Codes, acoustics, economics, power, PE HVAC&R area 4
The smallest area, 8–12 items, and the one that grew in October 2025: codes and standards, acoustics and vibration control, economic analysis and electrical concepts. Economics and electrical questions moved here from the old basic engineering practice section.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamPE Mechanical HVAC & Refrigeration
questions on the exam8–12
worked problems here15
areas, October 2025 outline4 of 4
Checked
What the area covers
Four topics in the specifications, in our words:
Codes and standards that govern HVAC and refrigeration design.
Acoustics and vibration control: sound control and absorption, attenuators, noise criteria, transmission and isolation.
Economic analysis of HVAC options.
Electrical concepts: power use, motor ratings, motor heat and current.
15 supportive knowledge problems
0 / 15 answered · 0 correct
Question 1 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A 50-hp fan motor has a service factor of 1.15 and operates continuously at 110% of its rated horsepower output. The motor nameplate efficiency is 92%. Calculate the true electrical input (kW) and the expected motor heat rejection (Btu/hr) to the surrounding space.
Given
Motor rating = 50 hp
Service factor = 1.15
Operating load = 110 % of rated output
Motor efficiency = 92 %
Find
Electrical input (kW) and motor heat to the space (Btu/h)
Solution
Pshaft = 50 × 1.10 × 0.746 = 41.03 kW
Pin = 41.03/0.92 = 44.6 kW
q = (44.6 − 41.03) × 3,412 = 12,174 Btu/hr
46.6 kW loads the motor to its full 1.15 service factor; the stem says it runs at 110%.
The input is right, but 152,176 Btu/hr counts all of it as motor heat; the shaft power leaves with the fan.
41.0 kW is the shaft output, not the electrical input, and its full value is again counted as motor heat.
44.6 kW and 12,174 Btu/hr: 55 hp of output (110% of 50 hp) is 41.03 kW at the shaft, ÷ 0.92 gives the input, and only the 3.57 kW difference is lost as heat at the motor.
40.5 kW uses the 50 hp rating; the motor is actually loaded to 110%, 55 hp.
Question 2 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
The Noise Reduction Coefficient (NRC) is computed as the average of sound absorption coefficients at 250, 500, 1000, and 2000 Hz. Why might this metric underrepresent a material’s performance in certain applications?
NRC averages only 250–2,000 Hz, so a material's weak absorption at 63–125 Hz, where mechanical-room rumble sits, never shows.
NRC stops at 2,000 Hz and leaves out 4,000 Hz, so it does not overweight high frequencies.
NRC is a plain arithmetic average of four absorption coefficients, rounded to 0.05.
NRC rates absorptive materials; a reflective surface simply has a low NRC.
Question 3 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Assembly
STC Rating
125 Hz TL (dB)
250 Hz TL (dB)
500 Hz TL (dB)
Wall A
45
32
38
44
Wall B
50
22
40
52
An engineer is evaluating two partition designs to isolate a mechanical room housing equipment that generates significant low-frequency noise, strictly dominated by the 125 Hz octave band. Based on the provided transmission loss data, which assembly will result in the lowest transmitted noise level, and what is the approximate difference in acoustic performance at the dominant frequency?
Given
Dominant noise band = 125 Hz
Wall A: STC 45, TL 32 dB at 125 Hz, 38 dB at 250 Hz, 44 dB at 500 Hz
Wall B: STC 50, TL 22 dB at 125 Hz, 40 dB at 250 Hz, 52 dB at 500 Hz
Find
Better wall and difference at the dominant band
Solution
TL125: A = 32 dB, B = 22 dB → A better by 10 dB despite the lower STC
STC is weighted toward speech frequencies and hides Wall B's weak 125 Hz performance.
Wall B is 10 dB worse, not better, at 125 Hz.
Wall A has the lower STC (45 vs 50); its advantage is 10 dB at 125 Hz, not STC.
Wall B leads only at 500 Hz (by 8 dB), a band that does not dominate this noise.
Wall A: at the dominant 125 Hz band it gives 32 dB of transmission loss against Wall B's 22 dB.
Question 4 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A private office situated directly beneath a VAV box and 15 feet of unlined rectangular sheet metal ductwork experiences excessive mechanical noise. The supplied air discharges through a diffuser located in a different room 20 feet farther down the corridor. The office features a standard acoustic tile ceiling. After installing internal acoustical lining inside the duct above the office, the noise level in the office remained virtually unchanged. What is the most likely path of the noise transmission and the recommended mitigation strategy?
Solution
Breakout TL rises with duct wall mass and stiffness (wrap, heavier gauge, round duct); internal lining attenuates sound inside the duct, not through its walls.
No outlet serves this office, and lining already failed to help; thicker lining does little for low-frequency breakout.
Turning vanes reduce turbulence in elbows; they do nothing for sound leaving straight duct walls.
The diffuser is in a different room 20 ft away, so its noise is not what this office hears.
The air is delivered to another room, so the noise in this office comes through the duct walls; adding mass outside the duct blocks it, which internal lining cannot.
Fiberglass inside the duct does not isolate vibration; that needs flexible connections and isolated hangers.
Question 5 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
An HVAC system is causing excessive noise in a private office space. Testing and balancing reports indicate that the branch volume damper serving the office is severely throttled to 15% open to reduce airflow, creating a localized pressure drop of 0.85 in. w.g. A sound attenuator installed at the air handling unit (AHU) discharge is verified to be performing to its specified dynamic insertion loss. Which of the following is the most likely cause of the persistent acoustic problem?
Solution
Damper/fitting regenerated sound rises steeply with velocity and pressure drop; balance with lower drops or add a terminal attenuator downstream.
The attenuator was verified to meet its specified insertion loss.
A damper 15% open with a 0.85 in. w.g. drop generates its own turbulence noise downstream of the attenuator, which cannot treat it.
Upstream breakout would be near the AHU, not at an office served through a heavily throttled branch damper.
Nothing in the reports points to vibration, while the severely throttled damper is a known noise source near the office.
The damper is burning off 0.85 in. w.g., so the branch has too much pressure, not too little.
Question 6 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A mechanical room contains three continuous-duty motors rated at 15 A, 20 A, and 40 A full-load current (FLA). An engineer initially sizes the feeder conductor ampacity at 93.75 A by applying a 125% multiplier to the sum of all three motors. According to NEC guidelines, what is the correct minimum feeder ampacity, and what specific error did the engineer make?
75 A is the plain sum; the code requires the extra 25% on the largest motor.
85 A: NEC 430.24 takes 125% of the largest motor's FLA (40 A × 1.25 = 50 A) plus 100% of the others (15 + 20).
125 A counts the largest motor twice: its 125% (50 A) is added to a 75 A sum that already includes its 40 A; NEC 430.24 adds 125% of the largest to 100% of the others only.
125% of the whole sum overstates the requirement; only the largest motor gets the 25% allowance.
There is no 115% factor for the smaller motors; they are counted at 100% of FLA.
Question 7 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Requirement
Specification
Wall Fire-Resistance Rating
2-hour
Duct Function
Supply Air
Smoke Control Integration
Part of engineered smoke-control system
System Status During Alarm
Operational for zone pressurization
A supply air duct penetrates a designated fire barrier. The parameters regarding the barrier and the duct's function are detailed in the table. According to NFPA 90A and NFPA 92 standards, select the appropriate damper assembly required for this specific penetration.
Solution
Barrier < 3 h → 1½ h fire damper; barrier ≥ 3 h → 3 h. Duct in an engineered smoke-control path → combination fire/smoke damper.
Fire dampers are rated 1½ or 3 hours, and a plain fire damper cannot be controlled as part of a smoke-control system.
A smoke damper alone has no fire rating, and the 2-hour barrier requires fire protection at the penetration.
There is no 2-hour damper rating class; dampers for barriers under 3 hours are rated 1½ hours.
3-hour dampers are for barriers rated 3 hours or more; this wall is rated 2 hours.
A barrier rated under 3 hours takes a 1½-hour fire damper, and because the duct serves smoke control it must also act as a smoke damper — a combination unit.
Question 8 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Octave band (Hz)
Fan SPL (dB)
Diffuser SPL (dB)
Return grille SPL (dB)
Limit (dB)
63
52
48
45
65
125
50
47
44
54
250
48
46
43
48
500
46
45
42
44
1000
44
43
40
41
2000
42
41
38
39
4000
40
39
36
37
A conference room has the project octave-band sound pressure limits listed in the table. The octave-band sound pressure levels from the supply fan, diffuser and return grille at the receiver are also given. Combine the three sources in each band and determine whether the design meets the limits. Identify any offending octave bands.
Given
Fan SPL (63–4000 Hz) = 52, 50, 48, 46, 44, 42, 40 dB
The combined level exceeds the limit in five bands, by up to 6.4 dB at 4000 Hz.
This adds the decibel values arithmetically (52 + 48 + 45 = 145 dB at 63 Hz); sound levels combine as energy, so 63 and 125 Hz are within limits.
Adding the three sources logarithmically gives 54.0, 52.4, 50.9, 49.4, 47.4, 45.4 and 43.4 dB; every band from 250 Hz up is over its limit.
This checks the fan alone, which is exactly at the 48 dB limit in the 250 Hz band; with the diffuser and grille added, 250 Hz reaches 50.9 dB.
The low bands have the most headroom (54.0 against 65 dB, 52.4 against 54 dB); the overages are in the mid and high bands.
Question 9 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Which type of vibration isolator would be most appropriate for a rooftop air handling unit?
Solution
fn ≈ 3.13/√δst (Hz, δ in in.) — a low fn needs large static deflection → springs
Neoprene mounts give about 0.25–0.5 in. of deflection, too little for a large AHU on a lightweight roof.
An inertia base adds mass and stability but still sits on springs; it is not the isolator itself.
Cork pads deflect only a fraction of an inch and isolate only high frequencies; they cannot handle low fan speeds on a roof.
Steel springs give 1–2 in. or more of static deflection, enough to isolate an AHU's low fan frequencies on a flexible roof structure.
Question 10 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
What is the annual cost of operating a 7.5 kW air conditioning unit that runs 1,200 hours per year if electricity costs $0.12 per kWh?
Given
Unit power = 7.5 kW
Operating hours = 1,200 h/yr
Electricity cost = $0.12 per kWh
Find
Annual operating cost ($)
Solution
cost = 7.5 kW × 1,200 h × $0.12/kWh = $1,080
$108 slips a decimal in the rate, using $0.012 per kWh.
$1,080: 7.5 kW × 1,200 h = 9,000 kWh at $0.12.
$9,000 is the 9,000 kWh of energy read as dollars; it was not multiplied by the rate.
$10,800 uses $1.20 per kWh, ten times the stated rate.
Question 11 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A chilled-water plant operates four identical pumps, each rated at 25 hp with a motor efficiency of 90% and running 6,000 hours per year at full load; baseline power draw is 20.7 kW per pump. A retrofit installs VFDs on all four pumps to run at 75% of full speed for 4,800 hours, with full speed for the remaining 1,200 hours. Assume pump power varies with the cube of speed, VFD efficiency is 95%, electricity costs $0.12 per kWh, and the four VFDs cost $38,000 installed. What is the simple payback period?
Given
Pumps = 4 × 25 hp, motor efficiency 90 %
Baseline draw = 20.7 kW per pump, 6,000 h/yr
With VFD: 75 % speed for 4,800 h, full speed for 1,200 h
1.38 years leaves out the 95% VFD efficiency, overstating the savings; the drive's losses apply at both speeds.
1.47 years: with VFDs each pump uses 9.19 kW for 4,800 h and 21.79 kW for 1,200 h, saving 215,716 kWh ($25,886) a year for the four.
2.02 years scales power with the square of speed; the stem gives the cube law.
4.04 years scales power linearly with speed, which badly understates the savings at 75% speed.
5.87 years compares the savings of one pump against the cost of all four VFDs.
Question 12 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
An HVAC control center uses a battery backup system that delivers 48 V DC and must supply 1,500 Wh of energy. If a maximum depth of discharge of 50% is prescribed, what is the minimum required battery capacity in Ampere-hours?
Given
Battery voltage = 48 V DC
Required energy = 1,500 Wh
Maximum depth of discharge = 50 %
Find
Minimum battery capacity (Ah)
Solution
C = E/(V·DoD) = 1,500 Wh/(48 V × 0.50) = 62.5 Ah
15.6 Ah multiplies by the 50% depth of discharge instead of dividing.
31.3 Ah is the usable charge alone; at 50% depth of discharge the battery must be twice that size.
62.5 Ah: 1,500 Wh at 48 V is 31.25 Ah of usable charge, and only half the battery may be discharged.
125.0 Ah applies the 50% depth of discharge twice.
Question 13 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Unit Designation
Cooling Capacity
Outdoor Air
Economizer
RTU-1
3.5 tons
600 cfm
Omitted
RTU-2
5.5 tons
1,500 cfm
Omitted
RTU-3
4.0 tons
800 cfm
Omitted
RTU-4
6.5 tons
1,800 cfm
Provided
Evaluate the proposed packaged rooftop units for a commercial facility located in Climate Zone 4A against standard prescriptive IECC economizer requirements. Assuming standard capacity thresholds strictly apply (systems ≥ 54,000 Btu/h require an economizer), which of the units listed in the table fails to comply?
Given
Climate zone = 4A
Economizer threshold = 54,000 Btu/h
RTU-1 = 3.5 tons, economizer omitted
RTU-2 = 5.5 tons, economizer omitted
RTU-3 = 4.0 tons, economizer omitted
RTU-4 = 6.5 tons, economizer provided
Find
Non-compliant unit
Solution
54,000 Btu/h ÷ 12,000 = 4.5 tons → units ≥ 4.5 tons need an economizer
RTU-2 (5.5 tons, none) fails; RTU-4 (6.5 tons) has one.
The 5.5-ton unit is 66,000 Btu/h, above the 54,000 Btu/h threshold, and has no economizer.
The 6.5-ton unit is above the threshold but has an economizer, so it complies.
4.0 tons is 48,000 Btu/h, still below 54,000 Btu/h, so no economizer is required.
3.5 tons is 42,000 Btu/h, below the threshold, so omitting the economizer is allowed.
Two units are below the threshold and the largest has an economizer; only one unit fails.
Question 14 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Zone
Area (ft2)
Occupancy
Rp (cfm/person)
Ra (cfm/ft2)
Office
1,000
20
5
0.06
Conference
500
25
5
0.06
An HVAC designer is evaluating a 100% outdoor air system serving two zones. The design outdoor condition is 95°F dry-bulb, and the indoor setpoint is 75°F in both zones. Assuming the system outdoor air intake equals the sum of the uncorrected zone breathing-zone outdoor airflows (ASHRAE 62.1), determine the sensible cooling load of the outdoor air using the standard sea-level equation. The zone data are given in the table.
Given
Outdoor design = 95 °F dry-bulb
Indoor setpoint = 75 °F
Office: 1,000 ft², 20 people, Rp = 5 cfm/person, Ra = 0.06 cfm/ft²
Conference: 500 ft², 25 people, Rp = 5 cfm/person, Ra = 0.06 cfm/ft²
340 Btu/hr is 1.08 × 315 without the 20°F temperature difference.
4,860 Btu/hr counts only the people component (225 cfm) and leaves out the area rate.
6,804 Btu/hr: the zones need 160 + 155 = 315 cfm of outdoor air, and 1.08 × 315 × 20°F = 6,804.
8,505 Btu/hr divides the airflow by a zone effectiveness of 0.8, although the stem asks for uncorrected breathing-zone flows.
28,350 Btu/hr uses the 4.5 total-heat factor with a dry-bulb difference; the sensible factor is 1.08.
Question 15 of 15 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
Octave Band
Forward Flow (+2000 fpm)
Reverse Flow (-2000 fpm)
No Flow (0 fpm)
63 Hz
15 dB
8 dB
10 dB
125 Hz
18 dB
12 dB
15 dB
250 Hz
25 dB
22 dB
24 dB
A large industrial exhaust fan is configured to draw air from a highly occupied manufacturing space through a long, straight return ductwork system. The face velocity of the air traveling across the selected acoustic silencer is precisely 2000 fpm. Sound generated by the exhaust fan propagates back through the same ductwork and into the occupied manufacturing space, creating a potential noise issue. Using the provided manufacturer performance table, determine the expected dynamic insertion loss of the silencer at the 63 Hz octave band for this specific installation scenario.
Given
Silencer face velocity = 2000 fpm
Application = exhaust fan; fan noise travels back to the space through the exhaust duct
63 Hz insertion loss: forward 15 dB, reverse 8 dB, no flow 10 dB
Find
Dynamic insertion loss at 63 Hz (dB)
Solution
Exhaust/return: sound against airflow → use the 'reverse flow' (−) dynamic insertion loss → 8 dB at 63 Hz.
8 dB: in an exhaust duct the fan noise travels toward the room against the airflow, so the reverse-flow (−2,000 fpm) rating applies.
10 dB is the no-flow (static) rating; at 2,000 fpm the flow direction changes the insertion loss.
12 dB is the reverse-flow value for the 125 Hz band, not 63 Hz.
15 dB is the forward-flow rating, for sound traveling with the air as in a supply duct.
18 dB is the forward-flow value at 125 Hz, the wrong band and the wrong flow direction.
Answer every question to see your score
Codes without the code book
Since this module comes with no design standards, code questions test what a document governs and what it requires in principle. Know which one answers which question:
ASHRAE 62.1: ventilation and acceptable indoor air quality.
ASHRAE 90.1: energy efficiency of buildings other than low-rise residential.
ASHRAE 15 and 34: safety of refrigeration systems, and refrigerant designations and safety classes.
NFPA 90A: installation of air-conditioning and ventilating systems, including fire and smoke dampers.
Model codes such as the International Mechanical Code and the International Energy Conservation Code, which take legal force only where a jurisdiction adopts them.
Common traps
Five mistakes outside the core
Decibels add on a log scale. Two equal sources give 3 dB more than one, not double the number:
L = 10 · log(∑ 10Li/10)
Sound power is not sound pressure. Lw belongs to the source; Lp depends on the room and the distance (ASHRAE Handbook—Fundamentals, Sound and Vibration).
Isolators can amplify. Transmissibility drops below one only when the forcing frequency exceeds √2 times the natural frequency; near resonance an isolator makes things worse. The natural frequency follows from static deflection:
fn ≈ 188/√δ (cycles per minute, δ in inches)
Simple payback ignores time. It counts neither interest nor the years after payback; compare options on present worth or annual cost (engineering economics).