Sub-topics Chemical characterization Sorption-Desorption - PowerPoint PPT Presentation

Sub-topics • Chemical characterization Sorption-Desorption Characteristics Determination of k d (The Distribution coefficient) • Thermal Characterization • Electrical Characterization

    2 K 1-D ADE C C C C    ρ d D v . . s.     i. dry η 2 t z t z       2 ρ C D C v C K i       dry s d   R 1    2   η t R z R z R = Retardation factor

The coefficient k d Also known as the partition(ing) or distribution coefficient Is a measure of sorption of contaminants to soils/rocks/admixtures (geomaterials). Defined as the ratio of the quantity of the sorbate sorbed per unit mass of solids (C s ) to the amount of the sorbate remaining in solution (C w ), at equilibrium. The reverse is true for desorption (leaching) process K d measurement, some issues Experimental conditions Measurement methodology Contaminant chemical characteristics Sorbents (particle size, geochemistry) Type (active/passive) and concentration of the sorbate

Partitioning Relationships C mg / Kg Solid   S K • Solid ↔ water d C mg / L Water W C s : Sorbate sorbed per unit mass of solids C w : Sorbate remaining in the solution, at equilibrium. 3 • Water ↔ vapor C mol / m air   g Henry' s law constant (H) 3 C mg / m water w Contaminant Concentration in geomaterials Total mass in unit volume of geomaterial : Volumetric water content θ w  .C s + θ w . C W + θ g . C g C T = : Volumetric vapour content θ g b θ = Saturation  Porosity (  ) If soil is saturated, θ g = 0 and θ w =   = bulk density of porous medium   C s +  C w b C T b

Determination of Sorption and Desoprtion Characteristics of Geomaterials Fail to simulate “ Geomaterial-Contaminant- 1. Batch Tests immobilizing agent Interaction ” in a realistic manner Fail to come up with recommendations regarding Generalized Isotherm Quite Time and Cost Intensive 2. Column Tests “ Low Hydraulic Conductivity ” Accelerated Physical Modeling Using a Geotechnical Centrifuge seems to be a viable option In situ field batch tests Field modelling tests k oc method k oc = k d .(100/OC) k oc = organic carbon normalized adsorption coefficient OC = percentage of organic carbon in the sample (g/g)

Available Methodologies ASTM (American Society of Testing and Materials). 1987. “ 24-hour Batch- Type Measurement of Contaminant Sorption by Soils and Sediments. ” In Annual Book of ASTM Standards, Water and Environmental Technology, Volume 11.04, pp. 163-167, Philadelphia, Pennsylvania. ASTM 1988. “Determining a Sorption Constant (k oc ) for an Organic Chemical in Soil and Sediments. ” In Annual Book of ASTM Standards, Water and Environmental Technology. Volume 11.04, pp. 731-737, Philadelphia, Pennsylvania. OCED TG 106, 2000, Determination of Soil Adsorption/Desorption Using a Batch Equilibrium Method. OECD: Organization for Economic Co-operation and Development

Factors Influencing Sorption and Desorption Characteristics  Specific-surface Area  Pozzolanic Activity (Lime Reactivity)  CEC Value  Liquid to Solid Ratio  pH of the Soil Solution  Buffer Capacity of the Sorbent  Temperature  Grain Size  Presence of other Ions  Ionic Strength  Organic Content & Fe- Mn Oxides  Carbonate Content

Batch Sorption Tests Geomaterial (in powder form) + Contaminant (in solution form) is allowed to interact for certain duration (with continuous stirring) Factors Influencing  Liquid to solid ratio (L/S = 10, 20, 50, 100, 200) Interaction time Variation of C e with interaction time 100 C i Initial concentration of contaminant L/S 10 20 50 100 200 80 C e Concentration of contaminant in solution after time interaction time is to be 60 recorded by sampling the solution C e (mg/l) frequently 40 C s Concentration of the contaminant sorbed 20      C C C (L/S) s i e 0 24 h 0.01 0.1 1 10 100 t s (h)

Sorption Characteristics (Isotherms) L/S 10 20 50 100 200 2.8 LR 4 mg/kg) Linear Isotherm (LR) 2.1 C s = K d · C e 1.4 C s (  10 0.7 0.0 0 1000 2000 3000 4000 5000 1.00 Langmuir Isotherm (LM) LM 0.75 C e /C s (kg/l) C 1 C   e e 0.50  C K b b d 0.25 s 0.00 0 1000 2000 3000 4000 5000 FH Freundlich Isotherm(FH) 5 10 C s (mg/kg)    -1 log(C ) log(K ) n log(C ) 3 s d e 10 1 10 -2 0 2 4 10 10 10 10 C e (mg/l)

Desorption Characteristics (Isotherms) C sl : the amount of contaminants present in the sorbate after desorption phenomena C el : the equilibrium concentration of contaminants    C C C (L/S) s s el l present in the solution after equilibration time C s : Concentration of the contaminant sorbed Variation of C el with leaching time 100 L/S Linear Isotherm (LR) 10 20 50 100 200   C K C sl dl el 10 Langmuir Isotherm (LM) C el (mg/l) C 1 C   el el  b b C K d sl l l l 1 Freundlich Isotherm (FH)    -1 log(C ) log(K ) n log(C ) 2 h sl dl l el 0.1 0.01 0.1 1 10 100 t l (h)

Relationship between CEC and K d CEC K d Principal minerals (meq/100 g) (ml/g) Quartz, Montmorillonite 49-57 3347-3580 Quartz, Orthoclase 6.4-6.6 1497-1530 10000 K d (ml/g) 1000 100 0 20 40 60 80 100 L/S

Some Important Relationships 3600 kd (ml/g) 3200 2800 12000 2400 0 5 10 15 20 25 10000 t (h) 8000 Kd (ml/g) 6000 4000 2000 0 3 4 5 6 7 8 9 pH

5 10 Material “Accelerated Physical Modeling of CS Sorption and Desorption Characteristics WC 4 10 of Geomaterials and Immobilizing Agents” IC EC (mS/cm) RSS Dali Naidu (2006) BSS FA-I 3 10 FA-II Material 2 CS WC IC RSS BSS FA-I FA-II 10 5 10 1 10 4 10 0 1 2 3 10 10 10 10 L/S EC (mS/cm) 1000 3 10 2 10 89 9.8 1 10 -1 0 1 2 3 4 -2 -1 0 1 2 3 10 10 10 10 10 10 10 10 10 10 10 10 Kd (l/kg) Kdl (l/kg)

Column Tests 55 The BTC Inner cylinder 1.0 Desorption starts @ 62 h Sorption 0.8 140 C t/ C 0 0.6 h 1 0.4 Middle cylinder sample L 0.2 Outer cylinder Porous disc 0.0 Base plate 10 0 20 40 60 80 100 120 140 160 t (h) 100 PV= V sol  [  L  (  d 2 /4)] -1 V sol is the volume of the solute passing through the sample L is the length of the sample,  is the porosity

    2 K 1-D ADE C C C C    ρ d D v . . s.     i. dry η 2 t z t z  ρ   K    2   dry C D C v C d i     R 1 s   η    2   t R z R z R = Retardation factor Centrifuge Modeling of Sorption/Desorption mechanisms         t t t y t t N mA mB mA       ln ln ln p mA A       t t t    mB mC mC        y , y and y y y y t t N t N t N ......... . AB   BC   AC   p mA A mB B mC C N N N B C C       ln ln ln       N N N A B A y sorption = 1 Y desorption = 0.5

Example Analysis No1 h L c 0 L CCL Transit Time Analysis Transit time, t: time required for solute concentration at the bottom of the barrier [e.g. c(L GCL , t) to reach a specified value relative to the source concentration [e.g., c(L GCL ,t)/c 0 = 0.5]

Advective Analysis t a = L/V s = n e L / K i Where: K CCL = hydraulic conductivity of compacted clay i CCL = hydraulic gradient across compacted clay n = porosity of compacted clay

Porosity: Use n e = n = 0.30 (case 1) n e = 0.01 (case 2) Use L CCL = 600 mm = 60 cm Typically: 10 -10 m/s < K CCL < 10 -8 m/s Use K CCL = 10 -9 m/s Hydraulic gradient considerations: For L CCL = 60 cm & h L = 30 cm, i = (h L + L CCL )/L CCL i = 1.5 t a = 0.3 x 0.6/ 10 -9 x 1.5 = 4 years = 0.01 x 0.6/10-9 x 1.5 = 0.13 years

Example Analysis No2 • A chemical waste is being discharged into a shallow injection well • Surrounding soil is sand with K = 6 x 10 -4 m/s and e = 0.9, i=0.006 • Water well 3 km downstream t required for the contaminants to travel from the source to the water well?

Porosity : n = e/(1+e) x 100% = (0.91/1.91) x 100 = 48 % V s = K i/n = 6 x 10 -4 x 0.006/0.48 = 0.075 10 -4 m/s t = L/Vs = 3000/0.075 = 13 years

Example analysis No3 Landfill is leaking leachate with a chloride concentration of 725 mg/l which enters an aquifer with the following properties: K = 3 x 10 -5 m/s, i=0.002, n e =0.23, D*=1 x 10 -9 m 2 /s Calculate the concentration of chloride in 1 year at a distance 15 m from the source

• Seepage velocity V s = K i/n e = 2.6 10 -7 m/s • Coefficient of longitudinal hydrodynamic dispersion a L = 0.00175 L 1.46 = 0.91 m • Coefficient of hydrodynamic dispersion D L D L = 0.91 x 2.6 10 -7 + 1 x 10 -9 = 2.4 10 -7 m 2 /s • Concentration C= 362.5 =[erfc (1.24) + exp (16.25) x erfc (4.22) = 30 mg/l