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.
Answers
Speed of wind on the upper surface of the wing, V1 = 70 m/s
Speed of wind on the lower surface of the wing, V2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m–3
According to Bernoulli’s theorem, we have the relation:
P1 + (1/2)ρV12 = P2 + (1/2)ρV22
P2 - P1 = (1/2) ρ (V12 - V22)
Where,
P1 = Pressure on the upper surface of the wing
P2 = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P2 - P1) A
= (1/2) ρ (V12 - V22) A
= (1/2) × 1.3 × [ 702 - 632 ] × 2.5
= 1512.87 N = 1.51 × 103 N
Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.
Answer:
ort by Riya67681 12.07.2019
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Speed of wind on the upper surface of the wing, V1 = 70 m/s
Speed of wind on the lower surface of the wing, V2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m–3
According to Bernoulli’s theorem, we have the relation:
P1 + (1/2)ρV12 = P2 + (1/2)ρV22
P2 - P1 = (1/2) ρ (V12 - V22)
Where,
P1 = Pressure on the upper surface of the wing
P2 = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P2 - P1) A
= (1/2) ρ (V12 - V22) A
= (1/2) × 1.3 × [ 702 - 632 ] × 2.5
= 1512.87 N = 1.51 × 103 N
Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.