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Sketching Streamlines and Lines of Constant Potential Example 1

This document provides examples of sketching streamlines and lines of constant potential for different fluid flow velocity fields. Example 1 shows an irrotational flow that can be described by a scalar potential. Example 2 shows a rotational rigid body flow that can be described by a stream function. Example 3 shows a non-irrotational simple shear flow that also uses a stream function. Diagrams of the streamlines and lines of constant potential/function are provided for each example.

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Prince Israel Eboigbe
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92 views6 pages

Sketching Streamlines and Lines of Constant Potential Example 1

This document provides examples of sketching streamlines and lines of constant potential for different fluid flow velocity fields. Example 1 shows an irrotational flow that can be described by a scalar potential. Example 2 shows a rotational rigid body flow that can be described by a stream function. Example 3 shows a non-irrotational simple shear flow that also uses a stream function. Diagrams of the streamlines and lines of constant potential/function are provided for each example.

Uploaded by

Prince Israel Eboigbe
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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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
u = ∇φ = ix + iy
∂x ∂y
= ux ix + uy iy
= xix − yiy . (2)

• Thus, ∂φ/∂x = x and ∂φ/∂y = −y from which


x2 y2
φ= + A(y), and φ = − + B(x), (3)
2 2
whence A(y) = −y 2 /2 and B(x) = x2 /2 so that

φ = (x2 − y 2 ). (4)
2
• Is the flow solenoidal?
∂ux ∂uy
∇·u = +
∂x ∂y
=  −  = 0. (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
∂ψ ∂ψ
ux = x = , uy = −y = − , (6)
∂y ∂x
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. (8)

1
• 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.

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


∂ux ∂uy
∇·u= + = 0. (10)
∂x ∂y

• So, we can express the velocity in terms of a stream function, ψ:

∂ψ Ωy 2
ux = = −Ωy ⇒ ψ = − + A(x)
∂y 2
∂ψ Ωx2
uy = − = Ωx ⇒ ψ = − + B(y), (11)
∂x 2
so, A(x) = −Ωx2 /2 and B(y) = −Ωy 2 /2 and ψ is given by
1
ψ = − Ω(x2 + y 2 ). (12)
2

• 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

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 repre-
sented using a scalar potential.

• Since ∂ux /∂x = ∂uy /∂y = 0, the flow is solenoidal, and so can be repre-
sented using a stream function, ψ.

• The stream function can be found as follows:


∂ψ y2
ux = = γy ⇒ ψ = γ + B(x)
∂y 2
∂ψ
uy = − = 0 ⇒ ψ 6= ψ(x) ⇒ B(x) = 0. (13)
∂x

• Thus, the stream function is given by

y2
ψ=γ . (14)
2

• We can use Eq. (14) to sketch the streamlines for this case of simple shear
flow, as shown in Fig. 3.

5
y

Figure 3: Lines of constant ψ for Example 3.

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