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Q10.14:
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 to be $1.3 kg m^{-3}$.

Solution :

Given Variables & Initial Setup

To determine the lift on the aeroplane wing, we identify the following parameters provided in the problem statement:

  • Velocity of air flow over the upper surface ($v_1$): $70\text{ m s}^{-1}$
  • Velocity of air flow over the lower surface ($v_2$): $63\text{ m s}^{-1}$
  • Area of the wing ($A$): $2.5\text{ m}^2$
  • Density of air ($\rho$): $1.3\text{ kg m}^{-3}$

Step 1: Applying Bernoulli’s Principle

According to Bernoulli’s Principle for streamline flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline [Per Bernoulli’s Equation: $P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$]. Assuming the wing is horizontal, the vertical height difference between the upper and lower surfaces is negligible ($\Delta h \approx 0$). Therefore, the equation simplifies to:

$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

Rearranging to find the pressure difference ($\Delta P = P_2 - P_1$):

$\Delta P = \frac{1}{2}\rho(v_1^2 - v_2^2)$

Step 2: Calculating the Pressure Difference

Substitute the given numerical values into the derived expression:

$\Delta P = \frac{1}{2} \times 1.3 \times (70^2 - 63^2)$

$\Delta P = 0.65 \times (4900 - 3969)$

$\Delta P = 0.65 \times 931$

$\Delta P = 605.15\text{ Pa}$

Step 3: Calculating the Total Lift

The lift force ($F_L$) acting on the wing is the product of the pressure difference and the surface area of the wing:

$F_L = \Delta P \times A$

$F_L = 605.15\text{ N m}^{-2} \times 2.5\text{ m}^2$

$F_L = 1512.875\text{ N}$

Visual Representation of Pressure Differential

v1 = 70 m/s v2 = 63 m/s Wing

Final Solution: The lift on the wing is $1512.875\text{ N}$.


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