The wind tunnel is designed so that the lower pressure outside the testing region B draws air out in order to reduce the boundary layer. Within region B there are 1000 holes, each 4 mm in diameter. If the pressure is adjusted so that the average velocity of the air through each hole is 10 m/s, determine the average velocity of the air exiting the tunnel at C. (Assume the air is incompressible).

the average velocity of the air exiting the tunnel at C is 20.8188 m/s

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

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explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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