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2 TWO-DIMENSIONAL FLOW
In this section we consider two-dimensional, incompressible, irrotational flow. The fluid velocity is in the x and y directions only, with components u and v which are functions of x and y.
2.1 Streamfunction and streamlines
For a two-dimensional flow the continuity equation (1.1) reduces to
0
u v x y
If we work with a variable ψ ( x , y ), defined by
u and v y x
∂ ψ ∂ψ = = − ∂ ∂
then the continuity equation is automatically satisfied, because
2 2 0
u v x y x y x y
∂ ∂ ∂ ψ ∂ψ
ψ is called the streamfunction. It is useful not only because it guarantees a solution of the continuity equation, but also because it helps us to visualise the flow.
A streamline is a curve in the flow field that is everywhere tangential to the fluid velocity. This definition means that the slope dy / dx of a streamline is given by
dy v / dx u x y
∂ ψ ∂ψ = = − ∂ ∂
After rearranging, this gives
dx dy 0 x y
∂ ψ ∂ψ
i.e. d ψ = 0 along a streamline.
Hence
(2.2)
Thus, if we know the function ψ ( x , y ), we can plot curves of constant ψ to determine the family of streamlines. It is customary to put arrows on the streamlines to indicate the direction of flow. This provides a useful way of visualising the flow. The conditions along a streamline and a solid body are the same - that is no flow through the streamline or the boundary. Thus any streamline in an inviscid flow can be considered as a solid boundary.
ψ = constant along a streamline
Figure 2 A family of streamlines.
Figure 2 shows a family of streamlines ψ = constant. It is clear that one of them is a closed streamline (i.e. encloses a finite area) and that is a particularly useful shape for a solid body. These streamlines describe flow around a wing at an angle of incidence of 10˚. Once we know the function ψ , we can evaluate the velocity field from u = ∂ ψ/∂ y , v = −∂ ψ/∂ x. The pressure then follows from Bernoulli's equation (1.7). The contours of
pressure coefficient ( ) 2
1 p 2
C = p − p ∞ / ρ U corresponding to the streamlines in Figure
2 are plotted below.
Figure 3: Contours of Cp
Question 2
A two-dimensional flow has a streamfunction ψ = Axy. Show that the flow is
irrotational and find the velocity potential. Plot sufficient streamlines to determine the flow pattern and also plot the equipotentials. Identify two practical situations where such a flow would occur.
Show that the isobars (curves of constant pressure) are circles with centres at the origin.
Question 3
Determine the streamfunction ψ ( r , θ) for a flow with velocity potential
φ = Ar 6 5^ cos 6 θ 5. Sketch the streamlines for 0 ≤ θ ≤ 5 π/3 and show that this can
describe flow onto a wedge of half-angle π/6. Find the variation of surface pressure with
distance from the apex of the wedge. Is the pressure gradient adverse or favourable?
2.2 Simple Flows
2.2.1 Uniform flow, u = ( U ,0)
Streamfunction:
u U and v 0 y x
integrate to give
ψ = Uy + C 1 where C 1 is an arbitraty constant (2.5)
Velocity potential:
u U and v 0 x y
integrate to give
φ = Ux + C 2 where C 2 is an arbitrary constant (2.6)
The integration constants C 1 and C 2 do not affect velocities or pressures in the flow and are usually set to zero. This arbitrary choice means that the streamline at y = 0 is chosen
as ψ = 0 and the equipotential at x = 0 is chosen as φ = 0.
y
x
φ = φ 1 φ^ =^ φ 2
U
2.2.2 Sources and sinks
x
y
r
Consider fluid flowing radially outward from the origin with a volume flow rate m (per unit z -length). Then, to satisfy conservation of mass,
(at radius 2 r^ )
m u r
π r
It is clearly a good idea to use polar co-ordinates here.
Streamfunction:
1 and 0 2 r
m u u r r r θ
integrate to give
m
= : the streamlines are radials (2.7)
Velocity potential:
1 and 0 r 2
m u u r r^ θ r
integrate to give
m
φ ln r
= : the equipotentials are circles. (2.8)
x
y
It is clear, then, that the flow has a stagnation point at S , a distance s upstream of the source where
m U
π s
= i.e. m = 2 π Us.
Adding the streamfunctions for a uniform flow and a source, we obtain
m
ψ Uy θ U y s θ
= + = + after substitution for m
The streamlines are given by the curves ψ = constant and are sketched below. In
particular, the streamline which goes through the stagnation point S (the position y = 0,
θ = π ) is
ψ = U ( y + s θ )= Us π
This streamline can represent a solid boundary, whose equation is y = s ( π − θ ). Note
that as x → ∞ , θ→ 0 and the half-height of the body is s π. Such a body is commonly
called a Rankine half-body.
π s
π s
ψ = π sU
S
The velocity at a position with polar co-ordinates ( r , θ) can be determined by
differentiating the streamfunction
s cos r
s sin u U v U r
The pressure field then follows from Bernoulli's equation (1.7). 2 2 2
s s p p U cos r (^) r
∞^ ρ θ
Misprint on handout
Contours of the nondimensional pressure coefficient, ( ) 2
p 2
C = p − p ∞ / ρ U , are
plotted below, together with the variation of Cp with angle θ along the body.
Cp = -0.
Cp = -0.
Cp = 0.
Cp = 0.
Figure 4: Contours of Cp for the Rankine half-body
There is a favourable pressure gradient along the body from the stagnation point ( Cp =1,
of course!) up to θ ≈ 63º ( Cp ≈ 0.59), after which there is a mild adverse pressure gradient
as x → ∞. However, this adverse pressure gradient is small and not sufficient to cause separation. We therefore conclude that this relatively simple flow provides useful information about the velocity and pressure field around the front part of a streamlined body, such as a bridge pier or strut placed in a uniform stream.
Figure 5: variation of Cp with angle for the Rankine half-body:
note θ = 0 corresponds to infinitely far downstream
Rankine Oval
Now let us consider the flow produced by the superposition of a uniform stream with a source and sink of equal strength.
a a
U
r
r 2
x
y
r 1
The streamfunction
m
ψ Uy θ θ
and the streamlines have the form shown below. In particular, the streamline ψ = 0 is
closed and can be used to represent a (fairly) streamlined cylindrical body - called a Rankine oval.
l (^) l
a (^) a
y
x
h
h
ψ = 0 + m − m
Question 5 Show that the half-length l and the half-height h of a Rankine oval are given by 1 (^2 ) 1 and 2
l m h m h tan a Ua a Ua a
= ^ + ^ = ^ −^ − ^
Hence show that m / Ua = 0.1 leads to a body of aspect ratio of about 20:1 and find the pressure coefficient on the body surface at the maximum thickness point. Note you do not need to solve these equations, just substitute appropriate values and show they satisfy the above equations approximately for a 20:1 aspect ratio.
Misprint on handout
Misprint on handout
So far we have derived just two solutions of Laplace's equation describing irrotational, two-dimensional flow (a uniform flow and a source/sink). Even these relatively simple flows have a number of practical applications. But we clearly need further solutions if we are to be able to investigate a wider range of flows. One way of generating new solutions
is to differentiate the ones we have already got. This works because, if φ 0 is a solution of
Laplace's equation, i.e. if
2
then differentiating this equation we find
2 0 0 x
The order of differentiation can be exchanged to show that
(^2 0 ) x
i.e. the derivative ∂ φ 0 /∂ x is also a solution of Laplace's equation. For example,
differentiating the velocity potential of a source, we find the new solution
( )
m mx ln r x (^) x y
Such a flow field is called a doublet and we will discuss it in more detail in Section 2.4.6.
An alternative and more versatile method for generating additional solutions of Laplace's equation is to use the powerful concept of complex potentials.
2.3 Complex potential
Suppose that F ( z ) is an analytic (i.e. a complex and differentiable) function of the complex variable z = x + iy. Then the chain rule of differentiation gives
1 F z dF dF x x dz dz
= = ×
while F z dF dF i y y dz dz
i.e.
F F i y x
If we expand the complex function F ( z ) into its real and imaginary parts by writing
F ( z ) = φ + i ψ, then the real and imaginary parts of (2.11) require
and y x y x
2.4 Simple flows by complex potentials
2.4.1 Uniform flow: F ( z ) = Uz****.
Then φ = Ux , ψ = Uy
u − iv = U i.e. u = U , v = 0
This is the uniform flow in the x -direction as in Section 2.2.1.
Question 6
What is the complex potential for uniform flow
of speed U at an angle α to the x -direction?
2.4.2 Stagnation flow F ( z ) =
Az^2.
Then 2 2
φ= A x − y ψ= Axy
u − iv = Az i.e. u = Ax , v = − Ay
This is the stagnation flow in Question 2.
It is possible to obtain a solution to this problem, which takes full account of viscosity. Although the method of solution is beyond the scope of this course, it is instructive to compare the viscous and inviscid solutions.
0.0^ 0.
Viscous (^) Inviscid
u / Ax
y ( A /ν)1/
Inviscid
Viscous
y ( A /ν)1/
v / ( A /ν)1/
Figure 6: Comparison between viscous and inviscid solutions for the velocity in a stagnation flow. Note that we can make the solutions agree if we shift the inviscid solution up – effectively shifting the wall up by a small amount.
y
x
U
First note that the region in which the viscous x -velocity differs from the inviscid value Ax is confined to a very small region close to the wall, specifically, to the region
y < 3( ν / A )^ 1 2, where ν is the kinematic viscosity. Similarly, the viscous y -velocity is
close to the inviscid value − Ay away from the wall. The most striking feature about the comparison between the y -velocities is that the best match between the viscous and inviscid flow solutions occurs if we shift up the inviscid solution by considering the wall
to be at a height 0.65( ν / A )1 2above the real surface. Note what a very small correction
this is. For example, in a flow of air with A = 100 s−^1 (this makes the speed in the x -direction 10 ms−^1 , 0.1m above the surface), this displacement distance
0.65( ν / A )^ 1 2= 0.25 mm.
The idea that viscous effects make the effective size of a body a bit bigger than its physical dimensions applies to all flows. It happens because within the boundary layer the tangential fluid velocity u is less than its inviscid value Us. The volume flow rate within a height h of the surface is therefore less than in an inviscid flow. By continuity this, in turn, has the effect of displacing streamlines outwards a small distance away from the surface.
h
d
U ∞
Us
u
Figure 7: The streamline initially at height d moves away from the surface as the boundary layer grows in order to keep the volume flux between the streamline and the wall constant.
The measure of the shift in the streamlines is the boundary-layer displacement thickness
δ*, which is defined as the distance by which the boundary would have to be displaced if
the flow were inviscid with the same volume flow at any section as the real flow.
Us y
u ( y )
Us
Real flow Hypothetical flow with displaced wall
i.e. the definition of the displacement thickness is that
φ = constant
ψ = constant
x
y
r
u θ
If we integrate u · dl around C 0 , a circle of radius r , we find that
0
2 , independent of C (^) 2 u dl r r r
∫.^ =^ ×^ = Γ
This is not surprising. In fact, we can prove that C
u dl
= Γ for any curve which
encloses the origin once. First note that, away from the origin, u = ∇ φ and hence the
vorticity ω = ∇ × u is zero. [Note that at the origin, φ does not exist because the angle θ
is not defined there and so ω is not necessarily zero there.] Recall that for any curve C
enclosing a surface S , Stokes theorem gives
c S
u dl = ω dS
Now ω is zero in the shaded region and so
0
C C
∫^ u dl^ =^ ∫ u dl = Γ
The quantity C
∫^ u dl. is an important parameter in flow fields. It is called the
circulation and will be discussed further in Section 2.6.
Returning to Stokes theorem, we have shown that for any contour C which encloses the origin:
· · C S
u dl = ω dS = Γ
C 0
dl
C 0
dl
u C
C
concentrated vorticity at the origin such that
ω. dS = Γ
We can directly interpret this result. The vorticity ω is zero except at the origin. At the
origin, it is so intense that, when integrated over a surface enclosing the origin, it is equal to Γ :
· S
ω dS = Γ
over any surface (no matter how small) enclosing the origin.
This concentration of vorticity is called a line vortex and Γ is the vortex strength.
Close to the vortex core, the flow velocities become large and viscous effects dominate in a real flow. However, the line vortex is useful to describe flows at reasonably large distances from the core.
Question 7 A crude approximation to a stationary hurricane is to model the flow field outside its core by the two-dimensional flow due to a superimposed line vortex and a line sink. Sketch the streamlines.
The wind velocity at a location A , 20 km away from the centre of a hurricane makes an angle of 65˚ with the radial towards the centre of the hurricane. The pressure at A is 550 Nm−^2 less than the pressure far away. What is the rate of volume influx of air per metre height (i.e. what is the strength of the apparent sink) and what is the circulation?
Hurricanes are far more complicated than this question implies. In particular, the energy to maintain a hurricane comes from the water vapour in moist warm air near the sea surface as it is condensed to form rain. A hurricane decays quickly once it is over land and the fluid mechanics are still not fully understood.
Line vortices move with the fluid
The development of the vorticity with time has a particularly simple form in an inviscid two-dimensional flow. To determine it, note that the two non-zero components of the momentum equation are
u u u 1 p u v
t x y ρ x
and
v v v 1 p u v
t x y ρ y
Pressure can be eliminated from these equations by subtracting ∂ / ∂ y of equation (2.14) from ∂ / ∂ x of equation (2.15):
v u v u u u u v u v t x y x x y y x y
∂ ^ ∂ ∂ ^ ∂ ^ ∂ ∂ ^ ∂ ^ ∂ ∂
−^ +^ +^ −^ +^ =
Question 8
The trailing vortex pattern left by an aircraft can, as a first approximation, be modelled by a pair of vortices of equal strength but opposite sign as shown below.
2 a
Show that the vortex pair moves downwards with a speed Γ / 4 π a.
In a frame of reference in which the vortices are at rest at (± a , 0), there is an oval-shaped streamline enclosing the vortices. Show that the equation of this streamline is
2 2 2 2
ln ( )
x a y x a x a y
Show that there are stagnation points on this streamline at x = 0, y = ± 3 a. Hence conclude that the pair of vortices carry with them an area of fluid, shaped in the form of an oval whose ratio of major to minor axes is about 1.21.
2.4.5 Corner and stagnation flows
Question 9 Sketch the streamlines for the complex potential F ( z ) = Az^ α^ for 3 2 1 3, 2, , and 2 3 2
α =. For which of these values of α does this potential represent a
physical flow? What is the variation of pressure along the surface θ = 0 , neglecting any
gravity effect?
We can use the results of Question 9 to determine the shape of the crest of a large amplitude wave on the sea surface. When a wave has the largest amplitude it can without breaking, its crest is observed to be in the form of a wedge, with the two faces
symmetrical. We will calculate the angle 2 β between the two faces of the crest.
sea surface
Schematic diagram of the crests of the largest amplitude breaking waves
In the frame of reference moving with the crest, the flow is steady and the sea surface is a streamline. This is the corner flow in Question 9, where you showed that at a distance r
from the corner u^2 + v^2^ = A^2^ α 2 r 2(^ α^ −1) where 2 β = π/ a. On the sea surface, the
pressure must be atmospheric. Using Bernoulli's equation shows that this requires
(^1) ( 2 2 ) constant 2
ρ u + v + ρ gz =
i.e. 2 2 2(^ 1)^ ( cos )
ρ A α r^ α^ −^ + ρ g h − r β independent of r.
For this to happen we require
2( α − 1) = 1 i.e. α= 3 / 2 and α^2 A^2 = 2 g cosβ
α = 3/2 implies crest angle 2 β = π/ α = 2 π/3, 120°
2.4.6 Doublet:
cos sin ( ) , , 2 2 2
F z z r r
[Note that this flow field can be obtained by differentiating that due to a source. See
equation (2.10).] The streamlines are curves, 2 2
y x + y
= constant i.e. they are circles
passing through the origin.
x
y
This source is called a doublet because it is produced by a source and sink pair of equal
strengths placed close together (see Appendix C). μ is the doublet source strength.
h^ β
r
z
Error on handout
