Non-steady State Continuous Flow

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

1 formulas

10 exam-style examples

~47 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Non-steady State Continuous Flow 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 non-steady state continuous flow describes physically and when it applies.
State every one of the 1 relation 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. Non-steady State Continuous Flow 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: non-steady state continuous flow.
Capstone Studio instructional photograph

Environmental Engineering — Non-steady State Continuous Flow: 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 1 relation 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

dx	Quantity produced by "dx = Dx + _n - k - Di x" — 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.

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

Completely mixed stream blending — Non-steady State Continuous Flow

A stream flowing 35.5 cfs at 21.5 mg/L receives a discharge of 12.0 cfs at 115.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 35.5 cfs, C₁ = 21.5 mg/L
Q₂ = 12.0 cfs, C₂ = 115.0 mg/L

Find

Mixed concentration C

Start with the thinking

Mass in equals mass out at steady state.
Weight by flow, never a simple average.

Step-by-step solution

Mass balance
$Q_{1} C_{1} + Q_{2} C_{2} = (Q_{1} + Q_{2}) C$
Loads
$35.5 (21.5) = 763.3; 12.0 (115.0) = 1, 380$
Total flow
$35.5 + 12.0 = 47.5 c f s$
Solve
$C = (763.3 + 1, 380) /47.5 = 45.12 m g / L$

Answer: C ≈ 45.1 mg/L

Why the other options are there

68.3 mg/L (unweighted average)
136.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 2

First-order removal in a CSTR versus a plug-flow reactor — Non-steady State Continuous Flow

A reactor of volume 378 m³ treats 1.55 m³/s carrying 59 mg/L of a contaminant that decays first-order with k = 0.05 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 = 378 m³
Q = 1.55 m³/s
C₀ = 59 mg/L
k = 0.05 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
$τ = 378/1.55 = 243.9 s = 0.068 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 59/ (1 + 0.05 \times 0.068) = 58.80 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 59 e^{(} - 0.05 \times 0.068) = 58.80 m g / L$
Comparison — plug flow removes 0.3% versus 0.3% for the CSTR

Answer: τ = 0.07 h; C_CSTR = 58.8 mg/L, C_PFR = 58.8 mg/L

Why the other options are there

58.8 mg/L (linear decay assumed)
58.8 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 3

Completely mixed stream blending — Non-steady State Continuous Flow (2)

A stream flowing 38.0 cfs at 7.5 mg/L receives a discharge of 10.5 cfs at 285.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 38.0 cfs, C₁ = 7.5 mg/L
Q₂ = 10.5 cfs, C₂ = 285.0 mg/L

Find

Mixed concentration C

Start with the thinking

Mass in equals mass out at steady state.
Weight by flow, never a simple average.

Step-by-step solution

Mass balance
$Q_{1} C_{1} + Q_{2} C_{2} = (Q_{1} + Q_{2}) C$
Loads
$38.0 (7.5) = 285.0; 10.5 (285.0) = 2, 993$
Total flow
$38.0 + 10.5 = 48.5 c f s$
Solve
$C = (285.0 + 2, 993) /48.5 = 67.58 m g / L$

Answer: C ≈ 67.6 mg/L

Why the other options are there

146.3 mg/L (unweighted average)
292.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 4

First-order removal in a CSTR versus a plug-flow reactor — Non-steady State Continuous Flow (2)

A reactor of volume 203 m³ treats 1.65 m³/s carrying 104 mg/L of a contaminant that decays first-order with k = 0.55 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 = 203 m³
Q = 1.65 m³/s
C₀ = 104 mg/L
k = 0.55 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
$τ = 203/1.65 = 123.0 s = 0.034 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 104/ (1 + 0.55 \times 0.034) = 102.1 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 104 e^{(} - 0.55 \times 0.034) = 102.1 m g / L$
Comparison — plug flow removes 1.9% versus 1.8% for the CSTR

Answer: τ = 0.03 h; C_CSTR = 102.1 mg/L, C_PFR = 102.1 mg/L

Why the other options are there

102.0 mg/L (linear decay assumed)
102.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 5

Completely mixed stream blending — Non-steady State Continuous Flow (3)

A stream flowing 3.0 cfs at 6.0 mg/L receives a discharge of 1.5 cfs at 75 mg/L. Find the fully mixed concentration.

Given

Q₁ = 3.0 cfs, C₁ = 6.0 mg/L
Q₂ = 1.5 cfs, C₂ = 75 mg/L

Find

Mixed concentration C

Start with the thinking

Mass in equals mass out at steady state.
Weight by flow, never a simple average.

Step-by-step solution

Mass balance
$Q_{1} C_{1} + Q_{2} C_{2} = (Q_{1} + Q_{2}) C$
Loads
$3.0 (6.0) = 18.0; 1.5 (75) = 112.5$
Total flow
$3.0 + 1.5 = 4.5 c f s$
Solve
$C = (18.0 + 112.5) /4.5 = 29.00 m g / L$

Answer: C ≈ 29.0 mg/L

Why the other options are there

40.5 mg/L (unweighted average)
81.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 6

First-order removal in a CSTR versus a plug-flow reactor — Non-steady State Continuous Flow (3)

A reactor of volume 4,506 m³ treats 1.85 m³/s carrying 230 mg/L of a contaminant that decays first-order with k = 0.55 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 = 4,506 m³
Q = 1.85 m³/s
C₀ = 230 mg/L
k = 0.55 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
$τ = 4506/1.85 = 2, 436 s = 0.677 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 230/ (1 + 0.55 \times 0.677) = 167.6 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 230 e^{(} - 0.55 \times 0.677) = 158.5 m g / L$
Comparison — plug flow removes 31.1% versus 27.1% for the CSTR

Answer: τ = 0.68 h; C_CSTR = 167.6 mg/L, C_PFR = 158.5 mg/L

Why the other options are there

144.4 mg/L (linear decay assumed)
167.6 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 7

Completely mixed stream blending — Non-steady State Continuous Flow (4)

A stream flowing 33.5 cfs at 19.5 mg/L receives a discharge of 17.5 cfs at 255.0 mg/L. Find the fully mixed concentration.

Given

Q₁ = 33.5 cfs, C₁ = 19.5 mg/L
Q₂ = 17.5 cfs, C₂ = 255.0 mg/L

Find

Mixed concentration C

Start with the thinking

Mass in equals mass out at steady state.
Weight by flow, never a simple average.

Step-by-step solution

Mass balance
$Q_{1} C_{1} + Q_{2} C_{2} = (Q_{1} + Q_{2}) C$
Loads
$33.5 (19.5) = 653.3; 17.5 (255.0) = 4, 463$
Total flow
$33.5 + 17.5 = 51.0 c f s$
Solve
$C = (653.3 + 4, 463) /51.0 = 100.3 m g / L$

Answer: C ≈ 100.3 mg/L

Why the other options are there

137.3 mg/L (unweighted average)
274.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 8

First-order removal in a CSTR versus a plug-flow reactor — Non-steady State Continuous Flow (4)

A reactor of volume 2,310 m³ treats 0.75 m³/s carrying 226 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 = 2,310 m³
Q = 0.75 m³/s
C₀ = 226 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
$τ = 2310/0.75 = 3, 080 s = 0.856 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 226/ (1 + 0.20 \times 0.856) = 193.0 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 226 e^{(} - 0.20 \times 0.856) = 190.5 m g / L$
Comparison — plug flow removes 15.7% versus 14.6% for the CSTR

Answer: τ = 0.86 h; C_CSTR = 193.0 mg/L, C_PFR = 190.5 mg/L

Why the other options are there

187.3 mg/L (linear decay assumed)
193.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 9

Series particulate control: overall collection efficiency and emission rate — Non-steady State Continuous Flow

A stack gas stream of 43 m³/s carries 13 g/m³ of particulate. It passes a cyclone at 75% efficiency followed by a fabric filter at 95.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 43 m³/s
C_in = 13 g/m³
η₁ = 0.75
η₂ = 0.950

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
$C_{1} = 13 (1 - 0.75) = 3.250 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 3.250 (1 - 0.950) = 0.1625 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.250 \times 0.0500 = 0.98750 = 98.750$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.1625 g / m^{3} \times 43 m^{3} / s \times 3600 s / h \div 1000 = 25.155 k g / h$

Answer: C_out = 0.1625 g/m³, η = 98.75%, emission = 25.16 kg/h

Why the other options are there

η = 170.0% (efficiencies added)
2,012 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

Example 10

Landfill volume required for a community's solid waste — Non-steady State Continuous Flow

A community of 78,385 people generates 3.0 kg/person/day of MSW and diverts 15% through recycling. Compacted in place at 596 kg/m³ with 20% additional daily cover volume, find the airspace needed for 18 years.

Given

Population = 78,385
Generation = 3.0 kg/cap/d
Diversion = 0.15
Compacted density = 596 kg/m³
Cover = 20%, design life 18 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
$78385 (3.0) (1 - 0.15) = 199, 882 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 199, 882/596 = 335.4 m^{3} / d a y$
Annual with cover
$335.4 \times 365 \times 1.20 = 146, 893 m^{3} / y r$
Design life
$146, 893 \times 18 = 2, 644, 073 m^{3}$

Answer: ≈ 146,893 m³/yr, or 2,644,073 m³ over 18 years

Why the other options are there

2,592,229 m³ (diversion and cover ignored)
1,313,223,098 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Non-steady State Continuous Flow

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

Non-steady State Continuous Flow contains 1 relation; 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.