(1) 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:

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:
w+u = u eese 22


Calculate in addition for this case the cross–flow angle )( = )( e  .

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, w, as function of the
ratio we/ue. Determine at which we/ue the maximum value of 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
0.2 0.22661 0.11405
0.3 0.32524 0.17091
0.4 0.41446 0.22749
0.5 0.49465 0.28356
0.6 0.56628 0.33889
0.7 0.62986 0.39319
0.8 0.68594 0.44616
0.9 0.73508 0.49751
1.0 0.77786 0.54692

1.1 0.81487 0.59411
1.2 0.84667 0.63883
1.3 0.87381 0.68085
1.4 0.89681 0.72000
1.5 0.91617 0.75616
1.6 0.93235 0.78924
1.7 0.94577 0.81925
1.8 0.95683 0.84619
1.9 0.96588 0.87017
2.0 0.97322 0.89130

2.5 0.99285 0.96058
3.0 0.99842 0.98851
3.5 0.99972 0.99733
4.0 0.99996 0.99951
4.5 1.00000 0.99993

(0)F” = 1.23259 (0)g = 0.57047