1) Definition (MCQ)
Q: What is the primary purpose of residential supply ductwork?
A. Remove indoor air to the outdoors
B. Deliver conditioned air from the air handler to the living spaces
C. Provide combustion air to the furnace
D. Balance static pressure only
Answer: B. Deliver conditioned air from the air handler to the living spaces.
Why: Supply ducts distribute cooled/heated air to rooms; returns bring air back.
2) Round-duct sizing (calculation)
Q: A bedroom requires 800 CFM. Target design face-velocity in the duct is 900 fpm. What round
duct diameter (inches) is required? Show steps.
Work & Answer:
1. Area (ft²) = CFM ÷ velocity = 800 ÷ 900 = 0.888888... ft².
2. For round duct: d (ft) = sqrt(4·Area / π).
4·Area = 4 × 0.888888... = 3.5555556.
Divide by π: 3.5555556 ÷ 3.14159265 = 1.1310.
sqrt(1.1310) = 1.0636 ft.
3. Convert to inches: 1.0636 ft × 12 = 12.763 in. → round to 12.8 in (practical: use 13 in round duct).
Answer: ≈12.8 in (use a 13″ round duct).
3) Static pressure budget (short calc)
Q: The fan external static pressure available = 0.50 in. w.c. Coil drop = 0.20 in. w.c., filter = 0.10 in.
w.c. What maximum static pressure remains for duct system losses?
Answer & Steps:
Remaining = 0.50 − 0.20 − 0.10 = 0.20 in. w.c. available for ducts and registers.
4) Equivalent length (practical)
Q: A flexible duct run is physically 25 ft. Manufacturer recommends adding 20 ft equivalent length for
fittings and turns. What total equivalent length do you use in friction charts?
Answer: 25 + 20 = 45 ft equivalent length.
Q: What is the primary purpose of residential supply ductwork?
A. Remove indoor air to the outdoors
B. Deliver conditioned air from the air handler to the living spaces
C. Provide combustion air to the furnace
D. Balance static pressure only
Answer: B. Deliver conditioned air from the air handler to the living spaces.
Why: Supply ducts distribute cooled/heated air to rooms; returns bring air back.
2) Round-duct sizing (calculation)
Q: A bedroom requires 800 CFM. Target design face-velocity in the duct is 900 fpm. What round
duct diameter (inches) is required? Show steps.
Work & Answer:
1. Area (ft²) = CFM ÷ velocity = 800 ÷ 900 = 0.888888... ft².
2. For round duct: d (ft) = sqrt(4·Area / π).
4·Area = 4 × 0.888888... = 3.5555556.
Divide by π: 3.5555556 ÷ 3.14159265 = 1.1310.
sqrt(1.1310) = 1.0636 ft.
3. Convert to inches: 1.0636 ft × 12 = 12.763 in. → round to 12.8 in (practical: use 13 in round duct).
Answer: ≈12.8 in (use a 13″ round duct).
3) Static pressure budget (short calc)
Q: The fan external static pressure available = 0.50 in. w.c. Coil drop = 0.20 in. w.c., filter = 0.10 in.
w.c. What maximum static pressure remains for duct system losses?
Answer & Steps:
Remaining = 0.50 − 0.20 − 0.10 = 0.20 in. w.c. available for ducts and registers.
4) Equivalent length (practical)
Q: A flexible duct run is physically 25 ft. Manufacturer recommends adding 20 ft equivalent length for
fittings and turns. What total equivalent length do you use in friction charts?
Answer: 25 + 20 = 45 ft equivalent length.
