Fluid Mechanics 3rd Year Mechanical Engineering Prof Brian Launder Lecture 10 The Equations of Motion for Steady Turbulent Flows

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Fluid Mechanics 3rd Year Mechanical Engineering Prof Brian Launder

  • Lecture 10

  • The Equations of Motion for Steady Turbulent Flows


  • To obtain a form of the equations of motion designed for the analysis of flows that are turbulent.

  • To understand the physical significance of the Reynolds stresses.

  • To learn some of the important differences between laminar and turbulent flows.

  • To understand why the turbulent kinetic energy has its peak close to the wall.

The strategy followed

  • We adopt the strategy ad-vocated by Osborne Reynolds in which the instantaneous flow propert-ies are decomposed into a mean and a turbulent part. (for the latter, Reynolds used the term sinuous).

  • We shall mainly use tensor notation for compactness. (Tensors hadn’t been invented in Reynolds’ time.)


  • We consider a turbulent flow that is incompressible and which is steady so far as the mean flow is concerned.

  • For most practical purposes one is interested only in the mean flow properties which will be denoted U, V, W (or Ui in tensor notation).

  • The instantaneous total velocity has components . (or )

  • So   

  • The difference between Ui and is denoted ui, the turbulent velocity:

  • NB the time average of ui is zero, i.e.

An important point to note

  • If a variable is a function of two independent variables, x and y, differential or integral operations on it with respect to x and y can be applied in any order.

  • Thus

  • So

Averaging the equations of motion

  • First, note that the instantaneous static pressure is likewise written as the sum of a mean and turbulent part:

  • The time average of , where the overbar denotes the time-averaging noted on the previous slide.

  • Treating the viscosity as constant, the time averaged value of the viscous term in the Navier-Stokes equations may be written:

  • But:

The continuity equation in turbulent flow

  • For a uniform density flow:

  • But …so

  • ..or

  • Thus, the fluctuating velocity also satisfies

  • or

The averaged momentum equation

  • From the averaging on Slide 6:

  • Convection Diffusion

  • This is known as the Reynolds Equation

  • Note that this is really three equations for i taking the value 1,2 and 3 in three orthogonal directions

  • Recall also that because the j subscript appears twice in the convection and diffusion terms, this implies summation, again for j=1,2, and 3.

  • Thus:

Boundary Layer form of the Reynolds Equation

  • The form of the Reynolds equation appropriate to a steady 2D boundary layer is taken directly from the laminar form with the inclusion of the same component of turbulent and viscous stress: i.e.

  • The accuracy of this boundary layer model is, for some flows, rather less than for the laminar flow case (i.e. the neglected terms are less “negligible”).

  • The form:

  • is a higher level of approximation.

Who was Osborne Reynolds?

  • Osborne Reynolds, born in Belfast - appointed in 1868 to the first full- time chair of engineering in England (Owens College, Manchester) at the age of 25.

  • Initially explored a wide range of physical phenomena: the formation of hailstones, the effect of rain and oil in calming waves at sea, the refraction of sound by the atmosphere…

  • …as well as various engineering works: the first multi-stage turbine, a laboratory-scale model of the Mersey estuary that mimicked tidal effects.

Entry into the details of fluid motion

  • By 1880 he had become fascinated by the detailed mechanics of fluid motion…..

  • ….especially the sudden transition between direct and sinuous flow which he found occurred when: UmD/  2000.

  • Submitted ms in early 1883 – reviewed by Lord Rayleigh and Sir George Stokes and published with acclaim. Royal Society’s Royal Medal in 1888.

Reynolds attempts to explain behaviour

  • In 1894 Reynolds presented orally his theoretical ideas to the Royal Society then submitted a written version.

  • This paper included “Reynolds averaging”, Reynolds stresses and the first derivation of the turbulence energy equation.

  • But this time his ideas only published after a long battle with the referees (George Stokes and Horace Lamb – Prof of Maths, U. Manchester)

Some features of the Reynolds stresses

  • The stress tensor comprises nine elements but, since it is symmetric ( ), only six components are independent since etc. or in Cartesian coordinates .

  • If turbulence is isotropic all the normal stresses (components where i=j) are equal and the shear stresses ( ) are zero. (Why??)

  • The presence of mean velocity gradients (whether normal or shear) makes the turbulence non-isotropic.

  • Non-isotropic turbulence leads to the transport of momentum usually orders of magnitude greater than that of molecular action.

More features of the Reynolds stresses

  • Turbulent flows unaffected by walls (jets, wakes) show little if any effect of Reynolds number on their growth rate (i.e. they are independent of ).

  • Turbulent flows (like laminar flows) obey the no-slip boundary condition at a rigid surface. This means that all the velocity fluctuations have to vanish at the wall.

  • So, right next to a wall we have to have a viscous sublayer where momentum transfer is by molecular action alone;

  • The presence of this sublayer means that growth rates of turbulent boundary layers will depend on Reynolds number.

Comparison of laminar and turbulent boundary layers

  • Laminar B.L.

  • Recall: The very steep near-wall velocity gradient in a turbulent b.l. reflects the damping of turbulence as the wall is approached

  • But why do turbulent velocity fluctuations peak so very close to the wall?

The mean kinetic energy equation

  • By multiplying each term in the Reynolds equation by Ui we create an equation for the mean kinetic energy:

  • The left side is evidently:

  • or, with KUi2 /2,

  • Re-organize the right hand side as:

  • A B C D E

  •  See next slide for physical meaning of terms

The “source” terms in the mean k.e eqn

  • A: Reversible working on fluid by pressure

  • B: Viscous diffusion of kinetic energy

  • C: Viscous dissipation of kinetic energy

  • D: Reversible working on fluid by turbulent stresses

  • E: Loss of mean kinetic energy by conversion to turbulence energy

A Query and a Fact

  • Question: How do we know that term E represents a loss of mean kinetic energy to turbulence?

  • Answer: Because the same term (but with an opposite sign) appears in the turbulent kinetic energy equation!

  • The mean and turbulent kinetic energy equations were first derived by Osborne Reynolds.

Boundary-layer form of mean energy equation

  • For a thin shear flow (U(y)) the mean k.e. equation becomes:

  • Consider a fully developed flow where the total (i.e. viscous + turbulent) shear stress varies so slowly with y that it can be neglected.

  • In this case, where does the conversion rate of kinetic energy reach a maximum?

Where is the conversion rate of mean energy to turbulence energy greatest?

  • This occurs where:

  • or where

  • or:

  • or, finally:

  • Thus, the turbulence energy creation rate is a maximum where the viscous and turbulent shear stresses are equal

The near wall peak in turbulence explained

  • The peak in turbulence energy occurs very close to the point where the transfer rate of mean energy to turbulence is greatest

  • This occurs where viscous and turbulent stresses are equal – i,e. within the viscosity affected sublayer!

  • Why the turbulent velocity fluctuations are so different in different directions will be examined in a later lecture.

Why is the normal stress perpendicular to the wall so much smaller than the other two?

  • Continuity for turbulent flow:

  • Apply this at y =0 (the wall)

  • But on this plane u=w=0 for all x and z

  • So, ; but u and w deriv’s w.r.t. y 0

  • Expand fluctuating velocities in a series:

  • But b1 must be zero (if )

  • So, while

  • Q: How does the shear stress vary for small y?

Extra slides

  • The following slides provide a derivation of the kinetic energy budget from the point of view of the turbulence.

  • They confirm the assertion made earlier that the term represents the energy source of turbulence.

  • We do not work through the slides in the lecture (Dr Craft will provide a derivation later) but the path parallels that for obtaining the mean kinetic energy.

The turbulence energy equation-1

  • Subtract the Reynolds equation from the Navier Stokes equation for a steady turbulent flow

  • This leads to:

  • Note the above makes use of since by continuity

The turbulence energy equation - 2

  • Multiply the boxed equation from the previous slide by and time average.

  • Note: where k is the turbulent

  • kinetic energy:

  • The viscous term is transformed as follows:

  •  turbulence energy dissipation rate

The turbulence energy equation - 3

  • After collecting terms and making other minor manipulations we obtain:

  • viscous turbulent diffusion generation dissipation

  • Note this is a scalar equation and each term has to have two tensor subscripts for each letter.

  • Repeat Q & A: How do we know that represents the generation rate of turbulence? Ans: The same term but with opposite sign appears in the mean kinetic energy equation.

A question for you

  • Compile a sketch of the mean kinetic energy budget for fully developed laminar flow between parallel planes.

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