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 $70\, m / s$ and $63 \,m / s$ respectively. What is the lift on the wing, if its area is $2.5\, m ^{2}$ ? Take the density of air to be $1.3\, kg / m ^{3}$.

• A. $5.1 \times 10^{2}\, N$
• B. $6.1 \times 10^{2}\, N$
• C. $1.6 \times 10^{3} \,N$
• D. $1.5 \times 10^{3} \,N$

Answer 1

Answer: (D)

Let the lower and upper surface of the wings of the aeroplane be at the same height $h$ and speeds of air on the upper and lower surfaces of the wings be $v_{1}$ and $v_{2}$.
Speed of air on the upper surface of the wing $v_{1}=70\, m / s$
Speed of air on the lower surface of the wings $v _{2}=63 \, m / s$
Density of the air $\rho=1.3\, kg / m ^{3}$
Area $ A=2.5\, m ^{2}$
According to Bernoulli's theorem,
$p_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h=p_{2}+\frac{1}{2} \rho v^{2}+\rho g h$
or $p_{2}-P_{1}=\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)$
$\therefore $ Lifting force acting on the wings,
$F=\left(p_{2}-p_{1}\right) \times A=\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right) \times A$
$\left[\because\right.$ Pressure $\left.=\frac{\text { Force }}{\text { Area }}\right]$
$=\frac{1}{2} \times 1.3 \times\left[(70)^{2}-(63)^{2}\right] \times 2.5$
$=\frac{1}{2} \times 1.3[4900-3969] \times 2.5$
$=\frac{1}{2} \times 1.3 \times 931 \times 2.5=1.51 \times 10^{3} \,N$