Load and Psychrometric Calculations

This is the chapter where you stop reciting and start calculating. CSLB's guide says some questions "require mathematical computation" and gives a calculator. The computations on the exam are almost always one of five kinds: a sensible, latent or total heat-flow calculation from airflow; an air-change or ventilation airflow calculation; a psychrometric property read or process; a conduction (U × A × ΔT) load; or a unit conversion. Master the five and the math is done.

2.1 The Units You Must Keep Straight

• BTU is a quantity of heat. BTU/hr is a rate. A ton of cooling is 12,000 BTU/hr.
• CFM is cubic feet of air per minute. A cubic foot of air is not a fixed weight: it depends on temperature and altitude. Standard air (sea level, about 70 °F) weighs about 0.075 lb per cubic foot.
• Dry-bulb (DB) temperature is what a regular thermometer reads. Wet-bulb (WB) is the temperature of a thermometer wrapped in a wet wick in moving air; it falls below dry-bulb by an amount that depends on how dry the air is.
• Humidity ratio is pounds of water vapor per pound of dry air. HVAC people usually express it in grains of moisture per pound of dry air, where 1 pound = 7,000 grains.
• Enthalpy (h) is the total heat in a pound of dry air plus its moisture, in BTU per pound of dry air, measured from an arbitrary reference. Only differences in enthalpy mean anything.

2.2 The Three Airflow Equations, and Where Their Constants Come From

Memorizing 1.08, 0.68 and 4.5 works until the exam changes the numbers on you or a choice looks wrong. Derive them once and you will never confuse them.

Sensible heat: Qs = 1.08 × CFM × ΔT (°F). Per minute, CFM cubic feet of standard air weighs CFM × 0.075 lb. Per hour that is 60 × 0.075 × CFM = 4.5 × CFM pounds. Air's specific heat is about 0.24 BTU per lb per °F. So Qs = 4.5 × 0.24 × CFM × ΔT = 1.08 × CFM × ΔT.

Total heat: Q_t = 4.5 × CFM × Δh (BTU/lb). It is the same mass flow, 4.5 × CFM pounds per hour, multiplied directly by the enthalpy change. The specific heat is already inside enthalpy, so no 0.24 appears.

Latent heat: QL = 0.68 × CFM × Δgrains. Mass flow is 4.5 × CFM lb of dry air per hour. Each grain of water removed is 1/7,000 lb. Condensing a pound of water vapor releases about 1,060 BTU. So QL = 4.5 × CFM × Δgrains × 1,060 ÷ 7,000 ≈ 0.68 × CFM × Δgrains. (Some references use latent heat values that give 0.69; use whatever the question states or your reference gives. The difference is about 1.5%.)

The memory aid. 1.08 goes with a temperature difference. 0.68 goes with a grains difference. 4.5 goes with an enthalpy difference. And sensible + latent = total.

Why mixing them is the classic trap. If you apply 1.08 to an enthalpy difference of 8 BTU/lb on 1,000 CFM you get 8,640, which is about a quarter of the true 36,000 (4.5 × 1,000 × 8). If you apply 4.5 to a dry-bulb difference you get a number roughly four times too large. Both wrong answers will be sitting in the option list, because the exam writer knows exactly which mistake you will make.

Altitude. All three constants assume sea-level air density. At altitude the air is thinner and carries less heat per cubic foot, so the constants shrink in proportion to density. Using the standard-atmosphere density relation, the density ratio is about 0.94 at 2,000 ft, 0.86 at 5,000 ft and 0.81 at 7,000 ft. At 5,000 ft the sensible constant becomes about 1.08 × 0.86 = 0.93. The practical consequence: a blower that moves a given CFM at altitude moves less mass, so it carries less heat, and a heat calculation done with sea-level constants overstates what the airflow can deliver.

2.3 Worked Calculations

Example A: sensible capacity. 1,800 CFM is cooled from 78 °F to 57 °F. Q_s = 1.08 × 1,800 × 21 = 40,824 BTU/hr.

Example B: total capacity from enthalpy. 1,000 CFM enters a coil at 30 BTU/lb and leaves at 22 BTU/lb. Q_t = 4.5 × 1,000 × 8 = 36,000 BTU/hr = 3 tons.

Example C: latent from grains. The same airstream loses 6 grains/lb. Q_L = 0.68 × 1,000 × 6 = 4,080 BTU/hr.

Example D: sensible, latent and total from one coil, cross-checked. A residential coil receives 1,800 CFM of air at 78 °F dry-bulb, 65 °F wet-bulb. Computing from the psychrometric equations at sea level, that air has a humidity ratio of about 71 grains/lb and an enthalpy of about 29.9 BTU/lb. It leaves at 57 °F dry-bulb, 56 °F wet-bulb: about 65 grains/lb and 23.8 BTU/lb.

• Sensible: 1.08 × 1,800 × (78 − 57) = 40,824 BTU/hr.
• Latent: 0.68 × 1,800 × (71 − 65) = about 7,300 BTU/hr (rounding grains to whole numbers).
• Sum: about 48,200 BTU/hr.
• Total by enthalpy: 4.5 × 1,800 × (29.9 − 23.8) = about 49,400 BTU/hr.

The two routes agree to within about 3%, which is as close as chart reading and rounded constants allow. The coil's sensible heat ratio is 40,824 ÷ 48,200 ≈ 0.85. (Your chart reading will differ by a grain or half a BTU/lb. The exam's choices are spread widely enough that this does not matter.)

Example E: the airflow that goes with a tonnage. A 3-ton system at the common residential design of about 400 CFM per ton moves 1,200 CFM. If the coil does 36,000 BTU/hr with an SHR of 0.75, sensible is 27,000 BTU/hr, and the required air temperature drop across the coil is ΔT = 27,000 ÷ (1.08 × 1,200) = 20.8 °F. This is the reason a normal residential split shows a temperature difference across the coil of roughly the high teens to low twenties, not 28 °F. If the system were pure sensible (SHR = 1), ΔT would be 36,000 ÷ (1.08 × 1,200) = 27.8 °F, which is what a dry-air process would show. The latent portion of the load takes up part of the capacity.

Example F: airflow versus latent removal. Slow a 3-ton system's blower from 1,200 to 900 CFM. The coil runs colder, so sensible capacity drops and latent capacity rises. That is how you buy dehumidification, and how you freeze a coil if you go too far. The design range of 350 to 450 CFM per ton is a convention, but the 2025 Energy Code's mandatory measures set a floor of 350 CFM per ton for ducted cooling in single-family buildings, and the Commission's duct leakage form instructions use 400 CFM per nominal ton to compute leakage percentages; the equipment's expanded performance data governs.

Example G: heating airflow from a furnace. A furnace is rated 90,000 BTU/hr input, 80% AFUE. Output = 72,000 BTU/hr. With a 50 °F temperature rise, CFM = 72,000 ÷ (1.08 × 50) = 1,333 CFM. The allowable rise comes from the equipment's rating plate and installation instructions, not from Manual J or the Title 24 forms.

Example H: finding temperature rise in the field. The same furnace measures 1,000 CFM of airflow. Rise = 72,000 ÷ (1.08 × 1,000) = 66.7 °F. If the rating plate's range tops out at 65 °F, the furnace is moving too little air. This is the calculation behind Chapter 11's furnace diagnosis.

2.4 Air Changes and Ventilation Airflow

CSLB's own sample question is an air-change calculation, so expect at least one.

Air changes per hour (ACH) tells you how many times per hour the full volume of a space is replaced. CFM = (volume in cubic feet × ACH) ÷ 60.

Example I. A parts warehouse bay is 90,000 ft³ and needs six air changes per hour. Required airflow = 90,000 × 6 ÷ 60 = 9,000 CFM. If each exhaust fan moves 1,500 CFM, you need 9,000 ÷ 1,500 = 6 fans. (This uses different numbers than CSLB's sample; the method is identical.)

Example J: going backwards. A 12-ft by 20-ft room with an 8-ft ceiling is 1,920 ft³. A 100 CFM bathroom-type exhaust fan produces 100 × 60 ÷ 1,920 = 3.1 ACH.

Rate-based ventilation. Codes also set outdoor air by a rate per person plus a rate per square foot of floor area, and the total is the sum. The rates themselves live in the adopted ventilation tables, which this book does not print, but the method is simple: outdoor air CFM = (people × CFM per person) + (floor area × CFM per square foot). Example: 30 people at a table rate of 5 CFM per person plus 3,000 ft² at 0.06 CFM per ft² gives 150 + 180 = 330 CFM (the rates here are illustrative, not code values).

Makeup air. Whatever you exhaust must be replaced. A kitchen hood exhausting 3,000 CFM requires roughly 3,000 CFM of makeup air from somewhere, or the building goes negative and pulls air backward through vents and flues. A return duct recirculates conditioned air; it does not replace air that left the building.

Ventilation is a load. Outdoor air that enters carries sensible and latent loads. Use the equations in 2.2 with the outdoor condition as the entering state. Example: 330 CFM of outdoor air at 95 °F DB entering a space held at 75 °F carries a sensible load of 1.08 × 330 × 20 = 7,128 BTU/hr before any latent load.

2.5 Reading the Psychrometric Chart

You will not be handed a chart in a closed-book exam without a question that tells you what to look for, but the exam does test the relationships the chart displays. Learn them as rules.

• Dry-bulb ≥ wet-bulb ≥ dew point. All three are equal only at saturation (100% relative humidity). The gap between dry-bulb and wet-bulb tells you how dry the air is.
• Dew point is the temperature at which the air's moisture begins to condense. It depends only on the moisture content (humidity ratio), not on dry-bulb. Warm air and cool air with the same humidity ratio share a dew point.
• Relative humidity is the moisture the air holds as a percentage of what it could hold at that dry-bulb. Warm the air without adding moisture and RH falls. Cool it and RH rises, reaching 100% at the dew point.
• Humidity ratio (grains/lb) is the true measure of how much water is in the air. Lines of constant humidity ratio run horizontal on the chart.
• Enthalpy lines slant down to the right, and they run almost parallel to the constant wet-bulb lines, which is why wet-bulb is such a good proxy for total heat content.

Four basic processes to recognize:

Cooling and dehumidifying in more detail. When a coil surface temperature is below the air's dew point, moisture condenses. The process line on the chart runs roughly toward the coil's apparatus dew point (ADP), the effective surface condition of the coil, but never all the way there, because some air bypasses the coil. The air leaves cooler and drier but not saturated in a real residential coil, typically at high relative humidity. The steeper the line, the more latent the process; the flatter, the more sensible.

Why an A-coil can "sweat" without dehumidifying. If the coil surface stays above the entering air's dew point, no moisture condenses. All the cooling is sensible. That is what a very high airflow does: it keeps the coil warmer. This is the mirror of Example F.

Mixing two airstreams. Return air and outdoor air mixed in a plenum blend in proportion to their airflow. Dry-bulb, humidity ratio and enthalpy of the mix are, to a good approximation, the CFM-weighted averages of the two streams. Example: 25% outdoor air at 95 °F DB, 75 °F WB (about 98.5 grains/lb, 38.3 BTU/lb) mixed with 75% return air at 75 °F DB, 63 °F WB (about 66.7 grains/lb, 28.4 BTU/lb):

• Mixed DB ≈ 0.25 × 95 + 0.75 × 75 = 80 °F.
• Mixed moisture ≈ 0.25 × 98.5 + 0.75 × 66.7 = 74.7 grains/lb.
• Mixed enthalpy ≈ 0.25 × 38.3 + 0.75 × 28.4 = 30.9 BTU/lb, which corresponds to a wet-bulb near 66 °F (an 80 °F DB mixed state at about 49% RH).

The exam version is usually: "A mixed-air temperature is 80 °F; the return is 75 °F, outdoors 95 °F; what fraction is outdoor air?" Fraction = (80 − 75) ÷ (95 − 75) = 25%.

Using the sensible heat ratio. A cooling load of 30,000 BTU/hr with 22,500 sensible has SHR = 0.75. Equipment is chosen so its coil SHR at the design conditions matches or exceeds the room's, otherwise the humidity runs high. A coil with SHR 0.85 on a room load with SHR 0.70 leaves the room humid.

Dew point and duct sweat. If a duct or pipe surface is colder than the dew point of the surrounding air, water condenses on it. A 55 °F supply duct in a 95 °F, 75 °F-wet-bulb attic (dew point near 67 °F) will sweat unless it is insulated and the insulation's vapor jacket is intact. This is why the exam links duct insulation, vapor barriers and sealing to humidity, not only to energy.

Key numbers and facts: psychrometrics - 1.08 with a temperature difference; 0.68 with grains; 4.5 with enthalpy. Sensible + latent = total. - 1.08 = 60 × 0.075 × 0.24. 4.5 = 60 × 0.075. 0.68 ≈ 4.5 × 1,060 ÷ 7,000. - 1 pound = 7,000 grains. 1 ton = 12,000 BTU/hr. - Altitude reduces the constants in proportion to air density (about 0.86 at 5,000 ft). - CFM = volume × ACH ÷ 60. - Dew point depends only on moisture content. DB ≥ WB ≥ DP, equal only at saturation. - Heating: horizontal right. Cooling: horizontal left. Cool + dehumidify: down and left. Humidify: up. - Mix fraction = (mixed − return) ÷ (outdoor − return). - SHR = sensible ÷ total. Lower airflow → colder coil → more latent.