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Exam Review

Key Facts: Taiwan Mechanical Engineering Technologist Examination Exam

6 papers

Constructed-response subjects, equally averaged.

Rules Articles 9–10 and Appendix 2

12 hours

Six two-hour sessions.

MOEX 2026 timetable

NTD 2,300

Standard paper-based senior-examination fee.

MOEX fee schedule

Review selected mechanical engineering concepts in English, then solve full official problems using each paper's language and calculator instructions.

Sample Taiwan Mechanical Engineering Technologist Examination Practice Questions

Try these sample questions to review concepts for the Taiwan Mechanical Engineering Technologist Examination exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1An ideal gas in a closed piston-cylinder assembly expands from an initial state of P1 = 500 kPa and V1 = 0.05 m³ to a final pressure of P2 = 100 kPa following a polytropic path PV^1.3 = C. What is the boundary work done by the gas during this process?
A.14.2 kJ
B.18.6 kJ
C.25.8 kJ
D.33.4 kJ
Explanation: For a polytropic process PV^n = C with n = 1.3, the final volume is V2 = V1 * (P1 / P2)^(1/n) = 0.05 * (500 / 100)^(1/1.3) ≈ 0.1724 m³. The boundary work is calculated as W_b = (P2 * V2 - P1 * V1) / (1 - n) = (100 * 0.1724 - 500 * 0.05) / (1 - 1.3) = (17.24 - 25.0) / (-0.3) ≈ +25.85 kJ.
2A proposed geothermal power heat engine operates between a heat source at 600°C and an environmental sink at 40°C. If the actual thermal efficiency of the plant is 60% of the maximum theoretical Carnot efficiency for these temperature limits, what is the actual thermal efficiency?
A.32.1%
B.38.5%
C.44.8%
D.56.0%
Explanation: Absolute temperatures are Th = 600 + 273.15 = 873.15 K and Tc = 40 + 273.15 = 313.15 K. The maximum Carnot efficiency is η_Carnot = 1 - (Tc / Th) = 1 - (313.15 / 873.15) ≈ 0.6414 (64.14%). The actual thermal efficiency is η_act = 0.60 * 64.14% ≈ 38.48% (38.5%).
3Air (ideal gas with Cp = 1.005 kJ/kg·K and gas constant R = 0.287 kJ/kg·K) is compressed from T1 = 300 K, P1 = 100 kPa to T2 = 450 K, P2 = 300 kPa. What is the specific entropy change (s2 - s1) of the air during this process?
A.+0.092 kJ/(kg·K)
B.-0.092 kJ/(kg·K)
C.+0.214 kJ/(kg·K)
D.-0.315 kJ/(kg·K)
Explanation: The change in specific entropy for an ideal gas with constant specific heats is Δs = Cp * ln(T2 / T1) - R * ln(P2 / P1). Substituting values: Δs = 1.005 * ln(450 / 300) - 0.287 * ln(300 / 100) = 1.005 * 0.405465 - 0.287 * 1.098612 = 0.40749 - 0.31530 = +0.0922 kJ/(kg·K).
4In a steam power plant operating on a superheated Rankine cycle, steam enters the turbine at 8 MPa and 500°C (h1 = 3399.5 kJ/kg, s1 = 6.7266 kJ/kg·K) and expands to a condenser pressure of 10 kPa (where sf = 0.6492 kJ/kg·K, sfg = 7.4996 kJ/kg·K, hf = 191.8 kJ/kg, hfg = 2392.1 kJ/kg). If the turbine has an isentropic efficiency of 85%, what is the actual turbine work output per kilogram of steam?
A.942 kJ/kg
B.1025 kJ/kg
C.1269 kJ/kg
D.1079 kJ/kg
Explanation: For isentropic expansion, s2s = s1 = 6.7266 kJ/kg·K. The vapor quality is x2s = (6.7266 - 0.6492) / 7.4996 = 0.8104. The isentropic enthalpy is h2s = 191.8 + 0.8104 * 2392.1 = 2130.28 kJ/kg. The ideal isentropic work is w_s = h1 - h2s = 3399.5 - 2130.28 = 1269.22 kJ/kg. The actual work output is w_act = η_t * w_s = 0.85 * 1269.22 ≈ 1078.8 kJ/kg (1079 kJ/kg).
5An ideal air-standard Brayton cycle operates with a compressor pressure ratio of rp = 8. Assuming constant specific heats with k = Cp / Cv = 1.4, what is the thermal efficiency of the cycle?
A.38.2%
B.41.5%
C.44.8%
D.51.2%
Explanation: For the ideal cold-air-standard Brayton cycle, eta = 1 - rp^[-(k-1)/k]. With rp = 8 and k = 1.4, eta = 1 - 8^(-0.2857) = 0.448, or 44.8%. This ideal relation assumes isentropic compression and expansion and constant specific heats.
6When comparing the theoretical thermal efficiencies of an ideal Otto cycle and an ideal Diesel cycle under identical compression ratios (r) and identical heat addition per unit mass:
A.The Diesel cycle is more efficient than the Otto cycle.
B.The Otto cycle is more efficient than the Diesel cycle.
C.Both cycles exhibit exactly equal thermal efficiency.
D.The relative efficiency depends entirely on the working fluid gas constant R.
Explanation: For the same compression ratio r, heat addition in the Otto cycle occurs at constant volume (reaching higher peak temperature and pressure), while heat addition in the Diesel cycle occurs at constant pressure (expansion occurs during heat addition, leading to a cutoff ratio rc > 1). Because [rc^k - 1] / [k * (rc - 1)] > 1 for rc > 1, the Otto cycle is inherently more efficient than the Diesel cycle for the same compression ratio.
7Atmospheric moist air is at a total barometric pressure of P = 101.325 kPa, dry-bulb temperature of 30°C, and relative humidity of 60%. The saturation pressure of water vapor at 30°C is 4.246 kPa. What is the humidity ratio (specific humidity ω) of this air in kg of water vapor per kg of dry air?
A.0.0160 kg/kg
B.0.0215 kg/kg
C.0.0261 kg/kg
D.0.0384 kg/kg
Explanation: The partial pressure of water vapor is Pv = φ * Psat(30°C) = 0.60 * 4.246 kPa = 2.5476 kPa. The humidity ratio is calculated using ω = 0.622 * [Pv / (P - Pv)] = 0.622 * [2.5476 / (101.325 - 2.5476)] = 0.622 * [2.5476 / 98.7774] ≈ 0.01604 kg H2O/kg dry air (16.0 g/kg).
8A composite plane wall consists of an inner layer of concrete (thickness L1 = 0.15 m, thermal conductivity k1 = 1.4 W/m·K) and an outer layer of insulation board (thickness L2 = 0.05 m, thermal conductivity k2 = 0.04 W/m·K). The inside surface is maintained at 22°C and the outside surface is maintained at -5°C. Under steady 1D conduction, what is the heat flux through the wall?
A.14.8 W/m²
B.16.5 W/m²
C.19.9 W/m²
D.25.2 W/m²
Explanation: The conductive thermal resistances per unit area are R1 = L1 / k1 = 0.15 / 1.4 ≈ 0.1071 m²·K/W and R2 = L2 / k2 = 0.05 / 0.04 = 1.250 m²·K/W. Total thermal resistance is R_total = 0.1071 + 1.250 = 1.3571 m²·K/W. The heat flux is q'' = ΔT / R_total = [22 - (-5)] / 1.3571 = 27 / 1.3571 ≈ 19.895 W/m² (19.9 W/m²).
9A solid copper sphere (density ρ = 8933 kg/m³, specific heat c = 385 J/kg·K, thermal conductivity k = 401 W/m·K) of diameter D = 50 mm is initially at Ti = 150°C. It is quenched in an oil bath maintained at T_inf = 30°C with an average convective heat transfer coefficient h = 120 W/m²·K. What is the time required for the sphere to cool to 50°C using the lumped capacitance method?
A.185 s
B.290 s
C.360 s
D.428 s
Explanation: For a sphere, the characteristic length is Lc = V / As = (π D³ / 6) / (π D²) = D / 6 = 0.050 / 6 ≈ 0.008333 m. The Biot number is Bi = h * Lc / k = 120 * 0.008333 / 401 ≈ 0.0025 << 0.1, validating the lumped model. The temperature response is (T - T_inf) / (Ti - T_inf) = exp[-t / τ], where the thermal time constant is τ = (ρ * c * Lc) / h = (8933 * 385 * 0.008333) / 120 ≈ 238.8 s. Then t = -τ * ln[(50 - 30) / (150 - 30)] = -238.8 * ln(20 / 120) = -238.8 * (-1.7918) ≈ 427.9 s (428 s).
10A straight rectangular fin of length L = 50 mm, thickness t = 2 mm, and unit width W = 1.0 m (W >> t) is attached to a plane surface at base temperature Tb = 120°C. The thermal conductivity of the fin material is k = 200 W/m·K and the convective coefficient to ambient air at T_inf = 20°C is h = 50 W/m²·K. Assuming an adiabatic fin tip, what is the rate of heat dissipation from the fin?
A.285 W
B.417 W
C.530 W
D.632 W
Explanation: Fin perimeter is P ≈ 2W = 2.0 m, cross-sectional area Ac = W * t = 0.002 m². The fin parameter is m = sqrt[h * P / (k * Ac)] = sqrt[50 * 2.0 / (200 * 0.002)] = sqrt(250) ≈ 15.811 m⁻¹. The dimensionless fin parameter product is m * L = 15.811 * 0.050 ≈ 0.7906. The base excess temperature is θb = Tb - T_inf = 120 - 20 = 100°C. For an adiabatic tip, q_fin = sqrt(h * P * k * Ac) * θb * tanh(m * L) = sqrt(50 * 2 * 200 * 0.002) * 100 * tanh(0.7906) = sqrt(40) * 100 * 0.6588 = 6.3245 * 100 * 0.6588 ≈ 416.7 W (417 W).

About the Taiwan Mechanical Engineering Technologist Examination Exam

Independent English-language MCQ study adaptation for 機械工程技師. Official papers assess extended written reasoning and calculations in Chinese, with English answers permitted where a paper explicitly allows them, as in the 2025 engineering mechanics paper. This bank is not an official translation, format simulation, or substitute for complete written solutions. Article 12's reciprocal route has separate provisions.

Exam sponsor: Ministry of Examination (考選部 / MOEX). The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

2026 day 1: 機械製造, 機動學與機械設計, 電工學（包括電機機械）. Day 2: 工程力學（包括靜力學、動力學與材料力學）, 熱力學與熱傳學（包括熱機）, 流體力學與流體機械. Designs and applied cases are written responses. The ordinary exam has no separate mandatory assignment, oral, interview, or hands-on practical stage.

Time Limit

12 testing hours: six 2-hour papers on November 14–15, 2026.

Passing Score

Article 10: mean at least 60 and no zero; if fewer than 16% qualify, the top-16% mechanism requires mean at least 50 and no zero, with cutoff ties included.

Exam / Certification Fees

NTD 2,300 standard registration fee.

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

17 study questions; 1/6 subject weight

Thermodynamics and Heat Transfer (熱力學與熱傳學，包括熱機)

Cycles, entropy, conduction, convection, radiation, refrigeration, and exchangers.

16 study questions; 1/6 subject weight

Electrical Engineering (電工學，包括電機機械)

Circuits, transformers, rotating machines, drives, and instrumentation.

17 study questions; 1/6 subject weight

Fluid Mechanics and Machinery (流體力學與流體機械)

Fluid forces, pipe flow, compressible flow, pumps, and turbines.

17 study questions; 1/6 subject weight

Kinematics and Machine Design (機動學與機械設計)

Mechanisms, gears, fatigue, shafts, bearings, springs, and joints.

17 study questions; 1/6 subject weight

Engineering Mechanics (工程力學，包括靜力學、動力學與材料力學)

Equilibrium, stress, deformation, motion, energy, and vibration.

16 study questions; 1/6 subject weight

Mechanical Manufacturing (機械製造)

Casting, forming, machining, welding, heat treatment, CNC, and tolerances.

Preparing for the Taiwan Mechanical Engineering Technologist Examination Exam

What You Need to Know

  • Passing score: Article 10: mean at least 60 and no zero; if fewer than 16% qualify, the top-16% mechanism requires mean at least 50 and no zero, with cutoff ties included.
  • Assessment: 2026 day 1: 機械製造, 機動學與機械設計, 電工學（包括電機機械）. Day 2: 工程力學（包括靜力學、動力學與材料力學）, 熱力學與熱傳學（包括熱機）, 流體力學與流體機械. Designs and applied cases are written responses. The ordinary exam has no separate mandatory assignment, oral, interview, or hands-on practical stage.
  • Time limit: 12 testing hours: six 2-hour papers on November 14–15, 2026.
  • Exam / certification fees: NTD 2,300 standard registration fee. Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
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Taiwan Mechanical Engineering Technologist Examination: Suggested Study Strategy

1State thermodynamic states, signs, and units before calculating.
2Check the assumptions behind ideal cycle and machine formulas.
3Work complete free-body diagrams and written derivations.
4Follow each official paper's language and calculator instructions.

Frequently Asked Questions

Who can apply?

Appendix 1 includes recognized junior-college-or-higher mechanical graduates and eligible graduates with six listed subjects totaling 18 credits. Legacy qualifying routes also exist. MOEX determines eligibility from credentials.

Is the exam available entirely in English?

The ordinary questions are Chinese. The 2025 engineering mechanics paper expressly allows Chinese or English answers; the other five require Chinese except for technical terms and formulas. Follow current paper instructions. Article 12's reciprocal foreign-license route is separate.

Does the 16% mechanism describe the pass rate?

No. It supplements the 60-average rule when fewer candidates qualify, with a minimum 50 average and no zero subject. It is not an observed statistic or upper quota.

Does this resource reproduce the official assessment?

No. It is independent English MCQ study practice, not an official translation, format simulation, or substitute for written calculations and design explanations. Passing the exam also does not complete every professional practice-license requirement.