Lin S.D. Water and Wastewater Calculations Manual

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solution:

These three sets of data have several unique characteristics and considera-

tions. Figure 1.7 demonstrates that S-shaped NBOD, and TBOD curves exist

at Lockport station (Illinois) that fit the mathematical formula represented

by Eq. (1.8). These curves usually occur at Lockport during cold weather peri-

ods, but not always.

Streams and Rivers 35

TABLE 1.10 Biochemical Oxygen Demand at 20ⴗC in the Kankakee River

Station sample: Kankakee 11

Date: 08/13/90

pH: 8.39

Temp: 23.70⬚C

Time, days TBOD, mg/L CBOD, mg/L NBOD, mg/L

0.78 0.81 0.44 0.37

2.79 1.76 1.13 0.63

3.52 2.20 1.26 0.93

6.03 2.42 1.62 0.79

6.77 2.83 1.94 0.89

7.77 3.20 2.22 0.98

8.76 3.54 2.49 1.06

9.49 3.83 2.81 1.02

10.76 4.19 3.09 1.10

13.01 4.85 3.76 1.10

13.43 5.08 3.94 1.14

14.75 5.24 4.02 1.22

15.48 5.57 4.42 1.15

17.73 6.64 5.30 1.17

19.96 7.03 5.85 1.19

Figure 1.7 BOD progressive curves for Lockport 18, January 16, 1990

(Butts and Shackleford, 1992).

Figure 1.8 illustrates Lockport warm weather BOD progression curves.

In comparisons with Figs. 1.7 and 1.8, there are extreme differences

between cold and warm weather BOD curves. The January NBOD

rep-

resents 75.3% of the TBOD

, whereas the September NBOD

contains

36 Chapter 1

Figure 1.8 BOD progressive curves for Lockport 36, September 26, 1990

(Butts and Shackleford, 1992).

Figure 1.9 BOD progressive curves for Kankakee 10, August 13, 1990

(Butts and Shackleford, 1992).

only 56.6% of the TBOD

. Furthermore, the September TBOD

is only

23.3% as great as the January TBOD

(see Table 1.9).

Figure 1.9 shows most tributary (Kankakee River) BOD characteris-

tics. The warm weather TBOD

levels for the tributary often approach

those observed at Lockport station (7.03 mg/L versus 8.92 mg/L), but the

fraction of NBOD

is much less (1.19 mg/L versus 5.05 mg/L) (Tables 1.9

and 1.10). Nevertheless, the TBOD

loads coming from the tributaries

are usually much less than those originating from Lockport since the trib-

utary flows are normally much lower.

8.4 Second-order reaction

In many cases, researchers stated that a better fit of BOD data can

be obtained by using a second-order chemical reaction equation, i.e. the

equation of a rotated rectangular hyperbola. A second-order chemical

reaction is characterized by a rate of reaction dependent upon the

concentration of two reactants. It is defined as (Young and Clark,

1965):

(1.29)

When applying this second-order reaction to BOD data, K is a con-

stant; C becomes the initial substance concentration; L

, minus the

BOD; y, at any time, t; or

(1.30)

rearranging

Integrating yields

(1.31)

2 y

52Kt

y5 y

y5 0

dsL

2 yd

t5 t

t5 0

2K dt

dsL

2 yd

52K dt

dsL

2 yd

5 KsL

2 yd

5 KC

Streams and Rivers 37

Multiplying each side of the equation by L

and rearranging,

(1.32)

(1.33)

(1.34)

where a is 1/KL

and b represents 1/L

. The above equation is in the form

of a second-order reaction equation for defining BOD data. The equa-

tion can be linearized in the form

(1.35)

in which a and b can be solved by a least-squares analysis. The simul-

taneous equations for the least-squares treatment are as follows:

at 1 b⌺t 2

⌺t

5 0

⌺

aa 1 bt 2

bt 5 0

⌺

aa 1 bt 2

b 5 0

5 a 1 bt

y 5

a 1 bt

y 5

1 1 KL

s1 1 KL

tdy 5 KL

y 5 KL

t 2 KL

y 5 KL

tsL

2 yd

2 y

52KL

2 y 2 L

2 y

52KL

1 2

2 y

52KL

38 Chapter 1

To solve a and b using five data points, t ⫽ 5, ⌺t ⫽ 15, and ⌺t

⫽ 55; then

Solving for b

(1.36)

Solving for a, by substituting b in the equation

(1.37)

a 5

⌺t

2 15bb

⌺t

2 15 3 0.1a

⌺t

3⌺t

5.5⌺t

1.5⌺t

5 1.1a

⌺t

b 2 0.3a

⌺t

b 5 0.10a

⌺t

3⌺t

bb 2 ca

⌺t

b 2 a

⌺t/y

bd 5 0

55 2 45

bb 2

⌺t

3⌺t

b 5 0

10b 5

⌺t

3⌺t

a 1 ba

b 2 a

⌺t/y

b 5 0

a 1 ba

b 2 a

⌺t

b 5 0

a 1 ba

⌺t

b 2 a

⌺t

b 5 0

a 1 ba

⌺t

b 2 a

⌺t/y

b 5 0

a⌺t 1 b⌺t

⌺t

5 0

Streams and Rivers 39

The velocity of reaction for the BOD curve is y/t. The initial reaction

is 1/a [in mg/(L ⭈ d)] and is the maximum velocity of the BOD reaction

denoted as v

The authors (Young and Clark, 1965) claimed that a second-order

equation has the same precision as a first-order equation at both 20⬚C

and 35⬚C.

Dougal and Baumann (1967) modified the second-order equation for

BOD predication as

(1.38)

where

⫽ 1/b (1.39)

K⬘⫽b/a (1.40)

This formula has a simple rate constant and an ultimate value of BOD.

It can also be transformed into a linear function for regression analy-

sis, permitting the coefficients a and b to be determined easily.

Example: The laboratory BOD data for wastewater were 135, 198, 216, 235,

and 248 mg/L at days 1, 2, 3, 4, and 5, respectively. Find K, L

, and v

for this

wastewater.

solution:

Step 1. Construct a table for basic calculations (Table 1.11)

Step 2. Solve a and b using Eqs. (1.36) and (1.37)

b 5 0.10a

⌺t

3⌺t

5 0.10s0.238167 2 3 3 0.068579d

5 0.003243

y 5

a 1 bt

t/bt

1 1

1/b

1 1

Krt

1 1

40 Chapter 1

TABLE 1.11 Data for Basic Calculations for Step 1

ty t/yt

1 135 0.007407 0.007407

2 198 0.010101 0.020202

3 216 0.013889 0.041667

4 235 0.017021 0.068085

5 248 0.020161 0.100806

Sum 0.068579 0.238167

Step 3. Calculate L

, K, v

, and K⬘ using Eqs. (1.38) to (1.40)

9 Determination of Reaeration Rate

Constant K

9.1 Basic conservation

For a stream deficient in DO but without BOD load, the classical for-

mula is

(1.41)

where dD/dt is the absolute change in DO deficit D over an increment

of time dt due to an atmospheric exchange of oxygen at air/water inter-

face; K

is the reaeration coefficient, per day; and D is DO deficit, mg/L.

52K

5 0.813 sper dayd

5 0.003243/0.003987

Kr 5 b/a

5 10.5 mg/sL

5 250.8 [mg/sL

dd]

5 1/0.003987 [mg/sL

dd]

5 1/a

5 0.00264 [per mg/sL

dd]

5 1/0.003987s308d

K 5 1/aL

5 308 smg/Ld

5 1/0.003243

5 1/b

a 5 1.1a

⌺t

b 2 0.3a

⌺t

5 1.1 3 0.068579 2 0.3 3 0.238167

5 0.003987

Streams and Rivers 41

Integrating the above equation from t

to t

gives

(1.42)

The K

values are needed to correct for river temperature according

to the equation.

(1.43)

where K

2(@T )

⫽ K

value at any temperature T ⬚C and K

2(@20)

⫽ K

value

at 20⬚C.

Example: The DO deficits at upstream and downstream stations are 3.55 and

2.77 mg/L, respectively. The time of travel in this stream reach is 0.67 days.

The mean water temperature is 26.5⬚C. What is the K

value for 20⬚C?

solution:

Step 1. Determine K

at 26.5⬚C

Step 2. Calculate K

2(@20)

therefore

5 0.33 sper dayd

2s@20d

5 0.37/s1.02d

26.5220

5 0.37/1.137

2s@Td

5 K

2s@20d

s1.02d

T220

5 0.37 sper dayd

52s20.248d/0.67

52aln

2.77

3.55

0.67

2s@26.5d

⌬t

2s@Td

5 K

2s@20d

s1.02d

T220

⌬t

5 D

52K

⌬t

52K

tln

52K

2 t

52K

dt

42 Chapter 1

9.2 From BOD and oxygen sag constants

In most stream survey studies involving oxygen sag equations (dis-

cussed later), the value of the reoxygenation constant (K

) is of utmost

importance. Under different conditions, several methods for determin-

ing K

are listed below:

I. K

may be computed from the oxygen sag equation, if all other

parameters are known; however, data must be adequate to support the

conclusions. A trial-and-error procedure is generally used.

II. The amount of reaeration (r

) in a reach (station A to station B)

is equal to the BOD exerted (L

⫺L

) plus oxygen deficiency from sta-

tion A to station B, (D

⫺ D

). The relationship can be expressed as

(1.44)

(1.45)

where r

is the amount of reaeration and D

is the mean (average) deficiency.

Example: Given L

for the upper and lower sampling stations are 24.6 and

15.8 mg/L, respectively; the DO concentration at these two stations are 5.35

and 5.83 mg/L, respectively; and the water temperature is 20⬚C; find the K

value for the river reach.

solution:

Step 1. Find D

, D

, and D

at 20⬚C

Step 2. Calculate r

and K

III. For the case, where DO ⫽ 0 mg/L for a short period of time and

without anaerobic decomposition (O’Connor, 1958),

(1.46)

max

5 K

5 sL

2 L

d 1 sD

2 D

5 s24.6 2 15.8d 1 s3.67 2 3.19d

5 9.28 smg/Ld

5 r

5 9.28/3.43

5 2.71 sper dayd

sat

at 208C 5 9.02 mg/L sTable 1.2d

5 9.02 2 5.35 5 3.67 smg/Ld

5 9.02 2 5.83 5 3.19 smg/Ld

5 s3.67 1 3.19d/2 5 3.43 smg/Ld

5 r

5 sL

2 L

d 1 sD

2 D

Streams and Rivers 43

where D

max

is the maximum deficit, mg/L, and C

is the saturation DO

concentration, mg/L. The maximum deficiency is equal to the DO satu-

ration concentration and the oxygen transferred is oxygen utilized by

organic matter.

Organic matter utilized ⫽ L

⫺ L

, during the time of travel t;

therefore

rate of exertion ⫽ (L

⫺ L

)/t ⫽ K

max

and

⫺ L

)/t ⫽ K

then

⫽ (L

⫺ L

)/C

t (1.47)

Example: Given that the water temperature is 20⬚C; L

and L

are 18.3

and 13.7 mg/L, respectively; and the time of travel from station A to station

B is 0.123 days. Compute K

solution: At 20⬚C,

IV. At the critical point in the river (O’Connor, 1958)

(1.48)

and

(1.49)

(1.50)

where K

is the deoxygenation rate in stream conditions, L

is the first-

stage ultimate BOD at critical point, and D

is the DO deficit at the crit-

ical point. These will be discussed in a later section.

V. Under steady-state conditions, at a sampling point (O’Connor, 1958)

(1.48)

5 0

5 K

5 0

5 9.02 smg/Ld

5 sL

2 L

d/C

5 s18.3 2 13.7d/s9.02 3 0.123d

5 4.15 sper dayd

44 Chapter 1