Design of a Complete-Mix Activated Sludge Reactor 13.3 Design of a Circular Settling Tank 13.10 Thickening of a Waste-Activated Sludge Using a Gravity-Belt Thickener 13.12 Design of an A
Trang 1P • A • R • T 3
ENVIRONMENTAL
CONTROL
Trang 3SECTION 13 WASTEWATER TREATMENT
AND CONTROL
Kevin D Wills, M.S.E., P.E.
Consulting Engineer Stanley Consultants, Inc.
Design of a Complete-Mix Activated
Sludge Reactor 13.3
Design of a Circular Settling Tank 13.10
Thickening of a Waste-Activated Sludge
Using a Gravity-Belt Thickener 13.12
Design of an Aerobic Digester 13.14
Design of an Aerated Grit Chamber
13.18
Design of Solid-Bowl Centrifuge for
Sludge Dewatering 13.20
Sizing of a Traveling-Bridge Filter 13.25
Design of a Rapid-Mix Basin and
Design of an Anaerobic Digester 13.43
Design of a Chlorination System for Wastewater Disinfection 13.46
Sanitary Sewer System Design 13.48
Selection of Sewage-Treatment Method
Calculation Procedure:
1. Compute the reactor volume
The volume of the reactor can be determined using the following equation derivedfrom Monod kinetics:
V r⫽
X (1 a ⫹k d c)
Trang 4where V r⫽Reactor volume (Mgal) (m3)
c ⫽Mean cell residence time, or the average time that the sludge remains
in the reactor (sludge age) For a complete-mix activated sludge cess,cranges from 5 to 15 days The design of the reactor is based
pro-oncon the assumption that substantially all the substrate (BOD) version occurs in the reactor, Acof 8 days will be assumed
con-Q ⫽Average daily influent flow rate (Mgd)⫽4.0 Mgd (15,140 m3/ d)
Y⫽Maximum yield coefficient (mg VSS / mg BOD5) For the activated
sludge process for domestic wastewater Y ranges from 0.4 to 0.8 A Y
of 0.6 mg VSS / mg BOD5will be assumed Essentially, Y represents
the maximum mg of cells produced per mg organic matter removed
S O⫽Influent substrate (BOD5) concentration (mg / L) ⫽240 mg / L
S ⫽Effluent substrate (BOD5) concentration (mg / L)⫽10 mg / L
X a⫽Concentration of microorganisms in reactor⫽Mixed Liquor VolatileSuspended Solids (MLVSS) in mg / L It is generally accepted that theratio MLVSS / MLSS⬇ 0.8, where MLSS is the Mixed Liquor Sus-pended Solids concentration in the reactor MLSS represents the sum
of volatile suspended solids (organics) and fixed suspended solids organics) For a complete-mix activated sludge process, MLSS rangesfrom 1,000 to 6,500 mg / L An MLSS of 4,500 mg / L will be assumed
(in-⫽⬎MLVSS⫽ (0.8)(4500 mg / L)⫽3600 mg / L
k d⫽Endogenous decay coefficient (d⫺1) which is a coefficient representingthe decrease of cell mass in the MLVSS For the activated sludge pro-
cess for domestic wastewater k d ranges from 0.025 to 0.075 d⫺1 A
value of 0.06 d⫺1will be assumed
2. Compute the hydraulic retention time
The hydraulic retention time () in the reactor is the reactor volume divided by the
influent flow rate: V r / Q Therefore, ⫽(0.83 Mgal) / (4.0 Mgd) ⫽0.208 days ⫽5.0 hours For a complete-mix activated sludge process, is generally 3–5 hours.Therefore, the hydraulic retention time is acceptable
3. Compute the quantity of sludge wasted
The observed cell yield, Y obs ⫽ Y/1 ⫹ k dc ⫽ 0.6/(1 ⫹ (0.06 d⫺1)(8 d))⫽ 0.41
mg / mg represents the actual cell yield that would be observed The observed cell
yield is always less than the maximum cell yield (Y).
The increase in MLVSS is computed using the following equation:
Trang 5This represents the increase of volatile suspended solids (organics) in the reactor.
Of course the total increase in sludge mass will include fixed suspended solids(inorganics) as well Therefore, the increase in the total mass of mixed liquor sus-pended solids (MLSS)⫽ P x(ss)⫽(3146 lb VSS / d) / (0.8)⫽3933 lb SS / d (1785.6
kg SS / d) This represents the total mass of sludge that must be wasted from thesystem each day
4. Compute the oxygen requirements based on ultimate carbonaceous oxygen demand (BODL)
The theoretical oxygen requirements are calculated using the BOD5 of the
waste-water and the amount of organisms (P x) wasted from the system each day If allBOD5were converted to end products, the total oxygen demand would be computed
by converting BOD5 to ultimate BOD (BODL), using an appropriate conversionfactor The ‘‘Quantity of Sludge Wasted’’ calculation illustrated that a portion ofthe incoming waste is converted to new cells which are subsequently wasted fromthe system Therefore, if the BODLof the wasted cells is subtracted from the total,the remaining amount represents the amount of oxygen that must be supplied tothe system From stoichiometry, it is known that the BODLof one mole of cells isequal to 1.42 times the concentration of cells Therefore, the theoretical oxygenrequirements for the removal of the carbonaceous organic matter in wastewater for
an activated-sludge system can be computed using the following equation:
lb O / d2 ⫽(total mass of BOD utilized, lb / d)L
⫺1.42 (mass of organisms wasted, lb / d)
Using terms that have been defined previously where f⫽ conversion factor forconverting BOD5to BODL(0.68 is commonly used):
lb / Mgal
Q(S O⫺S)冉8.34 mg / L 冊
ƒUsing the above quantities:
The food to microorganism ratio is defined as:
Trang 6F:M⫽S / O X a
where F:M is the food to microorganism ratio in d⫺1
F:M is simply a ratio of the ‘‘food’’ or BOD5 of the incoming waste, to theconcentration of ‘‘microorganisms’’ in the aeration tank or MLVSS Therefore, us-ing values defined previously:
(0.208 d)(3600 mg / L)
Typical values for F:M reported in literature vary from 0.05 d⫺1 to 1.0 d⫺1
depending on the type of treatment process used
A low value of F:M can result in the growth of filamentous organisms and isthe most common operational problem in the activated sludge process A prolifer-ation of filamentous organisms in the mixed liquor results in a poorly settlingsludge, commonly referred to as ‘‘bulking sludge.’’
One method of controlling the growth of filamentous organisms is through theuse of a separate compartment as the initial contact zone of a biological reactorwhere primary effluent and return activated sludge are combined This conceptprovides a high F:M at controlled oxygen levels which provides selective growth
of floc forming organisms at the initial stage of the biological process An F:M
ratio of at least 2.27 d⫺1in this compartment is suggested in the literature However,
initial F:M ratios ranging from 20–25 d⫺1have also been reported
The volumetric (organic) loading (V L) is defined as:
V L⫽S Q / V O r⫽S / O
V L is a measure of the pounds of BOD5applied daily per thousand cubic feet
of aeration tank volume Using values defined previously:
V L⫽(240 mg / L) / (0.208 d)⫽1154 mg / Ld⫽72 lb / 10 ft d (1.15 kg / Mm d)Volumetric loading can vary from 20 to more than 200 lb / 103ft3d (0.32 to 3.2
kg / Mm3d), and may be used as an alternate (although crude) method of sizingaeration tanks
6. Compute the waste activated sludge (WAS) and return activated sludge (RAS) requirements
Control of the activated sludge process is important to maintain high levels oftreatment performance under a wide range of operating conditions The principlefactors used in process control are (1) maintaining dissolved-oxygen levels in theaeration tanks, (2) regulating the amount of Return Activated Sludge (RAS), and(3) controlling the Waste Activated Sludge (WAS) As outlined previously in Part
5 ‘‘Compute the Food to Microorganism Ratio and the Volumetric Loading,’’ themost commonly used parameters for controlling the activated sludge process arethe F:M ratio and the mean cell residence time (c) The Mixed Liquor VolatileSuspended Solids (MLVSS) concentration may also be used as a control parameter.Return Activated Sludge (RAS) is important in maintaining the MLVSS concentra-tion and the Waste Activated Sludge (WAS) is important in controlling the meancell residence time (c)
The excess waste activated sludge produced each day (see step 3 ‘‘Compute theQuantity of Sludge Wasted’’) is wasted from the system to maintain a given F:M
or mean cell residence time Generally, sludge is wasted from the return sludge linebecause it is more concentrated than the mixed liquor in the aeration tank,
Trang 7FIGURE 2 Aeration tank mass balance.
FIGURE 1 Settling tank mass balance.
hence smaller waste sludge pumps are required The waste sludge is generallydischarged to sludge thickening and digestion facilities The alternative method ofsludge wasting is to withdraw mixed liquor directly from the aeration tank wherethe concentration of solids is uniform Both methods of calculating the waste sludgeflow rate are illustrated below
Use Figs 1 and 2 when performing mass balances for the determination ofRAS and WAS
X⫽Mixed Liquor Suspended Solids (MLSS) - see Part 1 ‘‘Compute theReactor Volume.’’
Q r⫽Return activated sludge pumping rate (Mgd)
X r⫽Concentration of sludge in the return line (mg / L) When lacking sitespecific operational data, a value commonly assumed is 8000 mg / L
Q e⫽Effluent flow rate (Mgd)
X e⫽Concentration of solids in effluent (mg / L) When lacking site specificoperational data, this value is commonly assumed to be zero
Q w⫽Wasted Activated Sludge (WAS) pumping rate from the reactor (Mgd)
⫽
Q w⬘ Waste Activated Sludge (WAS) pumping rate from the return line (Mgd)Other variables are as defined previously
Trang 8The actual amount of liquid that must be pumped to achieve process controldepends on the method used and the location from which the wasting is to beaccomplished Also note that because the solids capture of the sludge processingfacilities (i.e., thickeners, digesters, etc.) is not 100 percent and some solids arereturned, the actual wasting rate will be higher than the theoretically determinedvalue.
(a) Waste Activated Sludge (WAS) pumping rate from the return line If the
mean cell residence time is used for process control and the wasting is from thesludge return line (Fig 1), the wasting rate is computed using the following:
V X r
(Q X w ⬘ r⫹Q X ) e e
Assuming that the concentration of solids in the effluent from the settling tank
(X e) is low, then the above equation reduces to:
To determine the WAS pumping rate using this method, the solids concentration
in both the aeration tank and the return line must be known
If the food to microorganism ratio (F:M) method of control is used, the WASpumping rate from the return line is determined using the following:
(b) Waste Activated Sludge (WAS) pumping rate from the aeration tank If the
mean cell residence time is used for process control, wasting is from the aeration
tank (Fig 2), and the solids in the plant effluent (X e) are again neglected, then theWAS pumping rate is estimated using the following:
Trang 9lb SS / d) (1785.6 kg SS / d), and that either wasting method will achieve ac of 8days As can be seen, wasting from the aeration tank produces a much higher wasteflow rate This is because the concentration of solids in the bottom of the settlingtank (and hence the return line) is higher than in the aeration tank Consequently,wasting a given mass of solids per day is going to require a larger WAS pumpingrate (and larger WAS pumps) if done from the aeration tank as opposed to thereturn line The Return Activated Sludge (RAS) pumping rate is determined byperforming a mass balance analysis around either the settling tank or the aerationtank The appropriate control volume for either mass balance analysis is illustrated
in Fig 1 and 2 respectively Assuming that the sludge blanket level in the settling
tank remains constant and that the solids in the effluent from the settling tank (X e)are negligible, a mass balance around the settling tank (Fig 1) yields the followingequation for RAS pumping rate:
⫽5.0 Mgd (18,925 m / d)
As outlined above, the required RAS pumping rate can also be estimated byperforming a mass balance around the aeration tank (Fig 2) If new cell growth isconsidered negligible, then the solids entering the tank will equal the solids leavingthe tank Under conditions such as high organic loadings, this assumption may beincorrect Solids enter the aeration tank in the return sludge and in the influent flow
to the secondary process However, because the influent solids are negligible pared to the MLSS in the return sludge, the mass balance around the aeration tankyields the following equation for RAS pumping rate:
Q r 5.0 Mgd
Q 4.0 MgdRecirculation ratio can vary from 0.25 to 1.50 depending upon the type of ac-tivated sludge process used Common design practice is to size the RAS pumps sothat they are capable of providing a recirculation ratio ranging from 0.50 to 1.50
It should be noted that if the control volume were placed around the aerationtank in Fig 1 and a mass balance performed, or the control volume placed aroundthe settling tank in Fig 2 and a mass balance performed, that a slightly higher RAS
Trang 10pumping rate would result However, the difference between these RAS pumpingrates and the ones calculated above is negligible.
DESIGN OF A CIRCULAR SETTLING TANK
Domestic wastewater with an average daily flow of 4.0 Mgd (15,140 m3/ d) exitsthe aeration tank of a standard activated sludge treatment process Design a circularsettling tank to separate the sludge from the effluent The settling tank will work
in conjunction with the aeration tank Assume a peaking factor of 2.5
Calculation Procedure:
1. Determine the peak flow
Conventional examples of circular settling tank design utilize settling tests to sizethe tanks However, it is more common that settling facilities must be designedwithout the benefit settling tests When this situation develops, published values ofsurface loading and solids loading rates are generally used Because of the largeamount of solids that may be lost in the effluent if design criteria are exceeded,surface loading rates should be based on peak flow conditions Using a peaking
factor of 2.5, the daily peak flow (Q p) is:
3
Q p⫽2.5⫻4.0 Mgd⫽10.0 Mgd (37,850 m / d)
2. Find the settling tank surface area using surface loading criteria
The recommended surface loading rates (settling tank effluent flow divided by tling tank area) vary depending upon the type of activated sludge process used.However, surface loading rates ranging from 200 to 800 gal / day / ft2(8.09 to 32.4
set-L / m2d) for average flow, and a maximum of 1,000 gal / day / ft2 (40.7 L / m2d)for peak flow are accepted design values
The recommended solids loading rate on an activated sludge settling tank alsovaries depending upon the type of activated sludge process used and may be com-puted by dividing the total solids applied by the surface area of the tank Thepreferred units are lb / ft2h (kg / m2h) In effect, the solids loading rate represents
a characteristic value for the suspension under consideration In a settling tank offixed area, the effluent quality will deteriorate if solids loading is increased beyondthe characteristic value for the suspension Without extensive experimental workcovering all seasons and operating variables, higher rates should not be used fordesign The recommended solids loading rates vary depending upon the type ofactivated sludge process selected However, solids loading rates ranging from0.8 to 1.2 lb / ft2h (3.9 to 5.86 kg / m2h) for average flow, and 2.0 lb / ft2h (9.77
kg / m2h) for peak flow are accepted design values
For a Q pof 10.0 Mgd and a design surface loading rate of 1,000 gal / day / ft2atpeak flow, the surface area (A) of a settling tank may be calculated:
6
10 ⫻10 gal / day2
Trang 113. Find the settling tank surface area using solids loading criteria
The total solids load on a clarifier consists of contributions from both the influentand the return activated sludge (RAS) Assume the following (see ‘‘Design of AComplete-Mix Activated Sludge Reactor’’):
Max Solids (lb / d)⫽(Q p⫹Q )(X)(8.34 lb r L / mgMgal)
Using values given above;
Max Solids (lb / d)⫽(10 Mgd⫹5 Mgd)(4,500 mg / L)(8.34 lbL / mgMgal)
⫽562,950 lb / d
⫽23,456 lb / h (10,649 kg / h) of Suspended SolidsTherefore, using a solids loading rate of 2.0 lb / ft2h (9.77 kg / m2h) at peak flow,the surface area of a settling tank may be calculated:
23,456 lb / h
A⫽ 2 ⫽11,728 ft (1089.5 m )2.0 lb / ft h
In this case, the solids loading dominates and dictates the required settling tankarea
4. Select the number of settling tanks
Generally, more than one settling thank would be constructed for operational ibility Two tanks will be sufficient for this example Therefore, the surface area ofeach tank will be 11,728 ft2/ 2⫽5,864 ft2(544.8 m2) The diameter of each settlingtank is then 86.41 ft Use 87 ft (26.5 m) The total area of the two settling tanks
flex-is 11,889 ft2(110.4.5 m2) For an average flow rate of 4.0 Mgd (15,140 m3/ d) and
a RAS flow rate of 5.0 Mgd (18,925 m3), the total solids entering the settling tanks
at average daily flow is:
Solids (lb / d)⫽(4.0 Mgd⫹5.0 Mgd)(4,500 mg / L) (8.34 lbL / mgMgal)
⫽337,770 lb / d⫽14,074 lb / h (6389.6 kg / h)Therefore, the solids loading on the settling tanks at design flow is:
14,074 lb / h
Solids Loading⫽ 2 ⬇1.18 lb / ft h (5.77 kg / m h)
11,889 ftwhich is within the solids loading rate design criteria stated above
Related Calculations. Liquid depth in a circular settling tank is normally sured at the sidewall This is called the sidewater depth The liquid depth is a factor
mea-in the effectiveness of suspended solids removal and mea-in the concentration of thereturn sludge Current design practice favors a minimum sidewater depth of 12 ft(3.66 m) for large circular settling tanks However, depths of up to 20 ft (6.1 m)
Trang 12have been used The advantages of deeper tanks include greater flexibility in eration and a larger margin of safety when changes in the activated sludge systemoccur.
op-THICKENING OF A WASTE-ACTIVATED SLUDGE
USING A GRAVITY BELT THICKENER
A wastewater treatment facility produces 58,000 gal / day (219.5 m3/ d) of wasteactivated sludge containing 0.8 percent solids (8,000 mg / L) Design a gravity beltthickener installation to thicken sludge to 5.0 percent solids based on a normaloperation of 6 h / d and 5 d / wk Use a gravity belt thickener loading rate of 1,000
lb / h (454 kg / h) per meter of belt width Calculate the number and size of gravitybelt thickeners required, the volume of thickened sludge cake, and the solids capture
in percent
Calculation Procedure:
1. Find the dry mass of sludge that must be processed
Gravity belt thickening consists of a gravity belt that moves over rollers driven by
a variable speed drive unit The waste activated sludge is usually pumped from thebottom of a secondary settling tank, conditioned with polymer and fed into afeed / distribution box at one end The box is used to distribute the sludge evenlyacross the width of the moving belt The water drains through the belt as the sludge
is carried toward the discharge end of the thickener The sludge is ridged andfurrowed by a series of plow blades placed along the travel of the belt, allowingthe water released from the sludge to pass through the belt After the thickenedsludge is removed, the belt travels through a wash cycle
The 58,000 gal / day (219.5 m3/ d) of waste activated sludge contains
approxi-mately 3933 lb / d (1785.6 kg / d) of dry solids: See Design of a Complete-Mix
Activated Sludge Reactor, step 3—‘‘Compute the Quantity of Sludge Wasted,’’ and
step 6—‘‘Compute the WAS and RAS Requirements.’’
Based on an operating schedule of 5 days per week and 6 hours per day, thedry mass of sludge that must be processed is:
Weekly Rate: (3,933 lb / d)(7 d / wk)⫽27,531 lb / wk (12,499 kg / wk)
Daily Rate: (27,531 lb / wk) / (5 d / wk)⫽5506 lb / d (2499.7 kg / d)
Hourly Rate: (5506 lb / d) / (6 h / d)⫽918 lb / h (416.8 kg / h)
2. Size the belt thickener
Using the hourly rate of sludge calculated above, and a loading rate of 1,000 lb / hper meter of belt width, the size of the belt thickener is:
918 lb / hBelt Width⫽ ⫽0.918 m (3.01 ft)
1000 lb / hmUse one belt thickener with a 1.0 m belt width Note that one identical belt thickenershould be provided as a spare
Trang 13The thickened sludge flow rate (S) in gal / day (m3/ d) and the filtrate flow rate
(F) in gal / day (m3/ d) are computed by developing solids balance and flow balanceequations:
(a) Solids balance equation Solids in⫽solids out, which implies that: solids
in sludge feed⫽solids in thickened sludge⫹solids in filtrate Assume the ing:
follow-• Sludge feed specific gravity (s.g.)⫽1.01
• Thickened sludge s.g.⫽1.03
• Filtrate s.g.⫽1.0
• Suspended solids in filtrate⫽900 mg / L⫽0.09%
Therefore, the solids balance equation on a daily basis becomes:
5506 lb⫽(S, gal / day)(8.34 lb / gal)(1.03)(0.05)
⫹(F, gal / day)(8.34 lb / gal)(1.0)(0.0009) (1)
⇒5506 lb / d⫽0.4295(S)⫹0.0075(F)
(b) Flow balance equation Flow in ⫽ flow out, which implies that: influentsludge flow rate⫹washwater flow rate⫽thickened sludge flow rate⫹filtrate flowrate Daily influent sludge flow rate ⫽ (58,000 gal / day)(7 / 5) ⫽ 81,200 gal / day(307.3 m3/ d)
3. Compute the thickened sludge and filtrate flow rates
Washwater flow rate is assumed to be 16 gal / min (1.0 L / s) Washwater flow ratevaries from 12 gal / min (0.757 L / s) to 30 gal / min (1.89 L / s) depending on beltthickener size Therefore, with an operating schedule of 6 h / d (360 min / d) theflow balance equation on a daily basis becomes:
81,200 gal / day⫹(16 gal / min)(360 min / d)⫽S⫹F
⇒86,960 gal / day⫽S⫹F (2)Putting (EQ1) and (EQ 2) in matrix format, and solving for thickened sludge flow
rate (S) and filtrate flow rate (F):
F⫽75,458 gal / day (285.6 m / d) of filtrate
Therefore, the volume of thickened waste activated sludge exiting the gravity beltthickener is 11,502 gal / day (43.5 m3/ d) at 5.0 percent solids
4. Determine the solids capture
The solids capture is determined using the following:
Solids in Feed⫺Solids in Filtrate
Solids in FeedUsing values defined previously:
Trang 145506 lb / d⫺[(75,458 gal / day)(8.34 lb / gal)(1.0)(0.0009)]
The thickened sludge is generally pumped immediately to sludge storage tanks
or sludge digestion facilities If the thickener is operated 6.67 h / d, the thickenedsludge pumps (used to pump the thickened sludge to downstream processes) will
be sized based on the following thickened sludge flow rate:
12,823 gal / day
S⫽ ⫽32 gal / min (2.02 L / s)(6.67 h / d)(60 min / h)
Hydraulic (thickened sludge) throughput for a gravity belt thickener ranges from
25 (1.6 L / s) to 100 (6.3 L / s) gal / min per meter of belt width The filtrate flow of84,123 gal / day is generally returned to the head of the wastewater treatment facilityfor reprocessing
DESIGN OF AN AEROBIC DIGESTER
An aerobic digester is to be designed to treat the waste sludge produced by anactivated sludge wastewater treatment facility The input waste sludge will be12,823 gal / day (48.5 L / d) (input 5 d / wk only) of thickened waste activated sludge
at 5.0 percent solids—See Thickening of a Waste Activated Sludge Using a Gravity
Belt Thickener Assume the following apply:
1 The minimum liquid temperature in the winter is 15⬚C (59⬚F), and the maximumliquid temperature in the summer is 30⬚C (86⬚F)
2 The system must achieve a 40 percent Volatile Suspended Solids (VSS)
reduc-tion in the winter
3 Sludge concentration in the digester is 70 percent of the incoming thickened
sludge concentration
4 The volatile fraction of digester suspended solids is 0.8.
Calculation Procedure:
1. Find the daily volume of sludge for disposal
Factors that must be considered in designing aerobic digesters include temperature,solids reduction, tank volume (hydraulic retention time), oxygen requirements andenergy requirements for mixing
Because the majority of aerobic digesters are open tanks, digester liquid peratures are dependent upon weather conditions and can fluctuate extensively As
Trang 15tem-FIGURE 3 VSS reduction in aerobic digester vs liquid temperature
⫻sludge age (Metcalf & Eddy, Wastewater Engineering: Treatment,
Disposal, and Reuse, 3rd Ed., McGraw-Hill.)
with all biological systems, lower temperatures retard the process, whereas highertemperatures accelerate it The design of the aerobic digester should provide thenecessary degree of sludge stabilization at the lowest liquid operating temperatureand should supply the maximum oxygen requirements at the maximum liquid op-erating temperature
A major objective of aerobic digestion is to reduce the mass of the solids fordisposal This reduction is assumed to take place only with the biodegradable con-tent (VSS) of the sludge, although there may be some destruction of the inorganics
as well Typical reduction in VSS ranges from 40 to 50 percent Solids destruction
is primarily a direct function of both basin liquid temperature and sludge age, asindicated in Fig 3 The plot relates VSS reduction to degree-days (temperature⫻sludge age)
To ensure proper operation, the contents of the aerobic digester should be wellmixed In general, because of the large amount of air that must be supplied to meetthe oxygen requirement, adequate mixing is usually achieved However, mixingpower requirements should always be checked
The aerobic digester will operate 7 days per week, unlike the thickening cilities which operate intermittently due to larger operator attention requirements.The thickened sludge is input to the digester at 12,823 gal / day (48.5 L / d), 5 daysper week However, the volume of the sludge to be disposed of daily by the digesterwill be lower due to its operation 7 days per week (the ‘‘bugs’’ do not take the
fa-weekends off) Therefore the volume of sludge to be disposed of daily (Q i) is:
Q i⫽(12,823 gal / day)(5 / 7)⫽9,159 gal / day⫽1,224 ft / d (34.6 m / d)
2. Determine the required VSS reduction
The sludge age required for winter conditions is obtained from Fig 3 using theminimum winter temperature and required VSS reduction
To achieve a 40 percent VSS reduction in the winter, the degree-days requiredfrom Fig 3 is 475⬚Cd Therefore, the required sludge age is 475⬚Cd / 15⬚C ⫽31.7 days During the summer, when the liquid temperature is 30⬚C, the degree-days required is (30⬚C)(31.7 d)⫽ 951⬚Cd From Fig 3, the VSS reduction will
be 46 percent
Trang 16The total mass of solids processed by the digester will be 3,933 lb / d (1785.6
kg / d) which is the total mass of solids wasted from the treatment facility—See
Design of a Complete-Mix Activated Sludge Reactor, step 3 The total mass of VSS
input to the digester is:
3. Compute the volume of digested sludge
The volume of digested sludge is:
s.g.⫽specific gravity of digested sludge (assume s.g.⫽1.03)
% solids ⫽percent solids expressed as a decimal (incoming sludge: 5.0%)
Therefore, the volume of the digested sludge is:
V⫽ 3 ⫽832 ft / d⫽6223 gal / day (23.6 L / d)(62.4 lb / ft )(1.03)(0.05)
During the summer:
4. Find the oxygen and air requirements
The oxygen required to destroy the VSS is approximately 2.3 lb O2/ lb VSS (kg /kg) destroyed Therefore, the oxygen requirements for winter conditions are:(1258 lb VSS / d)(2.3 lb O / lb VSS)2 ⫽2893 lb O / d (1313.4 kg / d)2
The volume of air required at standard conditions (14.7 lb / in2and 68⬚F) (96.5kPa and 20⬚C) assuming air contains 23.2 percent oxygen by weight and the density
of air is 0.075 lb / ft3is:
Trang 172893 lb O / d2
Volume of Air⫽ 3 ⫽166,264 ft / d (4705.3 m / d)
(0.075 lb / ft )(0.232)For summer conditions:
• Winter: volume of air⫽166,264 ft3/ d/(0.1)(1,400 min / d)⫽1155 ft3/ min (32.7
m3/ min)
• Summer: volume of air⫽191,264 ft3/ d/(0.1)(1,440 min / d)⫽1328 ft3/ min (37.6
m3/ min)
To summarize winter and summer conditions:
5. Determine the aerobic digester volume
From the above analysis it is clear that the aerobic digester volume will be lated using values obtained under the winter conditions analysis, while the aerationequipment will be sized using the 1328 ft3/ min (37.6 m3/ min) air requirementobtained under the summer conditions analysis
calcu-The volume of the aerobic digester is computed using the following equation,assuming the digester is loaded with waste activated sludge only:
Q X i i
V⫽X(K P ⫹1 /)
d v c
where V⫽Volume of aerobic digester, ft3(m3)
Q i⫽Influent average flow rate to the digester, ft3/ d (m3/ d)
X i⫽Influent suspended solids, mg / L (50,000 mg / L for 5.0% solids)
X⫽Digester total suspended solids, mg / L
K d⫽Reaction rate constant, d⫺1 May range from 0.05 d⫺1 at 15⬚C (59⬚F)
to 0.14 d⫺1at 25⬚C (77⬚F) (assume 0.06 d⫺1at 15⬚C)
P v⫽Volatile fraction of digester suspended solids (expressed as a decimal)
⫽0.8 (80%) as stated in the initial assumptions
c⫽Solids retention time (sludge age), d
Trang 18Using values obtained above with winter conditions governing, the aerobic digestervolume is:
3(1,224 ft / d)(50,000 mg / L)
DESIGN OF AN AERATED GRIT CHAMBER
Domestic wastewater enters a wastewater treatment facility with an average dailyflow rate of 4.0 Mgd (15,140 L / d) Assuming a peaking factor of 2.5, size anaerated grit chamber for this facility including chamber volume, chamber dimen-sions, air requirement, and grit quantity
Calculation Procedure:
1. Determine the aerated grit chamber volume
Grit removal in a wastewater treatment facility prevents unnecessary abrasion andwear of mechanical equipment such as pumps and scrappers, and grit deposition inpipelines and channels Grit chambers are designed to remove grit (generally char-acterized as nonputrescible solids) consisting of sand, gravel, or other heavy solidmaterials that have settling velocities greater than those of the organic putresciblesolids in the wastewater
In aerated grit chamber systems, air introduced along one side near the bottomcauses a spiral roll velocity pattern perpendicular to the flow through the tank Theheavier particles with their correspondingly higher settling velocities drop to thebottom, while the rolling action suspends the lighter organic particles, which arecarried out of the tank The rolling action induced by the air diffusers is independent
of the flow through the tank Then non flow dependent rolling action allows theaerated grit chamber to operate effectively over a wide range of flows The heavierparticles that settle on the bottom of the tank are moved by the spiral flow of thewater across the tank bottom and into a grit hopper Screw augers or air lift pumpsare generally utilized to remove the grit from the hopper
The velocity of roll governs the size of the particles of a given specific gravitythat will be removed If the velocity is too great, grit will be carried out of thechamber If the velocity is too small, organic material will be removed with thegrit The quantity of air is easily adjusted by throttling the air discharge or usingadjustable speed drives on the blowers With proper adjustment, almost 100 percentgrit removal will be obtained, and the grit will be well washed Grit that is not wellwashed will contain organic matter and become a nuisance through odor emissionand the attraction of insects
Trang 19FIGURE 4 Aerated grit chamber (Metcalf & Eddy, Wastewater Engineering:
Treat-ment, Disposal, and Reuse, 3rd Ed., McGraw-Hill.)
Wastewater will move through the aerated grit chamber in a spiral path asillustrated in Fig 4 The rolling action will make two to three passes across thebottom of the tank at maximum flow and more at lesser flows Wastewater is in-troduced in the direction of the roll
At peak flow rate, the detention time in the aerated grit chamber should rangefrom 2 to 5 minutes A detention time of 3 minutes will be used for this example.Because it is necessary to drain the chamber periodically for routine maintenance,two redundant chambers will be required Therefore, the volume of each chamberis:
(peak flow rate, gal / day)(detention time, min)3
(7.48 gal / ft )(24 h / d)(60 min / h)Using values from above, the chamber volume is:
6(2.5)(4⫻10 gal / day)(3 min)
(7.48 gal / ft )(24 h / d)(60min / h)
2. Determine the dimensions of the grit chamber
Width-depth ratio for aerated grit chambers range from 1:1 to 5:1 Depths rangefrom 7 to 16 feet (2.1 to 4.87 m) Using a width-depth ratio of 1.2:1 and a depth
of 8 feet (2.43 m), the dimensions of the aerated grit chamber are:
Trang 20Width⫽(1.2)(8 ft)⫽9.6 ft (2.92 m)
3
2785 ftLength⫽(volume) / [(width)(depth)]⫽ ⫽36.3 ft (11.1 m)
(8 ft)(9.6 ft)Length-width ratios range from 3:1 to 5:1 As a check, length to width ratio forthe aerated grit chamber sized above is: 36.3 ft / 9.6 ft⫽3.78:1 which is acceptable
3. Determine the air supply required
The air supply requirement for an aerated grit chamber ranges from 2.0 to 5.0 ft3/min / ft of chamber length (0.185 to 0.46 m3/ minm) Using 5.0 ft3/ min / ft (0.46
m3/ minm) for design, the amount of air required is:
Air required (ft / min)⫽(5.0 ft / min / ft)(36.3 ft)⫽182 ft / min (5.2 m / min)
4. Estimate the quantity of grit expected
Grit quantities must be estimated to allow sizing of grit handling equipment such
as grit conveyors and grit dewatering equipment Grit quantities from an aeratedgrit chamber vary from 0.5 to 27 ft3/ Mgal (3.74 to 201.9 m3/ L) of flow Assume
a value of 20 ft3/ Mgal (149.5 m3/ L) Therefore, the average quantity of grit pected is:
Volume of grit (ft / d)⫽(20 ft / Mgal)(4.0 Mgd)⫽80 ft / d (2.26 m / d)Some advantages and disadvantages of the aerated grit chamber are listed below:
The same efficiency of grit removal is
possible over a wide flow range.
Power consumption is higher than other grit removal processes.
Head loss through the grit chamber is
minimal.
Additional labor is required for maintenance and control of the aeration system.
By controlling the rate of aeration, a grit
of relatively low putrescible organic
content can be removed.
Significant quantities of potentially harmful volatile organics and odors may
be released from wastewaters containing these constituents.
Preaeration may alleviate septic conditions
in the incoming wastewater to improve
performance of downstream treatment
units.
Foaming problems may be created if influent wastewater has surfactants present.
Aerated grit chambers can also be used
for chemical addition, mixing,
preaeration, and flocculation ahead of
Trang 21TABLE 1 Facility Capacity & Number of Centrifuges
Facility size, Mgd (m 3 / d)
Dewatering operation, h / d
Centrifuges operating ⫹ spare @ gal / min (L / s)
(Design Manual for Dewatering Municipal Wastewater Sludges, U.S EPA)
design parameters for the specification of a solid bowl centrifuge for dewateringthe sludge including: number of centrifuges, solids feed rate, percent solids recov-ery, dewatered sludge (cake) discharge rate, centrifugal force, polymer dosage, andpolymer feed rate Assume the following apply:
• Feed sludge is aerobically digested at 5.0 percent solids
• Dewatered sludge (cake) is to be 25 percent solids
• Centrate assumed to be 0.3 percent solids
• Polymer solution concentration is 25 percent
Calculation Procedure:
1. Select the number of centrifuges
The separation of a liquid-solid sludge during centrifugal thickening is analogous
to the separation process in a gravity thickener In a centrifuge, however, the appliedforce is centrifugal rather than gravitational and usually exerts 1,500 to 3,500 timesthe force of gravity Separation results from the centrifugal force-driven migration
of the suspended solids through the suspending liquid, away from the axis of tation The increased settling velocity imparted by the centrifugal force as well asthe short settling distance of the particles accounts for the comparatively high ca-pacity of centrifugal equipment
ro-Centrifuges are commonly used for thickening or dewatering Waste ActivatedSludge (WAS) and other biological sludges from secondary wastewater treatment
In the process, centrifuges reduce the volume of stabilized (digested) sludges tominimize the cost of ultimate disposal Because centrifuge equipment is costly andsophisticated, centrifuges are most commonly found in medium to large wastewatertreatment facilities
The capacity of sludge dewatering to be installed at a given facility is a tion of the size of a facility, capability to repair machinery on-site, and the avail-ability of an alternative disposal means Some general guidelines relating the min-imal capacity requirements are listed in Table 1 This table is based on theassumption that there is no alternative mode of sludge disposal and that the capacity
func-to sfunc-tore solids is limited
Using Table 1, the number of centrifuges recommended for a 4.0 Mgd (15,410
m3/ d) wastewater treatment facility is one operational ⫹ one spare for a total oftwo centrifuges
Trang 222. Find the sludge feed rate required
If the dewatering facility is operated 4 h / d, 7 d / wk, then the sludge feed rate is:
Sludge Feed Rate⫽(6,230 gal / day) / [(4 h / d)(60 / min / h)]
⫽26 gal / min (1.64 L / s)Although a 4 h / d operation is below that recommended in Table 1, the sludgefeed rate of 26 gal / min (1.64 L / s) is adequate for the size of centrifuge usuallyfound at a treatment facility of this capacity A longer operational day would benecessary if the dewatering facilities were operated only 5 days per week, or duringextended period of peak flow and solids loading
Assume a feed sludge specific gravity of 1.03 The sludge feed in lb / h iscalculated using the following equation:
(V)()(s.g.)(% solids)(60 min / h)
W s⫽ 7.48 gal / ft3
W s⫽Weight flow rate of sludge feed, lb / h (kg / h)
V⫽Volume flow rate of sludge feed, gal / min (L / s)
s.g.⫽specific gravity of sludge
% solids⫽percent solids expressed as a decimal
⫽density of water, 62.4 lb / ft3(994.6 kg / m3)
Using values obtained above, the sludge feed in lb / h is:
3(26 gal / min)(62.4 lb / ft )(1.03)(0.05)(60 min / h)
W s⫽ 7.48 gal / ft3
⫽670 lb / h of dry solids (304.2 kg / h)
3. Compute the solids capture
Since the solids exiting the centrifuge are split between the centrate and the cake,
it is necessary to use a recovery formula to determine solids capture Recovery isthe mass of solids in the cake divided by the mass of solids in the feed If thesolids content of the feed, centrate and cake are measured, it is possible to calculatepercent recovery without determining total mass of any of the streams The equationfor percent solids recovery is:
C S F⫺C C
R⫽100冉 冊冋 册F C S⫺C C
where R⫽Recovery, percent solids
C s⫽Cake solids, percent solids (25%)
F⫽Feed solids, percent solids (5%)
C c⫽Centrate solids, percent solids (0.3%)
Therefore, using values defined previously:
(5⫺0.3)
R⫽100(25 / 5)冋25⫺0.3册⫽95.14%
Trang 234. Determine the Dewatered Sludge Cake Discharge Rate
The dewatered sludge (cake) discharge rate is calculated using the following:Cake discharge rate (lb / h) dry solids⫽(sludge feed rate, lb / h)(solids recovery)
⫽(670 lb / h)(0.9514)⫽637.5 lb / h (289.4 kg / h) dry cake
The wet cake discharge in lb / h is calculated using the following:
Dry cake rate, lb / hWet cake discharge (lb / h)⫽ Cake % solids
637.5 lb / h
⇒Wet cake discharge (lb / h)⫽ 0.25
⫽2550 lb / h wet cake (1157.7 kg / h)The volume of wet cake, assuming a cake density of 60 lb / ft3 is calculated asfollows:
Wet cake rate, lb / h 2550 lb / h3
Volume of wet cake (ft / h)⫽Cake density, lb / ft3⫽ 60 lb / ft3
⫽42.5 ft / h (1.2 m / h) wet cakeFor a dewatering facility operation of 4 h / d, the volume of dewatered sludge cake
to be disposed of per day is:
(42.5 ft / h)(4 h / d)⫽170 ft / d⫽1272 gal / day (4.81 L / d)
5. Find the percent reduction in sludge volume
The percent reduction in sludge volume is then calculated using the following:
Sludge volume in⫺Sludge volume out
% Volume Reduction⫽ Sludge volume in
In most cases, a compromise is made between the process requirement andO&M considerations Operating at higher speeds helps achieve optimum perform-ance which is weighed against somewhat greater operating and maintenance costs.Increasing bowl speed usually increases solids recovery and cake dryness Todaymost centrifuges used in wastewater applications can provide good clarity and solidsconcentration at G levels between 1,800 and 2,500 times the force of gravity
Trang 246. Compute the centrifugal force in the centrifuge
The centrifugal acceleration force (G), defined as multiples of gravity, is a function
of the rotational speed of the bowl and the distance of the particle from the axis
of rotation In the centrifuge, the centrifugal force, G, is calculated as follows:
2(2N) R
32.2 ft / s
7. Find the polymer feed rate for the centrifuge
The major difficulty encountered in the operation of centrifuges is the disposal ofthe centrate, which is relatively high in suspended, non-settling solids The return
of these solids to the influent of the wastewater treatment facility can result in thepassage of fine solids through the treatment system, reducing effluent quality Twomethods are used to control the fine solids discharge and increase the capture Theseare: (1) increased residence time in the centrifuge, and (2) polymer addition Longerresidence time of the liquid is accomplished by reducing the feed rate or by using
a centrifuge with a larger bowl volume Better clarification of the centrate isachieved by coagulating the sludge prior to centrifugation through polymer addition.Solids capture may be increased from a range of 50 to 80 percent to a range of 80
to 95 percent by longer residence time and chemical conditioning through polymeraddition
In order to obtain a cake solids concentration of 20 to 28 percent for anaerobically digested sludge, 5 to 20 pounds of dry polymer per ton of dry sludgefeed (2.27 to 9.08 kg / ton) is required 15 lb / ton (6.81 kg / ton) will be used for thisexample Usually this value is determined through pilot testing or plant operatortrial and error
The polymer feed rate in lb / h of dry polymer is calculated using the following:
(polymer dosage, lb / ton)(dry sludge feed, lb / h)Polymer feed rate (lb / h)⫽
2000 lb / tonUsing values defined previously, the polymer feed rate is:
(15 lb / ton)(670 lb / h)Polymer feed rate (lb / h)⫽ 2000 lb / ton
⫽5.0 lb / h of dry polymer (2.27 kg / h)Polymer feed rate in gal / h is calculated using the following:
Trang 25polymer feed rate lb / h)Polymer feed rate (gal / h)⫽(8.34 lb / gal)(s.g.)(% polymer concentration)where s.g.⫽specific gravity of the polymer solution
% polymer concentration expressed as a decimal
Using values defined previously:
5.0 lb / hPolymer feed rate (gal / h)⫽(8.34 lb / gal)(1.0)(0.25)
⫽2.4 gal / h of 25% polymer solution (0.009 L / h)The polymer feed rate is used to size the polymer dilution / feed equipment requiredfor the sludge dewatering operation
Related Calculations. Selection of units for dewatering facility design is pendent upon manufacturer’s rating and performance data Several manufacturershave portable pilot plant units, which can be used for field testing if sludge isavailable Wastewater sludges from supposedly similar treatment processes but dif-ferent localities can differ markedly from each other For this reason, pilot planttests should be run, whenever possible, before final design decisions regarding cen-trifuge selection are made
de-SIZING OF A TRAVELING-BRIDGE FILTER
Secondary effluent from a municipal wastewater treatment facility is to receivetertiary treatment, including filtration, through the use of traveling bridge filters.The average daily flow rate is 4.0 Mgd (2778 gal / min) (15,140 m3/ d) and thepeaking factor is 2.5 Determine the size and number of traveling bridge filtersrequired
Calculation Procedure:
1. Determine the peak flow rate for the filter system
The traveling bridge filter is a proprietary form of a rapid sand filter This type offilter is used mainly for filtration of effluent from secondary and advanced waste-water treatment facilities In the traveling bridge filter, the incoming wastewaterfloods the filter bed, flows through the filter medium (usually sand and / or anthra-cite), and exits to an effluent channel via an underdrain and effluent ports locatedunder each filtration cell During the backwash cycle, the carriage and the attachedhood (see Fig 5) move slowly over the filter bed, consecutively isolating and back-washing each cell The washwater pump, located in the effluent channel, drawsfiltered wastewater from the effluent chamber and pumps it through the effluentport of each cell, forcing water to flow up through the cell thereby backwashingthe filter medium of the cell The backwash pump located above the hood drawswater with suspended matter collected under the hood and transfers it to the back-wash water trough During the backwash cycle, wastewater is filtered continuouslythrough the cells not being backwashed
Trang 26FIGURE 5 Traveling bridge filter.
Filtration in a traveling bridge filter is accomplished at a hydraulic loadingtypically in the range of 1.5 to 3.0 gal / min per square foot of filter surface (1.02
to 2.04 L / sm2) at average daily flow The maximum hydraulic loading used fordesign is typically 4.0 gal / min / ft2(2.72 L / sm2) at peak flow The peak hydraulicloading is used to size the traveling bridge filter
The peak flow for this treatment facility is calculated as follows:
Peak flow⫽(peaking factor)(average daily flow)⫽(2.5)(4.0 Mgd)
⫽10 Mdg⫽6944 gal / min (438.2 L / s)
Trang 272. Find the required filter surface area
Filter surface area required is calculated using:
peak flow (gal / min)2
Filter surface area required (ft )⫽Hydraulic loading (gal / minft )2
Using values from above:
6944 gal / min
Filter surface area required (ft )⫽ 2⫽1736 ft (161.3 m )
4.0 gal / minft
3. Determine the number of filters required
Standard filter widths available from various manufacturers are 8, 12, and 16 ft(2.44, 3.66, and 4.88 m) Using a width of 12 feet (3.66 m) and length of 50 feet(15.2 m) per filter, the area of each filter is:
Area of each filter⫽(12⬘)(50⬘)⫽600 ft (55.7 m )The number of filters required is 1736 ft2/ 600 ft2per filter ⫽ 2.89 Use 3 filtersfor a total filter area of 1800 ft2(167.2 m2)
It must be kept in mind that most state and local regulations stipulate that
‘‘rapid sand filters shall be designed to provide a total filtration capacity for themaximum anticipated flow with at least one of the filters out of service.’’ Therefore,
4 traveling bridge filters should be provided, each with filtration area dimensions
of 12⬘ wide⫻50⬘ long (3.66 m⫻15.2 m )
The media depth for traveling bridge filters ranges from 11ⴖ to 16ⴖ (27.9 to40.6 cm) Dual media may be used with 8ⴖ (20.3 cm) of sand underlying 8ⴖ(20.3cm) of anthracite
4. Find the hydraulic loading under various service conditions
The hydraulic loadings with all filters in operation is:
The amount of backwash water produced depends upon the quantity and ity of influent to the filter The backwash pumps are usually sized to deliver ap-
Trang 28qual-FIGURE 6 Process flow for direct filtration.
proximately 25 gal / min (1.58 L / s) during the backwash cycle Backwash water isgenerally returned to the head of the treatment facility for reprocessing
DESIGN OF A RAPID-MIX BASIN AND
FLOCCULATION BASIN
1.0 Mdg (3785 m3/ d) of equalized secondary effluent from a municipal wastewatertreatment facility is to receive tertiary treatment through a direct filtration processwhich includes rapid mix with a polymer coagulant, flocculation and filtration Sizethe rapid mix and flocculation basins necessary for direct filtration and determinethe horsepower of the required rapid mixers and flocculators
Calculation Procedure:
1. Determine the required volume of the rapid mix basin
A process flow diagram for direct filtration of a secondary effluent is presented inFig 6 This form of tertiary wastewater treatment is used following secondarytreatment when an essentially ‘‘virus-free’’ effluent is desired for wastewater rec-lamation and reuse
The rapid mix basin is a continuous mixing process in which the principleobjective is to maintain the contents of the tank in a completely mixed state Al-though there are numerous ways to accomplish continuous mixing, mechanicalmixing will be used here In mechanical mixing, turbulence is induced through theinput of energy by means of rotating impellers such as turbines, paddles, and pro-pellers
The hydraulic retention time of typical rapid mix operations in wastewatertreatment range from 5 to 20 seconds A value of 15 seconds will be used here.The required volume of the rapid mix basin is calculated as follows:
Volume (V )⫽(hydraulic retention time)(wastewater flow)
6(15 s)(1⫻10 gal / day)
86,400 s / d
2. Compute the power required for mixing
The power input per volume of liquid is generally used as a rough measure ofmixing effectiveness, based on the reasoning that more input power creates greater