16-Civ-B5 Water Supply and Wastewater Treatment · December 2015

Question 2 of 7: Total, Volatile and Inorganic Suspended Solids of a Sludge Sample

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 98-Civ-B5 Water Supply and Wastewater Engineering — National Examination, December 2015. Three hours; closed book, one aid sheet written on both sides; an approved calculator is permitted. Question 1 is compulsory and the candidate attempts any five of the remaining six, so 100 marks are written out of the 115 printed (Q1 = 25 marks, Q2 to Q7 = 15 marks each; Q2 splits 12 + 3). Every one of the seven questions is solved below, because this set is a study resource rather than an exam script.

Reference texts.

Metcalf & Eddy / AECOM, Wastewater Engineering: Treatment and Resource Recovery, 5th ed. — solids fractionation, nitrogen chemistry, sedimentation theory, anaerobic digestion.
Crittenden, Trussell, Hand, Howe & Tchobanoglous, MWH's Water Treatment: Principles and Design, 3rd ed. — coagulation and flocculation theory, softening, filtration, disinfection.
Davis & Cornwell, Introduction to Environmental Engineering, 5th ed. — treatment-train selection, hardness chemistry, residuals management.
Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design, 3rd ed. — mass balances, reaction kinetics, water-quality indicators.
CCME, Canadian Environmental Quality Guidelines (ammonia and receiving-water criteria) and Health Canada, Guidelines for Canadian Drinking Water Quality (treated-water objectives).
APHA/AWWA/WEF, Standard Methods for the Examination of Water and Wastewater — Method 2540 D/E for the solids series.

Check: representative design data. Questions 1, 3, 4 and 7 are discussion questions and print no numbers. Where a number appears in those answers it is a representative Canadian municipal value chosen by the solver to make the argument concrete; it is labelled as such at the point of use, and every one of them. The graded content of those questions is the reasoning, not the arithmetic.

Question 2: Total, Volatile and Inorganic Suspended Solids of a Sludge Sample (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single gravimetric solids determination on a sludge sample, following the three weighings of Standard Methods 2540 D and E.

Given data for the solids determination
Quantity	Symbol	Value
Sample volume filtered	$V$	20 mL = 0.020 L
Clean, dried filter paper	$W_1$	1.2345 g
Filter plus residue after drying at 105 °C	$W_2$	1.3456 g
Residue remaining after ignition at 550 °C	$W_3$	0.0234 g

Find. (a) The total suspended solids, the volatile suspended solids and the inorganic (fixed) suspended solids of the sludge, each expressed in mg/L; and (b) how the reported results would change had the muffle furnace been operated at 900 °C instead of 550 °C.

Check: unit typo in the source. The paper states the residue after ignition as “0.0234 mg/L”. That cannot be a concentration: it is one of three weighings, it is compared against masses of 1.2345 g and 1.3456 g, and a residue of 0.0234 mg/L against a TSS of over 5 000 mg/L would mean the sludge was 99.999 % organic, which no sludge is. The value is read here as a mass of 0.0234 g of ash, which is the only dimensionally consistent reading and which yields a volatile fraction of 79 %, exactly what a mixed municipal sludge gives. The exam's own note 2 invites the candidate to state such an assumption with the answer paper.

Approach. Every solids parameter in the series is a mass difference between two weighings, divided by the volume filtered: the 105 °C step separates suspended solids from water, and the 550 °C step separates the organic (volatile) fraction from the mineral (fixed) fraction of those same solids.

Figure 2.1 — What each weighing measures. The 105 °C weighing fixes the whole bar; ignition at 550 °C burns off the upper block and leaves the lower one, so VSS is obtained by difference and never weighed directly.

(a) TSS, VSS and inorganic suspended solids (12 marks)

Recover the mass of dry suspended solids retained on the filter. The filter gained mass only by the solids it captured, so $$m_{\mathrm{TSS}} = W_2 - W_1 = 1.3456 - 1.2345 = 0.1111\ \mathrm{g}$$
Convert that mass to a concentration on the filtered volume. Dividing by the 20 mL filtered and converting grams to milligrams, $$\mathrm{TSS} = \frac{m_{\mathrm{TSS}}}{V} = \frac{0.1111\ \mathrm{g}}{0.020\ \mathrm{L}}\times 10^{3}\ \frac{\mathrm{mg}}{\mathrm{g}} = \boxed{5555\ \mathrm{mg/L}}$$
Read the inorganic fraction directly from the ignition residue. Ignition at 550 °C destroys the organic matter and the cellulose of the filter itself; what survives is the mineral matter that came in with the sludge, so $$\mathrm{ISS} = \frac{W_3}{V} = \frac{0.0234\ \mathrm{g}}{0.020\ \mathrm{L}}\times 10^{3} = \boxed{1170\ \mathrm{mg/L}}$$ This quantity is also called the fixed suspended solids, FSS.
Obtain the volatile fraction by difference. Volatile suspended solids are never weighed directly — they are the mass that left during ignition: $$\mathrm{VSS} = \mathrm{TSS} - \mathrm{ISS} = 5555 - 1170 = \boxed{4385\ \mathrm{mg/L}}$$
Check the result against what a sludge should look like. The volatile fraction is $$\frac{\mathrm{VSS}}{\mathrm{TSS}} = \frac{4385}{5555} = 0.789 \;\;\text{or}\;\; 78.9\ \%$$ which sits squarely in the 70 to 85 % band expected of a mixed primary and waste activated sludge, and the two fractions sum back to the measured TSS. Had the sample been a digested sludge the ratio would have fallen to 55 to 65 %, and a chemically conditioned or lime-stabilised sludge would fall lower still.

(b) Effect of igniting at 900 °C instead of 550 °C (3 marks)

Standard Methods specifies 550 ± 50 °C for a reason, and it is not a tolerance for convenience. At that temperature organic matter is oxidised completely while the mineral matter is left intact. At 900 °C the mineral matter is no longer inert. Carbonates calcine,

$$\mathrm{CaCO_3} \xrightarrow{\ \sim 900\,{}^{\circ}\mathrm{C}\ } \mathrm{CaO} + \mathrm{CO_2}\uparrow$$

which alone removes 44.0 g of every 100.1 g of calcium carbonate present. Magnesium carbonate decomposes even lower, water of crystallisation is driven off hydrated salts, ammonium salts and some chlorides volatilise, and any calcium sulphate begins to break down. Every one of those losses is recorded by the balance as if it had been organic matter.

The consequences are specific and worth stating separately. TSS is unaffected, because it is fixed by the 105 °C weighing, which happens before the furnace is ever used. ISS is under-reported, because part of the mineral matter has been driven off as gas. VSS is over-reported by exactly the same amount, because it is obtained by difference — the error is not independent, it is transferred one-for-one from the fixed to the volatile column. If, say, half the 1170 mg/L of ash in this sample were calcium carbonate, the spurious loss is $0.5\times1170\times0.4397 = 257$ mg/L, so VSS would be reported as 4642 rather than 4385 mg/L, an error of 5.9 % on VSS and 22 % on ISS. There is a second, physical consequence: at 900 °C glass-fibre filters sinter and porcelain crucibles can spall, so part of the discrepancy may be loss of the apparatus itself rather than of the sample.

Because VSS is routinely used as the surrogate for active biomass in a digester or an activated-sludge tank, an inflated VSS propagates directly into an inflated estimate of biomass inventory, an under-estimated food-to-microorganism ratio and an over-estimated solids retention time. The test must be repeated at the correct temperature; the result cannot be corrected arithmetically, because the carbonate content of the ash is not known.

Question 2 — results
Quantity	Value
Mass of dry solids retained, $W_2-W_1$	0.1111 g
Total suspended solids, TSS	5555 mg/L
Inorganic (fixed) suspended solids, ISS	1170 mg/L
Volatile suspended solids, VSS	4385 mg/L
Volatile fraction, VSS/TSS	78.9 %
(b) Effect of 900 °C on TSS	No effect — set by the 105 °C weighing
(b) Effect of 900 °C on ISS	Under-reported (carbonates calcine, salts volatilise)
(b) Effect of 900 °C on VSS	Over-reported by the identical mass; about +257 mg/L, or +5.9 %, if half the ash were CaCO₃

Topic. The gravimetric solids series — total, volatile and fixed suspended solids — and the temperature dependence of the ignition step.

Key relations. $\mathrm{TSS} = (W_2-W_1)/V$; $\mathrm{ISS} = W_3/V$; $\mathrm{VSS} = \mathrm{TSS}-\mathrm{ISS}$; calcination of calcium carbonate with a mass loss of $M_{\mathrm{CO_2}}/M_{\mathrm{CaCO_3}} = 44.0/100.1 = 0.440$.

Why this works. The solids series is a sequence of thermal separations, each chosen so that one class of material is removed and the rest is not. Drying at 105 °C removes free and most bound water without decomposing organic matter, so the residue is genuinely “dry solids”. Igniting at 550 °C is hot enough to oxidise organic carbon to carbon dioxide and water but cool enough to leave carbonates, silicates and oxides untouched, so the residue is genuinely “mineral”. The whole method rests on that second window; move outside it and the classification stops meaning what its name says. VSS matters in practice because it is the cheapest available proxy for the mass of micro-organisms in a biological reactor, which is why every activated-sludge and digester calculation is written in terms of VSS rather than TSS.

Common pitfall. Two mistakes are almost universal on this question. The first is dividing by 20 instead of 0.020, or forgetting the gram-to-milligram conversion, which lands the answer three orders of magnitude out; the sanity check is that a sludge should report in thousands of mg/L, not in units or millions. The second is asserting that the 900 °C error affects “the results” in general without recognising that TSS is untouched and that the ISS and VSS errors are equal and opposite — that structure is the actual content of part (b).

Source. APHA/AWWA/WEF, Standard Methods, Method 2540 D and E; Metcalf & Eddy, Wastewater Engineering, 5th ed., Ch. 2 (constituents in wastewater, solids fractionation).

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