PE Mechanical Thermal and Fluid Systems Exam Prep
Start your PE Mechanical Thermal and Fluid Systems exam prep with this original, unofficial problem. All 36 problems, worked solutions, and a 12-week plan are free on this page, written to the NCEES specification effective October 2025.
Problem 1 · Conduction through a composite wall
TFS-HT-01 · Thermal/Fluid Principles: Heat transfer principles · Select one
A furnace wall is 0.20 m of firebrick (k = 1.0 W/m·K) backed by 0.10 m of insulation (k = 0.10 W/m·K). At steady state the inner surface is 900°C and the outer surface is 60°C. Treat it as one-dimensional conduction with perfect contact between the layers. What is the heat flux through the wall?
Show answer and solution
Answer: A. 700 W/m².
The layers are in series, so their resistances per unit area add: R = 0.20/1.0 + 0.10/0.10 = 0.20 + 1.00 = 1.20 m²·K/W. Heat flux q = ΔT/R = (900 − 60)/1.20 = 700 W/m². Notice that the thin insulation layer carries five-sixths of the total resistance.
Why the other choices miss:
- B (840 W/m²) ignores the brick and uses only the insulation's resistance.
- C (4,200 W/m²) ignores the insulation.
- D (5,040 W/m²) adds the layers' conductances as if they sat side by side (parallel) instead of stacked (series).
Try a change: the brick-to-insulation interface sits at 900 − 700 × 0.20 = 760°C. Interface temperatures fall out of the same series-resistance idea.
Principle: DOE-HDBK-1012/2-92, Vol. 2, Module HT-02, printed pp. 6, 9–10 (Fourier's law; series resistances; composite-wall example); OpenStax University Physics Vol. 2, §1.6 Mechanisms of Heat Transfer, Eq. 1.9. Topic: NCEES TFS specification, knowledge area 1A.
By Castleport Test Prep Editorial Team · Exam facts last verified October 8, 2026
Record your responses, then compare them with the worked solutions. This page does not automatically grade or save your answers.
Jump to: Topic ranges · What changed in 2025 · Your exam-day reference · More problems · 12-week plan
The exam at a glance
- 80 questions, computer-based, year-round at Pearson VUE test centers
- 8 hours of exam time inside a 9-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
- On screen you get the NCEES PE Mechanical Reference Handbook, and that's it. No design standards are supplied for PE Mechanical, and you can't bring your own references.
- Both SI and U.S. customary units; sea-level conditions unless a problem says otherwise
- Multiple choice plus multi-select, point-and-click, drag-and-drop, and fill-in-the-blank items, all scored right or wrong with no partial credit
- Results are pass/fail, typically 7–10 days after the exam
- First-time pass rate: 70% of 578 examinees (NCEES table updated July 2026)
Sources: NCEES PE Mechanical exam page; TFS exam specifications, effective October 2025, p. 1; NCEES computer-based testing page; NCEES Examinee Guide, May 2026, pp. 3 and 16.
What to study first: the 5 areas by official question range
Thermal/Fluid Principles has the largest published question range. Fluid Equipment and Distribution Systems and Power Systems and Components share the next-largest range. The published minima for those three areas total 53 questions, so they deserve substantial study time.
NCEES publishes a range for each knowledge area instead of the exact topic-area count. For year-round CBT exams, NCEES says its LOFT system gives examinees the same number of questions in the same topics while assembling different question sets of comparable difficulty. Use the published ranges to prioritize, not as a promise that your form will use a range midpoint.
| Rank | Knowledge area | Questions | What NCEES lists | Practice here |
|---|---|---|---|---|
| 1 | Thermal/Fluid Principles | 19–29 | Heat transfer (convection, conduction, radiation); thermodynamics (graphical processes, steam tables, Mollier diagrams, psychrometric charts, First and Second Laws); fluids (conservation of energy and mass, mixing, statics, Moody diagram, Mach number, shock properties, Bernoulli) | Problems 1–10 |
| 2 (tie) | Fluid Equipment and Distribution Systems | 17–26 | Piping components and connections (valves, fittings, joints, pressure vessels); fluid distribution (pipe flow, ductwork, friction factor, effective length, pressure drop, controls); pumps and fans (cavitation, curves, hydraulic power, series and parallel, efficiency, NPSH, potential energy storage); compressors, nozzles, and diffusers (efficiency, bleed air, isentropic flow) | Problems 11–19 |
| 2 (tie) | Power Systems and Components | 17–26 | Turbines (steam, gas, hydraulic, wind); boilers, steam generators, and waste heat recovery (heat rate, efficiency); condensers; power cycles (Rankine, Brayton, combined cycle, combined heat and power, combustion, internal combustion engines) | Problems 20–28 |
| 4 | Cooling/Heating Systems and Components | 12–18 | Heat exchangers (shell-and-tube, feedwater heaters); cooling towers (approach, drift, blowdown, makeup water); refrigeration cycles and heat pumps (vapor compression, absorption, thermal storage) | Problems 29–33 |
| 5 | Supportive Knowledge | 5–8 | Economic analysis; project planning and scheduling; electrical concepts; material and stress analysis; codes and standards | Problems 34–36 |
The ranges come from the NCEES TFS specification, pp. 1–2. The ranking is ours, and it says nothing about which topics are hardest. NCEES also notes that its listed examples aren't exclusive or exhaustive, so treat them as a floor for your review, not a fence.
One thing people miss: psychrometric charts sit in TFS area 1B. They sound like an HVAC topic, but they're on your list too.
What changed in October 2025, and can you still use an older book?
The format didn't change: 80 questions, a 9-hour appointment, and the handbook as your only reference. The topic outline did. The previous specification grouped everything into three big areas. The current one splits it into five.
| April 2020 specification (topics unchanged since 2017) | October 2025 specification |
|---|---|
| I. Principles, 28–44: basic engineering practice, fluid mechanics, heat transfer, mass balance, thermodynamics (including combustion), and supportive knowledge (pipe stress, joints, psychrometrics, codes) | 1. Thermal/Fluid Principles, 19–29. Psychrometric charts and Mollier diagrams are now named here. Combustion moved to power cycles, joints to piping components, and codes and economics to Supportive Knowledge. |
| II. Hydraulic and Fluid Applications, 21–33: pumps and fans, compressors, pressure vessels, control valves, actuators, connections, distribution systems | 2. Fluid Equipment and Distribution Systems, 17–26. Now also names ductwork, effective length, NPSH, nozzles and diffusers, bleed air, and potential energy storage. |
| III. Energy/Power System Applications, 21–33: turbines, boilers, IC engines, heat exchangers, cooling towers, condensers, cooling/heating, energy recovery, combined cycles | Split into 3. Power Systems and Components, 17–26 (now explicitly names hydraulic and wind turbines and combined heat and power) and 4. Cooling/Heating Systems and Components, 12–18 (heat exchangers, cooling towers, refrigeration and heat pumps, thermal storage) |
| (no separate area) | 5. Supportive Knowledge, 5–8: economics, project planning and scheduling, electrical concepts, material and stress analysis, codes and standards |
Sources: April 2020 specification, pp. 1–2; October 2025 specification, pp. 1–2. The row-by-row mapping is our reading of the two documents, not an NCEES crosswalk. Both specifications say their example lists are not exclusive or exhaustive, so a newly explicit example is not proof that the concept was absent from earlier exams.
So, can you use an older book? For the physics, yes. Steam tables and the Darcy equation didn't change. Here's what to do with it:
- Re-map its chapters to the five current areas so your hours follow the current ranges.
- Fill the gaps it may not cover: wind and hydraulic turbines, combined heat and power, ductwork, Gantt charts and critical path, electrical concepts, and thermal storage.
- Ignore any open-book advice. The computer-based PE Mechanical exam is closed book with an electronic reference, so tabbing a binder is wasted effort.
- Practice from the current handbook, not the book's own formula pages, since the handbook is what you'll have on screen.
The only reference you get: the PE Mechanical Reference Handbook
TFS supplies the NCEES PE Mechanical Reference Handbook as a searchable PDF. The Examinee Guide's format table lists no design standards for any PE Mechanical exam, so the handbook is your whole library (Examinee Guide, May 2026, pp. 10 and 16).
- Get it free: download the current handbook from your MyNCEES account and note the version shown there.
- No Ctrl+F: on exam day you search with the box on the left side of the reference window. You'll test on a 24-inch monitor, with room for the question and the handbook side by side (Examinee Guide, p. 10).
- Official practice exam: NCEES sells an 80-question PE Mechanical: Thermal and Fluid Systems practice exam with solutions, built to the October 2025 specification, as a $59.95 PDF (NCEES exam prep store). If you buy it, check the errata posted April 1, 2026.
A five-minute lookup drill. Do this with every problem you practice until it's automatic:
- Name the relationship in plain words, like "log mean temperature difference" or "net positive suction head."
- Search for that phrase in the handbook.
- Read the variable definitions and units beside the equation.
- Apply it.
- Write down the search words that worked.
Keep a lookup log so slow lookups get repeated until they're fast:
| Topic | Search words that worked | Where it was | Time to find | What slowed you down |
|---|---|---|---|---|
| (example) LMTD for a counterflow exchanger | "log mean" | Heat exchanger section | 2 min | Searched "LMTD" first |
36 original practice problems with worked solutions
These are original Castleport problems written to the NCEES TFS topic list. They aren't NCEES questions, and your score on them isn't a prediction of your exam result. This set isn't a full-length exam, and it doesn't cover every listed subtopic. The problems are roughly proportioned to the official ranges: 10, 9, 9, 5, and 3 across the five areas.
Most are standard multiple choice. Some use formats like those on the real exam: select-all-that-apply, enter-a-number, and matching. On the actual exam, every question, including those formats, is scored right or wrong with no partial credit (NCEES computer-based testing). We score our select-all and matching problems the same way.
Each problem gives you the property values it needs. On exam day, you'd pull those from the handbook's tables instead. Work each one before you open the solution, and when you miss, write down the first step that went wrong.
Thermal/Fluid Principles (10 problems)
Problem 1, conduction through a composite wall, is at the top of this page. Problems 2–10 continue here.
Problem 2 · Net radiation to a large room
TFS-HT-02 · Thermal/Fluid Principles: Heat transfer principles · Select one
A small gray surface (emissivity 0.80, area 0.50 m²) at 500 K sits inside a large room whose walls are at 300 K. Use σ = 5.67 × 10⁻⁸ W/m²·K⁴. What is the net radiant heat loss from the surface?
Show answer and solution
Answer: C. 1.23 kW.
Net exchange with large surroundings is P = εσA(Ts⁴ − Tsur⁴) = 0.80 × 5.67 × 10⁻⁸ × 0.50 × (500⁴ − 300⁴) = 1,234 W ≈ 1.23 kW.
Why the other choices miss:
- A (36 W) raises the temperature difference (200 K) to the fourth power instead of each temperature.
- B (60 W) uses Celsius temperatures (227°C and 27°C). Radiation laws need absolute temperature.
- D (1.42 kW) ignores the radiation arriving from the room walls.
Principle: OpenStax University Physics Vol. 2, §1.6 Mechanisms of Heat Transfer, "Radiation" subsection, net radiation rate. Topic: NCEES TFS specification, knowledge area 1A.
Problem 3 · Heat loss from a bare steam pipe
TFS-HT-03 · Thermal/Fluid Principles: Heat transfer principles · Enter a number
A bare steam line is 50 ft long with a 4.5-in. outside diameter. Its surface is at 300°F in 70°F air. Use a combined convection-and-radiation coefficient of 2.0 Btu/hr·ft²·°F applied as Q = hAΔT. Enter the heat loss in Btu/hr, rounded to the nearest 100.
Show answer and solution
Answer: 27,100 Btu/hr.
Surface area A = πDL = π × (4.5/12 ft) × 50 ft = 58.9 ft². Q = hAΔT = 2.0 × 58.9 × (300 − 70) = 27,096 ≈ 27,100 Btu/hr. Leaving the diameter in inches gives about 325,000 Btu/hr, twelve times too high. Convert lengths before you multiply.
Principle: DOE-HDBK-1012/2-92, Vol. 2, Module HT-02, printed p. 19 (Q = hAΔT). Topic: NCEES TFS specification, knowledge area 1A.
Problem 4 · Steam quality from a throttling calorimeter
TFS-TH-01 · Thermal/Fluid Principles: Thermodynamic principles · Select one
Wet steam at 2 MPa passes through a throttling valve and leaves at 100 kPa and 150°C, where h = 2,776.6 kJ/kg. At 2 MPa, use hf = 908.5 kJ/kg, hfg = 1,890.0 kJ/kg, and hg = 2,798.5 kJ/kg. Ignore heat loss and kinetic-energy change. What was the quality of the steam upstream?
Show answer and solution
Answer: B. 98.8%.
An adiabatic throttling valve does no work, so enthalpy stays constant: h₁ = h₂ = 2,776.6 kJ/kg. Then 2,776.6 = 908.5 + x(1,890.0), so x = 0.988, or 98.8%.
Why the other choices miss:
- A (66.8%) divides by hg instead of hfg. Quality multiplies hfg, the energy to vaporize.
- C (100%; it must have been saturated vapor) misses that wet steam can leave a throttle superheated. That's exactly why a throttling calorimeter works.
- D (It can't be found without the downstream entropy) treats the valve as isentropic. A throttle holds enthalpy constant, not entropy.
Principle: DOE-HDBK-1012/1-92, Vol. 1, Module HT-01, printed p. 29 (throttling process; isentropic process). Topic: NCEES TFS specification, knowledge area 1B.
Problem 5 · Second Law check on an efficiency claim
TFS-TH-02 · Thermal/Fluid Principles: Thermodynamic principles · Select one
A vendor claims a heat engine that takes heat from a 600 K source, rejects heat to a 300 K sink, and runs at 55% thermal efficiency. What is the right engineering response?
Show answer and solution
Answer: D. Reject it; the Carnot limit for these temperatures is 50%.
The Carnot efficiency is 1 − Tc/Th = 1 − 300/600 = 0.50. No engine working between two fixed-temperature reservoirs can beat a reversible engine, so a 55% claim violates the Second Law.
Why the other choices miss:
- A (Accept it; 55% is typical of modern plants) compares the claim with other plants instead of with the limit set by these two temperatures.
- B (Accept it only if the engine is reversible) gets it backward: even a reversible engine between these reservoirs reaches only 50%.
- C (Reject it; no heat engine can exceed 33%) invents a cap. One-third isn't a Second Law limit; the limit depends on the temperatures.
Principle: OpenStax University Physics Vol. 2, §4.5 The Carnot Cycle, Carnot efficiency and Carnot principle. Topic: NCEES TFS specification, knowledge area 1B.
Problem 6 · Adiabatic mixing of two air streams
TFS-TH-03 · Thermal/Fluid Principles: Thermodynamic principles · Enter a number
Two moist-air streams mix adiabatically at steady state. Stream 1 carries 3.0 kg/s of dry air at a humidity ratio of 0.010 kg water per kg dry air. Stream 2 carries 1.0 kg/s of dry air at 0.002 kg/kg. Enter the humidity ratio of the mixed stream, in kg/kg, to three decimals.
Show answer and solution
Answer: 0.008.
Water is conserved, so weight each humidity ratio by its dry-air flow: (3.0 × 0.010 + 1.0 × 0.002)/(3.0 + 1.0) = 0.032/4.0 = 0.008 kg/kg. A simple average of the two gives 0.006, which ignores that stream 1 carries three times as much air. On a psychrometric chart, the mixed state sits on the straight line between the two inlet states, closer to the bigger stream.
Principle: National Weather Service glossary, mixing ratio, humidity ratio, also called mixing ratio: water vapor per unit of dry air; OpenStax University Physics Vol. 1, §14.5 Fluid Dynamics, continuity (conservation of mass). Topic: NCEES TFS specification, knowledge area 1B.
Problem 7 · Which property stays constant?
TFS-TH-04 · Thermal/Fluid Principles: Thermodynamic principles · Match each item
Match each idealized process to the property it holds constant. One property is not used.
| Process | Property held constant |
|---|---|
| Adiabatic flow through a throttling valve | ______ |
| Reversible adiabatic expansion in an ideal turbine | ______ |
| Heating a gas sealed in a rigid tank | ______ |
Properties to choose from: Enthalpy, Entropy, Temperature, Volume.
Show answer and solution
Answer: Adiabatic flow through a throttling valve → enthalpy; Reversible adiabatic expansion in an ideal turbine → entropy; Heating a gas sealed in a rigid tank → volume; temperature is not used.
A throttle keeps enthalpy constant, so it plots as a horizontal line on a Mollier (h–s) diagram. A reversible adiabatic (isentropic) expansion keeps entropy constant, so it plots as a vertical line. A rigid tank can't change volume. Temperature is the trap: heating the tank raises it.
Principle: DOE-HDBK-1012/1-92, Vol. 1, Module HT-01, printed p. 29. Topic: NCEES TFS specification, knowledge area 1B.
Problem 8 · Differential manometer across an orifice
TFS-FL-01 · Thermal/Fluid Principles: Fluid principles · Select one
A water-over-mercury differential manometer across an orifice plate shows a 150-mm mercury deflection. Water fills both legs above the mercury, and the two pressure taps are at the same elevation. Use ρHg = 13,600 kg/m³, ρw = 1,000 kg/m³, and g = 9.81 m/s². What is the pressure difference across the orifice?
Show answer and solution
Answer: A. 18.5 kPa.
Write the hydrostatic balance down one leg and up the other. The water columns cancel except over the 150-mm deflection, so Δp = (ρHg − ρw)gh = 12,600 × 9.81 × 0.150 = 18,541 Pa ≈ 18.5 kPa.
Why the other choices miss:
- B (1.47 kPa) uses water's density alone.
- C (20.0 kPa) ignores the water column sitting on top of the mercury.
- D (21.5 kPa) adds the densities instead of subtracting.
Principle: OpenStax University Physics Vol. 1, §14.1 Fluids, Density, and Pressure, Eq. 14.4 (pressure at depth); OpenStax University Physics Vol. 1, §14.2 Measuring Pressure, manometers. Topic: NCEES TFS specification, knowledge area 1C.
Problem 9 · Pressure drop at a pipe reducer
TFS-FL-02 · Thermal/Fluid Principles: Fluid principles · Select one
Water (ρ = 1,000 kg/m³) flows at 0.030 m³/s through a horizontal pipe that reduces from 150-mm to 75-mm diameter. Upstream pressure is 200 kPa. Neglect losses. What is the pressure in the smaller pipe?
Show answer and solution
Answer: B. 178 kPa.
Continuity first: V₁ = Q/A₁ = 0.030/(π × 0.150²/4) = 1.70 m/s. Halving the diameter quarters the area, so V₂ = 6.79 m/s. Bernoulli on a horizontal line: Δp = ½ρ(V₂² − V₁²) = 500 × (46.1 − 2.9) = 21.6 kPa. So p₂ = 200 − 21.6 = 178 kPa.
Why the other choices miss:
- A (157 kPa) doubles the pressure drop.
- C (196 kPa) scales velocity by the diameter ratio (2×) instead of the area ratio (4×).
- D (222 kPa) adds the drop. Faster flow means lower static pressure, not higher.
Principle: OpenStax University Physics Vol. 1, §14.5 Fluid Dynamics, Eq. 14.14 (continuity); OpenStax University Physics Vol. 1, §14.6 Bernoulli's Equation, Eq. 14.16 (Bernoulli). Topic: NCEES TFS specification, knowledge area 1C.
Problem 10 · Normal shock in air
TFS-FL-03 · Thermal/Fluid Principles: Fluid principles · Select all that apply
Air at M = 2.0 (γ = 1.4) passes through a stationary normal shock. Select all that are true.
Show answer and solution
Answer: A, C and E. All correct choices, and only those, must be selected to earn credit.
Downstream Mach number: M₂² = [(γ − 1)M₁² + 2]/[2γM₁² − (γ − 1)] = (0.4 × 4 + 2)/(2.8 × 4 − 0.4) = 3.6/10.8, so M₂ = 0.577. The shock is adiabatic, so total temperature holds. It's irreversible, so total pressure drops and entropy rises (not falls). Static pressure jumps up across the shock (it doesn't fall).
Not correct: B (static pressure decreases); D (entropy decreases).
Principle: NASA Glenn Research Center, Normal Shock Wave Equations, normal shock relations. Topic: NCEES TFS specification, knowledge area 1C.
Fluid Equipment and Distribution Systems (9 problems)
Problem 11 · Head loss with fittings as equivalent length
TFS-FE-01 · Fluid Equipment and Distribution Systems: Fluid distribution systems · Select one
Water flows at 6.0 ft/s through 300 ft of pipe with an inside diameter of 0.3355 ft. The Darcy friction factor is f = 0.018. The valves and fittings add an equivalent length of 120 ft. Use g = 32.2 ft/s². What is the total friction head loss?
Show answer and solution
Answer: C. 12.6 ft.
Add the equivalent length to the straight pipe, then use Darcy: hf = f(L/D)(V²/2g) = 0.018 × (420/0.3355) × (6.0²/64.4) = 0.018 × 1,252 × 0.559 = 12.6 ft.
Why the other choices miss:
- A (3.6 ft) counts only the fittings' 120 ft.
- B (9.0 ft) leaves out the fittings' 120 ft.
- D (25.2 ft) drops the 2 in 2g.
Principle: DOE-HDBK-1012/3-92, Vol. 3, Module HT-03, printed p. 32 (Darcy equation) and p. 34 (equivalent length). Topic: NCEES TFS specification, knowledge area 2B.
Problem 12 · Valve selection
TFS-FE-02 · Fluid Equipment and Distribution Systems: Piping components and connections · Select all that apply
Select all statements that are correct for general piping practice.
Show answer and solution
Answer: B, D and E. All correct choices, and only those, must be selected to earn credit.
Gate valves are on/off valves; they aren't used to throttle. Globe valves throttle well, so they're the usual choice for regulating flow. A check valve has one job: stopping reverse flow. It doesn't throttle in either direction.
Not correct: A (a check valve can throttle flow in either direction); C (a gate valve is the preferred valve for fine flow control).
Principle: DOE-HDBK-1018/2-93, Vol. 2, Module ME-04 (Valves), printed pp. 10, 15, 35. Topic: NCEES TFS specification, knowledge area 2A.
Problem 13 · Reynolds number and flow regime
TFS-FE-03 · Fluid Equipment and Distribution Systems: Fluid distribution systems · Select one
Oil with a kinematic viscosity of 1.0 × 10⁻⁴ m²/s flows at 1.5 m/s in a pipe with a 50-mm inside diameter. Treating flow below Re = 2,000 as laminar, what are the Reynolds number and the regime?
Show answer and solution
Answer: D. 750; laminar.
Re = VD/ν = 1.5 × 0.050/(1.0 × 10⁻⁴) = 750. That's below 2,000, so the flow is laminar. On a Moody diagram, you'd read the friction factor from the laminar region, where roughness doesn't matter.
Why the other choices miss:
- A (75; laminar) is a decimal slip.
- B (750; turbulent) has the right number but the wrong regime.
- C (7.5 × 10⁵; turbulent) plugs the diameter in millimeters.
Principle: DOE-HDBK-1012/3-92, Vol. 3, Module HT-03, printed pp. 19–20 (Reynolds number; flow regimes). Topic: NCEES TFS specification, knowledge area 2B.
Problem 14 · NPSH available vs. required
TFS-FE-04 · Fluid Equipment and Distribution Systems: Pumps and fans · Select one
A pump draws 60°C water from an open tank at sea level. Use atmospheric pressure 101.3 kPa, vapor pressure 19.9 kPa, ρ = 983 kg/m³, and g = 9.81 m/s². The pump centerline is 3.0 m above the water surface, and suction-line losses are 1.5 m. The pump needs NPSHr = 4.5 m. Which statement is correct?
Show answer and solution
Answer: A. NPSHa ≈ 3.9 m; the NPSH requirement is not met.
Convert the pressure margin to head, then subtract lift and losses: NPSHa = (patm − pv)/(ρg) − zlift − hloss = 81,400/(983 × 9.81) − 3.0 − 1.5 = 8.44 − 4.5 = 3.94 m. That's below NPSHr = 4.5 m, so the stated NPSH requirement is not met and cavitation-related performance loss is a risk. Fixes include lowering the pump, cooling the water, or cutting suction losses.
Why the other choices miss:
- B (NPSHa ≈ 5.4 m; adequate margin) ignores the 3.0-m suction lift.
- C (NPSHa ≈ 6.9 m; adequate margin) ignores the 1.5-m suction losses.
- D (NPSHa ≈ 8.4 m; adequate margin) ignores both the lift and the losses.
Principle: DOE-HDBK-1012/3-92, Vol. 3, Module HT-03, printed pp. 48–49 (cavitation; NPSH). Topic: NCEES TFS specification, knowledge area 2C.
Problem 15 · Affinity laws for a speed change
TFS-FE-05 · Fluid Equipment and Distribution Systems: Pumps and fans · Select one
A centrifugal pump in a circulating loop with negligible static head draws 15 hp at 1,750 rpm. A drive slows it to 1,450 rpm. Estimate the new power.
Show answer and solution
Answer: B. 8.5 hp.
Power varies with the cube of speed: P₂ = 15 × (1,450/1,750)³ = 15 × 0.569 = 8.5 hp. The affinity laws hold well only when static head is small. With significant static head, find the new operating point on the system curve instead.
Why the other choices miss:
- A (7.1 hp) uses the fourth power of the speed ratio.
- C (10.3 hp) squares the speed ratio, which is the head law.
- D (12.4 hp) scales linearly, which is the flow law.
Principle: DOE Pumping Systems Tip Sheet #11, Adjustable Speed Pumping Applications, affinity laws and their static-head limit; DOE-HDBK-1012/3-92, Vol. 3, Module HT-03, printed p. 49 (pump laws). Topic: NCEES TFS specification, knowledge area 2C.
Problem 16 · Identical pumps in parallel
TFS-FE-06 · Fluid Equipment and Distribution Systems: Pumps and fans · Select all that apply
Two identical centrifugal pumps run in parallel on one system with a rising system curve. Select all that are true.
Show answer and solution
Answer: B, D and E. All correct choices, and only those, must be selected to earn credit.
Pumps in parallel share the same head and add their flows at that head. But the system doesn't stay at one head: more flow means more friction, so the operating point climbs the system curve and total flow ends up less than double. DOE calls the doubling assumption a common misconception. Adding heads at equal flow describes pumps in series.
Not correct: A (system flow doubles compared with one pump running); C (pumps in parallel add their heads at equal flow).
Principle: DOE-HDBK-1012/3-92, Vol. 3, Module HT-03, printed pp. 53–54 (pumps in parallel and series); DOE Pumping Systems Tip Sheet #8, Optimize Parallel Pumping Systems, Fig. 1. Topic: NCEES TFS specification, knowledge area 2C.
Problem 17 · Fan brake horsepower
TFS-FE-07 · Fluid Equipment and Distribution Systems: Pumps and fans · Select one
A fan moves 20,000 cfm against 4.0 in. w.g. of total pressure at 72% total efficiency. Using bhp = (cfm × total pressure in in. w.g.)/(6,362 × η), what is the brake horsepower?
Show answer and solution
Answer: C. 17.5 bhp.
bhp = 20,000 × 4.0/(6,362 × 0.72) = 80,000/4,581 = 17.5 bhp. The shaft must deliver more power than the air receives, so dividing by an efficiency below 1 should always make the number bigger.
Why the other choices miss:
- A (9.1 bhp) multiplies by efficiency instead of dividing.
- B (12.6 bhp) is air horsepower, before efficiency.
- D (24.3 bhp) divides by efficiency twice.
Principle: DOE, Improving Fan System Performance: A Sourcebook for Industry, p. 6 (fan total efficiency; constant 6,362). Topic: NCEES TFS specification, knowledge area 2C.
Problem 18 · Actual compressor exit temperature
TFS-FE-08 · Fluid Equipment and Distribution Systems: Compressors, nozzles, and diffusers · Select one
Air (γ = 1.4) is compressed from 100 kPa and 300 K to 800 kPa with an isentropic efficiency of 82%. Treat efficiency as the ideal temperature rise divided by the actual rise. What is the actual exit temperature?
Show answer and solution
Answer: C. 597 K.
Ideal exit: T₂s = 300 × 8^(0.4/1.4) = 300 × 1.811 = 543 K, an ideal rise of 243 K. Actual rise = 243/0.82 = 297 K, so T₂ = 597 K. A real compressor always runs hotter than an ideal one at the same pressure ratio.
Why the other choices miss:
- A (500 K) multiplies the ideal rise by efficiency instead of dividing.
- B (543 K) is the ideal (isentropic) exit temperature.
- D (1,325 K) uses an exponent of 1/γ instead of (γ − 1)/γ.
Principle: NASA Glenn Research Center, Compressor Thermodynamics, isentropic temperature–pressure relation; real compressor performs below ideal. Topic: NCEES TFS specification, knowledge area 2D.
Problem 19 · Stagnation temperature
TFS-FE-09 · Fluid Equipment and Distribution Systems: Compressors, nozzles, and diffusers · Enter a number
Air (γ = 1.4) flows at M = 2.0 with a static temperature of 300 K. Assuming adiabatic, isentropic flow, enter the stagnation (total) temperature in K.
Show answer and solution
Answer: 540 K.
T₀ = T[1 + (γ − 1)M²/2] = 300 × (1 + 0.2 × 4) = 300 × 1.8 = 540 K. This is the temperature the air would reach if brought to rest without heat transfer, which is why a diffuser's exit air is hotter than its inlet air.
Principle: NASA Glenn Research Center, Isentropic Flow Equations, Eq. 7, static-to-total temperature ratio. Topic: NCEES TFS specification, knowledge area 2D.
Power Systems and Components (9 problems)
Problem 20 · Steam turbine power
TFS-PS-01 · Power Systems and Components: Turbines · Select one
Steam enters a turbine at 20 kg/s with h₁ = 3,350 kJ/kg. Isentropic expansion to the exhaust pressure would give h₂s = 2,400 kJ/kg. Isentropic efficiency (actual enthalpy drop ÷ ideal enthalpy drop) is 85%. Neglect heat loss and changes in kinetic and potential energy. What is the power output?
Show answer and solution
Answer: B. 16.2 MW.
Actual enthalpy drop = 0.85 × (3,350 − 2,400) = 807.5 kJ/kg. Power = 20 × 807.5 = 16,150 kW ≈ 16.2 MW. For a turbine, efficiency shrinks the output; for a compressor, it grows the required input.
Why the other choices miss:
- A (13.7 MW) applies efficiency twice.
- C (19.0 MW) is the ideal output.
- D (22.4 MW) divides by efficiency, the compressor habit.
Principle: NASA Glenn Research Center, Power Turbine Thermodynamics, real turbine performance below ideal; DOE-HDBK-1012/1-92, Vol. 1, Module HT-01, printed p. 29 (isentropic process). Topic: NCEES TFS specification, knowledge area 3A.
Problem 21 · Wind turbine output
TFS-PS-02 · Power Systems and Components: Turbines · Select one
A wind turbine has an 80-m rotor diameter in a 10 m/s wind with ρ = 1.225 kg/m³. Its power coefficient Cp is 0.40. Estimate the power output.
Show answer and solution
Answer: A. 1.23 MW.
Swept area A = π(40)² = 5,027 m². Power in the wind = ½ρAV³ = 0.5 × 1.225 × 5,027 × 10³ = 3.08 MW. Captured power = 0.40 × 3.08 = 1.23 MW. Cp can't exceed the Betz limit of about 0.59, and doubling wind speed multiplies power by eight.
Why the other choices miss:
- B (0.12 MW) is a factor-of-ten slip.
- C (3.08 MW) leaves out Cp: that's the power in the wind, not the power captured.
- D (4.93 MW) uses the 80-m diameter as the radius.
Principle: Penn State AERSP 583, §2a.2 The Actuator Disk Model, power proportional to V³ and rotor area; Betz limit. Topic: NCEES TFS specification, knowledge area 3A.
Problem 22 · Heat rate to efficiency
TFS-PS-03 · Power Systems and Components: Boilers, steam generators, and waste heat recovery · Select one
A power plant's net heat rate is 9,500 Btu/kWh. Using 3,412 Btu per kWh of electricity, what is its thermal efficiency?
Show answer and solution
Answer: B. 35.9%.
One kWh of electricity is 3,412 Btu. Efficiency = 3,412/9,500 = 0.359, or 35.9%. A lower heat rate means a more efficient plant.
Why the other choices miss:
- A (2.78%) divides heat rate by 3,412 and reads the result as a percent.
- C (37.9%) uses 3,600 (kJ per kWh) in a Btu calculation.
- D (64.1%) is 1 minus the efficiency.
Principle: U.S. EIA, What is the efficiency of different types of power plants?, heat rate definition; 3,412 Btu per kWh. Topic: NCEES TFS specification, knowledge area 3B.
Problem 23 · Condenser cooling-water flow
TFS-PS-04 · Power Systems and Components: Condensers · Select one
A condenser rejects 150 MW to cooling water that warms by 10 K. Using cp = 4.18 kJ/kg·K, what cooling-water mass flow is needed?
Show answer and solution
Answer: A. 3,590 kg/s.
ṁ = Q/(cp·ΔT) = 150,000 kW/(4.18 × 10) = 3,589 kg/s ≈ 3,590 kg/s. Halving the allowed temperature rise would double the flow.
Why the other choices miss:
- B (359 kg/s) is off by a factor of ten.
- C (15,000 kg/s) leaves out cp.
- D (35,900 kg/s) is off by a factor of ten the other way.
Principle: DOE-HDBK-1012/2-92, Vol. 2, Module HT-02, printed p. 37 (heat-exchanger energy balance, ṁ·cp·ΔT). Topic: NCEES TFS specification, knowledge area 3C.
Problem 24 · Rankine cycle efficiency
TFS-PS-05 · Power Systems and Components: Power cycles · Select one
A simple Rankine cycle has these enthalpies (kJ/kg): pump inlet (saturated liquid) 192, pump exit 200, turbine inlet 3,400, turbine exit 2,300. What is the thermal efficiency?
Show answer and solution
Answer: B. 34.1%.
Net work = turbine work − pump work = (3,400 − 2,300) − (200 − 192) = 1,092 kJ/kg. Heat added in the boiler = 3,400 − 200 = 3,200 kJ/kg. Efficiency = 1,092/3,200 = 34.1%.
Why the other choices miss:
- A (32.1%) divides by the turbine-inlet enthalpy instead of the boiler heat input.
- C (34.4%) ignores pump work (1,100/3,200). Close, but net work is turbine work minus pump work.
- D (65.9%) is heat rejected over heat added.
Principle: OpenStax University Physics Vol. 2, §4.2 Heat Engines, Eq. 4.2 (efficiency = net work ÷ heat in). Topic: NCEES TFS specification, knowledge area 3D.
Problem 25 · Ideal Brayton cycle efficiency
TFS-PS-06 · Power Systems and Components: Power cycles · Select one
An ideal air-standard Brayton cycle (γ = 1.4) has a pressure ratio of 12. What is its thermal efficiency?
Show answer and solution
Answer: D. 50.8%.
Ideal Brayton efficiency = 1 − 1/PR^((γ − 1)/γ) = 1 − 1/12^0.2857 = 1 − 1/2.034 = 50.8%. Efficiency rises with pressure ratio, but turbine-inlet temperature limits how far designers can push it.
Why the other choices miss:
- A (49.2%) is 1/2.034, the reciprocal of the temperature ratio.
- B (63.0%) uses an exponent of 0.4 instead of (γ − 1)/γ.
- C (91.7%) is 1 − 1/PR, skipping the exponent.
Principle: MIT 16.Unified Thermodynamics notes, §3.7 Brayton Cycle, Eq. 3.10. Topic: NCEES TFS specification, knowledge area 3D.
Problem 26 · Combined-cycle efficiency
TFS-PS-07 · Power Systems and Components: Power cycles · Enter a number
A gas turbine converts 38% of its fuel energy to work. All of its rejected heat goes to a heat-recovery steam generator, and the steam bottoming cycle converts 30% of that rejected heat to work. Enter the overall combined-cycle efficiency in percent, to one decimal.
Show answer and solution
Answer: 56.6%.
The gas turbine makes 0.38 of the fuel energy as work and rejects 0.62. The steam cycle turns 30% of that 0.62 into work: 0.186. Total = 0.38 + 0.186 = 0.566, or 56.6%. Adding 38% and 30% gives 68%, which overstates it: the bottoming cycle only sees what the gas turbine rejected.
Principle: OpenStax University Physics Vol. 2, §4.2 Heat Engines, Eq. 4.2 (efficiency as work ÷ heat in), applied to each stage. Topic: NCEES TFS specification, knowledge area 3D.
Problem 27 · Stoichiometric air–fuel ratio
TFS-PS-08 · Power Systems and Components: Power cycles · Select one
Methane burns completely with stoichiometric air: CH₄ + 2(O₂ + 3.76 N₂) → CO₂ + 2H₂O + 7.52 N₂. Use molar masses of 28.97 g/mol for air and 16.04 g/mol for CH₄. What is the air–fuel ratio by mass?
Show answer and solution
Answer: D. 17.2.
Each mole of CH₄ needs 2 × 4.76 = 9.52 mol of air. Mass ratio = 9.52 × 28.97/16.04 = 17.2 kg of air per kg of fuel.
Why the other choices miss:
- A (2.0) is the moles of O₂ per mole of fuel.
- B (4.0) counts only the oxygen's mass.
- C (9.52) is the molar air–fuel ratio, not the mass ratio.
Principle: OpenStax Chemistry 2e, §4.3 Reaction Stoichiometry, stoichiometric factors from a balanced equation. Topic: NCEES TFS specification, knowledge area 3D.
Problem 28 · Waste heat recovery
TFS-PS-09 · Power Systems and Components: Boilers, steam generators, and waste heat recovery · Enter a number
Gas-turbine exhaust at 50 kg/s (cp = 1.1 kJ/kg·K) is cooled from 550°C to 150°C in a heat-recovery steam generator. Neglect losses. Enter the heat recovered in MW.
Show answer and solution
Answer: 22.0 MW.
Q = ṁ·cp·ΔT = 50 × 1.1 × (550 − 150) = 22,000 kW = 22.0 MW. A temperature difference in °C equals the same difference in K, so no conversion is needed.
Principle: DOE-HDBK-1012/2-92, Vol. 2, Module HT-02, printed p. 37 (ṁ·cp·ΔT energy balance). Topic: NCEES TFS specification, knowledge area 3B.
Cooling/Heating Systems and Components (5 problems)
Problem 29 · LMTD and flow arrangement
TFS-CH-01 · Cooling/Heating Systems and Components: Heat exchangers · Select one
A heat exchanger must transfer 400 kW with U = 500 W/m²·K. Hot fluid cools from 120°C to 70°C. Cold fluid warms from 20°C to 60°C. For a counterflow arrangement, what heat-transfer area is required?
Show answer and solution
Answer: A. 14.6 m².
In counterflow, the end temperature differences are 120 − 60 = 60 K and 70 − 20 = 50 K. LMTD = (60 − 50)/ln(60/50) = 54.8 K. Area = Q/(U × LMTD) = 400,000/(500 × 54.8) = 14.6 m². Counterflow needs less area than parallel flow for the same duty.
Why the other choices miss:
- B (13.3 m²) uses the larger end temperature difference (60 K) instead of the LMTD.
- C (16.0 m²) uses the smaller end temperature difference (50 K).
- D (20.5 m²) uses parallel-flow end differences (100 K and 10 K, LMTD 39.1 K).
Principle: DOE-HDBK-1012/2-92, Vol. 2, Module HT-02, printed pp. 3–4 (LMTD; Q = UAΔT) and p. 31 (counterflow vs. parallel flow). Topic: NCEES TFS specification, knowledge area 4A.
Problem 30 · Cooling-tower makeup water
TFS-CH-02 · Cooling/Heating Systems and Components: Cooling towers · Enter a number
A cooling tower circulates 5,000 gpm, and the water cools 12°F across the tower. Use these rules: evaporation is about 1% of circulating flow for each 10°F of cooling; blowdown = evaporation ÷ (cycles of concentration − 1); makeup = evaporation + blowdown, neglecting drift. The tower runs at 4 cycles. Enter the makeup flow in gpm.
Show answer and solution
Answer: 80 gpm.
Evaporation = 0.01 × (12/10) × 5,000 = 60 gpm. Blowdown = 60/(4 − 1) = 20 gpm. Makeup = 60 + 20 = 80 gpm. Dividing by cycles instead of (cycles − 1) gives 75 gpm.
Principle: EPA WaterSense at Work, §6.3 Cooling Towers, p. 6-9 (evaporation rule of thumb); FEMP/PNNL, Cooling Towers: Understanding Key Components, p. 4, Eqs. 1 and 6. Topic: NCEES TFS specification, knowledge area 4B.
Problem 31 · Cooling-tower water balance
TFS-CH-03 · Cooling/Heating Systems and Components: Cooling towers · Select all that apply
Select all that are true.
Show answer and solution
Answer: B, D and E. All correct choices, and only those, must be selected to earn credit.
Makeup replaces every water loss: evaporation, blowdown, and drift. Blowdown = evaporation/(cycles − 1), so more cycles means less blowdown. Cycles of concentration is about makeup divided by blowdown, not the other way around.
Not correct: A (blowdown equals evaporation divided by cycles of concentration); C (cycles of concentration equal blowdown divided by makeup).
Principle: EPA WaterSense at Work, §6.3 Cooling Towers, p. 6-9; FEMP/PNNL, Cooling Towers: Understanding Key Components, p. 4, Eqs. 1–7; FEMP Best Management Practice #10, Cooling Tower Management, makeup and cycles definitions. Topic: NCEES TFS specification, knowledge area 4B.
Problem 32 · Vapor-compression COP
TFS-CH-04 · Cooling/Heating Systems and Components: Refrigeration cycles and heat pumps · Select one
In an ideal vapor-compression refrigeration cycle, enthalpies are 400 kJ/kg at the evaporator exit, 440 kJ/kg at the compressor exit, and 250 kJ/kg at both the condenser exit and the evaporator inlet (after throttling). What is the refrigeration COP?
Show answer and solution
Answer: C. 3.75.
Refrigeration effect = 400 − 250 = 150 kJ/kg. Compressor work = 440 − 400 = 40 kJ/kg. COP = 150/40 = 3.75. The throttle holds enthalpy constant, which is why the condenser exit and evaporator inlet share 250 kJ/kg. For the same cycle as a heat pump, COP = 190/40 = 4.75, exactly 1 higher.
Why the other choices miss:
- A (0.27) inverts the ratio.
- B (4.75) is the heat-pump COP (190/40).
- D (5.75) counts compressor work twice: (190 + 40)/40.
Principle: OpenStax University Physics Vol. 2, §4.3 Refrigerators and Heat Pumps, Eqs. 4.3–4.4 (COP for refrigerators and heat pumps); DOE-HDBK-1012/1-92, Vol. 1, Module HT-01, printed p. 29 (throttling). Topic: NCEES TFS specification, knowledge area 4C.
Problem 33 · Chiller kW/ton to COP
TFS-CH-05 · Cooling/Heating Systems and Components: Refrigeration cycles and heat pumps · Enter a number
A chiller's full-load efficiency is 0.60 kW/ton. Using 1 ton of refrigeration = 3.517 kW, enter its COP to two decimals.
Show answer and solution
Answer: 5.86.
COP = cooling delivered ÷ power input = 3.517 kW/0.60 kW = 5.86. A lower kW/ton means a higher COP.
Principle: NIST SP 811, Appendix B.9 (heat flow rate conversions), ton of refrigeration = 12,000 Btu/h = 3,516.853 W; OpenStax University Physics Vol. 2, §4.3 Refrigerators and Heat Pumps, Eq. 4.3 (COP). Topic: NCEES TFS specification, knowledge area 4C.
Supportive Knowledge (3 problems)
Problem 34 · Net present value of a retrofit
TFS-SK-01 · Supportive Knowledge: Economic analysis · Select one
A variable-speed drive costs $40,000 installed and saves $9,000 a year, received at each year's end, for 10 years. There's no salvage value, and the discount rate is 8%. What is the net present value?
Show answer and solution
Answer: C. $20,400.
The uniform-series present-value factor is (P/A, 8%, 10) = [(1.08)¹⁰ − 1]/[0.08 × (1.08)¹⁰] = 6.710. Present value of savings = 9,000 × 6.710 = $60,391. NPV = 60,391 − 40,000 = $20,391 ≈ $20,400. Positive NPV means the drive pays for itself at an 8% cost of money.
Why the other choices miss:
- A (−$35,800) discounts only a single year-10 payment.
- B ($50,000) ignores the time value of money.
- D ($60,400) is the present value of the savings without subtracting the cost.
Principle: NIST Handbook 135, 2025 edition (Life-Cycle Costing Manual), §17.2.2.1 (uniform present value) and §6.1.1 (net savings). Topic: NCEES TFS specification, knowledge area 5A.
Problem 35 · Motor line current
TFS-SK-02 · Supportive Knowledge: Electrical concepts · Select one
A 50-hp, three-phase motor runs at full nameplate load on 460 V, with 93% efficiency and a 0.85 power factor. Use 1 hp = 0.7457 kW. What is the line current?
Show answer and solution
Answer: D. 59 A.
Input power = 50 × 0.7457/0.93 = 40.1 kW. Line current I = P/(√3 × V × pf) = 40,091/(1.732 × 460 × 0.85) = 59.2 A.
Why the other choices miss:
- A (50 A) leaves out power factor.
- B (55 A) ignores efficiency: nameplate horsepower is output, not input.
- C (103 A) uses the single-phase formula (no √3).
Principle: DOE, Determining Electric Motor Load and Efficiency, p. 4, Eq. 2 (input power from hp and efficiency); DOE-HDBK-1011/3-92, Vol. 3, Module ES-09, printed pp. 20–21 (three-phase power). Topic: NCEES TFS specification, knowledge area 5C.
Problem 36 · Thin-wall pressure vessel
TFS-SK-03 · Supportive Knowledge: Material and stress analysis · Select one
A cylindrical air receiver has a 1.2-m inside diameter and a 12-mm wall. It operates at 1.5 MPa gauge. Treat it as thin-walled. The steel's yield strength is 250 MPa. What is the factor of safety against yield, based on hoop stress?
Show answer and solution
Answer: D. 3.33.
Hoop stress σ = pr/t = 1.5 × 0.6/0.012 = 75 MPa. Factor of safety = 250/75 = 3.33. In a cylinder, hoop stress is twice axial stress, so hoop stress governs.
Why the other choices miss:
- A (0.30) inverts the ratio.
- B (1.67) uses the diameter where the radius belongs.
- C (6.67) uses the axial stress (pr/2t), which is half the hoop stress.
Principle: MIT OpenCourseWare 3.11, Pressure Vessels, Eqs. 2–3 (axial and hoop stress). Topic: NCEES TFS specification, knowledge area 5D.
How to read your results
Treat your percentage as feedback on these 36 problems, not an NCEES score. Three to ten problems per area is enough to show which methods you can't yet explain, but not enough to measure an area precisely.
NCEES counts correct answers, converts that count to a scaled score to even out differences between exam forms, and compares it with a passing standard set by subject-matter experts. It doesn't publish the passing score (Examinee Guide, p. 14). So no practice percentage converts to a pass or fail. The 70% first-time pass rate is the share of people who passed, not a score you need to hit.
Your 12-week TFS study plan
Example: 12 weeks at about 10 hours a week (120 hours). This is our planning example, weighted toward the areas with the biggest NCEES ranges. It's not an NCEES requirement, and no number of hours guarantees a pass. Keep every area in the plan, then shift hours toward whatever your error log says is weakest.
| Week | Focus | Do this | Finish the week with |
|---|---|---|---|
| 1 | Setup and baseline | Confirm your board approval and test date. Download the handbook. If you've already studied Problem 1, count it as review. Work the other 35 problems before opening any solution, then check them. | An error log and a list of the areas where you missed most |
| 2 | Heat transfer | Conduction and series resistance, convection, radiation, and the handbook's insulation and fin relations. Rework Problems 1–3. | Three lookup-log entries |
| 3 | Thermodynamics | Properties and steam tables, Mollier diagram, First and Second Laws, throttling, psychrometric chart processes. Rework Problems 4–7. | A quick T–s or h–s sketch for each process you studied |
| 4 | Fluid principles | Statics and manometers, continuity, Bernoulli, compressible flow, normal shocks. Rework Problems 8–10. | Shock and isentropic-flow tables located in the handbook |
| 5 | Piping and distribution | Darcy equation and Moody diagram, equivalent length, ductwork, valves and controls. Rework Problems 11–13. | A sketched system curve with an operating point |
| 6 | Pumps, fans, compressors, nozzles | NPSH, pump curves, affinity laws, series and parallel pumps, fan power, compressor and nozzle efficiency. Rework Problems 14–19. | A one-page pump-and-fan formula sheet with handbook search words |
| 7 | Turbines, boilers, waste heat | Steam, gas, hydraulic, and wind turbines; heat rate; heat-recovery steam generators. Rework Problems 20–22 and 28. | Every efficiency definition written in words |
| 8 | Power cycles and condensers | Rankine (including reheat and regeneration), Brayton, combined cycle, combined heat and power, combustion, IC engines, condenser duty. Rework Problems 23–27. | A labeled diagram for each cycle |
| 9 | Cooling and heating | LMTD and effectiveness, feedwater heaters, cooling towers, vapor-compression and absorption cycles, thermal storage. Rework Problems 29–33. | An energy-balance check on each problem |
| 10 | Supportive knowledge, then timed Block A | Economics, critical path and Gantt charts, motors and electrical, stress and safety factor, codes. Rework Problems 34–36. Then work Block A below, timed. | Time per problem logged |
| 11 | Long timed rehearsal | If you have the NCEES practice exam, sit it under exam conditions: two sections on one 8-hour budget, handbook and approved calculator only, first section submitted before you start the second. If not, work Blocks A and B back to back as this page's longest timed session; do not treat 36 repeated problems as a full exam simulation. | Results logged by area and error type |
| 12 | Targeted review | Rework every logged miss. Repeat your five slowest lookups. Confirm your appointment, ID, and calculator. | A short final repair list and a checked exam-day plan |
Timed mixed blocks. Each block has 18 problems, so give it 108 minutes (18 × 6). The problems are mixed so you have to recognize the method yourself, the way the exam makes you. These are problems you've already seen, so it's pacing and method practice, not a fresh full-length exam.
| Block | Problems, in this order | Time |
|---|---|---|
| A | 1, 11, 20, 29, 34, 2, 12, 21, 30, 3, 13, 22, 31, 4, 14, 23, 32, 5 | 108 minutes |
| B | 6, 15, 24, 33, 35, 7, 16, 25, 36, 8, 17, 26, 9, 18, 27, 10, 19, 28 | 108 minutes |
Adjusting the plan:
- Fewer hours a week? Stretch the calendar instead of cutting areas. At 6 hours a week, the same 120 hours takes about 20 weeks.
- Only six weeks? Combine weeks in pairs (1 and 2, 3 and 4, and so on) and keep both timed blocks. That's about 20 hours a week. If that isn't realistic, consider a later test date.
- Exam next week? Don't start a new system. Fix your two most repeated errors and your two slowest lookups, do one timed block, and check your logistics. You can reschedule a Pearson appointment at least 48 hours ahead for a $50 fee paid to Pearson (Examinee Guide, p. 6).
- Retaking? If you didn't pass, NCEES sends a diagnostic report comparing your performance in each knowledge area with the average of passing examinees. Weigh both the size of each gap and how many questions that area carries, give your top two areas a full week each, and then rerun the plan (Examinee Guide, pp. 14 and 20–21).
Copy this record and keep it with your notes:
My TFS prep record
Licensing board: ______ · Target test date: ______
Specification date: October 2025 · Handbook version (shown in MyNCEES): ______
Calculator model I'll bring: ______
Study hours I can really give per week: ______
Next topic and session date: ______
Turn each miss into your next study task
| What went wrong | Next step | How you know it's fixed |
|---|---|---|
| Picked the wrong method | Write one sentence on what the problem asks and which model fits, then rework a version with one input changed | You can say why the other method doesn't apply |
| Mixed up gauge and absolute pressure, or °C and K | Write the reference state next to every pressure and temperature before you substitute | The same slip doesn't show up on your next three problems |
| Mixed up unit systems | Write the unit chain on every line, and redo the problem in the other system | Units cancel to what the problem asks for |
| Put efficiency on the wrong side | Ask whether the real machine needs more input or gives less output than the ideal one | Your answer moves the right direction from the ideal value |
| Slow handbook lookup | Repeat the lookup drill and log the search words that worked | The same lookup takes under a minute |
| Arithmetic or rounding slip | Redo it from scratch, then check magnitude and sign | The check agrees without copying your first attempt |
Log lucky guesses too. A right answer you can't explain is a miss waiting to happen.
Exam-day rules that change how you practice
- About 6 minutes per question on average. That's 480 minutes ÷ 80 questions. It's a budget, not a per-question limit.
- Two sections, one clock. After roughly half the questions, you review and submit the first section, and you can't go back. Clear your flags before you submit. Time you save in the first section carries over.
- Breaks. The optional 50-minute break comes after the first submit. Unused break time doesn't extend your exam. Unscheduled breaks come out of your exam time, and you can't leave the building during them.
- Answer everything. There's no penalty for wrong answers, and some unscored pretest questions are mixed in where you can't spot them.
- One approved calculator. For 2026 test dates, NCEES approves the HP 33s, 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. Practice on the exact model you'll bring, and if your test date moves into 2027, check NCEES's 2027 list.
- Markers, not pencils. You get two reusable booklets and three markers for scratch work. Do some practice on a whiteboard or laminated sheet so it isn't new on exam day.
- Arrive 30 minutes early. The first and last names on your ID must match your appointment confirmation.
Sources: NCEES Examinee Guide, May 2026, pp. 8–12 and 14; NCEES 2026 approved-calculator list, p. 1.
Quick answers
Is TFS the right PE Mechanical exam for me? NCEES offers three PE Mechanical exams: HVAC and Refrigeration, Machine Design and Materials, and Thermal and Fluid Systems. Each has its own specification (NCEES PE Mechanical). Most candidates pick the one closest to their daily work. Open all three specifications and compare them before you register.
What's the pass rate? For the group NCEES reported in July 2026, 70% of 578 first-time examinees passed and 46% of 109 repeat examinees passed (NCEES). Those numbers describe groups of test takers, not your odds.
Is it open book? No. It's closed book with the electronic handbook (TFS specification, p. 1).
What score do I need? NCEES doesn't publish a passing score. It reports pass or fail (Examinee Guide, p. 14).
Do I register with NCEES or my state board? Check your licensing board's approval process first, then register and pay in MyNCEES. See NCEES exam registration or find your licensing board.
How soon can I retake it? Once per testing window (January–March, April–June, July–September, October–December), and no more than three times in any 12 months. Some boards are stricter. See NCEES retake rules.
Can I get testing accommodations? Yes, but you have to ask during registration. See NCEES exam accommodations.
Does passing give me a license? No. Your state or territorial licensing board issues the license, and the PE exam is one of its requirements.
Haven't taken the FE yet? Start with FE Mechanical exam prep.
Sources
Exam facts on this page were last verified October 8, 2026, against the NCEES sources below. We recalculated every practice problem and checked each one against its cited teaching source. They haven't been reviewed by NCEES or a licensed engineer.
NCEES
- PE Mechanical exam page (format, fee, references, pass rates)
- PE Mechanical: Thermal and Fluid Systems exam specifications, effective October 2025
- PE Mechanical: Thermal and Fluid Systems exam specifications, effective April 2020 (historical, for comparison)
- NCEES Examinee Guide, May 2026
- NCEES computer-based testing and alternative item types
- NCEES 2026 approved calculators (member-board memo, October 20, 2025)
- NCEES PE Mechanical: Thermal and Fluid Systems practice exam and its errata
Teaching sources for the practice problems
- U.S. DOE Fundamentals Handbooks: Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 1, Vol. 2, Vol. 3; Mechanical Science, Vol. 2; Electrical Science, Vol. 3
- U.S. DOE: Determining Electric Motor Load and Efficiency; Pumping Tip Sheet #11; Pumping Tip Sheet #8; Improving Fan System Performance
- NASA Glenn Research Center: normal shock, isentropic flow, compressor, and power turbine thermodynamics
- OpenStax University Physics: Vol. 1, §14.1, §14.2, §14.5, §14.6; Vol. 2, §1.6, §4.2, §4.3, §4.5; OpenStax Chemistry 2e, §4.3
- MIT 16.Unified thermodynamics notes, §3.7; MIT OpenCourseWare 3.11, Pressure Vessels; Penn State AERSP 583, actuator disk model
- U.S. EIA, power plant efficiency and heat rate; National Weather Service glossary, mixing ratio
- Cooling towers: EPA WaterSense at Work, §6.3; FEMP/PNNL cooling tower fact sheet; FEMP Best Management Practice #10
- NIST SP 811, Appendix B.9; NIST Handbook 135, 2025 edition
These teaching sources explain the methods. They aren't references supplied on the exam.
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