Consider a Venturi with a throat-to-inlet area ratio of 0.8, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level.
If the static pressure at the throat is 2100 lb/ft³, calculate the velocity of the airplane. Note that standard sea-level density and pressure are 1.23 kg/m³ (0.002377 slug/ft³) and 1.01 x 10⁵ N/m2 (2116lb/ft³), respectively.

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cjmejiab cjmejiab · Answer 1 · 2020-02-06T16:51:08+01:00

Defining the radius of area throat to inlet we would have that the proportional relationship would be [tex]\frac{A_2}{A_1}.[/tex]

This equation or relationship is obtained from continuity where

[tex]A_1V_1 = A_2V_2[/tex]

[tex]\frac{V_2}{V_1} = \frac{A_2}{A_1}= 0.8[/tex]

Now applying the Bernoulli equation between inlet and throat section we have,

[tex]\frac{p_1}{\rho g}+ \frac{v_1^2}{2g}+z_1=\frac{p_2}{\rho g}+ \frac{v_2^2}{2g}+z_2[/tex]

Here,

[tex]z_1 = z_2[/tex]

Then for a Venturi duct, the velocity of the airplane [tex]V_1[/tex] will be

[tex]V = \sqrt{\frac{2(p_1-p_2)}{\rho[(\frac{A_1}{A_2})^2-1]}}[/tex]

Our values are,

[tex]\frac{A_2}{A_1} = 0.8[/tex]

[tex]\rho = 0.002377slug/ft^3[/tex]

[tex]p_1 = 2116lb/ft^2[/tex]

[tex]p_2 = 2100lb/ft^2[/tex]

Replacing,

[tex]V= \sqrt{\frac{2(2116-2100)}{(0.002377)[(\frac{1}{0.8})^2-1]}}[/tex]

[tex]V = 154.7ft/s[/tex]

Therefore the velocity of the airplane is 154.7ft/s