Free PE Fire Protection Practice Test (40 Qs, 2027 Spec)

These 40 original, unofficial questions follow the NCEES PE Fire Protection specification that takes effect with the April 13, 2027 exam, and every one has a worked solution. No signup is required. Units switch between SI and U.S. customary the way the real exam does, so use the values given in each question.

Practice questions

Fire Protection Analysis (Questions 1–6)

Question 1 · Event-tree frequency

A storage room has an ignition frequency of 0.02 fires per year. On demand, the sprinkler system fails with probability 0.05 and the fire alarm fails with probability 0.10. Treat the two failures as independent. What is the annual frequency of a fire in which both the sprinklers and the alarm fail?

A. 1.0 × 10⁻⁴ per yearB. 1.0 × 10⁻³ per yearC. 3.0 × 10⁻³ per yearD. 1.7 × 10⁻² per year

Show answer and explanation

A. 1.0 × 10⁻⁴ per year. Multiply along the branch: 0.02 × 0.05 × 0.10 = 1.0 × 10⁻⁴ per year. Each branch of an event tree is conditional on the one before it, so independent probabilities multiply. Adding them is the classic slip.

B Stops after the sprinkler branch (0.02 × 0.05).
C Adds the failure probabilities (0.02 × 0.15) instead of multiplying.
D Calculates the branch where both systems work (0.02 × 0.95 × 0.90).

Fire Protection Analysis · 1A Fire risk assessment · PEFP-01. Method: Event-tree multiplication with the independence stated in the question.

Question 2 · ASET–RSET margin

A performance-based analysis predicts untenable conditions in a corridor at 15.0 minutes (the available safe egress time, ASET). Detection takes 60 s, alarm processing 30 s, pre-movement 3.0 min, and movement 4.0 min. What is the safety margin, ASET minus the required safe egress time (RSET), in minutes?

Enter a number in minutes, rounded to one decimal place.

Your answer (min)

Show answer and explanation

6.5 min. RSET = 1.0 + 0.5 + 3.0 + 4.0 = 8.5 min, so the margin is 15.0 − 8.5 = 6.5 min. Pre-movement is the piece people forget. It's real time, often minutes, spent noticing cues, checking what's going on, and getting ready to leave. Accepted here: 6.45 to 6.55 min.

11.0 Counts movement time only.
9.5 Leaves out pre-movement time.

Fire Protection Analysis · 1A Performance-based design · PEFP-02. Source: Kuligowski, The Process of Human Behavior in Fires, NIST TN 1632 (2009). Timeline arithmetic; pre-movement behavior per NIST TN 1632.

Question 3 · Two-zone model assumptions

Which are assumptions of a classic two-zone compartment fire model? Select all that apply.

Select all that apply. Credit requires every correct choice and no wrong ones.

A. The upper layer has one uniform temperature.B. The layer interface is at the same height everywhere in the room.C. Pressures are treated as hydrostatic; dynamic pressure effects are neglected.D. The model resolves three-dimensional velocity fields within each layer.E. Plume and doorway flows are calculated with algebraic correlations.

Show answer and explanation

A, B, C, E. A zone model idealizes a room as two well-mixed layers with a sharp interface between them. Pressures are hydrostatic, and the flows that move mass between layers (plumes, doorjets) come from algebraic correlations. That simplicity is why zone models run fast, and why they struggle with very large, tall, or oddly shaped spaces.

D False. Resolving three-dimensional flow is what computational fluid dynamics (CFD) models such as FDS do.

Fire Protection Analysis · 1B Fire models · PEFP-03. Source: Klote, NISTIR 5516, Method of Predicting Smoke Movement in Atria (NIST, 1994). NISTIR 5516, Section 3 (zone fire model concept).

Question 4 · CFD grid resolution

An analyst will model a 1,500 kW fire in FDS and wants D*/δx of about 10, where the characteristic fire diameter is D* = [Q / (ρ∞ cₚ T∞ √g)]^(2/5). Use ρ∞ = 1.204 kg/m³, cₚ = 1.005 kJ/kg·K, T∞ = 293 K, and g = 9.81 m/s². Which cell size δx is closest?

A. 0.05 mB. 0.11 mC. 0.23 mD. 0.45 m

Show answer and explanation

B. 0.11 m. D* = [1,500 / (1.204 × 1.005 × 293 × 3.132)]^0.4 = 1.13 m. For D*/δx ≈ 10, δx ≈ 0.11 m. D*/δx counts how many cells span the fire's characteristic diameter; too few and the plume isn't resolved.

A Gives D*/δx ≈ 23, finer than asked and about eight times as many cells as B.
C Gives D*/δx ≈ 5, near the coarse end of the 4–16 range used in U.S. NRC validation work.
D Gives D*/δx ≈ 2.5, below that range.

Fire Protection Analysis · 1B Fire models · PEFP-04. Source: FEMTC 2011 paper citing the FDS User's Guide D*/δx guidance and NRC 4–16 range. D*/δx resolution guidance from the FDS User's Guide as cited in the FEMTC 2011 paper.

Question 5 · UL 9540A

Which statement about UL 9540A is correct?

A. It is a product listing; a battery tested to UL 9540A needs no further evaluation.B. It is a test method that generates fire-propagation data, run in sequence from cell to module to unit to installation level.C. It tests only complete installations; cell data come from the battery manufacturer.D. Once testing is complete, it replaces NFPA 855's installation requirements.

Show answer and explanation

B. It is a test method that generates fire-propagation data, run in sequence from cell to module to unit to installation level. UL 9540A is a test method for thermal-runaway fire propagation. It starts with a single cell and moves to module, unit, and installation levels as needed. The data it produces feed installation decisions under codes such as NFPA 855, which is on the 2027 standards list.

A Confuses a test method with a certification.
C Reverses the sequence; cell-level testing comes first.
D Test data inform installation decisions; they don't replace the installation standard.

Fire Protection Analysis · 1C Energy storage systems · PEFP-05. Source: UL Solutions, Understanding UL 9540A, NFPA 855 and Large-Scale Fire Testing.

Question 6 · Battery off-gas dilution

Thermal runaway in a battery room releases flammable gas at a steady 0.30 m³/min. The gas mixture's lower flammable limit (LFL) is 6.0% by volume. The design keeps the well-mixed room concentration at or below 25% of the LFL. Ignoring the gas's own volume, what exhaust flow is required?

A. 1.25 m³/minB. 5.0 m³/minC. 20 m³/minD. 80 m³/min

Show answer and explanation

C. 20 m³/min. The allowed concentration is 0.25 × 6.0% = 1.5%. At steady state, concentration = release rate ÷ exhaust flow, so Q = 0.30 ÷ 0.015 = 20 m³/min.

A Multiplies the release by 25% instead of dividing by the allowed concentration.
B Dilutes only to the LFL itself (0.30 ÷ 0.06).
D Applies the 25% factor twice (0.30 ÷ 0.00375).

Fire Protection Analysis · 1C Energy storage systems · PEFP-06. Method: Steady-state dilution; the 25%-of-LFL design criterion is given in the question.

Fire Science (Questions 7–12)

Question 7 · Point-source radiation

A 2,000 kW pool fire has a radiative fraction of 0.30. Using the point-source model q″ = χᵣQ / (4πR²), at what distance does incident heat flux fall to 10 kW/m²?

A. 2.2 mB. 3.1 mC. 4.0 mD. 4.8 m

Show answer and explanation

A. 2.2 m. R = √(0.30 × 2,000 / (4π × 10)) = √4.77 = 2.2 m. The point-source model is a far-field approximation; close to a large fire it can understate flux, which is why solid-flame models exist.

B Uses 2πR², a hemisphere, instead of a sphere.
C Ignores χᵣ, treating all of the heat as radiated.
D Forgets the square root.

Fire Science · 2A Heat transfer · PEFP-07. Source: NUREG-1805 Supplement 1, Vol. 2, App. A (NRC, 2013). Chapter 5 point-source radiation spreadsheet (radiative fraction 0.30), App. A.

Question 8 · Blackbody emissive power

What is the emissive power of a blackbody surface at 800°C? Use σ = 5.67 × 10⁻¹¹ kW/m²·K⁴.

A. 23 kW/m²B. 60 kW/m²C. 75 kW/m²D. 75,000 kW/m²

Show answer and explanation

C. 75 kW/m². E = σT⁴ with absolute temperature: 5.67 × 10⁻¹¹ × (1,073.15)⁴ = 75 kW/m². The fourth power is why a few hundred degrees makes such a large difference in radiant exposure.

A Uses 800 instead of 1,073 K.
B Multiplies by an emissivity of 0.8 that the question didn't give.
D Uses σ in W/m²·K⁴ but labels the result kW/m².

Fire Science · 2A Heat transfer · PEFP-08. Method: Stefan–Boltzmann law.

Question 9 · Stoichiometric propane–air mixture

Propane burns as C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Air is 21% O₂ by volume (3.76 mol N₂ per mol O₂). What is the stoichiometric propane concentration in air, by volume?

A. 4.0%B. 4.2%C. 5.1%D. 16.7%

Show answer and explanation

A. 4.0%. Each mole of propane needs 5 mol O₂, which arrives with 5 × 3.76 mol N₂: 5 × 4.76 = 23.8 mol of air. The fuel fraction is 1 ÷ (1 + 23.8) = 4.0%.

B Divides by the air alone (1 ÷ 23.8), leaving the fuel out of the total.
C Mixes the fuel with nitrogen only (1 ÷ 19.8).
D Mixes the fuel with pure oxygen (1 ÷ 6).

Fire Science · 2B Fire chemistry · PEFP-09. Method: Stoichiometry with the air composition given in the question.

Question 10 · t-squared growth

A fire follows Q = αt² with the fast growth coefficient α = 0.04689 kW/s². How many seconds after effective ignition does it reach 2,000 kW? Round to the nearest second.

Enter a number in seconds, rounded to the nearest whole number.

Your answer (s)

Show answer and explanation

207 s. t = √(2,000 ÷ 0.04689) = 206.5 s, about 207 s. The fast growth time of 150 s is the time to reach 1,055 kW (1,000 Btu/s), not any fire size you choose. Accepted here: 205 to 208 s.

150 Uses the fast growth time, which is defined at 1,055 kW.
421 Uses the medium coefficient, 0.01127 kW/s².

Fire Science · 2C Fire dynamics · PEFP-10. Source: Klote, NISTIR 5516, Method of Predicting Smoke Movement in Atria (NIST, 1994). Table 2, typical fire growth constants.

Question 11 · Pool-fire heat release rate

A 2.0 m diameter gasoline pool burns at steady state. Use m″ = 0.055 kg/m²·s, ΔH_c,eff = 43,700 kJ/kg, kβ = 2.1 m⁻¹, and Q = m″ΔH_c,eff(1 − e^(−kβD))A. What is the heat release rate?

A. 6.6 MWB. 7.4 MWC. 7.6 MWD. 29.8 MW

Show answer and explanation

B. 7.4 MW. A = π(2.0)²/4 = 3.14 m², and 1 − e^(−4.2) = 0.985. Q = 0.055 × 43,700 × 0.985 × 3.14 = 7,440 kW ≈ 7.4 MW. The (1 − e^(−kβD)) term matters most for small pools; at 2 m it trims only about 1.5%.

A Uses kβ instead of kβD in the exponent.
C Drops the (1 − e^(−kβD)) term.
D Uses D² as the area instead of πD²/4.

Fire Science · 2C Fire dynamics · PEFP-11. Source: NUREG-1805 Supplement 1, Vol. 2, App. A (NRC, 2013). Chapter 3 pool-fire HRR equation and fuel property table, App. A.

Question 12 · Hot gas layer temperature (MQH)

A 500 kW fire burns in a gypsum-lined room with one door 0.9 m wide × 2.0 m high. The enclosing surface area, less the door, is 70 m², and the effective heat transfer coefficient hₖ is 0.039 kW/m²·K at the time of interest. Ambient is 25°C. Using ΔT = 6.85 [Q² / (A₀√H₀ · hₖA_T)]^(1/3), what is the upper-layer temperature?

A. 54°CB. 251°CC. 279°CD. 499°C

Show answer and explanation

B. 251°C. A₀√H₀ = 1.8 × 1.414 = 2.55. The denominator is 2.55 × 0.039 × 70 = 6.95. ΔT = 6.85 × (250,000 ÷ 6.95)^(1/3) = 226 K, so T = 25 + 226 = 251°C. Note the result is a temperature rise; add it to ambient.

A Uses Q instead of Q².
C Leaves out √H₀.
D Adds 273 to the rise, reporting kelvin as degrees Celsius.

Fire Science · 2C Enclosure fires · PEFP-12. Source: NUREG-1805 Supplement 1, Vol. 2, App. A (NRC, 2013). Chapter 2 hot gas layer (MQH method) and heat transfer coefficient, App. A.

Smoke Control Systems (Questions 13–15)

Question 13 · Atrium exhaust for a steady clear height

An atrium is designed for a steady 5,000 kW fire (convective fraction 0.7) and a 12 m clear height above the fuel. Ambient is 20°C. Use m = 0.071 Q_c^(1/3) z^(5/3) + 0.0018 Q_c (kg/s), an adiabatic smoke layer, cₚ = 1.0 kJ/kg·K, and ρ = 101,325 / (287T). What volumetric exhaust is needed?

A. 62 m³/sB. 71 m³/sC. 74 m³/sD. 85 m³/s

Show answer and explanation

B. 71 m³/s. Q_c = 3,500 kW. First check that the equation applies: mean flame height z_l = 0.166 Q_c^(2/5) = 4.3 m, below the 12 m clear height. Then m = 74.1 kg/s. The layer is 20 + 3,500 ÷ 74.1 = 67°C, so ρ = 1.037 kg/m³ and V = 74.1 ÷ 1.037 = 71 m³/s.

A Converts mass to volume with ambient-air density, which undersizes the fan.
C Reports the mass flow in kg/s as if it were m³/s.
D Uses the total heat release rate instead of the convective part.

Smoke Control Systems · 3B Calculations · PEFP-13. Source: Klote, NISTIR 5516, Method of Predicting Smoke Movement in Atria (NIST, 1994). Equations 14, 15, and 21–24 (simple plume equation from NFPA 92B).

Question 14 · Atrium smoke management

Which statements about atrium smoke management are correct? Select all that apply.

Select all that apply. Credit requires every correct choice and no wrong ones.

A. Where make-up air can reach the plume, its velocity should be limited (about 1 m/s, 200 fpm) so it doesn't disturb the plume.B. Passing a cold chemical-smoke test proves the system will perform in a real fire.C. A pre-stratified hot layer under the roof can keep smoke from reaching ceiling detectors; beam detectors across the plume path are a common answer.D. The simple plume equation is valid when the clear height is below the mean flame height.E. Assuming an adiabatic smoke layer is conservative when sizing exhaust volume.

Show answer and explanation

A, C, E. Fast make-up air near the plume deflects it and increases entrainment. Solar heating can trap a warm layer under an atrium roof that cooler smoke can't push through, so ceiling detectors may never see it. And an adiabatic layer is the hottest, least dense case, which gives the largest exhaust volume.

B False. Chemical smoke has little buoyancy and doesn't move like hot fire smoke.
D False. The plume equation applies above the mean flame height.

Smoke Control Systems · 3A Basis of design; 3C Testing · PEFP-14. Source: Klote, NISTIR 5516, Method of Predicting Smoke Movement in Atria (NIST, 1994). Sections 5.3, 6, 7, and 10.4.

Question 15 · Smoke filling time

An atrium has a ceiling 20 m above the fire and a uniform 3,600 m² cross-section. For a steady 2,000 kW fire, use t = (A/H²) · H^(4/3) · Q^(−1/3) · exp[(1.11 − z/H) / 0.28] in SI units. About how long until the first indication of smoke reaches 10 m above the fire?

A. 0.6 minB. 2.8 minC. 5.7 minD. 11.7 min

Show answer and explanation

C. 5.7 min. Check the limits first: A/H² = 9 (allowed range 0.9–14) and z/H = 0.5 (at least 0.2). Then t = 9 × 20^(4/3) × 2,000^(−1/3) × e^(2.18) = 343 s, about 5.7 min. This correlation estimates the first sign of smoke, not the idealized layer interface, which makes it conservative.

A Leaves out the A/H² factor.
B Uses z/H = 0.7, the time for smoke to reach 14 m.
D Uses z/H = 0.3, the time for smoke to reach 6 m.

Smoke Control Systems · 3B Calculations · PEFP-15. Source: Klote, NISTIR 5516, Method of Predicting Smoke Movement in Atria (NIST, 1994). Equations 17–18 and their stated range limits.

Fire Alarm and Signaling Systems (Questions 16–20)

Question 16 · Public-mode audibility

A space has an average ambient sound level of 60 dBA and a maximum of 68 dBA lasting more than 60 seconds. A horn produces 90 dBA at 10 ft. Ignoring reflections, the farthest occupant is 40 ft away. What level is required, and does the horn provide it there?

A. 75 dBA required; 78 dBA provided, so it complies.B. 75 dBA required; 66 dBA provided, so it fails.C. 73 dBA required; 84 dBA provided, so it complies.D. 83 dBA required; 78 dBA provided, so it fails.

Show answer and explanation

A. 75 dBA required; 78 dBA provided, so it complies.. Public-mode signals must be at least 15 dB above the average ambient or 5 dB above the maximum lasting 60 seconds, whichever is greater: max(75, 73) = 75 dBA. Sound from a point source drops about 6 dB per doubling of distance: 90 − 20 log(40/10) = 78 dBA.

B Subtracts 6 dB per 10 ft instead of per doubling of distance.
C Takes the smaller requirement and uses 10 log for distance.
D Adds 15 dB to the maximum instead of 5 dB.

Fire Alarm and Signaling Systems · 4B Design criteria; 4D Calculations · PEFP-16. Source: EC&M, Restricted audible mode operation notification (NFPA 72 Chapter 18 audibility). NFPA 72 Chapter 18 public-mode audibility rule; inverse-square attenuation.

Question 17 · Battery capacity

A fire alarm panel draws 1.2 A in standby and 4.5 A in alarm. Size the battery for 24 hours of standby followed by 5 minutes of alarm, then add the project's 20% safety margin. What capacity is required, in ampere-hours? Round to one decimal place.

Enter a number in ampere-hours, rounded to one decimal place.

Your answer (Ah)

Show answer and explanation

35.0 Ah. Standby: 1.2 × 24 = 28.8 Ah. Alarm: 4.5 × 5/60 = 0.375 Ah. Total 29.175 Ah × 1.2 = 35.0 Ah. The alarm load is large, but 5 minutes of it barely moves the total; standby dominates. Accepted here: 34.95 to 35.05 Ah.

29.2 Leaves out the margin.
61.6 Uses 5 hours of alarm instead of 5 minutes.

Fire Alarm and Signaling Systems · 4D Calculations · PEFP-17. Method: Durations and margin are given in the question.

Question 18 · Notification circuit voltage drop

A notification circuit carries 1.5 A to strobes lumped at its far end, 300 ft from the panel, on 14 AWG copper at 3.07 Ω per 1,000 ft. The design starting voltage is 20.4 V. What voltage reaches the appliances?

A. 19.0 VB. 17.6 VC. 21.2 VD. 18.5 V

Show answer and explanation

B. 17.6 V. Current goes out and back, so the loop is 600 ft: R = 0.600 × 3.07 = 1.84 Ω. Drop = 1.5 × 1.84 = 2.76 V, leaving 20.4 − 2.76 = 17.6 V. Compare that with the appliances' listed minimum operating voltage.

A Counts one conductor only (300 ft).
C Starts from 24 V nominal instead of the design starting voltage.
D Starts from 24 V and counts the drop twice.

Fire Alarm and Signaling Systems · 4D Calculations · PEFP-18. Method: Ohm's law with round-trip conductor length.

Question 19 · Smoke detector coverage

Spot-type smoke detectors with a 30 ft nominal spacing are installed on a smooth, flat ceiling. Under NFPA 72's alternative coverage rule, what is the greatest distance any point on the ceiling may be from the nearest detector?

A. 15 ftB. 21 ftC. 30 ftD. 42 ft

Show answer and explanation

B. 21 ft. All points on the ceiling must be within 0.7 times the selected spacing: 0.7 × 30 = 21 ft. A 21 ft circle just encloses a 30 × 30 ft square, which is why this rule lets you stretch spacing in narrow corridors.

A Half the spacing, the wall distance under the grid method.
C The center-to-center spacing.
D The full diagonal of a 30 ft square.

Fire Alarm and Signaling Systems · 4B Design criteria · PEFP-19. Source: National Training Center, The Point 7 Rule (NFPA 72 smoke detector spacing). NFPA 72 Chapter 17 smooth-ceiling spacing.

Question 20 · Heat detector response time

A heat detector has a response time index (RTI) of 100 (m·s)^½ and a 68°C rating. It is suddenly exposed to a steady 120°C ceiling jet moving at 2.0 m/s; ambient is 20°C. Neglect conduction losses. Using dT_d/dt = (√u / RTI)(T_g − T_d), when does it activate?

A. 33 sB. 34 sC. 46 sD. 71 s

Show answer and explanation

C. 46 s. This is a first-order response with time constant τ = RTI/√u = 100/1.414 = 70.7 s. For a step exposure, t = −τ ln[1 − (68 − 20)/(120 − 20)] = −70.7 × ln 0.52 = 46 s.

A Uses τ = RTI/u instead of RTI/√u.
B Assumes the detector heats at a constant rate (τ × 0.48).
D Reports the time constant itself.

Fire Alarm and Signaling Systems · 4D Detector response · PEFP-20. Method: First-order RTI response; the equation is given in the question.

Water-Based Fire Protection Systems (Questions 21–27)

Question 21 · Sprinkler end-head pressure

The most remote sprinkler (K = 5.6 gpm/psi^½) must deliver 0.20 gpm/ft² over its 130 ft² coverage area. What pressure is required at that sprinkler, in psi? Round to one decimal place.

Enter a number in psi, rounded to one decimal place.

Your answer (psi)

Show answer and explanation

21.6 psi. Flow = 0.20 × 130 = 26 gpm. From Q = K√P, P = (Q/K)² = (26/5.6)² = 21.6 psi. Accepted here: 21.55 to 21.65 psi.

4.6 Forgets to square Q/K.
5.4 Uses 0.10 gpm/ft².

Water-Based Fire Protection Systems · 5D Calculations · PEFP-21. Method: Q = K√P and density × area.

Question 22 · Hazen–Williams friction loss

500 gpm flows through 100 ft (equivalent length) of 3 in Schedule 40 steel pipe with an inside diameter of 3.068 in and C = 120. Using p = 4.52 Q^1.85 / (C^1.85 d^4.87) in psi per foot, what is the friction loss?

A. 27 psiB. 30 psiC. 38 psiD. 0.27 psi

Show answer and explanation

A. 27 psi. p = 4.52 × 500^1.85 / (120^1.85 × 3.068^4.87) = 0.270 psi/ft; × 100 ft = 27 psi. The d^4.87 term makes the actual inside diameter matter a lot.

B Uses the nominal 3 in instead of the 3.068 in actual inside diameter.
C Uses C = 100.
D Stops at the loss per foot.

Water-Based Fire Protection Systems · 5D Calculations · PEFP-22. Method: Hazen–Williams equation as given in the question.

Question 23 · Water supply volume

A design basis is 0.20 gpm/ft² over 1,500 ft², a 250 gpm hose allowance, and a 90-minute duration. Using density × area for the sprinkler flow (ignore hydraulic imbalance), what minimum stored volume does a tank need?

A. 27,000 galB. 33,000 galC. 49,500 galD. 66,000 gal

Show answer and explanation

C. 49,500 gal. Sprinkler flow = 0.20 × 1,500 = 300 gpm. With hose, 550 gpm × 90 min = 49,500 gal. A real hydraulic calculation will come out somewhat higher than density × area, because sprinklers farther from the remote end run at higher pressure.

A Leaves out the hose allowance.
B Uses 60 minutes.
D Uses 120 minutes.

Water-Based Fire Protection Systems · 5B Design criteria · PEFP-23. Method: Design criteria are given in the question.

Question 24 · Fire pump test curve

A centrifugal fire pump is rated 1,000 gpm at 100 psi net pressure. Its test curve shows 128 psi at churn, 101 psi at 1,000 gpm, and 62 psi at 1,500 gpm. Select all that apply.

Select all that apply. Credit requires every correct choice and no wrong ones.

A. The churn pressure is within the NFPA 20 limit.B. The 150%-flow point does not meet NFPA 20.C. NFPA 20 requires churn to be at least 140% of rated pressure.D. A 1,000 gpm pump can never supply a 1,400 gpm demand.E. Churn pressure plus maximum static suction pressure must not exceed the pressure rating of system components.

Show answer and explanation

A, B, E. Churn (no-flow) pressure may not exceed 140% of rated: 140 psi, so 128 psi is fine. At 150% of rated flow (1,500 gpm), the pump must still make at least 65% of rated pressure, 65 psi; 62 psi falls short.

C False. 140% is a maximum, not a minimum.
D False. A pump may be used up to 150% of its rated flow when pressure is adequate.

Water-Based Fire Protection Systems · 5C Principal components · PEFP-24. Source: ASPE, Engineering Considerations for Stationary Fire Pump Installations; Consulting-Specifying Engineer, Using fire pump speed as a design consideration. NFPA 20 centrifugal pump performance requirements as summarized in both articles.

Question 25 · Preaction system type

A records vault must keep water out of the sprinkler piping unless the detection system has operated and a sprinkler has opened. Which system fits?

A. Wet pipeB. Single-interlock preactionC. Double-interlock preactionD. Deluge

Show answer and explanation

C. Double-interlock preaction. A double-interlock preaction system admits water only when both events occur: the detection system operates and a sprinkler opens, which shows up as a loss of supervisory air pressure.

A Keeps water in the piping at all times.
B Admits water on detection alone.
D Uses open sprinklers and discharges over the whole area.

Water-Based Fire Protection Systems · 5A System selection · PEFP-25. Source: Viking, Preaction System Manual. NFPA 13 preaction system types as described in the manual.

Question 26 · Available flow from a hydrant test

A flow test reads 80 psi static and 60 psi residual while flowing 1,000 gpm. Using Q_R = Q_F (h_r / h_f)^0.54, what flow is available at 20 psi residual?

A. 553 gpmB. 1,810 gpmC. 2,330 gpmD. 3,000 gpm

Show answer and explanation

B. 1,810 gpm. h_r = 80 − 20 = 60 psi of drop; h_f = 80 − 60 = 20 psi. Q = 1,000 × (60/20)^0.54 = 1,000 × 1.81 = 1,810 gpm. Pressure drop grows faster than flow, so tripling the drop doesn't triple the flow.

A Inverts the pressure ratio.
C Uses an exponent of 0.77 instead of 0.54.
D Treats flow as proportional to pressure drop.

Water-Based Fire Protection Systems · 5B Water supply · PEFP-26. Source: City of Barrie, Fire Flow Test standard W501. Fire flow test calculation, item 7.

Question 27 · Pressure at the base of the riser

The most remote sprinkler needs 35 psi. Friction loss from the base of the riser to that sprinkler is 18 psi, and the sprinkler is 45 ft above the riser base. What pressure is required at the riser base?

A. 34 psiB. 53 psiC. 72 psiD. 98 psi

Show answer and explanation

C. 72 psi. Add the elevation head: 45 ft × 0.433 psi/ft = 19.5 psi. Total = 35 + 18 + 19.5 = 72.5 psi.

A Subtracts the elevation head.
B Ignores elevation.
D Adds 45 ft as if it were 45 psi.

Water-Based Fire Protection Systems · 5D Calculations · PEFP-27. Method: Static head of water, 0.433 psi per ft.

Special Hazard Systems (Questions 28–31)

Question 28 · Clean agent quantity

A 3,000 ft³ Class A surface-fire hazard is protected with HFC-227ea. For this calculation, use an extinguishing concentration of 6.7%. Use s = 1.885 + 0.0046T (ft³/lb) at T = 70°F and W = (V/s) · C / (100 − C). Apply NFPA 2001's Class A automatic-detection safety factor of 1.2. What agent weight is required?

A. 98 lbB. 109 lbC. 119 lbD. 130 lb

Show answer and explanation

C. 119 lb. Design concentration = 1.2 × 6.7 = 8.04%. s = 1.885 + 0.0046 × 70 = 2.207 ft³/lb. W = (3,000 ÷ 2.207) × 8.04 ÷ 91.96 = 119 lb. The (100 − C) term accounts for agent pushed out of the room as it discharges.

A Leaves out the safety factor.
B Uses C/100 instead of C/(100 − C).
D Uses the Class B safety factor of 1.3.

Special Hazard Systems · 6D Calculations · PEFP-28. Sources: NIST, FM-200 total flooding paper (agent quantity equation); NFPA 2001 public first-revision report, Section 7.2.2.1.2. The 6.7% extinguishing concentration is an exercise input; the 1.2 Class A automatic-detection safety factor is the standard-dependent principle being tested.

Question 29 · LFL of a gas mixture

A fuel gas is 50% methane (LFL 5.0%), 30% ethane (LFL 3.0%), and 20% propane (LFL 2.1%) by volume. Estimate the mixture's lower flammable limit with Le Chatelier's rule.

A. 2.1%B. 3.4%C. 3.8%D. 5.0%

Show answer and explanation

B. 3.4%. LFL_mix = 100 ÷ (50/5.0 + 30/3.0 + 20/2.1) = 100 ÷ 29.5 = 3.4%. The lower-LFL components pull the mixture down more than a simple average would.

A Uses the lowest component's LFL.
C Takes a weighted arithmetic average.
D Uses the largest component's LFL.

Special Hazard Systems · 6D Calculations · PEFP-29. Source: Zabetakis, Flammability Characteristics of Combustible Gases and Vapors, USBM Bulletin 627 (1965). Le Chatelier's mixing rule.

Question 30 · Foam concentrate for a fixed-roof tank

A 60 ft diameter fixed-roof tank is protected by Type II discharge outlets at 0.10 gpm/ft² of liquid surface for 55 minutes, using 3% concentrate. Ignore supplementary hose streams and pipe fill. How much foam concentrate is needed?

A. 254 galB. 467 galC. 933 galD. 15,600 gal

Show answer and explanation

B. 467 gal. Surface area = π(60)²/4 = 2,827 ft². Foam solution = 2,827 × 0.10 × 55 = 15,550 gal. At 3% proportioning, concentrate = 0.03 × 15,550 = 467 gal.

A Uses 30 minutes.
C Uses 6% concentrate.
D Reports foam solution, not concentrate.

Special Hazard Systems · 6B Design criteria · PEFP-30. Method: Rate, duration, and proportioning are given in the question.

Question 31 · Oxygen after an inert-gas discharge

An inert-gas system reaches a uniform 40% agent concentration in air that started at 20.9% oxygen. Ignoring leakage, what is the resulting oxygen concentration, in percent? Round to one decimal place.

Enter a number in percent, rounded to one decimal place.

Your answer (%)

Show answer and explanation

12.5%. The agent dilutes the air it mixes with: O₂ = 20.9 × (1 − 0.40) = 12.5%. That reduced oxygen level is how inert gases extinguish fires, and why occupied-space limits matter. Accepted here: 12.45 to 12.55%.

8.4 Calculates the share of oxygen displaced (20.9 × 0.40).
−19.1 Subtracts 40 percentage points from 20.9.

Special Hazard Systems · 6B Design criteria · PEFP-31. Source: NIST, Halon Options Technical Working Conference paper on inert-gas and halocarbon clean agents (2001). Oxygen after dilution, %O₂ = 21 − 21(c/100).

Passive and Structural Fire Protection (Questions 32–35)

Question 32 · Effective char depth

A glulam beam, 6.75 × 13.5 in, is exposed on three sides (bottom and both sides) for 1.5 hours. Use a nominal char rate βₙ = 1.5 in/hr, char depth a = βₙ t^0.813 (t in hours), and effective char depth = 1.2a. What residual width × depth remains for the structural check?

A. 3.15 × 11.7 inB. 1.74 × 11.0 inC. 4.25 × 11.0 inD. 2.25 × 11.25 in

Show answer and explanation

B. 1.74 × 11.0 in. a = 1.5 × 1.5^0.813 = 2.09 in; effective char = 1.2 × 2.09 = 2.50 in. Width loses char from both sides: 6.75 − 2(2.50) = 1.74 in. Depth loses it from the bottom only: 13.5 − 2.50 = 11.0 in. The 20% bump covers the heat-affected wood just behind the char, which is treated as having no strength.

A The 1-hour result (effective char 1.8 in).
C Chars only one side of the width.
D Uses a straight-line char of 1.5 × 1.5 = 2.25 in on the width only.

Passive and Structural Fire Protection · 7C Structural fire protection · PEFP-32. Source: WoodWorks, Using Char Methods to Demonstrate Fire Resistance of Exposed Wood Members (2024 NDS); STRUCTURE, Fire Protection of Mass Timber Connections (FDS/NDS char depth). NDS Chapter 16 char method; 1.5-hour effective char matches the NDS tabulated 2.5 in.

Question 33 · SFRM thickness for a substituted beam

A listed beam design requires 1.25 in of sprayed fire-resistive material (SFRM) on a beam with W/D = 0.82. A lighter beam with W/D = 0.62 is substituted. Using T₁ = T₂ (W₂/D₂ + 0.6) / (W₁/D₁ + 0.6), where subscript 1 is the substitute beam, what thickness does the substitute need (within the listing's limits)?

A. 1.07 inB. 1.25 inC. 1.45 inD. 1.65 in

Show answer and explanation

C. 1.45 in. T₁ = 1.25 × (0.82 + 0.6) / (0.62 + 0.6) = 1.25 × 1.42 / 1.22 = 1.45 in. A lighter section has less steel per inch of heated perimeter, so it heats faster and needs more protection.

A Swaps the two W/D terms.
B Assumes thickness carries over unchanged.
D Uses the plain W/D ratio without the 0.6 terms.

Passive and Structural Fire Protection · 7C Fire resistance equivalency · PEFP-33. Source: NFCA Guidance Bulletin 1004, UL Equation for Adjustment of SFRM Thickness. UL thickness adjustment equation.

Question 34 · Fire door test method

Which test method is used to evaluate a fire door assembly's ability to stop fire from passing from one room to the next?

A. ASTM E84 (UL 723), surface burning characteristicsB. NFPA 252 / UL 10C, fire tests of door assembliesC. ASTM E605, SFRM thickness and densityD. UL 9540A, battery thermal-runaway propagation

Show answer and explanation

B. NFPA 252 / UL 10C, fire tests of door assemblies. Fire doors are tested as assemblies under standards such as NFPA 252, UL 10B, or UL 10C. The test establishes a fire protection rating for the opening protective; the required door rating for a particular opening is then determined by the applicable code provisions for that wall and opening.

A Rates how fast flame spreads across an interior finish surface, not barrier performance.
C Measures thickness and density of sprayed fireproofing.
D Evaluates fire propagation in battery energy storage systems.

Passive and Structural Fire Protection · 7D Openings and penetrations · PEFP-34. Source: VT Industries, Flame Spread Classifications and Smoke Development (technical bulletin); ASTM E605/E605M, Thickness and Density of Sprayed Fire-Resistive Material (scope). Fire door test methods; ASTM E605 scope.

Question 35 · Interior finish class

An ASTM E84 test gives a wall panel a flame spread index of 30 and a smoke-developed index of 350. What interior finish class is it?

A. Class AB. Class BC. Class CD. Not classifiable; smoke is too high

Show answer and explanation

B. Class B. Flame spread index 26–75 is Class B. The smoke-developed limit of 450 applies to all three classes, so 350 passes.

A Requires a flame spread index of 0–25.
C Covers a flame spread index of 76–200.
D The smoke-developed limit is 450, not 350.

Passive and Structural Fire Protection · 7A Building materials · PEFP-35. Source: Massachusetts 780 CMR Chapter 8, Interior Finishes (classification by ASTM E84); VT Industries, Flame Spread Classifications and Smoke Development (technical bulletin). ASTM E84 interior finish classes.

Human Behavior and Evacuation (Questions 36–40)

Question 36 · Occupant load

A floor has a 1,200 ft² (net) conference room with tables and chairs and 6,000 ft² (gross) of offices. Use 15 ft² net per person for the conference room and 150 ft² gross per person for the offices. What is the total occupant load?

A. 120B. 140C. 211D. 231

Show answer and explanation

A. 120. Conference room: 1,200 ÷ 15 = 80. Offices: 6,000 ÷ 150 = 40. Total = 120. Watch the net/gross labels; they decide which area you divide.

B Uses 100 ft² gross for offices, the business factor NFPA 101 used before its 2018 edition.
C Uses 7 ft² net, a concentrated-assembly factor, for the conference room.
D Makes both of those errors.

Human Behavior and Evacuation · 8A Occupant load · PEFP-36. Source: MeyerFire, New occupant load factors coming to NFPA 101. Occupant load factors are given in the question; history of the business factor.

Question 37 · Required door width

360 occupants must leave a room through exit doors. Using a level-component capacity factor of 0.2 in per person, what total clear door width is required for capacity?

A. 72 inB. 108 inC. 170 inD. 1,800 in

Show answer and explanation

A. 72 in. 360 × 0.2 = 72 in. With 34 in clear doors, each serves 34 ÷ 0.2 = 170 people, so three doors are needed for capacity (two give only 340). Number-of-exits and remoteness rules are separate checks.

B Uses the stair factor, 0.3 in per person.
C Reports people per 34 in door, not inches.
D Divides by the factor instead of multiplying.

Human Behavior and Evacuation · 8A Arrangement and sizing · PEFP-37. Source: MeyerFire, Difference of IBC and NFPA 101 exit capacity factors. NFPA 101 capacity factors: 0.3 in per person for stairs, 0.2 in per person for level components in most occupancies.

Question 38 · Flow time through a doorway

600 people queue at a 44 in clear doorway. Use a boundary layer of 6 in on each side and a maximum specific flow of 24 persons per minute per foot of effective width. Ignoring travel time, how long does the flow take?

A. 6.8 minB. 9.4 minC. 12.5 minD. 25 min

Show answer and explanation

B. 9.4 min. Effective width = 44 − 2(6) = 32 in = 2.67 ft. Flow = 24 × 2.67 = 64 persons/min. Time = 600 ÷ 64 = 9.4 min. Edges of a doorway carry few people, which is why the hydraulic model subtracts a boundary layer.

A Uses the full 44 in clear width.
C Subtracts 12 in from each side (2.0 ft effective).
D Divides by the specific flow and ignores width.

Human Behavior and Evacuation · 8B Evacuation time · PEFP-38. Method: Effective-width hydraulic model; boundary layer and specific flow are given in the question.

Question 39 · Visibility through smoke

Smoke in a corridor has an optical density of 0.10 per meter (base 10). Using S = C/K, with K = 2.303 × optical density and C = 8 for a light-emitting exit sign, how far can occupants see the sign?

A. 13 mB. 35 mC. 80 mD. 30 m

Show answer and explanation

B. 35 m. K = 2.303 × 0.10 = 0.23 per meter. S = 8 ÷ 0.23 = 35 m. A light-reflecting sign (C ≈ 3) would be visible only about 13 m in the same smoke.

A Uses C = 3 for a light-reflecting sign.
C Skips converting optical density to the extinction coefficient.
D Makes both of those errors.

Human Behavior and Evacuation · 8C Tenability · PEFP-39. Source: Börger, Investigation of Smoke Characteristics by Photometric Measurements, FEMTC 2022; Waypoint-based approach to visibility (summary of Jin's sign-visibility constants), arXiv 2404.11439. Jin's visibility relationship and sign constants.

Question 40 · Occupant response

Which statements reflect research on occupant behavior in building fires? Select all that apply.

Select all that apply. Credit requires every correct choice and no wrong ones.

A. Occupants commonly seek information and confirm cues before they start moving.B. Occupants start moving immediately at the first alarm cue in nearly every case.C. Pre-evacuation time is part of RSET and can be a large share of it.D. Ambiguous cues and how occupants judge their own risk influence whether they act.E. Evacuation models predict behavior automatically, so pre-movement inputs aren't needed.

Show answer and explanation

A, C, D. Before people move, they perceive cues, interpret them, judge the risk, and decide. Along the way they seek information, mill, prepare, and warn others. All of that is pre-evacuation time.

B False. Delay activities such as information seeking and preparing are common.
E False. Models largely rely on user-supplied assumptions about behavior.

Human Behavior and Evacuation · 8C Psychological and physiological response · PEFP-40. Source: Kuligowski, The Process of Human Behavior in Fires, NIST TN 1632 (2009). Behavioral process phases and delay activities.

Review your answers

Count your results against all 40 questions: correct, wrong, and blank. If you worked only one area, score it against that area's own question count. A select-all question counts only if you picked every correct choice and nothing else; NCEES also scores every item right or wrong, with no partial credit (NCEES Examinee Guide, May 2026, p. 11).

This result describes your work on these 40 questions. It isn't an NCEES score, and it doesn't predict whether you'll pass. Three to seven questions per area is enough to point you somewhere, not enough to measure you.

What you do with the misses matters more than the total:

Rework each miss before you reread the solution. Cover the explanation and start again from the givens.
Find the first wrong step and name it. Most misses on this set fall into a few buckets: the wrong basis (Q instead of Q_c, gross instead of net area, nominal instead of actual pipe size), the wrong model (linear instead of square-law, average instead of log), a misremembered code criterion, units, or arithmetic.
Prove the fix on a fresh problem. Getting the same question right twice mostly shows you remember it.

Copy this log into your notes. Log lucky guesses too; a right answer you can't explain is a miss waiting to happen.

Review your answers
Question	Area	My answer	First wrong step	Cause	Fix in one line	Fresh problem done?
Example: Q13	Smoke Control	62 m³/s	Used ambient air density for hot smoke	Basis	Convert mass to volume at the smoke layer temperature

Two or more misses in one area is a reasonable signal to spend your next study block there before taking another set.

What this practice test covers

NCEES splits the 2027 exam into eight knowledge areas, each with a range of questions. We sized each area of this 40-question set from the midpoint of its official range (the midpoints add to 88; each area's share of 40 was rounded by largest remainder). Your real exam can land anywhere inside each range.

What this practice test covers
Knowledge area	Questions on the real exam (of 85)	Questions in this set
1. Fire Protection Analysis	10–15	6 (Q1–Q6)
2. Fire Science	10–15	6 (Q7–Q12)
3. Smoke Control Systems	6–9	3 (Q13–Q15)
4. Fire Alarm and Signaling Systems	8–12	5 (Q16–Q20)
5. Water-Based Fire Protection Systems	12–18	7 (Q21–Q27)
6. Special Hazard Systems	7–11	4 (Q28–Q31)
7. Passive and Structural Fire Protection	8–12	4 (Q32–Q35)
8. Human Behavior and Evacuation	9–14	5 (Q36–Q40)

Source: NCEES PE Fire Protection specification, effective April 2027, pp. 1–2.

Every area is sampled, but 40 questions can't reach every subtopic. Hazardous materials (NFPA 400), water mist, standpipes, dampers and firestopping, mass notification and emergency responder communication systems, and egress lighting and signage get little or no direct coverage here.

The real exam also uses point-and-click and drag-and-drop items. This set includes the two other alternative formats NCEES lists: select-all-that-apply (Q3, Q14, Q24, Q40) and enter-a-number (Q2, Q10, Q17, Q21, Q31).

Most questions here hand you the equation, constant, or design criterion they need, so you can work them anywhere. On exam day, finding that equation in the handbook or the right table in an NFPA standard is part of the job.

What changed for the April 2027 exam

The 2027 specification replaced the four broad areas used since 2018 with eight narrower ones and updated the list of standards NCEES supplies.

What changed for the April 2027 exam
Row label	Earlier specification (topics since 2018)	Starting April 13, 2027
Knowledge areas	4: Fire Protection Analysis; Fire Dynamics Fundamentals; Active and Passive Systems; Egress and Occupant Movement	8 (see the table above)
Energy storage systems	Not a listed topic	A listed subtopic, with NFPA 855-2023 supplied
NFPA 12 (carbon dioxide) and NFPA 25 (inspection, testing, and maintenance)	On the October 2022 standards list	Not on the 2027 list
Standard editions	For example, NFPA 13-2019, 72-2019, 101-2018 (October 2022 list)	NFPA 11-2024, 13-2022, 20-2022, 30-2024, 72-2022, 92-2024, 101-2024, 400-2022, 855-2023, 2001-2022
Questions and appointment	85 questions, 9.5 hours	Unchanged

Sources: 2027 specification, last page; earlier specification with October 2022 standards list.

NCEES scores answers that depend on a standard against the edition on its list, and solutions based on other editions don't get credit. If your study materials or office library use different editions, check the item you're relying on against the listed one.

How the real exam works

Length: 85 questions. The appointment is 9 hours 30 minutes: a 2-minute nondisclosure agreement, an 8-minute tutorial, 8 hours 30 minutes of exam time, and a 50-minute scheduled break. That's an average of 6 minutes per question.
Two sections: After roughly half the questions, you review and submit them, and you can't go back to that half.
Formats: Multiple choice plus alternative item types. Every question is scored right or wrong, with no partial credit. A few unscored pretest questions are mixed in, and you can't tell which.
References: Closed book. NCEES supplies the PE Fire Protection Reference Handbook and the listed NFPA standards on screen as searchable PDFs. You search with the search box on the reference; Ctrl+F doesn't work. You can't bring your own copies.
Units: Both SI and U.S. customary.
Calculator: One NCEES-approved model, or the on-screen TI-30XS.
Date, place, and cost: Offered once a year. The next date is April 13, 2027, at Pearson VUE test centers. The exam fee is $400, paid to NCEES; your licensing board may charge its own fee and decides whether you're eligible to sit.
Results: Usually 7–10 days, reported as pass or fail. If you don't pass, you get a diagnostic report by knowledge area.

Sources: NCEES PE Fire Protection page; 2027 specification; NCEES Examinee Guide, May 2026, pp. 3, 5, 8–11, 14, and 16.

Does your score here predict a pass?

No. NCEES counts your correct answers, converts them to a scaled score, and compares that with a passing standard set by subject-matter experts. It doesn't publish the passing score, so any "you need 70%" claim has no official basis (Examinee Guide, p. 14).

Which references should you practice with?

Download the current reference handbook from your MyNCEES account, and study from the exact NFPA editions on the 2027 list. NCEES supplies the standards during the exam but doesn't sell them; get study copies from NFPA (NCEES PE Fire Protection page). When you review a miss here, practice finding the equation or criterion in the handbook or standard: name it in plain words ("hot gas layer," "fire pump," "capacity factor"), search, then check the variable definitions and units.

Try a timed session

Work untimed first so you learn the methods. When you're ready, try all 40 in 240 minutes, which matches the exam's 6-minute average. That's an average, not a per-question limit. Practice the habits that matter on the real exam: skip and flag long problems, answer everything, and clear your flags before you "submit" the first 20.

PE Fire Protection pass rates

PE Fire Protection pass rates
Takers	Number	Pass rate
First-time	262	85%
Repeat	48	44%

Source: NCEES PE Fire Protection page, pass-rate table last updated July 2025, for examinees under NCEES member boards. These takers sat for the exam under the earlier specification, before the 2027 changes. They're group results, not your personal odds, and the table doesn't explain why first-time and repeat rates differ.

Quick answers

Is this the official NCEES practice exam? No. This is an independent, unofficial practice set. NCEES sells its own 85-question PE Fire Protection practice exam through MyNCEES; its listing shows a 2020 copyright, which is before the 2027 specification, and NCEES has posted errata for it (September 2025). If you use it, check the corrections and the standard editions its solutions rely on.

Are these real exam questions? No. We wrote them. Real exam questions are confidential, and every examinee agrees not to disclose them (Examinee Guide, p. 19).

Do I need to pass the FE first? Your state licensing board decides eligibility, not NCEES. Find yours in the NCEES board directory.

What if I don't pass? NCEES allows one attempt per testing window and no more than three in 12 months, and some boards are stricter. Because PE Fire Protection is offered once a year, a retake means the next annual date. See NCEES retake rules and how NCEES results work.

Sources and verification

Official sources

Technical sources for the practice questions

Fire dynamics and smoke: NUREG-1805 Supplement 1, Vol. 2 (U.S. NRC) · Klote, NISTIR 5516 · FDS grid-resolution guidance (FEMTC 2011)
Human behavior and tenability: Kuligowski, NIST TN 1632 · Börger, smoke visibility (FEMTC 2022) · Visibility constants summary (arXiv 2404.11439)
Special hazards: NIST clean agent papers (hold time, inert gas) · Zabetakis, USBM Bulletin 627 · UL Solutions on UL 9540A and NFPA 855
Water-based and alarm systems: ASPE on fire pumps · CSE on fire pump speed · Viking preaction manual · City of Barrie fire flow test standard · EC&M on NFPA 72 audibility · National Training Center on the 0.7 rule
Passive protection and egress: WoodWorks on NDS char methods · STRUCTURE on char depth · NFCA SFRM thickness bulletin · VT Industries on flame spread and door tests · Massachusetts 780 CMR Chapter 8 · ASTM E605 scope · MeyerFire on NFPA 101 capacity factors and occupant load factors

Last verified October 9, 2026. We checked the exam format, date, fee, specification, standards list, scoring, attempt rules, and pass-rate figures against the NCEES materials linked above. We independently recalculated all 40 answer keys and numeric acceptance ranges. Each source-dependent question names the technical source used for its underlying principle; questions marked "Method" rely on a calculation whose equation and inputs are stated in the question.

By Castleport Test Prep Editorial Team. This page was developed with AI-assisted research and editing; see how we use sources and AI assistance. Source and arithmetic checking is not professional engineering review.

Castleport Test Prep is an independent exam prep publisher, 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, and trademarks belong to their respective owners. These practice questions are original and unofficial.

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