Population Projection Equations

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

13 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 Population Projection Equations 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 population projection equations describes physically and when it applies.
State every one of the 13 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. Population Projection Equations 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: population projection equations.
Capstone Studio instructional photograph

Environmental Engineering — Population Projection Equations: 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 13 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

Linear Projection	Quantity produced by "Linear Projection = Algebraic Projection" — read its definition and unit from the handbook line directly above the equation.
Pt	Quantity produced by "Pt = P0 + k∆t" — read its definition and unit from the handbook line directly above the equation.
P0	Quantity produced by "P0 = population at time zero" — read its definition and unit from the handbook line directly above the equation.
k	Quantity produced by "k = growth rate" — read its definition and unit from the handbook line directly above the equation.
∆t	Quantity produced by "∆t = elapsed time in years relative to time zero" — read its definition and unit from the handbook line directly above the equation.
Log Growth	Quantity produced by "Log Growth = Exponential Growth = Geometric Growth" — read its definition and unit from the handbook line directly above the equation.
ln Pt	Quantity produced by "ln Pt = ln P0 + k∆t" — 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.

where
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

Population projection and design water demand — Population Projection Equations

A city of 67,532 grows at 1.2% per year. Project the population in 25 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 466 L/capita/day.

Given

P₀ = 67,532
i = 1.2%/yr
n = 25 yr
Per capita use = 466 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
$P = 67532 (1 + 0.012)^{2} 5 = 90, 996 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 67532 (1 + 0.012 \times 25) = 87, 792 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 90, 996 (466) /1000 = 42, 404 m^{3} / d$
Peak day
$Q_{p} e ak = 2.3 \times 42, 404 = 97, 530 m^{3} / d$

Answer: P = 90,996 (geometric) vs 87,792 (arithmetic); Q_avg = 42,404 m³/d, Q_peak = 97,530 m³/d

Why the other options are there

20,260 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 2

Population projection and design water demand — Population Projection Equations (2)

A city of 39,842 grows at 1.1% per year. Project the population in 10 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 403 L/capita/day.

Given

P₀ = 39,842
i = 1.1%/yr
n = 10 yr
Per capita use = 403 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
$P = 39842 (1 + 0.011)^{1} 0 = 44, 448 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 39842 (1 + 0.011 \times 10) = 44, 225 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 44, 448 (403) /1000 = 17, 913 m^{3} / d$
Peak day
$Q_{p} e ak = 2.7 \times 17, 913 = 48, 364 m^{3} / d$

Answer: P = 44,448 (geometric) vs 44,225 (arithmetic); Q_avg = 17,913 m³/d, Q_peak = 48,364 m³/d

Why the other options are there

4,383 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 3

Population projection and design water demand — Population Projection Equations (3)

A city of 17,018 grows at 2.9% per year. Project the population in 12 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 475 L/capita/day.

Given

P₀ = 17,018
i = 2.9%/yr
n = 12 yr
Per capita use = 475 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
$P = 17018 (1 + 0.029)^{1} 2 = 23, 982 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 17018 (1 + 0.029 \times 12) = 22, 940 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 23, 982 (475) /1000 = 11, 392 m^{3} / d$
Peak day
$Q_{p} e ak = 2.4 \times 11, 392 = 27, 340 m^{3} / d$

Answer: P = 23,982 (geometric) vs 22,940 (arithmetic); Q_avg = 11,392 m³/d, Q_peak = 27,340 m³/d

Why the other options are there

5,922 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 4

Population projection and design water demand — Population Projection Equations (4)

A city of 106,853 grows at 2.5% per year. Project the population in 24 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 352 L/capita/day.

Given

P₀ = 106,853
i = 2.5%/yr
n = 24 yr
Per capita use = 352 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
$P = 106853 (1 + 0.025)^{2} 4 = 193, 268 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 106853 (1 + 0.025 \times 24) = 170, 965 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 193, 268 (352) /1000 = 68, 030 m^{3} / d$
Peak day
$Q_{p} e ak = 2.9 \times 68, 030 = 197, 288 m^{3} / d$

Answer: P = 193,268 (geometric) vs 170,965 (arithmetic); Q_avg = 68,030 m³/d, Q_peak = 197,288 m³/d

Why the other options are there

64,112 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 5

Population projection and design water demand — Population Projection Equations (5)

A city of 157,001 grows at 2.4% per year. Project the population in 21 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 222 L/capita/day.

Given

P₀ = 157,001
i = 2.4%/yr
n = 21 yr
Per capita use = 222 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
$P = 157001 (1 + 0.024)^{2} 1 = 258, 346 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 157001 (1 + 0.024 \times 21) = 236, 130 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 258, 346 (222) /1000 = 57, 353 m^{3} / d$
Peak day
$Q_{p} e ak = 2.5 \times 57, 353 = 143, 382 m^{3} / d$

Answer: P = 258,346 (geometric) vs 236,130 (arithmetic); Q_avg = 57,353 m³/d, Q_peak = 143,382 m³/d

Why the other options are there

79,129 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 6

Population projection and design water demand — Population Projection Equations (6)

A city of 158,743 grows at 1.4% per year. Project the population in 18 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 350 L/capita/day.

Given

P₀ = 158,743
i = 1.4%/yr
n = 18 yr
Per capita use = 350 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
$P = 158743 (1 + 0.014)^{1} 8 = 203, 881 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 158743 (1 + 0.014 \times 18) = 198, 746 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 203, 881 (350) /1000 = 71, 359 m^{3} / d$
Peak day
$Q_{p} e ak = 2.3 \times 71, 359 = 164, 125 m^{3} / d$

Answer: P = 203,881 (geometric) vs 198,746 (arithmetic); Q_avg = 71,359 m³/d, Q_peak = 164,125 m³/d

Why the other options are there

40,003 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 7

Population projection and design water demand — Population Projection Equations (7)

A city of 104,765 grows at 1.7% per year. Project the population in 25 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 310 L/capita/day.

Given

P₀ = 104,765
i = 1.7%/yr
n = 25 yr
Per capita use = 310 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
$P = 104765 (1 + 0.017)^{2} 5 = 159, 676 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 104765 (1 + 0.017 \times 25) = 149, 290 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 159, 676 (310) /1000 = 49, 500 m^{3} / d$
Peak day
$Q_{p} e ak = 2.1 \times 49, 500 = 103, 949 m^{3} / d$

Answer: P = 159,676 (geometric) vs 149,290 (arithmetic); Q_avg = 49,500 m³/d, Q_peak = 103,949 m³/d

Why the other options are there

44,525 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 8

Population projection and design water demand — Population Projection Equations (8)

A city of 149,952 grows at 1.2% per year. Project the population in 16 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 295 L/capita/day.

Given

P₀ = 149,952
i = 1.2%/yr
n = 16 yr
Per capita use = 295 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
$P = 149952 (1 + 0.012)^{1} 6 = 181, 485 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 149952 (1 + 0.012 \times 16) = 178, 743 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 181, 485 (295) /1000 = 53, 538 m^{3} / d$
Peak day
$Q_{p} e ak = 2.5 \times 53, 538 = 133, 845 m^{3} / d$

Answer: P = 181,485 (geometric) vs 178,743 (arithmetic); Q_avg = 53,538 m³/d, Q_peak = 133,845 m³/d

Why the other options are there

28,791 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 9

Population projection and design water demand — Population Projection Equations (9)

A city of 72,435 grows at 2.9% per year. Project the population in 23 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 251 L/capita/day.

Given

P₀ = 72,435
i = 2.9%/yr
n = 23 yr
Per capita use = 251 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
$P = 72435 (1 + 0.029)^{2} 3 = 139, 798 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 72435 (1 + 0.029 \times 23) = 120, 749 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 139, 798 (251) /1000 = 35, 089 m^{3} / d$
Peak day
$Q_{p} e ak = 2.2 \times 35, 089 = 77, 197 m^{3} / d$

Answer: P = 139,798 (geometric) vs 120,749 (arithmetic); Q_avg = 35,089 m³/d, Q_peak = 77,197 m³/d

Why the other options are there

48,314 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

Example 10

Population projection and design water demand — Population Projection Equations (10)

A city of 100,059 grows at 3.3% per year. Project the population in 12 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 496 L/capita/day.

Given

P₀ = 100,059
i = 3.3%/yr
n = 12 yr
Per capita use = 496 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
$P = 100059 (1 + 0.033)^{1} 2 = 147, 727 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 100059 (1 + 0.033 \times 12) = 139, 682 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 147, 727 (496) /1000 = 73, 273 m^{3} / d$
Peak day
$Q_{p} e ak = 2.0 \times 73, 273 = 146, 545 m^{3} / d$

Answer: P = 147,727 (geometric) vs 139,682 (arithmetic); Q_avg = 73,273 m³/d, Q_peak = 146,545 m³/d

Why the other options are there

39,623 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Population Projection Equations

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

Population Projection Equations contains 13 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.