Question

Air is streaming past a horizontal airplane wing such that its speed is $90m{s}^{-1}$ at the lower surface and $120m{s}^{-1}$ over the upper surface. If the wing is 10 m long and has an average width of 2m, the difference of pressure on the two sides and the gross lift on the wing respectively, are (density of air $=1.3kg{m}^{-3})$

• A. 5 Pa, 900 N
• B. 95 Pa, 900 N
• C. 4095 Pa, 900 N
• D. 4095 Pa, 81900 N

Answer 1

Answer: (D)

4095 Pa, 81900 N

Applying Bernoulli's principle, we have
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2$
$\Rightarrow P_2-P_1=\frac{1}{2}\rho (v_1^2-v_2^2)\Rightarrow \Delta P=\frac{1}{2}\times 1.3\times ({120}^2-{90}^2)\Rightarrow \Delta P=0.65\times (120+90)\times (120-90)\Rightarrow \Delta P=0.65\times 210\times 30=4095Pa$
The difference in pressure provides the lift to the aeroplane. So
lift on the aeroplane=pressure difference $\times$ area of wings
$=\Delta P\times A=4095\times (10\times 2)=81900N$