Fluid Mechanics 3 rd Year Mechanical Engineering Prof Brian Launder

1. Fluid Mechanics3rd Year Mechanical EngineeringProf Brian Launder Lecture 10 The Equations of Motion for Steady Turbulent Flows

2. Objectives • 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.

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

4. Preliminaries • 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 turbulentvelocity: • NB the time average of uiis zero, i.e.

5. 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

6. 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:

7. The continuity equation in turbulent flow • For a uniform density flow: • But …so • ..or • Thus, the fluctuating velocity also satisfies or

8. The averaged momentum equation • From the averaging on Slide 6: Convection Diffusion This is known asthe 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, and3. • Thus:

9. 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.

10. 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. O

11. 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.

12. 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)

13. 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.

14. 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.

15. 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?

16. 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 KUi2 /2, • Re-organize the right hand side as:  A B C D E  See next slide for physical meaning of terms

17. 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

18. 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.

19. 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?

20. 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

21. 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.

22. 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?

23. 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.

24. 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

25. 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

26. 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.

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