Question

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–1and 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.

Answer 1

Speed of air on the upper surface of the wings (v1) = 70 m/s
speed of the air on the lower surface of the wings ( v2) = 63 m/s

density of air (d) = 1.3 Kg/m³
Area (A) = 2.5 m²

According to Bernoulli's theorem,
P1 + 1/2dv1² + dgh = P2 + 1/2dv2² + dgh
P2 - p1 = 1/2d(v1² - v2²)

So, lifting force = Pressure × A
= (P2 - P1) × A
= 1/2d(v1² - v2²) × A
= 1/2 × 1.3 ( 70² - 63²) × 2.5
= 1/2 × 1.3 × 931 × 2.5
= 1.51 × 10³ N

Answer 2

Speed of air on the upper surface of the wings (v1) = 70 m/s

speed of the air on the lower surface of the wings ( v2) = 63 m/s

density of air (d) = 1.3 Kg/m³

Area (A) = 2.5 m²

According to Bernoulli's theorem,

P1 + 1/2dv1² + dgh = P2 + 1/2dv2² + dgh

P2 - p1 = 1/2d(v1² - v2²)

So, lifting force = Pressure × A

= (P2 - P1) × A

= 1/2d(v1² - v2²) × A

= 1/2 × 1.3 ( 70² - 63²) × 2.5

= 1/2 × 1.3 × 931 × 2.5

= 1.51 × 10³ N

Read more on Brainly.in - https://brainly.in/question/1457838#readmore