Question

Air is streaming past a horizontal air plane wing such that its speed isover the upper surface is 120ms and at the lower surface is 90ms . If the density of air is1.3kgm³ find the difference in pressure between the top and bottom of the wing. If the wing is 10m long and has an average width of 2m, ​

Answer 1

Answer:

• Pressure Difference (ΔP) = 4095 Pascals.

Given:

1. Velocity at upper surface (v₁) = 120 m/s.
2. Velocity at Lower surface (v₂) = 90 m/s.
3. Density of air (ρ) = 1.3 Kg/m³.

Explanation:

$\rule{300}{1.5}$

This is an Application of Bernoullis theorem.

From Bernoullis theorem we Know,

$\large{\boxed{\bold{P + \dfrac{1}{2}\rho v^2 + \rho gh = Constant}}}$

$\bold{Here}\begin{cases} \rho \text{ denotes Density} \\ \text{v denotes Velocity} \\ \text{P denotes Pressure} \\ \text{h denotes Height} \\ \text{g denotes Acceleration due to gravity}\end{cases}$

Now, the Equation becomes,

$\large{\tt \hookrightarrow P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2 }$

But the Aeroplane is flying at a Same Altitude.

Therefore, h₁ = h₂ = h.

Substituting,

$\large{\tt \hookrightarrow P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh}$

$\large{\tt \hookrightarrow P_1 + \dfrac{1}{2}\rho v_1^2 + \cancel{\rho gh} = P_2 + \dfrac{1}{2}\rho v_2^2 + \cancel{\rho gh}}$

$\large{\hookrightarrow{\underline{\underline{\tt P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2 }}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

$\large{\tt \hookrightarrow P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2 }$

$\large{\tt \hookrightarrow P_2 - P_1 = \dfrac{1}{2}\rho v_1^2 - \dfrac{1}{2}\rho v_2^2}$

Let P₂ - P₁ = ΔP.

$\large{\tt \hookrightarrow \Delta P = \dfrac{1}{2}\rho v_1^2 - \dfrac{1}{2}\rho v_2^2}$

$\large{\tt \hookrightarrow \Delta P = \dfrac{1}{2} \rho \bigg[v_1^2 - v_2^2\bigg]}$

Substituting the values,

$\large{\tt \hookrightarrow \Delta P = \dfrac{1}{2} \times 1.3 \times \bigg[(120)^2 - (90)^2\bigg]}$

$\large{\tt \hookrightarrow \Delta P = \dfrac{1}{2} \times 1.3 \times (14400 - 8100)}$

$\large{\tt \hookrightarrow \Delta P = \dfrac{1}{2} \times 1.3 \times (6300)}$

$\large{\tt \hookrightarrow \Delta P = \dfrac{1}{\cancel{2}} \times 1.3 \times \cancel{6300}}$

$\large{\tt \hookrightarrow \Delta P = 1 \times 1.3 \times 3150}$

$\large{\tt \hookrightarrow \Delta P = 1 \times 13 \times 315}$

$\large{\tt \hookrightarrow \Delta P = 1 \times 4095}$

$\huge{\boxed{\boxed{\tt \Delta P = 4095 \: Pa}}}$

∴ The Pressure Difference is 4095 Pascals.

$\rule{300}{1.5}$