Inviscid Incompressible Flows
3.3 Two-Dimensional Flows
m + 2
3πa3ρ
dV0
dt .
This explains why the factor23πa3ρ in (3.2.43) is termed the virtual mass of the sphere.
3.3 Two-Dimensional Flows
Let us consider inviscid flow of an incompressible fluid past a rigid body of cylindrical shape whose cross-section is sketched in Figure 3.16; the generator of the body’s surface is a straight line perpendicular to the sketch plane. We shall assume that the cylinder is infinite in the spanwise direction.6We shall further assume that the velocity vector of the oncoming flow is perpendicular to the body’s generator. In this case, none of the fluid-dynamic functions is expected to be dependent on the spanwise coordinate z measured along the body’s generator, namely
∂
∂z = 0. (3.3.1)
Also, the spanwise velocity component w is expected to be zero, so that the velocity vector
V= (u, v, 0). (3.3.2)
Fluid flows that satisfy conditions (3.3.1) and (3.3.2) are termed two-dimensional.
V∞
u∞ v∞
S
x y
Fig. 3.16: A two-dimensional flow.
6In practice it is sufficient for the spanwise length of a cylinder to be much larger than the characteristic dimension of its cross-section.
3.3. Two-Dimensional Flows 153 In a two-dimensional flow, the vorticity vector ω has just one component, ωz, perpendicular to the plane of fluid motion. Indeed, using (3.3.1) and (3.3.2) in (1.4.15), we have If the flow considered is irrotational, i.e.
ωz= ∂v
∂x−∂u
∂y = 0, (3.3.3)
then there exists a velocity potential ϕ. It is related to the velocity field through the integral (3.2.2). In a two-dimensional flow, this integral assumes the form
ϕ(t, r) =
When dealing with three-dimensional flow (see Figure 3.9 on page 140), we demon-strated that the integral for ϕ(t, r) does not depend on a choice of contour C connecting points r0 and r. We shall now look at this result more closely. We took two contours C and C′ connecting r0 and r, and denoted the results of the integration along C and C′ by ϕ(t, r) and ϕ′(t, r), respectively. Then the difference between ϕ(t, r) and ϕ′(t, r) was calculated using Stokes’s theorem:
ϕ(t, r)− ϕ′(t, r) = however, relies on the assumption that the region occupied by the moving fluid is singly connected, i.e., for any closed contour eC, there exists a surface σ, resting on eC, that lies entirely inside the flow field. Two-dimensional flows past cylindrical bodies clearly do not belong to this category: one path connecting points r0 and r might go around one side of the body and the other might go around the other side, and the result of the integration in (3.3.4) cannot be proven to be the same.
Corresponding to this, we can subdivide all closed contours into two classes. In the first class are contours like C1 in Figure 3.17 that may be reduced to a point by a continuous deformation (shrinking). When the position vector r in the integral (3.3.4) makes a circle around such a contour, the potential ϕ(t, r) returns to its initial value.
In the second category are contours like C2. When travelling around such a contour, the potential increases by a value
∆ϕ = I
C2
V· dr = Γ, (3.3.5)
which coincides with the velocity circulation in the flow past a rigid body.
C1
C2
Fig. 3.17: Classification of closed contours in two-dimensional flows.
It is easily shown that the circulation Γ is a universal constant for a given two-dimensional flow, independent of the choice of a contour embracing the body. Indeed, let us consider two contours C2and C2′ as shown in Figure 3.18, and write
Γ = I
C2
V· dr, Γ′ = I
C2′
V· dr.
Then, using Stokes’s theorem, we will have I
C2
V· dr − I
C2′
V· dr = ZZ
σ
ωzds, (3.3.6)
with σ now denoting the region between C2 and C2′. Since in this region equation (3.3.3) holds, we can conclude that the two integrals on the left-hand side of (3.3.6) coincide with one another.
Thus, when dealing with two-dimensional irrotational flows past solid bodies, one can still introduce the velocity potential ϕ with the help of the integral (3.3.4). How-ever, in the general case, ϕ will not be a single-valued function, although the non-uniqueness in ϕ is rather simple. Each time an observer goes around the body, the value of ϕ at a given point r increases by the circulation Γ. In what follows, Γ will be assumed positive when the integral in (3.3.5) is calculated in the counter-clockwise direction.
C2
C2′
σ
Fig. 3.18: Two closed contours C2 and C2′ and the surface σ exploited in equa-tion (3.3.6).
3.3. Two-Dimensional Flows 155 Of course, each single-valued branch of the velocity potential ϕ still satisfies Laplace’s equation (3.2.7), which is written in two dimensions as
∂2ϕ
∂x2 +∂2ϕ
∂y2 = 0.
3.3.1 Stream function
Similar to the velocity potential ϕ(t, r), which has been introduced based on the zero-vorticity equation (3.3.3), we can use the continuity equation (3.1.1b), written in two dimensions as
∂u
∂x +∂v
∂y = 0, (3.3.7)
to introduce another scalar function ψ(t, r), termed the stream function, such that
∂ψ
∂x =−v, ∂ψ
∂y = u. (3.3.8)
For this purpose, we consider a vector field
A= (Ax, Ay, Az) such that
Ax=−v, Ay= u, Az= 0, (3.3.9)
Since neither u nor v depends on z,
curl A =
i j k
∂
∂x
∂
∂y 0
−v u 0 = k
∂u
∂x +∂v
∂y
= 0.
Hence, the vector field A is irrotational, and there exists a scalar function ψ(t, r) that, similar to (3.3.4), may be introduced using the integral
ψ(t, r) = Zr
r0
A· dr = Zr
r0
(−v dx + u dy). (3.3.10)
Variation of the upper limit of integration in (3.3.10) results in
∇ψ = A = (−v, u). (3.3.11)
Writing (3.3.11) in the coordinate-decomposition form
∂ψ
∂x = Ax=−v, ∂ψ
∂y = Ay= u proves the validity of equations (3.3.8).
The existence of the stream function relies solely on the continuity equation or, more precisely, on the form (3.3.7) it assumes in two-dimensional incompressible flows.
It does not matter if the flow is steady or unsteady, inviscid or viscous, irrotational or with non-zero vorticity, one can still introduce the stream function. In the case of irrotational flow, the stream function is easily seen to satisfy Laplace’s equation
∇2ψ = ∂2ψ
∂x2 +∂2ψ
∂y2 = 0, which is obtained by substituting (3.3.8) into (3.3.3).
In order to clarify the physical content of the stream function, let us consider two points M and M′ in the (x, y)-plane (see Figure 3.19) and calculate the fluid volume flux across a curve L joining M and M′ or, more precisely, the flux across an open surface swept out by translating curve L through unit distance in the z-direction. Reckoning the flux positive when it is in the direction shown by the arrow in Figure 3.19, we write
Q =− Z
L
V· ndl = − Z
L
(unx+ vny) dl. (3.3.12)
Here n = (nx, ny) is a unit vector normal toL and dl is the length of a small element of L. Notice that, unlike for the mass flux (1.6.3), the volume flux integral (3.3.12) does not involve the fluid density ρ.
Let us now introduce a unit vector τ = (τx, τy) tangent toL. It may be easily seen that
nx=−τy, ny= τx.
Hence, using (3.3.8), we can express the integral in (3.3.12) as Q =
Z
L
∂ψ
∂xτx+∂ψ
∂yτy
dl =
Z
L
∇ψ · τ dl =
Z
L
∇ψ · dr,
with dr being an increment of the position vector r alongL. We see that the volume flux across any curve joining two points in the flow field equals the difference between the values of ψ at these points:
Q = ψ(M′)− ψ(M). (3.3.13)
M
M′ n
τ Q
x y
L
Fig. 3.19: Calculation of volume flux Q through curveL between points M and M′.
3.3. Two-Dimensional Flows 157 We can now prove the following theorem.
Theorem 3.4 The stream function is constant along any streamline, and any line defined by the equation
ψ = const (3.3.14)
is a streamline.
Proof If line L in Figure 3.19 is a streamline, then, according to Definition 1.2 on page 29, the velocity vector is tangent toL, and therefore there is no fluid flux across L. Hence, Q = 0, and, using (3.3.13), we can conclude that for any two points M and M′ which belong to the same streamline, ψ(M′) = ψ(M ). This proves the first part of the theorem.
To prove that any line defined by equation (3.3.14) is a streamline, we note that for a line represented by equation (3.3.14), the unit vector normal to this line, n, may be calculated as
n= ∇ψ
|∇ψ| = 1
|∇ψ|
∂ψ
∂x,∂ψ
∂y
.
Let us consider the scalar product of the velocity vector V and the normal vector n:
V· n = 1
|∇ψ|
u∂ψ
∂x + v∂ψ
∂y
. (3.3.15)
Using (3.3.8) in (3.3.15), we see that V· n = 0, which proves that the velocity vector Vis perpendicular to n, and therefore is tangent to the line considered. ✷ Exercises 8
1. Consider a building in the form of a hemisphere that has an open ventilation window at the top of the roof. The entrance door, which is situated at point O on the ground level, is initially sealed and does not allow air to penetrate through it.
The building is exposed to wind, which is directed such that the entrance door finds itself at the front stagnation point; see Figure 3.20.
Fig. 3.20: Problem layout.
Show that the force that has to be applied to the door to open it outwards is F = 98ρV∞2A,
where ρ is the air density, V∞the wind speed, and A the area of the door surface.
Suggestion: You may assume that the ventilation window is small compared with the radius of the dome, and so is the door.
V∞
O
2. For a two-dimensional flow, the Euler equations are written as
∂u
∂t + u∂u
∂x+ v∂u
∂y = fx−1 ρ
∂p
∂x,
∂v
∂t + u∂v
∂x+ v∂v
∂y = fy−1 ρ
∂p
∂y,
∂u
∂x +∂v
∂y = 0.
Performing cross-differentiation of the x- and y-momentum equations, deduce a condition that should be imposed upon the body force f to ensure that the vorticity remains constant for each fluid particle as it travels through the flow field.
3. Prove that in a two-dimensional flow past a rigid cylindrical body the stream function ψ is a single-valued function, i.e., after making a circle around the closed contour C2 of Figure 3.17, ψ returns to its original value.
4. Assume that a flow of an incompressible fluid is steady and may be treated as inviscid. Assume further that the body force f is potential, i.e. there exists a scalar function U such that f =∇U. Show that under these conditions the gradient of the function H in the Bernoulli function (3.1.6) is related to the vorticity ω by the equation
∇H = V × ω. (3.3.16)
Suggestion: Use the Gromeko–Lamb form (3.1.3) of the Euler equations.
5. The streamline pattern in a two-dimensional flow is defined by equation (3.3.14), with different values of the stream function ψ corresponding to different stream-lines. Therefore, function H on the right-hand side of the Bernoulli equation (3.1.6) may be treated as a function of ψ. Starting with equation (3.3.16), prove
that dH
dψ =−ωz.
Hence, deduce that H is constant in an irrotational flow.
6. Suppose that a two-dimensional flow is irrotational and has velocity potential ϕ given by
ϕ = ax(x2− 3y2),
where a is a positive constant. Find the fluid volume flux across a curve connecting points M = (0, 0) and M′= (1, 1) in the (x, y)-plane.
7. The continuity equation for an incompressible fluid is written in spherical polar coordinates as
∂Vr
∂r +1 r
∂Vϑ
∂ϑ + 1 r sin ϑ
∂Vφ
∂φ +2Vr
r + Vϑ
r tan ϑ = 0. (3.3.17) Assume that the flow considered is axisymmetric with respect to the x-axis (see Figure 1.31 on page 84), and show that in this case equation (3.3.17) assumes the
3.3. Two-Dimensional Flows 159
form ∂
∂r r2sin ϑ Vr) + ∂
∂ϑ r sin ϑ Vϑ) = 0.
Based on this equation, introduce a scalar function ψ(r, ϑ), known as the Stokes stream function, such that
∂ψ
∂r =−r sin ϑ Vϑ, ∂ψ
∂ϑ = r2sin ϑ Vr, (3.3.18) and prove that the equation
ψ = const defines the streamlines in the flow.
Hint: You may use without proof the fact that gradient of ψ is written in spherical polar coordinates as
∇ψ = ∂ψ
∂rer+ 1 r sin ϑ
∂ψ
∂φeφ+1 r
∂ψ
∂ϑeϑ,
with er, eϑ, and eφ denoting the unit vectors in the radial, azimuthal, and merid-ional directions, respectively.
8. Combine (3.3.18) with the solution (3.2.30) for Vrand Vϑin the flow past a sphere, and show that in this flow
ψ = V∞
2
r2−a3
r
sin2ϑ.
9. Consider the steady inviscid flow of an incompressible fluid past a swept wing shown in Figure 3.21. The wing has an infinite span, and its profile at each cross-section OO′ remains the same, i.e. is independent of the spanwise coordinate z.
Write the Euler equations (3.1.1) and the impermeability (3.2.10) and free-stream (3.2.11) conditions in coordinate-decomposition form using Cartesian coordinates x, y, z with z-axis parallel to the wing generatrix.
V∞
χ
z O
O′
Fig. 3.21: Swept wing.
Prove that the flow in the (x, y)-plane perpendicular to the generatrix may be treated as two-dimensional.