Dissolved oxygenImportance Why is oxygen in water important?

Dissolved oxygen (DO) analysis measures the amount of gaseous oxygen (O2) dissolved in an aqueous solution. Oxygen gets into water by diffusion from the surrounding air, by aeration (rapid movement), and as a product of photosynthesis. DO is measured in standard solution units such as

milligrams O2 per liter (mg/L), millilitres O2 per liter (ml/L), millimoles O2 per liter (mmol/L), and moles O2 per cubic meter (mol/m3).

DO is measured by way of its oxidation potential with a probe that allows diffusion of oxygen into it.

The saturation solubility of oxygen in wastewater can be expressed asCs = (0.99)h/88 x 482.5/(T + 32.6)

For example, in freshwater in Ames at 350m and 20°C, O2 saturation is 8.8 mg/L. (Check for yourself, with = 1)

BODBiochemical oxygen demand or BOD is a procedure for determining the rate of uptake of dissolved oxygen by the organisms in a body of water BOD measures the oxygen uptake by bacteria in a water sample at a temperature of 20°C over a period of 5d in the dark. The sample is diluted with oxygen saturated de-ionized water, inoculating it with a fixed aliquot of microbial seed, measuring the (DO) and then sealing the sample to prevent further oxygen addition. The sample is kept at 20 °C for five days, in the dark to prevent addition of oxygen by photo-synthesis, and the dissolved oxygen is measured again. The difference between the final DO and initial DO is the BOD or, BOD5.

Once we have a BOD5 value, it is treated as just a concentration in mg/L

BOD can be calculated by:Diluted: ((Initial DO - Final DO + BOD of Seed) x Dilution Factor

BOD of seed (diluted activated sludge) is measured in a control: just deionized water without wastewater sample.

Significance: BOD is a measure of organic content and gives an indication on how much oxygen would be required for microbial degradation.

Oxygen depletion in streams

DO sag definitions

Cumulative oxygen supply + demandPlotting the two kinetic equations separately on a cumulative basis and adding these graphically produce the DO sag curve

Streeter-Phelps Model*

Mass Balance for the Model

Not a Steady-state situationrate O2 accum. = rate O2 in – rate O2 out + produced – consumed

rate O2 accum. = rate O2 in – 0 + 0 – rate O2 consumed

Kinetics Both reoxygenation and deoxygenation are 1st order

* Streeter, H.W. and Phelps, E.B. Bulletin #146, USPHS (1925)

Kinetics* for Streeter-Phelps Model

• Deoxygenation L = BOD remaining at any timedL/dt = Rate of deoxygenation equivalent to rate of BOD removaldL/dt = -k1L for a first order reaction

k1 = deoxygenation constant, f’n of waste type and temp.

*See Kinetics presentation if unfamiliar with the mathematical processing

kLdt

Ld

][ C

C

tdtk

L

dL0 0

ktkt eLLeL

Lorkt

L

L 000

ln

Developing the Streeter-Phelps

Rate of reoxygenation = k2D

D = deficit in D.O. k2 = reoxygenation constant*

2

3

21

)20(21

2

025.19.3

H

vk

T

Where– T = temperature of water, ºC– H = average depth of flow, m– ν = mean stream velocity, m/s

D.O. deficit = saturation D.O. – D.O. in the water

Typical values for k2 at 20 °C, 1/d (base e) are as follows:small ponds and back water 0.10 - 0.23sluggish streams and large lakes 0.23 - 0.35large streams with low velocity 0.35 - 0.46large streams at normal velocity 0.46 - 0.69swift streams 0.69 - 1.15rapids and waterfalls > 1.15

There are many correlations for this.The simplest one, used here, is from O’Connor and Dobbins, 1958

Combining the kinetics

OR

Net rate of change of oxygen deficiency, dD/dt

dD/dt = k1L - k2D

where L = L0e-k1t

dD/dt = k1L0e-k1t - k2D

Integration and substitution

tko

tktko eDeekk

LkD 221 )(

12

1

tkoc

tko

eLk

kD

DkeLkdt

dD

1

1

2

1

21 0

o

oc Lk

kkD

k

k

kkt

1

12

1

2

12

)(1ln

1

The last differential equation can be integrated to:

It can be observed that the minimum value, Dc is achieved when dD/dt = 0:

, since D is then Dc

Substituting this last equation in the first, when D = Dc and solving for t = tc:

Example: Streeter-Phelps

Wastewater mixes with a river resulting in aBOD = 10.9 mg/L, DO = 7.6 mg/LThe mixture has a temp. = 20 CDeoxygenation const.= 0.2 day-1

Average flow = 0.3 m/s, Average depth = 3.0 m

DO saturated = 9.1 mg/L• Find the time and distance downstream at which the

oxygen deficit is a maximum• Find the minimum value of DO

Solution…some values needed

• Initial Deficit Do = 9.1 – 7.6 = 1.5 mg/L (Now given, but could be calculated from proportional mix of river DO,

presumably saturated, and DO of wastewater, presumably zero)

• Estimate the reaeration constant:

k2 = 3.9 v½ (1.025T-20)½

H3/2

k2 = 3.9 x (0.3m/s)½ (1.02520-20)½

(3.0m)3/2 = 0.41 d-1

Solution…time and distance

days

Lk

kkDO

k

k

kkt

o

oc

67.2

9.102.0

)2.041.0(5.11

2.0

41.0ln

)2.041.0(

1

)(1ln

1

1

12

1

2

12

mdaysdayssmvtx cc 300,6967.2/400,86/3.0

Note that the effects will be maximized almost 70 km downstream

Solution…maximum DO deficiency

mg/L 1.3

e mg/L) 9.10(41.0

2.0 ))(2.67days(0.2day

2

1

1

1

tkoc eL

k

kD

The minimum DO value is 9.1-3.1 = 6 mg/L

Implication: DO probably not low enough for a fishkill, but if continued could lead to species differentiation and discourage sensitives species like trout.

Note that this BOD could have been calculated from mixing high-BOD wastewater with zero or near-zero BOD