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Unformatted text preview: Ch. VIII Potential Flow and Computational
Fluid Dynamics
Review of VelocityPotential Concepts
This chapter presents examples of problems and their solution for which the
assumption of potential flow is appropriate.
For low speed flows where viscous effects are neglected, the flow is irrotational
and V = ∇φ ∇×V=0 u= ∂φ
∂x v= ∂φ
∂y w= ∂φ
∂z The continuity equation , ∇ ⋅ V = 0 , now reduces to ∂2 φ ∂ 2φ ∂ 2φ
∇ V=
+
+
=0
∂ x2 ∂ y2 ∂ z2
2 The momentum equation reduces to Bernoulli’s equation: ∂φ P 1 2
+ + V + g z = const
∂t ρ 2
Review of Stream Function Concepts
For plane incompressible flow in xy coordinates a stream function exists such that u= ∂Ψ
∂y and v= − ∂Ψ
∂x The condition of irrotationality reduces to Lapace’s equation for Ψ and ∂2 Ψ ∂ 2 Ψ
+
=0
∂ x2 ∂ y2  v v VIII  1  eText Main Menu  Textbook Table of Contents  Study Guide Elementary PlaneFlow Solutions
Three useful planeflow solutions that are very useful in developing more complex
solutions are:
Uniform stream, iU, in the x direction: Ψ = Uy φ = Ux Line source or sink: Ψ = mθ φ = m ln r Line vortex: Ψ = − K ln r φ = Kθ In these expressions, the source strength, ‘m’ and vortex strength, ‘ K ‘, have the
2
dimensions of velocity times length, or [L /t].
If the uniform stream is written in plane polar coordinates, we have Ψ = U r sin θ Uniform stream, iU: φ = U r c os θ For a uniform stream moving at an angle, a , relative to the xaxis, we can write u = Uc os α = ∂Ψ ∂φ
=
∂y ∂x v = Usin α = − ∂ Ψ ∂φ
=
∂x ∂y After integration, we obtain the following expressions for the stream function and
velocity potential: Ψ = U (y cos α − x sin α ) φ = U (x cos α + y sin α ) Circulation
The concept of fluid circulation is very useful in the analysis of certain potential
flows, in particular those useful in aerodynamics analyses. Consider Figure 8.3
shown below:  v v VIII  2  eText Main Menu  Textbook Table of Contents  Study Guide We define the circulation, Γ , as
the counterclockwise line integral
of the arc length, ds times the
velocity component tangent to the
closed curve, C, e.g. Γ = ∫ V cos α d s = ∫ V ⋅ ds
c c Γ = ∫ (u dx + v dy + wdz )
c For most flows, this line integral around a closed path, starting and stoping at the
same point, yields Γ = 0. However,
for a vortex flow for which φ the integral yields Γ = 2πK = Kθ An equivalent calculation can by made by defining a circular path of radius r
around the vortex center to yield
2π Γ = ∫ vθ d s = ∫ 0 c K
r d φ = 2π K
r Superposition of Potential Flows
Due to the mathematical character of the equations governing potential flows, the
principle of superposition can be used to determine the solution of the flow which
results from combining two individual potential flow solutions.
Several classic examples of this are presented as follows:  v v VIII  3  eText Main Menu  Textbook Table of Contents  Study Guide 1. Source m at ( a,0) added to an equal sink at (+a, 0). 2a y
ψ = − m tan 2
x + y2 − a2
−1 1
(x + a) + y 2
φ = m ln
2
(x − a)2 + y 2
2 The streamlines and potential lines are two families of orthogonal circles (Fig.
4.13).
2. Sink m plus a vortex K, both at the origin. ψ = mθ − K ln r φ = m ln r + K θ The streamlines are logarithmic spirals swirling into the origin (Fig. 4.14). They
resemble a tornado or a bathtub vortex.
3. Uniform steam i U∞ plus a source m at the origin (Fig. 4.15), the Ranking half
body. If the origin contains a source, a plane halfbody is formed with its nose
to the left as shown below. If the origin contains a sink, m < 0, the halfbody
nose is to the right.. For both cases, the stagnation point is at a position
a = m / U∞ away from the origin.  v v VIII  4  eText Main Menu  Textbook Table of Contents  Study Guide Example 8.1
3 An offshore power plant cooling water intake has a flow rate of 1500 ft /s in water
30 ft deep as in Fig. E8.1. If the tidal velocity approaching the intake is 0.7 ft/s,
(a) how far downstream does the intake effect extend and (b) how much width of
tidal flow in entrained into the intake?
The sink strength is related to the volume flow,
Q and water depth by
3 1500 ft / s
m=
=
= 7.96 ft 2 / s
2π b
2 π 30 ft
Q The lengths a and L are given by
2 m 7.96 ft / s
a=
=
= 11.4 ft
U∞
0.7 ft / s
L = 2 π a = 2 π 11.4 ft = 71 ft
Flow Past a Vortex
Consider a uniform stream, U∞ flowing in the x direction past a vortex of strength
K with the center at the origin. By superposition the combined stream function is ψ = ψ stream + ψ vortex = U∞ r sin θ − K ln r
The velocity components of this flow are given by vr = 1 ∂ψ
= U∞ cos θ
r ∂θ vθ = − ∂ψ
K
= − U∞ sin θ +
∂r
r Setting v r and vθ = 0, we find the stagnation point at θ = 90û, r = a = K/ U∞
or (x,y) = (0,a).  v v VIII  5  eText Main Menu  Textbook Table of Contents  Study Guide An Infinite Row of Vortices
Consider an infinite row of vortices of equal strength K and equal spacing a. A
single vortex, i , has a stream function given by Fig. 8.7 Superposition of vortices
∞ ψ i = − K ∑ ln ri
i =1 This infinite sum can also be expressed as 1
1
2π y
2π x
ψ = − K ln cosh
− cosh
2 a
a 2 The resulting left and right flow above and below the row of vortices is given by u= ∂ψ
∂y =±
y >a πK
a  v v VIII  6  eText Main Menu  Textbook Table of Contents  Study Guide Plane flow past ClosedBody Shapes
Various types of external flows over a closedbody can be constructed by
superimposing a uniform stream with sources, sinks, and vortices.
Key Point: The body shape will be closed only if the net source of the outflow
equals the net sink inflow.
Two examples of this are presented below.
The Rankine Oval
A Rankine Oval is a cylindrical
shape which is long compared to
its height. It is formed by a
sourcesink pair aligned parallel to
a uniform stream.
The individual flows used to
produce the final result and the
combined flow field are shown in
Fig. 8.9. The combined stream
function is given by ψ = U∞ y − m tan −1 2a y
x2 + y2 − a2 or ψ = U∞ r sin θ + m(θ1 − θ 2 ) Fig. 8.9 The Rankine Oval The oval shaped closed body is the streamline, ψ = 0 . Stagnation points occur at
the front and rear of the oval, x = ± L, y = 0 . Points of maximum velocity and
minimum pressure occur at the shoulders, x = 0, y = ± h . Key geometric and
flow parameters of the Rankine Oval can be expressed as follows:  v v VIII  7  eText Main Menu  Textbook Table of Contents  Study Guide 2m L
= 1 + a U∞ a h/a
h
= cot
a
2 m / (U∞ a) 1/ 2 2 m / (U∞ a)
umax
=1 +
2
2
U∞
1+ h /a
As the value of the parameter m / (U ∞ a) is increased from zero, the oval shape
increases in size and transforms from a flat plate to a circular cylinder at the
limiting case of m / (U ∞ a) = ∞ .
Specific values of these parameters are presented in Table 8.1 for four different
values of the dimensionless vortex strength, K / (U ∞ a ).
Table 8.1 RankineOval Parameters m / (U∞ a) h/a L/h umax / U∞ 0.0
0.31
0.263
1.307
4.435
14.130
∞ 0.0
0.01
0.1
1.0
10.0
10.0
∞ L/a
1.0
1.10
1.095
1.732
4.458
14.177
∞ ∞
32.79
4.169
1.326
1.033
1.003
1.000 1.0
1.020
1.187
1.739
1.968
1.997
2.000 Flow Past a Circular Cylinder with Circulation
It is seen from Table 8.1 that as source strength m becomes large, the Rankine
Oval becomes a large circle, much greater in diameter than the sourcesink spacing
2a. Viewed, from the scale of the cylinder, this is equivalent to a uniform stream
plus a doublet. To add circulation, without changing the shape of the cylinder, we
place a vortex at the doublet center. For these conditions the stream function is
given by  v v VIII  8  eText Main Menu  Textbook Table of Contents  Study Guide r
a2 ψ = U∞ sin θ r − − K ln r
a
Typical resulting flows are shown in Fig. 8.10 for increasing values of nondimensional vortex strength K / U∞ a . Fig. 8.10 Flow past a cylinder with circulation for values of
K / U∞ a of (a) 0, (b) 1.0, (c) 2.0, and (d) 3.0
Again the streamline ψ = 0 is corresponds to the circle r = a. As the counterclockwise circulation Γ = 2 π K increases, velocities below the cylinder increase
and velocities above the cylinder decrease. In polar coordinates, the velocity
components are given by a2 1 ∂ψ
vr =
= U∞ cos θ 1 − 2 r
r ∂θ ∂ψ a2 K
vθ = −
= − U∞ sin θ 1 + 2 + ∂r
r r
For small K, two stagnation points appear on the surface at angles θ s or for which  v v VIII  9  eText Main Menu  Textbook Table of Contents  Study Guide sin θ s = K
2 U∞ a Thus for K = 0, θ s = 0 and 180o. For K / U∞ a = 1, θ s = 30 and 150o . Figure
8.10c is the limiting case for which with K / U∞ a = 2, θ s = 90o and the two
stagnation points meet at the top of the cylinder. The KuttaJoukowski Lift Theorem
The development in the text shows that from inviscid flow theory,
The lift per unit depth of any cylinder of any shape immersed in a
uniform stream equals to ρ U ∞ Γ where Γ is the total net circulation
contained within the body shape. The direction of the lift is 90o from
the stream direction, rotating opposite to the circulation.
This is the well known KuttaJoukowski lift theorem.  v v VIII  10  eText Main Menu  Textbook Table of Contents  Study Guide ...
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 Spring '08
 Sakar
 Fluid Dynamics

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