A flat plate is oriented parallel to a 15 m/s airflow at 20 o C and atmospheric pressure. The plate is 1 m long in the flow direction and 0.5 m wide. On one side of the plate, the boundary layer is tripped at the leading edge, and on the other side there is no tripping device. Find the total drag force on the plate.

0.506N

Explanation:

In this question, we are asked to calculate the total drag force on a plate which is oriented parallel to an air flow at a particular temperature and atmospheric pressure.

Please check attachment for complete solution, plate diagram and step-by-step explanation

F_d = 0.5063 N

Explanation:

The force due to shear stress is given as;

F_d = (C_f)•½•ρ•(V)²•B•L

Where;

C_f is total average shear stress coefficient on all sides of the plate

ρ is density

V is airflow speed

B is breadth of plate

L is length of plate

The density

and kinematic viscosity

of air at 20"C and atmospheric pressure

is 1.2 kg/m³ and 1.5 x 10^(-5) N-s/m2, respectively.

The Reynolds number based on the plate length is given as;

Re_L = (velocity x Length)/kinematic viscosity

Re_L = (15 x 1)/(1.5 x 10^(-5))

Re_L = 10^(6)

Now,The average shear stress coefficient on the "tripped" side of the plate is;

C_f = 0.074/(10^(6))^(1/5)) = 0.0047

The average shear stress coefficient on the "untripped" side of the plate is;

C_f = [0.53/(In²(0.06 x 10^(6)))] - (1520/(10^(6)) = 0.0028

Thus, we can plug in relevant values to find total drag force;

F_d = (0.0028 + 0.0047 )•½•(1.2) •(15)²•1•0.5

F_d = 0.0075 x 0.6 x 225 x 0.5 = 0.5063 N

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