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

Question 5 of 7: Biodegradable and Total COD from a Measured cBOD 5

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 5: Biodegradable and Total COD from a Measured cBOD₅ (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 measured five-day carbonaceous BOD and the non-biodegradable fraction of the organic matter it came from.

Given and assumed data
Quantity	Symbol	Value	Basis
Five-day carbonaceous BOD	$\mathrm{cBOD_5}$	300 mg/L	Given
Non-biodegradable fraction of the organic matter	$f_{nb}$	0.35	Given
First-order BOD rate constant, base e, at 20 °C	$k$	0.23 d⁻¹	Assumed — typical municipal wastewater
Test duration	$t$	5 d	Definition of BOD₅
Biodegradable COD equals ultimate carbonaceous BOD	$\mathrm{bCOD}=\mathrm{BOD_u}$	—	Assumed — both are the total oxygen equivalent of the biodegradable organics

Find. The biodegradable COD and the total COD of the wastewater, in mg/L.

Approach. The five-day bottle measures only the part of the biodegradable organic matter that is actually oxidised in five days, so the measurement is first extrapolated to the ultimate carbonaceous BOD, that ultimate value is identified with the biodegradable COD, and the given non-biodegradable fraction then scales it up to the total COD.

Figure 5.1 — How the three quantities relate. The total COD splits into a biodegradable and a non-biodegradable part; the biodegradable part equals the ultimate BOD, of which the five-day test captures 68.3 %. The non-biodegradable slice is invisible to any BOD test, however long it runs.

Extrapolate the five-day measurement to the ultimate carbonaceous BOD. BOD exertion is first order in the remaining biodegradable substrate, $$\mathrm{BOD}_t = \mathrm{BOD_u}\left(1-e^{-kt}\right)$$ so the fraction of the ultimate demand exerted in five days at $k = 0.23$ d⁻¹ is $$f_5 = 1-e^{-0.23\times 5} = 1-e^{-1.15} = 0.683$$
Invert for the ultimate demand. Rearranging, $$\mathrm{BOD_u} = \frac{\mathrm{cBOD_5}}{f_5} = \frac{300}{0.683} = 439.0\ \mathrm{mg/L}$$
Identify the ultimate BOD with the biodegradable COD. Both quantities are the total mass of oxygen required to oxidise the biodegradable organic matter completely — the BOD test does it biologically over an unbounded time, the COD test does it chemically in two hours. They therefore measure the same thing on the same material, and $$\mathrm{bCOD} = \mathrm{BOD_u} = \boxed{439\ \mathrm{mg/L}}$$
Scale up by the biodegradable fraction to reach the total COD. If 35 % of the organic matter is non-biodegradable, then 65 % is biodegradable, and since COD measures organic matter irrespective of whether organisms can use it, $$\mathrm{tCOD} = \frac{\mathrm{bCOD}}{1-f_{nb}} = \frac{439.0}{0.65} = \boxed{675\ \mathrm{mg/L}}$$
Recover the non-biodegradable residue and check the result. By difference, $$\mathrm{nbCOD} = 675.4-439.0 = 236\ \mathrm{mg/L}$$ and $236.4/675.4 = 0.350$, which closes exactly on the given fraction. Two ratios confirm the answer is physically sensible: $\mathrm{bCOD}/\mathrm{cBOD_5} = 1.46$, close to the value of about 1.5 that follows from any $k$ near 0.23 d⁻¹, and $\mathrm{tCOD}/\mathrm{cBOD_5} = 2.25$, within the 1.9 to 2.5 band reported for municipal wastewater. A ratio below about 1.8 would indicate a readily biodegradable waste; above about 3 it would indicate an industrial or already-treated effluent whose residual organics are largely refractory.

Check: the assumptions that carry the answer. Two assumptions do the work, and the exam explicitly invites them. First, $k = 0.23$ d⁻¹ base e at 20 °C, which is the standard textbook value for domestic wastewater; adopting the alternative convention $\mathrm{bCOD}=1.6\times\mathrm{cBOD_5}$ used by Metcalf & Eddy gives bCOD = 480 mg/L and tCOD = 738 mg/L instead, about 9 % higher, and a value of $k$ as low as 0.12 d⁻¹ would push bCOD to 665 mg/L. The answer is therefore only as good as the stated $k$, and on a real plant $k$ would be obtained by fitting a BOD time series rather than assumed. Second, the 0.35 non-biodegradable fraction is applied on a COD basis; applying it on a mass-of-organics basis would require the two fractions to have the same oxygen demand per unit mass, which is close but not exact. Both assumptions are stated with the answer, as note 2 of the paper requires.

Question 5 — results
Quantity	Value
Fraction of ultimate BOD exerted in 5 days, $1-e^{-kt}$	0.683
Ultimate carbonaceous BOD, BOD_u	439 mg/L
Biodegradable COD, bCOD	439 mg/L
Total COD, tCOD	675 mg/L
Non-biodegradable COD, nbCOD	236 mg/L
bCOD / cBOD₅	1.46
tCOD / cBOD₅	2.25

Topic. The relationship between the BOD and COD measures of organic strength, and the biodegradable/non-biodegradable partition of wastewater COD.

Key relations. First-order BOD exertion, $\mathrm{BOD}_t=\mathrm{BOD_u}(1-e^{-kt})$; the identity $\mathrm{bCOD}=\mathrm{BOD_u}$; the partition $\mathrm{tCOD}=\mathrm{bCOD}+\mathrm{nbCOD}$ with $\mathrm{nbCOD}/\mathrm{tCOD}=f_{nb}$.

Why this works. BOD and COD are both oxygen-equivalent measures of organic matter, but they use different oxidants and therefore see different material. The dichromate of the COD test oxidises nearly all organic carbon regardless of whether an organism could use it, so COD captures the refractory material as well; the micro-organisms of the BOD bottle oxidise only what they can metabolise, so BOD is blind to the refractory fraction no matter how long the bottle is incubated. The bridge between the two is the biodegradable COD, which both tests would report identically if the BOD test were run to completion. Everything in this question follows from placing that bridge correctly: extrapolate the BOD forward in time to remove the kinetic limitation, then divide by the biodegradable fraction to remove the substrate limitation. In design practice the same partition is what determines the effluent quality a biological plant can reach, because the non-biodegradable COD passes through untouched and sets the floor on effluent COD.

Common pitfall. The dominant error is to set bCOD equal to the measured cBOD₅ rather than to the ultimate BOD, which under-states every downstream result by about a third. A second error is to multiply by 0.65 instead of dividing by it, which computes a fraction of the biodegradable COD rather than the whole from which it came, and yields a total COD smaller than its own biodegradable part — an impossibility the candidate should catch immediately. A third is to forget that the test is carbonaceous BOD: had the plain BOD₅ been given for a nitrifying sample, part of the measured demand would be nitrogenous and would not belong in a carbon-based COD balance at all.

Source. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., Ch. 2 (characteristics of wastewater; BOD, COD and their interrelationships) and Ch. 8 (wastewater COD fractionation).

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Question 5 of 7: Biodegradable and Total COD from a Measured cBOD 5

Question 5: Biodegradable and Total COD from a Measured cBOD5 (15 marks)

Question 5: Biodegradable and Total COD from a Measured cBOD₅ (15 marks)