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

Question 6 of 7: TSS Removal in a Sedimentation Tank Depends on Surface Area, Not Depth

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 6: TSS Removal in a Sedimentation Tank Depends on Surface Area, Not Depth (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. An ideal rectangular sedimentation basin of length $L$, width $B$, depth $H$ and plan area $A_s = LB$, treating a flow $Q$ of a discrete (Type I) suspension. Two numerically identical basins of equal volume but different proportions are used below to make the result concrete.

Illustrative basins, sized by the solver for the demonstration
Quantity	Basin A	Basin B
Flow, $Q$	10 000 m³/d	10 000 m³/d
Plan area, $A_s$	500 m²	250 m²
Depth, $H$	3 m	6 m
Volume, $V=A_sH$	1500 m³	1500 m³
Particle density and water at 15 °C	$\rho_s=2650$ kg/m³, $\rho=999.1$ kg/m³, $\mu=1.139\times10^{-3}$ Pa s

Find. A mathematical demonstration that the removal efficiency depends on $A_s$ and not on $H$, and the consequence for a plant deciding between deepening an existing basin and building a second one alongside it.

Approach. Ideal-basin (Hazen) theory is set up by writing the two independent motions of a particle — horizontal transport with the bulk flow, vertical settling under gravity — and requiring that the particle reach the floor before it reaches the outlet; the depth then cancels algebraically, and Stokes' law converts the result into a particle size.

Figure 6.1 — Two basins of equal volume and equal detention time. The critical trajectory runs from the water surface at the inlet to the floor at the outlet. Halving the plan area doubles the overflow rate, so the smallest particle the basin captures grows by a factor of the square root of two.

Write the two independent motions. In the ideal basin the horizontal velocity is uniform over the cross-section and the vertical settling velocity is constant: $$v_h = \frac{Q}{BH}, \qquad v_s = \text{settling velocity of the particle class}$$ A particle entering at the water surface is the worst case, because it must fall the full depth $H$.
Impose the removal criterion. The particle is captured if the time available to cross the basin is at least the time needed to fall to the floor: $$t_{\mathrm{available}} = \frac{L}{v_h} \;\ge\; t_{\mathrm{required}} = \frac{H}{v_s}$$
Substitute the horizontal velocity and let the depth cancel. Putting $v_h = Q/(BH)$ into the inequality, $$\frac{L\,B\,H}{Q} \;\ge\; \frac{H}{v_s} \quad\Longrightarrow\quad \frac{LB\,\cancel{H}}{Q} \;\ge\; \frac{\cancel{H}}{v_s}$$ The depth appears once on each side and divides out. Rearranging leaves the criterion in terms of plan area alone: $$v_s \;\ge\; \frac{Q}{LB} = \frac{Q}{A_s} \;\equiv\; v_c$$
Name the result. The quantity $v_c = Q/A_s$ is the critical settling velocity, identical to the surface overflow rate, SOR. Every particle whose settling velocity equals or exceeds it is removed completely, whatever the depth of the tank: $$\boxed{v_c = \mathrm{SOR} = \frac{Q}{A_s}\quad\text{contains no }H}$$
Complete the removal function for the slower particles. A particle with $v_s < v_c$ is not removed if it enters at the surface, but it is removed if it enters low enough. Entering at depth $h$, it is captured when $h/v_s \le L/v_h$, that is when $h \le H\,v_s/v_c$. For a uniform inlet distribution the removed fraction of that class is $$R = \frac{h}{H} = \frac{v_s}{v_c} = \frac{v_s A_s}{Q}$$ which again contains no $H$. Integrating over the settling-velocity distribution of the suspension, $$R_{\mathrm{total}} = \left(1-F_{c}\right) + \frac{1}{v_c}\int_0^{v_c} v_s\,\mathrm{d}F$$ where $F_c$ is the mass fraction with $v_s < v_c$. Both terms are functions of $v_c = Q/A_s$ only, so the entire removal efficiency of the basin is a function of plan area and flow, and of nothing else.
Convert to a particle size with Stokes' law. In the laminar regime, $$v_s = \frac{g(\rho_s-\rho)d^2}{18\mu} \quad\Longrightarrow\quad d_{\min} = \sqrt{\frac{18\mu\,v_c}{g(\rho_s-\rho)}}$$ For Basin A, $v_c = 10\,000/500 = 20$ m/d $= 2.315\times10^{-4}$ m/s, giving $d_{\min} = 17.1\ \mu$m. For Basin B, $v_c = 10\,000/250 = 40$ m/d and $d_{\min} = 24.2\ \mu$m. Since $d_{\min}\propto\sqrt{v_c}$, halving the plan area lets particles $\sqrt{2}$ times larger escape. The Reynolds number at these velocities is $3\times10^{-3}$, comfortably within the Stokes regime.
Test the two basins on a real particle class and answer the design question. A 15 $\mu$m particle settles at 15.35 m/d. In Basin A it is removed to $15.35/20 = 76.8\ \%$; in Basin B, of identical volume and identical 3.6 h detention time, only to $15.35/40 = 38.4\ \%$. Building a second basin in parallel instead doubles the plan area to 1000 m², drops the overflow rate to 10 m/d, and removes that particle class completely. Deepening buys nothing; adding plan area buys everything.

The physical reason the algebra comes out this way is worth stating plainly. Making a tank deeper at fixed plan area does increase the detention time — but it increases the distance the particle must fall by exactly the same factor, so the extra time is consumed by the extra travel and nothing is gained. Making the tank larger in plan at fixed volume slows the horizontal velocity without changing the fall distance, and that is a real gain. The useful identity behind the whole argument is

$$\mathrm{SOR} = \frac{Q}{A_s} = \frac{Q}{V/H} = \frac{H}{V/Q} = \frac{H}{\tau}$$

which says that detention time only helps through the ratio $H/\tau$, and that ratio is fixed by the overflow rate no matter how the volume is arranged.

Check: where the ideal-basin result stops being the whole story. Depth is not irrelevant in a real tank, only absent from the ideal removal criterion. A minimum depth of about 3 m is needed to keep the horizontal velocity low enough that scour does not re-suspend settled solids (the scour criterion is a horizontal velocity below roughly 0.03 m/s for typical flocs), to accommodate the sludge blanket, and to damp inlet turbulence, density currents and wind-driven circulation. In a secondary clarifier treating flocculent activated sludge, depth also provides the thickening volume and the buffer that absorbs a peak solids load, so depth genuinely does affect performance there — for a different reason, and outside the Type I theory this question asks about. The particle density of 2650 kg/m³ assumed above is that of mineral silt; a biological floc at 1050 kg/m³ settles roughly thirty times more slowly for the same diameter.

Question 6 — results
Quantity	Basin A	Basin B
Volume	1500 m³	1500 m³
Detention time, $\tau=V/Q$	3.6 h	3.6 h
Critical settling velocity, $v_c=Q/A_s$	20 m/d	40 m/d
Smallest particle removed completely	17.1 $\mu$m	24.2 $\mu$m
Removal of a 15 $\mu$m class ($v_s=15.35$ m/d)	76.8 %	38.4 %
Governing criterion	$v_s\ge Q/A_s$ — contains no depth term
Partial-removal law	$R=v_s/v_c=v_sA_s/Q$ — also independent of depth
Design conclusion	A parallel basin halves the SOR to 10 m/d and removes the 15 $\mu$m class completely; deepening the existing basin changes nothing

Topic. Ideal (Hazen) sedimentation theory for discrete particles, the overflow-rate criterion, and its consequences for clarifier design.

Key relations. Critical settling velocity $v_c=Q/A_s=\mathrm{SOR}$; partial removal $R=v_s/v_c$; the identity $\mathrm{SOR}=H/\tau$; Stokes' law $v_s=g(\rho_s-\rho)d^2/18\mu$, so that $d_{\min}\propto\sqrt{\mathrm{SOR}}$.

Why this works. The result rests on the fact that the residence time available to a particle and the distance it must fall are both proportional to depth, so their ratio — which is what determines capture — is not. This is Hazen's insight of 1904, and it is the reason clarifiers are designed on overflow rate rather than on detention time, and the reason high-rate settlers work at all: a plate or tube settler inserted into an existing basin multiplies the effective plan area many times over within the same volume, and its performance is predicted correctly by exactly this analysis applied to the projected horizontal area of the plates. The same criterion, written as $Q/A_s$, reappears in grit chambers, in dissolved-air flotation (with the sign of the vertical velocity reversed) and in the design of stormwater settling ponds.

Common pitfall. The commonest error is to argue from detention time — “a deeper tank holds the water longer, so it must settle better” — which is exactly the intuition this question is set to demolish, and which the cancellation of $H$ refutes in one line. A second error is to conclude from the algebra that depth is unimportant in practice, and to propose an implausibly shallow tank; the scour, sludge-storage and hydraulic-stability requirements set an independent minimum depth that the ideal theory says nothing about. A third is to apply the Type I result to a secondary clarifier handling flocculent, hindered-settling activated sludge, where the solids loading rate and the thickening depth become co-equal design criteria.

Source. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., Ch. 5 (physical unit operations — Type I discrete settling and the overflow-rate concept); Crittenden et al., MWH's Water Treatment, 3rd ed., Ch. 10; original derivation after Hazen (1904).

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