Steady-State Reactor Parameters (Constant Density Systems)

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

10 formulas

10 exam-style examples

~60 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Steady-State Reactor Parameters (Constant Density Systems) 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 steady-state reactor parameters (constant density systems) describes physically and when it applies.
State every one of the 10 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. Steady-State Reactor Parameters (Constant Density Systems) 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: steady-state reactor parameters (constant density systems).
Capstone Studio instructional photograph

Environmental Engineering — Steady-State Reactor Parameters (Constant Density Systems): 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 10 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

Co	Quantity produced by "Co = initial concentration or influent concentration; Ct = final condition or effluent concentration." — read its definition and unit from the handbook line directly above the equation.
1 + kθ	Quantity produced by "1 + kθ" — read its definition and unit from the handbook line directly above the equation.
2kθ	Quantity produced by "2kθ" — 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.

Reaction Order r Ideal Batch Ideal Plug Flow Ideal CMFR
(C o – Ct ) (C o – Ct ) (C o – C t)
Zero –k
k k k
First –kC ln (C o C t ) ln (C o C t ) ( Co C t ) – 1
k k k
Second – kC2 ( Co C t ) – 1 ( Co C t ) – 1 ( Co C t ) – 1
kCo kCo kC t
Comparison of Steady-State Performance for Decay Reac ons of Different Order a
Equations for Ct
Reaction Order r Ideal Batch Ideal Plug Flow Ideal CMFR
t > Co/k 0
Second –kC2 Co Co
Time conditions are for ideal batch reactor only.
Davis, M.L., and S.J. Masten, Principles of Environmental Engineering and Science, 2nd ed., McGraw-Hill, 2004.

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 — Steady-State Reactor Parameters (Constant Density Systems)

A treatment tank holds 183,126 gal and treats 2.0 MGD. Find the hydraulic detention time.

Given

V = 183,126 gal
Q = 2.0 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
$2.0 \times 1 0^{6} = 2, 000, 000 g a l / d a y$
Substituting
$θ = 183126/2, 000, 000 = 0.0916 d a y$
Convert
$θ = 0.0916 (24) = 2.20 h r$

Answer: θ ≈ 2.20 hr

Why the other options are there

0.092 hr (day/hour conversion missed)
10.92 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 2

First-order removal in a CSTR versus a plug-flow reactor — Steady-State Reactor Parameters (Constant Density Systems)

A reactor of volume 4,222 m³ treats 1.10 m³/s carrying 82 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 = 4,222 m³
Q = 1.10 m³/s
C₀ = 82 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
$τ = 4222/1.10 = 3, 838 s = 1.066 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 82/ (1 + 0.20 \times 1.066) = 67.59 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 82 e^{(} - 0.20 \times 1.066) = 66.25 m g / L$
Comparison — plug flow removes 19.2% versus 17.6% for the CSTR

Answer: τ = 1.07 h; C_CSTR = 67.6 mg/L, C_PFR = 66.3 mg/L

Why the other options are there

64.5 mg/L (linear decay assumed)
67.6 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 3

Hydraulic detention time in a tank — Steady-State Reactor Parameters (Constant Density Systems) (2)

A treatment tank holds 182,320 gal and treats 4.4 MGD. Find the hydraulic detention time.

Given

V = 182,320 gal
Q = 4.4 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
$4.4 \times 1 0^{6} = 4, 400, 000 g a l / d a y$
Substituting
$θ = 182320/4, 400, 000 = 0.0414 d a y$
Convert
$θ = 0.0414 (24) = 0.99 h r$

Answer: θ ≈ 0.99 hr

Why the other options are there

0.041 hr (day/hour conversion missed)
24.13 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 4

First-order removal in a CSTR versus a plug-flow reactor — Steady-State Reactor Parameters (Constant Density Systems) (2)

A reactor of volume 4,820 m³ treats 1.05 m³/s carrying 328 mg/L of a contaminant that decays first-order with k = 0.10 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 = 4,820 m³
Q = 1.05 m³/s
C₀ = 328 mg/L
k = 0.10 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
$τ = 4820/1.05 = 4, 590 s = 1.275 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 328/ (1 + 0.10 \times 1.275) = 290.9 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 328 e^{(} - 0.10 \times 1.275) = 288.7 m g / L$
Comparison — plug flow removes 12.0% versus 11.3% for the CSTR

Answer: τ = 1.28 h; C_CSTR = 290.9 mg/L, C_PFR = 288.7 mg/L

Why the other options are there

286.2 mg/L (linear decay assumed)
290.9 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 5

Hydraulic detention time in a tank — Steady-State Reactor Parameters (Constant Density Systems) (3)

A treatment tank holds 192,250 gal and treats 0.6 MGD. Find the hydraulic detention time.

Given

V = 192,250 gal
Q = 0.6 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
$0.6 \times 1 0^{6} = 600, 000 g a l / d a y$
Substituting
$θ = 192250/600, 000 = 0.3204 d a y$
Convert
$θ = 0.3204 (24) = 7.69 h r$

Answer: θ ≈ 7.69 hr

Why the other options are there

0.320 hr (day/hour conversion missed)
3.12 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 6

First-order removal in a CSTR versus a plug-flow reactor — Steady-State Reactor Parameters (Constant Density Systems) (3)

A reactor of volume 1,706 m³ treats 0.85 m³/s carrying 268 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 = 1,706 m³
Q = 0.85 m³/s
C₀ = 268 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
$τ = 1706/0.85 = 2, 007 s = 0.558 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 268/ (1 + 0.55 \times 0.558) = 205.1 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 268 e^{(} - 0.55 \times 0.558) = 197.2 m g / L$
Comparison — plug flow removes 26.4% versus 23.5% for the CSTR

Answer: τ = 0.56 h; C_CSTR = 205.1 mg/L, C_PFR = 197.2 mg/L

Why the other options are there

185.8 mg/L (linear decay assumed)
205.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 7

Hydraulic detention time in a tank — Steady-State Reactor Parameters (Constant Density Systems) (4)

A treatment tank holds 81,387 gal and treats 5.2 MGD. Find the hydraulic detention time.

Given

V = 81,387 gal
Q = 5.2 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
$5.2 \times 1 0^{6} = 5, 200, 000 g a l / d a y$
Substituting
$θ = 81387/5, 200, 000 = 0.0157 d a y$
Convert
$θ = 0.0157 (24) = 0.38 h r$

Answer: θ ≈ 0.38 hr

Why the other options are there

0.016 hr (day/hour conversion missed)
63.89 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 8

First-order removal in a CSTR versus a plug-flow reactor — Steady-State Reactor Parameters (Constant Density Systems) (4)

A reactor of volume 2,874 m³ treats 0.95 m³/s carrying 392 mg/L of a contaminant that decays first-order with k = 0.50 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 = 2,874 m³
Q = 0.95 m³/s
C₀ = 392 mg/L
k = 0.50 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
$τ = 2874/0.95 = 3, 025 s = 0.840 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 392/ (1 + 0.50 \times 0.840) = 276.0 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 392 e^{(} - 0.50 \times 0.840) = 257.5 m g / L$
Comparison — plug flow removes 34.3% versus 29.6% for the CSTR

Answer: τ = 0.84 h; C_CSTR = 276.0 mg/L, C_PFR = 257.5 mg/L

Why the other options are there

227.3 mg/L (linear decay assumed)
276.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 9

Hydraulic detention time in a tank — Steady-State Reactor Parameters (Constant Density Systems) (5)

A treatment tank holds 109,875 gal and treats 5.5 MGD. Find the hydraulic detention time.

Given

V = 109,875 gal
Q = 5.5 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
$5.5 \times 1 0^{6} = 5, 500, 000 g a l / d a y$
Substituting
$θ = 109875/5, 500, 000 = 0.0200 d a y$
Convert
$θ = 0.0200 (24) = 0.48 h r$

Answer: θ ≈ 0.48 hr

Why the other options are there

0.020 hr (day/hour conversion missed)
50.06 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

Example 10

First-order removal in a CSTR versus a plug-flow reactor — Steady-State Reactor Parameters (Constant Density Systems) (5)

A reactor of volume 4,295 m³ treats 0.95 m³/s carrying 327 mg/L of a contaminant that decays first-order with k = 0.15 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 = 4,295 m³
Q = 0.95 m³/s
C₀ = 327 mg/L
k = 0.15 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
$τ = 4295/0.95 = 4, 521 s = 1.256 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 327/ (1 + 0.15 \times 1.256) = 275.2 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 327 e^{(} - 0.15 \times 1.256) = 270.9 m g / L$
Comparison — plug flow removes 17.2% versus 15.9% for the CSTR

Answer: τ = 1.26 h; C_CSTR = 275.2 mg/L, C_PFR = 270.9 mg/L

Why the other options are there

265.4 mg/L (linear decay assumed)
275.2 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

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

Steady-State Reactor Parameters (Constant Density Systems) contains 10 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.