Vadose Zone Penetration

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

5 formulas

10 exam-style examples

~55 min

All Environmental Engineering lectures

Learning objectives

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

This chapter section covers Vadose Zone Penetration 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 vadose zone penetration describes physically and when it applies.
State every one of the 5 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. Vadose Zone Penetration 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: vadose zone penetration.
Capstone Studio instructional photograph

Environmental Engineering — Vadose Zone Penetration: 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 5 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

D	Quantity produced by "D= A" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = volume of infiltrating hydrocarbon (m3)" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = area of spill (m2)" — read its definition and unit from the handbook line directly above the equation.
Rv	Quantity produced by "Rv = a constant reflecting the retention capacity of the soil and the viscosity of the product (see following table)" — 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
Typical Values of Rv
Soil Gasoline Kerosene Light Fuel Oil
Coarse Gravel 400 200 100
Gravel to Coarse Sand 250 125 62
Coarse to Medium Sand 130 66 33
Medium to Fine Sand 80 40 20
Fine Sand to Silt 50 25 12
A constant value representing capacity of soil and viscosity of product
Data from Shepherd, W. D. No date. Practical Geohydrological Aspects of Groundwater Contamination.
Dept. of Environmental Affairs, Houston: Shell Oil, as published in Underground Storage Tank
Corrective Action Technologies, U.S. Environmental Protection Agency, 1987, pp. 3-8 and 3-9, epa.gov.

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

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration

A water plant treating 37,373 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 3.8 mg/L, with 65 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 37,373 m³/d
Dose = 2.0 mg/L
Demand = 3.8 mg/L
t = 65 min
Target = 2.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$2.0 - 3.8 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(65) = 13.0 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 37373 m^{3} / d \div 1000 = 74.75 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .0 = 1.00 e + 4 or g ani s m s /100 m L$

Answer: Residual = 0.20 mg/L, CT = 13 mg·min/L, 74.7 kg Cl₂/day, survivors 1.0e+4/100 mL

Why the other options are there

CT = 130.0 (dose used instead of residual)
74,746 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 2

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (2)

A water plant treating 37,591 m³/d applies a chlorine dose of 5.5 mg/L against a demand of 1.4 mg/L, with 104 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 37,591 m³/d
Dose = 5.5 mg/L
Demand = 1.4 mg/L
t = 104 min
Target = 2.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$5.5 - 1.4 = 4.10 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 4.10(104) = 426.4 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$5.5 g / m^{3} \times 37591 m^{3} / d \div 1000 = 206.8 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .0 = 1.00 e + 4 or g ani s m s /100 m L$

Answer: Residual = 4.10 mg/L, CT = 426.4 mg·min/L, 206.8 kg Cl₂/day, survivors 1.0e+4/100 mL

Why the other options are there

CT = 572.0 (dose used instead of residual)
206,751 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 3

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (3)

A water plant treating 14,459 m³/d applies a chlorine dose of 5.5 mg/L against a demand of 1.8 mg/L, with 89 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 14,459 m³/d
Dose = 5.5 mg/L
Demand = 1.8 mg/L
t = 89 min
Target = 3.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$5.5 - 1.8 = 3.70 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 3.70(89) = 329.3 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$5.5 g / m^{3} \times 14459 m^{3} / d \div 1000 = 79.52 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{3} .0 = 1.00 e + 3 or g ani s m s /100 m L$

Answer: Residual = 3.70 mg/L, CT = 329.3 mg·min/L, 79.5 kg Cl₂/day, survivors 1.0e+3/100 mL

Why the other options are there

CT = 489.5 (dose used instead of residual)
79,525 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 4

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (4)

A water plant treating 6,066 m³/d applies a chlorine dose of 4.0 mg/L against a demand of 2.2 mg/L, with 28 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 6,066 m³/d
Dose = 4.0 mg/L
Demand = 2.2 mg/L
t = 28 min
Target = 2.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$4.0 - 2.2 = 1.80 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 1.80(28) = 50.4 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$4.0 g / m^{3} \times 6066 m^{3} / d \div 1000 = 24.26 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .0 = 1.00 e + 4 or g ani s m s /100 m L$

Answer: Residual = 1.80 mg/L, CT = 50 mg·min/L, 24.3 kg Cl₂/day, survivors 1.0e+4/100 mL

Why the other options are there

CT = 112.0 (dose used instead of residual)
24,264 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 5

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (5)

A water plant treating 29,370 m³/d applies a chlorine dose of 5.0 mg/L against a demand of 1.8 mg/L, with 43 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.5-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 29,370 m³/d
Dose = 5.0 mg/L
Demand = 1.8 mg/L
t = 43 min
Target = 2.5 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$5.0 - 1.8 = 3.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 3.20(43) = 137.6 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$5.0 g / m^{3} \times 29370 m^{3} / d \div 1000 = 146.9 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .5 = 3.16 e + 3 or g ani s m s /100 m L$

Answer: Residual = 3.20 mg/L, CT = 137.6 mg·min/L, 146.9 kg Cl₂/day, survivors 3.2e+3/100 mL

Why the other options are there

CT = 215.0 (dose used instead of residual)
146,850 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 6

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (6)

A water plant treating 30,905 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 4.0 mg/L, with 76 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 4.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 30,905 m³/d
Dose = 2.0 mg/L
Demand = 4.0 mg/L
t = 76 min
Target = 4.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$2.0 - 4.0 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(76) = 15.2 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 30905 m^{3} / d \div 1000 = 61.81 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{4} .0 = 1.00 e + 2 or g ani s m s /100 m L$

Answer: Residual = 0.20 mg/L, CT = 15 mg·min/L, 61.8 kg Cl₂/day, survivors 1.0e+2/100 mL

Why the other options are there

CT = 152.0 (dose used instead of residual)
61,810 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 7

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (7)

A water plant treating 26,642 m³/d applies a chlorine dose of 2.5 mg/L against a demand of 3.2 mg/L, with 83 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.5-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 26,642 m³/d
Dose = 2.5 mg/L
Demand = 3.2 mg/L
t = 83 min
Target = 3.5 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$2.5 - 3.2 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(83) = 16.6 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.5 g / m^{3} \times 26642 m^{3} / d \div 1000 = 66.61 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{3} .5 = 3.16 e + 2 or g ani s m s /100 m L$

Answer: Residual = 0.20 mg/L, CT = 17 mg·min/L, 66.6 kg Cl₂/day, survivors 3.2e+2/100 mL

Why the other options are there

CT = 207.5 (dose used instead of residual)
66,605 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 8

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (8)

A water plant treating 28,220 m³/d applies a chlorine dose of 6.0 mg/L against a demand of 1.4 mg/L, with 26 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 28,220 m³/d
Dose = 6.0 mg/L
Demand = 1.4 mg/L
t = 26 min
Target = 2.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$6.0 - 1.4 = 4.60 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 4.60(26) = 119.6 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$6.0 g / m^{3} \times 28220 m^{3} / d \div 1000 = 169.3 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .0 = 1.00 e + 4 or g ani s m s /100 m L$

Answer: Residual = 4.60 mg/L, CT = 119.6 mg·min/L, 169.3 kg Cl₂/day, survivors 1.0e+4/100 mL

Why the other options are there

CT = 156.0 (dose used instead of residual)
169,320 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 9

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (9)

A water plant treating 25,818 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 3.2 mg/L, with 20 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 4.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 25,818 m³/d
Dose = 2.0 mg/L
Demand = 3.2 mg/L
t = 20 min
Target = 4.0 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$2.0 - 3.2 = 0.20 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.20(20) = 4.0 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.0 g / m^{3} \times 25818 m^{3} / d \div 1000 = 51.64 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{4} .0 = 1.00 e + 2 or g ani s m s /100 m L$

Answer: Residual = 0.20 mg/L, CT = 4 mg·min/L, 51.6 kg Cl₂/day, survivors 1.0e+2/100 mL

Why the other options are there

CT = 40 (dose used instead of residual)
51,636 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

Example 10

Chlorine dose, CT value and daily chemical demand — Vadose Zone Penetration (10)

A water plant treating 10,431 m³/d applies a chlorine dose of 2.5 mg/L against a demand of 2.2 mg/L, with 37 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.5-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Given

Q = 10,431 m³/d
Dose = 2.5 mg/L
Demand = 2.2 mg/L
t = 37 min
Target = 2.5 log

Find

Residual, CT, kg/day of chlorine and surviving organisms

Start with the thinking

Residual = dose − demand; the residual, not the dose, drives disinfection credit.
Each log of removal divides the surviving organism count by ten.

Step-by-step solution

Formula
$r es i d u a l = d ose - d e man d$
Substituting
$2.5 - 2.2 = 0.30 m g / L$
Formula
$C T = C \times t$
Substituting — CT = 0.30(37) = 11.1 mg·min/L
Formula
$ma ss = d ose \times Q$
Substituting
$2.5 g / m^{3} \times 10431 m^{3} / d \div 1000 = 26.08 k g / d a y$
Log removal
$N = 1 0^{6} /1 0^{2} .5 = 3.16 e + 3 or g ani s m s /100 m L$

Answer: Residual = 0.30 mg/L, CT = 11 mg·min/L, 26.1 kg Cl₂/day, survivors 3.2e+3/100 mL

Why the other options are there

CT = 93 (dose used instead of residual)
26,078 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Vadose Zone Penetration

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

Vadose Zone Penetration contains 5 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.