Answer: the average velocity of the air exiting the tunnel at C is 20.8188 m/s
Step-by-step explanation:
From the image, Given that;
d₁ = 1.3 m, d₂ = 4 mm = 0.004 m (hole diameter), d₃ = 1.1m
v₁ = 15 m/s, v₂ = 10 m/s, v₃ = ?
we equate the volume flow rate between 1 and 2
A₁V₁ = 1000(A₂V₂) + Q
we know that the area is πd²/4
so
(πd₁²/4)V₁ = 1000((πd₂²/4)V₂) + Q
we substitute
((π(1.3)²/4)) × 15 = 1000((π(0.004)²/4)) × 10) + Q
19.9098 = 0.12566 + Q
Q = 19.9098 - 0.12566
Q = 19.7841 m³/s
now
Q = (πd₃²/4) V₃
we substitute
19.7841 = (π(1.1)²)/4) V₃
19.7841 = 0.9503V₃
V₃ = 19.7841 / 0.9503
V₃ = 20.8188 m/s
therefore the average velocity of the air exiting the tunnel at C is 20.8188 m/s