question_answer 14)

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 \[70\,m{{s}^{-1}}\] and \[63\,m{{s}^{-1}}\] respectively. What is the lift on the wing if its area is \[2.5\,{{m}^{2}}\]? Take the density of air is \[1.3\,kg\,\,{{m}^{-3}}\].

Answer:

Let \[{{Y}_{1}}:{{Y}_{2}}=1\cdot 8:1\] be the speeds on the upper and lower surfaces of the wing of aeroplane, and \[=150\times {{10}^{6}}N{{m}^{-2}},\] and \[\text{Y=}\frac{\text{Stress}}{\text{Strain}}\text{=}\frac{\text{150 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}}{\text{0 }\!\!\times\!\!\text{ 002}}\] be the pressures on upper and lower surfaces of the wing respectively. Then \[=7\cdot 5\times {{10}^{10}}N{{m}^{-2}}\] From Bernoulli's theorem, \[=300\times {{10}^{6}}N{{m}^{-2}}\] \[=3\times {{10}^{8}}N{{m}^{-2}}\] or \[\begin{align} & {{F}_{1}}=4+6=10kgf=10\times 9\cdot 8N; \\ & {{l}_{1}}=1\cdot 5m,\vartriangle {{l}_{1}}=?;2{{r}_{1}}=0\cdot 25cm \\ \end{align}\] \[\begin{align} & {{r}_{1}}=\left( 0\cdot 25/2 \right)cm=0\cdot 125\times {{10}^{-2}}m, \\ & {{Y}_{1}}=2\cdot 0\times {{10}^{11}}pa \\ \end{align}\]\[{{\text{F}}_{\text{2}}}\text{=6 }\!\!\times\!\!\text{ 0kgf=6 }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 8N;2}{{\text{r}}_{\text{2}}}\text{=0 }\!\!\times\!\!\text{ 25cm}\] This difference of pressure provides the lift to the aeroplane. So, lift on the aeroplane = pressure difference x area of wings \[\begin{align} & {{r}_{2}}=\left( 0\cdot 25/2 \right)cm=0\cdot 125\times {{10}^{-2}}m; \\ & {{Y}_{2}}=0\cdot 91\times {{10}^{11}}pa,{{l}_{2}}=1\cdot 0m,\vartriangle {{l}_{2}}=? \\ \end{align}\]\[{{Y}_{1}}=\frac{{{F}_{1}}\times {{l}_{1}}}{{{a}_{1}}\times \vartriangle {{l}_{1}}}=\frac{{{F}_{1}}\times {{l}_{1}}}{\pi r_{1}^{2}\times \vartriangle {{l}_{1}}}\]