D2 · HVAC and Refrigeration Distribution and Systems

Foundation lesson · 25 minute study session

Separate outdoor-air duty from recirculation

Recirculation moves air through a building; outdoor air dilutes contaminants and adds an environmental load. Filtration effectiveness depends on the contaminant and air path.

By the end of this lesson

Distinguish ventilation flow from supply flow.
Apply fully supplied ventilation criteria.
Calculate a simple well-mixed contaminant balance.

Use stated ventilation criteria

The exam specification includes codes and standards, but does not list separate design-standard PDFs. Do not assume a particular rate or edition merely because a textbook cites it. For a practice question using occupancy and area rates, supply those rates and any distribution effectiveness in the question, or verify a precise handbook reference. The supplied $R_p$ , $R_a$ and $E_z$ define this practice method; they are not asserted universal code values.

Ventilation and filtration serve different roles

Outside air can dilute a contaminant when its concentration is lower than indoors. A filter removes only contaminants it is capable of capturing. Particulate filtration does not automatically remove CO₂ or other gases. Leakage, bypass, source control and exhaust arrangements can change the result.

Write the balance

For one well-mixed zone at steady state with one pollutant source and no other removal process, generation equals outdoor-air flow times the indoor-outdoor concentration difference. Put generation and concentration into compatible units before solving. Recirculation with no effective pollutant removal does not change the steady dilution requirement.

A decision map for this lesson

Ventilation, dilution and filtration are different balances

Use the rates and applicability supplied by the question. A particle filter is not automatically a gas-removal device.

Choose the governing relationship and catch the wrong route
Condition or question	Governing decision	Error to avoid
A people-plus-area outdoor-air model is supplied	Calculate both terms, then apply the stated zone/system correction.	Replacing required outdoor air with recirculated cfm.
A well-mixed contaminant balance is specified	Use generation, outdoor background, allowable concentration rise and the stated removal paths.	Ignoring background concentration or mixing ppm with an unconverted volume fraction.
Media efficiency and bypass fraction are supplied	For equal inlet concentration and untreated bypass, use $\eta_{effective}=(1-b)\eta_{media}$ .	Applying the catalog media efficiency to air that bypasses the filter.
Occupant CO₂ is the concern	Check the specified ventilation or gas-removal mechanism.	Inferring CO₂ removal from particle efficiency or filter pressure drop.

On a narrow screen, scroll across the table to compare the decision and the error.

Relationships and their limits

Illustrative supplied-rate method

\dot V_{bz}=R_pP+R_aA;\qquad\dot V_{oz}=\dot V_{bz}/E_z

Use only the criteria and applicability supplied in the problem. This lesson does not assert these rates or an edition are in the handbook.

Simple dilution balance

G=\dot V_{oa}(C_{in}-C_{out})

Assumes steady, well-mixed conditions, compatible concentration units and no other removal.

Symbols and unit basis
Symbol	Meaning	Units or basis
$R_p$	Problem-supplied per-person rate	cfm/person
$P$	Design population	people
$R_a$	Problem-supplied area rate	cfm/ft²
$A$	Applicable area	ft²
$E_z$	Problem-supplied distribution effectiveness	dimensionless
$G$	Contaminant generation on compatible basis	e.g. ft³ contaminant/min
$\dot V_{bz}$	Breathing-zone flow under supplied method	cfm
$\dot V_{oz}$	Zone outdoor flow under supplied method	cfm
$\dot V_{oa}$	Outdoor airflow in dilution balance	cfm
$C_{in},\ C_{out}$	Indoor and outdoor concentration	compatible concentration units, e.g. volume fraction

One decision, fully worked

Apply the method.

An original practice problem specifies 6 cfm/person and 0.08 cfm/ft² for a 1,500-ft² zone with 30 people. It also specifies $E_z$ = 0.8 and instructs you to divide breathing-zone flow by $E_z$ . Find zone outdoor-air flow.

Use the stated rates; no external standard lookup is required.
Add the occupant contribution to the area contribution.
Apply the distribution factor to the complete breathing-zone result.

\dot V_{bz}=6(30)+0.08(1{,}500)=300\ \mathrm{cfm}

\dot V_{oz}=300/0.8=375\ \mathrm{cfm}

Decision

The supplied criteria require 375 cfm of zone outdoor air.

Independent check

The effectiveness below one increases the required flow. This is outside air, not the entire supply-air volume.

Specialist extension

Check bypass as well as filter-media efficiency

A filter rating describes removal through the media under its stated test conditions. A bypass path can reduce whole-system capture. For a supplied ideal two-path model with equal inlet concentration, effective removal is the treated-flow fraction times media efficiency. It is contaminant-specific; a particulate filter does not automatically remove a gas. A ventilation minimum and a pollutant-removal calculation answer different questions.

A new applied case

A system recirculates 1,200 cfm. Media removal efficiency is 0.90 for the stated particulate, but 15% of flow bypasses the media completely. Inlet concentration is the same on both paths.

\eta_{effective}=(1-0.15)(0.90)=0.765

Q_{clean}=1200(0.765)=918\ \mathrm{cfm}

Decision. Equivalent clean-air delivery is 918 cfm for this particulate and supplied model.

Cross-check. The result is less than the 1,080 cfm obtained by ignoring bypass. It does not replace the outdoor-air requirement.

Unit basis. Flow and equivalent clean-air delivery are cfm; bypass and removal fractions are dimensionless. This is the explicitly supplied mixing model.

NCEES PE Mechanical Reference Handbook v2.1; see the module reference path.

Open targeted practice in the study studio →

Transfer the method

HVAC and Refrigeration Distribution and Systems · HV-LC-D2

Try a new decision.

Select one

For this original problem only, use an outdoor-air criterion of 5 cfm/person plus 0.10 cfm/ft², then divide by a supplied distribution effectiveness of 0.80. A room has 40 people and 2,000 ft². The required outdoor-air flow is most nearly:

A320 cfmB400 cfmC500 cfmD200 cfm

A lesson check supports practice, not a readiness score. Your response stays in this page session and clears on reload.

Catch the common errors

Omitting the area or occupancy contribution.
Treating total supply cfm as all outdoor air.
Importing an unstated textbook rate or code edition.

Handbook navigation drill

Find it, then explain it.

Search: Filtration

Find filtration material and the AHU mixed-air section. Identify which terms concern contaminants and which concern air quantity.

Check your navigation

You can state what additional criteria a ventilation-rate question must supply.

Reference path: §9.3.2, p. 487; §9.3.12, p. 527; §3.1. These are printed handbook page numbers; section names and search terms remain the primary navigation method. See technical review status.

Theory sources

NCEES October 2025 HVAC specification, 2.B and 4.A; handbook v2.1, §§9.3.2 and 9.3.12.
Grondzik and Kwok, Mechanical and Electrical Equipment for Buildings, 12th ed., ch. 5 and §§12.13–12.14.
Lechner, Heating, Cooling, Lighting, 4th ed., §§16.17–16.19.

Worked examples and lesson prose are original ReasonSpan^™ material. Obtain authorized publications separately; this preview does not reproduce the supplied books or handbook.