Difference 69354

An airplane flies at a speed of 920 km/h at an altitude of 11 km above the surface of the Earth, where the air density is p = 0.36 kg/m³. Determine the pressure difference above and below the plane's wing if the top of the plane is 10% longer than the bottom.

Correct answer:

Δ =  2468.67 Pa

Step-by-step explanation:

$$\begin{gathered} v_1=920\text{km/h}\to \text{m/s}=920:3.6\text{m/s}=255.55556\text{m/s} \\ h=11\text{km}\to \text{m}=11\cdot 1000\text{m}=11000\text{m} \\ \rho =0.36{\text{kg/m}}^3 \\ g=9.81{\text{m/s}}^2 \\ q=100\%+10\%=1+\frac{10}{100}=1.1 \\ {v}_2=q\cdot v_1=\frac{11}{10}\cdot \frac{2300}{9}=\frac{11\cdot 2300}{10\cdot 9}=\frac{25300}{90}=\frac{2530}{9}\doteq 281.1111\text{m/s} \\ p=\rho \cdot g\cdot h=0.36\cdot 9.81\cdot 11000=\frac{194238}{5}=38847.6\text{Pa} \\ {p}_1+\rho \cdot g\cdot h+\frac{1}{2}\cdot \rho \cdot v_1^2=p_2+\rho \cdot g\cdot h+\frac{1}{2}\cdot \rho \cdot v_2^2 \\ {p}_1+\frac{1}{2}\cdot \rho \cdot v_1^2=p_2+\frac{1}{2}\cdot \rho \cdot v_2^2 \\ \Delta =p_2-p_1 \\ \Delta =\frac{1}{2}\cdot \rho \cdot (v_2^2-v_1^2)=\frac{1}{2}\cdot \rho \cdot ({\frac{2530}{9}}^2-{\frac{2300}{9}}^2)=\frac{7406}{3}\text{Pa}=2468.67\text{Pa} \end{gathered}$$

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