P • A • R • T3 ENVIRONMENTAL CONTROL Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. ENVIRONMENTAL CONTROL 13.3 SECTION 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 Flocculation Basin 13.28 Sizing a Polymer Dilution / Feed System 13.30 Design of a Trickling Filter Using NRC Equations 13.31 Design of a Plastic Media Trickling Filter 13.35 Sizing a Rotary-Lobe Sludge Pump 13.38 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 13.51 DESIGN OF A COMPLETE-MIX ACTIVATED SLUDGE REACTOR Domestic wastewater with an average daily flow of 4.0 Mgd (15,140 m 3 /d) has a five day Biochemical Oxygen Demand (BOD 5 ) of 240 mg /L after primary settling. The effluent is to have a BOD 5 of 10 mg /L or less. Design a complete-mix activated sludge reactor to treat the wastewater including reactor volume, hydraulic retention time, quantity of sludge wasted, oxygen requirements, food to microorganism ratio, volumetric loading, and WAS and RAS requirements. Calculation Procedure: 1. Compute the reactor volume The volume of the reactor can be determined using the following equation derived from Monod kinetics: ␪ QY(S Ϫ S) co V ϭ r X (1 ϩ k ␪ ) adc Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 13.4 ENVIRONMENTAL CONTROL where V r ϭ Reactor volume (Mgal) (m 3 ) ␪ c ϭ Mean cell residence time, or the average time that the sludge remains in the reactor (sludge age). For a complete-mix activated sludge pro- cess, ␪ c ranges from 5 to 15 days. The design of the reactor is based on ␪ c on the assumption that substantially all the substrate (BOD) con- version occurs in the reactor, A ␪ c of 8 days will be assumed. Q ϭ Average daily influent flow rate (Mgd) ϭ 4.0 Mgd (15,140 m 3 /d) Y ϭ Maximum yield coefficient (mg VSS/mg BOD 5 ). For the activated sludge process for domestic wastewater Y ranges from 0.4 to 0.8. A Y of 0.6 mg VSS/mg BOD 5 will be assumed. Essentially, Y represents the maximum mg of cells produced per mg organic matter removed. S O ϭ Influent substrate (BOD 5 ) concentration (mg/L) ϭ 240 mg/L S ϭ Effluent substrate (BOD 5 ) concentration (mg/L) ϭ 10 mg/L X a ϭ Concentration of microorganisms in reactor ϭ Mixed Liquor Volatile Suspended Solids (MLVSS) in mg/L. It is generally accepted that the ratio MLVSS/MLSS Ϸ 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 (in- organics). For a complete-mix activated sludge process, MLSS ranges from 1,000 to 6,500 mg /L. An MLSS of 4,500 mg/L will be assumed. ϭϾ MLVSS ϭ (0.8)(4500 mg / L) ϭ 3600 mg/L. k d ϭ Endogenous decay coefficient (d Ϫ 1 ) which is a coefficient representing the 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 Ϫ 1 .A value of 0.06 d Ϫ 1 will be assumed. Therefore: (8 d)(4.0 Mgd)(0.6 mg VSS/mg BOD )(240 Ϫ 10)mg/L 5 V ϭ r Ϫ 1 (3600 mg /L)(1 ϩ (0.06 d )(8d)) 33 ϭ 0.83 Mgal (110,955 ft ) (3140 m ) 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, ␪ ϭ (0.83 Mgal)/(4.0 Mgd) ϭ 0.208 days ϭ 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 ϭ Y/1 ϩ k d ␪ c ϭ 0.6/(1 ϩ (0.06 d Ϫ 1 )(8 d)) ϭ 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: P ϭ YQ(S Ϫ S)(8.34 lb /Mgal/ mg /L) x obs O where P x is the net waste activated sludge produced each day in (lb VSS/d). Using values defined above: mg VSS mg mg lb/ Mgal P ϭ 0.41 (4.0 Mgd) 240 Ϫ 10 8.34 ͩͪͩ ͪͩͪ x mg BOD L L mg/ L 5 ϭ 3146 lb VSS/ d (1428.3 kg VSS/ d) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL WASTEWATER TREATMENT AND CONTROL 13.5 This 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) ϭ P x(ss) ϭ (3146 lb VSS/d) / (0.8) ϭ 3933 lb SS / d (1785.6 kg SS/ d). This represents the total mass of sludge that must be wasted from the system each day. 4. Compute the oxygen requirements based on ultimate carbonaceous oxygen demand (BOD L ) The theoretical oxygen requirements are calculated using the BOD 5 of the waste- water and the amount of organisms (P x ) wasted from the system each day. If all BOD 5 were converted to end products, the total oxygen demand would be computed by converting BOD 5 to ultimate BOD (BOD L ), using an appropriate conversion factor. The ‘‘Quantity of Sludge Wasted’’ calculation illustrated that a portion of the incoming waste is converted to new cells which are subsequently wasted from the system. Therefore, if the BOD L of the wasted cells is subtracted from the total, the remaining amount represents the amount of oxygen that must be supplied to the system. From stoichiometry, it is known that the BOD L of one mole of cells is equal to 1.42 times the concentration of cells. Therefore, the theoretical oxygen requirements for the removal of the carbonaceous organic matter in wastewater for an activated-sludge system can be computed using the following equation: lb O / d ϭ (total mass of BOD utilized, lb /d) 2L Ϫ 1.42 (mass of organisms wasted, lb /d) Using terms that have been defined previously where f ϭ conversion factor for converting BOD 5 to BOD L (0.68 is commonly used): lb/ Mgal Q(S Ϫ S) 8.34 ͩ O ͪ mg/L lb O / d ϭϪ (1.42)(P ) 2 x ƒ Using the above quantities: (4.0 Mgd)(240 mg/ L Ϫ 10 mg /L)(8.34) lb O / d ϭϪ (1.42)(3146 lb /d) 2 0.68 ϭ 6816 lb O /d (3094.5 kg O /d) 22 This represents the theoretical oxygen requirement for removal of the influent BOD 5 . However, to meet sustained peak organic loadings, it is recommended that aeration equipment be designed with a safety factor of at least 2. Therefore, in sizing aeration equipment a value of (2)(6816 lb O 2 /d) ϭ 13,632 lb O 2 /d (6188.9 kg O 2 /d) is used. 5. Compute the food to microorganism ratio (F:M) and the volumetric loading (V L ) In order to maintain control over the activated sludge process, two commonly used parameters are (1) the food to microorganism ratio (F:M) and, (2) the mean cell residence time ( ␪ c ). The mean cell residence time was assumed in Part 1 ‘‘Compute Reactor Volume’’ to be 8 days. The food to microorganism ratio is defined as: Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL 13.6 ENVIRONMENTAL CONTROL F:M ϭ S / ␪ X Oa where F:M is the food to microorganism ratio in d Ϫ 1 . F:M is simply a ratio of the ‘‘food’’ or BOD 5 of the incoming waste, to the concentration of ‘‘microorganisms’’ in the aeration tank or MLVSS. Therefore, us- ing values defined previously: 240 mg /L Ϫ 1 F:M ϭϭ 0.321 d (0.208 d)(3600 mg/ L) Typical values for F:M reported in literature vary from 0.05 d Ϫ 1 to 1.0 d Ϫ 1 depending on the type of treatment process used. A low value of F:M can result in the growth of filamentous organisms and is the most common operational problem in the activated sludge process. A prolifer- ation of filamentous organisms in the mixed liquor results in a poorly settling sludge, commonly referred to as ‘‘bulking sludge.’’ One method of controlling the growth of filamentous organisms is through the use of a separate compartment as the initial contact zone of a biological reactor where primary effluent and return activated sludge are combined. This concept provides 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 Ϫ 1 in this compartment is suggested in the literature. However, initial F:M ratios ranging from 20–25 d Ϫ 1 have also been reported. The volumetric (organic) loading (V L ) is defined as: V ϭ SQ/V ϭ S / ␪ LOrO V L is a measure of the pounds of BOD 5 applied daily per thousand cubic feet of aeration tank volume. Using values defined previously: 33 3 V ϭ (240 mg /L)/ (0.208 d) ϭ 1154 mg /L ⅐ d ϭ 72 lb /10 ft ⅐ d (1.15 kg/ Mm ⅐ d) L Volumetric loading can vary from 20 to more than 200 lb /10 3 ft 3 ⅐ d (0.32 to 3.2 kg/Mm 3 ⅐ d), and may be used as an alternate (although crude) method of sizing aeration 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 of treatment performance under a wide range of operating conditions. The principle factors used in process control are (1) maintaining dissolved-oxygen levels in the aeration 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,’’ the most commonly used parameters for controlling the activated sludge process are the F:M ratio and the mean cell residence time ( ␪ c ). The Mixed Liquor Volatile Suspended 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 mean cell residence time ( ␪ c ). The excess waste activated sludge produced each day (see step 3 ‘‘Compute the Quantity 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 line because it is more concentrated than the mixed liquor in the aeration tank, Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL WASTEWATER TREATMENT AND CONTROL 13.7 FIGURE 2 Aeration tank mass balance. FIGURE 1 Settling tank mass balance. hence smaller waste sludge pumps are required. The waste sludge is generally discharged to sludge thickening and digestion facilities. The alternative method of sludge wasting is to withdraw mixed liquor directly from the aeration tank where the concentration of solids is uniform. Both methods of calculating the waste sludge flow rate are illustrated below. Use Figs. 1 and 2 when performing mass balances for the determination of RAS and WAS. X ϭ Mixed Liquor Suspended Solids (MLSS) - see Part 1 ‘‘Compute the Reactor Volume.’’ Q r ϭ Return activated sludge pumping rate (Mgd) X r ϭ Concentration of sludge in the return line (mg /L). When lacking site specific operational data, a value commonly assumed is 8000 mg/L. Q e ϭ Effluent flow rate (Mgd) X e ϭ Concentration of solids in effluent (mg / L). When lacking site specific operational data, this value is commonly assumed to be zero. Q w ϭ Wasted Activated Sludge (WAS) pumping rate from the reactor (Mgd) ϭ Q w Ј Waste Activated Sludge (WAS) pumping rate from the return line (Mgd) Other variables are as defined previously. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL 13.8 ENVIRONMENTAL CONTROL The actual amount of liquid that must be pumped to achieve process control depends on the method used and the location from which the wasting is to be accomplished. Also note that because the solids capture of the sludge processing facilities (i.e., thickeners, digesters, etc.) is not 100 percent and some solids are returned, the actual wasting rate will be higher than the theoretically determined value. (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 the sludge return line (Fig. 1), the wasting rate is computed using the following: VX r ␪ ϭ c (QX ϩ QX) w Ј ree Assuming that the concentration of solids in the effluent from the settling tank (X e ) is low, then the above equation reduces to: VX VX rr ␪ Ϸ ⇒ Q ϭ cw Ј QX ␪ X w Ј rcr Using values defined previously: (0.83 Mgal)(4500 mg/ L) 3 Q ϭϭ 0.0584 Mgd ϭ 58,400 gal /day (221 m /d) w Ј (8 d) (8000 mg /L) 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 WAS pumping rate from the return line is determined using the following: P ϭ QX(8.34 lb/Mgal / mg/L) x(ss) w Ј r Therefore: 3933 lb /d 3 Q ϭϭ 0.059 Mgd ϭ 59,000 gal /day (223.3 m /d) w Ј (8000 mg /L)(8.34) In this case, the concentration of solids in the sludge return line must be known. Note that regardless of the method used for calculation, if wasting occurs from the return line, the WAS pumping rate is approximately the same. (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 the WAS pumping rate is estimated using the following: VV rr ␪ Ϸ ⇒ Q Ϸ cw Q ␪ w c Using values defined previously: 0.83 Mgal 3 Q ϭϭ 0.104 Mgd ϭ 104,000 gal /day (393.6 m /d) w 8d Note that in case (a) or (b) above, the weight of sludge wasted is the same (3933 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL WASTEWATER TREATMENT AND CONTROL 13.9 lb SS /d) (1785.6 kg SS / d), and that either wasting method will achieve a ␪ c of 8 days. As can be seen, wasting from the aeration tank produces a much higher waste flow rate. This is because the concentration of solids in the bottom of the settling tank (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 pumping rate (and larger WAS pumps) if done from the aeration tank as opposed to the return line. The Return Activated Sludge (RAS) pumping rate is determined by performing a mass balance analysis around either the settling tank or the aeration tank. 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 following equation for RAS pumping rate: XQ Ϫ XQ rw Ј Q ϭ r X Ϫ X r Using values defined previously, the RAS pumping rate is computed to be: (4500 mg /L)(4.0 Mgd) Ϫ (8000 mg /L)(0.0584 Mgd) Q ϭ r 8000 mg /L Ϫ 4500 mg /L 3 ϭ 5.0 Mgd (18,925 m / d) As outlined above, the required RAS pumping rate can also be estimated by performing a mass balance around the aeration tank (Fig. 2). If new cell growth is considered negligible, then the solids entering the tank will equal the solids leaving the tank. Under conditions such as high organic loadings, this assumption may be incorrect. 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 com- pared to the MLSS in the return sludge, the mass balance around the aeration tank yields the following equation for RAS pumping rate: X(Q Ϫ Q ) w Q ϭ r X Ϫ X r Using values defined previously, the RAS pumping rate is computed to be: (4500 mg /L)(4.0 Mgd Ϫ 0.104 Mgd) 3 Q ϭϭ 5.0 Mgd (18,925 m / d) r 8000 mg /L Ϫ 4500 mg /L The ratio of RAS pumping rate to influent flow rate, or recirculation ratio ( ␣ ), may now be calculated: Q 5.0 Mgd r ␣ ϭϭ ϭ 1.25 Q 4.0 Mgd Recirculation 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 so that 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 aeration tank in Fig. 1 and a mass balance performed, or the control volume placed around the settling tank in Fig. 2 and a mass balance performed, that a slightly higher RAS Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL 13.10 ENVIRONMENTAL CONTROL pumping rate would result. However, the difference between these RAS pumping rates 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 m 3 /d) exits the aeration tank of a standard activated sludge treatment process. Design a circular settling 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 size the tanks. However, it is more common that settling facilities must be designed without the benefit settling tests. When this situation develops, published values of surface loading and solids loading rates are generally used. Because of the large amount 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 ϭ 2.5 ϫ 4.0 Mgd ϭ 10.0 Mgd (37,850 m / d) p 2. Find the settling tank surface area using surface loading criteria The recommended surface loading rates (settling tank effluent flow divided by set- tling tank area) vary depending upon the type of activated sludge process used. However, surface loading rates ranging from 200 to 800 gal / day/ft 2 (8.09 to 32.4 L/m 2 ⅐ d) for average flow, and a maximum of 1,000 gal /day/ft 2 (40.7 L /m 2 ⅐ d) for peak flow are accepted design values. The recommended solids loading rate on an activated sludge settling tank also varies 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. The preferred units are lb / ft 2 ⅐ h (kg/m 2 ⅐ h). In effect, the solids loading rate represents a characteristic value for the suspension under consideration. In a settling tank of fixed area, the effluent quality will deteriorate if solids loading is increased beyond the characteristic value for the suspension. Without extensive experimental work covering all seasons and operating variables, higher rates should not be used for design. The recommended solids loading rates vary depending upon the type of activated sludge process selected. However, solids loading rates ranging from 0.8 to 1.2 lb/ ft 2 ⅐ h (3.9 to 5.86 kg/m 2 ⅐ h) for average flow, and 2.0 lb / ft 2 ⅐ h (9.77 kg/m 2 ⅐ h) for peak flow are accepted design values. For a Q p of 10.0 Mgd and a design surface loading rate of 1,000 gal/ day /ft 2 at peak flow, the surface area (A) of a settling tank may be calculated: 6 10 ϫ 10 gal/ day 2 1000 gal /day/ ft ϭ ⇒ A 6 10 ϫ 10 gal/ day 22 A ϭϭ 10,000 ft (929 m ) 2 1000 gal /day/ ft Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. WASTEWATER TREATMENT AND CONTROL [...]... 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 7 feet (2.1 m) and a recirculation ratio of 2 Filter media will consist of rock Size both stages of the trickling filter assuming the efficiency (E ) of each stage is the same Calculation... 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 ϭ (12Ј)(50Ј) ϭ 600 ft2 (55.7 m2) The number of filters required is 1736 ft2 / 600 ft2 per filter ϭ 2.89 Use 3 filters for a total filter area of 1800... 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 5 to 20 seconds A value of 15 seconds will be used here The required volume of the rapid mix basin is... 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: Gϭ (2␲N)2R 32.2 ft / s2 where N ϭ Rotational speed of centrifuge (rev / s) R ϭ Bowl radius, ft (cm) The rotational speed and bowl diameter of the centrifuge will vary depending... 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 expected is: Volume of grit (ft3 / d) ϭ (20 ft3 / Mgal)(4.0 Mgd) ϭ 80 ft3 / d (2.26 m3 / d) Some advantages and disadvantages of the aerated grit chamber are listed below: Advantages The same efficiency of grit removal is possible over a wide flow range Head... volume of stabilized (digested) sludges to minimize the cost of ultimate disposal Because centrifuge equipment is costly and sophisticated, centrifuges are most commonly found in medium to large wastewater treatment facilities The capacity of sludge dewatering to be installed at a given facility is a function of the size of a facility, capability to repair machinery on-site, and the availability of an... lb / h wet cake (1157.7 kg / h) The volume of wet cake, assuming a cake density of 60 lb / ft3 is calculated as follows: Volume of wet cake (ft3 / h) ϭ Wet cake rate, lb / h 2550 lb / h ϭ Cake density, lb / ft3 60 lb / ft3 ϭ 42.5 ft3 / h (1.2 m3 / h) wet cake For a dewatering facility operation of 4 h / d, the volume of dewatered sludge cake to be disposed of per day is: (42.5 ft3 / h)(4 h / d) ϭ 170... The 58,000 gal / day (219.5 m3 / d) of waste activated sludge contains approximately 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, the dry mass of sludge that must be processed is:... the hourly rate of sludge calculated above, and a loading rate of 1,000 lb / h per meter of belt width, the size of the belt thickener is: Belt Width ϭ 918 lb / h ϭ 0.918 m (3.01 ft) 1000 lb / h ⅐ m Use one belt thickener with a 1.0 m belt width Note that one identical belt thickener should be provided as a spare Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)... d) 3 Compute the volume of digested sludge The volume of digested sludge is: Vϭ Ws (␳)(s.g.)(% solids) where V ϭ Sludge volume (ft3) (m3) Ws ϭ Weight of sludge (lb) (kg) ␳ ϭ density of water (62.4 lb / ft3) (994.6 kg / m3) s.g ϭ specific gravity of digested sludge (assume s.g ϭ 1.03) % solids ϭpercent solids expressed as a decimal (incoming sludge: 5.0%) Therefore, the volume of the digested sludge is: . subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS Downloaded from Digital Engineering Library. reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 13.4 ENVIRONMENTAL CONTROL