Standard Rate

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 Standard Rate 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 standard rate 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. Standard Rate 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: standard rate.
Capstone Studio instructional photograph

Environmental Engineering — Standard Rate: 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

Reactor Volume	Quantity produced by "Reactor Volume = 1 2 2 tr + V2ts" — 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.

V +V

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

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

A reactor of volume 3,235 m³ treats 1.65 m³/s carrying 300 mg/L of a contaminant that decays first-order with k = 0.60 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 = 3,235 m³
Q = 1.65 m³/s
C₀ = 300 mg/L
k = 0.60 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
$τ = 3235/1.65 = 1, 961 s = 0.545 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 300/ (1 + 0.60 \times 0.545) = 226.1 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 300 e^{(} - 0.60 \times 0.545) = 216.4 m g / L$
Comparison — plug flow removes 27.9% versus 24.6% for the CSTR

Answer: τ = 0.54 h; C_CSTR = 226.1 mg/L, C_PFR = 216.4 mg/L

Why the other options are there

202.0 mg/L (linear decay assumed)
226.1 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 2

Landfill volume required for a community's solid waste — Standard Rate

A community of 100,790 people generates 1.6 kg/person/day of MSW and diverts 15% through recycling. Compacted in place at 526 kg/m³ with 15% additional daily cover volume, find the airspace needed for 27 years.

Given

Population = 100,790
Generation = 1.6 kg/cap/d
Diversion = 0.15
Compacted density = 526 kg/m³
Cover = 15%, design life 27 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
$100790 (1.6) (1 - 0.15) = 137, 074 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 137, 074/526 = 260.6 m^{3} / d a y$
Annual with cover
$260.6 \times 365 \times 1.15 = 109, 386 m^{3} / y r$
Design life
$109, 386 \times 27 = 2, 953, 419 m^{3}$

Answer: ≈ 109,386 m³/yr, or 2,953,419 m³ over 27 years

Why the other options are there

3,021,401 m³ (diversion and cover ignored)
1,350,868,212 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 3

First-order removal in a CSTR versus a plug-flow reactor — Standard Rate (2)

A reactor of volume 3,402 m³ treats 0.65 m³/s carrying 54 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 = 3,402 m³
Q = 0.65 m³/s
C₀ = 54 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
$τ = 3402/0.65 = 5, 234 s = 1.454 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 54/ (1 + 0.25 \times 1.454) = 39.61 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 54 e^{(} - 0.25 \times 1.454) = 37.54 m g / L$
Comparison — plug flow removes 30.5% versus 26.7% for the CSTR

Answer: τ = 1.45 h; C_CSTR = 39.6 mg/L, C_PFR = 37.5 mg/L

Why the other options are there

34.4 mg/L (linear decay assumed)
39.6 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 4

Landfill volume required for a community's solid waste — Standard Rate (2)

A community of 33,583 people generates 2.9 kg/person/day of MSW and diverts 30% through recycling. Compacted in place at 686 kg/m³ with 25% additional daily cover volume, find the airspace needed for 28 years.

Given

Population = 33,583
Generation = 2.9 kg/cap/d
Diversion = 0.30
Compacted density = 686 kg/m³
Cover = 25%, design life 28 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
$33583 (2.9) (1 - 0.30) = 68, 173 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 68, 173/686 = 99.4 m^{3} / d a y$
Annual with cover
$99.4 \times 365 \times 1.25 = 45, 341 m^{3} / y r$
Design life
$45, 341 \times 28 = 1, 269, 557 m^{3}$

Answer: ≈ 45,341 m³/yr, or 1,269,557 m³ over 28 years

Why the other options are there

1,450,923 m³ (diversion and cover ignored)
696,733,068 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 5

First-order removal in a CSTR versus a plug-flow reactor — Standard Rate (3)

A reactor of volume 3,526 m³ treats 1.25 m³/s carrying 256 mg/L of a contaminant that decays first-order with k = 0.45 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 = 3,526 m³
Q = 1.25 m³/s
C₀ = 256 mg/L
k = 0.45 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
$τ = 3526/1.25 = 2, 821 s = 0.784 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 256/ (1 + 0.45 \times 0.784) = 189.3 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 256 e^{(} - 0.45 \times 0.784) = 179.9 m g / L$
Comparison — plug flow removes 29.7% versus 26.1% for the CSTR

Answer: τ = 0.78 h; C_CSTR = 189.3 mg/L, C_PFR = 179.9 mg/L

Why the other options are there

165.7 mg/L (linear decay assumed)
189.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 6

Landfill volume required for a community's solid waste — Standard Rate (3)

A community of 156,404 people generates 2.8 kg/person/day of MSW and diverts 30% through recycling. Compacted in place at 594 kg/m³ with 20% additional daily cover volume, find the airspace needed for 23 years.

Given

Population = 156,404
Generation = 2.8 kg/cap/d
Diversion = 0.30
Compacted density = 594 kg/m³
Cover = 20%, design life 23 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
$156404 (2.8) (1 - 0.30) = 306, 552 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 306, 552/594 = 516.1 m^{3} / d a y$
Annual with cover
$516.1 \times 365 \times 1.20 = 226, 043 m^{3} / y r$
Design life
$226, 043 \times 23 = 5, 198, 995 m^{3}$

Answer: ≈ 226,043 m³/yr, or 5,198,995 m³ over 23 years

Why the other options are there

6,189,280 m³ (diversion and cover ignored)
2,573,502,697 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 7

First-order removal in a CSTR versus a plug-flow reactor — Standard Rate (4)

A reactor of volume 3,989 m³ treats 0.05 m³/s carrying 252 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 = 3,989 m³
Q = 0.05 m³/s
C₀ = 252 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
$τ = 3989/0.05 = 79, 780 s = 22.161 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 252/ (1 + 0.15 \times 22.161) = 58.28 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 252 e^{(} - 0.15 \times 22.161) = 9.07 m g / L$
Comparison — plug flow removes 96.4% versus 76.9% for the CSTR

Answer: τ = 22.16 h; C_CSTR = 58.3 mg/L, C_PFR = 9.1 mg/L

Why the other options are there

-585.7 mg/L (linear decay assumed)
58.3 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 8

Landfill volume required for a community's solid waste — Standard Rate (4)

A community of 218,883 people generates 3.0 kg/person/day of MSW and diverts 20% through recycling. Compacted in place at 619 kg/m³ with 10% additional daily cover volume, find the airspace needed for 12 years.

Given

Population = 218,883
Generation = 3.0 kg/cap/d
Diversion = 0.20
Compacted density = 619 kg/m³
Cover = 10%, design life 12 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
$218883 (3.0) (1 - 0.20) = 525, 319 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 525, 319/619 = 848.7 m^{3} / d a y$
Annual with cover
$848.7 \times 365 \times 1.10 = 340, 736 m^{3} / y r$
Design life
$340, 736 \times 12 = 4, 088, 833 m^{3}$

Answer: ≈ 340,736 m³/yr, or 4,088,833 m³ over 12 years

Why the other options are there

4,646,402 m³ (diversion and cover ignored)
2,300,898,096 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 9

Hydraulic detention time in a tank — Standard Rate

A treatment tank holds 161,914 gal and treats 4.3 MGD. Find the hydraulic detention time.

Given

V = 161,914 gal
Q = 4.3 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
$4.3 \times 1 0^{6} = 4, 300, 000 g a l / d a y$
Substituting
$θ = 161914/4, 300, 000 = 0.0377 d a y$
Convert
$θ = 0.0377 (24) = 0.90 h r$

Answer: θ ≈ 0.90 hr

Why the other options are there

0.038 hr (day/hour conversion missed)
26.56 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

Example 10

Chronic daily intake, cancer risk and radioactive decay — Standard Rate

Drinking water contains 0.0240 mg/L of a carcinogen. An adult of 81 kg drinks 1.5 L/day. With a cancer slope factor of 0.57 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 25-year half-life remaining after 68 years.

Given

C = 0.0240 mg/L
IR = 1.5 L/d
BW = 81 kg
CSF = 0.57 (mg/kg·d)⁻¹
t½ = 25 yr, t = 68 yr

Find

CDI, lifetime risk and the remaining activity fraction

Start with the thinking

CDI normalises exposure to body weight so a dose-response slope can be applied.
A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

Formula
$C D I = (C \times I R) / B W$
Substituting — CDI = (0.0240 × 1.5)/81 = 4.444e-4 mg/kg·d
Formula
$R i s k = C D I \times C S F$
Substituting
$R i s k = 4.444 e - 4 (0.57) = 2.53 e - 4$
Interpretation — above the 10⁻⁶–10⁻⁴ risk range
Formula
$N / N_{0} = (½)^{(} t / t ½)$
Substituting
$(½)^{(} 68/25) = 15.177$

Answer: CDI = 4.44e-4 mg/kg·d, risk = 2.53e-4, 15.18% of the radionuclide remains

Why the other options are there

Risk = 0.01368 (body weight and intake ignored)
-36.0% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Standard Rate

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

Standard Rate 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.

Standard Rate

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

ReactorVolume=122tr+V2​ts

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section