Lin S.D. Water and Wastewater Calculations Manual

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Step 2. Compute ⌺y/ ⌺ty

Step 3. Find K

and L

On the 7-day time sequence graph (Fig. 1.5), enter the value 0.288 on the

⌺y/⌺ty scale, extend a horizontal line to the curve labeled ⌺y/⌺ty, and from

this point, follow a vertical line to the k

scale. A value of k

⫽ 0.264.

5 0.608 sper dayd

5 2.3026 3 k

5 2.3026 3 0.264

⌺y/⌺ty 5 635.2/2786 5 0.228

Streams and Rivers 25

Figure 1.4 ⌺y/L and ⌺y/⌺ty for various values of k

in a 5-day

sequence.

Extend the same vertical line to the curve labeled ⌺y/L

, obtaining a value

of 5.81. Since

The technique of the moment method for analyzing a set of BOD data con-

taining a lag phase is as follows:

1. Compute ⌺t, ⌺y, and ⌺ty

2. Compute t

and t

y and take the sum of the values of each quantity as ⌺t

and ⌺t

3. Compute the derived quantity:

in which n is the number of observations

⌺ty/⌺t 2 ⌺y/n

⌺st

yd/⌺t

2 ⌺y/n

5 109.3 smg/Ld

5 ⌺y/5.81 5 635.2/5.81

26 Chapter 1

Figure 1.5 ⌺y/L and ⌺y/⌺ty for various values of k

in a 7-day

sequence.

4. Enter the above quantity on appropriate curve to find k

5. Project to other curves for values and solve equations for C and L

. The

BOD equation with log phase is expressed as

(1.14b)

(1.14c)

in which t

is the lag period and

Example 2: With a lag phase BOD data, BOD values are shown in Table 1.8.

Find K

and L

, and complete an equation of the curve of best fit for the BOD

data.

C 5 10

y 5 L

[1 2 C10

]

y 5 L[1 2 10

st2t

]

Streams and Rivers 27

Figure 1.6 Curves for BOD computation with lag phase, based on

a 5-day time sequence.

solution:

Step 1. Compute ⌺t, ⌺y, ⌺ty, ⌺t

, and ⌺t

y (Table 1.8)

Step 2. Compute some quantities

and

Step 3. Enter in Fig. 1.6 on the vertical axis labeled

with the value 0.640, and proceed horizontally to the diagonal straight line.

Extend a vertical line to the axis labeled k

and read k

⫽ 0.173 per day (K

⫽

0.173 per day ⫻ 2.3026 ⫽ 0.398 per day). Extend the vertical line to curve labeled

⌺ty/⌺t 2 ⌺y/n

⌺t

y/⌺t

2 ⌺y/n

⌺ty/⌺t 2 ⌺y/n

⌺t

y/⌺t

2 ⌺y/n

113.13 2 91.8

125.14 2 91.8

5 0.640

⌺t

y/⌺t

5 6883/55 5 125.14

⌺y/n 5 459/5 5 91.8

⌺ty/⌺t 5 1697/15 5 113.13

28 Chapter 1

TABLE 1.8 Computations of ⌺t, ⌺y, ⌺ty, ⌺t

, and ⌺t

t y ty t

112 12 1 12

2 74 148 4 296

3 101 303 9 909

4 126 504 16 2016

5 146 730 25 3650

15 459 1697 55 6883

TABLE 1.7 Calculation of ⌺y and ⌺ty

ty ty

1 56.2 56.2

2 74.4 148.8

3 87.6 262.8

4 96.2 384.8

5 102.4 512.0

6 107.4 644.4

7 111.0 777.0

Sum 635.2 2786.0

and proceed horizontally from the intersection to the scale at the far right;

read

Step 4. Find CL

Step 5. Find L

, L

⫽ L for this case

Continue to the same vertical line to the curve labeled

and read horizontally on the inside right-hand scale

Then

Also

and lag period

Step 6. Write complete equation of the best fit for the data

y 5 177.6[1 2 e

20.398st20.787d

]

y 5 177.6[1 2 10

20.173st20.787d

]

5 0.787 sdaysd

0.173

log

1.368

log

C 5

242.94

177.6

5 1.368

5 177.6 smg/Ld

L 5 L

5 0.353s242.94d 1 91.8

L 2 ⌺y/n

5 0.353

L 2 ⌺y/n

CL 5

113.13 2 91.8

0.0878

5 242.94

⌺ty/⌺t 2 ⌺y/n

5 0.0878

Streams and Rivers 29

Logarithmic formula. Orford and Ingram (1953) reported that there is

a relationship between the observed BOD from domestic sewage and the

logarithm of the time of observation. If the BOD data are plotted against

the logarithm of time, the resultant curve is approximately a straight

line. The general equation is expressed as

⫽ m logt ⫹ b (1.22)

where m is the slope of the line and b is a constant (the intercept).

The general equation can be transformed by dividing each side by the

5-day BOD

intercept of the line to give

(1.22a)

⫽ s(M log t ⫹ B) (1.22b)

where s ⫽ BOD

intercept of the line

M ⫽ m/s, BOD rate parameter

B ⫽ b/s, BOD rate parameter

For domestic sewage oxidation at 20⬚C, the straight line through the

observed plotted points, when extrapolated to the logt axis, intercepts

the logt axis at 0.333 days. The general equation is

⫽ s(0.85 logt ⫹ 0.41) (1.23a)

where 0.41 is the BOD rate parameter for domestic sewage.

For any observed BOD curve with different oxidation time, the above

equation may be generalized as

⫽ S(log at ⫹ 0.41) (1.23b)

where S ⫽ BOD intercept (y-axis) of the line at 5/a days

⫽ 5-day BOD at the standardized

domestic sewage oxidation rate, when a ⫽ 1

⫽ strength factor

a ⫽ log t at x-axis intercept of normal domestic

sewage BOD curve divided by the x-axis intercept of

⫽

⫽ 1 for a standardized domestic sewage BOD

curve

0.333

x-axis intercept

5 M logt 1 B

logt 1

30 Chapter 1

The oxidation rate

(1.24a)

(1.24b)

For the logarithmic formula method, you need to determine the two

constants S and a. Observed BOD data are plotted on semilogarithmic

graph paper. Time in days is plotted on a logarithmic scale on the x-

axis and percent of 5-day BOD on a regular scale on the y-axis. The

straight line of best fix is drawn through the plotted points. The time

value of the x-axis intercept is then read, x

(for a standardized domes-

tic sewage sample, this is 0.333). The a value can be calculated by

a ⫽ 0.333/x

Example: At 20⬚C, S ⫽ 95 mg/L and x

⫽ 0.222 days. What is the BOD

equation and K

for the sample?

solution:

Step 1. Compute a

Step 2. Determine the simplified Eq. (1.23b)

Step 3. Compute the oxidation rate at t ⫽ 5 days, using Eq. (1.24a)

Step 4. Determine L

when t ⫽ 20 days from Step 2

5 158 smg/Ld

5 y

5 95s0.85 logs1.5 3 20d 1 0.41d

5 16.15[mg/sL

dd]

0.85S

0.85 3 95 mg/L

5 days

5 95s0.85 log 1.5t 1 0.41d

5 Ss0.85 log at 1 0.41d

5 1.5

a 5 0.333/x

5 0.333/0.222

0.85S

2.303t

5 k

2 y

0.85S

5 K

2 y

Streams and Rivers 31

Step 5. Compute K

Rapid methods. Sheehy (1960) developed two rapid methods for solv-

ing first-order BOD equations. The first method, using the “BOD slide

rule,” is applicable to the ratios of observations on whole day intervals,

from 1 to 8 days, to the 5-day BOD. The second method, the graphical

method, can be used with whole or fractional day BOD values.

The BOD slide rule consists of scales A and B for multiplication and

division. Scales C and D are the values of and plotted

in relation to values of k

t on the B scale. From given BOD values, the

value is easily determined from the BOD slide rule. Unfortunately,

the BOD slide rule is not on the market.

The graphical method is based on the same principles as the BOD slide

rule method. Two figures are used to determine k

values directly. One

figure contains k

values based on the ratios of BOD at time t (BOD

) to

the 5-day BOD for t less than 5 days. The other figure contains k

val-

ues based on the ratio of BOD

to BOD

at times t greater than 5 days.

The value of k

is determined easily from any one of these figures based

on the ratio of BOD

to BOD

8.3 Temperature effect on

A general expression of the temperature effect on the deoxygenation

coefficient (rate) is

(1.25)

where K

⫽ reaction rate at temperature T

, per day

⫽ reaction rate at temperature T

, per day

u ⫽ temperature coefficient

On the basis of experimental results over the usual range of river tem-

perature, u is accepted as 1.047. Therefore the BOD reaction rate at any

T in Celsius (working temperature), deviated from 20⬚C, is

(1.26)

1sTd

5 K

1s208Cd

3 1.047

sT220d

5 u

1 2 10

5 0.256 sper dayd

16.15 mg/sL

s158 2 95d mg/L

5 K

2 y

32 Chapter 1

(1.26a)

Thus

(1.27)

(1.28)

Example 1: A river water sample has k

⫽ 0.10 (one base 10) and L

⫽

280 mg/L at 20⬚C. Find k

and L

at temperatures 14⬚C and 29⬚C.

solution:

Step 1.

Using Eq. (1.26a)

at 14⬚C

at 29⬚C

Step 2. Find L

a(T)

, using Eq. (1.28)

at 14⬚C

5 246 smg/Ld

5 280s0.6 1 0.02 3 14d

as148Cd

5 L

as208Cd

s0.6 1 0.02T d

asTd

5 L

as208Cd

s0.6 1 0.02T d

5 0.15 sper dayd

5 0.10 3 1.047

1s298Cd

5 k

1s208Cd

3 1.047

s29220d

5 0.076 sper dayd

5 0.10 3 1.047

1s148Cd

5 k

1s208Cd

3 1.047

s14220d

1sTd

5 k

1s208Cd

3 1.047

sT220d

asTd

5 L

as208Cd

s0.6 1 0.2Td

asTd

5 L

as208Cd

[1 1 0.02sT 2 20d]

1sTd

5 k

1s208Cd

3 1.047

sT220d

Streams and Rivers 33

at 29⬚C

The ultimate BOD and K

values found in the laboratory at 20⬚C have to

be adjusted to river temperatures using the above formulas. Three types of

BOD can be determined: i.e. total, carbonaceous, and nitrogenous. TBOD

(uninhibited) and CBOD (inhibited with trichloromethylpyridine for nitrifi-

cation) are measured directly, while NBOD can be computed by subtracting

CBOD values from TBOD values for given time elements.

Example 2: Tables 1.9 and 1.10 show typical long-term (20-day) BOD

data for the Upper Illinois Waterway downstream of Lockport (Butts and

Shackleford, 1992). The graphical plots of these BOD progressive curves

are presented in Figs. 1.7 to 1.9. Explain what are their unique

characteristics.

5 330 smg/Ld

as298Cd

5 280s0.6 1 0.02 3 29d

34 Chapter 1

TABLE 1.9 Biochemical Oxygen Demand at 20ⴗC in the Upper Illinois

Waterway

Station sample: Lockport 18 Station sample: Lockport 36

Date: 01/16/90 Date: 09/26/90

pH: 7.03 pH: 6.98

Temp: 16.05⬚C Temp: 19.10⬚C

Time, TBOD, CBOD, NBOD, Time, TBOD, CBOD, NBOD,

days mg/L mg/L mg/L days mg/L mg/L mg/L

0.89 1.09 0.98 0.11 0.72 0.19 0.18 1.01

1.88 1.99 1.73 0.25 1.69 0.66 0.18 0.48

2.87 2.74 2.06 0.68 2.65 0.99 0.92 0.07

3.87 3.39 2.25 1.14 3.65 1.38 1.25 0.13

4.87 4.60 3.08 1.52 4.72 1.81 1.25 0.56

5.62 5.48 3.45 2.04 5.80 2.74 1.62 1.12

6.58 7.41 4.17 3.24 6.75 3.51 2.09 1.43

7.62 9.87 4.83 5.04 7.79 4.17 2.53 1.64

8.57 12.62 5.28 7.34 8.67 4.65 2.92 1.74

9.62 16.45 6.15 10.30 9.73 5.59 2.92 2.67

10.94 21.55 6.62 14.94 10.69 6.22 3.08 3.14

11.73 25.21 6.96 18.25 11.74 6.95 3.08 3.88

12.58 29.66 7.26 22.40 12.66 7.28 3.12 4.16

13.62 34.19 7.90 26.29 13.69 7.79 3.56 4.23

14.67 35.68 8.14 27.55 14.66 7.79 3.56 4.23

15.67 36.43 8.38 28.05 15.65 8.36 3.56 4.80

16.58 37.08 8.75 28.33 16.64 8.56 3.57 4.99

17.81 37.45 8.92 28.53 17.76 8.80 3.82 4.99

18.80 37.64 9.16 28.47 17.67 8.80 3.82 4.98

19.90 38.19 9.39 28.79 19.66 8.92 3.87 5.05

20.60 38.56 9.55 29.01