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PE HVAC exam: what the problems actually look like

Most PE HVAC prep shows you formulas. That is not what makes the exam hard. What makes it hard is that four plausible numbers sit in front of you, three of them are mistakes you could genuinely make, and you have about six minutes. Here are three real problems, worked in full.

· by the FE to PE Prep team

80

Questions

~6.5 min

Per question

4

Knowledge areas

PE Mech Handbook

Only reference

The anatomy of a PE HVAC question

Four things are true of nearly every question on this exam, and together they explain why engineers who know the material still run out of time.

  • It says “most nearly.” The options are close. You cannot eliminate three of them by inspection, which means you have to actually do the problem.
  • The wrong answers are engineered. A well-written distractor is not a random number — it is the answer you get from one specific, plausible error. Getting a clean-looking number is not evidence you are right.
  • The figure is part of the problem. Sometimes it carries the data — the numbered state points on a cycle diagram are how you know which enthalpy is which. Sometimes it fixes the configuration, and counterflow versus parallel flow changes which ends you are allowed to pair. Either way it is not decoration.
  • You have about six and a half minutes. NCEES gives you 9 hours for 80 questions, and that 9 hours has to absorb an 8-minute tutorial and your break — the scheduled break is deducted from your exam time, not added to the day. Six and a half minutes has to cover reading, handbook lookups and checking. Knowing the method is not enough if finding the property takes four.

Three real problems

These are drawn from our free PE HVAC set, with the figures we draw for every question. Try each one before you open the solution — and when you do open it, read the distractor notes, not just the answer.

1. Supply airflow from a sensible load

This is the most ordinary calculation in HVAC design, and it is on the exam because it is ordinary. The trap is not the algebra — it is the coefficient.

  1. 1
    medium
    A. Load components & supply airflow
    An office AHU cooling a perimeter zone must offset a room SENSIBLE load of Qs=60000Btu/hQ_s=60000\,\text{Btu/h}. The room is held at Troom=75°FT_{room}=75\,°\text{F} and the coil delivers supply air at Tsupply=55°FT_{supply}=55\,°\text{F} (see the AHU schematic). Using Qs=1.08V˙ΔTQ_s=1.08\,\dot V\,\Delta T, the required supply-air flow is most nearly:
    Figure for this questionOARAMAfiltercooling coilfanSA
    1. 667 cfm
    2. 3240 cfm
    3. 4410 cfm
    4. 2780 cfm
    Show the answer & worked solution

    Answer: D.2780 cfm

    ΔT=TroomTsupply=7555=20°F\Delta T=T_{room}-T_{supply}=75-55=20\,°\text{F}; V˙=Qs1.08ΔT=600001.08(20)=2778\dot V=\dfrac{Q_s}{1.08\,\Delta T}=\dfrac{60000}{1.08(20)}=2778 cfm. Distractors: used the LATENT coefficient 0.680.68 in place of the sensible 1.081.08 (44124412); used the total-enthalpy coefficient 4.54.5 (667667); and MULTIPLIED by 1.081.08 instead of dividing (1.08Qs/ΔT=32401.08\,Q_s/\Delta T=3240).

What the distractors teach: Three of the four options come from picking the wrong constant: 0.68 is the latent coefficient, 4.5 is the total-enthalpy coefficient, and one option multiplies by 1.08 instead of dividing. Every one of those is a mistake a competent engineer makes at minute 340 of an eight-hour exam.

2. Log-mean temperature difference, counterflow

A coil problem you could meet in any of three knowledge areas. It looks like bookkeeping until you notice which ends the problem wants you to pair.

  1. 2
    easy
    A. LMTD, effectiveness-NTU & coil duty
    A chilled-water cooling coil runs in pure COUNTERFLOW. The hot stream enters at Th,in=110T_{h,in}=110°F and leaves at Th,out=75T_{h,out}=75°F; the cold stream enters at Tc,in=45T_{c,in}=45°F and leaves at Tc,out=60T_{c,out}=60°F. The log-mean temperature difference is most nearly:
    Figure for this questionposition (area)Temperature (°F)hotcoldΔT1ΔT2
    1. 40 °F
    2. 34.1 °F
    3. 50 °F
    4. 39.2 °F
    Show the answer & worked solution

    Answer: D.39.2 °F

    Counterflow terminal differences: ΔT1=Th,inTc,out=11060=50\Delta T_1=T_{h,in}-T_{c,out}=110-60=50°F and ΔT2=Th,outTc,in=7545=30\Delta T_2=T_{h,out}-T_{c,in}=75-45=30°F. LMTD=ΔT1ΔT2ln(ΔT1/ΔT2)=5030ln(50/30)=39.15\text{LMTD}=\dfrac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}=\dfrac{50-30}{\ln(50/30)}=39.15°F. Distractors: used the ARITHMETIC mean (ΔT1+ΔT2)/2=40(\Delta T_1+\Delta T_2)/2=40°F (always larger than the log-mean); paired the ends as in PARALLEL flow, Th,inTc,inT_{h,in}-T_{c,in} and Th,outTc,outT_{h,out}-T_{c,out} (34.134.1°F); and took the LARGER terminal difference 5050°F alone (no averaging).

What the distractors teach: The arithmetic mean of the two terminal differences is 40 °F and it is offered to you. It is always larger than the log-mean, and it is right there as option A. Pairing the ends as though the flow were parallel gets you 34.1 °F, which is also offered.

3. Coefficient of performance from a cycle diagram

Here the figure is not decoration — the state points on the diagram are how you know which enthalpy is which. This is what most textbook problems strip out and the real exam does not.

  1. 3
    easy
    A. Vapor-compression cycle, COP & capacity
    A walk-in cooler operates the R-134a cycle shown (states 1→2→3→4) with h1=104h_1=104, h2=119h_2=119, h3=41.6h_3=41.6, and h4=41.6Btu/lbh_4=41.6\,\text{Btu/lb}. The coefficient of performance for cooling is most nearly:
    Vapor-compression refrigeration cycle plotted on a pressure–enthalpy (p–h) diagramp (psia)pcpecrit.1234h3=h4=41.6h1=104h2=119h (Btu/lb)
    1. 0.806
    2. 5.16
    3. 4.16
    4. 0.24
    Show the answer & worked solution

    Answer: C.4.16

    COP=h1h4h2h1=10441.6119104=62.415=4.16\text{COP}=\dfrac{h_1-h_4}{h_2-h_1}=\dfrac{104-41.6}{119-104}=\dfrac{62.4}{15}=4.16. Distractors: SWAPPED numerator and denominator, h2h1h1h4=0.24\dfrac{h_2-h_1}{h_1-h_4}=0.24 (work over effect); used the heating-COP form with the full drop h2h4h_2-h_4 in the numerator, h2h4h2h1=5.16\dfrac{h_2-h_4}{h_2-h_1}=5.16 (one greater than the cooling COP); and divided the effect by the condenser REJECTION instead of the work, h1h4h2h3=0.81\dfrac{h_1-h_4}{h_2-h_3}=0.81.

What the distractors teach: Invert the ratio and you get 0.24. Use the heating-COP form and you get 5.16, which is exactly one greater than the right answer — the kind of near-miss that looks like a rounding difference when you are checking your work quickly.

Where the questions come from

NCEES publishes exactly how many questions come from each knowledge area. It is the most useful study document you are not reading, because it tells you where your hours belong.

PE Mechanical HVAC and Refrigeration knowledge areas and question counts, per the NCEES specification
Knowledge areaQuestions
HVAC and Refrigeration Equipment and ComponentsThe largest block. Coils, chillers, towers, pumps and fans — sizing and rating them.24–36
HVAC and Refrigeration Distribution and SystemsDuctwork, piping, ventilation, controls, and refrigeration systems end to end.20–30
HVAC Loads and PsychrometricsLoads and air processes — including at 5,000 ft and at low temperature, not just sea level.18–27
Supportive KnowledgeCodes and standards, economics, and professional practice.8–12

If you want the pass-rate picture alongside this, the companion guide covers it: how hard is the PE Mechanical HVAC exam?

Common questions

What do PE HVAC exam questions actually look like?
Multiple choice, four options, phrased as 'most nearly' — the options are close enough that you cannot eliminate them by inspection. Many carry a figure (an AHU schematic, a cycle diagram, an equipment schedule). NCEES gives you 9 hours for 80 questions, including an 8-minute tutorial and an optional scheduled break that is deducted from your exam time — so roughly six and a half minutes per question, covering reading, handbook lookups and checking.
Are PE HVAC problems harder than FE problems?
They are longer and more layered rather than harder in the mathematics. An FE question usually tests one relationship. A PE HVAC question typically asks you to read a system, decide which relationship applies, pick the right coefficient or property, and then compute. The arithmetic is rarely the difficult part.
Can I use my own references in the PE HVAC exam?
No. NCEES specifies the exam as closed book with an electronic reference: the searchable PE Mechanical Reference Handbook is supplied inside the exam software and is your only reference. Unlike PE Civil, no design standards are supplied. That makes lookup speed a scoreable skill, so practising with that handbook open from day one is worth more than any additional textbook.
How many questions are on each PE HVAC topic?
The NCEES specification allocates 24-36 questions to Equipment and Components, 20-30 to Distribution and Systems, 18-27 to Loads and Psychrometrics, and 8-12 to Supportive Knowledge. Equipment and Components is the largest single block, which surprises engineers who assume psychrometrics dominates.

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Tell us how long you have. We'll budget your hours across every knowledge area using the official NCEES question weights — the heaviest areas get the most time, which is the opposite of how most people study.

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Work through seven more

Ten free PE HVAC problems in total — psychrometrics, coil duty, refrigeration cycles, equipment selection — each with the figure and the full worked solution. No signup.

Exam format, fees and specifications are published by NCEES and change over time — confirm current specifics at ncees.org. Independent study resource; not affiliated with, endorsed by, or sponsored by NCEES.