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PROBLEM 1.A – Viscosity

The temperature dependence of the dynamic viscosity μ of a gas is given by Sutherland's

expression:
3/ 2
µ T  T0 + S
=  
µ0  T0  T +S

where μ0 and T0 are reference values with μ0 = μ(T0); Sutherland's constant S has the dimension of
temperature and is characteristic of the gas (for air S = 111 K).

a. Show that the form of the above expression is independent of the choice of the reference
temperature, which can therefore be chosen arbitrarily.
Hint: choose a different reference temperature T1 and determine the expression for μ/μ1.

b. A convenient approximation of the viscosity relation is that according to a power law:

n
µ T 
= 
µ0  T0 

with n a constant. Determine how n relates to T0 and S, if this expression is to be a local

approximation of Sutherland's law for T near T0.

Calculate the corresponding value of n for the following cases:

- T0 = 288 K and S = 111 K (air at atmospheric conditions at 0 m SA)

- T0 very small (T0 << S)

- T0 very large (T0 >> S)

c. Investigate the accuracy of the power-law approximation by plotting μ/μ0 against T and
compare it to Sutherland's law. Use T0 = 288 K and S = 111 K and for the power law the
value of n that was calculated in part (b).
Determine the temperature interval in which the relative error introduced by the power-law
approximation remains within 5%.
(16) VISCOUS FLOWS - AE 4120 2

PROBLEM 2.C – Stagnation flow on an infinite swept wing

For the flow near the stagnation line of an infinite swept wing the numerical solution is given (see
table below) in the form of the nondimensional velocity profiles for the directions respectively
perpendicular and parallel to the stagnation line, with:

u / u e = F ′( η ) w / we = g( η )

a. Calculate the velocity profiles us/use and un/use, where the velocity vector has been
decomposed w.r.t. the flow direction just outside the boundary layer, for the case that we/ue =
0.5. Scale the components with the outer flow velocity use:

u se = u e2 + we2

Calculate in addition for this case the cross-flow angle ∆φ ( η ) = φ e − φ ( η ) .

Give a graphical representation of the results as profiles of η; also, plot the velocity
distributions in the form of a hodograph, i.e. us versus un.

b. Derive the general expression for the cross-flow angle at the wall, ∆ φ w , as function of the
ratio we/ue. Determine at which we/ue the maximum value of ∆ φ w occurs, and what its
value is.

Numerical solution:

η u/ue w/we η u/ue w/we η u/ue w/we

0.0 0.00000 0.00000

0.1 0.11826 0.05704 1.1 0.81487 0.59411 2.5 0.99285 0.96058
0.2 0.22661 0.11405 1.2 0.84667 0.63883 3.0 0.99842 0.98851
0.3 0.32524 0.17091 1.3 0.87381 0.68085 3.5 0.99972 0.99733
0.4 0.41446 0.22749 1.4 0.89681 0.72000 4.0 0.99996 0.99951
0.5 0.49465 0.28356 1.5 0.91617 0.75616 4.5 1.00000 0.99993
0.6 0.56628 0.33889 1.6 0.93235 0.78924
0.7 0.62986 0.39319 1.7 0.94577 0.81925
0.8 0.68594 0.44616 1.8 0.95683 0.84619
0.9 0.73508 0.49751 1.9 0.96588 0.87017
1.0 0.77786 0.54692 2.0 0.97322 0.89130

F" (0) = 1.23259 g ′(0) = 0.57047

(16) VISCOUS FLOWS - AE 4120 3

PROBLEM 3.F – Similar expansion boundary layer for β = –1

The Falkner-Skan equation describes the self-similar boundary layer solutions that are obtained for
pressure distributions that correspond to an external flow with u e ( x) ~ x m , and reads:
f ''' + f f '' + β (1 − f '2 ) = 0
2m
where β = and where the following scaling has been used:
m+1
u y m+1 ue x
= f '( η ) η=
ue x 2 ν

In addition to the common boundary conditions, the case of a similar solution with non-zero normal
velocity at the wall (suction or injection) has f(0) = f w ; where:
vw 2 ue x
fw = −
ue m +1 ν

For suction vw is negative, hence fw positive. For fw = 0 separation will occur for a certain negative
vale of β.
A result of the non-linear character of the Falkner-Skan equation is that for negative β not
every combination of β and fw allows a solution which possesses a boundary layer character (i.e.
with u asymptotically approaching ue), whereas other combinations may display multiple solutions.
This is illustrated here for the case β = –1.

a. Determine which (inviscid) expansion geometry corresponds to the case β = –1.

b. Show that for this special case β = –1 the Falkner-Skan equation can be analytically
integrated twice. Express the integration constants in the values which characterise the
solution at the wall (i.e. f w and f w '' ).

c. Show that, if the solution possesses a boundary layer character, from the behaviour of the
solution at large values of η the following relation can be derived:
f w" 2 = f w2 − 2
Hint: use the results of both integration steps!
Which values of fw allow boundary layer solutions?
Derive that the displacement thickness can be expressed in f w and f w '' .

d. The above shows that for certain values of f w two different solutions are possible. For one
of them f w '' is positive and for the other negative (the latter hence shows flow reversal near
the wall).
Determine both values of f w '' when f w = 2.25 and calculate the two corresponding velocity
profiles by a numerical integration (choose any method).

Note: in this case no iteration is required, as all boundary conditions at the wall are known!
(16) VISCOUS FLOWS - AE 4120 4

PROBLEM 4.C – Boundary layer development for an exponential external flow

The approximate boundary layer calculation method of the Pohlhausen/Thwaites type employs the
integral momentum equation in the form:

ue d θ 2 2due
= F( λ )= a − bλ where: λ= θ
ν dx ν dx

where λ features as a shape parameter of the velocity profile. With the linear relation between F and
λ, the momentum equation can be solved explicitly, giving the boundary layer development (in
terms of the momentum thickness) in an integral form.

N.B.: delay the substitution of specific values for a and b until indicated.

a. Show that a self-similar boundary layer results, when the external stream velocity varies as:
x m
ue ( x) = U0( )
L
Here U0 and L are velocity and length scales, and m is a constant.

- Determine the variation of θ with x and derive from this that λ is independent of x.

- Derive the expression which gives m as function of λ (and the constants a and b).
2m
- Derive also the relation between the Hartree parameter β = and λ.
m+1

b. Consider now the (non-similar) boundary layer which develops from x = 0 under the
influence of a pressure gradient corresponding to an exponentially accelerated external flow:

x
ue ( x) = U 0 exp( )
L

- Calculate the boundary layer development as function of x/L;

θ 2U 0
express the results in the dimensionless variables and λ
νL
- Determine at what value of x/L the momentum thickness reaches a maximum.

- Show that for large values of x/L the boundary layer develops towards similarity, and
determine what the value of λ is. Use the result of part (a) to determine the corresponding
value of m and β.

- Give a graphical representation of the boundary layer development on the interval

θ 2U 0
0≤x/L≤1; for a and b employ the values according to Thwaites. Plot both and λ as
νL
function of x/L.
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c. Next consider the (non-similar) boundary layer which develops from x = 0 with an
exponentially decelerated external flow:

x
u e ( x) = U 0 exp( − )
L

- Calculate the boundary layer development as function of x/L;

θ 2U 0
express the results in the dimensionless variables and λ
νL
- Determine where the method of Thwaites predicts separation to occur.

- Give a graphical representation of the boundary layer development on the interval 0≤x≤xsep;
θ 2U 0
for a and b employ the values according to Thwaites. Plot both and λ as function of
νL
x/L.

- Determine where separation would occur according to Stratford's prediction method.

(16) VISCOUS FLOWS - AE 4120 6

PROBLEM 5F – Heat transfer for the flat-plate boundary layer with suction

The momentum and heat transport in a boundary layer over a flat plate with suction, in neglection of
the effects of compression and dissipation, is governded by the follwoing equations:
∂u ∂u ∂ 2u
u +v =ν 2
∂x ∂y ∂y
∂T ∂T k ∂ 2T
u +v = 2
∂x ∂y ρ c p ∂y

with boundary condition, additional to the no-slip condition u ( x, 0) = 0 :

- prescribed normal velocity and temperature at the
= wall: v( x, 0) v= w ( x ); T ( x, 0) Tw ( x )
- constant velocity and temperature in the external flow: =
u ( x, ∞ ) U e ; = T ( x, ∞) T∞

a. Verify that a self-similar temperature solution is possible when the wall-normal velocity and
wall temperature vary according to vw ~ x −1/ 2 and ∆T= (Tw − T∞ ) ~ x n , with n constant.
Derive the differential equations that describe the self-similar velocity and temperature
solutions, together with the corresponding boundary conditions. Clearly show how the
conditions on vw ( x) and ∆T ( x) are reflected in constant values of the parameters in the
equation or the boundary conditions.

b. For the limiting case of strong suction, the flat-plate boundary layer transforms into the
“asymptotic suction boundary layer”. Derive the corresponding similarity equations for
velocity and temperature profile, using − yvw /ν as scaled normal coordinate.
N.B.: in this derivation two approaches can be followed, by applying the approximation of
strong suction either directly to the transformed equations found in (a), or by first simplifying
the original boundary layer (momentum and energy) equations, and then applying the
transformation.

c. Determine the solution for the velocity and temperature distribution (for arbitrary Prandtl
number) and plot the results for the following values of Pr: 0.5, 1 and 2.

- Compute the surface heat transfer and determine the Reynolds analogy factor s. What do you
observe regarding the effect of the Prandtl number on s for the suction boundary layer?
(16) VISCOUS FLOWS - AE 4120 7

PROBLEM 6.B – Instability and transition estimates of laminar boundary layers

A rough estimate of the location of the point of instability (critical point) or transition can be obtained
from semi-empirical correlations, such as e.g.:

- for the point of instability: Reθ ,crit exp(26.3 − 8 H )

= (after Wieghardt)
- for the point of transition: Reθ ,trans = 2.9(Re x,trans )0.4 (after Michel)

a. Apply these correlations to determine the point of instability and the point of transition for the
following self-similar boundary layer flows.
Give in both cases the values for Rex as well as Reθ.

-i- flat plate flow: =H 2.591

= θ 0.664 x Re x −1/ 2
-ii- stagnation point flow: =H 2.216
= θ 0.292 x Re x −1/ 2

b. Based on these estimates, what can you conclude about the effect of the pressure gradient on
the stability of a laminar boundary layer?

c. The velocity profile of the asymptotic suction boundary layer is given by the exponential
function:
u yv
1 − e −η
= − w
with: η =
ue ν

where vw is the normal velocity at the wall (for suction vw is negative).

Determine with the given correlation for the instability point, how large the suction velocity
−vw must be chosen, in order to keep the boundary layer on the margin of stability.
(16) VISCOUS FLOWS - AE 4120 8

PROBLEM 7.E – The Clauser plot

The so-called Clauser-plot technique is used to determine the wall shear stress in a turbulent boundary
layer from the mean velocity profile, in conditions where an accurate, direct determination of the
velocity gradient at the wall is not possible (which is the common situation for turbulent boundary
layers).

The table given below represents the measured velocity profile, such as determined e.g. by means of a
traverse with a pitot tube or a hot-wire probe. As a function of the distance y from the wall (measured
in mm), the velocity is given in non-dimensional form as u/Ue, where u is the local (mean) velocity in
the boundary layer, and Ue the velocity in the external flow.

a. Plot the velocity profile on a semi-logarithmic scale, with u/Ue versus Re y = yU e /ν .

b. Determine the value of the skin friction coefficient Cf by means of a curve fit of the law of the
wall to (the lower part of) the velocity profile.

c. Determine also the strength of the 'wake-component' of the velocity profile. For a measured
profile, this is defined as the maximum difference between u(y) and the law of the wall
expression.

Further data: ν = 15.0*10-6 m2/s

Ue = 10.521 m/s

y(mm) U/Ue y(mm) U/Ue

1 0.25 0.250 21 20.00 0.851

2 0.50 0.394 22 22.00 0.870
3 1.00 0.506 23 24.00 0.888
4 1.50 0.563 24 26.00 0.905
5 2.00 0.594 25 28.00 0.923
6 2.50 0.617 26 30.00 0.940
7 3.00 0.634 27 32.00 0.955
8 3.50 0.649 28 34.00 0.969
9 4.00 0.662 29 36.00 0.981
10 4.50 0.673 30 38.00 0.991
11 5.00 0.684 31 40.00 0.997
12 6.00 0.702 32 42.00 1.000
13 7.00 0.716 33 44.00 1.000
14 8.00 0.731 34 46.00 1.000
15 9.00 0.743
16 10.00 0.755
17 12.00 0.776
18 14.00 0.795
19 16.00 0.815
20 18.00 0.834
(16) VISCOUS FLOWS - AE 4120 9

PROBLEM 8.D – Production terms in the Reynolds Stress Equation

The production term Pij in the transport equation for the Reynolds stress component ui ' u j ' has the
following general form:

 ∂u ∂u j 
− u j ' uk ' i + ui ' uk '
Pij = 
 ∂xk ∂xk 

The production term P for the turbulent kinetic energy is related to this according to:

P11 + P22 + P33 ∂u

P= = −ui ' u j ' i
2 ∂x j

a. Completely expand the expressions for the six different production terms Pij (note that Pij is
symmetrical) under the assumption of a 2D mean flow. Replace the index notation by the
customary Cartesian notation, with: ( x1 , x2 , x3 ) → ( x, y, z ) and (u1 , u2 , u3 ) → (u , v, w) .

- Give also the corresponding expression for the production term P of the turbulent kinetic
energy.
NB: Only use the assumption that w and z-derivatives of mean flow properties are zero, but do
not introduce any simplification (symmetry arguments) regarding the possible impact on the
Reynolds stress components. Use the continuity equation for the mean flow, to replace ∂v / ∂y
by −∂u / ∂x .

- From symmetry arguments it follows that the Reynolds shear stress components u ' w ' and
v ' w ' are zero when the mean flow is 2D. Verify if this correspond to a zero value of the
production term for these components. Make the same assessment for the normal stress
component w '2 .

b. Determine how the different production terms further simplify upon introduction of these
symmetry arguments and under the additional assumption that the 2D mean flow has a
boundary layer character, so that v << u and ∂ / ∂x << ∂ / ∂y . Identify for each production
term the dominant contribution.

The wall region of a (2D, incompressible) turbulent boundary layer displays a universal structure,
which is known as the law of the wall. According to this concept, the properties of the flow can be
expressed uniquely in terms of the so-called wall scaling (also referred to as the use of 'wall units'). In
addition, it is commonly assumed that across the wall region the total shear stress is approximately
constant.

One element of this law of the wall is that the velocity profile of the mean flow satisfies an expression
of the following type:

u yv *
u+ = f ( y+ ) ; where:
= u+ = ; y+ with: v*
= τw / ρ
v* ν
(16) VISCOUS FLOWS - AE 4120 10

c. An accurate description of the velocity profile in the entire wall layer, including the viscous
region, is given by Spalding's implicit expression, which reads:

+ 2
+ + −κ B
 κu+ + (κ u ) (κ u + )3 
y = u +e e −1− κu − − 
 2 6 


- Show that in the wall region the production scales with v *4 /ν ; define the scaled production
as P + = Pν / v *4 . Derive that it can be expressed completely in terms of the gradient of the
mean velocity du + / dy + ; determine the expression for P + for large values of y+ (which
corresponds to the fully turbulent overlap region).

Hint: make also use of the assumption that the total shear stress is constant.

- Compute the variation of P + and plot this for the interval 0 ≤ y+ ≤ 50. In the calculations use
for the constants in the law of the wall the standard values κ = 0.41 and B = 5.0.

Hint: maintain in the derivations u+ as the independent variable, and make use of the relation
1
that: du + / dy + = +
dy / du +