Round → rectangular: equal area
Inputs: 10 in round, selected rectangle aspect 2:1.
Calculation: Preserve the 78.54 in² circle area and apply the 2:1 ratio.
Result: About 12.5 × 6.3 in equal-area rectangle.
Duct sizing and geometry
Convert round, rectangular, and flat-oval dimensions using the selected area or friction basis.
Convert round, rectangular and flat-oval duct dimensions on a selected equal-area or equal-friction basis, with the two equivalent diameters clearly distinguished.
Equal-area conversion preserves volume-flow velocity at a given CFM. Equal-friction conversion uses empirical shape relationships intended to compare straight-duct resistance at the same airflow.
Hydraulic diameter represents four times area divided by wetted perimeter. It is a different geometric quantity from the equal-friction round diameter and should not silently replace it in a sizing method.
Equal area: D = √(4A/π). Rectangular equal friction: De = 1.30(WH)^0.625/(W+H)^0.25. Flat-oval equal friction: De = 1.55A^0.625/P^0.25, A = πb²/4 + b(a−b), P = πb + 2(a−b). Hydraulic diameter: Dh = 4A/P.
D = equal-area round diameter; De = equal-friction round diameter; Dh = hydraulic diameter. W/H and a/b are the rectangular and oval full dimensions. A = section area and P = wetted perimeter. Use consistent length units within each shape relation; the form dimensions are inches.
Retain the selected conversion basis beside the dimensions. Equal-area, equal-friction and hydraulic diameters answer different questions and cannot be treated as interchangeable nominal sizes.
Inputs: 10 in round, selected rectangle aspect 2:1.
Calculation: Preserve the 78.54 in² circle area and apply the 2:1 ratio.
Result: About 12.5 × 6.3 in equal-area rectangle.
Inputs: 12 × 8 in rectangle.
Calculation: Equivalent circle diameter √(4×96/π).
Result: About 11.1 in equal-area round diameter.
Inputs: 16 in equivalent round diameter; selected rectangular width:height ratio 1.5:1.
Calculation: With W=rH, invert Equation 25: H = De/[1.30r^0.625/(r+1)^0.25], then W=rH.
Result: 18.018 × 12.012 in theoretical rectangular section; De=16 in.
Inputs: 28 × 14 in rectangular duct from ASHRAE Chapter 21 Table 9.
Calculation: Use Equation 25 with the clear inside dimensions.
Result: 21.327 in equivalent round diameter, matching the table’s rounded 21.3 in. Hydraulic diameter is a separate 18.667 in.
Inputs: 16 in round duct; fixed flat-oval minor axis 12 in.
Calculation: Set πD²/4 = πb²/4 + b(a−b), then solve a.
Result: Major axis 19.330 in; minor axis 12 in. This preserves area.
Inputs: 18 × 12 in flat oval with semicircular ends.
Calculation: A = π12²/4 + 12(18−12) = 185.097 in²; equal-area D = √(4A/π).
Result: Equal-area diameter 15.352 in; the separate equal-friction diameter is 15.254 in.
Inputs: 16 in equivalent round diameter; fixed flat-oval minor axis 12 in.
Calculation: Solve Equation 26 for major axis using the flat-oval area and perimeter equations, keeping the major axis at least 12 in.
Result: Major axis 19.662 in; minor axis 12 in; equal-friction De=16 in.
Inputs: 28 × 14 in flat-oval duct from ASHRAE Chapter 21 Table 9.
Calculation: Area 349.938 in² and perimeter 71.982 in give Equation 26 equivalent diameter.
Result: 20.702 in equivalent round diameter, matching the table’s rounded 20.7 in. Hydraulic diameter is a separate 19.446 in.
Equal area preserves average velocity when airflow is unchanged. Equal friction generally changes actual area, so its velocity needs a separate calculation.
The model uses two semicircular ends joined by straight sides. Enter the full major and minor axes specified by the fields, not a corner radius or the straight-side length.