Embed Size (px)
Citation preview
3A1 Incompressible Flow
BOUNDARY LAYER THEORY
T. B. Nickels 2007
1
Background
1 History
The field we now call fluid mechanics was, before 1905, split into two separate fields:Theoreticalhydrodynamics, which involved the theoretical analysis of inviscid fluids (which we have dis-cussed in the first part of 3A1), and Hydraulics, which was basically a set of empiricalrelations used by engineers in the design of fluid systems.M. J. Lighthill described the first as “The study of phenomena which could be proved, butnot observed” and the latter as “The study of phenomena which could be observed but notproved”. In other words, Theoretical Hydrodynamics was mathematically rigorous but didnot describe real phenomena and Hydraulics described real phenomena but had little if anytheoretical foundation.One particular and obvious failing of Theoretical hydrodynamics (potential flow), for exam-ple, is that it predicts zero drag on bodies such as spheres, cylinders and aerofoils.In 1905 Ludwig Prandtl united these two fields by introducing his famous BOUNDARYLAYER HYPOTHESIS. The essence of this hypothesis is that FAR from solid boundariesthe flow is well approximated by inviscid theory (the EULER EQUATIONS) and it is only ina thin region near the boundary (the BOUNDARY LAYER) that viscosity becomes impor-tant. He then developed BOUNDARY LAYER THEORY by making various assumptionsto simplify the full set of equations for this region close to the boundary.
2 Characterising the boundary layer
In IB you did an experiment in the small blue wind-tunnels in the hydraulics lab (it wasa very long time ago). In that experiment you measured the streamwise velocity near thesurface of a flat plate mounted in the wind-tunnel for two different cases: one in whichthe boundary layer was laminar and one in which it was turbulent. The latter case wasmade turbulent by the addition of a trip to the nose of the plate (essentially just a thinrod attached to the surface perpendicular to the oncoming flow). Some results are shown infigure 1. There are some obvious features of these results.
• The velocity drops from the free-stream value as the plate is approached down to zeroat the plate (of course you can never measure the speed right at the plate but you cansafely extrapolate the trend down to zero).
• The region where the velocity is changing is fairly thin relative to the size of thewind-tunnel.
• The laminar and turbulent velocity profiles are significantly different - the turbulentone has much larger gradients near the wall and extends further into the flow (isthicker).
The curve labelled “BLASIUS” refers to an exact solution for the laminar boundary layerthat we will derive later in this part of the course.Now before we go on to look at the boundary layer equations and solutions we might firstlook at how we can characterise, and parameterise the measured profiles.
2
Figure 1: IB Lab results
2.1 The skin-friction
Probably the most important boundary layer parameter for an engineer is the drag it exertson the wall. This is readily calculated using concepts from IB thermofluids. The shear stressin a Newtonian fluid is given by
τ = µ∂u
∂y. (1)
Now if we want to know the shear-stress acting on the wall (or, equivalently, that the wallexerts on the fluid) then we consider the value of the shear stress at the wall,
τw = µ∂u
∂y
∣∣∣∣∣y=0
. (2)
From this we can also come up with a non-dimensional skin-friction coefficient
C ′
f/2 = τw/(ρU21 ) (3)
where the dash is there for historical reasons and just means the local skin friction coefficientat some point along the plate as opposed to the overall skin friction coefficient Cf for thewhole plate (i.e. integrated over the whole plate length). If you look at figure 1 you cansee that the turbulent boundary layer has a much larger shear-stress at the wall than thelaminar one (look at the gradient at y = 0).
2.2 How thick is the boundary layer?
Since in some of the potential flow analysis we assume the solutions are valid outside theboundary layer then we might need to know the thickness of the boundary layer. The effect
3
y
u
Missing mass, momentum, energy(compared with the inviscid case)
U1
Inviscid solutionslip at the wall
U1-u(y)
u(y)
Figure 2: Boundary layer profile showing “missing” mass, momentum and energy
of viscosity on the mean velocity drops off but it is hard to say where it becomes negligible.One crude measure of thickness sometimes used in experiments is the “99% thickness”. Thisis the distance from the wall at which the velocity reaches 99% of its far-field or free-streamvalue. Whilst this value is useful it is also a bit arbitrary (why not the 98% thickness?) andit is also tricky to measure accurately since the velocity at this point changes very slowlywith distance from the wall.Another way to characterise the boundary layer is to compare the real flow with the flow wewould have in the inviscid case. In this situation there would be slip at the wall and hence noprofile. In comparison with this ideal flow we see that the mass flux is reduced (the velocitiesare less) and the momentum flux is also reduced (for the same reason). Further we couldsay that the kinetic energy flux is also reduced (we could extend this to other quantities).So we might say that in the boundary layer we have a mass flow deficit, a momentumdeficit and an energy deficit (when compared with the inviscid flow). The amount of thesedeficits is related to the thickness of the boundary in some way. We have already discusseddisplacement thickness (related to the mass flow deficit) in the first part of the course butwe will re-cap here.Mass flux - displacement thickness. If we want the inviscid flow to have the same mass fluxas the real flow then we need to remove a layer of fluid from the inviscid flow near the wall.We can work out the thickness of this layer (which we will call δ∗) by assuming that its massflux is equal to the amount missing when we compare the real flow and the ideal
ρU1δ∗ = ρ
∫∞
0(U1 − u)dy (4)
and hence
δ∗ =∫
∞
0
(
1 − u
U1
)
dy (5)
Note that this is an integral definition and so it is also much less sensitive to experimentalerror than the 99% thickness. We can also do the same for the momentum (we call thisthickness θ) and the energy (δE) i.e.
θ =∫
∞
0
(U1 − u)u
U21
dy =∫
∞
0
(
1 − u
U1
)u
U1
dy (6)
4
and
δE =∫
∞
0
(U21 − u2)u
U31
dy =∫
∞
0
(
1 −(u
U1
)2)
u
U1
dy (7)
Note that the momentum thickness and the energy thickness might be more appropriatelycalled the “missing-momentum thickness” and the “missing-energy thickness” since largervalues imply that there is more of the quantity missing relative to the inviscid flow. Sincethere is force acting on the fluid opposite to the flow (the skin-friction drag) then we mightexpect the momentum thickness to increase since momentum must be lost due to this externalforce (similarly energy is usually lost in the streamwise direction and so this thickness willusually increase). Note that, in the case of flows with pressure-gradients, there are additionalforces acting which can also change these “thicknesses” and they are not always acting inthe same direction as the skin-friction (just think about this for later in the lectures).
2.3 Shape factors
Another useful non-dimensional way of looking at boundary layer profiles is to considerthe ratios of these lengths. The most common parameter used is simply the ratio of thedisplacement thickness to the momentum thickness and is called the shape-factor (or form-
factor), H. So H = δ∗/θ. The reason it is called shape factor is that if one considers thedefinitions of the thicknesses given above then it may be noted that the momentum thicknessgives more weight to flow near the wall than the displacement thickness. Hence boundarylayers that have smaller velocities closer to the wall end up having a large shape-factor.In particular boundary layers close to separation have large shape-factors and hence thisparameter is often used as an indicator of whether separation is imminent.
5
Question 1.An experimentalist has measured the boundary layer velocity profile on a wing and found
that a useful curve-fit to the data is given by u/U1 = 1− e−5y/δ where y is the distance fromthe wall and δ is the 99% thickness of the boundary layer as measured.
a) Find the displacement, momentum and energy thicknesses of the boundary layer at thispoint in terms of δ.
b) She makes another measurement further downstream. Consider each thickness and statewhether it you think will be larger or smaller downstream - giving sensible reasons. Considerhow this will depend on where the measurements are taken on the aerofoil.
c) She decides a better curvefit for the flow near the wall is u/U1 = 16η0.926 − 15η, whereη = y/δ. She realises immediately that there is a difficulty with the behaviour of this functionas y → ∞ since it is not bounded. She also realises that the way to fix this is to integratefrom 0 to δ, rather than 0 to ∞ since it is a good fit in this range . Work out expressions forthe displacement and momentum thickness in this case. Why is this procedure (integratingonly up to δ) reasonable? Compare the results with those from part (a) to see the errorsinvolved in this approximation.
Note that this is a very common procedure since it allows for the use of functional forms thatare easy to analyse mathematically and only results in very small errors in the computedquantities.Answers: (a)δ/5, δ/10, δ/6, (b) All increase-faster in APG, (c) 0.193δ, 0.095δ
Question 2.In another experiment a researcher decides that a better fit to the velocity profile, in hisflow, is one in which the velocity gradient at the edge of the layer (y = δ99) is zero andU/U1 = 1 at y/δ99 = 1. This is often a reasonable approximation since the velocity typicallyasymptotes to the free-stream value extremely quickly. A suitable fit to the experimentalresults is U/U1 = a0(y/δ) − a1(y/δ)
2 − a2(y/δ)4.
a) Apply the boundary conditions in order to relate a2 and a1 to a0. Write down the expres-sion for U/U1. Note that the shape of the profile is now determined by only one parameter,a0 which can be varied to produce different profile shapes - try plotting the function forvarious values of 0 < a0 < 2.67 (this restriction gives physically plausible shapes).b) Consider the case where a0 = 2. This looks something like a zero pressure-gradientboundary layer (actually the shape is not a great fit to known solutions but that doesn’tmatter here). Find the displacement thickness and the momentum thickness in terms of δ.Find the shape-factor.c) Consider the case where a0 = 0. Find the displacement thickness and the momentumthickness in terms of δ. Find the shape-factor. What physical situation does this profilerepresent (consider the shear stress at the wall)?Answers: (a) a0η − (3a0/2 − 2)η2 − (1 − a0/2)η4, (b)δ∗ = δ/3, θ = 2δ/15, H = 5/2, (c)δ∗ = 8δ/15, θ = 8δ/63, H = 63/15 = 4.2, this is a separating (or separated) profile since thewall shear-stress is zero.
6
3 The evolution, or growth of boundary layers
Now we have some way of characterising a boundary layer given its velocity profile (eitherfrom experiment or theory) then we might ask how it develops along the surface of a body.Does it get thicker or thinner? How does the shear-stress at the wall (and hence the skin-friction) change and, on what does it depend. To answer this we need to look at theequations of motion, in particular the momentum and continuity equations are quite usefulin this regard. In general for a viscous flow these are represented by the Navier-Stokesequations and continuity. We are only interested in incompressible flow in this course so wewill assume constant density.
4 The Navier-Stokes Equations
see KUETHE & SCHETZE, Appendix B.8Liepman & Roshko 13.1–13.7Prandtl & Tietjens– Applied Aero-MechanicsSchlichting– Boundary Layer TheoryWhilst we have already derived these equations in the first part of 3A1, for completeness andconsistency we present them here again. Inviscid analysis suggests that a body in an inviscidfluid has zero drag. In order to resolve this paradox it is necessary to include viscous termsin the equations. This was first done by Navier in 1827 followed by other researchers butthe most elegant and complete form of the equations was derived by Stokes in 1845. Thisset of equations for the MOMENTUM of a fluid are known as the Navier-Stokes equationsand may be written
∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z=
−1
ρ
∂p
∂x+ ν
(
∂2u
∂x2+∂2u
∂y2+∂2u
∂z2
)
∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z=
−1
ρ
∂p
∂y+ ν
(
∂2v
∂x2+∂2v
∂y2+∂2v
∂z2
)
∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z=
−1
ρ
∂p
∂z+ ν
(
∂2w
∂x2+∂2w
∂y2+∂2w
∂z2
)
(8)
In addition to these equations we also have the equation for conservation of mass known influids as the CONTINUITY EQUATION which is given by
∂u
∂x+∂v
∂y+∂w
∂z= 0 (9)
It is often convenient to rewrite these equations in a more compact form using either VEC-TOR notation or TENSOR notation. In vector notation
∂u
∂t+ (u.∇)u =
−1
ρ∇p+ ν∇2u
∇.u = 0 (10)
7
and in tensor notation
∂ui
∂t+ uj
∂ui
∂xj=
−1
ρ
∂p
∂xi+ ν
∂2ui
∂xj∂xj
∂ui
∂xi= 0 (11)
where a repeated suffix denotes a summation over all values of the suffix (Einstein’s con-vention). These equations are generally accepted to be sufficient to fully describe the flowof any Newtonian fluid. It should be pointed out that while these equations appear simplethey are extremely difficult to solve since they are NON-LINEAR.
4.1 Two-dimensional flow
In most of the following analysis we will restrict ourselves to two-dimensional flows. In thiscase the Navier-Stokes equations become
∂u
∂t+ u
∂u
∂x+ v
∂u
∂y=
−1
ρ
∂p
∂x+ ν
(
∂2u
∂x2+∂2u
∂y2
)
∂v
∂t+ u
∂v
∂x+ v
∂v
∂y=
−1
ρ
∂p
∂y+ ν
(
∂2v
∂x2+∂2v
∂y2
)
∂u
∂x+∂v
∂y= 0 (12)
ASIDE: If ν → 0 then these reduce to the Euler equations. If we also specify STEADYFLOW (i.e. flow which does not change with time) then the time derivatives are zero (i.e.∂u/∂t = 0 and ∂v/∂t = 0) and if all boundary conditions are irrotational this reduces toPOTENTIAL FLOW. We can then define a stream function ψ and we have a solution ofLaplace’s equation ∇2ψ = 0 as we found in the first part of 3A1.We are, however, interested in the case when viscosity is important.Consider first the flow around a thin streamlined body, for example a thin aerofoil. Awayfrom the body the Euler equations apply and for steady flow the solution of Laplace’s equa-tion will give good results (and may be sufficient in many cases). In the region close to thesurface a boundary layer exists and we need to include the viscous terms. If the curvatureof the surface is small then we can neglect it and just use the full two-dimensional equationsas given above in x and y. Now if the boundary layer is thin we can calculate the velocityon the wall streamline, u1(x) using Euler and assume it is the same as the velocity at theouter edge of the boundary layer.Hence the boundary conditions for the solution of the equations are
y = 0 u = v = 0
y = ∞ u = u1(x) (13)
and in the free stream we know from Euler
u1∂u1
∂x=
−1
ρ
∂p
∂x(14)
8
δ
l
y
x
Figure 3: Schematic diagram of the boundary layer
(cf. Bernoulli 12u2
1 + p1/ρ = constant).So now we have set up the boundary conditions we need to see if we can simplify the equationsany further. In order to do this we first consider Prandtl’s ORDER OF MAGNITUDE argu-ment which may be used to simplify the full Navier-Stokes equations to a more manageableform in the case where the boundary layer is thin.
5 The boundary layer equations - Order of Magnitude
argument.
The essential part of this argument is to recognize that boundary layers are (in general) thinin comparison to their length of development (except perhaps right at the start of the body).Hence δ/l is small, where δ is the thickness of the boundary layer and l is the length overwhich it develops (see figure 3).Now we will examine the order of magnitude of all of the terms in the two-dimensionalNavier-Stokes equation in the region close to the boundary. By order of magnitude we meanthe size of the terms, which we will represent by O(), i.e. this means “is of the order ofmagnitude of”. This sort of argument is often called a scaling argument. In essence whatwe are doing is looking at how the various terms in the equation change when we changethe primary flow variables (such as the mean velocity, the size of the object we are studying,the viscosity of the fluid). To say that one variable scales with another quantity simplymeans that we expect it to increase proportionally when we increase that variable. To takean example, the skin friction drag on a body is proportional to its surface area so we mightsay that the skin friction drag scales with the surface area. This approach is very useful to ascientist or engineer since in this way we can determine which terms of an equation are likelyto be important under certain conditions and then simplify the equation (by dropping thoseterms that are likely to be insignificant). It should be noted that we do this all the time,often almost unconsciously, when, for example we ignore relativistic effects in the equationsof motion of a body, or neglect quantum effects. We expect these terms to be very smallin the situations we encounter most of the time. These are extreme examples (where theextra terms that would appear in the equations are extremely small) but in other situationsit is not as obvious which terms we can neglect. The order-of-magnitude analysis provides aformal procedure (in the less obvious cases) by which we can simplify the equations. Now wereturn to the two-dimensional Navier-Stokes equations and examine the order of magnitudeof the terms so we can consider in what circumstances we can neglect some. In particularwe are interested in the situation of a boundary layer developing on a wall. Starting from
9
left to right in the equations we find,
u1(x) = O(U∞)
∂u
∂x= O
(U∞
l
)
∂u
∂y= O
(U∞
δ
)
(15)
Now near the surface the slope of the streamlines must be O(δ/l) and therefore
v
U∞
= O
(
δ
l
)
v = O
(
U∞δ
l
)
(16)
(or we can get the same result using the continuity equation), show this. Also the secondorder terms are
∂2u
∂x2= O
(U∞
l2
)
∂2u
∂y2= O
(U∞
δ2
)
(17)
and the order of the other terms can be found by similar means.Substituting these into the momentum equations gives
O
(
U2∞
l
)
+ O
(
U2∞
l
)
= O
(
−1
ρ
∂p
∂x
)
+O(
ν(U∞
l2+U∞
δ2
))
O
(
U2∞δ
l2
)
+O
(
U2∞δ
l2
)
= O
(
−1
ρ
∂p
∂y
)
+O
(
ν
(
U∞δ
l3+U∞
lδ
))
(18)
Now in order to non-dimensionalise things we divide through by U 2∞/l This leads to
O(1) +O(1) = O
(
−∂(p/ρU2∞
)
∂(x/l)
)
+O
(
1U∞l
ν
)
+O
1
U∞lν
(
l
δ
)2
O
(
δ
l
)
+O
(
δ
l
)
︸ ︷︷ ︸
INERTIA FORCES
= O
(
−1
ρ
∂(p/ρU2∞
)
∂(y/l)
)
︸ ︷︷ ︸
PRESSURE GRADIENT FORCES
+O
(
δ
l
1U∞l
ν
)
+O
(
l
δ
1U∞l
ν
)
︸ ︷︷ ︸
V ISCOUS STRESS FORCES
(19)
Now, following Prandtl, we will examine what happens when U∞l/ν → ∞. Consider thefirst of the two momentum equations, bearing in mind that according to the boundary layerhypothesis δ << l and hence (l/δ)2 → ∞. The term in 1/Re goes to zero and the questionis what happens to the term
1
Re
(
l
δ
)2
(20)
as Re→ ∞ ??
10
Prandtl considered three cases:CASE (a)
1
Re
(
l
δ
)2
>> O(1) (21)
This leads to a balance between the pressure gradient forces and the viscous forces. Thissort of balance occurs in low Reynolds number flows where the inertia is small. HOWEVER,we are considering the case where Re→ ∞. One possible case where this might occur is inthe fully developed flow in a parallel pipe or duct. BUT there is a problem with this. Sincethe inertia terms have dropped out, then Reynolds number cannot be a ratio of inertia toviscous forces so we have a paradox. This case doesn’t make sense.CASE (b)
1
Re
(
l
δ
)2
<< O(1) (22)
Here we have a balance between pressure gradient forces and inertia forces. This case isO.K. but it simply represents the Euler equations. It suggests that viscous forces are notimportant, but we know that they must be in a boundary layer. So this solution is notuseful for our purposes. This is where the classical theorists went wrong. They assumedthat their inviscid solutions (eg. potential flow) should become more accurate at higherReynolds number and perhaps exact in the infinite Reynolds number limit. Experimentsshowed that this was not the case at all (bodies did not seem to approach a situation wherethey had zero drag, for example)CASE (c)
1
Re
(
l
δ
)2
= O(1) (23)
In this case inertia, pressure gradient and viscous forces are ALL equally important together.Consideration of this case led to the breakthrough of Prandtl. He recognised that, even inflows at extremely high Reynolds numbers there would be a region near the wall where thevelocity dropped to zero and hence, in this region, the local Reynolds number of the flowwould be low and so viscous forces could not be neglected. So, regardless of the externalReynolds number (eg. for a cylinder U∞D/ν, there is some region close to the surface,however small, where viscosity is important.Rearranging the previous equation leads to
δ
l= O
(
1√Re
)
. (24)
Note that this tells us how the boundary layer thickness varies with distance along a plate(how it grows). Substituting x, the distance from the leading edge for the length l andmultiplying across we find
δ = O
(√
νx
U∞
)
. (25)
Since it is always of this order then we can say that the boundary layer thickness scales inthis way - we will find the same result from an exact solution of the boundary layer equations
11
later on. Note also that (24) justifies our original assumption that δ/l is small - at least atreasonable Reynolds numbers. Note also that this Reynolds number is based on distancealong the plate so we would expect this assumption to break down very near the leadingedge where l is small and hence so is Re = U∞l/ν.We can substitute this back into our momentum equations to examine the consequences.
O(1) +O(1) = O
(
−∂(p/ρU2∞
)
∂(x/l)
)
+O(
1
Re
)
+O (1)
O
(
1√Re
)
+O
(
1√Re
)
= O
(
∂(p/ρU2∞
)
∂(y/l)
)
+O(
1
Re3/2
)
+O
(
1√Re
)
(26)
Now all the terms in inverse powers of Re go to zero as Re→ ∞ and hence ∂p/∂y = 0. Wecan now simplify the equations to
u∂u
∂x+ v
∂u
∂y= −1
ρ
∂p
∂x+ ν
∂2u
∂y2
0 = −∂p∂y
(27)
The second equation suggests that p 6= f(y) and hence p = p1(x)1so to summarize and
include this we finally have
u∂u
∂x+ v
∂u
∂y= −1
ρ
dp1
dx+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0 (28)
These equations are known as PRANDTL’S BOUNDARY LAYER EQUATIONS and theybecome asymptotically correct as Re → ∞. To solve them we need boundary conditionsand these are
y = 0 , u = v = 0
y = ∞ , u = u1(x)
u1du1
dx=
−1
ρ
dp1
dx(29)
These equations can be solved numerically or by series expansion for various cases and in thiscourse we will consider some exact solutions for laminar boundary layers - but a little later.
1This is useful when we consider using the potential flow solution for the outer flow to determine the
pressure along the wall in order to calculate the boundary layer. Since the pressure does not vary through
the boundary layer then we can use the pressure-gradient at the wall from the potential flow as the pressure-
gradient in the boundary layer equations
12
Question 3.
€
U∞
L
Hhsu
Let us consider the order of magnitude argument for a much simpler equation. Supposewe are examining the flow over a hill on the floor of a wind-tunnel. We are interested inthe pressure gradient along the surface of the hill and decide to use Bernoulli’s equation inits differentiated form. There is a potential energy term in the equation and we can writeBernoulli as
p +1
2ρu2 + ρgh = constant. (30)
Since we are interested in the pressure gradient then we take the derivative with respect tos, the distance along the surface (say, just to be careful, outside the boundary layer). Thisgives
∂p
∂s+ u
∂u
∂s+ ρg
∂h
∂s= 0, (31)
or
−∂p∂s
= ρu∂u
∂s+ ρg
∂h
∂s, (32)
which is, of course the Euler equation. Now the question is ”Is it safe to neglect the grav-ity term?” - if we can the problem becomes much simpler. We might also be interested inwhether we can do that in the real case we are modelling which might be a large hill in a field.We will consider the case where the hill is long and low so H/L is quite small (H/L << 1).The simplest way to decide on the importance of the terms is to estimate their size, or theirorder of magnitude.Perform an order of magnitude analysis of the equation and hence find under what condi-tions the contribution of the gravity term to the pressure-gradient may be neglected.Answer: It depends on the non-dimensional number gH/U 2
∞, if this is small we can neglect
the gravity term. Note that this number is a variation on the Froude number.
Before proceeding to find precise solutions of the equations we will look at the integralconstraints of the boundary layer as a whole. In other words we ignore the details of thevelocity profiles and instead integrate them up and see what we can say about the generalgrowth of the boundary layer based on the fact that the rate-of-change of momentum mustbe equal to the sum of the external forces. This should say something about, for example,how the (missing) momentum thickness changes along a plate. There are two, equivalentapproaches from slightly different perspectives - one proceeding from the boundary layerequations and the other from a straightforward control volume approach.
13
6 The Momentum Integral Equation
6.1 Integrating the boundary layer equations
There are many computational approaches which are based on approximations which yieldquick solutions of reasonable accuracy using the INTEGRATED form of the boundary layerequations (see Schichting).If we integrate the boundary layer equations from y = 0 to some point outside the boundarylayer y = h where h > δ and u = U1 at y = h then we get
∫ h
0(u∂u
∂x+ v
∂u
∂y− U1
dU1
dx)dy = −τo
ρ(33)
and using continuity we can put
v = −∫ y
0
∂u
∂xdy (34)
which leads to∫ h
0(u∂u
∂x− ∂u
∂y
∫ y
0
∂u
∂xdy − U1
dU1
dx)dy = −τo
ρ(35)
Now if we integrate BY PARTS and the second term becomes
∫ h
0
(
∂u
∂y
∫ y
0
∂u
∂xdy
)
dy = U1
∫ h
0
∂u
∂xdy −
∫ h
0u∂u
∂xdy
therefore,∫ h
0
(
2u∂u
∂x− U1
∂u
∂x− U1
dU1
dx
)
dy =−τoρ
∫ h
0
∂
∂x[u(U1 − u)] dy +
dU1
dx
∫ h
0(U1 − u)dy =
τoρ
(36)
Now the displacement and momentum thicknesses as derived earlier are,
δ∗ =∫
∞
0
(
1 − u
U1
)
dy
θ =∫
∞
0
u
U1
(
1 − u
U1
)
dy (37)
Now let h→ ∞τoρ
=d
dx(U2
1 θ) + δ∗U1dU1
dx(38)
and dividing through by U 21 and rearranging it is not difficult to show that we find
dθ
dx︸︷︷︸
inertia forces
+H + 2
U1θdU1
dx︸ ︷︷ ︸
Pressure gradient forces
=C ′
f
2︸︷︷︸
Skin friction forces
(39)
where
H =δ∗
θ= shape factor (40)
The boundary layer is often assumed to separate when H → 3 to 4 from empirical data.This equation can also be derived by considering forces on a small control volume.
14
6.2 The control volume approach (ab initio)
In order to find the momentum integral equation we consider forces on a small control volumewhich is part of the boundary layer. The rate of change of momentum of this control volumemust be balanced by the forces on the control volume (i.e. the wall stress and the pressuregradient forces).Consider a control volume
y
xA
B
C
D
E F
h
u
p
u+δu
p+δp
τw
δx
Edge of boundary layer
Figure 4: Control volume for derivation of momentum equation
MASS FLOW BALANCE Mass flow in through AE into control volume AEFD is
ρ∫ h
0udy (41)
Mass flow out of volume through DF is
ρ∫ h
0udy +
d
dx
[
ρ∫ h
0udy
]
.dx+O((dx)2) (42)
So the nett mass flow OUT OF the control volume AEFD is
d
dx
[
ρ∫ h
0udy
]
.dx +O((dx)2) (43)
So now we can calculate the mass-flow through the top, i.e. EF If Vh is the average velocityover EF then the rate of mass flow through EF is
ρVhdx (44)
and therefore, by continuity,
ρVh = − d
dx
[
ρ∫ h
0udy
]
dx+O(dx) (45)
So now we know the mass-flow through the sides of the control volume. We can now workout the momentum flux.
15
MOMENTUM BALANCE We can work out the net flux of x-momentum through the controlvolume and balance this with the x-direction forces acting on the control volume.
(Mx)DF − (Mx)AE =d
dx
[∫ h
0ρu2dy
]
dx +O((dx)2) (46)
(Mx)EF = ρVhU1dx = −U1d
dx
[
ρ∫ h
0udy
]
dx+O((dx)2) (47)
ForcesPressure force is x-direction on AEFD is −h.dp
Pressure force = −hdpdxdx+O((dx)2) (48)
Skin friction force is −τo.dxHence balancing the net forces with the net efflux of momentum
d
dx
[∫ h
0ρu2dy
]
dx− U1d
dx
[
ρ∫ h
0udy
]
dx = −h(
dp
dx
)
dx− τodx+O((dx)2) (49)
Now we divide through by dx and then let dx → 0 and hence the term O((dx)2) → 0 andthe equation becomes
d
dx
[∫ h
0ρu2dy
]
− U1d
dx
[
ρ∫ h
0udy
]
= −h(
dp
dx
)
− τo (50)
Now it only remains to rearrange this so we can express things in terms of δ∗ and θ whereas stated earlier
δ∗ =∫
∞
0(1 − u
U1
)dy (51)
and
θ =∫
∞
0
U
U1(1 − u
U1)dy (52)
hence we get
d
dx
[∫ h
0ρu(u− U1)dy
]
+dU1
dx
[∫ h
0ρudy
]
= hρU1dU1
dx− τo (53)
and rearranging this
d
dx
[∫ h
0ρu(u− U1)dy
]
+dU1
dx
[∫ h
0ρ(U1 − u)dy
]
= τo (54)
which leads to
ρU21
dθ
dx+ 2ρU1θ
dU1
dx+ ρU1Hθ
dU1
dx= τo (55)
and finally we get the von Karman momentum integral equation
dθ
dx+H + 2
U1θdU1
dx=
τoρU2
1
=C ′
f
2(56)
Some points to note are that
16
• This equation is valid for both laminar and turbulent flow
• It forms a good basis for approximate methods for solving boundary layer problems
• It also always serves as a good check on both calculations and experimental data
Question 4.A boundary layer develops with zero pressure-gradient. Its profile is measured at astreamwise location and it is found that a good fit to the function is given by U/U1 =1.54 ∗ (y/δ) − 0.61 ∗ (y/δ)4 for 0 < y < δ where δ is, say, the 99% thickness which is foundin this case to be 10mm. Another measurement is made 20mm downstream and the 99%thickness there is found to be 12mm. The same function also fits the profile well at thispoint.Use the momentum integral equation to estimate the value of the skin-friction coefficient(C ′
f) at this location. Comment on the likely accuracy of this estimate.Answer: C ′
f = 0.0257
Question 5.Uniform suction at the plate is sometimes used to control boundary layers - in some cases tomaintain laminar flow and in others to reduce the thickness of the boundary layer. Considerthe flow of a boundary layer over a porous surface through which flow is sucked such thatthe velocity normal to the plate is uniform and equal to −Vo.We will assume that there is no streamwise pressure-gradient in order to simplify the problem.Derive the momentum integral equation for this flow. Note that the boundary layer equationsare unchanged but the boundary condition at the wall is now different. In this case thecontinuity equation may be written as
v = −∫ y
0
∂u
∂xdy − Vo (57)
which ensures that at the wall the vertical velocity component is equal to the wall suctionvelocity. Note that if we differentiate both sides with respect to y we get back the usualcontinuity equation since Vo is constant everywhere.You can proceed either by integrating the boundary layer equations or by applying a controlvolume analysis. If you are keen do the problem using both methods.Answer: dθ/dx+ Vo/U1 = C ′
f/2
17
Question 6.The boundary layer equations apply to any long thin flow - that is any flow in which
gradients across the flow are much larger than those along the flow. There was no explicitmention of a wall in the derivation (the wall enters through the application of boundaryconditions when solving the equations). As a result it is possible to apply them to otherflows. Consider the 2D laminar wake behind a body (eg. an aerofoil) as shown. The wakemust have some deficit in velocity since some momentum is lost from the fluid due to thedrag of the body.
€
U1
u(y)
x
y flow
The equations are exactly the same as for the 2D boundary layer as given in (269). Thedefinition of the momentum thickness is the same here as for the boundary layer except thatthe integration is from y = −∞ and y = +∞.a) Derive the integral momentum equation in this case by integrating the boundary layerequation with respect to y between these limits and hence find an expression for dθ/dx. Notethat you can follow the basic procedure in the notes above. The end result is rather simpleand quite different from that for the boundary layer.b)Could you have predicted this result without doing any analysis - if so what is the reason-ing?c) Repeat the process using the control volume method (the ab initio approach).Answers: (a) dθ/dx = 0, (b) Yes
Laminar boundary layers
7 Similarity solutions
In this section some special exact solutions of the boundary layer equations will be consideredin which the shape of the profiles are invariant as the flow develops. Before we derive thesesolutions we will first see what we can find out about general constraints on the shape of thevelocity profiles that are given by the equations.
18
7.1 The shape of the velocity profile
In order to examine some constraints on the shape of the velocity profile we will assume itcan be expressed as a Taylor series expansion in the vicinity of y = 0 i.e. at the wall. Hencewe may write,
U(x, y) = a0(x)y + a1(x)y2 + a2(x)y
3 + a3(x)y4 + ... (58)
where the constants are functions of x. The first constant term (in y0)is obviously zero sincethe velocity goes to zero at the wall and so it has been dropped. Now we can substitute thisexpansion into the boundary layer equation and equate the coefficients of like powers of y.We will retain only terms of order lower than y3 - to simplify the mathematics and since itturns out that the higher order terms do not give much useful information( it helps to knowthe answer beforehand!)
a0da0
dxy2 − (1/2)a0
da0
dxy2 = −1
ρ
∂p
∂x+ ν(2a1 + 6a2y + 12a3y
2) (59)
Now equating the coefficients of like powers of y we find
a1 =1
2ρν
∂p
∂x(60)
a2 = 0 (61)
a3 =1
24νa0da0
dx. (62)
We also know that the shear stress at the wall is proportional to the velocity gradient at thewall i.e.
τw = µ∂U
∂y
∣∣∣∣∣y=0
= µa0 (63)
If we non-dimensionalise y with the local boundary layer thickness, ∆ and substitute backinto the expansion, then perform a little manipulation to put things in non-dimensional formwe find
U
U1=
1
2C ′
fRe∆ η +1
2Re∆
∆
ρU21
dp
dxη2 +
1
96C ′
fRe3∆
(
∆dC ′
f
dx− 2C ′
f
∆
ρU21
dp
dx
)
η4 + ... (64)
where η = y/∆ and Re∆ = U1∆/ν is the local Reynolds number based on the thickness.The main things that can be learned from this process are;
• There is no cubic term in the expansion and in zero pressure-gradient no quadraticterm
• The second derivative is balanced by the pressure-gradient
• The fourth order term depends on the gradient of the skin-friction - not just the localskin friction
• An adverse pressure-gradient reduces the velocity near the wall and a favourable doesthe opposite (not unexpected) - note you need to consider the term of order η4 to seethis clearly.
• Increasing Reynolds number increases the velocity near the wall
19
Another point that may be noted is that if all the coefficients of powers of η are constants(i.e. they do not vary with x) then the shape of the profile stays the same for all x - thisspecial case is known as a similarity solution and this constraint on the coefficients allows usto find some useful solutions of the boundary layer equations. Of course, with this limitedexpansion, ensuring constant coefficients does not necessarily guarantee that the shape isinvariant since higher order terms might vary with x - however it turns out that this expan-sion is sufficient. We examine true similarity solutions in the next two subsections of thenotes where we start by assuming that the complete shape does not vary with x.
Question 7.Using the results for the Taylor-series expansion (only the terms given) derive the conditionsnecessary for the shape of the profile given by U/U1 = f(y/∆) to be invariant with stream-wise distance x.
a) Use these conditions to find the variation of ∆ and C ′
f with x for the zero pressure-gradientcase.
b)In the case with pressure-gradient it is possible to do the same but the analysis is alittle tricky. The actual variations depend on the pressure-gradient. Instead show that(U1/Uo) = a(x/L)m and ∆/∆o = b(x/L)n give a constant shape when m and n are relatedin a particular way (Uo and ∆o are just some reference values used to make the expressionsnon-dimensional - they might be the values at some upstream location. L is also just a(constant) length used for non-dimensionalising the equations. a and b are constants). Statethe appropriate relationship between m and n (check that it is consistent with the zeropressure-gradient case). Note that the first two terms of the expansion are sufficient for part(b).
You may use the result that U1dU1/dx = −(1/ρ)dp/dx (from Bernoulli outside the layer) torelate the variation in U1 to the pressure-gradient.
Answer:(a) ∆ ∝√
νx/U1, C′
f ∝ 1/√
U1x/ν, (b) m + 2n− 1 = 0
7.2 The zero pressure-gradient boundary layer
Now that we have derived the integral constraints on the equations and we have some ideaof the factors affecting the overall growth of the boundary layer, we can look more closelyabout the actual shape of the boundary layer profiles. We already know something aboutthe shape from the Taylor series expansion, but here we will find the whole shape and theprecise variation of the integral parameters with streamwise distance. In order to do thiswe need to solve the boundary layer equations to find velocity profiles that satisfy thesedynamical equations. This is not that easy in general since we are solving a set of PDEs andthere are a wide variety of possibilities. There are various numerical techniques for doing sobut before considering them we will look at some analytical solutions which can lead us toa better understanding of the behaviour of the boundary layer and it develops and evolves.One way of simplifying the treatment of PDEs is to look for similarity solutions in which theboundary layer profiles develop in such a way that they maintain their shape whilst growing.These are the solutions we will consider in this section. The first and simplest case is the
20
U
y
U/U1
y/∆
1
4 x=1U1=4
x=2U1=5
x=3U1=6
Figure 5: Collapse of experimental data using similarity scaling
zero pressure-gradient boundary layer.In this case the pressure gradient is zero, so the equations we need to solve are
u∂u
∂x+ v
∂u
∂y= ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0 (65)
with boundary conditions
u = v = 0 at y = 0
u = U1 at y >> ∆
u = U1 at x = 0 (66)
where ∆ is a length scale which is proportional to δ and which will be defined later.Prandtl suggested that a SIMILARITY SOLUTION may be valid i.e. we assume that
u
U1
= F(y
∆
)
(67)
in other words the shape of the profiles stay the same when scaled with U1 and this ∆ soF (y/∆) is not a function of x. Now if we want to check this assumption we can plot u/U1
versus y/∆ for different x positions and different free-stream velocities: we find that theydo indeed seem to fall on a universal curve independent of U1, x and ν as shown in figure 5.This then reduces our PDE to an ODE since we are looking at a quantity, or quantitieswhich are only functions of one variable (y/∆ in this case).Now we can work out the terms in the boundary layer equations in terms of F . We will putη = y/∆ for convenience, and hence
u = U1F (η) (68)
and we note that the function F must have the following properties
F = 0 at η = 0
F = 1 at η = ∞ (69)
21
Now we can work out the terms in the equations in terms of F
∂u
∂x= U1F
′∂η
∂x(70)
where∂η
∂x=
−y∆2
d∆
dx. (71)
Note that the “dash” denotes differentiation of this function with respect to the one variableon which it depends, η (similarly two dashes represent differentiation twice - and so on).Now to find the terms in v we use continuity
v = −∫ y
0
∂u
∂xdy
=∫ y
0
d∆
dx
U1
∆ηF ′dy
= U1d∆
dx
∫ η
0ηF ′dη
= U1d∆
dx
(
[ηF ]η0 −[∫
Fdη]η
0
)
(72)
which leads to
v = U1d∆
dx
[
ηF −∫ η
0Fdη
]
(73)
Now in order to get rid of the integrals lets change the variable so that we only have D.E.srather than integro-differential equations. We can do this by writing
∫ η
0Fdη =
f
2
or F =f ′
2(74)
and hence we find that the terms are
u = U1f ′
2
∂u
∂x= −U1
∆
d∆
dxηf ′′
2
v = U1d∆
dx
(
ηf ′
2− f
2
)
∂u
∂y=
U1f′′
2∆
∂2u
∂y2=
U1f′′′
2∆2(75)
22
Now we can substitute these terms into the boundary layer equations to give;
−U21
∆
d∆
dxηf ′f ′′
4+U2
1
∆
d∆
dxηf ′f ′′
4− U2
1
∆
d∆
dx
ff ′′
4=νU1
∆2
f ′′′
2(76)
and hence
−U1
ν∆d∆
dx︸ ︷︷ ︸
function of x
=2f ′′′
ff ′′
︸ ︷︷ ︸
function of η
(77)
Now since x and η are independent variables then this equation is not possible unless bothsides are equal to a constant. By inspection this constant must be negative and hence wereduce this to two O.D.E.s
−U1
ν∆d∆
dx= −κ
2f ′′′
ff ′′= −κ (78)
where κ is a positive constant. The value of κ is arbitrary and its value only affects thedefinition of ∆ which we have not defined yet.What is IMPORTANT to note is that by assuming a SIMILARITY SOLUTION we havechanged the original P.D.E. into a pair of O.D.E.s.For convenience and for HISTORICAL reasons we use κ = 2.Now subsituting this in to the first equation we find
−U1
ν∆d∆
dx= −2
∫ ∆
0∆d∆ =
2ν
U1
∫ x
0dx
∆2
2=
2νx
U1
(79)
and hence we find an expression for ∆ i.e.
∆ = 2
√
νx
U1
(80)
which we actually indirectly found earlier by the order of magnitude argument. So we nowknow the way in which the boundary layer grows with streamwise distance, x.Now substituting κ = 2 into the second equation leads to
f ′′′ + ff ′′ = 0, (81)
which is called the Blasius equation. It is a 3rd order non-linear D.E. with boundary condi-tions
f ′ = 0 at η = 0
f ′ = 2 at η = ∞f = 0 at η = 0 (82)
23
The third condition can be derived from v = 0 using continuity. Now, as engineers, what wereally want is the local skin-friction coefficient since from this we can calculate drag. Thisis defined as
C ′
f =τo
12ρU2
1
(83)
and we know that
τ
ρ= ν
∂u
∂y
= νU1
2∆f ′′(η)
from which we find that at the wall (η = 0)
τoρ
=νU1f
′′(0)
2∆
=νU1
4
√
U1
νxf ′′(0)
=1
4
ν1/2U3/21
x1/2f ′′(0) (84)
and substituting into the expression for C ′
f we find
C ′
f =f ′′(0)
2
1
(Rex)1/2(85)
Now since f is a universal function of η then
f ′′(0)
2= a universal constant (86)
The equation we have just derived for the local skin friction coefficient is known as theBlasius Law of Skin Friction for Laminar Flow.The only thing we do not know in this expression is the value of the universal constantf ′′(0)/2. In order to find this we must solve the Blasius D.E.To do this we can use a series expansion method. Assume:
f = Co + C1η + C2η2
2!+ C3
η3
3!+ ...
f ′ = C1 + 2C2η
2!+ 3C3
η2
3!+ ... (87)
Now we know from boundary conditions that f(0) = 0 and f ′(0) = 0 and so Co = 0 andC1 = 0 immediately. Now we can substitute these expressions into the Blasius D.E. and wefind an expression as follows
C3 + C4η + (C2
2
2!+C5
2!)η2 + ... = 0 (88)
24
This series must converge for all η i.e. the left-hand side must be equal to zero for all η andhence all the coefficients of this series must be equal to ZERO. So
C3 = 0
C4 = 0
C22
2!+C5
2!= 0 (89)
and so on. When this is done we find that all non-zero coefficients in the expression for f(η)can be expressed in terms of C2 and hence we get
f =C2η
2
2!− C2
2η5
5!+
11C32η
8
8!− 375C4
2η11
11!+ ... (90)
so the problem reduces to finding a value of C2 which satisfies the outer boundary conditionas η → ∞.One simple approach is the SHOOTING METHOD. Here trial values of C2 which satisfy theboundary condtions are “launched” at η = 0 and their behaviour for large η is examined.The discrepancy from the desired solution is noted and the value is adjusted and a new trialis LAUNCHED.
C2=1
η
Correct boundary condition
f ’(inf) =2
f ’
2
C2=2
C2=1.328
Figure 6: Shooting method
The “wiggles” in the solution at very large η come from truncating the series. Blasius found(by hand!) that C2 = 1.32824 and since f ′′(0) = C2 then this leads to
C ′
f = 0.664(Rex)−1/2 (91)
and hence we know how the skin friction coefficient varies with streamwise distance, x. Nowwe haven’t defined ∆ in a simple physical way. Examining the Blasius solution found abovewe find that if δ is the 99% thickness of the layer then δ ≈ 2.5∆. The solution for the shapeof the profile is shown in figure 7.
25
0.0 1.0 2.0 3.0 4.0 η
0.0
0.5
1.0
1.5
U/U1
Figure 7: The Blasius solution
REFERENCES: Duncan, Thom and Young, Mechanics of Fluids, p.261 Schlichting, BoundayLayer Theory, p.121 Knudsen and Katz, Fluid Dynamics and Heat Transfer, p.253
NOTE: Schlichting uses a slightly different definition of η i.e. η = y√
U1/νx. Now we canlook at various cases of different pressure gradient and different wall shear stress.
7.3 The laminar boundary layer with pressure-gradient
In this case we use the full boundary layer equation
u∂u
∂x+ v
∂u
∂y= −1
ρ
dp1
dx+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0 (92)
Now to get a feel for the effect of pressure-gradient at the wall we consider what happens asy → 0. The inertia terms disappear since both u and v go to zero and hence we get
0 = −1
ρ
dp
dx+ ν
∂2u
∂y2(93)
and we also know by definition thatτ
ρ= ν
∂u
∂y(94)
(strictly there is a ν∂v/∂x term in this equation which is small according to the boundarylayer hypothesis.) and hence we find
dp
dx= µ
(
∂2u
∂y2
)
y=0
=
(
∂τ
∂y
)
y=0
(95)
26
and of course
τo = µ
(
∂u
∂y
)
y=0
(96)
Falkner and Skan Solution Now then suppose we have a pressure gradient acting. Falknerand Skan asked the question: What sort of pressure gradient will lead to a similarity solution?The answer they found is there is a similarity solution for the case where the free-streamvelocity varies as a simple power law of the form
U1 = axm (97)
This applies to flow over a wedge2. Now the equations we want to solve are
u∂u
∂x+ v
∂u
∂y= U1
dU1
dx+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0 (98)
and we proceed as before for the Blasius solution.Put u/U1 = F (η) where η = y/∆ hence u = U1F (η) but now U1 = φ(x).
∂u
∂x=dU1
dxF − ηF ′
U1
∆
d∆
dx(99)
Then we have to look at continuity, as before, to find v
v = −∫ η
0Fdη.∆
dU1
dx+ U1
d∆
dx
∫ η
0ηF ′dη (100)
and, as before, we put∫ η
0Fdη = f (101)
and hence∂u
∂x=dU1
dxf ′ − ηf ′′
U1
∆
d∆
dx(102)
and
v = −∆dU1
dxf + U1
d∆
dx[ηf ′ − f ] (103)
and as before∂u
∂y=U1f
′′
∆(104)
and∂2u
∂y2=U1f
′′′
∆2(105)
Then we can substitute this into the boundary layer equation leading to the complicatedexpression
U1f′
(
dU1
dxf ′ − ηf ′′
U1
∆
d∆
dx
)
+
(
−∆dU1
dxf + U1
d∆
dx[ηf ′ − f ]
)(
U1f′′
∆
)
= U1dU1
dx+νU1
∆2f ′′′
(106)
2You can convince yourself of this by working out V1 from the continuity equation and examining V1/U1
27
Now expanding this
U1dU1
dx(f ′)2−U2
1
∆
d∆
dxηf ′f ′′−U1
dU1
dxff ′′+
U21
∆
d∆
dxηf ′f ′′−U2
1
∆
d∆
dxff ′′ = U1
dU1
dx+νU1
∆2f ′′′ (107)
which becomes
∆2
ν
dU1
dx(f ′)2 − ∆2
ν
dU1
dxff ′′ − ∆U1
ν
d∆
dxff ′′ =
∆2
ν
dU1
dx+ f ′′′ (108)
Now we want f to be a universal function of η alone. This will be true if the coefficients ofthe terms in f are UNIVERSAL CONSTANTS. i.e. if
∆2
ν
dU1
dx= κ1
and∆U1
ν
d∆
dx= κ2 (109)
where κ1 and κ2 are some universal constants.We will first examine what happens if we assume that the first coefficient is a universalconstant and see if this is consistent with the second also being a universal constant.We return to the assumed form for the mean velocity variation
U1 = axm
dU1
dx= amxm−1
therefore
∆2
ν.amxm−1 = κ1
∆ =
√κ1ν
amxm−1(110)
We can now find d∆/dxd∆
dx=
√κ1ν
am
(1 −m
2
)
x1−m
2−1 (111)
and hence∆U1
ν
d∆
dx= κ1
1 −m
2m(112)
which is a constant for a given value of m as required. Now substituting back into theboundary layer equation
κ1(f′)2 − κ1ff
′′ − κ11 −m
2mff ′′ = κ1 + f ′′′ (113)
κ1(f′)2 − ff ′′κ1
2m+ 1 −m
2m= κ1 + f ′′′ (114)
κ1(f′)2 − ff ′′κ1
m + 1
2mff ′′ = κ1 + f ′′′ (115)
f ′′′ + κ1ff′′m+ 1
2m+ κ1[1 − (f ′)2] = 0 (116)
28
Now since κ1 is arbitrary then let
κ1
(m+ 1
2m
)
= 1 (117)
now the chosen value of κ1 will have an effect on the definition of ∆ as before. Also forhistorical reasons we will write κ1 = β Hence we find
f ′′′ + ff ′′ + β[1 − (f ′)2] = 0 (118)
with boundary conditions
η = 0 , f ′ = f = 0
η = ∞ , f ′ = 1 (119)
This is a third order ordinary differential equation which can be solved numerically or by aseries expansion method.The flow U1 = axm corresponds to the flow over the surface of a wedge of semi-angle πm
m+1
or πβ2
for m positive. When m = 0 we recover the Blasius solution i.e. a flat-plate with zero
πβ
Figure 8: Flow corresponding to Falkner-Skan solution
pressure gradient. Figure 9 shows solutions for several values of m from strong favourablepressure gradient (m = 1) to the case in which the boundary layer is on the verge ofseparation (m = −0.091). Separation will be discussed in the next section but we note herethat the point of separation (where the boundary is on the verge of separation, or has juststarted to separate) corresponds with zero skin-friction. Since the wall-shear stress is givenby τw = µ(∂u/∂y)|y=0 then this occurs when the gradient of the graph in the figure is zero(as it appears to be in the separating case). Comparing the gradients at the wall then itis obvious that the strong favourable pressure-gradient case has a larger wall shear-stressthan the other two (and hence a larger skin friction). A further obvious feature is that thefavourable pressure-gradient case has a larger area under the graph than the other two. Inaerodynamics jargon we would say that the profile is “fuller” or “more full”. This is justanother way to say that the skin friction is larger and a “fuller” profile will be more resistantto separation as a result (it can go further before the skin friction is reduced to zero).The m = 1 case is a special case of a favourable pressure-gradient where the flow correspondswith a plane stagnation point flow. In this case the boundary layer thickness is constant
29
0 1 2 3 4 5η
0
0.2
0.4
0.6
0.8
1
u/U1m=-0.091 (separating)
m=0
m=1 (Stag)
m=0 .33
m=-0.065
Figure 9: Falkner-Skan solutions for various pressure-gradients (plots courtesy of DavidDennis)
with x. This case was discussed in the first part of 3A1 where it was used to explain the ideaof the displacement thickness of the boundary layer. It is an EXACT SOLUTION of theNavier-Stokes equations and is also useful as a starting point for numerical schemes sincethis flow occurs at the nose of bodies where the boundary layer assumptions are not valid.This flow pattern is shown in figure 10 and the boundary layer profiles are shown in figure 11.
Figure 10: Plane stagnation point flow
7.4 The boundary layer as vorticity
Now that we have solved the boundary layer equations for some cases we can go on to lookat the flows in a slightly different manner. The vorticity in a two-dimensional flow is given
30
Figure 11: Boundary layer profiles in plane stagnation point flow
by
ω =∂v
∂x− ∂u
∂y. (120)
The boundary layer approximation suggests that the first term is very small relative to thesecond and so we will ignore it. If we now examine the Blasius solution we can differentiatethe velocity profile to find the vorticity. Simple inspection of the profile suggests what thiswill look like but it is plotted in figure 12.
0 1 2 3 4 5y/∆
0
0.1
0.2
0.3
0.4
0.5
ω∆/
U1
Figure 12: The Blasius vorticity distribution
We see that the boundary layer is simply a layer of vorticity near the wall (it dies off ratherrapidly at large distances from the wall). So an alternative way of looking at a boundarylayer is as a layer of vorticity. To examine how this evolves (in this laminar case) we can use
31
the vorticity equation which is, in 2D3,
Dω
Dt= ν
(
∂2ω
∂x2+∂2ω
∂y2
)
. (121)
The first term on the right is very small under the boundary layer approximation and wecan consider the material derivative on the left as being the rate of change with time whenmoving with the fluid - in this case we can just consider it a normal derivative, hence
∂ω
∂t= ν
∂2ω
∂y2. (122)
The solution to this equation is a Gaussian vorticity profile of the form
ω = ωoe−y2/4νt. (123)
(see the examples question in which you can derive this). A measure of the thickness of thelayer is δvort =
√4νt (which is the standard deviation of the Gaussian). Finally we ask how
we relate time in the moving frame to our fixed frame - moving at a velocity U1 in a time,t, we cover a distance of x = U1t metres and hence we can relate the time of the movingobserver to the distance along the plate t = x/U1. So then the thickness of this layer should
grow with distance as δvort =√
4νx/U1 - which is exactly the growth rate we worked outfor the Blasius layer using the similarity solution. Note that the actual shape of the Blasiusvorticity profile is not quite Gaussian due to the terms we have neglected. Nevertheless it is areasonable approximation for our purposes. This approach clearly illustrates the boundarylayer as a convection/diffusion problem - convection along the wall due to the flow anddiffusion out by the gradients.What is the effect of Reynolds number then on the boundary layer thickness? If we slowdown the flow (for example) then for a given distance the layer has a longer time to diffuseand hence we expect thicker boundary layers at low Reynolds number (similarly if we increasethe viscosity then the vorticity diffuses more quickly and we find the same result).We can extend this conceptual model to give a qualitative idea of the effect of pressure-gradients. Consider for example the case of an adverse pressure-gradient (APG). In thiscase the velocity along the wall is dropping with streamwise distance. Our vorticity stilldiffuses according to δ =
√4νt but when we want to convert this time into distance then
we need to use a decreasing convection velocity and hence in APGs the layer grows fasterwith respect to x (when compared with the ZPG case). A similar argument suggests thatboundary layers in favourable pressure-gradients grow more slowly with respect to x. It iseven possible to push these simple ideas a little further. Consider that the vorticity moveswith the free-stream velocity (or some fraction of it). Then at a given x location it has beentravelling for a time
t =∫ x
0
1
U1(x)dx (124)
and hence the thickness should be
∆ =
√
4ν∫ x
0
1
U1(x)dx. (125)
3bear in mind that the equation in three dimensions has vortex stretching and tilting terms not present
here
32
If we consider a power-law variation of external velocity (U ∝ xm as for the Falkner-Skansolutions) then it is not difficult to show that
∆ =
√
4ν
a(1 −m)xm−1. (126)
which is the same dependence on x as was found from the similarity solutions (the constantsare a little different). Note that the case of m = 1 (stagnation point) is difficult for thissimplified model since if we start from the origin then it takes infinite time to move awaydue to the zero velocity. Bear in mind that in this approach we have neglected variousadditional effects, such as the effect of the vertical component of velocity which can be veryimportant. For example in the case of suction the dominant convection of vorticity is suctiontowards the wall which is balance by the diffusion of vorticity away from the wall. Also inthe stagnation flow the boundary layer equations break down near the origin.So we have shown that the laminar boundary layer is a layer of vorticity at the wall. Thisconcept is often very useful when trying to understand the behaviour of boundary layers andis also used to model them. One way of doing this is to consider the boundary layer as a rowof line vortices along the surface - they represent, in a discrete fashion the continuous layerof vorticity. In particular the concept of separation is more intuitive when we consider thelayer of vorticity separating from the wall and in a model we might allow the line vortices toleave the surface and evolve freely in the outer flow (where they will move under the veloc-ities induced by the other line vortices in the flow. Note that there are more complicationsin correctly modelling a boundary layer in this manner but the concept is essentially the same.
33
Question 8.As we noted above the boundary layer can be considered to be a layer of vorticity near thewall. In this question we consider a layer of vorticity in an otherwise still fluid (apart from thevelocity induced by the vortex layer) - away from any boundary. We can use the similarity
method above to work out how the layer develops (we have already discussed the solution inthe notes but here we will derive it as an exercise in finding similarity solutions). Considera thin layer of vorticity in an otherwise irrotational flow. We consider that it extends toinfinity in the positive and negative x-directions. In this case it is indeed very long and thinand so the boundary layer approximation applies (exactly) since if it is uniform along itslength there are no gradients in that direction (they are not just small as they are for theBlasius solution). In this case the solution is governed by the vorticity equation of this form
∂ω
∂t= ν
∂2ω
∂y2. (127)
Now we look for a similarity solution where the shape of the vorticity profile remains thesame with time but it spreads out and the peak vorticity drops in some fashion i.e.
ω = ωof(y/δ) = ωof(η), (128)
where δ is some measure of the thickness of the layer and ωo is, say the peak vorticity at thecentre of the layer. We have introduced η = y/δ just to simplify the notation. Note thatthis is an unsteady problem and hence ωo and δ are both functions of time. Note also thatsince they are single values they are not functions of position, y. Now you can substitutethis function into the equation and find an ODE for the function f(y/δ). Note as a hintthat since f depends on δ then ∂f/∂t is not zero - you must use the chain rule eg. ∂f/∂t=(∂f/∂η)(dη/dt=(∂f/∂η)(dη/dδ)(dδ/dt) etc.a) Find the ODE for the function fb) Find the function f either by solving the ODE (not too hard) or by guessing the solutionand showing it satisfies the ODE.c) Find expressions for the variation of δ and ωo with time, tAnswers: see notes
34
Question 9.A two dimensional laminar jet issues from a slot into a still fluid. At some point downstreamthe centreline velocity is Uo and the width of the jet is δ (this might be measured by thepoint at which the velocity drops to half Uo for example). Assume that a similarity solutionof the form
U = Uof(y
δ
)
(129)
is applicable.a) Find the ODE for the velocity profile in this case. Note that the boundary layer equationshold for this flow.b) Find the condition on the variation of Uo and δ such that the shape of the profile isinvariant with streamwise distance.c) Using the result from (b) and the momentum integral equation (which is the same as forthe laminar wake in question 6) find the functional variation of Uo and δ with x. (Note thereis no streamwise pressure-gradient in this problem).
Answers: (a) f ′′ = a(f 2 − I1f′) − bI1f
′, where a = (δ2/ν) dUo/dx, b = (Uoδ/ν) dδ/dx,I1 =
∫ η0 fdη (b) a = constant, (c) Uo ∝ 1/x1/3, δ ∝ x2/3
Question 10.Consider the laminar wake from question 6. Assume a self-similar distribution of the velocitygiven by
U = U1 − Uo(x)f(η), (130)
where η = y/δ.It can be shown that for the zero pressure-gradient case (in which we are interested) asimilarity solution can only apply in the limit where U1/Uo → ∞. Since the wake spreadsand the deficit velocity drops off then this corresponds to a long way downstream. Considerthis case. (Note you cannot just say U = U1 since this corresponds to no wake at all).a) Find the ODE for the velocity profile f(η)(do not solve it) - retain only the largest termsin the limit.b) Find the variation of the thickness δ with x by assuming similarity applies.c) Use the momentum integral equation (derived earlier) with the result from (b) to find thevariation of Uo with x.Answers: (a) f ′′ + af ′ − bf = 0 where a = (δU1/ν) dδ/dx, b = (δU1/ν)(δ/Uo) dUo/dx,(b)δ ∝ √
x, (c)Uo ∝ 1/√x
7.5 The thermal and concentration boundary layers
The previous subsection considered the boundary layer as a layer of vorticity governed bythe vorticity equation, which, after application of the boundary layer approximations in 2Dis
Dω
Dt= ν
∂2ω
∂y2. (131)
This is interesting since the same equation applies for the diffusion of heat or of a passivescalar (a passive scalar is something that does not change the flow and is not a vector eg.concentration of dye in water) except that we replace ν with the appropriate diffusivity forthe quantity of interest. Hence we also have thermal boundary layers and concentration
35
boundary layers governed by the same equations. The equation is essentially a convection-diffusion equation and the evolution of all these three quantities is determined by how theyare moved around by the flow (convection) and how they diffuse out due to gradients (dif-fusion). So the equation for concentration of a passive scalar after applying the boundarylayer approximations is
Dc
Dt= Do
∂2c
∂y2, (132)
where Do is the diffusivity of the substance. In the case of the thermal boundary layer weconsider the case of flow over a heated plate. We will make the additional assumption thatthe temperature differences are not too large and so we don’t have to worry about buoyancyforces - we could make this more rigorous but for now just accept the assumption. In thiscase then we can write an equation for the convection/diffusion of heat
DT
Dt= α
∂2T
∂y2, (133)
where T is the local temperature of the fluid and α = λ/ρcp is the thermal diffusivity (λis the conductivity). We could use the same method of similarity to find solutions of theseequations but since the form of these equations is identical we can jump straight to thesolution. Then the thickness of the thermal boundary layer in laminar flow and developingin a zero pressure-gradient is simply given by
δT =√
4αx/U1 (134)
and we see immediately that the ratio of the thickness of the momentum boundary layer (or
layer of vorticity) to the thermal boundary layer is simply√
ν/α. The quantity under the
square-root is the Prandtl number. So we see that for high Prandtl number flows (such asliquids or oils) the thermal boundary layer is much thinner than the momentum boundarylayer and hence its behaviour is govern by the flow very near the wall (at very high Prandtlnumber it is immersed in the linear region right at the wall). In very low Prandtl numberflows it extends well beyond the momentum boundary layer and into the region of uniformflow outside. In air Prandtl number is of order one and so the layers are of comparablethickness.In the case of the concentration boundary layer the ratio of the momentum boundary layer
thickness to its thickness, δc is√
ν/Do. The quantity here under the square-root is calledthe Schmidt number. In water it has a value of about 1000 which means that momentum(and vorticity) diffuse about 30 times faster than dye in water.The thermal boundary layer equation can also be derived by considering the transport ofenthalpy. The left-hand-side represents the heat transferred into a small control volume bythe flow of fluid and the right-hand-side represents heat transfer from/to the small controlvolume due to conduction. Near the wall the convective terms become small (eg. U∂T/∂x)since the velocity goes to zero and hence the heat transfer (per unit area) from the wall issimply given by Q = λ∂T/∂y|y=0. The analogy with the shear-stress at the wall is clear andin that case we could think of the transfer of momentum from the wall as being due to thegradient of momentum (∂U/∂y). There are two usual wall boundary conditions consideredwhen solving this equation which are constant heat flux and constant temperature. Theadiabatic wall is one example of the first boundary condition in which the heat flux is zero.
36
y
€
T y( )
€
T∞
€
Tw
€
T y( )
€
T∞
€
Tw
€
T y( )
€
T∞
€
Tw
Hot wall Cold wall Adiabatic wall
gradient gradient gradient
Figure 13: Examples of thermal boundary layers
If it is not zero the wall can either be hot or cold and hence the heat transfer can be eitherto or from the wall - this implies that the gradient of the temperature at the wall can beof either sign. Some examples are shown in figure 13 to give some feel for what a thermalboundary layer might look like. Note the temperature gradients at the wall in each case.One further note regarding the thermal boundary layer equation is that it is linear and henceit is possible to produce additional solutions from known solutions by superposition. Thatis not true of the boundary layer equation (the momentum boundary layer).
8 Boundary layer separation
The Falkner-Skan solution produced some solutions for boundary layers subjected to pressure-gradients and we saw that, in one case (m = −0.091), the profile corresponded to a bound-ary layer on the verge of separation. This case has zero skin-friction and the point ofzero skin-friction is defined as point of separation. If we were to proceed to larger adversepressure-gradients then the flow near the wall would reverse and hence we would get negativeskin-friction. The problem with trying to analyse flows past separation is that the bound-ary layer approximation is no longer valid and hence the boundary layer equations are noteither. Once past separation we are forced back to using the Navier-Stokes equation sincegradients in both directions can be significant. It is also worth noting that separation inpractical applications also often leads to three-dimensional flow patterns in which even ourtwo-dimensional assumptions can break down (it is possible to have a purely two-dimensionalseparation in some cases but it is not common in real situations).Despite the complexity of separation we can, at least, examine some of the mechanismsinvolved and the features of the flow. First, for interest, return to the idea of the boundarylayer as a layer of vorticity. We have already touched on how this relates to separationearlier and here we just make some additional points. Consider the separating profile shownin figure 7.3. Without any calculations we can see that near the wall the gradient of the meanvelocity, ∂U/∂y is close to zero and hence the vorticity is almost zero right near the wall. Itobviously increases as we move further from the wall and then dies to zero at large distances.Hence in this case we have a layer of vorticity that is ”separated” from the wall by a regionof zero (or at least very small) vorticity. If the flow reverses near the wall then we haveopposite-signed vorticity right near the wall, then a region of almost zero vorticity and then
37
y
u
U1
u(y)
y
ω+
-
+- - - - - - - - - - - + +- - -
Conceptual model using small vortices
Figure 14: Separated boundary layer profile showing vorticity and a simple conceptual model
the vorticity from our original layer further out. It is often useful to keep this conceptualmodel in mind when trying to understand boundary layer behaviour. The conceptual modelgiven is not complete but it can provide a useful mental picture of separation and boundarylayers in general.
8.1 Separation criteria
When calculating boundary layers or just attempting to predict their behaviour it is usefulto have a criterion to determine where separation is likely. The obvious and most rigorousis to define the point where the skin-friction goes to zero but this is sometimes hard to dealwith mathematically or computationally. As a result various researchers have proposed othercriteria. One simple one is based on the shape-factor. This increases in adverse pressure-gradients and so it is possible to propose a single value at which separation occurs. A valueof around 3.7 is often used for laminar boundary layers (see Thwaites method later in thenotes). There are also other criteria which provide ways of estimating the separation pointfrom the external flow behaviour. None of these criteria is perfect since the separation pointdepends on the shape and the history of the boundary as it develops.
9 Arbitrary Pressure Gradients
The previous analytical solutions were for special cases where an analytical solution could befound. In the case of arbitrary pressure gradients then it is necessary to solve the equationsnumerically or by more complicated series methods. There are also approximate methodsthat make some assumptions about the profiles and reduce to much simpler solutions. We willfirst discuss an approximate method and then discuss the more general numerical approach.The approximate methods were derived at a time before widespread digital computing andwere designed so that hand calculations (albeit lengthy) are possible. They still have somenice features which are of use in understanding and dealing with simple problems. The aim of
38
such methods is not to give a detailed solution but to work out how the integral parametersof the flow (such as skin-friction, boundary layer thickness) vary as the flow develops.
9.1 Thwaites Method
(see Duncan, Thom and Young pp.273–278)In the general case of an arbitrary pressure-gradient we can no longer assume that thevelocity profile remains self-similar - the shape may change as the flow develops. If we areonly interested in the integral properties of the boundary layer (eg. how thick it is, thevalue of the skin-friction etc.) then we might instead make the next simplest assumptionthat is that the shape is no longer invariant but can be expressed in terms of one extraparameter - hence by giving a value to this parameter we determine the shape. This typeof method is classed as an integral method, since the aim is to solve the momentum integralequation for the variation of the integral parameters of the flow as it develops. Recall thatthe momentum integral equation relates three basic parameters (the momentum thickness, θ,the shape parameter, H, and the skin friction coefficient C ′
f) hence we need three equationsin total: the momentum integral equation provides one.In Thwaites method he defined a parameter, m, related to the shape of the profile as
m =θ2
U1
(
∂2u
∂y2
)
w
(135)
where the subscript w means at the wall (y = 0). This is directly related to the appliedpressure-gradient since at the wall
ν
(
∂2u
∂y2
)
w
=1
ρ
dp
dx= −U1
dU1
dx. (136)
so we can also write m as,
m = −θ2
ν
dU1
dx. (137)
He also defined a non-dimensional skin-friction parameter as
l =θ
U1
(
∂u
∂y
)
w
(138)
which is the thing that we might wish to find. So the situation is that we have a form
parameter, m which in some way uniquely determines the shape of the velocity profile and askin-friction parameter, l that is related to the skin-friction and must be a function only ofm since the skin friction depends only on the shape of the profile and the profile is uniquelydetermined, in some fashion, by m. Note also that the shape-factor H that appears in themomentum-integral equation must also just be a function of m since it is determined by theshape of the profile. Now if we use the momentum-integral equation we can find an equationthat relates l to m and H and the growth of the momentum thickness. If we knew l as afunction of m and H as a function of m then we could simply integrate the equation andfind how the momentum thickness and skin-friction vary with x for any pressure-gradient.The way Thwaites proceeded originally take the momentum integral equation and using theabove definitions with a little manipulation derive this equation
U1d
dx(θ2) = 2ν[m(H + 2) + l] = νL(m), (139)
39
where the L(m) = 2[m(H + 2) + l] which is, of course only a function of m. So we now findwe have an equation with an expression for the change of momentum thickness on the left (actually of θ2 - but we can find the growth rate from this) and on the right something thatis a function only of m. So if we can find the function on the RHS we can then go on topredict the growth of the boundary layer. To find this function Thwaites took the resultsfor various exact similarity solutions and plotted L(m) versus the value of m - he found thata good fit was given by L(m) = 0.45 + 6m - although there was some scatter especially foradverse pressure gradients (the reason for the scatter is that, in reality, the profiles cannotbe expressed simply in terms of one parameter, m - but it’s not a bad approximation to thetruth). With this (or any other) relationship that gives L in terms of m then the momentumintegral equation simplifies and , in the case of this fit, we find
U1d
dx(θ2) − 0.45ν + 6
dU1
dxθ2 = 0 (140)
which can be rearrangedd
dx(U6
1 θ2) = 0.45νU 5
1 (141)
and solved
[θ2]x1=
0.45ν
(U61 )x1
∫ x1
0U5
1dx (142)
where the particular quantities are evaluated at some point along the flow where x = x1.Now the way we might use this is to specify the variation of U1 with x in which we areinterested. This might come from solving the potential flow equations as we did in the firstpart of the course - this would prescribe the flow outside the boundary layer and hence thevariation of U1. Then we can substitute that into the above equation and integrate to findthe momentum thickness at any point of interest.Once the variation of θ has been evaluated then an additional equation is needed to solvefor the skin-friction variation and the variation of the shape factor (if required). Thwaitesproposed additional empirical relationships but perhaps the simplest are
H = 2.62 + 3.75m+ 5.24m2 −0.1 < m < 0 (143)
H = 2.472 +0.0147
0.107 −m0 < m < 0.1
These combined with the equation above provide sufficient information to solve for all therelevant parameters given an applied pressure-gradient or, equivalently, variation of externalvelocity(remember that we needed two additional equations which are the fit of L(m) andthe relationships above for H(m)). It is as well at this point to note the limitations of thissort of method. It assumes that all of the appropriate local parameters (such as skin-friction(l), momentum and shape factor) depend only on the local value of the pressure-gradientparameter m. This ignores the history of the boundary layer which may have undergone avariety of changes in external flow before reaching the point of interest. Only the integral ofthat history is taken into account. This is a fairly reasonable approximation for flows thatare slowly and smoothly developing as may be typical over normal aerofoil shapes.Note that the curve-fit of L(m) versus m was not precise and, as noted, showed somescatter. In order to provide a better fit, for the case of velocity profiles over aerofoils,Thwaites considered only the exact solutions that are most relevant to typical aerofoil flowsand tabulated the values of m, l(m) and H(m) which produce more accurate results than the
40
simple linear fit above (L(m) = 0.45+6m). A table of these values may be found in Duncan,Thom and Young, P. 275. The discussion and analysis in this section is based closely onthat provided in this reference.
Question 11.Suppose that for your flow of interest over the rear of an aerofoil of chord length C, say,(strongadverse pressure-gradients) a reasonably good fit to the known exact solutions for use inThwaites method is L(m) = 10m (it’s really not bad for 0.06 < m < 0.08). Suppose alsothat you have a measurement of the boundary layer at the start of the streamwise region ofinterest (x = xo) where the velocity outside the layer, U1(xo), is known as is the momentumthickness, θ(xo).a) What value of the shape factor, H does this fit for L(m) versus m predict will occur atseparation?b) Derive an expression for the momentum thickness, θ(x) in terms of the variation of theouter flow, U1(x) and the initial conditions (θ(xo), U1(xo)).c) Assume that the external flow varies as U1(x) = U1(xo)((x− xo)/C)−1/6. Find the varia-tion of θ(x) as a function of x and hence find the variation of the skin-friction coefficient asa function of x− x0/C and H.d) Using the result of part (c) and assuming a much simplified curve-fit forH i.e. H = 1+32mfind the location at which separation is predicted. Note the curve-fits here are chosen forsimplicity of analysis rather than accuracy. In order to achieve a concrete answer assumethat the initial Reynolds number U1(x0)θ(xo)/ν = 100 and θ(xo)/C = 0.01.
Answers: (a) 3, (b)m = 16
θ(xo)U1(xo)ν
θ(xo)C
(x−xo
C
)1/2, (c)C ′
f = 13
θ(xo)C
(x−xo
C
)−1/6
(3 − H),
(d)(x− xo)/C = 0.1406
9.2 Karman-Pohlhausen method
We will discuss the numerical approach (see Schlichting). According to Schlichting theboundary layer momentum equations with the the continuity equation can be put in theform
∂u
∂x= ν(
∂u
∂y
∫ y
0
1
u2
[
∂2u
∂y2− f(x)
]
dy +1
u
[
∂2u
∂y2− f(x)
]
) (144)
where1
ρ
dp
dx= νf(x) (145)
f(x) can be thought of as a “forcing function”. We imagine that f(x) is known by solvingthe Laplace equation for flow about the body, ignoring the boundary layers. We are mostlyinterested in the cases where this forcing is due to the pressure-gradient but in other situ-ations it might be due to some other force acting (eg. coriolis or electromagnetic or someother imposed forcing).Given this, the right hand side of the equation involves only derivatives and integrals withrespect to y. The left hand side has only derivatives of x.Assume curvature of body is sufficiently small so that we can flatten out the body’s surface.First we start with an initial profile which must be given. Usually it is expressed in termsof a series expansion. Now since we know the initial profile (i.e. we know u as a function ofy) and we know (1/ρ)dp/dx from potential flow theory then we know ∂u/∂x.
41
y
∆xxo
€
u yux
x y xo( ) ( , )+ ∂∂
∆
€
u y( )
Figure 15: Abitrary pressure gradient calculation
Now for the next profile at xo + ∆x then
u(xo + ∆x, y) = U(xo, y) +∂u
∂x(xo, y)∆x (146)
and so we can proceed along the wall and compute the rest of the boundary layer. Note: wemust be careful since
∂2u∂y2 − f(x)
u2(147)
must remain bounded for y → 0. Since
limy→0
∂2u∂y2 − f(x)
u2
=0
0. (148)
Note that this comes from the fact that the pressure-gradient at the wall is balanced by∂2u/∂y2 - consider the boundary layer equations as the wall is approached where the inertialterms go to zero since the velocity goes to zero. The problem is avoided as long as thenumerator goes to zero faster than the denominator.This approach is good as long as the boundary layer stays attached to the body (NO SEP-ARATION).Now for this first approximation we have assumed the boundary layer thickness was neg-liglible when calculating 1/ρdp/dx from potential flow. But now we know the boundary layerdisplacement thickness which effectively alters the shape of the body as seen by a potentialflow calculation. Hence we can now iterate and find the pressure gradient from potentialflow for our “new” body shape which is the body plus the displacement thickness of theboundary layer. Hence we can keep iterating until the solution converges to the requireddegree of accuracy.This approach is O.K. as long as the boundary layers remain attached. The good thing isthat if they remain attached in the computation then they will remain attached in real lifeso we know when the method breaks down.Once they do separate we must use the full Navier-Stokes equations for the separated regionand this region is usually unsteady for
Re =U1l
ν> 100 (149)
42
In order to compute this separated region we need to use one of the worlds largest super-computers at least and at high Reynolds number even this may not be sufficient. Fortunatelya lot of Heat Transfer problems occur at low Reynolds number and then we can use thismethod (eg chip cooling problems).
43
10 Turbulence
10.1 Transition to turbulence in boundary layers
Let us return to the IB boundary layer experiment. In that experiment we started witha smooth flat plate and measured the boundary layer at a certain speed (around 11 m/s).The boundary layer was laminar and the velocity profile matched the Blasius solution quitewell. Then a very small rod was added across the flow near the leading edge and the meanvelocity profile was measured again at the same speed. Despite the fact that the outer flowwas the same and the boundary conditions at the plate (in the region of interest) were thesame the profile was very different. What does this mean? Applying a small perturbation(or “kick”) to our Blasius boundary layer changed the solution completely. So then theBlasius boundary layer in the experiment was unstable to this perturbation. If it were stableit would have recovered and returned to the same solution. The resulting solution after theperturbation was a turbulent boundary layer which is very different in shape and is com-plicated and unsteady with the velocity fluctuating significantly with time (we measuredthat indirectly by listening to the flow with the stethoscope - the velocity fluctuations resultin pressure fluctuations which can be heard as sound). Despite the unsteady fluctuationswe could measure the time-averaged (mean) velocity profile since the time response of thepitot-tube and manometer are so slow they average out these fluctuations.
This instability of laminar boundary layers is quite common at higher Reynolds numbersand so turbulent boundary layers are quite common - so we need to understand somethingabout them.
First consider a rough argument for what is going on. When we add a perturbation to alaminar boundary layer then, mostly, the viscous stresses act to damp out this perturbation4,while the non-linear terms (the inertia forces) tend to increase it. As the Reynolds numberis increased then, by definition, the inertia forces become larger relative to the viscous forcesand hence the damping of the system is reduced. At some Reynolds number then the damp-ing will be not be sufficient to damp the perturbation and it will grow.
It is worth noting here that the Reynolds number of the boundary layer depends on thedistance along the plate (as well as the free-stream velocity). The reason for this is that,as the boundary layer gets thicker the velocity gradients are reduced - given a profile thatmaintains the same shape. As the velocity gradients are reduced the viscous stresses are re-duced and so is the damping. The result of this is that a boundary layer that starts laminarwill eventually become unstable at some distance down the plate and become turbulent -this process is known as transition. Figure 16 is a schematic diagram of this process showingthe critical length at which the boundary layer becomes turbulent.
Note also that it is difficult to precisely define this length since the transition starts witha very small perturbation (which might come from fluctuations in the outer flow, vibra-tions, surface imperfections) which grows to a point where it breaks down and becomes acomplicated three-dimensional motion. We could then put the transition point where the
4This is a simplification:viscosity can lead to destabilising influences but for our purposes this argument
is sufficiently accurate
44
xxtrans
Tollmein-Schlichting waves
Turbulence
U1
U1xtrans/ν 5 x 105
€
≈
Figure 16: Transition length for a boundary layer
perturbation first starts to grow (very difficult to measure), where it reaches a certain am-plitude or where the flow seems to be fully turbulent (difficult to define). Analytical studiescan define the location where the perturbation first starts to grow, whereas, for the engineerit is more useful to know where the flow is fully turbulent (or perhaps where the drag hasincreased significantly as it does when going from a laminar to a turbulent boundary layer).In order to find the location of the transition then it is necessary to examine the stability oflaminar boundary layers. Fortunately we have some solutions of the boundary layer equa-tions already such as the Blasius solution and the Falkner-Skan solutions (there are others).All we then need to do is to add a small perturbation to these solutions and see what hap-pens. If the perturbation is very small then we can linearise the non-linear boundary layerequations and find an equation for the evolution of the perturbation (does it grow or dieaway?).
This approach is known as the linearised theory of hydrodynamic stability which assumesvery small (strictly infinitessimal) two-dimensional disturbances (Orr-Sommerfeld equation).Whilst we cannot do the subject justice here the approach is conceptually straightforward.Take the boundary layer equations for a laminar boundary layer (eg. the Blasius solu-tion we derived earlier) and add a very, very small disturbance of the form, for exampleu′ = F (y)∗ei(α(x−ct)). Where α = 2π/l and l is the wavelength of the disturbance. So addingthis disturbance is like adding a wave onto the boundary layer with an amplitude that de-pends on time. Note that c is in general a complex number: the real part corresponds to thevelocity of the wave and the imaginary part to the growth rate of the disturbance. (Notethat we generally add a u′ and v′ perturbation by using a perturbation stream-function -this ensures continuity is satisfied - this is more complicated and we only want to get thegist of the idea here )
We add this on to the original solution and substitute it into the boundary layer equations.Since the perturbation is small it becomes possible to linearise the equations (basically, wethrow out terms that are higher order in our small perturbation) and extract an equationthat describes the growth of the disturbance for any given wavelength and Reynolds number.If the imaginary part of c (ci) found from solving this equation is greater than zero then thedisturbance grows exponentially, if it is less than zero it dies away.
45
Figure 17: Stability diagram for a boundary layer
Figure 17 shows contours of the growth rate of disturbances versus wavenumber (α the in-verse of the wavelength) and Reynolds number:this is for a zero pressure-gradient boundarylayer. The outermost contour is the neutral stability curve - i.e. the one for which the distur-bance neither grows nor decays. Any point inside this contour corresponds with an unstablecase - i.e. where the disturbance grows. The leftmost point on this contour corresponds tothe minimum stable Reynolds number, Re1 = 520 (in this case the Reynolds number on thediagram is based on δ1 = δ∗, the displacement thickness, and the free-stream velocity). Thewavenumber of this point - of about 0.3δ∗ corresponds with the most unstable wavelengthand hence the wave that starts the transition. Note that we have assumed that the boundarylayer is perturbed by infinitesimal waves of all wavelengths and the flow stability picks outthe first one with a non-negative growth rate. Also note that the growth rate of this firstunstable wave is very small (nominally zero) and so this point will be difficult to detect. Inpractical situations it is of more interest to know the point at which the flow is first turbulentwhich corresponds with waves with large growth rates and a range of wavelengths - this willoccur at higher Reynolds number. The numbers on the contours (ranging from 0 to 1.96)correspond to the amplification rates of the disturbances.
This analysis is analogous to the sort of analysis that is done in control theory where a sys-tem is perturbed with a given frequency disturbance and the response is noted. The classicalexample of this problem is the spring mass system with damping that becomes unstable forzero damping and at the resonant frequency. The boundary layer is a much more compli-cated system and is non-linear and hence the analysis is more complicated. Nevertheless, insimplistic terms, the Reynolds number may be considered to be an inverse damping param-eter since the viscous terms usually tend to damp out perturbations which is why the lowReynolds number cases are stable.
46
The first type of disturbance that just starts to grow is a two-dimensional wave on the bound-ary layer - these waves are known as Tollmein-Schlichting waves. Once they grow sufficientlynon-linear effects kick-in and the whole flow becomes complicated and finally turbulent.
In reality the disturbances are usually non-linear (i.e. they are not very small) and three-dimensional and in this case the behaviour is more complicated since the flow “bypasses”the usual route to transition. This is referred to as “Bypass transition”.
It is also of interest to note that laminar pipe flow and channel flow (Poiseulle flow) turn outto be stable according to linear stability analysis whereas in practice they become turbulentat high enough Reynolds numbers. The reason that they become turbulent is that they areunstable to finite disturbances (but stable to infinitesimal disturbances). A simple analogyto this situation is to compare trying to stand a sharpened pencil on its point (unstable toany disturbance however small) and standing it on its other flat end. In the second case itis stable to small disturbances but unstable to larger ones. This is really what happenedin the IB experiment. The laminar boundary layer in that case is stable for infinitesimalperturbations (otherwise it would become turbulent on its own) but it was unstable to thelarger perturbation given by the trip-wire (the rod at the start).
Question 12.Figure 17 shows the stability plot for a zero pressure-gradient boundary layer. The Reynoldsnumber for transition is shown as U1δ
∗/ν = 520.
a) Use the results from the Blasius analysis given earlier in the notes to express this as aReynolds number based on x (Rex = U1x/ν). Note:δ∗/∆ = 0.86 for the Blasius profile.
b) The usual engineering rule-of-thumb for zero pressure-gradient boundary layer transitionis Rex = 5 × 105. Compare this with the result from (a) and comment on the difference.Also convert the engineering value back to a value for δ∗U1/ν and look it up on the stabilityplot.
c) A laminar boundary layer develops in a zero pressure-gradient flow in a wind-tunnel whichhas been designed to be extremely low turbulence, acoustic noise and vibration (essentially sothere are no perturbations at all). A perturbation is added to the boundary layer such thatthe wavelength of the perturbation is l = 63δ∗. At approximately what Reynolds numberwill the boundary layer first become unstable. What Reynolds number would be a good en-gineering approximation for when the flow becomes turbulent (consider amplification rates)
Answers: (a) 9.14× 104 (b) It is larger since it is hard to detect the very start of transitionwhere the growth rate of the disturbances is close to zero - the engineering value correspondsto the point where the flow is definitely turbulent (rather than the initiation of the growth)- the point on the stability curves corresponds to a large growth rate (about 1).(c) AboutRedelta1
= 3000. To notice the transition then we need a decent amplification rate of sayabout 1 which makes the Reynolds number roughly 10,000.
47
10.1.1 Factors affecting transition
The area of transition and of stability theory for boundary layers is a broad field and isbeyond the scope of this course. We have touched on the ideas here so the basic approach isunderstood. What is more important from the point of an engineer is to understand whatfactors affect the transition and in what direction. The linear stability analysis provides amaximum Reynolds number for which the flow can remain laminar: in real flows transitionoften occurs well before this since real flows always contain small but finite perturbations.Due to the limited number of lectures available here we will merely list a few common factorsand their effects.
• Pressure-gradient; Adverse pressure-gradients make the b/l less stable and favourablethe opposite
• Free-stream turbulence; Leads to early transition (adds finite perturbations)
• Surface roughness - leads to early transition - tripping (again adding finite perturba-tions)
• Wall suction; tends to stabilise the boundary layer leading to later transition
10.2 Fully turbulent boundary layers
Turbulent flow is complex unsteady and three-dimensional and hence we must include time-dependence and three-dimensionality into the equations of motion.If we define the total derivative as
D
Dt=
∂
∂t+ u
∂
∂x+ v
∂
∂y+ w
∂
∂z(150)
then the Navier-Stokes equations can be written
Du
Dt=
−1
ρ
∂p
∂x+ ν∇2u
Dv
Dt=
−1
ρ
∂p
∂y+ ν∇2v
Dw
Dt=
−1
ρ
∂p
∂z+ ν∇2w (151)
These non-linear equations are extremely difficult to solve even on a computer. The reasonis that turbulent flows consist of a wide range of different sizes and frequencies of motions(large swirling motions to very small swirling motions). The ratio of the size of the largestmotion to the smallest increases with the Reynolds number. If we solve these on a gridthen it must be fine enough to resolve the smallest motions whilst the domain must be largeenough to resolve the largest motions. Hence the computational cost increases with theReynolds number and in fact approximately with the Reynolds number cubed.
In the case of boundary layers the highest Reynolds number that has so far been computedon a super-computer is about Rθ = U1θ/ν = 1410 (this was a zero pressure-gradient layer
48
t
U(t)
Uu’(t)
Figure 18: Definition of U
with simple boundary conditions). For comparison the turbulent boundary layer on an air-craft wing corresponds to about Rθ = 50000 - it is nowhere near feasible to computer evena zero pressure-gradient boundary layer at this Reynolds number using the biggest and bestsuper-computers available in the world today - it is also unlikely to be feasible in the nearfuture. In the case of the atmospheric boundary layer the Reynolds number is approximatelyRθ = 500000.
Faced with this difficulty of solving the exact equations (when research on turbulent bound-ary layers first started the use of computers to solve the equations was not even plausible,let alone feasible), researchers turned to statistical methods. The kinetic theory of gaseshad provided appropriate tools and had been very successful. It is not possible to solve theequations for the behaviour of the many billions of molecules in a volume of gas but it ispossible to predict their averaged behaviour in terms of, for example, the mean pressure andtemperature. The behaviour of turbulent flow is at least analogous in the sense that we canthink of it as a huge number small parcels of fluid moving around in an apparently randomfashion and we are interested in predicting some mean quantities (eg. the average drag on aplate caused by a turbulent boundary layer).
So instead of asking “what is the instantaneous value of the velocity at some point in theflow” we might instead ask “what is the time-average value of the velocity at that point andwhat is the standard deviation of the velocity from this average value”. To proceed in thismanner we consider splitting the variables into mean and fluctuating parts. We can do thissince, in our flows of interest the turbulence is a random, STATIONARY process (its meanvalues are independent of time) and hence we can break up the velocity components into amean value and a fluctuating component i.e.
u = U + u′(t) (152)
where U is the time average of u over a sufficiently long period so that it converges. If it isa stationary process then U is independent of time.
U =1
T
∫ t+T
tudt (153)
49
where T is sufficiently large (strictly T → ∞). It is a stationary process if we do the sameaveraging starting at another t and get the same result.So if we assume that the turbulence is a stationary random process we can write all thevariables in this form i.e.
u = U + u′(t)
v = V + v′(t)
w = W + w′(t)
p = P + p′(t)
ρ = ρ+ ρ′(t)
(154)
where we usually ignore the ρ term and assumed ρ is a constant for incompressible, lowMach number flow since
ρ′
ρ≈M2 (155)
If we substitute these terms into the Navier-Stokes equations and take time averages of theequations we can find the behaviour of the means and fluctuating components. We mustobserve the following rules:
∂u
∂x=
∂U
∂x
∫
uds =∫
Uds
u+ v = u+ v
u = U + u′ = U + u′ = u+ u′ (156)
and hence u′ = 0.Also additional rules
u.v = u.v = u.v
uu′ = u.u′ = 0 (157)
But it is important to note that while u′ = 0 and v′ = 0 u′v′ is not in general zero (this canoccur if u′ and v′ are 90o out of phase).
50
After we do this decompostion and then average the Navier-Stokes equations we get theREYNOLDS AVERAGED EQUATIONS
uj∂ui
∂xj=
−1
ρ
∂p
∂xi+ ν
∂2ui
∂xj∂xj− ∂u′iu
′
j
∂xj
∂ui
∂xi= 0
∂u′i∂xi
= 0 (158)
Note here that a repeated index denotes a summation hence, for example,
∂u′i∂xi
=∂u′
∂x+∂v′
∂y+∂w′
∂z(159)
where the three components of the velocity vector are u1, u2, u3 or u, v, w. Now for LAMINARflow we have
uj∂ui
∂xj=
−1
ρ
∂p
∂xi+ ν
∂2ui
∂xj∂xj
∂ui
∂xi= 0 (160)
and these equations are the same provided that the turbulence is stationary and the laminarflow is steady and we put
ui = ui (161)
except for the extra term
−∂u′
iu′
j
∂xj(162)
which represents the gradients of “stresses” due to the random turbulent motion. This termis a nine component tensor which is symmetric.
Question 13.Using the continuity equation and splitting the velocity components into mean and fluc-tuation parts show that for a turbulent flow both the mean velocity components and thefluctuating components satisfy the continuity equation separately as shown in (158) above.
Hinze and Schlichting have applied this to jets and boundary layers and have used an orderof magnitude analysis to find for boundary layers
u∂u
∂x+ v
∂u
∂y=
−1
ρ
dp
dx− ∂u′v′
∂y+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0 (163)
whereas for laminar flow
u∂u
∂x+ v
∂u
∂y= −dp1
dx+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0 (164)
51
Which is the same if u ≡ u and v ≡ v except for the extra turbulent stress term. If we definea new stress by
τ
ρ= −u′v′ + ν
∂u
∂y(165)
then the equations become
u∂u
∂x+ v
∂u
∂y=
−1
ρ
dp
dx+
∂
∂y
(
τ
ρ
)
∂u
∂x+∂v
∂y= 0
τ
ρ= −u′v′ + ν
∂u
∂y(166)
Unfortunately now we have 3 equations and 4 unknowns u, v, τ/ρ and u′v′, since we haveaveraged the equations and hence lost some information. In order to solve the equations wemust CLOSE them by finding an extra CLOSURE EQUATION. We will come back to thisin a later section.
Before going into this consider a physical explanation of this mysterious extra stress. Con-sider what u′v′ means. If this is negative then we have two cases; u′ is positive when v′ isnegative on average or u′ is negative when v′ is positive on average. Now if u′ is positive thena parcel of fluid is a bit faster than the mean and hence it has higher momentum. So whenu′v′ is negative this means that, on average higher momentum fluid moves down towards thewall and lower momentum fluid moves up away from the wall (it is best to think about thisfor a while and try waving your hands about to see what is going on).
Now why a stress? Consider for simplicity a control volume with the wall as its bottomboundary. The effect of this negative u′v′ is to increase the momentum in this control vol-ume by adding high momentum fluid and throwing out (the top) low momentum fluid. Thisincrease is effectively the same as would happen if we applied a net force to the controlvolume in the streamwise direction.
This is a simple (but correct) explanation. It also explains why the profiles for laminar andturbulent boundary layers look as they do. Consider the profiles from the IB experiment.In the turbulent case there is more momentum near the wall due to the redistribution ofmomentum by this Reynolds shear stress term. This also explains why turbulent boundarylayers are less likely to separate than laminar ones - the Reynolds shear stress redistributesthe momentum such that it is increased near the wall which compensates to some extent forthe drop in streamwise momentum near the wall due to the action of an adverse pressure-gradient. Put simply the pressure-gradient applies a backward force which reduces themomentum near the wall and the Reynolds stress constantly brings in new (positive) mo-mentum from the outer part of the flow.
52
Question 14.Integrate the boundary layer equation for the turbulent boundary layer (as shown in (163))from the wall to outside the layer and show that the momentum integral equation is thesame for a turbulent boundary layer as it is for a laminar boundary layer. The method isthe same as given earlier in the notes. If you are smart you might note that the equationis the same as for the laminar case except for the one extra term added on the right-handside that needs to be integrated up - so you might save some time and effort. Note that theReynolds shear stress goes to zero in the outer flow - outside the boundary layer there is noturbulence (also note that it must go to zero at the wall since the no-slip and no-penetrationconditions mean the fluctuations must go to zero - and the mean velocities too, of course!).
10.3 The mean velocity profile
The variation of the mean velocity is very important since once we have it we can work outthe momentum thickness, the shape factor and using the momentum-integral equation wecan then work out the wall shear stress and hence the skin-friction. This is a very importantquantity for an engineer - perhaps the most important.
The earliest researchers in the field simply used empirical curve-fits to experimental data asa way of solving the problem. The most famous is the one-seventh power-law profile wherethe mean velocity was assumed to vary as y1/7. This is a reasonable fit over most of theprofile but has the wrong behaviour very near the wall and at large distances from the wall(where it continues to increase without bound). The advantage of such a curve-fit is thatit is easy to deal with analytically and since it is integrated when finding the skin-frictionsmall errors don’t make a lot of difference to the answer. It is possible to derive the variationof the skin-friction with distance along the plate using this assumption - the full derivationis given in an Appendix to the notes.
One problem with the power-law is that there is no physics in its derivation and this be-comes obvious when we note that the best power law fit depends on the Reynolds number(sometimes 1/8 is better, sometimes 1/6 and so on). That indicates that there is somethingwrong somewhere.
In order to get a better understanding of the true shape we can apply some simple physicalarguments which we will then use for dimensional analysis so as to get some quantitativeresults.
Consider what happens very close to the wall. At the wall the velocity parallel to the wallmust go to zero due to the no-slip condition and the component normal to the wall must alsogo to zero due to the no-penetration condition. This is true for both laminar and turbulentboundary layers. This means that the fluctuating velocities must also go to zero at the walland hence very close to the wall the Reynolds shear stress must go to zero (in fact it is notdifficult to show that it must vary as y3 just by considering the Taylor series expansion of thefluctuations). The viscous stress, on the other hand depends only on the velocity gradientwhich does not go to zero at the wall but instead increases up to some finite value. Hencethere will be a region very close to the wall where the viscosity dominates and the equation
53
is the same as for a laminar boundary layer. This region is known as the laminar sublayer(or sometimes as the viscous sublayer). Its thickness depends on the Reynolds number andit is very thin at high Reynolds number.
As we move out from the wall then we would expect the Reynolds shear-stress (the turbu-lence stress) to increase while the viscous stress drops (the mean velocity gradient drops)and hence there will be some region where they are both important - this region is calledthe buffer region.
Further out as the viscous stress drops even further and the turbulence fluctuations increasethen we might expect a region where the viscous stress becomes negligible and hence themean flow might be independent of viscosity.As we move still further out the Reynolds shear stress drops and we might expect the outervariables such as the layer thickness and pressure-gradients to become important.
10.4 Some dimensional analysis and hand-waving
Using the physical arguments we have just discussed and some others we can do some di-mensional analysis in order to work out something about the shape of the mean velocityprofile.
Before making any assumptions we might look at the things on which the velocity profilemight depend. We will consider zero pressure-gradient flow in order to make the analysissimpler. So just consider all the relevant independent (or imposed) variables that affect theflow
u = f1(y, τo, ρ, ν, δ, U1), (167)
where f1 is some unknown function. Dimensional analysis would then lead us to four non-dimensional groups - which is not that helpful for our purposes. Now in order to simplify itwe look at the different regions discussed above and see what variables we can throw away.
Before starting, to simplify things note that the quantity√
τo/ρ has the dimensions of avelocity and since we are interested in velocities we can use this in place of τo. The “ve-locity” produced is known as the wall-shear velocity and here we will give it the symbol Uτ
hence Uτ =√
τo/ρ (in some literature the symbol used is u∗). In this way we also removedirect effects of the density since it only enters in modifying the kinematic viscosity and thewall-shear velocity.
Now consider a region quite close to the wall. If we are far enough away from the outeredge then we might guess that the local flow does not know anything directly about thefree-stream velocity or the boundary layer thickness (except in so far as they affect the shearstresses etc.). In this case we can write
u = f2(y, Uτ , ν), (168)
and we find only two non-dimensional quantities so this can be re-written as
u
Uτ
= f(yUτ
ν
)
, (169)
54
10 100 1000 10000z+
5
10
15
20
25
30
35
U+
10m/s (Rθ=2,583 & Κτ=1,047)10m/s (Rθ=9,256 & Κτ=3,436)10m/s (Rθ=20,013 & Κτ=7,611)20m/s (Rθ=37,542 & Κτ=14,379)30m/s (Rθ=52,103 & Κτ=19,956)40m/s (Rθ=68,151 & Κτ=26,376)Log-law (κ=0.41 & Α=5.0)
Figure 19: Law-of-the-wall scaling: Boundary layers scaled on wall-variables showing collapsein the inner region (note here z+ = y+) for a range of Reynolds numbers: zero pressure-gradient flow
where all the f functions are unknown and I am just using subscripts (f1, f2 etc.) to point outthat they are different. This famous expression was first derived by Prandtl and is known asthe “Law-of-the-wall”. This suggests that, close enough to the wall, all turbulent boundarylayer velocity profiles will collapse when scaled like this. It is an amazing result that is verywell supported by experimental data. Figure 191 shows profiles from a zero pressure-gradientboundary layer over a range of Reynolds numbers plotted in law-of-the-wall co-ordinates -the collapse is very good in the inner part but they start to diverge towards the outer edgedue to the effects of the outer flow variables (in particular δ).
The region where it applies covers the laminar sublayer, the buffer region and a significantpart of the flow further out. The region of applicability varies for different flows but it coversat least about the lower 10% of the thickness of the layer where all the significant actionis. As an aside, note that in the literature when something is non-dimensionalised using“wall-variables”(eg. Uτ , ν) then the quantity is superscripted with a plus. Hence the law ofthe wall might be written as u+ = f(y+).
Next lets go very close to the wall into the region of the viscous sublayer. To simplify thisfurther return to the boundary layer equation.
u∂u
∂x+ v
∂u
∂y= − ∂
∂y
(
τ
ρ
)
(170)
we have dropped the pressure-gradient term and based on our earlier argument the shearstress is just the viscous part - the Reynolds shear stress is negligible. To proceed further
55
note that as we go towards the wall the mean velocity components also go to zero and sothe left-hand side of the equation becomes negligible. We are then just left with
ν∂2u
∂y2= 0. (171)
Integrating once we find
ν∂u
∂y= constant, (172)
and at y = 0 this constant is just the wall-shear stress by definition. Integrating again wefind the profile is linear and may be written
u
Uτ
=yUτ
ν(173)
which is, of course, consistent with the law-of-the-wall as given in (191) but we have nowdefined the form of the function in some region very close to the wall. Empirically thisapplies well for a region of approximately y+ < 5. We sometimes say “for y is less than fivewall-units”. The quantity ν/Uτ has dimensions of a length and is often called a wall unit.The physical size of this quantity depends on the Reynolds number but just to give an idea,in a windtunnel with a flow of 30 m/s (not that fast) and say flow developing over a metrethen one wall-unit ≈ 15µm.
We will skip over the buffer region since in that region all the terms are important (the meanmomentum, the viscous stress and the Reynolds shear stress) and so it is hard to simplifythings further. Beyond this, however, but still within the region where the law-of-the-wallapplies, we expect that the viscous stress term becomes negligible in comparison with theReynolds stress (and the mean momentum). In this part then we would expect the flowto be independent of viscosity. Note that we need to be a bit careful. The velocity in thisregion cannot be independent of viscosity since the velocity at a given y relative to the walldepends on the change in velocity that occurs across the viscous sublayer and buffer regionplus the changes in velocity that occur locally which may be independent of viscosity. Sincethe change in velocity across the viscous sublayer and buffer layer depend on viscosity thenthe actual value of velocity further out must depend on viscosity.
We might seem to be stuck but re-reading the previous paragraph suggests a way forward.Whilst the actual velocity in this region depends on the viscosity (due to the increase verynear the wall) we might suggest that the changes that occur in the velocity within this regionare independent of the viscosity (since these are determined by local stresses). So then
∂u
∂y= f3(y, Uτ) (174)
this is consistent with the law-of-the-wall still but we have eliminated viscosity in the deriva-tive. We can proceed from here in two equivalent ways. The first and simplest is to note thatwe now have three variables and two fundamental quantities and so only one non-dimensionalnumber (Pi-group) which must therefore be constant
y
Uτ
∂u
∂y= A, (175)
56
where A is just some constant. Integrating gives
u
Uτ= A ln y +B (176)
where the constant B will depend on changes that occur in the viscous sublayer and bufferlayer. In order to fix the constant pick any point within the region with some fixed valueof yUτ/ν and therefore fixed u/Uτ and we find B = a1 − A ln(ν/Uτ ) where again a1 is justsome constant and so this can be written as
u
Uτ= A ln
(yUτ
ν
)
+ C (177)
The other, perhaps neater but equivalent way is to note that, from the law-of-the-wall
∂u
∂y=U2
τ
νf ′
(yUτ
ν
)
, (178)
where the dash denotes differentiation with respect to the argument. This is then substitutedinto 175 and integrated to find the function f in the law-of-the-wall:the same result follows.Note that, for historical reasons the constant A is usually replaced by 1/κ where the constantκ is referred to as the von-Karman constant. von Karman derived the log law in a differentmanner which resulted in the constant coming out in this manner. So we would usually seethe log-law written as
u
Uτ=
1
κln(yUτ
ν
)
+ C (179)
where κ is a universal constant with a value of about 0.4.
This expression is known as the logarithmic law for the mean velocity or, more affectionatelyas the log-law. Whilst there have been a variety of challenges to this in the past it is stronglysupported by data (see for example figure 191) and is at the very least an excellent fit todata over a wide range of Reynolds numbers (which has a sensible physical basis, as opposedto, say, the 1/7 power law). It applies in some part of the flow not too close to the wall, aswe have discussed, so we are beyond the viscous sublayer and buffer region - opinions differas to the number but a universal value of around y+ = yUτ/ν > 50 is a reasonable estimate.It also only applies when we are far enough from the outer edge of the layer and so for someuniversal value of y/δ which is usually taken as approximately y/δ < 0.1. The reason itis hard to precisely pin these numbers down is that we are considering a region where theviscous stresses are small enough to neglect and where the outer influences are also smallenough. These effects are always present and never exactly zero. As a final note since wenow have an estimate for where the log-law starts (y+ = 50) and where it ends (y/δ = 0.1)then we can work out the “length” of the log-law. The ratio of the outer limit to the inneris youter/yinner = 0.1δ/(50ν/Uτ ) = 0.002 ∗ δUτ/ν. So it depends on the Reynolds numberand gets longer for higher Reynolds number. Figure 191 shows this trend. On the figureKτ = δUτ/ν.
Now we turn to the outer part of the flow near the edge of the boundary layer. We postulatethat the flow out here does not know about the existence of the wall directly and what isimportant out here is the difference of the velocity in the layer from the free-stream velocity
57
(U1 − u(y)). We assume that the wall shear stress is still important since it determines thestresses further out from the wall (we would expect that if we doubled the wall shear-stressthen the stresses further out would also increase - it is only in this indirect sense that theflow out here “feels” the effect of the wall). This is an important assumption - the outerflow only feels the effect of the wall through the transmitted shear stress (we will returnto this later). Now we also not that out here the boundary layer thickness must enteras an important variable since the velocity will depend on how far we are from the edge.Dimensional analysis then gives
U1 − u(y) = f4(y, δ, Uτ , U1) (180)
which leads toU1 − u(y)
Uτ
= f4(y/δ, U1/Uτ ) (181)
but it is usual to drop the dependence on U1/Uτ since this is weak in zero pressure-gradientflow (this quantity varies only slowly with Reynolds number, or equivalently x)
U1 − u(y)
Uτ
= f4(y/δ) (182)
This expression is usually called the velocity defect law and is the outer flow’s equivalent ofthe law-of-the-wall. It is interesting that we can also derive the log-law by working from thisexpression and moving in. Recall to derive it before we started at the wall and moved outuntil we could neglect viscosity. In the outer part viscosity is already negligible (think aboutthe velocity gradient) but the boundary layer thickness is important. We might imagine thatas we move toward the wall and away from the edge then, far away from the edge (but stillwhere the viscous stress is neglible) then the boundary layer thickness would not be relevantany more - at least in terms of the changes in velocity so as before we could write
y
Uτ
∂u
∂y=
1
κ, (183)
and from (180)∂u
∂y=
−Uτ
δf ′
4(y/δ) (184)
then substituting and integrating to find f4(y/δ) gives
U1 − u(y)
Uτ
= −1
κln (y/δ) +D (185)
where D is some constant of integration. So we find we have another log law!! Also some-where in the middle not too close to the wall and also far from the edge. You may haveguessed that it is the same log-law just expressed in different variables. To show this take(179) and let y = δ - at this point then u = U1 so you have an expression for U1/Uτ . Subtract(179) from this and you have an expression for the defect. Figure 20 shows the collapse ofprofiles with velocity defect scaling. They collapse best in the outer region but then divergein the inner region as viscous effects become important.
Question 15.Show that the log-law for the outer flow is consistent with the log-law for the inner flow asdiscussed in the text above.
58
10-3 10-2 10-1 100
η
0
5
10
15
20
(U1-U
)/U
τ
10m/s (Rθ=2,583 & Κτ=1,047)10m/s (Rθ=9,256 & Κτ=3,436)10m/s (Rθ=20,013 & Κτ=7,611)20m/s (Rθ=37,542 & Κτ=14,379))30m/s (Rθ=52,103 & Κτ=19,956)40m/s (Rθ=68,151 & Κτ=26,376)Log_Law (κ=0.41 & Β=2.309)
Figure 20: Defect-law scaling: Boundary layers showing collapse in the outer region (notehere η = y/δ) for a range of Reynolds numbers: zero pressure-gradient flow
Finally we consider how to deal with the outer flow in combination with the law-of-the-wall. We have shown that the law-of-the-wall applies from the wall up to some value ofy/δ ≈ 0.1 where the effect of the outer flow and, in particular δ becomes important. Itbecomes logarithmic far enough from the wall but below this value. Beyond this point thewall-scaling no longer applies and the mean velocity deviates from the inner log-law solution.Coles(1956) suggested a simple way to account for this is to assume that this deviation is afunction of y/δ and hence that the whole profile could be written as
u(y)
Uτ= f
(yUτ
ν
)
+ φ(y
δ
)
. (186)
He called this deviation (φ) the wake function since its shape is similar to that of the wakeof a body. The different regions of the boundary layer are shown in figure 21. There is alot of information on the figure but don’t be deterred. Figure 191 shows velocity profiles forzero pressure-gradient boundary layers across a range of Reynolds numbers.Note that from this expression we can say something about the variation of the all-importantskin-friction coefficient. It is not difficult to show that C ′
f/2 = (Uτ/U1)2. Consider (189)
evaluated at the “edge” of the layer
U1
Uτ= f
(
δUτ
ν
)
+ φ(1). (187)
This can be rewritten as √√√√
2
C ′
f
= f
δU1
ν
√
C ′
f
2
+ φ(1). (188)
59
1 10 100 1000 10000yUτ/ν
0
10
20
30
U/Uτ
Boundary layer velocity profileU/Uτ=yUτ/νLogarithmic law
Wake
laminar sublayer
logarithmic region
Buffer region
y/δ = 0.110% of boundary layer thickness
Figure 21: Different regions of a turbulent boundary layer
which in the logarithmic part of the flow is√√√√
2
C ′
f
=1
κln
δU1
ν
√
C ′
f
2
+ C + φ(1). (189)
which is unfortunately and implicit expression for C ′
f and is in terms of the local Reynoldsnumber. It is possible to relate δ to the streamwise distance x through the momentumequation and after some simplification this leads to
√√√√
1
C ′
f
= a1 + a2 ln(RxC′
f) (190)
where Rx = U1x/ν and the a1 and a2 are just constants determined from experiments.This expression is sometimes called the Karman-Schoenherr law for skin-friction. The fullderivation is given in an appendix to these notes. It is not too difficult but takes a numberof steps. A comparison of this prediction from the above analysis with a large variety ofdata is shown in figure 22.
10.4.1 The effect of pressure-gradient
Since many flows of interest have an applied pressure-gradient then we need to consider howthings change with pressure-gradient. In this case there is an extra term in the boundarylayer equation. We will consider only moderate pressure-gradients which might be typical of,say, aerofoils and other common flows. We will also consider only moderately high Reynoldsnumbers.
60
Figure 22: Recent data showing C ′
f versus Rθ
In the inner part of the flow the pressure-gradient introduces and extra parameter which isthe non-dimensional pressure-gradient (non-dimensionalised on wall-variables) hence
u
Uτ= f
(
yUτ
ν,ν
ρU3τ
dp
dx
)
. (191)
It can be shown, however that at reasonable Reynolds number that this extra term becomesvery small due the being divided by U 2
τ and so for moderate to high Reynolds numbers andmoderate pressure-gradients this term is usually neglected. Hence the same law-of-the-wallapplies and all of the analysis above is still valid (we still get a log-law of the standard formand the inner region is not significantly modified). It is small relative to the other terms inthe equation (in particular the stresses).
In the outer region, however, this is not so. The viscous stress is negligible in the outerregion and the Reynolds shear stress is falling towards zero and so the pressure-gradient canbe important. Hence the defect law must be modified and we find
U1 − u(y)
Uτ= f4
(
y
δ,δ
ρU2τ
dp
dx
)
. (192)
Note that if we keep the non-dimensional pressure-gradient parameter constant then thedefect profiles will still collapse as for zero pressure-gradient layers but the shape will bedifferent. In the general case, however, the profiles still collapse in the inner region (usinglaw-of-the-wall scaling) but not in defect form.
Using the same logic as before Coles(1954) suggested then that this could be accounted forby allowing the wake function to vary with pressure-gradient. For simplicity (and based ondata) he suggested that the functional form would be the same but the strength of the wakepart (relative to the inner part) would change depending on the pressure-gradient this is
61
1 10 100 1000 10000 1e+05yUτ/ν
0
10
20
30
40
50
U/Uτ
Zero pressure-gradientAdverse pressure-gradientFavourable Pressure-gradient
Figure 23: The effect of pressure-gradients on the boundary layer profile
usually written asu(y)
Uτ
= f(yUτ
ν
)
+Π(x)
κW(y
δ
)
, (193)
where W (y/δ) is a universal function and Π is known as Coles’ wake factor which varieswith pressure-gradient (and distance along the plate if pressure-gradient is varying) - thefact that it is divided by κ is only a mathematical convenience for some other analysis.Coles found that the function W (y/δ) = 1 − cos(πy/δ) although this is only a curve-fit andobviously only works up to the edge of the layer (y/δ = 1) and not beyond. The effect ofpressure-gradients on the profiles are shown in figure 23 where it can be seen that the scalingstill applies in the inner region but the outer wake region is strongly affected. The valueof Π increases for adverse pressure-gradients and decreases for favourable ones. In stronglyfavourable pressure-gradients there is virtually no wake at all and the log law applies almostto the edge of the layer.
Note that the profile shape now depends on an extra parameter Π and so it is not possiblein general to find the skin-friction variation with x unless we know the function Π(x) - thiscan only come from measurements. In general we would know the pressure-gradient butif we could relate Π to the pressure-gradient parameter then we could solve the problem.There are various attempts to do this but all contain a large amount of empiricism andwork only in certain cases. In reality Π depends on a number of factors including thehistory of the boundary (what pressure-gradients it has undergone upstream of the locationof interest). Hence methods of evaluating the skin-friction in this manner are always justgood approximations for a particular situation.
62
Question 16.(a) A boundary layer has a profile given by
U1 − u(y)
Uτ= f(η) (194)
where η = y/δ. (a) Show that δ∗/δ = C1Uτ/U1, where δ∗ is the displacement thickness and
C1 =∫
∞
0f(η)dη (195)
(b) Show that θ/δ == C1Uτ/U1−C2(Uτ/U1)2 where θ is the momentum thickness. Note that
the definitions of momentum and displacement thickness are exactly the same for turbulentboundary layers as they are for laminar ones.
C2 =∫
∞
0(f(η))2dη (196)
(c) find an expression for the shape factor, H in terms of C1, C2, Uτ/U1.
(d) Consider the law-of-the wall expression evaluated at y = δ and show that this implies thatthe shape factor depends on Reynolds number (you need not derive the exact expression).
10.4.2 The effect of wall roughness
In many, if not most, engineering applications the surface over which the boundary layerdevelops is not perfectly smooth and may be very rough indeed. Now all real surfaces havesome roughness if you look close enough so the first question to consider is when the wallmay be considered smooth. It turns out that if the height of the protrusions on the surfaceis well within the laminar sublayer then the fluid flows smoothly over them and there is nosignificant effect on the overall flow. It is found that this occurs when the height of theroughness is nominally less than about 5 wall units. If we denote the roughness height byh then when h+ < 5. Note that the actual acceptable height will depend on the thicknessof the laminar sublayer which varies with Reynolds number so a flow that appears smoothto low Reynolds number boundary layer may well appear rough to one at higher Reynoldsnumber.
Once the roughness is larger than the laminar sublayer then we would expect it to introduceperturbations into the near wall flow and modify the region very near the wall significantly.This messes up our scalings very near the wall but we have noted already that the flowfurther out does not know what is going on at the wall except in-so-far as it detects a changein the stresses (which scale with the wall-shear stress). This prompted Townsend(1956) topropose his wall-similarity hypotheses. Consider the defect scaling. In this there are noextra parameters introduced by what is going on at the wall and hence the defect law willbe unchanged - the actual wall shear-stress increases but since this is accounted for in thescaling then the defect law still works perfectly well. Experiments confirm that this is thecase as long as the roughness elements are not so big as to extend well into the flow.
What about the near wall scaling? Our law-of-the-wall. The height, h, of the elements
63
imposes a new length scale and so we might write
U
Uτ
= f
(
yUτ
ν,hUτ
ν
)
(197)
We can also follow the same arguments as before to derive a log-law assuming that the heightof the roughness elements becomes unimportant far enough away from the wall (in terms ofthe velocity gradient), as is the viscosity. This has the form
U
Uτ=
1
κln(y
h
)
+ C(h+) + φ(y/δ) (198)
where the κ is the same as before but the C will depend on the flow near the wall henceC is a function of hUτ/ν = h+ since the wall shear stress will depend on the drag of theelements which is a combination of viscous drag and form drag. The wake function shouldbe unaffected since the outer flow is unaware of the details of the near wall flow.
If the roughness elements become larger then the roughness elements become like small bluffbodies. Just like the disk in the 3A1a experiment then their drag becomes independent ofthe Reynolds number and so even near the wall the viscous stresses become unimportant(relative to the form drag on the elements and the Reynolds shear stress). Hence our wall-unit length-scale ν/Uτ is no longer important. In this case then the law-of-the-wall becomes
U
Uτ
= f(y
h
)
(199)
and the log-law has the same form but in this case we would expect the constant, C, to bejust a universal constant (since there is no other non-dimensional parameter on which it candepend). In this case the wall-shear stress just depends on the form drag of the elements.We can again use a simple argument to look at the variation of the skin-friction coefficient.Again consider the log-law at the outer edge of the flow and we find
U1
Uτ=√
2/C ′
f =1
κln
(
δ
h
)
+ C
(
hU1
ν
C ′
f
2
)
+ φ(1). (200)
Consider a fixed height of roughness and increase the Reynolds number by increasing thespeed of the flow. The thickness of the boundary layer is only a weak function of Reynoldsnumber (it gets thinner as Re increases but we will neglect that to first order). As theReynolds number increases the sublayer gets thinner and thinner and so the roughness el-ements protrude further into the flow - h+ increases. Beyond a certain point, however, Cceases to depend on h+ and hence with δ/h constant the local skin-friction coefficient be-comes (roughly) constant. This is the situation that applies if we measure the drag of a plateof fixed length L and vary the Reynolds number - if we use a number of different roughnessheights then we can plot the drag of the plate as shown in figure 24.Now consider increasing the Reynolds number by keeping the velocity constant but moving
along the plate. In this case the boundary layer gets thicker and so the relative roughnessgets smaller as we move along the plate so the local skin friction coefficient drops for a fixedheight of roughness.
64
CD
U1L/ν
Large roughness
Decreasing roughness
Turbulent smooth curve
Laminar
Figure 24: Drag coefficient for a flat plate of length L (at zero incidence) with different levelsof roughness
10.5 Turbulent pipe flow
Since pipes are used quite frequently in engineering then turbulent pipe flow is of particularinterest. Flow in a pipe develops from the start of the pipe where a boundary layer formson the wall and grows with distance along the pipe, this might be initially turbulent orlaminar depending on the speed of the flow. After some distance the boundary layer growsuntil it reaches the centre and cannot grow anymore and so cannot develop with streamwisedistance. At some distance after this point the flow settles down to an internal equilibriumand it becomes independent of streamwise distance - the velocity profile (and the profilesof the other quantities eg. Reynolds stresses). This is known as fully-developed flow and,typically occurs after about 100 diameters from the start. This is the flow we will considerhere.
All of our previous arguments and dimensional analysis for the turbulent boundary layerstill apply in this case since none specifically referred to the geometry of the flow. The onlydifference is that in the pipe the “thickness” of the layer is fixed and equal to the radius ofthe pipe, for which we use the symbol a. Also the outer flow velocity U1 is now the velocityat the centre of the pipe. Hence, as before for boundary layers,
U
Uτ= f
(yUτ
ν
)
(201)
near the wall andU1 − U
Uτ= φ
(y
a
)
(202)
in the outer part - near the centre-line. And we can use the same arguments again to derivethe logarithmic law both in law-of-the-wall co-ordinates and in outer co-ordinates. Thesehave exactly the same form as before if we substitute a for δ and take U1 as the centre-linevelocity. Note that in this flow there is still a wake component but it must be independentof x since the mean velocity profile is independent of x. The wake component is, however,quite small since there is a favourable pressure-gradient acting in a pipe due to the frictionloss. To see this more clearly consider a control volume across the pipe. The momentum inmust equal the momentum out since the velocity profile does not change. Hence the sum
65
of the external forces is zero and so the force due difference in pressure across the controlvolume is balance by the force exerted by the wall shear-stress i.e. for a control volume oflength dx
τo2πadx = −dPdx
dxπa2 (203)
or, using the same definition of Uτ as before
a
2ρU2τ
dP
dx= constant. (204)
We should then not be surprised that the wake doesn’t change since the non-dimensionalpressure-gradient parameter based on the outer flow is a (negative) constant. Note alsothat since Uτ does not vary with x 5(and neither do a or ρ) then this implies a constantpressure-gradient - in other words the pressure varies linearly along the pipe.
11 The friction factor in a pipe
We can now determine the friction factor in a turbulent pipe flow. In this derivation thepipe flow is assumed to be fully-developed which is to say that the wall boundary layershave had enough development length to grow across the pipe and hence the velocity profileis self-similar (usually about 60 to 100 diameters). In a pipe we find that if the velocityprofiles are similar the problem reduces to a balance between the skin-friction forces and thepressure-gradient forces. Let us consider a cylindrical slug of fluid in the pipe.There is no inertia then
τo2πL = (p1 − p2)πa2 (205)
hence
τo =(p1 − p2)
L
(a
2
)
(206)
It is usual in pipe flow to use the bulk velocity, Ub which can be obtained using a stop-watchand a bucket. It is the average velocity over the pipe area i.e.
Ub =1
A
∫
AudA (207)
and from the definition of friction factor ,f ,
(p1 − p2)
Ld = f
1
2ρU2
b (208)
This leads tof
8=U2
τ
U2b
(209)
for a pipe. We can compare this to a boundary layer where
C ′
f
2=U2
τ
U21
(210)
5The wall shear stress depends on the gradient of the velocity at the wall which depends on the velocity
profile. In fully developed flow this does not change
66
and combining the two suggests that for a pipe flow
f
4=
τo12ρU2
b
(211)
Now we also know thatU1 − U
Uτ
= φ(y
a
)
(212)
and this leads to1
A
∫
A
U1
UτdA− 1
A
∫
A
U
UτdA =
1
A
∫
Aφ(y
a
)
dA (213)
Ub
Uτ− U1
Uτ=
1
A
∫
A−φ
(y
a
)
dA (214)
Now we can integrate across the pipe using annular strips. If we do this then
πa2 =∫ a
02π(y − a)d(y − a) (215)
soUb
Uτ
− U1
Uτ
=1
πa2
∫ a
02πa(1 − y
a)λ(
y
a)d(a(1 − y
a) (216)
where λ(y/a) is given by
λ(y
a) =
1
κln(
y
a) −W (1) +W (
y
a) (217)
and henceUb
Uτ− U1
Uτ= 2
∫ 1
0(1 − y
a)λ(
y
a)d(
y
a) = universal constant (218)
since the terms in the integral are all functions of y/a and the integral is definite. If W (1)and W (y/a) are considered to be small then it can be shown that
Ub
Uτ− U1
Uτ= − 3
2κ(219)
HINT: to do this integration you must know limx→0 x ln(x) which may be found froml’hospital’s rule.Now we want to find the friction factor so we must consider the region nearer the wall. Weknow that U = U1 at y = a and
U1
Uτ=
1
κln(aUτ
ν
)
+ A+W (1) (220)
and substituting this back we find
Ub
Uτ− 1
κln(aUτ
ν
)
− A−W (1) = − 3
2κ(221)
(retaining the H(1) term here). We also have a realtionship for the friction factor which is
Ub
Uτ
=
√
8
f(222)
67
1000 10000 1e+05 1e+06 1e+07 1e+08Re
0.01
0.02
0.03
0.04
0.05
f
Oregon Data - low temperature heliumPrinceton Data - very high pressure pipeCurvefit using theory
Figure 25: Friction factor relation from the recent paper of Smits et al.
hence √
8
f=
1
κln
(
Ubd
ν.Uτ
Ub
)
+ A+W (1) − 3
2κ− 1
κln(2) (223)
where d is the pipe diameter and Ubd/ν is the Reynolds number. Now substituting again forUb/Uτ in terms of the friction factor we get
√
8
f=
1
κln
(
Re.
√
8
f
)
+ A+W (1) − 3
2κ− 1
κln(2) (224)
which may be rewritten as√
8
f=
1
κln
(
Re.
√
8
f
)
+D (225)
where D is a universal constant. Schlichting gives this formula as
1√f
= 2.0 log10(Re√
f) − 0.8 (226)
which is called Prandtl’s smooth tube formula. Recent research using the combined re-sources of a 25 tonne pressurised (200 atmospheres) “super-pipe” at Princeton with a 30gliquid helium rig in Oregon have produced better estimates for the constants which givesan excellent fit to the data over nearly four orders of magnitude in Reynolds number (fromapproximately Re=2000 to Re=20,000,000). They find
1√f
= 1.93 log10(Re√
f) − 0.537 (227)
see figure 25The constants used generally are
κ = 0.41
A = 5.1. (228)
68
but note that the new constants given above suggest a value of κ = 0.422 for the super-pipedata. Note that this is still an active area of research. There is still some question as tothe correct values of the constants and whether they have the same asymptotic values indifferent flows (eg. pipes versus boundary layers). From a general engineering perspectivein most cases the errors incurred due to errors in κ are much smaller than the other errorsin design and manufacture (eg. pipe roughness, losses in imperfect joins etc.).
11.1 Rough pipe flow
Since most pipes in general use have some roughness then it is useful to consider the behaviourof rough pipes. Again the same arguments used for rough boundary layer flow and the samescalings apply when we replace δ by a and take U1 as the centreline velocity. It is alsopossible to calculate the skin friction for the rough pipe. Here we will consider just the casefor large roughness since it is easier. Using the arguments as before that when the elementsextend well beyond the sublayer their drag becomes independent of Reynolds number wefind the log law
u
Uτ=
1
κln(y/h) + C +W (y/δ) (229)
At the centreline this becomes
U1
Uτ=
1
κln(a/h) + C +W (1) (230)
where the term on the left is directly related to the inverse of the friction factor. Since,for a given pipe, everything on the right of the equation is constant then the friction-factorf becomes independent of Reynolds number. We already know how it behaves for smoothpipes from the above and in between the two we have some transition region between thatbehaviour and the constant value here. We can also note that increasing the relative rough-ness (h/a) reduces the right hand side and hence increases the friction factor (which is nottoo surprising). This information is summarised on a diagram known as the Moody diagram(figure 26) which shows the friction factor of a pipe versus Reynolds number for a rangeof different relative roughness values. Note also that the Reynolds number at which thefriction factor becomes constant varies with relative roughness. This is because the impor-tant parameter determining when we can neglect the viscous drag depends on the ratio ofthe roughness height to the sublayer thickness. At higher Reynolds number the sublayer isthinner and hence
12 Closure Hypotheses
So, as noted, we find ourselves short of one equation. The problem is to relate our newReynolds stress term somehow to the mean velocities - if we can do that then we can findsolutions. This is the Closure Problem much discussed in turbulence circles - in particularamong people trying to compute turbulent flows in practical situations. There are manydifferent closure hypotheses which have been used to solve the REYNOLDS AVERAGEDEQUATIONS.The simplest is the Boussinesq hypothesis. By analogy with LAMINAR flow Boussinesqsuggested the following relation.
69
2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 8 103
104
105
106
107 8
10
0.01
0.012
0.100
0.090
0.080
0.070
0.060
0.050
0.040
0.035
0.030
0.025
0.018
0.016
0.014
0.020
g
u
d
L
hf
2
2
⋅
=
ν
du=R e
pipe friction chartapplicable to circular pipes running full
© Glasgow College of Nautical Studies Faculty of Engineering
GCNS runs distance learning courses for the
Engineering Council Graduate Diploma.
website: www.glasgow-nautical.ac.uk
e-mail: [email protected]
Figure 26: Moody diagram
τTρ
= ε∂u
∂y(231)
where τT is the turbulent shear stress given by τT = −ρu′v′. The constant ε is called theEDDY VISCOSITY. In free shear flows such as jets and wakes ε is assumed to be a constantjust as ν is a constant - this gives results that are roughly right. In reality however ε is aproperty of the flow and not of the fluid and so its value can vary. In particular in the case ofboundary layers it is found that ε must vary with the distance from the wall to give resultsconsistent with experiments.The log-law has been derived by Squire(1947) using an eddy viscosity assumption.
τoρ
= U2τ = −u′v′ = ε
∂U
∂y(232)
Now the law of the wall suggests that ε is only a function of wall variables close to the walland furthermore beyond the viscous sub-layer then we would expect the viscous terms to benegliglible and hence ν should not be important, hence
ε = f(y, Uτ) (233)
Now dimensional analysis tells us that there is only one PI product and so
ε
yUτ
= constant = κ (234)
and substituing back into the equation for shear stress we find
U2τ = κyUτ
∂U
∂y(235)
70
and then we have an expression for ∂U/∂y which can be integrated and gives the same resultas before ( a log-law). In the outer part of the flow then we expect that the eddy viscosityshould depend on δ rather than y and hence
ε
δUτ= constant = B (236)
and thus the eddy viscosity is constant in the outer part of the layer. This approach has hadsome limited success. Its advantage is that it is mathematically and conceptually simple.Its disadvantage is that the physics is somewhat dubious - there is no reason to assume thatthe Reynolds shear-stress is proportional to the mean velocity gradient but it turns out thatthey are roughly similar. The Reynolds shear stress is large just outside the laminar sublayer(where it is approximately equal to the wall shear-stress and drops to zero outside the layer -just as the velocity gradient does. This does not, however, imply that they are proportionalto each other.
There are a variety of other ”turbulence models” in use. They all attempt to relate theturbulence quantities to the mean quantities in some fashion. They allow engineers to solvethe boundary layer equations using CFD methods. It is important to remember, however,that they are all based on extra assumptions about the physics which are no better than ourdimensional analysis and hand-waving arguments. As such they should be treated with careand not trusted unless they can be checked against experimental results.
71
13 Appendix:Power law profile used to calculate skin
friction - zero pressure-gradient flow.
Although we have found that a sensible physical functional form for the boundary layervelocity profile is logarithmic, before proceeding to use this fact to derive a skin-friction law,we shall consider an older and simpler approach to the problem. It was found quite early onin research that a reasonable empirical curve-fit to data is a 1/7th power law i.e.
u
Uτ= C1
(yUτ
ν
)1/7
, (237)
where the constant C1 actually varies with the Reynolds number but, let’s say, has a value ofaround 8.7 for this 1/7th power-law form at typical or reasonable Reynolds numbers and isa constant. There have been various variations on the value of the power but essentially it isfound that a power-law form like this fits data reasonably well 6. This power-law form is veryconvenient mathematically since we can easily integrate or differentiate it where necessary.Here we will derive the skin-friction variation for a general power-law profile of this form i.e.
u
Uτ= C1
(yUτ
ν
)1/n
, (238)
where for the 1/7th power-law the n would be 7. Note that the best fit to a typical profile willproduce a particular value of C1 for a given choice of n. It has been found empirically thatthe choice of n giving a good fit to data depends on the Reynolds number of the boundarylayer: higher Reynolds numbers are better fitted by larger values of n (for n = 7, C1 = 8.74,for n = 9 C1 = 10.6. This is a reflection of the fact that the profiles change with Reynoldsnumbers and we are trying to fit a universal function to them which is not appropriate -nevertheless the approach gives roughly reasonable results and is simple to analyse. It alsoshows - in a straightforward manner - the way in which skin-friction and boundary layergrowth can be derived from an assumed form of boundary layer profile. At the edge of thelayer this becomes
U1
Uτ= C1
(
δUτ
ν
)1/n
, (239)
Now consider the momentum equation
dθ
dx=C ′
f
2=U2
τ
U21
. (240)
Now we need the momentum thickness and so we integrate the profile up (from 0 to δ) andfind that
θ
δ=
n
(1 + n)(2 + n), (241)
which can then be substituted into the momentum integral equation along with (239 giving
n
(1 + n)(2 + n)
dδ
dx= (C1)
−2n/(n+1)
(
U1δ
ν
)−2/(n+1)
. (242)
6note, of course, that it does not have the correct asymptotic behaviour as y → ∞ but that is OK as long
as we avoid that region. Remember in question 1 in the notes where you showed that the integral properties
are not affected much as long as you cut off the integrals at some finite δ below which the fit to the profiles
is good
72
which we can now separate and integrate since we have only δ and x in the equation (andsome constants). We will assume that δ = 0 at x = 0. This then integrates to
δ
x= C ′
2R−2/(n+3)x , (243)
where the constant C ′
2 is given by
C ′
2 =
[
(2 + n)(n+ 3)
nC
−2n/(n+1)1
](n+1)/(n+3)
, (244)
and finally differentiating (243) and substituting into the momentum integral equation wecan find
C ′
f = C ′
3R−2/(n+3)x , (245)
whereC ′
3 = C ′
2 (2n/[(2 + n)(n+ 3)]) . (246)
Whilst the mathematics is a little tedious is it not horribly complicated. With the choiceof n = 7 then we find C ′
f = 0.0592R−1/5x and δ/x = 0.37R−1/5
x . These are useful roughexpressions which give reasonable estimates of the skin-friction and boundary layer growthfor zero pressure-gradient turbulent boundary layers.
73
Appendix 2:Skin friction derivation for a boundary layer
from the log law
SEE DUNCAN, THOM AND YOUNG P.320 to 323 The power-law fit for the velocityprofile of a turbulent boundary layer is, at best, a crude approximation. There is no physicsbehind the shape or scaling based on that approach. We have seen in previous sectionsthat there are sound scaling arguments that apply to the turbulent boundary layer andthese result in a logarithmic behaviour. The scaling arguments also show that there aretwo appropriate scaling regions in the flow and hence the analysis based on this theory isnecessarily more complicated. It is, however, more sound and also gives excellent agreementwith good experimental data. This is the only semi-theoretical solution known based onsimilarity laws. In the zero pressure-gradient case it is straightforward (if a little tricky) butin the case with pressure-gradient things become more difficult. It gives us an expression forhow the skin-friction varies based on the momentum integral equation and using the velocityprofile given earlier (i.e. the log-law plus a deviation function).So we have
dθ
dx=C ′
f
2(247)
U
Uτ
=1
κln(yUτ
ν
)
+ A +Π
κw(y
δ
)
(248)
U1 − U
Uτ
= −1
κln(y
δ
)
+Π
κ(w(1) − w(y/δ)) (249)
hence we can writeU1 − U
Uτ= f(y/δ) (250)
and we will define a new useful variable by
S =U1
Uτ
(251)
Now let us examine the displacement thickness
δ∗
δ=
∫ 1
0(1 − U/U1)dη
=Uτ
U1
∫ 1
0
U1 − U
Uτdη
=1
S
∫ 1
of(η)dη
=C1
S(252)
where as previously η = y/δ. This formula can also be used to give a precise definition of δsince δ∗ is know accurately (from an integral), hence
δ =δ∗S
C1(253)
74
and by a similar approach we can show that
θ
δ=C1
S− C2
S2(254)
where
C1 =∫ 1
0f(η)dη
C2 =∫ 1
0f 2(η)dη (255)
Now from the momentum integral equation
dθ
dx=C ′
f
2(256)
which can be written non-dimensionally as
dRθ
dRx
=C ′
f
2=
1
S2(257)
where Rθ = U1θ/ν and Rx = U1x/ν. Now at the edge of the layer we can write
κ[S − A− Π
κw(1)] = ln
(
δUτ
ν
)
(258)
orδUτ
ν= eκ[S−A−
Π
κw(1)] (259)
and this leads to
Rθ = SδUτ
ν(C1
S− C2
S2) (260)
and substituting
Rθ = (C1 −C2
S)eκ[S−A−
Π
κw(1)] (261)
But we also know thatdRθ
dRx=
1
S2(262)
from the momentum equation. For convenience we can put
A +Π
κw(1) = φ(1) (263)
and taking a derivative w.r.t. S
dRθ
dS= (
C2
S2+ κC1 −
κC2
S)eκ(S−φ(1)) (264)
Now apply the chain-ruledRx
dS=dRx
dRθ
.dRθ
dS(265)
and finally substituting in we find
75
dRx
dS= (C2 − κC2S + κC1S
2)eκ(S−φ(1)) (266)
integrating from the leading edge of the plate where x = 0 and S = 0 then we find
Rx = [C1S2 − (
2C1
κ+ C2)S +
2
κ(C1
κ+ C2)]e
κ(S−φ(1)) (267)
S is usually large so it is common to retain only the S2 terms and from this it is not difficultto show that the final skin friction law is
√√√√
1
C ′
f
= A′ +B′ log10(RxC′
f ) (268)
where A′ = 1.7, B′ = 4.15.
76
Appendix 3:Datacard
Module 3A1Boundary Layer Theory Data Card
Displacement thickness;
δ∗ =∫
∞
0
(
1 − u
U1
)
dy
Momentum thickness;
θ =∫
∞
0
(U1 − u)u
U21
dy =∫
∞
0
(
1 − u
U1
)u
U1dy
Energy thickness;
δE =∫
∞
0
(U21 − u2)u
U31
dy =∫
∞
0
(
1 −(u
U1
)2)
u
U1dy
H =δ∗
θ
Prandtl’s boundary layer equations (laminar flow);
u∂u
∂x+ v
∂u
∂y= −1
ρ
dp1
dx+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0
von Karman momentum integral equation;
dθ
dx+H + 2
U1θdU1
dx=
τoρU2
1
=C ′
f
2
Boundary layer equations for turbulent flow;
u∂u
∂x+ v
∂u
∂y=
−1
ρ
dp
dx− ∂u′v′
∂y+ ν
∂2u
∂y2
∂u
∂x+∂v
∂y= 0
77
