Water Quality Management in Rivers

Dissolved Oxygen Depletion

Dissolved Oxygen Depletion

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

Example (cont)

t

Lk

0

1ln tBOD

day

mg/L mg/L

5

158102

1ln

k

-1day 21.0k

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?

Example

202525 )056.1(20.0 -1day k

-1day 26.025 k

201010 )135.1(20.0 -1day k

-1day 056.010 k

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

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

Other Measures of Oxygen Demand

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

Water Quality Management in Rivers

Dissolved Oxygen Depletion

Dissolved Oxygen Sag Curve

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

Selecting System Boundaries

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

QQ

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

5. Calculate Critical time and DO

ad

dra

d

r

drc Lk

kkD

k

k

kkt 1ln

1

crcrcd tka

tktk

ar

adc eDee

kk

LkD