Lin S.D. Water and Wastewater Calculations Manual

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and

Step 2. Taking the average

The FS density for the water sample is reported as 2500 A/100 mL. Code A rep-

resents two or more filter counts within acceptable range and same dilution.

Note: Use two significant figures for all reported bacteria densities.

If more than one dilution is used, calculate the acceptable range results to

final reporting densities separately; then average for final recorded value.

Example 3: More than one dilution:

Volumes of 1.00, 0.30, 0.10, 0.03, and 0.01 mL generate TC colony counts of

TNTC (too numerous to count), 210, 78, 33, and 6. What is the TC density in

the water?

solution:

Step 1. In this case, only two volumes, 0.10 and 0.03 mL, produce TC colonies

in the acceptable limit. Calculate each MF density, separately,

and

Step 2. Compute the arithmetic mean of these two densities to obtain the

final recorded TC density

which would be recorded as 94,000 TC per 100 mL C. Code C stands for the

calculated value from two or more filter colony counts within acceptable range

but different dilution.

If all MF colony counts are below the acceptable range, all counts should

be added together. Also, all mL volumes should be added together.

Example 4: All results below the desirable range:

Volumes of 25, 10, 4, and 1 mL produced FC colony counts of 18, 10, 3, and

0, respectively. Calculate the FC density for the water sample.

100 mL

78,000 1 110,000

5 94,000

100 mL

33 3 100

0.03

5 110,000

100 mL

78 3 100

0.10

5 78,000

2400 1 2650

5 2525 [ reporting 2500A]

100 mL

53 3 100

5 2650

Streams and Rivers 115

solution:

which would be recorded as 77 FC per 100 mL B, where code B represents

results based on colony counts outside the acceptable range.

If colony counts from all membranes are zero, there is no actual calcula-

tion possible, even as an estimated result. If there had been one colony on the

membrane representing the largest filtration volume, the density is esti-

mated from this filter only.

If all MF counts are above the acceptable limit, calculate the density as

shown in Example 4.

Example 5: Filtration volumes 40, 15, 6, and 2 mL produced FC colony

counts of 1, 0, 0, and 0, respectively. Estimate the FC density for the sample.

solution:

which would be recorded as 2.5 FC per 100 mL K1, where K1 means less than

2.5/100 mL using one filter to estimate the value.

Example 6: Membrane filtration volumes of 0.5, 1.0, 3.0, and 10 mL produced

FC counts of 62, 106, 245, and TNTC, respectively. Compute the FC density.

solution:

which would be reported as 9200 FC per 100 mL B.

If bacteria colonies on the membrane are too numerous to count, use a

count of 200 colonies for the membrane filter with the smallest filtration vol-

ume. The number at which a filter is considered TNTC is when the colony

count exceeds 200 or when the colonies are too indistinct for accurate counting.

Example 7: Given that filtration volumes of 2, 8, and 20 mL produce TC

colony counts of 244, TNTC, and TNTC, respectively, what is the TC density

in the sample?

solution:

Using the smallest filtration volume, 2 mL in this case, and TC for 200 per

filter, gives a better representative value. The estimated TC density is

100 mL

200 3 100

5 10,000 L

100 mL

s62 1 106 1 245d 3 100

0.5 1 1 1 3

5 9180 sreporting 9200Bd

100 mL

1 3 100

5 2.5

5 77.5

100 mL

s18 1 10 1 3 1 0d 3 100

25 1 10 1 4 1 1

116 Chapter 1

Code L, for bacterial data sheet description, has to be changed to greater

than; i.e. all colony counts greater than acceptable range, calculated as when

smallest filtration volume had 200 counts.

If some MF colony counts are below and above the acceptable range and

there is a TNTC, use the useable counts and disregard any TNTCs and its

filtration volume.

Example 8: Assume that filtration volumes 0.5, 2, 5, and 15 mL of a sam-

ple produced FC counts of 5, 28, 67, and TNTC. What is the FC density?

solution:

The counts of 5 and TNTC with their volume are not included for

calculation.

Bacterial standards. Most regulatory agencies have stipulated stan-

dards based on the bacterial density for waters. For example, in Illinois,

the Illinois Department of Public Health (IDPH) has promulgated the

indicator bacteria standards for recreational-use waters as follows:

■

A beach will be posted “Warning—Swim At Your Own Risk” when

bacterial counts exceed 1000 TC per 100 mL or 100 FC per 100 mL.

■

A beach will be closed when bacterial densities exceed 5000 TC per

100 mL or 500 FC per 100 mL in water samples collected on two con-

secutive days.

The Illinois Pollution Control Board (IEPA, 1999) has adopted rules

regarding FC limits for general-use water quality standards applicable

to lakes and streams. The rules of Section 302.209 are

a. During the months May through October, based on a minimum of five

samples taken over not more than a 30-day period, FCs (STORET number

31616) shall not exceed a geometric mean of 200 per100 mL, nor shall

more than 10% of the samples during any 30-day period exceed 400 per

100 mL in protected waters. Protected waters are defined as waters that,

due to natural characteristics, aesthetic value, or environmental signifi-

cance, are deserving of protection from pathogenic organisms. Protected

waters must meet one or both of the following conditions:

(1) They presently support or have the physical characteristics to support

primary contact.

(2) They flow through or adjacent to parks or residential areas.

5 [reporting 1400 B]

100 mL

s28 1 67d 3 100

2 1 5

5 1357

Streams and Rivers 117

Example: FC densities were determined weekly for the intake water from

a river. The results are listed in Table 1.22. Calculate the 30-day moving geo-

metric mean of FC densities and complete the table.

solution:

Step 1. Check the minimum sampling requirement: Five samples were col-

lected in less than a 30-day period

Step 2. Calculate geometric means (M

) from each consecutive five samples

Bacterial die-off in streams.

There have been many studies on the die-

off rate of bacteria in streams. Most of the work suggests that Chick’s

law, one of the first mathematics formulations to describe die-off curves,

remains quite applicable for estimating the survival of pathogens and

nonpathogens of special interest in stream sanitation investigations.

Chick’s law is

N ⫽ N

⫺kt

(1.125)

logN/N

⫽⫺kt (1.126)

where N ⫽ bacterial density at time t, days

⫽ bacterial density at time 0

k ⫽ die-off or death rate, per day

5 2

42 3 20 3 300 3 50 3 19

5 47

118 Chapter 1

TABLE 1.22 Fecal Coliform Densities

FC density/100 mL

Date, 1997 Observed Moving geometric mean

5 May 42

12 May 20

19 May 300

26 May 50

2 June 19 47

9 June 20 41

16 June 250 68

23 June 3000 110

30 June 160 140

7 July 240 220

14 July 20 220

21 July 130 200

28 July 180 110

Before Chick’s law is applied, the river data are generally transformed

to bacterial population equivalents (BPE) in the manner proposed by

Kittrell and Furfari (1963). The following expressions were used for TC

and FC data:

(1.127)

(1.128)

Here the FC : TC ratio is assumed to be 0.964, as reported by Geldreich

(1967).

Average stream flows and geometric means for each stream sampling

station are used for BPE calculation. The relationship of BPE versus

time of travel are plotted on semilog paper for both TC and FC.

Reasonably good straight lines of fit can be developed. The decay or

death rates for TC and FC can be estimated from the charts.

Example: TC density in the effluent is 9800 per 100 mL. Its densities are

determined to be 5700, 2300, and 190 TC/100 mL, respectively, after the time

of travel for 8, 24, and 60 h. Determine the die-off rate of TC in the stream.

solution:

Step 1. Calculate k for each stream reach using Eq. (1.126); t ⫽ 1/3, 1.0, and

2.5 days

Similarly

Also

The average

Note: This problem also can be solved by graphic method. Plot the data on

semilogarithmic paper and draw a straight line. The slope of the line is the

die-off rate.

k 5 0.673sper dayd

k 5 0.685 sper dayd

2 ks2.5d 5 logs190/9800d 521.712

k 5 0.629 sper dayd

2 ks1d 5 log s2300/9800d 520.629

k 5 0.706 sper dayd

2ks1/3d 5 logs5700/9800d 520.235

2kt 5 logsN/N

BPE 5 Qscfsd 3 FC/100 mL 3 6.1 3 0.964 3 10

BPE 5 Qscfsd 3 TC/100 mL 3 6.1 3 10

Streams and Rivers 119

18.3 Macroinvertebrate biotic index

The macroinvertebrate biotic index (MBI) was developed to provide a

rapid stream-quality assessment. The MBI is calculated at each stream

station as a tool to assess the degree and extent of wastewater dis-

charge impacts. The MBI is an average of tolerance rating weighted by

macroinvertebrate abundance, and is calculated from the formula

(IEPA, 1987):

(1.129)

where n

⫽ number of individuals in each taxon i

⫽ tolerance rating assigned to that taxon i

N ⫽ total number of individuals in the sediment sample

Most macroinvertebrate taxa known to occur in Illinois have been

assigned a pollution tolerance rating, ranging from 0 to 11 based on lit-

erature and field studies. A 0 is assigned to taxa found only in unaltered

streams of high water quality, and an 11 is assigned to taxa known to

occur in severely polluted or disturbed streams. Intermediate ratings are

assigned to taxa that occur in streams with intermediate degrees of

pollution or disturbance. Appendix A presents a list of these tolerance

ratings for each taxon.

Example: The numbers of macroinvertebrates in a river sediment sample

are 72, 29, 14, 14, 144, 445, 100, and 29 organisms/m

for Corbicula, Perlesta

placida, Stenonema, Caenis, Cheumatopsyche, Chironomidae, Stenelmis,

and Tubificidae, respectively. The tolerance values for these organisms are,

respectively, 4, 4, 4, 6, 6, 6, 7, and 10 (Appendix A). Compute MBI for this

sample.

solution:

> 6.0

5 5.983

3 6 1 100 3 7 1 29 3 10d

847

s72 3 4 1 29 3 4 1 14 3 4 1 14 3 6 1 144 3 6 1 445

MBI 5

⌺

n51

d/N

N 5 72 1 29 1 14 1 14 1 144 1 445 1 100 1 29 5 847

n 5 8

MBI 5

⌺

i51

d/N

120 Chapter 1

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