Sizing the Duct System and Return Path for a 5-Ton Office TI
A 2,200 ft² open-office tenant improvement gets one new 5-ton (60,000 Btu/h) packaged RTU rated 2,000 cfm at 0.90 in. w.c. external static. Manual J: 52,000 Btu/h total, 38,000 Btu/h sensible at 95°F outdoor, 75°F / 50% RH indoor. External airstream components: 10 kW electric heat kit 0.18 in. w.c., supply diffusers 0.10, return grille 0.07. Index supply path: 55 ft trunk plus 30 ft branch, with 70 ft equivalent length for fittings. Ducted return: 20 ft of duct plus 15 ft equivalent for fittings. A 12 x 12 ft conference room has a 300 cfm diffuser and one 36 in. x 80 in. door with a 3/4 in. undercut; the owner refuses visible transfer grilles in the corridor. The blower has taps for 1,600 / 1,800 / 2,000 cfm.
1. What design friction rate (in. w.c. per 100 ft) is available for this duct system? Show the arithmetic and state any assumptions about the index run.
Read the worked answer
Available static for ductwork
Start with the unit's rated external static and subtract every component that sits in the airstream outside the cabinet:
- Heat kit: 0.18
- Supply diffusers: 0.10
- Return grille: 0.07
- Total deductions: 0.35 in. w.c.
Available static = 0.90 − 0.35 = 0.55 in. w.c.
Total effective length of the index run
The index (longest) circuit is the one with the highest total pressure loss, which here includes both supply and return because the return is ducted:
- Supply trunk 55 + branch 30 + return duct 20 = 105 ft measured
- Fittings equivalent = 70 (supply) + 15 (return) = 85 ft
- TEL = 190 ft total (105 + 85)
Friction rate
FR = available static × 100 / TEL = 0.55 × 100 / 190 = 0.29 in. w.c. per 100 ft (about 0.29–0.30)
Assumptions worth stating in the field: friction rate must be calculated on the longest path, not an average; flex duct, volume dampers, and any fire dampers must be included in the equivalent lengths; and if the return is a plenum or open ceiling cavity rather than ducted, the return length drops out and the friction rate rises (here it would go to about 0.37). That is exactly the kind of assumption that should be confirmed before the shop cuts metal, because a 0.08 in. w.c. error changes every duct size on the sheet.
2. The plan is drawn at 2,000 cfm. Is that the right airflow for the calculated load, and which blower tap would you set? Justify with numbers.
Read the worked answer
Nominal check
2,000 cfm ÷ 5 tons = 400 cfm/ton - nominal, so nothing is obviously wrong at nameplate conditions.
Sensible check
Sensible capacity = 1.08 × cfm × ΔT. At 2,000 cfm and a 20°F coil ΔT:
1.08 × 2,000 × 20 = 43,200 Btu/h sensible
The load is 38,000 Btu/h sensible, so the unit has roughly 14% sensible margin. That sounds fine until you compare total capacity to total load: 60,000 available vs 52,000 required is about 15% oversized.
The judgment call
An oversized unit at full airflow satisfies sensible load but satisfies latent load only by accident, and short cycling makes it worse. The lever we have is airflow:
- At 1,800 cfm: 1.08 × 1,800 × 20 = 38,880 Btu/h sensible - almost exactly the calculated sensible load, with the remaining capacity going to latent removal.
- At 1,600 cfm: 34,560 Btu/h sensible - below the load on a design day, and 320 cfm/ton is low enough to risk a cold, wet coil and icing in mild weather.
Answer: set the 1,800 cfm tap (360 cfm/ton). It matches sensible load, improves latent removal and run time, and stays safely above the ~350 cfm/ton floor. Then verify in the field with a supply/return ΔT and a wet-bulb split rather than trusting the tap label, because a restrictive duct system can leave you well below the nominal cfm even on the correct tap.
3. The conference room door has only a 3/4 in. undercut and the owner will not allow corridor transfer grilles. Quantify the return-air problem and propose a solution that respects that constraint.
Read the worked answer
Quantify the existing path
Undercut free area = 36 in. × 0.75 in. = 27 in² = 27 / 144 = 0.1875 ft²
Velocity through the undercut at 300 cfm = 300 / 0.1875 = 1,600 fpm
That is far beyond a reasonable return-path velocity (a practical ceiling is roughly 300–500 fpm, well under 0.05 in. w.c. of pressure drop). The result is a whistling door, the door standing open, and the room pressurizing so the 300 cfm supply never actually arrives. Doubling the undercut to 1.5 in. only gets to 54 in² and about 800 fpm - still too high, so trimming the door is not a fix.
Required free area
At 400 fpm: 300 / 400 = 0.75 ft² = 108 in² net free area. A stamped grille is typically 60–70% free area, so gross grille size is about 108 / 0.65 ≈ 166 in² per grille, or two roughly 10 x 10 in. grilles with comparable net area.
Solution that keeps grilles out of the corridor
Install a jump duct / ceiling transfer path from the conference room ceiling into the adjacent open-office return plenum or return duct: a short length of lined or insulated flex (or sheet metal) with a grille at each end, each sized to the ~110 in² net free area computed above, with the room-end grille located high on a wall or in the ceiling away from the supply diffuser. This gives the 300 cfm a low-velocity path without a corridor penetration, and it must be lined or the transfer path itself becomes the noise source. Document the net free area on the drawing so the balancer can confirm the room holds its design cfm at the door closed.