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Grit Chambers...
( a ) Weir velocity control :
A proportional weir, a Parshall flume or a Venturi flume may be used for control of water velocity in grit chambers.
Illustrations of these control devices are contained in figure given below.
Proportional weir design formula ;
Other symbols are as indicated in figure given above - a and all dimensions are expressed in feet.
Parshall flume design formula ;
In the illustration ( figure given above - b ), the depth of the liquid in the grit chamber above the grit space is
the same as H a , the depth in the Parshall flume ahead of the throat. The velocity through a channel of the
illustrated cross-section can be determined from the following formula ;
Table C-2 illustrates the variations in velocity through channels of different sizes used in conjunction with Parshall
flumes of different sizes discharging various quantities under free-flow conditions. The quantities and values of H are
taken from Table C-2. Table C-3 can serve as a guide in selecting the size of Parshall a flume and the number and
approximate size of channels to suit a specified set of flow conditions.
Venturi flume design formula ;
( b ) Example :
Assuming that the diameter of the inlet sewer is 2 feet, that the average rate of sewage flow is 1.67 cfs, that the maximum
rate of flow and as high as practicable over the entire range of flows. Assume a depth in the effluent channel of 1.5 feet
at maximum rate of flow. Then the required width (w) would be 5 / (1.5 x 2) = 1 ft. 8 in. The designs for various types of
control sections and grit chamber cross-sections are as follows ;
( 1 ) Proportional weir : As the depth (h) in the chamber above the weir crest cannot exceed 2 feet without submerging the
crown of the inlet sewer, let h = 1.75 for first trial. Determine W by substitution in equation C-1. Then ;
W = ( 5 cfs ) / [ ( 1 fps ) ( 1.75 ft ) ] = 2.86 ft
The design value of h = ( 5 cfs ) / [ ( 1 fps ) ( 2.86 ft ) ] = 1.77 ft
To prevent appreciable divergence of the velocity from 1 fps when the flows are low, the depth (d) of the rectangular
section of the weir opening should be minimum practicable. As indicated by equations C-2 and C-3, this depth is a function
of b and x. The breadth (b) is limited by the width of the effluent channel (in this case, 1 ft. 8 in.) and x is limited by
the size of solids to be passed (in this case, about 3 inches, assuming that a bar screen will be ahead of the grit
chamber). Try d 0.15 and solve for b in equation 2. Then ;
therefore, b = 1.53 (satisfactory). For various values of y, values of y/d are determined and the corresponding values
of x/b are taken from the table; the values of x are then determined as follows ;
The above tabulation indicates that the trial design is satisfactory.
( 2 ) Bar screen : Assumptions ; Maximum daily flow = 4 mgd ; Maximum storm flow = 7 mgd ; Maximum allowable velocity
through bar rack for maximum daily flow = 2 fps ; Then, using the design procedure in the preceding paragraph ; Maximum
daily flow = 4 × 1.547 = 6.188 cfs ; Maximum storm flow = 7 × 1.547 = 10.829 cfs.
Since Q = Av, 6.188 = A × 2, and the net area A through the bars is 3.094 sq. ft. For a maximum allowable velocity through
the bar rack of 3 fps during maximum storm flow, the net area through the bars must be 10.829/3 = 3.61 sq. ft. The gross
area will be based on the larger of the two net areas (in this case, 3.61 sq.ft.). A rack consisting of 2-inch × 5/16-inch
bars, spaced to provide clear openings of 1 inch, has an efficiency of 0.768, yielding ;
Gross area = ( 3.61 ) / ( 0.768 ) = 4.70 sq.ft
( 3 ) Wastewater depth : The channel width in this case might be established at 3 feet; in which case, the water depth
would be 4.70/3.0 = 1.57 feet. This is a theoretical water depth which may be affected by subsequent plant units. For
instance, a grit chamber may follow the screen chamber and be of such design that a head may build up, the effect of which
would be observed in the screen chamber. This is particularly true if the sewage flow by pipeline from the screen chamber
to the grit chamber and the flow in the grit chamber are subject to controlled velocity. A screen chamber may be followed
immediately by a wet well; in which case the sewage depth, especially at the low flows, would be less than that computed.
In fact, it is sometimes necessary to increase the sewage depth by installing stop planks across the channel behind the
screen. The planks serve as a weir, and the head built over this impromptu weir further serves to increase the sewage
depth in the channel. The head loss through a bar rack is computed from the formula ;
If the screen is half plugged with screening, leaves and other debris ; from Q = ( A ) ( V ), the area is directly
proportional to the velocity. In other words, if the area is cut in half, the velocity must double. The head loss,
therefore, is ;
The increase in head loss is over one-half foot as the screen becomes half plugged. The need for accurate control of the
cleaning cycle and protection against surge loads is thus demonstrated.
( 4 ) Parshall flume : From inspection of table C-13, it is readily obvious that by using two grit channels with bottom
width of 0.75 ft and with sides sloping at an angle whose cotangent is 0.67, a Parshall flume with a throat width of 0.75
ft would control the velocity within the specified limits. The table of discharge for Parshall flumes (table C-2) indicates
that for flows of 0.67, 1.67 and 5.0 cfs, the values of H are 0.34, 0.67 a and 1.38, respectively. By appropriate
substitution in equation C-4, the corresponding velocities are found to be 0.91, 1.04 and 1.08 fps.
( 5 ) Venturi flume : In this case, design for D = 2.0. Then W, H, b and d are determined by substituting (in eqs C-S
through C-8, respectively) as follows ;
For various values of Q, the values of h and v (determined from eqs C-9 and C-10) are as follows ;
This tabulation indicates that for values of Q varying between 0.67 cfs and 5.0 cfs, the velocity in the grit chamber would
not vary more than 10 percent from 1.0 fps and that the design would be satisfactory in this respect.
( c ) Discussion :
The text states that for proper operation of a proportional weir, complete free-flow conditions must exist below the weir
crest. Therefore, in this case, the depth of the effluent channel below the weir crest would have to be 1.5 ft, the maximum
liquid depth in the effluent channel. With a Parshall flume (according to the text), the permissible submergence is 65
percent of H or, in this case, 0.65 × 1.38 a = 0.9 ft. The effluent channel floor would have to be 1.5 - 0.9 = 0.6 ft below
the floor of the approach to the Parshall flume. For a Venturi flume (according to the text), the head loss to be provided
for should not be less than H/3 or, in this case, 2.35/3 = 0.78 ft. The crest could, therefore, be submerged 2.0-0.78 =
1.22 ft and the effluent channel floor could be 1.5-1.22 = 0.28 ft below the crest. However, to satisfy the design with
respect to d, the depth of the effluent channel would have to be 0.35 ft below the crest. The foregoing analysis indicates
that the three devices would require head losses of 1.5, 0.60 and 0.35 ft, respectively. Consideration of these losses and
local conditions would determine which of the three designs should be used.
Mechanical Flocculation...
( a ) Design requirements and criteria :
Design a mechanical flocculator as a part of chemical precipitation to handle a flow rate of 4 mgd. The following conditions
apply ; Temperature = 20 O C ( 68 O F ) ; Paddle-tip speed = 1.2 fps ; Coefficient of drag of paddles
= 1.8.
( b ) Calculations and results :
( 1 ) Using a detention time of 20 minutes, determine tank volume. Volume = flow rate × detention time ;
( 2 ) Using a depth of 10 ft, determine tank dimensions. Width is usually set by standard sludge removal equipment sizes. A
width of 12 ft will be used here.
A tank size of 12 ft x 62 ft would be appropriate here and would produce a satisfactory length-to-width ratio of about
5.4:1.
( 3 ) A major design factor in flocculator design is mean velocity gradient (G), measured in ft/sec ft. A typical value of
30 ft/sec ft will be used in this case.
( 4 ) The theoretical power requirement is calculated by using the following formula ;
Convert this to horsepower ;
Hp = ( P ) ( 1 Hp / 550 ft . lb / sec ) = ( 140.4 ) / ( 550 ) = 0.26 ; use 0.3 Hp
( 5 ) Determine the paddle area from the following formula ;
Sedimentation...
( a ) Design requirements and criteria :
Design a sedimentation unit to provide settling for a sewage flow rate of 4 mgd, with suspended solids concentration of 300
mg/L. The following conditions apply ; Surface loading rate = 600 gpd/sq ft ; Suspended solids removal = 60 % ; Sludge
solids content = 4 % ; Sludge specific density = 1.02.
( b ) Calculations and results :
( 1 ) Calculate total tank surface area ;
Surface area = ( Flow rate ) / ( Surface loading rate ) = ( 4,000,000 gpd ) / ( 600 gpd / sq ft ) = 6,666.7 ;
use 6,670 sq ft.
( 2 ) Using a depth of 8 ft, calculate total volume ;
V = ( 8 ) ( 6,670 ) = 53,360 cu ft.
( 3 ) This volume can be divided among three rectangular tanks (in parallel), 20 ft wide and 120 ft long, with a
satisfactory length-to-width ratio of 6:1. Two circular tanks (in parallel), 35 ft in diameter, would also be suitable.
This will provide flexibility of operation during routine or emergency maintenance.
( 4 ) Calculate weir length requirement, assuming 3 rectangular tanks and allowable weir loading rate of 15,000
gpd / linear ft.
Weight removed = ( 4 mgd ) ( 300 mg / L ) ( 0.60 ) = 6,000 lb / day ; therefore, 1,500 lbs are removed per 1 mpd
flow.
( 6 ) Calculate sludge volume, assuming a specific gravity of 1.02 and a moisture content of 96 % ( 4 % solids ) ;
Sludge volume = ( 6,000 lb / day ) / [ ( 1.20 ) ( 62.4lb / cu ft ) ( 0.04 ) ] = 2,360 cu ft / day ( @ 4 mgd ) =
17,700 gpd
( 7 ) Sludge handling in this example consists of removing sludge manually from settling tank sludge hopper, using a
telescoping drawoff pipe which discharges the sludge into a sump from which it is removed by a sludge pump (or pumps).
Assume that the sludge will be wasted every 8 hours and pumps for 0.50 - hour to the digester.
Sludge sump capacity = ( Daily sludge volume ) / ( Number of wasting periods per day ) = ( 2,360 cu ft ) / ( 3 ) =
787 cu ft ( 5,900 gal )
Increase capacity 10 percent to compensate for scum removal volumes ;
Sludge pumping capacity = ( Sludge and scum volume / wasting period ) / ( 30 minutes pumping / wasting period ) =
( 6,500 ) / ( 30 min ) = 217 ; use 220 gpm
Chemical Precipitation...
( a ) Design requirements and criteria :
Calculate the sludge production, using chemical addition in primary sedimentation. Assume that addition of 60 lbs of
ferrous sulfate and 700 lbs / mil gal of lime yields 70 percent suspended solids removal under the following conditions ;
Flow rate = 4 mgd ; Suspended solids concentration = 300 mg / L.
The reactions that will occur are as follows ;
( b ) Calculations and results :
All interim calculations are computed on the basis of a flow volume of 1 mil gal.
( 1 ) Determine the weight of suspended solids removed ;
Solids weight = ( 0.70 ) ( 300 mg / L ) [ ( 8.34 lb / mil gal ) / ( mg / L ) ] = 1,750 lb / mil gal
( 2 ) Determine weight of ferric hydroxide formed from ferrous sulfate ;
( 3 ) Determine weight of CaCO 3 formed in reacting with SO 4 hardness ;
Single Stage Stone - Media Trickling Filters...
( a ) Design requirements and criteria :
Design a trickling filter to treat 2 mgd of primary settled effluent under the following conditions ; Raw wastewater
BOD 5 = 250 mg / L ; Primary clarifier BOD 5 removal efficiency 30 percent ; Required effluent
BOD 5 = 30 mg / L ; Design temperature = 20 O C ; Design without and with recirculation.
( b ) Calculations and results :
( 1 ) Design without recirculation ;
Primary treated effluent BOD = ( 250 ) ( 1 - 0.30 ) = 175 mg / L
Since the design temperature is 20 O C, no temperature correction is required.
BOD loading to filter = ( 2.0 mgd ) ( 175 mg / L ) [ ( 8.34 lb / mil gal ) / ( mg / L ) ] = 2,930 lb / day
Assume a practical filter depth of 6 ft for stone-media filters. Apply NRC formula ;
( 2 ) Design with 1 : 1 recirculation ;
( 3 ) Design with 1 : 2 recirculation ;
( c ) Pumps :
Recirculation pumps will be sized to provide constant rate recirculation. Vertical-shaft, single suction units with motors
mounted on top of the pumps, installed in a dry well or on an upper floor, will be used. Each pump will be provided with its
individual pipe connection to the wet well.
Two Stage Stone - Media Trickling Filters...
( a ) Design requirements and criteria :
Design a two-stage trickling filter to treat 3.0 mgd of primary settled effluent, assuming the following conditions ;
Raw wastewater BOD 5 = 250 mg/L ; Primary clarified BOD 5 removal efficiency = 30 percent ;
Required effluent BOD 5 = 30 mg/L ; Design temperature = 20 O C.
( b ) Calculations and results :
- Primary clairfier effluent BOD 5 = ( 250 ) ( 1 - 0.3 ) = 175 mg/L
- Overall trickling filter efficiency = ( 175 - 30 ) / ( 175 ) = 0.828 = 82.8 %
- Filter depth = 6 ft
- Recirculation = 1 : 2
- Assuming that the first stage filter efficiency = 75 percent
- Overall efficiency = 0.828 = ( E 1 ) + ( E 2 ) ( 1 - E 1 ) =
0.75 + ( E 2 ) ( 1 — 0.75 ) = 0.31
- Recirculation factor F = ( 1 + R ) / ( 1 + 0.10 x R ) 2 = ( 1 + 2 ) / ( 1 + 0.10 x 2 ) 2 =
2.08
- Organic loading to first stage filter, W = ( 3.0 ) ( 8.34 ) ( 250 ) ( 1 — 0.30 ) = 4,379 ; use 4,380 lb/day
- Now, using the NRC formula ;
The design of a two—stage system requires careful economic considerations to insure that minimal filter volume is required
and that proper recirculation rates are selected to optimize filter volume and pumping requirements, resulting in the
lowest—cost, most flexible system.
Plastic Media Trickling Filters...
( a ) Design requirements and criteria :
Design a plastic media trickling filter to treat 0.2 mgd of primary settled effluent, assuming the following conditions ;
Raw wastewater BOD = 250 mg/L ; Primary clarifier BOD removal efficiency = 30 percent ; Required final effluent BOD = 25
mg/L ; Winter design temperature = 10 O C.
( b ) Calculations and results :
The formula presented for plastic media trickling filters states the following ; L E and L O in
units of mg / L ; K 20 as 1 / day ; D as ft and Q as gpm / sq . ft.
Plastic media manufacturers should be consulted to determine the exact proportions of filter media that provide for a
minimum of 5,720 sq ft area and 12 ft depth.
Activated Sludge, Closed - Loop Reactor...
( a ) Design requirements and criteria :
The following design criteria are for an oxidation ditch with horizontal-shaft rotor aerators, using a single- channel,
oval configuration and miltiple units in parallel, operated in an extended aeration mode with nitrification.
Influent requirements ;
- Q, avg daily flow = 1.0 mgd
- Q, peak flow = 2.0 mgd
- BOD 5 = 250 mg/L
- TSS = 250 mg/L
- VSS = 200 mg/L
- TKN = 25 mg/L (all NH 3 - N)
- P = 8 mg/L
- pH = 6.5 to 8.5
- Minimum temp = 10 O C
- Maximum temp = 25 O C
- Use 20 lb BOD/1,000 cu ft—day
- V = ( 1,042.5 lb/day ) / ( 20 lb/1,000 cu ft-day ) = 52,125 cu ft
- Hydraulic detention time = [ ( 52,125 ) ( 7.48 ) ( 24 ) ] / ( 0.5 × 10 6 ) = 18.75 hr
( c ) Calculation of rotor requirements :
- Rotor mixing requirement = 16,000 gal/ft of rotor; recommended by a manufacturer for maintaining a channel velocity of
1.0 fps
- Length of rotor = [ ( 52,125 ) ( 7.48 ) ] / ( 16,000 ) = 24.4 ; use 25 ft
- Oxygenation requirement = 2.35 lb O 2 /lb BOD-recommended for domestic sewage. Assume the following operating
conditions ; RPM = 72 ; Immersion = 8 in
- Oxygenation = 3.75 lb O 2 /hr-ft ( from figure given below ), using a 42-in diameter MAGNA rotor
- Length of rotor = [ ( 1,042.5 ) ( 2.35 ) ] / [ ( 24 ) ( 3.75 ) ] = 27.2 ; use 28 ft
- Use two rotors per each unit oxidation ditch or 2- x 14-ft length rotor
- Theoretical oxygen transfer requirement, lb O 2 /hr-ft = [ ( 1,042.5 ) ( 2.35 ) ] / [ ( 24 ) ( 28 ) ] =
3.65 lb O 2 /hr-ft
- Actual immersion = 8 in
- Power requirement = 0.84 kW/ft of rotor ( from figure given below ), using 42-in diameter MAGNA rotor
- Size all motors to allow at least 1.5 inch above the 8-in actual immersion to allow for peak flows
- Motor horsepower required at ( 8 + 1.5 ) = 9.5 in = [ ( 0.99 ) ( BHP ) ( 14 ) ( 1.34 ) ] / ( 0.95 ) = 19.6 hp
- Use standard horsepower, or 20 hp each
( e ) Channel sizing :
- Ditch liquid volume = 52,125 cu ft
( f ) Final clarifier :
( g ) Check for nitrificaiton in oxidation ditch :
- Find MLVSS concentration required at standard conditions ( 20 O C, pH = 8.5 ) to completely nitrify ammounia
shown in figure given below
- MLVSS = ( 1 / 0.0075 ) ( 25 / 18.75 ) = 177.8 mg/L ( where 0.0075 mg NH 3 -N/mg MLVSS-hr is the rate of
nitrification given in figure shown above )
- Change MLVSS concentration at standard conditions to design conditions [ 10 O C, pH = 6.5 to 8.5
( assume pH is 7.0 ) ], using figures for temperature and pH corrections given below
( h ) Check size of final clarifier with solids loading :
- A weir on each oxidation ditch is provided and sized so that the head differential over the weir between the maximum and
minimum flow rates is less than 1.25 in
- Maximum flow rate = Q PEAK + maximum recirculation rate
- Minimum flow rate = Q MIN + recirculation rate
- For average daily flows between 200,000 gal/day to 1 mgd, the length (L) of the overflow weir is ;
- L = [ ( 3.5 ) ( 347.2 ) ] / ( 102 ) = 11.9 ; use 12 ft
( j ) Calculate the return sludge flow rate, Q R :
- The equipment manufacturer recommends the following ;
- Minimum pumping capacity = ( 0.25 ) ( Q AVG ) = ( 0.25 ) ( 347.2 gpm ) = 87 gpm
- Maximum pumping capacity = ( 1.0 ) ( Q AVG ) = ( 1.0 ) ( 347.2 gpm ) = 347 gpm
- 2 pumps, each with 44 to 174 gpm capacity
( k ) Calculate amount of sludge for disposal :
- The amount of sludge wasted is the same as in the vertical-shaft aerator example since the influent and effluent
characteristics and the MLVSS concentrations are identical
- Px = 683.2 lb/day of VSS from 2 oxidation ditches
- Total sludge 933.4 lb/day dry solids from 2 oxidation ditches
- 2 pumps, each with a capacity of 120 gpm for a 52-min/day wasting schedule as in the vertical-shaft aerator example
( l ) Sizing of sludge drying beds :
- BOD population equivalent = [ ( 1.0 ) ( 250 ) ( 8.34 ) ] / ( 0.17 ) = 12,265 persons
- Drying bed area = ( 12,265 ) ( 1 sq ft/cap ) = 12,265 sq ft
- Use 10 drying beds, each at 35 ft x 35 ft
- Total area = 12,250 sq ft
Microstrainer...
( a ) Design requirements and criteria :
Determine the solids loaidng on a microstrainer, treating a secondary effluent containing 15 mg/L suspended solids, at a
flow rate of 0.8 mgd.
( b ) Calculations and results :
( 1 ) Determine screen surface area, using a hydraulic loading of 600 gal/sq ft/hr ;
- Surface area = [ ( 800,000 ) ( 1 / 24 ) ] / ( 600 ) = 55.6 sq ft ; use 60 sq ft
- Assuming of the drum surface area is submerged: a drum, 5 ft dia x 3 ft wide, would provide 32 sq ft of submerged screen
area
( 2 ) The solids loading is determined as follows ;
- Solids loading in lb/mil gal = SS concentration in mg/L x 8.34
- 15 mg/L x 8.34 = 125 lb/mil gal
- 125 lb/mil gal x 0.8 mgd = 100 lb/day
- Solids loading = ( 100 lb/day ) / ( 32.0 sq ft ) = 3.1 lb/sq ft/day
- This loading is high and might require solids removal pretreatment if screen clogging occurs
- The required horsepower is taken from the following tabulation
Multi - Media Filtration...
( a ) Design requirements and criteria :
Determine the size of a multi-media gravity filter used in treating a secondary effluent, assuming a hydraulic loading of
5 gpm/sq ft and a flow rate of 0.7 mgd.
( b ) Calculations and results :
( 1 ) Surface area = [ ( 700,000 gal/day ) ( 1 day/1,440 min ) ] / ( 5 gal/min/sq ft ) = 97.2 sq ft ; use 100 sq ft ;
- Provide two 50 sq ft filter units
( 2 ) Depth is determined by the filter media layer depth. A common media depth would be 18 in coal and 6 in sand ;
( 3 ) Backwash pumping ;
- Per filter unit = filter surface areas x backwash rate = 50 sq ft x 15 gpm/sq ft = 750 gpm
- Provide two 750 gpm pumps (includes 1 standby) for each filter unit
( 4 ) Backwash water storage tank volume ;
- Vol = wash rate x waste time period = 750 gpm x 8 min = 6,000 gal/filter unit (mimimum)
- Backwash water may be obtained from the chlorine contact basin if its design volume is, as a minimum, slightly greater
than the required washwater volume
( 5 ) Backwash wastewater is to be returned to the primary settling tanks and the return rate cannot exceed 15 percent of
the design flow ;
- Maximum pumping rate = 0.15 x 700,000 gpd = 105,000 gpd = 73 gpm
- Provide two 70-gpm waste pumps (includes 1 standby) and provide a wastewater storage sump equivalent to backwash water
volume or 6,000 gallons
Activated Carbon Adsorption...
( a ) Design requirements and criteria :
- Q = 1.5 mgd = 1,042 gpm
- Hydraulic loading = 5 gpm/sq ft (determined from pilot unit or estimated)
- Contact time = 30 minutes (determined from laboratory carbon isotherms)
- Secondary effluent BOD = 25 mg/L
- Effluent BOD required = 10 mg/L
- lb BOD removed/lb carbon applied = 0.2 lb BOD rem'd / lb carbon (determined laboratory carbon isotherms)
- Carbon type = Calgon Filtrasorb 300 (determined from laboratory carbon isostherms)
( b ) Calculations and results :
( 1 ) Surface area ;
( 2 ) Bed depth required at 30 min contact time ;
( 3 ) Regeneration of carbon ;
Phosphorus Removal...
( a ) Design requirements and criteria :
Determine what post-secondary mineral treatment for phosphorus removal will meet an effluent requirement of 1 mg/L as P.
Assume the following conditions apply :
- Wastewater flow = 0.5 mgd
- Influent phosphorus concentration = 12 mg/L
- Treatment will be post secondary
- Influent alkalinity = 300 mg/L as CaCO 3
( b ) Calculations and results :
( 1 ) Determine lb/day of phosphorus incoming ;
( 2 ) Determine total dosage rate for mineral addition, using an Al:P weight ratio of 2:1 and an Fe:P ratio of 3:1 ;
Nitrification - Denitrification...
( a ) Design requirements and criteria :
Design a nitrification-denitrification system to treat 20 mg/L of influent nitrogen. Use a separate stage configuration
with suspended growth denitrification. Assume the following conditions apply ;
- Wastewater flow rate = 1.2 mgd
- Temperature = 15 O C
- Operating pH = 7.8
- Nitrification : Influent NH 3 - N concentration = 20 mg/L ; MLVSS = 2,000 mg/L ; Recycle = 100 % of average
flow
- There is no waste sludge
- Denitrification : MLVSS = 2,000 mg/L ; Recycle = 75 % ; Detention time = 2.5 hr
( b ) Calculations and results :
( 1 ) Determine loading ;
( 2 ) Determine tank volume. Use figure given above to estimate the volumetric NH 3 - N loading at
15 O C with an MLVSS concentration of 2,000 mg/L ;
From figure given below, get the performance efficiency and adjust taqnk volume accordingly ; at pH = 7.8, the
efficiency is 88 percent.
Tank volume = ( 12,120 cu ft ) / ( 0.88 ) = 13,773 cu ft
( 3 ) Determine detention time ;
Detention time = [ ( 13,773 cu ft ) ( 7.48 gal/hr ) ] / [ ( 1,200,000 gal/day ) ( 1 day/24 hr ) ] =
2.06 hr ; use 2.1 hr
( 4 ) Set diffused air supply at 1 scf/gal wastewater treated ;
( 5 ) Determine clarifier volume, using an overflow rate of 800 gpd/sq ft and a tank depth of 12 ft ;
- Surface area = ( 1,200,000 gpd ) / ( 800 gpd/sq ft ) = 1,500 sq ft
- Volume = ( 12 ft ) ( 1,500 sq ft ) = 18,000 cu ft
- 1 tank, 20 ft wide by 80 ft long, would be suitable
( 6 ) Assuming 90 percent NH 3 - N reduction in the nitrification stage, determine NO 3 - N and
loading applied to the denitrification stage ( 0.90 ) ( 200 lb/day NH 3 - N ) = 180 lb/day
NO 3 - N
Methanol dose = ( 180 lb/day NH 3 - N ) ( 3 lb methanol / lb NO 3 - N ) = 540 lb methanol/day
( 8 ) Determine tank volume, using detention time ;
Volume = ( 1,200,000 gal/day ) ( 1 day/24 hr ) ( 2.1 hr ) ( 1 cu ft/7.48 gal ) = 14,037 cu ft ; use 14,040
Anaerobic Sludge Digestion...
( a ) Design requirements and criteria :
Determine the digester volume, and gas yield and heat requirement for the anaerobic digestion of combined activated sludge
and primary sludge. Assume the following conditions apply ;
- Sludge amount = 1,200 lb/day
- Sludge solids after thickening = 3%
- Detention time = 15 days
- Volatile matter reduction = 60%
- Temperature = 80 O F (in digester)
- Gas yield rate = 15 cu ft/lb VSS destructed
- Volatile sludge content = 75%
( b ) Calculations and results :
( 1 ) Determine digester volume ;
- Sludge volume = ( 1,200 lb/day ) / [ ( 8.34 ) ( 0.03 ) ] = 4,796 ; use 4,800 gal/day = ( 4,800 ) / ( 7.48 ) =
642 cu ft/day
- Digester volume = ( 642 cu ft/day ) ( 15 days ) = 9,630 cu ft
( 2 ) Determine gas yield ;
Gas yield = ( 15 cu ft/lb VSS destroyed ) ( 1,200 lb/day ) ( 0.75 ) ( 0.60 ) = 8,100 cu ft/day = 5.6 cfm
( 3 ) Using a circular tank 20 feet deep and 26 feet in diameter, determine the heat requirement. Given ;
( 4 ) Determine heat supplied by utilizing sludge gas, assuming a heat value of 600 Btu/cu ft ;
Heat from sludge gas = ( 8,100 cu ft/day ) ( 600 Btu/cu Ft ) = 4,860,000 Btu/day
Aerobic Sludge Digestion...
( a ) Design requirements and criteria :
Determine digester volume and air requirements for the aerobic digestion of activated sludge after thickening. Assume the
following conditions apply ;
- Sludge quantity = 1,200 lb/day
- Sludge concentration = 3 percent
- Detention time = 20 days
- Air supply requirement = 30 cfm/l,000 cu ft digester volume
( b ) Calculations and results :
( 1 ) Determine digester volume ;
- Sludge volume = ( 1,200 lb/day ) / [ ( 8.34 ) ( 0.03 ) ] = 4,796 gal day = ( 4,796 gal day ) ( 1 cu ft / 7.48 gal ) =
641 cu ft/day
- Digester volume = ( 641 cu ft/day ) ( 20 day ) = 12,820 cu ft
( 2 ) Determine air requirement ;
Air required = ( 12,820 cu ft ) ( 30 cfm/1,000 cu ft ) = 384.6 cfm ; use 385 cfm
(This also satisfies mixing requirements.)
Sludge Pumping...
( a ) Design requirements and criteria :
Determine the horsepower and pressure requirements for pumping sludge from a settling tank to a thickener. Assume the
following conditions apply ;
- Sludge is pumped at 6 fps
- 150 ft of 8" pipe is used
- Thickener is 10 ft above settling tank (elev diff = 10')
- Sludge specific gravity = 1.02
- Sludge moisture content = 95 percent
- Pressure at inlet side of pumps = 3 psi
- Coefficient of friction = 0.01
( b ) Calculations and results :
( 1 ) Calculate head loss, using Manning's Equation, to give friction loss ; F (for water) ;
( 2 ) Calculate total pumping head ;
- H = 10 ft + 15 ft = 25 ft of sludge
- Add 3 ft to this account for losses due to valves, elbows, etc.
- Total H = 29 ft of sludge
(3) Assuming a pump efficiency of 60 percent, calculate horsepower requirement ;
Gravity Sludge Thickener...
( a ) Design requirements and criteria :
Size a gravity thickener tank to handle an activated sludge. Assume the following conditions apply ;
- Amount of sludge to thickener = 1,200 lb/day
- Sludge solids content = 1 percent (i.e., 10,000 mg/L)
- Solids loading for thickener = 8 lbs/sq ft/day
- Tank depth = 12 ft
( b ) Calculations and results :
( 1 ) Determine surface area ;
SA = ( 1,200 lb/day ) / ( 8 lb/sq ft/day ) = 150 sq ft
Design a vacuum filter to dewater 10,000 gpd digested sludge containing 5 percent solids. Assume the following
conditions ;
- Specific resistance, r = 3 x 10 7 sec 2 /g
- Required vacuum = 25 in Hg
- Weight of filtered solids per unit volume of filtrate = 4 lb/cu ft
- Cycle time, 0 (from manufacturer) 6 min
- Form time (from manufacturer) 3 min
( b ) Calculations and results :
These values are typical values for digested sludge. For design purposes, it is preferred to use values based on pilot or
laboratory studies.
- Compute the total solids in the sludge ( S ) = ( 0.024 ) ( 8.34 ) ( 50,000 ) = 10,000 lb/day = 417 lb/hr
- Required filter area ( A ) = ( 417 ) / ( 5.56 ) = 75 sq ft
- Therefore ; filter diameter = 4 ft and filter length = 6 ft
- If specific resistance is unknown for a specific application, use loading rates as follows ;
Determine the capacity of a chlorinator for an activated sludge wastewater treatment plant with an average flow of 2 mgd.
The peaking factor for the treatment plant is 2.5. The average required chlorine dosage is 8 mg/L and the maximum required
chlorine dosage is 20 mg/L. EPA regulations require 30 min contact time at peak hour conditions.
( b ) Calculations and results :
( 1 ) Determine capacity of the chlorinator at peak flow ;
- lb Cl 2 /day = ( 20 mg/L ) ( 8.34 lb/gal ) ( 2 mgd ) ( 2.5 ) = 834 lb/day
- Use four 250 lb/day units
( 2 ) Estimate the daily consumption of chlorine ;
It should be noted that the total unit capacity is about 6 times the average needed chlorine. This is to cover the peak
hydraulic flow of wastewater and to cover a wider range of dosage of chlorine that might be needed under unfavorable
conditions. Space requirements for the 4 units and for storage of chlorine tanks is estimated to be about 400 sq ft.
( 3 ) Chlorine contact tank. Required volume for 30 min contact time ;
Baffling is used to construct a more regular tank shape and to prevent short circuiting. Using a flow channel width of 15.3
ft and 4 side-by-side plug flow compartments, the overall tank width is ;
- The length is ; ( 4 ) ( 15.3 ) = 61.2 ft
- The tank dimensions are therefore ; 61.2 ft x 38.3 ft x 8 ft (6 ft SWD)
Mound Systems...
( a ) Design requirements and criteria :
Munitions storage guard house. Maximum capacity: 9 persons.
Area is underlain by coarse sand and irregular rocks. Bedrock is over 50' below site. Soil excavations reveal no hard-pan,
but water table is only 2' below ground.
- Slope = 5 percent
- Percolation rate = 55 min/in at 24 in
- Manifold length = 125 ft
- Septic tank anticipated volume = 2,000 gal
( b ) Calculations and results :
( 1 ) Choose a location : Using soil maps, boreholes and soil pits, choose a location free of trees with no clay lenses or
large boulders in the soil profile. Choose mound site before building the housing.
( 2 ) Calculate waste load : Consider fill occupancy. Whoever is on guard lives there, so anticipate 9 berthing areas.
150 gal/day/room x 9 rooms = 1,350 gal/day
( 3 ) Select fill material: When quality fill is not readily available, the mound system is excessively expensive.
Therefore, utilize the soil maps to locate areas of medium sand. If the sand is not >25 percent coarse sand and <5 percent
fine sand, then it should bbe "up—graded" by screening out fines and materials greater than 2 mm in size, using a trommel.
Final stockpiled material should have an infiltration rate of 1.0 to 1.5 gal/ft 2 /day.
( 4 ) Size the absorption area : Use filtration rate of 1.20 gal/ft 2 /day. Absorption area required = 1,350
gal/day 1.20 gal/ft 2 /day = 1,124 ft 2 . Use a conservative value of 1,200 ft 2 .
Since the area is permeable and of constant slope, a bed system could be used (trenches might be used if the topography
required following a contour or was very narrow). Two long narrow beds, A = 10" wide; each bed will be 60" long (B).
( 5 ) Mound height : At upslope side, minimum depth = 1 ft. At downslope side, depth = E = D + slope (A). Side depth
= E = 1 + 0.05 (25) = 2.25. Bed depth (F) = 1 ft minimum with at least 8 in of aggregate (not limestone) beneath the
distribution piping. Bed total thickness = 1 ft. Cap should be a minimum of 6 in of subsoil and 6 in of reclaimed topsoil.
To provide drainage, use 2.0" at centerline.
( 6 ) Mound length and width ;
( 7 ) Calculate basal area : On a sloping site, the basal area is that area under the slope of the bed; while on level
sites, the whole area of the mound may be used. Since this is a sloped area ;
- Basal area ( BA ) = ( B ) ( A + I ) = ( 60 ) [ ( 2 x 10 ) + 5 + 15 ] = 2,400 ft 2
- Basal area required ( BAR ) = Daily flow infiltration capacity of soil
Typical design loading rates are ;
Therefore, BAR = 900 gal/day x 0.74 gal/ft 2 /day = 666. Since BA is lower than BAR, the slopes could be
shortened. However, if slope is cut to 2:1, erosion may set in. Therefore, size is acceptable at 2,400 ft since this
figure is conservative by a factor of 4.
( 8 ) Distribution system : For this size system, manifold pipe should be 2"; laterals, 1" PVC. The manifold pipe should
run from the pumping chamber (or dosing siphon) to the center of the mound to an "elbow" and then through a "reducer"
(from 2" to 1"), thence through a "tee" to the 6 lateral distribution pipes below. The manifold pipe may be run above or
below the beds but, in either case, must be designed so that wastewater will drain from the laterals and the manifold
between doses. Table given below is used to determine perforation spacing and pipe diameter. In this case, laterals of
1-in PVC are 30 ft long so that a 30-in spacing of 7/32-in holes is chosen.
( 10 ) Dosing volume/pump selection : To be certain that all laterals are quickly and completely filled upon each dosing,
dosing volume is estimated at 10 times void volume, or 350 gallons. Controls, pump size and septic tanks should be chosen
based upon conventional engineering practices and local codes. The design should include a high pressure cutoff to prevent
over—pressurization of the system, a pump chamber with appropriate control, and a pump pedestal to protect the pump from
settled solids. Use figure to check flow rate against mound absorption area. Dosing volume should be about 1/5 th of the
septic tank capacity. In this case, the septic tank is about 2,000 gallons. Since each dose will be 350 gallons, the tank
will always remain 80 percent full, allowing settlement of solids and pumping (or siphoning) of relatively clear effluent
to the mound. The pump must be sized to deliver 2 ft of head at the distal end of laterals after due consideration of ;
( a ) elevation difference between pump and lateral invert and ( b ) friction losses in pump, manifolds and lateral. In
this case, a pump might supply 2.5 ft of head at the supply end to overcome friction losses. A pump capable of delivering
2.5 ft of head at 120 gals/mm would pump for about 3 min to deliver the 350-gal dose. After installation, pressure, timing
and dose rate will be adjusted prior to final burial of the piping. Each dose must fill all laterals to about 2 ft head to
assure even distribution. However, pressure should not be too great as pipes may burst.
( 11 ) When the system is fully laid out with pipes cemented in place, it should be tested and dosing volume adjusted as
necessary to assure that all holes are operating, that all pipes are full at each dosing, and that there is no excess
pressure. Pressure head at end of laterals should be about 2 ft when in full operation at time of pump cutoff. Pipes are
then covered with more gravel, then protected by woven plastic fabric covered by straw. Finally, subsoil and then topsoil
are carefully spread over the top, and the shaped mound is seeded to grass. Do not neglect to ditch the edges of the mound
to channel runoff around the mound.
