Critical oxygen deficit

On the DO sag curve a minimum concentration occurs at some point, along a stream. If the Streeter–Phelps equation is differentiated with respect to time, and set equal to zero, the time at which the minimum DO occurs is expressed by

${\displaystyle t_{crit}={\frac {1}{k_{2}-k_{1}}}\ln {\left[{\frac {k_{2}}{k_{1}}}\left({1-{\frac {D_{a}(k_{2}-k_{1})}{L_{a}k_{1}}}}\right)\right]}}$

To find the value of the critical oxygen deficit, ${\displaystyle D_{crit}}$, the Streeter–Phelps equation is combined with the equation above, for the critical time, ${\displaystyle t_{crit}}$. Then the minimum dissolved oxygen concentration is

${\displaystyle DO_{crit}=DO_{sat}-D_{crit}}$

Mathematically it is possible to get a negative value of ${\displaystyle DO_{crit}}$, even though it is not possible to have a negative amount of DO in reality.^[3]

The distance traveled in a river from a given point source pollution or waste discharge downstream to the ${\displaystyle DO_{crit}}$ (which is the minimum DO) is found by

Surface plot depicting the dissolved oxygen (DO) concentration in a river. DO is shown on the vertical axis, with the along-stream and cross-stream directions on the x and y axes, respectively. A continuous input of biological material is added to the river at x = 75 m, y = 15 m, beginning at t = 0.

${\displaystyle x_{crit}=vt_{crit}}$

where ${\displaystyle v}$ is the flow velocity of the stream. This formula is a good approximation as long as the flow can be regarded as a plug flow (turbulent).

Estimation of reaeration rate

Several estimations of the reaeration rate exist, which generally follow the equation

${\displaystyle k_{2}=Kv^{a}H^{-b}}$

where

• ${\displaystyle K}$ is a constant.
• ${\displaystyle v}$ is the flow velocity [m/s].
• ${\displaystyle H}$ is the depth [m].
• ${\displaystyle a}$ is a constant.
• ${\displaystyle b}$ is a constant.

The constants depend on the system to which the equation is applied, i.e. the flow velocity and the size of the stream or river. Different values are available in the literature.

The software "International Hydrological Programme" applies the following equation derived on the basis of values used in published literature^[4]

${\displaystyle k_{2}=2.148v^{0.878}H^{-1.48}}$

where

• ${\displaystyle k_{2}}$ ${\displaystyle [d^{-1}]}$.
• ${\displaystyle v}$ is the average flow velocity [m/s].
• ${\displaystyle H}$ is the average depth of flow in the river [m].

Temperature correction

Both the deoxygenation rate, ${\displaystyle k_{1}}$ and reaeration rate, ${\displaystyle k_{2}}$ can be temperature corrected, following the general formula.^[2]

${\displaystyle k=k_{20}\theta ^{(T-20)}}$

where

• ${\displaystyle k_{20}}$ is the rate at 20 degrees Celsius.
• θ is a constant, which differs for the two rates.
• ${\displaystyle T}$ is the actual temperature in the stream in degC.

Normally θ has the value 1.048 for ${\displaystyle k_{1}}$ and 1.024 for ${\displaystyle k_{2}}$. An increasing temperature has the most impact on the deoxygenation rate, and results in an increased critical deficit (${\displaystyle D_{crit}}$), and ${\displaystyle x_{crit}}$ decreases. Furthermore, a decreased ${\displaystyle DO_{sat}}$ concentration occurs with increasing temperature, which leads to a decrease in the DO concentration.^[2]

Mixing of rivers

When two streams or rivers merge or water is discharged to a stream it is possible to determine the BOD and DO after mixing assuming steady state conditions and instantaneous mixing. The two streams are considered as dilutions of each other thus the initial BOD and DO will be ^[4]

${\displaystyle L_{a}={\frac {L_{s}Q_{s}+L_{b}Q_{b}}{Q_{s}+Q_{b}}}}$

and

${\displaystyle DO_{0}={\frac {DO_{s}Q_{s}+DO_{b}Q_{b}}{Q_{s}+Q_{b}}}}$

where

• ${\displaystyle L_{a}}$ is the initial concentration of BOD in the river downstream of the mixing, also called BOD(0). The unit of ${\displaystyle L_{a}}$ is ${\displaystyle {\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}}$.
• ${\displaystyle L_{b}}$ is the background BOD of the concentration in the river ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle L_{s}}$ is the BOD of the content of the merging river ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle DO_{0}}$ is the initial concentration of the dissolved oxygen in the river downstream of the conjoining point ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle DO_{b}}$ is the background concentration of the dissolved oxygen content in the river ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle DO_{s}}$ is the background concentration of the dissolved oxygen content in the merging river ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle Q_{b}}$ is the flow in the river upstream from the mixing point ${\displaystyle [{\tfrac {m^{3}}{s}}]}$.
• ${\displaystyle Q_{s}}$ is the flow in the merging river upstream from the mixing point ${\displaystyle [{\tfrac {m^{3}}{s}}]}$.

Numerical approach

Nowadays it is possible to solve the classical Streeter–Phelps equation numerically by use of computers. The differential equations are solved by integration.

Background BOD

Background BOD or benthic oxygen demand is the diffuse source of BOD represented by the decay of organic matter that has already settled on the bottom. This will give rise to a constant diffuse input thus the change in BOD over time will be

${\displaystyle {\frac {dL}{dt}}=-k_{1}L+L_{b}}$

where

• ${\displaystyle k_{1}}$ is the rate for oxygen consumption by BOD, usually in ${\displaystyle d^{-1}}$.
• ${\displaystyle L}$ is the BOD from organic matter in the water ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle L_{b}}$ is the background BOD input ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.

Sedimentation of BOD

Sedimented BOD does not directly consume oxygen and this should therefore be taken into account. This is done by introducing a rate of BOD removal combined with a rate of oxygen consumption by BOD. Giving a total rate for oxygen removal by BOD ^[2]

${\displaystyle k_{r}=k_{1}+k_{3}}$

where

• ${\displaystyle k_{1}}$ is the rate of oxygen consumption by BOD, usually in ${\displaystyle d^{-1}}$.
• ${\displaystyle k_{3}}$ is the rate of settling of BOD, usually in ${\displaystyle d^{-1}}$.

The change in BOD over time is described as

${\displaystyle {\frac {dL}{dt}}=-k_{r}L}$

where ${\displaystyle L}$ is the BOD from organic matter in the water ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.

${\displaystyle k_{r}}$ is typically in the range of 0.5-5 ${\displaystyle d^{-1}}$.^[2]

Sediment oxygen demand

Measurement of SOD using an oxygen electrode. A: Water, B: Sediment, C: Box, D: Oxygen electrode.

Oxygen can be consumed by organisms in the sediment. This process is referred to as sediment oxygen demand (SOD). Measurement of SOD can be undertaken by measuring the change of oxygen in a box on the sediment (benthic respirometer).

The change in oxygen deficit due to consumption by sediment is described as

${\displaystyle {\frac {dD}{dt}}=-k_{2}D+{\frac {S}{H}}}$

where

• ${\displaystyle H}$ is the depth of the river [m]
• ${\displaystyle S}$ is the SOD ${\displaystyle [{\tfrac {g}{m^{2}d}}]}$
• D is the saturation deficit ${\displaystyle [{\tfrac {\mathrm {g} }{\mathrm {m} ^{3}}}]}$.
• ${\displaystyle k_{2}}$ is the reaeration rate [${\displaystyle d^{-1}}$].

The range of the SOD is typically in the range of 0.1 – 1 ${\displaystyle {\tfrac {g}{m^{2}d}}}$ for a natural river with low pollution and 5 – 10 ${\displaystyle {\tfrac {g}{m^{2}d}}}$ for a river with moderate to heavy pollution.^[2]

Nitrification

Ammonium is oxidized to nitrate under aerobic conditions

NH₄⁺ + 2O₂ → NO₃⁻ + H₂O + 2H⁺

Ammonium oxidation can be treated as part of BOD, so that BOD = CBOD + NBOD, where CBOD is the carbonaceous biochemical oxygen demand and NBOD is nitrogenous BOD. Usually CBOD is much higher than the ammonium concentration and thus NBOD often does not need to be considered. The change in oxygen deficit due to oxidation of ammonium is described as

${\displaystyle {\frac {dD}{dt}}=k_{N}N-k_{2}D}$

where

• D is the saturation deficit.
• ${\displaystyle k_{N}}$ is the nitrification rate ${\displaystyle [d^{-1}]}$.
• ${\displaystyle N}$ is ammonium-nitrogen concentration.

The range of ${\displaystyle k_{N}}$ is typically 0.05-0.5 ${\displaystyle d^{-1}}$.^[2]

Photosynthesis and respiration

Photosynthesis and respiration are performed by algae and by macrophytes. Respiration is also performed by bacteria and animals. Assuming steady state (net daily average) the change in deficit will be

${\displaystyle {\frac {dD}{dt}}=-k_{2}D+(R+P)_{avg}}$

where

• ${\displaystyle R}$ is the respiration ${\displaystyle [{\tfrac {mg}{Ld}}]}$.
• ${\displaystyle P}$ is the photosynthesis ${\displaystyle [{\tfrac {mg}{Ld}}]}$.

Note that BOD only includes respiration of microorganisms e.g. algae and bacteria and not by macrophytes and animals.

Due to the variation of light over time, the variation of the photosynthetic oxygen can be described by a periodical function over time, where time is after sunrise and before sunset^[2]

${\displaystyle P(t)=P_{max}sin\left({\frac {\pi }{f}}+(t-t_{s})\right)}$

where

• ${\displaystyle P(t)}$ is the photosynthesis at a given time ${\displaystyle [{\tfrac {mg}{Ld}}]}$.
• ${\displaystyle P_{max}}$ is the daily maximum of the photosynthesis ${\displaystyle [{\tfrac {mg}{Ld}}]}$.
• ${\displaystyle f}$ is the fraction of day with sunlight, usually ${\displaystyle {\tfrac {1}{2}}}$ day.
• ${\displaystyle t_{s}}$ is the time at which sun rises ${\displaystyle [d]}$.

The range of the daily average value of primary production ${\displaystyle (P-R)}$ is typically 0.5-10 ${\displaystyle {\tfrac {mg}{Ld}}}$.^[2]