Water treatment and distribution

This chapter covers the following:

• Population modeling
• Mass balance and calculations
• Steady-state reactor parameters (constant density systems)
• Specific gravity for a solids slurry
• NRC trickling filter performance
• Activated sludge

Population modeling

Population modeling helps you estimate current and future population based on trends such as births, deaths, and migration. In environmental and water resources engineering, population projections matter because design flows and loading rates for water supply and wastewater systems depend directly on how many people the system will serve.

Different projection methods fit different situations:

• Linear (arithmetic) growth: adds a constant number of people per unit time
• Exponential (geometric) growth: grows by a constant fraction per unit time
• Logistic (decreasing-rate-of-increase) growth: growth slows as the population approaches a saturation value
• Ratio/correlation methods: tie local growth to growth in a larger reference region

Linear projection = algebraic projection

Use this model when the population increases by about the same number of people each year.

$$P_t=P_0+k\Delta t$$

where

• $P_t$ = population at time $t$
• $P_0$ = population at time zero
• $k$ = growth rate
• $\Delta t$ = elapsed time in years relative to time zero

Log growth = exponential growth = geometric growth

Use this model when the population grows proportionally to its current size (a constant percentage-type growth).

$$P_t=P_0{e}^{k\Delta t}$$

$$\ln P_t=\ln P_0+k\Delta t$$

where

• $P_t$ = population at time $t$
• $P_0$ = population at time zero
• $k$ = growth rate
• $\Delta t$ = elapsed time in years relative to time zero

Percent growth

This form is often used when growth is applied in discrete periods (for example, year-by-year compounding).

$$P_t=P_0(1+k{)}^n$$

where

• $P_t$ = population at time $t$
• $P_0$ = population at time zero
• $k$ = growth rate
• $n$ = number of periods

Ratio and correlation growth

This approach assumes the community’s population stays in a constant ratio to a larger region’s population.

$$\frac{P_2}{P_{2R}}=\frac{P_1}{P_{1R}}=k$$

where

• $P_2$ = projected population
• $P_{2R}$ = projected population of a larger region
• $P_1$ = population at last census
• $P_{1R}$ = population of larger region at last census
• $k$ = growth ratio constant

Decreasing-rate-of-increase growth

Use this model when growth slows over time as the population approaches a maximum (saturation) value.

$$P_t=P_0+(S-P_0)(1-e^{-k(t-t_0)})$$

where

• $P_t$ = population at time $t$
• $P_0$ = population at time zero
• $k$ = growth rate constant
• $S$ = saturation population
• $t,t_0$ = future time, initial time

Mass balance and calculations

A mass balance applies conservation of mass to a defined control volume. The basic idea is:

• mass can enter
• mass can leave
• mass can be created or destroyed by reactions (or other sources/sinks)
• whatever is left shows up as accumulation inside the control volume

Mass balances are used throughout environmental engineering for reactor analysis, mixing problems, and pollutant fate and transport under both steady-state and dynamic conditions.

Mass balance

$$\frac{dM}{dt}=\frac{d{M}_{in}}{dt}-\frac{d{M}_{out}}{dt}\pm r$$

$$M=CQ=CV$$

Continuity equation

$$Q=vA$$

where

• $M$ = mass
• $M_{in}$ = mass in
• $M_{out}$ = mass out
• $r=k{C}^n$ = reaction rate

$$k=\text{reaction rate constant}\left(\frac{1}{{\text{(concentration units)}}^{n-1}\cdot \text{time}}\right)$$

• $n$ = order of reaction
• $C$ = concentration (mass/volume)
• $Q$ = flow rate
• $V$ = volume
• $v$ = velocity
• $A$ = cross-sectional area of flow

Unit conversion formula

This conversion is commonly used to compute mass loading from concentration and flow.

$$M(\text{lb/day})=C(\text{mg/L})\times Q(\text{MGD})\times 8.34[\text{lb-L/(mg-MG)}]$$

where

• $\text{MGD}$ = million gallons per day

FE pitfall: carry full precision through kt-based reactor calculations and round only at the final answer - early rounding in $C_0/C_t$ or $kt$ compounds through the exponential and CMFR expressions below. Also confirm your unit system before substituting: match concentration and flow units (mg/L, MGD, cfs, ft³) in the mass-loading equation above, and don’t mix overflow rates given in gpd/ft² with values needed in m/d in the clarifier tables later in this chapter.

Steady-state reactor parameters (constant density systems)

In steady-state reactor analysis, key properties (such as flow rate and concentration at a given location) don’t change with time. For constant density systems, this assumption simplifies the governing equations and lets you relate:

• detention (retention) time
• reaction kinetics (reaction order and rate constant)
• effluent concentration

These relationships are used for ideal reactor models such as the completely mixed flow reactor (CMFR) and the plug flow reactor (PFR).

Steady-state retention times ($t$) for decay reactions of different order

Each table below is organized by reaction order (rows) and reactor type (columns): find the row matching your reaction order, then read across to the column matching your reactor - ideal batch, ideal plug flow, or ideal CMFR.

Equations for mean retention times ($t$)

Reaction order	$r$	Ideal batch	Ideal plug flow	Ideal CMFR
Zero${}^{a,b}$	$-k$	$\frac{C_0-C_t}{k}$	$\frac{C_0-C_t}{k}$	$\frac{C_0-C_t}{k}$
First	$-kC$	$\frac{\ln (C_0/C_t)}{k}$	$\frac{\ln (C_0/C_t)}{k}$	$\frac{(C_0/C_t)-1}{k}$
Second	$-k{C}^2$	$\frac{(C_0/C_t)-1}{k{C}_0}$	$\frac{(C_0/C_t)-1}{k{C}_0}$	$\frac{C_0-C_t}{k{C}_t^2}$

${}^a$ Expressions are valid for $kt\le C_0$; otherwise $C_t=0$ ${}^b{C}_0$ = initial concentration or influent concentration; $C_t$ = final condition or effluent concentration.

Steady-state performance for decay reactions of different order

Equations for $C_t$

Reaction order	$r$	Ideal batch	Ideal plug flow	Ideal CMFR
Zero${}^b$	$-k$	$C_0-kt$	$C_0-kt$	$C_0-kt$
First	$-kC$	$C_0[\exp (-kt)]$	$C_0[\exp (-kt)]$	$\frac{C_0}{1+kt}$
Second	$-k{C}^2$	$\frac{C_0}{1+kt{C}_0}$	$\frac{C_0}{1+kt{C}_0}$	$\frac{(4kt{C}_0+1{)}^{1/2}-1}{2kt}$

${}^b{C}_0$ = initial concentration or influent concentration; $C_t$ = final condition or effluent concentration.

Note: Here $\frac{C_t}{C_0}=1-X$, where $X$ is the fractional conversion - the moles of material reacted per mole fed.

Example: First-order CMFR effluent concentration

A completely mixed flow reactor (CMFR) treats a waste stream with an influent concentration $C_0=120$ mg/L using a first-order reaction with $k=0.25$ day⁻¹. The hydraulic retention time is $t=4$ days. What is the steady-state effluent concentration?

From the performance table above, the ideal CMFR expression for a first-order reaction is:

$$C_t=\frac{C_0}{1+kt}$$

Substituting the given values:

$$C_t=\frac{120}{1+(0.25)(4)}=\frac{120}{2}=60\text{mg/L}$$

Answer: $60$ mg/L

Specific gravity for a solids slurry

For a solids slurry, specific gravity captures the combined effect of:

• how much solid is present
• the density of the solid particles
• the liquid phase properties

You’ll use slurry specific gravity in hydraulic calculations, pump selection, and process design for systems that handle suspended or settled solids.

$$S=\frac{W_w+W_s}{(W_w/1.0)+(W_s/S_s)}$$

where

• $S$ = specific gravity of wet sludge
• $W_w$ = weight of water (lb or kg)
• $W_s$ = weight of dry solids (lb or kg)
• $S_s$ = specific gravity of dry solids

Volume of waste sludge

$$V=\frac{W_s}{(s/100)\cdot \rho S}=\frac{W_s}{\left(\frac{100-p}{100}\right)\cdot \rho S}$$

The two forms are equivalent because solids content and water content add to 100%: $s=100-p$.

where

• $V$ = volume of sludge, ft³ (gal or m³)
• $W_s$ = weight of dry solids (lb or kg)
• $s$ = solids content (%)
• $\rho$ = unit weight of water (62.4 lb/ft³ or 1,000 kg/m³)
• $S$ = specific gravity of wet sludge
• $p$ = water content (%), where $p=100-s$

NRC trickling filter performance

The NRC (National Research Council) trickling filter performance equations are empirical relationships used to estimate organic removal efficiency under different loading conditions. They incorporate key factors such as:

• influent substrate concentration
• hydraulic loading
• filter depth
• recirculation

Because they’re straightforward to apply, NRC equations are commonly used for preliminary design and performance checks of conventional trickling filter systems.

Single-stage or first-stage rock filter

$$E_t=\frac{100}{1+0.0561\sqrt{\frac{W}{VF}}}$$

where

• $E_t$ = efficiency of BOD removal at 20°C
• $W$ = BOD loading (lb/day)
• $V$ = volume of filter media (10³ ft³)
• $F$ = recirculation factor

Recirculation factor

$$F=\frac{1+R}{(1+R/10{)}^2}$$

Example: NRC trickling filter efficiency

A single-stage rock trickling filter treats a BOD loading of $W=800$ lb/day. The filter media volume is $V=40$ (that is, $40\times 10^3{\text{ft}}^3$), and the recirculation ratio is $R=2$. What BOD removal efficiency does the NRC equation predict at 20°C?

Step 1: find the recirculation factor.

$$F=\frac{1+2}{(1+2/10{)}^2}=\frac{3}{(1.2{)}^2}=\frac{3}{1.44}=2.08$$

Step 2: substitute into the NRC efficiency equation.

$$E_t=\frac{100}{1+0.0561\sqrt{\frac{W}{VF}}}=\frac{100}{1+0.0561\sqrt{\frac{800}{(40)(2.08)}}}=\frac{100}{1+0.0561(3.10)}=\frac{100}{1.174}=85.2\%$$

Answer: $85.2\%$ BOD removal

Activated sludge

The activated sludge process is a biological treatment method that uses suspended microorganisms to remove organic matter (and, in some configurations, nutrients). Aeration and mixing keep the biomass in suspension and provide oxygen so microorganisms can metabolize biodegradable pollutants. The biomass forms flocs that are later separated from the treated effluent by sedimentation.

Common design and operating concepts include mixed liquor suspended solids, sludge age, food-to-microorganism ratio, and oxygen demand.

Food-to-microorganism (F/M) ratio

The food-to-microorganism ratio compares the organic loading entering the aeration basin to the biomass available to consume it. This is the FE Reference Handbook’s activated-sludge loading equation:

$$\frac{F}{M}=\frac{Q_0{S}_0}{VX}$$

where $Q_0$ = influent flow rate, $S_0$ = influent BOD concentration, $V$ = aeration basin volume, and $X$ = biomass concentration (MLSS or MLVSS) in the basin.

Example: Food-to-microorganism (F/M) ratio

An activated sludge aeration basin has a volume of $100,000$ ft³. The basin receives an influent flow of $Q_0=5$ MGD with a BOD concentration of $S_0=150$ mg/L, and the mixed liquor volatile suspended solids concentration is $X=2,500$ mg/L. What is the F/M ratio?

First convert the basin volume to million gallons (MG) so it matches the units of $Q_0$, using $1{\text{ft}}^3=7.48$ gal:

$$V=100,000{\text{ft}}^3\times 7.48\frac{\text{gal}}{{\text{ft}}^3}=748,000\text{gal}=0.748\text{MG}$$

Then apply the F/M equation:

$$\frac{F}{M}=\frac{Q_0{S}_0}{VX}=\frac{(5)(150)}{(0.748)(2,500)}=\frac{750}{1,870}=0.40{\text{day}}^{-1}$$

Answer: $0.40$ day⁻¹

Sludge production equation

$$X=\frac{{\theta }_{c}Y(S_0-S_e)}{\theta (1+k_d{\theta }_c)}$$

Steady-state mass balance around secondary clarifier

$$(Q_0+Q_R)X=Q_e{X}_e+Q_R{X}_r+Q_w{X}_w$$

Solids residence time

$${\theta }_c=\frac{VX}{Q_w{X}_w+Q_e{X}_e}$$

Sludge volume/day

$$Q_s=\frac{M(100)}{{\rho }_s(\%\text{ solids})}$$

Sludge volume index (SVI)

$$SVI=\frac{\text{Sludge volume after settling (mL/L)}\times 1,000}{\text{MLSS (mg/L)}}$$

where

• $k_d$ = microbial death rate (day⁻¹)
• $S_e$ = effluent BOD or COD (kg/m³)
• $S_0$ = influent BOD or COD (kg/m³)
• $X$ = biomass concentration (MLSS or MLVSS, kg/m³)
• $X_e$ = effluent suspended solids (kg/m³)
• $X_r$ = recycle suspended solids (kg/m³)
• $X_w$ = waste sludge suspended solids (kg/m³)
• $\theta$ = hydraulic residence time
• ${\theta }_c$ = solids retention time
• $Y$ = yield coefficient (kg biomass/kg BOD)
• $M$ = sludge production rate (dry weight basis)
• ${\rho }_s$ = density of solids
• $Q_e$ = effluent flow rate
• $Q_R$ = recycle flow rate
• $Q_w$ = waste flow rate
• $R$ = recycle ratio = $Q_R/Q_0$

Example: Solids retention time (SRT)

An aeration basin holds $V=1.5$ MG at a biomass concentration of $X=2,500$ mg/L. The plant discharges effluent at $Q_e=5$ MGD carrying $X_e=20$ mg/L of suspended solids, and wastes sludge at $Q_w=0.06$ MGD with $X_w=8,000$ mg/L. What is the solids retention time?

$${\theta }_c=\frac{VX}{Q_w{X}_w+Q_e{X}_e}=\frac{(1.5)(2,500)}{(0.06)(8,000)+(5)(20)}=\frac{3,750}{480+100}=\frac{3,750}{580}=6.5\text{days}$$

Answer: $6.5$ days

Clarifier design equations

Overflow rate = hydraulic loading rate:

$$v_0=\frac{Q}{A_{\text{surface}}}$$

Terminal settling velocity ($v_c$): velocity at which the smallest particle is 100% removed.

Hydraulic residence time:

$$\theta =\frac{V}{Q}$$

Two more clarifier parameters show up on the reference sheet but rarely drive a numeric calculation on their own: weir loading rate, $\text{WOR}=Q/\text{weir length}$, and horizontal velocity, $v_h=Q/A_{\text{cross-section}}$.

where

• $Q$ = flow rate
• $A_{\text{surface}}$ = surface (plan-view) area of the clarifier
• $A_{\text{cross-section}}$ = flow cross-sectional area normal to the horizontal flow
• $V$ = tank volume

Typical primary clarifier efficiency percent removal

Overflow rates (gpd/ft²)	1,200	1,000	800	600
Overflow rates (m/d)	48.9	40.7	32.6	24.4
Suspended solids	54%	58%	64%	68%
BOD₅	30%	32%	34%	36%

Note: gpd = gallons per day