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5.3 Duct Geometry and Worked Airflow Calculations

Key Takeaways

  • The volumetric airflow continuity equation Q = A × V (CFM = Area in sq ft × Velocity in FPM) is the fundamental governing law for fluid flow in HVAC distribution ducts.

  • Duct cross-sectional area calculations must always convert inch dimensions to square feet by dividing by 144, and internal acoustic duct liner thickness must be subtracted twice from external dimensions.

  • A calculation must use the internal flow dimensions at the measurement plane, accounting for liner or obstructions when applicable.

  • Worked results should be checked against duct size, measured pressure, design flow, and instrument range before reporting.

Last updated: August 2026

Duct Geometry and Worked Airflow Calculations

A traverse produces velocity at a defined plane; converting it to airflow requires the net internal area at that same plane. Record whether dimensions came from a drawing, label, or field measurement and whether they are outside or clear inside dimensions.

5. Duct Geometric Calculations and Net Internal Area

To compute total volumetric flow (Q=A×VQ = A \times V), technicians must calculate the precise net cross-sectional area of ductwork. External dimensions must be adjusted for internal insulation (acoustic liner).

   Rectangular Duct             Round Spiral Duct          Flat Oval Duct
  +------------------+             .--------.             .--------------.
  |  |============|  |           /  /======\ \           /  /==========\  \
  |  |   Net H    |  |          |  |  Net D | |         (  ( Net a  Net A ) )
  |  |============|  |           \  \======/ /           \  \==========/  /
  +------------------+             '--------'             '--------------'
         Net W

1. Rectangular Ducts

A (sq ft)=W (in.)×H (in.)144A\text{ (sq ft)} = \frac{W\text{ (in.)} \times H\text{ (in.)}}{144}
  • Internal Liner Deduction: Acoustic duct liner lines all four interior walls. Subtract twice the liner thickness (tlinert_{liner}) from each nominal dimension:
Wnet=Wnominal−(2×tliner),Hnet=Hnominal−(2×tliner)W_{net} = W_{nominal} - (2 \times t_{liner}), \quad H_{net} = H_{nominal} - (2 \times t_{liner}) Anet (sq ft)=[Wnominal−2tliner]×[Hnominal−2tliner]144A_{net}\text{ (sq ft)} = \frac{[W_{nominal} - 2t_{liner}] \times [H_{nominal} - 2t_{liner}]}{144}

2. Round Ducts

A (sq ft)=π×D24×144=π×D2576≈D2183.35A\text{ (sq ft)} = \frac{\pi \times D^2}{4 \times 144} = \frac{\pi \times D^2}{576} \approx \frac{D^2}{183.35}
  • Or in terms of radius r=D/2r = D/2:
A (sq ft)=π×r2144A\text{ (sq ft)} = \frac{\pi \times r^2}{144}
  • Internal Liner Deduction:
Dnet=Dnominal−(2×tliner)D_{net} = D_{nominal} - (2 \times t_{liner}) Anet (sq ft)=π×(Dnominal−2tliner)2576A_{net}\text{ (sq ft)} = \frac{\pi \times (D_{nominal} - 2t_{liner})^2}{576}

3. Flat Oval Ducts

Flat oval ductwork consists of a rectangular center section flanked by two semicircular ends:

  • Let aa = Minor internal axis (height/depth in inches)
  • Let AmajorA_{major} = Major internal axis (width in inches)
A (sq ft)=(π×a24)+[a×(Amajor−a)]144A\text{ (sq ft)} = \frac{\left( \frac{\pi \times a^2}{4} \right) + \left[ a \times (A_{major} - a) \right]}{144}

6. Step-by-Step Worked Field Numerical Examples

Example 1: Standard Air Calculation in a Rectangular Supply Main

Scenario: A technician performs an equal-area Pitot-tube traverse on a 30 in.×16 in.30\text{ in.} \times 16\text{ in.} unlined rectangular supply duct at standard conditions. Each point's velocity pressure has been converted to velocity before averaging, producing a representative mean velocity of 1,878.5 FPM1,878.5\text{ FPM}. Calculate total airflow.

  1. Calculate internal cross-sectional area (AA): A=30×16144=480144=3.333 sq ftA = \frac{30 \times 16}{144} = \frac{480}{144} = 3.333\text{ sq ft}
  2. Use the point-by-point mean velocity: Vavg=1,878.5 FPMV_{avg} = 1,878.5\text{ FPM}
  3. Calculate volumetric flow (QQ): Q=A×Vavg=3.333 sq ft×1,878.5 FPM=6,261 CFMQ = A \times V_{avg} = 3.333\text{ sq ft} \times 1,878.5\text{ FPM} = \mathbf{6,261\text{ CFM}}

Do not average velocity pressures and then take one square root; the square-root conversion must be performed at each equal-area point before velocities are averaged.


Example 2: Non-Standard Air Density Correction at High Altitude and Heating Temperature

Scenario: A discharge heating air duct measuring 36 in.×20 in.36\text{ in.} \times 20\text{ in.} is traversed on a project in Denver, CO (elevation 5,280 ft5,280\text{ ft}, measured barometric pressure 24.80 in. Hg24.80\text{ in. Hg}). The air temperature in the duct is 125∘F125^\circ\text{F}. The measured velocity pressure is Vp=0.18 in. w.g.V_p = 0.18\text{ in. w.g.} Calculate the actual airflow in CFM.

  1. Calculate Internal Area (AA): A=36×20144=720144=5.00 sq ftA = \frac{36 \times 20}{144} = \frac{720}{144} = 5.00\text{ sq ft}
  2. Calculate Density Correction Factor (KdK_d): Kd=(530460+125)×(24.8029.92)=(530585)×0.82888=0.90598×0.82888=0.7509K_d = \left( \frac{530}{460 + 125} \right) \times \left( \frac{24.80}{29.92} \right) = \left( \frac{530}{585} \right) \times 0.82888 = 0.90598 \times 0.82888 = \mathbf{0.7509}
  3. Calculate Actual Density (ρactual\rho_{actual}): ρactual=0.075×0.7509=0.05632 lb/ft3\rho_{actual} = 0.075 \times 0.7509 = 0.05632\text{ lb/ft}^3
  4. Calculate Density-Corrected Velocity (VV): V=4005×0.180.7509=4005×0.23971=4005×0.48960=1961 FPMV = 4005 \times \sqrt{\frac{0.18}{0.7509}} = 4005 \times \sqrt{0.23971} = 4005 \times 0.48960 = \mathbf{1961\text{ FPM}} (Note: The uncorrected standard formula would yield 4005×0.18=1699 FPM4005 \times \sqrt{0.18} = 1699\text{ FPM}, creating an uncorrected error of 13.4%13.4\%!)
  5. Calculate Total Actual Airflow (QQ): Q=5.00 sq ft×1961 FPM=9805 CFMQ = 5.00\text{ sq ft} \times 1961\text{ FPM} = \mathbf{9805\text{ CFM}}

Example 3: Round Spiral Duct with Internal Acoustic Liner

Scenario: A 22 in.22\text{ in.} nominal diameter round supply duct is lined internally with 1.0 in.1.0\text{ in.} fiberglass acoustic insulation. An average traverse velocity of 1350 FPM1350\text{ FPM} is measured under standard air conditions. Determine the actual airflow.

  1. Determine Net Internal Diameter (DnetD_{net}): Dnet=22 in.−(2×1.0 in.)=20.0 in.D_{net} = 22\text{ in.} - (2 \times 1.0\text{ in.}) = 20.0\text{ in.}
  2. Calculate Net Cross-Sectional Area (AnetA_{net}): Anet=π×(20.0)2576=3.14159×400576=1256.64576=2.182 sq ftA_{net} = \frac{\pi \times (20.0)^2}{576} = \frac{3.14159 \times 400}{576} = \frac{1256.64}{576} = 2.182\text{ sq ft}
  3. Calculate Volumetric Flow Rate (QQ): Q=Anet×V=2.182 sq ft×1350 FPM=2946 CFMQ = A_{net} \times V = 2.182\text{ sq ft} \times 1350\text{ FPM} = \mathbf{2946\text{ CFM}}
Test Your Knowledge

A 16-inch nominal diameter round supply duct is fitted with a 1.0-inch internal acoustic fiberglass liner. If the average measured velocity across the net cross-section is 1,600 FPM under standard conditions, what is the volumetric airflow in CFM?

A

2,234 CFM

B

1,710 CFM

C

1,955 CFM

D

1,422 CFM

Test Your Knowledge

During cold weather testing of an outside air intake duct, air temperature is measured at 20°F with barometric pressure at 29.92 in. Hg, resulting in an air density of 0.0828 lb/ft³. If the measured velocity pressure is 0.36 in. w.g., what is the actual air velocity?

A

2,403 FPM

B

2,612 FPM

C

2,286 FPM

D

2,150 FPM

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