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Water Quality Lecture 13: Introduction to Environmental Engineering.

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Presentation on theme: "Water Quality Lecture 13: Introduction to Environmental Engineering."— Presentation transcript:

1 Water Quality Lecture 13: Introduction to Environmental Engineering

2 Measures of Water Quality What is clean water? Is D.C.’s drinking water clean? –What are some parameters that determine water quality? Dissolved oxygen – oxygen probe and oxygen meter Biochemical Oxygen Demand Solids Nitrogen – ammonia Bacteriological quality

3 Biochemical Oxygen Demand BOD –Measurement of the rate at which microbes degrading organic matter use oxygen One can calculate the theoretical oxygen demand If you know the chemical formula – for example 1.67 x M glucose –C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O –6 x 1.67moles/L x 32 g/mole x 1000 mg/g = 321 mg/L BOD is used because wastewater isn’t one organic molecule

4 Biochemical Oxygen Demand Standard BOD Test Run in the absence of light 20  C for 5 days, defined as BOD 5 Standard 300 mL volume BOD bottle –BOD ultimate, L, BOD greater than 20 days BOD = I –F I = initial DO, mg/L F = final DO, mg/L

5 Biochemical Oxygen Demand What if the F is =0? Dilution of the sample is required when F = 0. If F = 0 we don’t know how much DO would have been used. BOD = (I – F)D D = dilution

6 Biochemical Oxygen Demand Dilution by how much? The proper dilution amount is rarely known before the test. A proper BOD test requires a change of DO of 2 mg/L And more than 2 mg/L remaining at the end of the test The dilution required is calculated as:

7 Biochemical Oxygen Demand Seeding –The addition of active microorganisms that take up oxygen May be required in samples that do not have their own If seeding is necessary, any BOD that is contributed by the seed must be subtracted

8 Biochemical Oxygen Demand BOD with dilution and seed BODt = BOD at time t, mg/L I, F = initial and final DO of sample and seed, mg/L I’, F’ = initial and final DO of seed water, mg/L X = seeded dilution water in sample bottle, mL Y = seeded dilution in bottle with only seed dilution water mL D = dilution of sample

9 Biochemical Oxygen Demand Example Standard BOD test with a 1:30 dilution with seeded dilution water is run. Both bottles begin at saturation, 9.2 mg/L. After five days, the bottle with waste has a DO of 2 mg/L, while the DO of the seed = 8 mg/L. Find the BOD 5.

10 Biochemical Oxygen Demand Typical BOD values –Wastewater BOD 5  250 mg/L –Effluent BOD 5 from wastewater < 30 mg/L –Industrial wastewater = 30,000 mg/L –US Engineers often speak in terms of lbs BOD 5 /day Mass rate load of a plant to the receiving stream

11 Biochemical Oxygen Demand The BOD test occurs in a bottle, –Thus the mass balance is Rate of DO accum. = rate of DO consumed dz/dt = -r z = dissolved oxygen, mg/L Assume that it is a first-order reaction

12 Biochemical Oxygen Demand –As O 2 is used, the amount still to be used is z, the amount already used is y L = y + z L= ultimate demand

13 Biochemical Oxygen Demand Variables y = BOD at any time t L = ultimate BOD k 1 = deoxygenation constant –Example Find the BOD5 for a waste with an ultimate BOD = 282 mg/L and a k 1 = /day

14 Nitrogenous Oxygen Demand Nitrogenous BOD Organic matter + O 2  bacteria + NH 3 + CO 2 + et.al. 2NH 3 + 3O 2  2NO H + + 2H 2 O 2NO O 2  2 NO 3 - Two moles of O 2 per mole of NH 3 BOD ult = a (BOD 5 ) + b(KN) –KN = Kjeldahl nitrogen (measure of organic nitrogen and ammonia) –a and b are constants, a = 1.2 & b = 4.0 for example

15 Homework Chapter 8 Problems: 3, 7, 16, 22


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