What do PE HVAC exam questions actually look like?

Mostly multiple choice with four options, usually phrased 'most nearly', and 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 8 hours of exam time for 80 questions, inside a 9-hour appointment that also holds the nondisclosure agreement, an 8-minute tutorial and a 50-minute scheduled break. The break does not come out of your exam time, so you have 6 minutes per question on average, 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 practicing 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.

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 six minutes on average. Here are three practice problems, worked in full.

Updated October 6, 2026 · by the FE → PE Prep team

Questions

6 min

Per question

Knowledge areas

PE Mech Handbook

Only reference

The anatomy of a PE HVAC question

Four things shape most questions 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.
Often, 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 six minutes a question. NCEES gives you 8 hours of exam time for 80 questions. The nondisclosure agreement, an 8-minute tutorial and a 50-minute scheduled break fill the rest of the 9-hour appointment, so none of them comes out of your exam time. Six minutes has to cover reading, handbook lookups and checking. Knowing the method is not enough if finding the property takes four.

Three practice problems

These are drawn from our free PE HVAC set, figures included. Try each one before you open the solution. When you do open it, read the distractor notes, not just the answer.

1. Supply airflow from a room cooling load

This is the most ordinary calculation in HVAC design, and it is on the exam because it is ordinary. The load calculation hands you a total load and a sensible heat ratio; only the sensible part sets the airflow.

1
medium
A. Load components & supply airflow
An office AHU cooling a perimeter zone at a sea-level site must offset a total room cooling load of $75{,}000\,\text{Btu/h}$ with a sensible heat ratio of $0.80$ . The room is held at $75\,°\text{F}$ , and the supply air enters the room at $55\,°\text{F}$ (see the AHU schematic). The required supply-air flow (cfm) is most nearly:
Figure — drag or pinch to move around
1. 833 cfm
2. 2730 cfm
3. 3410 cfm
4. 4340 cfm
Show the answer & worked solution
Answer: B.2730 cfm
The supply air is sized on the room SENSIBLE load; the latent part is removed at the coil. With SHR $=$ sensible/total (§7.1), $q_s=\text{SHR}\,q_t=0.80(75{,}000)=60{,}000$ Btu/h. $\Delta t=t_{room}-t_{supply}=75-55=20\,°\text{F}$ , and the §9.1.7 sea-level sensible relation $q_s=1.10\,Q\,\Delta t$ gives $Q=\dfrac{q_s}{1.10\,\Delta t}=\dfrac{60{,}000}{1.10(20)}=2{,}727$ cfm. Distractors: paired the total load with the total-heat factor $4.5$ but a temperature difference; $4.5$ multiplies an enthalpy difference ( $833$ ); sized the air on the TOTAL load, $75{,}000/[1.10(20)]$ , though the latent part is removed at the coil and does not set the airflow ( $3{,}409$ ); and divided by the latent factor per grain, $4840/7000=0.691$ , instead of the sensible $1.10$ ( $4{,}339$ ).

What the distractors teach

Take the sensible share first: 0.80 × 75,000 = 60,000 Btu/h, then 60,000 ÷ (1.10 × 20) = 2,730 cfm. Size the air on the whole 75,000 Btu/h and you get 3,410 cfm, which means an oversized fan and an overcooled room; the latent load is removed at the coil and does not set the airflow. The other two wrong options come from the coefficient. 0.691 (4,840 divided by 7,000) is the latent factor per grain of moisture, and dividing by it gives 4,340. 833 pairs the total load with 4.5, the total-heat factor, which belongs with an enthalpy difference, not a temperature difference. The PE Mechanical Reference Handbook prints 1.10 as its sea-level sensible factor. Many textbooks use 1.08, which the handbook does not print, so practice with 1.10.

2. Counterflow heat exchanger: sizing the surface

A heat-recovery exchanger to size. Three of the four end temperatures are given; the fourth comes from an energy balance, and only then can the log-mean be taken.

2
medium
A. LMTD, effectiveness-NTU & coil duty
In a heat-recovery plate heat exchanger, $40\,\text{gpm}$ of process water is cooled from $140$ °F to $65$ °F by $75\,\text{gpm}$ of domestic water entering at $50$ °F. The two streams flow in counterflow, both are water, and the overall heat-transfer coefficient is $U=500\,\text{Btu/(hr-ft}^2\text{-°F)}$ . The required heat-transfer area is most nearly:
Figure — drag or pinch to move around
1. 60 ft²
2. 92.3 ft²
3. 103 ft²
4. 200 ft²
Show the answer & worked solution
Answer: C.103 ft²
Duty from the process water (§9.1.7, standard water): $\dot Q=500\,(40)(140-65)=1{,}500{,}000$ Btu/hr. The domestic water absorbs the same duty, so $T_{Co}=T_{Ci}+\dfrac{\dot Q}{500\,(75)}=50+40=90$ °F. Counterflow terminal differences (§5.5.3): $T_{Hi}-T_{Co}=140-90=50$ °F and $T_{Ho}-T_{Ci}=65-50=15$ °F, so $\Delta T_{lm}=\dfrac{50-15}{\ln(50/15)}=29.07$ °F. With $F=1$ for counterflow (§5.5.1), $A=\dfrac{\dot Q}{U F\,\Delta T_{lm}}=\dfrac{1{,}500{,}000}{500\,(1)(29.07)}=103$ ft². Distractors: used the ARITHMETIC mean of the terminal differences, $(50+15)/2=32.5$ °F, in place of the log-mean ( $92.3$ ft²); assumed the domestic water warms as much as the process water cools, $T_{Co}=50+75=125$ °F, ignoring the different flows, which leaves $15$ °F at both ends ( $200$ ft²); and used the LARGER terminal difference, $50$ °F, as the driving difference ( $60$ ft²).

What the distractors teach

The process water gives up 500 × 40 × 75 = 1,500,000 Btu/hr, so the 75 gpm of domestic water leaves at 90 °F and the end differences are 50 °F and 15 °F. Their log-mean, (50 − 15)/ln(50/15) = 29.1 °F, sizes the surface at 103 ft². Their arithmetic mean, 32.5 °F, is about 12% above the log-mean and sizes the surface at 92.3 ft², which is offered. Assuming the domestic water warms by the same 75 °F the process water cools leaves 15 °F at both ends and 200 ft². And the larger end difference on its own, 50 °F, gives 60 ft².

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. The question also asks for the COP the owner pays for, on the motor's electrical input.

3
medium
A. Vapor-compression cycle, COP & capacity
A food-plant cold room is served by an open-drive compressor operating the R-134a cycle shown (states 1→2→3→4), with $h_1=104$ , $h_2=119$ , $h_3=41.6$ , and $h_4=41.6\,\text{Btu/lb}$ . The compressor's electric motor has an efficiency of $0.90$ . The cooling coefficient of performance based on the electrical input to the motor is most nearly:
Figure — drag or pinch to move around
1. 3.74
2. 4.16
3. 6.24
4. 12.8
Show the answer & worked solution
Answer: A.3.74
From the diagram's enthalpies (§4.5.2), the cycle COP is $\text{COP}_{ref}=\dfrac{h_1-h_4}{h_2-h_1}=\dfrac{104-41.6}{119-104}=\dfrac{62.4}{15}=4.16$ , cooling per unit of SHAFT work. The motor needs $W_{el}=W_{shaft}/\eta_m$ (§1.11.1), so on the electrical input $\text{COP}_{el}=\text{COP}_{ref}\,\eta_m=4.16(0.90)=3.74$ . With an open drive the motor loss leaves as heat to the machine room and never reaches the refrigerant, so $h_2$ is unchanged. Distractors: stopped at the cycle COP, $4.16$ , leaving out the motor loss that the electrical input has to cover; left $h_4$ out of the numerator, $\dfrac{h_1}{h_2-h_1}\,\eta_m=6.24$ ; and multiplied the electrical COP by $3.412$ Btu/h per W, $12.77$ , which turns a COP into an EER (§9.6.2), not a dimensionless COP.

What the distractors teach

Read the cycle first: (104 − 41.6) ÷ (119 − 104) = 62.4 ÷ 15 = 4.16, the cooling per unit of shaft work. The open-drive motor is 90% efficient, so the electrical input is the shaft work divided by 0.90, and the COP on that input is 4.16 × 0.90 = 3.74. Stop at the cycle and you pick 4.16. It is the most tempting wrong answer because it is a correct calculation of a different quantity. Leave h4 out of the numerator and you get 6.24. Multiply by 3.412 and you have turned a COP into an EER, 12.8.

Where the questions come from

NCEES publishes a question range for each knowledge area in its exam specification. 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 area	Questions
HVAC 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, acoustics and vibration control, economic analysis, and electrical concepts such as motor ratings, power use and amperage.	8–12

If you want the pass-rate picture alongside this, the companion guide covers it: how hard is the PE Mechanical HVAC exam? And for turning these question counts into a study calendar: how to prepare for the PE Mechanical HVAC exam.

Common questions

What do PE HVAC exam questions actually look like?: Mostly multiple choice with four options, usually phrased 'most nearly', and 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 8 hours of exam time for 80 questions, inside a 9-hour appointment that also holds the nondisclosure agreement, an 8-minute tutorial and a 50-minute scheduled break. The break does not come out of your exam time, so you have 6 minutes per question on average, 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 practicing 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.

Build your study plan

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.

Work through seven more

Ten free PE HVAC problems in total, covering loads, heat exchangers, refrigeration cycles, cooling towers, ductwork and controls, each with a full worked solution and, where the problem needs one, its figure. No signup.

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