14.4 Applied HVAC/R Trade Math & Unit Conversions

14.4 Applied HVAC/R Trade Math & Unit Conversions

[!IMPORTANT] Calculator rules matter here. Prov permits only a simple four-function calculator in the testing room. Scientific calculators, phones, tablets, and programmable devices are prohibited, and formula sheets or stickers may not be added to a reference book. Every calculation on this exam must be reachable with add, subtract, multiply, and divide — which is exactly why the constants below are worth memorizing rather than deriving.

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Capacity and Energy Conversions

The ton definition is not merely a trade convention in Arkansas — A.C.A. § 17-33-101(12) defines "one ton" as twelve thousand (12,000) BTUH in the licensing statute itself, and § 17-33-101(7) defines horsepower as 746 watts. That is what ties the Class B ceiling of 25 tons cooling and 15 horsepower refrigeration to real nameplate data.

[!TIP] Worked conversion — a license-scope question in disguise. A packaged unit is rated 288,000 BTU/hr of cooling. May a Class B contractor install it? 288,000 ÷ 12,000 = 24 tons. The Class B per-unit ceiling under 17 CAR § 261-103(b)(1) is 25 tons, so yes — 24 tons is within scope. Change the rating to 312,000 BTU/hr and it becomes 26 tons, which exceeds the ceiling and requires a Class A contractor.

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The Three Air-Side Heat Equations

These are the most-used formulas in the trade. All three assume standard air at sea level, 70°F, and a density of 0.075 lb/ft³.

Sensible Heat

Q(sensible) = 1.08 × CFM × ΔT

where Q is in BTU/hr and ΔT is the dry-bulb temperature difference in °F.

The constant 1.08 is derived: 60 min/hr × 0.075 lb/ft³ × 0.24 BTU/lb·°F = 1.08.

[!TIP] Worked example. An air handler moves 1,600 CFM. Entering air is 78°F dry bulb, leaving air is 58°F. ΔT = 78 − 58 = 20°F Q = 1.08 × 1,600 × 20 = 34,560 BTU/hr sensible

Latent Heat

Q(latent) = 0.68 × CFM × Δgr

where Δgr is the humidity ratio difference in grains of moisture per pound of dry air.

[!TIP] Worked example. The same 1,600 CFM coil takes air from 78 grains to 58 grains. Δgr = 20 grains Q = 0.68 × 1,600 × 20 = 21,760 BTU/hr latent

Total Heat

Q(total) = 4.5 × CFM × Δh

where Δh is the enthalpy difference in BTU per pound of dry air, read from the psychrometric chart.

[!TIP] Worked example. Entering enthalpy 30.5 BTU/lb, leaving enthalpy 23.5 BTU/lb, at 1,600 CFM. Δh = 7.0 BTU/lb Q = 4.5 × 1,600 × 7.0 = 50,400 BTU/hr total

Cross-check: sensible 34,560 + latent 21,760 = 56,320, against the 50,400 total from enthalpy. The discrepancy signals that the assumed state points are not internally consistent — which is exactly how a field technician catches a bad measurement. On the exam, total = sensible + latent always holds, and the sensible heat ratio (SHR) = sensible ÷ total.

Sensible Heat Ratio

SHR = Q(sensible) ÷ Q(total)

A residential system in Arkansas's humid climate typically designs near 0.75–0.80. A low SHR means a large share of the load is moisture removal; a high SHR (0.90+) is typical of dry climates or high-outdoor-air commercial applications with separate dehumidification.

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The Water-Side Heat Equation

Q = 500 × GPM × ΔT

where Q is BTU/hr, GPM is gallons per minute, and ΔT is the water temperature difference in °F.

The constant 500 is derived: 60 min/hr × 8.33 lb/gal × 1.0 BTU/lb·°F = 499.8 ≈ 500.

[!TIP] Worked example — sizing a hydronic circulator. A boiler must deliver 240,000 BTU/hr across a 20°F design temperature drop (180°F supply, 160°F return). GPM = Q ÷ (500 × ΔT) = 240,000 ÷ (500 × 20) = 24 GPM

Halve the design ΔT to 10°F and the required flow doubles to 48 GPM — which is why a wider ΔT lets a designer use smaller pipe and a smaller pump. This inverse relationship between ΔT and flow is a favorite exam question.

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Airflow, Area, and Velocity

CFM = Velocity (fpm) × Area (sq ft), therefore Velocity = CFM ÷ Area.

Rectangular duct area (sq ft) = (W × D) ÷ 144, with W and D in inches.

Round duct area (sq ft) = (π × d²) ÷ 576, with d in inches. (This is πd²/4 in square inches, divided by 144.)

[!TIP] Worked example — rectangular. A 24 × 12 duct carries 1,800 CFM. Area = (24 × 12) ÷ 144 = 288 ÷ 144 = 2.0 sq ft Velocity = 1,800 ÷ 2.0 = 900 fpm

[!TIP] Worked example — round. A 14-inch round duct carries 1,200 CFM. Area = (3.1416 × 14 × 14) ÷ 576 = 615.75 ÷ 576 = 1.069 sq ft Velocity = 1,200 ÷ 1.069 = 1,123 fpm

Fan Laws

For a single fan in a fixed system, changing speed changes performance predictably:

• Airflow varies directly with speed: CFM₂ = CFM₁ × (RPM₂ ÷ RPM₁)
• Static pressure varies with the square of speed: SP₂ = SP₁ × (RPM₂ ÷ RPM₁)²
• Brake horsepower varies with the cube of speed: BHP₂ = BHP₁ × (RPM₂ ÷ RPM₁)³

[!TIP] Worked example. A fan delivers 4,000 CFM at 800 RPM against 1.0 in. w.c., drawing 2.0 BHP. The balancer increases speed to 1,000 RPM. Ratio = 1,000 ÷ 800 = 1.25 CFM = 4,000 × 1.25 = 5,000 CFM SP = 1.0 × 1.25² = 1.0 × 1.5625 = 1.56 in. w.c. BHP = 2.0 × 1.25³ = 2.0 × 1.953 = 3.91 BHP

A 25% airflow increase demanded a 95% increase in motor power. This is why "just speed up the fan" so often trips the motor overload.

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Electrical Formulas

The three-phase constant 1.732 is the square root of 3.

[!TIP] Worked example — electric heat. A 10 kW electric resistance heater operates at 240 V single phase (power factor 1.0 for a resistive load). Amps = 10,000 W ÷ 240 V = 41.7 A Heat output = 10 kW × 3,412 = 34,120 BTU/hr

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Areas, Volumes, and Air Changes

• Rectangular area = length × width
• Circle area = π × r² = πd² ÷ 4
• Triangle area = (base × height) ÷ 2
• Room volume = length × width × height
• Air changes per hour (ACH) = (CFM × 60) ÷ room volume in ft³
• CFM required for a target ACH = (volume × ACH) ÷ 60

[!TIP] Worked example. A commercial kitchen measures 30 ft × 24 ft with a 10 ft ceiling and requires 15 air changes per hour. Volume = 30 × 24 × 10 = 7,200 ft³ CFM = (7,200 × 15) ÷ 60 = 1,800 CFM

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Offsets: Right-Triangle Constants

Duct and pipe offsets are 45°, 30°, or 60° right triangles. Rather than using trigonometry — impossible on a four-function calculator — the trade uses fixed constants:

Travel is the length of the diagonal piece; run is the horizontal distance consumed; offset is the vertical displacement.

[!TIP] Worked example. A refrigerant line must offset 18 inches around a beam using 45° fittings. Travel = 18 × 1.414 = 25.45 inches — the center-to-center length of the diagonal piece. Run = 18 × 1.000 = 18 inches of horizontal space consumed. Deduct the fitting take-out from the travel to get the cut length of tubing.

Using 30° fittings instead: travel = 18 × 2.000 = 36 inches, run = 18 × 1.732 = 31.2 inches. The shallower angle has lower pressure loss but needs far more room — the trade-off the exam is testing.

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Percentages, Efficiency, and Mixed Air

Efficiency = output ÷ input

[!TIP] Worked example. A furnace has an input of 100,000 BTU/hr and an AFUE of 95%. Output = 100,000 × 0.95 = 95,000 BTU/hr To deliver 100,000 BTU/hr instead, required input = 100,000 ÷ 0.95 = 105,263 BTU/hr. Note the difference: dividing by the efficiency sizes the equipment; multiplying finds the output of equipment already selected.

Mixed-air temperature is a weighted average by airflow:

MAT = [(OA CFM × OA temp) + (RA CFM × RA temp)] ÷ Total CFM

[!TIP] Worked example. 2,000 total CFM with 500 CFM of outdoor air at 95°F and 1,500 CFM of return air at 75°F. MAT = [(500 × 95) + (1,500 × 75)] ÷ 2,000 = (47,500 + 112,500) ÷ 2,000 = 160,000 ÷ 2,000 = 80°F

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Common Exam Traps

• Trap: 1.08 vs. 4.5 vs. 0.68. 1.08 pairs with dry-bulb ΔT (sensible), 0.68 with grains (latent), 4.5 with enthalpy (total). Using the wrong constant with the right numbers is the most common failure.
• Trap: 500 vs. 1.08. 500 is the water constant; 1.08 is the air constant. Never cross them.
• Trap: Round duct area. Divide by 576 when the diameter is in inches, not by 144.
• Trap: Fan law exponents. Airflow is linear, pressure is squared, horsepower is cubed.
• Trap: Efficiency direction. Multiply input by efficiency to get output; divide required output by efficiency to size input.
• Trap: The 1.732 factor. It belongs only in three-phase calculations.
• Trap: Offset constants. 45° travel is offset × 1.414; 30° travel is offset × 2.000. Candidates reverse them because 30° "sounds smaller."