Area 1 is 18–27 items: load calculations, and the processes you trace on a psychrometric chart when air is heated, cooled, humidified or dried. The October 2025 outline asks for those processes at three conditions: sea level, 5,000 ft and low temperature.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamPE Mechanical HVAC & Refrigeration
questions on the exam18–27
worked problems here20
areas, October 2025 outline1 of 4
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What the area covers
Three topics in the specifications, in our words:
Heating and cooling loads: conduction and solar gain through the envelope, infiltration and outdoor air, people, lighting and equipment, for single rooms and whole buildings.
Heating and cooling processes: sensible heating and cooling, mixing of airstreams, coil processes and evaporative cooling, each at sea level, at 5,000 ft and at low temperature.
Humidification and dehumidification: adding moisture with water or steam, removing it at a coil or by other means, under the same three conditions.
20 loads and psychrometrics problems
0 / 20 answered · 0 correct
Question 1 of 20 · HVAC Loads and Psychrometrics · 18–27 items
Parameter
Design value
Elevation
5,000 ft
Air density
0.063 lb/ft3
Specific heat
0.24 Btu/(lb·°F)
Ventilation flow
2,050 cfm
Outdoor temperature
95 °F
Indoor temperature
75 °F
Calculate the actual design sensible cooling load for the ventilation air using rigorous density-based calculations, ignoring standard sea-level rule-of-thumb approximations like Qs = 1.08 · CFM · ΔT.
0.62 MBH drops the 60 min/h conversion, so it is the load per minute of airflow, not per hour.
31.24 MBH applies the density correction twice: the 0.063 lb/ft³ result scaled again by 0.063/0.075.
37.20 MBH uses the site density: 60 × 0.063 × 2,050 × 0.24 × 20 = 37,195 Btu/h.
44.28 MBH is the sea-level 1.08 factor, which builds in 0.075 lb/ft³ and overstates the load at 5,000 ft by about 19%.
45.10 MBH uses a rounded 1.1 factor — a sea-level shortcut that is even further from the 5,000-ft density.
Question 2 of 20 · HVAC Loads and Psychrometrics · 18–27 items
A mechanical designer computes the required cooling capacity for a dedicated outdoor air system (DOAS) that dehumidifies hot, humid outdoor air down to a 55°F dew point. The calculation is flagged during a technical peer review. The reviewer notes that the error involves using sensible-only formulas (1.08 × CFM × ΔT) for processes that change humidity without accounting for latent energy and ΔW. Which of the following best describes the resulting impact on the system design?
Solution
Dehumidifying air needs ṁa·Δh = sensible + latent; 1.08·cfm·ΔT gives only the sensible part.
qT = 4.5 · cfm · Δh (sea level)
Airflow in a DOAS is set by the ventilation requirement, not by the coil load formula, so it is not inflated by this error.
Oversized would need a calculation that adds load; dropping the latent term only removes load, so capacity comes out short.
The bypass factor is a property of coil geometry and face velocity; a load-formula error does not change it.
Undersized: the sensible-only formula leaves out the latent energy of the condensed moisture, which for humid outdoor air can match or exceed the sensible part.
Nothing in the error reduces airflow; the mistake is in the energy the coil must remove, not in the air quantity.
Question 3 of 20 · HVAC Loads and Psychrometrics · 18–27 items
A constant-enthalpy humidifier conditions 12,500 cfm of standard air (ρ = 0.075 lb/ft3). The air enters at 75°F DB and a humidity ratio of 0.0060 lb/lb. The humidifier adds moisture until the leaving air reaches a humidity ratio of 0.0115 lb/lb. Calculate the required make-up water flow rate in gallons per hour (gph) to maintain this process. Assume the density of water is 8.33 lb/gal.
37.1 gph: 56,250 lb/h of dry air picks up 0.0055 lb/lb, or 309.4 lb/h of water, divided by 8.33 lb/gal.
40.5 gph uses the entering humidity ratio 0.0060 as the moisture added instead of the rise to 0.0115.
77.7 gph uses the leaving humidity ratio 0.0115 alone, as if the entering air were bone dry.
118.2 gph adds the two humidity ratios (0.0115 + 0.0060) instead of taking their difference: 56,250 × 0.0175/8.33.
309.4 is the water rate in lb/h; it is labeled gph without dividing by 8.33 lb/gal.
Question 4 of 20 · HVAC Loads and Psychrometrics · 18–27 items
An office conference room has an area of 1,200 ft² and an occupancy of 30 people. Per ASHRAE Standard 62.1, the breathing-zone outdoor air requirement uses Rp = 5 cfm/person and Ra = 0.06 cfm/ft². The zone air distribution effectiveness is Ez = 0.8. Calculate the required zone outdoor airflow (cfm), accounting for the distribution effectiveness adjustment.
178 cfm multiplies Vbz by 0.8 instead of dividing, which lowers the outdoor air when mixing is poor.
188 cfm drops the area component and corrects only the people rate, 150/0.8.
222 cfm is Vbz = 5 × 30 + 0.06 × 1,200 before the effectiveness correction.
240 cfm applies Ez only to the area term, 150 + 72/0.8; the whole Vbz is divided by Ez.
278 cfm: the breathing-zone 222 cfm divided by Ez = 0.8, because imperfect mixing needs more outdoor air at the diffuser.
Question 5 of 20 · HVAC Loads and Psychrometrics · 18–27 items
A meeting room with 20 occupants has a ventilation requirement of 15 cfm per person. The outdoor air is at 95°F dry-bulb and 78°F wet-bulb (W ≈ 0.0168 lb/lb), and the room is maintained at 75°F dry-bulb and 50% relative humidity (W ≈ 0.0093 lb/lb). At sea level, what is the approximate total ventilation cooling load (sensible plus latent)?
Given
Occupants = 20
Ventilation rate = 15 cfm per person
Outdoor air = 95 °F DB, 78 °F WB, W ≈ 0.0168 lb/lb
17,400 Btu/h is sensible 6,480 plus latent 4,840 × 300 × 0.0075 ≈ 10,890 for the 300 cfm of outdoor air.
10,900 Btu/h is only the latent part from the humidity-ratio difference; the 20°F dry-bulb difference is left out.
6,480 Btu/h is only the sensible part, 1.08 × 300 × 20; the outdoor air is also far more humid than the room.
870 Btu/h uses 15 cfm — one person's ventilation — instead of 20 × 15 = 300 cfm.
Question 6 of 20 · HVAC Loads and Psychrometrics · 18–27 items
A cooling coil at sea level operates with an apparatus dew point of 50°F. If the coil’s bypass factor is reduced to 0.10 and the entering air is at 90°F DB, what is the leaving air dry-bulb temperature according to the bypass equation?
Given
Apparatus dew point ADP = 50°F
Bypass factor BF = 0.10
Entering air dry-bulb = 90°F
Find
Leaving air dry-bulb (°F)
Solution
T2 = ADP + BF·(T1 − ADP)
T2 = 50 °F + 0.10 × (90 − 50) °F = 54 °F
About 86°F treats 0.10 as the contact factor, as if only 10% of the air touched the coil.
About 54°F: 10% of the air bypasses at 90°F and 90% leaves at the 50°F ADP, so 50 + 0.10 × 40.
About 50°F is the apparatus dew point itself, which only a coil with zero bypass would deliver.
About 46°F subtracts the bypass term from the ADP; bypassed air is warmer, so it raises the leaving temperature.
Question 7 of 20 · HVAC Loads and Psychrometrics · 18–27 items
Time Period
Outdoor Temp (°F)
Duration (hours)
00:00 - 08:00
30
8
08:00 - 16:00
50
8
16:00 - 24:00
40
8
An intermittent mechanical ventilation system provides 2,400 CFM of outdoor air but operates on a 25% duty cycle (15 minutes per hour). The indoor setpoint is continuously maintained at 70°F. Using the tabulated 24-hour outdoor temperature profile, calculate the total average daily sensible heating energy required to temper this ventilation air.
19,440 is the average hourly rate after the duty cycle (1.08 × 2,400 × 30 × 0.25); it was never multiplied by 24 hours.
155,520 Btu/day averages the three periods' energies instead of adding them, so it covers only eight hours.
466,560 Btu/day: the continuous daily energy of 1,866,240 Btu times the 25% duty cycle.
1,866,240 Btu/day is the energy if the fan ran all 24 hours; the 25% duty cycle is not applied.
7,464,960 Btu/day divides by the 0.25 duty cycle instead of multiplying, quadrupling the continuous energy.
Question 8 of 20 · HVAC Loads and Psychrometrics · 18–27 items
During the psychrometric design of a dedicated outdoor air system for a space with a heavily latent cooling load, the calculated sensible heat ratio (SHR) line is plotted originating from the desired room condition. If this plotted SHR line fails to intersect the saturation curve, what design strategy is typically mandated?
Solution
SHR line too steep to meet the saturation curve → cool and dehumidify to the required W, then reheat sensibly (often with recovered or condenser heat).
Evaporative cooling adds moisture, the opposite of what a latent-dominated space needs.
A colder coil removes more moisture but also delivers air too cold for the small sensible load; without reheat the space overcools.
Bypassed outdoor air carries its full moisture into the supply, which works against the latent load.
More air at the same leaving state still has the coil's SHR; it overcools the space before the moisture is removed.
Overcool to the dew point the latent load needs, then reheat to the supply temperature the sensible load allows; no single coil leaving state lies on the steep SHR line.
Question 9 of 20 · HVAC Loads and Psychrometrics · 18–27 items
A dedicated outdoor air system in a dry climate manufacturing facility utilizes a direct evaporative cooler equipped with rigid media. The system has a tested saturation effectiveness of 0.80 and operates continuously. The facility's control sequence requires a maximum supply air dry-bulb temperature of 68.0°F to offset internal heat gains effectively. The historical climate bin data for the primary operating season is provided in the comprehensive table below. Carefully evaluate the unit's cooling performance across these varying environmental conditions to determine the total number of annual operating hours that supplemental mechanical cooling is strictly needed.
Bin
Mean DB (°F)
Mean WB (°F)
Annual Hours
1
105.0
68.0
150
2
102.0
66.0
300
3
95.0
62.0
450
4
88.0
58.0
600
5
82.0
56.0
800
Given
Saturation effectiveness ε = 0.80
Maximum supply dry-bulb = 68.0 °F
Bin 1: 105.0 °F DB, 68.0 °F WB, 150 h
Bin 2: 102.0 °F DB, 66.0 °F WB, 300 h
Bin 3: 95.0 °F DB, 62.0 °F WB, 450 h
Bin 4: 88.0 °F DB, 58.0 °F WB, 600 h
Bin 5: 82.0 °F DB, 56.0 °F WB, 800 h
Find
Annual hours needing mechanical cooling (h)
Solution
Tout = Tdb − ε(Tdb − Twb)
Bin 1: 105.0 − 0.80 × 37.0 = 75.4 °F; bin 2: 73.2 °F; bin 3: 68.6 °F; bin 4: 64.0 °F; bin 5: 61.2 °F
hours = 150 + 300 + 450 = 900 h
150 h counts only bin 1, whose 68.0°F wet-bulb is already at the limit, as if only bins with a wet-bulb at or above 68°F needed help; at 0.80 effectiveness bins 2 and 3 also leave above 68.0°F (73.2 and 68.6°F).
450 h counts only bins 1 and 2; bin 3 leaves at 68.6°F, which still exceeds the 68.0°F limit.
900 h: bins 1–3 leave the cooler at 75.4, 73.2 and 68.6°F, all above 68.0°F; bins 4 and 5 reach 64.0 and 61.2°F.
1,400 h is the time the cooler meets 68°F on its own (bins 4–5), the complement of what was asked.
2,300 h is every bin in the table; bins 4 and 5 are met by evaporative cooling alone.
Question 10 of 20 · HVAC Loads and Psychrometrics · 18–27 items
In a cooling load calculation for a multi-zone building, what is the primary purpose of the block load calculation compared to individual zone calculations?
Solution
Zone loads size terminals and branch ducts; the block (coincident) load sizes central equipment.
Energy-code compliance is checked against envelope and equipment requirements; a block load is a sizing step, not a code test.
Block load sizes central plant and air handlers: zones peak at different hours, so the simultaneous building peak is below the sum of zone peaks.
Ventilation is set by ASHRAE 62.1 rates per person and per area, not by a cooling block load.
Block load is not an average of zone conditions; it is the coincident peak of the whole building at one hour.
Question 11 of 20 · HVAC Loads and Psychrometrics · 18–27 items
An air handling unit mixes two varying air streams. The outdoor air stream has a flow rate of 2,000 cfm at 95°F dry-bulb and 78°F wet-bulb (enthalpy = 41.6 Btu/lb, humidity ratio = 0.0168 lb/lb). The return air stream has a flow rate of 8,000 cfm at 75°F dry-bulb and 50% relative humidity (enthalpy = 28.1 Btu/lb, humidity ratio = 0.0093 lb/lb). Assuming standard air density for both streams, calculate the enthalpy of the mixed air stream.
Given
Outdoor air = 2,000 cfm, 95 °F DB / 78 °F WB, h = 41.6 Btu/lb, W = 0.0168 lb/lb
Return air = 8,000 cfm, 75 °F DB / 50 % RH, h = 28.1 Btu/lb, W = 0.0093 lb/lb
30.80 Btu/lb: with equal densities, 0.20 × 41.6 + 0.80 × 28.1 weights each stream by its share of the flow.
31.48 Btu/lb takes the outdoor fraction as 2,000/8,000 = 0.25, its ratio to return air instead of to the 10,000 cfm total: 0.25 × 41.6 + 0.75 × 28.1.
34.85 Btu/lb is the plain average of the two enthalpies, ignoring that return air is four times the outdoor flow.
38.90 Btu/lb swaps the weights, giving outdoor air 80% of the mix.
41.60 Btu/lb is the outdoor-air enthalpy alone, as if the unit ran on 100% outdoor air.
Question 12 of 20 · HVAC Loads and Psychrometrics · 18–27 items
Parameter
Indoor Design
Outdoor Design
Dry-Bulb Temp
75°F
95°F
Humidity Ratio
0.0095 lb/lb
0.0175 lb/lb
Enthalpy
28.1 Btu/lb
38.6 Btu/lb
A building envelope has a total area of 20,000 sq ft. The normalized air leakage rate is 0.15 cfm per sq ft of envelope area. Using the psychrometric data provided in the table and a standard latent heat conversion factor of 4,840 for humidity ratio differences, calculate the design latent infiltration load.
Saturation pressures from the steam tables (FE Reference Handbook, Thermodynamics).
13%: the vapor pressure stays at 0.17 psia while saturation pressure rises from 0.34 psia at 68°F to 1.28 psia at 110°F.
31% scales 50% by 68/110 in °F; relative humidity follows saturation pressure, which is strongly nonlinear in temperature.
46% scales 50% by the absolute-temperature ratio 528/570 °R, which again ignores how fast saturation pressure grows.
50% assumes relative humidity is conserved; sensible heating conserves the humidity ratio, so relative humidity falls.
Question 14 of 20 · HVAC Loads and Psychrometrics · 18–27 items
When warm, moist air is passed over a chilled water coil with a surface temperature significantly below the air's initial dew point, what is the resulting path of the primary air on a standard psychrometric chart?
Solution
Cooling and dehumidifying lowers both dry-bulb and humidity ratio; the line runs from the entering state toward the coil ADP.
Straight down would remove moisture at constant dry-bulb, as a desiccant process with reheat might; a cold coil also drops dry-bulb.
A horizontal line is sensible cooling only, which happens when the coil surface stays above the air's dew point.
Up and to the right is heating with humidification, the opposite direction to a chilled-water coil.
Moving along a wet-bulb line is adiabatic (evaporative) cooling, which adds moisture rather than removing it.
Down and to the left: dry-bulb falls (left) while moisture condenses on the coil (down) at the same time.
Question 15 of 20 · HVAC Loads and Psychrometrics · 18–27 items
A controls engineer proposes a sequence of operations for an air handler using an indirect evaporative cooler. The sequence modulates the cooler's staging based on the assumption that the absolute minimum temperature the primary air can be cooled to is its initial dew-point temperature. What is the fundamental flaw in this control logic?
Solution
Tp,out = Tp,in − ε(Tp,in − Twb,sec)
The floor is the secondary-stream wet-bulb, not the primary-air dew point.
The secondary air is cooled by evaporation toward its wet-bulb, and the primary air can only be cooled toward that temperature across the heat exchanger.
An apparatus dew point belongs to a chilled cooling coil; an indirect evaporative cooler has no such coil surface.
The sensible heat ratio describes the space load; it does not set the lowest temperature a heat exchanger can deliver.
The secondary air dry-bulb would be the limit of a dry heat exchanger; wetting the secondary side is what lets the primary air go below it.
This is the engineer's flawed assumption; the primary air gains no moisture and never approaches its own dew point.
Question 16 of 20 · HVAC Loads and Psychrometrics · 18–27 items
An engineer is sizing a makeup air unit for a facility located at 6,000 feet above sea level. The engineer reads the humidity ratio from a standard sea-level psychrometric chart (14.696 psia), using the entering dry-bulb temperature and relative humidity. A senior reviewer notes that at the site (11.78 psia) W = 0.622·pw/(p − pw) must use the local barometric pressure; the latent load is computed from the dry-air mass flow (lb/h). Based on the table below, what is the primary consequence of using the sea-level chart?
Property Parameter
Sea Level (14.696 psia)
6,000 ft (11.78 psia)
Saturation pressure at 75°F
0.430 psia
0.430 psia
Saturation humidity ratio
0.0188 lb/lb
0.0236 lb/lb
Solution
W = 0.622 · pw/(p − pw)
Ws = 0.622 × 0.430/(11.78 − 0.430) = 0.0236 lb/lb vs 0.0188 at 14.696 psia
Same t and RH → higher W at altitude → a sea-level chart understates the moisture to remove.
Overestimating would need the sea-level chart to read W too high; lower barometric pressure raises W, so the error runs the other way.
Sensible heat depends on dry-bulb change and mass flow; the humidity-ratio error in the table does not raise it.
The table shows a humidity-ratio deviation; the sensible load is not what the omitted pressure correction distorts.
Dew point depends on vapor pressure; the error is in the humidity ratio derived from it, and reading too little moisture would not give a high dew point.
Underestimating latent load: at 11.78 psia the same dry-bulb and RH carry more moisture per pound of dry air (0.0236 vs 0.0188 at saturation), so a sea-level chart reads W too low.
Question 17 of 20 · HVAC Loads and Psychrometrics · 18–27 items
Which of the following factors would have the LEAST impact on the cooling load calculation for a commercial office space?
Solution
Internal gains (people, equipment, lights) and solar gain through glass dominate office cooling loads.
Computers and other plug loads are a large, continuous internal gain in offices.
Occupants add sensible and latent gain per person; in an office this is a major internal load.
Glazing area and orientation set solar gain, often the largest single envelope load in a commercial space.
Interior paint color changes almost nothing: interior walls see no direct sun, so absorptance there barely moves the load.
Question 18 of 20 · HVAC Loads and Psychrometrics · 18–27 items
What is the approximate internal heat gain from 25 people engaged in moderately active office work in a conference room?
Per-person rates from ASHRAE Fundamentals, heat gain from occupants.
450 Btu/h is one person's total gain; it was never multiplied by the 25 occupants.
5,000 Btu/h counts only the latent 200 Btu/h per person; the sensible part is the larger share.
6,250 Btu/h counts only the sensible 250 Btu/h per person and drops the latent moisture gain.
11,250 Btu/h: 25 people at about 250 Btu/h sensible plus 200 Btu/h latent each, the ASHRAE value for moderately active office work.
Question 19 of 20 · HVAC Loads and Psychrometrics · 18–27 items
What is the daily cooling load contribution from lighting in a 20,000 ft² retail space with a lighting power density of 1.5 W/ft² that operates for 12 hours?
Given
Floor area A = 20,000 ft2
Lighting power density = 1.5 W/ft2
Operating time = 12 h
Find
Daily lighting cooling load (ton-hours)
Solution
P = 20,000 ft2 × 1.5 W/ft2 = 30 kW
E = 30 kW × 3,412 Btu/kWh × 12 h = 1,228,320 Btu
E = 1,228,320 ÷ 12,000 = 102 ton-h
1,228 is the daily energy in thousands of Btu; it was divided by 1,000 rather than 12,000 Btu per ton-hour.
102 ton-hours: 30 kW × 3,412 Btu/h per kW for 12 h is 1,228,320 Btu, divided by 12,000 Btu per ton-hour.
30 divides 360 kWh by 12 directly, skipping the 3,412 Btu/kWh conversion.
9 is the instantaneous load in tons (30 kW ÷ 3.517 kW/ton); it has not been multiplied by the 12 operating hours.
Question 20 of 20 · HVAC Loads and Psychrometrics · 18–27 items
An HVAC engineer is calculating the cooling load for an IT server room. Which of the following factors would contribute MOST to the total load?
Solution
Equipment heat ≈ electrical input (W × 3.412 Btu/h per W) dominates a data-room load.
Lighting is around 1 W/ft², a small fraction of the equipment density in a server room.
Server rooms usually have little or no glazing, and solar gain is small next to tens of watts per square foot of equipment.
IT equipment turns nearly all its electrical input into heat around the clock, typically most of a server room's load.
Envelope conduction is minor in an interior server room compared with the equipment heat.
Answer every question to see your score
Sea level, 5,000 ft and below freezing
Problems are set at sea level unless they say otherwise, and the familiar airflow constants carry that assumption inside them:
1.08 = 60 min/h × 0.075 lb/ft3 × 0.24 Btu/(lb·°F)
At 5,000 ft the standard atmosphere is about 12.2 psia instead of 14.7, so a cubic foot of air holds roughly 17% less mass at the same temperature. Rebuild the constant from the site density instead of reusing 1.08, and do the same for the latent constant. The humidity ratio depends on the total pressure as well as on the vapor pressure, so a chart drawn for sea level gives the wrong humidity ratio and enthalpy for the same dry bulb and relative humidity. Use a chart or the equations for the site pressure (standard atmosphere and the equations: ASHRAE Handbook—Fundamentals, Psychrometrics).
Low temperature is new in the October 2025 outline. Below 32 °F the moisture in saturated air is in equilibrium with ice, the values of humidity ratio get very small, and the normal-range chart stops at freezing; read a low-temperature chart or the tables.
Common traps
Five chart and load mistakes
Grains or pounds. The latent constant is 0.68 with humidity ratio in grains per pound and 4,840 with pounds per pound; 7,000 grains make a pound. A mixed pair misses by a factor of 7,000.
Heat gain is not cooling load. The radiant part of solar, lighting and people gains is absorbed by the structure and reaches the air hours later, so the load peaks after the gain does (ASHRAE Handbook—Fundamentals, Nonresidential Cooling and Heating Load Calculations).
Enthalpy is not read along the wet-bulb line. On the chart the two families of lines are close but not parallel; take enthalpy from its own scale.
The wrong design day. Cooling design pairs a dry bulb with its mean coincident wet bulb; dehumidification design uses the dew-point condition with its own coincident dry bulb. Sizing for latent load on the dry-bulb day misses the humid case (ASHRAE Handbook—Fundamentals, Climatic Design Information).
Credits at the heating design point. Design heating loads are taken without solar and internal gains: the design hour is a cold night with the building empty.
How this area links to the rest
Moist-air properties start as thermodynamics; the property relations are explained once on FE Mechanical thermodynamics, and this page owns their HVAC use. Outdoor air rates and filtration belong to distribution and systems, coils and evaporative coolers as machines to equipment and components. In the exam, though, a single problem often starts with a load and ends at a coil.