Answer to Question #162649 in Mechanics | Relativity for Dorcas

Question

Air flows through a heating duct with a square cross-section with 9-inch sides at a speed of 5.7 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.

Answer 1

By the equation of continuity

(Rate of volumetric flow rate is constant)

$A_1V_1 = A_2V_2 \\ (9 \times 9) \times 5.7 = (13 \times 13) \times V_2 \\ 461.7 = 169 \times V_2 \\ V_2 = \frac{461.7}{169} = 2.73 \;ft/s$

Answer: 2.73 ft/s