A liquid (ρ = 1000 kg/m3; μ = 2 × 10–2 N ∙ s/m2; v = 2 × 10–5 m2/s) flows tangentially p

A liquid (ρ = 1000 kg/m³; μ = 2 × 10^–2 N ∙ ^s/m²; v = 2 × 10^–5 m²/s) flows tangentially past a flat plate with total length of 4 m (parallel to the flow direction), a velocity of 1 m/s, and a width of 1.5 m. What is the skin friction drag (shear force) on one side of the plate?

Expert Solution

Given that

ρ = 1000 kg/m3, µ = 2 × 10-2 Ns/m2,

v = 2 × 10-5 m2/s, v = 1m/s

⇒ = Reynolds number

Rex = Vx/υ = 1m/s × 2m/2 × 10-5m2/s = 105.

Rex = 105

⇒ The boundary layer is laminar.

⇒ yRex0.5/x = (0.0008m)(105)0.5/2m = 0.126

Use figure to obtain u/U0

u/U0 ≈ 0.15;

u = 0.15 m/s.

Rex = 105

u = 0.15 m/s.

Junaid answered 2 years ago

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