Question

In a test experiment on a model airplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are $70m{s}^{-1}$ and $63m{s}^{-1}$ respectively. What is the lift on the wing if its area is $2.5$ m${}^2$ ? Take the density of air to be $1.3kg{m}^{-3}$.

Answer 1

Speed of wind on the upper surface of the wing, $V_1=70m/s$
Speed of wind on the lower surface of the wing, $V_2=63m/s$
Area of the wing, $A=2.5m^2$
Densityof air, $\rho =1.3kg/m^3$
According to Bernoullis theorem, we have the relation:

$P1+\left(1/2\right)\rho \left(V_1^2\right)=P2+\left(1/2\right)\rho \left(V_2^2\right)$

$P2-P1=\left(1/2\right)\rho (V_1^2-V_2^2)$
Where,
$P1=$ Pressure on the upper surface of the wing
$P2=$ Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing $=(P2-P1)A$
$=\left(1/2\right)\rho (V_1^2-V_2^2)A$
$=\left(1/2\right)1.3[{70}^2-{63}^2]2.5$
$=1512.87$
$N=1.51\times {10}^{3}N$
Therefore, the lift on the wing of the aeroplane is $1.51\times {10}^{3}N$.