Biochemical Measures of Water Quality Introduction to DO

Biochemical Measures of Water Quality
Introduction to DO
DO = Dissolved oxygen
Aquatic aerobic organisms need oxygen to survive
Maximum amount in clean water is about 9 mg/L.
DO varies with temperature, salinity, elevation, and turbulence (mixing).
Measurement of DO
DO is measured with an electronic instrument.
Special probe senses DO, sends signal to the instrument.
Concentration is given in units of mg/L





Dissolved Oxygen Problem Situations
Effect of Temperature on DO
o Hot water from a power plant decreases DO level in the receiving water
Effect of Salinity on DO
o Salt from roads and irrigated fields enters streams causes DO level to decrease
Effect of Elevation
o DO in Rio Ruidoso about 7 mg/L
o DO in Pecos River at Artesia is about 8 mg/L
o Conclusion: Ruidoso's treated wastewater must be cleaner than Artesia's
Effect of Turbulence on DO
o A stream with good mixing will replenish DO quickly
o A slow, sluggish stream (or a lake) will replenish DO slowly
DO Requirements by Fish
Trout 4-5 mg/L
Bass 3-4 mg/L
Carp 2-3 mg/L
Catfish 1-2 mg/L
Dissolved Oxygen in Conclusion
Saturation DO decreases as temperature increases
Saturation DO decreases as salinity increases
Saturation DO decreases as elevation increases
Introduction to BOD
Organic pollutants are food for bacteria
o Proteins, fats, carbohydrates, etc.
Need oxygen to consume food
HCOH + O2  CO2 + H2O
BOD = Biochemical oxygen demand.
Amount of oxygen required by bacteria to degrade a waste.
A gross indicator of water pollution
Surrogate test for total organic pollution in water


BOD Test Background
It takes no more than 5 days for any stream in England to reach the ocean
5 Days Chosen by British Scientists around 1860.
BOD Test Description
If waste is very strong, it is diluted with a known amount of water.
If waste does not contain bacteria, they are added (seeded).
Samples Conditions:
Test is run for 5 days
Temperature is kept at 20° C
Bottles are kept in the dark.
Analyzed by measuring the DO at the beginning and end of a time period, usually 5 days.
Example Calculation:
o Waste is diluted 1:100
o Initial DO = 9 mg/L, Final = 3 mg/L
o BOD = 100*(9-3) = 600 mg/L

Typical BOD5 Values
Automotive anti-freeze (ethylene glycol): 600,000 mg/L
Meat packing waste: 5,000 mg/L
Domestic wastewater: 300 mg/L
Wastewater treatment plant effluent <30 mg/L

The amount of oxygen required to biodegrade the waste (BOD) is illustrated in the following plot. This plot shows that the total amount of food remaining, L, decreases with time from an initial Lo value. The amount of oxygen needed to degrade the waste, y, increases asymptotically to its ultimate value, Lo. To simplify equations, let y equal the BOD
the BOD Curve

At any given time the initial amount of food, Lo, must be equal to the sum of food consumed, y, plus food remaining, L. Therefore,

Solving of L,

The amount of food remaining (expressed as oxygen) L decreases exponentially with time. The food consumption rate (dL/dt) is a function of the amount of food available at any given time.

Where k1 represents the biodegradation rate constant. The following graph shows the effect of varying k1 constants. The yellow curve represents a readily biodegradable waste. The blue curve is a recalcitrant waste and the magenta one is one that is intermediate between the two extremes.

Separating variables leads to:

Integrating and substituting limits of integration yields:

Solving for L,

Since L = Lo - y it is possible to solve for y to obtain the equation for the BOD curve,

DO and BOD in Conclusion
Bacteria use DO to degrade pollutants:
DO Level decreases
Pollutant is completely or partially degraded to CO2
Oxygen is replenished by:
o Reaeration (i.e., mixing)
o Photosynthesis (algae in water)