AOS 104 Fundamentals of Air and Water Pollution

Transcription

1 AOS 104 Fundamental of Air and Water Pollution Dr. Jeffrey Lew AIM: jklew888 MS Grade Homework 150 pt 2 Mierm 300 pt Final Exam Total 550 pt 1000 pt 2 Homework There will be 6 homework et Homework i due by the end of buine (~ 5 pm) on the due date Late homework will receive partial credit a outlined in the yllabu You are encouraged to work together and dicu approache to olving problem, but mut turn in your own work 3

2 Lecture Note 4 Topic of the coure Calculating pollution concentration Type of water pollution Health effect from water pollution Type of air pollution Health effect of air pollution Urban air pollution bad ozone Effect of pollution on acidity (ph) acid rain Stratopheric air pollution depletion of good ozone Global climate change 5 Introduction Meaurement of Concentration Liquid Air Converion Involving the Ideal Ga Law Material Balance Model Baic Steady-tate Model With Conervative Pollutant Reidence Time Steady and Non-teady Model With Nonconervative Pollutant 6

3 Unit of Meaurement Both SI and Britih unit ued Be able to convert between thee two tandard Example Quantity SI Britih Length m ft Volume m 3 ft 3 Power watt BTU/hr Denity kg/m 3 lb/ft 3 7 Concentration The amount of a pecified ubtance in a unit amount of another ubtance Uually, the amount of a ubtance diolved in water or mixed with the atmophere Can be expreed a... Ma/Ma g/kg, lb/ton, ppmm, ppbm Ma/Volume g/l, µg/m 3 Volume/Volume Volume/Ma Mole/Volume ml/l, ppmv, ppbv L/kg molarity, M, mol/l 8 Liquid Concentration of ubtance diolved in water are generally given a ma per unit volume. e.g., milligram/liter (mg/l) or microgram/ liter (µg/l) Concentration may alo be expreed a a ma ratio, for example: 7 ma unit of ubtance A per million ma unit of ubtance B i 7 ppmm. 9

4 Example 1 23 µg of odium bicarbonate are added to 3 liter of water. What i the concentration in µg/l and in ppb (part per billion) and in mole/l? 23 µg 3 L = 7.7 µg L To find the concentration in ppb we need the weight of the water. 10 Standard aumption: denity of water i 1g/ml or 1000 g/l (at 4 C) g 1 L 1 L 1000 g = = = 7.7 = 7.7 ppbm 9 billion For denity of water i 1000 g/l, 1 μg/l = 1 ppbm 1 mg/l = 1 ppmm 11 Sometime, liquid concentration are expreed a mole/l (M). e.g., concentration of odium bicarbonate (NaHCO 3 ) i 7.7 µg/l g Molecular Weight = [ (3 16) ] mol = 84 g mol Molar Concentration = 7.7µg 1 L 1 mol 84 g 1 g µg = mol L 12

5 Air Gaeou pollutant ue volume ratio: ppmv, ppbv Or, ma/volume concentration ue m 3 for volume Example 2 A car i running in a cloed garage. Over 3 minute, it expel 85 L of CO. The garage i 6 m 5 m 4 m. What i the reulting concentration of CO? Aume that the temperature in the room i 25 C. 13 Solution: The volume of CO i 85 L and Volume of room = 6 m 5 m 4 m = 120 m 3 = L Concentration = 85 L L = 708 ppmv = = Example 2 Cont. Intead of 85 L of CO, let ay 3 mole of CO were emitted. We need to find the volume occupied by three mole of CO. IDEAL GAS LAW: PV= nrt P = Preure (atm) V = Volume (L) n = Number of mole R = Ideal Ga Contant = L atm K 1 mol 1 T = Temperature (K) 15

6 To remember the unit it All Lover Mut Ki (atm, L, mol, K) Thi law tell you how a ga repond to a change in it phyical condition. Change P, V or T, and the other adjut 16 The Ideal Ga Law PV = nrt o P 1 V 1 = nrt 1 P 2 V 2 = nrt 2 etc. Can rearrange to get the equality, P 1 V 1 T 1 = nr = P 2 V 2 T 2 17 The ideal ga law alo tell you that: at 0 C (273 K) and 1 atm (STP), 1 mole occupie 22.4 L at 25 C (298 K) (about room temperature) and 1 atm, 1 mole occupie 24.5 L 18

7 Back to Example 2 Volume of CO = ( 3 mol) 24.5 L mol = 73.5 L Volume of room = 6 m 5 m 4 m Concentration = = 120 m 3 = L 73.5 L L = = = 613 ppmv 19 Material Balance Expree Law of Conervation of Ma Material balance can be applied to many ytem organic, inorganic, teady-tate, financial, etc. 20 Baic equation of material balance Input = Output Decay + Accumulation (eq. 1.11) Input, output, etc. are uually given a rate, but may alo be quantitie (i.e. mae). Thi equation may be written for the overall ytem, or a erie of equation may be written for each component and the equation olved imultaneouly. Steady-tate (or equilibrium), conervative ytem are the implet Accumulation rate = 0, decay rate = 0 21

8 Example 3 Problem 1.7 Agricultural dicharge containing 2000 mg/l of alt i releaed into a river that already ha 400 ppm of alt. A town downtream need water with <500 ppm of alt to drink. How much clean water do they need to add? Maximum recommended level of alt for drinking water = 500 ppm. Brackih water have > 1500 ppm alt. Saline water have > 5000 ppm alt. Sea water ha 30,000 34,000 ppm alt m 3 / 400 ppm Q m 3 / C ppm S+R m 3 / 500 ppm S m 3 / C ppm R m 3 / 0 ppm 5.0 m 3 / 2000 mg/l What i the concentration in flow Q? What flow of clean water (R) mut be mixed to achieve 500 ppm? (What i the ratio of R/S?) 23 To implify, we can break thi into two ytem firt (remembering that 1 ppmm = 1 mg/l) 25.0 m 3 / 400 ppm Q m 3 / x ppm 5.0 m 3 / 2000 mg/l Baic equation: input = output ( 25.0 m3 400 ppmm) m ppmm Q = Total flow = 25.0 m3 ( ) = ( Q m3 C ppmm) m3 m3 = 30.0 Thu, C = the alt concentration at the takeout from the river = 667 ppmm 24

9 Solve the other part of the ytem: S+R m 3 / 500 ppm S m 3 / 667 ppm R m 3 / 0 ppm S m3 ( ) m3 667 ppmm + R m3 0 ppmm = S + R 500 ppmm 667S = 500S + 500R R S = = Reidence Time Lifetime or reidence time of ubtance amount / rate of conumption Lifetime of Earth petroleum reource: 1.0 x J / 1.35 x J/yr = 74 year ANWR ha ~5.7 to barrel of oil; bet gue i The US conume 19 million barrel/day of oil. How much time doe thi give u? 26 The reidence time may be defined for a ytem in teady-tate a: Stock (material in ytem) Flow rate (in or out) Reidence time in a lake: The average time water pend in the lake Some water may pend year in the lake Some may flow through in a few day Depend on mixing. 27

10 In the firt approximation, conider only tream flow in and tream flow out. T = M/Fin = M/Fout For thi very imple teady tate ytem, we calculate the reidence time Ex. The volume of a lake fed by a tream flowing at m 3 /day i m 3. What i the reidence time of the water in the lake? m 3 = 430 day m day 28 More Material Balance What if a ubtance i removed by chemical, biological or nuclear procee? The material i till in teady-tate if it concentration i not changing. Steady-tate for a non-conervative pollutant: We now need to include the decay rate in our material balance expreion: Input rate = Output rate + Decay rate 29 Aume decay i proportional to concentration ( 1 t order decay ). dc = kc where k = reaction rate coefficient, in unit of 1/time. C = concentration of pollutant Separate variable and integrate: C dc t = k C 0 C 0 30

11 Solution: ln( C ) ln( C 0 ) = kt kt 0 ln( C ) ln ( C ) = ln C 0 C 0 Take the exponential of each ide: C = C 0 e kt = kt For a particular ytem (i.e., a lake), we can write a total ma decay rate (ma/time), that we can compare with the input and output rate: k ha unit of 1/time C ha unit of ma/volume V ha unit of volume = kcv ma removal rate Thu the decay rate = kcv (ma/time) 31 Material Balance Equation Steady tate with decay Input rate = Output rate + kcv Example 4 A lake with a contant volume of 10 7 m 3 i fed by a clean tream at a flow of 50 m 3 /. A factory dump 5.0 m 3 / of a non-conervative pollutant with a concentration of 100 mg/l into the lake. The pollutant ha a reaction (decay) rate coefficient of 0.4/day (= ). Find the teady-tate concentration of the pollutant in the lake. 32 Start by drawing a diagram of the problem tatement. 5 m 3 / 100 mg/l 50 m 3 / 0 mg/l 10 7 m 3??? K= 0.4 /day = Solution: Aume the volume of the lake i contant, o the outflow i equal to the inflow, or Water outflow rate = 55 m 3 / For the pollutant: Input rate = Output rate + Decay rate 33

12 Variation Non-teady ituation Conider a lake that initially had zero concentration of the pollutant, and then a pollutant wa introduced. How i the concentration changing with time? (i.e., a tranient phenomenon not teady-tate) Step function repone Ma balance: Accumulation rate = Input rate Output rate Decay rate V dc = S QC kcv 34 Eventually ytem reache a teady-tate concentration, C( ) (i.e., when dc/ = 0) C ( ) = C = S Q + kv Concentration a a function of time (before teady-tate i reached) i given by the tranient equation: dc = S V QC V kc 35 which can be rearranged to give: dc = k + Q C S Q + kv So we can ubtitute for C : dc = k + Q C C [ ] To integrate, we imply the C C term: y = C C dy = dc 36

13 dy = k + Q y a familiar, eparable differential equation (k+q/v i a contant!), with a olution of the form: y = y 0 e k + Q t Subtituting and rearranging, C ( t ) = C + C 0 C where y 0 = C o C ( )exp k + Q t At t = 0, exp = 1 t =, exp = 0 37 C ( t ) = C + ( C 0 C )exp k + Q V t What i the general behavior of thi equation? At time = 0, the exponential term goe to 1 o C = C(0) At time =, exp goe to 0 C = C Concentration C C0 Time 38 Example 5 Bar with volume of 500 m 3 Freh air enter at a rate of 1000 m 3 /hr Bar i clean when it open at 5 PM Formaldehyde i emitted at 140 mg/hr after 5 PM by moker k = the formaldehyde removal rate coeff. = 0.40/hr What i the concentration at 6 PM? ( ) = C ( 0) C C t ( )exp k + Q t + C Solution Firt we need C 39

14 Q = 1000 m 3 /hr; V = 500 m 3 ; S=140 mg/hr; k = 0.40 /hr S C = Q + kv = mg/hr m 3 /hr + 0.4/hr 500 m 3 C = mg/m 3 ( ) For the concentration at 6 PM, one hour after the bar open, ubtitute known value into: C ( t ) = ( C 0 C )exp k + Q t + C C ( t ) = mg m 3 exp m 3 hr + hr 500.0m 3 t C ( t ) = mg 1 e 2.4t C 1 hr m 3 ( ) ( ) = mg ( ) = mg/m 3 m 3 1 e mg m m 3 / 100 mg/l 50 m 3 / 0 mg/l 10 7 m 3 k = 0.4 /day 1 m 3 / 42 mg/l Pollutant B??? New factory open, dicharging pollutant B. If k i 0.4/day, what i the concentration after 2 day of dicharge? S = 1 m L mg m day 42 L = mg day Q = ( ) m L L m day = day 41 V = 10 7 m 3 Q = L/day S = mg/day k = 0.4/day C = C ( t ) = C + C 0 C ( )exp k + Q t C = S Q + kv 42

15 Firt tep: C = S Q + KV = mg day L day day 1010 L = mg/l 43 Now ue tep function to find concentration at 2 day C ( t ) = C + C 0 C C0 = 0; C t ( )exp k + Q t ( ) = 1 exp k + Q V = 1 exp = ( 0.829)0.41 mg L day t C m day m 3 2 day 0.41 mg L = 0.34 mg/l 44