Upload
others
View
10
Download
0
Embed Size (px)
Citation preview
Chapter.6 Differential Analysis of Fluid Flow
Copyright Β© The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter. 6Differential Analysis of Fluid Flow
Potential Flow
μ΄λκ·Ό
Mκ΄ 517νΈ[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
Chapter.6 Differential Analysis of Fluid Flow
Inviscid Flow: Eulerβs equations of motion
β’ 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:
β’ For an inviscid flow in which all the shearing stresses are zero, and the normal stresses are replaced by βp, the Navier-Stokes Equations reduce to Eulerβs equations
β’ In Cartesian coordinates:
βπ = ππ₯π₯ = ππ¦π¦ = ππ§π§
ππ β π΅π β 0 = πππ½
ππ‘+ π Β· π» π½
πππ₯ βππ
ππ₯= π
ππ’
ππ‘+ π’
ππ’
ππ₯+ π£
ππ’
ππ¦+ π€
ππ’
ππ§
πππ¦ βππ
ππ¦= π
ππ£
ππ‘+ π’
ππ£
ππ₯+ π£
ππ£
ππ¦+ π€
ππ£
ππ§
πππ§ βππ
ππ§= π
ππ€
ππ‘+ π’
ππ€
ππ₯+ π£
ππ€
ππ¦+ π€
ππ€
ππ§
Chapter.6 Differential Analysis of Fluid Flow
The Bernoulli equation derived from Eulerβs equations
β’ Steady, incompressible, inviscid, along a streamline
Inviscid Flow: Eulerβs equations of motion
π
π+
π2
2+ ππ§ = ππππ π‘
π1
πΎ+
π12
2π+ π§1 =
π2
πΎ+
π22
2π+ π§2
ππ β π΅π = πππ½
ππ‘+ π Β· π» π½ = π π Β· π» π½
π½ Β· π» π½ =π
2π» π½ Β· π½ - π½ x (π»x π½)
ππ β π΅π Β· ππ =π
2ππ» π½ Β· π½ Β· ππ β ππ½ x (π»x π½) Β· ππ
dot product
Β· ππ
Vector identity
β΅ π β«½ ππ πππππ π π‘ππππππππ
β΅ π»π Β· ππ = dπβ πππ§ βππ
πβ π
1
2π2 = 0
& incompressible
Chapter.6 Differential Analysis of Fluid Flow
The Bernoulli equation derived from Eulerβs equations
β’ The Bernoulli equation can also be derived, starting from Eulerβs equations. For inviscid, incompressible fluids, we end up with the same equation
β’ 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 βheadβ form by dividing each term by π so that
β’ The Bernoulli equation is restricted to the following:
- Inviscid flow- Steady flow- Incompressible flow- Flow along a streamline
Inviscid Flow: Eulerβs equations of motion
π
π+
π2
2+ ππ§ = ππππ π‘
π1
πΎ+
π12
2π+ π§1 =
π2
πΎ+
π22
2π+ π§2
Chapter.6 Differential Analysis of Fluid Flow
The Bernoulli equation for steady, incompressible, irrotational flow
Inviscid Flow: Eulerβs equations of motion
ππ β π΅π = πππ½
ππ‘+ π Β· π» π½ = π π Β· π» π½
π½ Β· π» π½ =π
2π» π½ Β· π½ - π½ x (π»x π½)
ππ β π΅π Β· ππ =π
2ππ» π½ Β· π½ Β· ππ β ππ½ x (π»x π½) Β· ππ
dot product
Β· ππ
Vector identity
β΅ πππππ‘ππ‘πππππ
β΄ π ππ‘ππ ππππ πππ πππ πππ‘βπ β πππ§ βππ
πβ π
1
2π2 = 0
Chapter.6 Differential Analysis of Fluid Flow
The Irrotational Flow and corresponding Bernoulli equation
β’ 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:
β’ But it can now be applied between any two points in the flow field, not limited to applications along a streamline.
Inviscid Flow: Eulerβs equations of motion
π» Γ π = 0
π1
πΎ+
π12
2π+ π§1 =
π2
πΎ+
π22
2π+ π§2
Various regions of flow around bodies
Various regions of flow through channels
Chapter.6 Differential Analysis of Fluid Flow
The Velocity Potential
β’ For an irrotational flow:
so we have
β’ It follows that in this case the velocity components can be expressed in terms of a scalar function π π₯, π¦, π§, π‘ , called velocity potential, as
In vector form
Inviscid Flow: Eulerβs equations of motion
π» Γ π½ =ππ€
ππ¦β
ππ£
ππ§ π +
ππ’
ππ§β
ππ€
ππ₯ π +
ππ£
ππ₯β
ππ’
ππ¦ π = 0
ππ€
ππ¦=
ππ£
ππ§,
ππ’
ππ§=
ππ€
ππ₯,
ππ£
ππ₯=
ππ’
ππ¦
π’ =ππ
ππ₯, π£ =
ππ
ππ¦, π€ =
ππ
ππ§
π½ = π»π
Chapter.6 Differential Analysis of Fluid Flow
β’ 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.
β’ And therefore for incompressible, irrotational flow, it follows that
β’ The velocity potential satisfies the Laplace equation.In Cartesian coordinates:
In cylindrical coordinates:
Inviscid Flow: Eulerβs equations of motion
π» β π½ = 0
π»2π = 0
π2π
ππ₯2+
π2π
ππ¦2+
π2π
ππ§2= 0
1
π
π
πππ
ππ
ππ+
1
π2
π2π
ππ2+
π2π
ππ§2= 0
Satisfy Laplace egn !!Linear & homogeneousβ€ superposition allowed
Chapter.6 Differential Analysis of Fluid Flow
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 π1
Solution 2 π2 Solution 3 π3 = π1 +π2
π’ =ππ
ππ¦, π£ = β
ππ
ππ₯
π£π =1
π
ππ
ππ, π£π = β
ππ
ππ
x
y
π£ =ππ
ππ¦
π’ =ππ
ππ₯
x
y
ΞΈr
π
π£π =ππ
ππ
π£π =1
π
ππ
ππ
π’ =ππ
ππ₯, π£ =
ππ
ππ¦
π£π =ππ
ππ, π£π =
1
π
ππ
ππ
π
Chapter.6 Differential Analysis of Fluid Flow
β’ 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,
β’ If we now impose the condition of irrotationality for 2D, it follows
and in terms of the stream function
π’ =ππ
ππ¦, π£ = β
ππ
ππ₯
ππ’
ππ¦=
ππ£
ππ₯
π
ππ¦
ππ
ππ¦=
π
ππ₯β
ππ
ππ₯
π2π
ππ₯2 +π2π
ππ¦2 = 0Satisfy Laplace egn !!=> Superposition allowed
6.5 Some Basic, Plane Potential Flows
Chapter.6 Differential Analysis of Fluid Flow
β’ Thus, for a plane irrotational flow we can use either the velocity potential or the stream function - both must satisfy Laplaceβs 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 π (called equipotential lines) are orthogonal to lines of constant π (streamlines) at all points where they intersect. Recall that two lines are orthogonal if the product or their slopes is β 1, as illustrated by this figure
β’ Along streamlines π = const:
β’ Along equipotential lines π = const.
ππ¦
ππ₯=
π£
π’Along π=const
ππ =ππ
ππ₯ππ₯ +
ππ
ππ¦ππ¦ = π’ππ₯ + π£ππ¦ = 0
ππ¦
ππ₯= β
π’
π£Along π=const
x
a
ba
b
y
π
πΓ β
π
π= β1
Lines of constant π lines of constant π(equipotential lines) (stream lines)
Flow net: a family of streamlines andequipotential lines (see Fig. 6.15)
6.5 Some Basic, Plane Potential Flows
slope
β΅ π β«½ ππ πππππ π π‘ππππππππ
ππ =ππ
ππ₯ππ₯ +
ππ
ππ¦ππ¦ = βπ£ππ₯ + π’ππ¦ = 0
Chapter.6 Differential Analysis of Fluid Flow
Uniform flow at angle Ξ± with the π₯ axis
y
x
π = π1
π = π1 π = π2
π
π = π2
π = π3
π = π4
π’ = π =ππ
ππ₯=
ππ
ππ¦
π£ = 0 =ππ
ππ¦= β
ππ
ππ₯
β΄ π = ππ₯ +c = f(x)
π = ππ¦ = f(y)
Set zero (arbitrary constant)
6.5 Some Basic, Plane Potential Flows
Chapter.6 Differential Analysis of Fluid Flow
Uniform flow at angle Ξ± with the π₯ axis
Velocity potential:
Stream function:
Velocity components:
π = π π₯ cos πΌ + π¦ sin πΌ
π = π π¦ cos πΌ β π₯ sin πΌ
π’ = π cos πΌ , π£ = π sin πΌ
y
x
Ξ±
π
π = π1
π = π2
π = π3
π = π4
π = π1
π = π2
6.5 Some Basic, Plane Potential Flows
π’ = π cosπΌ =ππ
ππ₯=
ππ
ππ¦
π£ = π sin πΌ =ππ
ππ¦= β
ππ
ππ₯
π = ππ₯ cosπΌ + π π¦
πβ² π¦ =ππ
ππ¦= π sin πΌ
π π¦ = ππ¦ sin πΌ + π
Chapter.6 Differential Analysis of Fluid Flow
6.5.2 Source and sink (π > 0 source; π < 0 sink)
6.5 Some Basic, Plane Potential Flows
For the conservation of mass
or
and (purely radial flow)
2ππ π£π = π (the volume rate of flow) = const
π£π =π
2ππ
π£π = 0
β΄ππ
ππ= ππ =
π
2ππ,
1
π
ππ
ππ= ππ = 0
β΄ π =π
2πππ π
1
π
ππ
ππ= ππ =
π
2ππ, β
ππ
ππ= ππ = 0 β΄ π =
π
2ππ
source flow ; π > 0sink flow ; π < 0magnitude of m; the strength
From the stream function for the source
x
y
π = ππππ π‘
π£π
π = ππππ π‘
ΞΈ
r
Chapter.6 Differential Analysis of Fluid Flow
6.5.2 Source or sink (π > 0 for source; π < 0 for sink)
Velocity potential:
Stream function:
Velocity components:
π =π
2πln π
π =π
2ππ
π£π =π
2ππ, π£π = 0
y
x
r
ΞΈ
π = constantπ = constant
π£π
6.5 Some Basic, Plane Potential Flows
πππ‘β πππ’π β πππππ πππ πππ‘βππππππ
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
= 4ππππ 2π
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
- Fully developed 2-dimensioanl channel flow- Height h, and both plates are fixed- πpβππ₯<0 and constant- Steady, incompressible, 2-D (in x-y plane)
6.5.2 Source and Sink
cf)
h
u(y)
x
y
π’ =1
2π
ππ
ππ₯π¦2 β βπ¦ and π£ = 0
ππ
ππ¦= π’ =
1
2π
ππ
ππ₯(π¦2 β βπ¦)
integration
π =1
2π
ππ
ππ₯
π¦3
3β
βπ¦2
2+ π(π₯)
0 = π = βππ
ππ₯= βπβ²(π₯) β΄ π π₯ = π
π =1
2π
ππ
ππ₯
π¦3
3β
βπ¦2
2+ π ,
π =1
2π
ππ
ππ₯
π¦3
3β
βπ¦2
2, ππ‘ππ = β
1
12π
ππ
ππ₯β3
Derive the Stream function π along the dashed line.For simplicity, letβs put π = 0 on the bottom wall, but what is π on the top wall?
integration
B.C.; along y=0, π = 0 so that β΄ c = 0
at y = h
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
6.5.3. Vortex (concentric circles)
; Interchange the velocity potential π & stream function π for the source.
π = πΎπ π = βπΎ ln π , πΎ=const
π£π =1
π
ππ
ππ= β
ππ
ππ=
πΎ
π
π£π = 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
π£π = ππ
π£π =πΎ
π
π β€ π0
π > π
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
πππππ‘ππ‘πππππ
Irrotational flow
Chapter.6 Differential Analysis of Fluid Flow
- For an irrotational flow, so thatπ = π»π π β ππ = π»π β ππ = ππ
π€ = π
ππ = 0
π β π
π€ = 0
2π πΎ
ππππ = 2ππΎ
- π€ (Circulation) ; the line integral of the tangential component of the velocity taken around a closed curve in the flow field
π€ = π
π β π π
if there are singularities enclosed within the curve.
ex) the free vortex with π£π = πΎ π
Chapter.6 Differential Analysis of Fluid Flow
Free vortex (π€ > 0 counterclockwise; π€ < 0 clockwise)
Velocity potential:
Stream function:
Velocity components:
π =π€
2ππ
π = βπ€
2πln π
π£π = 0, π£π =π€
2ππ
y
r
ΞΈ
π = constant
π = constant
π£π
6.5.3. Vortex (concentric circles)
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
6.5.4. Doublet ; Combination of a source and sink
The source β sink pair
π = βπ
2π(π1 β π2)
π‘ππ β2ππ
π= π‘ππ(π1 β π2)
x
y
π2
π1r
π
π2 π π1
=π‘πππ1 β π‘πππ2
1 + π‘πππ1π‘πππ2
π‘πππ1 =ππ πππ
ππππ π β π
π‘πππ2 =ππ πππ
ππππ π + π
=2πππ πππ
π2 β π2
β΄ π = βπ
2ππ‘ππβ1
2πππ πππ
π2 β π2
2πππ πππ
π(π2 β π2)
π = βπ
2π
2πππ πππ
π2 β π2= β
ππππ πππ
π(π2 β π2)
For small values of
π
mβ β€ 2ππ/mβ β€ 2πππ πππ
π2βπ2 β
Chapter.6 Differential Analysis of Fluid Flow
Doublet (with strength πΎ = ππ/π)
Velocity potential:
Stream function:
Velocity components:
π =πΎ πππ π
π
π = βπΎ π ππ π
π
π£π = βπΎ πππ π
π2 , π£π = βπΎ π ππ π
π2
6.5.4. Doublet ; Combination of a source and sink
Doublet ; letting the source and sink approach one another (π β 0) while increasing the strength (m ββ)
β΄ π = βπΎπ πππ
π, πΎ =
ππ
π(strength of
doublet)
π =πΎπππ π
π
Streamlines for a doublet
ππ =1
π
ππ
ππ, ππ = β
ππ
ππ
π£π =ππ
ππ, π£π =
1
π
ππ
ππ
Chapter.6 Differential Analysis of Fluid Flow
Doublet (with strength πΎ = ππ/π)
Stream function: π = βπΎ π ππ π
π
6.5.4. Doublet ; Combination of a source and sink
x
y
π1π2 = βπΎπ π ππ π = βπΎπ¦
π1 = βπΎπ¦
(π₯2+π¦2)
π₯2 + π¦2 +πΎ
π1
π¦ +πΎ
2π1
2 =πΎ
2π1
2
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
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
ΞΈ
Stagnation point
b
Ο = Οππ
Οπ
Οπ
6.6 Superposition of Basic, Plane Potential Flows
At the stagnation point (π₯ = βπ;π = π)
π = ππ’ππππππ ππππ€ + ππ ππ’πππ
= πππ πππ +π
2ππ
π = πππππ π +π
2πππ π
π£π = 0 β π =π
2ππ
β΄ π = π 2ππ β΄ ππ π‘πππππ‘πππ =π
2ππ
π=π=
π
2= πππ
(Stream function of uniform flow)
β΄ π = π(π¦πππ πΌ β π₯π πππΌ)
If πΌ = 0Β° and π¦ = ππ πππ then π = ππ π πππSee (6.97~98)
ππ =1
π
ππ
ππ=
π
2ππfor the source only
π = π
π : uniform flow
Chapter.6 Differential Analysis of Fluid Flow
6.6.1 Source in a Uniform Stream-Half-Body
β΄ ππ π‘πππππ‘πππ = πππ = πππ πππ + πππ (Streamline passing through SP)
π =π(π β π)
π πππ[0 β€ π β€ 2π]
π£π =1
π
ππ
ππ= ππππ π +
π
2ππ
π£π = βππ
ππ= βππ πππ β΄ π2 = ππ
2 + ππ2 = π2 +
πππππ π
ππ+ (
π
2ππ)2
π = π 2ππ
π2 = π2(1 + 2π
ππππ π +
π2
π2)
or
See the figures above- The width of the half-body ~ 2ππ
π¦ = ππ πππ = π π β π
π¦πππ₯ = ππ π β 0 2π
from eq (6-100)
as or
- To get the velocity components at any points,
and
and since
π£π =1
π
ππ
ππ, π£π = β
ππ
ππ
π£π =ππ
ππ, π£π =
1
π
ππ
ππ
eq (6-100)
Chapter.6 Differential Analysis of Fluid Flow
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.
The flow around a half-body: (a) superposition of a source and a uniform flow; (b) replacement of streamline π = πππ with solid boundary to form half-body.
Velocity potential:
Stream function:
Velocity components:
π = ππ cos π +π
2πln π
π = ππ π ππ π +π
2ππ
π£π = ππππ π +π
2ππ, π£π = βπ sin π
y
x
b
U
Stagnation
point
Source
r
ΞΈ
Stagnation point
b
Ο = Οππ
Οπ
Οπ
Chapter.6 Differential Analysis of Fluid Flow
6.6.2 Source & Sink in a Uniform Stream β Rankine ovals
- Given the reference pressure π0 and velocity π, you can obtain the pressure at any point from BE (Bernoulli eqn)
- Important Note!!
Inviscid flow β neglecting viscosityβ the fluid βslipsβ on the bodyβ the tangential velocity on the body is not β Zeroββ βSlipβ condition ! applied for viscous flow
If flow separation doesnβt occur, itβs reasonable!!
- Rankine ovals (an oval shape) ; a uniform flow + a source + a sink
Β· Whatβs the egn for streamline passing through SP.Β· The width of the Ranking ovalsΒ· Velocity at any points
π0 +1
2ππ2 = π +
1
2ππ2
Given reference Solved above
π = ππ’ππππππ ππππ€ + ππ ππ’πππ + ππ πππ
Equal strength
= πππ πππ +π
2π(π2 β π1)
a a
ΞΈ1
ΞΈ3
ΞΈ2m
?
Chapter.6 Differential Analysis of Fluid Flow
Stagnation point
b
Ο = Οππ
Οπ
Οπ
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
β’ Rankine ovals are formed by combining a source and sink with a uniform flow.
β’ The flow around a Rankine oval: (a) superposition of source-sink pair and a uniform flow; (b) replacement of streamline π = 0 with solid boundary to form Rankine oval.
Velocity potential:
Stream function:
Body half length:
Body half width: from y axis-π=0 intersection
π = ππ πππ π βπ
2πππ π1 β ππ π2
π = ππ sin π βπ
2ππ‘ππβ1
2ππ sin π
π2 β π2= ππ¦ β
π
2ππ‘ππβ1
2ππ¦
π₯2 + π¦2 β π2
π =ππ
ππ+ π2
1/2from stag point (π£π = π£π=0) at y=0 or π= π
β =β2 β π2
2ππ‘ππ
2ππβ
π
SourceSink
r π1
π1
π2
π2 π
x
yU
Stagnation
PointStagnation
PointΟ =0
h
h
+m -m
π΅ π΅
a a
a a
6.6.2 Source & Sink in a Uniform Stream β Rankine ovals
(6.105)
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
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)
π = πππ πππ βπΎπ πππ
π= π β
πΎ
π2 ππ πππ
π = πππππ π +πΎπππ π
π
π = πππ π‘πππ‘ at π = π
π = 0 for π = π ππ π βπΎ
π2 = 0
πΎ = ππ2
β΄ π = ππ 1 βπ2
π2 π πππ
β΄ π = ππ 1 +π2
π2 πππ π see Figure 6.26
Velocity components?
Chapter.6 Differential Analysis of Fluid Flow
6.6.2 Flow around a Circular Cylinder
The velocity components
On the surface of the cylinder (π = π)
The pressure distribution on the cylinder surface from the Bernoulli equation
The resultant force (per unit length)
π£π =ππ
ππ=
1
π
ππ
ππ= π 1 β
π2
π2πππ π
π£π =1
π
ππ
ππ= β
ππ
ππ= βπ 1 +
π2
π2 π πππ
π£π = 0, π£ππ = β2ππ πππ
π0 +1
2ππ2 = ππ +
1
2ππ£ππ
2
β΄ ππ = π0 +1
2ππ2(1 β 4π ππ2π)
πΉπ₯ = β 0
2π
ππ πππ ππππ
πΉπ¦ = β 0
2π
ππ π ππππππ
; drag
; liftdβ²Alembertβ²s 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
Chapter.6 Differential Analysis of Fluid Flow
6.6.2 Flow around a Circular Cylinder
- Adding a free vortex to the flow around a cylinder
and
- The tangential velocity, π£ΞΈ, on the surface of the cylinder (π = π)
Resemble the flow in a uniform stream with a rotating cylinder
π€ ; - the circulation and the vortex strength- can change the streamlines patterns
Letβs determine the location of stagnation points(π = ππ π‘ππ where π£ΞΈ = 0 in above egn)
π = ππ 1 βπ2
π2 π πππ βπ€
2πln π
π = ππ 1 +π2
π2 πππ π +π€
2ππ
π£ππ = βππ
πππ=π
= β2ππ πππ +π€
2ππ
π ππππ π‘ππ =π€
4πππ
1)
2) ; some other location on the surface
3) ; located away from the cylinder
π€ = 0; ππ π‘ππ = 0 ππ π
π€
4πππβ€ 1
π€
4πππ> 1
see Figure 6.27
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
6.6.2 Flow around a Circular Cylinder
- To get the force per unit length on the cylinder.consider ππ from the Bernoulli egn
or
ex) ; π + tive & π€ + tive Fy downward
π + tive & π€ - tive Fy upward
π0 +1
2ππ2 = ππ +
1
2π(β2ππ πππ +
π€
2ππ)2
ππ = π0 +1
2ππ2 1 β 4π ππ2π +
2π€π πππ
πππβ
π€2
4π2π2π2
β΄ πΉπ₯ = β π
2π
ππ πππ ππππ = 0
πΉπ¦ = β 0
2π
ππ π ππππππ = βπππ€
Fy -tive
(lift)
β΅ π πππ & π ππ2π πππ π π¦ππππ‘πππ π€ππ‘ π = π/2
Chapter.6 Differential Analysis of Fluid Flow
Flow around a Circular Cylinder
β’ A doublet combined with a uniform flow can be used to represent flow around a circular cylinder.
Velocity potential:
Stream function:
Velocity components:
π = ππ πππ π +πΎ πππ π
π
π = ππ sin π βπΎ sin π
π
π£π = π 1 βπ1
π2cos π , π£π = βπ 1 +
π2
π2sin π
U
2U
a
π =0r
ΞΈ
6.6 Superposition of Basic, Plane Potential Flows
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
Chapter.6 Differential Analysis of Fluid Flow
Assignments (Chapter 6) β 6th Edition
β’ P6.4, P6.6, P6.8, P6.9, P6.10, P6.14, P6.16, P6.18
β’ P6.24, P6.27, P6.28, P6.32, P6.34, P6.36, P6.37, P6.39, P6.41
β’ P6.45, P6.46, P6.47, P6.49, P6.53, P6.54, P6.55
β’ P6.62, P6.64, P6.67, P6.70, P6.80, P6.84
β’ P6.92, P6.93, P6.95, P6.101, P6.105
Assignments (Chapter 6) β 7th Edition
β’ P6.4, P6.6, P6.8, P6.9, P6.10, P6.13, P6.15, P6.17
β’ P6.24, P6.28, P6.29, P6.33, P6.35, P6.36, P6.37, P6.39, P6.41
β’P6.46, P6.47, P6.49, P6.55, P6.56, P6.57
β’ P6.66, P6.70, P6.75, P6.77, P6.85, P6.88
β’ P6.94, P6.95, P6.97, P6.104, P6.108