Question

A smooth flat plate 2.0 m wide and 2.5 m long is towed in oil (rd = 0.8) at a velocity of 1.5 m/s along its length. find the thickness of the boundary layer and shear stress at the center?

Answer 1

pls provide coefficient of viscosity ???

Answer 2

solution.

Given : B = 2 m; L = 2.5 m; Sp. gravity = 0.8; U = 1.5 m/s, voil = 10–4 m2/s.

Thickness of boundary layer, δL:

ReL = –4

1.5 2.5 37500

10

UL

v

×

= =

Since ReL is less than 5 × 105, so the boundary layer at the trailing edge is laminar, and Blasius

equation gives,

δx = 5

x

x

Re

At the trailing edge x = L,

∴ δL = 5 2.5 0.0645 m

37500

× = = 64.55 mm (Ans.)

Shear stress at the trailing edge, τL:

According to Blasius average co-efficient of drag (C*

D) is given by:

C*

D = 0.664 0.664 0.664

37500x LRe Re

= =

∴ τL = * 21

2

DC U× ρ

= 0.664 1

237500 × × (0.8 × 1000) × 1.52 = 3.086 N/m2 (Ans.)