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Rooftop unit supply duct check

Air distribution · duct velocity, equivalent round and velocity pressure

Up on the roof, a packaged rooftop unit delivers 2,000 cfm down through the curb into an 18 in × 12 in rectangular main supply duct (216 in² of free area), which runs exposed under the deck serving an open shop floor of 24,000 ft³. The balancer is at the duct with a drill, a pitot tube and a manometer, about to put in the traverse holes — but wants the paper numbers first, because a traverse without a prediction is just a reading. Two translations stand between the sheet metal and the answers: the duct checked against the friction chart — which is drawn for round duct — and the velocity pressure the pitot traverse should expect at this flow. Take standard air at 1.2014 kg/m³. One duct, four numbers — velocity, equivalent round, velocity pressure, air changes — and each one answers to a different trade.

2,000 cfmRTU18 × 12 in1,333 fpm24,000 ft³ · 5 ACH

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • V̇ = 2,000 cfm — Supply airflow
  • A = 216 in² — Duct free area (18 × 12 in)
  • a = 18 in — Duct side a
  • b = 12 in — Duct side b
  • ρ = 1.2014 kg/m³ — Standard air density
  • V_room = 24,000 ft³ — Shop floor volume
Determine
  1. (a)the average velocity in the duct
  2. (b)the equivalent round duct diameter for the friction chart
  3. (c)the velocity pressure a pitot tube should read
  4. (d)the air changes per hour the unit gives the space
Step 1 of 4(a) · solve for Flow velocity

Velocity is the first check on any duct: 2,000 cfm through 1.5 ft² is 1,333 fpm (6.77 m/s) — fine for a shop supply duct, loud in an office ceiling. The slip to avoid is using the sheet-metal nominal with an internal liner in place: an inch of liner all round shrinks 18 × 12 to 16 × 10 and adds 35% velocity.

AvQ
Rearranged for v
v=QAv = \tfrac{Q}{A}
Your values, in your units
v=(2,000 cfm)(216 in2)v = \tfrac{\left( 2{,}000\ \text{cfm} \right)}{\left( 216\ \text{in}^{2} \right)}
Converted to base units
v=(56,633.7 L/min)(0.139355 m2)v = \tfrac{\left( 56{,}633.7\ \text{L/min} \right)}{\left( 0.139355\ \text{m}^{2} \right)}
Answer
v=6.7733 m/sv = 6.7733\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Volumetric Flow Rate (Q = Av) solver →

Step 2 of 4(b) · solve for Equivalent round diameter

Friction charts are drawn for round duct, so the rectangle has to become the round duct with the SAME friction at the SAME flow — Huebscher's 1.30(ab)^0.625/(a+b)^0.25, which lands at 16.0 in. Note it is NOT the equal-area circle (that would be 16.6 in): equal friction, not equal area, is the promise.

baDe
Rearranged for Dₑ
De=1.30(ab)0.625(a+b)0.25D_e = 1.30 \tfrac{(ab)^{0.625}}{(a+b)^{0.25}}
Your values, in your units
De=1.30((18 in)×(12 in))0.625((18 in)+(12 in))0.25D_e = 1.30 \tfrac{(\left( 18\ \text{in} \right) \times \left( 12\ \text{in} \right))^{0.625}}{(\left( 18\ \text{in} \right) + \left( 12\ \text{in} \right))^{0.25}}
Converted to base units
De=1.30((0.4572 m)×(0.3048 m))0.625((0.4572 m)+(0.3048 m))0.25D_e = 1.30 \tfrac{(\left( 0.4572\ \text{m} \right) \times \left( 0.3048\ \text{m} \right))^{0.625}}{(\left( 0.4572\ \text{m} \right) + \left( 0.3048\ \text{m} \right))^{0.25}}
Answer
De=406 mmD_e = 406\ \text{mm}

Carried onward at full precision, not this rounded figure.

Open the Equivalent Round Duct Diameter solver →

Step 3 of 4(c) · solve for Dynamic pressure

The pitot traverse reads this directly: q = ½ρv² = 27.6 Pa, about 0.111 in w.g. — the field rule VP = (V/4,005)² gives the same number, because 4,005 is just ½ρ in fpm-and-inches clothing. It scales with velocity SQUARED, which is why a traverse can resolve small flow changes.

vρq
Rearranged for q
q=12ρv2q = \tfrac{1}{2} \rho v^{2}
6.7733 m/scarried from step 1
Your values, in your units
q=12 (1.2014 kg/m3) (6.77333 m/s)2q = \tfrac{1}{2} \, \left( 1.2014\ \text{kg/m}^{3} \right) \, \left( 6.77333\ \text{m/s} \right)^{2}
Answer
q=27.559 Paq = 27.559\ \text{Pa}

Carried onward at full precision, not this rounded figure.

Open the Dynamic Pressure (q = ½ρv²) solver →

Step 4 of 4(d) · solve for Air changes per hour

Finally the ventilation yardstick: 2,000 cfm is 120,000 ft³/h into a 24,000 ft³ room — 5 air changes per hour, a healthy shop-floor figure. Mind that this is SUPPLY air changes; codes that ask for outdoor-air ACH want this number times the outdoor-air fraction, not this number.

VV̇ACH
Rearranged for ACH
ACH=3600 V˙Vroom\mathrm{ACH} = \tfrac{3600 \, \dot{V}}{V_{room}}
Your values, in your units
ACH=3600×(2,000 cfm)(24,000 ft3)\mathrm{ACH} = \tfrac{3600 \times \left( 2{,}000\ \text{cfm} \right)}{\left( 24{,}000\ \text{ft}^{3} \right)}
Converted to base units
ACH=3600×(56,633.7 L/min)(679,604 L)\mathrm{ACH} = \tfrac{3600 \times \left( 56{,}633.7\ \text{L/min} \right)}{\left( 679{,}604\ \text{L} \right)}
Answer
ACH=5 1/h\mathrm{ACH} = 5\ \text{1/h}

Carried onward at full precision, not this rounded figure.

Open the Air Changes per Hour (ACH) solver →

Answer

The duct runs at about 1,333 fpm, behaves like a 16.0 in round duct on the friction chart, should show 0.111 in w.g. of velocity pressure on the pitot, and gives the shop 5 air changes per hour.

Why this order

These four numbers are one duct seen by four different people. The velocity is the acoustician's and the balancer's number — it decides noise and sets the pitot reading. The equivalent round is the designer's, because every friction chart and ductulator is drawn for round duct and a rectangle has more wetted perimeter per unit area, so Huebscher's exponents (0.625 and 0.25, fitted in 1948) find the round duct with equal friction rather than equal area. The velocity pressure is the measurement number: a pitot tube cannot read flow, only ½ρv², so the traverse recovers v from q and then multiplies back through the area — this chain run in reverse. And ACH is the code official's number, tying the whole airside to the room it serves. The cross-checks all live in trade units: 1,333 fpm is (1,333/4,005)² = 0.111 in w.g. ✓, and 18 × 12 in tables list 16.0 in equivalent round ✓.

The errors worth naming: liner. An inch of acoustic liner turns 18 × 12 into 16 × 10, cuts the area 26% and raises velocity 35% and velocity pressure 83% — a balancer who traverses a lined duct against the sheet-metal dimensions chases a phantom flow deficit all afternoon. Second, the equal-area shortcut: √(4ab/π) gives 16.6 in, and a friction chart read at 16.6 instead of 16.0 under-predicts the loss about 10% — small on one fitting, compounding over a hundred metres of duct. Third, the 4,005 constant is standard air only: at altitude or in a hot exhaust the density drops, and the honest ½ρv² this chain computes is what keeps the traverse honest when the rule of thumb quietly is not.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.