Answer:
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
