BIOREMEDIATION EXAM - 1999 SOLUTION
Point values are shown in [brackets]. List assumptions used and show units.
1. [2] Write a balanced stoichiometry for the aerobic bio-mineralization of ortho-xylene.
C6H4(CH3)2 + 10.5 O2 -> 8 CO2 + 5 H2O
2. [2] The Michaelis-Menten biokinetics for aerobic bacteria degrading o-xylene are: K = 1 g/g-d, Ks =
1 mg/L, and net yield = 0.4 g VSS/g o-xylene. If the initial biomass concentration is 10 mg VSS/L, and
50 mg/L o-xylene is completely biodegraded, what is the final biomass concentration (in mg VSS/L)?
Y * (So - Sf) = biomass growth; 0.4 g VSS/g oX (50 mg/L oX) = 20 mg/L VSS
10 mg/L VSS initial + 20 mg/L VSS growth = 30 mg/L VSS final
3. [4] If no other nitrogen source is available, how many grams of nitrate-N would be consumed per
gram of ortho-xylene biomineralized under aerobic conditions (for the bacteria in question 2)?
of the dry cell weight, 50% is carbon, 14% is nitrogen.
Use the nitrate-N for cell material. 1 g oX biomineralized = 0.4 g VSS
assume that only carbon was combusted away: 0.4 g C * 14 g N / 50 g C = 0.112 g NO3-N
OR assume VSS = 0.8 * total dry mass; then 0.4 g VSS/0.8 = 0.5 g dry mass * 0.5 g C/g dry
0.25 g C * 14/50 = 0.07 g NO3-N
4. [4] How many mg/L of nitrate-N would be consumed as an electron acceptor to biodegrade 50
mg/L o-xylene if no oxygen is available?
C8H10 + x NO3- + x H+ -> 8 CO2 + (x/2) N2 + ((10-x) / 2) H2O
O: 3x = 16 + (10-x)/2 => x = 6
C8H10 + 6 NO3- + 6 H+ -> 8 CO2 + 3 N2 + 2 H2O
50 mg/L oX * 1 mol/106 g * 6 mol / 1 mol * 14 g/mol =39.6 mg/L NO3-N
5. [2] Write a balanced stoichiometry for aerobic co-metabolic biodegradation of chloroform (CHCl3).
CHCl3 + 0.25 O2 + 1.5 H2O -> CO2 + 3 HCl
6. [5] What are 6 co-substrates for aerobic cometabolic TCE degradation?
methane, ammonia, propane, isoprene, butane, toluene, phenol
For 2 of these, give 1 advantage and 1 disadvantage (compared to the alternate co-substrates).
methane: gas, non-toxic, fastest rate TCE deg., low solubility, potentially explosive;
intermediate toxicity
phenol: high solubility, potential to avoid intermediate toxicity (high Tc), toxic itself,
toluene: toxic to humans to a low effluent conc. allowed; supports a range of microbes
7. [6] What are the three primary purposes of the additives used in composting?
rich organics for co-substrates
bulking agent (better oxygen supply)
fast degrading organics produce heat to increase bioavailability and biokinetic rates
nutrients (N, P)
list the purpose(s) of each of the following:
cow manure - co-substrate; heat generation
straw - bulking
crushed oyster shells - buffer pH changes
8. [4] List 4 different types of ex-situ biological reactors used to treat contaminated gases.
biofilter biotrickling filter bioscrubber suspended-growth reactor
9. [4] Given the following data from a 4 cm diameter x 10 cm tall soil core: weight = 218.9 g, dry weight = 217.14 g,
ignited (post 550°C) weight = 216.88 g. When the final (post ignited) soil was added into a beaker containing 100 mL
of water, the final water level in the beaker was 190 mL. Calculate the soil porosity.
pb = (1-n) ps; pb = dry weight / total cylinder volume; ps = ignited weight / water volume displaced
217.14 g / 126.66 cm3 = (1-n) (216.88 g / 90 cm3)
1.714 = (1-n) 2.41 n = 0.289
10. [8] Given the following kinetic coefficients for a pure culture of aerobic bacteria: maximum specific benzene
degradation rate 5 g benz/g VSS-d; benzene half-saturation coefficient 0.1 mg/L; net yield 0.5 g VSS/g benz; substrate
inhibition coefficient 100 mg benz/L; endogenous decay rate 0.01 d
-1
; maximum specific growth rate on xylene 4 d
-1
;
xyl half-saturation concentration 1 mg/L; substrate inhibition coefficient 500 mg xylene/L; net yield 0.4 g VSS/g xyl.
a. Find the initial benzene degradation rate with 100 mg/L VSS and 20 mg/L benzene
dS/dt = K S X / (Ks + S + S^2/Ki)
dS/dt = 5 g/g-d * 20 mg/L * 100 mg/L / (0.1 mg/L + 20 mg/L + 20 mg/L^2 / 100 mg/L)
dS/dt = 10000 mg/L-d / 24.1 = 415 mg benz / L soln -d
b. the initial xylene degradation rate with 200 mg/L VSS, 20 mg/L benzene, and 10 mg/L xylene
dS/dt = K S X / (Ks (1 + B/Ks-b) + S + S^2/Ki)
µm / Y = K => K = 4 d
-1
/ 0.4 g VSS/g oX = 10 g oX/g VSS-d
dS/dt = 10 g/g-d * 10 mg/L * 200 mg/L / (1 mg/L (1 + (20 mg/L/0.1 mg/L)) + 10 mg/L + 100
mg/L/500)
dS/dt = 20000 mg/L-d / 211.2
dS/dt = 94 mg oXylene / L soln -d
11. [3] List three compounds that can be degraded by anaerobic bacteria but can not be degraded by
any known aerobic bacteria under any conditions (including potential co-substrate availability).
tetrachloroethene, hexachloroethane, carbon tetrachloride, >6 Cl-PCB, hexachlorobenzene
12. [6] List three methods that you could use to prove that the bacteria present in a soil sample have
the capability to degrade BTEX compounds (be a specific as possible).
plate on agar with BTEX present as the sole carbon source, look for growth
use a DNA probe to look for TOL, TMO, TOM, TBU, TOD coding sequence in DNA
use an enzyme assay to look for above listed enzymes
use a lab test in closed flask, add radiolabeled BTEX, look for
14
CO
2
production; compare to
“killed” controls
13. [6] The following compounds are all in a gas stream bubbled into a 2 m deep liquid reactor.
Cmpd Inlet Gas Conc, mg/L MW Solubility, mg/L H Kow, L/L
A 10 100 1000 0.2 1000
B 10 120 500 0.2 100
C 10 110 2000 0.1 10
a. If bioactivity maintains the liquid concentration of all compounds below 0.05 mg/L, which
compound (A, B, or C) will have the lowest concentration in the effluent gas?
predominant problem is mass transfer of compound into liquid out of the gas
CMPD C has the lowest Henry’s and therefore greatest desire to partition into liquid
b. If bioactivity maintains the liquid concentration of cmpd A at 0.2 mg/L, cmpd B at 2 mg/L,
and cmpd C at 20 mg/L, which will have the lowest concentration in the effluent gas?
Here, the high liquid conc prevents a high “driving force” to get the compound to partition out
of the gas and into the liquid. “H * Cl” term is highest for C (0.1*20 = 2) so it will transfer the
worst. H * Cl is the lowest for compound A (0.2*0.2 = 0.04) so it will transfer the best and have the
lowest concentration in the effluent gas.
14. [9] Given that an FBR at steady-state is treating 100 m
3
/min contaminated water with inlet
concentration of 100 mg/L toluene and an effluent concentration 0.1 mg/L toluene. If the
biodegradation kinetics of toluene are 1st order below 5 mg/L toluene with k = 10 L/mg biomass-d:
a. What is the total biomass concentration in the reactor (in g biomass)?
Mass Balance TOLUENE:
100 mg/L * 100 m3/min = 0.1 mg/L * 100 m3/min + 10 L/mg-d (BIO) 0.1 mg/L tol
1 E7 mg/min - 1 E4 mg/min = 1(BIO) /d
9.99 E6 mg/min * 60*24 min/d = BIO /d
BIO = 1.43856 E10 mg biomass = 1439 kg bio
b. This reactor contains 1500 kg of sand packing, the packed bed volume is 1 m
3
with
porosity of 0.4, and the reactor is operated with a recycle rate that causes 200% bed expansion.
What is the final depth of reactor needed, assuming that the reactor has a cross-sectional area of 1 m
2
?
packed bed 1 m3 + expansion of 1 m3(2) = 3 m3 total expanded bed
add 10% safety factor => 3 * 1.1 = 3.3 m3 /1 m2 = 3.3 m deep
c. What is the HRT in your reactor?
HRT = Vl / Ql = (3.3 m3 total - 0.6 m3 solid) / 100 m3/min = 0.027 min
15. [6] A soil slurry reactor operated with an HRT of 50 days has an inlet concentration of 1000 mg
TOC/L slurry, the solids content in the slurry is 10% by volume, and the effluent concentration is 50
mg TOC/L solids and 10 mg TOC/L liquid. (assume soil solids have a density of 2 g/mL)
a. If the biodegradation rate is first order with respect to the liquid concentration, write an equation(s)
to calculate the biodegradation rate constant.
MASS BALANCE TOC
100 mg TOC/L * Qsl = 50 mg TOC/L solids (0.1 L solids/L slurry) Qsl +
10 mg TOC/L liq (0.9 L lig/L slurry) Qsl + k (10 mg/L) VOL
dividing both sides by slurry flowrate (Qsl): 100 = 5 + 9 + 10 K HRT
b. If the system is at equilibrium, what is the soil:water partitioning coefficient of the TOC?
Kd = Csoil / Cwater = 50 mg TOC/L solids / 10 mg TOC/L liq
Kd = 5 L liq / L solids
16. [5] Given the results below from gas samples taken from 8 ft deep in the vadose zone at 3 sites.
SITE 1 SITE 2 SITE 3
sample % O2 % CO2 HC,
mg/L
% O2 % CO2 HC,
mg/L
% O2 % CO2 HC,
mg/L
1 18 1 0 3 1 15 8 3 1
2 5 8 10 1 1 20 16 1 0
3 6 7 8 2 1 10 4 8 3
4 7 6 6 2 1 5 1 10 10
5 6 7 8 15 1 0 2 9 8
HC = hyrdocarbons; assume that the average H of the HCs is 0.1.
At which sites would you further investigate the potential for remediation by bioventing? BRIEFLY justify.
NOT site 1, since not oxygen limited (O2 conc > 5% everywhere)
SITE 2 since oxygen is depleted significantly where there is HC comtamination compared to bkg
SITE 3 since oxygen is depleted <5% at some high HC contaminated areas, and more than bkg
What are 2 further tests that you would conduct in the field prior to implementing full-scale bioventing?
on site OUR test
in situ air permeability test
17. [4] Given a bioventing well screened from 5 ft below ground to 15 ft below ground. The soil porosity (coarse
gravel) is 0.5; the soil is 10% water saturated, 10% NAPL saturated. You operate an air injection well at 10 cfm which
gives a radius of influence of 50 ft. What is soil gas exchange rate in pore volumes per day?
pore volume of gas within radius of influence = b * pi * r
2
* n-air
10 ft * pi * (50 ft)^2 * (0.5*0.8)
31416 ft3 = 1 pore volume
pore volumes / day = 10 cfm / 31416 cf/pore vol = 0.0003 pore vol / min = 0.46 pore vol/day
18. [9] A site is contaminated with jet fuel that has penetrated the vadose zone and is now floating on
the water table and dissolving into the unconfined aquifer. Background groundwater concentrations of
sulfate, Fe
+3
, nitrate and oxygen are 50 mg/L, 50 mg/L, 50 mg/L, and 8 mg/L, respectively.
a. Under conditions of natural attenuation, most of the jet fuel (by mass) likely be degraded
by bacteria using which of the available electron acceptors?
utilization factors: 3 mg O2/mg HC; nitrate 4.9, iron 21.5, sulfate 4.7
Therefore, sulfate!
Also, remember on the “pie graph” from lecture than methanogenesis was 39%, sulfate
reduction 29%, nitrate reduction 14%, aerobic 10%, iron 8% (as an average over sites)
=> also credit if you said nitrate (more preferred for energy and almost as much available)
b. If there is a drinking water well downgradient that is at risk, in which scenario below (a) or
(b) would natural attenuation be more likely to attenuate the plume before it reaches the well, or no
difference predicts (assume that all non-listed site parameters are the same for A and B):
OPTION A OPTION B your answer + briefly why
b1 groundwater velocity 1
m/yr
groundwater velocity
10 m/y
Option A better chance to attenuate
faster GW velocity gives less time for
attenuation and bioreactions;
b2 the foc of the soil is
0.1%
the foc of the soil is
1%
B better chance to attenuate
higher foc = more sorption, slower plume
movement (bigger Retardation)
b3 longitudinal
hydrodynamic
dispersion is 1 m
2
/d
longitudinal
hydrodynamic
dispersion is 10 m
2
/d
A; less dispersion will slow the movement
ahead of the “Average” velocity of the
plume
b4 transverse
hydrodynamic
dispersion is 0.1 m
2
/d
transverse
hydrodynamic
dispersion is 1 m
2
/d
B; more dilution and “sideways”
movement of plume
b5 10 kg of fuel spilled 100 kg fuel spilled A; less source, greater chance that is will
be attenuated sooner, maybe before plume
hits receptor
19. [8] To determine whether natural attenuation would be sufficient to protect a down-gradient groundwater well from a
plume of chloroform contamination, would you be more likely to use an instantaneous biodegradation model based on
electron acceptors or a first-order model? justify your answer.
CF only degrades aerobically cometabolically; it is also a “slow degrading” compound
anaerobically. Therefore, first-order model probably better than instantaneous (also more typical to
model chlorinated cmpds with 1st order and fuel hydrocarbons with inst. model)
What tests could you run to provide data to support this claim regarding the appropriate kinetic model?
look for production of by-products
look for slow tailing decrease in concs with distance (inst model more blocky)
look for depletion in electron acceptors/donors slowly down the plume rather than “sudden”
blocky change
try to fit Bioscreen model to the current plume
What other data would you gather (in the field and/or the lab) to support a case for natural attenuation to EPA?
bacteria naturally present in the site soil can degrade CF (in the lab)
CF-degrading bacteria active on site (DCM, CM, Cl- by-products)
gene probes?
run a risk model to prove enough attenuation occurring for minimal risk to receptors
20. [3] The addition of a non-ionic surfactant into the groundwater injected into a soil column (foc 2%, porosity 0.3,
5% saturated with PCB NAPL) failed to increase the biodegradation removal of PCBs from the column (compared to a
control without surfactant addition). List 3 possible causes for the lack of enhanced PCB biodegradation.
surfactant was toxic to the bacteria
PCB in the micelle not available for biodegradation
biodegradation of PCB not bio-availability limited (limited by electron acceptor or donor,
nutrients, etc)
higher PCB solubility was toxic to the bacteria
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