Unformatted text preview: CVEN3502 Workshop
Water & Wastewater Characterisation
SOLUTIONS QUESTION 1 (Adapted from Mines, Chapter 3, Q3).
A wastewater sample was analysed to characterise the solids content. An evaporating dish was
initially weighed to obtain the tare mass and then reweighed after evaporation of a 150 ml sample
and then again after ignition in a furnace at 550oC. Then a Whatman “ashless” filter was weighed
to obtain the tare mass and then reweighed after filtration of a 150 ml sample and then again after
ignition in a furnace at 550oC. Use the measurements below to determine the total solids (TS),
volatile solids (VS), dissolved solids (DS), total suspended solids (TSS), and total volatile
suspended solids (TVSS).
Tare mass of evaporating dish = 24.3520 g
Mass of evaporating dish plus residue after evaporation @ 105°C = 24.3970 g
Mass of evaporating dish plus residue after ignition @ 550°C = 24.3850 g
Mass of Whatman “ashless” filter and tare = 1.5103 g
Mass of Whatman “ashless” filter and tare after drying @ 105°C = 1.5439 g
Residue on Whatman “ashless” filter and tare after ignition @ 550°C = 0.024 g
SOLUTION: . . . . / . . . . / . . . / 1 . . . . 0.300 0.224 / 0.076 / . 0 0.224 0.160 . / 0.064 / QUESTION 2
A water sample contains Ca2+ (20 mg/L), Na+ (15 mg/L) and Mg2+ (10 mg/L). The molar mass of Ca2+ =
40 g/mol, Na+ = 23 g/mol, and Mg2+ = 24 g/mol. The molar mass of carbonate ion (CO32-) = 60 g/mol.
Calculate the total hardness (TH) as CaCO3 for this water sample.
Molar concentration of Ca2+ = 0.020 g/L / 40 g/mol = 5.0 x 10-4 mol/L
Molar concentration of Mg2+ = 0.010 g/L / 24 g/mol = 4.2 x 10-4 mol/L.
(note: Na+ does not contribute to hardness)
Total Molar concentration of hardness species: 9.2 x 10-4 mol/L.
Total concentration of hardness as CaCO3 = concentration x molar mass of CaCO3.
= 9.2 x 10-4 mol/L x (40+60) g/mol = 0.092 g/L = 92 g/L QUESTION 3
A 100 ml sample of water is titrated with 0.02N H2SO4. The initial pH is 9.8 and 16.6 ml acid is required
to reach the pH 4.5 endpoint. Determine the following:
a) Hydrogen ion concentration [H+]
b) Hydroxide ion concentration [OH-]
c) Total alkalinity expressed as mg/L CaCO3. 2 SOLUTION:
a) [H+] = 10-pH = 10-9.8 = 1.6 x 10-10 mol/L. b) [OH-].[H+] = kw = 10-14.
[OH-] = 10-14/(1.6 x 10-10) = 6.3 x 10-5 mol/L.
c) . . . . ∗ = 166 mg/L CaCO3. QUESTION 4
An engineer working at a water treatment plant identified cryptosporidium in raw drinking water at
a concentration of 1 oocyst in 10L of water. A tolerable risk concentration has been identified as 4
x 10-5 oocysts/L. Use Table 4.9 from the Australian Guidelines for Water Recycling (see lecture
slides) to determine the following:
a) Is chlorination likely to be sufficient treatment?
b) Would dual media filtration followed by chlorination be sufficient?
c) Suggest an alternative water treatment strategy. SOLUTION:
a) Initial concentration = 1 in 10L = 0.1 oocyst/L = 10-1 oocyst/L.
Chlorination provides 0- 0.5 log removal of cryptosporidium.
So final concentration after chlorination would be (at least) 10-1 x 10-0.5 = 10-1.5 oocyst/L. This is
much higher than the tolerable dose of 4 x 10-5 oocysts/L. So chlorination alone would not be
sufficient. b) Dual media filtration provides 1.5-2.5 log removal of cryptosporidium.
So final concentration after dual media filtration followed by chlorination would be (at least) 10-1 x
10-0.5 x 10-2.5 = 10-4 oocyst/L. This is still higher than the tolerable dose of 4 x 10-5 oocysts/L. c) In order to achieve an exposure less than the tolerable dose, removal by a factor of at least:
10-1 / 4 x 10-5 = 2500 is required. In log-terms, this is 3.4 log removals.
From Table 4.9, there are many combinations that could achieve 3.4 log removals. Examples
include chlorination plus ultraviolet light; or one of the membrane filtration options (each >6.0 log
removal). 3 ...
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