
19.0d
26. For a given set of heterotrophic biokinetic coefficients, calculate the effluent soluble
substrate concentration (Se).
K s 1 + kd c overall
Se =
(7.197)
Yk - kd - 1
c overall
mg
1 + 0.05d -1 19 d
L
g VSS
0.6
3.6 d -1 - 0.05d -1
g BOD5
60
Se =
19 d
3.0
-1
mg
L
27. For a given set of autotrophic biokinetic coefficients, calculate the effluent
ammonium nitrogen concentration using the following equation:
+
NH 4 - N e =
K NS 1 + kd
NS
c design
YNS k NS - kd
c design
NS
(7.198)
-1
mg
1 + 0.04 d -1 9.5d
L
g VSS
0.15
2.1d -1 - 0.04 d -1
g NH +4 -N
0.57
+
NH 4 - N e =
9.5d
+
NH 4 - N e = 0.49
mg
L
83
-1
28. The nitrogen utilized in synthesis neglecting the small amount of nitrogen synthesized by
the nitrifiers is calculated using Equation and assuming FN = 0.12:
Y Si Se FN
N syn =
+ X e FN
(7.199)
1+ kd c overall
0.60
N syn =
g VSS
mg
mg
200
3.0
0.12
g BOD5
L
L
mg TSS
+ 5
-1
L
1+ 0.05d 19d
N syn = 7.7
0.75
mg VSS
0.12
mg TSS
mg N
L
29. Determine the amount of nitrogen to be oxidized using Equation (7.200).
NO = TKN 0 - NH 4 N - N syn
e
mg
mg
mg
mg
0.49
7.7
21.8
L
L
L
L
30. Calculate the volume of the first oxic zone necessary to achieve nitrification for a given
temperature, pH, and DO as follows:
Q c design
Y Si - S e
Voxic1 =
+ XL
(7.201)
X
1 + kd c design
NO = 30
m3
9.5d
d
mg
3500
L
38, 000
Voxic1 =
0.6
mgVSS
g TSS
mg
mg
200
- 3.0
mg BOD5 0.75 gVSS
L
L
-1
1 + 0.05d 9.5d
+ 50
mg
L
Voxic1 = 16, 200 m3
First-Stage Anoxic Zone Calculations
11. Estimate the nitrate nitrogen concentration in the mixed liquor recycle returned to the
first stage anoxic zone using the following equation:
NO(Q)
(7.202)
N=
(Qmlr +Qr +Q)
mg
21.8
L
N=
m3
38, 000
d
3
38, 000
m
d
3
3 38, 000
m
d
3
38, 000
m
d
4.4
mg
L
12. Estimate the nitrate equivalence of dissolved oxygen in the mixed liquor recycle as
follows:
84
NO3
NO3
N
eq 1
= DO
mlr
mg
N = 0.5
eq 1
L
NO3
N
eq1
0.35
g NO3 - N
g O2
m3
38,000
3
d
g NO3 - N
0.35
g O2
= 20.0
1 kg
1000 g
Qmlr
(7.203)
1 kg
1000 g
kg
d
13. Calculate the mass of nitrates to be removed in the first stage anoxic zone using the
following equation and assuming that the nitrate nitrogen concentration in the effluent is
1.0 mg/L as N:
1 kg
(7.204)
NOR1 = Qr NO3 e + Qmlr (N)
1000 g
m3 mg
m3
mg
1
+ 38,000
3 4.4
d
L
d
L
kg
NOR1 = 540
d
1 kg
1000 g
NOR1 = 38,000
14. The total mass of nitrates to be removed in the first stage anoxic zone is calculated as
follows:
(7.205)
TNOR1 = NOR1 + NO3 N
eq1
kg
kg
kg
+ 20.0
560
d
d
d
15. The volume of the 1st-stage anoxic zone can be determined using Equation (7.209).
TNOR1 = 540
Vanoxic1
TNOR1 1000 g kg
0.03 Q Si 1.06
0.029 X 1.06
Vanoxic1
0.029 3500
Vanoxic1
T 20
0.03 38, 000 m3 d
560 kg d 1000 g kg
mg
L
T 20
1.06
200
mg
1.06
L
15 20 C
15 20 C
5140 m3
Second-Stage Anoxic Zone Calculations
1. Calculate the quantity of nitrates to be removed in the 2nd-stage anoxic zone using
Equation (7.210). Assume an effluent nitrate concentration of 1.0 mg/L.
NOR2 =
Q NO - NO3 - N
1000 g/kg
85
e
- NOR1
(7.210)
m3
d
38, 000
NOR2 =
mg
mg
- 1.0
kg
L
L
- 540
1000 g/kg
d
21.8
250
kg
d
2. The nitrate equivalence of dissolved oxygen in the mixed liquor from the 1st-stage oxic
zone (1.0 mg/L) is calculated as:
g NO3 - N
1 kg
NO3 N = Q + Qr DOml 0.35
eq 2
g O2
1000 g
NO3
N
eq 2
NO3
= 38,000
N
eq 2
m3
m3
+ 38,000
d
d
= 27
1.0
mg
L
0.35
g NO3 - N
g O2
1 kg
1000 g
kg
d
3. The total mass of nitrates to be removed in the 2nd-stage anoxic zone is calculated using
Equation (7.212).
TNOR2 = NOR2 + NO3 N
eq2
kg
kg
kg
+ 27
277
d
d
d
4. The specific denitrification rate in the 2nd-stage anoxic zone is calculated using Equation
(7.213) corrected for ambient temperature (Burdick et al., 1982):
0.706
SDNR2 = 0.12( c ) -overall
[1.02] T - 20
TNOR2 = 250
SDNR2 = 0.12 19d
-0.706
[1.02]15- 20
0.0136d -1
5. The volume of the 2nd-stage anoxic zone is calculated as:
Vanoxic2 =
Vanoxic2 =
TNOR2 (1000 g/kg)
X SDNR2
(7.214)
277 kg/d (1000 g/kg)
mg
3500
0.0136 d -1
L
5820m3
Second Oxic Zone or Re-aeration Zone Calculations
The 2nd-stage oxic zone volume is calculated as follows assuming a detention time of 45
minutes:
Voxic2 =
1 hr
60 min
86
1d
24 hr
(Q)
(7.215)
1 hr
60 min
Voxic2 = 45 min
1d
24 hr
38,000
m3
d
1190 m3
Check the Overall Design
7. First, calculate the total volume of the biological system using the following equation:
Vtotal = Vanoxic1
Voxic1
Vanoxic2
Voxic2
Vtotal = 5140m3 16, 200m3 5820m3 1190m3
28,350m3
8. Determine the total quantity of sludge produced as follows:
Px =
0.60
Px =
Y Si - S e
1 + kd
+XL
1 kg
1000 g
Q
c overall
g VSS
g TSS
mg
mg
200
- 3.0
g BOD 5 0.75g VSS
L
L
-1
1+ 0.05d 19 d
Px = 4970
+ 50
(7.217)
mg
L
38, 000
m3
d
1 kg
1000 g
kg
d
9. Compute the new overall MCRT using Equation (7.218).
c overall
c overall
=
=
Vtotal
X
Px (1000 g/kg)
28,350 m3 3500 g m3
% Difference
kg
4970
d
g
1000
kg
20.0d 19.0d
100%
20.0d
20.0 d
5% OK
Since the difference between the original and calculated value for the overall MCRT is within 5
percent, there it is not necessary to repeat the procedure
Oxygen Requirements
Oxygen is required to meet both the carbonaceous and nitrogenous demand. The total kilograms
of oxygen required can be estimated using the following equations:
87
(7.219)
O 2 = CBOD + NOD - DOC
CBOD =
Q ( 1-1.42 Y ) S i - S e
CBOD = 38,000
1.42 0.05d -1
+ 1.42 k d
1 kg
1000 g
m3
g VSS
mg
mg
(1 1.42 0.6
) 200
3.0
d
g BOD5
L
L
2625
mg
16,200 m3
L
m3
NOD = 38,000
4.57
d
DOC = Q 2.86
21.8
mg
L
gO 2
g NO - N
gO 2
m3
2.86
d
g NO3- - N
1 kg
1000 g
1 kg
1000 g
NO - NO3
3
21.8
O2 = CBOD + NOD - DOC = 4130
1 kg
1000 g
1 kg
1000 g
NOD = Q 4.57 NO
DOC = 38,000
X Voxic1
(7.220)
3790
e
mg
mg
- 1.0
L
L
kg
gd
1 kg
1000 g
(7.222)
1 kg
1000 g
kg
kg
kg
3790
2260
d
d
d
2260
5660
kg
d
kg
d
Alkalinity Requirements
The effluent alkalinity can be calculated as follows from the 4-stage anoxic/oxic process:
ALK e = ALKo - 7.14 NO + 3.57 NO - NO3 - N
ALK e = 200
(7.223)
e
mg
mg
mg
mg
- 7.14 21.8
+ 3.57 21.8
- 1.0
L
L
L
L
119
mg
L
33. A 4-stage Bardenpho process is to be designed to meet a 5/5/3/ mg/L effluent standard for
BOD5, TSS, and TN, respectively. The influent wastewater characteristics, mixed liquor
parameters, and biokinetic coefficients to be used in design are presented in the tables below.
Assume that the influent ammonium concentration is equal to the TKN concentration, RAS flow
= 1Q and MLR flow = 4Q, DO in the RAS and MLR recycle is 1.0 mg/L and 1.0 mg/L entering
the second anoxic zone.
Wastewater characterization
Influent
Q=
Mixed liquor
56,775 m3/d
DO =
88
2.0 mg/L
pH =
7.2
MLSS =
3,500 mg/L
Minimum
temperature
18°C
MLVSS =
0.70 MLSS
TKN =
30 mg/L
KDO =
0.5 mg/L
Alkalinity =
250 mg/L as CaCO3
MLVSS =
2,625 mg/L
BOD5 =
200 mg/L
TSS =
167 mg/L
Inerts (XL) =
50 mg/L
Heterotrophic biokinetic coefficients at 20°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.06 d-1
k=
5.0 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 20°C
Coefficient
Coefficient
YNS =
0.15 g VSS/g NH+4 -N
(kd)NS =
0.05 d-1
kNS =
3.0 d-1
(Ks)NS =
0.74 mg/L NH+4 -N
Heterotrophic biokinetic coefficients at 18°C
Coefficient
Coefficient
Y=
0.6 g VSS/g BOD5
kd =
0.055 d-1
k=
4.4 d-1
Ks =
60 mg/L BOD5
Nitrosomonas biokinetic coefficients at 18°C
Coefficient
YNS =
Coefficient
0.15 g VSS/g NH+4 -N
(kd)NS =
89
0.046 d-1
2.6 d-1
kNS =
(Ks)NS =
0.67 mg/L NH+4 -N
A temperature correction coefficient ( ) of 1.04 should be used for correcting both the
heterotrophic and Nitrosomonas kd values, whereas a of 1.07 should be used for correcting both
the heterotrophic and Nitrosomonas k values. The heterotrophic Ks value is corrected with a of
1.00 and the Nitrosomonas Ks is corrected with a value of 1.053. All temperature correction
coefficients were obtained from Metcalf and Eddy (2003). Use a safety factor of 2.5 and peaking
factor of 1.2 and assume that Fn = 0.10.
Solution
First Oxic Zone Calculations
1. Calculate the maximum growth rate of Nitrosomonas for the ambient temperature of
DO
18°C.
(7.192)
1-0.833 7.2 - pH
MAX NS = YNS k NS
K DO + DO
MAX
=
NS
0.15gVSS
gNH +4 -N
2.6d-1
2 mg L
0.5mg L +2 mg L
1-0.833 7.2 - 7.2
0.31d
1
2. Correct the Nitrosomonas decay coefficient for temperature.
kd
NS T C
kd
kd
NS 18 C
NS 20 C
0.05d -1
1.04
T 20 C
1.04
NS 20 C
18
20 C
0.046 d -1
3. Calculate the minimum MCRT required for nitrification.
( max )NS NH 4 N
1
0
=
- kd NS
K NS + NH 4 N
c min
(7.193)
0
1
=
c
min
c
min
0.31d
-1
30 mg L
0.67 mg L + 30 mg L
- 0.046d -1
1
= 3.8d
0.26 d -1
4. Calculate the design MCRT based on a safety factor of 2.5 and peaking factor of 1.2 as
follows:
(7.194)
c design =
c min SF PF
c design
= 3.8d 2.5 1.2
90
11.4d
5. Estimate the overall MCRT by multiplying the design MCRT by a factor ranging from
1.8 to 2.0, assume a value of 2.0.
c overall =
c design 1.8 to 2.0
c overall
= 11.4d 2.0
22.8d
6. For a given set of heterotrophic biokinetic coefficients, calculate the effluent soluble
substrate concentration (Se).
K s 1 + kd c overall
Se =
(7.197)
Yk - kd - 1
c overall
60
Se =
22.8d
mg
1 + 0.055d -1 22.8d
L
g VSS
0.6
4.4 d -1 - 0.055d -1
g BOD5
2.3
-1
mg
L
7. For a given set of autotrophic biokinetic coefficients, calculate the effluent
ammonium nitrogen concentration using the following equation:
K NS 1 + kd
+
NH 4 - N e =
NS
c design
YNS k NS - kd
c design
NS
(7.198)
-1
mg
1 + 0.046 d -1 11.4 d
L
g VSS
0.15
2.6 d -1 - 0.046 d -1
+
g NH 4 -N
0.67
+
NH 4 - N e =
11.4 d
+
NH 4 - N e = 0.35
-1
mg
L
8. The nitrogen utilized in synthesis neglecting the small amount of nitrogen synthesized by
the nitrifiers is calculated using Equation and assuming FN = 0.10:
Y Si Se FN
N syn =
+ X e FN
(7.194)
1+ kd c overall
0.60
N syn =
g VSS
mg
mg
200
2.3
0.10
g BOD5
L
L
mg TSS
+ 5
-1
L
1+ 0.055d 22.8d
N syn = 5.6
0.70
mg N
L
9. Determine the amount of nitrogen to be oxidized using Equation (7.200).
91
mg VSS
0.10
mg TSS
NO = TKN 0 - NH 4
N
e
- N syn
(7.200)
mg
mg
mg
mg
0.35
5.6
24.1
L
L
L
L
10. Calculate the volume of the first oxic zone necessary to achieve nitrification for a given
temperature, pH, and DO as follows:
Q c design
Y Si - S e
Voxic1 =
+ XL
(7.201)
X
1 + kd c design
NO = 30
m3
56, 775
11.4 d
d
Voxic1 =
mg
3500
L
0.6
mgVSS
g TSS
mg
mg
200
- 2.3
mg BOD5 0.70 gVSS
L
L
-1
1 + 0.055d 11.4 d
+ 50
mg
L
Voxic1 = 28,500 m3
First-Stage Anoxic Zone Calculations
16. Estimate the nitrate nitrogen concentration in the mixed liquor recycle returned to the
first stage anoxic zone using the following equation:
NO(Q)
(7.202)
N=
(Qmlr +Qr +Q)
24.1
N=
mg
L
56, 775
3
56, 775
m
d
m3
d
3
4
56, 775
m
d
3
56, 775
m
d
4.0
mg
L
17. Estimate the nitrate equivalence of dissolved oxygen in the mixed liquor recycle as
follows:
g NO3 - N
1 kg
(7.203)
NO3 N = DO mlr 0.35
Qmlr
eq 1
g O2
1000 g
NO3
mg
N = 1.0
eq 1
L
NO3
N
eq1
g NO3 - N
0.35
g O2
= 80.0
m3
56,775
4
d
1 kg
1000 g
kg
d
18. Calculate the mass of nitrates to be removed in the first stage anoxic zone using the
following equation and assuming that the nitrate nitrogen concentration in the effluent is
1.0 mg/L as N:
92
NOR1 = Qr NO3
e
1 kg
1000 g
+ Qmlr (N)
(7.204)
m3 mg
m3
mg
1
+ 56,775
4 4.0
d
L
d
L
kg
NOR1 = 965
d
1 kg
1000 g
NOR1 = 56,775
19. The total mass of nitrates to be removed in the first stage anoxic zone is calculated as
follows:
(7.205)
TNOR1 = NOR1 + NO3 N
eq1
kg
kg
kg
+ 80.0
1050
d
d
d
20. The volume of the 1st-stage anoxic zone can be determined using Equation (7.209) .
TNOR1 = 965
TNOR1 1000 g kg
Vanoxic1
0.03 Q Si 1.06
0.029 X 1.06
Vanoxic1
T 20
0.03 56, 775 m3 d
1050 kg d 1000 g kg
mg
1.06
L
8270 m3
0.029 3500
Vanoxic1
T 20
200
mg
1.06
L
18 20 C
18 20 C
Second-Stage Anoxic Zone Calculations
6. Calculate the quantity of nitrates to be removed in the 2nd-stage anoxic zone using
Equation (7.210). Assume an effluent nitrate concentration of 1.0 mg/L.
NOR2 =
Q NO - NO3 - N
1000 g/kg
56, 775
NOR2 =
m3
d
e
- NOR1
mg
mg
- 1.0
kg
L
L
- 965
1000 g/kg
d
24.1
347
kg
d
7. The nitrate equivalence of dissolved oxygen in the mixed liquor from the 1st-stage oxic
zone (1.0 mg/L) is calculated as:
g NO3 - N
1 kg
NO3 N = Q + Qr DOml 0.35
(7.211)
eq 2
g O2
1000 g
NO3
N
NO3
m3
m3
+ 56,775
eq 2
d
d
kg
N = 40
eq 2
d
= 56,775
93
1.0
mg
L
0.35
g NO3 - N
g O2
1 kg
1000 g
8. The total mass of nitrates to be removed in the 2nd-stage anoxic zone is calculated using
Equation (7.212).
TNOR2 = NOR2 + NO3 N
eq2
kg
kg
kg
+ 40
387
d
d
d
9. The specific denitrification rate in the 2nd-stage anoxic zone is calculated using Equation
(7.213) corrected for ambient temperature (Burdick et al., 1982):
0.706
SDNR2 = 0.12( c ) -overall
[1.02] T - 20
TNOR2 = 347
-0.706
SDNR2 = 0.12 22.8d
[1.02]18- 20
0.01268d-1
10. The volume of the 2nd-stage anoxic zone is calculated as:
TNOR2 (1000 g/kg)
Vanoxic2 =
X SDNR2
Vanoxic2 =
387 kg/d (1000 g/kg)
mg
3500
0.01268d -1
L
(7.214)
8720m3
Second Oxic Zone or Re-aeration Zone Calculations
The 2nd-stage oxic zone volume is calculated as follows assuming a detention time of 45
minutes:
Voxic2 =
Voxic2 = 45 min
1 hr
60 min
1 hr
60 min
1d
24 hr
1d
24 hr
(7.216)
(Q)
56,775
m3
d
1770 m3
Check the Overall Design
10. First, calculate the total volume of the biological system using the following equation:
(7.216)
Vtotal = Vanoxic1 Voxic1 Vanoxic2 Voxic2
Vtotal = 8270 m3 28,500 m3 8720 m3 1770 m3
47, 260 m3
11. Determine the total quantity of sludge produced as follows:
Px =
Y Si - S e
1 + kd
+XL
Q
c overall
94
1 kg
1000 g
(7.217)
0.60
Px =
g VSS
g TSS
mg
mg
200
- 2.3
g BOD 5 0.70 g VSS
L
L
-1
1+ 0.055d 22.8d
Px = 7110
+ 50
mg
L
56, 775
m3
d
1 kg
1000 g
kg
d
12. Compute the new overall MCRT using Equation (7.218).
c overall
c overall
% Difference
=
Vtotal
X
Px (1000 g/kg)
47, 260 m3 3500 g m3
=
kg
7110
d
23.3d 22.8d
23.3d
g
1000
kg
100%
2.2% 5% OK
23.3d
Since the difference between the original and calculated value for the overall MCRT is within 5
percent, there it is not necessary to repeat the procedure
Oxygen Requirements
Oxygen is required to meet both the carbonaceous and nitrogenous demand. The total kilograms
of oxygen required can be estimated using the following equations:
O 2 = CBOD + NOD - DOC
CBOD =
Q ( 1-1.42 Y ) S i - S e
CBOD = 56,775
1.42 0.055d -1
(7.219)
+ 1.42 k d
X Voxic1
1 kg
(7.220)
1000 g
m3
g VSS
mg
mg
(1 1.42 0.6
) 200
2.3
d
g BOD5
L
L
2450
mg
28,500 m3
L
1 kg
1000 g
NOD = Q 4.57 NO
NOD = 56,775
m3
4.57
d
DOC = Q 2.86
24.1
1 kg
+
1000 g
mg
L
gO 2
1 kg
1000 g
1 kg
1000 g
NO - NO3
3
g NO - N
95
(7.221)
6250
e
kg
gd
1 kg
1000 g
(7.222)
gO 2
m3
2.86
d
g NO3- - N
mg
mg
1 kg
kg
- 1.0
3750
L
L
1000 g
d
kg
kg
kg
kg
O2 = CBOD + NOD - DOC = 7,110
6250
3750
9, 610
d
d
d
d
Alkalinity Requirements
DOC = 56,775
24.1
The effluent alkalinity can be calculated as follows from the 4-stage anoxic/oxic process:
ALK e = ALKo - 7.14 NO + 3.57 NO - NO3 - N
ALK e = 250
(7.223)
e
mg
mg
mg
mg
- 7.14 24.1
+ 3.57 24.1
- 1.0
L
L
L
L
160
mg
L
34. A batch settling column analysis on sludge from an extended aeration activated sludge
process is presented below. Determine the diameter (ft) of two secondary clarifiers operating in
parallel if the design flow (Q) is 10 MGD, return activated sludge (RAS) flow (Qr) is 10 MGD,
the MLSS concentration is 3,000 mg/L, and the underflow solids concentration is 12%.
SS
(mg/L)
1,000
3,000
4,500
5,500
7,500
10,000
12,500
15,000
V
(fph)
15.00
10.50
7.00
5.00
1.60
0.50
0.25
0.13
Solution Part A
Pot solids flux versus solids concentration.
96
2.50
2.00
Solids Flux, lb/(h-ft2)
GL = 1.70
1.50
1.00
0.50
0.00
0
2000
4000
6000
8000
10000
12000
14000
16000
Solids Concentration, mg/L
At a underflow solids concentration of 12000 mg/L, the limiting solids flux is 1.70 lb/(hft2).
Solution Part B
A
Q Qr MLSS
GL
10 10 MGD 3000 mg/L 8.34
A
1.7
12, 265ft 2
2
D 88 90 ft
A
6312ft 2
lb
h ft 2
D2
4
lb
MG mg L
1d
24 h
35. A batch settling column analysis on sludge from an oxidation ditch is presented below.
Determine the diameter (m) of two secondary clarifiers operating in parallel if the total flow (Q
+Qr ) to the clarifiers is 56,775 m3/d at a MLSS concentration of 3,500 g/m3, and the underflow
solids concentration is 10,000 g/m3.
97
SS
(g/m3)
1,000
3,000
4,500
5,500
7,500
10,000
12,500
V
(mph)
2.5
1.5
0.6
0.25
0.11
0.05
0.03
Solution Part A
Pot solids flux versus solids concentration.
5.00
4.50
4.00
Solids Flux, kg/m2-h
3.50
3.00
GL = 2.8
2.50
2.00
1.50
1.00
0.50
0.00
0
2000
4000
6000
8000
10000
12000
14000
Solids Concentration, g/m3
At a underflow solids concentration of 10,000 g/m3, the limiting solids flux is 2.80 kg/(hm2).
Solution Part B
98
A
Q Qr MLSS
GL
56, 775
A
m3
d
3500 g/m3
kg
2.8
h m2
2956 m 2
A
2
D
43.4 m
1478m
1d
24 h
1kg
1000 g
D2
4
2
36. A circular secondary clarifier processes a total wastewater flow (Q +Qr) at 7,570 m3/d at a
MLSS concentration of 2,800 g/m3. Given the following batch settling column analysis and a
limiting solids flux of 7.0 kg/(m2h) determine the following:
a) Solids underflow concentration (g/m3)
b) Diameter of the secondary clarifiers (m)
c) Clarifier overflow rate (m/h)
SS
(g/m3)
1,500
2,600
3,950
5,420
6,950
9,000
12,000
V
(m/h)
5.00
3.25
2.00
1.00
0.55
0.25
0.13
Solution Part A
Pot solids flux versus solids concentration.
99
9.00
8.00
Solids Flux, kg/h-m2
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.00
0
2000
4000
6000
8000
10000
12000
14000
Solids Concentration, g/m3
The underflow solids concentration at a limiting solids flux rate of 7.0 kg/m2h is equal to
13,200 g/m3. See the solids flux plot above.
Solution Part B
Q Qr MLSS
GL
A
7,570
A
m3
d
7.0
A 126m 2
D
2800 g/m3
kg
h m2
D2
4
1d
24 h
1kg
1000 g
12.7 m
Solution Part C
Q Qr MLSS Qr X r
7570
Qr
m3
g
2800 3
d
m
1606
Qr 13, 200
g
m3
m3
d
100
m3
Q Qr
7570
d
3
m
m3
Q 7570
1606
d
d
V0
Q
A
5964 m3 d
12.7 m
4
2
5964
47.1
m3
d
m 1d
d 24 h
1.96
m
h
37. A batch settling column analysis on sludge from a complete-mix activated sludge process is
presented below. A circular secondary clarifier processes a total flow of 2.0 MGD (Q +Qr) at a
MLSS concentration of 3,200 mg/L. Determine the following assuming a limiting solids flux
value of 1.2 lb/(ft2h):
a) Solids concentration in the underflow (mg/L)
b) Underflow rate (gpd)
c) Diameter of the secondary clarifiers (ft)
d) Overflow rate (gpd/ft2)
SS
(mg/L)
1,000
1,200
2,000
3,800
6,100
8,200
10,000
11,000
V
(fph)
19.5
19
12
3.5
1.1
0.5
0.3
0.2
Solution Part A
Pot solids flux versus solids concentration.
101
1.60
1.40
Solids Flux, lb/h-ft2)
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0
2000
4000
80009200 mg/L10000
6000
12000
Solids Concentration, mg/L
The underflow solids concentration at a limiting solids flux of 1.2 lb/ft2h is equal to 9,200 mg/L.
See the solids flux plot above.
Solution Part B
Q Qr MLSS Qr X r
2.0 MGD 3200
Qr
mg
L
Qr 9, 200
mg
L
0.69 MGD
Solution Part C
A
Q Qr MLSS
GL
2.0MGD 3200 mg/L
A
1.2
A 1853ft 2
D
48.6
8.34 lb
MG mg L
lb
h ft 2
D2
4
50 ft
102
1d
24 h
Solution Part C
V0
Q
A
2.0 MGD 0.69 MGD
50 ft
4
106gal
MG
667
2
gpd
ft 2
38. Two circular secondary clarifiers are to be designed to treat an average daily flow (ADF) of
10.0 MGD from a CMAS process. The peak:average flow ratio is 2.5. The MLSS at ADF is
3,500 mg/L and the return activated sludge (RAS) flow is 6.0 MGD. During the peak flow
condition, the MLSS concentration will be lowered to 2,500 mg/L and the RAS flow increased to
10.2 MGD. Use overflow rates of 600 and 1,500 gpd/ft2 at average and peak flows. Solids
loading rates of 25 and 50 ppd/ft2 are to be used at average and peak loading conditions,
respectively. Determine the diameter of the clarifiers (ft).
Determine clarifier area based on overflow rate.
As
Q
V0
As
Q
V0
10 106 gpd
600 gpd ft 2
1.67 104 ft 2
2.5 10 106 gpd
1500 gpd ft 2
1.67 104 ft 2
Next, determine clarifier area based on solids loading rate.
As
Q Qr MLSS
SLR
As
10
6 MGD 3500
mg
L
25lb d ft
2
2.5 10 10.2 MGD
Q Qr MLSS
SLR
4
8.34 lb
MG mg L
2500
50 lb d ft
2
mg
L
1.87 104 ft 2
8.34 lb
MG mg L
2
The largest area is 1.8710 ft ; determine the diameter as follows.
1.87 104 ft 2
2
D 109 ft 110 ft
As
9.35 103 ft 2
103
D2
4
1.47 104 ft 2
39. Two square secondary clarifiers will be used to treat an average daily flow (ADF) of 5,000
m3/d from a high-purity oxygen activated sludge process with a MLSS concentration of 6,000
mg/L. At peak flow (10,000 m3/d), the MLSS concentration is reduced to 4,500 g/m3. The RAS
flow at average and peak flows are 3,000 and 6,000 m3/d, respectively. Additional design
criteria include overflow rates of 30 and 70 m3/(dm2) and solids loading rates of 100 and 240
kg/(dm2), respectively at average and peak loading conditions. Determine the diameter (m) of
the clarifiers.
Determine clarifier area based on overflow rate.
As
Q
V0
5000 m3 /d
30 m3 d m 2
167 m 2
As
Q
V0
10, 000m3 /d
70 m3 d m 2
143m 2
Next, determine clarifier area based on solids loading rate.
As
As
Q Qr MLSS
SLR
Q Qr MLSS
SLR
5000 3000
m3
d
6000
100 kg d m
10, 000 6000
m3
d
g
m3
1kg
1000 g
2
4500
240 kg d m
g
m3
1kg
1000 g
2
480 m 2
300 m 2
The largest area is 480 m2, determine the diameter as follows.
As
D
D 2 480 m 2
4
2
17.5 m
240 m 2
40. Two circular secondary clarifiers are designed to treat wastewater from a high-rate trickling
filter system. The average and peak flow rates are 10 and 25 MGD, respectively. A
recirculation flow of 2Q, based on average daily flow will be used at all times. The
overflow rates at average and peak flow will be 400 and 1,000 gpd/ft2, respectively.
Determine:
a) The diameter (ft) of the clarifiers if the total flow is applied to the secondary
clarifiers.
104
b) The diameter (ft) of the clarifiers when direct recirculation around the
clarifiers is used.
Solution Part A
As
As
Q
V0
Q
V0
106gal
10 20 MGD
MG
2
400 gpd ft
75, 000 ft 2
106gal
MG
2
1000 gpd ft
25 20 MGD
45, 000 ft 2
Clarifier diameter for Part A is determined below.
As
D
D 2 75, 000 ft 2
4
2
218.5 230 ft
37,500 ft 2
Solution Part B
For direct recirculation the areas are calculated as follows:
As
As
Q
V0
Q
V0
106gal
MG
400 gpd ft 2
10 MGD
106gal
MG
1000 gpd ft 2
25, 000 ft 2
25 MGD
D2
As
4
D 126.2
25, 000 ft 2
2
130 ft
25, 000 ft 2
12,500 ft 2
41. Two existing chlorine contact basins (CCBs) at a secondary WWTP have the following
dimensions: 7 ft70 ft15 ft. The average daily flow (ADF) and peak hour flow (PHF) to the
basins are 5 MGD and 10 MGD, respectively. Will the CCBs meet a detention time of 30
minutes at ADF or 15 minutes at PHF with one unit out of service? If not, what could be done?
105
Solution
The detention time at ADF with one basin in service is calculated below.
V
Q
7 ft×70ft×15ft
5 106 gpd
7.48gal
ft 3
24 h
d
60 min
= 15.8 min @ ADF
h
The detention time at ADF with two basins in service is calculated below.
2
V
Q
7 ft×70ft×15ft
5 106 gpd
7.48gal
ft 3
24 h
d
60 min
= 31.7 min @ ADF
h
The detention time at PHF with one basin in service is calculated below.
V
Q
7 ft×70ft×15ft
10 106 gpd
7.48gal
ft 3
24 h
d
60 min
= 7.9 min @ PHF
h
The detention time at PHF with two basins in service is calculated below.
2
V
Q
7 ft×70ft×15ft
10 106 gpd
7.48gal
ft 3
24 h
d
60 min
= 15.8 min @ PHF
h
To meet the detention time requirements, both basins must be in operation at all times. A third
CCB with the same dimensions of 7 ft70 ft15 ft should be constructed to provide reliability
and redundancy, especially if the wastewater flow rate increases.
42. Primary sludge contains 4% dry solids that are 75% volatile.
Determine:
a) The specific gravity (Ss) of the dry solids.
b) The specific gravity (Ssl) of the sludge.
c) The volume (ft3) of the thickened sludge, if 1,000 lb of dry solids are thickened
to 7% in a gravity belt thickener.
Solution Part A
Estimate the specific gravity of the dry solids knowing that 75% are volatile and 25% are fixed
using Equation (7.252). Assume that the specific gravity of the volatile (organic) and fixed
(mineral) solids are 1.0 and 2.5, respectively and the total mass of the dry solids is 100 kg.
Therefore, 0.70 100 = 70 kg of volatile solids and 30 kg of fixed solids. Substitute the
appropriate values into Equation (7.252).
106
Ms
Ss
100 lb
Ss
Ss
Mf
Mv
Sv
Sf
75lb
1.0
25lb
2.5
1.18
Solution Part B
Calculate the specific gravity of the sludge using Equation and the specific gravity of the dry
solids of 1.18. Assume that the total mass of the sludge is 100 lb; therefore, the mass of dry
solids is 100 lb 0.04 = 4 lb and the mass of water is 100 lb – 4 lb = 96 lb.
M sl
Ssl
Mw
Sw
100 lb
S sl
S sl
Ms
Ss
96 lb
1.0
(7.252)
4 lb
1.18
1.01
Solution Part C
Estimate the volume of the sludge after thickening to 6% solids (dry weight basis) by gravity belt
thickening. Use Equation (7.257). The density of water is equal to 12.4 lb/ft3 at 4ºC and is the
value generally used in sludge calculations.
Vsl
Vsl
Ms
S sl Ps
1000 lb
62.4 lb m
1.01
ft 3
(7.257)
227 ft 3
0.07
43. Waste activated sludge contains 1% dry solids that are 80% volatile.
Determine:
a) The specific gravity (Ss) of the dry solids.
b) The specific gravity (Ssl) of the sludge.
107
c) The volume (m3) of the thickened sludge if 60,000 kg of dry solids is thickened
to 6% in a gravity belt thickener.
Solution Part A
Estimate the specific gravity of the dry solids knowing that 80% are volatile and 20% are fixed
using Equation (7.252). Assume that the specific gravity of the volatile (organic) and fixed
(mineral) solids are 1.0 and 2.5, respectively and the total mass of the dry solids is 100 kg.
Therefore, 0.80 100 = 80 kg of volatile solids and 20 kg of fixed solids. Substitute the
appropriate values into Equation (7.252).
Ms
Ss
Mv
Sv
100 kg
Ss
Ss
Mf
Sf
80 kg
1.0
20 kg
2.5
1.14
Solution Part B
Calculate the specific gravity of the sludge using Equation (7.253) and the specific gravity of the
dry solids of 1.14. Assume that the total mass of the sludge is 100 kg; therefore, the mass of dry
solids is 100 kg 0.01 = 1 kg and the mass of water is 100 kg – 1 kg = 99 kg.
M sl
Ssl
Mw
Sw
100 kg
S sl
Ssl
Ms
Ss
(7.2453)
99 kg 1 kg
1
1.14
1.00
Solution Part C
Estimate the volume of the sludge after thickening to 6% solids (dry weight basis) by gravity belt
thickening. Use Equation (7.257). The density of water is equal to 1000 kg/m3 at 4ºC and is the
value generally used in sludge calculations.
Vsl
Ms
S sl Ps
(7.257)
108
Vsl
60, 000 kg
1000 kg
1.00
0.06
m3
1000 m3
44. Ten thousand gallons of primary sludge at 5% solids is blended with 5,500 gallons of WAS
at 1.2% solids.
Determine:
a) The solids concentration (%) of the blended sludge.
b) The final volume (ft3) of the blended sludge if it is thickened to 8% using a
thickening centrifuge. Assume the specific gravity of the blended sludge is 1.0
Solution A
C
10,000gal 5% + 5500gal 1.2%
10,000gal +5500gal
3.65% solids
Solution B
Ms
10,000gal 5×10,000
Ms
mg
L
5500gal 1.2×10,000
mg
L
8.34lbm
MG mg L
MG
106gal
4720 lb m
The volume is determined using Equation (7.257).
Vsl
Ms
S sl Ps
1.0
4720 lb m
62.4 lb m
0.08
ft 3
946 ft 3
45. Five thousand cubic meters of primary sludge at 6% solids is blended with 3,500 m3 of WAS
at 0.8% solids.
Determine:
a) The solids concentration (%) of the blended sludge.
b) The final volume (m3) of the blended sludge if it is thickened to 7% using a
GBT. Assume the specific gravity of the blended sludge is 1.03.
Solution A
C
5000 m3 6% + 3500m3
0.8%
5000 m3 +3500 m3
Solution B
109
3.86% solids
Ms
5000 m3 6×10,000
Ms
g
m3
3500 m3 0.8×10,000
g
m3
kg
1000g
3.28 105 kg
From Equation (7.257), the volume is determined as follows:
Vsl
3.28 105 kg
1000 kg
1.03
0.07
m3
Ms
S sl Ps
4550 m3
46. One thousand cubic meters per day of primary sludge at 3% solids concentration and
specific gravity of 1.03 is to be thickened to 5% solids in a gravity thickener. The solids
concentration in the supernatant is estimated to be 700 g/m3.
Determine:
a) The diameter of the thickener (m).
b) Thickened sludge flow (m3/d).
Solution A
Ms
m3
g
1000
3×10,000 3
d
m
1.03
kg
1000 g
Assume a solids loading rate of 120 kg/(m2d) from Table 7.25.
As
Ms
SLR
As
258 m 2
3.09 104 kg/d
kg
120 2
m d
D2
4
D 18.1m
Solution B
Qin
Qunderflow
Qsupernatant
Qsupernatant
1000
m3
d
1000
m3
d
Qunderflow
Perform materials balance on solids as follows:
110
258 m 2
3.09 104
kg
d
M s in
M s underflow
Qin Conc S
M s supernatant
Qunderflow Conc S Qsupernatant Conc S
S = specific gravity of species
3.09 104
kg
d
Qunderflow 5 10,000
3.09 104
kg
d
50 Qunderflow
0.7 Qsupernatant
3.09 104
kg
d
50 Qunderflow
0.7 1000
Qunderflow
Qsupernatant
1000
613
m3
d
g
m3
1kg
1000g
m3
d
Qsupernatant 700
g
m3
1kg
1000g
Qunderflow
m3
d
Qunderflow 1000
m3
d
613
m3
d
387
m3
d
47. Fifty thousand gallons per day of primary sludge at 3.5% solids concentration and specific
gravity of 1.05 is to be thickened to 5.5% solids in a gravity thickener. The solids concentration
in the supernatant is estimated to be 600 mg/L.
Determine:
a) The diameter of the thickener (ft).
b) Thickened sludge flow (gpd).
Solution A
Ms
50, 000
gal
mg
3.5×10,000
d
L
1.05
8.34lb
MG mg L
Assume a solids loading rate of 25 lb/(ft2d) from Table 7.25.
111
15,325
lb
d
As
Ms
SLR
As
613ft 2
D
28 30 ft
15,325 lb/d
25lb
ft 2 d
613ft 2
D2
4
Solution B
Qin
Qunderflow
Qsupernatant
Qsupernatant
0.05 MGD
0.05 MGD Qunderflow
Perform materials balance on solids as follows:
M s in
M s underflow
Qin Conc S
M s supernatant
Qunderflow Conc S Qsupernatant Conc S
S = specific gravity of species
15,325
lb
d
Qunderflow 5.5 10, 000
15,325
lb
d
Qsupernatant
15,325
lb
d
mg
L
8.34lb
MG mg L
4.59 105 Qunderflow
Qsupernatant 600
mg
L
8.34lb
MG mg L
5004 Qsupernatant
0.05MGD- Qunderflow
4.59 105 Qunderflow
5004 0.05MGD- Qunderflow
Qunderflow
0.033 MGD
Qsupernatant
0.05 MGD Qunderflow
0.05 MGD 0.033 MGD
0.017 MGD
48. A gravity belt thickener (GBT) is designed to thicken 10,000 pound per day of anaerobically
digested primary and WAS at a 3.0% solids concentration to 6% solids. The GBT will operate 5
days per week for 8 hours each day. The SS concentration in the filtrate is 900 mg/L. Use the
112
following design criteria to design the GBT: hydraulic loading rate = 100 gpm/m; solids loading
rate = 1000 lb/(m·h); and washwater rate = 25 gpm/meter of belt.
Solution
Calculate the daily solids loading rate to the GBT.
10,000
lb 7 d
d wk
1wk
5d
1d
lb
lb
= 1750
or 14,000
8h
h
d
Determine the sludge volume entering the GBT using Equation (7.257) and assuming the
specific gravity of the wet sludge (Ssl) is equal to 1.0.
Vsl
MS
S sl
=
Ps
14,000 lb/d
= 5.6 104 gal/d or 117 gpm
3
1.00 8.34 lb ft 0.03
Draw a schematic diagram so that a flow and materials balance can be performed.
Washwater
0 % Solids
Sludge
cake
Sludge
feed
GBT
3 % Solids
14,000 lb/d
6 % Solids
900 mg/L
Filtrate
Determine the belt width of the GBT based on hydraulic loading.
belt width
117 gpm
1.2 m
100 gpm m
Determine the belt width of the GBT based on solids loading.
belt width
1750 lb h
1000 lb
m h
1.75 m
2.0 m
113
Use two, 2.0-m belts for redundancy.
Determine the sludge cake and filtrate flows by performing a flow and materials balance around
the GBT. Cake and filtrate flows are denoted as QC and QF, respectively.
Qsludge
Qwashwater
QF
gpm
2.0m = 50gpm
m
Qwashwater
25
117 gpm
50 gpm
167 gpm
QF
167
QF
QC
8h
d
MG
106 gal
QC
g 60 min
min
h
0.0802 MGD
QF
QC
QF
QC
0.0802 MGD
QC
0.0802 MGD
Perform a materials balance on solids. Recall that a 1% solids concentration is equal to 10,000
mg/L assuming the specific gravity is equal to 1.0.
M sludge
MC
MF
14,000 lb/d QC 6 10,000
14,000 lb/d 5 105 QC
mg
L
8.34lb
MG mg L
QF
0.0802 MGD
900
mg
L
8.34lb
MG mg L
7510 QF
14, 000 lb/d 7510 0.0802 MGD QC
QC
QF
5.0 105 QC
0.0272 MGD or 57 gpm
QC
0.0802 MGD
0.0272 MGD
0.053MGD or 110 gpm Now,
calculate the solids in the filtrate.
0.053MGD 900
mg
L
8.34lb
MG mg L
398
lb
d
The percent capture through the GBT can be calculated using the following equation:
114
% Capture =
% Capture
solids in feed - solids in filtrate
solids in feed
14, 000 lb/d 398lb/d 100
14, 000 lb/d
100
(7.258)
97.2 %
A schematic diagram showing the complete materials balance is shown below:
Washwater
0 % Solids
50 gpm
Sludge
cake
Sludge
feed
GBT
1 % Solids
14,000 lb/d
117 gpm
6 % Solids
13,602 lb/d
57 gpm
900 mg/L
398 lb/d
110 gpm
Filtrate
49. A gravity belt thickener (GBT) is designed to thicken 600 m3/d of WAS at a 0.8% solids
concentration to 5% solids. The GBT will operate 6 days per week for 7 hours each day. The SS
concentration in the filtrate is 1,200 g/m3. Use the following design criteria to design the GBT:
hydraulic loading rate = 800 Lpm/m; solids loading rate = 200-600 kg/(m·h) ; and washwater
rate = 95 Lpm/meter of belt.
Solution
Calculate the daily solids loading rate to the GBT.
m3 7 d
600
d wk
1wk
6d
0.8 10,000
g
m3
1kg
kg
kg
= 5600
or 800
1000g
d
h
Draw a schematic diagram so that a flow and materials balance can be performed.
115
Washwater
0 % Solids
Sludge
cake
Sludge
feed
GBT
0.8 % Solids
5600 kg/d
5 % Solids
1200 g/m3
Filtrate
Determine the belt width of the GBT.
600
belt width
m3 7 d
d wk
1667 Lpm
800 Lpm m
1wk
6d
1000 L
m3
1d
7h
1h
60 min
1667 Lpm or 700
m3
d
Use two, 2.0-m GBTs, one for redundancy.
2.1 m
Check the actual solids loading rate (SLR) to see if it meets the design criteria.
solids loading rate
800 lb/h of solids
2.0 m
400
kg
m h
This is acceptable since the SLR design criteria ranges from 200 to 600 kg/(m h).
Determine the sludge cake and filtrate flows by performing a flow and materials balance around
the GBT. Cake and filtrate flows are denoted as QC and QF, respectively.
Qsludge
Qwashwater
Qwashwater
95
QF
Lpm
2.0 m = 190 Lpm
m
1667 Lpm 190 Lpm
1860 Lpm
1857
QC
QF
L 60 min
min
h
QF
QC
QC
7h
d
m3
1000 L
116
780
m3
d
780
m3
d
QC
780
QF
QC
m3
d
QF
Perform a materials balance on solids. Recall that a 1% solids concentration is equal to 10,000
mg/L assuming the specific gravity is equal to 1.0.
M sludge
MC
MF
5600 kg/d QC 5 10,000
5600 kg/d
50 QC
QF
m3
780
d
1kg
1000g
QC
50 QC
QF 1200
g
m3
1kg
1000g
1.2 QF
5600 kg/d 1.2 780
QC
g
m3
m3
d
m3
95.5
or
d
m3
96
d
m3
96
d
m3
684
d
Now, calculate the solids in the filtrate.
m3
g
684
1200 3
d
m
1kg
1000g
820
kg
d
The percent capture through the GBT can be calculated using Equation (7.258).
% Capture =
% Capture
solids in feed - solids in filtrate
solids in feed
5600 kg/d 820 kg/d 100
5600 kg/d
100
85.3 %
A schematic diagram showing the complete materials balance is shown below:
117
(7.258)
Washwater
0 % Solids
80 m3/d
Sludge
cake
Sludge
feed
GBT
m3/d
700
0.8% Solids
5600 kg/d
96 m3/d
5% Solids
4800 kg/d
684 m3/d
0.8% Solids
820 kg/d
Filtrate
50. 57,000 gallons of waste activated sludge at a 1.2 % solids concentration is to be processed
daily by an aerobic digester. The volatile solids content is 75% and the minimum design
operating temperature is 12ºC. A temperature correction coefficient ( ) of 1.06 will be used for
adjusting the digestion time for temperature variations. Assume an oxygen demand of 2.0
pounds of oxygen per pound of volatile solids destroyed and a 50% reduction in volatile solids.
Determine the following:
a)
b)
c)
d)
e)
Design hydraulic detention time ( ) or digestion time in (days).
Aerobic digester volume in (ft3).
Volatile solids loading rate to the aerobic digester in (lb/d ft3).
Quantity of oxygen required (lb/d) to stabilize the waste activated sludge.
Total air required (ft3/min) assuming 0.075 lb air/ft3, 0.23 lb O2/lb air and the diffusers
have a transfer efficiency of 6.0%.
Solution Part A
Select a hydraulic detention time of 16 days from Table 7.32 and correct it for the minimum
design operating temperature of 12°C using Equation (7.269).
tT
C
t20
C
1.06
20
T C
16d 1.06
20
12
25.5 d
Solution Part B
The aerobic digester volume is calculated by multiplying the hydraulic detention time by the
volume of sludge fed to the digester as:
V
t Qsludge
25.5 d 57,000
gal
d
Solution Part C
Calculate the volatile solids loading to the digester.
118
ft 3
7.48gal
1.94 105 ft 3
VS loading 57, 000
VS loading
gal 1MG
d 106gal
8.34 lb
MG mg L
0.75 1.2 10, 000
mg
L
4280 lb/d
The volatile solids loading rate to the digester is determined as:
VS loading to digester
4280 lb/d VS
1.94 105 ft 3
0.022
lb VS
Less than 0.04 but should work.
ft 3 d
Solution Part D
Oxygen required to stabilize the VS in the secondary sludge (WAS) is estimated as:
O2 to stabilize VS in WAS
4280
lb VS
0.50
d
2.0
lb O2
lb BOD
4280
lb O2
d
Solution Part E
Finally, the total air required (ft3/(min·1000 ft3) to stabilize the WAS is estimated as follows
assuming 0.075 lb air/ft3 , 0.23 lb O2/lb air, and the diffuser transfer efficiency is 6%.
4280 lb O 2 d
air required
0.075 lb air ft 3
0.06
0.23lb O 2
lb air
1d
24 h
1h
ft 3
= 2870
Chec
60 min
min
k to see if the mixing requirements have been met.
air required
1000 ft 3
2870 ft 3 / min 1000 ft 3
1.94 105 ft 3 1000 ft 3
14.8ft 3 / min
1000 ft 3
20-40 ft 3 / min
Additi
1000 ft 3
onal air must be supplied to keep the sludge in suspensions. The air requirement must be
increased as follows:
30ft 3 / min
1.94 105 ft 3
3
1000 ft
5820
ft 3
min
51. An aerobic digester is to be designed to stabilize 600 kg/d of primary sludge at 5% solids
and 75% volatiles along with 400 kg/d of waste-activated sludge at 1% solids and 75% volatiles.
The volume of the combined sludge that will be fed to the digester is 104 m3/ d. The minimum
design operating temperature is 18°C. A temperature correction coefficient ( ) of 1.06 should be
used to correct the detention or digestion time for temperature. The primary sludge contains 210
119
kg of BOD with an oxygen demand of 1.9 kg of oxygen per kg of BOD, whereas use an oxygen
demand of 2.0 kg of oxygen per kg of VS destroyed for the WAS. Assume that 90% of the BOD
contained in the primary sludge is destroyed and that 50% of the volatiles solids in the WAS is
destroyed.
Determine the following:
f)
g)
h)
i)
j)
k)
l)
Design hydraulic detention time ( ) or digestion time in days.
Aerobic digester volume in (m3).
Volatile solids loading rate to the aerobic digester in (kg/d m3).
Quantity of oxygen required to stabilize the primary sludge in kg/d.
Quantity of oxygen required (kg/d) to stabilize the secondary sludge (WAS).
Total quantity of oxygen required (kg/d) to stabilize the combined sludge.
Total air required (m3/min) if there are 1.2 kg of air per cubic meter of air and the
atmosphere consists of 23% oxygen by weight (0.23 kg O2/kg air) and the diffusers have
a transfer efficiency of 4.0%.
Solution Part A
Select a hydraulic detention time of 19 days from Table 7.32 and correct it for the minimum
design operating temperature of 18°C using Equation (7.269).
t18 C
t20
C
1.06
20
18 C
19d 1.06
20
18
21.3 d
Solution Part B
The volume of primary sludge is:
Vsl
Ms
S sl Ps
600 kg
1000 kg
1.0
m3
12 m3
0.05
The volume of WAS is:
Vsl
Ms
S sl Ps
400 kg
1000 kg
1.0
m3
Total sludge volume 12
m3
d
40 m3
0.01
40
m3
d
52
m3
d
The solids concentration in the blended sludge is determined as follows:
C
12 m3 d 5% + 40 m3 d 1.0%
12 m3 d + 40 m3 d
120
1.92% solids
The aerobic digester volume is calculated by multiplying the hydraulic detention time by the
volume of sludge fed to the digester as:
V
t Qsludge
m3
21.3 d 52
d
1100 m3
Solution Part C
Calculate the volatile solids loading to the digester.
VS loading
600 kg/d 400 kg/d 0.75
750 kg/d
The volatile solids loading rate to the digester is determined as:
VS loading to digester
750 kg/d VS
1100 m3
0.68
kg VS
Meets criteria in Table 7.32.
m3 d
Solution Part D
Oxygen required to stabilize the BOD in the primary sludge is estimated as:
O2 to stabilize primary BOD
210
kg BOD
d
1.9
kg O2
kg BOD
0.90
359
kg O2
d
Solution Part E
Oxygen required to stabilize the VS in the secondary sludge (WAS) is estimated as:
O2 to stabilize VS in WAS
400
kg TS
d
0.75
kg VS
kg TS
2.0
kg O2
kg BOD
0.50
300
kg O2
Sol
d
ution Part F
The total oxygen required to stabilize the combined primary and secondary sludge is determined
below:
total O2 required 359 300 659
kg O2
d
Solution Part G
Finally, the total air required (m3/(min·1000 m3) to stabilize the combined sludge is estimated as
follows assuming 1.2 kg air/m3 , 0.23 kg O2/kg air, and the diffuser transfer efficiency is 4%.
121
659 kg O 2 d
air required
1.2 kg air m3
0.04
0.23kg O 2
kg air
1d
24 h
1h
m3
= 41.5
Check
60 min
min
to see if the mixing requirements have been met.
air required
1000 m3
41.5 m3 / min 1000 m3
1100 m3
1000 m3
37.7 m3 / min
1000 m3
60 m3 / min
1000 m3
This does not meet the mixing requirements; therefore, additional air must be supplied beyond
process requirements.
65 m3 / min
1000 m3
3
1100 m
m3
71.5
min
52. Calculate the anaerobic digester capacity (m3) required for conventional single-stage
anaerobic digestion given the following data: raw sludge production = 1000 kg/d; VS in raw
sludge = 72%; moisture content of raw sludge = 94%; solids content of raw sludge = 5% solids;
digestion period = 30 days; VS Reduction = 52%; storage time required = 90 days; percent solids
in thickened digested sludge = 5.5%, and specific gravity of digested sludge = 1.02.
Solution
Calculate the volume of raw sludge fed to the digester using Equation (7.257) assuming Ssl is 1.0.
The density of water is 1000 kg/m3 = 62.4 lbm/ft3 = 8.34 lbm/gal.
Ps
1.0 Pw
Vsl
1.0 0.94 0.06
MS
S sl Ps
(7.248)
1000 kg/d
1.0 1000 kg m3 0.05
m3
20.0
d
V1
Perform a materials balance on the mass of solids entering and exiting the digester. First,
determine the volatile and fixed solids in the raw sludge fed to the digester. Recall that total
solids = volatile solids + fixed solids.
Mass of volatile solids in raw sludge = 0.72 (1000 kg/d) = 720 kg/d
Mass of fixed solids in raw sludge = Total Solids – Volatile Solids = 1000 - 720 = 280 kg/d
Next, calculate the volatile and total solids quantities remaining after digestion. Digestion of the
solids causes a 52% reduction in the volatile solids and the fixed solids are assumed to remain
unchanged.
122
Mass of volatile solids remaining after digestion = (1-0.52)
720 kg/d = 345.6 kg/d
Mass of total solids remaining after digestion = VS + FS = 345.6 + 280 = 625.6 kg/d
Next, determine the volume of digested sludge that accumulates daily in the tank assuming that
the sludge thickens to 5.5% solids using Equation (7.257) and that the specific gravity of the
digested sludge is 1.03.
Vsl
MS
Ssl Ps
625.6 kg/d
1.02 1000 kg m3 0.055
11.2
m3
d
V2
Finally, the volume of the conventional, single-stage anaerobic digester is determined from
Equation (7.270).
V
V1
V2
2
T1
V2 T2
(20.0 m3 d +11.2 m3 d )
11.2 m3
30d
2
d
1480 m3
90d
53. Calculate the anaerobic digester capacity (ft3) required for conventional single-stage
anaerobic digestion given the following data: 40,000 gallons per day of combined primary sludge
and WAS at 3.0% solids; volatile solids content = 75%; digestion period = 30 days; VS
Reduction = 55%; storage time required = 100 days; percent solids in thickened digested sludge
= 6.5%, and specific gravity of digested sludge = 1.04.
Solution
Calculate the volume of raw sludge fed to the digester using Equation (7.257) assuming Ssl is 1.0.
The density of water is 1000 kg/m3 = 62.4 lbm/ft3 = 8.34 lbm/gal.
Ms
40,000
Vsl
MS
S sl Ps
gal 1MG
d 106 gal
3 10,000
mg
L
10,000 lb/d
1.0 62.4 lb m ft 3 0.03
8.34lb
MG mg L
5340
10,000
lb
d
ft 3
d
Perform a materials balance on the mass of solids entering and exiting the digester. First,
determine the volatile and fixed solids in the raw sludge fed to the digester. Recall that total
solids = volatile solids + fixed solids.
Mass of volatile solids in raw sludge = 0.75 (10,000 lb/d) = 7500 lb/d
Mass of fixed solids in raw sludge = Total Solids – Volatile Solids = 10,000 - 7500 = 2500 lb/d
123
Next, calculate the volatile and total solids quantities remaining after digestion. Digestion of the
solids causes a 55% reduction in the volatile solids and the fixed solids are assumed to remain
unchanged.
Mass of volatile solids remaining after digestion = (1-0.55)
7500 lb/d = 3375 lb/d
Mass of total solids remaining after digestion = VS + FS = 3375 + 2500 = 5875 lb/d
Next, determine the volume of digested sludge that accumulates daily in the tank assuming that
the sludge thickens to 7% solids using Equation (7.257) and that the specific gravity of the
digested sludge is 1.03.
Vsl
MS
S sl Ps
5875 lb/d
1.04 62.4 lb m ft 3 0.065
1390
ft 3
d
V2
Finally, the volume of the conventional, single-stage anaerobic digester is determined from
Equation (7.270).
V
V1
V2
2
T1
(5340ft 3 d +1390 ft 3 d )
1390ft 3
30d
2
d
V2 T2
100d
2.40 105 ft 3
54. Determine the volume of a first-stage, high-rate anaerobic digester based on a maximum
volatile solids loading rate of 2.5 kg VS/(m3d) and a detention time of 18 days. 700 cubic
meters per day of combined primary and waste activated sludge at 3.6% solids and a volatile
content of 71% are fed to the digester.
Solution Part A
Determine the volume of the digester using both methods; use the larger of the two volumes for
the final volume of the digester.
First calculate the mass of volatile solids as follows:
Ms
700
m3
kg
d 1000g
3.6 10,000
g
m3
0.71
17,900
lb VS
d
Next, calculate the volume of the anaerobic digester based on volatile solids loading.
kg VS
d
kg VS
2.5 3
m d
17,900
V
7160 m3
Solution part b
124
Next calculate the volume of the digester based on detention time:
V
Q
m3
18d 700
d
12,600 m3
Use the larger of the two areas for the final design; therefore, the volume of the digester should
be 12,600 m3 .
55. Determine the volume of a first-stage, high-rate anaerobic digester based on a maximum
volatile solids loading rate of 0.15 lb VS/(ft3d) and a detention time of 16 days. 20,000 gallons
of raw primary sludge at 5% solids and a volatile content of 73% are fed to the digester.
Solution Part A
Determine the volume of the digester using both methods; use the larger of the two volumes for
the final volume of the digester.
First calculate the volume of the anaerobic digester based on volatile solids loading.
Ms
20,000
gal 1MG
d 106 gal
lb VS
d
lb VS
0.15 3
ft d
5 10,000
mg
L
8.34lb
MG mg L
0.73
6090
lb VS
d
6090
V
40,600 ft 3
Solution part b
Next calculate the volume of the digester based on detention time:
V
Q
gal
16d 20,000
d
ft 3
7.48gal
42,800ft 3
Use the larger of the two areas for the final design; therefore, the volume of the digester should
be 42,800 ft3 .
56. An activated sludge WWTP produces an average of 100,000 gallons per day of primary
sludge at 5% solids and 250,000 gallons per day of WAS at 1.1% solids. If the WAS is
thickened to 6.0% solids by dissolved air flotation and then combined with the primary sludge
prior to dewatering, what is the quantity (lb/d)and solids concentration of the blended sludge (%
solids)? Determine the belt width (m) of a belt filter press (BFP) if the BFP produces a sludge
125
cake of 22% solids. Assume that the BRP operates 5 days a week for 8 hours each day. The
solids in the filtrate is assumed to be 1,800 mg/L and the addition of polymer for conditioning
contributes negligible weight. Assume a washwater rate of 25 gpm/meter of belt and hydraulic
loading rate= 45gpm/meter of belt.
Solution Part A
Determine the quantity of dry sludge that enters the BFP.
M PS
M Total
100,000
gal 1MG
d 106 gal
MWAS
250,000
M PS
MWAS
gal 1MG
d 106 gal
41, 700
5 10,000
mg
L
1.1 10,000
lb TS
lb TS
22,935
d
d
8.34lb
MG mg L
mg
L
41,700
8.34lb
MG mg L
64, 635
lb TS
d
22,935
lb TS
d
lb TS
d
The volume of the thickened WAS to be blended with the primary sludge is determined as
follows using Equation (7.257). Recall that the density of water = 1000 kg/m3 = 62.4 lbm/ft3 =
8.34 lbm/gal.
Vsl
MS
S sl Ps
22,935 ppd
1.0 8.34 lb m gal 0.06
45,833 gpd
The solids concentration in the blended sludge is calculated as follows:
C
45,833gal d 6% + 100, 000gal d 5.0%
45,833gal d +100, 000gal d
5.3% solids
The quantity of solids that enter the BFP on a daily basis is calculated below.
M Total
64,635
lb TS 7 d
d
wk
1wk
5d
90,500
lb TS
d
The volume of blended sludge fed to the BFP is determined as follows.
Vsl
MS
S sl Ps
90,500 ppd
1.0 8.34 lb m gal
mg
5.3 10,000
L
126
0.205 MGD
Vsl
0.205 MGD
106 gal
MG
1d
8h
1h
60 min
427 gpm
The solids loading rate should vary between 800-1200 lb/(hm) and hydraulic loading rate should
range from 40-50gpm/meter. Use a hydraulic loading rate of 45gpm/meter to determine the
width of the BFP.
Belt width =
427 gpm
45gpm/meter
Go with five, 2-meter Belt Filter Presses
9.5m
Check the solids loading rate.
SLR
90,500lb/d 1d
5 2m
24 h
377 lb
h m
The SLR is on the low side but is acceptable. The design of the BFP is controlled in this case by
the hydraulic loading rate.
Qwashwater
25
gpm
10.0 m = 150gpm
m
Solution Part B
Draw a schematic of the process before completing material balances.
Washwater
250 gpm
0% Solids
Sludge
Sludge
Belt
filter press
Centrifuge
cake
feed
90,500 lb/d
427 gpm
5.3% Solids
22% Solids
Filtrate
1800 mg/L
Solution Part E
Perform a materials balance on solids and flows.
127
M sludge
M filtrate
M cake
90,500
lb
mg
Qfiltrate 22 10, 000
d
L
90,500
lb
1.50 104 Qfiltrate 1.83 106 Qcake
d
Qsludge
8.34 lb
MG mg L
Qwashwater
Qfiltrate
427 gpm
250 gpm
677 gpm
Qfiltrate
677 gpm
60min
h
677 gpm
Qfiltrate
Qcake 1800
mg
L
8.34 lb
MG mg L
Qcake
Qfiltrate
Qcake
Qcake
MG
106 gal
Qfiltrate
0.325 MGD
8h
d
0.325MGD
Qcake
Qcake
lb
1.50 104 0.325 MGD Qcake 1.83 106 Qcake
d
lb
90,500
1.50 104 Qcake 4875 1.83 106 Qcake
d
Qcake 0.0471MGD
90,500
Qfiltrate
M filtrate
0.325 MGD
Qcake
0.28MGD 1800
% Capture =
% Capture =
0.325 MGD 0.0471 0.278 MGD
mg
L
8.34lb
MG mg L
solids in feed - solids in filtrate
4200
100
solids in feed
90,500lb d - 4200lb d
90,500lb d
128
100
95.4%
lb
d
Washwater
0.12 MGD
0% Solids
Sludge
Sludge
Belt
filter press
Centrifuge
cake
feed
90,500 lb/d
0.205 MGD
5.3% Solids
Filtrate
86,300 lb/d
0.047 MGD
22% Solids
4200 lb/d
0.28 MGD
1800 mg/L
57. A 2.0 m belt filter press dewaters 100 gpm of anaerobically digested sludge at a solids
content of 6.5%. Four pounds of polymer are added per ton of sludge, and the liquid polymer
flow rate is 6.4 gallons per minute. The washwater flow rate is 30 gpm per meter of belt. The
solids concentrations in the filtrate and cake are 1,800 mg/L and 30%, respectively. The BFP
operates 16 h/d and 7 days per week. Perform a materials balance around the BFP and calculate
the following:
a) Hydraulic loading rate, gpm/m.
b) Solids loading rate, lb/(mh).
c) Solids capture (%).
Solution Part A
Determine the hydraulic loading rate on the BFP neglecting the polymer dosage.
Qsludge
100 gpm
gpm
= 50
2m
m
Solution Part B
Determine the quantity of dry sludge that enters the BFP.
129
M sludge 100
gal 60 min
min
h
5.20 104
M Polymer
16 h
d
1MG
106 gal
lb TS 4lb polymer
d
ton
6.5 10,000
1ton
2000 lb
104
mg
L
8.34lb
MG mg L
5.20 104
lb polymer
d
Solution Part C
Perform a materials balance on flows and solids as follows. Draw a schematic of the process
before completing material balances.
Washwater
Polymer
60 gpm
0% Solids
6.4 gpm
104 lb/d
Sludge
Sludge
Belt
filter press
Centrifuge
cake
feed
30% Solids
5.20104 lb/d
100 gpm
5.3% Solids
Filtrate
1800 mg/L
Qsludge
Qwashwater
Qwashwater
Qpolymer
Qfiltrate
Qcake
30 gpm 2 m = 60 gpm
100 gpm
60 gpm 6.4 gpm Qfiltrate
166.4 gpm Qfiltrate
Qcake
Qfiltrate
0.16 MGD - Qcake
M sludge
M polymer
M filtrate
0.24 MGD
M cake
lb
lb
104
M filtrate
d
d
lb
5.20 104
M filtrate M cake
d
5.20 104
5.20 104
lb
d
5.20 104
lb
1.50 104 Q filtrate
d
Q filtrate 1800
mg
L
Qcake
M cake
8.34lb
MG mg L
2.50 106 Qcake
130
Qcake 30 10,000
mg
L
8.34lb
MG mg L
lb TS
d
lb
lb
1.50 104
0.16 MGD - Qcake
d
d
0.02 MGD
5.20 104
Qcake
Qfiltrate
0.16 MGD - Qcake
2.50 106 Qcake
0.16 MGD 0.02 MGD = 0.14 MGD
The quantity of solids that exit the BFP on a daily basis is calculated below.
M cake
0.02 MGD 30 10,000
M filtrate
0.14 MGD 1800
mg
L
mg
L
8.34lb
MG mg L
8.34lb
MG mg L
5.00 104
2.10 103
lb TS
d
lb TS
d
Calculate the % capture using Equation (7.258).
% Capture =
% Capture =
solids in feed - solids in filtrate
100
solids in feed
78, 216.4lb d - 3153lb d
100
78, 216.4lb d
96%
Draw a schematic to show the final flows and solids loading rates into and out of the belt filter
press.
Washwater
0.12 MGD
0% Solids
Sludge
Sludge
Belt
filter press
Centrifuge
cake
feed
90,500 lb/d
0.205 MGD
5.3% Solids
Filtrate
4200 lb/d
0.279 MGD
1800 mg/L
131
86,300 lb/d
0.046 MGD
22% Solids