Air flows through a heating duct with a square cross-section with 9-inch sides at a speed of 6.1 ft/s. Just before reaching an outlet in the floor of a room, the duct widens to assume a square cross-section with sides equal to 13 inches. Compute the speed of the air flowing into the room (in ft/s), assuming that we can treat the air as an incompressible fluid.

2.9237 ft/s

Explanation:

Given the data in the question;

A₁ = 9-inch × 9-inch = 81 in² = 81 / 144 = 0.5625 ft²

V₁ = 6.1 ft/s

A₂ = 13 in × 13 in = 169 in² = 1.17361 ft²

v₂ = ?

using the the equation if continuity

( Rate of volumetric flow is constant )

A₁V₁ = A₂V₂

we substitute

0.5625 ft² × 6.1 ft/s = 1.17361 ft² × V₂

3.43125 ft³/s = 1.17361 ft² × V₂  

V₂  = 3.43125 ft³/s  / 1.17361 ft²

V₂ = 2.9237 ft/s

Therefore, the speed of the air flowing into the room is 2.9237 ft/s

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