Lin S.D. Water and Wastewater Calculations Manual

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Step 3. Determine a and b

Step 4. Determine degree of BOD removal

For plant A

1 – a ⫽ 1 ⫺ 0.61 ⫽ 0.39 i.e. 39% removal efficiency

For plant B

1 – b ⫽ 1 – 0.29 ⫽ 0.71 i.e. 71% removal efficiency

Step 5. Determine cost ratio

15.3 Method 3

The principle of this method is to load the river to the utmost so as to

minimize the total amount of wastewater treatment. The proportion of

and P

can be computed by either Method 1 or Method 2. The cost of

treatment can be divided in proportion to population.

There are two cases of initial BOD loading:

Case 1. L

⬎ L (L is the BOD load of plant A influent)

■

Let plant A discharge L without a treatment.

■

Have plant B treat an amount equal to L – L

■

City A is assessed for P

(L – L

■

City B is assessed for P

(L – L

Plant A

Plant B

1 2 a

1 2 b

0.39

0.71

1.82

0.55

a 5 2.1 3 0.29 5 0.61

b 5 4837/16,441 5 0.29

4837 5 11,137b 1 5304b

4837 5 0.912 3 s1.1bd 3 5815 1 b 3 5304

5 paL

1 bL

a 5 2.1b

a/b 5 P

5 0.677/0.323 5 2.1/1

Streams and Rivers 85

Case 2. L

⬍ L

■

Let plant A discharge L

and treat L – L

■

Have plant B treat an amount equal to (L – L

) – pL

■

Then the total treatment will remove (L – L

) ⫹ (L – L

– pL

) ⫽

2L – (1 ⫹ p)L

– L

■

City A is assessed for P

[2L – (1 ⫹ p)L

– L

■

City B is assessed for P

[2L – (1 ⫹ p)L

– L

Example 3: Using data listed in Examples 1 and 2, determine the cost ratio

assessed to plants A and B by Method 3.

solution:

Step 1. Select case

Since L

⫽ 5815 lb/d ⬍ L ⫽ 6910 lb/d

Case 2 will be applied

Step 2. Determine amounts needed to be treated

For plant A

For plant B

Total treatment ⫽ 1095 ⫹ 1074 ⫽ 2169 (lb/d)

Step 3. Compute cost assessments

For Example 2,

For plant A

Cost ⫽ 0.677 ⫻ treatment cost of 2169 lb/d

For plant B

Cost ⫽ 0.323 ⫻ treatment cost of 2169 lb/d

5 0.323

5 0.677

5 1074 slb/dd

L 2 L

2 pL

5 6910 2 4837 2 0.912 3 1095

5815 5 lb/d discharged without treatment

6910 2 5815 5 1095 slb/dd treated

86 Chapter 1

16 Velz Reaeration Curve (a Pragmatic

Approach)

Previous sections discuss the conceptual approach of the relationships

between BOD loading and DO in the stream. Another approach for

determining the waste-assimilative capacity of a stream is a method

first developed and used by Black and Phelps (1911) and later refined and

used by Velz (1939). For this method, deoxygenation and reoxygenation

computations are made separately, then added algebraically to obtain the

net oxygen balance. This procedure can be expressed mathematically as

(1.110)

where DO

net

⫽ dissolved oxygen at the end of a reach

⫽ initial dissolved oxygen at beginning of a reach

rea

⫽ dissolved oxygen absorbed from the atmosphere

used

⫽ dissolved oxygen consumed biologically

All units are in lb/d or kg/d.

16.1 Dissolved oxygen used

The dissolved oxygen used (BOD) can be computed with only field-

observed dissolved oxygen concentrations and the Velz reoxygenation

curve. From the basic equation, the dissolved oxygen used in pounds per

day can be expressed as:

The terms DO

and DO

net

are the observed field dissolved oxygen val-

ues at the beginning and end of a subreach, respectively, and DO

rea

the reoxygenation in the subreach computed by the Velz curve. If the dis-

solved oxygen usage in the river approximates a first-order biological

reaction, a plot of the summations of the DO

used

versus time of travel can

be fitted to the equation:

(1.111)

and L

and K

are determined experimentally in the same manner as

for the Streeter–Phelps method. The fitting of the computed values to

the equations must be done by trial and error, and a digital computer

is utilized to facilitate this operation.

16.2 Reaeration

The reaeration term (DO

rea

) in the equation is the most difficult to deter-

mine. In their original work, Black and Phelps (1911) experimentally

used

5 L

[1 2 exps2K

td]

used

5 sDO

2 DO

net

d 1 DO

rea

sin lb/dd

net

5 DO

1 DO

rea

2 DO

used

Streams and Rivers 87

derived a reaeration equation based on principles of gas transfer of dif-

fusion across a thin water layer in acquiescent or semiquiescent system.

A modified but equivalent form of the original Black and Phelps equa-

tion is given by Gannon (1963):

(1.112)

where R ⫽ percent of saturation of DO absorbed per mix

⫽ initial DO in percent of saturation

K ⫽ p

am/4L

in which m is the mix or exposure time in

hours, L is the average depth in centimeters, and a is the

diffusion coefficient used by Velz

The diffusion coefficient a was determined by Velz (1947) to vary with

temperature according to the expression:

⫽ a

(1.1

T⫺20

) (1.113)

where a

⫽ diffusion coefficient at T ⬚C

⫽ diffusion coefficient at 20⬚C

when the depth is in feet,

⫽ 0.00153

when the depth is in centimeters,

⫽ 1.42

Although the theory upon which this reaeration equation was devel-

oped is for quiescent conditions, it is still applicable to moving, and

even turbulent, streams. This can be explained in either of two ways

(Phelps, 1944):

(1) Turbulence actually decreases the effective depth through which

diffusions operate. Thus, in a turbulent stream, mixing brings lay-

ers of saturated water from the surface into intimate contact with

other less oxygenated layers from below. . . The actual extent of such

mixing is difficult to envision, but it clearly depends upon the fre-

quency with which surface layers are thus transported. In this

change-of-depth concept, therefore, there is a time element involved.

(2) The other and more practical concept is a pure time effect. It is

assumed that the actual existing conditions of turbulence can be

replaced by an equivalent condition composed of successive periods

of perfect quiescence between which there is instantaneous and

R 5 100a

1 2 B

100

bs81.06dae

29K

225K

88 Chapter 1

complete mixing. The length of the quiescent period then becomes the

time of exposure in the diffusion formula, and the total aeration per

hour is the sum of the successive increments or, for practical purposes,

the aeration period multiplied by the periods per hour.

Because the reaeration equation involves a series expansion in its

solution, it is not readily solved by desk calculations. To facilitate the

calculations, Velz (1939, 1947) published a slide-rule curve solution to

the equation. This slide-rule curve is reprinted here as Fig. 1.11 (for its

use, consult Velz, 1947). Velz’s curve has been verified as accurate by two

independent computer checks, one by Gannon (1963) and the other by

Butts (1963).

Although the Velz curve is ingenious, it is somewhat cumbersome to

use. Therefore, a nomograph (Butts and Schnepper, 1967) was developed

for a quick, accurate desk solution of R at zero initial DO(B

in the

Black and Phelps equation). The nomograph, presented in Fig. 1.12, was

constructed on the premise that the exposure or mix time is character-

ized by the equation

M ⫽ (13.94)log

(H) – 7.45 (1.114)

in which M is the mix time in minutes and H is the water depth in feet.

This relationship was derived experimentally and reported by Gannon

(1963) as valid for streams having average depths greater than 3 ft. For

Streams and Rivers 89

Figure 1.11 Standard reoxygeneration curve (Butts et al., 1973).

90 Chapter 1

Figure 1.12 Oxygen absorption nomograph (Butts and Schnepper, 1967).

an initial DO greater than zero, the nomograph value is multiplied by

(1 – B

) expressed as a fraction because the relationship between B

and

the equation is linear.

The use of the nomograph and the subsequent calculations required

to compute the oxygen absorption are best explained by the example that

follows:

Example: Given that H ⫽ 12.3 ft; initial DO of stream ⫽ 48% of saturation;

T ⫽ 16.2⬚C; DO load at saturation at 16.2⬚C ⫽ 8800 lb; and time of travel in

the stream reach ⫽ 0.12 days. Find the mix time (M) in minutes; the percent

DO absorbed per mix at time zero initial DO (R

); and the total amount of oxy-

gen absorbed in the reach (DO

rea

solution:

Step 1. Determine M and R

From Fig. 1.12, connect 12.3 ft on the depth-scale with 16.2⬚C on the

temperature scale. It can be read that:

M ⫽ 27.5 min on the mix timescale, and

⫽ 0.19% on the R

-scale

Step 2. Compute DO

rea

Since R

is the percent of the saturated DO absorbed per mix when the ini-

tial DO is at 100% deficit (zero DO), the oxygen absorbed per mix at 100%

deficit will be the product of R

and the DO load at saturation. For this exam-

ple, lb/mix per 100% deficit is

However, the actual DO absorbed (at 48% saturation) per mix is only

Because the amount absorbed is directly proportional to the deficit, the num-

ber of mixes per reach must be determined:

The time of travel

The mix time

M ⫽ 27.5 min

Consequently, the number of mixes in the reach ⫽ 173/27.5 ⫽ 6.28.

5 173 min

5 0.12 days 3 1440 min/d

t 5 0.12 days

16.72 3 [1 2 s48/100d] 5 8.7slb/mix per 100% deficitd

8800 lb 3

0.19

100

5 16.72 slb/mix per 100% deficitd

Streams and Rivers 91

Reaeration in the reach is therefore equal to the product of the DO absorbed

per mix ⫻ the number of mixes per reach. DO

rea

is computed according to the

expression:

(1.115)

17 Stream DO Model (a Pragmatic

Approach)

Butts and his coworkers (Butts et al., 1970, 1973, 1974, 1975, 1981;

Butts, 1974; Butts and Shackleford, 1992) of the Illinois State Water

Survey expanded Velz’s oxygen balance equation. For the pragmatic

approach to evaluate the BOD/DO relationship in a flowing stream, it

is formulated as follows:

(1.116)

where DO

⫽ net dissolved oxygen at the end of a stream reach

⫽ initial dissolved oxygen at the beginning of a reach

⫽ dissolved oxygen consumed biologically within a reach

⫽ dissolved oxygen derived from natural reaeration

within a reach

⫽ dissolved oxygen derived from channel dams,

tributaries, etc.

Details of computation procedures for each component of the above

equation have been outlined in several reports mentioned above.

Basically the above equation uses the same basic concepts as employed

in the application of the Streeter–Phelps expression and expands the

Velz oxygen balance equation by adding DO

. The influences of dams and

tributaries are discussed below.

17.1 Inﬂuence of a dam

Low head channel dams and large navigation dams are very impor-

tant factors when assessing the oxygen balance in a stream. There are

certain disadvantages from a water quality standpoint associated

with such structures but their reaeration potential is significant

during overflow, particularly if a low DO concentration exists in the

5 DO

2 DO

1 DO

5 54.6 slbd

5 8.7 3 6.28

5 a1 2

100

0.19

100

173

27.5

bs8800d

rea

5 a1 2

% DO saturation

100

bsDO saturation loadd

92 Chapter 1

upstream pooled waters. This source of DO replenishment must be

considered.

Procedures for estimating reaeration at channel dams and weirs have

been developed by researchers in England (Gameson, 1957; Gameson

et al., 1958; Barrett et al., 1960; Grindrod, 1962). The methods are eas-

ily applied and give satisfactory results. Little has been done in devel-

oping similar procedures for large navigation and power dams. Preul and

Holler (1969) investigated larger structures.

Often in small sluggish streams, more oxygen may be absorbed by

water overflowing a channel dam than in a long reach between dams.

However, if the same reach were free flowing, this might not be the

case; i.e. if the dams were absent, the reaeration in the same stretch of

river could conceivably be greater than that provided by overflow at a

dam. If water is saturated with oxygen, no uptake occurs at the dam

overflow. If it is supersaturated, oxygen will be lost during dam over-

flow. Water, at a given percent deficit, will gain oxygen.

The basic channel dam reaeration formula takes the general form

(1.117)

where r ⫽ dissolved oxygen deficit ratio at temperature T

q ⫽ water quality correction factor

b ⫽ weir correction factor

T ⫽ water temperature, ⬚C

h ⫽ height through which the water falls, ft

The deficit dissolved oxygen ratio is defined by the expression

(1.118)

where C

⫽ dissolved oxygen concentration upstream of the dam, mg/L

⫽ dissolved oxygen concentration downstream of the

dam, mg/L

⫽ dissolved oxygen saturation concentration, mg/L

⫽ dissolved oxygen deficit upstream of the dam, mg/L

⫽ dissolved oxygen deficit downstream of the dam, mg/L

Although Eqs. (1.117) and (1.118) are rather simplistic and do not

include all potential parameters which could affect the reaeration of

water overflowing a channel dam, they have been found to be quite reli-

able in predicting the change in oxygen content of water passing over a

dam or weir (Barrett et al., 1960). The degree of accuracy in using the

equations is dependent upon the estimate of factors q and b.

For assigning values for q, three generalized classifications of water

have been developed from field observations. They are q ⫽ 1.25 for clean

or slightly polluted water; q ⫽ 1.0 for moderately polluted water; and

q ⫽ 0.8 for grossly polluted water. A slightly polluted water is one in

r 5 sC

2 C

d/sC

2 C

d 5 D

r 5 1 1 0.11qbs1 1 0.46T dh

Streams and Rivers 93

which no noticeable deterioration of water quality exists from sewage

discharges; a moderately polluted stream is one which receives a sig-

nificant quantity of sewage effluent; and a grossly polluted stream is one

in which noxious conditions exist.

For estimating the value of b, the geometrical shape of the dam is

taken into consideration. This factor is a function of the ratio of weir coef-

ficients W of various geometrical designs to that of a free weir where

W ⫽ (r – 1)/h (1.119)

Weir coefficients have been established for a number of spillway types,

and Gameson (1957) in his original work has suggested assigning b val-

ues as follows:

Spillway type b

Free 1.0

Step 1.3

Slope (ogee) 0.58

Slopping channel 0.17

For special situations, engineering judgment is required. For example,

a number of channel dams in Illinois are fitted with flashboards during

the summer. In effect, this creates a free fall in combination with some

other configuration. A value of b, say 0.75, could be used for a flashboard

installation on top of an ogee spillway (Butts et al., 1973). This combi-

nation would certainly justify a value less than 1.0, that for a free weir,

because the energy dissipation of the water flowing over the flashboards

onto the curved ogee surface would not be as great as that for a flat sur-

face such as usually exists below free and step weirs.

Aeration at a spillway takes place in three phases: (1) during the fall;

(2) at the apron from splashing; and (3) from the diffusion of oxygen due

to entrained air bubbles. Gameson (1957) has found that little aeration

occurs in the fall. Most DO uptake occurs at the apron; consequently, the

greater the energy dissipation at the apron, the greater the aeration.

Therefore, if an ogee spillway is designed with an energy dissipater, such

as a hydraulic jump, the value of b should increase accordingly.

Example: Given C

⫽ 3.85 mg/L, T ⫽ 24.8⬚C, h ⫽ 6.6 ft (1.0 m), q ⫽ 0.9,

and b ⫽ 0.30 (because of hydraulic jump use 0.30).

Find DO concentration below the dam C

solution:

Step 1. Using dam reaeration equation to find r

r ⫽ 1 ⫹ 0.11qb(1 ⫹ 0.046T )h

⫽ 1 ⫹ 0.11 ⫻ 0.9 ⫻ 0.30(1 ⫹ 0.046 ⫻ 24.8) ⫻ 6.6

⫽ 1.42

94 Chapter 1