Question

Experiments were conducted in a wind tunnel with a wind speed of 50 km/hr on a flat plate of size 2 m long and 1 m wide. The density of air is 1.15 kg/m3, the co-efficient of lift and drag are 0.75 and 0.15 respectively. Determine (i) the lift force, (ii) the drag force, (iii) the resultant force, (iv) direction of resultant force and (v) the power exerted by air on plate

Answer 1

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Experiments were UNIT -8 FLUID MECHANICS .... wind tunnel with a wind speed of 50km/hr on a flat plate of size 2m long and 1m wide. The density of air is 1.15kg/ .

Answer 2

The lift force is 166.404 N, the drag force is 33.28 N, the resultant force is 169.67 N, the direction of the resultant force is 78.69°, and the power exerted by the air on the plate is 462.26 W.

Given,

Area of plate = 2 m²

Velocity of air = 50 km/hr = 13.89 m/s

Density of air, ρ = 1.15 kg/m3

Value of CD = 0.15 and CL = 0.75

To Find,

(i) the lift force, (ii) the drag force, (iii) the resultant force, (iv)the direction of the resultant force, and (v) the power exerted by the air on the plate.

Solution,

(i) Lift Force (FL)

FL = CL×A×ρ×V2/2 = 0.75×2×1.15×13.892/2 = 166.404 N

(ii) Drag force (FD)

FD = CD×A×ρ×U2/2 = 0.15×2×1.15×13.892/2 = 33.28 N

(iii) Resultant force (FR)

FR = √(F²D+F²L) = √33.28²+166.404² = 169.67 N

(iv) The direction of resultant force ( θ)

The direction of the resultant force is given by.

tanθ = FL/FD = 166.38/33.275 = 5

θ = tan⁻¹5 = 78.69°.

(v) Power exerted by the air on the plate

Power = Force in the direction of motion × Velocity

Power = FD×U N m/s = 33.280×13.89 W = 462.26 W

Hence, the lift force is 166.404 N, the drag force is 33.28 N, the resultant force is 169.67 N, the direction of the resultant force is 78.69°, and the power exerted by the air on the plate is 462.26 W.

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