Free PE Mechanical Thermal and Fluid Systems Practice Test
40 original practice questions for the NCEES PE Mechanical: Thermal and Fluid Systems exam. These questions are unofficial, written by Castleport Test Prep rather than NCEES, and free to use with no signup.
Practice questions
Thermal/fluid principles (TFS-01 to TFS-11)
TFS-01 · Heat transfer: conduction and convection
A steel steam line has a 168 mm outside diameter and an outer pipe-surface temperature of 180°C. It's covered with 50 mm of insulation with k = 0.050 W/(m·K). The combined outside surface coefficient is 10 W/(m²·K), and the surrounding air is at 20°C. At steady state, ignoring contact resistance, the heat loss per meter of pipe is closest to:
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Answer and explanation
Answer: B. The two resistances act in series, so add them. r₁ = 0.084 m and r₂ = 0.134 m.
- Insulation: R = ln(r₂/r₁)/(2πk) = ln(1.595)/(2π × 0.050) = 1.487 m·K/W.
- Outside film: R = 1/(2πr₂h) = 1/(2π × 0.134 × 10) = 0.119 m·K/W.
q′ = (180 − 20)/(1.487 + 0.119) = 99.7 ≈ 100 W/m.
Common misses:
- C: Drops the outside film resistance and counts the insulation alone.
- A: Puts the outside diameter over the inside radius inside the log.
- D: Counts only the outside film, as if there were no insulation.
Topic: Thermal/Fluid Principles, 1A Heat transfer principles. Principle: FAMU-FSU EML3015 lecture notes, “Insulation” slides; MIT Unified Engineering, §16.5.1, Eq. 16.36.
They cover all five knowledge areas of the specification effective October 2025, and every question comes with a worked solution and the reason each wrong answer is wrong.
How this set works: 40 questions in NCEES knowledge-area order, roughly in proportion to the official question ranges. Most have one correct answer. Three are select-all, six ask you to enter a number, and one asks you to match values to terms. NCEES uses these formats along with standard multiple choice. As on the exam, there's no partial credit: a select-all question counts only if you pick every correct choice and nothing else (NCEES computer-based testing). At the exam's average pace of 6 minutes per question (480 minutes ÷ 80), the full set takes about 4 hours. That roughly matches one half of the real exam, since NCEES has you submit about half the questions before the scheduled break. Use g = 9.81 m/s² (32.2 ft/s²) and water at 1,000 kg/m³ unless a question gives other values. Assume sea-level conditions unless a question says otherwise, which is the same rule the NCEES specification sets.
TFS-02 · Heat transfer: radiation
Two large, parallel gray plates face each other across a vacuum. Plate 1 is at 600 K and plate 2 is at 400 K. Both have an emissivity of 0.80. Using σ = 5.67 × 10⁻⁸ W/(m²·K⁴), what is the net radiant heat flux between them?
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Answer and explanation
Answer: B. For two large parallel gray surfaces, q/A = σ(T₁⁴ − T₂⁴)/(1/ε₁ + 1/ε₂ − 1).
The blackbody exchange is σ(T₁⁴ − T₂⁴) = 5,897 W/m². The emissivity term is 1.25 + 1.25 − 1 = 1.5. Divide: 5,897/1.5 = 3,931 W/m².
Common misses:
- D: Treats both plates as blackbodies.
- C: Multiplies by a single emissivity.
- A: Multiplies by ε₁ε₂. That product isn't the exchange factor for two facing plates.
Topic: Thermal/Fluid Principles, 1A Heat transfer principles. Principle: MIT Unified Engineering, §19.3, Eq. 19.3.
TFS-03 · Heat transfer: forced convection
Water is being heated in a smooth tube with a 25 mm inside diameter. The flow is fully developed turbulent, with Re = 50,000 and Pr = 4.0. The water's thermal conductivity is 0.62 W/(m·K). Using the Dittus–Boelter correlation, the convection coefficient is closest to:
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Answer and explanation
Answer: C. The fluid is being heated, so the Prandtl exponent is n = 0.4.
Nu = 0.023 Re^0.8 Pr^0.4 = 0.023 × 5,743 × 1.741 = 230.
h = Nu·k/D = 230 × 0.62/0.025 = 5,704 W/(m²·K).
Common misses:
- B: Uses n = 0.3, the exponent for a fluid being cooled.
- A: Leaves out the Prandtl-number term.
- D: Leaves the diameter in millimeters.
Topic: Thermal/Fluid Principles, 1A Heat transfer principles. Principle: University of Waterloo ECE 309, Ch. 6, Internal Flow §2, slide 12.
TFS-04 · Thermodynamics: throttling and steam quality
Wet steam from a 1.0 MPa line is bled through a throttling calorimeter to 100 kPa. It leaves at 120°C, which is superheated at that pressure. Use these values: at 100 kPa and 120°C, h = 2,716.6 kJ/kg; at 1.0 MPa, h_f = 762.5, h_fg = 2,014.6, and h_g = 2,777.1 kJ/kg. Neglect heat loss and kinetic-energy changes. What is the steam quality in the line?
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Answer and explanation
Answer: B. An adiabatic throttle does no work, so enthalpy doesn't change: h₁ = h₂ = 2,716.6 kJ/kg.
At 1.0 MPa, x = (h − h_f)/h_fg = (2,716.6 − 762.5)/2,014.6 = 0.970.
Common misses:
- A: Divides by h_g instead of h_fg.
- C: Computes h₂/h_g, which isn't a quality.
- D: Gives the moisture fraction, 1 − x.
Topic: Thermal/Fluid Principles, 1B Thermodynamic principles. Principle: DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 1, Module HT-01, p. 29 (throttling). Property values in the stem come from the IAPWS-IF97 steam formulation. Your handbook's tables may differ in the last digit.
TFS-05 · Thermodynamics: psychrometric mixing
Two moist-air streams mix adiabatically at constant pressure. Stream 1 carries 2.0 kg/s of dry air at a humidity ratio of 0.010 kg/kg. Stream 2 carries 1.0 kg/s of dry air at 0.004 kg/kg. Enter the humidity ratio of the mixed stream, in kg water per kg dry air, to three decimal places.
Enter a number in kg/kg dry air.
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Answer and explanation
Answer: 0.008 kg/kg dry air. Both the dry air and the water are conserved, so the mixed humidity ratio is the dry-air-weighted average:
ω₃ = (2.0 × 0.010 + 1.0 × 0.004)/3.0 = 0.008.
On a psychrometric chart, the mixed state sits on the straight line between the two inlet states. It lands two-thirds of the way toward stream 1, because stream 1 carries twice the dry air.
Common misses:
- 0.007: The plain average of the two humidity ratios. It ignores that stream 1 carries twice the dry air.
Topic: Thermal/Fluid Principles, 1B Thermodynamic principles. Principle: MIT OCW 4.42J, “Water Vapor” notes, p. 15, Eq. 26c.
TFS-06 · Thermodynamics: second law
Heat flows steadily at 100 kW through a rod from a reservoir at 500 K to a reservoir at 300 K. The rod's own state doesn't change. What is the rate of entropy generation?
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Answer and explanation
Answer: A. Add the entropy changes of the two reservoirs. The hot one loses Q̇/T_H and the cold one gains Q̇/T_L:
Ṡ_gen = 100/300 − 100/500 = 0.333 − 0.200 = 0.133 kW/K.
It comes out positive, as any real heat transfer across a finite temperature difference must.
Common misses:
- D: Flips the sign. Negative entropy generation would violate the second law.
- C: Counts only the cold reservoir.
- B: Counts only the hot reservoir.
Topic: Thermal/Fluid Principles, 1B Thermodynamic principles. Principle: MIT Unified Engineering, §5.5, “Heat transfer between two heat reservoirs”.
TFS-07 · Thermodynamics: closed system, first law
A sealed rigid tank holds 2.0 kg of air. The air is heated from 300 K to 400 K. For air, c_v = 0.718 kJ/(kg·K) and c_p = 1.005 kJ/(kg·K). How much heat is added?
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Answer and explanation
Answer: B. A rigid tank can't do boundary work, so all the heat goes into internal energy:
Q = ΔU = m·c_v·ΔT = 2.0 × 0.718 × 100 = 143.6 kJ.
Common misses:
- C: Uses c_p, which belongs to a constant-pressure process.
- A: Uses the gas constant, R = 0.287 kJ/(kg·K), instead of c_v.
- D: Multiplies by the final temperature instead of the temperature change.
Topic: Thermal/Fluid Principles, 1B Thermodynamic principles. Principle: MIT Unified Engineering, §2.4.1, Specific heats.
TFS-08 · Fluid principles: statics (manometer)
A mercury U-tube manometer (SG 13.6) is connected across an orifice in a horizontal water line. Water fills both legs above the mercury. The manometer reads 150 mm. What is the pressure difference across the orifice?
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Answer and explanation
Answer: B. The water standing above the mercury pushes back in both legs, so use the density difference:
Δp = (ρ_Hg − ρ_w)gh = (13,600 − 1,000)(9.81)(0.150) = 18,540 Pa ≈ 18.5 kPa.
Common misses:
- C: Ignores the water column above the mercury.
- A: Uses the density of water alone.
- D: Adds the two densities.
Topic: Thermal/Fluid Principles, 1C Fluid principles. Principle: Al-Mustaqbal University, fluid mechanics lecture, §2.6, Eq. 2.8 (manometer pressure balance).
TFS-09 · Fluid principles: shock properties
Air (γ = 1.4) at Mach 2.00 passes through a normal shock. Enter the static-pressure ratio across the shock, p₂/p₁, to two decimal places.
Enter a number.
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Answer and explanation
Answer: 4.50. p₂/p₁ = [2γM₁² − (γ − 1)]/(γ + 1) = (2 × 1.4 × 4 − 0.4)/2.4 = 4.50. A normal-shock table gives the same value.
Common misses:
- 0.58: The downstream Mach number, not the pressure ratio.
- 0.72: The stagnation-pressure ratio, p₀₂/p₀₁. It drops across a shock because the shock is irreversible.
Topic: Thermal/Fluid Principles, 1C Fluid principles. Principle: NASA Glenn Research Center, “Normal Shock Wave Equations”.
TFS-10 · Fluid principles: hydrostatic force
A vertical rectangular gate is 2.0 m wide and 3.0 m tall. Its top edge is 1.0 m below the free water surface. Atmospheric pressure acts on both sides. What is the resultant hydrostatic force on the gate?
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Answer and explanation
Answer: C. Use the depth of the gate's centroid: h_c = 1.0 + 1.5 = 2.5 m.
F = ρg·h_c·A = 1,000 × 9.81 × 2.5 × 6.0 = 147 kN.
The force acts at the center of pressure, below the centroid: 2.5 + I_c/(h_c·A) = 2.5 + 4.5/15 = 2.80 m below the surface.
Common misses:
- A: Uses the depth of the top edge.
- D: Uses the depth of the bottom edge.
- B: Measures the centroid from the top of the gate and forgets the 1.0 m of water above it.
Topic: Thermal/Fluid Principles, 1C Fluid principles. Principle: LibreTexts, Tatum, Classical Mechanics, §16.6, Eq. 16.6.1.
TFS-11 · Fluid principles: conservation of mass and energy (mixing)
Hot water at 2.0 kg/s and 80°C mixes adiabatically with cold water at 15°C. The mixed stream leaves at 40°C. Assume a constant specific heat. Enter the cold-water mass flow, in kg/s, to one decimal place.
Enter a number in kg/s.
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Answer and explanation
Answer: 3.2 kg/s. The heat the hot stream gives up equals the heat the cold stream picks up:
2.0(80 − 40) = ṁ(40 − 15), so ṁ = 80/25 = 3.2 kg/s.
Common misses:
- 1.25: Swaps the two temperature differences.
Topic: Thermal/Fluid Principles, 1C Fluid principles. Principle: MIT Unified Engineering, §2.5.2, Eq. 2.11 (steady-flow energy equation).
Fluid equipment and distribution systems (TFS-12 to TFS-20)
TFS-12 · Pumps: NPSH available
A pump draws 60°C water from an open tank at sea level (101.325 kPa). The pump centerline is 3.0 m above the liquid surface, and the suction-line losses are 1.2 m. At 60°C, ρ = 983.2 kg/m³ and the vapor pressure is 19.95 kPa. What is the NPSH available?
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Answer and explanation
Answer: A. NPSHa = (p_atm − p_v)/(ρg) − suction lift − suction losses
= (101,325 − 19,950)/(983.2 × 9.81) − 3.0 − 1.2 = 8.44 − 4.2 = 4.24 m.
Common misses:
- C: Forgets the vapor pressure, the term that makes hot water hard to pump.
- B: Forgets the suction losses.
- D: Adds the suction lift instead of subtracting it.
Topic: Fluid Equipment and Distribution Systems, 2C Pumps and fans. Principle: Hydraulic Institute, Pump Principles, Eq. 1.D.6 (NPSHA); DOE Fundamentals Handbook, Vol. 3, “Cavitation,” p. 48. Water properties in the stem come from the IAPWS-IF97 formulation.
TFS-13 · Pumps: parallel operation and system curves
Each of two identical pumps follows H = 50 − 2,000Q², with H in m and Q in m³/s. The system curve is H = 20 + 3,000Q². What is the operating flow with both pumps running in parallel?
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Answer and explanation
Answer: B. In parallel, the two pumps split the flow at the same head. The combined curve is H = 50 − 2,000(Q/2)² = 50 − 500Q².
Set it equal to the system curve: 50 − 500Q² = 20 + 3,000Q², so 3,500Q² = 30 and Q = 0.0926 m³/s.
That's only about 20% more than one pump alone, because the system curve rises steeply.
Common misses:
- A: One pump running alone.
- D: Assumes parallel pumps double the flow.
- C: The result for the same two pumps in series. With this friction-heavy system, series happens to deliver more flow.
Topic: Fluid Equipment and Distribution Systems, 2C Pumps and fans. Principle: DOE Fundamentals Handbook, Vol. 3, “Centrifugal Pumps in Parallel,” pp. 53–54.
TFS-14 · Fluid distribution: major and minor losses
Water flows at 2.5 m/s through 60 m of pipe with a 100 mm inside diameter. The friction factor is 0.020. The fittings and valves have a combined loss coefficient ΣK = 6.5. What is the total head loss?
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Answer and explanation
Answer: D. The velocity head is V²/2g = 6.25/19.62 = 0.319 m.
- Major (pipe friction): f(L/D)(V²/2g) = 0.020 × 600 × 0.319 = 3.82 m
- Minor (fittings): ΣK(V²/2g) = 6.5 × 0.319 = 2.07 m
Total = 5.89 m.
Common misses:
- C: Counts the pipe friction only.
- A: Counts the fittings only.
- B: Uses V instead of V² in the velocity head.
Topic: Fluid Equipment and Distribution Systems, 2B Fluid distribution systems. Principle: DOE Fundamentals Handbook, Vol. 3, p. 32, Eq. 3-14; p. 34, Eqs. 3-15 and 3-16.
TFS-15 · Fans: fan laws (USCS)
A fan at 1,200 rpm delivers 10,000 cfm at 2.00 in. w.g. and draws 5.0 bhp. In the same duct system, with the fan laws applying, what power does it need to deliver 12,000 cfm?
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Answer and explanation
Answer: D. Flow scales with speed, so the speed must rise by 12,000/10,000 = 1.2, to 1,440 rpm.
Power scales with the cube of speed: 5.0 × 1.2³ = 8.64 bhp.
Pressure scales with the square, so it rises to 2.88 in. w.g.
Common misses:
- B: Scales power linearly with speed.
- C: Scales power with the square of speed, which is the pressure law.
- A: Gives the new pressure in in. w.g., not the power.
Topic: Fluid Equipment and Distribution Systems, 2C Pumps and fans. Principle: DOE Improving Fan System Performance sourcebook, §1, “Rotational Speed,” p. 5.
TFS-16 · Piping components: valve flow coefficient (USCS)
A control valve passes 100 gpm of water (SG 1.0) with a 9.0 psi pressure drop. Enter the valve's flow coefficient, C_v, to one decimal place.
Enter a number.
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Answer and explanation
Answer: 33.3. Q = C_v√(ΔP/SG), so C_v = 100/√(9.0/1.0) = 100/3 = 33.3.
Common misses:
- 11.1: Divides by ΔP instead of its square root.
- 300: Multiplies by the square root instead of dividing.
Topic: Fluid Equipment and Distribution Systems, 2A Piping components and connections. Principle: ISA, Baumann, Control Valve Primer, Ch. 5, p. 21 and Eq. 5-4.
TFS-17 · Nozzles: exit velocity
Air expands through an adiabatic nozzle from 600 K, with negligible inlet velocity, down to 480 K. Take c_p = 1.005 kJ/(kg·K). What is the exit velocity?
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Answer and explanation
Answer: C. With no heat and no work, the whole enthalpy drop becomes kinetic energy:
V₂ = √[2c_p(T₁ − T₂)] = √(2 × 1,005 J/(kg·K) × 120 K) = 491 m/s.
Common misses:
- A: Leaves c_p in kJ, so the answer is off by a factor of √1,000.
- B: Drops the 2.
- D: Uses the inlet temperature instead of the temperature drop.
Topic: Fluid Equipment and Distribution Systems, 2D Compressors, nozzles, and diffusers. Principle: MIT Unified Engineering, §2.5.2, Eq. 2.11, and §2.5.4.2 (flow through a nozzle).
TFS-18 · Compressors: isentropic efficiency
Air enters a compressor at 300 K and is compressed through a pressure ratio of 6. Take γ = 1.4. The compressor's isentropic efficiency is 0.82. What is the actual exit temperature?
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Answer and explanation
Answer: C. The ideal exit temperature is T₂s = 300 × 6^(0.4/1.4) = 500.6 K.
Compressor efficiency is ideal work over actual work, so the real temperature rise is larger: 200.6/0.82 = 244.6 K.
T₂ = 300 + 244.6 = 545 K.
Common misses:
- B: The isentropic exit temperature.
- A: Multiplies the rise by η, which is turbine logic.
- D: Divides the absolute exit temperature, not the rise, by η.
Topic: Fluid Equipment and Distribution Systems, 2D Compressors, nozzles, and diffusers. Principle: University of Waterloo ME 354, Ch. 8, “Compressor and Turbine Efficiencies”.
TFS-19 · Pumps: cavitation and NPSH
Select all that apply. A pump draws water from an open tank. Which changes reduce the NPSH available and raise the risk of cavitation?
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Answer and explanation
Answers: A, B, C.
A, B, and C each subtract from NPSHa:
- Warmer water has a higher vapor pressure.
- More suction lift takes away static head.
- A partly closed suction valve adds suction friction.
As on the exam, this counts as correct only if you pick exactly these three.
Common misses:
- D: A larger pipe lowers suction velocity and friction, so it raises NPSHa.
- E: A flooded suction turns lift into positive static head, which also raises NPSHa.
Topic: Fluid Equipment and Distribution Systems, 2C Pumps and fans. Principle: DOE Fundamentals Handbook, Vol. 3, “Cavitation,” p. 48; Hydraulic Institute, Pump Principles, Eq. 1.D.6 (NPSHA).
TFS-20 · Fluid distribution: laminar pipe flow
Oil with a kinematic viscosity of 1.0 × 10⁻⁴ m²/s flows at 1.2 m/s through 50 m of pipe with a 100 mm inside diameter. What is the friction head loss?
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Answer and explanation
Answer: C. Check the flow regime before choosing f. Re = VD/ν = 1.2 × 0.1/(1.0 × 10⁻⁴) = 1,200, so the flow is laminar.
For laminar flow, Poiseuille's law written in Darcy form gives f = 64/Re = 0.0533.
h_f = f(L/D)(V²/2g) = 0.0533 × 500 × 1.44/19.62 = 1.96 m.
Common misses:
- B: Assumes a typical turbulent f = 0.02 without checking Re.
- A: Uses the Fanning factor, 16/Re, in the Darcy equation.
- D: Leaves out g. 19.2 is the loss in J/kg, not a head in meters.
Topic: Fluid Equipment and Distribution Systems, 2B Fluid distribution systems. Principle: OpenStax University Physics Vol. 1, §14.7, Eqs. 14.19–14.20 (Poiseuille's law and Reynolds number); DOE Fundamentals Handbook, Vol. 3, Eq. 3-14.
Power systems and components (TFS-21 to TFS-30)
TFS-21 · Turbines: steam turbine output
Steam flows through a turbine at 25 kg/s. The inlet enthalpy is 3,450 kJ/kg, and the isentropic exit enthalpy is 2,300 kJ/kg. The turbine's isentropic efficiency is 0.85, and the generator's efficiency is 0.97. What is the electrical output?
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Answer and explanation
Answer: A. Actual turbine work = ṁη_t(h₁ − h₂s) = 25 × 0.85 × 1,150 = 24,438 kW.
Electrical output = 24,438 × 0.97 = 23,700 kW.
Common misses:
- B: Stops at shaft power and skips the generator.
- C: Applies the generator efficiency but skips the turbine efficiency.
- D: Divides by turbine efficiency, which is compressor logic.
Topic: Power Systems and Components, 3A Turbines. Principle: University of Waterloo ME 354, Ch. 8, “Compressor and Turbine Efficiencies”.
TFS-22 · Power plants: heat rate (USCS)
A power plant delivers 250 MW net from a fuel energy input of 650 MW. Enter its net heat rate, in Btu/kWh, to the nearest 10.
Enter a number in Btu/kWh.
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Answer and explanation
Answer: 8,870 Btu/kWh. Efficiency η = 250/650 = 0.3846.
Heat rate = 3,412 Btu/kWh ÷ η = 8,870 Btu/kWh.
Common misses:
- 9,360: The heat rate in kJ/kWh (3,600 ÷ η), labeled with the wrong unit.
- 1,312: Multiplies by η instead of dividing.
Topic: Power Systems and Components, 3B Boilers, steam generators, and waste heat recovery. Principle: U.S. EIA FAQ, “What is the efficiency of different types of power plants?”.
TFS-23 · Boilers: efficiency
A boiler makes 50 kg/s of steam at 3,300 kJ/kg from feedwater at 900 kJ/kg. It burns 3.2 kg/s of fuel with a heating value of 44,000 kJ/kg. What is the boiler efficiency?
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Answer and explanation
Answer: C. Useful heat = ṁ(h_steam − h_feedwater) = 50 × 2,400 = 120,000 kW.
Fuel input = 3.2 × 44,000 = 140,800 kW.
Efficiency = 120,000/140,800 = 85.2%.
Common misses:
- D: Uses the steam enthalpy without subtracting the feedwater's, which gives an impossible result above 100%.
- B: Uses the feedwater enthalpy as the output.
- A: Gives the losses, 1 − η, not the efficiency.
Topic: Power Systems and Components, 3B Boilers, steam generators, and waste heat recovery. Principle: U.S. DOE, Steam System Survey Guide, §4.2, Eqs. 9–10 (boiler efficiency).
TFS-24 · Condensers: duty and cooling-water flow
Steam enters a condenser at 30 kg/s with an enthalpy of 2,400 kJ/kg. It leaves as saturated liquid at 192 kJ/kg. The cooling water warms by 10 K, with c_p = 4.18 kJ/(kg·K). What cooling-water flow is required?
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Answer and explanation
Answer: A. Condenser duty = 30 × (2,400 − 192) = 66,240 kW.
Cooling water: ṁ = 66,240/(4.18 × 10) = 1,585 kg/s.
Common misses:
- B: Ignores the enthalpy the condensate still carries out.
- C: Leaves out c_p.
- D: Leaves out the temperature rise.
Topic: Power Systems and Components, 3C Condensers. Principle: MIT Unified Engineering, §2.5.2, Eq. 2.11.
TFS-25 · Condensers: surface area (LMTD)
A condenser rejects 50 MW. Steam condenses at a constant 45°C, and the cooling water warms from 25°C to 35°C. The overall heat-transfer coefficient is 3,000 W/(m²·K). What heat-transfer area is required?
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Answer and explanation
Answer: C. The temperature differences at the two ends are 45 − 25 = 20 K and 45 − 35 = 10 K.
LMTD = (20 − 10)/ln(20/10) = 14.43 K.
A = Q/(U × LMTD) = 50 × 10⁶/(3,000 × 14.43) = 1,155 m².
Common misses:
- B: Uses the arithmetic mean temperature difference, 15 K.
- D: Uses the smaller end difference.
- A: Uses the larger end difference.
Topic: Power Systems and Components, 3C Condensers. Principle: MIT Unified Engineering, §18.5, Eqs. 18.28 and 18.36.
TFS-26 · Power cycles: Rankine with a real turbine
A Rankine cycle has turbine inlet steam at 8 MPa and 500°C (h₃ = 3,399.5 kJ/kg) and a 10 kPa condenser. The isentropic turbine-exit enthalpy is h₄s = 2,130.2 kJ/kg. Saturated liquid leaves the condenser at h₁ = 191.8 kJ/kg with v₁ = 0.00101 m³/kg. The pump is isentropic, and the turbine's isentropic efficiency is 0.85. What is the cycle's thermal efficiency?
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Answer and explanation
Answer: A. - Pump work: v₁ΔP = 0.00101 × (8,000 − 10) = 8.07 kJ/kg, so h₂ = 199.9 kJ/kg.
- Turbine work: 0.85 × (3,399.5 − 2,130.2) = 1,078.9 kJ/kg.
- Heat added: q_in = 3,399.5 − 199.9 = 3,199.6 kJ/kg.
η = (1,078.9 − 8.07)/3,199.6 = 33.5%.
Common misses:
- B: The ideal cycle, with an isentropic turbine.
- C: Divides the ideal efficiency by 0.85.
- D: The Carnot efficiency between 773 K and 319 K (saturation at 10 kPa). That's an upper limit, not this cycle's efficiency.
Topic: Power Systems and Components, 3D Power cycles. Principle: FAMU-FSU EML 4304, Rankine cycle slides, “Energy Analysis” and “Thermal Efficiency”. Steam properties in the stem come from the IAPWS-IF97 formulation.
TFS-27 · Power cycles: Brayton
Using cold-air properties with k = 1.4, what is the thermal efficiency of an ideal air-standard Brayton cycle with a pressure ratio of 10?
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Answer and explanation
Answer: A. η = 1 − 1/r_p^((k−1)/k) = 1 − 10^(−0.2857) = 1 − 0.518 = 48.2%.
Common misses:
- B: Reports the 0.518 term itself instead of 1 minus it.
- C: Uses 1/k as the exponent.
- D: Calculates 1 − 1/r_p and drops the exponent.
Topic: Power Systems and Components, 3D Power cycles. Principle: MIT Unified Engineering, §3.7.1, Eq. 3.10.
TFS-28 · Power cycles: combustion and excess air
Propane (C₃H₈, molar mass 44.1) burns completely with 20% excess air. Treat air as O₂ + 3.76 N₂ with a molar mass of 28.97. What is the air-fuel ratio by mass?
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Answer and explanation
Answer: C. Stoichiometric combustion: C₃H₈ + 5(O₂ + 3.76N₂).
With 20% excess air, the fuel gets 6.0 × 4.76 = 28.56 mol of air per mol.
A/F = 28.56 × 28.97/44.1 = 18.8.
Common misses:
- B: The stoichiometric ratio, with no excess air.
- D: The molar ratio, not the mass ratio.
- A: Counts only the mass of the oxygen.
Topic: Power Systems and Components, 3D Power cycles. Principle: University of Waterloo ME 354, Çengel problem 15-80 solution (balanced propane combustion).
TFS-29 · Turbines: wind turbine power
A wind turbine has an 80 m rotor diameter. The wind speed is 10 m/s, the air density is 1.225 kg/m³, and the power coefficient is 0.40. How much power does it extract?
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Answer and explanation
Answer: A. Swept area A = π(40)² = 5,027 m².
P = ½ρAV³C_p = 0.5 × 1.225 × 5,027 × 1,000 × 0.40 = 1.23 MW.
Common misses:
- B: Uses the Betz limit, 0.593, instead of the stated C_p.
- C: Gives the total power in the wind, as if C_p were 1.
- D: Uses the diameter as the radius.
Topic: Power Systems and Components, 3A Turbines. Principle: Queen Mary University of London, “Power in the wind” and “Betz limit” notes.
TFS-30 · Power cycles: cycle improvements
Select all that apply. Which statements about power cycles are true?
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Answer and explanation
Answers: A, C, D.
A, C, and D are true.
Common misses:
- B: Backward. Lowering condenser pressure raises efficiency.
- E: A regenerator only helps while the turbine exhaust is hotter than the compressor discharge. At high pressure ratios that stops being true.
Topic: Power Systems and Components, 3D Power cycles. Principle: FAMU-FSU EML 4304, Rankine slides on improving efficiency, reheat, and regeneration; MIT Unified Engineering, §8.6; MIT Unified Engineering, §8.7 (combined cycle); Mines Paris, “Gas turbine cycles and variants”.
Cooling/heating systems and components (TFS-31 to TFS-37)
TFS-31 · Heat exchangers: effectiveness–NTU
In a counterflow heat exchanger, the hot stream has a heat-capacity rate of 2.0 kW/K and enters at 150°C. The cold stream has 4.0 kW/K and enters at 30°C. NTU, based on the smaller heat-capacity rate, is 1.5. What is the heat-transfer rate?
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Answer and explanation
Answer: B. The capacity ratio is C_r = 2.0/4.0 = 0.5. For counterflow:
ε = [1 − e^(−NTU(1−C_r))]/[1 − C_r·e^(−NTU(1−C_r))] = 0.528/0.764 = 0.691.
q = ε·C_min·(T_h,in − T_c,in) = 0.691 × 2.0 × 120 = 166 kW.
Common misses:
- A: Uses the parallel-flow relation.
- C: The maximum possible heat transfer, which assumes ε = 1.
- D: Multiplies by C_max instead of C_min.
Topic: Cooling/Heating Systems and Components, 4A Heat exchangers. Principle: Al-Mustaqbal University, heat transfer lecture, Table 10-3 (ε–NTU relations); MIT Unified Engineering, §18.5.3.
TFS-32 · Heat exchangers: closed feedwater heater
A closed feedwater heater warms 100 kg/s of feedwater from 600 to 800 kJ/kg. Extraction steam enters at 2,800 kJ/kg and leaves as drain at 700 kJ/kg. The heater is adiabatic. What extraction-steam flow does it need?
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Answer and explanation
Answer: C. Each kilogram of steam gives up 2,800 − 700 = 2,100 kJ. The feedwater needs 100 × 200 = 20,000 kW.
ṁ_steam = 20,000/2,100 = 9.52 kg/s.
Common misses:
- A: Credits the steam with its full 2,800 kJ/kg and ignores what the drain carries out.
- B: Subtracts the feedwater inlet enthalpy from the steam enthalpy.
- D: Uses the drain enthalpy alone.
Topic: Cooling/Heating Systems and Components, 4A Heat exchangers. Principle: MIT Unified Engineering, §2.5.2, Eq. 2.11.
TFS-33 · Cooling towers: range and approach
Match the values. A cooling tower receives hot water at 35°C and returns cold water at 29°C. The ambient wet-bulb temperature is 24°C, and the dry-bulb temperature is 32°C. Choose the correct value for each term. Values: 3 K, 5 K, 6 K, 11 K.
| Term | Your value (3 K, 5 K, 6 K, or 11 K) |
|---|---|
| Range | |
| Approach |
Show answerHide answer
Answer and explanation
Answer: Range = 6 K; Approach = 5 K.
Range = 6 K. Range is hot water minus cold water: 35 − 29.
Approach = 5 K. Approach is cold water minus the entering wet bulb: 29 − 24.
Evaporation can't cool water below the wet-bulb temperature, which is why approach, not range, tells you how hard the tower is working.
Common misses:
- 3 K: Measures cold water against the dry bulb.
- 11 K: Hot water minus wet bulb, which isn't a standard tower term.
Topic: Cooling/Heating Systems and Components, 4B Cooling towers. Principle: SPX Cooling Technologies (Marley), Cooling Tower Fundamentals, pp. 14 and 16.
TFS-34 · Cooling towers: makeup water (USCS)
A cooling tower evaporates 60 gpm and runs at 4 cycles of concentration. Drift is negligible. What is the makeup-water rate?
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Answer and explanation
Answer: C. Blowdown B = E/(C − 1) = 60/3 = 20 gpm.
Makeup = evaporation + blowdown = 60 + 20 = 80 gpm.
Common misses:
- B: Uses E/C for blowdown.
- A: Forgets blowdown entirely.
- D: Multiplies evaporation by the cycles of concentration.
Topic: Cooling/Heating Systems and Components, 4B Cooling towers. Principle: SPX Cooling Technologies (Marley), Cooling Tower Fundamentals, p. 31, Formulas 2 and 5 (with drift = 0); U.S. DOE FEMP, Cooling Towers fact sheet, Eqs. 1 and 6.
TFS-35 · Refrigeration: vapor-compression cycle (mixed units)
A 10-ton vapor-compression system has these refrigerant enthalpies: leaving the evaporator, h₁ = 400 kJ/kg; leaving the compressor, h₂ = 430 kJ/kg; leaving the condenser, h₃ = 260 kJ/kg. The refrigerant is throttled, so h₄ = h₃. What compressor power does it need? (1 ton = 12,000 Btu/h = 3.517 kW.)
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Answer and explanation
Answer: C. Capacity = 10 × 3.517 = 35.17 kW.
The refrigeration effect is h₁ − h₄ = 140 kJ/kg, so ṁ = 35.17/140 = 0.251 kg/s.
Compressor power = ṁ(h₂ − h₁) = 0.251 × 30 = 7.54 kW. (COP = 140/30 = 4.67.)
Common misses:
- B: Sizes the mass flow from the condenser's enthalpy change instead of the evaporator's.
- D: Treats a ton as 12 kW.
- A: Treats a ton as 1 kW.
Topic: Cooling/Heating Systems and Components, 4C Refrigeration cycles and heat pumps. Principle: WPI ES 3001, Ch. 10 slides, “The Vapor-Compression Refrigeration Cycle,” Eqs. 10.3, 10.5, 10.6 (ton = 200 Btu/min).
TFS-36 · Chillers: kW/ton and COP (USCS ↔ SI)
A chiller runs at 0.60 kW/ton. Enter its COP, to two decimal places.
Enter a number.
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Answer and explanation
Answer: 5.86. One ton of refrigeration is 3.517 kW of cooling, so COP = 3.517/0.60 = 5.86. A 3.516 conversion gives the same 5.86.
Common misses:
- 1.67: Takes 1/0.60, as if a ton were 1 kW.
- 20: Takes 12/0.60, mixing kBtu/h with kW.
Topic: Cooling/Heating Systems and Components, 4C Refrigeration cycles and heat pumps. Principle: Bureau of Energy Efficiency (India), Ch. 9, §9.3 and §9.5.3.
TFS-37 · Refrigeration: LiBr–water absorption chillers
Select all that apply. Which statements about single-effect lithium bromide–water absorption chillers are true?
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Answer and explanation
Answers: A, B, E.
A, B, and E are true. DOE lists single-stage absorption chillers at a COP of roughly 0.70–0.79.
Common misses:
- C: Backward. Absorption COPs are well below those of electric chillers.
- D: DOE describes lithium bromide–water units for chilling at about 40°F and above. Ammonia–water systems handle colder duty.
Topic: Cooling/Heating Systems and Components, 4C Refrigeration cycles and heat pumps. Principle: U.S. DOE, Absorption Chillers for CHP Systems fact sheet, pp. 1–3, Table 2.
Supportive knowledge (TFS-38 to TFS-40)
TFS-38 · Economics and electrical concepts: annual energy cost
A pump needs 60 kW of shaft power. Its motor is 93% efficient, it runs 6,000 hours a year, and electricity costs $0.12/kWh. What is the annual energy cost?
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Answer and explanation
Answer: C. The motor's electrical input is larger than its output: 60/0.93 = 64.5 kW.
Annual cost = 64.5 × 6,000 × 0.12 = $46,452.
Common misses:
- B: Ignores the motor's losses.
- A: Multiplies by efficiency, which makes the input smaller than the output.
- D: Uses 8,760 hours instead of the stated 6,000 run hours.
Topic: Supportive Knowledge, 5A/5C Economic analysis; electrical concepts. Principle: U.S. DOE, Determining Electric Motor Load and Efficiency, p. 8, Eq. 7.
TFS-39 · Project planning: critical path
A plant shutdown has five activities:
- A: 3 days, can start immediately
- B: 4 days, after A
- C: 7 days, after A
- D: 2 days, after B
- E: 5 days, after both C and D
What is the minimum project duration, and which path is critical?
Show answerHide answer
Answer and explanation
Answer: B. Total each path through the network:
- A–B–D–E: 3 + 4 + 2 + 5 = 14 days
- A–C–E: 3 + 7 + 5 = 15 days
The longest path sets the duration: 15 days, A–C–E. Activities B and D each have 1 day of float.
Common misses:
- C: Adds every activity as if they all ran one after another.
- A: Picks the shorter path.
- D: Pairs the right duration with the wrong path.
Topic: Supportive Knowledge, 5B Project planning and scheduling. Principle: MIT OCW ESD.36, Lecture 2, slides 18–20, 29, and 31.
TFS-40 · Material and stress analysis: pressure piping
A steam header has a 300 mm inside diameter and a 10 mm wall. It carries 4.0 MPa internal pressure, and the material's yield strength is 250 MPa. Using the thin-wall hoop stress based on the inside diameter, what is the factor of safety against yield?
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Answer and explanation
Answer: C. Hoop stress σ = pD/(2t) = 4.0 × 300/20 = 60 MPa.
Factor of safety = 250/60 = 4.17.
This is a thin-wall estimate for practice, not a piping-code wall-thickness calculation.
Common misses:
- D: Uses the axial stress, pD/(4t) = 30 MPa, which is half the hoop stress.
- B: Uses pD/t.
- A: Inverts the ratio.
Topic: Supportive Knowledge, 5D Material and stress analysis. Principle: Purdue ME 323, “Pressure Vessels,” Summary.
Score your set and find your weak areas
Score on these 40 practice questions: ____ / 40 correct.
Give one point for each correct answer. A select-all question counts only if your choices match the full correct set. A numeric answer must be in the requested unit and rounding. For the matching question, both values must be right.
Your score describes how you did on these 40 questions. It isn't an NCEES score and doesn't predict a pass. NCEES converts the number you get right to a scaled score, compares it with a standard set by subject-matter experts, and doesn't publish the passing score (NCEES Examinee Guide, May 2026, p. 14).
Mark your misses by knowledge area. Count any question you got right only by guessing as a miss.
| NCEES knowledge area | Questions on the real exam (of 80) | Questions here | Your misses |
|---|---|---|---|
| 1 Thermal/Fluid Principles | 19–29 | TFS-01 to TFS-11 (11) | |
| 2 Fluid Equipment and Distribution Systems | 17–26 | TFS-12 to TFS-20 (9) | |
| 3 Power Systems and Components | 17–26 | TFS-21 to TFS-30 (10) | |
| 4 Cooling/Heating Systems and Components | 12–18 | TFS-31 to TFS-37 (7) | |
| 5 Supportive Knowledge | 5–8 | TFS-38 to TFS-40 (3) | |
| Total | 80 | 40 |
The ranges come from the NCEES Thermal and Fluid Systems specification, effective October 2025. Our counts are an editorial split that fits inside half of each range. They aren't an NCEES exam form. With 3 to 11 questions per area, your tally tells you where to look next, not how well you know a whole area.
What your pattern means
| What you see | What it likely means on this set | What to do next |
|---|---|---|
| Misses cluster in one or two areas | Those sampled topics need repair | Start there, beginning with the area that has the larger official range |
| Misses spread across every area | Shared setup habits are slipping (units, energy balances, efficiency direction) | Rework untimed from a blank page before adding timed sets |
| Lots of right answers you couldn't explain | Luck is hiding gaps | Count those guesses as misses and log them |
| Right answers but slow | Setup or lookup speed is the bottleneck | Run the handbook drill below, then short timed sets |
| Fast, with many misses | Speed isn't your problem yet | Slow down and write the governing equation before any numbers |
Log each miss by its first wrong step
A score tells you that something went wrong. The log tells you why, and the why is what you study next.
| Question | Area | First wrong step | Error type | Fix | Fresh problem to try | Retry date |
|---|---|---|---|---|---|---|
| Example: TFS-18 | 2 | Multiplied the temperature rise by η | Wrong model (efficiency direction) | Write down whether ideal or actual work is larger before applying η | A new compressor problem with a different pressure ratio | (when done) |
Error types to choose from:
- concept gap
- wrong equation or model
- handbook lookup
- property or table lookup
- unit conversion
- algebra or arithmetic
- state-point or diagram setup
- misread condition
- pacing
- lucky guess
Three steps that fix most misses:
- Rework the question from a blank page. Name the unknown and its units, then write the governing relationship before you touch a number.
- Find the relationship in your handbook. Note the search term that found it.
- Try a fresh problem on the same idea. Getting this exact question right a second time mostly proves you remember the answer.
PE Mechanical Thermal and Fluid Systems exam facts
| Detail | Current answer |
|---|---|
| Questions | 80 |
| Exam time | 8 hours. The 9-hour appointment adds a 2-minute nondisclosure agreement, an 8-minute tutorial, and a 50-minute scheduled break |
| When and where | Year-round, computer-based, at NCEES-approved Pearson test centers |
| Specification | Effective beginning October 2025, five knowledge areas |
| Reference | Closed book. NCEES supplies the PE Mechanical Reference Handbook on screen; no design standards are supplied for PE Mechanical |
| Units | SI and U.S. Customary; sea-level conditions unless a question says otherwise |
| Question types | Multiple choice plus alternative item types: multiple correct, point and click, drag and drop, and fill in the blank |
| Fee | $400 per attempt, paid to NCEES; your licensing board may charge its own fee |
| Results | Pass or fail, typically 7–10 days after the exam |
Sources: NCEES PE Mechanical page; TFS specification, effective October 2025, p. 1; NCEES Examinee Guide, May 2026, pp. 3, 10–11, 14, and 16.
Pass rate: In the NCEES table updated July 2026, 70% of 578 first-time Thermal and Fluid Systems examinees passed, and 46% of 109 repeat examinees passed (NCEES PE Mechanical page, pass-rate table). For year-round exams, NCEES reports the January–June group in its July update. These are group rates, not your personal odds.
Scoring: Your result is based on the number of questions you answer correctly, with nothing taken off for wrong answers. Every question, including select-all and the other alternative types, is scored right or wrong with no partial credit. A few unscored pretest questions are mixed in, and you can't tell which ones they are. If you don't pass, NCEES sends a diagnostic report showing your performance in each knowledge area on a 0–15 scale (Examinee Guide, pp. 11, 14, and 20).
Practice the way the exam works
- Search the handbook, not your notes. The handbook is the only reference you get. It appears on screen as a searchable PDF, and you search it with the box on the left side. Ctrl+F doesn't work (Examinee Guide, p. 10). Download the current version from your MyNCEES account and practice with it.
- Run a five-minute lookup drill. Pick a problem and name the relationship you need in plain words, such as "net positive suction head" or "log mean temperature difference." Search for it, check the variable definitions and units, then solve. Write down the search term that worked.
- Use the calculator you'll bring. You may bring one NCEES-approved calculator, and an on-screen TI-30XS is also available (Examinee Guide, p. 8). Check your model against the calculator policy linked at the bottom of the NCEES exams page.
- Plan for the locked first half. After about half the questions, you review and submit them, and you can't go back to them. Clear your flags before you submit (Examinee Guide, p. 11).
- Never leave a question blank. Wrong answers cost nothing.
- Know what this page doesn't simulate. It covers single-answer, select-all, numeric-entry, and matching questions. The real exam also uses point-and-click items on a graphic, and its questions aren't calibrated to ours.
After this set
- Work out a study plan. Our PE exam prep guide has a 12-week plan, a first 60-minute session, and a method for choosing what to study first.
- Ready to book? See how NCEES exam registration works. If you need testing accommodations, read how to request NCEES accommodations first, because you request them during registration.
- After the exam: NCEES exam results and the NCEES retake policy.
Common questions
Is 70% the passing score? No. The 70% figure is the share of first-time Thermal and Fluid Systems examinees who passed in NCEES's July 2026 table. NCEES doesn't publish a passing score, because it varies slightly with exam difficulty (Examinee Guide, p. 14).
Are these questions as hard as the real ones? We can't promise that. They follow the current specification's topics and use realistic setups. They haven't been calibrated against NCEES questions, though, so treat your score as feedback on these 40 items only.
My prep book uses different topic names. Is it out of date? Possibly. The specification before October 2025 grouped topics as Principles, Hydraulic and Fluid Applications, and Energy/Power System Applications (April 2020 specification). The engineering in an older book can still be sound. Remap its chapters to the five current areas above, and fill in anything the current list adds.
Do I have to pass the FE first, or have four years of experience? Your state licensing board decides who may sit for the PE. NCEES designs the PE for engineers with at least four years of post-college experience, but eligibility rules are the board's (NCEES PE Mechanical page; Examinee Guide, p. 2). Find your board in the NCEES member board directory.
Sources and verification
Last verified: October 8, 2026. We checked the exam format, fee, specification and its effective date, item types, scoring rules, reference and calculator rules, and pass rates against the NCEES documents listed below. We re-solved all 40 questions, including the arithmetic behind every wrong answer, and checked each underlying principle against the teaching source linked under its question. Steam and water property values given in the questions were calculated from the IAPWS-IF97 formulation.
Official sources:
- NCEES: PE Mechanical
- NCEES PE Mechanical: Thermal and Fluid Systems CBT specification, effective October 2025 (PDF)
- NCEES Examinee Guide, May 2026 (PDF)
- NCEES: Computer-based testing and alternative item types
- NCEES exams page, including the calculator policy
The teaching source for each question appears directly beneath it. For how we check exam facts, see our methodology. Found an error? The corrections page explains how to report it.
AI-assisted tools were used to develop and check this content. This is editorial source checking, not professional engineering review.
Written by the Castleport Test Prep Editorial Team.
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), Pearson VUE, or any state engineering licensing board. The practice questions on this page are original and unofficial; they are not NCEES exam questions. Exam and credential names identify their subjects, and trademarks belong to their respective owners. This resource doesn't guarantee an exam result or determine eligibility or licensure.