boundary layer equation

Boundary Layer Equations

Prof. Rohit GoyalProfessor, Department of Civil Engineering

Malaviya National Institute of Technology Jaipur

E-Mail: rgoyal_jp@yahoo.com

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Topics Covered

Different Boundary Layer Thickness Nominal Thickness Displacement Thickness Momentum Thickness Energy Thickness

Equations for different BL thickness Boundary Layer Equations

Assumptions

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Nominal Thickness (δ)

Nominal thickness of the boundary layer is defined as the thickness of zone extending from solid boundary to a point where velocity is 99% of the free stream velocity (U)

This is arbitrary, especially because transition from 0 velocity at boundary to the U outside the boundary takes place asymptotically.

It is based on the fact that beyond this boundary, effect of viscous stresses can be neglected.

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Other Definitions of BL Thickness

Many other definitions of boundary layer thickness has been introduced at different times and provide important concepts based on mathematical calculations and logic

These definitions are Displacement Thickness (δ*) Momentum Thickness (θ) Energy Thickness (δe)

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Displacement Thickness

Presence of boundary layer introduces a retardation to the free stream velocity in the neighborhood of the boundary

This causes a decrease in mass flow rate due to presence of boundary layer

A “velocity defect” of (U-u) exists at a distance y along y axis

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Displacement Thickness

Displacement thickness may be thought of as the distance (measured perpendicular to the boundary) with which the boundary may be imagined to have been shifted such that the actual flow rate would be the same as that of an ideal fluid (with slip) flowing around the displaced boundary

This may be imagined in as explained in figures on next page

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Velocity Distribution

U

SolidBoundary

Equivalent Flow Rate

U

Velocity Defect

VelocityDefect

δ *

Ideal FluidFlow

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Eqn. for Displacement Thickness

By equating the flow rate for velocity defect to flow rate for ideal fluid

If density is constant, this simplifies to

δ* would always be smaller than δ

( )∫ −=δ

ρδρ0

* dyuUU

∫

−=

δδ

0

* 1 dyU

u

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Momentum Thickness

Retardation of flow within boundary layer causes a reduction in the momentum flux too

So similar to displacement thickness, the momentum thickness (θ) is defined as the thickness of an imaginary layer in free stream flow which has momentum equal to the deficiency of momentum caused to actual mass flowing inside the boundary layer

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Eqn. for Momentum Thickness

By equating the momentum flux rate for velocity defect to that for ideal fluid

If density is constant, this simplifies to

θ would always be smaller than δ* and δ

( ) ( )∫ −=δ

ρθρ0

2 uUudyU

∫

−=

δθ

01 dyU

u

U

u

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Graphical Representation

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Energy Thickness

Similarly Energy thickness (δe) is defined as the thickness of an imaginary layer in free stream flow which has energy equal to the deficiency of energy caused to actual mass flowing inside the boundary layer

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Eqn. for Energy Thickness

By equating the energy transport rate for velocity defect to that for ideal fluid

If density is constant, this simplifies to

( ) ( )∫ −=δ

ρδρ0

222

2

1

2

1uUudyU e

∫

−=

δδ

0 2

2

1 dyU

u

U

ue

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Boundary Layer Assumptions

Following assumptions are made for the analysis of the boundary layer It is assumed (also observed to great extend) that

Reynolds number of flows are large and the thickness of boundary layer are small in comparison with any characteristic dimension of the boundary

The boundary is streamlined so that the flow pattern and pressures determined by ideal flow theory are accurate

It is possible to treat the flow at constant density and isothermal conditions prevail so that viscosity is also constant

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Approximations Made

Using these assumptions following approximations are made The pressure does not vary across any given

section of the boundary layer. So pressure determined by ideal fluid theory at the edge of boundary holds within the boundary layer also

Since flow is essentially parallel so that the shear stress are solely determined by Newton’s law of viscosity τ=µ(∂u/∂y)

Compared with thin boundary layer, the gentle curvature of the boundary has practically no influence on the flow properties

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Coordinate System

Last approximations allows us to choose coordinate system with x-axis along the curved body and y axis along normal to boundary as shown below

Strictly speaking this coordinate system is curvilinear but is expected to behave like a rectangular system in the thin region of the boundary layer

Continuity Equation

Only steady two dimensional flow is considered for simplicity

Continuity Equation in 2D is ∂u/∂x+∂v/∂y = 0

Where u and v are velocity components in x and y axes

Momentum Equation

Since velocity component in y direction is negligibly small so momentum equation is considered only in x direction

Considering a small control volume of sides ∆x and ∆y and thickness in the third direction as unity is considered as shown on next page.

Control Volume

Summation of Forces

Neglecting component of gravity in x-direction, only pressure and shear stress as shown on control volume are considered

There would be a negative shear stress on lower face because layer below is trying to retard the motion of particles within control volume. Similarly shear stress on top surface would be positive

xyy

xyxx

ppypFx ∆

∆

∂∂++∆−∆

∆

∂∂+−∆=∑ τττ

Forces in x-Direction

So total force in x direction

Using Newton’s law of viscosity

This would be equal to change in rate of momentum in x-direction

xyy

xyxx

ppypFx ∆

∆

∂∂++∆−∆

∆

∂∂+−∆=∑ τττ

yxy

u

x

pFx ∆∆

∂∂+

∂∂−=∑ 2

2

µ

Change in rate of momentum

From Left and Right Faces

Mass entering the left face = ρu∆y Momentum entering the left face =

ρu2∆y Momentum leaving right face

= ρu2∆y + ∂(ρu2∆y)/∂x ∆x = ρ(u2+ ∂u2/∂x ∆x)∆y

From Top and Bottom faces

Mass entering from bottom face = ρv∆x Momentum entering the bottom face =

(ρv∆x)u Momentum leaving top face

= ρuv∆x + ∂(ρuv∆x)/∂y ∆y = ρ(uv+ ∂(uv)/∂y ∆y)∆x

Net momentum in x-direction

( )

xuvyuxyy

uvuvyx

x

uu ∆−∆−∆

∆

∂∂++∆

∆

∂∂+ ρρρρ 2

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Net Rate of Momentum

Net momentum in x-direction simplifies to

Using continuity equation ∂v/∂y=-∂u/∂x So the net rate of momentum

yxy

uv

y

vu

x

uu ∆∆

∂∂+

∂∂+

∂∂

2ρ

yxy

uv

x

uu ∆∆

∂∂+

∂∂ρ

Prandtl BL Equation

Equating net rate of momentum to forces in x-direction

Which simplifies to

This is also referred as Prandtl BL Eqn.

yxy

u

x

pyx

y

uv

x

uu ∆∆

∂∂+

∂∂−=∆∆

∂∂+

∂∂

2

2

µρ

2

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y

u

x

p

y

uv

x

uu

∂∂+

∂∂−=

∂∂+

∂∂ υ

ρ