Question

The difference in the pressure between the air just over the wing $P_1$ and that under the wing $P_2$ is:

• A. $2.8\times {10}^3$ Pa
• B. $0.723\times {10}^2$ Pa
• C. $1.46\times {10}^4$ Pa
• D. Zero

Answer 1

The correct option is C $1.46\times {10}^4$ Pa

Compression of the streamlines means that the stream tube above the wing
has a smaller cross-sectional area than that in front of the plane and from the continuity equation , the velocity v' of the air must therefore be greater above the wing.
Area $A^{\prime }=\frac{8}{10}A$ $\Rightarrow \frac{A}{A^{\prime }}=\frac{10}{8}$
From continuity equation,
$Av=A^{\prime }{v}^{\prime }$
$\therefore v^{\prime }=\frac{A}{A^{\prime }}v=\frac{10}{8}\times 200=250m{s}^{-1}$
The greater velocity (v',as obtained above) implies lower pressure than the normal pressure of the air in front of the plane. Given that the flow-lines under the wing are not compressed at all. The pressure under the wing is just the normal pressure of the air in front of the wing.

From Bernoulli's equation , with both points 1 (for p1) and 2 (for p2) at effectively same elevation.
$P_a+\frac{1}{2}{\rho }_a{v}_1^2=P_b+\frac{1}{2}{\rho }_b{v}_2^2$
Given:
$v_2=200m{s}^{-1}$
$\therefore P_2-P_1=\frac{1}{2}{\rho }_a(v_1^2-v_2^2)=\frac{1}{2}\times 1.3({250}^2-{200}^2)=1.46\times {10}^{4}Pa$