Question

In a test experiment on a model aeroplane 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 1.3 kg $m^{–3}$.

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, $A=2.5m^2$
Density of air, $\rho =1.3kg{m}^{–3}$
According to Bernoulli’s theorem, we have the relation:
$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2{P}_2-P_1=\frac{1}{2}\rho (V_1^2-V_2^2)$
Where,
$P_1$ = Pressure on the upper surface of the wing
$P_2$= 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 $=(P_2-P_1)A$
$=\frac{1}{2}\rho (V_1^2-V_2^2)A=\frac{1}{2}1.3\left(\right(70{)}^3-\left(63{)}^2\right)\times 2.5$
=1512.87
$=1.51\times {10}^{3}N$
Therefore, the lift on the wing of the aeroplane is $1.51\times {10}^{3}N.$