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The minimum unobstructed total area of the outdoor and return air ducts to a forced-air warm-air furnace shall be a minimum of ____ sq in per 1000 BTU/h output rating capacity of the furnace.

a. 1
b. 2
c. 4
d. 6

User Saiwing
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1 Answer

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Final answer:

The volume of air that needs to be moved in the house is 715 m³, and the flow rate is 0.793 m³/s. The cross-sectional area of the air duct is 0.071 m². The average speed of air in the duct is 11.17 m/s.

Step-by-step explanation:

To calculate the average speed of air in the duct, we need to know the volume of air to be moved and the cross-sectional area of the duct. The volume of the house is provided as a rectangular solid with dimensions 13.0 m wide, 20.0 m long, and 2.75 m high. To find the volume of air to be moved, we multiply these dimensions giving:

V = 13.0 m × 20.0 m × 2.75 m = 715 m³

Since the heater needs to move this air every 15 minutes, we divide the volume by the time in seconds to find the flow rate:

Flow rate = 715 m³ / (15 min × 60 sec/min) = 0.793 m³/s

To calculate the cross-sectional area of the duct, we use the diameter (0.300 m) to find the radius (0.150 m) and apply the area formula for a circle (A = πr²):

A = π × (0.150 m)² = 0.071 m²

Lastly, to find the average speed of air (v), we use the relationship between flow rate (Q), speed (v), and cross-sectional area (A): Q = v × A. Solving for v gives us:

v = Q / A = 0.793 m³/s / 0.071 m² = 11.17 m/s

Therefore, the average speed of air in the duct is 11.17 m/s.

User Carlos Santillan
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