Castleport Test Prep

PE Fire Protection Exam Prep

Start your PE Fire Protection exam prep with this original, unofficial problem, written to the new NCEES specification for the April 13, 2027 exam. All 36 problems, worked solutions, and a dated study plan to exam day are free on this page.

Problem 1 · Flame height from a pool fire

FS-01 · Fire Science, 2C · Select one

A 1.2 m diameter pool fire burns steadily with a total heat release rate (HRR) of 1,500 kW. Using Heskestad's flame height correlation, L = 0.235 Q^(2/5) − 1.02 D (L and D in m, Q in kW), what is the mean flame height, closest to?

Show answer and solution

Answer and solution

Answer: B. 3.2 m.

Q^(2/5) = 1,500^0.4 = 18.64, so 0.235 × 18.64 = 4.38 m. Subtract 1.02 × 1.2 = 1.22 m: L ≈ 3.16 m. The 4.4 m answer drops the −1.02D term, which is what shortens the flames of a wide fire. The 5.6 m answer adds that term instead of subtracting it. The 17.7 m answer uses an exponent of 3/5 instead of 2/5. The correlation is empirical and uses SI units, so keep it in kW and meters.

Try a change: Keep 1,500 kW but double the diameter to 2.4 m. The flame drops to about 1.9 m. The same heat spread over a wider base burns lower.

Source: NUREG-1824 Vol. 3 (NRC/EPRI, 2007), §3.4, Eq. 3-14. Topic: NCEES PE Fire Protection specification, April 2027, 2C.

By Castleport Test Prep Editorial Team · Exam facts last verified October 9, 2026

What changed for 2027 · 2027 topic ranges · All 36 problems · Plan to April 13

The exam at a glance

  • One date: Tuesday, April 13, 2027, at Pearson VUE test centers. Fire Protection is offered once a year; NCEES encourages candidates to reserve a seat early after registration and licensing-board approval.
  • 85 questions with 8.5 hours of exam time inside a 9.5-hour appointment (2-minute nondisclosure agreement, 8-minute tutorial, 50-minute scheduled break).
  • $400 exam fee paid to NCEES. Your licensing board may charge its own application fee.
  • References on screen: the NCEES PE Fire Protection Reference Handbook plus 10 NFPA standards at specific editions (listed below). You can't bring your own copies.
  • Units: both SI and U.S. customary.
  • Question types: multiple choice plus alternative item types (multiple correct, point and click, drag and drop, fill in the blank), all scored right or wrong.
  • Results: pass or fail, typically 7–10 days after the exam.
  • Pass rates: 85% of 262 first-time takers and 44% of 48 repeat takers, in the Fire Protection row NCEES labels "July 2025." No newer row had been posted when we checked.

Sources: NCEES PE Fire Protection exam page; PE Fire Protection specification, effective April 2027, p. 1; NCEES Examinee Guide, May 2026, pp. 3, 11, 14, and 16.

What changed for the April 2027 exam

NCEES rebuilt the topic outline into eight areas, added energy storage systems (with NFPA 855 as a supplied standard), dropped NFPA 12 and NFPA 25 from the reference list, and moved every remaining standard to a newer edition. The question count and appointment length didn't change. The current NCEES specification is explicitly effective beginning April 2027 and lists the standards that apply to that version (April 2027 specification).

Where the old topics went

Where the old topics went
Through April 2026 (topics in effect since October 2020)April 2027 specification
Fire Protection Analysis: 17–26 questions (hazard and risk analysis, data interpretation, fire tests, design fire selection, codes)1. Fire Protection Analysis: 10–15. Now names performance-based design, fire models, and energy storage systems
Fire Dynamics Fundamentals: 11–172. Fire Science: 10–15. Heat transfer, fire chemistry, fire dynamics
Active and Passive Systems: 31–47, split into the six rows belowNo longer one area
Water-Based Fire Protection Systems: 9–145. Water-Based Fire Protection Systems: 12–18
Special Hazard Systems: 4–66. Special Hazard Systems: 7–11. Adds oxygen reduction and ignitable liquid drainage floor assemblies; names LFL calculations
Detection, Alarm, and Signaling Systems: 6–94. Fire Alarm and Signaling Systems: 8–12. Names mass notification and emergency responder communication systems
Smoke Control Systems: 4–63. Smoke Control Systems: 6–9. Adds inspection, testing, and commissioning
Explosion Protection and Prevention Systems: 2–3No standalone area. Flammable limits now appear under Special Hazard calculations
Passive Building Systems: 6–97. Passive and Structural Fire Protection: 8–12. Names structural fire analysis, char, and fire resistance equivalency
Egress and Occupant Movement: 11–178. Human Behavior and Evacuation: 9–14. Prescriptive egress, performance-based egress, tenability, and decision making

Sources: specification effective April 2027, pp. 1–2; previous specification and October 2022 standards list, pp. 1–4.

NFPA editions supplied on exam day

NFPA editions supplied on exam day
NFPA standardBefore April 2027April 2027
11, Low-, Medium-, and High-Expansion Foam20162024
12, Carbon Dioxide Extinguishing Systems2018Removed
13, Installation of Sprinkler Systems20192022
20, Installation of Stationary Pumps for Fire Protection20192022
25, Inspection, Testing, and Maintenance of Water-Based Systems2020Removed
30, Flammable and Combustible Liquids Code20182024
72, National Fire Alarm and Signaling Code20192022
92, Standard for Smoke Control20182024
101, Life Safety Code20182024
400, Hazardous Materials Code20192022
855, Installation of Stationary Energy Storage SystemsNot listed2023 (new)
2001, Clean Agent Fire Extinguishing Systems20182022

NCEES scores code-based answers against the listed edition, and its specification says solutions based on other standards won't receive credit. Study the editions above, even if your jurisdiction or employer uses a newer one. NCEES does not provide the design standards for study, so arrange lawful access to the listed editions before you rely on code-navigation practice. (April 2027 specification, standards page)

Studying from an older book or practice exam?

Many underlying engineering methods still transfer, but code-dependent requirements, topic coverage, and reference editions can change. If your material was written before the 2027 specification:

  • Add: energy storage systems and NFPA 855, fire model assumptions and limits, and smoke control inspection, testing, and commissioning.
  • Drop as code study: NFPA 12 (carbon dioxide systems) and NFPA 25 (inspection, testing, and maintenance). Neither is supplied anymore.
  • Redo your code lookups in the 2022 and 2024 editions listed above. Section numbers and tables move between editions.

NCEES's PE Fire Protection practice-exam listing still showed the ©2020 version when checked October 9, 2026, and says it was produced from the specification that took effect in October 2018. That means it predates the April 2027 specification. NCEES also says practice exams are normally updated about six months before revised specifications take effect, so check the current listing in MyNCEES before relying on it for 2027 coverage (NCEES practice exam listing). Corrections to current practice exams are posted on the NCEES errata page.

What to study first: the eight areas by NCEES question range

Water-based systems have the highest published question range. The other seven areas still matter, and the ranges are not a personal study-time prescription.

What to study first: the eight areas by NCEES question range
Knowledge areaNCEES question rangeWhat NCEES listsPractice here
5. Water-Based Fire Protection Systems12–18System selection; design criteria; principal components; hydraulics, equipment sizing, supportsProblems 11–16
1. Fire Protection Analysis10–15Performance-based design and fire risk assessment; fire models; energy storage systemsProblems 6–10
2. Fire Science10–15Heat transfer; fire chemistry; fire dynamicsProblems 1–5
8. Human Behavior and Evacuation9–14Prescriptive egress; performance-based egress; psychological and physiological responseProblems 17–21
4. Fire Alarm and Signaling Systems8–12System selection; design criteria; principal components; voltage drop, battery, notification, detector responseProblems 22–25
7. Passive and Structural Fire Protection8–12Materials; compartmentalization; structural fire protection; openings and penetrationsProblems 26–29
6. Special Hazard Systems7–11System selection; design criteria; principal components; LFL, concentrations, flow ratesProblems 30–33
3. Smoke Control Systems6–9Basis of design; plume/venting/pressure/exhaust calculations; inspection, testing, commissioningProblems 34–36

The ranges and topic labels above are NCEES's. Use them to understand exam breadth, then use your error log to decide where your own study time belongs. NCEES also says the listed examples are not exclusive or exhaustive. (April 2027 specification, pp. 1–2)

Use the ranges together with your own results. Spend the most time where an area is both large and weak for you. If your day job is all sprinklers, Area 5 may need a quick review while alarm, smoke control, and ESS need real hours.

Three things deserve early attention:

  • Energy storage systems are new to the 2027 outline. Materials built to the pre-2027 specification may not cover them, and NFPA 855-2023 is now in the supplied standards list.
  • Ten standards are supplied, so finding things counts. Unlike the PE Mechanical exams, Fire Protection gives you code text. Fast, accurate lookups are a skill to practice, not something to leave for exam day.
  • Both unit systems appear. Many fire dynamics correlations are SI-only, while hydraulics and smoke control often show up in U.S. units. Problems 1–5, 7, 19, and 30–31 use SI; most of the rest use U.S. customary units.

How to study with the handbook and the 10 standards

Run every study session through four steps:

  1. Learn. Review the concept and the assumptions behind each formula. Ask what each correlation is for and where it stops working.
  2. Find. Locate the relation in the NCEES handbook or the provision in the supplied standard. Write down the section and the search words that worked.
  3. Solve. Work fresh problems before you look at any solution.
  4. Log. Record every miss and every lucky guess by cause (see the error log).

On exam day, the handbook and standards appear as searchable PDFs. You search with a box on the left side of the screen; Ctrl+F doesn't work (Examinee Guide, p. 10). NCEES posts a video on searching the on-screen references. You can download the current handbook free from your MyNCEES account.

A lookup drill. Pick 10 prompts from the week's area. Give yourself 60 to 90 seconds each to find the governing equation, table, or section. Search with one distinctive term, not the whole question. Before you use anything, read its scope and variable definitions. Log a failed lookup separately from a concept error. (The 60–90 seconds is our drill target, not an NCEES rule.)

Start with lookups like these, and record where you found each one:

How to study with the handbook and the 10 standards
Find thisWhere to lookRecord
Flame height, plume, and ceiling jet correlationsNCEES handbookUnits the correlation expects and whether it uses total or convective HRR
Hazen-Williams, K-factor, and elevation relationsNCEES handbook; NFPA 13-2022Units and the C value convention
Sprinkler design criteria and hose stream allowancesNFPA 13-2022The chapter and table for each hazard or storage type
Fire pump acceptance and performanceNFPA 20-2022The flow and pressure points tested
Secondary power and notification requirementsNFPA 72-2022Standby and alarm durations; audibility rules
Smoke control design, plumes, and testingNFPA 92-2024Plume equations in both unit systems; testing frequencies
Egress capacity, occupant load, and travel distanceNFPA 101-2024Capacity factors and occupancy-specific exceptions
Energy storage system requirementsNFPA 855-2023How the standard uses UL 9540A test data
Clean agent design concentrations and quantitiesNFPA 2001-2022Flooding equations and safety factors
Foam, flammable liquids, and hazardous materialsNFPA 11-2024, 30-2024, 400-2022Application rates, classification tables, quantity limits

36 original practice problems with worked solutions

These are original Castleport problems written to the April 2027 Fire Protection topic list. They aren't NCEES questions, they aren't a full-length exam, and they don't cover every subtopic. Every code-like value you need is given in the problem, so you can solve each one without the standards. On the real exam, you'll pull those values from the supplied editions yourself.

Most are single-answer multiple choice. Four ask you to enter a number, two are select-all-that-apply, and one is a matching item. NCEES also uses point-and-click and drag-and-drop items, which this page doesn't reproduce. On the real exam, every question is scored right or wrong, with no partial credit (Examinee Guide, p. 11).

Format clarification: The 36 problems on this page are 30 single-answer multiple-choice items, four numeric-entry items, and two select-all-that-apply items. There is no separate matching exercise in this set.

Work each problem before you open the solution. When you miss one, write down the first step that went wrong.

Fire Science (Problems 1–5)

What this area tests most often:

  • Heat release rate is the master variable. Flame height, plume flow, radiation, and detector response all start from it.
  • Check total versus convective HRR. Plume equations often use the convective part; radiation uses the radiative fraction.
  • Correlations are empirical. Each has units it expects and a range where it applies. Ceiling jets switch equations at r/H = 0.18.
  • Growth rates matter as much as peaks. A t-squared fire's time to a given size drives detection and egress timelines.

Problem 1 is at the top of this page.

Problem 2 · Radiant heat flux to a nearby target

FS-02 · Fire Science, 2A · Select one

A fire with a total HRR of 2,000 kW has a radiative fraction of 0.35. Treat it as a point source. What radiant heat flux reaches a target 5 m from the fire, facing it?

Show answer and solution

Answer and solution

Answer: C. 2.2 kW/m².

q″ = χr Q ÷ (4πR²) = 0.35 × 2,000 ÷ (4π × 5²) = 700 ÷ 314.2 ≈ 2.2 kW/m². The 6.4 kW/m² answer leaves out the radiative fraction, as if every kilowatt left as radiation. The 11.1 kW/m² answer uses R instead of R². The 8.9 kW/m² answer drops the 4 in 4πR². The point-source model is a far-field tool; it loses accuracy within a few fire diameters of the flames.

Try a change: Move the target to 2.5 m. Halving the distance multiplies the flux by four, to about 8.9 kW/m².

Source: NUREG-1824 Vol. 3 (NRC/EPRI, 2007), §3.5.1, Eq. 3-17; CFAST 7.7.7 Technical Reference Guide (NIST TN 1889v1), §1.1 and §1.5. Topic: NCEES PE Fire Protection specification, April 2027, 2A.

Problem 3 · Time for a fast t-squared fire

FS-03 · Fire Science, 2C · Enter a number

A design fire grows as a t-squared fire at the “fast” rate, defined as reaching 1,054 kW at 150 s. Enter the time, in seconds, for the fire to reach 500 kW.

Your answer (s) · accepted tolerance ±2 s

Accepted tolerance: ±2 s.

Show answer and solution

Answer and solution

Answer: 103 s.

For a t-squared fire, Q = Q_ref × (t/t_ref)², so t = 150 × √(500 ÷ 1,054) ≈ 103 s. An answer of 71 s scales linearly, which ignores the square. The same 500 kW takes about 207 s at the medium rate (1,054 kW at 300 s).

Source: CFAST 7.7.7 User's Guide (NIST TN 1889v2), §6.3, Eq. 6.2, p. 28. Topic: NCEES PE Fire Protection specification, April 2027, 2C.

Problem 4 · Heat release from oxygen consumption

FS-04 · Fire Science, 2B · Select one

A burning item consumes oxygen at 0.10 kg/s. Using the oxygen consumption constant of 13.1 MJ released per kg of O₂ consumed, what is the heat release rate?

Show answer and solution

Answer and solution

Answer: D. 1.31 MW.

HRR = 0.10 kg/s × 13.1 MJ/kg = 1.31 MJ/s = 1.31 MW. The 13.1 MW and 131 kW answers are decimal slips. The 1.31 kW answer treats MJ as kJ. The constant is nearly the same for most organic fuels, which is why calorimeters can measure HRR from oxygen depletion without knowing exactly what is burning.

Source: FDS 6.11.1 User's Guide (NIST SP 1019), §13, Eq. 13.2; CFAST 7.7.7 Technical Reference Guide (NIST TN 1889v1), p. 13 (Huggett). Topic: NCEES PE Fire Protection specification, April 2027, 2B.

Problem 5 · Ceiling jet temperature at a detector

FS-05 · Fire Science, 2C · Select one

A 1,000 kW fire burns under a flat, unobstructed 4 m ceiling. Alpert's ceiling-jet correlations (SI) are ΔT = 16.9 Q^(2/3) ÷ H^(5/3) for r/H ≤ 0.18, and ΔT = 5.38 (Q/r)^(2/3) ÷ H for r/H > 0.18 (ΔT in °C, Q in kW, H and r in m). What is the gas temperature rise at a detector 3 m from the plume centerline?

Show answer and solution

Answer and solution

Answer: A. 65 °C.

First pick the region: r/H = 3 ÷ 4 = 0.75, which is greater than 0.18, so use the ceiling-jet form. ΔT = 5.38 × (1,000 ÷ 3)^(2/3) ÷ 4 ≈ 5.38 × 48.1 ÷ 4 ≈ 65 °C. The 168 °C answer uses the plume-region equation, which only applies directly above the fire. The 26 °C answer divides by H^(5/3) in the ceiling-jet equation. The 71 °C answer swaps r and H.

Try a change: A detector right above the fire (r/H ≤ 0.18) would see about 168 °C. That is why spacing and radial distance matter so much for response time.

Source: NISTIR 6470 (NIST, 2000), §1.2, Eqs. 7–8, p. 4. Topic: NCEES PE Fire Protection specification, April 2027, 2C.

Fire Protection Analysis (Problems 6–10)

What this area tests most often:

  • Energy storage systems are new to this outline. Know what UL 9540A test data tell you and how NFPA 855 uses them.
  • Every fire model has assumptions. Zone models assume two uniform layers; CFD needs adequate grid resolution.
  • Performance-based design compares two times. Available safe egress time (ASET) must exceed required safe egress time (RSET), with margin.
  • Risk math depends on independence. Common-cause failures change the answer.

Problem 6 · What a UL 9540A test report evaluates

FPA-01 · Fire Protection Analysis, 1C · Select one

A battery energy storage system design review includes a UL 9540A test report. Which statement best describes what UL 9540A is intended to evaluate?

Show answer and solution

Answer and solution

Answer: B. Thermal-runaway and fire-propagation behavior in battery energy storage systems.

UL Solutions describes UL 9540A as a test method for evaluating thermal-runaway fire propagation in battery energy storage systems. Its test levels examine behavior from cells and modules through larger system and installation conditions; the exact sequence and requirements can change by edition. Structural fire resistance, sprinkler-density selection, and occupant egress are separate design questions.

Source: UL Solutions, UL 9540A test method overview. Topic: NCEES PE Fire Protection specification, April 2027, 1C.

Problem 7 · Characteristic fire diameter for a CFD model

FPA-02 · Fire Protection Analysis, 1B · Select one

You are setting up a Fire Dynamics Simulator (FDS) model with a 1,000 kW design fire. Ambient air: T∞ = 293 K, ρ∞ = 1.204 kg/m³, cp = 1.005 kJ/(kg·K), g = 9.81 m/s². What is the characteristic fire diameter, D* = [Q ÷ (ρ∞ cp T∞ √g)]^(2/5)?

Show answer and solution

Answer and solution

Answer: C. 0.96 m.

ρ∞ cp T∞ √g = 1.204 × 1.005 × 293 × 3.132 = 1,110. Q ÷ 1,110 = 0.901, and 0.901^0.4 ≈ 0.96 m. The 1.51 m answer leaves out √g. The 0.24 m and 0.10 m answers are D/4 and D/10, which are grid cell sizes, not D itself. Modelers compare cell size to D to judge how well a fire is resolved. The current FDS guide warns against treating any ratio from past validation work as an automatic minimum, so check results with a grid sensitivity study.

Source: FDS 6.11.1 User's Guide (NIST SP 1019), §6.3.6, Eq. 6.2, p. 41. Topic: NCEES PE Fire Protection specification, April 2027, 1B.

Problem 8 · Where a two-zone model fits poorly

FPA-03 · Fire Protection Analysis, 1B · Select all that apply

A two-zone fire model such as CFAST uses simplified compartment-layer and radiation submodels. In which of these situations should you be especially cautious about those simplifications? Select all that apply.

Show answer and solution

Answer and solution

Answer: A, B and D.

A, B and D. The two-layer compartment approximation is weakest in geometries such as long corridors and tall shafts where conditions vary strongly with position. Separately, CFAST's point-source radiation treatment is a poor near-field approximation within a few fire diameters of the flames. The room-like office in C, with a clearly stratified layer, is much closer to the model's intended compartment representation.

Scored all-or-nothing, like NCEES multiple-correct items: you need A, B, and D and nothing else.

Source: CFAST 7.7.7 Technical Reference Guide (NIST TN 1889v1), §1.1 and §1.5. Topic: NCEES PE Fire Protection specification, April 2027, 1B.

Problem 9 · RSET and the safety margin

FPA-04 · Fire Protection Analysis, 1A · Select one

In a simplified performance-based egress calculation, treat these phases as sequential after ignition: detection 60 s, notification 30 s, pre-movement 120 s, and travel 150 s. Your analysis predicts untenable conditions in the egress path at 480 s. What are the required safe egress time (RSET) and the margin (ASET − RSET)?

Show answer and solution

Answer and solution

Answer: D. RSET 360 s; margin 120 s.

Under the sequential-phase assumption stated in the problem, RSET = 60 + 30 + 120 + 150 = 360 s. The available safe egress time (ASET) is 480 s, so the margin is 120 s. The 240 s answer leaves out pre-movement. The 150 s answer counts only travel. The 480 s answer confuses ASET with RSET. In an actual analysis, define the time components and any overlap consistently in the design basis before comparing RSET with ASET.

Source: NIST TN 1623 (Bukowski, 2009), Part 2, Performance Metrics. Topic: NCEES PE Fire Protection specification, April 2027, 1A.

Problem 10 · Two protection systems failing together

FPA-05 · Fire Protection Analysis, 1A · Select one

In a fault tree, an event requires both the automatic detection system (probability of failure on demand 0.10) and the sprinkler system (0.05) to fail. Assuming the failures are independent, what is the probability that both fail on demand?

Show answer and solution

Answer and solution

Answer: A. 0.005.

An AND gate multiplies independent probabilities: 0.10 × 0.05 = 0.005. The 0.15 answer adds them, which is the rough OR-gate approximation. The 0.145 answer is the probability that at least one fails, 1 − (0.90 × 0.95). The 0.075 answer averages them. Watch the independence assumption: a shared water supply or shared power source creates a common-cause failure and makes the real probability higher.

Source: Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 1A.

Water-Based Fire Protection Systems (Problems 11–16)

What this area tests most often:

  • Think of hydraulics as a pressure budget: end-head pressure, plus friction, plus elevation, compared with what the supply delivers.
  • Density × area gives flow; K-factor turns flow into pressure.
  • Read supply curves with pressure drops, not residual pressures.
  • Add the hose stream allowance before you compare demand to supply.
  • Fire pump tests check points on a curve, not just the rated point.

Problem 11 · End-head pressure from density and area

WB-01 · Water-Based Fire Protection Systems, 5D · Select one

The hydraulically most remote sprinkler must deliver 0.20 gpm/ft² over its 130 ft² coverage area. The sprinkler has K = 5.6 gpm/psi^½. What end-head pressure is required, closest to?

Show answer and solution

Answer and solution

Answer: B. 21.6 psi.

Flow first: q = 0.20 × 130 = 26 gpm. Then q = K√p gives p = (q/K)² = (26 ÷ 5.6)² ≈ 21.6 psi. The 4.6 psi answer forgets to square. The 0.83 psi answer divides q by K². The 10.6 psi answer uses K = 8.0, a different sprinkler. A real design also checks the minimum operating pressure in NFPA 13-2022 and the sprinkler's listing.

Source: FM Global Data Sheet 3-0, Hydraulics of Fire Protection Systems (2010). Topic: NCEES PE Fire Protection specification, April 2027, 5D.

Problem 12 · Friction loss in a feed main

WB-02 · Water-Based Fire Protection Systems, 5D · Select one

A flow of 500 gpm passes through 100 ft (equivalent length) of 4 in. Schedule 40 steel pipe with an inside diameter of 4.026 in. and a Hazen-Williams C of 120. Using p = 4.52 Q^1.85 ÷ (C^1.85 d^4.87), in psi per foot, what is the friction loss?

Show answer and solution

Answer and solution

Answer: D. 7.2 psi.

The formula gives 0.0718 psi per foot. Multiply by 100 ft: about 7.2 psi. The 0.072 psi answer stops at the per-foot value. The 72 psi answer is a decimal slip. The 4.8 psi answer uses C = 150; the problem says 120, and a lower C means a rougher pipe and more friction.

Try a change: Double the flow to 1,000 gpm and the loss rises by 2^1.85, to about 26 psi. Friction grows much faster than flow.

Source: FM Global Data Sheet 3-0, Hydraulics of Fire Protection Systems (2010). Topic: NCEES PE Fire Protection specification, April 2027, 5D.

Problem 13 · Pressure lost to elevation

WB-03 · Water-Based Fire Protection Systems, 5B · Select one

The highest sprinkler branch line is 60 ft above the fire pump discharge gauge. How much pressure does elevation alone cost?

Show answer and solution

Answer and solution

Answer: C. 26.0 psi.

A column of water costs 0.433 psi per foot: 60 × 0.433 ≈ 26.0 psi. The 138.6 psi answer divides by 0.433 instead of multiplying. The 13.0 psi answer halves the factor. The 60.0 psi answer treats feet as psi.

Source: FM Global Data Sheet 3-0, Hydraulics of Fire Protection Systems (2010). Topic: NCEES PE Fire Protection specification, April 2027, 5B.

Problem 14 · Reading a hydrant flow test

WB-04 · Water-Based Fire Protection Systems, 5B · Enter a number

A hydrant flow test gives a static pressure of 80 psi and a residual pressure of 60 psi while flowing 1,000 gpm. Using Q_R = Q_F × (H_R ÷ H_F)^0.54, where H is the pressure drop from static, enter the flow available at a residual pressure of 40 psi, to the nearest 10 gpm.

Your answer (gpm) · accepted tolerance ±10 gpm

Accepted tolerance: ±10 gpm.

Show answer and solution

Answer and solution

Answer: 1,450 gpm.

The test drop is H_F = 80 − 60 = 20 psi. The drop at 40 psi residual is H_R = 80 − 40 = 40 psi. Q_R = 1,000 × (40 ÷ 20)^0.54 = 1,000 × 1.454 ≈ 1,450 gpm. An answer of 2,000 gpm scales linearly. An answer near 800 gpm uses the residual pressures instead of the pressure drops.

Source: City of Oxnard, Fire-Flow Test Worksheet (rev. 2009). Topic: NCEES PE Fire Protection specification, April 2027, 5B.

Problem 15 · Does the fire pump pass?

WB-05 · Water-Based Fire Protection Systems, 5C · Select one

A fire pump is rated 1,000 gpm at 100 psi net. On the acceptance test, net pressure at churn (no flow) is 118 psi, and at 1,500 gpm it is 68 psi. Apply two criteria: at 150% of rated flow, net pressure must be at least 65% of rated; and churn pressure must not exceed 140% of rated. Is the pump acceptable on these two criteria?

Show answer and solution

Answer and solution

Answer: A. Yes — both criteria are met.

65% of 100 psi is 65 psi, and the pump makes 68 psi at 1,500 gpm. 140% of 100 psi is 140 psi, and churn is 118 psi. Both pass. The two “No” choices misread the curve: pressure is supposed to drop below rated at overload and rise above rated at churn. The rated point matters on a real test, but the question asks only about these two criteria. On the exam, take the criteria from NFPA 20-2022.

Source: FM Global Data Sheet 3-7, Fire Protection Pumps (2012, rev. 2025), §2.4 and §3.1.5. Topic: NCEES PE Fire Protection specification, April 2027, 5C.

Problem 16 · Demand versus supply with hose allowance

WB-06 · Water-Based Fire Protection Systems, 5B · Select one

A sprinkler system needs 1,100 gpm at 52 psi at the base of the riser. The design adds a 250 gpm hose stream allowance at the same point. The water supply is the one from Problem 14 (static 80 psi; 60 psi residual at 1,000 gpm). Which statement is correct?

Show answer and solution

Answer and solution

Answer: B. Demand is 1,350 gpm; the supply at 52 psi is only about 1,200 gpm, so it falls short.

Total demand is 1,100 + 250 = 1,350 gpm. Supply at 52 psi: the drop is 80 − 52 = 28 psi, so Q = 1,000 × (28 ÷ 20)^0.54 ≈ 1,200 gpm. That's short of 1,350 gpm. The first choice forgets the hose allowance. The third reads the supply at 40 psi instead of the 52 psi the system needs. The fourth subtracts the hose allowance. Compare demand and supply at the same point and pressure, and apply any safety margin the design basis requires.

Source: City of Oxnard, Fire-Flow Test Worksheet (rev. 2009); Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 5B.

Human Behavior and Evacuation (Problems 17–21)

What this area tests most often:

  • Occupant load and egress capacity are separate checks. One counts people; the other sizes the components.
  • Effective width is narrower than clear width in flow calculations.
  • Pre-movement time is often the biggest unknown in an evacuation timeline.
  • Tenability uses dose, which depends on both concentration and exposure time.

Problem 17 · Stair egress capacity

HB-01 · Human Behavior and Evacuation, 8A · Enter a number

A stair has a clear width of 44 in. Using a capacity factor of 0.3 in. per person (given), enter the number of occupants the stair can serve.

Your answer (persons) · enter a whole number

Accepted tolerance: ±0 persons.

Show answer and solution

Answer and solution

Answer: 146 persons.

44 ÷ 0.3 = 146.7. Round down to 146 people, because a stair can't serve a fraction of a person beyond its width. The problem supplies the 0.3 in./person factor; do not treat that number as a universal code value. On the 2027 exam, use the applicable factor from the supplied NFPA 101-2024 or from the problem statement.

Source: 2025 California Building Code (based on the 2024 IBC), §1005.3.1. Topic: NCEES PE Fire Protection specification, April 2027, 8A.

Problem 18 · Occupant load from a load factor

HB-02 · Human Behavior and Evacuation, 8A · Select one

A 6,000 ft² office floor uses an occupant load factor of 100 ft² per person, gross (given). What is the occupant load?

Show answer and solution

Answer and solution

Answer: D. 60.

Occupant load = area ÷ load factor = 6,000 ÷ 100 = 60 people. The 600 answer divides by 10. The 6 answer divides by 1,000. The 100 answer reports the load factor itself. Check whether the factor is gross or net: a net factor applies only to the usable floor area.

Source: Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 8A.

Problem 19 · Flow time through a corridor

HB-03 · Human Behavior and Evacuation, 8B · Select one

300 people leave through a 1.5 m wide corridor. Assume a 0.2 m boundary layer along each wall and a specific flow of 1.3 persons per second per meter of effective width (both given). Ignoring queuing upstream, how long does it take for everyone to pass?

Show answer and solution

Answer and solution

Answer: B. 210 s.

People don't use the full width, so subtract a boundary layer on each side: 1.5 − (2 × 0.2) = 1.1 m effective width. Flow = 1.3 × 1.1 = 1.43 persons/s, and 300 ÷ 1.43 ≈ 210 s. The 154 s answer uses the full 1.5 m. The 178 s answer subtracts only one boundary layer. The 256 s answer subtracts 0.6 m.

Source: NIST TN 1623 (Bukowski, 2009), Part 2, Performance Metrics; Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 8B.

Problem 20 · Fractional effective dose for CO

HB-04 · Human Behavior and Evacuation, 8C · Select one

An occupant is exposed to a steady 1,000 ppm of carbon monoxide for 30 minutes. Using FED_CO = 2.764 × 10⁻⁵ × C^1.036 × t (C in ppm, t in min), is incapacitation predicted (FED ≥ 1)?

Show answer and solution

Answer and solution

Answer: C. Yes — FED is about 1.06.

1,000^1.036 = 1,282, so FED = 2.764 × 10⁻⁵ × 1,282 × 30 ≈ 1.06. That reaches 1, so incapacitation is predicted. The 0.35 answer is the 10-minute dose. The 0.83 answer drops the 1.036 exponent. The 30 answer forgets the coefficient. Dose accumulates over time, which is why tenability depends on both concentration and duration. More susceptible people can be affected at lower doses.

Source: FDS 6.11.1 User's Guide (NIST SP 1019), §22.10.21 (FED); NIST TN 1644 (2009), p. 6. Topic: NCEES PE Fire Protection specification, April 2027, 8C.

Problem 21 · Shortening pre-movement time

HB-05 · Human Behavior and Evacuation, 8C · Select one

Which change is most likely to shorten occupants' pre-movement time in a public building?

Show answer and solution

Answer and solution

Answer: A. Replace the bell-only signal with clear, informative voice messages.

Research on public buildings found that an alarm signal alone often doesn't prompt people to leave. Clear voice messages that say what is happening and what to do get faster reactions, and live messages do better than prerecorded ones. A louder bell helps audibility but still doesn't tell people what's going on. Exit signs help with wayfinding once people move, not with the decision to start. Corridor length has nothing to do with the delay before people begin moving.

Source: Proulx, “How to initiate evacuation movement in public buildings,” Facilities 17 (1999), pp. 333–334 (NRC Canada). Topic: NCEES PE Fire Protection specification, April 2027, 8C.

Fire Alarm and Signaling Systems (Problems 22–25)

What this area tests most often:

  • Battery size = standby + alarm, then the margin the design calls for.
  • Voltage drop uses the round-trip conductor length.
  • Sound drops about 6 dB per doubling of distance in a free field, less indoors.
  • Circuit loading uses worst-case device current.

Problem 22 · Secondary power battery size

FA-01 · Fire Alarm and Signaling Systems, 4D · Select one

A fire alarm panel draws 0.8 A in standby and 3.2 A in alarm. The project requires 24 hours of standby followed by 5 minutes of alarm, plus a 20% capacity margin (all given). What is the minimum battery capacity?

Show answer and solution

Answer and solution

Answer: C. 23.4 Ah.

Standby: 0.8 A × 24 h = 19.2 Ah. Alarm: 3.2 A × (5 ÷ 60) h = 0.27 Ah. Total 19.47 Ah, times 1.2 for the margin = 23.4 Ah. The 19.5 Ah answer leaves out the margin. The 35.2 Ah answer treats 5 minutes as 5 hours, and 42.2 Ah does that and adds the margin. On the exam, standby and alarm durations come from NFPA 72-2022 or the problem itself.

Source: Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 4D.

Problem 23 · Voltage drop on a notification circuit

FA-02 · Fire Alarm and Signaling Systems, 4D · Select one

A notification appliance circuit carries 2.0 A to appliances lumped at the end of a 300 ft run of 12 AWG wire (2.0 Ω per 1,000 ft of conductor, given). The panel puts out 20.4 V at the end of battery life. What voltage reaches the last appliance?

Show answer and solution

Answer and solution

Answer: B. 18.0 V.

Current goes out and back, so use 600 ft of conductor: 0.6 × 2.0 = 1.2 Ω. Drop = 2.0 A × 1.2 Ω = 2.4 V, leaving 20.4 − 2.4 = 18.0 V. The 19.2 V answer uses only the one-way length. The 15.6 V answer doubles the round trip again. The 20.4 V answer ignores wire resistance. Lumping the load at the end is conservative; spread-out appliances drop less.

Source: Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 4D.

Problem 24 · Sound level at a greater distance

FA-03 · Fire Alarm and Signaling Systems, 4B · Select one

An outdoor horn measures 84 dBA at 10 ft. Assuming free-field, point-source behavior, what level should you expect at 40 ft?

Show answer and solution

Answer and solution

Answer: A. 72 dBA.

In a free field, sound from a point source drops about 6 dB each time the distance doubles. 10 → 20 → 40 ft is two doublings, so 84 − 12 = 72 dBA. The 78 dBA answer uses 3 dB per doubling. The 21 dBA answer divides 84 by 4. The 60 dBA answer subtracts 24 dB. Indoors, reflections usually make the drop smaller.

Source: OSHA Technical Manual, Sec. III Ch. 5 (Noise), §II.B.6. Topic: NCEES PE Fire Protection specification, April 2027, 4B.

Problem 25 · Appliances on one circuit

FA-04 · Fire Alarm and Signaling Systems, 4C · Select one

A notification circuit is rated 3.0 A. Each horn/strobe draws 0.12 A at its worst-case operating voltage (given). How many appliances can the circuit carry?

Show answer and solution

Answer and solution

Answer: D. 25.

3.0 ÷ 0.12 = 25 appliances. The arithmetic uses the worst-case current supplied in the stem. In an actual design, use the listed appliance current and the applicable circuit and power-supply requirements rather than assuming this simplified division is the only check.

Source: Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 4C.

Passive and Structural Fire Protection (Problems 26–29)

What this area tests most often:

  • Calculated fire resistance methods have limits. Know them before you apply the equation.
  • Wood loses section to char; the effective char depth adds a weakened layer.
  • Lighter steel heats faster, so it needs thicker protection.
  • Firestop ratings measure different things: flame passage versus temperature rise.

Problem 26 · Effective char depth for a timber member

PS-01 · Passive and Structural Fire Protection, 7C · Select one

For exposed wood members, use a nominal char rate βn = 1.5 in./h, char depth a_char = βn × t^0.813 (t in hours), and effective char depth a_eff = 1.2 × a_char. What is the effective char depth for a 1.5-hour exposure?

Show answer and solution

Answer and solution

Answer: C. 2.5 in..

a_char = 1.5 × 1.5^0.813 = 1.5 × 1.39 = 2.09 in. Multiply by 1.2: a_eff ≈ 2.5 in. The 2.7 in. answer uses linear time (1.5 × 1.5 × 1.2). The 2.1 in. answer stops at the char depth. The 2.25 in. answer is linear and skips the 1.2. The 1.2 factor accounts for the heated, weakened zone just behind the char. AWC limits this method to ratings of 2 hours or less.

Source: AWC Technical Report 10 (2021), Eqs. 1.4-8 to 1.4-10, p. 11. Topic: NCEES PE Fire Protection specification, April 2027, 7C.

Problem 27 · Adjusting spray-applied fireproofing thickness

PS-02 · Passive and Structural Fire Protection, 7C · Select one

A beam tested with 1.0 in. of spray-applied fire-resistive material (SFRM) has W1/D1 = 0.80. A substitute beam has W2/D2 = 0.50. Using h2 = h1 × [(W1/D1) + 0.6] ÷ [(W2/D2) + 0.6], what SFRM thickness does the substitute beam need?

Show answer and solution

Answer and solution

Answer: D. 1.27 in..

h2 = 1.0 × (0.80 + 0.6) ÷ (0.50 + 0.6) = 1.4 ÷ 1.1 ≈ 1.27 in. A lighter section (lower W/D) heats faster, so it needs more protection. The 1.60 in. answer ignores the 0.6 terms. The 0.79 in. answer inverts the ratio. The 1.00 in. answer assumes nothing changes.

Source: FM Global Data Sheet 1-21, Fire Resistance of Building Assemblies, §4.1, Eq. 1. Topic: NCEES PE Fire Protection specification, April 2027, 7C.

Problem 28 · Limits on the SFRM substitution equation

PS-03 · Passive and Structural Fire Protection, 7C · Select all that apply

For the SFRM thickness equation in Problem 27, which conditions must hold for it to apply? Select all that apply.

Show answer and solution

Answer and solution

Answer: A, B and C.

A, B and C. The equation has a floor on how light the substitute section can be, a minimum thickness, and a minimum rating. D is false: the equation exists largely to adjust protection for sections lighter than the tested one, as Problem 27 shows. Restrained ratings carry extra conditions, and intumescent or mastic coatings need their own testing.

Scored all-or-nothing: A, B, and C, and nothing else.

Source: FM Global Data Sheet 1-21, Fire Resistance of Building Assemblies, §4.1, Eq. 1. Topic: NCEES PE Fire Protection specification, April 2027, 7C.

Problem 29 · Which firestop rating covers temperature rise?

PS-04 · Passive and Structural Fire Protection, 7D · Select one

A firestop system for a pipe penetration must limit temperature rise on the unexposed side, not just keep flames from passing. Under ASTM E814 testing, which rating addresses that?

Show answer and solution

Answer and solution

Answer: B. T rating.

ASTM E814 bases the F rating on flame occurrence on the unexposed side. The T rating adds temperature rise on the unexposed side. A flame spread index is a surface-burning property used for interior finish, not penetrations. The wall's own rating covers the assembly, not the firestop system sealing the hole through it.

Source: ASTM E814-23a, §5.2 (Significance and Use). Topic: NCEES PE Fire Protection specification, April 2027, 7D.

Special Hazard Systems (Problems 30–33)

What this area tests most often:

  • Halocarbon agents and inert gases use different flooding equations.
  • Specific vapor volume depends on temperature, so read the design temperature carefully.
  • Mixture flammable limits aren't simple averages.
  • Ventilation design for flammable vapors works backward from a target fraction of the LFL.

Problem 30 · Clean agent quantity

SH-01 · Special Hazard Systems, 6D · Select one

A 100 m³ room at 20 °C is protected by FK-5-1-12 at a 4.5% design concentration. The agent's specific vapor volume is S = 0.0664 + 0.000274 T (m³/kg, T in °C). Using W = (V/S) × C ÷ (100 − C), how much agent is needed?

Show answer and solution

Answer and solution

Answer: B. 65.6 kg.

S = 0.0664 + 0.000274 × 20 = 0.07188 m³/kg. W = (100 ÷ 0.07188) × (4.5 ÷ 95.5) ≈ 65.6 kg. The 62.6 kg answer uses C/100, which ignores how the agent displaces air as it is added. The 0.34 kg answer multiplies by S. The 71.0 kg answer evaluates S at 0 °C instead of 20 °C. On the exam, design concentration and agent data come from NFPA 2001-2022 or the problem.

Source: Honeywell FK-5-1-12 System Manual HSN-MNL-F2-R4 (2024), §3.1.10. Topic: NCEES PE Fire Protection specification, April 2027, 6D.

Problem 31 · Inert gas quantity

SH-02 · Special Hazard Systems, 6D · Select one

An inert gas system must reach a 40% concentration in a 100 m³ room. Take S = 0.70 m³/kg at the design temperature (given). Using X = (V/S) × ln[100 ÷ (100 − C)], how much agent is needed?

Show answer and solution

Answer and solution

Answer: C. 73.0 kg.

ln(100 ÷ 60) = 0.511, so X = (100 ÷ 0.70) × 0.511 ≈ 73.0 kg. The 31.7 kg answer uses log₁₀ without the 2.303 conversion. The 57.1 kg answer treats the concentration linearly (0.40). The 130.9 kg answer uses ln(100 ÷ 40). Inert gases use the logarithmic form because a mix of agent and air vents out as the concentration builds.

Source: Honeywell IG-541 System Manual HSN-MNL-IG541-R3 (2024), §4.1.9. Topic: NCEES PE Fire Protection specification, April 2027, 6D.

Problem 32 · Lower flammable limit of a gas mixture

SH-03 · Special Hazard Systems, 6D · Enter a number

A fuel gas is 60% methane (LFL 5.0%), 30% propane (LFL 2.1%), and 10% butane (LFL 1.8%) by volume (LFLs given). Using Le Chatelier's mixing rule, enter the mixture's lower flammable limit in vol%, to two decimal places.

Your answer (vol%) · accepted tolerance ±0.02 vol%

Accepted tolerance: ±0.02 vol%.

Show answer and solution

Answer and solution

Answer: 3.14 vol%.

1/LFL_mix = 0.60/5.0 + 0.30/2.1 + 0.10/1.8 = 0.120 + 0.143 + 0.056 = 0.318, so LFL_mix ≈ 3.14%. An answer of 3.81% weights the LFLs linearly by volume fraction, which overstates the limit.

Source: Liekhus et al., J. Loss Prev. Process Ind. 13 (2000), Eq. 1 (NIOSH). Topic: NCEES PE Fire Protection specification, April 2027, 6D.

Problem 33 · Ventilation to stay below 25% of the LFL

SH-04 · Special Hazard Systems, 6B · Select one

Propane vapor is released at 0.2 ft³/min into a room. The design goal is to keep the concentration at or below 25% of the LFL (propane LFL 2.1%, given). Assuming perfect mixing, what ventilation rate is needed?

Show answer and solution

Answer and solution

Answer: A. 38 cfm.

Target concentration = 0.25 × 0.021 = 0.00525. Ventilation = release rate ÷ target = 0.2 ÷ 0.00525 ≈ 38 cfm. The 9.5 cfm answer designs to 100% of the LFL. The 0.38 cfm answer uses 2.1 instead of 0.021. The 152 cfm answer applies the 25% factor twice. Real rooms don't mix perfectly, so designs add a mixing factor on top.

Source: OSHA Technical Manual, Sec. III Ch. 3 (Ventilation), general (dilution) ventilation; Worked from first principles using only the inputs given. Topic: NCEES PE Fire Protection specification, April 2027, 6B.

Smoke Control Systems (Problems 34–36)

What this area tests most often:

  • Plume mass flow sets the exhaust rate needed to hold a smoke layer at a given height.
  • Convert mass to volume at smoke temperature, not ambient.
  • Pressurization has a floor and a ceiling: enough pressure to hold back smoke, not so much that people can't open the doors.
  • The 2027 outline adds inspection, testing, and commissioning to this area.

Problem 34 · Plume mass flow in an atrium

SC-01 · Smoke Control Systems, 3B · Select one

An atrium design fire has a total HRR of 5,000 Btu/s; take the convective HRR as Qc = 0.7Q. For the axisymmetric plume (U.S. units): limiting flame height zl = 0.533 Qc^(2/5); above it, m = 0.022 Qc^(1/3) z^(5/3) + 0.0042 Qc (m in lb/s, Qc in Btu/s, z in ft). What is the plume mass flow at a smoke layer interface 40 ft above the fuel?

Show answer and solution

Answer and solution

Answer: D. 171 lb/s.

Qc = 0.7 × 5,000 = 3,500 Btu/s. zl = 0.533 × 3,500^0.4 ≈ 13.9 ft, so 40 ft is above the flames and the main equation applies. m = 0.022 × 3,500^(1/3) × 40^(5/3) + 0.0042 × 3,500 ≈ 156 + 15 ≈ 171 lb/s. The 197 lb/s answer uses total HRR instead of convective. The 156 lb/s answer drops the second term. The 111 lb/s answer uses the below-flame-height form. These are the NFPA 92 axisymmetric plume equations; the exam supplies NFPA 92-2024.

Source: Klote, ASHRAE AtriumCalc Report v1.1 (2015), Eqs. 2–4 (same as NFPA 92 axisymmetric plume). Topic: NCEES PE Fire Protection specification, April 2027, 3B.

Problem 35 · Volumetric exhaust rate

SC-02 · Smoke Control Systems, 3B · Select one

To hold the smoke layer at 40 ft, exhaust must match the 171 lb/s plume mass flow from Problem 34. Smoke density at the exhaust is 0.068 lb/ft³ (given). Using V = 60 m ÷ ρ, what volumetric exhaust rate is needed?

Show answer and solution

Answer and solution

Answer: C. 151,000 cfm.

V = 60 × 171 ÷ 0.068 ≈ 151,000 cfm. The 137,000 cfm answer uses ambient air density (0.075 lb/ft³); hot smoke is lighter, so the same mass takes more volume. The 2,500 cfm answer forgets the 60 s/min conversion. The 700 cfm answer multiplies by density instead of dividing.

Source: Klote, ASHRAE AtriumCalc Report v1.1 (2015), Eq. 14. Topic: NCEES PE Fire Protection specification, April 2027, 3B.

Problem 36 · Door opening force in a pressurized stair

SC-03 · Smoke Control Systems, 3A · Select one

A stair pressurization system holds 0.10 in. w.c. across a 3 ft × 7 ft door. The knob is 3 in. from the latch edge (d = 0.25 ft), and the door closer needs 10 lb. Using F = Fdc + 5.2 W A ΔP ÷ [2(W − d)] (F in lb, W and d in ft, A in ft², ΔP in in. w.c.), what force is needed to open the door?

Show answer and solution

Answer and solution

Answer: B. 16.0 lb.

Pressure part: 5.2 × 3 × 21 × 0.10 ÷ [2 × (3 − 0.25)] = 32.76 ÷ 5.5 ≈ 6.0 lb. Add the 10 lb closer force: about 16.0 lb. The 6.0 lb answer leaves out the closer. The 15.5 lb answer uses W instead of W − d. The 11.1 lb answer leaves out the 5.2 constant. This is why pressurization has an upper limit as well as a lower one: too much pressure and people can't open the door.

Source: 2025 California Fire Code (based on the 2024 IFC), §909.6.2; Consulting-Specifying Engineer, “Smoke control design considerations” (2016). Topic: NCEES PE Fire Protection specification, April 2027, 3A.

How to read your results

Treat your percentage as feedback on these 36 problems, not an NCEES score or a prediction of passing. Three to six problems per area can show which methods you can't yet explain, but they can't measure an area precisely.

NCEES counts correct answers, converts the count to a scaled score to adjust for differences between exam forms, and compares it with a standard set by subject-matter experts. It doesn't publish the passing score, so no practice percentage converts into an exam result (Examinee Guide, p. 14).

Your plan to April 13, 2027

Assumptions: you start the week of October 12, 2026, study about 8 focused hours a week (roughly 200 hours in all), and work full time. The plan works area by area, largest first, then switches to mixed and timed practice. This is our editorial plan, not an NCEES requirement, and no number of hours guarantees a pass.

Your plan to April 13, 2027
Week ofFocusDo thisFinish the week with
Oct 12Setup and baselineConfirm your board's approval process and register in MyNCEES. Download the 2027 specification and the handbook. Work all 36 problems before opening any solution.An error log and a ranked list of weak areas
Oct 19Get your referencesArrange lawful study access to the 10 standards at the editions in the table above. Read each scope chapter and table of contents.You can say what each standard governs
Oct 26 – Nov 9 (3 weeks)Water-based systemsNFPA 13 design criteria, hydraulic calculations, NFPA 20 pumps, water supplies. Rework Problems 11–16.A hand hydraulic calculation for a small tree system
Nov 16 – Nov 23 (2 weeks)Fire scienceHeat transfer, combustion, HRR, plumes, ceiling jets, enclosure fires. Rework Problems 1–5.Each correlation located in the handbook, with its units
Nov 30Fire protection analysisPerformance-based design process, fire models, risk; NFPA 855 and UL 9540A. Rework Problems 6–10.A one-page summary of how ESS hazards are evaluated
Dec 7 – Dec 14 (2 weeks)Human behavior and evacuationNFPA 101 occupant load, capacity, travel distance; timed egress; tenability. Rework Problems 17–21.Capacity and flow-time calculations in both unit systems
Dec 21 – Dec 28Light weeksTwo to four hours: retry earlier misses on fresh problems.An up-to-date error log
Jan 4 – Jan 11 (2 weeks)Fire alarm and signalingNFPA 72 design, battery, voltage drop, audibility, ECS and MNS. Rework Problems 22–25.Battery and voltage-drop calculations in under 6 minutes each
Jan 18 – Jan 25 (2 weeks)Passive and structuralFire resistance, compartmentation, openings and penetrations, char, SFRM. Rework Problems 26–29.A list of each method's limits
Feb 1 – Feb 8 (2 weeks)Special hazardsNFPA 2001, 11, 30, and 400; flammable limits. Rework Problems 30–33.Agent quantity calculations in both unit systems
Feb 15Smoke controlNFPA 92 plumes, exhaust, pressurization, testing and commissioning. Rework Problems 34–36.A plume-to-exhaust calculation without notes
Feb 22 – Mar 1 (2 weeks)Mixed practice and lookupsMixed sets, so you choose the method yourself. Daily lookup drills across all 10 standards.Lookup times logged by standard
Mar 8Set up the rehearsalIf a current 2027-aligned 85-question practice source is available, use it; otherwise assemble fresh mixed sets totaling 85 questions without treating them as an official full-length exam. Set out your calculator, references, and scratch-work setup. Block the day.Rehearsal day on the calendar
Mar 15Full rehearsalWork 85 questions in 8.5 hours. Submit the first half before starting the second. Use only the handbook, the 10 standards, your approved calculator, and a scratch-work setup that approximates the test center.Results by area and error type
Mar 22 – Mar 29 (2 weeks)Fix the leaksSpend most of the time on areas that cost the most questions. Check each fix on a problem you haven't seen.A short repair list based on work you actually did
Apr 5Taper and logisticsShort mixed sets. Confirm your appointment, ID, route, and calculator. Nothing new the day before.An exam-day plan
Apr 13Exam day

A typical 8-hour week: two 90-minute sessions on concepts and handbook relations, two 60-minute problem sessions, two 30-minute lookup drills, one 90-minute mixed or timed set, and 30 minutes updating the error log. Move the blocks around work and family; keep the mix.

Timed blocks you can run with this page: Block A is the odd-numbered problems (1, 3, 5 … 35) and Block B the even-numbered ones (2, 4, 6 … 36). Each has 18 problems, so give it 18 × 6 = 108 minutes. You'll have seen these before, so treat them as pacing practice, not a fresh exam.

Starting later?

The same plan compresses if you hold the order and raise the weekly hours:

  • Starting in mid-January (12 weeks, about 14 hours a week): Jan 18 setup and references · Jan 25 – Feb 1 water-based · Feb 8 fire science · Feb 15 analysis and ESS · Feb 22 human behavior · Mar 1 alarm · Mar 8 passive and special hazards · Mar 15 smoke control and mixed sets · Mar 22 full rehearsal · Mar 29 fix the leaks · Apr 5 taper.
  • Fewer than 8 weeks: don't try to cover everything evenly. Work all 36 problems, rank your weak areas, put most hours into the biggest weak areas, and keep daily lookup drills and at least one timed mixed set. Be honest about whether April 13 is realistic for you.
  • Strong in one specialty? Keep a short review there and don't skip it entirely. Move the saved hours to areas you don't touch at work.
  • More time than this? Add a mixed-review week after every two areas rather than lingering in your favorite topics.

What to study next

When you're not sure where to spend the next session, use this order:

  1. Stay inside the 2027 specification.
  2. Rank weak areas from your error log, not from how a topic feels.
  3. Break ties by the area's question range and how often the same mistake repeats.
  4. Retest each fix on a fresh problem.
  5. Stop polishing a strong niche while a large area is still weak.

Turn each miss into your next study task

Turn each miss into your next study task
What went wrongNext stepHow you know it's fixed
Didn't know the concept or modelGo back to the concept and its assumptions before doing more problemsYou can explain why the method applies
Couldn't find the provision or relationRepeat the lookup; record the section and the search words that workedThe same lookup is faster next time
Used the wrong edition or wrong standardRe-find it in the edition listed for 2027You cite the right edition from memory
Picked the wrong equation or region (total vs. convective HRR, plume vs. ceiling jet)Write one sentence on what's being asked and which form applies, then rework with one input changedYou can say why the other form doesn't fit
Units or unit systemRedo it with units on every lineUnits cancel to what the question asks for
Missed a given assumption in the stemUnderline every given value and condition, then reworkNothing in the stem goes unused
Arithmetic or signRedo it from scratch, then check magnitude and directionThe check agrees without copying your first try
Ran out of timeNote where the time went; practice flagging and returning in a timed blockYour average moves toward 6 minutes

Error log fields: problem or source · date · area · minutes spent · your answer · correct or lucky guess · first wrong step · error type (from the table) · where you found the fix (standard, edition, section, search words) · date you'll retry on a fresh problem.

If your exam is next week

Don't start anything new. Pick your two most repeated errors and your two slowest lookups, fix those, and run one timed block. Confirm your appointment, ID, and calculator. If you need to move your appointment, you must reschedule at least 48 hours ahead, and Pearson charges $50 (Examinee Guide, p. 6). Fire Protection is offered once a year, so moving off April 13, 2027 means waiting for the next annual date, which NCEES hadn't posted when we checked.

Exam-day rules that change how you practice

  • About 6 minutes per question on average. That's 510 minutes ÷ 85 questions. It's a budget, not a per-question limit.
  • Two sections, one clock. After roughly half the questions, you review and submit them, and you can't go back. The sections aren't timed separately, so time you save early carries forward.
  • Breaks: the optional 50-minute break comes after you submit the first section. Unused break time doesn't add exam time, and unscheduled breaks come out of your exam time.
  • No penalty for wrong answers, so answer everything. Some questions are unscored pretest items that look like all the others.
  • One approved calculator. For 2026 exams, NCEES approved the HP 33s and HP 35s, any Casio fx-115 or fx-991 model, and any TI-30X or TI-36X model; an on-screen TI-30XS is also available. NCEES reviews the list each year, and the 2027 list hadn't been posted when we checked, so confirm it before April.
  • Scratch work: you get two reusable booklets and three markers, not paper and pencil. Practice some problems on a whiteboard so it isn't new on exam day.
  • Arrive 30 minutes early, and make sure the first and last names on your appointment confirmation match your ID.

Sources: NCEES Examinee Guide, May 2026, pp. 6, 8–12, and 14; NCEES 2026 calculator list, p. 1.

Quick answers

Are NFPA 12 and NFPA 25 still on the exam?

Not as supplied references. Neither appears on the standards list for exams beginning April 2027 (specification, standards page). Carbon dioxide still appears among the special hazard system types NCEES names, so know how CO₂ systems work even though NFPA 12 isn't supplied.

What's the pass rate?

In the Fire Protection row NCEES labels "July 2025," 85% of 262 first-time takers passed, as did 44% of 48 repeat takers (NCEES). Those numbers describe a past group of examinees on the old specification, not your personal odds or a readiness target.

What score do I need?

NCEES doesn't publish a passing score, and results are reported only as pass or fail. Claims that you need a specific percentage aren't based on anything official (Examinee Guide, p. 14).

I failed in April 2026. How do I use my diagnostic report?

The report compares your performance in each knowledge area, on a 0–15 scale, with the average of passing examinees (Examinee Guide, pp. 20–21). Your report uses the old four-area outline, so map each weak area onto the 2027 areas with the table above before you plan. Then confirm each weakness with fresh problems.

Do I need the FE first, or four years of experience?

Your state licensing board decides who can sit for the PE. NCEES designs the exam for engineers with at least four years of post-college experience, but whether you can test earlier is your board's rule. Find yours in the NCEES board directory. If you still need the FE, start with FE exam prep.

Does passing make me a licensed PE?

No. Your licensing board issues the license. Passing the exam is one of its requirements.

Do I need a prep course?

No course is required. Whatever you use, check that it matches the April 2027 specification and the NFPA editions listed above.

How do registration, results, retakes, and accommodations work?

You register and pay in MyNCEES after checking your board's approval process. You can attempt an exam once per testing window and no more than three times in 12 months, and some boards are stricter; for Fire Protection, the once-a-year date is the practical limit. Accommodations must be requested during registration. See NCEES exam registration, NCEES exam results, NCEES retake rules, and NCEES exam accommodations.

Comparing Fire Protection with other PE exams? See PE exam prep.

Sources

Exam facts on this page were last verified October 9, 2026, against the NCEES sources below. Every practice item was re-solved during this audit, calculation-based answers were independently recomputed, and consequential source/method mappings were rechecked against the cited material available to us. The problems have not been reviewed by NCEES or by a licensed fire protection engineer. This guide was developed with AI-assisted research and editing; see how Castleport uses sources and AI assistance.

NCEES

Teaching sources for the practice problems

These teaching sources explain the methods. They are not the references supplied on the exam, and the California code sections are cited only for equations and factors that match the national model codes they're based on.

Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES) or the National Fire Protection Association (NFPA). Exam, credential, and standard names identify their subjects; trademarks belong to their respective owners. The practice problems are original and unofficial, and nothing on this page guarantees a passing result or a license.

PE Fire Protection Exam Prep: 2027 Plan + 36 Problems