(1) 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 me xxu ~)(, and reads: 2”’ ” (1 ‘ )f + f f + f = 0
where 1
2
m+
mβ = and where the following scaling has been used: ‘
e
u = f ( )u
1
2
eu xy m+η = x
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 wf(0) = f; where: 2
1
w e
w
e
v u xf = u m
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. wf and ”wf).
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: 2 22w wf “ = f
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 wf and ”wf.
d. The above shows that for certain values of wf two different solutions are possible. For one
of them ”wf is positive and for the other negative (the latter hence shows flow reversal near
the wall).
Determine both values of ”wf when wf = 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!
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 me xxu ~)(, and reads: 2”’ ” (1 ‘ )f + f f + f = 0
where 1
2
m+
mβ = and where the following scaling has been used: ‘
e
u = f ( )u
1
2
eu xy m+η = x
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 wf(0) = f; where: 2
1
w e
w
e
v u xf = u m
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. wf and ”wf).
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: 2 22w wf “ = f
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 wf and ”wf.
d. The above shows that for certain values of wf two different solutions are possible. For one
of them ”wf is positive and for the other negative (the latter hence shows flow reversal near
the wall).
Determine both values of ”wf when wf = 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!
