BOD lab report

LAB

: ENVIRONMENTAL ENGINEERING

EXPERIMENT

: BIOCHEMICAL OXYGEN DEMAND (BOD)

1.0 OBJECTIVE To measures the strength of the water sample ( water, wastewater, etc) based on the amount of oxygen needed to stabilize the organic matter in the sample. 2.0 THEORY Biochemical Oxygen Demand is a common, environmental procedure for determining theextent to which oxygen within a sample can support microbial life. This method is popular in many environmental laboratories analyzing waste water, compost, sludge, and soil samples. When a measurement is made of all oxygen consuming materials in a sample, the result is termed “Total Biochemical Oxygen demand “( TBOD ) , or often just simply “ Biochemical Oxygen Demand “ (BOD). Because the test is performed over a five day period, it is often referred to as a “five Day BOD “, or a BOD5. In addition, this procedure is only suitable for samples void of serious matrix interferences. To gain a broader appreciation of oxygen demand, additional avenues of interest may be explored including CBOD (carbonaceous oxygen demand), COD (chemical oxygen demand), and TOC (total organic carbon).Because of complications measuring this ultimate BOD (BOD u), BODu is usually extrapolated from laboratory 5-day BOD bottle tests BODt = BODu ( 1 – e-kt ) In many biological treatment plants, the facility effluent large numbers of nitrifying organisms which are developed during the treatment process. These organisms can exert an oxygen demand as they convert nitrogenous compounds (ammonia and organic nitrogen) to more stable forms (nitrites and nitrates). At least part of this oxygen demand is normally measured in a five day BOD. Sometimes it is advantageous to measure just the oxygen demand exerted by organic (carbonaceous) compounds, excluding the oxygen demand exerted by the nitrogenous compounds. To accomplish this, the nitrifying organisms can be inhibited from using oxygen by the addition of a nitrification inhibitor to the samples. The result is termed “Carbonaceous Biochemical Oxygen Demand”or BOD.

Generally, high BOD indicates a high content of easily degradable organic material in sample, and low BOD indicates a low volume of organic material substances which are difficult to break down.

3.0 EQUIPMENT & REAGENTS

       

6 units BOD bottles – 300ml 100ml beaker 100ml graduated cyclinder 1unit pipettes (ml) DO meter pH meter Air pump Incubator, capable of maintaining 20 ± 1°C

Nutrient Buffer solution 1. Phosphate Buffer - Dissolve 8.5g potassium di hydrogen phosphate (KH2PO4), are dissolved in approximately 500ml distilled water, 21.75g di potassium hydrogen phosphate (K2HPO4) are added, followed by 33.4g di sodium hydrogen phosphate (Na2HPO4) and 1.7g ammomium chloride (NH4CI). Adjust pH to 7.2 if necessary with either 1 NH 2SO4 or NaOH. Dilute to 1liter 2. Magnesium sulfate - 22.5g magnesium sulphate (MgSO4.7H2O) is diddolved in distrilled water and the solution made up to 1liter. 3. Calcium chloride - 27.5g of anhydrous calcium chloride (CaCI2) are dissolved in distilled water and the solution made up to 1liter. 4. Ferric chloride -

0.25g FeCI3.6H2O is dissolved in distilled water and the solution made up to 1liter.

BOD Dilution water

Add 1ml of each of the 4 nutrient buffer solutions to 1liter of deionized water and aerate for at least 1 hr before conducting BOD test to ensure the DO concentration in the dilution water is at least 7.5 mg/L.

Chemical for pH adjustment 1N of sulfuric acid, H2SO4 or 1N sodium hydroxide, NaOH

4.0 BOD5 MEASUREMENT PROCEDURES 4.1 DETERMINATION OF SAMPLE SIZE : Once a general range for the BOD of a sample has been determined, the dilutions can be established which will ensure that at least one dilution will meet the criteria for valid BOD results. The following procedure can be used to calculate volumes for sample dilution from the estimated BOD. For example, suppose the estimated BOD of an influent sample is 400 mg/L and assume the DO of saturated dilution water is 8.0 mg/L. since the criteria for most valid results states that the DO depletion at the end of five days incubation should be at least 2.0 mg/L and the residual DO at least 1.0 mg/L , the formulas to calculate the minimum and maximum estimated dilution are as follows: A. mL sample added to BOD bottle = ( minimum allowable depletion, mg/L x Volume of BOD bottle, mL ) / estimated BOD , mg/L. B.

ml sample added to BOD bottle = ( maximum allowed depletion , mg/l x Volume of BOD bottle, ml ) / estimated BOD, mg/l.

NOTE: Those sample dilutions which deplete less 2 mg/L, or have a final DO of less than 1 mg/L would not be use in the calculation of the average sample BOD.

Table 1: Volume of sample for dilution

Expected BOD

Volume of sample to be diluted

Range

to 300 ml in BOD bottle (ml)

(mg O2/L) 0-7 7 - 21 12 - 42 30- 105 60 - 210 120 - 420 300 - 1050 600 - 2100 1200 - 4200 3000 - 10500 6000 - 21000

300 100 50 20 10 5 2 1 0.5 0.2 0.1

To determine the value of the BOD in mg/l, use the following formula:

When dilution water is not seeded:

D1 – D2 BOD5,mg/L = P

D1 = DO of diluted sample immediately after preparation, mg/L, D2 = DO of diluted sample after 5 d incubation at 20oC, mg/L, P =reciprocal of dilution factor, D= volume of BOD bottle/volume of sample used in mL

NOTE: The samples must undergo dilution if they have a BOD > 7, otherwise all the dissolved oxygen will be used up before 5 days have elapsed. [Note: raw domestic wastewater typically has a BOD of around 300 mgO2/l] 4.2 PRE-TREATMENT SAMPLES. 1) That contain caustic alkalinity or acidity 2) Caustic alkalinity or acidity can prevent bacteria from growing during the course of the BOD test. 3) To prevent this, samples which have pH values higher than pH 8.0 or lower than pH 6.0 must be neutralized to pH 7.0 before test is performed. NOTE :Neutralized samples must be seeded for the BOD test. Procedure for neutralizing samples: 1.

Pour 50 mL of samples into a 100 mL beaker.

2.

Measure the pH of the sample using a pH meter. If the pH is out of the range of pH 6.0 to pH 8.0 countinue with steps 3 – 6, otherwise perform the BOD test on the untreated sample.

3.

Add 1 N sulfiris acid if the sample is alkaline, or 1N sodium hydroxide if the sample is acidic, intil the pH reaches 7.0.

4.

Calculate the amount of sulfuric acid or sodium hydroxide needed to neutralize 1000ml of the sample.

5.

Add the calculated amount of acid or base to the sample.

6.

Repeat steps 1-5 until the pH test shows pH 7.0.

7.

Calculation the amount of 1 N sodium hydroxide or 1 N sulfuric acid needed to neutralized the sample to pH 7.0 using the following formula: mL needed = ( mL acid or base used x mL total test sample ) mL sample portion used for neutralization. For example, suppose 1.3 mL of 1 N NaOH are use to neutralize 50 mL of sample to pH 7.0 calculate the volume of NaOH to be added to neutralize the sample as follows:

mL 1 N NaOH needed = ( 1.3 mL x 1000 mL ) /50 mL = 1300 / 50 = 26ml

4,3 BOD5 PROCEDURES 1. Label separately 4 unit of 300ml BOD bottles (1 blank, 3 samples) 2.

Pipette appropriate volumes into bottles. Fill all the bottles to the top with dilution water.

3. Measure the initial DO concentration using DO meter. 4. Place all the bottles into the BOD incubator, this is set at 20 oC and is dark to prevent the growth of algae. Leave for 5 days. 5. After 5 days, remove the samples from the incubator and measure the final DO content of each. NOTE: The dilution water blanks are used only to check the quality of the dilution water. If the quality of the water is good and free from impurities, the depletion of DO should NOT be less than 0.2 mg/l. in any event, do not use the depletion obtained as blank correction.

5.0 RESULT AND CALCULATION 1. Type of sample: FKAAS river 2. Initial pH sample: 7.01 3. Volume of sample used: 50ml Volume of

Volume of

Sample

incubation

sample

BLANK 1 2 3

bottle, ml 300 300 300 300

size, ml 0 50 50 50

Initial DO,

Final DO,

mg/l

mg/l

7.48 7.46 7.46 7.47

7.42 4.74 4.81 4.9

DO depletion, mg/l 0.06 2.72 2.65 2.57

6.0 DATA ANALYSIS i.

Show all the calculation and state if any of the data needs to be discarded.

BOD,mg/L= D1 – D2 P

Blank = 7.48 – 7.42 = 0.06 the value is less than 0.2 mg/L, so not calculation for average BOD.

sample 1 = 7.46–4.74 = 16.32 mg/L ( 50 / 300 ) Sample 2 = 7.46 – 4.81 = 15.9 mg/L ( 50 / 300 )

BODx, mg/l 16.32 15.9 15.42

Sample 3 = 7.47 – 4.9 = 15.42 mg/L ( 50 / 300 )

ii.

Does your BOD dilution watershow ’toxic effect’? The sample is not showing any toxic effect, it is because the waste water, took from UTHM is not from a toxic source. And the pH value of the sample is nearly neutral and after our group use 5 mL of wastewater, the value of BOD obtained is slightly higher than other group. It means that the sample doesn’t show any toxic effect.

iii.

Could you rely on your BOD results? Why? Yes, because the average BOD result is compared with the effluent standard (DOE, Malaysia).

iv.

Determine the pollution level of samples based on the BOD results. The BOD5 for 5ml are more than 100 ml/g, which is not in the range of Standard A and Standard B. Standard A (20mg/l) is the indicator for drinking water while Standard B (50 mg/l) is the indicator for inland water quality. Hence, we can say that the water sample is quite polluted because the location of the sample is the UTHM waste water. The area is polluted because usually people waste. The organic material come from the waste will consume the oxygen in the water due to the microbes’ activities. The consequence is the water polluted and the BOD 5 reading is high.

v.

Differentiate between seeded and unseeded BOD. Seeding is a process of adding live bacteria and microorganism to a sample. If the samples tested contain materials which could kill or injure the microorganisms, the condition must be corrected and healthy active organisms added.

7.0 DISCUSSION The BOD5 for 50ml are more than 100 ml/g, which is not in the range of Standard A and Standard B. Standard A (20mg/l) is the indicator for drinking water while Standard B (50 mg/l) isthe indicator for inland water quality. Hence, we can say that the water sample is quite polluted because the location of the sample is the FKAAS river. This area is polluted because usually by natural waste that come from plants and animals. The consequence is the water polluted and the BOD5 reading is high.

8.0 CONCLUSION The water source at FKAAS river has been polluted due to the natural waste from animal and plant. The BOD5 reading is out of the Standard A and B. The recommendation to solve the problemis to apply water treatment to the water source and to prevent the pollution of water.

APPENDIX ENVIRONMENTAL QUALITY ACT 1974 (ENVIRONMENTAL QUALITY (SEWAGE AND INDUSTRIAL EFFLUENTS) REGULATIONS 1978 [Regulation 8 (1), 8 (2), 8 (3)] PARAMETER LIMITS OF EFFLUENT OF STANDARDS A AND B Parameter

Temperature

Unit

C

pH Value

Standard A

B

40

40

6.0 - 9.0

5.5 - 9.0

BODs at 20ºC

mg/1

20

50

COD

mg/1

50

100

Suspended Solids

mg/1

50

100

Mercury

mg/1

0.005

0.005

Cadmium

mg/1

0.01

0.02

Chromium, Hexavaient

mg/1

0.05

0.05

Arsenic

mg/1

0.05

0.10

Cyanide

mg/1

0.05

0.10

Lead

mg/1

0.10

0.5

Chromium, Trivalent

mg/1

0.20

1.0

Copper

mg/1

0.20

1.0

Manganese

mg/1

0.20

1.0

Nickel

mg/1

0.20

1.0

Tin

mg/1

0.20

1.0

Zinc

mg/1

1.0

1.0