Lin S.D. Water and Wastewater Calculations Manual

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⫽ stoichiometric ratio of oxygen uptake per unit of algae

respired, mg/mg

m ⫽ algal growth rate coefficient, per day

r ⫽ algal respiration rate coefficient, per day

Ag ⫽ algal mass concentration, mg/L

13 Natural Self-Puriﬁcation in Streams

When decomposable organic waste is discharged into a stream, a series

of physical, chemical, and biological reactions are initiated and there-

after the stream ultimately will be relieved of its pullutive burden. This

process is so-called natural self-purification. A stream undergoing self-

purification will exhibit continuously changing water quality charac-

teristics throughout the reach of the stream.

Dissolved oxygen concentrations in water are perhaps the most impor-

tant factor in determining the overall effect of decomposable organic

matters in a stream. It is also necessary to maintain mandated DO lev-

els in a stream. Therefore, the type and the degree of wastewater treat-

ment necessary depend primarily on the condition and best usage of the

receiving stream.

Recently the US Environmental Protection Agency has revised the

water quality model QUALZE for total maximum daily loads to rivers

and streams (US EPA, 1997). Interested readers may refer to an excel-

lent example of total maximum daily load analysis in Appendix B of the

US EPA technical guidance manual.

13.1 Oxygen sag curve

The dissolved oxygen balance in a stream which is receiving waste-

water effluents can be formulated from a combination of the rate of oxy-

gen utilization through BOD and oxygen transfer from the atmosphere

into water. Many factors involved in this process are discussed in the

previous sections. The oxygen sag curve (DO balance) is as a result of

DO added minus DO removed. The oxygen balance curve or oxygen pro-

file can be mathematically expressed (Streeter–Phelps, 1925) as previ-

ously discussed:

(1.8b)

where k

and k

are, respectively, deoxygenation and reoxygenation rates

to the base 10 which are popularly used. Since k

is determined under lab-

oratory conditions, the rate of oxygen removed in a stream by oxidation

may be different from that under laboratory conditions. Thus, a term k

2 k

s10

2 10

d 1 D

3 10

Streams and Rivers 55

is often substituted. Likewise, the rate of BOD removal in a stream may

not equal the deoxygenation rate in a laboratory bottle and the oxidation

rate in a stream, so the term k

is used to reflect this situation. Applying

these modified terms, the oxygen sag equation becomes

(1.69)

If deposition occurs, k

will be greater than k

; and if scour of sediment

organic matter occurs, k

will be less then k

Computation of organic waste load-capacity of streams or rivers may

be carried out by using Eq. (1.69) (the following example), the Thomas

method, the Churchill–Buckingham method, and other methods.

Example: Station 1 receives a secondary effluent from a city. BOD loading at

station 2 is negligible. The results of two river samplings at stations 3 and 4

(temperature, flow rates, BOD

, and DO) are adopted from Nemerow (1963) and

are shown below. Construct an oxygen sag curve between stations 3 and 4.

Flow Q

BOD

DO, mg/L DO deficit

Station Temp ⬚C cfs mg/L lb/d Measure Saturated mg/L lb/d

(1) (2) (3) (4) (5) (6) (7) (8) (9)

3 20.0 60.1 36.00 11662 4.1 9.02 4.92 1594

17.0 54.0 10.35 3012 5.2 9.61 4.41 1284

Mean 7337 1439

4 20.0 51.7 21.2 5908 2.6 9.02 6.42 1789

16.5 66.6 5.83 2093 2.8 9.71 6.91 2481

Mean 4000 2135

solution:

Step 1. Calculation for columns 5, 7, 8, and 9

Col. 5: lb/d ⫽ 5.39 ⫻ col. 3 ⫻ col. 4

Col. 7: taken from Table 1.2

3.6 miles

v=1 mph

2 k

s10

2 10

d 1 D

3 10

56 Chapter 1

Col. 8: mg/L ⫽ col. 7 ⫺ col. 6

Col. 9: lb/d ⫽ 5.39 ⫻ col. 3 ⫻ col. 8

Note: 5.39 lb/d ⫽ 1 cfs ⫻ 62.4 lb/ft

⫻ 86,400 s/d ⫻ (1 mg/10

mg)

Step 2. Determine ultimate BOD at stations 3 (L

) and 4 (L

)

Under normal deoxygenation rates, factor of 1.46 is used as a multiplier to

convert BOD

to the ultimate first-stage BOD.

Step 3. Compute deoxygenation rate k

Step 4. Compute average BOD load , average DO deficit , and the dif-

ference in DO deficit ⌬D in the reach (Stations 3 and 4) from above

Step 5. Compute reaeration rate k

f 5 k

5 5.9/1.76 5 3.35

5 5.90 sper dayd

5 1.76 3

8276

1787

1392

2.303 3 0.15 3 1787

5 k

⌬D

2.303⌬tD

5 1392slb/dd

⌬D 5 s1789 1 2481d 2 s1594 1 1284d

5 1787 lb/d

s1439 1 2135d lb/d

5 8276 lb/d

L 5

1 L

d 5

s10,712 1 5840d lb/d

5 1.76 sper dayd

⌬t

log

0.15 days

log

10,712

5840

⌬t 5

3.6 miles

1.0 mph 3 24 h/d

5 0.15 days

5 4000 lb/d 3 1.46 5 5840 lb/d

5 7337 lb/d 3 1.46 5 10,712 lb/d

Streams and Rivers 57

Step 6. Plot the DO sag curve

Assuming the reaction rates k

⫽ k

) and k

remain constant in the reach

(stations 3–4), the initial condition at station 3:

Using Eq. (1.69)

When t ⫽ 0.015

D values at various t can be computed in the same manner. We get D values

in lb/d as below. It was found that the critical point is around t ⫽ 0.09 days.

The profile of DO deficit is depicted as below (Fig. 1.10):

⫽ 1439 D

0.09

⫽ 2245

0.015

⫽ 1745 D

0.105

⫽ 2227

0.03

⫽ 1960 D

0.12

⫽ 2190

0.045

⫽ 2104 D

0.135

⫽ 2137

0.06

⫽ 2192 D

0.15

⫽ 2073

0.075

⫽ 2235

0.015

1.76 3 10,712

5.9 2 1.76

s10

21.7630.015

2 10

25.930.015

d 1 1439 3 10

25.930.015

5 1745 slb/dd

2 k

s10

2 10

d 1 D

3 10

5 10,712 lb/d

5 1439 lb/d

58 Chapter 1

Figure 1.10 Profile of DO deficit.

Thomas method. Thomas (1948) developed a useful simplification of the

Streeter-Phelps equation for evaluating stream waste–assimilative

capacity. His method presumes the computation of the stream deoxy-

genation coefficient (k

) and reoxygenation coefficient (k

) as defined in

previous sections (also example). He developed a nomograph to calcu-

late the DO deficit at any time, t (DO profile), downstream from a

source of pollution load. The nomograph (not shown, the interested

reader should refer to his article) is plotted as D/L

versus k

t for var-

ious ratios of k

. In most practical applications, this can be solved

only by a tedious trial-and-error procedure. Before the nomograph is

used, k

, k

, D

, and L

must be computed (can be done as in the pre-

vious example). By means of a straightedge, a straight line (isopleth)

is drawn, connecting the value of D

at the left of the point repre-

senting the reaeration constant ⫻ time of travel (k

t) on the appropri-

ate k

curve (a vertical line at k

t and intercept on the k

curve).

The value D

is read at the intersection with the isopleth. Then, the

value of the deficit at time t(D

) is obtained by multiplying L

with the

interception value.

Nemerow (1963) claimed that he had applied the Thomas nomograph

in many practical cases and found it very convenient, accurate, and

timesaving. He presented a practical problem-solving example for an

industrial discharge to a stream. He illustrated three different DO pro-

files along a 38-mile river: i.e. (1) under current loading and flow con-

ditions; (2) imposition of an industrial load at the upstream station

under current condition; and (3) imposition of industrial load under 5-

year low-stream conditions.

Churchill–Buckingham method. Churchill and Buckingham (1956)

found that the DO values in streams depend on only three variables:

temperature, BOD, and stream flow. They used multiple linear corre-

lation with three normal equations based on the principle of least

squares:

(1.70)

(1.71)

(1.72)

where y ⫽ DO dropped in a reach, mg/L

⫽ BOD at sag, mg/L

⫽ temperature, ⬚C

⫽ flow, cfs (or MGD)

a, b, and c ⫽ constants

⌺x

y 5 ax

1 bx

1 cx

⌺x

y 5 ax

1 bx

1 cx

⌺x

y 5 ax

1 bx

1 cx

Streams and Rivers 59

By solving these three equations, the three constants a, b, and c can be

obtained. Then, the DO drop is expressed as

(1.73)

This method eliminates the often questionable and always cumbersome

procedure for determining k

, k

, and k

. Nemerow (1963) reported

that this method provides a good correlation if each stream sample is

collected during maximum or minimum conditions of one of the three

variables. Only six samples are required in a study to produce practi-

cal and dependable results.

Example: Data obtained from six different days’ stream surveys during

medium- and low-flow periods showed that the DO sag occurred between sta-

tions 2 and 4. The results and computation are as shown in Table 1.12.

Develop a multiple regression model for DO drop and BOD loading.

solution:

Step 1. Compute elements for least-squares method (see Table 1.13)

Step 2. Apply to three normal equations by using the values of the sum or

average of each element

229.84a ⫹ 326.7b ⫹ 122.81c ⫽ 39.04

326.7a ⫹ 474.42b ⫹ 179.77c ⫽ 56.98

122.81a ⫹ 179.77b ⫹ 92.33c ⫽ 23.07

Solving these three equations by dividing by the coefficient of a:

a ⫹ 1.421b ⫹ 0.534c ⫽ 0.1699

a ⫹ 1.452b ⫹ 0.550c ⫽ 0.1744

a ⫹ 1.464b ⫹ 0.752c ⫽ 0.1879

y 5 ax

1 bx

1 cx

1 d

60 Chapter 1

TABLE 1.12 DO Sag between Stations 2 and 4

Observed DO, mg/L DO drop, BOD @ Temperature, Flow,

mg/L sag, mg/L °C 1000 cfs

Station 2 Station 4 yx

7.8 5.7 2.1 6.8 11.0 13.21

8.2 5.9 2.3 12.0 16.5 11.88

6.1 4.8 1.9 14.8 21.2 9.21

6.0 3.2 2.8 20.2 23.4 6.67

5.5 2.3 3.2 18.9 28.3 5.76

6.2 2.9 3.3 14.3 25.6 8.71

Sum 15.6 87.0 126.0 55.44

Mean 2.6 14.5 21.0 9.24

Subtracting one equation from the other:

0.043b ⫹ 0.218c ⫽ 0.0180

0.012b ⫹ 0.202c ⫽ 0.0135

Dividing each equation by the coefficient of b:

b ⫹ 5.07c ⫽ 0.4186

b ⫹ 16.83c ⫽ 1.125

Subtracting one equation from the other:

11.76c ⫽ 0.7064

c ⫽ 0.0601

Substituting c in the above equation:

Similarly

Check on to the three normal equations with values of a, b, and c:

The preceding computations yield the following DO drop:

Y ⫽⫺0.0241x

⫹ 0.1139x

⫹ 0.0601x

Step 3. Compute the DO drop using the above model

The predicted and observed DO drop is shown in Table 1.14.

d 5 Y 2 sax

1 bx

1 cx

5 39.04 2 s2229.84 3 0.0241 1 326.7 3 0.1139 1 122.81 3 0.0601d

5 0

a 5 0.1699 2 1.421 3 0.1139 2 0.534 3 0.0601

520.0241

b 1 5.07 3 0.0601 5 0.4186

b 5 0.1139

Streams and Rivers 61

TABLE 1.13 Elements for Least-Squares Method

4.41 14.28 23.10 27.74 46.24 74.8 89.83 121.00 145.31 174.50

5.29 27.60 37.95 27.32 144.00 198.0 142.56 272.25 176.02 141.13

3.61 28.12 40.28 17.50 219.04 313.76 136.31 449.44 195.25 84.82

7.84 56.56 65.52 18.68 408.04 472.68 134.74 547.56 196.08 44.49

10.24 60.48 90.56 18.43 357.21 534.87 108.86 800.89 163.01 33.18

10.89 47.19 84.48 28.74 204.49 366.08 124.55 655.36 222.98 75.86

Sum 42.28 234.23 341.89 138.42 1379.02 1960.19 736.85 2846.50 1078.64 553.99

Mean 7.05 39.04 56.98 23.07 229.84 326.70 122.81 474.42 179.77 92.33

It can be seen from Table 1.14 that the predicted versus the observed DO

drop values are reasonably close.

Step 4. Compute the allowable BOD loading at the source of the pollution

The BOD equation can be derived from the same least-squares method by cor-

relating the upstream BOD load (as x

) with the water temperature (as x

flow rate (as x

), and resulting BOD (as z) at the sag point in the stream.

Similar to steps 1 and 2, the DO drop is replaced by BOD at station 4 (not

shown). It will generate an equation for the allowable BOD load:

z ⫽ ax

⫹ bx

⫹ cx

⫹ d

Then, applying the percent of wastewater treatment reduction, the BOD in

the discharge effluent is calculated as x

. Selecting the design temperature

and low-flow data, one can then predict the BOD load (lb/d or mg/L) from the

above equation. Also, applying z mg/L in the DO drop equation gives the pre-

dicted value for DO drop at the sag.

13.2 Determination of

The value k

can be determined for a given reach of stream by deter-

mining the BOD of water samples from the upper (A) and lower (B) ends

of the section under consideration:

(1.74)

(1.75a)

(1.75b)

slogL

2 log L

log

52k

62 Chapter 1

TABLE 1.14 Predicted and Observed DO Drop

Dissolved oxygen drop, mg/L

Sampling date Calculated Observed

1 1.93 2.1

2 2.35 2.3

3 2.61 1.9

4 2.62 2.8

5 3.07 3.2

6 3.04 3.3

While the formula calls for first-stage BOD values, any consistent BOD

values will give satisfactory results for k

Example 1: The first-stage BOD values for river stations 3 and 4 are 34.6

and 24.8 mg/L, respectively. The time of travel between the two stations is

0.99 days. Find k

for the reach.

solution:

Example 2: Given

⫽ 32.8 mg/L, L

⫽ 24.6 mg/L

⫽ 3.14 mg/L, D

⫽ 2.58 mg/L

t ⫽ 1.25 days

Find k

and k

solution:

Step 1. Determine k

using Eqs. (1.44) and (1.45)

Reaeration

Mean deficit

Step 2. Compute k

slog L

2 log L

1.25

slog 32.8 2 log 24.6d

5 0.1sper dayd

5 s3.14 1 2.58d/2 5 2.86

8.76

2.86

5 3.06 sper dayd or k

3.06

2.303

5 1.33 sper dayd

5 sL

2 L

d 1 sD

2 D

5 s32.8 2 24.6d 1 s3.14 2 2.58d

5 8.76 smg/Ld

slog L

2 log L

0.99

slog 34.6 2 log 24.8d

5 0.15 sper day for base 10d

5 2.3026k

5 0.345 per day for log base e

Streams and Rivers 63

13.3 Critical point on oxygen sag curve

In many cases, only the lowest point of the oxygen sag curve is of inter-

est to engineers. The equation can be modified to give the critical value

for DO deficiency (D

) and the critical time (t

) downstream at the crit-

ical point. At the critical point of the curve, the rate of deoxygenation

equals the rate of reoxygenation:

(1.76)

Thus

(1.77a)

(1.77b)

where L

⫽ BOD remaining.

Use K

if deoxygenation rate at critical point is different from K

Since

then

(1.78a)

(1.78b)

Let

(1.79)

(1.80)D

3 e

f 5

3 10

3 e

5 L

3 e

5 K

2 K

5 0

64 Chapter 1