Lin S.D. Water and Wastewater Calculations Manual

Подождите немного. Документ загружается.

require nitrogen as an energy source. The two-step nitrification can be

expressed as

(1.87)

and

(1.88)

(1.89)

Overall 2 moles of O

are required for each mole of ammonia; i.e. the N : O

ratio is 14 : 64 (or 1 : 4.57). Typical domestic wastewater contains 15 to 50 mg/L

of total nitrogen. For this example, effluent NH

⫽ 2.3 mg/L as N; therefore,

the ultimate nitrogenous BOD is

⫽ 2.3 mg/L ⫻ 4.57 ⫽ 10.51 (mg/L)

Step 6. Calculate DO deficit immediately downstream of the wastewater load

From Table 1.2, the DO saturation value at 26⬚C is 8.02 mg/L.

The initial DO deficit D

Step 7. Compute D

and DO values at various times t

Using

(1.90)

and at 26⬚C

⫽ 1.29 per day

⫽ 0.33 per day

⫽ 0.39 per day

⫽ 12.24 mg/L

⫽ 10.51 mg/L

⫽ 1.92 mg/L

⫽ 1.0 day

2 K

2 e

d 1 D

2 K

st2t

2 e

st2t

]

5 8.02 mg/L 2 6.10 mg/L 5 1.92 mg/L

5 6.10 smg/Ld

DO 5

13.31 3 2 1 78.64 3 6.8

13.31 1 78.64

1 2O

1 2H

1 H

Nitrobacter

Nitrosomonas

1 2H

1 H

Streams and Rivers 75

Fot t ⱕ 1.0 day:

When t ⫽ 0.1 days

⫽ 4.2075(0.9675 ⫺ 0.8790) ⫹ 1.6876

⫽ 2.060 (mg/L)

0.2

⫽ 2.17 mg/L and so on

For t ⬎ 1.0 day

⫽ 4.2075(0.6956 ⫺ 0.2420) ⫹ 0.4646

⫹ 4.5543(0.9618 ⫺ 0.8790)

⫽ 2.75 (mg/L)

1.2

⫽ 3.04 mg/L and so on

Step 8. Produce at table for DO sag

Table 1.15 gives the results of the above calculations for DO concentrations

in the stream at various locations.

Step 9. Explanation

From Table 1.15, it can be seen that minimum DO of 5.60 mg/L occurred at

t ⫽ 0.8 days due to carbonaceous demand. After this location, the stream

starts recovery. However, after t ⫽ 1.0 day, the stream DO is decreasing due

to nitrogenous demand. At t ⫽ 2.0, it is the critical location (22.22 miles below

the outfall) with a critical DO deficit of 3.83 mg/L and the DO level in the

water is 4.19 mg/L.

5 1.11 miles

Distance 5 0.463 mph 3 24 h/d 3 0.1 days

When t 5 0.1 days

5 0.463 mph

V 5 0.681 ft/s 5 0.681 ft/s 3 3600 s/h 3

1 mile

5280 ft

0.39 3 10.51

1.29 2 0.39

[

20.39s1.121d

2 e

21.29s1.121d

]

1.1

5 4.2075se

20.3331.1

2 e

21.2931.1

d 1 1.92e

21.2931.1

0.1

0.33 3 12.24

1.29 2 0.33

20.3330.1

2 e

21.2930.1

d 1 1.92e

21.2930.1

76 Chapter 1

13.4 Simpliﬁed oxygen sag computations

The simplified oxygen sag computation was suggested by Le Bosquet and

Tsivoglou (1950). As shown earlier at the critical point

(1.48)

and (1.49)

Let the lb/d of first-stage BOD be C

(1.91)

5.39Q

5 5.39L

Cslb/dd 5

smg/Ld Q scfsd 3 62.38 slb/cfd 3 86,400 ss/dd

1,000,000

5 K

5 0

Streams and Rivers 77

TABLE 1.15 DO Concentrations at Various Locations

Distance below DO deficit, Expected DO,

∗

t, days outfall, miles mg/L mg/L

0.0 0.00 1.92 6.10

0.1 1.11 2.06 5.96

0.2 2.22 2.17 5.85

0.3 3.33 2.26 5.74

0.4 4.44 2.32 5.70

0.5 5.55 2.37 5.64

0.6 6.67 2.40 5.62

0.7 7.78 2.41 5.61

0.8 7.89 2.42 5.60

0.9 10.00 2.41 5.61

1.0 11.11 2.40 5.62

1.1 12.22 2.75 5.27

1.2 13.33 3.04 4.98

1.3 14.44 3.27 4.75

1.4 15.55 3.45 4.57

1.5 16.67 3.59 4.43

1.6 17.78 3.69 4.33

1.8 18.89 3.81 4.21

2.0 22.22 3.83 4.19

2.1 23.33 3.81 4.21

2.2 24.44 3.79 4.23

∗

Expected DO ⫽ saturated DO (8.02 mg/L) ⫺ DO deficit

where C

⫽ constant

If D

⫽ 0, then D

is a function of L

, or

(1.92)

where C

, C

⫽ constants

(1.93)

Since K

⫽ K

(1.94)

Hence the minimum allowable DO concentration at the critical point

(DO

) is the DO saturation value (S, or DO

sat

) minus D

. It can be writ-

ten as

(1.95)

(1.95a)

From the above equation, there is a linear relationship between DO

and 1/Q. Therefore, a plot of DO

versus 1/Q will give a straight line,

where S is the y-intercept and C

is the slope of the line. Observed val-

ues from fieldwork can be analyzed by the method of least squares to

determine the degree of correlation.

14 SOD of DO Usage

The SOD portion of DO usage is calculated using the following formula:

(1.96)

where DO

sod

⫽ oxygen used per reach, mg/L

G ⫽ the SOD rate, g/(m

⭈ d)

t ⫽ the retention time for the reach

H ⫽ the average water depth in the reach, ft

All biological rates, including BOD and SOD, have to be corrected for

temperature using the basic Arrhenius formula:

(1.97)R

5 R

T2A

sod

3.28Gt

5 S 2 C

5 S 2 D

5 L

DQ 5 C

D 5 L

3 C

78 Chapter 1

where R

⫽ biological oxygen usage rate at a temperature, T ⬚C

⫽ biological oxygen usage rate at ambient or standard

temperature (20⬚C as usual), A⬚C

u ⫽ proportionality constant, 1.047 for river or stream

Example: A stream reach has an average depth of 6.6 ft. Its SOD rate

was determined as 3.86 g/(m

⭈ d) at 24⬚C. The detention time for the reach

is 0.645 days. What is the SOD portion of DO usage at the standard tem-

perature of 20⬚C?

solution:

Step 1. Determine SOD rate G at 20⬚C

Step 2. Calculate DO

sod

15 Apportionment of Stream Users

In the United States, effluent standards for wastewater treatment

plants are generally set for BOD, total suspended solids (TSS), and

ammonia-nitrogen (NH

-N) concentrations. Each plant often should

meet standards of 10–10–12 (mg/L for BOD, TSS, and NH

-N, respec-

tively), right or wrong. It is not like classically considering the maximum

use of stream-assimilative capacity. However, in some parts of the world,

maximum usage of stream natural self-purification capacity may be a

valuable tool for cost saving of wastewater treatment. Thus, some con-

cepts of apportionment of self-purification capacity of a stream among

different users are presented below.

The permissible load equation can be used to compute the BOD load-

ing at a critical point L

, as previously stated (Eq. (1.84); Thomas, 1948):

(1.98)log L

5 log D

1 c1 1

2 k

a1 2

0.418

dloga

5 1.03 smg/Ld

3.28 3 3.21 3 0.645

6.6

sod

3.28Gt

5 G

/s1.047d

5 3.86/1.2017 5 3.21[g/sm

dd]

5 G

s1.047

24220

Streams and Rivers 79

where L

⫽ BOD load at the critical point

⫽ DO deficit at the critical point

⫽ DO deficit at upstream pollution point

⫽ river BOD removal rate to the base 10, per day

⫽ reoxygenation rate to the base 10, per day

also may be computed from limited BOD concentration and flow.

Also, if city A and city B both have the adequate degree of treatment to

meet the regulatory requirements, then

(1.99)

where p ⫽ part or fraction of L

remaining at river point B

a ⫽ fraction of L

discharged into the stream

b ⫽ fraction of L

discharged into the stream

⫽ BOD load at point A, lb/d or mg/L

⫽ BOD load at point B, lb/d or mg/L

If L

⫽ L

⫽ paL ⫹ bL (1.99a)

A numerical value of L

is given by Eq. (1.98) and a necessary relation-

ship between a, b, p, L

, and L

is set up by Eq. (1.99). The value p can

be computed from

Since

Therefore

(1.100)

(1.100a)

The apportionment factors a and b can be determined by the following

methods.

p 5 10

log p 52k

log

52k

t 5 log sfraction remainingd

log p 52k

5 paL

1 bL

80 Chapter 1

15.1 Method 1

Assume: City A is further upstream than city B, with time of travel t

given, by nature, a larger degree of stream purification capacity. The

time from city B to the lake inlet is also t. Since

(1.101)

The proportion removed by the stream of BOD added by city A is

(1.102)

(1.102a)

The proportion removed by the stream of BOD added by city B is

(1.102b)

Then the percent amount removed BOD from cities A and B, given as

and P

, respectively, are

(1.103)

(1.104)

Calculate the degree of BOD removal proportion:

(1.105)

a 5 s1 1 pdb

s1 1 pd/s2 1 pd

1/s2 1 pd

1 1 p

2 1 p

1 2 p

s1 2 p

d 1 s1 2 pd

s1 2 pd

s1 2 pds1 1 p 1 1d

1 1 p

2 1 p

s1 2 pds1 1 pd

s1 2 pds1 1 p 1 1d

1 2 p

s1 2 p

d 1 s1 2 pd

s1 2 pds1 1 pd

s1 2 pds1 1 pd 1 s1 2 pd

5 1 2 10

5 1 2 p

5 1 2 10

t32

5 1 2 10

22k

5 1 2 10

5 1 2 p

y 5 L

s1 2 10

Streams and Rivers 81

Solving Eqs. (1.99a) and (1.105) simultaneously for a and b:

(1.106)

(1.107)

The required wastewater treatment plant efficiencies for plants A and

B are 1 – a and 1 – b, respectively. The ratio of cost for plant A to plant

B is (1 – a) to (1 – b).

Example 1: Assume all conditions are as in Method 1. The following data

are available: k

⫽ 0.16 per day; t ⫽ 0.25 days; flow at point C, Q

⫽ 58 MGD

L ⫽ 6910 lb/d; and regulation required L

ⱕ 10 mg/L. Determine the appor-

tionment of plants A and B, required plant BOD removal efficiencies, and

their cost ratio.

solution:

Step 1. Compute allowable L

Step 2. Compute p

Step 3. Calculate a and b by Eqs. (1.106) and (1.107)

5 0.49

5 0.70 3 0.697

a 5

1 1 p

1 1 p 1 p

b 5

4837

6910

1 1 0.912

1 1 0.912 1 0.912

5 0.912

p 5 10

5 10

20.1630.25

5 4837 slb/dd

5 10 mg/L 3 8.34 lb/d MGD

mg/L 3 58 MGD

1 1 p

1 1 p 1 p

a 5 s1 1 pdb

b 5

1 1 p 1 p

5 ps1 1 pdb 1 b 5 sp 1 p

1 1db

5 pa 1 b

5 paL 1 bL

82 Chapter 1

Step 4. Determine percent BOD removal required

For plant A:

1 – a ⫽ 1 – 0.49 ⫽ 0.51 i.e. 51% removal needed

For plant B:

1 – b ⫽ 1 – 0.255 ⫽ 0.745 i.e. 74.5% removal needed

Step 5. Calculate treatment cost ratio

15.2 Method 2

Determine a permissible BOD load for each city as if it were the only

city using the river or stream. In other words, calculate maximum BOD

load L

for plant A so that dissolved oxygen concentration is main-

tained above 5 mg/L (most state requirements), assuming that plant B

does not exist. Similarly, also calculate maximum BOD load L

for plant

B under the same condition. It should be noted that L

may be larger

than L

and this cannot be the case in Method 1.

The BOD removed by the stream from that added

at point A: referring to Eq. (1.102a)

(1 – p

at point B: referring to Eq. (1.102b)

(1 – p)L

The percentage removed is

at point A,

(1.108)

at point B,

(1.109)

s1 2 pdL

s1 2 p

1 s1 2 pdL

s1 2 p

1 s1 2 pdL

Cost, plant A

Cost, plant B

1 2 a

1 2 b

0.51

0.745

0.68

1.46

5 0.255

b 5

1 1 p 1 p

b 5 0.70 3

2.744

Streams and Rivers 83

The degree of BOD removal is

(1.105)

Solve Eqs. (1.99), (1.105), (1.108), and (1.109) for a and b. Also determine

(1 – a) and (1 – b) for the required wastewater treatment efficiencies and

cost ratios as (1 – a)/(1 – b).

Example 2: All conditions and questions are the same as for Example 1,

except using Method 2.

solution:

Step 1. Compute allowable L

and L

For point B, using Eq. (1.100)

For point A,

Step 2. Calculate P

and P

, using Eq. (1.108)

5 1 2 0.677 5 0.323

5 0.677

978.4

978.4 1 466.8

s1 2 0.912

d5815

s1 2 0.912

d5815 1 s1 2 0.912d5304

s1 2 p

1 s1 2 pdL

5 5815 slb/dd

5 4837 lb/d/0.8318

5 10

20.1630.2532

5 10

20.08

5 0.8318

5 5304 lb/d

5 4837/0.912

5 L

/0.912 sL

5 L

in this cased

p 5

5 10

20.04

5 0.912

log p 5 log

52k

t 520.16 3 0.25 520.04

84 Chapter 1