Overview, Chemistry Review, Kinetic Reaction RatesLecture No. 21. OverviewA. UsesB. QuantitiesC. Sources of Supply/PollutantsD. Treatment Plants/Pollutants2. Chemistry Review (water at standard conditions)A. ConversionsB. Chemical Review1.) Dilutions2.) Solute and solvent3.) Concentration in physical units4.) Concentration in chemical units5.) Reactions involving standard solutions6.) Formulas(see references nos. 1,2 and 3)C. Water Demand3. Kinetic Reaction RatesA. GeneralB. Zero Order ReactionC. 1st Order ReactionD. Second Order ReactionE. Half Lives, t1/2F. SummaryG. Determination of OrderH. Thermodynamics4. Reaction SummaryA. StoichiometryB. KineticsC. ThermodynamicsOverview, Chemistry Review, Kinetic Reaction RatesLecture No. 21. OverviewA. Uses-drinking and culinary (potable water)-cleaning, laundering, bathing-gardening, filling swimming pools-industrial purposes-etc.B. Quantities-average day = 150gpcd @ 40-60psi-maximum day = 2-3 times average day-hydraulic design (distribution system) = maximum day + fire flow demand-treatment plant design = maximum dayC. Sources of Supply/Pollutants-rainwater/pollen, particulate matter, gases-surface water/minerals, salts, organics, biologicals-groundwater/soluble material-man adds/treated and untreated wastes, body contaminants, oil spills etc.D. Treatment Plants/Pollutants-disinfection only-filtration/color, turbidity-iron and manganese removal-softening/hardness removalOverview, Chemistry Review, Kinetic Reaction Rates, Page No.22. Chemistry Review (water at standard conditions)A. Conversions-specific gravity = 1-specific weight = 62.4lbs/ft3(function of temperature)-1 = 7.48gal/ft3-1 = 8.34lb/gal-1 = 7000grains/1lb.-lb/day = Xmg/l x 8.34(lb/MG)/(mg/l) x Q(MGD)-1 = 1%/10,000mg/l-1 = 325,829gal/acre-ft-1 = 43,560ft2/1acre-1 = 17.1(mg/l)/(grain/gal)Typical Treatment Plant Flow ScenariosFiltration(surface waters)Iron and Manganese Removal(ground water)SofteningrapidmixflocculationsettlingfiltrationdisinfectionstoragesystemsystemsystemaerationsettlingfiltrationdisinfectiondisinfectionstoragestoragerapidmixflocculationsettlingrecarbonationOverview, Chemistry Review, Kinetic Reaction Rates, Page No.3-1 = 1cfs/.646MGD-1 = 1cfs/448.8gpm-1 = MGD/694.4gpm-ExampleGiven: A 50MGD filtration plant chlorinates the finished water at the rate of 5mg/lFind: How many 1-ton cylinders chlorine cylinders must be ordered each month?lb/day = mg/l x 8.34(lb/MG)/(mg/l) x Q(MGD)lb/day = 5mg/l x 8.34(lb/MG)/(mg/l) x 50MGDlb/day = 2085usage per month = 2085lb/day x 365days/12monthsusage per month = 63,419lbs/month1-ton cylinders = 63,419lbs/month x 1cylinder/1tons x 1ton/2000lbs= 31.7cylinders use 32NOTE: Since the requirement for chlorine is relatively great, larger scale deliveries such as railroad tank cars may be in order.Given: A town of 28,000 uses water at the rate of 150gpcd.Find: Annual consumption in acre-ft = 28,000cap x 150gal/(capxday) x365days/year x 1acre-ft/325,829gal = 4705acre-ft/yearGiven: 1 acre-footFind: Show that the conversion factor to gallons is accurate.1 acre-ft x 43,560ft2/1acre = 43,560ft3 43,560ft3 x = 325,829 gallonsThis number is often round up to 326,000, depending on the accuracy required.Given: A circular tank holds 1MG and is 36’high.Find: The diameter.V = 1,000, 000 gallons x V = 133,690ft3 The volume also equals the volume of a cyclinder:V = D2 x h = D2 x 36ft = 133,690ft3 D = 68.78ftNote: In the environmental engineering business, tank size in gallons and cubic feet are interchangeable.Given: The Weymouth Water Treatment Plant is 600MGDFind: What is this flow in cfs and gpm1.)cfs600 MGD x1cfs/.646MGD928.79cfs2.)gpm600MGD x 694.4 gpm/MGD416,640gpmOverview, Chemistry Review, Kinetic Reaction Rates, Page No.4B. Chemical Review1.) Dilutions- Dilutions are made by combining two solutions, perhaps a sample and dilution water, the latter being reasonably pure water. A 3:1 dilution refers to 3 total parts, the number to the left of the colon, of which 1 part, the number to the right of the colon, is the sample that is being diluted. A 3:1 solution, therefore consists of 2 parts water and 1 part sample which add up to 3 total parts.- Dilutions are often performed because the full strength sample is off-scale meaning that it is so strong that it can not be measured with the technique or device being used. In the above case, the final answer would be multiplied by 3 in order to achieve the correct results.- Example. Given:A iron sample was prepared and placed in the spectrophotometer and proved to be off -scale at which point the original liquid was diluted 10:1. A value of 3.6 mg/l was subsequently read from the spectrophotometer.Find: 1.) How was the dilution made 2.) How much iron is in the sample.1.) The dilution was made by taking a 100ml graduated cylinder and pouring 10 ml of the iron sample in it. Then the cylinder was filled to the 100 ml mark by adding 90 ml of dilution (uncontaminated) water and the two quantities were thoroughly mixed. Any 10:1 may be used, e.g. 1 ml and 9 ml or 10,000ml and 90,000ml. Consider the sample needed to do the testing the difficulty of mixing.2.) The final answer is the spectrophotometer reading times the dilution factor, in this case, 3.6mg/l x 10 = 36.0 mg/l.2.) Solute and solvent- solute-dissolved substance(sugar)- solvent-substance in which the solute is dissolved(coffee)- solution-solute plus solvent3.) Concentration in physical units- weight of solute per unit volume of solution(mg/l), e.g. 15.8g of NaCl per liter of solution- % composition, e.g. 3%H2O2 is 100g of solution, that is, 3g H2O2 + 97gH2O = 100g-weight of solute per weight of solution(ppm) e.g. 97g of KCl in 100g of water4.) Concentration in chemical units- Molarity-The number of moles of the solute contained in 1 liter of solution. Note: A mole of atoms of any element is that amount ofOverview, Chemistry Review, Kinetic Reaction Rates, Page No.5substance containing the same number of atoms as exactly 12g of pure carbon-12, Avogadro's number, 6.023 x 1023 .The mass in grams of a mole of atoms of any element is equal to its atomic weight, e.g. .5 molar(.5M) solution of H2SO4 contains 49.03g of H2SO4 per liter of solution.H 1 x 2 = 2.0S 32.064 x 1 = 32.06O 16.0 x 4 = 64.0 98.06 = 1mole-g.5M = .5(98.06) = 49.03Morality = number of moles of solute/number of liters of solution- Normality-The number of gram equivalent weights of solute containedin 1 liter of solution. Note: Equivalent weight is that