Exercises__WS14_10..11.14-en.

Transcrição

Exercises__WS14_10..11.14-en.
Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Task 1: Risk analysis: Transport time„Contaminant source – Receptor“ in groundwater
Processes: Convection + Adsorption = Advection
Contaminant: Benzene (KOC = 38)
Distance „Contaminant source – Receptor“: 1 km
Hydraulic gradient: 0.01
Sediment: medium sand kf = 10-3 m/s = 86.4 m/d
Solid density s= 2.65 kg/l
Total porosity = 0.35
Effective Porosiy = 0.2
Content of organic carbon = 0.01
Solution:
1. Darcy-law  Darcy-velocity
2. effective Porosity  mean travel velocity
3. Adsorption  Retardation  Particle velocity  Time
Task 2 – 5: Equilibrium mass partitioning of DNAPL-contaminant TCE in different phases
At the aquifer basis in a aquitard wanne(?) a TCE-Pool has been formed. The extensions in xand y-direction amount 10m and the hight 1m.
Task 2: Calculate the TCE-mass within the pool (pure TCE-Phase).
The aquifer consists of medium sand with a total porosity of 0.35.
The density of TCE amounts 1.46 kg/l.
Task 3: The TCE – Phase stays in solution equilibrium with groundwater. How much TCE can
maximal dissolve into the water phase?
The geometric dimensions of both adjacent phases are the same, namely 10m x 10m x 1m and the
the porosity (= 0.35), too. The maximal solution of TCE in water amounts 1100 mg/l.
Task 4: How much TCE can be adsorbed at the solid matrix?
Solid density ρs = 2.65 kg/l
Organic carbon content of the sediment: fOC = 0.004 mg-OC/mg-sed
KOC = 18.2 l/kg
Hint: Calculate the mass of the solid matrix (medium sand) that is contained in the control
volume. With this you can calculate the adsorbed TCE – mass.
Task 5: Because of the high volatility (VOC) a large part of the TCE will be degass into the
above-lying unsaturated zone. How much TCE will be degass into the gas phase ? Assume the
same geometry as for the water phase.
Henry-coefficient of TCE HTCE = 0.417
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Time scales
Task 6: How many pore volumes (= PV) have to be exchanged over the TCE-Pool, till the natural
groundwater flow has completely dissolved the TCE-Pool ?
(partitioning equilibrium is assumed !)
Task 7: What time (convective time scale for Pool-dissolution) will pass, if the mean velocity
v = 1m/d ?
Task 8: What time (= tp&t) is necessary to pump the contaminated water bulk (= 1 pore volume) with
5 pumping wells; each operates with a pumping rate of 100 m3/d ?
Task 9: How much kg TCE will be released diffusively by the TCE-Pool during 1 day?
Hint:
Diffusive mass flux:
QTCE = AjTCE,
Dimension: kg/d
Phase boundary:
A = A
mass flux density (1. Fick’s law): jTCE = DTCECTCE /
Diffusion coefficient:
Diffusion layer thickness:
Concentration difference:
DTCE = 1.0510-8 m2/s
  1mm
CTCE = 1100 mg/l
Task 10: How much TCE will be released diffusively during tp&t ?
Task 11: Which concentration corresponds to the released mass?
Time scales:
Zeitskala: Bioremdiation
Übungsaufgabe 12A: TCE (Aliphat)
Eine TCE-Fahne hat eine transversale Ausdehnung von 10m und eine vertikale Ausdehnung von
1m. Die natürliche Grundwassergeschwindigkeit beträgt 1m/d. Die TCE-Concentration beträgt
500mg/l.
Stellen Sie die stöchiometrische Reaktionsgleichung für den aeroben Umsatz von TCE zu
CO2 und H2O auf!
Berechnen Sie den massenstöchiometrischen Faktor: f = O2Masse/TCE-Masse.
Übungsaufgabe 12B: Benzen (Aromat)
a) Stellen Sie die stöchiometrische Reaktionsgleichung für den aeroben Umsatz von Benzen zu
CO2 und H2O auf!
b) Berechnen Sie den massenstöchiometrischen Faktor: f = O2Masse/Benzen-Masse.
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Übungsaufgabe 13: Es soll A) eine Sauerstoffgas-Injektion und B) eine GelöstSauerstoffinjektion über Injektionsbrunnen durchgeführt werden. Die Gelöst-Concentration
beträgt unter Atmosphärendruck (pO2 = 1atm) ca. 50 mg/l. Zeigen Sie mit dem Henry-Gesetz,
dass sich diese bei einer Injektionstiefe von 10m unter GW-Spiegel verdoppelt.
Übungsaufgabe 14: Um das Strömungsfeld nicht wesentlich zu stören, können Sie mit einer
Injektionsrate von 100 Liter/Tag arbeiten. Bestimmen Sie die Anzahl N und die Anordnung der
Injektionsbrunnen A) für eine Sauerstoffgasinjektion und B) für eine Gelöst-Gasinjektion.
Übungsaufgabe 15: Bestimmen Sie die optimale Mischstrecke LDisp, wenn die transversale
Dispersivity αT = 0.1m beträgt.
Lösung:
transversale Dispersionsstrecke: LT2 = DT⋅t, LT = 1m
transversaler Dispersionskoeffizient: D = α ⋅v
T
T
Übungsaufgabe 16: Bestimmen Sie die Ausdehnung des aeroben Reaktionsraumes für die in
situ-Sanierungsmassnahme, wenn das Sanierungsziel 25 mg/l beträgt. Der aerobe Bioabbau wird
durch eine Kinetik 1. Ordnung beschrieben; die Halbwertszeit von TCE beträgt 30 Tage.
Die Contaminantfahne befindet sich im steady-state.
Übungsaufgabe 17: Leiten Sie das hydarulische Kriterium für eine reaktive Wand ab!
Übungsaufgabe 18: Die gesamte Contaminantmasse MTCE soll in der reaktiven Wand /
abstromseitigen Reaktionsraum adsorbiert, chemisch umgewandelt bzw. biologisch abgebaut
werden. Leiten Sie dafür dass Massenbilanzkriterium ab!
Übungsaufgabe 19: Ein Trägergasstrom wird mittels 3 Air-sparging-Brunnen in den Aquifer
gepumpt und danach mit einer SVE-Einrichtung wieder abgesaugt. Der kontaminierte GWKörper besitzt ein Volumen von 35 m3 (= VP = Porenvolumen). Die Concentration von TCE
beträgt 500 mg/l und der dimensionslose Henry-Koeffizient ist 0.42.
Mit jedem Injektionsbrunnen können Sie 100 Liter Trägergas pro Stunde injizieren.
Wie lange dauert eine vollständige Sanierung mittels Stripping ?
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Übungsaufgabe 20
Sie sollen für einen Toluol-Schadensfall eine geeignete Sanierungsmaßnahme vorschlagen. Aus
einer Risikoanalyse wissen Sie, dass die NA-Prozesse nicht ausreichen, um das Sanierungsziel,
d.h. eine Zielconcentration von 1% der maximal lösbaren Concentration Cmax = 515 mg/L, am
Rezeptorstandort zu erreichen. Da Toluol gut aerob abbaubar ist, entscheiden Sie sich für eine
Sauerstoffinjektion, d.h. für eine reaktive Sauerstoffwand.
Die Ausdehnung der Contaminantfahne im Anstrom der Reaktiven Wand (RW) und die
hydraulischen Parameter sind:
Parameter der Contaminantfahne:
Breite der Fahne
LF = 30 m
mittlere Länge der Fahne Lx = 200 m
Aquifermächtigkeit
bF = 10 m
Porosity
 = 0.2
hydraul. Gradient
i = 0.01
hydraul. Conductivity
kf = 10-3 m/s
Parameter der Reaktiven Wand/ des Reaktionsraumes:
Breite der Wand
BRW = 10 m
Länge des Reaktionsraumes LRW = 10 m
Aquifermächtigkeit
bRW = 10 m
Porosity
 = 0.2
Im Abstrom der reaktiven Wand (Reaktionsraum) findet ein mikrobieller Abbau von Toluol statt.
In erster Näherung können Sie die Abbaukinetik als Kinetik 1. Ordnung beschreiben und
annehmen, dass nach einer Einlauf-Phase sich ein „steady-state“ einstellt. Aus der Vorlesung
„Bioremediation“ wissen Sie, dass die Ratenkonstante k1 proportional zur ElektronenakzeptorConcentration ist:
k1  A  [O2 ]w
(*)
mit A = 0.4 L/(mg·Tag).
Welche mittlere Sauerstoffconcentration müssen Sie im Reaktionsraum gewährleisten, damit Ihr
Sanierungsziel erreicht wird?
Hinweis: Bestimmen Sie zuerst k1 und dann über die Formel (*) die mittlere
Sauerstoffconcentration.
Task 21: Pump and Treat: Best case scenario (Domenico & Schwarz D 20.1 und D20.2)
A) A contaminant plume covers a x-y-area of 41490 m2 and extends through the whole aquifer (B
= 16.7m). The porosity amounts 0.3. How long takes P&T with 3 pumping wells (Q = 0.13
m3/min) ?
B) If one assumes that the contaminant with a Kd-value of 0.2 L/kg will be adsorbed at the solid
matrix (s = 2.65 kg/L), to which factor changes the in A) calculated Clean-up-time ?
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Tasks with EXCEL-worksheets
Task 22: Risk-Analysis for the SAFIRA-test site Leuna
Sketch:
z
10
m
z = z2-z1
Gasphase
0.5
m
LNAPL
0
Parameter:
Benzene-concentration C0 = 1780 mg/L
effective porosity eff = 0.3
hydraulic conductivity kf = 0.001 m/s
Potential difference h = h1 – h2 = 0.1m
L = 100m
longitudinal dispersivity = 10m
Dg,eff = 3.8  10-6 m2/s
HBz = 0.24
50
m
100
m
5
x
Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Questions:
1. Which Benzene mass reaches the atmosphere at the third day?
(cross section: 10 m x 10 m)? Transport path: soil air – atmosphere –man.
You can calculate the flux density both analytically and numerically
(Finite Differences: j = - Dg,eff C/z, z = 0.01m; see sketch).
2. Which Benzene mass reaches the groundwater at the 10th day?
(groundwater body: 50m × 10m × 0.5 m)?
3. After how many days the MAK-Concentration (10 micro-g/L) is exceeded within the drinking
well SAFLEU2?
Technical Guide Concentration (TRK) = 3.25 micro-g/L
EU-Guide value = 0.001 micro-g/L
Tasks: Answer the questions with an EXCEL-work-sheet!
Hints:
EXCEL 2010:
1. ERFC() = GAUSSFKOMPL() = 1 – GAUSSFEHLER()
2. Calculation options: click “automatically”
EXCEL 2003:
3. Activate the list of available Add-Ins, click the control box Analyse-functions and
click OK.
4. For small arguments there is an error #ZAHL! Ignore!
5. Options -> calculations -> automatically
Task 23: Risk-Analysis for the SAFIRA-test site Bitterfeld
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Processes
The contaminant Monochlorbenzene (MCB = ClBz = DNAPL) migrates due to leakages during
industrial operation/production as phase into the groundwater and will be hindered and adsorbed
by the lignite seam (Aquitard). There a DNAPL-Pool is formed.
Process A): MCB dissolved into the above flowing groundwater and will be transported by the
groundwater flow. In the nearness of the lignite the sediment will contain a high content of
organic carbon foc, and hence a high degree of adsorption.
Process B): MCB will migrate through the lignite seam by diffusion and gets into the tertiary
aquifer.
Estimate the risk (time scales) for both processes!
Questions:
1. Which MCB/TCE-mass dissolved maximally into the groundwater body (100m × 100m × 1m)
that lies above the MCB/TCE-Pool ?
2. After which time reaches MCB/TCE (Climit = 10 micro-g/L) the river ‚Mulde‘ with and without
adsorption at lignite seam?
Hint: Since the lignite leads to a higher foc only over a partial distance (section) of 2 km, you can
calculate the time difference for the transport along the lignite, i.e. the retardation takes place
only on the lignite section.
3. Which MCB/TCE-mass gets into the atmosphere at the first day after reaching the limit value
Climit at the “Fuhne”-wetland (cross section: 100 m x 1 m) ?
4. After which time MCB reaches the tertiary aquifer with a breakthrough concentration of 10
micro-g/L due to diffusion through the lignite seam? The lignite seam has a mean thickness of
about 6m?
Tasks: Answer the questions with an EXCEL-work-sheet!
Sketch:
z
100 m
GW-Körper
foc = 0.01
DNAPL-Pool
Kohle
0
1m
foc = 0
foc = 0.4
2 km
4 km
x
Parameter:
Contaminant: Monochlorbenzene (Abbreviation: MCB or in sketch ClBz)
max
Maximal MCB-Dissolved-Concentration CMCB
= 500 mg/L
Geometry „MCB-Pool“: 100m × 100m × 1m
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
distance SAFBIT2 – SAFBIT3 = 4 km
Quartary Aquifer:
Content organic carbon in the aquifer obove the lignite seam : foc = 0.01
Total Porosity
tot = 0.35
effective Porosity
eff = 0.3
solid density
s = 2.65 kg/l
hydraulic conductivity
kf = 0.005 m/s
Potential difference
h = h2 – h3 = 4m
Transport
longitudinal dispersivity = 10 m
lignite/coal seam
Length
Lcoal = 2 km
Thickness
Dcoal = 6 m
Total Porosity
tot = 0.6
solid density
s = 1.2 kg/l
Content organic carbon of the lignite seam foc = 0.4
KOC = 136 l/kg
-9
2
molecular diffusion coefficent of MCB in water DwMCB
, 0 = 2×10 m/s
MCB
effectiver diffusion coefficient DwMCB
, eff  Dw, 0 / R
(R – retardation coefficient due to adsorption)
Übungsaufgabe 24: Pump and Treat: Single-well-capture zone
A) Erstellen Sie ein EXCEL-file zur Berechnung der Single-Well-Capture Zone !
B) Die Dimension der Pachtfläche, z.B. Oxywall-Leuna, beträgt 100m x 50m und die
Fahnenbreite 50m. Sie sollen einen Pumpbrunnen bei minimalen Betriebskosten installieren! Sie
müssen sicherstellen, dass Sie nur GW Ihrer Pachtfläche bewegen, d.h. die Capture Zone muß
vollständig in Ihrer Pachtfläche liegen.
Gegeben:
kf = 10-4 m/s (Feinsand)
i = 0.1 (starker hydraulischer Gradient)
B = 5 m (Aquifermächtigkeit)
Bestimmen Sie die Position (x-Koordinate) des Pumpbrunnens und dessen Pumprate!
Übungsaufgabe 25: Bioremediation
Am Safira-Standort Bitterfeld befinden sich grosse Mengen chlorierter Aliphaten (PCE: Per- oder
Tetrachlorethen, TCE, DCE, VC) und Aromaten (HCH, DCB, MCB). Am ‚Hot Spot: Grube
Antonie’ liegt PCE in Phase vor und das vorbeiströmende GW befindet sich im lokalen
max
max
Gleichgewicht mit dem Hot Spot: C PCE
= 155 mg/L (Vergleiche mit CTCE
= 1100 mg/L!).
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Gegeben:
Porosity = 0.3
kf = 10-3 m/s (Mittelsand)
i = 0.001
A = 10 m2 (durchflossene Cross section)
kPCE = 0.001 1/d
F = mTCE/mPCE : massenstöchiometrischer Faktor
A) Infolge mikrobieller Dechlorierung wird PCE zu TCE abgebaut. Berechnen Sie die
Wegstrecke nach der die PCE-Concentration kleiner als 1 mg/L ist. Verwenden Sie die steadystate-Näherung !
Lösung: 1460 m
B) Berechnen Sie den steady-state-Concentrationsverlauf von TCE! Variieren Sie die TCERatenkonstanten (0.5, 0.1, 0.05, 0.01, 0.0011, 0.001, 0.00099, 0.0005 1/d) und diskutieren Sie
die Singularität bei TCE = 0.001!
C) Welche PCE-Masse wird innerhalb eines Tages umgesetzt ?
Hinweis: Integrieren Sie über Wegstrecke, die eine Wasservolumen x A während Zeit t
zurücklegt! Differenz zu M0 ergibt abgebaute Masse.
D) Welche TCE-Masse entsteht während eines Tages ?
Aufgabe 26 Modulare Bioremediation-Technologie
Sie wissen, dass Sanierungstechnologien für kontaminierte Böden (= unsaturated zone) und für
kontaminierte Grundwässer (= saturated zone) modular aufgebaut sind. In Abb. 1 ist ein
modulares Sanierungskonzept dargestellt.
A) Diskutieren Sie die hydraulischen Sanierungsmaßnahmen und den Verlauf des
Grundwasserspiegels!
--> P&T, Absenkung des GW-Spiegels
B) Charakterisieren Sie die Bioremediation-Sanierungsmaßnahme! Nennen Sie
Elektronenakzeptor und Elektronendonator! Begründen Sie dies mit dem Partialladungskonzept,
wenn der Contaminant Benzen ist!
C) Welches Redoxpotenzial stellt sich infolge der Sanierung ein?
D) Welche physiko-chemische Sanierungsmaßnahme wird durchgeführt?
--> Air stripping, Air sparging: Henry-Gesetz
E) Welche Injektionen (Phasen?) werden durchgeführt?
F) Charakterisieren Sie die jeweiligen Prozesse, die in den einzelnen Sanierungsmodulen
ausgenutzt werden, um die organischen Contaminants hoher Volatilität (VOCs) aus dem Boden
und dem Grundwasser zu beseitigen!
--> SAN1-Modul: P&T; SAN2-Modul: techn.-gesteuerte BR; SAN3-Modul: physiko-chemisch,
Air stripping
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Abbildung 1 Bioremediation
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Übungsaufgabe 27: Natural Attenuation
A) Betrachten Sie den aeroben Abbau von TCE. Stellen Sie stöchiometrische Reaktionsgleichung
auf, welche die Massenbilanz und Ladungsbilanz erfüllt!
B) Betrachten Sie den aeroben Abbau von Benzen. Stellen Sie stöchiometrische
Reaktionsgleichung auf und berechnen Sie den stöchiometrischen Faktor!
C) Berechnen Sie die Benzen-Masse, die sich je Liter nach 50 Tagen aerob umgesetzt hat,
dabei soll das GW als ruhend betrachtet werden. Vergleichen Sie den Umsatz, welcher sich
infolge einer Monod-Kinetik ergibt mit dem Umsatz einer 1.order-Kinetik und erstellen Sie dazu
ein EXCEL-file !
Gegeben:
[O2] = 10 mg/L
[Bz] = 500 mg/L
M = 1 mg/L
kmax = 1.56 1/Tag
KS = 10 mg/L
KO2 = 1 mg/L
ÜA 28 Case-Study 1: PCE-Schadensfall „Auensee-Leipzig“
In der unmittelbaren Nähe des Erholungsgebietes „Auensee-Leipzig“ wurde über ca. 40 Jahre
eine chemische Reinigung mit chlorierten Lösungsmittel (TCE, PCE) betrieben. Durch Leckagen
im Entsorgungssystem sind diese Lösungsmittel in den Untergrund gelangt. Die geologische
Struktur wird von oben nach unten wie folgt charakterisiert:
Obere Schicht: mittelsandiger bis feinkiesiger oberer quartärer Grundwasserleiter (GWL)
Mittlere Schicht: Braunkohle (BK)-Schicht, die aber nicht durchgängig zusammenhängend ist;
sondern eher als stückige BK im mittelsandigen Sediment charakterisiert werden kann
Untere Schicht: mittelsandiger, unterer, tertiärer GWL
Basis: Rubelton-Aquitard.
A) Charakterisieren Sie eine mögliche Contaminantverteilung im Querschnitt vom oberen GWL
bis zum Rupelton-Aquitard.
B) Welche Sanierungsmaßnahmen halten Sie für geeignet und welche für ungeeignet (jeweils 2)?
Begründen Sie Ihre Vorschläge!
C) Wichtig für Ihre Entscheidung, welche Sanierungsmaßnahme geeignet ist, ist die Zeit, die Sie
bis zum Eintreffen des Contaminants am Rezeptor zur Verfügung haben.
Führen Sie eine Risikoanalyse durch, indem Sie die Zeit „Quelle-Rezeptor“ berechnen (in
Tagen!):
A) für den oberen GWL und
B) für die BK.
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Verwenden Sie dazu den Isohypsenplan (siehe Abb. 2)!
Zur Vereinfachung nehmen Sie einen konstanten hydraulische Gradienten an. Diesen berechnen
Sie an der Bilanzebene!
D) Ihre in B) aufgezählten Sanierungsmaßnahmen müssen die notwendige Bedingung des
stöchiometrischen Massenumsatzes erfüllen (ankommende Contaminantfracht muss zu x%
entsprechend Ihrem Sanierungsziel umgesetzt werden). Folglich benötigen Sie zur
Dimensionierung Ihrer Sanierungsmaßnahme die Contaminantfracht.
Berechnen Sie die spezifische Contaminantfracht (in mg/Tage) für die in Abbildung 1
dargestellte Bilanzebene entlang der Hauptstrombahn (durchflossene Fläche A = 1 m2)
A) für den oberen GWL und
B) für die BK.
E) Die PCE-Kontamination in der BK-Schicht ist um einen Faktor 15 höher. Leitet sich daraus
auch ein 15mal höheres Risiko für den Auensee ab. Begründen Sie Ihre Antwort!
Parmeter
Contaminant: PCE
Log(Koc) = 2.42 (Dimension von Koc [kg/L])
A = 1 m2 ,
Abstand Quelle-Rezeptor = 500 m
Feststoffdichte s = 2.5 kg/L
Tabelle 1: Parmeter für oberen GWL und BK-Schicht.
Oberer GWL
BK-Schicht
PCE [mg/L]
10
150
kf [m/s]
3 × 103
5 × 106
fOC
2.46 × 103
1.62 × 102
12
Tot Porosity
0.35
0.4
Eff. Porosity
0.3
0.3
Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Abbildung 2 Isohypsenplan für Schadensfall „Auensee-Leipzig“:
Blauer Pfeil entspricht Hauptstrombahn.
Rote dicke Linie: Bilanzebene
Maßstab: schwarzer Pfeil (obere linke Ecke) entspricht 50 m in der Realität.
50 m
m
Naherholungsgebie
500
Übungsaufgabe 29: Case Study 2 „Bitterfeld-Bayer-Riegel“: P&T – 4-Brunnen-Gallerie
Im Anstrom einer Investorfläche (Bayer-AG) auf dem ehemaligen Gelände des
Chemiekombinates Bitterfeld soll der Contaminantstrom (Concentration ca. 1000 mg/L)
hydraulisch mit einer 4-Brunnen-Gallerie abgefangen werden. Die Contaminantconcentrationen
in den Beobachtungsbrunnen P4 (700,0) und P5(700,1000) (alle Längenangaben in Meter!) sind
unterhalb 1 mg/L.
Die Koordinaten der Eckpunkte der Investorfläche sind: (0,0), (0,1000), (-1000,0), (-1000,1000).
Bestimmen sie die x- und y-Koordinaten jedes Pumpbrunnens und die Pumpleistung mit Hilfe
des Four-well-Capture-Zone Diagramms (siehe Vorlesungsskript). Der Abstand der Brunnen
beträgt 100 m.
Im Anstrom der Investorfläche haben Sie 3 Beobachtungsbrunnen, P1, P2, P3, zur Verfügung.
Sie messen die folgenden Grundwasserhöhen in Bezug zur Geländeoberkante (!): h1 = 5 m, h2 =
5 m, h3 = 4.132 m. Die Koordinaten der Beobachtungsbrunnen sind: P1(250,200), P2(250,800),
P3(500,500).
Die Grundwassergeschwindigkeit bestimmen Sie mit Hilfe des hydraulischen Dreiecks.
Die hydraulischen Eigenschaften des Aquifers sind:
Mächtigkeit 10 m
Effektive Porosity 0.3
Hydraulische Conductivity 0.001 m/s.
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Task 30: Case Study 3 "Dimensioning of modular reactive walls"
For groundwater contamination, caused by (1,2)-di-chlorobenzene (DCB: 2 Cl-atoms substituting
two hydrogen atoms), you have to dimension a modular, two-step reactive wall. The length of the
lease area is L. You know that with increasing degree of chlorination aerobic degradation
becomes very slow.
Therefore, you decide to carry out a reductive dechlorination of DCB to mono-chlorobenzene (=
MCB) using a reactive iron wall. Down-gradient, directly into the effluent from the iron wall, you
inject oxygen (reactive oxygen wall), since MCB is well aerobically degradable and can be
mineralized completely to CO2 and H2O. For simplicity, you can assume that the oxygen wall
(aerobic reaction zone) is connected directly to the reactive iron wall!
Hint: Make yourself clear the geometry with a sketch, before you start your calculation!
The steady-state solution, you must apply for each reactive wall separately, where the left edge of
each wall has always to be chosen as x = 0 (This is also logical!).
Parameters:
Length of the lease area L = 500 m
Width of the lease area = 10 m
More of the reactive iron wall = 10m
Width of the reactive iron wall = 10 m
Aquifer thickness = 15 m
Contaminant plume width = 10 m
Groundwater velocity located upstream of the iron wall u = 1 m / day
DCB concentration located upstream: CDCB(x = 0) = 140 mg / L
Porosity of the aquifer = 0.3
Porosity of the iron wall = 0.4
Molar mass of chlorine is 35 g / mol
A) The reactivity of the reactive iron wall (rate constant kIron [1/day]?) should be dimensioned
in such a manner that 90% of the influent DCB will be de-chlorinated. Use the steady-state
solution. In the application of the hydraulic criterion (equal sign!) use the relationship between
volumetric flux density and flow velocity.
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
B) How long is the aerobic reaction zone?
The effluent MCB-contaminant load (= mass flux =QMCB = CMCB  Qw, convective approach)
from the iron wall has to be completely degraded within the lease area. What oxygen mass per
unit time (= QO2) you have to inject?
Hints:
1. The water flow rate Qw, you can easily calculate.
2. To calculate CMCB (MCB-concentration at the output of the iron wall that is the input
concentration of the oxygen wall), you first have to calculate the reductively dechlorinated DCBmols per liter and then to take into account the stoichiometry of the reductive dechlorination:
x Mole DCB  x Mole MCB!
3. First, establish the stoichiometric equation for aerobic degradation of MCB and calculate the
stoichiometric factor: fO2 = O2-mass / MCB-mass, so that you can meet the required
stoichiometry condition for the complete degradation, i.e. MCB each molecule gets x oxygen
molecules!
C) The reactivity of the downstream aerobic reaction zone is kO2 = 6.0  103 1/day. What goal
concentration you reach at the end of the aerobic reaction zone, i.e. at the end of the lease area?
Task 31 Case-Study 4: Dimensioning of modular Reactive Walls
GAC-Wall/ORC-Wall
A) GAC-Wand
A DNAPL pool consisting of PCE, TCE and DCE is a continuous source of groundwater
contamination. The contaminant plume should be cleaned up through a 2-stage modular reactive
wall. The reactive walls extend over the entire aquifer thickness.
In a first step a reactive adsorption wall consisting of granular activated carbon (GAC) should
bind the chlorinated aliphatics. The width of the GAC-wall is adjusted to the width of the
contaminant plume. The flow rate of groundwater is small compared to the adsorption rate, so
that you can assume partitioning equilibrium between the aqueous phase and adsorptive-bound
chlorinated aliphatics. The technical Kd values (L/kg) of activated carbon are:
Log(Kd,PCE) = 2.19
Log(Kd,TCE) = 2.97
Log(Kd,DCE) = 0.046
Determine the PCE, TCE, and DCE concentration at the exit of the GAC-wall!
Note: Make yourself clear the geometry with a sketch before you start your calculation!
Because of the mass balance, the inflowing contaminant concentration C win,i (mg/L) is equal to the
sum of adsorptive-bound contaminant concentration Cads,i (mg/kg) and outflowing contaminant
concentration C wout,i (mg/L) (mg/L) (i = PCE, TCE, DCE)! Think about between what
concentrations partitioning equilibrium (PartEqu) holds. The contaminant flux always needs a
certain amount of time (relaxation time) before PartEqu is establisehd. In the mass balance
equation, all concentrations must have the same dimension, namely mg/L. Therefore, you must
multiply Cads,i by a factor b/, where b is the GAC-bulk density and  the porosity.
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
B) ORC-Wall
You know that the sorption capacity decreases with the degree of chlorination. Consequently,
most of the DCE will pass the GAC-wall. However, since the reactivity of aerobic microbial
degradation increases with decreasing degree of chlorination, you install directly to the GACwall, a ORC-wall (ORC-oxygen releasing compound). Dimension the ORC wall so that the
released oxygen per day is sufficient for complete aerobic degradation of DCE, i.e. which
oxygen mass must release the ORC-wall per day?
Parameters:
Width of the GAC-Wall = 12 m
Width of the ORC-Wand = 12 m
Width of the contaminant plume = 12 m
Aquifer thickness = 13 m
Groundwater velocity upstream of the GAC-Wall u = 1 m/Tag
Porosity of the Aquifer = 0.4
Porosity of the GAC-Wall = 0.4
GAC-solid density = 0.6 kg/L
Porosity of the ORC-Wall = 0.4
Concentrations upstream of the GAC-wall:
PCE-Concentration = 140 mg/L
TCE-Concentration = 840 mg/L
DCE-Concentration = 600 mg/L
Mol-Masse von Chlor ist 35 g/Mol
Ü-Aufgabe 32 Case-Study 5:„Auensee“ : Gekoppelte Wasserstoff-Sauerstoff-Gaswand
In der unmittelbaren Nähe des Erholungsgebietes „Leipzig- Auensee“ wurde über ca. 40 Jahre
eine chemische Reinigung mit chlorierten Lösungsmittel (PCE, TCE) betrieben. Durch Leckagen
im Entsorgungssystem sind diese Lösungsmittel in den Untergrund gelangt. Der kontaminierte
Grundwasserstrom führt direkt zum Auensee. Mit Hilfe einer Gekoppelten WasserstoffSauerstoff-Gaswand (ED-EA-Technologie) soll der Grundwasserstrom gereinigt werden, bevor
er den Auensee erreicht. Dimensionieren Sie die beiden reaktiven Gaswände!
Für eine erste Abschätzung, soll angenommen werden, dass sich die hydraulische Conductivity
des Aquifers durch die Gasinjektionen nicht ändern soll, was natürlich nicht richtig ist!
Hinweis: Machen Sie sich mit einer Skizze die Geometrie klar, bevor Sie anfangen zu rechnen!
Wasserstoff-Gaswand: Reduktive Dechlorierung von PCE zu DCE
32A) Wie viel Norm-Liter gasförmigen Wasserstoff (1 Mol Wasserstoff-Gas entspricht 22.4
Norm-Liter) müssen Sie für die vollständige reduktive Dechlorierung von PCE zu DCE pro Tag
injizieren und unter welchem Druck?
Hinweis: Stellen Sie zuerst die Reaktionsgleichung auf!
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Exercises for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
32B) Bestimmen Sie mit Hilfe des Partialladungskonzeptes Elektronendonator und
Elektronenakzeptor!
Sauerstoff-Gaswand: Aerober mikrobieller Abbau von DCE
32C) Wie viel Norm-Liter gasförmigen Sauerstoff (1 Mol Sauerstoff-Gas entspricht 22.4 NormLiter) müssen Sie für den vollständigen aeroben Abbau pro Tag injizieren und unter welchem
Druck?
Hinweis: Stellen Sie zuerst die Reaktionsgleichung auf!
Aus 6A) kennen Sie quasi die DCE-Concentration am Ende des anaeroben Reaktionsraumes (=
Eingangsconcentration für Sauerstoffwand!), wenn Sie eine vollständige Dechlorierung (PCE 
DCE) voraussetzen.
32D) Bestimmen Sie mit Hilfe des Partialladungskonzeptes Elektronendonator und
Elektronenakzeptor!
32E) Welche DCE-Concentration wird nach 10 m des aeroben Reaktionsraumes erreicht, wenn
die Ratenkonstante für den aeroben Abbau kO2 = 0.23 1/day beträgt. Es wird steady-state
vorausgesetzt.
Parameter
Grundwasserspiegel unter GOK (= Geländeoberkante) = 5 m
Tiefe der Gasinjektion = 15 m unter GOK
Breite der Wasserstoff-Wand = 15 m
Länge des anaeroben Reaktionsraumes = 20 m
Breite der Sauerstoff-Wand = 15 m
Länge des aeroben Reaktionsraumes = 20 m
Aquifermächtigkeit = 10 m
Breite der Contaminantfahne = 15 m
Grundwassergeschwindigkeit im Anstrom der Wasserstoff-Wand u = 1 m/Tag
Porosity des Aquifers = 0.35
PCE-Concentration im Anstrom der Wasserstoff -Wand = 140 mg/L
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Übungsaufgaben: Darcy-Gesetz - REV
Übungsaufgabe 33:
33-1: Berechnen Sie den hydraulischen Druck pw und die hydraulische Conductivity Kf bei x1
= 1 m (h1 = 5.2 m) und bei x2 = 2 m (h1 = 5.0 m) für A) Grobsand (k = 10-9 m2, eff = 0.3) und
für B) Sandstein (k = 10-14 m2, eff = 0.25). Die Dichte von Wasser beträgt 1 kg/L und die
dynamische Viskosität ca. 10-3 Pa s.
33-2: Welcher Druck herrscht direkt an GW-Oberfläche?
33-3: Welcher Druck herrscht 10 m unterhalb der GW-Oberfläche?
33-4: Berechnen Sie die mittlere GW-Geschwindigkeit für A) und B)!
Übungsaufgabe 34: Calculate the permeability and the mean capillary radius for different
porous media, if the hydrualic conductivity is given: fine sand: 10-4 m/s, medium sand: 10-3
m/s, coarse sand: 10-2 m/s. The correction factor C = 0.01.
Übungsaufgabe 35: The hydraulic gradient is 0.001. Calculate the Darcy flux density for
medium sand. The porosity is 0.35. What is the mean velocity?
Übungsaufgabe 36: Using Hagen-Poiseuille-law and assume that the porous media can be
described by a homogeneous capillary bundle model, what is the radius of the capillary?
Übungsaufgabe 37: Write Darcy's law for all 3 directions. Do not forget that the hydrostatic
pressure is ρwgz. The isotropic hydraulic gradient is 0.001. Calculate the Darcy velocity for all
3 directions for fine sand.
Tasks: Darcy-Law – Pore scale
Übungsaufgabe 38: Porous media is described by a capillary bundle model with a lognormal
(pore size) radius distribution. The mean radius is 0.3 mm and the SDV is 0.1 mm. Calculate
the Darcy flux density for a hydraulic gradient of 0.01. Approximate the parabolic velocity
profile by the mean velocity.
Übungsaufgaben*: Air sparging
Übungsaufgabe 39: 1D- Horizontal gas bubble flow
Below an aquitard layer you inject air for short time pulse Δt = 10 s with an injection rate of 1
L/min. The porosity is 0.3. Calculate the radius of the injected gas bubble. Assume that the
bubble is rigid (no deformation) and that only one bubble is formed that travels in positive xdirection. The cross section at the injection point can be estimated as 25 cm2.
Calculate the maximal travel distance of the gas bubble by applying Newton's second law and
Stokes law for the friction force: Ff = 6πηvR.
A) The dynamic viscosity is given by the rigid gas bubble
B) The dynamic viscosity is given by the adjacent water phase
C) Increase the injection rate by the factor 10 and repeat all calculations.
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Übungsaufgabe 40: 2D- gas bubble flow
The point of injection is within the homogenous aquifer 10m below the groundwater surface.
Calculate the path of the injected gas bubbles (because of symmetry 2 bubbles will be formed
one in negative x-direction and the second in positive x-direction) for Q1 = 1 L/min and Q2 =
10 L/min. Calculate the breakthrough time and the radius of influence.
Task 41: Partitioning equilibrium: LNAPL-Multi-component phase – Raoult’s law
a) A LNAPL-Mixed phase contains the following Mol-fractions: X-Benzene = 0.25, XToluene = 0.35, X-Ethylbenzene = 0.15, X-Xylene = 0.15. Calculate the maximal
concentrations for Benzene, Toluene, Ethylbenzene and Xylene within the water phase!
Calculate also the corresponding partial pressures of the gas phase, if partitioning equilibrium
holds.
b) Weisen Sie nach, dass in Abbildung: Henry‘s law - Raoult‘s law (Folie 47)
kein Verteilungsgleichgewicht besteht!
Beachten Sie, dass foc maximal gleich 1 sein kann (Koc =38)!
Task 42: Modelling with BIOSCREEN
BIOSCREEN uses an approximate solution (see West et al. (2007) Vol. 45, No. 2, GROUND
WATER, p. 126–135, download from UFZ-webpage). Since BIOSCREEN is used by
engineers and scientists around the world for designing remediation technologies, it is
important to estimate the absolute error (= Abs(Cexact - CBioscreen)). Estimate the absolute error
for the concentration profile of the risk analyses of LNAPL-scenario (Task 22). Compare the
exact 1D-concentration of the advection-dispersion equation with the centerline concentration
for t = 1 year and a modeling area of 500m by 20m. Since no reactive processes are
considered (no biodegradation) you have to set the rate constant zero!
In order to compare the BIOSCREEN solution and the 1D-exact solution you have to set the
transversal and vertical dispersion zero or a very small value, e.g. 0.001.
The original BIOSCREEN software is based on American Units: Length is measured in foots
(1 foot is 0.3048 m).
There is no biodegradation (large half-life time, e.g. 100 years) and the background
concentration of other species, like DO, NO3, etc. is zero.
Parameter
The geometry of the source (pure Benzene phase) is 20m x 20m x 5m.
Hydraulic gradient 0.01
Hydraulic conductivity kf = 10-3 m/s
Solid density s= 2.65 kg/L
Total porosity = Effective porosity = 0.3
Contaminant Benzene (KOC = 38)
Total Benzene mass = 500 t
Content of organic carbon = 0.01
Longitudinal dispersivity δx = αx = 10 m
Download the software (EXCEL-file) and manual from UFZ-webpage.
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Hint: DATA-input in BIOSCREEN can be done directly or indirectly! For indirect data input
you gave the hydraulic gradient and the mean velocity is calculated by BIOSCREEN!
To use the indirect data input you have to push the button “Restore Formulas for Vs ..”!
Use as model area plume length and lateral source extension in foot (1 foot = 0.3048 m)!
Task 43: Bioremediation case study: BTEX-MTBE contamination at Leuna test site
A) General task, test site description, field data
The chemical plant for gasoline production at the Leuna test site operated up to 1990, for
about 50 years. The last 20 years MTBE was used in gasoline instead of lead. Through the
SAFIRA-investigation a large BTEX- and MTBE-contamination of the groundwater were
detected. The lateral dimensions of the source are 100 m by 100 m and the thickness is about
3 m. The total source mass M0 of the different components could be estimated between 100
and 800 t. The maximal concentration at x = 0 (= C0, source output into groundwater) is
equal to the mixed solubility for a specific component that is given by Raoult’s law!
From upgradient field measurements the following ED- and EA-concentrations were obtained
at the plume centerline (see Table 2 and 3).
The aquifer is a medium-sandy aquifer.
The estimated plume length (Lp, unit = ft) is 600 ft.
The fraction of organic carbon foc = 0.02.
The soil bulk density is 1.855 kg/L.
Hint: DATA-input in BIOSCREEN can be done directly or indirectly! For indirect data input
you gave the hydraulic gradient and the mean velocity is calculated by BIOSCREEN!
To use the indirect data input you have to push the button “Restore Formulas for Vs ..”!
Use as model area plume length and lateral source extension in foot (1 foot = 0.3048 m)!
Make sure that your BIOSCREEN field data table has 11 columns at the right x-positions
shown in Table 2 and 3!
Important Note: According to the BIOSCREEN manual the widths of the source zone sections
add to the total width, e.g. 4 sections: 25m + 25m + 25m + 25m = 100 m
Table 1 Pure solubilities of source components.
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Table 2 Electrondonor concentrations measured at plume centerline at Leuna test site in
mg/L.
Table 3 Electron acceptor background concentrations measured at plume centerline at Leuna
test site in mg/L.
Table 4 Mass stoichiometric coefficients (= utilization factor) for the BTEX-group.
B) Additional parameters of the LEUNA-test site
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
B1) Transport data for Dispersive-Advective Flow
Hydraulic gradient: 0.005.
Hydraulic conductivity is 0.001 m/s.
Effective Porosity
Typical values
Clay
0.01 - 0.20
Sandstone
0.005 - 0.10
Silt
0.01 - 0.30
Unfract. Limestone 0.001- 0.05
Fract. Granite
0.00005 - 0.01
Fine Sand
0.10 - 0.30
Medium Sand 0.15 - 0.30
Coarse Sand
0.20 - 0.35
Gravel
0.10 - 0.35
Organic Carbon Partition Coefficient (Koc, unit: L/kg)
Typical values:
Benzene: 38 L/kg
Ethylbenzene: 95 L/kg
Toluene: 135 L/kg
Xylene: 240 L/kg
MTBE: 11.2 L/kg
Fraction Organic Carbon (foc) = 0.02
Estimated Plume Length (Lp, unit = ft): 600 ft
typical values: For BTEX plumes, 50 - 500 ft. For chlorinated solvents, 50 to 1000 ft.
Dispersivities (unit = ft)
Longitudinal Dispersivity (alpha_x):
Transverse Dispersivity (alpha_y):
Vertical Dispersivity (alpha_z):
alpha_y = 0.10 alpha_x
alpha_z is very low (i.e. 1 × 10-99 ft)
B2) Source zone data
Source half-life time t1/2,source = ln(2)/ksource = 0.693/ksource ( unit = year)
ksource = source rate constant (unit = 1/year)
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Figure 1 Definition of source zone data.
Description:
The Domenico (1987) model assumes the source is infinite, i.e. the source concentrations are
constant. In BIOSCREEN, however, an approximation for a declining source concentration
has been added. Note that this is an experimental relationship, and it should be applied with
caution. The declining source term is based on the following assumptions:
• There is a finite mass of organics in the source zone present as a free-phase
or residual NAPL. The NAPL in the source zone dissolves slowly as fresh
groundwater passes through.
• The change in source zone concentration can be approximated as a firstorder
decay process. For example, if the source zone concentration "half-life"
is 10 years and the initial source zone concentration is 1 mg/L, then the
source zone concentration will be 0.5 mg/L after 10 years, and 0.25 mg/L after 20
years.
Definition:
t1/2,source = (0.693 * M0 ) / (Q * C0)
Q = Groundwater flow through source zone (L/yr)
C0 = Effective source zone concentrationat t = 0 (mg/L)
B3) Field data: Biodegradation kinetics
First-Order Decay Coefficient (lambda, kbio) unit: 1/day, 1/year
Definition: Rate coefficient describing first-order decay process for dissolved constituents.
The first-order decay coefficient equals 0.693 divided by the half-life of the contaminant in
groundwater. In BIOSCREEN, the first-order decay process assumes that the rate of
biodegradation depends only on the concentration of the contaminant and the rate coefficient.
For example, consider 3 mg/L benzene dissolved in water in a beaker (= Becher). If the halflife of the benzene in the beaker is 728 days, then the concentration of benzene 728 days from
now will be 1.5 mg/L (ignoring volatilization and other losses).
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
TASK:
Source zone characterization of Multi-component LNAPL (BTEX, MTBE)
43-1. Calculate the mixed solubilities for MTBE and BTEX-components using Raoults law!
The mol-fractions are given in Table 1. The density of MTBE is 0.74 kg/L.
43-2. Determine the total source mass by trial-and-error curve fitting and using the MTBEfield data (see Table 2). In presence of BTEX the MTBE behaves like a conservative tracer,
i.e. no biodegradation! For the best fit the standard deviation (can be calculated in EXCEL)
must be a minimum! Try only values M0 = n × 100 t, where n is a natural number!
The steady-state MTBE-plume is reached after 0.5 years.
43-3. The rate constant of the source zone can be calculated through the following formula:
𝑘𝑠 =
𝑄𝑤 ∙𝐶0
𝑀0
,
where Qw is the groundwater flow rate [ft3/year].
TASK: Characterization of the MTBE/BTEX-plume
43-4. Calculate the MTBE plume mass!
The time dependence of the source mass is
𝑀(𝑡) = 𝑀0 × exp⁡(−𝑘𝑠 ∙ 𝑡).
The difference mass M(t) = M0 – M(t) is released into the aquifer, i.e. the plume mass. Part
of this mass is adsorbed and the other part is dissolved into the plume. Assuming partitioning
equilibrium you can calculate the plume mass Mplume and the adsorbed mass Mads. Compare
the plume mass with the BIOSCREEN result using a summation over the concentration array!
The steady-state plume for MTBE is reached after 0.5 years.
43-5. Determine the steady-state plume mass for Benzene and Toluene-component without
biodegradation by following the same calculation steps as for MTBE!
The steady-state plume for Benzene is reached after 1 year.
The steady-state plume for Toluene is reached after 4 year.
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
TASK:
Biodegradation: Utilization factors, Biodegradation capacity, Instantaneous
reactions
Figure 2 Typical plume centerline concentration profile of EDs (= BTEX) and EAs (O2,
NO3, SO4, CO2, and Fe(3+). Note that Fe(2+) and CH4 are the by-products of anaerobic
biodegradation.
43-6A. Calculate the mass stoichiometric coefficient (or utilization factor) for the BTEXgroup for aerobic biodegradation. Average the 4 utilization factors and compare your result
with the result in Table 4.
43-6B. Calculate the mass stoichiometric coefficient (or utilization factor) for the Benzene for
nitrate reduction and sulfate reduction. Average the 4 utilization factors and compare your
results with the result in Table 4.
43-6C. Calculate the mass stoichiometric coefficient (or utilization factor) for MTBE for
aerobic biodegradation and nitrate reduction.
43-7. Calculate the intrinsic Biodegradation capacity for Benzene for the whole LEUNA test
site. Use the utilization factors from Table 4 and the following general formula:
Biodegradation capacity = ∑𝐸𝐴𝑠 𝐶𝐸𝐴 /𝑈𝐹(𝐵𝑧, 𝐸𝐴).
CEA means the averaged concentration of the electron acceptor. The sum runs over all EAs!
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Figure 3 Instantaneous reactions and biodegradation capacity (BC).
43-8. Determine the Benzene and Toluene mass that is removed or mineralized by intrinsic
biodegradation assuming instantaneous reactions!
Calculate the mass flux at source-plume interface (x = 0) and 300 ft downgradient (x = 300
ft).
TASK: Biodegradation: First order kinetics
43-9. Determine the Benzene mass that is removed or mineralized by aerobic biodegradation
during Direct oxygen Gas Injection (DGI). Oxygen gas is injected in a gallery of 10 lances
along the source-groundwater interface. The biodegradation rate constant is 1 1/year.
Calculate the mass flux at source-plume interface (x = 0) and 300 ft downgradient (x = 300
ft). Simulation time is 1 year.
TASK 44: Designing and Dimensioning of Reactive Walls using
BIOSCREEN
Test site description, field data
The chemical plant for gasoline production at the Leuna test site operated up to 1990, for
about 50 years. The last 20 years MTBE was used in gasoline instead of lead. Through the
SAFIRA-investigation a large BTEX- and MTBE-contamination of the groundwater were
detected. The lateral dimensions of the source are 100 m by 100 m and the thickness is about
3 m. The total source mass M0 of the different components could be estimated between 100
and 800 t. The maximal concentration at x = 0 (= C0, source output into groundwater) is
equal to the mixed solubility for a specific component that is given by Raoult’s law!
From upgradient field measurements the following ED- and EA-concentrations were obtained
at the plume centerline (see Table 2 and 3).
The aquifer is a medium-sandy aquifer.
The estimated plume length (Lp, unit = ft) is 600 ft.
The fraction of organic carbon foc = 0.02.
The soil bulk density is 1.855 kg/L.
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Homework for the course: „Remediation of contaminated sites: Innovative groundwater remediation methods“
Hint: DATA-input in BIOSCREEN can be done directly or indirectly! For indirect data input
you gave the hydraulic gradient and the mean velocity is calculated by BIOSCREEN!
To use the indirect data input you have to push the button “Restore Formulas for Vs ..”!
Use as model area plume length and lateral source extension in foot (1 foot = 0.3048 m)!
Make sure that your BIOSCREEN field data table has 11 columns at the right x-positions
shown in Table 2 and 3(see TASK 43)!
Important Note: According to the BIOSCREEN manual the widths of the source zone sections
add to the total width, e.g. 4 sections: 25m + 25m + 25m + 25m = 100 m
ORC-Wall
300 ft down-gradient of the source-groundwater interface (x = 0) a reactive ORC-wall should
be installed for stimulating the aerobic degradation of Benzene. The geometry of the RW is
(Lx, Ly, Lz). ORC-reactive materials, for instance Permanganate, are cost-intensive materials.
Therefore you have to optimize the costs. The objective is that the concentration at the left
and right boundary (y1 = Ly/2 and y2 = -Ly/2) of the RW is less or equal to 15% of the
Centerline-Benzene concentration. The thickness of the ORC-Wall is Lx = LRW =1 m.
Use for dimensioning the steady-state BIOSCREEN plume shape!
The Benzene source mass can be calculated by trial-and-error fit (see TASK 43-5). The
steady-state plume for Benzene is reached after 1 year.
A) Dimensioning of the ORC-RW
Determine the reactive area ARW = Ly x Lz, i.e. lateral extension of the RW and its height Lz.
B) Designing of the ORC-RW
Determine the reactivity of the RW (rate constant k [1/day] using the mass balance criterion
(1) of a RW (see lectures RW(1)). You have to use the exact advective-dispersive mass flux
and not(!) the simple convective approximation, since the error of 20% can lead to serious
contamination down-gradient of the RW! The exact mass flux can be obtained from
BIOSCREEN.
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