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Analysis and Design

9% of exam

Mass balance, hydraulic and solids (sediment/sludge) loading, and hydraulic flow measurement with weirs, flumes, and meters.

4 concepts

A. Mass balance

Steady-State Mass Balance

Control-volume bookkeeping — accumulation = in − out ± reaction — that, killed at steady state, solves blending, splitting, and CSTR/PFR reactor problems across the WRE exam.

Almost every quantitative question in water-resources and environmental engineering is a mass balance wearing a costume. Mix two streams, size a lagoon, find an effluent concentration, check a chlorine dose, balance a clarifier — each one is the same idea: draw a boundary, account for everything that crosses it, and account for whatever is created or destroyed inside. Points are lost not because the algebra is hard but because the control volume is drawn loosely, a reaction term is dropped, or a concentration in

\text{mg/L}

is multiplied by a flow in

\text{MGD}

without the

8.34

that makes the units close. The PE Civil Reference Handbook states the governing relation compactly in

\S6.7

Water Quality (Mass Conservation and Continuity); this concept is about wielding it fluently. Master it and a large block of the Analysis-and-Design questions collapses into one disciplined procedure.

The control-volume statement

Pick a boundary — a tank, a reach of river, a pipe junction, an entire plant — and write the conservation law for the constituent of interest. The rate of accumulation of mass inside the boundary equals the rate in, minus the rate out, plus whatever a reaction produces (or minus what it consumes). Mass inside is

M = CV

(concentration times volume), and each transport term is a flow times a concentration,

\dot m = QC

. Getting the sign of the reaction term right — generation positive, decay negative — is half the battle.

\frac{dM}{dt} = \dot m_{in} - \dot m_{out} \pm \dot m_{rxn}, \qquad M = CV,\quad \dot m = QC

Kill the accumulation: steady state

"Steady state" is the single most useful simplification on the exam: when nothing inside the control volume is changing with time,

dM/dt = 0

, and the balance reduces to in = out ± reaction. Unless a problem explicitly describes filling, draining, a slug load, or a startup transient, assume steady state. For a non-reactive (conservative) constituent the reaction term also vanishes, leaving simply

\sum Q_{in}C_{in} = \sum Q_{out}C_{out}

— the workhorse behind every blending and splitting calculation.

0 = \dot m_{in} - \dot m_{out} \pm \dot m_{rxn} \;\Longrightarrow\; \sum Q_iC_i\big|_{in} = \sum Q_jC_j\big|_{out} \;(\text{conservative})

Blending and splitting streams

When two or more streams merge, a conservative steady-state balance gives the flow-weighted average concentration directly. The total flow out is the sum of the inflows (a volumetric balance), and the mass out is the sum of

Q_iC_i

. Splitting is the same law run backward: a stream divided into branches keeps the parent concentration in every branch, but apportions the flow. Always run the flow balance and the mass balance as two separate equations — they are independent and together they close the problem.

C_{mix} = \frac{\sum_i Q_i C_i}{\sum_i Q_i}

Reactions: order and rate

Inside the boundary a constituent may decay. The reaction rate is

r = kC^n

, where

n

is the order. A zero-order reaction (

n=0

) proceeds at a constant rate independent of concentration (

r=k

, units of

\text{mg/L}\cdot\text{d}^{-1}

); a first-order reaction (

n=1

) is proportional to the amount present (

r=kC

, with

k

in

\text{d}^{-1}

). BOD exertion, disinfection die-off, and most natural-attenuation processes are modeled first-order. The order dictates how the reaction term enters the balance, so identify it before you substitute.

r = kC^{n}, \qquad \dot m_{rxn} = rV = kC^{n}V

The CSTR (complete-mix) reactor

In a continuous-stirred tank reactor the contents are perfectly mixed, so the concentration everywhere inside — and therefore in the effluent — equals

C

. The reaction is evaluated at that single exit concentration. Defining the hydraulic detention time

\tau = V/Q

, the steady-state balance for a first-order decay rearranges to a clean result: the higher

k\tau

is, the more is removed. The handbook gives the underlying relation

C_0 V_{CSTR}/F_{A0}

form in

\S6.1.5.10

Flow Reactors; in concentration terms it is the expression below.

Q C_0 = Q C + kCV \;\Longrightarrow\; \frac{C}{C_0} = \frac{1}{1+k\tau}, \qquad \tau = \frac{V}{Q}

The PFR (plug-flow) reactor and why it wins

In a plug-flow reactor fluid moves through as discrete "plugs" that do not mix longitudinally; concentration falls continuously along the length, so the reaction always sees a higher local concentration than the fully-diluted CSTR exit. Integrating the first-order decay over the residence time gives an exponential decay in

k\tau

. For the same

\tau

,

k

, and influent, a PFR always removes more than a single CSTR — the gap widens as

k\tau

grows. A useful bridge:

N

equal CSTRs in series approach PFR behavior as

N\to\infty

, with

C_N/C_0 = (1+k\tau_i)^{-N}

.

\frac{C}{C_0}\bigg|_{PFR} = e^{-k\tau}, \qquad \frac{C}{C_0}\bigg|_{N\,CSTR} = \frac{1}{(1+k\tau_i)^{N}}

Make the units close: the 8.34 factor

Concentrations live in

\text{mg/L}

but design loads live in

\text{lb/day}

or

\text{kg/d}

, and the conversion is the most-used number in the WRE exam. In USCS,

M\,(\text{lb/day}) = C\,(\text{mg/L}) \times Q\,(\text{MGD}) \times 8.34

, where

8.34

is

\text{lb}

per million gallons per

\text{mg/L}

(the weight of a gallon of water). In SI it is simpler still:

\text{mg/L} = \text{g/m}^3

, so

M\,(\text{kg/d}) = C\,(\text{g/m}^3)\times Q\,(\text{m}^3/\text{d})/1000

. The handbook lists this in

\S6.7

as the concentration-to-mass conversion.

M\,(\text{lb/day}) = C\,(\text{mg/L}) \times Q\,(\text{MGD}) \times 8.34\ \tfrac{\text{lb}}{\text{Mgal}\cdot\text{mg/L}}

Exam strategy

Before touching numbers, do four things: (1) draw the boundary and label every stream crossing it with both

Q

and

C

; (2) ask whether it is steady state — if yes, strike

dM/dt

; (3) ask whether the constituent is conservative or reactive, and if reactive, name the order; (4) fix one unit system and convert everything into it first. For reactor problems, decide CSTR vs PFR from the mixing description ('complete-mix', 'aerated lagoon'

\to

CSTR; 'long channel', 'chlorine contact tank'

\to

PFR) and compute

\tau = V/Q

as your first move. If the answer wants a load, finish with the

8.34

(or

/1000

in SI). A quick sanity check — the mixed concentration of two streams must lie between the two inlet concentrations — catches most arithmetic slips.

Key equations

General mass balance

\frac{dM}{dt} = \dot m_{in} - \dot m_{out} \pm \dot m_{rxn}

Accumulation = in − out ± reaction. $M=CV$ (mass in the control volume); each transport rate $\dot m = QC$ . Foundation of every WRE balance ( $\S6.7$ ).

Steady-state, conservative

\sum Q_i C_i\big|_{in} = \sum Q_j C_j\big|_{out}

Continuity (volumetric balance)

Q = vA, \qquad \sum Q_{in} = \sum Q_{out}

Flow-weighted blend concentration

C_{mix} = \dfrac{\sum_i Q_i C_i}{\sum_i Q_i}

Reaction rate

r = kC^{n}

$n$ = reaction order (0 or 1 on the exam); $k$ = rate constant ( $\text{mg/L}\cdot\text{d}^{-1}$

Hydraulic detention time

\tau = \theta = \dfrac{V}{Q}

Mean residence time in a reactor or tank; $V$ = volume, $Q$

CSTR, first-order

\dfrac{C}{C_0} = \dfrac{1}{1+k\tau}

CSTR, zero-order

C = C_0 - k\tau

Complete-mix reactor with constant-rate (zero-order) removal. Valid only while $C \ge 0$ .

PFR, first-order

\dfrac{C}{C_0} = e^{-k\tau}

Steady-state effluent of an ideal plug-flow reactor with first-order decay. Removes more than a CSTR at equal

N CSTRs in series, first-order

\dfrac{C_N}{C_0} = \dfrac{1}{(1+k\tau_i)^{N}}

Mass-to-load conversion (USCS)

M\,(\text{lb/day}) = C\,(\text{mg/L})\, Q\,(\text{MGD})\,(8.34)

$8.34$ = lb per Mgal per mg/L. The most-used conversion in WRE; $Q$

Mass-to-load conversion (SI)

M\,(\text{kg/d}) = \dfrac{C\,(\text{g/m}^3)\, Q\,(\text{m}^3/\text{d})}{1000}

Worked examples

Blending a discharge into a river

Problem. A river flows at $12\,\text{MGD}$ carrying a conservative pollutant at $1.5\,\text{mg/L}$ . A treated discharge of $2.0\,\text{MGD}$ at $25\,\text{mg/L}$ joins it and fully mixes. What is the pollutant concentration just downstream?

Solution. Draw the control volume around the mixing point. Conservative, steady state, so mass in = mass out. Flow balance:

Q_{mix} = 12 + 2.0 = 14\,\text{MGD}

. Mass balance:

Q_{mix}C_{mix} = (12)(1.5) + (2.0)(25) = 18 + 50 = 68\ \tfrac{\text{mg}}{\text{L}}\cdot\text{MGD}

C_{mix} = \frac{(12)(1.5)+(2.0)(25)}{12+2.0} = 4.86\,\text{mg/L}

CSTR versus PFR at the same detention time

Problem. An aerated lagoon and a long plug-flow channel each provide a detention time of $\tau = 5\,\text{d}$ and treat a waste with influent BOD $C_0 = 200\,\text{mg/L}$ that decays first-order at $k = 0.30\,\text{d}^{-1}$

Effluent load in lb/day and kg/d

Problem. A treatment plant discharges $6.0\,\text{MGD}$ of effluent with a BOD of $18\,\text{mg/L}$ . Express the BOD mass loading in lb/day and in kg/d.

Solution. USCS:

M = C\,Q\,(8.34) = (18)(6.0)(8.34) = 901\,\text{lb/day}

Common pitfalls

•Dropping the reaction term. A conservative balance (in = out) is only valid when nothing decays or is generated. If the problem names BOD decay, disinfection, or settling, you owe a $\pm kC^nV$ term — leaving it out overstates the effluent.
•Forgetting the 8.34. Multiplying $\text{mg/L}\times\text{MGD}$ gives a number with no useful units. The $8.34\ \text{lb}/(\text{Mgal}\cdot\text{mg/L})$ is mandatory in USCS load problems; $Q$ must be in MGD, not gpd or cfs.
•Mixing unit systems mid-problem. Combining $\text{mg/L}$ with $\text{m}^3/\text{d}$ then applying $8.34$ double-converts. Pick SI or USCS at the start and put every quantity in it before substituting.
•Confusing CSTR and PFR. The complete-mix form $C_0/(1+k\tau)$ and the plug-flow form $C_0 e^{-k\tau}$
•Evaluating the CSTR reaction at the inlet. In a CSTR the reaction rate uses the EXIT (bulk) concentration, not $C_0$ . Using $C_0$ inflates the predicted removal.
•Not running the flow balance separately. The volumetric balance ( $\sum Q_{in}=\sum Q_{out}$ ) and the mass balance are two equations. Skipping the flow balance and dividing mass by the wrong total flow is a classic blending error.
•Treating a transient as steady state. Filling tanks, slug loads, and startup all have $dM/dt\neq0$ . Only strike the accumulation term when the problem says or implies steady conditions.

References

NCEES PE Civil Reference Handbook — §6.7 Water Quality (Mass Conservation and Continuity)
NCEES PE Civil Reference Handbook — §6.1.5.10 Flow Reactors, Steady State (CSTR/PFR)
Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery — reactor kinetics and mass-balance fundamentals

B. Hydraulic loading

Hydraulic Loading and Overflow Rates

Surface overflow rate, hydraulic loading rate, detention time, and weir loading — the four geometry-and-flow ratios that size and check every clarifier, basin, and filter.

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C. Solids loading

Solids Loading, Flux and State-Point Analysis

Solids loading rate, gravity and underflow solids flux, the state-point method for secondary clarifiers, and the recycle/underflow solids balance that sets RAS concentration.

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D. Hydraulic flow measurement

Flow Measurement: Weirs, Flumes and Meters

Rectangular and V-notch weir laws, the Parshall flume, and orifice/venturi differential meters — head-discharge relationships and the discharge coefficients that calibrate them.

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When $dM/dt=0$ and no reaction: mass in equals mass out. Use for blending, splitting, and tracer balances.

Flow balance, independent of the mass balance. $Q$ = volumetric flow, $v$ = velocity, $A$ = cross-sectional area.

Concentration after streams merge (conservative, steady state). Result must fall between the lowest and highest inlet concentration.

for

n{=}0

,

\text{d}^{-1}

for

n{=}1

);

C

= concentration.

= flow. Keep units consistent (e.g. days).

Steady-state effluent of a complete-mix reactor with first-order decay. Reaction evaluated at exit (= bulk) concentration.

\tau

.

$N$ equal-volume complete-mix tanks, each with detention $\tau_i$ ; approaches PFR as $N\to\infty$ ( $\S6.1.5.10$ ).

MUST be in MGD.

Uses $\text{mg/L}=\text{g/m}^3$ . No 8.34 needed in SI — divide by 1000 to reach kg.

.

C_{mix} = \dfrac{68}{14} = 4.86\,\text{mg/L}

. Sanity check: the result lies between

1.5

and

25\,\text{mg/L}

and sits near the larger flow's value — correct. Final answer:

C_{mix} = 4.86\,\text{mg/L}

.

. Find the effluent BOD from each, modeled as an ideal CSTR and an ideal PFR.

Solution. Both are first-order, steady state. Compute

k\tau = (0.30)(5) = 1.5

. CSTR (complete-mix):

C = \dfrac{C_0}{1+k\tau} = \dfrac{200}{1+1.5} = \dfrac{200}{2.5} = 80.0\,\text{mg/L}

. PFR (plug-flow):

C = C_0\,e^{-k\tau} = 200\,e^{-1.5} = 200(0.2231) = 44.6\,\text{mg/L}

. Sanity check: at equal

\tau

the PFR removes more (effluent

44.6 < 80.0

), as expected. Final answers:

C_{CSTR}=80.0\,\text{mg/L}

,

C_{PFR}=44.6\,\text{mg/L}

.

\frac{C_{CSTR}}{C_{PFR}} = \frac{C_0/(1+k\tau)}{C_0\,e^{-k\tau}} = \frac{e^{k\tau}}{1+k\tau}=\frac{4.482}{2.5}=1.79

. Convert:

901/2.205 = 409\,\text{kg/d}

. SI cross-check:

Q = 6.0\,\text{MGD} = 6.0(3785.4) = 22{,}712\,\text{m}^3/\text{d}

, so

M = (18\,\text{g/m}^3)(22{,}712)/1000 = 409\,\text{kg/d}

— the two routes agree. Final answer:

901\,\text{lb/day} \approx 409\,\text{kg/d}

.

M = (18)(6.0)(8.34) = 901\,\text{lb/day} = 409\,\text{kg/d}

are not interchangeable; using the CSTR equation for a contact tank under-predicts removal. Read the mixing description.