Question

A liquid flows tangentially past a flat plate of dimensions total length of 4 m (parallel to the flow direction) and width of 1.5 m. fluid velocity is 1 m/s (length wise). What is the skin friction drag (shear force) on both sides of the plate? liquid density = 1000 kg/m³, dynamic viscosity = 2 x 10-2 N*s/m².

Answer 1

Sol: Information provided in the question:

Length of plate (L) = 4 m

Width of plate (B) = 1.5 m

fluid velocity is (Uo) = 1 m/s (length wise)

liquid density (ρ)= 1000 kg/m³

dynamic viscosity () = 2 x 10^-2 N*s/m²

=> We have to find the skin friction drag (shear force) on both sides of the plate :

Step (1) : We have to find the value of Reynolds number (R_e) so to decide the type of flow:

R_e = (ρ x Uo x L ) /   

R_e = (1000 x 1 x 4 ) / 2 x 10^-2

R_e = 200000 = 2 x 10⁵

so we can see that R_e = 200000 = 2 x 10⁵ < 5x10⁵ that is Laminar

Step (2): Now we calculate local Shear stress coefficient (C_f)

(C_f) = 1.33 / (R_e )^1/2

(C_f) = 1.33 / ( 2 x 10⁵)^1/2

(C_f) = 2.9739 x 10^-3

Step (3) : Now the skin friction drag (shear force) :

F_s = C_f x B x L x ρ x Uo² / 2

F_s = 2.9739 x 10^-3x 1.5 x 4 x 1000 x 1² / 2

F_s = 8.921 N (On one side of plate)

Hence the skin friction drag (shear force) on both sides of the plate : 2 x  F_s = 2 x 8.921 N = 17.843 N