Energy Sources and Conversion Processes

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Energy Sources and Conversion Processes 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 energy sources and conversion processes describes physically and when it applies.
State every one of the 0 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. Energy Sources and Conversion Processes 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: energy sources and conversion processes.
Capstone Studio instructional photograph

Environmental Engineering — Energy Sources and Conversion Processes: 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

ENERGY SOURCES
BIOMASS PHOTOSYNTHESIS SUN PHOTO-
FUELS CL VOLTAICS
SOLAR WIND, HYDRO,
THERMAL WAVES, TIDAL
OCEAN
THERMAL
BATTERIES, FUEL CELLS
ENERGY FORMS
ELECTROCHEMICAL
CHEMICAL
HEAT MECHANICAL ELECTRICITY
WORK
NUCLEAR
FISSION,
FUSION
ENERGY SOURCES
FOSSIL FUELS NUCLEAR GEOTHERMAL TO END USES:
FUELS RESIDENTIAL
INDUSTRIAL
GAS, OIL, COAL TRANSPORTATION
Tester, Jefferson W., Elizabeth M. Drake, Michael J. Driscoll, Michael W. Golay, and William A. Peters, Sustainable Energy: Choosing Among Options, MIT Press, 2012, p.12.

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

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

Example 1

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes

A reactor of volume 1,966 m³ treats 0.80 m³/s carrying 101 mg/L of a contaminant that decays first-order with k = 0.30 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 = 1,966 m³
Q = 0.80 m³/s
C₀ = 101 mg/L
k = 0.30 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
$τ = 1966/0.80 = 2, 458 s = 0.683 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 101/ (1 + 0.30 \times 0.683) = 83.83 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 101 e^{(} - 0.30 \times 0.683) = 82.30 m g / L$
Comparison — plug flow removes 18.5% versus 17.0% for the CSTR

Answer: τ = 0.68 h; C_CSTR = 83.8 mg/L, C_PFR = 82.3 mg/L

Why the other options are there

80.3 mg/L (linear decay assumed)
83.8 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 2

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (2)

A reactor of volume 4,642 m³ treats 1.20 m³/s carrying 350 mg/L of a contaminant that decays first-order with k = 0.30 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 = 4,642 m³
Q = 1.20 m³/s
C₀ = 350 mg/L
k = 0.30 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
$τ = 4642/1.20 = 3, 868 s = 1.075 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 350/ (1 + 0.30 \times 1.075) = 264.7 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 350 e^{(} - 0.30 \times 1.075) = 253.6 m g / L$
Comparison — plug flow removes 27.6% versus 24.4% for the CSTR

Answer: τ = 1.07 h; C_CSTR = 264.7 mg/L, C_PFR = 253.6 mg/L

Why the other options are there

237.2 mg/L (linear decay assumed)
264.7 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 3

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (3)

A reactor of volume 3,931 m³ treats 1.85 m³/s carrying 213 mg/L of a contaminant that decays first-order with k = 0.40 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 = 3,931 m³
Q = 1.85 m³/s
C₀ = 213 mg/L
k = 0.40 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
$τ = 3931/1.85 = 2, 125 s = 0.590 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 213/ (1 + 0.40 \times 0.590) = 172.3 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 213 e^{(} - 0.40 \times 0.590) = 168.2 m g / L$
Comparison — plug flow removes 21.0% versus 19.1% for the CSTR

Answer: τ = 0.59 h; C_CSTR = 172.3 mg/L, C_PFR = 168.2 mg/L

Why the other options are there

162.7 mg/L (linear decay assumed)
172.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 4

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (4)

A reactor of volume 4,830 m³ treats 1.85 m³/s carrying 322 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 = 4,830 m³
Q = 1.85 m³/s
C₀ = 322 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
$τ = 4830/1.85 = 2, 611 s = 0.725 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 322/ (1 + 0.50 \times 0.725) = 236.3 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 322 e^{(} - 0.50 \times 0.725) = 224.1 m g / L$
Comparison — plug flow removes 30.4% versus 26.6% for the CSTR

Answer: τ = 0.73 h; C_CSTR = 236.3 mg/L, C_PFR = 224.1 mg/L

Why the other options are there

205.2 mg/L (linear decay assumed)
236.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 5

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (5)

A reactor of volume 4,326 m³ treats 1.75 m³/s carrying 173 mg/L of a contaminant that decays first-order with k = 0.40 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 = 4,326 m³
Q = 1.75 m³/s
C₀ = 173 mg/L
k = 0.40 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
$τ = 4326/1.75 = 2, 472 s = 0.687 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 173/ (1 + 0.40 \times 0.687) = 135.7 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 173 e^{(} - 0.40 \times 0.687) = 131.4 m g / L$
Comparison — plug flow removes 24.0% versus 21.5% for the CSTR

Answer: τ = 0.69 h; C_CSTR = 135.7 mg/L, C_PFR = 131.4 mg/L

Why the other options are there

125.5 mg/L (linear decay assumed)
135.7 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 6

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (6)

A reactor of volume 2,804 m³ treats 0.45 m³/s carrying 318 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 = 2,804 m³
Q = 0.45 m³/s
C₀ = 318 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
$τ = 2804/0.45 = 6, 231 s = 1.731 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 318/ (1 + 0.10 \times 1.731) = 271.1 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 318 e^{(} - 0.10 \times 1.731) = 267.5 m g / L$
Comparison — plug flow removes 15.9% versus 14.8% for the CSTR

Answer: τ = 1.73 h; C_CSTR = 271.1 mg/L, C_PFR = 267.5 mg/L

Why the other options are there

263.0 mg/L (linear decay assumed)
271.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 7

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (7)

A reactor of volume 3,256 m³ treats 2.00 m³/s carrying 179 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 = 3,256 m³
Q = 2.00 m³/s
C₀ = 179 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
$τ = 3256/2.00 = 1, 628 s = 0.452 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 179/ (1 + 0.10 \times 0.452) = 171.3 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 179 e^{(} - 0.10 \times 0.452) = 171.1 m g / L$
Comparison — plug flow removes 4.4% versus 4.3% for the CSTR

Answer: τ = 0.45 h; C_CSTR = 171.3 mg/L, C_PFR = 171.1 mg/L

Why the other options are there

170.9 mg/L (linear decay assumed)
171.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 8

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (8)

A reactor of volume 1,513 m³ treats 0.65 m³/s carrying 370 mg/L of a contaminant that decays first-order with k = 0.25 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 = 1,513 m³
Q = 0.65 m³/s
C₀ = 370 mg/L
k = 0.25 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
$τ = 1513/0.65 = 2, 328 s = 0.647 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 370/ (1 + 0.25 \times 0.647) = 318.5 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 370 e^{(} - 0.25 \times 0.647) = 314.8 m g / L$
Comparison — plug flow removes 14.9% versus 13.9% for the CSTR

Answer: τ = 0.65 h; C_CSTR = 318.5 mg/L, C_PFR = 314.8 mg/L

Why the other options are there

310.2 mg/L (linear decay assumed)
318.5 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 9

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (9)

A reactor of volume 3,460 m³ treats 1.90 m³/s carrying 324 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 = 3,460 m³
Q = 1.90 m³/s
C₀ = 324 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
$τ = 3460/1.90 = 1, 821 s = 0.506 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 324/ (1 + 0.50 \times 0.506) = 258.6 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 324 e^{(} - 0.50 \times 0.506) = 251.6 m g / L$
Comparison — plug flow removes 22.3% versus 20.2% for the CSTR

Answer: τ = 0.51 h; C_CSTR = 258.6 mg/L, C_PFR = 251.6 mg/L

Why the other options are there

242.1 mg/L (linear decay assumed)
258.6 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

Example 10

First-order removal in a CSTR versus a plug-flow reactor — Energy Sources and Conversion Processes (10)

A reactor of volume 3,559 m³ treats 1.30 m³/s carrying 100 mg/L of a contaminant that decays first-order with k = 0.25 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 = 3,559 m³
Q = 1.30 m³/s
C₀ = 100 mg/L
k = 0.25 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
$τ = 3559/1.30 = 2, 738 s = 0.760 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 100/ (1 + 0.25 \times 0.760) = 84.03 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 100 e^{(} - 0.25 \times 0.760) = 82.69 m g / L$
Comparison — plug flow removes 17.3% versus 16.0% for the CSTR

Answer: τ = 0.76 h; C_CSTR = 84.0 mg/L, C_PFR = 82.7 mg/L

Why the other options are there

81.0 mg/L (linear decay assumed)
84.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Energy Sources and Conversion Processes

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

Energy Sources and Conversion Processes contains 0 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.