Mechanical Properties of Fluids
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Question : 14
Total: 31
In a test experiment on a model aeroplane in a wind tunnel, theflow speeds on the upper and lower surfaces of the wing are 70 m s – 1 63 m s – 1 2.5 m 2 ? 1.3 k g m – 3
Solution:
Let v 1 and v 2 P 1 and P 2
Here,v 1 = 70 m s − 1 ;
v 2 = 63 m s − 1 ;
ρ = 1.3 k g m − 3
The level of the upper and lower surfaces of the wings from the ground may be taken same.
∴ h 1 = h 2
area of wing,A = 2.5 m 2
Thus from Bernoulli’s Theorem,
P 1 + ρ g h 1 +
ρ v 1 2 = P 2 + ρ g h 2 +
ρ v 2 2 P 2 − P 1 =
ρ ( v 1 2 − v 2 2 ) . . . ( i )
This pressure difference provides the lift to the aeroplane.
If F be the lift on the wing, then
F = ( P 2 − P 1 ) × A =
ρ ( v 1 2 − v 2 2 ) × A =
× 1.3 × ( 70 2 − 63 2 ) × 2.5
=
× 1.3 × 931 × 2.5 = 1512.9 N
= 1.5 × 10 3 N
Here,
The level of the upper and lower surfaces of the wings from the ground may be taken same.
area of wing,
Thus from Bernoulli’s Theorem,
This pressure difference provides the lift to the aeroplane.
If F be the lift on the wing, then
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