DESIGN OFA RAPID-MIX BASIN AND FLOCCULATION BASIN

DESIGN OFA RAPID-MIX BASIN AND FLOCCULATION BASIN 1.0 Mdg (3785 m3/d) of equalized secondary effluent from a municipal wastewater treatment facility is to receive tertiary treatment through a direct filtration process which includes rapid mix with a polymer coagulant, flocculation and filtration Size the rapid mix and flocculation basins necessary for direct filtration and determine the horsepower of the required rapid mixers and flocculators Calculation Procedure: Determine the required volume of the rapid mix basin A process flow diagram for direct filtration of a secondary effluent is presented in Fig This form of tertiary wastewater treatment is used following secondary treatment when an essentially "virus-free" effluent is desired for wastewater reclamation and reuse The rapid mix basin is a continuous mixing process in which the principle objective is to maintain the contents of the tank in a completely mixed state Although there are numerous ways to accomplish continuous mixing, mechanical mixing will be used here In mechanical mixing, turbulence is induced through the input of energy by means of rotating impellers such as turbines, paddles, and propellers The hydraulic retention time of typical rapid mix operations in wastewater treatment range from 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) V= (I* s)(l x 10* gal/day) 86,400 s/d = 1?4 m 24 ft3 (0>68 m3) Compute the power required for mixing The power input per volume of liquid is generally used as a rough measure of mixing effectiveness, based on the reasoning that more input power creates greater turbulence, and greater turbulence leads to better mixing The following equation is used to calculate the required power for mixing: G= [-^ V /^ Polymer Addition From Secondary Treatment To Disinfection Rapid Mix (Coagulation) Flocculation FIGURE Process flow for direct filtration Filtration where G = Mean velocity gradient (s l ) P = Power requirement (ft-lb/s) (kW) fju = Dynamic viscosity (lb-s/ft2) (Pa-s) V= Volume of mixing tank (ft3) (m3) G is a measure of the mean velocity gradient in the fluid G values for rapid mixing operations in wastewater treatment range from 250 to 1,500 s"1 A value of 1,000 s"1 will be used here For water at 6O0F (15.50C), dynamic viscosity is 2.36 x IQ-5 lb-s/ft2 (1.13 x 10~3 Pa-s) Therefore, the required power for mixing is computed as follows: P = G2IJiV= (1,000 s~1)2(2.36 x 10-5 Ib-s/ft2)(24 ft3) = 566 ft-lb/s = 1.03 horsepower (0.77 kW) Use the next largest motor size available = 1.5 horsepower (1.12 kW) Therefore, a 1.5 horsepower (1.12 kW) mixer should be used Determine the required volume and power input for flocculation The purpose of flocculation is to form aggregates, or floes, from finely divided matter The larger floes allow a greater solids removal in the subsequent filtration process In the direct filtration process, the wastewater is completely mixed with a polymer coagulant in the rapid mix basin Following rapid mix, the flocculation tanks gently agitate the wastewater so that large "floes" form with the help of the polymer coagulant As in the rapid mix basins, mechanical flocculators will be utilized For flocculation in a direct filtration process, the hydraulic retention time will range from to 10 minutes A retention time of minutes will be used here Therefore, the required volume of the flocculation basin is F= (Sn1In)(I x 1O^ gal/day) l,440mm/d = a l ? f t l G values for flocculation in a direct filtration process range from 20 to 100 s"1 A value of 80 s"1 will be used here Therefore, the power required for flocculation is P = G2VV= (80 s~1)2(2.36 x 10-5)(743 ft3) = 112 ft-lb/s = 0.2 horsepower (0.15 kW) Use the next largest motor size available = 0.5 horsepower (0.37 kW) Therefore, a 0.5 horsepower (0.37 kW) flocculator should be used It is common practice to taper the energy input to flocculation basins so that floes initially formed will not be Broken as they leave the flocculation facilities In the above example, this may be accomplished by providing a second flocculation basin in series with the first The power input to the second basin is calculated using a lower G value (such as 50 s"1) and hence provides a gentler agitation Related Calculations If the flows to the rapid mix and flocculation basin vary significantly, or turn down capability is desired, a variable speed drive should be provided for each mixer and flocculator The variable speed drive should be controlled via an output signal from a flow meter immediately upstream of each respective basin It should be noted that the above analysis provides only approximate values for mixer and flocculator sizes Mixing is in general a "black art," and a mixing manufacturer is usually consulted regarding the best type and size of mixer or flocculator for a particular application SIZING A POLYMER DILUTION/ FEED SYSTEM 1.0 Mgd (3,785 m3/d) of equalized secondary effluent from a municipal wastewater treatment facility is to undergo coagulation and flocculation in a direct filtration process The coagulant used will be an emulsion polymer with 30 percent active ingredient Size the polymer dilution/feed system including: the quantity of dilution water required, and the amount of neat (as supplied) polymer required Calculation Procedure: Determine the daily polymer requirements Depending on the quality of settled secondary effluent, organic polymer addition is often used to enhance the performance of tertiary effluent filters in a direct filtration process: see Design of a Rapid Mix Basin and Flocculation Basin Because the chemistry of the wastewater has a significant effect on the performance of a polymer, the selection of a type of polymer for use as a filter aid generally requires experimental testing Common test procedures for polymers involve adding an initial polymer dosage to the wastewater (usually part per million, ppm) of a given polymer and observing the effects Depending upon the effects observed, the polymer dosage should be increased or decreased by 0.5 ppm increments to obtain an operating range A polymer dosage of ppm (2 parts polymer per x 106 parts wastewater) will be used here In general, the neat polymer is supplied with approximately 25 to 35 percent active polymer, the rest being oil and water As stated above, a 30 percent active polymer will be used for this example The neat polymer is first diluted to an extremely low concentration using dilution water, which consists of either potable water or treated effluent from the wastewater facility The diluted polymer solution usually ranges from 0.005 to 0.5 percent solution The diluted solution is injected into either a rapid mix basin or directly into a pipe A 0.5 percent solution will be used here The gallons per day (gal/day) (L/d) of active polymer required is calculated using the following: Active polymer (gal/day) = (wastewater flow, Mgd) x (active polymer dosage, ppm) Using the values outlined above: Active polymer = (1.0 Mgd)(2 ppm) = gal/day active polymer (pure polymer) = 0.083 gal/hr (gal/h) (0.31 L/h) Find the quantity of dilution water required The quantity of dilution water required is calculated using the following: ^M , ,„, active polymer, gal/h Dilution water (gal/h) = ————^—^——*—.—% solution used (as a decimal) Therefore, using values obtain above: Dilution water = — —— = 16.6 gal/h (62.8 L/h) 0.005 Find the quantity of neat polymer required The quantity of neat polymer required is calculated as follows: , ,„ N active polymer, gal/h Neat polymer (gal/h) = — : ^-7 TT-T % active polymer in emulsion as supplied Using values obtained above: Neat polymer = 0'083 gal/h = 0.277 gal/h (1.05 L/h) This quantity of neat polymer represents the amount of polymer used in its "as supplied" form Therefore, if polymer is supplied in a 55 gallon (208.2 L) drum, the time required to use one drum of polymer (assuming polymer is used 24 h/d, d/wk) is: 55 gal Time required to use one drum of polymer = ————— = 200 h = days 0.277 gal/h DESIGN OFA TRICKLING FILTER USING THE NRC EQUATIONS A municipal wastewater with a flow rate of 1.0 Mgd (3,785 m3/d) and a BOD5 of 240 mg/L is to be treated by a two stage trickling filter system The effluent wastewater is to have a BOD5 of 20 mg/L Both filters are to have a depth of feet (2.1 m) and a recirculation ratio of Filter media will consist of rock Size both stages of the trickling filter assuming the efficiency (E) of each stage is the same Calculation Procedure: Find the efficiency of the trickling filters The modern trickling filter, shown in Fig 7, consists of a bed of highly permeable medium to which microorganisms are attached and through which wastewater is percolated or trickled The filter media usually consists of either rock or a variety of plastic packing materials The depth of rock varies but usually ranges from to feet (0.91 to 244 m) Trickling filters are generally circular, and the wastewater is distributed over the top of the bed by a rotary distributor Filters are constructed with an underdrain system for collecting the treated wastewater and any biological solids that have become detached from the media This underdrain system is important both as a collection unit and as a porous structure through which air can circulate The collected liquid is passed to a settling tank where the solids are separated from the treated wastewater In practice, a portion of the treated wastewater is recycled to dilute the strength of the incoming wastewater and to maintain the biological slime layer in a moist condition The organic material present in the wastewater is degraded by a population of microorganisms attached to the filter media Organic material from the wastewater is absorbed onto the biological slime layer As the slime layer increases in thickness, the microorganisms near the media face lose their ability to cling to the media surface The FIGURE Cutaway view of a trickling filter (Metcalf & Eddy, Wastewater Engineering: Treatment, Disposal, and Reuse, 3rd Ed., McGraw-Hill.) liquid then washes the slime off the media, and a new slime layer starts to grow The phenomenon of losing the slime layer is called "sloughing" and is primarily a function of the organic and hydraulic loading on the filter Two possible process flow schematics for a two-stage trickling filter system are shown in Fig The NRC equations for trickling filter performance are empirical equations which are primarily applicable to single and multistage rock systems with recirculation The overall efficiency of the two-stage trickling filter is calculated using: Overall efficiencyJ = influent BOD5 - effluent BOD5 — x 100 influent BOD5 Using the influent and effluent BOD5 values presented in the problem statement, the overall efficiency is: Overall efficiency = 24° "ff ~ ™ mg/L x I00 = 91.7% 240 mg/L Recycle Influent Primary Clarifier Recycle First-Stage Filter Second-Stage Filter Recycle Influent Primary Clarifier First-Stage Filter Effluent Clarifier Recycle Clarifier Second-Stage Filter FIGURE Two-stage trickling filter process flow schematics Effluent Clarifier Also, overall efficiency = El+ E2(I -E1), and E1=E2 where E1 = The efficiency of the first filter stage, including recirculation and settling (%) E2 = The efficiency of the second filter stage, including recirculation and settling (%) Substituting E1 for E2, setting up as a quadratic equation, and solving for E1: E\ - 2E1 + 0.917 = O = E1 = 0.712 or 71.2% Therefore, the efficiency of each trickling filter stage is 71.2 percent Analyze the first stage filter For a single stage or first stage rock trickling filter, the NRC equation is 100 El= 1+0.0561 /E V VF where W= BOD5 loading to the filter, Ib/d (kg/d) V= Volume of the filter media, 103ft3(m3) F= Recirculation factor 2a Compute the recirculation factor of the filter Recirculation factor represents the average number of passes of the influent organic matter through the trickling filter The recirculation factor is calculated using F= 1+ * [1 + (fl/10)]2 where R = Recirculation ratio = QJQ Qr = Recirculation flow Q = Wastewater flow Using values from above, the recirculation factor is F =_JL±2 = 208 [1+(2/1O)] 2b Compute the BOD5 loading for the first stage filter The BOD5 loading for the first stage filter is calculated using W - (Influent BOD5, mg/L)(Wastewater flow, Mgd)(8.34 Ib/Mgal/mg/L) Using values from above, the BOD5 loading for the first stage filter is W= (240 mg/L)(1.0 Mgd)(8.34) - 2,002 Ib BOD5/d (908.9 kg/d) 2c Compute the volume and diameter of the first stage filter Therefore, the volume of the first stage trickling filter is calculated as follows: 71.2 = 1QQ — 1+0.0561 /2,2002 Ib/d V F(2.08) V= 18.51 103 ft3 (523.8 m3) Using the given depth of feet (2.1 m), and a circular trickling filter, the area and diameter of the first stage filter are Area=^^ depth 18.51 xl 3ft3 = > 4 f t ft diameter=5g-02 ft(177m) Analyze the second stage filter The BOD5 loading for the second stage trickling filter is calculated using W = (I-E1)W where W = BOD5 loading to the second stage filter W=(I- 0.712)(2,002 Ib/d) = 577 Ib BOD5/d (261.9 kg/d) The NRC equation for a second stage trickling filter is ^IT2 _ | 100 0.0561 W 1-E1 ^VF Using terms defined previously and values calculated above, the volume of the second stage trickling filter is 100 E2= | 0.0561 /577 Ib/d 1-0.712 V F(2.08) V= 64.33 103 ft3 (1.82 m3) The area and diameter of the second stage filter are Area = 64'33 * 1^ ft3 = 9,190 ft2 diameter - 108.17 ft (32.97 m) ft Compute the BOD5 loading and hydraulic loading to each filter The BOD5 (organic) loading to each filter is calculated by dividing the BOD5 loading by the volume of the filter in 103 ft3 (m3): First stage filter: BOD5 loading = Jffig^ = 108.2 -^ (1.74kg/m3-d) Second stage filter: BOD5 loading = J™^ = 8.97 -^ (0.14kg/m3-d) BOD5 loading for a first stage filter in a two stage system typically ranges from 60 to 120 lb/103 ft3-d (0.96 to 1.93 kg/m3-d) The second stage filter loading typically ranges from to 20 lb/103 ft3-d (0.08 to 0.32 kg/m3-d) The hydraulic loading to each filter is calculated as follows: (1 + R)(0 Hydraulic loading = — —T-TT (Area)(l ,440mm/d) First stage filter: Hydraulic loading y = (1+2)(1 x 106 gal/day) —: —— (2,644 ft2)( 1,440 min/d) = 0.79 gal/min-ft2 (0.54 L/s-m2) Second stage filter: Hydraulic loading = J (1+2)(1 x 106 gal/day) —: : (9,190 ft2)( ,440 mm/d) = 0.23 gal/min-ft2 (0.156 L/s-m2) Hydraulic loading for two stage trickling filter systems typically ranges from 0.16 to 0.64 gal/min-ft2 (0.11 to 0.43 L/s-m2) Related Calculations In practice, the diameter of the two filters should be rounded to the nearest ft (1.52 m) to accommodate standard rotary distributor mechanisms To reduce construction costs, the two trickling filters are often made the same size When both filters in a two stage trickling filter system are the same size, the efficiencies will be unequal and the analysis will be an iterative one DESIGN OFA PLASTIC MEDIA TRICKLING FILTER A municipal wastewater with a flow of 1.0 Mgd (694 gal/min) (3,785 m3/d) and a BOD5 of 240 mg/L is to be treated in a single stage plastic media trickling filter without recycle The effluent wastewater is to have a BOD5 of 20 mg/L Determine the diameter of the filter, the hydraulic loading, the organic loading, the dosing rate, and the required rotational speed of the distributor arm Assume a filter depth of 25 feet (7.6 m) Also assume that a treatability constant (k20/2o) °f 0.075 (gal/min)0 5/ft2 was obtained in a 20 foot (6.1 m) high test filter at 2O0C (680F) The wastewater temperature is 3O0C (860F) Calculation Procedure: Adjust the treatability constant for wastewater temperature and depth Due to the predictable properties of plastic media, empirical relationships are available to predict performance of trickling filters packed with plastic media However, the treatabil- ity constant must first be adjusted for both the temperature of the wastewater and the depth of the actual filter Adjustment for temperature The treatability constant is first adjusted from the given standard at 2O0C (680F) to the actual wastewater temperature of 3O0C (860F) using the following equation: ^30/2O = £20/20 0r~2° where ^30720 = Treatability constant at 3O0C (860F) and 20 foot (6.1 m) filter depth Ar20720= Treatability constant at 2O0C (680F) and 20 foot (6.1 m) filter depth = Temperature activity coefficient (assume 1.035) T = Wastewater temperature Using above values: £30/20 = (0.075 (gal/min)0 VfF)(1.035)30-20 = 0.106 (gal/min)0 5/ft2 Adjustment for depth The treatability constant is then adjusted from the standard depth of 20 feet (6.1 m) to the actual filter depth of 25 feet (7.6 m) using the following equation: £30/25 = £30/20(^1/^2)* where &30725 = Treatability constant at 3O0C (860F) and 25 foot (7.6 m) filter depth £30/20 = Treatability constant at 3O0C (860F) and 20 foot (6.1 m) filter depth D1 = Depth of reference filter (20 feet) (6.1 m) D2 = Depth of actual filter (25 feet) (7.6 m) jc = Empirical constant (0.3 for plastic medium filters) Using above values: £30/25 = (0.106 (gal/mm)0-5/ft2)(20/25)°-3 = 0.099 (gal/min)0 5/ft2 (0.099 (L/s)° 5/m2) Size the plastic media trickling filter The empirical formula used for sizing plastic media trickling filters is ^=exphftD(a,n ^i where Se = BOD5 of settled effluent from trickling filter (mg/L) S1 = BOD5 of influent wastewater to trickling filter (mg/L) &20 = Treatability constant adjusted for wastewater temperature and filter depth = (£30/25) D= Depth of filter (ft) Qv = Volumetric flowrate applied per unit of filter area (gal/min-ft2) (L/s-m2) = QIA Q = Flowrate applied to filter without recirculation (gal/min) (L/s) A= Area of filter (ft2) (m2) n = Empirical constant (usually 0.5) Rearranging and solving for the trickling filter area (A): / -In(S6IS1) \ i/n ™ (kw2s)D ) Using values from above, the area and diameter of the trickling filter are A = 694 gal/min ( ^^ft J^ =6"'6 ff 29 ft (9>1 m) ' Calculate the hydraulic and organic loading on the filter The hydraulic loading (QIA ) is then calculated: Hydraulic loading = 694 gal/min/699.6 ft2 - 0.99 gal/min-ft2 (0.672 L/s-m2) For plastic media trickling filters, the hydraulic loading ranges from 0.2 to 1.20 gal/min-ft2 (0.14 to 0.82 L/s-m2) The organic loading to the trickling filter is calculated by dividing the BOD5 load to the filter by the filter volume as follows: Organic loading = (1.0 Mgd)(240 mg/L)(8.34 lb-L/mg-Mgal) 3 (699.6 ft )(25 ft)(10 fWlOOO ft ) = 114 To^Fd (557kg/m2'd) For plastic media trickling filters, the organic loading ranges from 30 to 200 Ib/103 ft -d (146.6 to 977.4 kg/m2-d) Determine the required dosing rate for the filter To optimize the treatment performance of a trickling filter, there should be a continual and uniform growth of biomass and sloughing of excess biomass To achieve uniform growth and sloughing, higher periodic dosing rates are required The required dosing rate in inches per pass of distributor arm may be approximated using the following: Dosing rate = (organic loading, lb/103 ft3-d)(0.12) Using the organic loading calculated above, the dosing rate is: Dosing rate = (114 lb/103 ft3-d)(0.12) = 13.7 in/pass (34.8 cm/pass) Typical dosing rates for trickling filters are listed in Table To achieve the typical dosing rates, the speed of the rotary distributor can be controlled by (1) reversing the location of some of the existing orifices to the front of the distributor arm, (2) adding reversed deflectors to the existing orifice discharges, and (3) by operating the rotary distributor with a variable speed drive Determine the required rotational speed of the distributor The rotational speed of the distributor is a function of the instantaneous dosing rate and may be determined using the following: = " L6(6r) (WPK) Vel dty = ° 26 gal/min (7.48 gal/tfX(^4X3 in/12 in/ft)')(60 s/min) = L2 *" (0'37 "^ Using Fig 11 with a velocity of 1.2 ft/s (0.37 m/s) and a solids content of percent, the multiplication factor (k) is 12 Therefore, the dynamic head loss for this system when pumping percent solids is: Dynamic Head Loss5o/0 = (Dynamic Head Losswater)(&) Dynamic Head Loss5o/0 = (4.12')(12) = 49.44 feet (15.1 m) Use 50 feet of head loss (15.2 m) The total head loss is the sum of the static head and the dynamic head Ioss5o/0 Therefore, the total head loss (Total Dynamic Head or TDH) for the system is 10'+ 50'= 60 feet (18.3m) This translates to a discharge pressure on the pump of TDH = (60')/(2.31 ft/psi) = 26 IMn2 (179.1 kPa) Therefore, the design condition for the rotary lobe pump is 26 gal/min (1.64 L/s) at 26 lb/in2 (179 IkPa) Choose the correct pump for the application At this point, a rotary lobe pump manufacturers catalog is required in order to choose the correct pump curve for this application This is accomplished by choosing a pump performance curve that meets the above design condition An example of a manufacturers curve that satisfies the design condition is shown in Fig 12 Plotting a horizontal line from 26 gal/min (1.64 L/s) on the left to the 26 lb/in2 (179.1 kPa) pressure line and reading down gives a pump speed of approximately 175 rpm This means that for this pump to deliver 26 gal/min (1.64 L/s) against a pressure of 26 lb/in2 (179.1 kPa), it must operate at 175 rpm The motor horsepower required for the rotary lobe pump is calculated using the following empirical formula (taken from the catalog of Alfa Laval Pumps, Inc of Kenosha, WI): W r (NKSf) (?)(#/)! HP= ^- [(0.043X^)(P)+ ^- + f^] Hp = Motor horsepower Af= Pump speed (rpm) q = Pump displacement, gal/100 revolutions (L/100 revs) P = Differential pressure or TDH, psi, (kPa) Sf= Factor related to pump size which is calculated using: Sf= (#)(0.757) + Nf= Factor related to the viscosity of the pumped liquid which is calculated using Nf= 2.2 ^/Viscosity (cp) The pump speed was found in Fig 12 to be 175 rpm; the pump displacement is taken from the pump curve and is 25 gal/100 revs (94.6 L/100 revs); the differential pressure or U.S GPM (m3/hr) CAPACITY Test Medium: top Displacement: U.S gal/1 OO rev 25 Port Size: inches (76.2 mm) Speed -RPM FIGURE 12 Pump performance curve (A If a Laval Pumps, Inc.) TDH was calculated above to be 26 lb/in2 (179.1 kPa); Sf= (25 gal/100 revs)(0.757) + = 21.925 Determine the pump horsepower The viscosity of a sewage sludge is dependent upon the percent solids and may be found using Table Using Table 3, a percent solids sludge has a viscosity of 1,050 cp Therefore: Nf= 2.2 Vl,050 cp = 22.36 Using the values outlined above, the motor horsepower is calculated: HP iir K043*25gai/io°revs)(26p s i ) + id*—- + ^u I 175 rpm r (175 rpm)(21.925) (25)(22.36) = 1.3hp(0.97kW) The resulting horsepower of 1.3 hp (0.97 kW) is at best an informed estimate of the horsepower which will be absorbed at the pump shaft during pumping It includes no service factor or margin for error, neither does it allow for inefficiency in power transmission systems TABLE Viscosity of Sewage Sludge vs Percent Solids Percent solids Viscosity in centipoise (cp) 10 80 250 560 1,050 1,760 10 2,750 4,000 5,500 7,500 (Courtesy Alfa Laval Pumps, Inc.) Find the installed horsepower of the pump Actual installed horsepower will need to be greater than this calculated horsepower to allow for the random torque increases due to large solids and to provide a prudent safety factor Therefore, the minimum installed motor horsepower is calculated as follows: Installed Motor hp F= (1.2)(Calculatedhp) —— -——^ x 100 Motor Drive Efficiency (%) Assuming a motor efficiency of 90 percent, the installed motor horsepower is Installed Motor hp = (L2)(1 hp) x 100 = 1.73 hp (1.29 kW) Q^ 90/o Use the next largest motor size available, which is 2.0 hp (1.49 kW) Therefore, the rotary lobe sludge pump will have a 2.0 hp (1.49 kW) motor To summarize, the requirements for a rotary lobe pump for this application are: design condition of 26 gal/min (1.64 L/s) at 26 lb/in2 (179.1 kPa), operating at 175 rpm with a 2.0 hp (1.49 kW) motor Often a variable speed drive is provided so that the pump rpm (hence flow rate) may be adjusted to suit varying sludge flow rate demands at the downstream dewatering facility Also, a second pump is generally provided as a spare DESIGN OF AN ANAEROBIC DIGESTOR A high rate anaerobic digestor is to be designed to treat a mixture of primary and waste activated sludge produced by a wastewater treatment facility The input sludge to the digester is 60,000 gal/day (227.1 m3/d) of primary and waste activated sludge with an average loading of 25,000 Ib/d (11,350 kg/d) of ultimate BOD (BODL) Assume the yield coefficient (Y) is 0.06 Ib VSS/lb BODL (kg/kg), and the endogenous coefficient (kd) is 0.03 d"1 at 350C (950F) Also assume that the efficiency of waste utilization in the digester is 60 percent Compute the digester volume required, the volume of methane gas produced, the total volume of digester gas produced, and the percent stabilization of the sludge Calculation Procedure: Determine the required digester volume and loading Anaerobic digestion is one of the oldest processes used for the stabilization of sludge It involves the decomposition of organic and inorganic matter in the absence of molecular oxygen The major applications of this process ar,e in the stabilization of concentrated sludges produced from the treatment of wastewater In the anaerobic digestion process, the organic material is converted biologically, under anaerobic conditions, to a variety of end products including methane (CH4) and carbon dioxide (CO2) The process is carried out in an airtight reactor Sludge, introduced continuously or intermittently, is retained in the reactor for varying periods of time The stabilized sludge, withdrawn continuously or intermittently from the reactor, is reduced in organic and pathogen content and is nonputrescible In the high rate digestion process, as shown in Fig 13, the contents of the digester are heated and completely mixed For a complete-mix flow through digester, the mean cell residence time (0C) is the same as the hydraulic retention time (B) In the United States, the use and disposal of sewage sludge is regulated under 40 CFR Part 503 promulgated February 1993 The new regulation replaces 40 CFR Part 257—the original regulation governing the use and disposal of sewage sludge, in effect since 1979 The new regulations state that "for anaerobic digestion, the values for the mean-cellresidence time and temperature shall be between 15 days at 350C (950F) to 550C (1310F) and 60 days at 2O0C (680F)." Therefore, for an operating temperature of 350C (950F), a mean cell residence time of 15 days will be used The influent sludge flow rate (0 is 60,000 gal/day = 8,021 fVVd (226.9 m3/d) The digester volume Vrequired is computed using V = 0CQ V= (15