14.1 Water quality

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14. Environmental engineering

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Water quality

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This chapter covers the following:

Ground and surface water quality, basic water chemistry
Biochemical oxygen demand (BOD)
BOD exertion
Kinetic temperature corrections
Stream modeling - Streeter Phelps
Oxygen saturation
Dilution purification of wastewater streams

Basic water chemistry

Groundwater tends to run low in dissolved oxygen and turbidity but high in hardness and dissolved minerals; surface water sees more reaeration and organic loading, so it runs higher in BOD and turbidity with more variable DO. The next chapter, Water treatment and distribution, covers how these differences drive treatment design.

Definitions

Hardness: The concentration of dissolved $C a^{2 +}$ and $M g^{2 +}$ , expressed as an equivalent concentration of calcium carbonate ( $C a C O_{3}$ ).
Alkalinity: Water’s capacity to neutralize acids, from dissolved bicarbonate, carbonate, and hydroxide ions, also expressed as $m g / L$ as $C a C O_{3}$ .
Equivalent weight: A substance’s molecular weight divided by the equivalents ( $n$ ) it contributes: $EW = \frac{MW}{n}$ .

To compare ions on a common basis, convert to an equivalent concentration of $C a C O_{3}$ ( $E W_{C a C O_{3}} = 50 m g / m e q$ ):

$Concentration as C a C O_{3} = concentration \times \frac{E W _{C a C O_{3}}}{E W _{substance}}$

Example: phosphate concentration as CaCO₃

A sample contains $31 m g / L$ of $P O_{4}^{3 -}$ (molecular weight $= 95 g / m o l$ , contributing $n = 3$ equivalents). Find the equivalent concentration as $C a C O_{3}$ .

$E W_{P O_{4}^{3 -}} = \frac{95}{3} = 31.7 m g / m e q$

$Concentration as C a C O_{3} = 31 \times \frac{50}{31.7} = 48.9 m g / L as C a C O_{3}$

Answer: $\approx 48.9 m g / L$ as $C a C O_{3}$

For $a A + b B ⇌ c C + d D$ , the equilibrium constant expression is:

$K = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

Example: equilibrium constant

For the reaction $A + 2 B ⇌ C$ , equilibrium concentrations are $[A] = 0.5 m o l / L$ , $[B] = 0.2 m o l / L$ , and $[C] = 0.1 m o l / L$ . Find $K$ .

$K = \frac{[ C ]}{[ A ] [ B ] ^{2}} = \frac{0.1}{( 0.5 ) ( 0.2 ) ^{2}} = \frac{0.1}{0.02} = 5$

Answer: $K = 5$

Biochemical oxygen demand (BOD)

Biochemical oxygen demand (BOD) measures how much dissolved oxygen microorganisms need to biologically oxidize biodegradable organic matter under aerobic conditions. In practice, BOD is most commonly reported as the 5-day test at 20°C, written as $BOD_{5}$ .

Because BOD reflects the amount of biodegradable organic pollution present, it’s widely used in:

wastewater treatment design
regulatory compliance
evaluating impacts on receiving waters

The three equations below cover the setups you’ll see: the basic equation applies when the dilution water is unseeded, the seeded version applies when seed organisms are added to the dilution water, and $BO D_{t}$ applies when you need the BOD exerted at some incubation time $t$ other than the standard 5 days.

Basic BOD testing/sampling equation

$BOD (mg/L) = \frac{D _{1} - D _{2}}{P}$

Example: BOD5 dilution test

A $10 m L$ sample is diluted to $300 m L$ with unseeded water. DO is $8.6 m g / L$ at time zero and $4.1 m g / L$ after 5 days at 20°C. Find $BO D_{5}$ .

$P = \frac{10}{300} = 0.0333$

$BOD = \frac{8.6 - 4.1}{0.0333} = 135 m g / L$

Answer: $BO D_{5} \approx 135 m g / L$

When the dilution water is seeded

$BOD (mg/L) = \frac{( D _{1} - D _{2} ) - ( B _{1} - B _{2} ) f}{P}$

Where:

$D_{1}$ = dissolved oxygen of diluted sample immediately after preparation (mg/L)
$D_{2}$ = dissolved oxygen of diluted sample after 5-day incubation at 20°C (mg/L)
$B_{1}$ = dissolved oxygen of seed control before incubation (mg/L)
$B_{2}$ = dissolved oxygen of seed control after incubation (mg/L)
$f$ = fraction of seeded dilution water volume in sample to volume of seeded dilution water in seed control
$P$ = fraction of wastewater sample volume to total combined volume

$BO D_{t}$

$BOD_{t}$ is the BOD exerted after $t$ days of incubation, based on the difference between the blank and the sample, scaled by the dilution factor.

$BOD_{t} = (D O_{b, t} - D O_{s, t}) \times dilution factor$

Where:

$D O_{b, t}$ = dissolved oxygen concentration in blank after $t$ days in incubation (mg/L)
$D O_{s, t}$ = dissolved oxygen concentration in sample after $t$ days in incubation (mg/L)

Dilution factor:

$Dilution factor = \frac{volume of bottle or diluted sample}{volume of undiluted sample}$

Notice that the dilution factor is just the reciprocal of $P$ ( $dilution factor = 1/ P$ ), so multiplying by the dilution factor gives the same result as dividing by $P$ in the basic BOD equation above - the $BO D_{t}$ form simply substitutes the blank/sample DO readings for $D_{1} / D_{2}$ .

BOD exertion

BOD exertion describes how oxygen demand is applied over time as microorganisms decompose organic matter. The demand isn’t used up instantly; it increases gradually as biochemical reactions proceed. Under ideal conditions, this behavior is modeled with first-order kinetics.

$BOD_{t} = L_{0} (1 - e^{- k t})$

Where:

$BOD_{t}$ = amount of BOD exerted at time $t$ (mg/L)
$k$ = BOD decay rate constant, base $e$ (day⁻¹)
$k = 2.30 K$
$K$ = rate constant, base 10 (day⁻¹)
$L_{0}$ = ultimate BOD (mg/L)
$t$ = time (days)

Watch out: Exponentials like $BO D_{t} = L_{0} (1 - e^{- k t})$ and the Streeter-Phelps deficit equation need base- $e$ constants. Convert base-10 $K$ first: $k = 2.30 K$ .

Kinetic temperature corrections

Reaction rates in water quality processes depend strongly on temperature. To model conditions at temperatures other than the standard reference of 20°C, you adjust the rate constant using a temperature correction.

$k_{T} = k_{20} θ^{T - 20}$

Where:

$T$ = temperature of interest (°C)
$k_{T}$ = BOD rate constant at the temperature of interest (day⁻¹)
$k_{20}$ = BOD rate constant determined at 20°C (day⁻¹)
$θ$ = temperature coefficient
BOD ( $k$ ): $θ = 1.135$ for $T = 4$ to $20° C$ , or $θ = 1.056$ for $T = 21$ to $30° C$
Reaeration ( $k_{r}$ ): $θ = 1.024$

Stream modeling - Streeter-Phelps

The Streeter-Phelps model describes dissolved oxygen (DO) changes in a stream after an organic waste discharge. It combines two competing processes:

oxygen depletion from BOD exertion (deoxygenation)
oxygen replenishment from the atmosphere (reaeration)

The model predicts the DO deficit downstream, including the critical point where DO reaches its minimum before recovering toward saturation. The saturation concentration $D O_{s a t}$ is covered in the next section.

A two-part diagram showing BOD oxygen consumption over time and the resulting dissolved oxygen sag curve (Streeter-Phelps model) with initial deficit, critical point, and recovery to saturation downstream of discharge. — Stream modeling with Streeter Phelps

$D = \frac{k _{d} L _{a}}{k _{r} - k _{d}} [e^{- k_{d} t} - e^{- k_{r} t}] + D_{a} e^{- k_{r} t}$

$t_{c} = \frac{1}{k _{r} - k _{d}} ln (\frac{k _{r}}{k _{d}} (1 - D_{a} \frac{k _{r} - k _{d}}{k _{d} L _{a}}))$

$D O = D O_{sat} - D$

Where:

$D$ = dissolved oxygen deficit (mg/L)
$D O$ = dissolved oxygen concentration (mg/L)
$D_{a}$ = initial dissolved oxygen deficit in mixing zone (mg/L)
$D O_{sat}$ = saturated dissolved oxygen concentration (mg/L)
$k_{d}$ = deoxygenation rate constant, base $e$ (day⁻¹)
$k_{r}$ = reaeration rate constant, base $e$ (day⁻¹)
$L_{a}$ = initial ultimate BOD in mixing zone (mg/L)
$t$ = time (days)
$t_{c}$ = time at which minimum dissolved oxygen occurs (days)

Example: Streeter-Phelps deficit

A river has deoxygenation rate $k_{d} = 0.2 d a y^{- 1}$ and reaeration rate $k_{r} = 0.4 d a y^{- 1}$ (both already base $e$ ), initial ultimate BOD $L_{a} = 20 m g / L$ , and initial deficit $D_{a} = 2 m g / L$ immediately after mixing. Find the dissolved oxygen deficit $D$ at $t = 2$ days downstream.

$D = \frac{( 0.2 ) ( 20 )}{0.4 - 0.2} [e^{- (0.2) (2)} - e^{- (0.4) (2)}] + (2) e^{- (0.4) (2)}$

$D = 20 [0.6703 - 0.4493] + 2 (0.4493) = 4.42 + 0.90 = 5.32 m g / L$

Answer: $D \approx 5.3 m g / L$

Oxygen saturation

Oxygen saturation is the maximum dissolved oxygen concentration water can hold when it’s in equilibrium with the atmosphere. It depends mainly on:

temperature
atmospheric pressure
salinity

Saturation is a key reference value because it sets the upper limit for DO and helps define the oxygen deficit used in stream modeling. Henry’s law describes the underlying relationship:

$D O_{sat} = K_{H} \cdot P_{O_{2}}$

Where:

$K_{H}$ = Henry’s law constant (moles/L·atm)
$P_{O_{2}}$ = partial pressure of oxygen (atm)

On the FE exam, look up $D O_{sat}$ from the DO saturation table (about $9.1 m g / L$ at $20° C$ ) rather than computing it from Henry’s law.

$D O_{stream}$ is the same $D O_{sat}$ used in this chapter’s deficit equation ( $D = D O_{sat} - D O$ ); there’s no separate formula to learn.

Dilution purification of wastewater streams

Dilution and natural purification control what happens to wastewater after it enters a receiving water.

Dilution lowers pollutant concentrations through mixing with cleaner water.
Purification reduces pollutants through processes such as biodegradation, sedimentation, and reaeration.

Together, these processes determine a stream’s assimilative capacity and help guide discharge permit limits.

If untreated or partially treated sewage is instantly mixed upon discharge into a large water body, the resulting parameter levels (e.g., temperature, DO, BOD, suspended solids) can be estimated using a weighted average. If the mixed conditions meet water quality standards, pretreatment may not be required.

Consider a wastewater flow rate $Q_{w}$ with ultimate BOD $L_{w}$ and dissolved oxygen $D O_{w}$ mixing with a river with flow rate $Q_{r}$ , ultimate BOD $L_{r}$ , and dissolved oxygen $D O_{r}$ . Each parameter below follows the same weighted-average pattern; keep $Q_{w}$ and $Q_{r}$ in matching flow units.

The initial (immediately after mixing) ultimate BOD of the river-wastewater mix is:

$L_{o} = \frac{Q _{w} L _{w} + Q _{r} L _{r}}{Q _{w} + Q _{r}}$

Initial dissolved oxygen immediately after mixing $D O_{o}$ is given by:

$D O_{o} = \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Temperature immediately after mixing $T_{o}$ is given by:

$T_{o} = \frac{Q _{w} T _{w} + Q _{r} T _{r}}{Q _{w} + Q _{r}}$

Initial oxygen deficit after mixing $D_{o}$ is given by:

$D_{o} = D O_{s a t} - \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Example: river-wastewater mixing

$Q_{w} = 2 MG D$ with $D O_{w} = 2.0 m g / L$ enters a river at $Q_{r} = 18 MG D$ with $D O_{r} = 8.5 m g / L$ . Find the DO immediately downstream (flows already share units).

$D O_{o} = \frac{( 2 ) ( 2.0 ) + ( 18 ) ( 8.5 )}{2 + 18} = \frac{157}{20} = 7.85 m g / L$

Answer: $D O_{o} = 7.85 m g / L$

Biochemical oxygen demand (BOD)

Measures oxygen needed by microbes to oxidize organic matter
Key for wastewater treatment design, compliance, and impact assessment
Main formula: $BOD = \frac{D _{1} - D _{2}}{P}$
- Adjust for seeded dilution: $BOD = \frac{( D _{1} - D _{2} ) - ( B _{1} - B _{2} ) f}{P}$
$BOD_{t}$ after $t$ days: $(D O_{b, t} - D O_{s, t}) \times dilution factor$

BOD exertion

Describes time-dependent oxygen demand as organics decompose
Modeled by first-order kinetics: $BOD_{t} = L_{0} (1 - e^{- k t})$
- $L_{0}$ = ultimate BOD, $k$ = decay rate constant
$k = 2.30 K$ (base 10 to base $e$ conversion)

Kinetic temperature corrections

Reaction rates increase with temperature
Adjust rate constant: $k_{T} = k_{20} θ^{T - 20}$
- $θ = 1.024$ for reaeration
- $k_{20}$ = rate at 20°C, $T$ = temperature of interest

Stream modeling - Streeter Phelps

Predicts DO changes after waste discharge
- Deoxygenation (BOD exertion) vs. reaeration (atmospheric O $_{2}$ )
DO deficit: $D = \frac{k _{d} L _{a}}{k _{r} - k _{d}} [e^{- k_{d} t} - e^{- k_{r} t}] + D_{a} e^{- k_{r} t}$
Critical point (minimum DO): $t_{c} = \frac{1}{k _{r} - k _{d}} ln (\frac{k _{r}}{k _{d}} (1 - D_{a} \frac{k _{r} - k _{d}}{k _{d} L _{a}}))$
$D O = D O_{sat} - D$

Oxygen saturation

Maximum DO water can hold at equilibrium with atmosphere
- Depends on temperature, pressure, salinity
$D O_{sat} = K_{H} \cdot P_{O_{2}}$
Oxygen deficit: $D = D O_{stream} - D O$

Dilution purification of wastewater streams

Dilution: lowers pollutant concentration by mixing
Purification: pollutant reduction via biodegradation, sedimentation, reaeration
Initial mixed values (after discharge):
- Ultimate BOD: $L_{o} = \frac{Q _{w} L _{w} + Q _{r} L _{r}}{Q _{w} + Q _{r}}$
- DO: $D O_{o} = \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$
- Temperature: $T_{o} = \frac{Q _{w} T _{w} + Q _{r} T _{r}}{Q _{w} + Q _{r}}$
- Oxygen deficit: $D_{o} = D O_{s a t} - \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Next | 14.2 Water treatment and distribution

Water quality

This chapter covers the following:

Ground and surface water quality, basic water chemistry
Biochemical oxygen demand (BOD)
BOD exertion
Kinetic temperature corrections
Stream modeling - Streeter Phelps
Oxygen saturation
Dilution purification of wastewater streams

Basic water chemistry

Definitions

Hardness: The concentration of dissolved $C a^{2 +}$ and $M g^{2 +}$ , expressed as an equivalent concentration of calcium carbonate ( $C a C O_{3}$ ).
Alkalinity: Water’s capacity to neutralize acids, from dissolved bicarbonate, carbonate, and hydroxide ions, also expressed as $m g / L$ as $C a C O_{3}$ .
Equivalent weight: A substance’s molecular weight divided by the equivalents ( $n$ ) it contributes: $EW = \frac{MW}{n}$ .

To compare ions on a common basis, convert to an equivalent concentration of $C a C O_{3}$ ( $E W_{C a C O_{3}} = 50 m g / m e q$ ):

$Concentration as C a C O_{3} = concentration \times \frac{E W _{C a C O_{3}}}{E W _{substance}}$

Example: phosphate concentration as CaCO₃

A sample contains $31 m g / L$ of $P O_{4}^{3 -}$ (molecular weight $= 95 g / m o l$ , contributing $n = 3$ equivalents). Find the equivalent concentration as $C a C O_{3}$ .

$E W_{P O_{4}^{3 -}} = \frac{95}{3} = 31.7 m g / m e q$

$Concentration as C a C O_{3} = 31 \times \frac{50}{31.7} = 48.9 m g / L as C a C O_{3}$

Answer: $\approx 48.9 m g / L$ as $C a C O_{3}$

For $a A + b B ⇌ c C + d D$ , the equilibrium constant expression is:

$K = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

Example: equilibrium constant

For the reaction $A + 2 B ⇌ C$ , equilibrium concentrations are $[A] = 0.5 m o l / L$ , $[B] = 0.2 m o l / L$ , and $[C] = 0.1 m o l / L$ . Find $K$ .

$K = \frac{[ C ]}{[ A ] [ B ] ^{2}} = \frac{0.1}{( 0.5 ) ( 0.2 ) ^{2}} = \frac{0.1}{0.02} = 5$

Answer: $K = 5$

Biochemical oxygen demand (BOD)

Because BOD reflects the amount of biodegradable organic pollution present, it’s widely used in:

wastewater treatment design
regulatory compliance
evaluating impacts on receiving waters

Basic BOD testing/sampling equation

$BOD (mg/L) = \frac{D _{1} - D _{2}}{P}$

Example: BOD5 dilution test

A $10 m L$ sample is diluted to $300 m L$ with unseeded water. DO is $8.6 m g / L$ at time zero and $4.1 m g / L$ after 5 days at 20°C. Find $BO D_{5}$ .

$P = \frac{10}{300} = 0.0333$

$BOD = \frac{8.6 - 4.1}{0.0333} = 135 m g / L$

Answer: $BO D_{5} \approx 135 m g / L$

When the dilution water is seeded

$BOD (mg/L) = \frac{( D _{1} - D _{2} ) - ( B _{1} - B _{2} ) f}{P}$

Where:

$D_{1}$ = dissolved oxygen of diluted sample immediately after preparation (mg/L)
$D_{2}$ = dissolved oxygen of diluted sample after 5-day incubation at 20°C (mg/L)
$B_{1}$ = dissolved oxygen of seed control before incubation (mg/L)
$B_{2}$ = dissolved oxygen of seed control after incubation (mg/L)
$f$ = fraction of seeded dilution water volume in sample to volume of seeded dilution water in seed control
$P$ = fraction of wastewater sample volume to total combined volume

$BO D_{t}$

$BOD_{t}$ is the BOD exerted after $t$ days of incubation, based on the difference between the blank and the sample, scaled by the dilution factor.

$BOD_{t} = (D O_{b, t} - D O_{s, t}) \times dilution factor$

Where:

$D O_{b, t}$ = dissolved oxygen concentration in blank after $t$ days in incubation (mg/L)
$D O_{s, t}$ = dissolved oxygen concentration in sample after $t$ days in incubation (mg/L)

Dilution factor:

$Dilution factor = \frac{volume of bottle or diluted sample}{volume of undiluted sample}$

BOD exertion

$BOD_{t} = L_{0} (1 - e^{- k t})$

Where:

$BOD_{t}$ = amount of BOD exerted at time $t$ (mg/L)
$k$ = BOD decay rate constant, base $e$ (day⁻¹)
$k = 2.30 K$
$K$ = rate constant, base 10 (day⁻¹)
$L_{0}$ = ultimate BOD (mg/L)
$t$ = time (days)

Watch out: Exponentials like $BO D_{t} = L_{0} (1 - e^{- k t})$ and the Streeter-Phelps deficit equation need base- $e$ constants. Convert base-10 $K$ first: $k = 2.30 K$ .

Kinetic temperature corrections

$k_{T} = k_{20} θ^{T - 20}$

Where:

$T$ = temperature of interest (°C)
$k_{T}$ = BOD rate constant at the temperature of interest (day⁻¹)
$k_{20}$ = BOD rate constant determined at 20°C (day⁻¹)
$θ$ = temperature coefficient
BOD ( $k$ ): $θ = 1.135$ for $T = 4$ to $20° C$ , or $θ = 1.056$ for $T = 21$ to $30° C$
Reaeration ( $k_{r}$ ): $θ = 1.024$

Stream modeling - Streeter-Phelps

The Streeter-Phelps model describes dissolved oxygen (DO) changes in a stream after an organic waste discharge. It combines two competing processes:

oxygen depletion from BOD exertion (deoxygenation)
oxygen replenishment from the atmosphere (reaeration)

$D = \frac{k _{d} L _{a}}{k _{r} - k _{d}} [e^{- k_{d} t} - e^{- k_{r} t}] + D_{a} e^{- k_{r} t}$

$t_{c} = \frac{1}{k _{r} - k _{d}} ln (\frac{k _{r}}{k _{d}} (1 - D_{a} \frac{k _{r} - k _{d}}{k _{d} L _{a}}))$

$D O = D O_{sat} - D$

Where:

$D$ = dissolved oxygen deficit (mg/L)
$D O$ = dissolved oxygen concentration (mg/L)
$D_{a}$ = initial dissolved oxygen deficit in mixing zone (mg/L)
$D O_{sat}$ = saturated dissolved oxygen concentration (mg/L)
$k_{d}$ = deoxygenation rate constant, base $e$ (day⁻¹)
$k_{r}$ = reaeration rate constant, base $e$ (day⁻¹)
$L_{a}$ = initial ultimate BOD in mixing zone (mg/L)
$t$ = time (days)
$t_{c}$ = time at which minimum dissolved oxygen occurs (days)

Example: Streeter-Phelps deficit

A river has deoxygenation rate $k_{d} = 0.2 d a y^{- 1}$ and reaeration rate $k_{r} = 0.4 d a y^{- 1}$ (both already base $e$ ), initial ultimate BOD $L_{a} = 20 m g / L$ , and initial deficit $D_{a} = 2 m g / L$ immediately after mixing. Find the dissolved oxygen deficit $D$ at $t = 2$ days downstream.

$D = \frac{( 0.2 ) ( 20 )}{0.4 - 0.2} [e^{- (0.2) (2)} - e^{- (0.4) (2)}] + (2) e^{- (0.4) (2)}$

$D = 20 [0.6703 - 0.4493] + 2 (0.4493) = 4.42 + 0.90 = 5.32 m g / L$

Answer: $D \approx 5.3 m g / L$

Oxygen saturation

Oxygen saturation is the maximum dissolved oxygen concentration water can hold when it’s in equilibrium with the atmosphere. It depends mainly on:

temperature
atmospheric pressure
salinity

Saturation is a key reference value because it sets the upper limit for DO and helps define the oxygen deficit used in stream modeling. Henry’s law describes the underlying relationship:

$D O_{sat} = K_{H} \cdot P_{O_{2}}$

Where:

$K_{H}$ = Henry’s law constant (moles/L·atm)
$P_{O_{2}}$ = partial pressure of oxygen (atm)

On the FE exam, look up $D O_{sat}$ from the DO saturation table (about $9.1 m g / L$ at $20° C$ ) rather than computing it from Henry’s law.

$D O_{stream}$ is the same $D O_{sat}$ used in this chapter’s deficit equation ( $D = D O_{sat} - D O$ ); there’s no separate formula to learn.

Dilution purification of wastewater streams

Dilution and natural purification control what happens to wastewater after it enters a receiving water.

Dilution lowers pollutant concentrations through mixing with cleaner water.
Purification reduces pollutants through processes such as biodegradation, sedimentation, and reaeration.

Together, these processes determine a stream’s assimilative capacity and help guide discharge permit limits.

The initial (immediately after mixing) ultimate BOD of the river-wastewater mix is:

$L_{o} = \frac{Q _{w} L _{w} + Q _{r} L _{r}}{Q _{w} + Q _{r}}$

Initial dissolved oxygen immediately after mixing $D O_{o}$ is given by:

$D O_{o} = \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Temperature immediately after mixing $T_{o}$ is given by:

$T_{o} = \frac{Q _{w} T _{w} + Q _{r} T _{r}}{Q _{w} + Q _{r}}$

Initial oxygen deficit after mixing $D_{o}$ is given by:

$D_{o} = D O_{s a t} - \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Example: river-wastewater mixing

$Q_{w} = 2 MG D$ with $D O_{w} = 2.0 m g / L$ enters a river at $Q_{r} = 18 MG D$ with $D O_{r} = 8.5 m g / L$ . Find the DO immediately downstream (flows already share units).

$D O_{o} = \frac{( 2 ) ( 2.0 ) + ( 18 ) ( 8.5 )}{2 + 18} = \frac{157}{20} = 7.85 m g / L$

Answer: $D O_{o} = 7.85 m g / L$

Achievable FE Civil

14. Environmental engineering

Our FE Civil course is now in "early access" - get 50% off for a limited time.

Water quality

10 min read

Font

Discuss

Feedback

This chapter covers the following:

Ground and surface water quality, basic water chemistry
Biochemical oxygen demand (BOD)
BOD exertion
Kinetic temperature corrections
Stream modeling - Streeter Phelps
Oxygen saturation
Dilution purification of wastewater streams

Basic water chemistry

Definitions

Hardness: The concentration of dissolved $C a^{2 +}$ and $M g^{2 +}$ , expressed as an equivalent concentration of calcium carbonate ( $C a C O_{3}$ ).
Alkalinity: Water’s capacity to neutralize acids, from dissolved bicarbonate, carbonate, and hydroxide ions, also expressed as $m g / L$ as $C a C O_{3}$ .
Equivalent weight: A substance’s molecular weight divided by the equivalents ( $n$ ) it contributes: $EW = \frac{MW}{n}$ .

To compare ions on a common basis, convert to an equivalent concentration of $C a C O_{3}$ ( $E W_{C a C O_{3}} = 50 m g / m e q$ ):

$Concentration as C a C O_{3} = concentration \times \frac{E W _{C a C O_{3}}}{E W _{substance}}$

Example: phosphate concentration as CaCO₃

A sample contains $31 m g / L$ of $P O_{4}^{3 -}$ (molecular weight $= 95 g / m o l$ , contributing $n = 3$ equivalents). Find the equivalent concentration as $C a C O_{3}$ .

$E W_{P O_{4}^{3 -}} = \frac{95}{3} = 31.7 m g / m e q$

$Concentration as C a C O_{3} = 31 \times \frac{50}{31.7} = 48.9 m g / L as C a C O_{3}$

Answer: $\approx 48.9 m g / L$ as $C a C O_{3}$

For $a A + b B ⇌ c C + d D$ , the equilibrium constant expression is:

$K = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

Example: equilibrium constant

For the reaction $A + 2 B ⇌ C$ , equilibrium concentrations are $[A] = 0.5 m o l / L$ , $[B] = 0.2 m o l / L$ , and $[C] = 0.1 m o l / L$ . Find $K$ .

$K = \frac{[ C ]}{[ A ] [ B ] ^{2}} = \frac{0.1}{( 0.5 ) ( 0.2 ) ^{2}} = \frac{0.1}{0.02} = 5$

Answer: $K = 5$

Biochemical oxygen demand (BOD)

Because BOD reflects the amount of biodegradable organic pollution present, it’s widely used in:

wastewater treatment design
regulatory compliance
evaluating impacts on receiving waters

Basic BOD testing/sampling equation

$BOD (mg/L) = \frac{D _{1} - D _{2}}{P}$

Example: BOD5 dilution test

A $10 m L$ sample is diluted to $300 m L$ with unseeded water. DO is $8.6 m g / L$ at time zero and $4.1 m g / L$ after 5 days at 20°C. Find $BO D_{5}$ .

$P = \frac{10}{300} = 0.0333$

$BOD = \frac{8.6 - 4.1}{0.0333} = 135 m g / L$

Answer: $BO D_{5} \approx 135 m g / L$

When the dilution water is seeded

$BOD (mg/L) = \frac{( D _{1} - D _{2} ) - ( B _{1} - B _{2} ) f}{P}$

Where:

$D_{1}$ = dissolved oxygen of diluted sample immediately after preparation (mg/L)
$D_{2}$ = dissolved oxygen of diluted sample after 5-day incubation at 20°C (mg/L)
$B_{1}$ = dissolved oxygen of seed control before incubation (mg/L)
$B_{2}$ = dissolved oxygen of seed control after incubation (mg/L)
$f$ = fraction of seeded dilution water volume in sample to volume of seeded dilution water in seed control
$P$ = fraction of wastewater sample volume to total combined volume

$BO D_{t}$

$BOD_{t}$ is the BOD exerted after $t$ days of incubation, based on the difference between the blank and the sample, scaled by the dilution factor.

$BOD_{t} = (D O_{b, t} - D O_{s, t}) \times dilution factor$

Where:

$D O_{b, t}$ = dissolved oxygen concentration in blank after $t$ days in incubation (mg/L)
$D O_{s, t}$ = dissolved oxygen concentration in sample after $t$ days in incubation (mg/L)

Dilution factor:

$Dilution factor = \frac{volume of bottle or diluted sample}{volume of undiluted sample}$

BOD exertion

$BOD_{t} = L_{0} (1 - e^{- k t})$

Where:

$BOD_{t}$ = amount of BOD exerted at time $t$ (mg/L)
$k$ = BOD decay rate constant, base $e$ (day⁻¹)
$k = 2.30 K$
$K$ = rate constant, base 10 (day⁻¹)
$L_{0}$ = ultimate BOD (mg/L)
$t$ = time (days)

Watch out: Exponentials like $BO D_{t} = L_{0} (1 - e^{- k t})$ and the Streeter-Phelps deficit equation need base- $e$ constants. Convert base-10 $K$ first: $k = 2.30 K$ .

Kinetic temperature corrections

$k_{T} = k_{20} θ^{T - 20}$

Where:

$T$ = temperature of interest (°C)
$k_{T}$ = BOD rate constant at the temperature of interest (day⁻¹)
$k_{20}$ = BOD rate constant determined at 20°C (day⁻¹)
$θ$ = temperature coefficient
BOD ( $k$ ): $θ = 1.135$ for $T = 4$ to $20° C$ , or $θ = 1.056$ for $T = 21$ to $30° C$
Reaeration ( $k_{r}$ ): $θ = 1.024$

Stream modeling - Streeter-Phelps

The Streeter-Phelps model describes dissolved oxygen (DO) changes in a stream after an organic waste discharge. It combines two competing processes:

oxygen depletion from BOD exertion (deoxygenation)
oxygen replenishment from the atmosphere (reaeration)

$D = \frac{k _{d} L _{a}}{k _{r} - k _{d}} [e^{- k_{d} t} - e^{- k_{r} t}] + D_{a} e^{- k_{r} t}$

$t_{c} = \frac{1}{k _{r} - k _{d}} ln (\frac{k _{r}}{k _{d}} (1 - D_{a} \frac{k _{r} - k _{d}}{k _{d} L _{a}}))$

$D O = D O_{sat} - D$

Where:

$D$ = dissolved oxygen deficit (mg/L)
$D O$ = dissolved oxygen concentration (mg/L)
$D_{a}$ = initial dissolved oxygen deficit in mixing zone (mg/L)
$D O_{sat}$ = saturated dissolved oxygen concentration (mg/L)
$k_{d}$ = deoxygenation rate constant, base $e$ (day⁻¹)
$k_{r}$ = reaeration rate constant, base $e$ (day⁻¹)
$L_{a}$ = initial ultimate BOD in mixing zone (mg/L)
$t$ = time (days)
$t_{c}$ = time at which minimum dissolved oxygen occurs (days)

Example: Streeter-Phelps deficit

A river has deoxygenation rate $k_{d} = 0.2 d a y^{- 1}$ and reaeration rate $k_{r} = 0.4 d a y^{- 1}$ (both already base $e$ ), initial ultimate BOD $L_{a} = 20 m g / L$ , and initial deficit $D_{a} = 2 m g / L$ immediately after mixing. Find the dissolved oxygen deficit $D$ at $t = 2$ days downstream.

$D = \frac{( 0.2 ) ( 20 )}{0.4 - 0.2} [e^{- (0.2) (2)} - e^{- (0.4) (2)}] + (2) e^{- (0.4) (2)}$

$D = 20 [0.6703 - 0.4493] + 2 (0.4493) = 4.42 + 0.90 = 5.32 m g / L$

Answer: $D \approx 5.3 m g / L$

Oxygen saturation

Oxygen saturation is the maximum dissolved oxygen concentration water can hold when it’s in equilibrium with the atmosphere. It depends mainly on:

temperature
atmospheric pressure
salinity

Saturation is a key reference value because it sets the upper limit for DO and helps define the oxygen deficit used in stream modeling. Henry’s law describes the underlying relationship:

$D O_{sat} = K_{H} \cdot P_{O_{2}}$

Where:

$K_{H}$ = Henry’s law constant (moles/L·atm)
$P_{O_{2}}$ = partial pressure of oxygen (atm)

On the FE exam, look up $D O_{sat}$ from the DO saturation table (about $9.1 m g / L$ at $20° C$ ) rather than computing it from Henry’s law.

$D O_{stream}$ is the same $D O_{sat}$ used in this chapter’s deficit equation ( $D = D O_{sat} - D O$ ); there’s no separate formula to learn.

Dilution purification of wastewater streams

Dilution and natural purification control what happens to wastewater after it enters a receiving water.

Dilution lowers pollutant concentrations through mixing with cleaner water.
Purification reduces pollutants through processes such as biodegradation, sedimentation, and reaeration.

Together, these processes determine a stream’s assimilative capacity and help guide discharge permit limits.

The initial (immediately after mixing) ultimate BOD of the river-wastewater mix is:

$L_{o} = \frac{Q _{w} L _{w} + Q _{r} L _{r}}{Q _{w} + Q _{r}}$

Initial dissolved oxygen immediately after mixing $D O_{o}$ is given by:

$D O_{o} = \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Temperature immediately after mixing $T_{o}$ is given by:

$T_{o} = \frac{Q _{w} T _{w} + Q _{r} T _{r}}{Q _{w} + Q _{r}}$

Initial oxygen deficit after mixing $D_{o}$ is given by:

$D_{o} = D O_{s a t} - \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Example: river-wastewater mixing

$Q_{w} = 2 MG D$ with $D O_{w} = 2.0 m g / L$ enters a river at $Q_{r} = 18 MG D$ with $D O_{r} = 8.5 m g / L$ . Find the DO immediately downstream (flows already share units).

$D O_{o} = \frac{( 2 ) ( 2.0 ) + ( 18 ) ( 8.5 )}{2 + 18} = \frac{157}{20} = 7.85 m g / L$

Answer: $D O_{o} = 7.85 m g / L$

Biochemical oxygen demand (BOD)

Measures oxygen needed by microbes to oxidize organic matter
Key for wastewater treatment design, compliance, and impact assessment
Main formula: $BOD = \frac{D _{1} - D _{2}}{P}$
- Adjust for seeded dilution: $BOD = \frac{( D _{1} - D _{2} ) - ( B _{1} - B _{2} ) f}{P}$
$BOD_{t}$ after $t$ days: $(D O_{b, t} - D O_{s, t}) \times dilution factor$

BOD exertion

Describes time-dependent oxygen demand as organics decompose
Modeled by first-order kinetics: $BOD_{t} = L_{0} (1 - e^{- k t})$
- $L_{0}$ = ultimate BOD, $k$ = decay rate constant
$k = 2.30 K$ (base 10 to base $e$ conversion)

Kinetic temperature corrections

Reaction rates increase with temperature
Adjust rate constant: $k_{T} = k_{20} θ^{T - 20}$
- $θ = 1.024$ for reaeration
- $k_{20}$ = rate at 20°C, $T$ = temperature of interest

Stream modeling - Streeter Phelps

Predicts DO changes after waste discharge
- Deoxygenation (BOD exertion) vs. reaeration (atmospheric O $_{2}$ )
DO deficit: $D = \frac{k _{d} L _{a}}{k _{r} - k _{d}} [e^{- k_{d} t} - e^{- k_{r} t}] + D_{a} e^{- k_{r} t}$
Critical point (minimum DO): $t_{c} = \frac{1}{k _{r} - k _{d}} ln (\frac{k _{r}}{k _{d}} (1 - D_{a} \frac{k _{r} - k _{d}}{k _{d} L _{a}}))$
$D O = D O_{sat} - D$

Oxygen saturation

Maximum DO water can hold at equilibrium with atmosphere
- Depends on temperature, pressure, salinity
$D O_{sat} = K_{H} \cdot P_{O_{2}}$
Oxygen deficit: $D = D O_{stream} - D O$

Dilution purification of wastewater streams

Dilution: lowers pollutant concentration by mixing
Purification: pollutant reduction via biodegradation, sedimentation, reaeration
Initial mixed values (after discharge):
- Ultimate BOD: $L_{o} = \frac{Q _{w} L _{w} + Q _{r} L _{r}}{Q _{w} + Q _{r}}$
- DO: $D O_{o} = \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$
- Temperature: $T_{o} = \frac{Q _{w} T _{w} + Q _{r} T _{r}}{Q _{w} + Q _{r}}$
- Oxygen deficit: $D_{o} = D O_{s a t} - \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Next | 14.2 Water treatment and distribution

Water quality

This chapter covers the following:

Ground and surface water quality, basic water chemistry
Biochemical oxygen demand (BOD)
BOD exertion
Kinetic temperature corrections
Stream modeling - Streeter Phelps
Oxygen saturation
Dilution purification of wastewater streams

Basic water chemistry

Definitions

Hardness: The concentration of dissolved $C a^{2 +}$ and $M g^{2 +}$ , expressed as an equivalent concentration of calcium carbonate ( $C a C O_{3}$ ).
Alkalinity: Water’s capacity to neutralize acids, from dissolved bicarbonate, carbonate, and hydroxide ions, also expressed as $m g / L$ as $C a C O_{3}$ .
Equivalent weight: A substance’s molecular weight divided by the equivalents ( $n$ ) it contributes: $EW = \frac{MW}{n}$ .

To compare ions on a common basis, convert to an equivalent concentration of $C a C O_{3}$ ( $E W_{C a C O_{3}} = 50 m g / m e q$ ):

$Concentration as C a C O_{3} = concentration \times \frac{E W _{C a C O_{3}}}{E W _{substance}}$

Example: phosphate concentration as CaCO₃

A sample contains $31 m g / L$ of $P O_{4}^{3 -}$ (molecular weight $= 95 g / m o l$ , contributing $n = 3$ equivalents). Find the equivalent concentration as $C a C O_{3}$ .

$E W_{P O_{4}^{3 -}} = \frac{95}{3} = 31.7 m g / m e q$

$Concentration as C a C O_{3} = 31 \times \frac{50}{31.7} = 48.9 m g / L as C a C O_{3}$

Answer: $\approx 48.9 m g / L$ as $C a C O_{3}$

For $a A + b B ⇌ c C + d D$ , the equilibrium constant expression is:

$K = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$

Example: equilibrium constant

For the reaction $A + 2 B ⇌ C$ , equilibrium concentrations are $[A] = 0.5 m o l / L$ , $[B] = 0.2 m o l / L$ , and $[C] = 0.1 m o l / L$ . Find $K$ .

$K = \frac{[ C ]}{[ A ] [ B ] ^{2}} = \frac{0.1}{( 0.5 ) ( 0.2 ) ^{2}} = \frac{0.1}{0.02} = 5$

Answer: $K = 5$

Biochemical oxygen demand (BOD)

Because BOD reflects the amount of biodegradable organic pollution present, it’s widely used in:

wastewater treatment design
regulatory compliance
evaluating impacts on receiving waters

Basic BOD testing/sampling equation

$BOD (mg/L) = \frac{D _{1} - D _{2}}{P}$

Example: BOD5 dilution test

A $10 m L$ sample is diluted to $300 m L$ with unseeded water. DO is $8.6 m g / L$ at time zero and $4.1 m g / L$ after 5 days at 20°C. Find $BO D_{5}$ .

$P = \frac{10}{300} = 0.0333$

$BOD = \frac{8.6 - 4.1}{0.0333} = 135 m g / L$

Answer: $BO D_{5} \approx 135 m g / L$

When the dilution water is seeded

$BOD (mg/L) = \frac{( D _{1} - D _{2} ) - ( B _{1} - B _{2} ) f}{P}$

Where:

$D_{1}$ = dissolved oxygen of diluted sample immediately after preparation (mg/L)
$D_{2}$ = dissolved oxygen of diluted sample after 5-day incubation at 20°C (mg/L)
$B_{1}$ = dissolved oxygen of seed control before incubation (mg/L)
$B_{2}$ = dissolved oxygen of seed control after incubation (mg/L)
$f$ = fraction of seeded dilution water volume in sample to volume of seeded dilution water in seed control
$P$ = fraction of wastewater sample volume to total combined volume

$BO D_{t}$

$BOD_{t}$ is the BOD exerted after $t$ days of incubation, based on the difference between the blank and the sample, scaled by the dilution factor.

$BOD_{t} = (D O_{b, t} - D O_{s, t}) \times dilution factor$

Where:

$D O_{b, t}$ = dissolved oxygen concentration in blank after $t$ days in incubation (mg/L)
$D O_{s, t}$ = dissolved oxygen concentration in sample after $t$ days in incubation (mg/L)

Dilution factor:

$Dilution factor = \frac{volume of bottle or diluted sample}{volume of undiluted sample}$

BOD exertion

$BOD_{t} = L_{0} (1 - e^{- k t})$

Where:

$BOD_{t}$ = amount of BOD exerted at time $t$ (mg/L)
$k$ = BOD decay rate constant, base $e$ (day⁻¹)
$k = 2.30 K$
$K$ = rate constant, base 10 (day⁻¹)
$L_{0}$ = ultimate BOD (mg/L)
$t$ = time (days)

Watch out: Exponentials like $BO D_{t} = L_{0} (1 - e^{- k t})$ and the Streeter-Phelps deficit equation need base- $e$ constants. Convert base-10 $K$ first: $k = 2.30 K$ .

Kinetic temperature corrections

$k_{T} = k_{20} θ^{T - 20}$

Where:

$T$ = temperature of interest (°C)
$k_{T}$ = BOD rate constant at the temperature of interest (day⁻¹)
$k_{20}$ = BOD rate constant determined at 20°C (day⁻¹)
$θ$ = temperature coefficient
BOD ( $k$ ): $θ = 1.135$ for $T = 4$ to $20° C$ , or $θ = 1.056$ for $T = 21$ to $30° C$
Reaeration ( $k_{r}$ ): $θ = 1.024$

Stream modeling - Streeter-Phelps

The Streeter-Phelps model describes dissolved oxygen (DO) changes in a stream after an organic waste discharge. It combines two competing processes:

oxygen depletion from BOD exertion (deoxygenation)
oxygen replenishment from the atmosphere (reaeration)

$D = \frac{k _{d} L _{a}}{k _{r} - k _{d}} [e^{- k_{d} t} - e^{- k_{r} t}] + D_{a} e^{- k_{r} t}$

$t_{c} = \frac{1}{k _{r} - k _{d}} ln (\frac{k _{r}}{k _{d}} (1 - D_{a} \frac{k _{r} - k _{d}}{k _{d} L _{a}}))$

$D O = D O_{sat} - D$

Where:

$D$ = dissolved oxygen deficit (mg/L)
$D O$ = dissolved oxygen concentration (mg/L)
$D_{a}$ = initial dissolved oxygen deficit in mixing zone (mg/L)
$D O_{sat}$ = saturated dissolved oxygen concentration (mg/L)
$k_{d}$ = deoxygenation rate constant, base $e$ (day⁻¹)
$k_{r}$ = reaeration rate constant, base $e$ (day⁻¹)
$L_{a}$ = initial ultimate BOD in mixing zone (mg/L)
$t$ = time (days)
$t_{c}$ = time at which minimum dissolved oxygen occurs (days)

Example: Streeter-Phelps deficit

A river has deoxygenation rate $k_{d} = 0.2 d a y^{- 1}$ and reaeration rate $k_{r} = 0.4 d a y^{- 1}$ (both already base $e$ ), initial ultimate BOD $L_{a} = 20 m g / L$ , and initial deficit $D_{a} = 2 m g / L$ immediately after mixing. Find the dissolved oxygen deficit $D$ at $t = 2$ days downstream.

$D = \frac{( 0.2 ) ( 20 )}{0.4 - 0.2} [e^{- (0.2) (2)} - e^{- (0.4) (2)}] + (2) e^{- (0.4) (2)}$

$D = 20 [0.6703 - 0.4493] + 2 (0.4493) = 4.42 + 0.90 = 5.32 m g / L$

Answer: $D \approx 5.3 m g / L$

Oxygen saturation

Oxygen saturation is the maximum dissolved oxygen concentration water can hold when it’s in equilibrium with the atmosphere. It depends mainly on:

temperature
atmospheric pressure
salinity

Saturation is a key reference value because it sets the upper limit for DO and helps define the oxygen deficit used in stream modeling. Henry’s law describes the underlying relationship:

$D O_{sat} = K_{H} \cdot P_{O_{2}}$

Where:

$K_{H}$ = Henry’s law constant (moles/L·atm)
$P_{O_{2}}$ = partial pressure of oxygen (atm)

On the FE exam, look up $D O_{sat}$ from the DO saturation table (about $9.1 m g / L$ at $20° C$ ) rather than computing it from Henry’s law.

$D O_{stream}$ is the same $D O_{sat}$ used in this chapter’s deficit equation ( $D = D O_{sat} - D O$ ); there’s no separate formula to learn.

Dilution purification of wastewater streams

Dilution and natural purification control what happens to wastewater after it enters a receiving water.

Dilution lowers pollutant concentrations through mixing with cleaner water.
Purification reduces pollutants through processes such as biodegradation, sedimentation, and reaeration.

Together, these processes determine a stream’s assimilative capacity and help guide discharge permit limits.

The initial (immediately after mixing) ultimate BOD of the river-wastewater mix is:

$L_{o} = \frac{Q _{w} L _{w} + Q _{r} L _{r}}{Q _{w} + Q _{r}}$

Initial dissolved oxygen immediately after mixing $D O_{o}$ is given by:

$D O_{o} = \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Temperature immediately after mixing $T_{o}$ is given by:

$T_{o} = \frac{Q _{w} T _{w} + Q _{r} T _{r}}{Q _{w} + Q _{r}}$

Initial oxygen deficit after mixing $D_{o}$ is given by:

$D_{o} = D O_{s a t} - \frac{Q _{w} D O _{w} + Q _{r} D O _{r}}{Q _{w} + Q _{r}}$

Example: river-wastewater mixing

$Q_{w} = 2 MG D$ with $D O_{w} = 2.0 m g / L$ enters a river at $Q_{r} = 18 MG D$ with $D O_{r} = 8.5 m g / L$ . Find the DO immediately downstream (flows already share units).

$D O_{o} = \frac{( 2 ) ( 2.0 ) + ( 18 ) ( 8.5 )}{2 + 18} = \frac{157}{20} = 7.85 m g / L$

Answer: $D O_{o} = 7.85 m g / L$

Water quality

Basic water chemistry

Biochemical oxygen demand (BOD)

Basic BOD testing/sampling equation

When the dilution water is seeded

BODt​

BOD exertion

Kinetic temperature corrections

Stream modeling - Streeter-Phelps

Oxygen saturation

Dilution purification of wastewater streams

Water quality

Basic water chemistry

Biochemical oxygen demand (BOD)

Basic BOD testing/sampling equation

When the dilution water is seeded

BODt​

BOD exertion

Kinetic temperature corrections

Stream modeling - Streeter-Phelps

Oxygen saturation

Dilution purification of wastewater streams

Water quality

Basic water chemistry

Biochemical oxygen demand (BOD)

Basic BOD testing/sampling equation

When the dilution water is seeded

BODt​

BOD exertion

Kinetic temperature corrections

Stream modeling - Streeter-Phelps

Oxygen saturation

Dilution purification of wastewater streams

Water quality

Basic water chemistry

Biochemical oxygen demand (BOD)

Basic BOD testing/sampling equation

When the dilution water is seeded

BODt​

BOD exertion

Kinetic temperature corrections

Stream modeling - Streeter-Phelps

Oxygen saturation

Dilution purification of wastewater streams

$BO D_{t}$

$BO D_{t}$

$BO D_{t}$

$BO D_{t}$