Hydrologic Processes Important for Contaminant Transport

Transcription

1 Hydrologic Processes Important for Contaminant Transport Quantity of Water available for transport- mass flow Degree of dispersion-variability in pore size distribution Greater is the variability, greater is the degree of dispersion and thus highly likely to appear earlier in ground water

2 Contaminant Traits Important for Water solubility Water Quality Adsorption/Desorption to and from soils Diffusive properties Degradation Volatilization Sinks and sources

3 Solubility of a Contaminant Most inorganic contaminants are readily soluble in water Most handbooks of Physics and Chemistry give solubility of various contaminants

4 Solubility of various Inorganic Compounds (g/100g of water) Substance 0 C 10 C 20 C 30 C g/100g H 2 O AlCl NH 4 NO CaSO 4. 2H 2 O HgCl

5 Retardation Factor or Coefficient R=1+(ρ b K d /θ) where R=Retardation factor, ρ b =bulk density, g cm - 3 ; K d =adsorption or distribution coefficient, cm 3 g - 1 ; θ =volumetric water content, cm 3 cm - 3. Higher the bulk density, higher is the retardation factor Higher the adsorption coefficient, higher is the retardation factor. Lower is the water content, higher is the retardation factor.

6 Distribution (Adsorption) Coefficient Adsorption Isotherm is the equilibrium relationship between adsorbed concentration (C( d ) and the dissolved concentration (C( l ). The slope of the line is called distribution coefficient. Linear adsorption isotherm, C d =K d C l Adsorbed Concentration, C d Linear Adsorption Isotherm Slope=K d Dissolved Concentration, C l

7 How do we get Adsorption Isotherm Adsorbed Concentration, C d K d1 Tightly adsorbed K d2 Lightly adsorbed Dissolved Concentration, C l

8 Calculation of Retardation factor Calculate the retardation factor of Atrazine (K d =1.6 cm 3 g -1 ) in a clay loam soil (ρ( b =1.3 g cm - 3 ) at θ=0.20 cm 3 cm -3. R=(1+(1.3 x 1.6/0.2)) =11.4 Units= g cm -3 x cm 3 g -1 /cm 3 cm Unitless cm -3

9 Types of Adsorption Coefficients Freundlich Isotherm C d =K f C l 1/n where n<1 Jury et al, Soil Physics, John Wiley & sons

10 Distribution Coefficients Compound Atrazine Bromacil DBCP DDT Lindane Phorate K d, cm 3 g

11 Distribution Coefficient, K d Higher is the value of K d, higher is the degree of adsorption i.e. less is available in solution phase and thus less likely to move through soil and contaminate the ground water.

12 Freundlich Isotherm = 1 n n 1 K f Cl 1 dc dc d l C d =K f C l 1/n R( c) = 1+ 1 n ρ C b θ 1 n l 1 dc dc d l At lower concentrations, the retardation coefficient is higher and at higher concentrations the retardation coefficient is lower. This means there is less likelihood of completely cleaning up a site.

13 Mass Flow T 1 T 2 T 3 T 4 T 5 T 6 T 7

14 Dispersion T 1 T 2 T 3 T 4 T 5 T 6 T 7

15 Mass Flow + Dispersion Red Piston Flow Real soil Blue Time or Pore Volume

16 Diffusion of Sugar in a Container Water Sugar Cube After a while all water in the container will be sugary. That is because sugar is diffusing from higher concentration near the cube to lower concentrations away due to random molecular motion.

17 Diffusion of Smoke in Room Room Air Smoke Bomb After a while the whole room will be filled with smoke. This is because smoke is moving (diffusing) from higher concentration near the smoke bomb to area of lower concentration away.

18 Random Molecular Motion- Diffusion

19 Diffusion T 1 T 2 T 3 T 4 T 5 T 6 T 7

20 Mass Flow + Dispersion+ Diffusion Red Piston Flow Advection + Dispersion + Diffusion Blue Time or Pore Volume Advection + Dispersion Diffusion is only important when water is not flowing

21 Diffusion in hot and Cold Water Hot Cold

22 Concepts of Advection and Dispersion

23 C/C Breakthrough Curves They refer to the relationship of leachate concentration from a soil column vs. time or pore volume. Piston Flow Real soil C/C Real soil Piston flow Time or Pore Volume Time or Pore Volume

24 What does the breakthrough curve tells us It tells us how fast the solute is moving towards the bottom of the soil column. How long will it take for a solute to appear at the bottom. Will a remediation strategy be effective in minimizing the transport of contaminant downward-laboratory experiment For the piston type of flow a contaminant appears at the bottom of the soil column in one pore volume In real soils, there is gradual appearance or disappearance of the contaminant from the bottom of the soil column

25 What is Pore Volume Pore volume refers to the combined volume of all the pores in a soil column of given dimension. A soil column 30 cm long, 5 cm diameter (volume=589 cm 3 ) and bulk density of 1.5 g/ cm 3 has pore volume of =porosity x vol.=(1-(1.5/2.65)) (1.5/2.65)) x 589 =256 cm 3. 1 PV=256 cm 3

26 Effect of Retardation Factor on Solute Breakthrough Higher is the retardation factor slower is the appearance of the contaminant at the water table. Higher is the retardation factor, longer it will take to flush all the chemical out of the soil.

27 Reasons for skewness in breakthrough curves (BTC) There are variety of pores Some very large, some very small Contaminant travels faster through large pores and slower through small pores Wider the distribution of pores, greater is the skewness in BTC At one pore volume half of the contaminant is out.

28 Animation of Dispersion

29 Dispersion of a Slug of Chemical

30 Contaminant Flux Refers to amount of contaminant transport per unit area per unit time, g cm -2 t -1. Processes contributing to contaminant flux Mass flow of water-advection Hydrodynamic dispersion Contaminant diffusion Sinks for a contaminant or sources of a contaminant

31 Mass Flow of Contaminants Advection Mass flow of contaminant is due to mass flow of water. Contaminant flux due to mass flow of water (q c ) = flux of water moving thru the soil (q) x the concentration of the contaminant in the moving water (C( l ) (mg/cm 2 /t q c =q x C l /t )= (cm 3 /cm 2 /t) x (mg/l)

32 Calculations of Mass Flow If 2 cm of water of infiltrates into the soil in 10 minutes, calculate the amount of chloride that will move into the soil with water if the chloride concentration of the infiltrating water is 2 mg/l. q c = 2 cm 3 cm -2 /10 minutes x 60 minutes h - 1 x 2 mg/l=12 cm 3 cm -2 h -1 x 2 mg L -1 x 1 L/ 1000 cm 3 =0.024 mg cm -2 h -1

33 Hydrodynamic Dispersion Soils are made up of various size pores. Water velocity through a pore depends upon its size This variation in pore velocity causes some contaminant to move faster and some contaminant to move slower. This variation in contaminant movement (due to variation in pore water velocity) is called hydrodynamic dispersion. Hydrodynamic dispersion is one of the major reasons for skewness in BTC.

34 Dispersion Cartoon Fette, C.W. r, Applied Hydrogeology

35 Contaminant Diffusion Chemicals move from higher concentration to lower concentration even when water is not moving. This movement is called contaminant diffusion. Heavier the contaminant molecule slower is the diffusion flux Bigger the concentration gradient greater is the diffusion flux Temperature, density, viscosity are other factors that affect contaminant diffusion

36 Sinks or Sources Plant uptake, nitrate are taken up by plant (sink) and microbes (sink or source) Precipitation (sink)- CaCO 3, CaSO 4 or dissolution (source)- CaCO 3, CaSO 4 Mineralization (source)-no 3 ; dentrification (sink)-no 3 ; volatilization (sink)-gasoline; degradation into daughter products (sink)- atrazine

37 Contaminant Flux Contaminant Flux= Mass Flux (Advection)+dispersion Flux+ Diffusion Flux+ Sink or source terms If dispersion and diffusion are absent and there are no sink or source terms then, contaminant transport is due to advection and more like piston flow. With dispersion and diffusion, there will be some early appearance of contaminant and it will take long time to displace all the contaminant from the soil.

38 Concepts of Advection and Dispersion

39 Piston Flow and Dispersion Effect 1.0 Piston Flow 1.0 Piston flow Real soil Real soil C/C C/C Time or Pore Volume Time or Pore Volume Relative Concentration of 0.5 is called the center of mass because half of the mass has moved out of the soil column and half is still in the soil column

40 Breakthrough Time (t( b ) Refers to the time when the center of mass of the contaminant front reaches a given depth. Center of mass-when 50% of the solute has left the soil and 50% is still remaining in the soil. If we know the velocity at which contaminant front is moving then we can calculate the breakthrough time for a given depth. t b =L/V; L=Depth, V= Contaminant front velocity.

41 Pore Water Velocity Pore water velocity refers to the velocity at which water front is moving through soil. Infiltration rate refers to the rate at which water is entry. Darcy s s flux refers to an equivalent infiltration rate at which water is moving through the soil. q=2 cm h -1 means 2 cm of water had moved past a given depth in one hour. It does not mean that water has moved two cm in the soil in one hour. This 2 cm of water has moved through both soil particles as well as through soil pores.

42 Pore Water Velocity Since this refers to velocity of water front in soil, we need to know how much of the volume is occupied by pores. Pore water velocity can be calculated from infiltration rate or Darcy s s flux. Pore water velocity= infiltration rate/water filled porosity Pore water velocity=darcy s s flux/water filled porosity

43 Pore Water Velocity What is the pore water velocity (V) given Darcy s flux in a saturated soil is 5 cm h -1? Saturated water content of the soil=0.50 cm 3 cm -3. V= q/θ s =5 cm h -1 /0.5 cm 3 cm -3 =10 cm h -1 This means although water flux was 5 cm h -1, the wetting front moved 10 cm in one hour because water was moving only through the pore spaces (50%) and not through the soil particles.

44 Pore Water Velocity What is the pore water velocity (V) given Darcy s s flux in a unsaturated soil is 2 cm h -1? Unsaturated water content of the soil=0.25 cm 3 cm -3 (water filled porosity). V= q/θ=2 cm h -1 /0.25 cm 3 cm -3 =8 cm h -1 This means although water flux was 2 cm h -1, the wetting front moved 8 cm in one hour because water was moving only through the 25% of the bulk volume. Remaining 25% of the bulk volume was empty and water did not move through 50% of the bulk volume occupied by soil particles.

45 Examples of Breakthrough Time How long will it take for a non-reactive contaminant to show up at 100 cm depth if the Darcy s s flux is 5 cm h -1 and the soil was saturated, θ s =0.50 cm 3 cm -3. V= q/θ s =5 cm h -1 /0.5 cm 3 cm -3 =10 cm h -1 t b =L/V=100 cm/10 cm h -1 =10 hours

46 Examples of Breakthrough Time How long will it take for a reactive contaminant (R=11.4) to show up at 100 cm depth if the Darcy s s flux is 5 cm h -1 the soil was saturated, θ s =0.50 cm 3 cm -3. V= q/θ s =5 cm h -1 /0.5 cm 3 cm -3 =10 cm h -1 V R =V/R=10 cm h -1 /11.4=0.877 cm h -1 t b =L/ V R =100 cm/0.877 cm h -1 = 114 hours

47 Examples of Distance to Travel Calculate the time it will take for a non-reactive solute to reach water table at 10 m depth if the Darcy s s flux in an unsaturated soil (θ=0.25( cm 3 cm -3 ) was 2 cm h -1. V= q/θ=2 cm h -1 /0.25 cm 3 cm -3 =8 cm h -1 V=Distance/time; Time=Distance/velocity Time= 10 m x 100 cm m -1 /8 cm h -1 =125 hours

48 Examples of Time to Breakthrough Calculate the time it will take for a reactive contaminant (R=11.4) to reach water table at 10 m depth if the Darcy s s flux in an unsaturated soil (θ=0.25 cm 3 cm -3 ) was 2 cm h -1. V= q/θ=2 cm h -1 /0.25 cm 3 cm -3 =8 cm h -1 V R =V/R=8 cm h -1 /11.4=0.70 cm h -1 V R =Distance/time; Time=Distance/velocity Time= 10 m x 100 cm m -1 /0.7 cm h -1 = hours

49 Examples of Distance to Travel Calculate the distance travelled by a non- reactive solute in six days if the Darcy s s flux in an unsaturated soil (θ=0.25( cm 3 cm -3 ) was 2 cm h q cmh 1 V 8 x = = = cmh cm cm θ s L = V x. t = 8cmh 1 x 6days x24h = 1152cm = 11.52m

50 Examples of Distance Travelled Calculate the distance travelled by a reactive contaminant (R=11.4) in 60 days if the Darcy s s flux in an unsaturated soil (θ=0.25 cm 3 cm -3 ) was 2 cm h -1. V x q = = h θ 0.25cm cm s 2 1 cmh 1 = 8cm 3 3 L = V R. t = Vxt R = 8cmh x60days x 24hr = 1010cm = 10.1m

51 Effect of Flow Velocity Plume movement at low ground water velocity Source Xylene Benzene MTBE (Methyl tertiary-butyl ether) Plume movement represents high ground water velocity MTBE is gasoline additive, highly soluble

52 Convective-dispersive Flow Pulse Condition C = C 2 0 erfc L 2 v t x D t L + exp v L x D L L + v t x erfc 2 D t L C=Solute concentration at any time mg/l C 0 =Initial solute concentration, mg/l L=Length of the flow path, m V x =Average linear velocity, m/day T=Time, day D L =longitudinal dispersion coefficient Fetter, applied hydrogeology, 2001 Prentice Hall

53 Solution of Convective- dispersive Solute Flow Spreading of a solute slug due to convection and dispersion. With time the peak concentration is less but it spreads over much bigger distance. This is also the solution of previous equation. Fetter, 2001 Applied Hydrogeology, Prentice Hall

54 Spread of a Contaminant Development of a plume From a continuous source. Dot density indicates solute concentration. Travel of a contaminant Slug from a one-time point source Fetter, Applied Hydrogeology. Prentice Hall

55 Computations A landfill is leaking leachate with chloride concentration of 725 mg/l which enters an aquifer whose properties are as follows: Hydraulic conductivity=3.0x10-3 cm/s dh/l=0.002, Porosity=0.23 D*=1x10-9 m 2 /s Compute Cl concentration in 1 y at a distance of 15 m.

56 Computations I Flux=3.0x10-5 x0.002=6.0x10-8 Pore water velocity (V( x )=6.0 x10-8/0.23=2.6 x10-7 D L =a L x V x +D*; a L =0.83 (log L) a L =0.83 (log 15m) =1.23 m D L =(1.23 x 2.6x10-7 )+1x10-9 m 2 /s=3.2x year=1y x 60s/min x 1440 min/day x 365 days/y=3.15 x 10 7 s Fetter, Applied Hydrogeology, Prentice Hall

57 Fetter, Applied Hydrogeology. Fetter, Applied Hydrogeology. Prentice Hall Prentice Hall Computations II Computations II + + = exp x x x x x x erfc x x x x x x x x x erfc C + + = t D t v L erfc D L v t D t v L erfc C C L x L x L x 2 exp 2 2 0

58 Error Function and Error Function Complementary all%20213/table%20of%20error%20func tion%20values.pdf dsp.elet.polimi.it/fondstiol/comperrfnc.pdf erfc (x)=1-erf(x) and erfc(-x)=1+erf(x)

59 Computations III C = erfc + exp 12.9 erfc 6.3 ( ) [ erfc( 1.08) exp12.9 ( ) ( 3.68) ] C = erfc Zero C = 362.5x0.127= 46mg/ L Fetter, Applied Hydrogeology. Prientice Hall

60 Summary of Time of Travel or Distance to Travel Find the Flux, cm 3 /cm 2 /t Find pore water velocity (V)=Flux/water filled porosity Velocity of retarding fluid (V R )= V/R Time to breakthrough in piston transport=length of the column or depth to water table/v R Distance a chemical has traveled in a given time=v R x t

61 Summary on Flux Long term flux (D) =P+I-ET ET-RO because we assume ΔS S is zero Short term flux=infiltration rate=p+i-ro Short term flux through soil, you use Darcy s s law=q=k s or K unsat (ΔH/L) For groundwater aquifers, you calculate flux using Darcy s s law=q=k s (ΔH/L)

62 Summary on Concentration Prediction Relative Concentration (C/C0) of non-retarding contaminant in soils at 1 PV will be 0.5 (center of mass) Relative concentration of retarding contaminant in soils at R x PV will be 0.5 (center of mass) Concentration of a contaminant in piston transport is same as we added (C 0 ) at 1 PV. Concentration of a contaminant in ground water under pulse conditions is given by long equation.

63 Chemical and Biological Degradation Many contaminants degrade over time due to chemical or biological reactions. Most often their degradation can be described by an exponential function: M t = M 0 e -μt where M 0 is the mass of contaminant at time zero, M t is the mass at time t, μ is the rate of degradation, t is time.

64 Example of Degradation Calculate the mass of acetone remaining in soil after 24 hours if the degradation rate is hr -1. Initial mass=100 g M t = M 0 e -μt x 24 M t =100 e =50 g In other words, ½ of the Acetone is degraded in 24 hours. Therefore, 24 hours will also be the half-life life of acetone in this soil.

65 First Order Degradation for Various Half-lives lives Half-Life, days m, d

66 Breakthrough Curves with First Order Decay One effect of decay is that it removes the long tail of the BTC. Also, the total mass showing up at the bottom is also less.

67 K d and Half-Life of Common Pesticides Compounds K d, cm 3 g -1 τ 1/2, days Atrazine Bromacil DBCP DDT Lindane Phorate

68 Degradation and Half-Life Higher is the degradation rate shorter is the half-life. life. Atrazine has higher degradation rate (smaller half-life) life) than DDT which has lower degradation rate (higher half life).

69 Example of Contaminant Higher the degradation rate, less is the total amount of contaminant that will appear in groundwater (area under the curve). Higher is the degradation rate less time it will take to displace all contaminant out of the soil (See the time at μ=0 rate). Degradation

70 Chemical Decay in Piston Flow Model In piston flow, there is no dispersion or diffusion The contaminant moves with a velocity V=J w /θ and reaches a distance L in t b =L/V. If chemical was added as a front of concentration C 0 at t=0 it will have a concentration C 0 e -μt b =C 0 e -μl/v. If the contaminant was added as a pulse of mass then the mass that passes z=l will be M 0 e - μl/v.

71 Piston Flow Breakthrough Curve with Degradation 1.0 No degradation μ=0.02 Slight degradation C/C0 0.5 μ=0.04 Moderate degradation μ=0.06 High degradation 1.0 Pore Volume

72 Preferential Flow Preferential flow leads to by passing the filtering capacity of the soil. Preferential flow is caused by: Presence of macropores Soil heterogeneities Differences in properties of resident and invading fluids

73 Macropore Flow Macropore flow results due to the presence of earthworm macropores,, cracks between pedons,, and root channels (tap root system) Earthworm macropores could be mm in diameter. More prevalent in no-till or manured soils. Generally effective when they are open at the soil surface or open into water table or groundwater.

74 Examples of BTC with Preferential Flow Preferential flow leads to early appearance of contaminant in the groundwater. Preferential flow also leads to deeper penetration of the contaminant into the soil profile. Jury et al. Soil Physics

75 Q 4 πr ρ g L + d w = 8 Flow η L Through a Tube-Poiseuille Poiseuille s Law Q = π r 4 8 ρ ν w g L + L d Where Q= water flux, cm 3 /t; r=tube radius (cm); L is the length of the tube (cm); d=depth of ponded water (cm), ρ w =density of water (1.0 g cm -3 ); Water viscosity (0.01 g cm -1 s -1 ); Acceleration due to gravity (981 cm/sec/sec).

76 Macropore and Root Channel Earthworm Channel Root Channel & Soil Discontinuities

77 Type of Earthworms Lumricus rubellus Lumbricus terrestris Aprrectodea tuberculata

78

79 Poiseuille Poiseuille Law Law + = L d L g r Q w η ρ π = L d L g r Q w η ρ π = L d L g r Q w η ρ π = + = = L d L g r r L d L g r A Q q w w η ρ π η ρ π L r d

80 Calculations on Flow Through a Q = Worm Hole π 4 r ρ g w 8η L + L Calculate the breakthrough time through a worm hole that is 1 mm in diameter, 100 cm long and under 10 cm of pressure head. Q=[ x (0.05) 4 x 981 x ( )]/[8 x 100 x 0.01]= cm 3 s -1 Water velocity (V( h ) through worm hole=q/πr 2 =0.265/[ x (0.05) 2 ]=33.7 cm s -1 t b =L/ V h =100/33.7=2.97 s d

81 Volatilization of Contaminants Some organic chemicals move in vapor phase in addition to liquid phase. The quantity of contaminant that moves in vapor phase depends upon its ability to partition into the vapor phase from its solution phase. The relationship describing a contaminant s vapor phase concentration (C g ) to its liquid phase concentration (C( l ) is called Henry s law. C g =K H C l where K H =Henry s s constant

82 K d, Half-life, life, and Henry s Constant for Common Pesticides Compound Atrazine Bromacil DBCP DDT Lindane Phorate K d, cm 3 g -1 τ 1/2, d K H, cm cm 3 of water cm -3 of air 2.5 x x x x x x 10-4

83 Henry s s Constant, K H Higher is the value of Henry s s constant, greater is the ability of the contaminant to convert to vapor phase. For the same solution concentration, DBCP (K H =8.3 x 10-3 ) will have higher concentration in vapor phase followed by Lindane (1.3 x 10-4 ) and then Atrazine (2.5 x 10-7 ).

84 Volatilization of Contaminants Some chemicals have tendency to volatilize. These chemicals can move out of the soil in vapor phase. The process controlling their movement in vapor phase is diffusion. Volatilization Flux=-D a g [C [C soil surface -C air ]/d D a g =Diffusion coefficient of the gas in open air; C soil surface =concentration of gas at the soil surface; C air =concentration of gas in open atmosphere; d=stagnant layer depth.

85 Volatilization of Contaminants

86 Contaminant Flux in Soil in Vapor Phase Flux=-D s g [C 1 -C 2 ]/L where D S g is diffusion coefficient of a contaminant in soil Lets say D S g =0.5 cm 2 /day Then flux=0.5 (1-0.5)/1=0.25 mg/cm 2 /day The units are: (cm 2 /day)x (mg/cm 3 )/(cm) =mg/cm 2 /day Soil surface A 1 cm B C 1 =1 mg/cm 3 C 2 =0.5 mg/cm 3

87 Indices of Pesticide Movement Pesticide movement depends upon all three factors Its ability to adsorb on the soil particles Its ability to degrade Its ability to volatilize Higher is the ability of a pesticide to adsorb, degrade and volatilize, lower will be its amount appearing in the groundwater. Based on these three factors, several different indices have been developed to characterize a pesticide s s ability to move in soils

88 Type of Breakthrough Curves For anion there may be early appearance of the contaminant at the bottom of the column. This is is due to anion exclusion. Clay are negatively charged and thus repel the anions making them appear earlier.

89 Breakthrough Curves This refers to a relationship between the contaminant concentration that is coming out at the bottom of a soil column vs. time or the quantity of water applied. The shape of the curve tells us how a contaminant will behave in a given soil.

90 Effect of Dispersion on the Breakthrough Curve Skewness in the breakthrough curves reflects the degree of dispersion. Higher is the dispersion coefficient, more skewness in the breakthrough curve. If no dispersion is present, solute transport is like a piston flow.

91 What is Breakthrough Curve It refers to the relationship of leachate concentration vs. time or pore volume.