# chapter 8a

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Applications Water Quality

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Measures of Water Quality Some of the Most basic and Important Measures Dissolved Oxygen Biochemical Oxygen Demand Solids Nitrogen Bacteriological

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Dissolved Oxygen (DO) Typically Measured by DO probe and Meter Electrochemical Half Cell Reaction

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Biochemical Oxygen Demand (BOD) Amount of oxygen used by microorganisms to decompose organic matter in a water Theoretical BOD can be determined by balancing a chemical equation in which all organic matter is converted to CO 2 Calculate the theoretical oxygen demand of 1.67 x 10 -3 moles of glucose (C 6 H 12 O 6 ): C 6 H 12 O 6 + O 2  CO 2 + H 2 O general, unbalanced eqn C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O 1.67x 10 -3 moles glucose/L x 6 moles O 2 / mole glucose x 32 g O 2 /mole O 2 = 0.321 g O 2 /L = 321 mg O 2 /L

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BOD Test Dark 20 o C Time Standard – 5 days Ultimate

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BOD = I - F I = Initial DO F = Final DO If all the DO is used up the test is invalid, as in B above To get a valid test dilute the sample, as in C above. In this case the sample was diluted by 1:10. The BOD can then be calculated by: BOD = (I – F) D D = dilution as a fraction D = volume of bottle/(volume of bottle – volume of dilution water) BOD = (8 – 4) 10 = 40 mg/L

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For the BOD test to work microorganisms have to be present. Sometimes they are not naturally present in a sample so we have to add them. This is called “seeding” a sample If seed is added you may also be adding some BOD. We have to account for this in the BOD calculation: BOD = [(I – F) – (I’ – F’)(X/Y)]D Where: I’ = initial DO a bottle with only dilution water and seed F’ = final DO of bottle with only dilution water and seed X = amount of seeded dilution water in sample bottle, ml Y = amount of seeded dilution water in bottle with only seeded dilution water

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Example Calculate the BOD 5 of a sample under the following conditions. Seeded dilution water at 20 o C was saturated with DO initially. After 5 days a BOD bottle with only seeded dilution water had a DO of 8 mg/L. The sample was diluted 1:30 with seeded dilution water. The sample was saturated with DO at 20 o C initially. After five days the DO of the sample was 2 mg/L. Since a BOD bottle is 300 ml a 1:30 dilution would have 10 ml sample and 290 ml seeded dilution water. From the table, at 20 o C, DO sat = 9.07 mg/L BOD 5 = [(9.07 – 2) – (9.07 – 8)(290/300)] 30 = 174 mg/L

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If we do a mass balance on the BOD bottle: dz/dt = -r Where: z = dissolved oxygen necessary for the microorganisms to decompose the organic matter If r is first order: dz/dt = -k 1 t Separate the variables and integrate: z = z 0 e -k1 t Z is defined as the amount of oxygen still to be used by the microorganisms to degrade the waste. If we define y to be the amount of oxygen which already been used to degrade the waste: L = z + y L = ultimate demand for O 2

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So: z = L - y By substitution: L – y = z 0 e -k1 t But z o = L, so: y = L – Le -k1 t Or y = L(1-e -k1 t )

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y = L(1-e -k1 t ) Rearranging: (t/y) 1/3 = (1/(k 1 L) 1/3 ) + (k 1 2/3 /(6 L 1/3 )) t y = b + m x So: k 1 = 6 (slope/intercept) L = 1/(6 (slope)(intercept) 2 )

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Example From a BOD test we have the following data: Slope = 0.545 Intercept = 0.021 k 1 = 6(0.021/0.545) = 0.64 day -1 L = 1/[6(0.021)(0.545) 2 ] = 26.7 mg/L

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Nitrogenous Oxygen Demand

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Solids Total Solids Residue on evaporation at 103 o C TS = (W ds – W d )/V Where: W ds = weight of dish plus solids after evaporation W d = weight of dish alone V = volume of sample

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Total Solids can be divided into two fractions: Suspended Solids Dissolved Solids Dissolved solids are the solids that can pass through a glass fiber filter with a 0.45 micro pore size Suspended solids are the solids that can not pass through a glass fiber filter with a 0.45 micron pore opening

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Suspended solids SS = (F df – F d )/ V Where: F df = weight of the Filter plus dry filtered solids F d = weight of the clean, dry filter V = volume of sample

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Volatile and Fixed Solids Volatile solids are the solids that are volatilized at 600 o C Fixed solids are the solids that remain after heating to 600 o C Generally the volatile solids are considered to be the organic fraction of the solids. Volatile Solids = Total Solids – Fixed Solids

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Nitrogen Organic Nitrogen Ammonia Nitrates + Nitrites (NO 3 - , NO 2 - ) Organic Nitrogen + Ammonia = Kjeldahl Nitrogen

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Microbiological Measurements Waterborne Pathogens Salmonella (typhoid fever) Shigella (bacillary dysentery) Hepatitus virus Entameoba Histolytica (ameobic dysentery) Giardia Lamblia (“bever fever”) Cryptosporidium Indicator Organisms coliforms MPN test 