Mechanics of Fluids

316 Boundary layers, wakes and other shear layers
To sum up, the principal characteristics of the turbulent boundary layer
on a flat plate with zero pressure gradient and moderate Reynolds number
are these:
1. The thickness of the boundary layer increases approximately as the 4 5
power of the distance from the leading edge (compared with the power
1
2 for a laminar layer).
2. The shear stress τ0 at the boundary is approximately inversely proportional
to the fifth root of the local Reynolds number (compared with the
square root for a laminar layer).
3. The total friction drag is approximately proportional to 9 5 power of the
velocity of the main flow and 4 3
5 power of the length (compared with 2
and 1
2 powers respectively for the laminar layer).
Example 8.5 A train is 100 m long, 2.8 m wide and 2.75 m high.
The train travels at 180 km · h −1 through air of density 1.2 kg · m −3
and kinematic viscosity 1.5 × 10 −5 m 2 · s −1 . You may assume that
the frictional drag of the train is equivalent to the drag of a turbulent
boundary layer on one side of a flat plate of length l = 100 m and
breadth b = 8.3 m. Taking the constant in equation 8.24 equal to
0.37, calculate
(a) the boundary layer thickness at the rear of the train
(b) the frictional drag acting on the train, DF
(c) the power required to overcome the frictional drag.
Solution
(a) um = 180 km · h−1 × 103 m · km −1
3600 s · h−1 50 m · s −1
Substituting in equation 8.24, with x = l
� �1/5 ν
δ = 0.37 l
um
4/5 �
(1.5 × 10
= (0.37)
−5 ) m2 · s−1 50 m · s−1 = 0.73 m
(b) Rel = uml
ν = 50 m · s−1 × 100 m
(1.5 × 10−5 ) m2 = 3.333 × 108
· s−1 From equation 8.25
CF = 2F
ϱu 2 m l = 0.074(Re l) −1/5
� 1/5
(100 m) 4/5