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 m²? Take the density of air to be 1.3 kg m⁻³.
The lift on the wing is approximately 1513 N.
Given:
- Speed on upper surface, v₂ = 70 m s⁻¹
- Speed on lower surface, v₁ = 63 m s⁻¹
- Area of wing, A = 2.5 m²
- Density of air, ρ = 1.3 kg m⁻³
Step 1: Calculate pressure difference According to Bernoulli's principle, the pressure difference between the upper and lower surfaces of the wing is given by: ΔP = (ρ/2)(v₂² - v₁²) ΔP = (1.3/2)(70² - 63²) ΔP = 0.65 × (4900 - 3969) ΔP = ΔP = 605.15 N m⁻²
Step 2: Calculate the lift force The lift on the wing is the product of pressure difference and wing area: Lift = ΔP × A Lift = Lift = 1512.875 N 1513 N
Therefore, the lift on the wing is approximately 1513 N.
Explanation
This question applies Bernoulli's principle to calculate dynamic lift on an aircraft wing. The context explains that when air flows over an aerofoil, the streamlines crowd together above the wing, causing higher flow speed on top compared to the bottom. According to Bernoulli's principle, higher speed means lower pressure. The pressure difference between lower and upper surfaces creates an upward lift force. The formula ΔP = (ρ/2)(v₂² - v₁²) is explicitly derived in Example 9.7 of the textbook context. Multiplying this pressure difference by the wing area gives the total lift force.
Solution Steps
Step 1: Identify given values - v₂ = 70 m s⁻¹, v₁ = 63 m s⁻¹, A = 2.5 m², ρ = 1.3 kg m⁻³
Step 2: Apply Bernoulli's principle formula for pressure difference: ΔP = (ρ/2)(v₂² - v₁²)
Step 3: Calculate v₂² - v₁² = 4900 - 3969 = 931
Step 4: Calculate ΔP = (1.3/2) × 931 = 605.15 N m⁻²
Step 5: Calculate lift force: Lift = ΔP × A = = 1512.875 N 1513 N