chapter. 6 differential analysis of fluid flow potential flow

Chapter.6 Differential Analysis of Fluid Flow

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Chapter. 6Differential Analysis of Fluid Flow

Potential Flow

이동근

M관 517호[email protected]

Reference:1. Fluid Mechanics, Frank M White, 6th Edition, 2008, McGraw Hill2. Munson et al., Fundamentals of Fluid Mechanics, 5th Edition, 2006, John Wiley &Sons, Inc


Inviscid Flow: Euler’s equations of motion

• Flow fields in which the shearing stresses are zero are said to be inviscid, nonviscous, or frictionless. For fluids in which there are no shearing stresses the normal stress at a point is independent of direction:

• For an inviscid flow in which all the shearing stresses are zero, and the normal stresses are replaced by –p, the Navier-Stokes Equations reduce to Euler’s equations

• In Cartesian coordinates:

−𝑝 = 𝜎𝑥𝑥 = 𝜎𝑦𝑦 = 𝜎𝑧𝑧

𝜌𝒈 − 𝜵𝑝 − 0 = 𝜌𝜕𝑽

𝜕𝑡+ 𝑉 · 𝛻 𝑽

𝜌𝑔𝑥 −𝜕𝑝

𝜕𝑥= 𝜌

𝜕𝑢

𝜕𝑡+ 𝑢

𝜕𝑢

𝜕𝑥+ 𝑣

𝜕𝑢

𝜕𝑦+ 𝑤

𝜕𝑢

𝜕𝑧

𝜌𝑔𝑦 −𝜕𝑝

𝜕𝑦= 𝜌

𝜕𝑣

𝜕𝑡+ 𝑢

𝜕𝑣

𝜕𝑥+ 𝑣

𝜕𝑣

𝜕𝑦+ 𝑤

𝜕𝑣

𝜕𝑧

𝜌𝑔𝑧 −𝜕𝑝

𝜕𝑧= 𝜌

𝜕𝑤

𝜕𝑡+ 𝑢

𝜕𝑤

𝜕𝑥+ 𝑣

𝜕𝑤

𝜕𝑦+ 𝑤

𝜕𝑤

𝜕𝑧


The Bernoulli equation derived from Euler’s equations

• Steady, incompressible, inviscid, along a streamline


𝑝

𝜌+

𝑉2

2+ 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡

𝑝1

𝛾+

𝑉12

2𝑔+ 𝑧1 =

𝑝2

𝛾+

𝑉22

2𝑔+ 𝑧2

𝜌𝒈 − 𝜵𝑝 = 𝜌𝜕𝑽

𝜕𝑡+ 𝑉 · 𝛻 𝑽 = 𝜌 𝑉 · 𝛻 𝑽

𝑽 · 𝛻 𝑽 =𝟏

2𝛻 𝑽 · 𝑽 - 𝑽 x (𝛻x 𝑽)

𝜌𝒈 − 𝜵𝑝 · 𝑑𝒓 =𝟏

2𝜌𝛻 𝑽 · 𝑽 · 𝑑𝒓 − 𝜌𝑽 x (𝛻x 𝑽) · 𝑑𝒓

dot product

· 𝑑𝒓

Vector identity

∵ 𝑉 ⫽ 𝑑𝒓 𝑎𝑙𝑜𝑛𝑔 𝑠𝑡𝑟𝑒𝑎𝑚𝑙𝑖𝑛𝑒

∵ 𝛻𝜙 · 𝑑𝒓 = d𝜙− 𝑔𝑑𝑧 −𝑑𝑝

𝜌− 𝑑

1

2𝑉2 = 0

& incompressible


The Bernoulli equation derived from Euler’s equations

• The Bernoulli equation can also be derived, starting from Euler’s equations. For inviscid, incompressible fluids, we end up with the same equation

• It is often convenient to write the Bernoulli equation between two points (1) and (2) along a streamline and to express the equation in the “head” form by dividing each term by 𝑔 so that

• The Bernoulli equation is restricted to the following:

- Inviscid flow- Steady flow- Incompressible flow- Flow along a streamline


𝑝

𝜌+

𝑉2

2+ 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡

𝑝1

𝛾+

𝑉12

2𝑔+ 𝑧1 =

𝑝2

𝛾+

𝑉22

2𝑔+ 𝑧2


The Bernoulli equation for steady, incompressible, irrotational flow


𝜌𝒈 − 𝜵𝑝 = 𝜌𝜕𝑽

𝜕𝑡+ 𝑉 · 𝛻 𝑽 = 𝜌 𝑉 · 𝛻 𝑽

𝑽 · 𝛻 𝑽 =𝟏

2𝛻 𝑽 · 𝑽 - 𝑽 x (𝛻x 𝑽)

𝜌𝒈 − 𝜵𝑝 · 𝑑𝒓 =𝟏

2𝜌𝛻 𝑽 · 𝑽 · 𝑑𝒓 − 𝜌𝑽 x (𝛻x 𝑽) · 𝑑𝒓

dot product

· 𝑑𝒓

Vector identity

∵ 𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙

∴ 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑝𝑎𝑡ℎ𝑠− 𝑔𝑑𝑧 −𝑑𝑝

𝜌− 𝑑

1

2𝑉2 = 0


The Irrotational Flow and corresponding Bernoulli equation

• If we make one additional assumption-that the flow is irrotational the analysis of inviscid flow problems is further simplified. The Bernoulli equation has exactly the same form at that for inviscid flows:

• But it can now be applied between any two points in the flow field, not limited to applications along a streamline.


𝛻 × 𝑉 = 0

𝑝1

𝛾+

𝑉12

2𝑔+ 𝑧1 =

𝑝2

𝛾+

𝑉22

2𝑔+ 𝑧2

Various regions of flow around bodies

Various regions of flow through channels


The Velocity Potential

• For an irrotational flow:

so we have

• It follows that in this case the velocity components can be expressed in terms of a scalar function 𝜙 𝑥, 𝑦, 𝑧, 𝑡 , called velocity potential, as

In vector form


𝛻 × 𝑽 =𝜕𝑤

𝜕𝑦−

𝜕𝑣

𝜕𝑧 𝒊 +

𝜕𝑢

𝜕𝑧−

𝜕𝑤

𝜕𝑥 𝒋 +

𝜕𝑣

𝜕𝑥−

𝜕𝑢

𝜕𝑦 𝒌 = 0

𝜕𝑤

𝜕𝑦=

𝜕𝑣

𝜕𝑧,

𝜕𝑢

𝜕𝑧=

𝜕𝑤

𝜕𝑥,

𝜕𝑣

𝜕𝑥=

𝜕𝑢

𝜕𝑦

𝑢 =𝜕𝜙

𝜕𝑥, 𝑣 =

𝜕𝜙

𝜕𝑦, 𝑤 =

𝜕𝜙

𝜕𝑧

𝑽 = 𝛻𝜙


• The velocity potential is a consequence of the irrotationality of the flow field, whereas the stream function is a consequence of conservation of mass. It is to be noted, however, that the velocity potential can be defined for a general three-dimensional flow, whereas the stream function is restricted to two-dimensional flow.

• And therefore for incompressible, irrotational flow, it follows that

• The velocity potential satisfies the Laplace equation.In Cartesian coordinates:

In cylindrical coordinates:


𝛻 ∙ 𝑽 = 0

𝛻2𝜙 = 0

𝜕2𝜙

𝜕𝑥2+

𝜕2𝜙

𝜕𝑦2+

𝜕2𝜙

𝜕𝑧2= 0

1

𝑟

𝜕

𝜕𝑟𝑟

𝜕𝜙

𝜕𝑟+

1

𝑟2

𝜕2𝜙

𝜕𝜃2+

𝜕2𝜙

𝜕𝑧2= 0

Satisfy Laplace egn !!Linear & homogeneous⤍ superposition allowed


6.5 Some Basic, Plane Potential Flows

Laplace egn ; a linear & homogeneous partial differential equation

For simplicity, only plane (2-D) flows

(in Cartesian coord.)

(in Cylindrical coord.)

A stream function for plane flow is

or

Solution 1 𝜙1

Solution 2 𝜙2 Solution 3 𝜙3 = 𝜙1 +𝜙2

𝑢 =𝜕𝜓

𝜕𝑦, 𝑣 = −

𝜕𝜓

𝜕𝑥

𝑣𝑟 =1

𝑟

𝜕𝜓

𝜕𝜃, 𝑣𝜃 = −

𝜕𝜓

𝜕𝑟

x

y

𝑣 =𝜕𝜙

𝜕𝑦

𝑢 =𝜕𝜙

𝜕𝑥

x

y

θr

𝑉

𝑣𝑟 =𝜕𝜙

𝜕𝑟

𝑣𝜃 =1

𝑟

𝜕𝜙

𝜕𝜃

𝑢 =𝜕𝜙

𝜕𝑥, 𝑣 =

𝜕𝜙

𝜕𝑦

𝑣𝑟 =𝜕𝜙

𝜕𝑟, 𝑣𝜃 =

1

𝑟

𝜕𝜙

𝜕𝜃

𝑉


• For potential flow, basic solutions can be simply added to obtain more complicated solutions because of the major advantage of Laplace equation that it is a linear PDE. For simplicity, only plane (two-dimensional) flows will be considered. Since we can define a stream function for plane flow,

• If we now impose the condition of irrotationality for 2D, it follows

and in terms of the stream function

𝑢 =𝜕𝜓

𝜕𝑦, 𝑣 = −

𝜕𝜓

𝜕𝑥

𝜕𝑢

𝜕𝑦=

𝜕𝑣

𝜕𝑥

𝜕

𝜕𝑦

𝜕𝜓

𝜕𝑦=

𝜕

𝜕𝑥−

𝜕𝜓

𝜕𝑥

𝜕2𝜓

𝜕𝑥2 +𝜕2𝜓

𝜕𝑦2 = 0Satisfy Laplace egn !!=> Superposition allowed



• Thus, for a plane irrotational flow we can use either the velocity potential or the stream function - both must satisfy Laplace’s equation in two dimensions. It is apparent from these results that the velocity potential and the stream function are somehow related. It can be shown that lines of constant 𝜙 (called equipotential lines) are orthogonal to lines of constant 𝜓 (streamlines) at all points where they intersect. Recall that two lines are orthogonal if the product or their slopes is – 1, as illustrated by this figure

• Along streamlines 𝜓 = const:

• Along equipotential lines 𝜙 = const.

𝑑𝑦

𝑑𝑥=

𝑣

𝑢Along 𝜓=const

𝑑𝜙 =𝜕𝜙

𝜕𝑥𝑑𝑥 +

𝜕𝜙

𝜕𝑦𝑑𝑦 = 𝑢𝑑𝑥 + 𝑣𝑑𝑦 = 0

𝑑𝑦

𝑑𝑥= −

𝑢

𝑣Along 𝜙=const

x

a

ba

b

y

𝑎

𝑏× −

𝑏

𝑎= −1

Lines of constant 𝜙 lines of constant 𝜓(equipotential lines) (stream lines)

Flow net: a family of streamlines andequipotential lines (see Fig. 6.15)


slope

∵ 𝑉 ⫽ 𝑑𝒓 𝑎𝑙𝑜𝑛𝑔 𝑠𝑡𝑟𝑒𝑎𝑚𝑙𝑖𝑛𝑒

𝑑𝜓 =𝜕𝜓

𝜕𝑥𝑑𝑥 +

𝜕𝜓

𝜕𝑦𝑑𝑦 = −𝑣𝑑𝑥 + 𝑢𝑑𝑦 = 0


Uniform flow at angle α with the 𝑥 axis

y

x

𝜓 = 𝜓1

𝜙 = 𝜙1 𝜙 = 𝜙2

𝑈

𝜓 = 𝜓2

𝜓 = 𝜓3

𝜓 = 𝜓4

𝑢 = 𝑈 =𝜕𝜙

𝜕𝑥=

𝜕𝜓

𝜕𝑦

𝑣 = 0 =𝜕𝜙

𝜕𝑦= −

𝜕𝜓

𝜕𝑥

∴ 𝜙 = 𝑈𝑥 +c = f(x)

𝜓 = 𝑈𝑦 = f(y)

Set zero (arbitrary constant)



Uniform flow at angle α with the 𝑥 axis

Velocity potential:

Stream function:

Velocity components:

𝜙 = 𝑈 𝑥 cos 𝛼 + 𝑦 sin 𝛼

𝜓 = 𝑈 𝑦 cos 𝛼 − 𝑥 sin 𝛼

𝑢 = 𝑈 cos 𝛼 , 𝑣 = 𝑈 sin 𝛼

y

x

α

𝑈

𝜓 = 𝜓1

𝜓 = 𝜓2

𝜓 = 𝜓3

𝜓 = 𝜓4

𝜙 = 𝜙1

𝜙 = 𝜙2


𝑢 = 𝑈 cos𝛼 =𝜕𝜙

𝜕𝑥=

𝜕𝜓

𝜕𝑦

𝑣 = 𝑈 sin 𝛼 =𝜕𝜙

𝜕𝑦= −

𝜕𝜓

𝜕𝑥

𝜙 = 𝑈𝑥 cos𝛼 + 𝑓 𝑦

𝑓′ 𝑦 =𝜕𝜙

𝜕𝑦= 𝑈 sin 𝛼

𝑓 𝑦 = 𝑈𝑦 sin 𝛼 + 𝑐


6.5.2 Source and sink (𝑚 > 0 source; 𝑚 < 0 sink)


For the conservation of mass

or

and (purely radial flow)

2𝜋𝑟 𝑣𝑟 = 𝑚 (the volume rate of flow) = const

𝑣𝑟 =𝑚

2𝜋𝑟

𝑣𝜃 = 0

∴𝜕𝜙

𝜕𝑟= 𝜈𝑟 =

𝑚

2𝜋𝑟,

1

𝑟

𝜕𝜙

𝜕𝜃= 𝜈𝜃 = 0

∴ 𝜙 =𝑚

2𝜋𝑙𝑛 𝑟

1

𝑟

𝜕𝜓

𝜕𝜃= 𝜈𝑟 =

𝑚

2𝜋𝑟, −

𝜕𝜓

𝜕𝑟= 𝜈𝜃 = 0 ∴ 𝜓 =

𝑚

2𝜋𝜃

source flow ; 𝑚 > 0sink flow ; 𝑚 < 0magnitude of m; the strength

From the stream function for the source

x

y

𝜓 = 𝑐𝑜𝑛𝑠𝑡

𝑣𝑟

𝜙 = 𝑐𝑜𝑛𝑠𝑡

θ

r


6.5.2 Source or sink (𝑚 > 0 for source; 𝑚 < 0 for sink)

Velocity potential:

Stream function:


𝜙 =𝑚

2𝜋ln 𝑟

𝜓 =𝑚

2𝜋𝜃

𝑣𝑟 =𝑚

2𝜋𝑟, 𝑣𝜃 = 0

y

x

r

θ

𝜙 = constant𝜓 = constant

𝑣𝑟


𝑏𝑜𝑡ℎ 𝑒𝑞𝑢𝑖 − 𝑙𝑖𝑛𝑒𝑠 𝑎𝑟𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙


= 4𝑟𝑐𝑜𝑠2𝜃


- Fully developed 2-dimensioanl channel flow- Height h, and both plates are fixed- 𝑑p∕𝑑𝑥<0 and constant- Steady, incompressible, 2-D (in x-y plane)

6.5.2 Source and Sink

cf)

h

u(y)

x

y

𝑢 =1

2𝜇

𝑑𝑝

𝑑𝑥𝑦2 − ℎ𝑦 and 𝑣 = 0

𝜕𝜓

𝜕𝑦= 𝑢 =

1

2𝜇

𝑑𝑝

𝑑𝑥(𝑦2 − ℎ𝑦)

integration

𝜓 =1

2𝜇

𝑑𝑝

𝑑𝑥

𝑦3

3−

ℎ𝑦2

2+ 𝑔(𝑥)

0 = 𝜈 = −𝜕𝜓

𝜕𝑥= −𝑔′(𝑥) ∴ 𝑔 𝑥 = 𝑐

𝜓 =1

2𝜇

𝑑𝑝

𝑑𝑥

𝑦3

3−

ℎ𝑦2

2+ 𝑐 ,

𝜓 =1

2𝜇

𝑑𝑝

𝑑𝑥

𝑦3

3−

ℎ𝑦2

2, 𝜓𝑡𝑜𝑝 = −

1

12𝜇

𝑑𝑝

𝑑𝑥ℎ3

Derive the Stream function 𝜓 along the dashed line.For simplicity, let’s put 𝜓 = 0 on the bottom wall, but what is 𝜓 on the top wall?

integration

B.C.; along y=0, 𝜓 = 0 so that ∴ c = 0

at y = h


6.5.3. Vortex (concentric circles)

; Interchange the velocity potential 𝜙 & stream function 𝜓 for the source.

𝜙 = 𝐾𝜃 𝜓 = −𝐾 ln 𝑟 , 𝐾=const

𝑣𝜃 =1

𝑟

𝜕𝜙

𝜕𝜃= −

𝜕𝜓

𝜕𝑟=

𝐾

𝑟

𝑣𝑟 = 0

(rotational vortex)

(irrotational vortex)

ex) the motion of a liquid contained in a tank that is rotated about this axis

ex) the swirling motion of the water as it drains from a bathtub

𝑣𝜃 = 𝜔𝑟

𝑣𝜃 =𝐾

𝑟

𝑟 ≤ 𝑟0

𝑟 > 𝑟

and

Forced vortex ; the fluid is rotating as a rigid body

Free vortex ; irrotational motion

Combined vortex ; a forced vortex as a central core and a free vortex outside the core

zr

zr

r

r

FrFFzr

iiri

r

1Fx

F

r

1)rF(

rr

1F

if

r

1i

r

ff

0

0K0zr

iiri

r

1Vx

zr

𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙

Irrotational flow

https://www.youtube.com/watch?v=glfm3NMMxh0


- For an irrotational flow, so that𝑉 = 𝛻𝜙 𝑉 ∙ 𝑑𝑠 = 𝛻𝜙 ∙ 𝑑𝑠 = 𝑑𝜙

𝛤 = 𝑐

𝑑𝜙 = 0

𝜞 ≠ 𝟎

𝛤 = 0

2𝜋 𝐾

𝑟𝑟𝑑𝜃 = 2𝜋𝐾

- 𝛤 (Circulation) ; the line integral of the tangential component of the velocity taken around a closed curve in the flow field

𝛤 = 𝑐

𝑉 ∙ 𝑑 𝑠

if there are singularities enclosed within the curve.

ex) the free vortex with 𝑣𝜃 = 𝐾 𝑟


Free vortex (𝛤 > 0 counterclockwise; 𝛤 < 0 clockwise)

Velocity potential:

Stream function:


𝜙 =𝛤

2𝜋𝜃

𝜓 = −𝛤

2𝜋ln 𝑟

𝑣𝑟 = 0, 𝑣𝜃 =𝛤

2𝜋𝑟

y

r

θ

𝜓 = constant

𝜙 = constant

𝑣𝜃

6.5.3. Vortex (concentric circles)


6.5.4. Doublet ; Combination of a source and sink

The source – sink pair

𝜓 = −𝑚

2𝜋(𝜃1 − 𝜃2)

𝑡𝑎𝑛 −2𝜋𝜓

𝑚= 𝑡𝑎𝑛(𝜃1 − 𝜃2)

x

y

𝑟2

𝑟1r

𝑎

𝜃2 𝜃 𝜃1

=𝑡𝑎𝑛𝜃1 − 𝑡𝑎𝑛𝜃2

1 + 𝑡𝑎𝑛𝜃1𝑡𝑎𝑛𝜃2

𝑡𝑎𝑛𝜃1 =𝑟𝑠𝑖𝑛𝜃

𝑟𝑐𝑜𝑠𝜃 − 𝑎

𝑡𝑎𝑛𝜃2 =𝑟𝑠𝑖𝑛𝜃

𝑟𝑐𝑜𝑠𝜃 + 𝑎

=2𝑎𝑟𝑠𝑖𝑛𝜃

𝑟2 − 𝑎2

∴ 𝜓 = −𝑚

2𝜋𝑡𝑎𝑛−1

2𝑎𝑟𝑠𝑖𝑛𝜃

𝑟2 − 𝑎2


𝜋(𝑟2 − 𝑎2)

𝜓 = −𝑚

2𝜋


𝑟2 − 𝑎2= −

𝑚𝑎𝑟𝑠𝑖𝑛𝜃

𝜋(𝑟2 − 𝑎2)

For small values of

𝑎

m↑ ⤍ 2𝜋𝜓/m↓ ⤍ 2𝑎𝑟𝑠𝑖𝑛𝜃

𝑟2−𝑎2 ↓


Doublet (with strength 𝐾 = 𝑚𝑎/𝜋)

Velocity potential:

Stream function:


𝜙 =𝐾 𝑐𝑜𝑠 𝜃

𝑟

𝜓 = −𝐾 𝑠𝑖𝑛 𝜃

𝑟

𝑣𝑟 = −𝐾 𝑐𝑜𝑠 𝜃

𝑟2 , 𝑣𝜃 = −𝐾 𝑠𝑖𝑛 𝜃

𝑟2


Doublet ; letting the source and sink approach one another (𝑎 → 0) while increasing the strength (m →∞)

∴ 𝜓 = −𝐾𝑠𝑖𝑛𝜃

𝑟, 𝐾 =

𝑚𝑎

𝜋(strength of

doublet)

𝜙 =𝐾𝑐𝑜𝑠𝜃

𝑟

Streamlines for a doublet

𝜈𝑟 =1

𝑟

𝜕𝜓

𝜕𝜃, 𝜈𝜃 = −

𝜕𝜓

𝜕𝑟

𝑣𝑟 =𝜕𝜙

𝜕𝑟, 𝑣𝜃 =

1

𝑟

𝜕𝜙

𝜕𝜃


Doublet (with strength 𝐾 = 𝑚𝑎/𝜋)

Stream function: 𝜓 = −𝐾 𝑠𝑖𝑛 𝜃

𝑟


x

y

𝜓1𝑟2 = −𝐾𝑟 𝑠𝑖𝑛 𝜃 = −𝐾𝑦

𝜓1 = −𝐾𝑦

(𝑥2+𝑦2)

𝑥2 + 𝑦2 +𝐾

𝜓1

𝑦 +𝐾

2𝜓1

2 =𝐾

2𝜓1

2


6.6.1 Source in a Uniform Stream-Half-Body

Flow around a half-body is obtained by the addition of a source to a uniform flow.

y

x

b

U

Stagnation

point

Source

r

θ

Stagnation point

b

ψ = π𝑏𝑈

π𝑏

π𝑏

6.6 Superposition of Basic, Plane Potential Flows

At the stagnation point (𝑥 = −𝑏;𝜃 = 𝜋)

𝜓 = 𝜓𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑓𝑙𝑜𝑤 + 𝜓𝑠𝑜𝑢𝑟𝑐𝑒

= 𝑈𝑟𝑠𝑖𝑛𝜃 +𝑚

2𝜋𝜃

𝜙 = 𝑈𝑟𝑐𝑜𝑠𝜃 +𝑚

2𝜋𝑙𝑛 𝑟

𝑣𝑟 = 0 → 𝑈 =𝑚

2𝜋𝑏

∴ 𝑏 = 𝑚 2𝜋𝑈 ∴ 𝜓𝑠𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 =𝑚

2𝜋𝜃

𝜃=𝜋=

𝑚

2= 𝜋𝑏𝑈

(Stream function of uniform flow)

∴ 𝜓 = 𝑈(𝑦𝑐𝑜𝑠𝛼 − 𝑥𝑠𝑖𝑛𝛼)

If 𝛼 = 0° and 𝑦 = 𝑟𝑠𝑖𝑛𝜃 then 𝜓 = 𝑈𝑟 𝑠𝑖𝑛𝜃See (6.97~98)

𝜈𝑟 =1

𝑟

𝜕𝜓

𝜕𝜃=

𝑚

2𝜋𝑟for the source only

𝒓 = 𝒃

𝑈 : uniform flow



∴ 𝜓𝑠𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 = 𝜋𝑏𝑈 = 𝑈𝑟𝑠𝑖𝑛𝜃 + 𝑏𝑈𝜃 (Streamline passing through SP)

𝑟 =𝑏(𝜋 − 𝜃)

𝑠𝑖𝑛𝜃[0 ≤ 𝜃 ≤ 2𝜋]

𝑣𝑟 =1

𝑟

𝜕𝜓

𝜕𝜃= 𝑈𝑐𝑜𝑠𝜃 +

𝑚

2𝜋𝑟

𝑣𝜃 = −𝜕𝜓

𝜕𝑟= −𝑈𝑠𝑖𝑛𝜃 ∴ 𝑉2 = 𝜈𝑟

2 + 𝜈𝜃2 = 𝑈2 +

𝑈𝑚𝑐𝑜𝑠𝜃

𝜋𝑟+ (

𝑚

2𝜋𝑟)2

𝑏 = 𝑚 2𝜋𝑈

𝑉2 = 𝑈2(1 + 2𝑏

𝑟𝑐𝑜𝑠𝜃 +

𝑏2

𝑟2)

or

See the figures above- The width of the half-body ~ 2𝜋𝑏

𝑦 = 𝑟𝑠𝑖𝑛𝜃 = 𝑏 𝜋 − 𝜃

𝑦𝑚𝑎𝑥 = 𝑏𝜋 𝜃 → 0 2𝜋

from eq (6-100)

as or

- To get the velocity components at any points,

and

and since

𝑣𝑟 =1

𝑟

𝜕𝜓

𝜕𝜃, 𝑣𝜃 = −

𝜕𝜓

𝜕𝑟

𝑣𝑟 =𝜕𝜙

𝜕𝑟, 𝑣𝜃 =

1

𝑟

𝜕𝜙

𝜕𝜃

eq (6-100)



Flow around a half-body is obtained by the addition of a source to a uniform flow.

The flow around a half-body: (a) superposition of a source and a uniform flow; (b) replacement of streamline 𝜓 = 𝜋𝑏𝑈 with solid boundary to form half-body.

Velocity potential:

Stream function:


𝜙 = 𝑈𝑟 cos 𝜃 +𝑚

2𝜋ln 𝑟

𝜓 = 𝑈𝑟 𝑠𝑖𝑛 𝜃 +𝑚

2𝜋𝜃

𝑣𝑟 = 𝑈𝑐𝑜𝑠𝜃 +𝑚

2𝜋𝑟, 𝑣𝜃 = −𝑈 sin 𝜃

y

x

b

U

Stagnation

point

Source

r

θ

Stagnation point

b

ψ = π𝑏𝑈

π𝑏

π𝑏


6.6.2 Source & Sink in a Uniform Stream – Rankine ovals

- Given the reference pressure 𝑃0 and velocity 𝑈, you can obtain the pressure at any point from BE (Bernoulli eqn)

- Important Note!!

Inviscid flow → neglecting viscosity→ the fluid “slips” on the body→ the tangential velocity on the body is not “ Zero”→ “Slip” condition ! applied for viscous flow

If flow separation doesn’t occur, it’s reasonable!!

- Rankine ovals (an oval shape) ; a uniform flow + a source + a sink

· What’s the egn for streamline passing through SP.· The width of the Ranking ovals· Velocity at any points

𝑃0 +1

2𝜌𝑈2 = 𝑃 +

1

2𝜌𝑉2

Given reference Solved above

𝜓 = 𝜓𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑓𝑙𝑜𝑤 + 𝜓𝑠𝑜𝑢𝑟𝑐𝑒 + 𝜓𝑠𝑖𝑛𝑘

Equal strength

= 𝑈𝑟𝑠𝑖𝑛𝜃 +𝑚

2𝜋(𝜃2 − 𝜃1)

a a

θ1

θ3

θ2m

?


Stagnation point

b

ψ = π𝑏𝑈

π𝑏

π𝑏


• Rankine ovals are formed by combining a source and sink with a uniform flow.

• The flow around a Rankine oval: (a) superposition of source-sink pair and a uniform flow; (b) replacement of streamline 𝜓 = 0 with solid boundary to form Rankine oval.

Velocity potential:

Stream function:

Body half length:

Body half width: from y axis-𝜓=0 intersection

𝜙 = 𝑈𝑟 𝑐𝑜𝑠 𝜃 −𝑚

2𝜋𝑙𝑛 𝑟1 − 𝑙𝑛 𝑟2

𝜓 = 𝑈𝑟 sin 𝜃 −𝑚


2𝑎𝑟 sin 𝜃

𝑟2 − 𝑎2= 𝑈𝑦 −

𝑚


2𝑎𝑦

𝑥2 + 𝑦2 − 𝑎2

𝑙 =𝑚𝑎

𝜋𝑈+ 𝑎2

1/2from stag point (𝑣𝜃 = 𝑣𝑟=0) at y=0 or 𝜃= 𝜋

ℎ =ℎ2 − 𝑎2

2𝑎𝑡𝑎𝑛

2𝜋𝑈ℎ

𝑚

SourceSink

r 𝑟1

𝜃1

𝑟2

𝜃2 𝜃

x

yU

Stagnation

PointStagnation

Pointψ =0

h

h

+m -m

𝓵 𝓵

a a

a a

6.6.2 Source & Sink in a Uniform Stream – Rankine ovals

(6.105)


6.6.3 Flow around a Circular Cylinder

; a Uniform flow + a Doublet

In order to get the streamlines around a circular cylinder,

(doublet strength)

𝜓 = 𝑈𝑟𝑠𝑖𝑛𝜃 −𝐾𝑠𝑖𝑛𝜃

𝑟= 𝑈 −

𝐾

𝑟2 𝑟𝑠𝑖𝑛𝜃

𝜙 = 𝑈𝑟𝑐𝑜𝑠𝜃 +𝐾𝑐𝑜𝑠𝜃

𝑟

𝜓 = 𝑐𝑜𝑠𝑡𝑎𝑛𝑡 at 𝑟 = 𝑎

𝜓 = 0 for 𝑟 = 𝑎 𝑖𝑓 𝑈 −𝐾

𝑎2 = 0

𝐾 = 𝑈𝑎2

∴ 𝜓 = 𝑈𝑟 1 −𝑎2

𝑟2 𝑠𝑖𝑛𝜃

∴ 𝜙 = 𝑈𝑟 1 +𝑎2

𝑟2 𝑐𝑜𝑠𝜃 see Figure 6.26

Velocity components?



The velocity components

On the surface of the cylinder (𝑟 = 𝑎)

The pressure distribution on the cylinder surface from the Bernoulli equation

The resultant force (per unit length)

𝑣𝑟 =𝜕𝜙

𝜕𝑟=

1

𝑟

𝜕𝜓

𝜕𝜃= 𝑈 1 −

𝑎2

𝑟2𝑐𝑜𝑠𝜃

𝑣𝜃 =1

𝑟

𝜕𝜙

𝜕𝜃= −

𝜕𝜓

𝜕𝑟= −𝑈 1 +

𝑎2

𝑟2 𝑠𝑖𝑛𝜃

𝑣𝑟 = 0, 𝑣𝜃𝑠 = −2𝑈𝑠𝑖𝑛𝜃

𝑃0 +1

2𝜌𝑈2 = 𝑃𝑠 +

1

2𝜌𝑣𝜃𝑠

2

∴ 𝑃𝑠 = 𝑃0 +1

2𝜌𝑈2(1 − 4𝑠𝑖𝑛2𝜃)

𝐹𝑥 = − 0

2𝜋

𝑃𝑠𝑐𝑜𝑠𝜃𝑎𝑑𝜃

𝐹𝑦 = − 0

2𝜋

𝑃𝑠𝑠𝑖𝑛𝜃𝑎𝑑𝜃

; drag

; liftd′Alembert′s paradox;In the potential theory, the drag and the lift are both zero in a uniform stream. However, there is a significant drag developed on a cylinder when it is placed in a moving fluid.

see Figure 6.27



- Adding a free vortex to the flow around a cylinder

and

- The tangential velocity, 𝑣θ, on the surface of the cylinder (𝑟 = 𝑎)

Resemble the flow in a uniform stream with a rotating cylinder

𝛤 ; - the circulation and the vortex strength- can change the streamlines patterns

Let’s determine the location of stagnation points(𝜃 = 𝜃𝑠𝑡𝑎𝑔 where 𝑣θ = 0 in above egn)

𝜓 = 𝑈𝑟 1 −𝑎2

𝑟2 𝑠𝑖𝑛𝜃 −𝛤

2𝜋ln 𝑟

𝜙 = 𝑈𝑟 1 +𝑎2

𝑟2 𝑐𝑜𝑠𝜃 +𝛤

2𝜋𝜃

𝑣𝜃𝑠 = −𝜕𝜓

𝜕𝑟𝑟=𝑎

= −2𝑈𝑠𝑖𝑛𝜃 +𝛤

2𝜋𝑎

𝑠𝑖𝑛𝜃𝑠𝑡𝑎𝑔 =𝛤

4𝜋𝑈𝑎

1)

2) ; some other location on the surface

3) ; located away from the cylinder

𝛤 = 0; 𝜃𝑠𝑡𝑎𝑔 = 0 𝑜𝑟 𝜋

𝛤

4𝜋𝑈𝑎≤ 1

𝛤

4𝜋𝑈𝑎> 1

see Figure 6.27



- To get the force per unit length on the cylinder.consider 𝑃𝑠 from the Bernoulli egn

or

ex) ; 𝑈 + tive & 𝛤 + tive Fy downward

𝑈 + tive & 𝛤 - tive Fy upward

𝑃0 +1

2𝜌𝑈2 = 𝑃𝑠 +

1

2𝜌(−2𝑈𝑠𝑖𝑛𝜃 +

𝛤

2𝜋𝑎)2

𝑃𝑠 = 𝑃0 +1

2𝜌𝑈2 1 − 4𝑠𝑖𝑛2𝜃 +

2𝛤𝑠𝑖𝑛𝜃

𝜋𝑎𝑈−

𝛤2

4𝜋2𝑎2𝑈2

∴ 𝐹𝑥 = − 𝑎

2𝜋

𝑃𝑠𝑐𝑜𝑠𝜃𝑎𝑑𝜃 = 0

𝐹𝑦 = − 0

2𝜋

𝑃𝑠𝑠𝑖𝑛𝜃𝑎𝑑𝜃 = −𝜌𝑈𝛤

Fy -tive

(lift)

∵ 𝑠𝑖𝑛𝜃 & 𝑠𝑖𝑛2𝜃 𝑎𝑟𝑒 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑤𝑟𝑡 𝜃 = 𝜋/2


Flow around a Circular Cylinder

• A doublet combined with a uniform flow can be used to represent flow around a circular cylinder.

Velocity potential:

Stream function:


𝜙 = 𝑈𝑟 𝑐𝑜𝑠 𝜃 +𝐾 𝑐𝑜𝑠 𝜃

𝑟

𝜓 = 𝑈𝑟 sin 𝜃 −𝐾 sin 𝜃

𝑟

𝑣𝑟 = 𝑈 1 −𝑎1

𝑟2cos 𝜃 , 𝑣𝜃 = −𝑈 1 +

𝑎2

𝑟2sin 𝜃

U

2U

a

𝜓 =0r

θ

6.6 Superposition of Basic, Plane Potential Flows


Assignments (Chapter 6) – 6th Edition

• P6.4, P6.6, P6.8, P6.9, P6.10, P6.14, P6.16, P6.18

• P6.24, P6.27, P6.28, P6.32, P6.34, P6.36, P6.37, P6.39, P6.41

• P6.45, P6.46, P6.47, P6.49, P6.53, P6.54, P6.55

• P6.62, P6.64, P6.67, P6.70, P6.80, P6.84

• P6.92, P6.93, P6.95, P6.101, P6.105

Assignments (Chapter 6) – 7th Edition

• P6.4, P6.6, P6.8, P6.9, P6.10, P6.13, P6.15, P6.17

• P6.24, P6.28, P6.29, P6.33, P6.35, P6.36, P6.37, P6.39, P6.41

•P6.46, P6.47, P6.49, P6.55, P6.56, P6.57

• P6.66, P6.70, P6.75, P6.77, P6.85, P6.88

• P6.94, P6.95, P6.97, P6.104, P6.108

chapter. 6 differential analysis of fluid flow potential flow

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