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Consider a low speed open circuit subsonic wind tunnel with an inlet to throat area ratio of 12. The tunnel is turned on and the pressure difference between the inlet (the settling chamber) and the test section is read as a height difference of 10 cm on a U-tube mercury manometer. (The density of liquid mercury is 1.36 x 10-4 kg/m3 ). Calculate the velocity of the air in the test section.

Respuesta :

Answer:

velocity = 147.57 m/s

Explanation:

given data

inlet to throat area ratio = 12

height difference Δh = 10 cm = 0.1 m

density of liquid mercury = 1.36 × [tex]10^{4}[/tex] kg/m³

solution

we get here weight of mercury that is express as

weight of mercury = density of liquid mercury ×  g    .................1

weight of mercury = 1.36 × [tex]10^{4}[/tex] × 9.8

weight of mercury = 1.33 × [tex]10^{5}[/tex]  N/m²  

and

area ratio is

[tex]\frac{a1}{a2}[/tex]  = 12

so velocity of air in the test section will be

velocity = [tex]\sqrt{\frac{2\times w \times \triangle h}{\rho (1-(area\ ration)^2)}}[/tex]      .......................1

put here value and we get

velocity = [tex]\sqrt{\frac{2\times 1.33\times 10^5 \times 0.1}{1.23 (1-(\frac{1}{12})^2)}}[/tex]    

velocity = 147.57 m/s

The velocity of air in the test section is; v = 147.802 m/s

Velocity of air

We are given;

  • inlet to throat area ratio; a1/a2 = 12
  • height difference Δh = 10 cm = 0.1 m
  • density of liquid mercury; ρ = 1.36 × 10⁴ kg/m³

Thus, weight of mercury is;

W = ρg

W = 1.36 × 10⁴ × 9.8

W = 13.328 × 10⁴ N/m²

Formula for the velocity of the air in the test section is;

v = √[(2ρgΔh)/(ρ_air × (1 - a2/a1)]

Where ρ_air is density of air = 1.23 kg/m³

Thus;

v = √[(2 × 1.36 × 10⁴ × 9.8 × 0.1)/(1.23 × (1 - 1/12)]

Solving this gives us;

v = 147.802 m/s

Read more on Velocity at; https://brainly.com/question/4931057

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