Ultrafiltration

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

7 formulas

10 exam-style examples

~59 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Ultrafiltration within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what ultrafiltration describes physically and when it applies.
State every one of the 7 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Ultrafiltration is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: ultrafiltration.
Capstone Studio instructional photograph

Environmental Engineering — Ultrafiltration: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 7 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

ε	Quantity produced by "ε = membrane porosity" — read its definition and unit from the handbook line directly above the equation.
r	Quantity produced by "r = membrane pore size" — read its definition and unit from the handbook line directly above the equation.
∆P	Quantity produced by "∆P = net transmembrane pressure" — read its definition and unit from the handbook line directly above the equation.
µ	Quantity produced by "µ = viscosity" — read its definition and unit from the handbook line directly above the equation.
δ	Quantity produced by "δ = membrane thickness" — read its definition and unit from the handbook line directly above the equation.
Jw	Quantity produced by "Jw = volumetric flux (m/s)" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

fr 2 # D P
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hydraulic detention time in a tank — Ultrafiltration

A treatment tank holds 64,519 gal and treats 3.9 MGD. Find the hydraulic detention time.

Given

V = 64,519 gal
Q = 3.9 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$3.9 \times 1 0^{6} = 3, 900, 000 g a l / d a y$
Substituting
$θ = 64519/3, 900, 000 = 0.0165 d a y$
Convert
$θ = 0.0165 (24) = 0.40 h r$

Answer: θ ≈ 0.40 hr

Why the other options are there

0.017 hr (day/hour conversion missed)
60.45 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 2

Chemical feed rate from dosage — Ultrafiltration

A plant treating 0.5 MGD requires a 2.0 mg/L dose. What is the daily chemical demand in pounds?

Given

Q = 0.5 MGD
Dose = 2.0 mg/L
8.34 lb/gal

Find

lb/day of chemical

Start with the thinking

The 8.34 factor converts mg/L·MGD to lb/day.
Adjust for purity if the product is not 100% active.

Step-by-step solution

Feed
$l b / d a y = Q (M G D) \times d ose (m g / L) \times 8.34$
Substituting
$l b / d a y = 0.5 (2.0) (8.34)$
Evaluate — 8.3 lb/day
At 65% available strength — 12.8 lb/day of product

Answer: ≈ 8 lb/day

Why the other options are there

1.0 lb/day (8.34 omitted)
0.12 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 3

Converting ppm to mg/m³ — Ultrafiltration

A stack gas contains 42.5 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 42.5 ppm
MW = 28 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 42.5 (28) /24.45$
Evaluate — 48.67 mg/m³
Reverse check
$pp m = 48.67 (24.45) /28 = 42.5 ✓$

Answer: ≈ 48.7 mg/m³

Why the other options are there

37.1 mg/m³ (ratio inverted)
53.1 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 4

First-order removal in a CSTR versus a plug-flow reactor — Ultrafiltration

A reactor of volume 496 m³ treats 1.35 m³/s carrying 394 mg/L of a contaminant that decays first-order with k = 0.20 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Given

V = 496 m³
Q = 1.35 m³/s
C₀ = 394 mg/L
k = 0.20 h⁻¹

Find

τ, CSTR effluent and PFR effluent

Start with the thinking

The mass balance for steady state is: in − out − reaction = 0.
For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

Formula
$τ = V / Q$
Substituting
$τ = 496/1.35 = 367.4 s = 0.102 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 394/ (1 + 0.20 \times 0.102) = 386.1 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 394 e^{(} - 0.20 \times 0.102) = 386.0 m g / L$
Comparison — plug flow removes 2.0% versus 2.0% for the CSTR

Answer: τ = 0.10 h; C_CSTR = 386.1 mg/L, C_PFR = 386.0 mg/L

Why the other options are there

386.0 mg/L (linear decay assumed)
386.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 5

Activated sludge F/M ratio, aeration time and sludge production — Ultrafiltration

An activated sludge plant treats 3,813 m³/d with an influent BOD of 163 mg/L, effluent BOD 17 mg/L, MLSS 2,112 mg/L, and an aeration basin volume of 5,467 m³. With Y = 0.55 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 10 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 3,813 m³/d
S₀ = 163 mg/L, S = 17 mg/L
X = 2,112 mg/L, V = 5,467 m³
Y = 0.55, k_d = 0.06 d⁻¹, SRT = 10 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 3813 (163) / (5467 \times 2112) = 0.054 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 5467/3813 = 1.4338 d = 34.41 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.55 (3813) (163 - 17) / (1 + 0.06 \times 10) m g / d$
Evaluate
$P_{x} = 191.4 k g V S S / d a y$

Answer: F/M = 0.05 d⁻¹, τ = 34.4 h, sludge = 191.4 kg/d

Why the other options are there

F/M = 294.3 (basin volume omitted)
P_x = 306.2 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 6

Circular clarifier overflow rate, detention time and Stokes settling — Ultrafiltration

A circular clarifier 34 m in diameter and 3.0 m deep treats 19,531 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 100 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 19,531 m³/d
D = 34 m, depth = 3.0 m
d_p = 100 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (34)^{2} = 907.9 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 19531/907.9 = 21.51 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 724/19531 = 3.35 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 19531/(π × 34) = 182.9 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (1.0 e - 4)^{2} / (18 \times 0.001) = 777.0 m / d$
Compare
$v_{s} = 777.0 m / d > S O R = 21.51 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 21.5 m/d, τ = 3.3 h, weir loading = 182.9 m³/m·d, v_s = 777.0 m/d

Why the other options are there

SOR = 61.0 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 7

Series particulate control: overall collection efficiency and emission rate — Ultrafiltration

A stack gas stream of 27 m³/s carries 12 g/m³ of particulate. It passes a cyclone at 79% efficiency followed by a fabric filter at 97.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 27 m³/s
C_in = 12 g/m³
η₁ = 0.79
η₂ = 0.975

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 12 (1 - 0.79) = 2.520 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 2.520 (1 - 0.975) = 0.0630 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.210 \times 0.0250 = 0.99475 = 99.475$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0630 g / m^{3} \times 27 m^{3} / s \times 3600 s / h \div 1000 = 6.124 k g / h$

Answer: C_out = 0.0630 g/m³, η = 99.48%, emission = 6.12 kg/h

Why the other options are there

η = 176.5% (efficiencies added)
1,166 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 8

Landfill volume required for a community's solid waste — Ultrafiltration

A community of 243,616 people generates 1.6 kg/person/day of MSW and diverts 40% through recycling. Compacted in place at 526 kg/m³ with 25% additional daily cover volume, find the airspace needed for 14 years.

Given

Population = 243,616
Generation = 1.6 kg/cap/d
Diversion = 0.40
Compacted density = 526 kg/m³
Cover = 25%, design life 14 yr

Find

Annual and total landfill airspace

Start with the thinking

Only the landfilled fraction consumes airspace — diverted material is subtracted first.
Daily cover soil is real volume and must be added to the waste volume.

Step-by-step solution

Formula
$l an df i l l e d ma ss = p o p u l a t i o n \times r a t e \times (1 - d i v er s i o n)$
Substituting
$243616 (1.6) (1 - 0.40) = 233, 871 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 233, 871/526 = 444.6 m^{3} / d a y$
Annual with cover
$444.6 \times 365 \times 1.25 = 202, 859 m^{3} / y r$
Design life
$202, 859 \times 14 = 2, 840, 025 m^{3}$

Answer: ≈ 202,859 m³/yr, or 2,840,025 m³ over 14 years

Why the other options are there

3,786,700 m³ (diversion and cover ignored)
1,195,082,650 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 9

Chlorine dose, CT value and daily chemical demand — Ultrafiltration

A water plant treating 29,076 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 1.8 mg/L, with 105 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 29,076 m³/d
Dose = 2.0 mg/L
Demand = 1.8 mg/L
t = 105 min
Target = 3.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$2.0 - 1.8 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(105) = 21.0 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 29076 m^{3} / d \div 1000 = 58.15 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{3} .0 = 1.00 e + 3 or g ani s m s /100 m L$

Answer: Residual = 0.20 mg/L, CT = 21 mg·min/L, 58.2 kg Cl₂/day, survivors 1.0e+3/100 mL

Why the other options are there

CT = 210.0 (dose used instead of residual)
58,152 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Example 10

Population projection and design water demand — Ultrafiltration

A city of 65,884 grows at 1.9% per year. Project the population in 24 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 488 L/capita/day.

Given

P₀ = 65,884
i = 1.9%/yr
n = 24 yr
Per capita use = 488 L/cap/d

Find

Projected population and design flows

Start with the thinking

Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

Formula (geometric)
$P = P_{0} (1 + i)^{n}$
Substituting
$P = 65884 (1 + 0.019)^{2} 4 = 103, 505 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 65884 (1 + 0.019 \times 24) = 95, 927 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 103, 505 (488) /1000 = 50, 510 m^{3} / d$
Peak day
$Q_{p} e ak = 2.3 \times 50, 510 = 116, 174 m^{3} / d$

Answer: P = 103,505 (geometric) vs 95,927 (arithmetic); Q_avg = 50,510 m³/d, Q_peak = 116,174 m³/d

Why the other options are there

30,043 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ultrafiltration

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Ultrafiltration contains 7 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.