Biological Oxygen Demand (BOD5) - Chapter 3 PDF

Biological Oxygen Demand (BOD5) Oxygen Demand It is a measure of the amount of “reduced” organic and inorganic matter in a water and wastewater. Measured in several ways: BOD - Biochemical Oxygen Demand COD - Chemical Oxygen Demand ThOD - Theoretical Oxygen Demand Bichemical Oxygen Demand (BO...

Biological Oxygen Demand (BOD5) Oxygen Demand It is a measure of the amount of “reduced” organic and inorganic matter in a water and wastewater. Measured in several ways: BOD - Biochemical Oxygen Demand COD - Chemical Oxygen Demand ThOD - Theoretical Oxygen Demand Bichemical Oxygen Demand (BOD5) BOD determines the amount of dissolved oxygen needed by aerobic organisms in a wastewater body to break the organic material present in the given wastewater sample at certain temperature over a specific period of time, Usually time is taken is 5 days, The temperature is 20 0C. Biochemical oxygen demand measures the amount of oxygen used by bacteria to metabolize organic material in wastewater, . micro-organisms Organic matter + O2 ⇒ H2O + CO2 ➢ Five-day BOD (BOD5) is performed at 20°C in dark over five days, with added seed (bacteria), nutrients and oxygen. Expressed as mg/L. ➢ Nutrients: Iron, magnesium & calcium salts and phosphate buffer. BOD test is a long-established test BOD determination involves the measurement of the dissolved oxygen used by microorganisms in the biochemical oxidation of organic matter in the waste. BOD test results used to: ➢ to determine approximate quantity of Oxygen that will be required to stabilize organic matter in waste. ➢ to determine the size of wastewater treatment facilities. ➢ to measure the efficiency of some treatment processes. BOD test There are three different cases to carry BOD test: ✓ Case 1: Sample only – no dilution ✓ Case 2: Sample with dilution ✓ Case 3: Sample with seed L or BOD remaining BOD - loss of biodegradable organic matter (oxygen demand) BOD Bottle Lo Lo-Lt = BODt Lt Time BOD Bottle BOD Bottle BOD Bottle BOD Bottle Case 1: Sample Only – No Dilution BOD = DOi – DOf where DOi = the initial dissolved oxygen (DO) of the wastewater, DOf = final dissolved oxygen (DO) of wastewater 5 days later. Case 2: Sample with Dilution Case 3: Sample with seed ➢Organic matter transferred with the seed, • Seeding provides micro-organisms to oxidize organic matter, – Not required for − municipal wastes, biologically treated effluents, surface water samples, – Some industrial wastes may require seeding. ➢ Settle sewage can be used as a source of seeding, BOD test procedure In the standard BOD test (see Fig. a), a small sample of the wastewater to be tested is placed in a BOD bottle (volume = 300 ml). The bottle is then filled with dilution water saturated in oxygen and containing the nutrients required for biological growth. Before the bottle is stoppered, the oxygen concentration in the bottle is measured (see Fig. b). After the bottle is incubated for 5 days at 20 oC, the dissolved oxygen concentration is measured again. The BOD of the sample is the difference in the dissolved oxygen concentration values, expressed in milligram per liter, divided by the decimal fraction of sample used. The computed BOD value is known as the 5-day, 20 oC biochemical oxygen demand After incubation, the dissolved oxygen of the sample is measured and BOD is calculated using Eq. (*) or (**). D1 Where, D1- Dissolved oxygen of diluted sample immediately after preparation, mg/L D2- Dissolved oxygen of diluted sample after 5-day incubation at 200C, mg/L B1- Dissolved oxygen of seed control before incubation, mg/L B2- Dissolved oxygen of seed control after incubation f - fraction of seeded dilution water volume in sample to volume of seeded dilution water in seed control. P- fraction of wastewater sample volume to total combined volume. Example 1: Determination of BOD from laboratory Data. The following information is available for a seeded 5-day BOD test conducted on a wastewater sample. 15 ml of the waste sample was added directly into a 300 ml BOD incubation bottle. The initial DO of the diluted sample was 8.8 mg/L and the final DO after 5 days was 1.9 mg/L. The corresponding initial and final DO of the seeded dilution water was 9.1 and 7.9, respectively. What is the 5-day BOD (BOD5) of the wastewater sample? Solution 5-BOD can be determined as follows: BOD cont. Theoretically, an infinite time is required to complete the biological oxidation of organic matter. For practical purposes, the rxn may be considered complete in 20 days. It has been found by experience that a reasonably large percentage of total BOD exerted in 5 days: Modeling of BOD Reactions The rate of BOD oxidation (exertion) is modeled based on the assumption that the amount of organic material remaining at any time t is governed by a first-order function as given below: K1 = first-order reaction rate constant 1/d UBOD = ultimate carbonaceous BOD, mg/L t = time, d Thus BOD exerted up to time t is given by BODt = UBOD –BODr =UBOD – UBOD (e-K1t) BODt =UBOD(1- e-K1t) (1) This equation (1) is the standard expression used to define the BOD for wastewater. The value of K1 for untreated wastewater is generally about 0.12 to 0.46 1/d, with a typical value about 0.26 1/d. If k1 at 200 C is equal to 0.23 1/d, the 5-day oxygen demand is about 68% of the ultimate first-stage demand. The temperature at which the BOD of a wastewater sample is determined is usually 200 C. It is possible, however, to determine the reaction constant k at a temperature other than 200 C using the following relationship Equation (1) along with Eq. (2), makes it possible to convert test results from different time periods and temperatures to the standard 5-day 200 C test, as illustrated in the following example. Example 2: Calculation of BOD Determine the 1-day BOD and ultimate first-stage 5-day BOD for a wastewater whose 5-day 200C BOD is 200 mg/L. The reaction constant k (base e)= 0.23 d-1. What would have been the 5-day BOD if the test had been conducted at 250C? Solution 1) Ultimate carbonaceous BOD 200 = UBOD (1- e-0.23X5) 200= UBOD (1- 0.316) UBOD = 293 mg/l Solution 2) 1-day BOD Solution BOD contains of: ➢ Carbonaceous Biochemical Oxygen Demand. ➢ Nitrogenous Biochemical Oxygen Demand. Carbonaceous Biochemical Oxygen Demand (CBOD) The Carbonaceous Biochemical Oxygen Demand (CBOD) represents that portion of oxygen demand involved in the conversion of organic carbon to carbondioxide, water and energy, Nitrification in the BOD test Noncarbonaceous matter, such as ammonia is produced during the hydrolysis of proteins. It is known that a number of bacteria are capable of oxidizing ammonia to nitrite and subsequently to nitrate. The generalized reactions are as follows: Conversion of ammonia to nitrite (Nitrosomonas bacterium): Conversion of nitrite to nitrate (Nitrobacter bacterium): Overall conversion of ammonia to nitrate (Nitrobacter bacterium): NH3 + 2 O2 HNO3 + H2O The oxygen demand associated with the oxidation of ammonia to nitrate is called the nitrogenous biochemical oxygen demand (NBOD) Problem: A secondary effluent has a total BOD5 of 45 mg/L. A BOD5 test is run on the same secondary effluent at the same time using a nitrification “inhibitor”. Using the data for this test as given below, calculate the nitrogenous BOD (assume a 300 ml BOD bottle is used to run this test). Sample size Initial DO, mg/L Final DO, mg/L 30 mL 8.9 6.0 50 mL 8.8 4.2 Solution a) Case I (30ml sample) b) Case II (50ml sample) Limitations of BOD5 The limitations of BOD are as follows 1) A high concentration of active, acclimated seed bacteria is required. 2) Pretreatment is needed when Dealing with toxic wastes, and the effects of nitrifying organisms must be reduced. 3) Only the biodegradble organics are measured 4) The relatively long period of time required to obtain test results. BOD Example This result was obtained for a BOD test on a wastewater sample. The sample was diluted by a factor of 20 prior to the test. What is the BOD5 ? BOD5 = (8 - 1.7)*20 = 126 mg/L Chemical Oxygen Demand COD The COD test is used to measure the oxygen equivalent of the organic material in wastewater that can be oxidized chemically using dichromate in an acid solution Some differences between BOD and COD 1) Many organic substances which are difficult to oxidized biologically, such as lignin, can be oxidized chemically, 2) Certain organic substances may be toxic to the microorganisms used in the BOD test, From an operational standpoint, one of the main advantages of the COD test is that it can be completed in about 2.5 h, compared with 5 or more days for the BOD test. Rabid COD test that takes only about 15 min has been developed. Ex. 3. Calculation of ThOD Solution 1) Balanced reaction for the carbonaceous oxygen demand 2) Balanced reaction for the nitrogenous oxygen demand 3) Determine ThOD Problem1: . Calculate the B.O.D. of a reference sample given that 1mL of seed material was used in the reference sample. (300 mL BOD bottles were used) Volume Used Initial D.O. Final D.O. Seed Material 5 mL 8.6 mg/L 5.8 mg/L Reference Sample 5 mL 8.7 mg/L 4.8 mg/L Reference Solution Organics Seed Bacteria And Organics 5 mL 1 1 mL 5 mL 2 Example Seeded BOD Calculation. Calculate the B.O.D. of a reference sample given that 1mL of seed material was used in the reference sample. (300 mL BOD bottles were used) Seed Material Reference Sample Volume Used 5 mL 5 mL Initial D.O. 8.6 mg/L 8.7 mg/L Final D.O. 5.8 mg/L 4.8 mg/L D1 - D2 BOD X 300 mL = seeded Sample Volume, mL D1 = Depletion of Reference Sample D1 = 8.7 mg/L - 4.8 mg/L D1 = 3.9 mg/L Example Seeded BOD Calculation. Calculate the B.O.D. of a reference sample given that 1mL of seed material was used in the reference sample. (300 mL BOD bottles were used) Seed Material Volume Used Initial D.O. Final D.O. 5 mL 8.6 mg/L 5.8 mg/L Reference Sample 5 mL 8.7 mg/L 4.8 mg/L D1 - D2 BOD X 300 mL = seeded Sample Volume, mL D2 = Depletion of Seed (In Seeded Sample) 8.6 – 5.8 D2 = 5 D2 = 2.8 = 0.56 5 Example Seeded BOD Calculation. Calculate the B.O.D. of a reference sample given that 1mL of seed material was used in the reference sample. (300 mL BOD bottles were used) Seed Material Volume Used Initial D.O. Final D.O. 5 mL 8.6 mg/L 5.8 mg/L Reference Sample 5 mL 8.7 mg/L 4.8 mg/L D1 - D2 BOD X 300 mL = seeded Sample Volume, mL 3.9 mg/L - 0.56 mg/L X 300 mL BOD = 5 mL 3.34 mg/L X 300 mL BOD = 5 mL BOD = 200.4 mg/L Example #2 Seeded BOD Calculation. Calculate the B.O.D. of an de-chlorinated effluent sample given that 1mL of seed material was used in the sample. (300 mL BOD bottles were used) Seed Material Sample Volume Used Initial D.O. Final D.O. 10 mL 8.3 mg/L 5.6 mg/L 200 mL 8.5 mg/L 3.7 mg/L D1 - D2 BOD X 300 mL = seeded Sample Volume, mL D1 = Depletion of Reference Sample D1 = 8.5 mg/L - 3.7 mg/L D1 = 4.8 mg/L Example #2 Seeded BOD Calculation. Calculate the B.O.D. of an de-chlorinated effluent sample given that 1mL of seed material was used in the sample. (300 mL BOD bottles were used) Volume Used Initial D.O. Final D.O. Seed Material Sample 10 mL 8.3 mg/L 5.6 mg/L 200 mL 8.5 mg/L 3.7 mg/L D1 - D2 BOD X 300 mL = seeded Sample Volume, mL D2 = Depletion of Seed D2 = (In Seeded Sample) 8.3 – 5.6 10 D2 = 2.7 10 = 0.27 Example #2 Seeded BOD Calculation. Calculate the B.O.D. of an de-chlorinated effluent sample given that 1mL of seed material was used in the sample. (300 mL BOD bottles were used) Seed Material Volume Used Initial D.O. Final D.O. 10 mL 8.3 mg/L 5.6 mg/L Sample 200 mL 8.5 mg/L 3.7 mg/L D1 - D2 BOD X 300 mL = seeded Sample Volume, mL 4.8 mg/L - 0.27 mg/L X 300 mL BOD = 200 mL 4.53 mg/L X 300 mL BOD = 200 mL BOD = 6.8 mg/L 1. Calculate the B.O.D. given: D.O. IN = 7.0 mg/L D.O. OUT (5-day) = 3.5 mg/L Vol. Sample in B.O.D. bottle = 15 mL Depletion, mg/L X 300 mL Sample Volume, mL BOD = BOD = 7.0 mg/L - 3.5 mg/L 15 mL BOD = 3.5 mg/L 15mL BOD = 70 mg/L X 300 mL X 300 mL