Biological Treatment

Transcription

1 Biological Treatment Biological treatment is the degradation of organic waste by the action of microorganisms. Biotransformation This refers to the simplification of an organic compound to a daughter compound, or partial degradation. Mineralization This is the complete breakdown of organic molecules into cellular mass, carbon dioxide, water, and inert inorganic residuals. Advantages Gentle reaction temperature and pressure Equipment is simple Disadvantages Long reaction time Large space is required ome wastes are non-biodegradable (pretreatment may be required) 128

2 Basic Microbiology Energy and ubstrate ources Any living matter requires energy and carbon for growth and maintenance The organic waste serves as a source of carbon and energy Heterotrophic bacteria are the most commonly used organisms involved in biological treatment Fungi and higher living forms can be effective in specific application 129

3 Enzymatic Processes Enzymes are large protein molecules composed primarily of amino acids twisted into complex shapes by peptide links and hydrogen bonding Enzymes function as catalysts so they lower the reaction activation energy Enzymes are specific to substrates Different enzymes may be required in a sequential manner to degrade organic waste to successively simple compounds 130

4 ubstrate Biodegradability Biodegradable compounds Recalcitrant or refractory compounds resist degradation Persistent compounds degradation is slow In general, the following conditions are associated with recalcitrant or persistent compounds: Halogenation Large number of halogens Highly branched Low solubility in water Atomic charge difference To determine the biodegradability, we can use tests such as 5-day BOD, ultimate oxygen demand, respirometry studies. 131

5 Determination of BOD dilution to cover different ranges of BOD seeding with a bacterial culture (saprophytic and other micro-organisms and some autotrophic bacteria) that has acclimated to the organic matter or other materials in the wastewater incubation period of five days at the constant temperature of 20 o C measure dissolved oxygen before and after incubation Non-seeded BOD (mg / l) = D 1 D 2 P D1 D2 -B1 B2 f eeded BOD (mg / l) = P D1 = dissolved oxygen of diluted sample immediately after preparation, mg/l D2 = dissolved oxygen of diluted sample after 5 days incubation at 20 o C, mg/l P = decimal volumetric fraction of sample used B1 = dissolved oxygen of seed control before incubation, mg/l B2 = dissolved oxygen of seed control after incubation, mg/l f = ratio of seed in sample to seed in control = (% seed in D1)/ (% seed in B1) 132

6 133

7 134

8 Example: A wastewater sample is diluted by a factor of 10 using seeded dilution water. If the following results are obtained, determine the 5-day and 6-day BOD. The ratio of seed in sample to seed in control, f = 1. Dissolved oxygen, mg/l Time (day) Diluted sample eeded sample P 0.1, 10 5-day BOD: f ( D ) ( ) ( / ) 1 D2 B1 B2 f BOD mg L P ( ) ( )(1) BOD5 59.3( mg / L) day BOD: ( D ) ( ) ( / ) 1 D2 B1 B2 f BOD mg L P ( ) ( )(1) BOD6 61.9( mg / L)

9 The kinetics of the BOD reaction can be formulated with first-order reaction kinetics, for practical purposes: dlt dt kl t where Lt is the amount of the first-stage BOD remaining in the water at time t and k is the reaction rate constant. This equation can be integrated as ln Lt 0 t kt or Lt kt e L where L or BODL is the BOD remaining at time t=0 (i.e., the total or ultimate first-stage BOD initially present). The amount of BOD remaining at time t is kt Lt L( e ) and y, the amount of BOD that has been exerted at any time t, is yt L Lt L( 1 e kt ) Note that the 5-day BOD equals 5k y5 L L5 L( 1 e ) For polluted water and wastewater, k (base e) is around 0.2 ( ) day

10 Example: Determine the 1-day BOD and ultimate first-stage BOD for a wastewater whose 5-day, 20 o C BOD is 200 mg/l. The reaction constant k (base e) = 0.23 day -1. olution: 1. Determine ultimate BOD kt L L( e ) y t 5 L L L 1 e L=293 mg/l L(1 e Determine 1-day BOD k y1 L L1 L(1 e ) = 293(1-e ) = 60 mg/l 5k ) 137

11 3. Nitrogenous Biochemical Oxygen Demand (NBOD) - the oxygen demand associated with the oxidation of ammonia to nitrate. Carbonaceous Biochemical Oxygen Demand (CBOD) - suppressed BOD. Elimination of the interference of nitrifying bacteria by pretreatment or by the use of inhibitory agents. Limitation in the BOD Test A high concentration of active, acclimated seed bacteria is required Pretreatment is needed when dealing with toxic waste, and the effects of nitrifying organisms must be reduced Only biodegradable organics are measured An arbitrary, long period of time is required to obtain results Inhibition and Toxicity An organic substance that is biodegradable at one concentration can become persistent at higher concentrations by inhibiting the growth of the microbial culture. At even higher concentrations, the substance can become toxic to the culture. 138

12 Engineering Factors Electron Acceptor The catabolic reactions involve a transfer of electrons from the waste to an electron acceptor, the biological process by which this occurs is termed respiration. aerobic respiration - bacteria utilize oxygen as the terminal acceptor of electrons removed from oxidized organic compounds. anaerobic process - biotreatment occurs in the absence of oxygen, producing methane. The terminal electron acceptors can be: Nitrates - reduced to nitrogen ulfates - reduced to hydrogen sulfide Carbon dioxide - reduced to methane Moisture Biodegradation requires moisture for two reasons: For cellular growth because cellular tissue is 75-80% moisture As a medium for movement of the microorganisms to the substrate, or vice versa. 139

13 Temperature Temperature has a major influence on growth rate. There is an optimum temperature for cell growth. ph Enzyme activity depends on ph. Total Dissolved olids (TD) TD should not exceed 40,000 mg/l TD should not vary by more than a factor of 2 over a period of a few days. Nutrients Organic carbon (TOC) Phosphorus (P) Nitrogen (N) TOC:N:P ratio is around 20:5:1 Reactor Design and Operation Flow equalization Mixing regime olids retention time Hydraulic retention time Other Factors (carbon source...) 140

14 Growth Kinetics The performance of biological waste treatment can be measured by the rate at which microorganisms metabolize the waste which, in return, is directly related to their rate of growth. Bacteria reproduce by binary fission, and the time required to divide is referred to as generation or doubling time, which can be as little as 15 to 20 minutes. Batch Culture A small inoculation of bacterial cells is introduced to a batch liquid containing a high concentration of a single substrate. The pure culture is then incubated under optimum conditions. The pattern for population growth of such a culture follows five phases: Lag phase - Cells require a period to adapt to the new environment before beginning their culture; pre-adapted cells need a much shorter lag phase. Exponential-growth phase - The biomass grows exponentially as cells divide at a constant rate which is the maximum growth rate for the established conditions. 141

15 tationary phase - Upon approaching exhaustion of the substrate, the number of dyeing cells offsets the number of new cells. Death phase - Without the addition of substrate, dyeing cells outnumber the new cells. The rate can be as great as the inverse of the exponential growth phase. Endogenous phase - In some cases, the rate of population decline slows as dead cells supply the substrate needed for new cells. 142

16 Kinetics of Batch Culture The rate at which cells divide is the specific growth rate or. It is measured as generations per unit time, or the inverse of the doubling time. Without die-off, the bacterial cells accumulate in the batch culture at an exponential rate: dx X (1) dt where X = concentration of biomass = specific growth rate (time -1 ) t = time 143

17 Monod Model The specific growth rate is related to the substrate concentration by the Monod equation: max (2) Ks where max = maximum specific growth rate = concentration of substrate in solution Ks = half-velocity constant (substrate concentration at which the specific growth rate is one-half max) 144

18 ubstituting Eq. (2) into (10) yields dx X max (3) dt Ks The ratio of the increase in cellular mass to the decrease in mass of substrate is termed the growth yield coefficient, Y, which can be expressed as: dx dx / dt Y (4) d d / dt ubstituting (5) into (4) gives d X max dt Y Ks (5) The term max /Y is referred to as the degradation rate constant or maximum rate of substrate removal per unit weight of biomass, k. Therefore, d kx dt Ks (6) and dx YkX (7) dt Ks In reality, not all cells are in the growth phase, a portion of them are in a maintenance or death phase. The net yield in biomass is thus the hypothetical amount of new cell formation [Eq. (7)] less any reduction due to death and maintenance. The reduction can be grouped and 145

19 defined as endogenous decay. Inclusion of decay in Eq. (7) gives the following: dx YkX dt K bx s (8) where b = endogenous decay constant (time -1 ) The basic equations form the foundation for any detailed analysis of the kinetics of biological treatment systems. There are four important kinetic parameters which can be determined only through treatability experiments. 146

20 olids Retention Time (RT) This is also referred to as mean cell residence time or c. A longer value of c results in more efficient degradation, smaller reactor size, and lower cost. If the RT drops below the cell regeneration time, biomass will wash out faster than it forms new cells. Without complete data, conceptual screening studies for activated sludge use a rule of thumb of an RT of about days with X not exceeding 5000 mg/l. A higher value of X will likely result in the clarifier failing. XV c ( RT ) Rate of biomass wastage (9) where V = volume of reactor 147

21 Example. A 5000 liter bioreactor operates at a biomass concentration of 2000 mg/l, measured as mixed liquor volatile suspended solids (MLV), and treats 10,000 liters per day of liquid waste containing 1000 mg/l of total organic carbon (TOC). The suspended solids are separated in a clarifier following the bioreactor with recycle of separated sludge. The recycle flow rate is 5000 liters/day. Each day, 300 liters of recycle are wasted. The effluent from the clarifier contains 40 mg/l MLV. What is the solids retention time (RT), which is defined as XV ( RT ) where c Rate of biomass X = concentration of biomass V = volume of reactor wastage The rule of thumb is that the normal RT is of about days. If the calculated RT is too short, how can you increase it to the required time? 148

22 olution. The conceptual flow and solids balance for a system with recycle is shown above. RT is calculated for a recycle system as follows: XV c ( RT ) (10) Q QW X E QW XU X = 2000 mg/l; V = 5000 liters; Q = 10,000 liters/day; = 1000 mg TOC/L; QU = 5000 liters/day; QW = 300 liters/day. From the solids balance around the clarifier, we have Q QU QW X Q QW X E QU XU olving for the underflow sludge concentration, XU, Q Q U QW X Q QW X X E U Q U (10, )(2000) (10, )(40) mg/l Now solving for RT, XV c ( RT ) Q Q X Q X W (2000)(5000) 4.7 days (10, )(40) (300)(5800) This is a very short RT, probably too short for efficient treatment. To increase the RT, we recycle all the sludge (i.e., QW = 0), the new RT is XV (2000)(5000) c ( RT ) 25 days QX (10,000)(40) E E W U 149

23 Application of Growth and ubstrate Removal Kinetics to Biological Treatment Microorganism and substrate mass balance A mass balance for the microorganisms in the completemix reactor shown below can be written as: Q, 0, X0 Q,, X X, V, Accumulation = Inflow Outflow + Net growth dx dt where r g V QX 0 QX Vr g (11) K max s X bx Yr su bx (12) dx X V QX QX V max 0 bx (13) dt Ks If it is assumed that the concentration of microorganisms in the influent can be neglected and that steady state conditions prevails (dx/dt=0), the above equation can be simplified to yield Q 1 max b (14) V Ks where = hydraulic retention time, V/Q. 150

24 The term 1/ corresponds to the net specific growth rate. In this case there is no biomass recycled so is the same as c, the mean cell residence time. XV XV V c ( RT ) Rate of biomass wastage QX Q Performing a substrate balance similar to the microorganism mass balance, we have d kx V Q0 Q V (15) dt Ks At steady state (d/dt=0), the resulting equation is kx 0 (16) Ks where = V/Q. Effluent microorganism and substrate concentrations By rearranging the substrate mass balance equation (eq. 16) it yields 0 (17) K kx s 151

25 ubstitute this equation into the biomass balance equation (eq. 14) to give 1 0 max b (18) kx olving for X b max b kx Y X max 0 0 Yobs0 k1 b 1 b (19) ubstitute eq. (19) into eq. (16) yields k Y 0 0 K 1 b K s s max 0 1 b K 1 s b Yk b 1 Ks1 b b1 max (20) The observed yield, Yobs, is given as Y Y obs (21) 1 b 152

26 Complete mix with recycle: CENG 4710 Environmental Control, Prof. X. Hu Reactor ettling tank (clarifier) Q, 0 Qe,, Xe X0 = 0 X, V, Qr, Xr, Qw, Xr, The mean hydraulic retention time for the reactor is = V/Q. The mean cell residence time is XV XV c ( RT ) Rate of biomass wastage QwX r Qe X which is different from. Mass balance: Accumulation = Inflow Outflow + Net growth dx dt where r g V Qw X r Qe X e Vrg QX 0 (22) Yr bx (23) su where rsu is the substrate consumption rate and Y is the yield. e 153

27 By assuming that the concentration of microorganisms in the influent is zero and that steady state conditions prevails (dx/dt=0), the above equation can be simplified to yield QwX r Qe X e 1 r Y su b (24) VX c X The term rsu is determined using the following expression: Q r su 0 0 (25) V where (0 - )=mass concentration of substrate utilized 0 = substrate concentration in influent = substrate concentration in effluent The mass concentration of microorganisms X in the reactor can be obtained by substituting Eq. (25) into Eq. (24) and solving for X. 1 Y 0 b c X Y b b c X c c c Y ( 0 ) X (26) (1 b ) Performing a substrate balance, c 154

28 155 ) (1 ) ( ) (1 ) ( 0 max 0 max max 0 c c s c c s s su b K b Y K Y X K Y Y X r ) (1 max c c s b K ) (1 1 ) (1 1 ) (1 max 1 max max c c c b c c c s b b b b K c The effluent substrate concentration is found to be equal to 1 1 max b b K c c s (27) The observed yield is c obs b Y Y 1 (28)

29 Toxic Inhibition Many hazardous wastes can inhibit their own degradation at increased concentrations. The specific growth rate at lower concentrations increases in a linear relationship. However, as the concentration of the toxic substance increases further, the specific growth rate slows, eventually reaching a threshold concentration above which growth ceases. To take into account the toxic inhibition, the Monod model must be modified, and the Haldane equation comes: d kx (29) dt 2 Ks Ki where Ki = inhibition coefficient (mass/volume) The maximum, or *, occurs at concentration *. 156

30 Minimum ubstrate Concentration The minimum sufficient substrate would occur when formation of new biomass equals loss by endogenous decay: YkminX bx (30) Ks min olving for min: bk s min (31) Yk b For aerobic treatment systems, min is in the range of 0.1 to 1.0 mg/l. In contrast, for the specific conditions of many contaminated sites, some toxic organics are considered to be harmful to receptors at lower concentrations as low as 0.01 mg/l. In such cases, it is necessary to induce cometablism or add a primary substrate. Otherwise, effective biodegradation may not occur. Mathematical Approximation for Hazardous Waste Cases For the typical hazardous waste application, the target organic chemical is at a very low concentration, the numerical value of is much smaller than Ks, i.e., << Ks. Hence, eq. (6) becomes: 157

31 d k X (32) dt Ks If the biomass concentration is further assumed to be approximately constant, integration of eq. (32) gives: ln k K Xt (33) 0 s where t = duration of treatment 0 = initial substrate concentration The constant X can be grouped with the two degradation coefficients as follows: k k' X (34) Ks where k = first order degradation rate constant (time -1 ) The contaminant concentration after treatment for the time interval t is: e k ' 0 t (35) In most cases the waste degradation rates are reported in terms of half-life: ln k' (36) t0.5 t0.5 where t0.5 = half-life (days) 158

32 159

33 Conventional Treatment Pretreatment: Equalization - to dampen/modulate hydraulic surges and variable organic loadings in continuous flow systems. Chemical Treatment - typically to precipitate toxic metals, if present, but could involve other steps such as breaking of emulsions. Physical eparation - sedimentation of metallic precipitates, removal of floating materials, etc. Conditioning - typically to supply nutrients and to adjust ph to optimum range. Biodegradation: After pretreatment, the liquid waste flows into the bioreactor where the dissolved organics are metabolized by the biomass with a resulting growth of cellular mass. 160

34 uspended Growth ystems CENG 4710 Environmental Control, Prof. X. Hu The effluent washes the suspended biomass out of the bioreactor to a separation step where the biomass is separated from the treated effluent, typically by sedimentation. ubsequent steps (e.g., filtration) may also be utilized dependent upon the regulatory limitations on the quality of the effluent. A major portion of the settled biomass (i.e., sludge) is returned to the bioreactor to maintain the proper solids retention time (i.e., sludge age) and the proper ratio between substrate and acclimated microorganisms (i.e., food:biomass ratio). 161

35 Recycling increases the solids retention time beyond the simple hydraulic retention time. This speeds degradation and reduces the size of the needed reactor. Excess sludge must be removed for treatment and disposal. The typical sludge age of an effective system is about 20 to 30 days, and no system should have a sludge age outside the range of 10 to 50 days. A lower sludge age would tend to result in system washout, instability, or filamentous organism growth. A higher sludge age could produce non-settling, pinpoint floc. An additional consideration is that the biomass ML should exceed 1000 mg/l to promote proper flocculation. 162

36 Powdered activated carbon may be added to the bioreactor to remove organic compounds not metabolized by the biomass. The largest application of this technology is the 40-MGD activated sludge system operated by Du Pont s Chambers works in Deepwater, NJ, UA. uspended growth systems are best suited to wastes containing moderate to high concentrations of organics, as high as 5000 mg/l of TOC. Low concentrations of organics do not yield enough growth to provide a properly flocculated biomass. This produces poor final settling of biomass such that new cells cannot replace those lost through washout and endogenous decay. 163

37 Example: Biomass calculations. An industrial plant generates a process wastewater stream which has averaged 440 gal/min and 4,800 mg/l of COD over the past three years. The regulatory agency has stated that the organic chemical constituents are hazardous and must be reduced to 100 mg/l measured as COD. The stormwater runoff from the site also contains the organic chemicals of concern, and all runoff from an annual average rainfall of 60 inches must also be treated. The industrial plant conducted some studies and found the following: Rainfall/runoff measurements over three months reported a total of 12 inches of rainfall over that period which yielded 3,300,000 gallons of runoff containing 300 mg/l of COD. Treatability tests of the process wastewater reported the following data: BOD:COD:: 1:2 Biomass production rate (Y) = 0.7 mg/mg BOD removed Endogenous decay rate (b) = 0.05 day -1 The plant is evaluating the feasibility of treating the wastewater and runoff in a full-scale completely mixed, suspended growth system with solids recycle. Based on the above information, could a sufficient biomass concentration be maintained to prevent settling problems? 164

38 olution. Calculate average runoff: Runoff = (3.3 x 10 6 gal/ 12 in) x (60 in/year) = 16.5 x 10 6 gal/year = 45,200 gal/day Calculate average daily flow: Flow = (440 gal/min x 1440 min/day) + 45,200 gal/day = 633,600 gal/day + 45,200 gal/day = 678,800 gal/day Calculate average BOD of influent. (633, mg/l) (45, mg/l) Ave. COD 678,800 = 4500 mg/l Avg. BOD = 0.5 x 4,500 mg/l = 2250 mg/l Estimate suspended biomass (ML): The steady-state equations include a number of interdependent variables. Use of these equations may necessitate assuming values for select variables. These assumed values represent starting points from which refinements may be necessary. For this example, it is first necessary to assume a sludge age (c) and hydraulic detention time () as follows: c = 50 days = 7 days 165

39 With this starting point, the suspended biomass can be calculated from equation (19) for this type of system: Y X c 1 bc = 3100 mg/l This is a reasonable level for the suspended solids in a reactor. It is within the range defined by a minimum of 1000 mg/l and a maximum of 5000 mg/l as discussed earlier. 166

40 Process Design and Control Relationships The term (-rsu/x) is known as the specific substrate utilization rate, U, which is calculated as r Q U su 0 0 X X V X Hence, the net specific growth rate, 1/c, is 1 r Y su b YU b c X To determine the specific utilization ratio U, the substrate utilized and the mass of microorganisms effective in this utilization must be known. The substrate utilized can be evaluated by determining the difference between the influent and the effluent COD or BOD5. The mean cell-residence time, c, is often used as the treatment control parameter. To control the growth rate of microorganisms and hence their degree of waste stabilization, a specified percentage of the cell mass in the system must be wasted each day. For example, if it is determined that a c of 10 days is needed for a desired treatment efficiency, then 10 percent of the total cell mass is wasted from the system per day. 167

41 A term closed related to the specific utilization rate U and commonly used in practice as a design and control parameter is known as the foodmicroorganism ratio (F/M), which is defined as F / M 0 X The terms U and F/M are related by the process efficiency (E) as follows ( F / M ) E U E where E = process efficiency, percent 0 = influent substrate concentration = effluent substrate concentration Example: Activated-sludge process analysis. An organic waste having a soluble BOD5 of 250 mg/l is to be treated with a complete-mix activated sludge process. The effluent BOD5 is to be equal to or less than 20 mg/l. Assume that the temperature is 20 o C, the flowrate is 5.0 Mgal/day, and that the following conditions are applicable. 168

42 1. Influent volatile suspended solids to reactor are negligible. 2. Return sludge concentration = 10,000 mg/l of suspended solids = 8,000 mg/l volatile suspended solids. 3. Mixed-liquor volatile suspended solids (MLV) = biomass concentration = 3,500 mg/l = 0.80 x total ML. 4. Mean cell-residence time c = 10 days. 5. Hydraulic regime of reactor = complete mix. 6. Kinetic coefficients, Y = 0.65 lb cells/lb BOD5 utilized b =0.06 d It is estimated that the effluent will contain about 25 mg/l of biological solids, of which 80 percent is volatile and 65 percent is biodegradable. Assume that the biodegradable biological solids can be converted from ultimate BOD demand to a BOD5 demand using the factor 0.68 [e.g., BOD K value = 0.1 d -1 (base 10)]. 8. Waste contains adequate nitrogen, phosphorus, and other trace nutrients for biological growth. 169

43 olution From the following equation, it is known that the BODL of one mole of cells is equal to 1.42 times the concentration of ultimate BOD. C H NO 113 5(32) cells O2 5CO2 H2O NH energy 1. Estimate the soluble BOD5 in the effluent. Effluent BOD5 = influent soluble BOD5 escaping treatment + BOD5 of effluent biological solids. obs pollutan t cell 20 = + 25(0.65)(1.42)(0.68) = 4.3 mg/l soluble BOD5 The biological treatment efficiency based on soluble BOD5 would be E s (100) 98% 250 The overall plant efficiency would be E overall (100) 92% Compute the reactor volume. The volume of the reactor can be determined by: 170

44 XV YQ c ( 0 ) (1 b ) (3,500mg / L)( V Mgal) V = 1.4 Mgal c 0.65(5Mgal/d)(10 d)( ) (1 (0.06/ d)(10d)) 2. Compute the sludge-production rate on a mass basis. The observed yield is Y 0.65 Yobs 1 b (10) c The biomass production rate is Biomass production, lb V/ld = (Yobs lb/lb)[(0 - ) mg/l][(q Mgal/d) [8.34 (lb/mgal)/(mg/l)] (:unit conversion) = 0.406( )(5)(8.34) = 4,160 lb V/d 171

45 3. Compute the biomass-wasting rate if wasting is accomplished from the reactor, as shown in (a): or from the recycle line, as shown in (b): Also assume that Qe = Q and the V in the effluent is equal to 20 mg/l (0.80 x 25 mg/l). 172

46 (a) Determine the wasting rate from the reactor. XV c Q X Q X w e e 10 Q w (3,500 mg/l)(1.4 Mgal) (3,500 mg/l) (5 Mgal/d)(20 mg/l) Qw = Mgal/day (b) Determine the wasting rate from the recycle line. XV c Q X Q X w r e e 10 Q w (3,500 mg/l)(1.4 Mgal) (8,000 mg/l) (5 Mgal/d)(20 mg/l) Qw = Mgal/day Note that in either case, the weight of sludge wasted is the same (4,160 lb V/d), and that either wasting method will achieve a c of 10 days for the system. 173

47 5. Compute the recirculation ratio using a suspended solids mass balance around the reactor neglecting the suspended solids in the influent. Aerator V conc = 3,500 mg/l Return V conc = 8,000 mg/l 3,500(Q + Qr) = 8,000(Qr) Qr/Q = R = Compute the hydraulic retention time for the reactor. HRT = V/Q = 1.4 Mgal/(5 day) = day = 6.7 hr 7. Check the specific substrate utilization rate, the food-to-microorganism ratio, and the volumetric loading rate (a) The specific substrate utilization rate is 0 ( ) mg/l U X (0.28 day) (3,500 mg/l) mg BOD utilized mg MLV day 174

48 (b) The food-to-microorganism ratio is 250 mg/l F / M 0 X (0.28 d) (3,500 mg/l) mg BOD utilized mg MLV day (c) The volumetric loading rate is ( mg/l)(q Mgal/day) 250(5) mg / L VLR V Mgal 1.4 day 175