Lin S.D. Water and Wastewater Calculations Manual

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determined at any specific reach as the channel volume of the reach

divided by the flow as follows:

(1.3)

where t ⫽ time of travel at a stream reach, days

V ⫽ stream reach volume, ft

or m

Q ⫽ average stream flow in the reach, ft

/s(cfs) or m

86,400 ⫽ a factor, s/d

Example: The cross-sectional areas at river miles 62.5, 63.0, 63.5, 64.0,

64.5, and 64.8 are, respectively, 271, 265, 263, 259, 258, and 260 ft

at a sur-

face water elevation. The average flow in 34.8 cfs. Find the time of travel for

a reach between river miles 62.5 and 64.8.

solution:

Step 1. Find average area in the reach

Step 2. Find volume

Step 3. Find t

5 Dissolved Oxygen and Water Temperature

Dissolved oxygen (DO) and water temperature are most commonly in

situ monitored parameters for surface waters (rivers, streams, lakes,

reservoirs, wetlands, oceans, etc.). DO concentration in milligrams per

5 1.06 days

3,190,000 ft

34.8 ft

/s 3 86,400 s/d

t 5

86,400

5 3,190,000 ft

V 5 262.7 ft

3 12,144 ft

5 12,144 ft

5 2.3 miles 3 5280

mile

Distance of the reach 5 s64.8 2 62.5d miles

5 262.7 ft

Average area 5

s271 1 265 1 263 1 259 1 258 1 260d ft

t 5

86,400

Streams and Rivers 5

liter (mg/L) is a measurement of the amount of oxygen dissolved in water.

It can be determined with a DO meter or by a chemical titration method.

The DO in water has an important impact on aquatic animals and

plants. Most aquatic animals, such as fish, require oxygen in the water

to survive. The two major sources of oxygen in water are from diffusion

from the atmosphere across the water surface and the photosynthetic

oxygen production from aquatic plants such as algae and macrophytes.

Important factors that affect DO in water (Fig. 1.1) may include water

temperature, aquatic plant photosynthetic activity, wind and wave mix-

ing, organic contents of the water, and sediment oxygen demand.

Excessive growth of algae (bloom) or other aquatic plants may provide

very high concentration of DO, so called supersaturation. On the other

hand, oxygen deficiencies can occur when plant respiration depletes

oxygen beyond the atmospheric diffusion rate. This can occur especially

during the winter ice cover period and when intense decomposition of

organic matter in the lake bottom sediment occurs during the summer.

These oxygen deficiencies will result in fish being killed.

6 Chapter 1

Figure 1.1 Factors affecting dissolved oxygen concentration in

water.

5.1 Dissolved oxygen saturation

DO saturation (DO

sat

) values for various water temperatures can be com-

puted using the American Society of Civil Engineers’ formula (American

Society of Civil Engineering Committee on Sanitary Engineering

Research, 1960):

sat

⫽ 14.652 ⫺ 0.41022T ⫹ 0.0079910T

⫺ 0.000077774T

(1.4)

where DO

sat

⫽ dissolved oxygen saturation concentration, mg/L

T ⫽ water temperature, ⬚C

This formula represents saturation values for distilled water (b ⫽ 1.0)

at sea level pressure. Water impurities can increase the saturation level

(b ⬎ 1.0) or decrease the saturation level (b ⬍ 1.0), depending on the

surfactant characteristics of the contaminant. For most cases, b is

assumed to be unity. The DO

sat

values calculated from the above formula

are listed in Table 1.2 (example: DO

sat

⫽ 8.79 mg/L, when T ⫽ 21.3⬚C)

for water temperatures ranging from zero to 30⬚C (American Society of

Engineering Committee on Sanitary Engineering Research, 1960).

Example 1: Calculate DO saturation concentration for a water temperature

at 0, 10, 20, and 30⬚C, assuming b ⫽ 1.0.

solution:

(a) At T ⫽ 0⬚C

sat

⫽ 14.652 ⫺ 0 ⫹ 0 ⫺ 0

⫽ 14.652 (mg/L)

(b) At T ⫽ 10⬚C

sat

⫽ 14.652 – 0.41022 ⫻ 10 ⫹ 0.0079910 ⫻ 10

– 0.000077774 ⫻ 10

⫽ 11.27 (mg/L)

sat

⫽ 14.652 – 0.41022 ⫻ 20 ⫹ 0.0079910 ⫻ 20

– 0.000077774 ⫻ 20

⫽ 9.02 (mg/L)

(d) At T ⫽ 30⬚C

sat

⫽ 14.652 ⫺ 0.41022 ⫻ 30 ⫹ 0.0079910 ⫻ 30

⫺ 0.000077774 ⫻ 30

⫽ 7.44 (mg/L)

The DO saturation concentrations generated by the formula must be cor-

rected for differences in air pressure caused by air temperature changes and

Streams and Rivers 7

for elevation above the mean sea level (MSL). The correction factor can be cal-

culated as follows:

(1.5)

where f ⫽ correction factor for above MSL

A ⫽ air temperature, ⬚C

E ⫽ elevation of the site, feet above MSL

Example 2: Find the correction factor of DO

sat

value for water at 620 ft above

the MSL and air temperature of 25⬚C. What is DO

sat

at a water temperature

of 20⬚C?

f 5

2116.8 2 s0.08 2 0.000115AdE

2116.8

8 Chapter 1

TABLE 1.2 Dissolved Oxygen Saturation Values in mg/L

Temp,

⬚C 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0 14.65 14.61 14.57 14.53 14.49 14.45 14.41 14.37 14.33 14.29

1 14.25 14.21 14.17 14.13 14.09 14.05 14.02 13.98 13.94 13.90

2 13.86 13.82 13.79 13.75 13.71 13.68 13.64 13.60 13.56 13.53

3 13.49 13.46 13.42 13.38 13.35 13.31 13.28 13.24 13.20 13.17

4 13.13 13.10 13.06 13.03 13.00 12.96 12.93 12.89 12.86 12.82

5 12.79 12.76 12.72 12.69 12.66 12.62 12.59 12.56 12.53 12.49

6 12.46 12.43 12.40 12.36 12.33 12.30 12.27 12.24 12.21 12.18

7 12.14 12.11 12.08 12.05 12.02 11.99 11.96 11.93 11.90 11.87

8 11.84 11.81 11.78 11.75 11.72 11.70 11.67 11.64 11.61 11.58

9 11.55 11.52 11.49 11.47 11.44 11.41 11.38 11.35 11.33 11.30

10 11.27 11.24 11.22 11.19 11.16 11.14 11.11 11.08 11.06 11.03

11 11.00 10.98 10.95 10.93 10.90 10.87 10.85 10.82 10.80 10.77

12 10.75 10.72 10.70 10.67 10.65 10.62 10.60 10.57 10.55 10.52

13 10.50 10.48 10.45 10.43 10.40 10.38 10.36 10.33 10.31 10.28

14 10.26 10.24 10.22 10.19 10.17 10.15 10.12 10.10 10.08 10.06

15 10.03 10.01 9.99 9.97 9.95 9.92 9.90 9.88 9.86 9.84

16 9.82 9.79 9.77 9.75 9.73 9.71 9.69 9.67 9.65 9.63

17 9.61 9.58 9.56 9.54 9.52 9.50 9.48 9.46 9.44 9.42

18 9.40 9.38 9.36 9.34 9.32 9.30 9.29 9.27 9.25 9.23

19 9.21 9.19 9.17 9.15 9.13 9.12 9.10 9.08 9.06 9.04

20 9.02 9.00 8.98 8.97 8.95 8.93 8.91 8.90 8.88 8.86

21 8.84 8.82 8.81 8.79 8.77 8.75 8.74 8.72 8.70 8.68

22 8.67 8.65 8.63 8.62 8.60 8.58 8.56 8.55 8.53 8.52

23 8.50 8.48 8.46 8.45 8.43 8.42 8.40 8.38 8.37 8.35

24 8.33 8.32 8.30 8.29 8.27 8.25 8.24 8.22 8.21 8.19

25 8.18 8.16 8.14 8.13 8.11 8.10 8.08 8.07 8.05 8.04

26 8.02 8.01 7.99 7.98 7.96 7.95 7.93 7.92 7.90 7.89

27 7.87 7.86 7.84 7.83 7.81 7.80 7.78 7.77 7.75 7.74

28 7.72 7.71 7.69 7.68 7.66 7.65 7.64 7.62 7.61 7.59

29 7.58 7.56 7.55 7.54 7.52 7.51 7.49 7.48 7.47 7.45

30 7.44 7.42 7.41 7.40 7.38 7.37 7.35 7.34 7.32 7.31

SOURCE: American Society of Civil Engineering Committee on Sanitary Engineering

Research, 1960

solution:

Step 1. Using Eq. (1.5)

Step 2. Compute DO

sat

From Example 1, at T ⫽ 20⬚C

sat

⫽ 9.02 (mg/L)

With an elevation correction factor of 0.977

sat

⫽ 9.02 mg/L ⫻ 0.977 ⫽ 8.81 (mg/L)

5.2 Dissolved oxygen availability

Most regulatory agencies have standards for minimum DO concentra-

tions in surface waters to support indigenous fish species in surface

waters. In Illinois, for example, the Illinois Pollution Control Board stip-

ulate that dissolved oxygen shall not be less than 6.0 mg/L during at least

16 h of any 24-h period, nor less than 5.0 mg/L at any time (IEPA, 1999).

The availability of dissolved oxygen in a flowing stream is highly vari-

able due to several factors. Daily and seasonal variations in DO levels

have been reported. The diurnal variations in DO are primarily induced

by algal productivity. Seasonal variations are attributable to changes in

temperature that affect DO saturation values. The ability of a stream

to absorb or reabsorb oxygen from the atmosphere is affected by flow fac-

tors such as water depth and turbulence, and it is expressed in terms

of the reaeration coefficient. Factors that may represent significant

sources of oxygen use or oxygen depletion are biochemical oxygen

demand (BOD) and sediment oxygen demand (SOD). BOD, including

carbonaceous BOD (CBOD) and nitrogenous BOD (NBOD), may be the

product of both naturally occurring oxygen use in the decomposition of

organic material and oxygen depletion in the stabilization of effluents

discharged from wastewater treatment plants (WTPs). The significance

of any of these factors depends upon the specific stream conditions. One

or all of these factors may be considered in the evaluation of oxygen use

and availability.

5 0.977

2116.8 2 47.8

2116.8

2116.8 2 s0.08 2 0.000115 3 25d620

2116.8

f 5

2116.8 2 s0.08 2 0.000115AdE

2116.8

Streams and Rivers 9

6 Biochemical Oxygen Demand Analysis

Laboratory analysis for organic matter in water and wastewater

includes testing for biochemical oxygen demand, chemical oxygen

demand (COD), total organic carbon (TOC), and total oxygen demand

(TOD). The BOD test is a biochemical test involving the use of micro-

organisms. The COD test is a chemical test. The TOC and TOD tests are

instrumental tests.

The BOD determination is an empirical test that is widely used for

measuring waste (loading to and from wastewater treatment plants),

evaluating the organic removal efficiency of treatment processes, and

assessing stream assimilative capacity. The BOD test measures: (1) the

molecular oxygen consumed during a specific incubation period for the

biochemical degradation of organic matter (CBOD); (2) oxygen used to

oxidize inorganic material such as sulfide and ferrous iron; and (3)

reduced forms of nitrogen (NBOD) with an inhibitor (trichloromethylpyri-

dine). If an inhibiting chemical is not used, the oxygen demand measured

is the sum of carbonaceous and nitrogenous demands, so-called total

BOD or ultimate BOD.

The extent of oxidation of nitrogenous compounds during the 5-day

incubation period depends upon the type and concentration of micro-

organisms that carry out biooxidation. The nitrifying bacteria usually

are not present in raw or settleable primary sewage. These nitrifying

organisms are present in sufficient numbers in biological (secondary)

effluent. A secondary effluent can be used as “seeding” material for an

NBOD test of other samples. Inhibition of nitrification is required for a

CBOD test for secondary effluent samples, for samples seeded with sec-

ondary effluent, and for samples of polluted waters.

The result of the 5-day BOD test is recorded as carbonaceous bio-

chemical oxygen demand, CBOD

, when inhibiting nitrogenous oxygen

demand. When nitrification is not inhibited, the result is reported as

BOD

(incubation at 15⬚C for 5 days).

The BOD test procedures can be found in Standard Methods for the

Examination of Water and Wastewater (APHA, AWWA, and WEF, 1995).

When the dilution water is seeded, oxygen uptake (consumed) is

assumed to be the same as the uptake in the seeded blank. The differ-

ence between the sample BOD and the blank BOD, corrected for the

amount of seed used in the sample, is the true BOD. Formulas for cal-

culation of BOD are as follows (APHA, AWWA, and WEF, 1995):

When dilution water is not seeded:

(1.6)BOD, mg/L 5

2 D

10 Chapter 1

When dilution water is seeded:

(1.7)

where D

, D

⫽ DO of diluted sample immediately after

preparation, mg/L

, D

⫽ DO of diluted sample after incubation at 20⬚C, mg/L

P ⫽ decimal volumetric fraction of sample used; mL of

sample/300 mL for Eq. (1.6)

⫽ DO of seed control before incubation, mg/L

⫽ DO of seed control after incubation, mg/L

f ⫽ ratio of seed in diluted sample to seed in seed control

P ⫽ percent seed in diluted sample/percent seed in seed

control for Eq. (1.7)

If seed material is added directly to the sample and to control bottles:

f ⫽ volume of seed in diluted sample/volume of seed in seed control

Example 1: For a BOD test, 75 mL of a river water sample is used in the

300 mL of BOD bottles without seeding with three duplications. The initial

DO in three BOD bottles read 8.86, 8.88, and 8.83 mg/L, respectively. The

DO levels after 5 days at 20⬚C incubation are 5.49, 5.65, and 5.53 mg/L,

respectively. Find the 5-day BOD (BOD

) for the river water.

solution:

Step 1. Determine average DO uptake

Step 2. Determine P

Step 3. Compute BOD

BOD

3.30 mg/L

0.25

5 13.2 smg/Ld

P 5

300

5 0.25

5 3.30smg/Ld

[s8.86 2 5.49d 1 s8.88 2 5.65d 1 s8.83 2 5.53d]

x 5

⌺sD

2 D

BOD, mg/L 5

2 D

d 2 sB

2 B

Streams and Rivers 11

Example 2: The wastewater is diluted by a factor of 1/20 using seeded con-

trol water. DO levels in the sample and control bottles are measured at 1-day

intervals. The results are shown in Table 1.3. One milliliter of seed material

is added directly to diluted and to control bottles. Find daily BOD values.

solution:

Step 1. Compute f and P

Step 2. Find BODs using Eq. (1.7)

Day 1:

Similarly

Day 2:

5 74.4 smg/Ld

BOD

s7.98 2 4.13d 2 s8.25 2 8.12d1

0.05

5 57.2 smg/Ld

s7.98 2 5.05d 2 s8.25 2 8.18d1

0.05

BOD

2 D

d 2 sB

2 B

P 5

5 0.05

f 5

1 mL

5 1.0

12 Chapter 1

TABLE 1.3 Change in Dissolved Oxygen and Biochemical Oxygen Demand

with Time

Dissolved oxygen, mg/L

Time, Diluted Seeded BOD,

days sample control mg/L

0 7.98 8.25 —

1 5.05 8.18 57.2

2 4.13 8.12 74.4

3 3.42 8.07 87.6

4 2.95 8.03 96.2

5 2.60 7.99 102.4

6 2.32 7.96 107.4

7 2.11 7.93 111.0

For other days, BOD can be determined in the same manner. The results

of BODs are also presented in the above table. It can be seen that BOD

for

this wastewater is 102.4 mg/L.

7 Streeter–Phelps Oxygen Sag Formula

The method most widely used for assessing the oxygen resources in

streams and rivers subjected to effluent discharges is the Streeter–Phelps

oxygen sag formula that was developed for the use on the Ohio River in

1914. The well-known formula is defined as follows (Streeter and

Phelps, 1925):

(1.8a)

(1.8b)

where D

⫽ DO saturation deficit downstream, mg/L or lb

(DO

sat

– DO

) at time t

t ⫽ time of travel from upstream to downstream, days

⫽ initial DO saturation deficit of upstream water,

mg/L or lb

⫽ ultimate upstream BOD, mg/L

e ⫽ base of natural logarithm, 2.7183

⫽ deoxygenation coefficient to the base e, per day

⫽ reoxygenation coefficient to the base e, per day

⫽ deoxygenation coefficient to the base 10, per day

⫽ reoxygenation coefficient to the base 10, per day

In the early days, K

or K

and k

or k

were used classically for values

based on e and 10, respectively. Unfortunately, in recent years, many

authors have mixed the usage of K and k. Readers should be aware of

this. The logarithmic relationships between k and K are K

⫽ 2.3026k

and K

⫽ 2.3026k

; or k

⫽ 0.4343K

and k

⫽ 0.4343K

The Streeter–Phelps oxygen sag equation is based on two assump-

tions: (1) at any instant, the deoxygenation rate is directly proportional

to the amount of oxidizable organic material present; and (2) the reoxy-

genation rate is directly proportional to the DO deficit. Mathematical

expressions for assumptions (1) and (2) are

(1.9)

5 K

2 L

2 k

[10

2 10

] 1 D

2 K

2 e

] 1 D

Streams and Rivers 13

and

(1.10)

where ⫽ the net rate of change in the DO deficit, or the absolute

change of DO deficit (D) over an increment of time dt

due to stream waste assimilative capacity affected by

deoxygenation coefficient K

and due to an atmospheric

exchange of oxygen at the air/water interface affected by

the reaeration coefficient K

⫽ ultimate upstream BOD, mg/L

⫽ ultimate downstream BOD at any time t, mg/L

Combining the above two differential equations and integrating between

the limits D

, the initial upstream sampling point, and t, any time of flow

below the initial point, yields the basic equation devised by Streeter and

Phelps. This stimulated intensive research on BOD, reaction rates, and

stream sanitation.

There are some shortcomings in the Streeter–Phelps equation. The two

assumptions are likely to generate errors. It is assumed that: (1) wastes

discharged to a receiving water are evenly distributed over the river’s

cross section; and (2) the wastes travel down the river as a plug flow with-

out mixing along the axis of the river. These assumptions only apply

within a reasonable distance downstream. Effluent discharge generally

travels as a plume for some distance before mixing. In addition, it is

assumed that oxygen is removed by microbial oxidation of the organic

matter (BOD) and is replaced by reaeration from the surface. Some fac-

tors, such as the removal of BOD by sedimentation, conversion of sus-

pended BOD to soluble BOD, sediment oxygen demand, and algal

photosynthesis and respiration are not included.

The formula is a classic in sanitary engineering stream work. Its

detailed analyses can be found in almost all general environmental engi-

neering texts. Many modifications and adaptations of the basic equation

have been devised and have been reported in the literature. Many

researches have been carried out on BOD, K

, and K

factors. Illustrations

for oxygen sag formulas will be presented in the latter sections.

8 BOD Models and

Computation

Under aerobic conditions, organic matter and some inorganics can be

used by bacteria to create new cells, energy, carbon dioxide, and residue.

The oxygen used to oxidize total organic material and all forms of nitro-

gen for 60 to 90 days is called the ultimate BOD (UBOD). It is common

that measurements of oxygen consumed in a 5-day test period called

52K

14 Chapter 1