Castleport Test Prep

Free PE Mechanical HVAC and Refrigeration Practice Test

These 40 original, unofficial questions follow the NCEES PE Mechanical: HVAC and Refrigeration specification effective October 2025, and every one has a worked solution. Use U.S. customary units and sea-level conditions unless a question says otherwise.

Practice questions

HVAC Loads and Psychrometrics (Questions 1–10)

Question 1 · Supply airflow after fan heat

A room has a sensible cooling load of 48,600 Btu/h and is held at 75°F. Air leaves the cooling coil at 53°F, and a downstream supply fan raises it 2°F. Ignore duct heat gain. Using q = 1.08 × cfm × ΔT, what supply airflow is required?

Answer for Question 1: Supply airflow after fan heat

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Answer and explanation

A. 2,250 cfm. Air reaches the room at 53 + 2 = 55°F, so ΔT = 75 − 55 = 20°F. Airflow = 48,600 ÷ (1.08 × 20) = 2,250 cfm. Size room airflow with the temperature of the air that actually enters the room. Using the colder coil-leaving temperature makes every cfm look like it removes more heat than it does. The 1.08 is 0.075 lb/ft³ × 0.24 Btu/lb·°F × 60 min/h, so it applies to standard air.

  • B Ignores the fan heat (ΔT = 22°F).
  • C Counts the fan heat twice (ΔT = 18°F).
  • D Leaves out ΔT, so the units aren't even cfm.

HVAC Loads and Psychrometrics · 1B Heating/Cooling Processes · PEHVAC-01. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, Thermal Components, sensible heat of air (1.08 basis); U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, HT-02 Eq. 2-15, Q = ṁcpΔT.

Question 2 · Envelope conduction loss

A building has 2,000 ft² of gross exterior wall, which includes 200 ft² of glazing. Opaque wall U = 0.05, glazing U = 0.30, and the 2,000 ft² roof has U = 0.03 (all Btu/h·ft²·°F). Indoor 70°F, outdoor 10°F. Ignoring infiltration and every other load, what is the conduction heat loss?

Answer for Question 2: Envelope conduction loss

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Answer and explanation

B. 12,600 Btu/h. Net opaque wall = 2,000 − 200 = 1,800 ft². UA = 1,800(0.05) + 200(0.30) + 2,000(0.03) = 90 + 60 + 60 = 210 Btu/h·°F. Loss = 210 × 60°F = 12,600 Btu/h.

  • A Uses the gross wall area, so the glazing area is counted twice.
  • C Leaves out the roof.
  • D Gives the glazing the wall's U-value.

HVAC Loads and Psychrometrics · 1A Heating/Cooling Loads · PEHVAC-02. Principle: U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, HT-02 pp. 9–10 (resistance and overall U); Eq. 2-3.

Question 3 · Humidity ratio from relative humidity

Air is at 75°F dry bulb, 50% relative humidity, and 14.696 psia. Saturation pressure of water at 75°F is 0.430 psia. Using W = 0.622 pw/(p − pw), what is the humidity ratio?

Answer for Question 3: Humidity ratio from relative humidity

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Answer and explanation

C. 0.0092 lb/lb. Vapor partial pressure pw = 0.50 × 0.430 = 0.215 psia. W = 0.622 × 0.215 ÷ (14.696 − 0.215) = 0.00924 lb water/lb dry air, about 64.7 grains/lb. The 0.622 is the ratio of the molar masses of water and dry air.

  • A Drops the 0.622.
  • B Uses saturation pressure, which is the 100% RH answer.
  • D Reports the vapor pressure as if it were W.

HVAC Loads and Psychrometrics · 1C Humidification/Dehumidification Processes · PEHVAC-03. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, Humidity parameters, Eq. (8) and molar-mass ratio 0.621945.

Question 4 · Latent ventilation load

1,000 cfm of outdoor air at W = 0.0160 lb/lb is brought into a space held at W = 0.0093 lb/lb. Using q_L = 4,840 × cfm × ΔW, what latent load does this air add?

Answer for Question 4: Latent ventilation load

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Answer and explanation

D. 32,400 Btu/h. q_L = 4,840 × 1,000 × (0.0160 − 0.0093) = 32,428 ≈ 32,400 Btu/h. The 4,840 is roughly 0.075 lb/ft³ × 60 min/h × 1,076 Btu/lb, where 1,076 is an approximate heat of vaporization.

  • A Uses the room humidity ratio by itself.
  • B Uses the outdoor humidity ratio by itself.
  • C Uses the grains-based constant (0.68) with ΔW in lb/lb.

HVAC Loads and Psychrometrics · 1A Heating/Cooling Loads · PEHVAC-04. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, Thermal Components, standard-air basis; ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, moist-air properties.

Question 5 · Mixed air and coil load

Return air at 6,000 lb dry air/h and h = 28 Btu/lb mixes adiabatically with outdoor air at 2,000 lb dry air/h and h = 38 Btu/lb. The mixture leaves the cooling coil at h = 23 Btu/lb. Neglect condensate enthalpy and casing losses. What is the total coil load?

Answer for Question 5: Mixed air and coil load

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Answer and explanation

C. 5.00 tons. Mix on a dry-air mass basis: h_mix = (6,000 × 28 + 2,000 × 38) ÷ 8,000 = 30.5 Btu/lb. Coil load = 8,000 × (30.5 − 23) = 60,000 Btu/h = 5.00 tons.

  • A Uses the return-air enthalpy alone.
  • B Averages 28 and 38 without weighting by mass (33 Btu/lb).
  • D Multiplies an hourly mass flow by 60 again.

HVAC Loads and Psychrometrics · 1B Heating/Cooling Processes · PEHVAC-05. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, adiabatic mixing of two moist airstreams; NIST Guide to the SI (SP 811), Appendix B.9, ton of refrigeration = 12,000 Btu/h.

Question 6 · Sensible heat at 5,000 ft

A heating coil at a 5,000-ft site raises 2,000 cfm (actual) of air by 20°F. Barometric pressure there is 12.23 psia; at sea level (14.696 psia) the sensible constant is 1.08. Treating air as an ideal gas at the same temperature, what is the coil output?

Answer for Question 6: Sensible heat at 5,000 ft

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Answer and explanation

A. 36,000 Btu/h. At the same temperature, density scales with pressure. Corrected constant = 1.08 × (12.23/14.696) = 0.899. q = 0.899 × 2,000 × 20 ≈ 35,950, about 36,000 Btu/h. The same cfm moves about 17% less air mass at this altitude.

  • B Keeps the sea-level constant.
  • C Divides by the pressure ratio instead of multiplying.
  • D Applies the correction twice.

HVAC Loads and Psychrometrics · 1B Heating/Cooling Processes (5,000-ft elevation) · PEHVAC-06. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, Table 1, standard atmospheric data (12.23 psia at 5,000 ft); NCEES PE Mechanical: HVAC and Refrigeration specification, effective October 2025, 1B lists 5,000-ft elevation processes.

Question 7 · Water removed by a cooling coil

4,000 cfm enters a cooling coil with specific volume 14.0 ft³/lb dry air and W = 0.0115 lb/lb. It leaves at W = 0.0080 lb/lb. There's no bypass or leakage, and all condensate drains. How much water does the coil remove, in lb/h? Round to the nearest 0.1.

Enter a number in lb/h, rounded to one decimal place.

Answer for Question 7: Water removed by a cooling coil

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Answer and explanation

60.0 lb/h. Dry-air flow = 4,000 × 60 ÷ 14.0 = 17,142.9 lb/h. Water removed = 17,142.9 × (0.0115 − 0.0080) = 60.0 lb/h. Use the specific volume you're given; that's what converts cfm at this state to dry-air mass. Accepted here: any entry from 59.95 up to (not including) 60.05.

  • 63.0 Uses the standard 4.5 factor instead of the given specific volume.
  • 1.0 Forgets to multiply by 60 min/h (that's lb/min).

HVAC Loads and Psychrometrics · 1C Humidification/Dehumidification Processes · PEHVAC-07. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, moisture balance on dry-air basis.

Question 8 · Coincident peak load

Three zones are served by one plant. At 10:00 their cooling loads are 3, 4, and 2 tons. At 14:00 they're 5, 3, and 2 tons. These two hours govern the design, and the loads include all allowances. What plant capacity is needed for these states?

Answer for Question 8: Coincident peak load

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Answer and explanation

D. 10 tons. Add the zones hour by hour, then take the largest total: 9 tons at 10:00 and 10 tons at 14:00, so 10 tons. Zones don't peak at the same time, so the plant needs the largest simultaneous sum, not the sum of each zone's peak.

  • A Adds each zone's own peak (5 + 4 + 2), which never happen together.
  • B Uses the 10:00 total.
  • C Sizes for the largest single zone.

HVAC Loads and Psychrometrics · 1A Heating/Cooling Loads · PEHVAC-08. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 18 Nonresidential Cooling and Heating Load Calculations, §1.1 Terminology: largest hourly sum of simultaneous zone loads.

Question 9 · Cooling then reheat

Air goes from 80°F, W = 0.012 through a cooling coil to 55°F, W = 0.008, then through a dry sensible reheat coil to 65°F, W = 0.008, all at the same pressure. Select all that apply.

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

Answer for Question 9: Cooling then reheat

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Answer and explanation

A, C, D. W falls from 0.012 to 0.008 across the cooling coil, so it condenses water (A). Dry reheat adds sensible heat only, so W stays 0.008 and enthalpy rises (D). Same W at the same pressure means the same vapor pressure and the same dew point (C). Relative humidity drops during reheat, but that isn't moisture removal.

  • B is false. Lower relative humidity after reheat is not removed water; W is unchanged.
  • E is false. W stays at 0.008; dry bulb rising doesn't add moisture.

HVAC Loads and Psychrometrics · 1C Humidification/Dehumidification Processes · PEHVAC-09. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, humidity ratio and dew point definitions; sensible heating at constant W.

Question 10 · Direct evaporative cooling

A direct evaporative cooler receives air at 95°F dry bulb, 65°F wet bulb. Its saturation effectiveness, (T_db,in − T_db,out)/(T_db,in − T_wb,in), is 85%. What leaving dry-bulb temperature should you expect?

Answer for Question 10: Direct evaporative cooling

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Answer and explanation

B. 69.5°F. T_out = 95 − 0.85 × (95 − 65) = 69.5°F. The process is close to adiabatic saturation, so wet bulb stays near 65°F while humidity ratio climbs.

  • A Applies (1 − ε) to the temperature drop.
  • C Assumes a perfect 100%-effective cooler.
  • D Takes 85% of the entering dry bulb.

HVAC Loads and Psychrometrics · 1C Humidification/Dehumidification Processes · PEHVAC-10. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, adiabatic saturation and thermodynamic wet bulb.

HVAC and Refrigeration Distribution and Systems (Questions 11–21)

Question 11 · Total duct system loss

At design flow, 200 ft of straight duct loses 0.08 in. wg per 100 ft. The fittings all reference V = 2,000 fpm and have a combined loss coefficient K = 2.4. The filter loses 0.45 in. wg and the coil 0.70 in. wg. Velocity pressure is VP = (V/4,005)². Ignoring anything else, what is the total pressure loss?

Answer for Question 11: Total duct system loss

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Answer and explanation

D. 1.91 in. wg. Straight duct: 2 × 0.08 = 0.16. VP = (2,000/4,005)² = 0.249, so fittings = 2.4 × 0.249 = 0.60. Total = 0.16 + 0.60 + 0.45 + 0.70 = 1.91 in. wg. A loss coefficient multiplies velocity pressure; it isn't a pressure by itself.

  • A Adds K = 2.4 as if it were in. wg.
  • B Uses one velocity pressure instead of K × VP.
  • C Counts only 100 ft of straight duct.

HVAC and Refrigeration Distribution and Systems · 2A Air Distribution Systems and Ductwork · PEHVAC-11. Principle: U.S. DOE/AMCA, Improving Fan System Performance, p. 8, Δp = C(V/1,097)²ρ; 1,097/√0.075 ≈ 4,005.

Question 12 · Zone outdoor airflow

An office zone has 25 people and 2,500 ft². Use Rp = 5 cfm/person, Ra = 0.06 cfm/ft², and zone air distribution effectiveness Ez = 0.8. What is the zone outdoor airflow, Voz?

Answer for Question 12: Zone outdoor airflow

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Answer and explanation

A. 344 cfm. Breathing-zone airflow Vbz = Rp·Pz + Ra·Az = 5(25) + 0.06(2,500) = 275 cfm. Voz = Vbz ÷ Ez = 275 ÷ 0.8 = 344 cfm. Poorer distribution means you must deliver more outdoor air to get the same air to the breathing zone.

  • B Stops at Vbz.
  • C Multiplies by Ez instead of dividing.
  • D Counts only the people component.

HVAC and Refrigeration Distribution and Systems · 2B Air Quality and Ventilation · PEHVAC-12. Principle: WBDG/UFC 3-410-01, ASHRAE Standard 62.1 IAQ Process Flow Chart, §6.2.1.1 and §6.2.1.3 (ASHRAE 62.1 Ventilation Rate Procedure).

Question 13 · Pump head for a closed loop

A completely filled closed chilled-water loop serves an 80-ft-tall building. Common supply and return piping loses 18 ft of water at full flow. Three parallel terminal branches need 22, 30, and 26 ft at their design flows; the easier branches will be balanced. What minimum pump differential head is required?

Answer for Question 13: Pump head for a closed loop

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Answer and explanation

C. 48 ft. Pump head = common losses + the most demanding parallel path = 18 + 30 = 48 ft. The other branches get 8 and 4 ft of balancing loss. In a closed, filled loop the water going up is balanced by water coming down, so the pump doesn't supply static lift.

  • A Adds the 80-ft building height.
  • B Adds all three parallel branches as if they were in series.
  • D Leaves out the common piping.

HVAC and Refrigeration Distribution and Systems · 2C Fluid Distribution Systems and Piping · PEHVAC-13. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, closed systems: pumps do not provide static lift; U.S. DOE/Hydraulic Institute, Improving Pumping System Performance, 2nd ed., p. 8, closed-loop systems are friction-dominated.

Question 14 · Chilled-water flow rate

A cooling coil removes 240,000 Btu/h using water that enters at 44°F and leaves at 56°F. Using gpm = q ÷ (500 × ΔT), what water flow is needed, in gpm? Round to the nearest 0.1.

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

Answer for Question 14: Chilled-water flow rate

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Answer and explanation

40.0 gpm. gpm = 240,000 ÷ (500 × 12) = 40.0. The 500 comes from about 8.33 lb/gal × 60 min/h × 1 Btu/lb·°F, so it's for water. A glycol mix needs its own density and specific heat. Accepted here: any entry from 39.95 up to (not including) 40.05.

  • 18,518.5 Uses 1.08, the standard-air constant, in place of 500.

HVAC and Refrigeration Distribution and Systems · 2C Fluid Distribution Systems and Piping · PEHVAC-14. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, Heat transferred to or from water, ≈500 constant.

Question 15 · Steam coil condensate

A steam coil delivers 400,000 Btu/h. Steam enters saturated and leaves as saturated condensate. Latent heat at the coil pressure is 945 Btu/lb. How much condensate does the coil produce?

Answer for Question 15: Steam coil condensate

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Answer and explanation

B. 423 lb/h. Only latent heat is released, so ṁ = 400,000 ÷ 945 = 423 lb/h. This is the steady condensate rate from the stated coil load; actual trap and return-system selection also depends on operating and startup conditions.

  • A Divides by total vapor enthalpy (about 1,160 Btu/lb).
  • C Uses a 1,000 Btu/lb rule of thumb.
  • D Gives the per-minute value but calls it per hour.

HVAC and Refrigeration Distribution and Systems · 2C Fluid Distribution Systems and Piping (steam/condensate) · PEHVAC-15. Principle: U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, HT-02 p. 36, Q = ṁΔh.

Question 16 · Control valve Cv

A coil branch needs 30 gpm of water. The branch has 10 psi available, and the piping and coil use 6 psi at that flow, so the rest is allocated to the control valve. With Cv = Q√(SG/Δp) and SG = 1, what Cv is required?

Answer for Question 16: Control valve Cv

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Answer and explanation

B. 15.0. Valve pressure drop = 10 − 6 = 4 psi. Cv = 30 × √(1/4) = 15.0.

  • A Gives all 10 psi to the valve.
  • C Uses the 6-psi coil and piping drop.
  • D Divides by 4 instead of its square root.

HVAC and Refrigeration Distribution and Systems · 2D Control Concepts · PEHVAC-16. Principle: the Cv relationship is given in the question.

Question 17 · Removing control offset

A proportional-only discharge-air controller settles 2°F below setpoint under a steady load. Which change removes the steady-state offset?

Answer for Question 17: Removing control offset

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Answer and explanation

D. Add integral action. Integral action keeps adding up error until the error is zero, so the output settles at setpoint.

  • A Higher gain shrinks the offset but doesn't remove it, and can make the loop hunt.
  • B Derivative reacts to how fast the error changes, not to a steady error.
  • C A wider deadband accepts more error.

HVAC and Refrigeration Distribution and Systems · 2D Control Concepts · PEHVAC-17. Principle: Åström & Murray, Feedback Systems (FBSwiki), PID Control, PID Control: integral action gives zero steady-state error.

Question 18 · Supply-air temperature reset

A VAV zone has a fixed sensible cooling load. The room is 75°F and q = 1.08 × cfm × (T_room − T_supply). Supply air temperature is reset from 55°F to 60°F. Select all that apply.

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

Answer for Question 18: Supply-air temperature reset

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Answer and explanation

A, C, D. ΔT drops from 20°F to 15°F, so cfm rises by 20/15 = 1.33 (A). A reheating zone starts from warmer air, so it needs less reheat (C). Fan power grows with airflow, roughly with the cube on a fixed system (D).

  • B is false. For a fixed load, a smaller ΔT means more air, not less.
  • E is false. A warmer coil leaving temperature generally removes less moisture.

HVAC and Refrigeration Distribution and Systems · 2D Control Concepts (temperature reset) · PEHVAC-18. Principle: U.S. DOE/AMCA, Improving Fan System Performance, pp. 4–5, fan laws; ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, sensible heat of air.

Question 19 · Primary-secondary mixing

A primary loop delivers 60 gpm at 44°F to a common pipe with negligible loss. The secondary loop draws 80 gpm, and its return water is 56°F. At steady state with equal water properties, what is the secondary supply temperature?

Answer for Question 19: Primary-secondary mixing

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Answer and explanation

C. 47°F. The secondary pulls 20 gpm more than the primary supplies, so 20 gpm of secondary return flows backward through the common pipe and mixes into the supply. T = (60 × 44 + 20 × 56) ÷ 80 = 47°F.

  • A Averages 44 and 56 without weighting by flow.
  • B Reverses the flows (60 at 56°F, 20 at 44°F).
  • D Assumes the secondary still gets 44°F water despite the shortfall.

HVAC and Refrigeration Distribution and Systems · 2C Fluid Distribution Systems and Piping · PEHVAC-19. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, compound (primary-secondary) pumping and common pipe; U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, energy balance.

Question 20 · Refrigerant mass flow

An ideal R-134a cycle runs at 20°F evaporating and 100°F condensing. h₁ = 106.0 Btu/lb (saturated vapor leaving the evaporator), h₂ = 118.8 Btu/lb (after isentropic compression), and h₃ = h₄ = 45.1 Btu/lb (saturated liquid, then throttled). What refrigerant flow does a 50-ton load need?

Answer for Question 20: Refrigerant mass flow

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Answer and explanation

A. 164 lb/min. Refrigerating effect = h₁ − h₄ = 60.9 Btu/lb. One ton is 200 Btu/min, so flow = 50 × 200 ÷ 60.9 = 164 lb/min. For reference, COP = 60.9 ÷ (118.8 − 106.0) = 4.76.

  • B Uses h₂ − h₄, which is condenser heat per pound.
  • C Divides by compressor work per pound.
  • D Gives lb/h but labels it lb/min.

HVAC and Refrigeration Distribution and Systems · 2E Refrigeration and Refrigeration Systems · PEHVAC-20. Principle: NIST Guide to the SI (SP 811), Appendix B.9, ton of refrigeration = 12,000 Btu/h; U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, p. 36, Q = ṁΔh.

Question 21 · Refrigerant safety class

R-454B is classified A2L under ASHRAE Standard 34. Select all that apply.

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

Answer for Question 21: Refrigerant safety class

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Answer and explanation

A, B, D. The letter is toxicity: A lower, B higher. The number is flammability: 1 no flame propagation, 2 lower flammability, 2L a subclass that burns very slowly, 3 highly flammable. A2L refrigerants are flammable, just mildly.

  • C is false. A2L refrigerants can burn; class 1 is the no-flame-propagation class.
  • E is false. B is the higher-toxicity class, not a flammability class.

HVAC and Refrigeration Distribution and Systems · 2E Refrigeration and Refrigeration Systems (refrigerants) · PEHVAC-21. Principle: ASHRAE Handbook—Refrigeration (2026), Ch. 1 Halocarbon and Hydrocarbon Refrigeration Systems, Standard 34 toxicity/flammability classes and R-454B = A2L.

HVAC Equipment and Components (Questions 22–35)

Question 22 · Cooling tower blowdown

A cooling tower evaporates 12 gpm and runs at 4 cycles of concentration. Treat blowdown as all non-evaporative losses. What blowdown rate is needed, in gpm? Round to the nearest 0.1.

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

Answer for Question 22: Cooling tower blowdown

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Answer and explanation

4.0 gpm. Blowdown = evaporation ÷ (cycles − 1) = 12 ÷ 3 = 4.0 gpm. Make-up = evaporation + blowdown = 16 gpm. More cycles means less blowdown and make-up, until scaling and corrosion limit you. Accepted here: any entry from 3.95 up to (not including) 4.05.

  • 3.0 Divides by cycles instead of (cycles − 1).
  • 16.0 Gives the make-up flow instead.

HVAC Equipment and Components · 3A Cooling Towers and Fluid Coolers · PEHVAC-22. Principle: U.S. DOE FEMP, Cooling Towers: Understanding Key Components, p. 3, Eqs. 1 and 6.

Question 23 · Condenser water flow from range and approach

A chiller's evaporator load is 40 tons, and its compressor adds 38 kW to the refrigerant. All of that heat goes to the tower. Water enters the tower at 95°F, the entering-air wet bulb is 76°F, and the tower's approach (leaving water minus wet bulb) is 6°F. Using 500 × gpm × range, what condenser water flow is needed? Use 1 kW = 3,412 Btu/h.

Answer for Question 23: Condenser water flow from range and approach

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Answer and explanation

D. 93.8 gpm. Heat rejected = 40 × 12,000 + 38 × 3,412 = 609,656 Btu/h. Leaving water = 76 + 6 = 82°F, so the range (entering minus leaving water) = 95 − 82 = 13°F. Flow = 609,656 ÷ (500 × 13) = 93.8 gpm.

  • A Rejects only the evaporator load.
  • B Uses the 6°F approach as if it were the range.
  • C Uses 95 − 76 = 19°F as the range.

HVAC Equipment and Components · 3A Cooling Towers and Fluid Coolers · PEHVAC-23. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 40 Cooling Towers, definitions of range and approach; NIST Guide to the SI (SP 811), Appendix B.9, ton and kW conversions.

Question 24 · Boiler fuel flow

A gas boiler has 2,000 MBH input and 82% efficiency. The natural gas heating value is 1,030 Btu/ft³. What gas flow does the boiler burn?

Answer for Question 24: Boiler fuel flow

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Answer and explanation

B. 1,942 ft³/h. Fuel flow follows input, not output: 2,000,000 ÷ 1,030 = 1,942 ft³/h. Output is 0.82 × 2,000 = 1,640 MBH.

  • A Divides the output by the heating value.
  • C Divides the input by efficiency a second time.
  • D Gives the per-minute value but calls it per hour.

HVAC Equipment and Components · 3B Boilers and Furnaces · PEHVAC-24. Principle: U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, energy balance.

Question 25 · Counterflow area with LMTD

In a counterflow plate heat exchanger, the hot side goes from 200°F to 120°F and the cold side from 100°F to 110°F. The duty is 1,000,000 Btu/h and U = 250 Btu/h·ft²·°F. What heat transfer area is required?

Answer for Question 25: Counterflow area with LMTD

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Answer and explanation

C. 86.0 ft². Counterflow end differences: 200 − 110 = 90°F and 120 − 100 = 20°F. LMTD = (90 − 20) ÷ ln(90/20) = 46.5°F. Area = 1,000,000 ÷ (250 × 46.5) = 86.0 ft².

  • A Uses the arithmetic mean of 55°F, which overstates ΔT when the end differences are this far apart.
  • B Uses parallel-flow end differences (100 and 10°F).
  • D Uses the largest end difference, 90°F.

HVAC Equipment and Components · 3C Heat Exchangers · PEHVAC-25. Principle: U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, HT-02 p. 3 Eq. 2-2 and pp. 36–37.

Question 26 · Heat pump with resistance backup

At the outdoor design condition, a building needs 90,000 Btu/h of heat. The heat pump supplies 66,000 Btu/h at a heating COP of 2.75. Electric resistance supplies the rest at COP = 1. Use 1 kW = 3,412 Btu/h. What is the combined electrical demand?

Answer for Question 26: Heat pump with resistance backup

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Answer and explanation

A. 14.07 kW. Heat pump input = 66,000 ÷ 2.75 = 24,000 Btu/h. Resistance makes up 90,000 − 66,000 = 24,000 Btu/h at COP 1. Total = 48,000 ÷ 3,412 = 14.07 kW. The system's overall COP is 90,000 ÷ 48,000 = 1.875, not 2.75.

  • B Runs the whole load at the heat pump's COP.
  • C Leaves out the resistance backup.
  • D Converts the whole thermal load to kW at COP 1.

HVAC Equipment and Components · 3D Condensers/Evaporators (heat pumps) · PEHVAC-26. Principle: NIST Guide to the SI (SP 811), Appendix B.9, Btu/h–watt conversion; U.S. DOE Fundamentals Handbook, Thermodynamics, Heat Transfer, and Fluid Flow, Vol. 2, energy balance.

Question 27 · Chiller kW/ton to COP

A chiller is rated at 0.55 kW/ton. What is its COP?

Answer for Question 27: Chiller kW/ton to COP

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Answer and explanation

D. 6.39. One ton is 3.517 kW of cooling, so COP = 3.517 ÷ 0.55 = 6.39.

  • A Takes 1 ÷ 0.55.
  • B That's EER (12 ÷ 0.55, in Btu/W·h), a different unit.
  • C Divides the other way.

HVAC Equipment and Components · 3D Condensers/Evaporators (chillers) · PEHVAC-27. Principle: NIST Guide to the SI (SP 811), Appendix B.9, ton of refrigeration = 3.516853 kW.

Question 28 · Ice storage mass

An ice storage tank must supply 2,000 ton-hours using latent heat only. The latent heat of fusion is 144 Btu/lb. How much ice is needed?

Answer for Question 28: Ice storage mass

Show answer and explanation

Answer and explanation

C. 166,700 lb. 2,000 ton-h × 12,000 Btu/ton-h = 24,000,000 Btu. Ice = 24,000,000 ÷ 144 = 166,667 lb.

  • A Uses 1,000 Btu per ton-hour.
  • B Reports Btu as pounds.
  • D Uses 288 Btu/lb.

HVAC Equipment and Components · 3D Condensers/Evaporators (thermal storage) · PEHVAC-28. Principle: NIST Guide to the SI (SP 811), Appendix B.9, ton of refrigeration = 12,000 Btu/h.

Question 29 · Fan laws

A fan delivers 10,000 cfm at 900 rpm and draws 8.0 bhp. On the same system, airflow must rise to 11,000 cfm. What new brake horsepower should you expect?

Answer for Question 29: Fan laws

Show answer and explanation

Answer and explanation

B. 10.6 bhp. On the same system, flow scales with speed, so speed rises 10% to 990 rpm. Power scales with speed cubed: 8.0 × 1.1³ = 10.6 bhp. Pressure rises by 1.1² = 1.21.

  • A Scales power linearly.
  • C Uses the square law, which is for pressure.
  • D Inverts the square ratio.

HVAC Equipment and Components · 3E Pumps, Compressors, and Fans (laws) · PEHVAC-29. Principle: U.S. DOE/AMCA, Improving Fan System Performance, pp. 4–5, fan laws.

Question 30 · Pump operating point

A pump curve is H = 60 − 0.001Q², and an open system's curve is H = 10 + 0.004Q², with H in ft and Q in gpm. At what flow will the pump operate, in gpm? Round to the nearest 0.1.

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

Answer for Question 30: Pump operating point

Show answer and explanation

Answer and explanation

100.0 gpm. The pump runs where the curves cross: 60 − 0.001Q² = 10 + 0.004Q², so 50 = 0.005Q² and Q = 100.0 gpm. Head there is 50 ft. Accepted here: any entry from 99.95 up to (not including) 100.05.

  • 109.5 Drops the system's 10-ft static head.
  • 129.1 Subtracts the coefficients instead of adding them.

HVAC Equipment and Components · 3E Pumps, Compressors, and Fans (curves) · PEHVAC-30. Principle: U.S. DOE/Hydraulic Institute, Improving Pumping System Performance, 2nd ed., Appendix C tip sheet: pump and system curve intersection.

Question 31 · Pump motor input

A pump moves 300 gpm of water against 80 ft of total head. Pump efficiency is 75% and motor efficiency 92%. What is the motor's electrical input?

Answer for Question 31: Pump motor input

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Answer and explanation

A. 6.55 kW. Fluid hp = 300 × 80 × 1.0 ÷ 3,960 = 6.06 hp. Brake hp = 6.06 ÷ 0.75 = 8.08 hp. Input = 8.08 × 0.7457 ÷ 0.92 = 6.55 kW.

  • B Converts fluid horsepower to kW.
  • C Skips the motor efficiency.
  • D Leaves the answer in horsepower.

HVAC Equipment and Components · 3E Pumps, Compressors, and Fans · PEHVAC-31. Principle: U.S. DOE/Hydraulic Institute, Improving Pumping System Performance, 2nd ed., p. 9, fluid power = HQ(s.g.)/3,960; U.S. DOE, Determining Electric Motor Load and Efficiency, p. 3, Eq. 2.

Question 32 · NPSH available

A pump draws 160°F water from an open tank at 14.7 psia. The water's vapor pressure is 4.75 psia and its density 61.0 lb/ft³. The tank surface is 5 ft above the impeller centerline, and suction friction is 3 ft. What is the net positive suction head available?

Answer for Question 32: NPSH available

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Answer and explanation

D. 25.5 ft. Pressure head above vapor pressure = (14.7 − 4.75) × 144 ÷ 61.0 = 23.5 ft. Add 5 ft of static head and subtract 3 ft of friction: NPSHA = 25.5 ft. Compare that with the pump's NPSHR; DOE notes a common design rule of keeping NPSHA about 25% above NPSHR.

  • A Ignores vapor pressure.
  • B Treats the 5 ft as a suction lift.
  • C Adds friction instead of subtracting it.

HVAC Equipment and Components · 3E Pumps, Compressors, and Fans (NPSH) · PEHVAC-32. Principle: U.S. DOE/Hydraulic Institute, Improving Pumping System Performance, 2nd ed., pp. 20–21 and 38, cavitation and NPSH.

Question 33 · Cooling coil capacity

A coil cools 8,000 cfm of standard air from 80°F db/67°F wb (h ≈ 31.5 Btu/lb) to 55°F db/54°F wb (h ≈ 22.6 Btu/lb). Using q = 4.5 × cfm × Δh, what is the coil load?

Answer for Question 33: Cooling coil capacity

Show answer and explanation

Answer and explanation

B. 26.7 tons. q = 4.5 × 8,000 × (31.5 − 22.6) = 320,400 Btu/h = 26.7 tons. The 4.5 is 0.075 lb/ft³ × 60 min/h.

  • A Counts only sensible heat (1.08 × 8,000 × 25), missing the latent part.
  • C Uses the entering enthalpy instead of the difference.
  • D Reports MBH as tons.

HVAC Equipment and Components · 3F Cooling and Heating Coils · PEHVAC-33. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 13 Hydronic Heating and Cooling, total heat of air (4.5 basis); ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, moist-air enthalpy.

Question 34 · Heat recovery plus heating coil

A sensible-only heat recovery device has equal stream heat-capacity rates and 70% effectiveness. Outdoor air enters at 10°F and exhaust at 70°F. The 2,000 cfm of outdoor air must then be heated to 65°F; use 1.08 for this air. Ignoring fan heat, leakage, and frost, what heating does the downstream coil need?

Answer for Question 34: Heat recovery plus heating coil

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Answer and explanation

C. 28,080 Btu/h. Recovered outdoor air temperature = 10 + 0.70 × (70 − 10) = 52°F. Coil heat = 1.08 × 2,000 × (65 − 52) = 28,080 Btu/h.

  • A Ignores the recovery device (55°F rise).
  • B Applies (1 − ε), giving 28°F leaving the device.
  • D Heats to the 70°F exhaust temperature instead of 65°F.

HVAC Equipment and Components · 3G Energy Recovery · PEHVAC-34. Principle: ASHRAE Handbook—HVAC Systems and Equipment (2024), Ch. 26 Air-to-Air Energy Recovery Equipment, Effectiveness, Eqs. (1), (2a), (4a), and ideal assumptions.

Question 35 · Coil capacity screen

At the design air and water conditions, a coil must provide at least 72,000 Btu/h sensible and 24,000 Btu/h latent. Four candidates list total/sensible capacity in kBtu/h: A 100/64, B 105/80, C 100/82, D 90/65. Which candidate meets both requirements?

Answer for Question 35: Coil capacity screen

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Answer and explanation

B. Candidate B. Latent = total − sensible: A 36, B 25, C 18, D 25 kBtu/h. Only B meets both 72 sensible and 24 latent. This is a capacity screen; a real selection also checks leaving conditions, rows, and water-side performance.

  • A Has plenty of latent but only 64 sensible.
  • C Meets sensible, but its latent is only 18.
  • D Short on sensible (65) and total (90 vs. 96).

HVAC Equipment and Components · 3F Cooling and Heating Coils (selection) · PEHVAC-35. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 1 Psychrometrics, sensible and latent components.

Supportive Knowledge (Questions 36–40)

Question 36 · Net present value of an upgrade

An upgrade costs $18,000 more up front. It saves $4,000/yr in energy but adds $500/yr in maintenance, both at year-end for 10 years. At 6%, P/A = 7.3601. Ignore taxes, escalation, and salvage. What is the incremental net present value?

Answer for Question 36: Net present value of an upgrade

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Answer and explanation

D. +$7,760. Net annual savings = 4,000 − 500 = $3,500. Present value = 3,500 × 7.3601 = $25,760. NPV = 25,760 − 18,000 = +$7,760.

  • A Ignores the added maintenance.
  • B Skips discounting (10 × 3,500 − 18,000).
  • C Treats the first cost as an annual cost.

Supportive Knowledge · 4C Economic Analysis · PEHVAC-36. Principle: NIST Guide to the SI (SP 811), Appendix B.9, standard present-worth factor P/A = [(1+i)ⁿ − 1]/[i(1+i)ⁿ]; value given in stem.

Question 37 · Three-phase motor current

A 25-hp, 460-V three-phase motor runs at rated load with 92% efficiency and a 0.86 power factor. What is the line current?

Answer for Question 37: Three-phase motor current

Show answer and explanation

Answer and explanation

A. 29.6 A. Input = 25 × 0.7457 ÷ 0.92 = 20.26 kW. I = 20,264 ÷ (√3 × 460 × 0.86) = 29.6 A.

  • B Leaves out √3.
  • C Leaves out power factor.
  • D Leaves out efficiency.

Supportive Knowledge · 4D Electrical Concepts · PEHVAC-37. Principle: U.S. DOE, Determining Electric Motor Load and Efficiency, p. 3, Eqs. 1–2.

Question 38 · Motor heat into the airstream

A 10-hp fan motor, 90% efficient, runs at rated load. Both the motor and fan sit in the airstream. How much heat do they add to the air?

Answer for Question 38: Motor heat into the airstream

Show answer and explanation

Answer and explanation

B. 28,300 Btu/h. With the motor in the airstream, all electrical input ends up as heat in the air: 10 × 2,544 ÷ 0.90 = 28,270 ≈ 28,300 Btu/h. If the motor sat outside the airstream, only the shaft power, 25,440 Btu/h, would reach the air.

  • A Counts shaft power only (motor outside the airstream).
  • C Counts only the motor's losses.
  • D Multiplies by efficiency instead of dividing.

Supportive Knowledge · 4D Electrical Concepts (heat output) · PEHVAC-38. Principle: NIST Guide to the SI (SP 811), Appendix B.9, hp = 745.7 W; Btu/h = 0.2930711 W.

Question 39 · Combining sound levels

Three sources measured at the same point produce 80 dB, 80 dB, and 83 dB. Assume they're uncorrelated. What is the combined level?

Answer for Question 39: Combining sound levels

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Answer and explanation

D. 86 dB. Add the energy, not the decibels: L = 10 log₁₀(10⁸ + 10⁸ + 10^8.3) = 10 log₁₀(4.0 × 10⁸) = 86.0 dB. Two equal sources add about 3 dB.

  • A Adds decibels arithmetically.
  • B Averages them.
  • C Keeps only the loudest source.

Supportive Knowledge · 4B Acoustics and Vibration Control · PEHVAC-39. Principle: ASHRAE Handbook—Fundamentals (2025), Ch. 8 Sound and Vibration, Combining Sound Levels, Eq. (10); OSHA Technical Manual, Sec. III Ch. 5 Noise, §II.B.5, decibels are logarithmic.

Question 40 · Spring isolation efficiency

A 900-rpm fan sits on spring isolators with 1.0 in. of static deflection. Ignoring damping, about what isolation efficiency do you get?

Answer for Question 40: Spring isolation efficiency

Show answer and explanation

Answer and explanation

C. 95%. Natural frequency fn = (1/2π)√(g/δ) = 3.13 ÷ √1.0 = 3.13 Hz. Forcing frequency = 900/60 = 15 Hz, so the ratio r = 4.79. Transmissibility = 1 ÷ (r² − 1) = 0.046, about 95% isolation. Isolation only starts once r is above √2; real isolators have some damping, so use the manufacturer's data for selection.

  • A Uses 1 − 1/r.
  • B Reports the transmissibility as the efficiency.
  • D Would only be true if r were below √2.

Supportive Knowledge · 4B Acoustics and Vibration Control · PEHVAC-40. Principle: OpenStax University Physics Vol. 1, §15.1 Simple Harmonic Motion, mass-spring natural frequency.

Review your answers

Count your results against all 40 questions: number correct, number wrong, and number left blank. If you worked only one topic area, score that area against its own question count. A "select all" question counts only if you picked every correct choice and nothing else, which is how NCEES scores alternative item types too.

This result describes your work on these 40 questions. It isn't an NCEES score and it doesn't predict whether you'll pass.

The useful part is what you do with the misses:

  1. Rework each miss before rereading the solution. Cover the explanation and try again from the givens.
  2. Find the first wrong step and name it. Most misses here fall into a few buckets: wrong boundary (coil vs. room, input vs. output), wrong basis (dry-air mass, actual vs. standard air, absolute vs. gauge), units, a wrong model (square law vs. cube law, LMTD vs. average), or arithmetic.
  3. Prove the fix on a fresh problem. Getting the same question right the second time mostly shows you remember it.

Copy this log into your notes:

Review your answers
QuestionMy answerFirst wrong stepCauseFix in one lineFresh problem done?
Example: Q12,045 cfmUsed 53°F coil air instead of 55°F room supplyBoundarySize room airflow at the air that enters the room

Log lucky guesses too. A right answer you can't explain is a miss waiting to happen.

Use the completed log to build your next study block around the mistakes that repeat most often.

What this practice test covers

NCEES groups the exam into four knowledge areas, each with a range of questions. We sized each area of this 40-question set from the midpoint of its official range. Your real exam can land anywhere inside each range.

What this practice test covers
Knowledge areaQuestions on the real exam (of 80)Questions in this set
1. HVAC Loads and Psychrometrics18–2710 (Q1–Q10)
2. HVAC and Refrigeration Distribution and Systems20–3011 (Q11–Q21)
3. HVAC Equipment and Components24–3614 (Q22–Q35)
4. Supportive Knowledge8–125 (Q36–Q40)

Source: NCEES HVAC and Refrigeration specification, effective October 2025, pp. 1–2.

Every area is sampled, but 40 questions can't hit every subtopic. Food storage, VRF systems, and codes interpretation get little or no direct coverage here.

The real exam also uses point-and-click and drag-and-drop items. This set includes the two other alternative formats NCEES lists: select-all-that-apply (Q9, Q18, Q21) and enter-a-number (Q7, Q14, Q22, Q30).

Constants used in these questions

These shortcuts save time, but each one carries an assumption. When a problem gives different properties, use those.

Constants used in these questions
ConstantUsed forWhere it comes from
1.08Sensible heat of air, Btu/h = 1.08 × cfm × ΔT (°F)0.075 lb/ft³ × 0.24 Btu/lb·°F × 60 min/h
4,840Latent heat of air, Btu/h = 4,840 × cfm × ΔW (lb/lb)0.075 × 60 × about 1,076 Btu/lb
4.5Total heat of air, Btu/h = 4.5 × cfm × Δh (Btu/lb)0.075 × 60
500Water, Btu/h = 500 × gpm × ΔTabout 8.33 lb/gal × 60 min/h × 1 Btu/lb·°F
4,005Velocity pressure, in. wg = (fpm/4,005)²standard air at 0.075 lb/ft³
3,960Pump fluid horsepower = gpm × ft × SG ÷ 3,960unit conversion to hp
6,362Fan horsepower = cfm × in. wg ÷ (6,362 × efficiency)unit conversion to hp
12,000Btu/h per ton of refrigeration (3.517 kW)definition
3,412Btu/h per kWunit conversion
2,544Btu/h per horsepower (0.7457 kW)unit conversion

The air constants are for standard air at sea level. At altitude, scale them by the density ratio, as in Q6. Sources: ASHRAE Handbook—HVAC Systems and Equipment, Ch. 13 (air and water constants and their property assumptions); DOE fan sourcebook, pp. 6 and 8; DOE pump sourcebook, p. 9; NIST SP 811, App. B.9.

How the real exam works

  • Length: 80 questions. The appointment is 9 hours: a 2-minute nondisclosure agreement, an 8-minute tutorial, 8 hours of exam time, and a 50-minute scheduled break. That's an average of 6 minutes per question.
  • Two sections: After roughly half the questions, you review and submit them, and you can't go back to that half.
  • Formats: Multiple choice plus alternative item types. Every question is scored right or wrong, with no partial credit. A few unscored pretest questions are mixed in, and you can't tell which.
  • References: Closed book. You get the NCEES PE Mechanical Reference Handbook on screen as a searchable PDF, using its search box (Ctrl+F doesn't work). The Mechanical exams supply no separate design standards.
  • Conditions: U.S. customary units, sea level unless a question says otherwise.
  • Calculator: One NCEES-approved model, or the on-screen TI-30XS.
  • Cost and timing: $400 to NCEES per attempt, offered year-round at Pearson VUE test centers. Your state board may charge its own fee and decides whether you're eligible to sit.
  • Results: Usually 7–10 days, reported as pass or fail. If you don't pass, you get a diagnostic report by knowledge area.

Sources: NCEES Mechanical exam page; NCEES Examinee Guide, May 2026, pp. 3, 8–11, 14, and 16; NCEES computer-based testing.

Does your score here predict a pass?

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

Which references can you use on exam day?

Only the reference handbook NCEES puts on screen. Download the current version from your MyNCEES account and practice with it, because the version you study should match the one on your exam. These 40 questions give you every property and constant you need, so you can work them without it. On the real exam, practice finding things with the handbook's search box: name the relationship in plain words ("affinity," "log mean," "NPSH," "psychrometric"), search, then check the variable definitions and units.

Try a timed session

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

PE HVAC and Refrigeration pass rates

PE HVAC and Refrigeration pass rates
CohortFirst-time takersFirst-time pass rateRepeat takersRepeat pass rate
January–June 202679971%22940%

Source: NCEES PE pass-rate table, updated July 2026, for examinees under NCEES member boards. These are group results, not your personal odds, and the table does not establish why first-time and repeat pass rates differ.

Quick answers

Is this the official NCEES practice exam? No. This is an independent, unofficial practice set. NCEES provides its own exam-prep materials through MyNCEES.

Are these real exam questions? No. We wrote them. Real exam questions are confidential, and every examinee agrees not to disclose them.

Do I need to pass the FE first? Your state licensing board decides, not NCEES. Find yours in the NCEES board directory. If you still need the FE, try our free FE Mechanical practice test.

What if I fail? You can take an NCEES exam once per testing window and no more than three times in 12 months; some boards are stricter. See NCEES retake rules and how NCEES results work.

Sources and verification

Official sources

Technical sources for the practice questions

Last verified October 8, 2026. We checked the exam format, fee, specification, scoring, reference, retake rules, and pass-rate figures against the current NCEES materials linked above. We also independently recalculated all 40 answers and numeric acceptance ranges and rechecked the cited technical principles used by the set.

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

Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify their subjects, and trademarks belong to their respective owners. These practice questions are original and unofficial.

Free PE Mechanical HVAC and Refrigeration Practice Test (40 Qs)