60 problems from all four areas of the October 2025 outline, sized by the middle of each published range, so equipment (24–36 items on the exam) is the biggest block and supportive knowledge the smallest. Every problem is in USCS units. Pick an answer and the solution opens, each option explained.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamPE Mechanical HVAC & Refrigeration
practice problems60
questions on the exam80
exam time8 h
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60 PE HVAC&R practice problems
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Question 1 of 60 · HVAC Loads and Psychrometrics · 18–27 items
An air handling unit uses direct steam injection to humidify 8,000 cfm of supply air (ρ = 0.075 lb/ft3). The air enters the humidification section at 60°F DB with a humidity ratio of 0.0040 lb/lb. The target leaving condition requires a humidity ratio of 0.0085 lb/lb. If the steam is supplied at a pressure where its latent heat of vaporization is 1,000 Btu/lb, calculate the required steam mass flow rate in lb/hr.
0.162 divides the vapor mass by the 1,000 Btu/lb latent heat; a mass balance needs no latent heat, which is extra data here.
2.70 is the rate per minute: the 600 lb/min dry-air flow times 0.0045 was never multiplied by 60.
144 lb/hr uses the entering humidity ratio 0.0040 as if it were the moisture to add.
162 lb/hr: 36,000 lb/h of dry air must gain 0.0045 lb of water per lb, and all of it comes from the injected steam.
306 lb/hr uses the leaving humidity ratio 0.0085 alone, ignoring the moisture the air already carries.
Question 2 of 60 · HVAC Loads and Psychrometrics · 18–27 items
A single-story commercial building measures 150 ft by 100 ft with a 12 ft ceiling height. The infiltration rate is estimated at 0.5 Air Changes per Hour (ACH). If the outdoor design temperature is 95°F and the indoor setpoint is 75°F, calculate the sensible cooling load from infiltration.
Given
Building = 150 ft × 100 ft × 12 ft
Infiltration = 0.5 ACH
Outdoor design temperature = 95 °F
Indoor setpoint = 75 °F
Find
Sensible infiltration cooling load (Btu/h)
Solution
V = 150 × 100 × 12 = 180,000 ft3
Q = ACH · V/60 = 0.5 × 180,000/60 = 1,500 cfm
qs = 1.08 × 1,500 × (95 − 75) = 32,400 Btu/hr
32,400 Btu/hr: 0.5 air change per hour of a 180,000 ft³ building is 1,500 cfm, times 1.08 × 20°F.
64,800 Btu/hr ignores the 0.5 factor and uses one air change per hour, 3,000 cfm: 1.08 × 3,000 × 20.
129,600 Btu/hr inverts the air-change rate to 2 per hour; 0.5 ACH is one building volume every two hours.
135,000 Btu/hr puts the 4.5 total-heat factor in front of a dry-bulb difference; 4.5 belongs with an enthalpy difference.
1,944,000 Btu/hr uses 90,000 ft³/h as if it were cfm; the hourly volume has to be divided by 60.
Question 3 of 60 · HVAC Loads and Psychrometrics · 18–27 items
An engineer sizes a heating coil for a high-altitude facility based strictly on the volumetric flow rate requirement of the space at standard conditions, entirely ignoring local air density changes. Which of the following describes the most likely consequence of this specific design flaw?
Solution
q = ρ · Q · 60 · cp · ΔT
At 5,000 ft ρ ≈ 0.062 lb/ft3 vs 0.075 at sea level: the same cfm carries about 17 % less mass, so heat delivered drops in proportion.
Face velocity is cfm divided by face area; the design cfm is unchanged, so velocity matches the design.
Insufficient capacity: the thinner air carries less mass per cfm, so the same airflow delivers less heat to the space than the sea-level sizing assumed.
Pressure drop at a fixed velocity scales with density, so the thinner air gives a lower drop, not a higher one.
Lower density reduces mass flow per cfm, which lowers delivered heat; nothing in the error inflates capacity.
Fan power at a given cfm falls with air density, so a high-altitude fan draws less power, not more.
Question 4 of 60 · HVAC Loads and Psychrometrics · 18–27 items
Parameter
Value
Outdoor air (OA) flow rate
2,000 cfm
Outdoor air dry-bulb
95 °F
Return air (RA) flow rate
8,000 cfm
Return air dry-bulb
75 °F
An air handling unit mixes outdoor air and return air before cooling it through a chilled water coil. The system parameters are provided in the table. The local barometric pressure is standard sea-level (14.696 psia). The cooling coil has an apparatus dew point (ADP) of 50 °F and a bypass factor of 0.15. Determine the final dry-bulb temperature of the air leaving the cooling coil.
54.4 °F: the mix enters at 79 °F, and 15% of it bypasses the 50 °F ADP, so 50 + 0.15 × 29.
55.3 °F uses 85 °F, the plain average of 95 and 75 °F, instead of weighting the mix by the 2,000 and 8,000 cfm flows.
56.8 °F applies the bypass factor to the 95 °F outdoor air, as if no return air were mixed in.
61.9 °F applies the bypass factor to the whole 79 °F mixed temperature instead of its excess over the ADP: 50 + 0.15 × 79.
74.7 °F treats 0.15 as the contact factor: only 15% of the air is cooled, so the coil barely changes the 79 °F mix.
Question 5 of 60 · HVAC Loads and Psychrometrics · 18–27 items
Face Velocity (fpm)
Saturation Effectiveness (%)
300
92
400
88
500
85
600
81
Air enters a direct evaporative cooling unit at 95°F dry-bulb and 65°F wet-bulb. The unit processes an airflow of 10,000 cfm and has a media face area of 20 square feet. Using the manufacturer's media performance data provided in the table, determine the expected dry-bulb temperature of the air leaving the cooler.
65.0°F is the entering wet-bulb, the limit of a cooler with 100% saturation effectiveness.
67.4°F uses the 92% effectiveness at 300 fpm, a slower face velocity than this unit runs.
68.6°F uses the 88% effectiveness at 400 fpm; the face velocity here is 500 fpm.
69.5°F: 10,000 cfm over 20 ft² is 500 fpm, where the table gives 85%, and 95 − 0.85 × 30 = 69.5°F.
70.7°F uses the 81% effectiveness at 600 fpm instead of the 500 fpm row.
Question 6 of 60 · HVAC Loads and Psychrometrics · 18–27 items
A two-stage evaporative cooling system processes 10,000 CFM of outside air. Stage 1 is an indirect evaporative cooler with a sensible cooling effectiveness of 65%. Stage 2 is a direct evaporative cooler with an effectiveness of 85%. The entering outside air is at 100°F DB and 70°F WB. The secondary scavenging air for the indirect cooler is also at 70°F WB. The wet-bulb temperature of the primary air leaving Stage 1 is 64°F. Determine the final dry-bulb temperature of the air leaving Stage 2.
Given
Airflow = 10,000 cfm
Indirect stage effectiveness = 65 %
Direct stage effectiveness = 85 %
Outside air = 100 °F DB, 70 °F WB
Secondary air wet-bulb = 70 °F
Wet-bulb leaving stage 1 = 64 °F
Find
Dry-bulb leaving stage 2 (°F)
Solution
T1 = 100 − 0.65 × (100 − 70) = 80.5 °F
T2 = 80.5 − 0.85 × (80.5 − 64) = 66.5 °F
80.5°F is the air leaving stage 1; the direct stage has not been applied.
74.5°F applies only the direct stage to the 100°F outdoor air and skips the indirect precooling.
71.6°F runs stage 2 toward the original 70°F wet-bulb; sensible cooling in stage 1 lowered the primary air's wet-bulb to 64°F.
69.4°F applies the direct stage to the 100°F outdoor dry-bulb instead of the 80.5°F air leaving stage 1: 100 − 0.85 × (100 − 64).
66.5°F: stage 1 brings the air to 80.5°F; stage 2 then closes 85% of the gap to the primary air's new 64°F wet-bulb.
Question 7 of 60 · HVAC Loads and Psychrometrics · 18–27 items
An air handling unit mixes return air and outside air. The properties are measured as shown in the table below. Calculate the final mixed air enthalpy using the exact mass-based energy balance equation:
hmix = (ṁ1 h1 + ṁ2 h2)/(ṁ1 + ṁ2)
Stream
Flow Rate
Enthalpy (Btu/lb)
Specific Volume (cu ft/lb)
Outside Air
2,500 cfm
34.0
14.0
Return Air
7,500 cfm
24.0
13.5
Given
Outside air = 2,500 cfm, h = 34.0 Btu/lb, v = 14.0 ft3/lb
Return air = 7,500 cfm, h = 24.0 Btu/lb, v = 13.5 ft3/lb
31.5 Btu/lb swaps the stream shares, giving outdoor air 75% of the mix.
29.0 Btu/lb is the plain average of 34.0 and 24.0, ignoring that return air is three times the outdoor flow.
26.6 Btu/lb multiplies each cfm by its specific volume instead of dividing, which overweights the lighter outdoor air.
26.5 Btu/lb weights by cfm directly, which assumes both streams have the same density; the table gives 14.0 and 13.5 ft³/lb.
26.4 Btu/lb: each cfm is divided by its own specific volume, giving 178.6 and 555.6 lb/min of dry air before weighting.
Question 8 of 60 · HVAC Loads and Psychrometrics · 18–27 items
An industrial facility has an aggregate total thermal capacity of 1,200,000 Btu/°F. The building envelope overall heat transfer coefficient–area product (ΣUA) is 8,000 Btu/(hr·°F). The continuous infiltration introduces an additional thermal conductance equivalent to 4,000 Btu/(hr·°F). Calculate the building's overall thermal time constant τ to assess its thermal lag in response to temperature changes.
0.01 hours is conductance divided by capacity, the inverse of the time constant (units of 1/h).
69 hours is 0.693τ, the time to close half of a temperature step; the time constant is the 63% point.
100 hours: 1,200,000 Btu/°F of capacity over the combined 12,000 Btu/(hr·°F) of envelope and infiltration conductance.
150 hours uses only the 8,000 Btu/(hr·°F) envelope conductance and leaves out infiltration.
300 hours uses only the 4,000 Btu/(hr·°F) infiltration conductance and leaves out the envelope.
Question 9 of 60 · HVAC Loads and Psychrometrics · 18–27 items
A critical manufacturing space requires a sensible heat ratio (SHR) of exactly 0.75 to maintain strict environmental conditions. Using the manufacturer's coil performance data below, at which chilled water (CHW) flow rate does the coil deliver this SHR?
CHW Flow (GPM)
Sensible Capacity (MBH)
Latent Capacity (MBH)
30
150.0
40.0
40
165.0
55.0
50
172.0
68.0
60
175.5
71.5
Given
Required SHR = 0.75
30 gpm: sensible 150.0 MBH, latent 40.0 MBH
40 gpm: sensible 165.0 MBH, latent 55.0 MBH
50 gpm: sensible 172.0 MBH, latent 68.0 MBH
60 gpm: sensible 175.5 MBH, latent 71.5 MBH
Find
Chilled water flow giving SHR 0.75 (gpm)
Solution
SHR = qs/(qs + qL)
SHR(40 gpm) = 165.0/(165.0 + 55.0) = 0.750
30 gpm: 0.789; 50 gpm: 0.717; 60 gpm: 0.711
70 gpm is outside the table; SHR keeps falling with flow, so extrapolating moves further from 0.75.
60 gpm gives 175.5/247.0 = 0.711, the lowest SHR in the table.
50 gpm gives 172.0/240.0 = 0.717, more latent removal than the space needs and a lower SHR.
30 gpm gives 150.0/190.0 = 0.789; too little latent capacity for the required 0.75.
Question 10 of 60 · HVAC Loads and Psychrometrics · 18–27 items
A multi-purpose classroom has a variable occupancy schedule throughout the day. Using the occupancy data and per-person latent heat gain rates provided below, calculate the peak latent load (Btu/h) for coil sizing purposes. Assume the latent moisture gain rate is 200 Btu/h per person during high-activity periods and 150 Btu/h per person during low-activity periods.
Time Period
Number of Occupants
Activity Level
8:00–10:00 AM
30
High
10:00 AM–12:00 PM
20
Low
12:00–2:00 PM
15
Low
2:00–4:00 PM
35
High
4:00–6:00 PM
10
Low
Given
8–10 AM: 30 people, high activity
10 AM–12 PM: 20 people, low
12–2 PM: 15 people, low
2–4 PM: 35 people, high
4–6 PM: 10 people, low
Latent gain: 200 Btu/h per person high, 150 Btu/h low
7,000 Btu/h: the 2–4 PM period, 35 people at the 200 Btu/h high-activity rate, is the largest of the five.
6,500 Btu/h averages the two high-activity periods (6,000 and 7,000 Btu/h); a coil is sized for the larger of them.
6,000 Btu/h is the 8–10 AM period (30 × 200); a later period has more people at the same rate.
5,250 Btu/h applies the low-activity 150 Btu/h rate to the 35-person period.
3,950 Btu/h is the average of the five period loads; a coil is sized for the peak, not the average.
Question 11 of 60 · HVAC Loads and Psychrometrics · 18–27 items
A building was designed with a cooling load of 50 tons based on a design outdoor condition of 95°F. If the actual outdoor temperature only reaches 85°F, approximately what percentage of the design capacity would be required under these conditions, assuming a linear relationship and a constant indoor setpoint of 75°F?
200% inverts the ratio, 20°F/10°F; the load falls with the smaller temperature difference, so the fraction is below 1.
100% assumes the load does not change with outdoor temperature, which contradicts the linear relationship given in the stem.
89% is the ratio of outdoor temperatures, 85/95; load follows the indoor–outdoor difference, not the outdoor reading.
50%: the indoor–outdoor difference falls from 20°F to 10°F, and the stated linear model scales load with that difference.
Question 12 of 60 · HVAC Loads and Psychrometrics · 18–27 items
At 5,000 ft elevation (barometric pressure 12.23 psia), an air stream has a dry-bulb temperature of 75°F and a relative humidity of 50%. At sea level (14.696 psia), the same dry-bulb temperature and relative humidity give a humidity ratio of 0.0092 lb/lb. Estimate the humidity ratio at 5,000 ft.
About 0.0077 lb/lb applies the pressure ratio upside down, 12.23/14.696, lowering W where it should rise.
About 0.0092 lb/lb reads the sea-level chart unchanged; humidity ratio depends on barometric pressure.
About 0.0111 lb/lb: the vapor pressure is the same at the same t and RH, but it now divides into a smaller dry-air pressure, raising W by about 1.2 times.
About 0.0133 lb/lb applies the 1.2 pressure correction twice.
Question 13 of 60 · HVAC Loads and Psychrometrics · 18–27 items
An engineer is evaluating the regeneration heat requirement for a solid desiccant wheel operating at elevated temperatures. High-moisture regeneration exhaust air leaves the wheel at 160°F. The engineer calculates the sensible regeneration exhaust energy using the standard dry-air specific heat of 0.24 Btu/(lb·°F). What is the primary thermodynamic error in this calculation approach?
Solution
cp,moist ≈ 0.24 + 0.444·W Btu/(lbda·°F)
At high W the vapor term is not negligible; use moist-air enthalpy h = 0.24t + W(1,061 + 0.444t).
The moisture driven off the wheel raises the stream's specific heat: water vapor is about 0.444 Btu/(lb·°F), so moist-air cp = 0.24 + 0.444W, and 0.24 alone understates the energy.
That transfer is on the process side of the wheel; the question is about the regeneration exhaust energy.
The desiccant itself does not vaporize; the heat of adsorption belongs to the water, and it is not the sensible-term error described.
An energy balance on a mass basis does not use density, so a density assumption is not the error in this calculation.
Matrix carry-over heats the process side; it does not change the specific heat used on the regeneration exhaust.
Question 14 of 60 · HVAC Loads and Psychrometrics · 18–27 items
Performance Metric
Conventional Sub-Cool & Reheat
Desiccant DOAS
Supply Air Dew Point
42°F
42°F
Cooling Coil Leaving Temp
42°F
55°F
Reheat Energy Required
High
Minimal
Regeneration Thermal Source
N/A
Waste Heat Available
An engineer is evaluating a Dedicated Outdoor Air System (DOAS) for an indoor pool facility with continuously high latent loads. Two system architectures are proposed: a conventional sub-cool and reheat cycle, and a DOAS integrating a solid desiccant wheel. Based on the system performance matrix, what is the primary thermodynamic advantage of selecting the desiccant DOAS strategy over the conventional refrigeration cycle?
Solution
Conventional: cool all air to the 42 °F dew point, then reheat. Desiccant DOAS: the wheel removes moisture, the coil stops at 55 °F, regeneration uses waste heat.
Both columns show a 42°F supply dew point; the advantage is how that dew point is reached.
Adsorption releases heat, so a desiccant wheel warms the process air rather than precooling it.
Both loads still have to be calculated; the system only splits which component handles each one.
The table still shows a cooling coil in the desiccant DOAS, leaving at 55°F; sensible cooling is still needed.
The desiccant does the moisture removal, so the coil leaves at 55°F instead of 42°F and the supply still reaches a 42°F dew point with little reheat.
Question 15 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A contractor selects an oversized two-way control valve for a heating coil that requires a maximum flow of 60 gpm. The chosen valve has a fully open Cv of 100. A differential pressure controller maintains a constant 9 psi drop across the valve at all times. Based on the manufacturer's flow characteristic table below, at approximately what percentage of stem travel will this valve operate during maximum design load, and what is the primary controllability risk associated with this selection?
Stem travel at design load (%) and controllability risk
Solution
Cv,req = Q/√ΔP = 60/√9 = 20 → 20/100 = 20 % of Cv
Table: 20 % of Cv at 40 % travel → oversized valve, poor low-load control (hunting).
40% open: 60 gpm at 9 psi needs Cv = 20, which is 20% of the valve's 100 and falls at 40% stroke; at part load it works near its seat and hunts.
The stroke is right, but choked flow needs a pressure drop near the vapor-pressure limit; 9 psi on a heating coil is far from it.
20% is the fraction of Cv, not of stem travel; the table puts 20% of Cv at 40% travel, and a 9 psi hot-water drop does not cavitate.
60% travel gives 50% of Cv, which would pass 150 gpm at 9 psi — far more than needed; with a constant 9 psi drop, authority is not the problem.
80% travel corresponds to 80% of Cv, four times the Cv of 20 the load requires.
Question 16 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
An AHU economizer is failing to meet the actual building cooling load. The mixed-air temperature controller is commanded to maintain a 55°F setpoint. The controller commands the outdoor air damper to modulate, and the damper feedback signal matches the command perfectly. The BAS displays a mixed-air temperature of 55°F. However, a temporary diagnostic logger reveals the actual mixed-air temperature is persistently 61°F. What is the primary diagnostic conclusion?
Solution
Loop satisfied on its own signal + actuator feedback = command + independent logger disagrees → sensor bias.
A tuning error would show as hunting or offset on the displayed value; the BAS shows the setpoint held steadily.
The loop holds 55°F on the signal it sees and the damper follows its command, so the 6°F gap to the logger points to the sensor.
Damper feedback matches the command, so the actuator is moving the damper as told.
A high-limit lockout would close the damper to minimum, yet the damper is modulating on command.
Question 17 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
Pressure (psia)
Tsat (°F)
hf (Btu/lb)
hg (Btu/lb)
30
15.3
15.0
105.0
150
105.9
45.0
118.0
A vapor-compression refrigeration cycle operates using the refrigerant properties shown in the table. The system provides a cooling capacity of 29,700 Btu/hr. The evaporator pressure is 30 psia and the condenser pressure is 150 psia. The compressor suction vapor is superheated to 30.3 °F. The liquid entering the expansion valve is subcooled to 95.9 °F. The enthalpy of the refrigerant at the compressor discharge is measured to be 130.0 Btu/lb.
Assume the specific heat of superheated vapor is 0.20 Btu/(lb · °R) and the specific heat of subcooled liquid is 0.30 Btu/(lb · °R). What is the required compressor power in Btu/hr?
9,900 Btu/hr: h1 = 108.0 and h3 = 42.0 Btu/lb give 66 Btu/lb of refrigeration effect, 450 lb/hr of flow, and 22 Btu/lb of compression.
10,371 Btu/hr ignores liquid subcooling: with h3 = 45.0 Btu/lb the refrigeration effect drops to 63 Btu/lb and flow rises to 471 lb/hr.
11,250 Btu/hr takes compression from saturated vapor (105.0 Btu/lb) instead of the superheated suction at 108.0 Btu/lb.
12,375 Btu/hr ignores both superheat and subcooling, using saturated values at both ends of the cycle.
Question 18 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A centrifugal fan draws 8,000 CFM of standard air through a return duct and discharges it through a supply duct. The return duct has a static pressure of -1.50 in. w.g. and an average air velocity of 1,200 fpm at the fan inlet. The supply duct has a static pressure of 2.20 in. w.g. and an average air velocity of 1,600 fpm at the fan discharge. What is the total pressure developed by the fan?
Given
Airflow = 8,000 cfm standard air
Inlet static pressure = −1.50 in. w.g.
Inlet velocity = 1,200 fpm
Discharge static pressure = 2.20 in. w.g.
Discharge velocity = 1,600 fpm
Find
Fan total pressure (in. w.g.)
Solution
VP = (V/4,005)2: VPo = 0.160, VPi = 0.090 in. wg
FTP = (2.20 + 0.160) − (−1.50 + 0.090) = 3.77 in. wg
3.61 in. wg is the fan static pressure (total pressure minus outlet velocity pressure), not the total pressure asked for.
3.70 in. wg is the static pressure rise only; the velocity pressures at inlet and outlet are left out.
3.77 in. wg: total pressure out (2.20 + 0.16) minus total pressure in (−1.50 + 0.09).
3.86 in. wg adds the outlet velocity pressure but forgets to subtract the inlet velocity pressure.
3.95 in. wg adds both velocity pressures; the inlet velocity pressure belongs to the inlet total pressure and is subtracted.
Question 19 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
Pressure (psia)
hf (Btu/lb)
hfg (Btu/lb)
15
181.2
969.7
50
250.2
924.2
200
355.5
843.3
A facility cascades saturated steam condensate through two successive flash tanks to recover low-pressure steam. Initially, 10,000 lb/hr of saturated condensate at 200 psia enters a medium-pressure flash tank maintained at 50 psia. The remaining saturated liquid drains and flows into a low-pressure flash tank maintained at 15 psia. Based on the provided thermodynamic properties, what is the total combined mass flow of flash steam produced across both stages?
631 lb/hr is only the second-stage flash from the low-pressure tank.
1,139 lb/hr is only the first stage; the liquid draining to the 15 psia tank flashes again.
1,770 lb/hr: 1,139 lb/hr flashes at 50 psia, and the 8,861 lb/hr of liquid left flashes another 631 lb/hr at 15 psia.
1,797 lb/hr flashes 200 psia straight to 15 psia in one step; two stages yield slightly less because the first flash takes its steam at 50 psia.
1,851 lb/hr applies the second-stage fraction to the full 10,000 lb/hr instead of the liquid left after the first flash.
Question 20 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A hydronic cooling loop has a design flow rate of 250 gpm. The piping, coils, and fittings account for a dynamic pressure drop of 60 ft of head, excluding the control valve. An engineer specifies a two-way control valve with a fully open flow coefficient (Cv) of 65. Assuming the fluid is water with a specific gravity of 1.0, what is the calculated fully open valve authority?
0.198 adds 14.79 psi to 60 ft without converting head to psi.
0.246 divides the valve's 14.79 psi by the 60 ft of circuit head, mixing units and leaving the valve out of the denominator.
0.363: the open valve drops (250/65)² = 14.79 psi, against 60 ft / 2.31 = 25.97 psi in the rest of the circuit.
0.570 is valve drop over the rest of the circuit; authority divides by the total including the valve.
0.637 is the circuit's share of the total drop, the complement of the valve authority.
Question 21 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
An engineer is verifying the superheat of an evaporator charged with refrigerant blend R-407A. The suction pressure measured at the evaporator outlet is 45 psig. The actual temperature of the suction line at the same location is measured at 38°F. The saturation properties for R-407A are provided in the table below.
Calculate the actual operating superheat of the evaporator.
11.0°F: for a zeotropic blend superheat is measured from the dew point, 27.0°F at 45 psig, where the last liquid has boiled off.
15.5°F reads the dew point from the 40 psig row instead of the measured 45 psig.
16.3°F uses the midpoint of bubble and dew temperatures, an average used for evaporator rating, not for superheat.
17.5°F reads the bubble point from the 50 psig row (20.5°F) and measures from the bubble point instead of the dew point at 45 psig.
21.5°F measures from the 16.5°F bubble point, where boiling begins; that overstates the vapor-only superheat by the glide.
Question 22 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A facility has three interconnected zones (A, B, and C). Assume that the only internal air transfers occur sequentially: air transfers from Zone A directly into Zone B, and air transfers from Zone B directly into Zone C. Given the steady-state mechanical supply, return, and exhaust flow rates for each zone in the table below, determine the unknown interzone transfer airflow rate from Zone B to Zone C. Assume standard air density.
100 cfm balances zone B without the 200 cfm it receives from zone A.
200 cfm is the transfer from A into B, not the one from B into C.
300 cfm: zone A's 200 cfm surplus enters B, and B then has 1,800 + 200 − 1,500 − 200 = 300 cfm to pass on; zone C's balance confirms it.
500 cfm leaves out zone B's own 200 cfm of exhaust.
700 cfm adds zone B's exhaust instead of subtracting it as an outflow.
Question 23 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A testing, adjusting, and balancing (TAB) technician is performing a proportional balancing procedure on a variable volume system with four terminal zones. All branch dampers are initially fully open. The technician measures the initial actual airflow and compares it to the design airflow, as listed in the table below. According to the proportional balancing method, one zone must be identified as the index terminal and its damper must remain fully open while the other branches are progressively throttled. What is the initial flow ratio of the index terminal for this system?
Zone
Design CFM
Initial Actual CFM
Zone 1
500
600
Zone 2
800
720
Zone 3
400
480
Zone 4
300
240
Given
Zone 1: design 500 cfm, actual 600 cfm
Zone 2: design 800 cfm, actual 720 cfm
Zone 3: design 400 cfm, actual 480 cfm
Zone 4: design 300 cfm, actual 240 cfm
Find
Initial flow ratio of the index terminal (%)
Solution
ratio = actual/design: 1.20, 0.90, 1.20, 0.80 → index = zone 4 at 80 %
80%: zone 4 (240 of 300 cfm) has the lowest ratio, so it is the index terminal and stays fully open.
90% is zone 2; it is short of design but not the shortest, so it gets throttled toward the index ratio.
100% is the final target ratio after the fan is adjusted, not the initial ratio of the index terminal.
120% belongs to zones 1 and 3, the branches with the most excess flow, which are throttled first.
125% inverts zone 4's ratio (300/240); flow ratio is actual over design.
Question 24 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A 24 in. × 24 in. square ceiling grille is operating in a commercial return air system and is drawing in 800 cfm of air. The manufacturer's catalog lists the effective area factor (Ak) for this specific grille design as 2.50 ft2. What is the effective face velocity at the grille?
Given
Grille size = 24 in × 24 in
Airflow = 800 cfm
Effective area Ak = 2.50 ft2
Find
Effective face velocity (fpm)
Solution
v = Q/Ak = 800 cfm ÷ 2.50 ft2 = 320 fpm
80 fpm multiplies the 4 ft² nominal area by 2.50, treating Ak as a factor instead of an area.
200 fpm uses the 4 ft² nominal face area; the blades and frame leave only 2.50 ft² open.
320 fpm: 800 cfm through the 2.50 ft² effective area that the catalog gives for this grille.
2,000 fpm multiplies cfm by area instead of dividing.
Question 25 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
Diffuser Model
Total Pressure Drop at 500 CFM
Existing
0.10 in. w.g.
New
0.25 in. w.g.
A terminal branch duct serves a single diffuser that requires 500 CFM. The branch has a balancing damper that is currently throttled, creating a 0.30 in. w.g. pressure drop. The existing diffuser has a total pressure drop of 0.10 in. w.g. at 500 CFM. An engineer replaces the existing diffuser with a new model that has a total pressure drop of 0.25 in. w.g. at 500 CFM. Assuming the main duct pressure upstream of the branch remains constant, what must the new pressure drop across the balancing damper be to maintain exactly 500 CFM, and how must the damper be adjusted relative to its original position?
Given
Diffuser airflow = 500 CFM
Damper pressure drop = 0.30 in. w.g.
Existing diffuser drop = 0.10 in. w.g.
New diffuser drop = 0.25 in. w.g.
Find
New damper pressure drop (in. w.g.) and adjustment
Solution
ΔPbranch = 0.30 + 0.10 = 0.40 in. w.g.
ΔPdamper,new = 0.40 − 0.25 = 0.15 in. w.g. → open the damper
0.05 in. w.g. subtracts the new diffuser drop from the old damper drop and loses the old diffuser's 0.10.
Leaving the damper alone puts 0.55 in. w.g. of resistance on a 0.40 in. w.g. branch, so flow falls below 500 cfm.
0.55 in. w.g. adds the new diffuser drop to the old damper drop; closing the damper would cut flow further.
0.45 in. w.g. adds the 0.15 increase to the damper; the damper has to give up that pressure, not take more.
0.15 in. w.g.: the branch still has 0.40 in. w.g. available, the new diffuser takes 0.25, so the damper must open to drop only 0.15.
Question 26 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
A chilled water system follows the formula Q = 500 × gpm × ΔT. If a system needs to remove 720,000 Btu/h with a ΔT of 8°F, what is the required chilled water flow rate in gpm?
Given
Heat removal Q = 720,000 Btu/h
Water temperature difference Δt = 8 °F
Find
Chilled water flow (gpm)
Solution
gpm = Q/(500·ΔT) = 720,000/(500 × 8) = 180 gpm
3 gpm divides the load per minute (12,000 Btu/min) by 500 × 8; the 500 constant already turns gpm into lb/h, so the load must be in Btu/h.
60 is the load in tons (720,000/12,000), not a water flow.
90 gpm uses a 1,000 constant instead of 500; the 500 comes from 8.33 lb/gal × 60 min/h × 1 Btu/lb·°F.
180 gpm: 720,000 Btu/h divided by 500 × 8°F.
Question 27 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
How does intercooling in a multi-stage vapor compression cycle affect the specific work input compared to a single-stage compression?
Solution
w ∝ Tin·[(p2/p1)(k−1)/k − 1] — lower Tin at each stage → lower total work
Added equipment raises first cost, not the thermodynamic work per pound of refrigerant.
Compression work depends on inlet temperature; removing heat between stages lowers it.
Intercooling changes the gas state between stages, which directly lowers the work per pound.
Cooler gas at each stage inlet is denser and needs less work per pound for the same pressure ratio, so total work falls.
Question 28 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
What is the primary advantage of a water-side economizer compared to an air-side economizer?
Solution
Air-side: limited by outdoor enthalpy/humidity. Water-side: limited by outdoor wet-bulb via the tower; outdoor air stays out.
An air-side economizer uses only fan energy and is usually the more efficient of the two when outdoor air is suitable.
The tower and plate exchanger add water treatment and cleaning, so maintenance is usually higher.
Water-side economizers need heat exchangers, valves and controls, usually costing more than air-side dampers.
A water-side economizer works off the cooling tower, so it depends on outdoor wet-bulb and never brings humid outdoor air into the building.
Question 29 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
What is the primary function of a thermostatic steam trap in a steam heating system?
Solution
Thermostatic element: open when cool (condensate/air below saturation), closed at steam temperature.
Steam pressure is set by the boiler controls and pressure-reducing valves, not by a trap.
Removing condensate helps avoid water hammer, but a trap does not prevent it by itself; slope, drip legs and drainage do that.
Steam flow to equipment is regulated by control valves; a trap passes condensate, not modulated steam.
Every steam trap's job is to pass condensate while holding back live steam; a thermostatic trap opens on the lower temperature of condensate and air, so it vents air well at startup.
Question 30 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
In a multiplex refrigeration system for food storage, several compressors are connected to common suction and discharge manifolds. How does this arrangement improve operation?
Solution
Parallel rack: one suction group, N compressors → capacity steps + N−1 redundancy.
Manifolding compressors does not halve the charge, which is set mainly by the evaporators, condenser and piping.
Speed is set by the motor and drive; a manifold does not raise compressor speed.
The compressors share common suction and discharge pressures, so they are coupled, not isolated.
Shared manifolds let the rack stage compressors to follow load and keep refrigeration going on the remaining units when one is down.
Question 31 of 60 · HVAC Equipment and Components · 24–36 items
A condensate pump is installed 4 ft below a condensate receiver operating at atmospheric pressure. The suction line is 2-in. Schedule 40 steel pipe (inside diameter 2.067 in.), 12 ft long, with two 90° elbows (K = 0.9 each) and one gate valve (K = 0.2); the friction factor is f = 0.020. The pump handles 200°F condensate at 50 gpm. The vapor pressure at 200°F is 11.526 psia, atmospheric pressure is 14.7 psia, and the water density at 200°F is 60.1 lb/ft³. The pump manufacturer specifies NPSHr = 8 ft. What is the available NPSH (NPSHa), and does it meet a required margin of 3 ft above NPSHr?
Given
Pump = 4 ft below the receiver; receiver at atmospheric pressure
Suction pipe = 2 in. Sch 40, ID 2.067 in., 12 ft long
Fittings = two 90° elbows K = 0.9 each, gate valve K = 0.2
Condensate = 200 °F, 50 gpm
Vapor pressure = 11.526 psia
Atmospheric pressure = 14.7 psia
Friction factor f = 0.020
Density = 60.1 lb/ft3
NPSHr = 8 ft
Required margin = 3 ft
Find
NPSHa (ft) and whether the 3-ft margin is met
Solution
V = (50/448.8)/(π/4 × (2.067/12)2) = 4.78 ft/s, V2/2g = 0.355 ft
10.40 ft: 35.22 ft of atmosphere plus 4 ft of static head, minus 27.62 ft of vapor pressure and 1.20 ft of suction losses; that is only 2.40 ft above NPSHr.
37.40 ft uses the vapor pressure of 60°F water; the condensate is at 200°F.
11.60 ft leaves out the 1.20 ft of pipe and fitting losses, which is what pushes the margin under 3 ft.
38.02 ft forgets the vapor pressure; at 200°F it cancels most of the atmospheric head.
2.40 ft subtracts the 4 ft of static head; the pump sits below the receiver, so the head adds.
Question 32 of 60 · HVAC Equipment and Components · 24–36 items
What is the primary advantage of using a flooded evaporator rather than a direct expansion (DX) evaporator in a large chiller?
Solution
Flooded: all tubes wetted → high h, close approach; costs: large charge, level and oil management.
Flooded evaporators need level control (float or level sensor) and oil-return provisions; distribution is not simpler overall.
Every tube in a flooded shell is bathed in boiling liquid, so the refrigerant-side coefficient is high and the approach temperature small.
Flooded evaporators are the type that use low-side floats or level controls.
A flooded shell holds a large pool of liquid; low charge is the DX evaporator's advantage.
Question 33 of 60 · HVAC Equipment and Components · 24–36 items
A cooling tower circulates 400 gpm with a range of 12°F. The tower has a drift rate of 0.002% of the circulating flow, and the plant targets 4.0 cycles of concentration (COC). Using the evaporation heuristic E ≈ 0.001 × gpm × range, compute the total makeup water flow (gpm).
Given
Circulating flow = 400 gpm
Range = 12 °F
Drift = 0.002 % of circulation
Cycles of concentration = 4.0
Evaporation ≈ 0.001 × gpm × range
Find
Total makeup flow (gpm)
Solution
E = 0.001 × 400 × 12 = 4.8 gpm, D = 0.00002 × 400 = 0.008 gpm
B + D = E/(COC − 1) = 1.6 gpm → B = 1.592 gpm
M = E + B + D = 6.40 gpm (9,216 gal/day)
4.80 gpm is evaporation alone; dissolved solids must also leave with blowdown and drift.
6.00 gpm takes blowdown as E/COC = 1.2 gpm instead of E/(COC − 1).
6.40 gpm: evaporation 4.8 gpm plus 1.6 gpm of blowdown and drift together, since E/(COC − 1) already covers both concentrated-water losses.
7.20 gpm reads the drift as 0.2% of circulation (0.8 gpm) and then adds it on top of E/(COC − 1) = 1.6 gpm, which already includes drift: 4.8 + 1.6 + 0.8.
64.00 gpm uses 0.01 instead of 0.001 in the evaporation heuristic, making everything ten times too large.
Question 34 of 60 · HVAC Equipment and Components · 24–36 items
What is the primary purpose of water treatment in a cooling tower system?
Treatment keeps surfaces near their clean coefficient; it cannot raise performance above clean-tube design.
Evaporation concentrates minerals and the warm, aerated water breeds bacteria; treatment keeps scale, corrosion and biofilm (including Legionella) in check.
Cooling tower water is not an antifreeze solution; freeze protection comes from basin heaters and drain-down, not treatment chemicals.
Range is set by the heat load and water flow; chemicals do not change it.
Question 35 of 60 · HVAC Equipment and Components · 24–36 items
Which VRF (variable refrigerant flow) arrangement is required for simultaneous heating and cooling in different zones?
Solution
Simultaneous heating/cooling = heat-recovery VRF (3-pipe with branch selectors, or 2-pipe with a BC controller).
Isolation valves are for service; they shut units off but cannot change their mode.
A heat-recovery VRF system routes hot gas and liquid through branch selector or BC-controller boxes so some zones heat while others cool.
A heat-pump VRF system switches the whole condensing unit between heating and cooling; zone thermostats cannot make it do both at once.
A distributor splits refrigerant among coil circuits; it does not let zones run in opposite modes.
Question 36 of 60 · HVAC Equipment and Components · 24–36 items
What is the principal advantage of a refrigerant-to-water brazed plate heat exchanger over a shell-and-tube design for a water-cooled condensing unit?
Solution
BPHE: compact, high h, low charge; limits: no mechanical cleaning, fouling and freeze sensitivity.
Brazed plate units are more vulnerable to freeze damage because the small channels rupture when water freezes.
Thin corrugated plates give very high coefficients, so a brazed plate unit does the same duty in a fraction of the volume of a shell-and-tube.
Narrow plate channels foul and plug more readily than shell-and-tube passages.
A brazed plate pack is sealed and cannot be opened, so only chemical cleaning in place is possible.
Question 37 of 60 · HVAC Equipment and Components · 24–36 items
A 400-ton chiller rejects 5,400,000 Btu/h to a cooling tower at 95°F entering and 85°F leaving water. The site design wet-bulb is 80°F, while the tower's catalog rating of 380 tower tons is at 78°F wet-bulb. The manufacturer's correction factor for an 80°F wet-bulb is 0.92 (capacity at site = 0.92 × rated). Using 15,000 Btu/h per tower ton, what rated capacity does the job require, and is the 380-ton tower adequate?
331 tons multiplies by 0.92 instead of dividing, as if the warmer wet-bulb helped the tower.
391 tons: 5,400,000/15,000 = 360 tower tons at site, and only 92% of a rating is available at 80°F wet-bulb, so 360/0.92 is needed.
489 tons combines the 12,000 Btu/h ton with the correction; the duty in tower tons is 360, not 450.
360 tons is the duty at site conditions; it ignores that the tower delivers only 92% of its rating at the higher wet-bulb.
450 tons divides by 12,000 Btu/h, the refrigeration ton; tower tons are based on 15,000 Btu/h of heat rejection.
Question 38 of 60 · HVAC Equipment and Components · 24–36 items
Parameter
Scenario A (non-condensing)
Scenario B (condensing)
Flue gas temperature
385°F
125°F
Measured O2 (%)
5.5
4.0
Estimated stack loss (%)
12.5
3.2
Water dew point of flue gas
~135°F
~135°F
A consulting engineer is evaluating two boiler venting scenarios for a commercial building retrofit. Both use natural gas and deliver the same heating output, and both are connected to the existing Type B (galvanized steel) vent. Based on the operating data in the table, which assessment and recommendation are correct?
Solution
Flue temperature vs water dew point (~135 °F for natural gas): A 385 °F > 135 °F → dry; B 125 °F < 135 °F → condensing, acidic condensate → corrosion-resistant vent + neutralization.
Type B is listed for non-condensing gas appliances at A's temperature, and a trap does not protect galvanized steel from acid.
This has the dew-point logic backwards: gas above its dew point stays dry, gas below it condenses.
B's 3.2% stack loss shows latent heat is being recovered, and the vent material still has to change.
Condensing does recover latent heat, but the acidic condensate corrodes galvanized Type B vent; inspection does not make it acceptable.
Scenario B's 125°F flue is below the 135°F dew point, so acidic condensate forms; it needs PVC, CPVC or AL29-4C venting and a neutralizer, while A stays dry.
Question 39 of 60 · HVAC Equipment and Components · 24–36 items
A water-cooled chiller will use brackish water with high chloride content in its condenser. Which factor is most critical when selecting the condenser tube material?
Solution
Aggressive water → choose metallurgy (90/10 CuNi, titanium, duplex) for corrosion resistance, then check conductivity and cost.
The cheapest tube that fails early costs more over the chiller's life; corrosion resistance governs.
Chlorides pit copper and many stainless grades; cupronickel or titanium tubes are chosen for corrosion resistance first.
Copper conducts best but corrodes quickly in brackish water; a slightly lower conductivity costs far less than tube failures.
A thicker wall delays a leak but does not stop pitting; the material choice decides tube life.
Question 40 of 60 · HVAC Equipment and Components · 24–36 items
In a multi-circuit DX cooling coil, how does the design of the refrigerant distributor influence coil capacity and defrost performance?
Solution
Distributor = equal pressure drop to each circuit → equal feed → full coil use, even frost, complete defrost.
A coil below 32°F still collects frost however well it is fed; defrost is still needed.
The distributor's job is splitting flow evenly; uneven feed directly affects how the coil frosts and defrosts.
Maldistribution leaves part of the coil underused, so capacity drops directly.
An even split keeps every circuit active; a starved circuit runs high superheat and frosts unevenly, which cuts capacity and leaves ice after defrost.
Question 41 of 60 · HVAC Equipment and Components · 24–36 items
A design engineer calculates the expected leaving water temperature of a cooling tower serving a 500-ton chiller. The local design dry-bulb temperature is 95°F and the wet-bulb is 78°F. The engineer uses an approach specification of 10°F and computes the tower leaving water temperature as 95 − 10 = 85°F. Identify the error and give the correct leaving water temperature.
105°F adds the approach to the dry-bulb; an evaporative tower cools toward the wet-bulb, not above the dry-bulb.
Range is entering minus leaving water; the 10°F given is the approach, and 85°F remains wrong.
The error is named right, but 68°F subtracts the approach from the wet-bulb; leaving water is always above the wet-bulb.
Approach is measured from the wet-bulb, the limit an evaporative tower can approach, so 78 + 10 = 88°F.
85°F subtracts the approach from the dry-bulb, which has no role in tower approach.
Question 42 of 60 · HVAC Equipment and Components · 24–36 items
An engineering intern submits the following solution for an energy recovery ventilator (ERV) problem: "Given outdoor air at 95°F dry-bulb, 78°F wet-bulb (enthalpy 41.6 Btu/lb) and return air at 75°F, 50% RH (enthalpy 28.1 Btu/lb), with 2,000 cfm of outdoor air and a device sensible effectiveness εs = 0.70, the enthalpy recovered is Δh = εs(hoa − hra) = 0.70 × (41.6 − 28.1) = 9.45 Btu/lb. Then total recovery is Q = 2,000 cfm × 9.45 Btu/lb = 18,900 Btu/h." Identify the errors in this solution.
Solution
q = 4.5 · cfm · εt · (hoa − hra)
With the total effectiveness: q = 4.5 × 2,000 × εt × 13.5 Btu/lb.
The enthalpies 41.6 and 28.1 Btu/lb match the stated conditions; the sensible effectiveness was applied to an enthalpy difference, not a temperature change.
The effectiveness error is real, but cfm × Btu/lb does not give Btu/h; the 4.5 factor is missing.
Both the effectiveness type and the units are wrong, so the 18,900 Btu/h result is not valid.
The unit error is real, but a sensible effectiveness says nothing about moisture transfer, so it cannot be applied to enthalpy.
Total heat needs the total (enthalpy) effectiveness, and cfm must become dry-air mass flow (× 0.075 × 60, i.e., 4.5 × cfm × Δh).
Question 43 of 60 · HVAC Equipment and Components · 24–36 items
A gasketed-plate heat exchanger is specified for a heating application with a design flow rate of 180 gpm. The manufacturer states a pressure drop of 6.5 psi at this flow. The rest of the system (piping and fittings) adds 22 ft of head loss. The fluid is water at 140°F with a specific weight of 61.5 lb/ft³. Calculate the brake horsepower the pump must deliver. Pump efficiency is 80% and motor efficiency is 91%.
Given
Flow = 180 gpm
Exchanger pressure drop = 6.5 psi
Other system head loss = 22 ft
Specific weight γ = 61.5 lb/ft3
Pump efficiency = 80 %
Motor efficiency = 91 %
Find
Pump brake horsepower (hp)
Solution
hHX = 6.5 × 144/61.5 = 15.2 ft → H = 22 + 15.2 = 37.2 ft
1.24 hp treats 6.5 psi as 6.5/61.5 ft, dropping the 144 in²/ft² conversion.
1.67 hp is the water horsepower; the 80% pump efficiency has not been applied.
2.09 hp: 6.5 psi is 15.2 ft of 140°F water, so the pump makes 37.2 ft at 180 gpm with SG 0.986 and 80% efficiency.
2.29 hp also divides by the 91% motor efficiency; that is electrical input, not the brake horsepower at the pump shaft.
3.32 hp counts the 22 ft of system loss twice.
Question 44 of 60 · HVAC Equipment and Components · 24–36 items
Flow (gpm)
Head (ft)
0
100
40
95
80
80
120
55
160
20
Two identical centrifugal pumps operate in parallel to serve a hydronic system. The system head curve is defined by Hsys = 10 + 0.002 Q2, where Q is the total system flow in gpm and Hsys is in feet. A single pump's performance at its base speed of 1,750 RPM is given in the table. If both pumps are operated at identical reduced speeds via VFDs to provide a total combined system flow of 120 gpm, what is the required operating speed of each pump?
pump: H = 100 − 0.003125·Q2 at 1,750 RPM → scaled: H = 100r2 − 0.003125·Q2
38.8 = 100r2 − 11.25 → r = 0.7075 → N = 1,238 RPM
750 RPM scales speed by a flow ratio read along the full-speed curve; affinity laws apply along a system curve through the origin, which this one is not.
1,090 RPM scales only the shutoff head to 38.8 ft (√0.388) and ignores the flow term of the pump curve.
1,238 RPM: each pump carries 60 gpm at the system's 38.8 ft, and the scaled curve 100(N/1,750)² − 0.003125 × 60² meets that point when N/1,750 = 0.707.
1,602 RPM sizes one pump for the full 120 gpm; in parallel each pump carries half the flow at the common head.
1,750 RPM is full speed; two pumps at full speed would push the system well past 120 gpm.
Question 45 of 60 · HVAC Equipment and Components · 24–36 items
Equipment
System Type
Rated Approach
Unit A
Closed-Circuit Cooler
9.5°F
Unit B
Open Cooling Tower
4.0°F
Unit C
Plate-and-Frame HX
3.5°F
A critical cooling loop requires a maximum leaving fluid temperature of 86.0°F while rejecting heat to a 78.0°F ambient wet-bulb environment. The engineer evaluates two fully-isolated system options:
Option 1: Unit A (Closed-Circuit Fluid Cooler)
Option 2: Unit B (Open Cooling Tower) coupled with Unit C (Plate-and-Frame HX)
Based on the vendor approach data in the table, determine which system meets the leaving fluid requirement and identify the corresponding combined approach temperature.
Given
Required leaving fluid = 86.0 °F
Ambient wet-bulb = 78.0 °F
Closed-circuit cooler approach = 9.5 °F
Open tower approach = 4.0 °F
Plate-and-frame HX approach = 3.5 °F
Find
Option that meets the requirement and its combined approach (°F)
8.0°F is the allowed approach, not the fluid cooler's rated 9.5°F.
The closed-circuit cooler's 9.5°F approach would deliver 87.5°F, above the 86.0°F limit.
9.5°F exceeds the allowed 8.0°F, so the fluid cooler alone cannot hold 86.0°F.
4.0°F counts only the tower; the plate exchanger adds its own 3.5°F approach in series.
Option 2: the tower's 4.0°F approach plus the plate exchanger's 3.5°F gives 7.5°F, inside the 8.0°F allowed (86 − 78).
Question 46 of 60 · HVAC Equipment and Components · 24–36 items
A chiller rated at 100 tons operates at 60% load and draws 77 kW at that load. What is its part-load efficiency in kW/ton?
Given
Rated capacity = 100 tons
Part load = 60 %
Power input = 77 kW
Find
Part-load efficiency (kW/ton)
Solution
kW/ton = 77 kW/(0.60 × 100 tons) = 1.28 kW/ton
0.46 kW/ton scales the power by 60% as well, applying the part-load fraction to the measured kW.
0.77 kW/ton divides by the 100-ton rating instead of the 60 tons the chiller is producing.
1.28 kW/ton: 77 kW divided by the 60 tons actually produced.
9.35 is the EER (12/1.28), a different efficiency metric, not kW per ton.
Question 47 of 60 · HVAC Equipment and Components · 24–36 items
Parameter
Baseline Burner
Low-NOx Retrofit
Combustion Efficiency
80.0%
75.0%
NOx Emissions
0.12 lb/MMBtu
0.04 lb/MMBtu
A facility is evaluating a low-NOx burner retrofit for a natural-gas boiler with an output capacity of 60 MMBtu/hr operating 8,000 hours per year at full load. Natural gas costs $6.00 per MMBtu. Using the provided performance data, calculate the expected annual increase in fuel costs resulting from the retrofit.
$144,000 takes 5 percentage points of the output-basis cost ($2,880,000); efficiency divides the output, so input changes by more.
$192,000 takes 5 percentage points of the retrofit fuel cost; the increase is the difference of the two inputs.
$240,000: fuel input rises from 60/0.80 = 75 to 60/0.75 = 80 MMBtu/hr, and 5 × 8,000 h × $6.00 = $240,000.
$3,600,000 is the baseline's total fuel cost (75 MMBtu/hr), not the increase.
$3,840,000 is the retrofit's total fuel cost (80 MMBtu/hr), not the increase.
Question 48 of 60 · HVAC Equipment and Components · 24–36 items
Stream
Enthalpy (Btu/lb)
Boiler Water
355.0
Blowdown Water
125.0
Makeup Water In
28.0
Makeup Water Out
85.0
A boiler system utilizes a continuous blowdown heat recovery heat exchanger to preheat incoming makeup water. The blowdown mass flow rate is 4,500 lb/hr. The heat exchanger has a heat transfer efficiency of 85% to account for ambient losses. Using the thermodynamic properties provided in the table, calculate the actual energy recovered and transferred to the makeup water.
879,750 Btu/hr: 4,500 lb/hr of blowdown cools from 355.0 to 125.0 Btu/lb, releasing 1,035,000 Btu/hr, of which 85% reaches the makeup water.
1,035,000 Btu/hr is the heat given up by the blowdown before the 85% exchanger efficiency.
1,217,647 Btu/hr divides the 1,035,000 Btu/hr by 0.85 instead of multiplying; losses reduce the heat that reaches the makeup water.
1,357,875 Btu/hr applies the 85% efficiency to the full blowdown enthalpy rather than to the drop across the exchanger.
1,597,500 Btu/hr uses the full 355.0 Btu/lb enthalpy of the blowdown, as if it left at 0 Btu/lb.
Question 49 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A rooftop HVAC unit includes a 15-hp, 460-V, three-phase compressor motor. The unit nameplate lists the compressor FLA = 21 A, MCA = 26.25 A, and MOCP = 40 A. The VFD manufacturer specifies that the drive's rated output is 25 A and recommends upstream branch-circuit protection not to exceed 35 A to prevent nuisance tripping during transient inrush and to protect internal drive components. The local AHJ requires documentation of OCPD selection rationale per NEC Article 430 and manufacturer instructions. Which of the following represents the most appropriate OCPD selection and required justification for this installation?
Solution
OCPD ≤ min(MOCP 40 A, drive max 35 A) and ≥ MCA 26.25 A → 35 A; conductors per MCA.
The VFD's 35 A maximum is more restrictive than the unit nameplate's 40 A MOCP, and NEC 110.3(B) requires following the listed equipment's instructions.
45 A exceeds both the 40 A MOCP and the drive's 35 A maximum.
Nothing conflicts: the lower of the two maximums governs, so selection need not wait for the AHJ.
25 A is below the 26.25-A MCA, so the breaker would trip under normal running current.
The drive manufacturer's maximum is part of its listing and instructions, so it is not advisory.
Question 50 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
An HVAC contractor projects annual energy savings of 35,000 kWh. If electricity costs $0.10 per kWh, determine the annual dollar savings and convert the energy saving to Btu (1 kWh = 3,412 Btu).
Given
Annual energy reduction = 35,000 kWh
Electricity cost = $0.10/kWh
Conversion 1 kWh = 3,412 BTU
Find
Annual dollar savings and energy saving (BTU)
Solution
$ = 35,000 × $0.10 = $3,500
E = 35,000 kWh × 3,412 Btu/kWh = 1.19 × 108 Btu
1.19 × 105 uses 3.412, the kBtu-per-kWh factor, and labels the result Btu.
$3,500 and 1.19 × 108 Btu: 35,000 kWh × $0.10, and 35,000 × 3,412 = 119,420,000 Btu.
1.03 × 10¹ divides kWh by 3,412 instead of multiplying.
$350 slips a decimal in the rate, using $0.01 per kWh.
Question 51 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A campus with its own on-site boiler plant burns 2,500 therms of natural gas per month (1 therm = 100,000 Btu); the plant efficiency is 80%. The campus has 150,000 ft² of building area. What is the site energy use intensity (EUI), based on the fuel purchased, in kBtu/ft²/month?
The on-site plant is part of the site, so its 80 % efficiency is not applied.
0.17 uses 10,000 Btu per therm instead of 100,000.
1.33 multiplies by the 80% plant efficiency, giving the heat the plant delivers; a site EUI based on fuel purchased counts the gas itself.
1.67: the plant is on the campus, so site energy is the fuel purchased, 2,500 therms × 100,000 Btu = 250,000 kBtu, over 150,000 ft².
2.08 divides by the 80% efficiency, as if the gas were the useful output and more fuel were burned.
Question 52 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
An ERV system has an incremental first cost of $25,000. It is rated to recover 150,000 Btu/hr of gross heating energy. The unit operates for 4,000 hours annually during the heating season. The facility's heating energy cost is $12.00 per MMBtu. The added static pressure introduces a parasitic fan penalty of 3 kW, which runs continuously during the same 4,000 operating hours. Electricity costs $0.15 per kWh. What is the simple payback period for this ERV installation?
1.79 years credits heat recovery for all 8,760 hours of the year instead of the 4,000-hour heating season.
2.78 years adds the $1,800 fan cost to the savings instead of subtracting it.
3.47 years uses the gross $7,200 heating savings and ignores the fan penalty.
3.73 years prices the 12,000 kWh of fan energy at the $12/MMBtu heating rate (40.9 MMBtu, $491) instead of $0.15/kWh ($1,800).
4.63 years: 600 MMBtu of recovered heat is worth $7,200, less $1,800 for 12,000 kWh of fan energy, so $25,000 ÷ $5,400.
Question 53 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A medical facility is evaluating a proposed upgrade to its primary air-handling unit controls. Based on the system modifications shown below, which statement best characterizes the expected changes to the system's life-cycle cost profile?
Control Parameter
Baseline Sequence
Proposed Sequence Upgrade
Airflow Delivery
Constant volume delivery
Variable frequency drive (VFD)
Ventilation Rate
Fixed 20% outdoor air
Demand-controlled (CO2 sensors)
Maintenance Cycle
Standard filter replacement
Added sensor calibration routine
Solution
LCC = first cost + PW(energy) + PW(O&M): VFD + DCV → energy ↓, first cost ↑, O&M ↑ (sensor calibration).
The table adds a sensor calibration routine, so the O&M profile changes.
VFD fan control and CO2-based ventilation cut fan and outdoor-air energy, while the new sensors add calibration work.
First cost does rise, but O&M rises too because of the added calibration.
Sensor power is a few watts, tiny next to the fan and outdoor-air conditioning savings.
Adding drives and sensors raises first cost, and the calibration routine adds maintenance.
Question 54 of 60 · HVAC Loads and Psychrometrics · 18–27 items
Relative Humidity (%)
Silica Gel Capacity (%)
Molecular Sieve Capacity (%)
10
5
18
30
15
20
50
28
21
70
38
22
A pharmaceutical facility requires deep dehumidification to maintain a space relative humidity of 10% at 70°F. The engineer is comparing two desiccant materials for the application. Based on the provided non-linear adsorption isotherm values, which operational characteristic distinguishes the molecular sieve from the silica gel for this specific low-humidity requirement?
Solution
Read the isotherms at the design condition: 10 % RH → molecular sieve 18 % vs silica gel 5 % of dry weight. A desiccant for deep drying must hold moisture at low RH.
At 10% RH the molecular sieve still holds 18% of its weight in water against 5% for silica gel, which is what deep drying to 10% RH needs.
Above 50% RH the molecular sieve adds only one point (21% to 22%); its isotherm is flattest there.
Silica gel holds less at 10% and 30% RH (5 and 15% against 18 and 20%); it leads only at 50% RH and above.
Silica gel climbs from 5% to 38% across the table; it is the strongly humidity-dependent material of the two.
The table shows the opposite: the molecular sieve rises only from 18% to 22% between 10% and 70% RH, an almost flat isotherm.
Question 55 of 60 · HVAC and Refrigeration Distribution and Systems · 20–30 items
Filter Stage
E1 Efficiency (%)
E2 Efficiency (%)
E3 Efficiency (%)
Stage 1 (Pre-filter)
30
50
80
Stage 2 (Secondary)
60
80
90
Stage 3 (Final)
95
99
99.9
A critical environment AHU relies on three sequential filtration stages to control particulate matter. Using the provided manufacturer efficiencies for each stage, calculate the overall single-pass removal efficiency of the system for particles in the E1 size range.
Given
Stage 1 E1 efficiency = 30 %
Stage 2 E1 efficiency = 60 %
Stage 3 E1 efficiency = 95 %
Find
Overall single-pass E1 removal efficiency (%)
Solution
P = (1 − 0.30)(1 − 0.60)(1 − 0.95) = 0.014
η = 1 − P = 98.60 %
17.10% multiplies the three efficiencies (0.30 × 0.60 × 0.95); filters in series multiply penetrations, 1 − 0.70 × 0.40 × 0.05.
61.67% averages the three efficiencies; filters in series multiply penetrations.
72.00% combines only the first two stages and leaves out the final filter.
95.00% is the final stage alone; the upstream stages also remove part of what the final filter would see.
98.60%: the three stages pass 0.70 × 0.40 × 0.05 = 1.4% of E1 particles, so 98.6% is removed.
Question 56 of 60 · HVAC Equipment and Components · 24–36 items
A central plant is replacing a 500-ton open cooling tower with a 500-ton closed-circuit fluid cooler to isolate the condenser water loop. Both units will operate at the same full-load design conditions, rejecting identical amounts of heat to the atmosphere. Which of the following statements correctly compares the expected full-load evaporative water loss of the new closed-circuit cooler to the original open tower?
Solution
E ≈ q/hfg ≈ 0.001 × gpm × range (same q → about the same E)
The sealed coil isolates the process fluid, but a separate spray water circuit still evaporates over the coil.
Approach affects the leaving fluid temperature, not the amount of water evaporated for a given heat load.
Pump heat is a tiny addition to the load; evaporation follows the total heat rejected.
At full wet design load the coil surface stays wetted and sensible rejection is a small share in both units.
Both reject the same heat mostly by evaporating spray water; the closed coil keeps the process loop sealed but not the spray water.
Question 57 of 60 · HVAC Equipment and Components · 24–36 items
A chilled water air handling unit is experiencing a loss of sensible cooling capacity due to waterside fouling of the cooling coil. In response, the building automation system's PID loop fully opens the chilled water control valve, leading to a drastically reduced waterside delta-T (often called "low delta-T syndrome"). If the facility operator manually overrides the chilled water supply temperature setpoint, lowering it from 44°F to 40°F instead of addressing the fouling directly, what is the most likely immediate system-level impact?
Solution
Lower CHWS → lower evaporator saturation temperature → higher lift → higher kW/ton (≈ 1–2 % per °F); the fouling stays unsolved.
A setpoint change does not command the pumps; constant-speed primary pumps keep their flow, and variable secondary pumps may even slow down.
A 40°F supply setpoint drags the evaporating temperature down, raising the pressure ratio the compressor must make, so kW/ton rises.
Colder supply water makes the coil surface colder, which increases latent removal.
Valve authority depends on pressure drops, not on water temperature; the valve is already wide open.
Colder water lowers the apparatus dew point; bypass factor depends on coil geometry and face velocity.
Question 58 of 60 · HVAC Equipment and Components · 24–36 items
An open-drive reciprocating compressor circulates a refrigerant mass flow rate of 200 lb/hr. Thermodynamic cycle analysis determines that the ideal isentropic compression work is 15 Btu/lb. The compressor has an isentropic efficiency of 0.75 and a volumetric efficiency of 0.85. The electrical motor driving the compressor has an efficiency of 0.90.
Calculate the required electrical motor input power in kW. Assume 1 kW = 3,412 Btu/hr.
Given
Refrigerant flow = 200 lb/hr
Isentropic work = 15 Btu/lb
Isentropic efficiency = 0.75
Volumetric efficiency = 0.85
Motor efficiency = 0.90
1 kW = 3,412 Btu/hr
Find
Motor input power (kW)
Solution
Wshaft = ṁ·ws/ηs = 200 × 15/0.75 = 4,000 Btu/hr
Pin = 4,000/(0.90 × 3,412) = 1.302 kW
0.593 kW multiplies by the efficiencies instead of dividing, so the input comes out below the ideal work.
0.879 kW is the ideal isentropic power, before any efficiency.
1.055 kW multiplies the 4,000 Btu/hr shaft power by the 0.90 motor efficiency instead of dividing: 4,000 × 0.90/3,412.
1.172 kW is the compressor shaft power; the 90% motor efficiency has not been applied.
1.302 kW: 200 lb/hr × 15 Btu/lb ÷ 0.75 is 4,000 Btu/hr of shaft work, ÷ 0.90 motor efficiency is 4,444 Btu/hr of electricity.
Question 59 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A newly installed rectangular supply duct system must be tested to verify compliance with SMACNA Leakage Class 4. The duct section being tested is 48 in. × 24 in. and 60 ft long, and it operates at a design static pressure of 2.0 in. w.g. The allowable leakage is F = CL·P0.65, where F is the allowable leakage in cfm per 100 ft² of duct surface, CL is the leakage class and P is the test pressure in in. w.g. The measured leakage at 2.0 in. w.g. is 14 cfm. Determine the allowable leakage for this section and whether the installation passes.
22.6 cfm uses half the perimeter (48 + 24 in.), so the surface area is halved.
28.8 cfm drops the pressure term and uses only the class number per 100 ft².
6.3 cfm is the allowance per 100 ft² of surface; the section has 720 ft², so the total allowance is 7.2 times larger.
57.6 cfm uses P to the first power instead of 0.65, overstating the allowance.
45.2 cfm: F = 4 × 2.00.65 = 6.28 cfm per 100 ft², and the duct has 12 ft × 60 ft = 720 ft² of surface; 14 cfm is well under it.
Question 60 of 60 · Supportive Knowledge (HVAC and Refrigeration) · 8–12 items
A supply fan generates the octave-band sound power levels shown in the table below. Two dissipative silencers are installed in series in the supply duct. Their insertion losses (IL) at each octave band are also provided. Calculate the resulting octave-band sound power levels after both silencers and determine whether the design meets a target room-level insertion loss of 25 dB at 500 Hz.
Octave Band (Hz)
Source Level (dB)
Silencer 1 IL (dB)
Silencer 2 IL (dB)
125
92
8
6
250
95
12
9
500
98
15
12
1000
94
14
11
2000
89
13
10
Given
Fan source level at 500 Hz = 98 dB
Silencer 1 IL at 500 Hz = 15 dB
Silencer 2 IL at 500 Hz = 12 dB
Target insertion loss at 500 Hz = 25 dB
Find
Net insertion loss at 500 Hz and whether target is met
Solution
ILtotal(500 Hz) = 15 + 12 = 27 dB ≥ 25 dB → meets the target; Lw = 98 − 27 = 71 dB
71 dB is the sound power left after the silencers, not an insertion loss, so it cannot be compared with a 25 dB target.
The second silencer attenuates what the first passes on, so its 12 dB counts too.
No self-noise data are given, and self-noise adds a level downstream; it does not subtract from insertion loss.
Averaging the two silencers treats them as alternatives; in series each one removes its own share.
Yes: for silencers in series the insertion losses add band by band, 15 + 12 = 27 dB, above the 25 dB target.
Answer every question to see your score
How this set is built
Weighted by midpoints. The specifications publish a range per area, so the set takes the middle of each: 15 problems on loads and psychrometrics, 17 on distribution and systems, 21 on equipment and 7 on supportive knowledge.
Units as on the exam. All values are in USCS, and a problem states its elevation when it is not at sea level. Problems from SI sources were left out rather than converted.
A simpler format. Every problem here has one correct option. The exam itself uses multiple choice + AITs (alternative item types such as drag-and-drop); the table below lists them.
This set cannot predict a pass. The exam turns your number of right answers into a scaled score, and the cut is not published, so a percentage here does not translate into it.
Pace for six minutes a problem
80 questions in 8 h is six minutes each, and the 50 min scheduled break splits the exam in two: once you submit the first half you cannot reopen it. A long look-up you cannot finish is better flagged and revisited before you submit that half.
The exam behind the set
Checked Oct 29, 2026. Sources are numbered and listed under Source notes. Timing, scoring and retake rules: exam facts.
Fact
Detail
Source
Checked
Questions
80 questionsIncludes unscored pretest questions that are not marked.
NCEES: PE Mechanical exams · NCEES: PE Mechanical HVAC and Refrigeration specifications, October 2025 (PDF) · NCEES Examinee Guide, May 2026 (PDF) [1][2][3]
Question types
Multiple choice plus alternative item types (multiple correct, point and click, drag and drop, fill in the blank)No partial credit on alternative items.
NCEES: PE Mechanical exams · NCEES Examinee Guide, May 2026 (PDF) [1][3]
NCEES: PE Mechanical exams · NCEES Examinee Guide, May 2026 (PDF) [1][3]
Exam time
8 hours
NCEES: PE Mechanical exams · NCEES Examinee Guide, May 2026 (PDF) [1][3]
Scheduled break
50 minutes, after about half the questionsYou review and submit the first half before the break and cannot return to it; unscheduled breaks run on the exam clock.
NCEES: PE Mechanical exams · NCEES Examinee Guide, May 2026 (PDF) [1][3]
Units
USCS onlySea-level conditions unless a problem says otherwise.
NCEES: PE Mechanical HVAC and Refrigeration specifications, October 2025 (PDF) [2]
Reference supplied
PE Mechanical reference handbook, on screen as a searchable PDFThe current version is free in your MyNCEES account; NCEES sells no printed copy; on exam day search with the box on the left (Ctrl+F does not work); no personal copies.
NCEES: PE Mechanical exams · NCEES Examinee Guide, May 2026 (PDF) [1][3]
Passing score
Not published (a scaled score against a minimum ability level)The number of correct answers is converted to a scaled score; wrong answers cost nothing; first-time and repeat takers are graded to the same standard.
The exam owner's practice exam is a separate product
The council sells its own e-book: 80 questions, $59.95 as of Oct 29, 2026, written to the October 2025 specifications. Our set is written separately and borrows nothing from it or from any exam form. Use the e-book once, timed, close to exam day; use the area pages here for the repetitions before it.
No. Exam questions are confidential, and nothing here comes from a form. It is written to the October 2025 outline, in USCS units.
Is there a target percentage on this set?
No. The exam's cut score is not published, so any target would be invented. Use the score here to find the weakest of the four areas, then work that area's page.
Why are there no SI problems?
The exam has none: USCS only. Practicing conversions you will never need costs time you could spend on look-ups.
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