Sketching streamlines and lines of constant
potential
Example 1
Sketch the streamlines and lines of constant potential for a flow described by the
velocity field u = ux ix + uy iy = xix − yiy . This is a 2D velocity, so that uz = 0.
• Is the flow irrotational? The vorticity is given by (see equation sheet for
the formula for curl)
∂uz ∂uy
∂ux ∂uz
∂uy ∂ux
ω =∇∧u =
−
ix +
−
iy +
−
iz
∂y
∂z
∂z
∂x
∂x
∂y
= 0,
(1)
whence the flow is irrotational.
• Hence, u can be written in terms of a scalar potential φ (see p. 2 of Lecture
1):
∂φ
∂φ
ix +
iy
∂x
∂y
= ux ix + uy iy
= xix − yiy .
u = ∇φ =
(2)
• Thus, ∂φ/∂x = x and ∂φ/∂y = −y from which
y2
x2
+ A(y), and φ = − + B(x),
2
2
whence A(y) = −y 2 /2 and B(x) = x2 /2 so that
φ = (x2 − y 2 ).
2
φ=
(3)
(4)
• Is the flow solenoidal?
∂ux ∂uy
+
∂x
∂y
= − = 0.
∇·u =
(5)
Thus the flow is solenoidal.
• Since ∇ · u = 0, ∇ · ∇ ∧ Ψ with Ψ = ψiz (see p. 2 of Lecture 1).
• So the velocity field is given by
ux = x =
∂ψ
∂ψ
, uy = −y = − ,
∂y
∂x
(6)
which upon integration of the first and second equations, respectively gives
ψ = xy + A(x), and ψ = xy + B(y),
(7)
hence, A = B = 0 and the stream function is given by
ψ = xy.
1
(8)
• It is possible to use Eqs. (4) and (8) to sketch lines of constant φ and ψ,
respectively. These lines are sketched in Fig. 1 and the flow is said to be
extensional. Note that these lines are orthogonal.
y
x
Figure 1: Lines of constant φ and ψ for Example 1.
2
Example 2
Sketch the streamlines for a flow described by the velocity field u = −Ωyix +Ωxiy .
• Using the same methods as in Example 1, we can determine that
ω = 2Ωiz .
(9)
Thus, the flow is not irrotational since the vorticity, ω 6= 0.
• The flow is solenoidal since
∇·u=
∂ux ∂uy
+
= 0.
∂x
∂y
(10)
• So, we can express the velocity in terms of a stream function, ψ:
∂ψ
Ωy 2
= −Ωy ⇒ ψ = −
+ A(x)
∂y
2
∂ψ
Ωx2
= −
= Ωx ⇒ ψ = −
+ B(y),
∂x
2
ux =
uy
(11)
so, A(x) = −Ωx2 /2 and B(y) = −Ωy 2 /2 and ψ is given by
1
ψ = − Ω(x2 + y 2 ).
2
(12)
• It is possible to use Eq. (12) to sketch the streamlines associated with the
flow, which in this case corresponds to rigid body rotation, as shown in Fig.
2.
3
y
x
Figure 2: Lines of constant ψ for Example 2.
4
Example 3
Sketch the streamlines for a flow described by the velocity field u = γyix .
• Using the same procedures as in the previous two examples, ω = −γiz .
Thus, the flow is not irrotational and the velocity field can not be represented using a scalar potential.
• Since ∂ux /∂x = ∂uy /∂y = 0, the flow is solenoidal, and so can be represented using a stream function, ψ.
• The stream function can be found as follows:
ux =
uy = −
y2
∂ψ
= γy ⇒ ψ = γ + B(x)
∂y
2
∂ψ
= 0 ⇒ ψ 6= ψ(x) ⇒ B(x) = 0.
∂x
(13)
• Thus, the stream function is given by
ψ=γ
y2
.
2
(14)
• We can use Eq. (14) to sketch the streamlines for this case of simple shear
flow, as shown in Fig. 3.
5
y
x
Figure 3: Lines of constant ψ for Example 3.
6