1. Applications Water Quality

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

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

4. 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

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

6. 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) DD = dilution as a fraction D = volume of bottle/(volume of bottle – volume of dilution water) BOD = (8 – 4) 10 = 40 mg/L

7. 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

8. 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

9. 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 + yL = ultimate demand for O 2

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

11. 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 )

12. 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

13. Nitrogenous Oxygen Demand

14. 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

15. 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

16. 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

17. 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

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

19. 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