Deformation and voracity
Each of the expressions given below represents a two–dimensional in compressible flow field, with
velocity components u and v in resp. the x– and y–direction of a Cartesian coordinate frame:
(1) u = Ax v = –Ay
(2) u = Ay v = Ax
Here A is a constant, with dimension of [velocity/length]. Assume that A has a positive value.
a. Show that both velocity fields are irrotational and satisfy the (incompressible) continuity
equation. Determine for each flow field the shape of the streamlines en sketch the flow
pattern (indicate the flow direction).
b. Determine for each of these flow fields how a square fluid element is transported and deformed by the flow over a small time interval Δt. Make a clear sketch for each case. Determine also for each case which (viscous) normal and tangential stresses work on the sides of the fluid element. Indicate these stresses in the sketch, noting the proper direction of the stresses and their relative magnitude.
c. Show that the flow field of (2) is identical to that of (1) and can be obtained by a rotation of the coordinate frame over an angle of 45o.
Compare the stress situations that have been derived for each case above, and use this to determine the consequence which the orientation of the fluid element with respect to the flow field has on the stress situation.
d. As the velocity field is irrotational (potential flow), it satisfies the flow equations for both inviscid and viscous flow, and the pressure can be obtained from Bernoulli’s relation (see
White, section 2–10).
– Compute the pressure field p(x,y) and determine the shape of the isobars in the flow.
– Compute explicitly the components of the gradient of the viscous stress tensor ij j / x;
use this result to explain how the “inviscid potential flow” can also satisfy the viscous flow
equations, even though the viscous stresses are not zero.
(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
