In general, when a fluid flows over a stationary surface
, e.g. the flat plate, the bed of a river, or the wall of a pipe, the fluid touching the surface is brought to rest by the shear stress
to at the wall. The region in which flow adjusts from zero velocity at the wall to a maximum in the main stream of the flow is termed the boundary layer
. The concept of boundary layers is of importance in all of viscous fluid dynamics and also in the theory of heat transfer.
Basic characteristics of all laminar and turbulent boundary layers are shown in the developing flow over a flat plate. The stages of the formation of the boundary layer are shown in the figure below:
Boundary layers may be either laminar, or turbulent depending on the value of the Reynolds number.
The Reynolds number is the ratio of inertia forces to viscous forces and is a convenient parameter for predicting if a flow condition will be laminar or turbulent. It is defined as
in which V is the mean flow velocity, D a characteristic linear dimension, ρ fluid density, μ dynamic viscosity, and ν kinematic viscosity.
For lower Reynolds numbers, the boundary layer is laminar and the streamwise velocity changes uniformly as one moves away from the wall, as shown on the left side of the figure. As the Reynolds number increases (with x) the flow becomes unstable and finally for higher Reynolds numbers, the boundary layer is turbulent and the streamwise velocity is characterized by unsteady (changing with time) swirling flows inside the boundary layer.
Transition from laminar to turbulent boundary layer occurs when Reynolds number at x exceeds Rex ~ 500,000. Transition may occur earlier, but it is dependent especially on the surface roughness. The turbulent boundary layer thickens more rapidly than the laminar boundary layer as a result of increased shear stress at the body surface.
The external flow reacts to the edge of the boundary layer just as it would to the physical surface of an object. So the boundary layer gives any object an “effective” shape which is usually slightly different from the physical shape. We define the thickness of the boundary layer as the distance from the wall to the point where the velocity is 99% of the “free stream” velocity.
To make things more confusing, the boundary layer may lift off or “separate” from the body and create an effective shape much different from the physical shape. This happens because the flow in the boundary has very low energy (relative to the free stream) and is more easily driven by changes in pressure.
Special reference: Schlichting Herrmann, Gersten Klaus. Boundary-Layer Theory, Springer-Verlag Berlin Heidelberg, 2000, ISBN: 978-3-540-66270-9
A long thin flat plate is placed parallel to a 1 m/s
stream of water at 20°C
. Assume that kinematic viscosity of water at 20°C is equal to 1×10-6 m2/s
At what distance x from the leading edge will be the transition from laminar to turbulent boundary layer (i.e. find Rex ~ 500,000).
In order to locate the transition from laminar to turbulent boundary layer, we have to find x at which Rex ~ 500,000.
x = 500 000 x 1×10-6 [m2/s] / 1 [m/s] = 0.5 m
We define the thickness
of the boundary layer as the distance from the wall to the point where the velocity is 99% of the “free stream” velocity. For laminar boundary layers
over a flat plate, the Blasius solution
of the flow governing equations gives:
where Rex is the Reynolds number based on the length of the plate.
For a turbulent flow the boundary layer thickness is given by:
This equation was derived with several assumptions. The turbulent boundary layer thickness formula assumes that the flow is turbulent right from the start of the boundary layer.
Consider a water flow (20°C) at v = 0.1 m/s
past a flat plate 1 m long. Compute the boundary layer thickness in the middle of the plate. Assume that kinematic viscosity of water at 20°C is equal to 1×10-6 m2/s
The Reynolds number for the middle of the plate is equal to:
ReL/2 = 0.1 [m/s] x 0.5 [m] / 1×10-6 [m2/s] = 50 000
This satisfies the laminar conditions. The boundary layer thickness is therefore equal to:
δ ≈ 5.0 x 0.5 / (50 000)½ = 0.011 m