water quality management in rivers. dissolved oxygen depletion
TRANSCRIPT
Biochemical Oxygen Demand Measurement
• Take sample of waste; dilute with oxygen saturated water; add nutrients and microorganisms (seed)
• Measure dissolved oxygen (DO) levels over 5 days• Temperature 20° C• In dark (prevents algae from growing)• Final DO concentration must be > 2 mg/L • Need at least 2 mg/L change in DO over 5 days
Example
• A BOD test was conducted in the laboratory using wastewater being dumped into a Lake. The samples are prepared by adding 3.00 mL of wastewater to the 300.0 mL BOD bottles. The bottles are filled to capacity with seeded dilution water.
Example : Raw Data
Time(days)
Dilutedsample
DO (mg/L)
Blank SeededSample DO
(mg/L)0 7.95 8.151 3.75 8.102 3.45 8.053 2.75 8.004 2.15 7.955 1.80 7.90
Example : Calculations
• What is the BOD5 of the sample?
• Plot the BOD with respect to time.
sample diluted
the of ionsconcentrat DO final and initial DO ,DO
(blank)water diluted seeded
the of ionsconcentrat DO final and initial B ,B
volumetotal sample/ volumefactor dilutionPP
P)])(1B(B-)DO[(DOBOD
fi
fi
fifim
Example : Time – Concentration Plot
0
100
200
300
400
500
600
700
0 1 2 3 4 5 6
time (days)
BO
D (
mg
/L)
Modeling BOD as a First-order Reaction
0123456
0 10 20 30time (days)
Co
nc. (m
g/L
)
Organic matter oxidized
Organic matter remaining
Modeling BOD Reactions
• Assume rate of decomposition of organic waste is proportional to the waste that is left in the flask.
demandoxygen uscarbonaceo ultimate theis where
:yieldsequation thisSolving
)(timeconstant rate BOD the
tafter timeleft demandoxygen ofamount where
1-
o
ktot
t
tt
L
e LL
k
L
- kLdt
dL
Ultimate BOD
0123456
0 10 20 30time (days)
Co
nc
. (m
g/L
)
Lt
Lo
Lo- Lt BOD exerted
L remaining
BODt
Ultimate Biochemical Oxygen Demand
Lt = amount of O2 demand left in sample at time, t
Lo = amount of O2 demand left initially (at time 0, no DO demand has been exerted, so BOD = 0)
At any time, Lo = BODt + Lt (that is the amount of DO demand used up and the amount of DO that could be used up eventually)Assuming that DO depletion is first order
BODt = Lo(1 - e-kt)
Example
• If the BOD5 of a waste is 102 mg/L and the BOD20 (corresponds to the ultimate BOD) is 158 mg/L, what is k?
kteL 10tBOD
kteL
0
1 tBOD
ktL
0
1ln tBOD
Biological Oxygen Demand: Temperature Dependence
• Temperature dependence of biochemical oxygen demand As temperature increases, metabolism
increases, utilization of DO also increases
kt = k20T-20
= 1.135 if T is between 4 - 20 oC = 1.056 if T is between 20 - 30
oC
Example
The BOD rate constant, k, was determined empirically to be 0.20 days-1 at 20 oC.
What is k if the temperature of the water increases to 25 oC?
What is k if the temperature of the water decreases to 10 oC?
Nitrogenous Oxygen Demand
• So far we have dealt only with carbonaceous demand (demand to oxidize carbon compounds)
• Many other compounds, such as proteins, consume oxygen
• Mechanism of reactions are different
Nitrogenous Oxygen Demand
• Nitrification (2 step process)2 NH3 + 3O2 2 NO2
- + 2H+ + 2H2O
2 NO2- + O2 2 NO3
-
– Overall reaction:NH3 + 2O2 NO3
- + H+ + H2O
• Theoretical NBOD = N /gO g 4.57 14
16 x 4
oxidized nitrogen of gramsused oxygen of grams
2
Nitrogenous oxygen demand
• Untreated domestic wastewater
ultimate-CBOD = 250 - 350 mg/Lultimate-NBOD = 70 - 230 mg/L
Total Kjeldahl Nitrogen (TKN) = total concentration of organic and ammonia nitrogen in wastewater: 15 - 50 mg/L as N
Ultimate NBOD 4.57 x TKN
Chemical Oxygen Demand
• Chemical oxygen demand - similar to BOD but is determined by using a strong oxidizing agent to break down chemical (rather than bacteria)
• Still determines the equivalent amount of oxygen that would be consumed
• Value usually about 1.25 times BOD
Mass Balance Approach
• Originally developed by H.W. Streeter and E.B. Phelps in 1925
• River described as “plug-flow reactor”
• Mass balance is simplified by selection of system boundaries
• Oxygen is depleted by BOD exertion
• Oxygen is gained through re-aeration
Steps in Developing the DO Sag Curve
1. Determine the initial conditions
2. Determine the re-aeration rate from stream geometry
3. Determine the de-oxygenation rate from BOD test and stream geometry
4. Calculate the DO deficit as a function of time
5. Calculate the time and deficit at the critical point
Initial Mixing
Qr = river flow (m3/s)DOr = DO in river (mg/L)Lr = BOD in river (mg/L)
Qmix = combined flow (m3/s)DO = mixed DO (mg/L)La = mixed BOD (mg/L)
Qw = waste flow (m3/s)DOw = DO in waste (mg/L)Lw = BOD in waste (mg/L)
1. Determine Initial Conditions
a. Initial dissolved oxygen concentration
b. Initial dissolved oxygen deficit
where D = DO deficit (mg/L)
DOs = saturation DO conc. (mg/L)
rw
rrww
DOQDOQDO
mix
rrwwsa Q
DOQDOQDOD
DODOD s
1. Determine Initial Conditions
DOsat is a function of temperature. Values can be found in Table.
c. Initial ultimate BOD concentration
rw
rrwwa QQ
LQLQL
2. Determine Re-aeration Rate
a. O’Connor-Dobbins correlation
where kr = reaeration coefficient @ 20ºC (day-1)
u = average stream velocity (m/s)
h = average stream depth (m)
b. Correct rate coefficient for stream temperature
where Θ = 1.024
2/3
2/19.3
h
ukr
2020,
Trr kk
Determine the De-oxygenation Rate
a. rate of de-oxygenation = kdLt
where kd = de-oxygenation rate coefficient (day-1)
Lt = ultimate BOD remaining at time (of travel downstream) t
b. If kd (stream) = k (BOD test)
and
tkt
deLL 0
tkd
deLk 0tiondeoxygenta of rate
3. Determine the De-oxygenation Rate
c. However, k = kd only for deep, slow moving streams. For others,
where η = bed activity coefficient (0.1 – 0.6)d. Correct for temperature
where Θ = 1.135 (4-20ºC) or 1.056 (20-30ºC)
h
ukkd
2020,
Trr kk
4. DO as function of time
• Mass balance on moving element
• Solution is
DkLkdt
dDrtd
tka
tktk
dr
adt
rrd eDeekk
LkD