Ratio and Correlation Growth

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

6 formulas

10 exam-style examples

~57 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Ratio and Correlation Growth 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 ratio and correlation growth describes physically and when it applies.
State every one of the 6 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. Ratio and Correlation Growth 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: ratio and correlation growth.
Capstone Studio instructional photograph

Environmental Engineering — Ratio and Correlation Growth: 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 6 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

P2R	Quantity produced by "P2R = P1R = k" — read its definition and unit from the handbook line directly above the equation.
P2	Quantity produced by "P2 = projected population" — read its definition and unit from the handbook line directly above the equation.
P1	Quantity produced by "P1 = population at last census" — read its definition and unit from the handbook line directly above the equation.
P1R	Quantity produced by "P1R = population of larger region at last census" — read its definition and unit from the handbook line directly above the equation.
k	Quantity produced by "k = growth ratio constant" — 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.

P2 P1
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

Least-squares slope for a settlement trend

Settlement readings (t, s) are (1, 4), (2, 7), (3, 9) and (4, 12) mm. Fit s = a + bt by least squares.

Given

n = 4
Σt = 10, Σs = 32, Σts = 95, Σt² = 30

Find

a and b

Start with the thinking

Compute the sums before touching the formulas.
Slope first; the intercept follows from the means.

Step-by-step solution

Slope — b = (nΣts − ΣtΣs)/(nΣt² − (Σt)²)
Numerator
$4 (95) - (10) (32) = 380 - 320 = 60.0$
Denominator
$4 (30) - 100 = 20.0$
Slope
$b = 60.0/20.0 = 3.00 mm p er p er i o d$
Intercept
$a = s ˉ - b t ˉ = 8.00 - 3.00 (2.50) = 0.500 mm$

Answer: s = 0.500 + 3.00t (mm)

Why the other options are there

b = 0.333 (ratio inverted)
a = 8.00 (mean reported as intercept)

Reference: FE Reference Handbook — Probability and Statistics — Linear regression

Example 2

Population projection and design water demand — Ratio and Correlation Growth

A city of 157,231 grows at 1.6% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 207 L/capita/day.

Given

P₀ = 157,231
i = 1.6%/yr
n = 26 yr
Per capita use = 207 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
$P = 157231 (1 + 0.016)^{2} 6 = 237, 561 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 157231 (1 + 0.016 \times 26) = 222, 639 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 237, 561 (207) /1000 = 49, 175 m^{3} / d$
Peak day
$Q_{p} e ak = 2.7 \times 49, 175 = 132, 773 m^{3} / d$

Answer: P = 237,561 (geometric) vs 222,639 (arithmetic); Q_avg = 49,175 m³/d, Q_peak = 132,773 m³/d

Why the other options are there

65,408 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 3

Population projection and design water demand — Ratio and Correlation Growth (2)

A city of 118,315 grows at 1.0% per year. Project the population in 13 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 490 L/capita/day.

Given

P₀ = 118,315
i = 1.0%/yr
n = 13 yr
Per capita use = 490 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
$P = 118315 (1 + 0.010)^{1} 3 = 134, 654 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 118315 (1 + 0.010 \times 13) = 133, 696 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 134, 654 (490) /1000 = 65, 980 m^{3} / d$
Peak day
$Q_{p} e ak = 1.9 \times 65, 980 = 125, 362 m^{3} / d$

Answer: P = 134,654 (geometric) vs 133,696 (arithmetic); Q_avg = 65,980 m³/d, Q_peak = 125,362 m³/d

Why the other options are there

15,381 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 4

Population projection and design water demand — Ratio and Correlation Growth (3)

A city of 90,512 grows at 3.0% per year. Project the population in 17 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 495 L/capita/day.

Given

P₀ = 90,512
i = 3.0%/yr
n = 17 yr
Per capita use = 495 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
$P = 90512 (1 + 0.030)^{1} 7 = 149, 603 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 90512 (1 + 0.030 \times 17) = 136, 673 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 149, 603 (495) /1000 = 74, 053 m^{3} / d$
Peak day
$Q_{p} e ak = 2.7 \times 74, 053 = 199, 944 m^{3} / d$

Answer: P = 149,603 (geometric) vs 136,673 (arithmetic); Q_avg = 74,053 m³/d, Q_peak = 199,944 m³/d

Why the other options are there

46,161 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 5

Population projection and design water demand — Ratio and Correlation Growth (4)

A city of 42,437 grows at 3.3% per year. Project the population in 13 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 406 L/capita/day.

Given

P₀ = 42,437
i = 3.3%/yr
n = 13 yr
Per capita use = 406 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
$P = 42437 (1 + 0.033)^{1} 3 = 64, 722 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 42437 (1 + 0.033 \times 13) = 60, 642 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 64, 722 (406) /1000 = 26, 277 m^{3} / d$
Peak day
$Q_{p} e ak = 2.6 \times 26, 277 = 68, 320 m^{3} / d$

Answer: P = 64,722 (geometric) vs 60,642 (arithmetic); Q_avg = 26,277 m³/d, Q_peak = 68,320 m³/d

Why the other options are there

18,205 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 6

Population projection and design water demand — Ratio and Correlation Growth (5)

A city of 97,528 grows at 1.0% per year. Project the population in 14 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 396 L/capita/day.

Given

P₀ = 97,528
i = 1.0%/yr
n = 14 yr
Per capita use = 396 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
$P = 97528 (1 + 0.010)^{1} 4 = 112, 106 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 97528 (1 + 0.010 \times 14) = 111, 182 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 112, 106 (396) /1000 = 44, 394 m^{3} / d$
Peak day
$Q_{p} e ak = 2.6 \times 44, 394 = 115, 424 m^{3} / d$

Answer: P = 112,106 (geometric) vs 111,182 (arithmetic); Q_avg = 44,394 m³/d, Q_peak = 115,424 m³/d

Why the other options are there

13,654 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 7

Population projection and design water demand — Ratio and Correlation Growth (6)

A city of 116,224 grows at 3.0% per year. Project the population in 17 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 215 L/capita/day.

Given

P₀ = 116,224
i = 3.0%/yr
n = 17 yr
Per capita use = 215 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
$P = 116224 (1 + 0.030)^{1} 7 = 192, 101 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 116224 (1 + 0.030 \times 17) = 175, 498 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 192, 101 (215) /1000 = 41, 302 m^{3} / d$
Peak day
$Q_{p} e ak = 2.9 \times 41, 302 = 119, 775 m^{3} / d$

Answer: P = 192,101 (geometric) vs 175,498 (arithmetic); Q_avg = 41,302 m³/d, Q_peak = 119,775 m³/d

Why the other options are there

59,274 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 8

Population projection and design water demand — Ratio and Correlation Growth (7)

A city of 71,844 grows at 1.9% per year. Project the population in 23 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 358 L/capita/day.

Given

P₀ = 71,844
i = 1.9%/yr
n = 23 yr
Per capita use = 358 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
$P = 71844 (1 + 0.019)^{2} 3 = 110, 764 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 71844 (1 + 0.019 \times 23) = 103, 240 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 110, 764 (358) /1000 = 39, 653 m^{3} / d$
Peak day
$Q_{p} e ak = 1.9 \times 39, 653 = 75, 341 m^{3} / d$

Answer: P = 110,764 (geometric) vs 103,240 (arithmetic); Q_avg = 39,653 m³/d, Q_peak = 75,341 m³/d

Why the other options are there

31,396 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 9

Population projection and design water demand — Ratio and Correlation Growth (8)

A city of 139,969 grows at 3.3% per year. Project the population in 26 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 277 L/capita/day.

Given

P₀ = 139,969
i = 3.3%/yr
n = 26 yr
Per capita use = 277 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
$P = 139969 (1 + 0.033)^{2} 6 = 325, 567 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 139969 (1 + 0.033 \times 26) = 260, 062 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 325, 567 (277) /1000 = 90, 182 m^{3} / d$
Peak day
$Q_{p} e ak = 2.2 \times 90, 182 = 198, 400 m^{3} / d$

Answer: P = 325,567 (geometric) vs 260,062 (arithmetic); Q_avg = 90,182 m³/d, Q_peak = 198,400 m³/d

Why the other options are there

120,093 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

Example 10

Population projection and design water demand — Ratio and Correlation Growth (9)

A city of 121,647 grows at 2.0% per year. Project the population in 22 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 388 L/capita/day.

Given

P₀ = 121,647
i = 2.0%/yr
n = 22 yr
Per capita use = 388 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
$P = 121647 (1 + 0.020)^{2} 2 = 188, 064 p eo pl e$
Formula (arithmetic)
$P = P_{0} (1 + in)$
Substituting
$P = 121647 (1 + 0.020 \times 22) = 175, 172 p eo pl e$
Formula
$Q_{a} v g = P \times p er c a p i t a u se$
Substituting
$Q_{a} v g = 188, 064 (388) /1000 = 72, 969 m^{3} / d$
Peak day
$Q_{p} e ak = 2.9 \times 72, 969 = 211, 609 m^{3} / d$

Answer: P = 188,064 (geometric) vs 175,172 (arithmetic); Q_avg = 72,969 m³/d, Q_peak = 211,609 m³/d

Why the other options are there

53,525 (growth increment reported as population)
Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Ratio and Correlation Growth

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

Ratio and Correlation Growth contains 6 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.