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ENVENG 4110/7110 Solutions
Q1
(a) Solubility is that of the pure substance multiplied by its molar fraction
TPH Benzene Toluene Ethylbenzene Xylenes
Mass Fraction (%) 100 1.1 2.6 1.1 5.2
Molar Mass (g/mol) 102.5 78.1 92.2 106.2 106.2
Solubility (mg/L) 1780 515 152 198
Mass in 1 mol of petrol (g) 102.5 1.1275 2.665 1.1275 5.33
Moles and molar fraction 1 0.01444 0.02890 0.01062 0.05019
Effective Solubility (mg/L) 25.70 14.88 1.614 9.938
(10 marks)
(b) C = PM/(RT) = 183 x 0.0781 / (8.314 x 293.15) = 0.005864 kg/m3
The initial mass flux Fo = 1.32 x 10-6 x 0.005864 / 2.5 = 3.096 x 10-9 kg/m2/s
The initial mass of benzene per unit area is 0.2 x 750 x 0.25 x 0.011 = 0.4125 kg/m2
Let m = mass per unit horizontal area as a function of time
dm/dt = -3.096 x 10-9(m/mo) or dm/m = -(3.096 x 10-9 / 0.4125) dt = -7.5055 x 10-9 dt
Solution ln (m/mo) = -7.5055 x 10-9 t
Let time for m = mo /2 be t1/2 = - ln (0.5) / 7.5055 x 10-9 = 9.235 x 107 seconds
t1/2 = 2.928 years
(10 marks)
(c) Remediation: first remove as much free-floating product as possible by using skimmer
pumps in an interception trench. It is usually necessary to excavate and remove highly
affected soil to landfill. Monitored natural attenuation is then a possible option depending
on the availability of electron acceptors and the appropriate bacteria. This can effectively
oxidise the hydrocarbons to carbon dioxide and water. If there are low levels of dissolved
oxygen in the groundwater the addition of extra oxygen from hydrogen peroxide or
magnesium peroxide can accelerate bio-oxidation. Adding nutrients (e.g. nitrate and
phosphate) can often be effective in increasing biological treatment. Thermal desorption
coupled with a vapour collection/treatment system is another option, depending on the
permeability of the soil. Pneumatic or hydraulic fracturing can be used if the soil is
insufficiently permeable but with risk to foundations, if any.
(5 marks)
Q2
(a) v = Darcy velocity = flow per unit area perpendicular to the flow (m/s)
K = hydraulic conductivity (m/s)
h = total head (m) = height to water table in unconfined aquifers and piezometric height
for confined aquifers, both relative to a suitable datum
x = horizontal distance (m)
dh/dx = total head gradient (no units)
Darcy Equation may be generalised as a tensor equation Vi = j Kij h/xj
Darcy’s equation fails for turbulent flows and time-varying flows
(7 marks)
(b) Take datum at base of aquifer (30 m below GL)
r1 = 5.3 m r2 = 15.7 m
h1 = 24.9 – 1.25 = 23.65 m h2 = 24.9 – 0.31 = 24.59 m
Q = 100 L/min = 0.1 m3/min = 0.1 x 60 x 24 m3/d = 144 m3/d
b = 30 - 12 = 18 m
Substitution in formula gives K = 1.471 m/d
For drawdown at the well use r2 = 0.05 m and h2 as the unknown
h=19.613 m
Drawdown = 24.9 – 19.613 = 5.287 m
(12 marks)
(c)
Q = L b K dh/dx
L = Q / (K b dh/dx) = 144 / (1.471 x 18 x 0.005) = 1088 m
(6 marks)
L
Q3
(a) ON of Na2CrO4 = 0 because it is a neutral molecule. Four oxygen atoms in a compound have ON =
4 x (-2) = -8. Two sodium atoms in a compound have ON = 2 x (+1) = + 2.
Thus ON of chromium = 0 - 2 + 8 = +6. This is chromium’s highest oxidation state and the species is
referred to as chromium (VI) or hexavalent chromium.
Chromium (VI) is highly toxic and a known carcinogen, whereas chromium (III) is an essential
nutrient and only toxic in large doses. Chromium causes severe skin rashes with dermal contact and
can ulcerate the nasal passages and cause lung cancer if inhaled. In the digestive tract stomach
irritation and ulcers have been caused. Anaemia has also been found in humans. In animals,
stomach tumours and possibly cancers in other organs have been caused by ingestion of Cr(VI). Male
fertility in humans is also affected adversely. In the body Cr(VI) is reduced to Cr(III) and eventually
excreted.
There is a risk that in the desert region the lagoon will dry out by evaporation and wind-blown soil
may carry to the town and be inhaled with risk of cancer and nasal irritation. Seepage downwards in
the unconfined surface aquifer will lead to chromium contamination which will be transported
towards the town as the groundwater flows. If the town dwellers use bore water from the surface
aquifer for drinking, food preparation and washing there is risk of ingestion and dermal contact
exposure. Use of bore water in swimming pools will also give rise to exposure by dermal contact and
ingestion
(8 marks)
(b) Using Darcy’s law: V = - K dh/dx
K = 0.00012 cm/s = 0.00012 x 3600 x 24 = 10.368 cm/d = 0.10368 m/d
Seepage velocity Vs = V/n, where n = porosity. Vs =0.002 x 0.10368 / 0.25 = 8.294x10-4 m/d
Time to reach the nearby town is 1500 / 8.294x10-4 = 1.808x106 days = 4955 years
(8 marks)
( c) One option is to reduce the Cr(VI) to Cr(III) by bacterial action. For this to occur a suitable
substrate must be provided as a carbon source (e.g. molasses, lactic acid) plus essential nutrients
(Nitrate and phosphate). These must be injected into the plume upstream of the town. Pump and
treat is also possible with extraction wells and reinjection wells. Chemical treatment could be to
reduce Cr(VI) by applying the extracted groundwater to zero-valent, elemental iron under acid
conditions to produce iron oxide and Cr(III).
2Fe + 2CrO42- + H2O + 4H+ = 2Fe(OH)3 + Cr2O3
Ferrous iron can also be used. Once in the trivalent state the chromium can be precipitated as a
hydroxide by raising the pH by adding an alkali. Another technique is to use ion exchange resins:
first stage a strong acidic resin to produce chromic acid followed by a weak basic resin to adsorb the
chromic acid.
(9 marks)
Q4
(a) Molar mass of C2Cl4 = 2 x 12.01 + 4 x 35.45 = 165.82 g/mol
For PCE, = (PM)/(RT) = (101325 x 0.16582)/(8.314 x 293.15) = 6.8937 kg/m3
For Air, = (PM)/(RT) = (101325 x 0.029)/(8.314 x 293.15) = 1.2056 kg/m3
Thus PCE is much denser than air
(5 marks)
(b) Density of liquid PCE is 1.622 times denser than water and is a DNAPL. PCE will tend to sink
through the aquifer until it meets an impermeable layer (or aquitard).
(3 marks)
(c) Take 1 Litre of soil : Air space = 0.28 L, water space = 0.15 L
Concentration of PCE in water Cw = 150 mg/L = (1000 x 0.15)/165.82 = 0.9046 mol/m3
Mass of PCE in water = 22.5 mg
Vapour Pressure PCE = HCw = 0.0259 x 0.9046 = 0.023429 atm
Concentration PCE in air = 0.023429 x 6.8937 = 0.1615 kg/m3 = 0.1615 g/L = 161.5 mg/L
Mass of PCE in air = 161.5 x 0.28 = 45.224 mg
Mass of soil = 1.5 kg
Kd = FocKoc = 0.02 x 245 L/kg = 4.9 L/kg
Mass of PCE adsorbed on soil = 1.5 x 4.9 x 150 mg = 1102.5 mg
Total PCE = 22.5 + 45.224 + 1102.5 = 1170.224 mg
Saturation value for PCE = 1170.224/1.5 = 780.15 mg/kg
(17 marks)
Q5
(a) Mass TCE (mg) per gram of GAC = qe = 30.9 x (5.0 – 0.01)2/3 = 90.232 mg/g
Mass of GAC = 50 kg = 50,000 g
Mass of TCE = 50,000 x 90.232 mg = 4511582 mg
Number of litres = 4511582/5 = 902316 L = 902.3 m3
(7 marks)
(b) TOC = 50 mg/L = 0.05/(12.01 x 7) = 5.9474 x 10-4 mol/L
1 mole of toluene requires 9 moles of oxygen gas
At 20 oC and 1 atm, 1 mole of air occupies 24.04 L and so the
oxygen concentration = 0.21/24.04 = 0.008735 mol/L
Oxygen required for 1 L of influent = 5.9474 x 10-4 x 9 = 5.35266 x 10-3 mol/L
Air required (100% efficiency) = 5.35266 x 10-3 / 0.008735 = 0.6128 litres of air per litre of
influent. At 20% efficiency we require 0.6128/0.2 = 3.064 litres of air per litre of influent.
(7 marks)
(c) Equation for ion exchange: 2 R-SO3H + Ni2+ = (R-SO3)2Ni = 2 H+
Assuming 100% efficiency the effluent will be sulphuric acid (dissociated in water).
Neutralisation of pH by adding an alkali such as Ca(OH)2.
(3 marks)
(d) Nickel 50 mg/L equivalent to 0.05 / 58.71 = 0.0008516 mol/L. The chemical equivalent is
twice this because Nickel is divalent. Chemical equivalent = 0.0017032 eq/L
The volume of resin = x 0.62 x 1.7 = 1.923 m3 = 1923 L, giving a capacity of 1.8 x 1923 =
3461.4 equivalents. Litres of influent treated = 3461.4 / 0.0017032 L = 2.0323 ML
Surface loading rate = Q/A = 10 m/h and so Q = 10 x x 0.62 = 11.31 m3/h = 11310 L/h
Time to saturate =2.0323 ML / 0.01131 ML/h = 179.7 hours = 7.49 days
(8 marks)
Q6 Trains from the mid-west travel through Adelaide taking products, including
containers of CCA liquid (chromated copper arsenate, wood preservative) to port in Port
Adelaide. In late March 2018, there is a derailment near the site where Fizzydrinks
Pty Ltd manufactures soft drinks and, unknown to Fizzydrinks, the ground water near
it is contaminated. Fizzydrinks uses bore water for the manufacture of its soft drinks
and in the two weeks after the derailment, numerous members of the public fall ill.
The SA Dept of Health investigates and links the illness with the consumption of the
soft drink from Fizzydrinks.
A public authority is called onto the Fizzydrinks site, tests the various ingredients of
soft drink and discovers the contaminant in the bore water.
In your answers please reference the relevant legislation.
(a) The national policy on site contamination talks about the states having a
nationally consistent approach to site contamination. What is the policy called
and what else does it say?
(3 marks)