BOD and Oxygen Saturation

LECTURE 19

LECTUREOVERYIEW: After a brief introduction to the dissolved oxygen problem, I describe how BOD serves as a means to quantify the oxygen demand of a wastewater. Simple mass balances are developed for the BOD of a batch system. Following n brief review of Ilenry's Inw. Ialso present a general over- view of the saturntiot~of oxygcri in water.

In Lec. I. I stated that engineers originally became involved in water-quality modeling to assess the impact of sewage on receiving waters. We will now study the tnodels that were developed to solve this problem. To plxe these models in context, I'll first describe the cycle of organic production and decomposition that occurs in the biosphere.

19.1 THE ORGANIC PRODUCTION/DECOMPOSITlON CYCLE As tlepicted in Fig. 19.1 the biosphere can be viewed as a cycle of life and death. Powered by the sun. autotrophict organisms (primarily plants) convert simple inor- sanic nutrients into more complex organic molecules. In photosyrrtlresis. solar energy is *stored as chemical energy in the organic molecules. In addition oxygen is liberilted and carbon dioxide is consonicd. The organic matter then serves ils an energy source for heterotrophic organisms (bacteria arid animals) in the reverse processes of respiraliorr and decomposilio:~.

! 'The [ern) nrtrorroplric refers ro or~nrrisrnsIikc pl:inls !hat do nor clc.prnd on o~llcrorpaninnis for nurrition. i In corilrnsr heterorroplric orgnnis~iicconsicl of n~l~rllnl~nrttl 111os1 I~acleriathat whsisl on organic nlatlcr. 348 P~RTIV Dissolved Oxygen and Pathogens LEcrvRE IP BOD and Oxygen Saturation 339

Hlgh energy

t _t RIVER

Salurallon

-h Solar PRODUCTION CTMOSPHERE\DECOMPOSITION Chemlcal Ir energy (plants) energy E __t - Recovery zone I cl0

DIOXIDE t Crltlcal concentratlon I (decornposltlon = reaeratlon) Low energy \GI/ Decomposlllc n Reaeration FIGURE 19.1 dominates dominates The natural cycle of organic production and decomposition. FIGURE 19.2 The dissolved oxygen "sag" that occu-s below sewage discharges into streams. These return the organic matter to the simpler inorganic state. During breakdown, oxygen is consurned and carbon dioxide is liberated. The cycle can bc represented in chemical terms by the following simple expres- solid organic matter. This has two impacts. First, the solid matter makes the water sion: turbid. Thus light cannot penetrate and plant growth is suppressed. Some of the solids settle downstream fro111 the semuge outfall and create sludge beds that can emit noxious odors (Fig. 19.30). Second, the organic niatter provides food for het- respirarion (19.1) erotrophic organisms. Consequently the right side of the cycle in Fig. 19.1 becomes Carbon 'Water Sugar Oxygen dominant. Large populations of decomposer organis~llsbreak down the organic dioxide matter in tlie water and in the process deplete the dissolved oxygen. In addition According to this reversible reaction, carbon dioxide and water are used to syllthesize decomposition of the organic matter takes place in the sludge bed and a sediment organic matter (the sugar glucose) and to create oxygen inthe forward photosynthesis oxygen demand supplements the decay in the water. reaction. Conversely the or:ariic inatter is broken down and oxygen is consumed in As oxygen levels drop. atmospheric oxygen enters the water to compensate for the reverse recpiratioli alld rfecoriiposition reactions. the oxygcli deficit. At first the oxygen consumptio~~in the water and to the sedi- , The chemistry of the prodi~ctionldeconipositio~icycle is far more coniplicatetl ments dwarfs this renerntio~t.However. as the organic lnatter is assimilated and the ' than Eq. 19.1. For exalnple many different organic compounds are created arid bro- oxygen levels tlrop, there will come a point at which the depletion and the reaer- ken down in the process. In adtlition other elements beyond carbon, hytlrogen, and ation will be in balance. At this point the lowest or "criticnl" level of oxygen will , oxygen are involved. In Inter lcctil~csI'll present a rnore colilplete representation. be reached. Beyond this point reaeration dontinates and oxygen levels begin to rise. However. Eq. 19.1 provides ;I st:~rtingpoint lor our elforts to quantify tlie process. In this recovery zone the water becomes clearer beci~useniuch ol tlie solid niatter from the discharge will have settled. In addition inorganic nutrients liberated during the decomposition process will be high. Conscql~entlythe recovery zone will often 19.2 THE DISSOLVED OXYGEN SAG be do~ninatedby the growth of plants. Thus the left side of the cycle from Fig. 19.1 Now that we have the big picture. let's link the lifeldeath cycle with the environment can become o;~erenipliasized. in a stre;itii below a \rastewa[rr discharge (Fig. 19.2). If the stream is origillally un- Beyond the chemical changes. sewage also lends to sigliiticant effects on the polluted. dissolved oxygcl~lc\.cls above [he discharge will be near saturation. The biota. As in Fig. 19.3b and c. molds and bac~eriadominate near the discharge. In inrroduction of tile untreated sewage will elevate the levels of both tlissolved and addition the bacteria thernselves provide a food source for a succession of organisms LECTURE IV BOD and Oxygen Saturation 35 1

consisting of ciliates. rotifers. and crustaceans. The diversity of higher organisms decrea.;es drastically in the degradation and active decomposition zones below the discharge. r\t the same time the total number of organisms increases (Fig. 19.3d __3 River I and e).As recovery ensues. these trends are reversed. All the chernical interactions form the characteristic dissolved oxygen "sag" mg"- Actlve shown in Fig. 19.2. The key feature of the sag is the critical or minimum concentra- Clean water da"on Recovery decomposltlon tion. The location and magnitude of this critical concentration depends on a number of factors, including the size of the loading, the stream's flow and morphometry, water temperature. etc. The goal of the models formulated below will be to simulate the sag as 3 funcrion of ~l~esefactors.

19.3 EXPERIMENT The lir5t step in motleling the DO sag will be to characterize the strength of the wastewater. To tlo this we will focus on the respiration/decomposition portion of the life/de;lth cycle. In terms of the general chemical representation from Eq. 19.1. " this is 300 Sewage bacterla I \ no. m~-I respiralion c6fII2o6 + 602 A 6c02+ 6H20 (19.2) Clllaler no. rnL-1 Rotlferr and no. mL-I (xlM)) crurtaceanr Now let's irnagirie a closed-batch experiment to investigate how this reaction no. mL-I (~25,000) affects a simple environri~er~t(recall Lec. 2). Suppose that you placed a quantity of SrtFibr illto a hottlc of wntcr will1 1111 irlilinl oxygen content of oo. Yo11 nlso ntltl n smnll anlourit of bacteria nntl stopper 111e bottle. Assi~mingthat deconiposition proceeds as a lirst-order reaction. a riiass hnlnrlce for the glucose can be written as

where g = ~lucosecorlcentration (mg-glucose L-I) and kl = the decomposition rate in the bottle (d-I). Il'tlie illitin1 level of glucose is go, this equation can be solved for Sludge worms 1 (el 8 = Roe-kll (19.4) Aquatlc Insects Aqualic Insects (x 25) Next a mass b;llance can be writterl for oxygen,

- -4 0 4 8 12 where o = oxysen co~icer~tr:~tion(mgO L-I) and r,,, = the stoichionletric ratio of Travel time (d) oxygen co11si1111etlto glr~cosetlccomposetl (mgO ~ng-glucose-').Equa~ion 19.4 can be substituted into Eq. 19.5 to give FIGURE 19.3 The changes in the biota below a sewage treatment plan1 eHlenl (redrawn from Bartsch and lngram 1977),

Ifthe initial level of oxygen is 00, tllis equation can be solved for 352 PART IV Dtssolved Oxygen and Pathogens I-ECTIIRE IV BOD and Oxygen Sa~uration 353 I.! ., Accortlicig to this equation. the bottle will originally have a dissolved oxygen level of 00. Thereafter the oxygen will decrease exponentially and asymptotically approach a level of i Oxygen level In bottle

Oxygen equivalents consumed during Before proceeding further, let's work an example. 1 EXAMPLE 19.1. OXYGEN DEPLETION INA CLOSED BATCH SYSTEM. You place 2 mg of glucose in a 250-mL bottle. After adding a small quantity of bacteria. you I! ,. -. fill the remainder of the volume with water arid sropper the bottle. The inlt~alcoricentra- tion of oxygen is 10 mg L-I. If glucose decomposes at a rate of 0. I d-I. determirie the I oxygen concentration as a function of time in this closed batch systern. i i 0 10 20 30 40 50 Solution: Firx we must deterniine the initial concentration of glucose. Time (d) FIGURE E19.1

Nexr we must determine the ratio of mass of oxygen consumed per rnass of glucose decomposed. Using the stoichiometry of Eq. 19.2 we can calculate '0. . 19.4 BIOCHEMICAL OXYGEN DEMAND Tlie experiment outlined at the end of the last section was intended to show how the decomposition process could be modeled for a simple batch system. At face ., , value a similar approach could be used to model how sewage would affect oxygen .- Therefore rota1 clrcornpotition of the glucose would consttlne tlie following amount of levels. However. such an approach woultl be probleniatic because, as mentioned I'xygcll: previously, sewage is not coniposetl of silnpie sugar. Thus to ri_eorously apply the ro,go = 1.0667 x 8 = 8.5333 nig L-' approach, we would have to characterize the concentrations of the myriad organic conlpounds in each sewage sample. We would firrther need to determine the stoi- Consequently the oxygen level in the bottle will ultimately approach (Eq. 19.8) cliioliietry of the decomposition for each reaction. Finally each of the compounds I - could decompose at a different rate. Obviously such a rigorous approach would be !. : .:. , ; i o -. 10 - 8.5333 = 1.4667 mg L-' i~iipractical. 'rI (.I In the early days of modeling. the constraints on such an approach were even Equation 19.7 can then be used to compute the oxygen level as a function of time. more severe because of the limitations of the technology for characterizing organic . ..I. ... . Some results. along with the glucose levels (expressed as oxygen equivalents) computed compounds. As a consequence the lirst water-quality analysts took an empirical . fl.<. with Eq. 19.4 are displayed below: approach and simply disregarded the composition of the sewage. As in our simple experiment from Example 19.1. the analysts introduced some sewage into a Time batch reactor and merely measured how lnucli oxygen was consumed. The resulting Id) qirantity was dubbed biocltetnicnl o.rygett detnnttd or BOD. In ternis of our siniple niodel we can now deline a new variable L (mg0L-I) that is the amount of oxidizable organic matter remaining in the bottle expressed as oxygen equivalents. A mass balance for L for the batch system can be written as

Ifthe initial level is Lo, this equation can be solved for I. A graph of the rcsrllts can also be generated: 354 PART IV Dissolved Oxygen and Parhogens .) LECTURE IY BOD and Oxygen Saturation 355 Note that the oxygen consumed during the decomposition process can be de- tined as Thus by disregarding the exact composition of the organic matter. we avoid the ne- cessity of characterizing the orgiinic matter and its stoichiometry relative to oxygen. or substituting Eq. 19.10, Finally it should be noted that although the exact co~iipositionof the organic matter in wastewater is not characterized. the organic carbon content can be measured directly. In these cases Eq. 19.16 can be reformulated to estimate the BOD where y = BOD tmgO L-I). We can now see that the value can be defined as on the basis of organic carbon content, as in either the initial concentration of oxidizable organic matter (expressed in oxygen Lo = rucCorB (19.17) units) or as the ultimate BOD. This notion is reinforced by Fig. 19.4. which shows both Eq. 19.10 and 19.12. wt :re C,,, = organic carbon concentration of the wastewater (mgC L-I) and r,,, = Next a mass balance can be written for oxygen, ratio of mass of oxygen consumed per mass of carbon assimilated (mgO mgc-I). Again. Eq. 19.2 provides a basis for estimating the ratio

If the initial level of oxygen is 00, this equation can be solved for Finally it should be noted that along with the deconlposition of carbonaceous matter. an additio~ialoxygen demand is exerted due to the oxidation of ammonia to nitrate in the process called nitrijcation. The oxygen demand due to nitrification According to this equation the bottle will originally have a dissolved oxygen is sornetirnes referred to as nitrogerrorrs BOD or NBOD to distinguish it from the level of 00. Thereafter the oxylen will decrease exponentially and asynlptotically carboriaccous BOD or CBOD described above. In Lec. 23 we will describe nitrifica- approach a level of tion and NBOD in detail.

Notice that this development is identical to the glucose experiment. In fact the 19.5 BOD MODEL FOR A STREAM glucose experinierit could be modeled in tenns of BOD by substituting Now we can take our rnodel and apply it to a stream below a sewage treatment plant. When we do this we must now consider that, in addition to decomposition, the BOD can also be re~iiovedby sedimentation. Therefore the mass balance for a constant- flow, constant-geometry channel can be written as

where k, = total removal rate (d-I), which is coniposerl of both decomposition and settling, k, = kd + k, ( 19.20) where kd = deco~npositionrate in the stream (d-I) a~irlk, = settling removal rate (d-I). Recognize that the deco~npositionrate represents the sariie type of process as ! i the kl from the bottle experiment (recall Eqs. 19.3 and 19.9). We have used a different subscript in Eq. 19.20 to highlight that deco~npositionin a natural environment such as a river will generally be different from that in a bottle. In addition note that the settling rate is related to more fundamental parameters by Tlme (d)

FIGURE 19.4 The value L, can be defined as either the initial concentration of oxidizable organic matter or as Ihe ultimate BOD. where v, = BOD settling velocity (m d-I) and H = water depth (m). 356 PART IV Dissolved Oxygen and Pathogens LEcrvRE 19 BOD and Oxygen Satura~ion 357

At steady-state. Eq. 19.19 becomes kr InL = Inh- -r ( 19.25) U ' Consequently if the simple model holds. a plot of In L versus .r/U (that is. travel time) should yield a straight line with o slope of k,. Ifcomplete mixing is assumed at the location of the discharge, an initial concentra- Figure 19.5 shows a typical pattern that might be observed in a stream receiving tion can be calculated as the flow-weighted average (recall Eq. 9.42) of the loading untreated sewage. Note how the rate is higher immediately below the discharge. (subscript w) and the BOD in the river upstream of the discharge (subscript r), Several factors underlie such a pattern. In particular the higher rates are usually caused by the fast degradation of readily decomposable organics and the settling of, sewage particulates.

Using this value as an initial condition. we can solve Eq. 19.22 for

kr 19.6 BOD LOADINGS, CONCENTRATIONS, AND RATES L = be-77-' 1 19.24) Before proceeding to other aspects of DO modeling. let's review some of the param- Thus the BOD is reduced by decomposition and settling as it is carried downstream. eter.; that relate to bioche~iiicaloxygen demand. Aside from modeling the distribution of BOD below a point source in a plug-flow stream with constant parameters. Eq. 19.24 provides a framework for estimating the removal rate in such systems. To do this, the natural logarithm can be taken, 19.6.1 BODS(5-Day BOD) As in Table 19.1, typical values for the BOD bottle decay rate range from 0.05 to 0.5 d-I. with a geometric mean of about 0. I5 d-I. This information can be used to estimate a 95% response time for the bottle test as rs5 = 310.15 = 20 d. Because such a long measilrernent period is unacceptable, water-quality analysts early on ___C River ;~tloptcda 5-tl:\y tcsl. Although shortening the incubation time to 5 d makes the test practical, we must then have a nieans to extrapolate the 5-d result to the ultimate BOD level. This is usually done by perforniing a long-tern1 BOD to estimate the decay rate. If the first- order decornposition model holds, Eq. 19. I2 can be used to compilte

where y5 = 5-day BOD. Tahle 19.1 includes typical values for the ratio of 5-day to ultimate BOD. I \ Deep stream 19.6.2 BOD Loadings and Concentrations Table 19.3- provides typical valr~esof flow rate and BOD for raw sewage from both the United States and developing countries. In general the Row rate for the United

Tt\RI,E 19.1 FIGURE 19.5 Typical val~~esof the BOD bottle decomposition Plot of BOO downstream from a point source of untreated sewage rate for various levels of treatment. BOD, is the into a plug-flow river having constant hydrogeometric characteristics. r~ltiniateBOI). Values here are for CBOD In the initial stretch, high BOO removal rates will occur due to settling and fast decomposition of easily degradable organic matter. Further Trent nletlt k,(ZO"C) H01),1001). downstream the lower removal rates will occur as the more refractory Ut~trented 0.35 (0.20-0.50) 0.83 organic matter degrades at a slower rate. Prim:~ry 0.20 (0.10-0.30) 0.63 Activarcd sludge 0.075 (0.05 -0.10) 0.3 1 ,. 358 P~RTIV Di~solvedOxygen and Pathogens LE~RE19 BOD and Oxygen Saturation 359

TABLE 19.2 10 Typical loading rates for untreated domestic sewage - -,.--.-..-.. *.-.-...... -- . . . .".-.*...,.-IL,._..-.. - --\ Range of f observations Per-capila flow rote Per-cnpitn CllOD CBOD concentrotion (m' capita d ') (m' capita d.') fmg L-')

United Stares 0.57 ( 150)' 125 (0.275)' 220 Developing countries 0.19 (50)' 60 (0.132)' 320

States is higher because higher water use typically acco~iipariiesa higher staridard of living. Per capita generation rate of BOD is also higher because of garbage 0.1 1 10 FIGURE 19.7 In-stream decomposition rate disposals and other accouterments of a developed economy like that of the United H (m) versus depth (Bowie el al. 1985). States. The average concentration of the developing countries is generally higher because the lower water use in these countries outweighs the higher per capita BOD contribution for the United States. As might be expected. the bottle rate can rarely be directly applied to rivers because the bottle environment is not a good representation of the river. In fact on- ly in deep. slow rivers would the two converge. In most other rivers, environmental 19.6.3 BOD Removal Rates factors tend to make removal higher than for tlie bottle. The primary causes of this increase are settling and bed effects. The bottle BOD decomposition rate provides a first estimate of the removal rate in natural waters. As listed in Table 19.1. the rates depend on the degree of treatment Settling. Settling effects relate to the fact that for sewage with a significant of the sewage prior to discharge. Raw sewage is a mixture of compounds ranging fraction of organic solids. the total removal rate instreams is a combination of settling from easily decomposable sups to refractory substances that take longer to break and deconiposition (recall Eqs. 19.20 and 19.2 1), down. Because waste treatriient tends to selectively retiiove tlie former. BOD bottle I rates tenti to he lower for treated sewage. "3 k, = k,, + - (19.27) I I-I Using some typical settling velocities. we can see (Fig. 19.6) that the settling effect can be particularly significant for raw sewage in shallow streams (that is. < 1 m).

Bed effects. All other things equal, attached bacteria generally are more effec- tive decomposers than free-floating bacteria. Bottom decomposition can be param- eterized as a mass-transfer flux of BOD. Thus in a way similar to settling, bottom decotnposition becomes more pronounced in sliallower systems because the effect beco~ilesmore significant relative to the volu~iictricdecomposition in the water. This trend, which is displayed in Fig. 19.7, has been fit by the equations (Hy- droscicrice 197 1 ) k,, = 0.3 (3°,43JOsHs8ft (,IY.28) kd = 0.3 H >8fta Thus up to about 2.4 rn (8 ft) the rate decreases with depth. Above 2.4 m the rate approaches a constant value that is typical of bottle rates. Finally. BOD decomposition rates car1 be extrapolated to other temperatures US- FIGURE 19.6 irig 2.44 = with 1.047. Plot of total removal rate versus stream depth for BOD that is 50% in settleable Eq. (k kloeT-20). 0 = form. A range of settling veloctties is deptcted. Note that a decomposition rate of In sutnniary BOD removal rates tend to increase with temperature and tend 0.35 3.' is used. to hc higher immetliately downstrearii frorn point sources. The latter effect is more LECNRE19 BOD 1 .360 PART IV Dissolved Oxygen and Pathogens and Oxygen Saturation 361 pronounced for untreated wastewater. In addition enhanced settling and bed effects TABLE 19.3 means that shallower systems typically exhibit higher BOD removal rates than Henry's constants for some gases commonly encountered deeper waters. in water-quality modeling (modified from Kavanaugh and Trussell 1980)

Henry's constant (20°C) 19.7 HENRY'S LAW AND THE IDEAL GAS LAW Compound Formula IDImenslonless) (atm mJ mole-') I If a beaker of gas-free distilled water is opened to the atmosphere, gaseous com- Methane CHI 44.4 1.55 X I@ pounds such as oxygen, carbon dioxide. and nitrogen cross the air-water interface Oxygen o? 32.2 7.74 x 10-I 1 and enter into solution (Fig. 19.8). The process will continue until an equilibrium is I Nilrogen N? 28.4 6.84 x 10-I i ) established between the partiai pressure of the gas in the atmosphere and the concen- Carbon dioxide CO? . 1.13 2.72 x lo-? tration in the aqueous phase. This equilibrium is quantified by Henry's law, which Hydrogen sulfide H?S 0.386 9.27 x lo-' i can be represented as Sulfur dioxide SOz 0.0284 6.84 x lo-' .I Ammonia Nti, 0.000569 1.37 X to-' t: where H, = Henry's constant fatm m3 mole-') + p = partial pressure (atm) c = water concentration (mole m-)) where cg and cl are the gas and liquid concentrations, respectively (mole rn-'). Henry's constants for some gases commonly encountered inwater-quality modeling Notice that Henry's law specifies that at equilibrium the ratio of the gaseous to the water concentration will be maintained at a constant value. The water concen- are summarized in Table 19.3. + tration for a particular gaseous level is referred to as the sntrtrntion concentration. Equation 19.29 can also be represented in dimensionless form by invoking the ideal gas law In the following example we determine the saturation concentration for oxygen. J EXAbIP1,E 19.2. IIENRY'S I.A\V AND OXYGEN SATURATION. Determine the sarttration concentration of oxygen in water at ZO°C. Note that clean. dry air near sea level is cornposed of approximately 20.958 oxygen by volunle. :; where R = universal gas constant l8.206 X atm m3 (K mole)-'] and Tfl = b absolute temperature (K). Thus the itleal gas law provides a means to express the Sol~rtio~i:Assrtrlling ~liatDalton's law holds. the partinl pressttre of oxygen can be com- partial pressure in concentration units of moles ni-'. Substituting Eq. 19.30 into puted as 19.29 and rearranging yields the dinlensionless Henry's constant 1) = 0.2095( 1 atm) = 0.3095 atrn ? This value can then be substituted into Eq. 19.29 along with the value of tlenry'sconstant - from Table 19.3 to yield C, p 0.2095 = - = ---- - 0.2707 mole 111-' H, 0.774

(17' - g-oxygen c. = 0.2707 nlole 111- ' = 8.66 tng I.-' niole

19.8 DISSOLVED OXYGEN SATURATION FIGURE 19.8 As calculatetl in the previous example. the saturation concentration of oxygen in a The closed system in (a) is undersaturated with oxygen. When nati~ralwnter is on the order of I0 mg L-'. In general several environmental fac- it is opened to the atmosphere (b), oxygen comes into solution until an equilibrium (c) is reached. Henry's law provides the tors can affect this value. From the perspective of water-quality modeling, the most means lo quantify this equilibrium condition. iniportant of these are; 362 PART IV Dissolved Oxygen and Pathogens LE~RE19 BOD and Oxygen Saturation 363

19.8.2 Salinity Effect The following equation can be used to establish the dependence of saturation on salinity (APHA 1992): i

1.0754 X 10' + 2. I407 X I@ \n o,, = Ino,, - s I T't T,Z I I where o,, = saturation concentration of dissolved oxygen in saltwater at L atm (mg L-I) and S = salinity (g L-' = parts per thousand. ppt, sometimes given as %o). Salinity can be related to chloride concentration by the following approximation:

S = 1.80655 x Chlor ( 19.35)

/ where Chlor = chloride concentration (ppt). The higher the salinity, the less oxygen Salt water (S = 35 ppt) can be held by water (Fig. 19.9).

EXAhlPLE 19.3. OXYGEN SATURATION FOR AN ESTUARY. Determine the saturation for an estuary with a temperature of 20°C and a salinity of 25 ppt. Solution: Equation 19.32 can be used to compute

FIGURE 19.9 Relationship of oxygen saturation in water to temperature and salinity.

Temperature Salinity This corresponds to a freshwater saturation value of Partial pressure variations due to elevation o,, = e' '"' = 9.092 rng L-' Sevcral empirically derived equations have been developed to predict how these Equation 19.34 can be used to correct for salinity, factors influence saturation. These are reviewed in the following sections.

1.0754 X 10' + 2.1407 X ID' In o,, = 2.207 - 25 = 2.060 293.15 292.15? 19.8.1 Temperature Effect which corresponds to a saltwater saturation value of The following equation can be used to establish the dependence of oxygen saturation on teniperature (APHA 1992): o,, = P''~ = 7.816 Ing L-'

Thus the saltwater value is about 86% of the freshwater value.

! 19.8.3 Pressure Effect i i wliere o,, = saturation concentratiori of clissolved oxygen in fresh water at I atni The followin_g equatiotl can be used lo establish the dependence of saturation on (n~pL-I) and Ta = absolute temperature (K). Remember that pressure (APHA 1992): Ta = T + 273.15 ( 19.33) where T = temperature ("C). Accordi~igto this equation, saturation decreases with increasing temperature. As displayed in Fig. 19.9, freshwater concentration ranges from about 14.6 nig L-I at O°C to 7.6 mg L-I at 30°C. .361 PART IV Dissolved Oxygen and Pathogens LECWRE 19 BOD and Oxygen Saturation 365 iL PROBLEMS - I 19.1. Use separation of variables to solve Eq. 19.6 for Eq. 19.7. I I I I I I I 1 19.2. A tanker truck careens off the road and dumps 30.000 L of glucose syrup into a small mountain lake. The concentration of the syrup is 100 g-glucose L-'. Sea level (a) Compute the grams of CBOD spilled. (b) Determine the lake's saturation concentration of oxygen (T = IOOC; elev = 11.000 ft.). 19.3. A lake with a bay has the following characteristics:

Lake Bay

Mean depth 8 m 3 rn I Surface area 1.6 x 106 m2 0.4 x 106 m' 4.8 km (3 miles) I Inflow 50.000 mJ d-' 4800 mJ d'' Inflow BOD concentrarion 0 mg L-' 57 rng L-I i I I I I I A subdivision housing 1000 people is planned that will discharge raw sewage into the bay. Each individual contributes about 0.568 m' capita-' d-' of wastewater and FIGURE 19.10 Relationship of oxygen saturation in water to temperature and elevation above 113.4 g capita-' d-' of carbonaceous biochemical oxygen demand. sea level. (a) The bav inflow has a chloride concentrations of 50 mg L-I. The lake and the bay have chloride concentrations of 5 and 10 mg L-'. respectively. Determine the bulk diffusion coefficient between the lake and the bay. (b) If the BOD decays at a rate of 0.1 d-I and settles at a rate of 0.1 m d-'. deter- where p = atmospheric pressure (atm) mine the steady-state BOD concentration of the lake and bay in mg L-'. Make o,, = saturation concentration of dissolved oxygen at p (mg L-I) this determination with and without the subdivision. I oSl = saturation coticentration of dissolved oxygen at I atm (mg L-I) I p,, = partial pressure of water vapor (atm) 19.4, The following data have been collected for dissolved and total BOD below a point source of untreated sewage into a stream. Use this data to estimate the BOD removal p,., can he calculated by I rates (k,, k,. and kd) for the river. The vtlocity and depth are 6600 m d-' and 2 m. I I respectively. 3840.70 216,961 1 Inp,., = 11.8571 ------Ta T,Z .r fkm) 1 The parnmcter 0 cnri be coriiputed as Dissolved (mg L- ') Total (lng L-I)

.r (km) I Dissolvcd (mg L-') Notice that this fortnula is written in tenns of temperature in degrees Celsius rather Total (nig L-') than Kelvin. Zisoti et nl. ( 1975) Iiave developetl n handy cipproxitnatio~ibased on clev;ition. ! 19.5. Detern~inethe saturation cor~cer~trationsat ZOdC for (ti) nitrogen gas (78.1% by volume o,, = o,l [I - 0.1 148 X elev(km)] 11 9.39) i in the atmosphere) and (b)carbon dioxide (0.03 14%). 19.6. A flow of 2 crns with a 5-d BOD of I0trig L-' is tliscllarged from an activated-sludge or in English units. treatment plant to a stream with a flow of 5 crns and zero BOD. Stream characteristics are k,lo = 0.2 d-'. cross-sectional area = 25 m', and T = 28°C. o,, = o,~[I - 0.000035 X elev(ft)] ( 19.40) (a) What is the concentration of BOD nt the mixing point? (B) How far below the effluent will lie stream BOD concentration fall to 5% of its where elev = elevntion above sea level. As displayed in Fig. 19.10 this relationship original value? indicates that as pressure decreases at higher elevations, the saturation drops.