bod and its kinetic aspects

8/19/2019 BOD and its kinetic aspects

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Biochemical Oxygen Demand - BOD

• Measurement of amount of oxygen used by bacteria to

metabolize organic material in water

organic material + bacteria + O2→ CO2 + H2O

Five-day BOD (BOD5) is performed at 20°C in dark over

five days, with added seed (bacteria), nutrients* andoxygen. Expressed as mg/L.

*Iron, magnesium & calcium salts and phosphate buffer


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Uses of BOD Test

Used as a measurement of waste strength

- is an effluent discharge limitation parameter in mill permits

Used as basis for designing & evaluating biological

treatment systems

-to determine amount of Oxygen needed to stabilize organic

matter in waste

-to determine size of wastewater treatment facilities

-to measure efficiency of some treatment processes

BOD

time. d

B O D

BOD exerted, BODt

BODr

UBOD

BODt

BOD remaining, BODr

Ultimate BOD

BOD Exertion & Remaining

Metcalf & Eddy, 4th edition


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Limitations of BOD5

• Only 60-70% of soluble organic matter is metabolized in

five days

• Inorganic oxidation may also occur:

1) Inorganic oxidation of sulfur or iron:

S-- + 2 O2→ SO4-- or 4Fe++ + 3O2→ 2Fe2O3

2) Nitrification may also occur if ammonia is present

Conversion of ammonia to nitrate:NH3 + 2 O2→ HNO3 + H2O

Details of Nitrification

Nitrifying bacteria can oxidize ammonia, common to many waste

streams, resulting in nitrogenous oxygen demand (NBOD).

Conversion of ammonia to nitrite

(Nitrosomonas bacterium):

NH3 + 3/2 O2 → HNO2 + H2O

Conversion of nitrite to nitrate

(Nitrobacter bacterium):

HNO2 + 1/2 O2 → HNO3


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Nitrogenous and Carbonaceous BOD

Nitrifying bacteria can be suppressed by chemical treatment,

pasteurization, or chlorination followed by dechlorination


Functional Analysis of BOD test: interrelationship of organic waste, bacteria, and oxygen

consumed (Metcalf & Eddy, 4th Edition)

Waste is

‘Stabilized’


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Functional Analysis of BOD test: Organic waste remaining and Oxygen consumed

(Metcalf & Eddy, 4th Edition)

Reaction Order Form Units Comment on Rate

Zero-order rate = k 1/timeConcentration has no

effect

First-order rate = kC Concentration/timeDirectly proportional

to concentration

Second-order rate = kC2Concentration x

Concentration/time

Proportional to second

power of concentration

Reaction order can be determined by summingthe exponents on the concentration terms in an equation

Oxygen consumed is a 1st order reaction:

r r

BODk dt

dBODrate 1

BOD (mg/L) is a concentration term.

BOD remaining decreases with time, as oxygen is consumed.


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In effluents from biological treatment processes,

k1 normally varies from 0.12 to 0.46, 1/d

Average is 0.23/d

(k1 (base e) = K1(base 10)*2.303)


Reaction rates are affected by temperature

k is affected by temperature (T)

van’t Hoff-Arrhenius (‘uh-REE-nee-us’) relationship:

k1T= k120

θ(T-20)

θ = 1.056 for temperatures between 20º - 30ºC

EXAMPLE:

If river temperature is 25°C, what will k1 be in that environment?

Find k1 at 25°C if k1 at 20°C = 0.23

k1 = k120(1.056)(T-20)

= 0.23(1.056)(25-20)

= 0.30/d

How can we determine k1 experimentally?


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Add 1 ml/L Nutrients:

Phosphate Buffer Solution

Magnesium Sulfate Solution

Calcium Chloride Solution

Ferric Chloride Solution

Experiment to Determine BOD Kinetics

Incubate at 20° C, take D.O. readings daily for ~10 days

How much of the effluent sample should be

added to the dilution water, to make up each set?


9/12

Requirements for valid kinetic experiment:

Daily residual D.O. of at least 1 mg/L

Day 2 D.O. decrease of about 2 mg/L

Day 2 tends to have the highest D.O. decrease.

need to calculate BOD exerted from day 2 to day 3

to know what dilution to use when preparing sample

where:

t = time, days

BODr = amount of BOD remaining at time t

k1 = 1st order reaction rate, 1/day

Integrating the above expression between the limits of UBOD

and BODt and t = 0 and t = t yields:

BODr = UBOD(e-k1t)

Where UBOD = total or ultimate carbonaceous BOD, mg/L

The amount of BOD exerted at time t

(what gets regulated) is:

BODt = UBOD – BODr= UBOD – UBOD(e-k1t)

= UBOD(1 – e-k1t)

r r BODk

dt

dBODrate 1

BOD Kinetics


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Now we can find BOD2

Given: BOD5 = 20 mg/L, t = 20°C , k1 = 0.23/d

Find: BOD2

Solution:

First, find UBOD:

BOD5 = UBOD(1 – e-k1t)

20 = UBOD(1 – e-0.23 x 5)

= UBOD(1 – 0.3166)

= UBOD * 0.683

UBOD = 29.3 mg/L

Next, find BOD2:

BODt = UBOD(1 – e-k1t)

BOD2 = 29.3(1 – e-0.23 x2

)= 29.3(1 – 0.6313)

= 10.8 mg/L

BODt= UBOD(1 – e-k1t)

UBOD = 29.3 mg/L

Find BOD3:

BODt = UBOD(1 – e-k1t)

BOD3 = 29.3(1 – e-0.23 x3)

= 29.3(1 – 0.5016)

= 14.6 mg/L

the BOD exerted between days 2 and 3 is

BOD3 – BOD2 = 14.6 – 10.8 = ~4 mg/L

BODt

= UBOD(1 – e-k1t)

Recall requirements for valid kinetic experiment:

Daily: residual D.O. of at least 1 mg/L

D.O. decrease of about 2 mg/L

Day 2 tends to have the highest D.O. decrease.

need to calculate BOD change from day 2 to day 3


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Instructions: Fits data to equation y

= L(1 - e-kt)

Enter up to 50 pairs of time, BOD

(y) data in columns A and B. Enter

an initial guess for k in cell B7 and a

confidence level in D5.

Click the "Find k & L" button.

Results appear in cells B7 and D7.

If a "failed to converge" message

appears, try a new guess for k that

is more than an order of magnitudehigher or lower than the original k.

Recalculate for new data and/or

confidence level.

0

500

1000

1500

2000

2500

3000

3500

4000

4500

0 5 10 15

B O D

Time

Model Fit to Data

Fujimoto graphical method to find UBOD (L)

BODt= UBOD(1 – e-k1t)

Then find k using:

Prepare table of BODtvs BODt+1

Plot BODt vs BODt+1,

connect with linear trendline.

Use option to display equation

on chart.

Add line of slope 1.

Intersection of two lines is

UBOD.

Use equation on chart to

calculate UBOD where X=Y.

y = 0.8003x + 943.08

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

5500

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500

B O D @ t +

1 , m g / L

BOD @ t, mg/L

Fujimoto Plott, days BOD, mg/L BOD @ t + 1

0 0 366

1 366 1431

2 1431 2269

3 2269 3513

6 3513 3605

7 3605 3720

8 3720 3840

9 3840 3885

10 3885 3970

14 3970

bod and its kinetic aspects

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