Lin S.D. Water and Wastewater Calculations Manual

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or at this point

D ⫽ K

L (1.51)

The value of D can be obtained from a field measurement. The value of

L can be obtained by measuring oxygen uptake from a sample taken

from a given point on a river. Thus, K

can be computed from the above

equation.

9.3 Empirical formulas

The factors in the Streeter–Phelps equation has stimulated much

research on the reaeration rate coefficient K

. Considerable controversy

exists as to the proper method of formulas to use. Several empirical

and semi-empirical formulas have been developed to estimate K

, almost

all of which relate stream velocity and water depth to K

, as first pro-

posed by Streeter (1926). Three equations below are widely known and

employed in stream studies:

O’Connor and Dobbins (1958) (1.52)

Churchill et al. (1962) (1.53)

Langbein and Durum (1967) (1.54)

where K

⫽ reaeration rate coefficient, per day

V ⫽ average velocity, ft/s (fps)

H ⫽ average water depth, ft

On the basis of K

-related physical aspects of a stream, O’Connor

(1958) and Eckenfelder and O’Connor (1959) proposed K

formulas as

follows:

For an isotropic turbulence (deep stream)

(1.55)

For a nonisotropic stream

(1.56)

where D

⫽ diffusivity of oxygen in water, ft

H ⫽ average depth, ft

480D

0.5

0.25

1.25

V d

0.5

2.3H

1.5

7.63V

1.33

11.57V

0.969

1.673

13.0V

0.5

1.5

Streams and Rivers 45

S ⫽ slope

V ⫽ velocity of flow, fps ⫽ B (HS) (nonisotropic if B ⬍ 17;

isotropic if B ⬎ 17)

The O’Connor and Dobbins equation, Eq. (1.52), is based on a theory

more general than the formula developed by several of the other investi-

gators. The formula proposed by Churchill et al. (Eq. (1.53)) appears to be

more restrictive in use. The workers from the US Geological Survey

(Langbein and Durum, 1967) have analyzed and summarized the work of

several investigators and concluded that the velocities and cross section

information were the most applicable formulation of the reaeration factor.

Example: A stream has an average depth of 9.8 ft (3.0 m) and velocity of flow

of 0.61 ft/s (0.20 m/s). What are the K

values determined by the first three

empirical formulas in this section (Eqs. (1.52) to (1.54))? What is K

at 25⬚C

of water temperature (use Eq. (1.54) only).

solution:

Step 1. Determine K

at 20⬚C

(a) by the O’Conner and Dobbins formula (Eq. (1.52))

(b) by the Churchill et al. formula (Eq. (1.53))

Step 2. For a temperature of 25⬚C

9.4 Stationary ﬁeld monitoring procedure

Larson et al. (1994) employed physical, biochemical, and biological fac-

tors to analyze the K

value. Estimations of physical reaeration and the

2s@25d

5 K

2s@20d

s1.02d

T220

5 0.22s1.02d

25220

5 0.24 sper dayd

7.63 3 0.61

s9.8d

1.33

5 0.22 sper dayd

11.57 3 s0.61d

0.969

s9.8d

1.673

5 0.16 sper dayd

13.0V

0.5

1.5

13.0 3 s0.61d

0.5

s9.8d

1.5

5 0.33 sper dayd

46 Chapter 1

associated effects of algal photosynthetic oxygen production (primary

productivity) are based on the schematic formulations as follows:

Physical aeration ⫽ ambient DO ⫺ light chamber DO ⫹ SOD

Algal productivity:

Gross ⫽ light chamber DO ⫺ dark chamber DO

Net ⫽ light chamber DO at end ⫺ light chamber DO at beginning

The amount of reaeration (REA) is computed with observed data from

DataSonde as

REA ⫽ C

⫺ C

⫺ POP ⫺ PAP ⫹ TBOD ⫹ SOD (1.57)

where REA ⫽ reaeration (⫽D

⫺ D

⫽ dD), mg/L

and C

⫽ observed DO concentrations in mg/L at t

and t

respectively

POP ⫽ net periphytonic (attached algae) oxygen

production for the time period (t

⫺ t

⫽⌬t), mg/L

PAP ⫽ net planktonic algae (suspended algae) oxygen

production for the time period t

⫺ t

, mg/L

TBOD ⫽ total biochemical oxygen demand (usage) for the

time period t

⫺ t

, mg/L

SOD ⫽ net sediment oxygen demand for the time period

⫺ t

, mg/L

The clear periphytonic chamber is not used if periphytonic productivity/

respiration is deemed insignificant, i.e. POD ⫽ 0. The combined effect

of TBOD – PAP is represented by the gross output of the light chamber,

and the SOD is equal to the gross SOD chamber output less than dark

chamber output.

The DO deficit D in the reach is calculated as

(1.58)

where S

and S

are the DO saturation concentration in mg/L at t

and

, respectively, for the average water temperature (T ) in the reach.

The DO saturation formula is given in the preview section (Eq. (1.4)).

Since the Hydrolab’s DataSondes logged data at hourly intervals, the

DO changes attributable to physical aeration (or deaeration) are avail-

able for small time frames, which permits the following modification of

the basic natural reaeration equation to be used to calculate K

values

(Broeren et al., 1991)

(1.59)

⌺

i51

i11

2 R

d 52K

2 ⌺

i51

1 S

i11

1 C

i11

D 5

1 S

1 C

Streams and Rivers 47

where R

⫽ an ith DO concentration from the physical aeration

DO-used “mass diagram” curve

i⫹1

⫽ a DO concentration 1 hour later than R

on the physical

reaeration DO-used curve

⫽ an ith DO saturation concentration

i⫹1

⫽ a DO saturation concentration 1 hour later than S

⫽ an ith-observed DO concentration

i⫹1

⫽ an observed DO concentration 1 hour later than C

Note: All units are in mg/L.

Thus, the reaeration rate can be computed by

(1.60)

Both the algal productivity/respiration (P/R) rate and SOD are biolog-

ically associated factors which are normally expressed in terms of grams

of oxygen per square meter per day (g/(m

⭈ d)). Conversion for these areal

rates to mg/L of DO usage for use in computing physical aeration (REA)

is accomplished using the following formula (Butts et al., 1975):

(1.61)

where U ⫽ DO usage in the river reach, mg/L

G ⫽ SOD or P/R rate, g/(m

⭈ d)

t ⫽ time of travel through the reach, days

H ⫽ average depth, ft

Example: Given that C

⫽ 6.85 mg/L, C

⫽ 7.33 mg/L, POP ⫽ 0, at T ⫽ 20⬚C,

gross DO output in light chamber ⫽ 6.88 mg/L, dark chamber DO output ⫽

5.55 mg/L, gross SOD chamber output ⫽ 6.15 mg/L, and water temperature

at beginning and end of field monitoring ⫽ 24.4 and 24.6⬚C, respectively,

⫺ t

⫽ 2.50 days, calculate K

at 20⬚C.

solution:

Step 1. Determine REA

TBOD

⫺ PAP ⫽ gross ⫽ 6.88 mg/L ⫺ 5.55 mg/L ⫽ 1.33 mg/L

SOD ⫽ 6.15 mg/L ⫺ 5.55 mg/L ⫽ 0.60 mg/L

From Eq. (1.57)

REA ⫽ C

⫺ C

⫺ POP ⫹ (⫺PAP ⫹ TBOD) ⫹ SOD

⫽ 7.33 ⫺ 6.85 ⫺ 0 ⫹ 1.33 ⫹ 0.60

⫽ 2.41 mg/L

U 5

3.28Gt

REA

Dst

2 t

48 Chapter 1

Step 2. Calculate S

, S

, and D

at T ⫽ 24.4⬚C

⫽ 14.652 ⫺ 0.41022(24.4) ⫹ 0.007991(24.4)

⫺ 0.00007777(24.4)

⫽ 8.27 (mg/L) or from Table 1.2

at T ⫽ 24.6⬚C, use same formula

⫽ 8.24 mg/L

Note: The elevation correction factor is ignored, for simplicity.

From Eq. (1.58)

Step 3. Compute K

with Eq. (1.60)

10 Sediment Oxygen Demand

Sediment oxygen demand (SOD) is measure of the oxygen demand char-

acteristics of the bottom sediment which affects the dissolved oxygen

resources of the overlying water. To measure SOD, a bottom sample is

especially designed to entrap and seal a known quantity of water at the

river bottom. Changes in DO concentrations (approximately 2 h, or until

the DO usage curve is clearly defined) in the entrapped water are

recorded by a DO probe fastened in the sampler (Butts, 1974). The test

temperature should be recorded and factored for temperature correction

at a specific temperature.

SOD curves can be plotted showing the accumulated DO used (y-axis)

versus elapsed time (x-axis). SOD curves resemble first-order car-

bonaceous BOD curves to a great extent; however, first-order kinetics

are not applicable to the Upper Illinois Waterway’s data. For the most

part, SOD is caused by bacteria reaching an “unlimited” food supply.

Consequently, the oxidation rates are linear in nature.

REA

Dst

2 t

2.41 mg/L

1.17 mg/L 3 2.50 days

5 0.82 per day

D 5

1 S

1 C

s8.27 1 8.24 2 6.85 2 7.33d

5 1.17 smg/Ld

Streams and Rivers 49

The SOD rates, as taken from SOD curves, are in linear units of mil-

ligrams per liter per minute (mg/(L ⭈ min)) and can be converted into

grams per square meter per day (g/(m

⭈ d)) for practical application. A

bottle sampler SOD rate can be formulated as (Butts, 1974; Butts and

Shackleford, 1992):

(1.62)

where SOD ⫽ sediment oxygen demand, g/(m

⭈ d)

S ⫽ slope of stabilized portion of the curve, mg/(L ⭈ min)

V ⫽ volume of sampler, L

A ⫽ bottom area of sampler, m

Example: An SOD sampler is a half-cylinder in shape (half-section of steel

pipe). Its diameter and length are ft (14 in) and 2.0 ft, respectively.

(1) Determine the relationship of SOD and slope of linear portion of usage

curve.

(2) What is SOD in g/(m

⭈ d) for S ⫽ 0.022 mg/(L ⭈ min)?

solution:

Step 1. Find area of the SOD sampler

Step 2. Determine the volume of the sampler

Step 3. Compute SOD with Eq. (1.62)

5 195S ignoring the volume of two union connections

1400 3 S 3 30.27

3 0.217

SOD 5

1440SV

5 30.27 L

V 5

3 3.14 3 s7/12 ftd

3 s1.0 ftd 3 s28.32 L/ft

V 5

5 0.217 m

A 5 1

sftd 3 2.0 sftd 3 0.0929 sm

/ft

SOD 5

1440 SV

50 Chapter 1

Note: If two union connections are considered, the total volume of water con-

tained within the sampler is 31.11 L. Thus, the Illinois State Water Survey’s

SOD formula is

SOD

⫽ 206.6S

Step 4. Compute SOD

10.1 Relationship of sediment

characteristics and SOD

Based on extensive data collections by the Illinois State Water Survey,

Butts (1974) proposed the relationship of SOD and the percentage of

dried solids with volatile solids.

The prediction equation is

SOD ⫽ 6.5(DS)

⫺0.46

(VS)

0.38

(1.63)

where SOD ⫽ sediment oxygen demand g/(m

⭈ d)

DS ⫽ percent dried solids of the decanted sample by weight

VS ⫽ percent volatile solids by weight

Example: Given DS ⫽ 68% and VS ⫽ 8%, predict the SOD value of this

sediment.

solution:

10.2 SOD versus DO

An SOD–DO relationship has been developed from the data given by

McDonnell and Hall (1969) for 25-cm deep sludge. The formula is

SOD/SOD

⫽ DO

0.28

(1.64)

where SOD ⫽ SOD at any DO level, g/(m

⭈ d)

SOD

⫽ SOD at a DO concentration of 1.0 mg/L, g/(m

⭈ d)

DO ⫽ dissolved oxygen concentration, mg/L

SOD 5 6.5sDSd

20.46

sVSd

0.38

6.5 3 s8d

0.38

s68d

0.46

6.5 3 2.20

6.965

5 2.05 [g/sm

dd]

SOD 5 206.6 3 0.022

5 4.545 [g/sm

dd]

Streams and Rivers 51

This model can be used to estimate the SOD rate at various DO con-

centration in areas having very high benthic invertebrate populations.

Example: Given SOD

⫽ 2.4 g/(m

⭈ d), find SOD at DO concentrations of

5.0 and 8.0 mg/L.

solution:

Step 1. For DO

⫽ 5.0 mg/L, with Eq. (1.64)

Step 2. For DO

⫽ 8.0 mg/L

Undisturbed samples of river sediments are collected by the test lab-

oratory to measure the oxygen uptake of the bottom sediment: the

amount of oxygen consumed over the test period is calculated as a zero-

order reaction (US EPA, 1997):

(1.65)

where dC/dt ⫽ rate change of oxygen concentration, g O

/(m

⭈ d)

SOD ⫽ sediment oxygen demand, g O

/(m

⭈ d)

H ⫽ average river depth, m

11 Organic Sludge Deposits

Solids in wastewaters may be in the forms of settleable, flocculation or

coagulation of colloids, and suspended. When these solids are being car-

ried in a river or stream, there is always a possibility that the velocity

of flow will drop to some value at which sedimentation will occur. The

limiting velocity at which deposition will occur is probably about 0.5 to

0.6 ft/s (15 to 18 cm/s).

Deposition of organic solids results in a temporary reduction in the

BOD load of the stream water. Almost as soon as organic matter is

deposited, the deposits will start undergoing biological decomposition,

SOD

5 2.4 3 8

0.28

5 2.4 3 1.79

5 4.30 smg/Ld

SOD

5 SOD

3 DO

0.28

5 2.4 3 5

0.28

5 2.4 3 1.57

5 3.77 smg/Ld

52 Chapter 1

which results in some reduction in the DO concentration of the water

adjacent to the sediment material. As the deposited organic matter

increases in volume, the rate of decomposition also increases. Ultimately,

equilibrium will be established. Velz (1958) has shown that at equilib-

rium the rate of decomposition (exertion of a BOD) equals the rate of

deposition.

In order to properly evaluate the effects of wastewater on the oxygen

resources of a stream, it may be necessary to account for the contribu-

tion made by solids being deposited or by sediments being scoured and

carried into suspension.

From field observations, Velz (1958) concluded that enriched sedi-

ments will deposit and accumulate at a stream velocity of 0.6 ft/s (18 cm/s)

or less, and resuspension of sediments and scouring will occur at a flow

velocity of 1.0 to 1.5 ft/s (30 to 45 cm/s). Velz (1958) has reported that

the effect of sludge deposits can be expressed mathematically. The accu-

mulation of sludge deposition (L

) is

(1.66)

where L

⫽ accumulation of BOD in area of deposition, lb/d

⫽ BOD added, lb/d

k ⫽ rate of oxidation of deposit on base 10

t ⫽ time of accumulation, days

Example: Given (A) k ⫽ 0.03; t ⫽ 2 days

(B) k

⫽ 0.03; t ⫽ 30 days

⫽ 0.03; t ⫽ 50 days

Find the relationship of L

and P

solution for A: (with Eq. (1.66))

This means that in 2 days, 188% of daily deposit (P

) will have been accu-

mulated in the deposit area, and 13% of P

will have been oxidized. The rate

of utilization is about 6.9% per day of accumulated sludge BOD.

5 1.88P

0.0691

s0.13d

2.303 3 0.03

s1 2 10

20.0332

2.303k

s1 2 10

2kt

2.303k

s1 2 10

2kt

Streams and Rivers 53

solution for B:

In 30 days, 1265% of daily deposit will have been accumulated and the rate

of utilization is about 87.4% of daily deposit.

solution for C:

In 50 days, 1403% of daily deposit will have been accumulated and the rate

of utilization is about 97% of daily deposit (98% in 55 days, not shown). This

suggests that equilibrium is almost reached in approximately 50 days, if

there is no disturbance from increased flow.

12 Photosynthesis and Respiration

Processes of photosynthesis and respiration by aquatic plants such as

phytoplankton (algae), periphyton, and rooted aquatic plants (macro-

phytes) could significantly affect the DO concentrations in the water col-

umn. Plant photosynthesis consumes nutrients and carbon dioxide

under the light, and produces oxygen. During dark conditions, and dur-

ing respiration, oxygen is used. The daily average oxygen production and

reduction due to photosynthesis and respiration can be expressed in

the QUALZE model as (US EPA, 1997):

(1.67)

(1.68)

where dC/dt ⫽ rate of change of oxygen concentration, mg O

/(L ⭈ d)

P ⫽ average gross photosynthesis production, mg O

/(L ⭈ d)

R ⫽ average respiration, mg O

/(L ⭈ d)

⫽ stoichiometric ratio of oxygen produced per unit of

algae photosynthesis, mg/mg

5 sa

m 2 a

rdAg

5 P 2 R

0.0691

s1 2 10

20.03350

0.0691

s0.97d

5 14.03P

0.0691

s1 2 10

20.03330

0.0691

s0.874d

5 12.65P

54 Chapter 1