ENVS4450

Environmental Modeling

Chapter 7:

Dissolved Oxygen Sag Curves in Streams

Quote

“[Mathematics] The handmaiden of the Sciences”

-Eric Temple Bell

Concepts

•Introduction

•Input sources

•Mathematical Model

•Sensitivity analysis

•Limitations

Case Study: Any Stream, Anywhere

•Every stream has inputs oforganic waste

–Spreads disease

–Consumes DO ondecomposition

•Ancient communities builtnear flowing water

e.g. NY City, London,W. Europe

Case Study: Any Stream, Anywhere

The Problem: D.O. < BOD

Sewage treatment begins

Chemical process:

MO’s consume DO

Physical process:

Re-aeration byatmosphere

Meadows et al., 2004

Introduction

•Modeling the effects of release of oxidizable organic matter into aflowing body of water

–DO = chemical measurement of dissolved oxygen (mg L-1)

–BOD = total DO needed to oxidize organic matter in a water sample= change from initial DO at saturation to amount after 5 days

BOD

time

Introduction

•Standard of living ~ adequate water andwastewater treatment

Human Risks

•Challenge of preventing rapid spead of disease

e.g. typhoid fever (bacteria), hepatitis (viruses),cryptosporidosis (protozoa)

•Removed by sand filtration andchlorination/ozonation

Aquatic Risks

•Aerobic organisms depend on DO

•8-12 mg L-1

•Affected by temperature and salt

Inc. salt dec. DO

Wipple and Wipple (1911)

The Streeter-Phelps Equation

without trmt:

with trmt:

Organic matter is oxidized, stream re-aerates

End

•Review

Basic Input Sources

•Parameters for S-P equation:

–Wastewater: Flow rate, temperature, DO, BOD

•BOD measured in lab – DO measured after several days (flat portionof curve)

•The following material and model iscovered in:

CHEM3500/3550

Basic Input SourcesSewage Treatment Plants

•Remove turbidity, oxidizable organic matter, and pathogens

–Turbidity – settling tanks and filters

–Organic matter – trickling filters, activated sludge

–Pathogens – filtration, chloination, ozonation

ftp://ftp.wiley.com/public/sci_tech_med/pollutant_fate/

Basic Input SourcesSewage Treatment Plants

•Prelininary - screening of large materials

•Primary - sedimentation - settling tanks

•Secondary - biological aeration – tricklingfilters, activated sludge - metabolizes andflocculates dissolved organics

•Tertiary – e.g. P removal

http://www.waterencyclopedia.com/Tw-Z/Wastewater-Treatment-and-Management.html

Basic Input Sources

•Wastewater Treatment PlantModel

Movie

1. Wastewater Treatment and Discharge (2000)2. Wastewater Generation and Collection (2000)3. Our Urban Environment: Water Quality (2000)

End

•Review

Mathematical Model

Take a river: What parameters and processes would be important indeveloping a model for the oxidation of organic waste?

our model river: draw in parameters

Ultimate BODLof mix

Stream DO deficit

Consumption DO by MO’s

Re-aeration by atmosphere

Amount DO consumed

The Streeter-Phelps Equation

D = k’BODL [exp(-k’(x/v) – exp(-k2’(x/v))] + D0exp(-k2’(x/v))

k2’ – k’

where: D = DO concentration deficit (value below saturation) (mg L-1),k’2= the re-aeration constant (in d-1),BODL= the ultimate BOD (in mg L-1),k’= the BOD rate constant for oxidation (d-1),x = distance downstream from the point source (km),v = average water velocity (km d-1)Do= initial oxygen deficit of mixed stream and wastewater (mg L-1)

Consumption by MO’s

Re-aeration by atmos. O2

D is not the remaining DO content but the amount of original DOconsumed…must be subtracted from original DO without BOD waste

The Streeter-Phelps Equation

DO at a given distance below the input:

The Streeter-Phelps Equation

•k2’ = first-order rate constant for re-aeration

•Eact measurements are difficult, get from tables:

The Streeter-Phelps Equation

•BODL = ultimate BOD or maximum O2 required to oxize thewaste sample

•Determined from 5 day BOD test or using equation:

BODL = BOD5

1 – exp(-k’(x/v))

•Where k’ is obtained from a 20 day BOD experiment

•D0 = DO level in the stream upstream from input - initial DO ofstream-waste mixture

The Streeter-Phelps Equation

Zone of Clean Water (Zone 1)

Zone of Degradation (Zone 2)Zone of Active Decomposition (Zone 3)

Zone of Recovery (Zone 4)

Zone of Cleaner Water (Zone 5)

Algae, fungi, protozoa, worms,larger planst die

Gray/black, H2S, CH4, NH3productions,

Minimum D = criticaldissolved oxygen = Dc

The Streeter-Phelps Equation

tc = 1 ln k2’ 1 – D0(k2’-k’)

k2’ – k’ k’ k’ BODL

and xc = vtc

Critical DO concentration, Dc:

Dc = k’ BODL exp(-k’(xc/v))

k2’

Problem

1. Determine Dc and its location.

2. Estimate the 20 °C BOD5 of a sample taken at xc.

3. Plot the curve.

Example Problem: A city discharges 25 million gallons per day of domesticsewage into a stream whose typical rate of flow is 250 cubic feet per second.The velocity of the stream is appoximately 3 miles per hour. The temperatureof the sewage is 21 °C, while that of the stream is 15 °C. The 20 °C BOD5 ofthe sewage is 180 mg/L, while that of the stream is 1.0 mg/L. The sewagecontains no DO, but the stream is 90% saturated upstream of the discharge.At 20 °C, k’ is estimated to be 0.34 per day while k2’ is 0.65 per day.

1.Determine DO in stream before discharge (=upstream DO):

Saturation conc. at 15 °C = 10.2 mg/L

Upstream is 90% saturated = 10.2 mg/L x 0.90 = 9.2 mg/L

2.Determine mixture, T, DO, and BOD using mass balance:

Flow rate stream: = 250 ft3/s = 612 x 106 L/d

Flow rate sewage: 25 x 106 gallons/d = 94.8 x 106 L/d

Temperature of mixture:

T = stream input + sewage input – output effect

0 = (stream flow)(stream temp.) + (sewage flow)( sewage temp) – (mix flow)(mix temp)

0 = (612 x 106 L/d)(15 °C) + (94.8 x 106 L/d)(20 °C) – (612 x 106 L/d + 94.8 x 106 L/d)Tmix

Tmix = (612 x 106 L/d)(15 °C) + (94.8 x 106 L/d)(20 °C) = 15.7 °C

(612 x 106 L/d +94.8 x 106 L/d)

DO in mixture

Net change in DO = Stream input + Sewage output – Output

0 = (stream flow)(stream DO) + (sewage flow)(sewage DO) – (mix flow)(mix DO)

0 = (612 x 106 L/d)(9.2 mg/L) + (94.8 x 106 L/d)(0.0) - (612 x 106 L/d + 94.8 x 106 L/d)(Domix)

DOmix = (612 x 106 L/d)(9.2 mg/L) + (94.8 x 106 L/d)(0.0 mg/L)

(612 x 106 L/d + 94.8 x 106 L/d)

= 7.97 mg/L

BOD5 of mixture:

Net change in BOD5 = BOD5 = Stream input + Sewage output – Output

0 = (stream flow)(stream BOD5) + (sewage flow)(sewage BOD5) – (mix flow)(mix BOD5)

0 = (612 x 106 L/d)(1.0 mg/L) + (94.8 x 106 L/d)(80 mg/L) - (612 x 106 L/d + 94.8 x 106L/d)(BOD5)

BOD5mixture = (612 x 106 L/d)(1.0 mg/L) + (94.8 x 106 L/d)(80 mg/L) = 25.0 mg/L

(612 x 106 L/d + 94.8 x 106 L/d)

BODL of mixture (at 20 °C)

BODL = BOD5= 25.0 mg/L= 30.6 mg/L

1 – exp(-k’(x/v) 1 – exp(-0.34/d)(5 d)

3.Correct rate constants to 15.7 °C

k’ = 0.34(1.135)15.7-20 = 0.197 d-1

k2’ = 0.65(1.024)15.7-20 = 0.587 d-1

4. Determine tc and xc:

D0 = (initial stream O2 - O2 of mixture)

= (9.2 – 7.97) = 1.23 mg O2 L-1

4. Determine tc and xc:

tc = 1 ln k2’ 1 – D0(k2’-k’)

k2’ – k’ k’ k’ BODL

= 2.42 d

xc = vtc = 3 mi/h x 24 h/d x 2.42 d = 174.2 mi = 280 km

5. Determine Dc:

5. Determine Dc:

V = 3 mi/h = 72 mi/d

Dc = k’ BODL exp(-k’(xc/v)

k2’

= 0.197 d-1(30.6 mg/L) exp(-(0.197 d-1)(174.2 mi / 72 mi d-1)))

0.587 d-1

= 6.37 mg L-1

The DO will be depressed 6.37 mg L-1 from saturation.Minimum DO = 9.2 mg L-1 - 6.37 mg L-1 = 2.83 mg L-1

6. Determine BOD5 at critical point, xc:

BOD5 = BODL exp(-k’(x/v))

= (30.6 mg L-1) exp(-0.197 d-1)(174.2 mi)/(72 mi d-1) = 19.0 mg L-1

20 °C BOD5 = BOD5 [1 – exp(-k’)(5)]

= 19.0 mg L-1 [1 – exp(-0.34 d-1)(5 d)] = 15.5 mg L-1

Easier method

•Use Fate!!!

•Much easier than by hand

End

•Review

Sensitivity Analysis

Limitations

•It uses average re-aeration rates of the stream (problem in alternatingriffle and pool areas)

•Sedimentation is not allowed in the basic model, but can beincorporated with additional experimental data

Remediation

•Problems are:-Eutrophication

-Odors

-Low/no D.O.

-Aquatic death

-Microbes/Pathogens

•Source removal! (install treatment plant)

including BOD, NO3-, NH3/NH4+, PO43-

removal, but you still will have organic rich

sediments for some time

•Time (flowing aquatic systems can be very resilient)

•Notice the difference between the recovery of a biodegradable pollutantversus nonbiodegradable!

End

•Review