streeter phelphs equation

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Environmental Modeling

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 of

organic waste

– Spreads disease

– Consumes DO on

decomposition

• Ancient communities built

near flowing water

e.g. NY City, London,

W. Europe

Chemical process:

MO’s consume DO

Physical process:

Re-aeration by

atmosphere

Case Study: Any Stream, Anywhere

The Problem is D.O. < BOD

Sewage treatment begins

Meadows et al., 2004

Introduction

• Modeling the effects of release of oxidizable organic matter into a

flowing body of water

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

– BOD is the total DO needed to oxidize organic matter in a water

sample = change from initial DO at saturation to amount after 5

days

• Standard of living ~ adequate water and wastewater treatment

Human Risks

• Challenge of preventing rapid spead of disease

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

• Removed by sand filtration and chlorination/ozonation

Aquatic Risks

• Aerobic organisms depend on DO

• 8-12 mg L-1

• Affected by temperature and salt

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 portion

of curve)

Basic Input Sources Sewage 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 Sources Sewage Treatment Plants

• Prelininary - screening of large materials

• Primary - sedimentation - settling tanks

• Secondary - biological aeration – trickling

filters, activated sludge - metabolizes and

flocculates dissolved organics

• Tertiary – e.g. P removal

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

Basic Input Sources

• Wastewater Treatment Plant

Model

Mathematical Model

Take a river: What parameters and processes would be important in

developing 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 DO

consumed…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

• Exact measurements are difficult, get from tables:

The Streeter-Phelps Equation

• BODL = ultimate BOD or maximum O2 required to oxize the

waste 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 of

stream-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, NH3

productions,

Minimum D = critical

dissolved 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 domestic

sewage 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 temperature

of the sewage is 21 °C, while that of the stream is 15 °C. The 20 °C BOD5 of

the sewage is 180 mg/L, while that of the stream is 1.0 mg/L. The sewage

contains 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 10 6 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 10 6 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 106

L/d)(BOD5)

BOD5mixture = (612 x 10 6 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:

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

Sensitivity Analysis

Limitations

• It uses average re-aeration rates of the stream (problem in alternating

riffle and pool areas)

• Sedimentation is not allowed in the basic model, but can be

incorporated 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

+, PO4 3-

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 pollutant versus nonbiodegradable!

End

• Review