100+ Free Registered Public Utility Engineer — HVAC (China) Practice Questions

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2026 Statistics

Key Facts: Registered Public Utility Engineer — HVAC (China) Exam

MOHURD & MOHRSS

Joint National Statutory Examination Authorities

PRC Registered Public Utility Engineer Regulations

60%

Standard Passing Benchmark on Professional Exam Subjects

MOHRSS National Professional Qualification Standard

Single Sitting

Score Requirement for Professional Knowledge + Case Analysis

National Professional Qualification Examination Center

GB 50736 / 51251 / 55015

Primary National HVAC, Smoke, and Energy Codes Tested

MOHURD National Engineering Standards

3 Years

Mandatory Registration Renewal Cycle with Continuing Education

MOHURD National Engineer Registration Center

The China Registered Public Utility Engineer (HVAC) Examination is the official national professional licensure credential for HVAC, building services, and thermal fluid engineers. Administered by MOHURD and MOHRSS, it assesses mastery of psychrometric processes, heating and cooling load calculations, hydronic systems, smoke control (GB 51251), cleanrooms (GB 50073), and building energy efficiency (GB 55015 / GB 50736).

Sample Registered Public Utility Engineer — HVAC (China) Practice Questions

Try these sample questions to test your Registered Public Utility Engineer — HVAC (China) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1According to moist air thermodynamics and GB 50736-2012, moist air enthalpy $h$ (in $\text{kJ/kg dry air}$) at atmospheric pressure is calculated as $h = 1.005 t + d(2501 + 1.86 t)$, where $t$ is dry-bulb temperature ($^\circ\text{C}$) and $d$ is moisture content (humidity ratio in $\text{kg/kg dry air}$). What is the specific enthalpy of moist air at $t = 25^\circ\text{C}$ with a moisture content of $d = 0.010\text{ kg/kg}$ ($10\text{ g/kg}$)?

A.25.13 kJ/kg

B.50.60 kJ/kg

C.75.32 kJ/kg

D.100.25 kJ/kg

Explanation: Using the standard moist air enthalpy formula $h = 1.005 t + d(2501 + 1.86 t)$: 1) Sensible enthalpy of dry air = $1.005 \times 25 = 25.125\text{ kJ/kg}$. 2) Enthalpy of water vapor = $d \times (2501 + 1.86 \times 25) = 0.010 \times (2501 + 46.5) = 0.010 \times 2547.5 = 25.475\text{ kJ/kg}$. 3) Total moist air enthalpy $h = 25.125 + 25.475 = 50.60\text{ kJ/kg dry air}$.

2In HVAC psychrometrics, the Sensible Heat Ratio (SHR / 显热比) represents the proportion of sensible cooling load to total cooling load ($SHR = Q_s / Q_t$). If an air-conditioned space has a sensible cooling load $Q_s = 42\text{ kW}$ and a latent cooling load $Q_l = 18\text{ kW}$, what is the space Sensible Heat Ratio (SHR)?

A.0.30

B.0.43

C.0.70

D.1.43

Explanation: Total cooling load $Q_t = Q_s + Q_l = 42\text{ kW} + 18\text{ kW} = 60\text{ kW}$. The Sensible Heat Ratio is defined as $SHR = \frac{Q_s}{Q_t} = \frac{42\text{ kW}}{60\text{ kW}} = 0.70$. On the psychrometric chart, $SHR$ determines the slope of the space condition line ($\\epsilon = \Delta h / \Delta d = 2501 / (1 - SHR)$ approx).

3An air handling unit (AHU) adiabatically mixes an outdoor air stream (State W: $L_w = 2000\text{ m}^3/\text{h}$, $t_w = 35^\circ\text{C}$, $h_w = 85.0\text{ kJ/kg}$) with a return air stream (State N: $L_n = 6000\text{ m}^3/\text{h}$, $t_n = 25^\circ\text{C}$, $h_n = 50.0\text{ kJ/kg}$). Assuming constant air density, what are the mixed air dry-bulb temperature $t_m$ and enthalpy $h_m$ at the mixing point M?

A.$t_m = 27.5^\circ\text{C}, h_m = 58.75\text{ kJ/kg}$

B.$t_m = 30.0^\circ\text{C}, h_m = 67.50\text{ kJ/kg}$

C.$t_m = 28.5^\circ\text{C}, h_m = 62.50\text{ kJ/kg}$

D.$t_m = 26.5^\circ\text{C}, h_m = 55.00\text{ kJ/kg}$

Explanation: Total mixed air volume $L_m = L_w + L_n = 2000 + 6000 = 8000\text{ m}^3/\text{h}$. The outdoor air fraction is $x_w = 2000 / 8000 = 0.25$ (25%), and return air fraction is $x_n = 6000 / 8000 = 0.75$ (75%). 1) Mixed temperature $t_m = x_w t_w + x_n t_n = (0.25 \times 35) + (0.75 \times 25) = 8.75 + 18.75 = 27.5^\circ\text{C}$. 2) Mixed enthalpy $h_m = x_w h_w + x_n h_n = (0.25 \times 85.0) + (0.75 \times 50.0) = 21.25 + 37.50 = 58.75\text{ kJ/kg}$.

4A cooling coil receives mixed air at state $t_1 = 28^\circ\text{C}, h_1 = 60.0\text{ kJ/kg}$ and discharges treated supply air at state $t_2 = 16^\circ\text{C}, h_2 = 40.0\text{ kJ/kg}$. The apparatus dew point (ADP / 机器露点) of the coil is $t_{adp} = 12^\circ\text{C}, h_{adp} = 34.0\text{ kJ/kg}$. What is the coil bypass factor ($BF$ / 旁通系数) and contact factor ($CF$ / 接触系数)?

A.$BF = 0.25, CF = 0.75$

B.$BF = 0.33, CF = 0.67$

C.$BF = 0.50, CF = 0.50$

D.$BF = 0.15, CF = 0.85$

Explanation: The coil bypass factor $BF$ is defined by the ratio of actual unachieved cooling/dehumidification to the theoretical maximum: $BF = \frac{t_2 - t_{adp}}{t_1 - t_{adp}} = \frac{16 - 12}{28 - 12} = \frac{4}{16} = 0.25$ (or using enthalpy: $\frac{40.0 - 34.0}{60.0 - 34.0} = \frac{6}{26} \approx 0.231$, with temperature-based $BF = 0.25$). The contact factor (or efficiency) is $CF = 1 - BF = 1 - 0.25 = 0.75$ (75%).

5When moist air is heated sensibly across a hot-water heating coil without adding or removing moisture ($d = \text{constant}$), how do the relative humidity ($\phi$) and dew point temperature ($t_d$) change?

A.Relative humidity increases; dew point temperature increases

B.Relative humidity decreases; dew point temperature remains constant

C.Relative humidity decreases; dew point temperature decreases

D.Relative humidity remains constant; dew point temperature decreases

Explanation: In sensible heating: 1) Moisture content $d$ remains constant, so the partial pressure of water vapor $p_v$ remains unchanged. 2) Because dew point temperature $t_d$ is solely a function of $p_v$, $t_d$ remains constant. 3) Dry-bulb temperature $t$ increases, which drastically increases the saturated vapor pressure $p_{v,sat}(t)$. Consequently, relative humidity $\phi = p_v / p_{v,sat}(t)$ decreases.

6In a direct evaporative cooling process (direct spray with recirculated water without external heat input), what is the thermodynamic trajectory of the moist air state on the psychrometric chart?

A.Along a constant dry-bulb temperature line ($t = \text{const}$)

B.Along an approximate constant wet-bulb temperature (isenthapic $h \approx \text{const}$) line with decreasing dry-bulb temperature and increasing humidity

C.Along a constant relative humidity line ($\phi = \text{const}$)

D.Along a constant moisture content line ($d = \text{const}$)

Explanation: When water recirculates without external heating or cooling, the water temperature reaches the entering air wet-bulb temperature ($t_w$). As water evaporates, the latent heat of vaporization is supplied entirely by the sensible heat of the air. Therefore, the air dry-bulb temperature drops, moisture content increases, and the process proceeds along the constant wet-bulb / approximate isenthalpic line ($h \approx \text{const}$).

7When saturated dry steam at $100^\circ\text{C}$ is injected into an air stream for humidification (isothermal humidification / 等温加湿), what is the primary characteristic of the air state transformation on the psychrometric chart?

A.Dry-bulb temperature drops substantially while moisture content increases

B.Dry-bulb temperature remains nearly constant (slight increase) while moisture content and enthalpy increase

C.Enthalpy remains strictly constant while dry-bulb temperature rises

D.Relative humidity decreases while moisture content increases

Explanation: Steam carries both latent heat and its own sensible heat. Because the specific heat of steam is small compared to the latent heat of vaporization ($r_0 \approx 2501\text{ kJ/kg}$), injecting dry steam increases moisture content $d$ and total enthalpy $h$ with negligible change in dry-bulb temperature (typically $\Delta t < 0.5-1.0^\circ\text{C}$). The process direction angle $\\epsilon = \Delta h / \Delta d \approx 2600-2700\text{ kJ/kg}$.

8A reverse Carnot refrigeration cycle operates between an evaporating temperature of $t_e = 5^\circ\text{C}$ and a condensing temperature of $t_c = 35^\circ\text{C}$. What is the theoretical maximum Coefficient of Performance ($COP_{Carnot}$)?

A.8.27

B.9.27

C.7.27

D.0.14

Explanation: Absolute temperatures in Kelvin: $T_e = 5 + 273.15 = 278.15\text{ K}$, $T_c = 35 + 273.15 = 308.15\text{ K}$. Theoretical Carnot refrigeration COP is: $COP_{Carnot} = \frac{T_e}{T_c - T_e} = \frac{278.15}{308.15 - 278.15} = \frac{278.15}{30} = 9.2717 \approx 9.27$.

9In a single-stage vapor compression refrigeration cycle, introducing liquid refrigerant subcooling (过冷) before the expansion valve without changing evaporating or condensing pressures will produce which of the following thermodynamic effects?

A.Decreases specific refrigeration capacity and decreases COP

B.Increases specific refrigeration capacity, decreases vapor quality after throttling, and increases COP

C.Increases compressor discharge temperature and increases compression work

D.Decreases condensing heat rejection and leaves COP unchanged

Explanation: Subcooling the liquid refrigerant before throttling shifts the entering state of the expansion valve to the left on the p-h diagram. This reduces vapor flashing during expansion (lower vapor quality $x$), directly increases specific refrigeration effect ($q_0 = h_1 - h_4$), leaves specific compression work ($w = h_2 - h_1$) unchanged, and consequently increases the cycle COP ($COP = q_0 / w$).

10At standard atmospheric pressure ($P = 101.325\text{ kPa}$), moist air has a partial pressure of water vapor $p_v = 1.70\text{ kPa}$. Using the ideal gas humidity ratio equation $d = 0.622 \frac{p_v}{P - p_v}$, what is the moisture content $d$ of this air?

A.$0.0053\text{ kg/kg dry air}$

B.$0.0106\text{ kg/kg dry air}$ ($10.6\text{ g/kg}$)

C.$0.0168\text{ kg/kg dry air}$

D.$0.0212\text{ kg/kg dry air}$

Explanation: Using the moisture content formula: $d = 0.622 \times \frac{p_v}{P - p_v} = 0.622 \times \frac{1.70}{101.325 - 1.70} = 0.622 \times \frac{1.70}{99.625} = 0.622 \times 0.017064 = 0.01061\text{ kg/kg dry air} = 10.61\text{ g/kg dry air}$.

About the Registered Public Utility Engineer — HVAC (China) Exam

The National Qualification Examination for Registered Public Utility Engineers — HVAC (全国勘察设计注册公用设备工程师（暖通空调）执业资格考试) is administered under the MOHURD and MOHRSS survey-and-design framework. Its Foundation and Professional stages cover thermodynamics, psychrometrics, heating, air conditioning and refrigeration, ventilation, smoke control, and heat and cold sources.

Assessment

Foundation Examination: Public Fundamentals and Specialized Fundamentals objective papers. Professional Examination: two Professional Knowledge sessions and two Professional Case Analysis sessions over two days, totaling 200 and 100 points respectively.

Time Limit

Foundation: 8 hours total (4 hrs AM + 4 hrs PM); Professional: 12 hours total across 2 days (3 hrs AM + 3 hrs PM per day)

Passing Score

Foundation: 132/240 points (55%); Professional Knowledge: 120/200 points (60%); Professional Case Analysis: 60/100 points (60%)

Exam Fee

Set by the provincial examination authority; consult the current registration notice (Ministry of Housing and Urban-Rural Development (MOHURD) & Ministry of Human Resources and Social Security (MOHRSS))

Registered Public Utility Engineer — HVAC (China) Exam Content Outline

18%

Thermodynamics & Psychrometrics

Covers moist air thermodynamics and psychrometric state transformations: dry-bulb, wet-bulb, and dew-point temperatures, relative humidity $\phi$, enthalpy $h = 1.005 t + d(2501 + 1.86 t)$, mixing of two air streams, heating, surface cooling and dehumidification, adiabatic evaporative cooling, steam humidification, Sensible Heat Ratio ($SHR = Q_s / Q_t$), apparatus dew point (ADP), and cooling coil bypass factor ($BF$).

22%

Heating Engineering & Load Calculations

Focuses on GB 50736 building heat loss and heating systems: building envelope basic heat loss $Q = F K (t_n - t_{wn}) \alpha$, correction factors for orientation, height, and wind, infiltration heat loss calculations, radiator selection and surface area sizing ($Q = K F \Delta T$), low-temperature hot water radiant floor heating (max supply water temp 60°C, floor surface temperature limits), district heating network hydraulic calculations, pipe friction/local resistance, and expansion vessel sizing ($V_p = \alpha \Delta t V_c$).

26%

Air Conditioning Systems & Cooling Load Calculations

Encompasses cooling load calculations under GB 50736 (envelope solar radiation, harmonic reaction / transfer function methods, internal occupant, lighting, and equipment loads), Air Handling Unit (AHU) coil sizing, Constant Air Volume (CAV) and Variable Air Volume (VAV) systems, Fan Coil Unit plus Dedicated Outdoor Air System (FCU + DOAS), Variable Refrigerant Flow (VRF), chilled water piping network layout (direct vs reverse return), chilled water pump head calculation, and chiller energy metrics (COP and IPLV).

20%

Ventilation, Smoke Control & Cleanroom Technology

Addresses ventilation physics, building fire smoke safety, and clean environment engineering: natural ventilation neutral pressure plane, minimum outdoor air exchange rates under GB 50736, GB 51251 staircase and anteroom mechanical pressurization air volume calculation ($L_s = L_1 + L_2$), smoke exhaust air volume calculation per fire compartment, duct friction and dynamic resistance, fan affinity laws, and ISO 14644 / GB 50073 cleanroom airflow patterns, room air changes, and multi-stage HEPA filtration.

14%

Cold & Heat Sources and Building Energy Conservation

Details central cooling and heating plants and statutory energy standards: water-cooled centrifugal and screw chillers, air-source heat pumps (ASHP), ground-source heat pumps (GSHP / GB 50366), lithium bromide absorption chillers, ice and chilled water thermal energy storage, boiler plant design, and compliance with mandatory energy codes GB 55015-2021 and GB 50189-2015.

How to Pass the Registered Public Utility Engineer — HVAC (China) Exam

What You Need to Know

Passing score: Foundation: 132/240 points (55%); Professional Knowledge: 120/200 points (60%); Professional Case Analysis: 60/100 points (60%)
Assessment: Foundation Examination: Public Fundamentals and Specialized Fundamentals objective papers. Professional Examination: two Professional Knowledge sessions and two Professional Case Analysis sessions over two days, totaling 200 and 100 points respectively.
Time limit: Foundation: 8 hours total (4 hrs AM + 4 hrs PM); Professional: 12 hours total across 2 days (3 hrs AM + 3 hrs PM per day)
Exam fee: Set by the provincial examination authority; consult the current registration notice

Keys to Passing

Work through all 100 available questions
Review every answer and explanation
Track weak areas and revisit them
Use our AI tutor for tough concepts

Registered Public Utility Engineer — HVAC (China) Study Tips from Top Performers

1Master Psychrometric Chart Processes and Calculations: Practice fast computation of moist air enthalpy $h = 1.005 t + d(2501 + 1.86 t)$, mixing point coordinates ($t_m = (L_w t_w + L_n t_n)/(L_w + L_n)$), sensible heat ratio line slope, apparatus dew point (ADP), and supply air state determination.

2Memorize Mandatory Heating Design Limits: Pay close attention to GB 50736 radiant floor heating temperature limits (supply water $\le 60^\circ\text{C}$, ideal $35^\circ\text{C}-45^\circ\text{C}$, supply/return $\Delta t \le 10^\circ\text{C}$, floor surface temperature limits: $29^\circ\text{C}$ for occupied zones, $32^\circ\text{C}$ for bathrooms, $35^\circ\text{C}$ for perimeter areas).

3Internalize GB 51251 Smoke Control Formulas: Practice the calculation of positive pressure air volume for pressurized stairwells and anterooms ($L_s = L_1 + L_2$), door opening velocity criteria ($v \ge 0.7\text{ m/s}$ or $1.0\text{ m/s}$), and smoke exhaust rate calculations based on fire compartment area ($60\text{ m}^3/(\text{h}\cdot\text{m}^2)$) or smoke reservoir plume dynamics.

4Execute Hydronic & Pumping Calculations: Master Darcy-Weisbach friction calculations, specific frictional resistance $R_m$, equivalent length method for local fittings, expansion tank volume ($V_p = \alpha \Delta t V_c$), and pump affinity laws ($Q \propto n$, $H \propto n^2$, $P \propto n^3$).

5Apply Energy Conservation Standards (GB 55015 / GB 50189): Understand minimum chiller COP/IPLV thresholds, heat recovery requirements when outdoor air exhaust $\ge 3000\text{ m}^3/\text{h}$ with large enthalpy differences, energy consumption monitoring, and variable speed pumping controls.

6Structure Multi-Step Case Calculations: In open-book case problems, clearly state the applicable GB code clause, list design parameters, calculate intermediate values (such as mass flow rates $\dot{m} = Q / (c_p \Delta t)$), and double-check unit conversions (kW to W, m³/h to m³/s, kPa to Pa).

Frequently Asked Questions

What is the Registered Public Utility Engineer (HVAC) qualification and what is its practice scope?

The Registered Public Utility Engineer (Heating, Ventilation and Air Conditioning) (注册公用设备工程师(暖通空调)) is a national statutory professional engineering qualification in China governed by MOHURD and MOHRSS. Licensed HVAC engineers possess statutory authority to independently lead, execute, calculate, stamp, and take legal responsibility for HVAC, building thermal environment, smoke control, cleanroom, and energy conservation design documentation across civil and industrial building projects.

What is the format and structure of the HVAC examination?

The qualification has a closed-book Foundation Examination with Public and Specialized Fundamentals, followed by a two-day open-book Professional Examination with Professional Knowledge and Professional Case Analysis sessions. The professional totals are 200 points for Knowledge and 100 for Case Analysis.

What are the passing scores and score validity rules?

The current passing thresholds are 132/240 for Foundation, 120/200 for Professional Knowledge, and 60/100 for Professional Case Analysis. Both professional parts must be passed in the same examination year.

What key national codes and standards are most heavily tested?

The most critical national codes tested include GB 50736 (Design Code for Heating, Ventilation and Air Conditioning of Civil Buildings), GB 51251 (Technical Standard for Smoke Management Systems in Buildings), GB 55015 (General Code for Building Energy Conservation and Renewable Energy Utilization), GB 50189 (Design Standard for Energy Efficiency of Public Buildings), GB 50016 (Code for Fire Protection Design of Buildings), and GB 50073 (Design Code for Cleanroom).

Why is this OpenExamPrep practice bank presented in English?

This practice module provides 100 high-yield multiple-choice questions in English designed to build conceptual and quantitative calculation mastery for bilingual engineering practitioners, international HVAC engineers, and candidates preparing for Chinese engineering standards. It meticulously preserves official Chinese terminology, standard code designations (e.g., GB 50736, GB 51251, GB 55015), formula variables, and statutory criteria in parentheses for seamless cross-referencing.