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Jianming Yang
Chapters 3&4
1
Fall 2012
Chapter 3: Integral Relations for a Control Volume
Chapter 4: Differential Relations for Fluid Flow
1 Basic Physical Laws of Fluid Mechanics
Laws of mechanics are written for a system, i.e., a fixed amount of matter.
1) Conservation of mass:
dM
dt
=0
2) Conservation of momentum: F = Ma =
3) Conservation of angular momentum:
d(Mu)
dHG
dt
dt
= MG
dE
4) Conservation of energy: = Q̇ − Ẇ
dt
∆E = heat added to system
− work done by system
δQ
5) Second law of thermodynamics: dS ≥
T
The entropy of an isolated system can only
increase.
And equations of state are needed for a complete description.
058:0160
Jianming Yang
Chapters 3&4
2
Fall 2012
2 Reynolds Transport Theorem
In fluid mechanics we are usually interested in a region of space, i.e., control volume
(CV), rather than individual masses or particular systems. Therefore, we need to
transform conservation laws from a system analysis to a control volume analysis. This is
accomplished through the use of Reynolds transport theorem (RTT).
(Actually RTT was derived in thermodynamics for control volume forms of conservation
laws for mass, momentum, and energy, but not in general form or referred as RTT)
The system extensive thermodynamic properties, which depend on mass, can be written
as 𝐵sys = (𝑀, 𝑀𝐮, 𝐸). Then conservation laws are of form
𝑑
𝑑𝑡
Therefore,
𝑑𝐵sys
𝑑𝑡
(𝑀, 𝑀𝐮, 𝐸 ) = RHS
needs to be related to changes in CV.
Recall, definition of corresponding system intensive thermodynamic properties
𝛽=
𝑑𝐵sys
𝑑𝑀
= (1, 𝐮, 𝑒)
are independent of mass. This requires a relationship between
∫𝐶𝑉 𝛽𝑑𝑚 = ∫𝐶𝑉 𝛽𝜌𝑑𝑉.
𝑑𝐵sys
𝑑𝑡
and
𝑑𝐵𝐶𝑉
𝑑𝑡
with 𝐵𝐶𝑉 =
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Jianming Yang
Chapters 3&4
3
Fall 2012
For an arbitrarily deforming moving CV, the absolute fluid velocity:
𝐮 = 𝐮𝑠 + 𝐮𝑟
where 𝐮𝑠 is the velocity of the control surface defining CV and 𝐮𝑠 = 𝐮𝑠 (𝒓, 𝑡), and the
relative velocity 𝐮𝑟 = 𝐮𝑟 (𝒓, 𝑡)
Arbitrary fixed control volume
𝑑𝐵𝑠𝑦𝑠
lim
𝑑𝑡
∆𝑡→0
(𝐵𝐶𝑉 )𝑡+∆𝑡 −(𝐵𝐶𝑉 )𝑡
∆𝑡
∆𝑡→0
lim
∆𝑡→0
= lim
∆𝐵𝑡+∆𝑡
∆𝑡
(𝐵𝐶𝑉 +∆𝐵)𝑡+∆𝑡 −(𝐵𝐶𝑉 )𝑡
∆𝑡
Arbitrarily deforming moving control volume
= lim
(𝐵𝐶𝑉 )𝑡+∆𝑡 −(𝐵𝐶𝑉 )𝑡
∆𝑡→0
= time rate of change of 𝐵 in CV =
∆𝑡
𝑑𝐵𝐶𝑉
𝑑𝑡
+ lim
∆𝑡→0
=
𝑑
∫
𝑑𝑡 𝐶𝑉
= net outflux of 𝐵 from CV across CS = ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
∆𝐵𝑡+∆𝑡
∆𝑡
𝛽𝜌𝑑𝑉
058:0160
Jianming Yang
Chapters 3&4
4
Fall 2012
General form RTT for deforming moving control volume
𝑑𝐵sys
=
𝑑𝑡
𝑑
∫
𝑑𝑡 𝐶𝑉
𝛽𝜌𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
Special Cases:
1) Non-deforming CV moving at constant velocity
𝑑𝐵sys
𝑑𝑡
= ∫𝐶𝑉
𝜕
(𝛽𝜌)𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
𝜕𝑡
2) Fixed CV
𝑑𝐵sys
𝑑𝑡
= ∫𝐶𝑉
𝜕
𝜕𝑡
(𝛽𝜌)𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮 ∙ 𝐧𝑑𝐴
Gauss’ Theorem: ∫𝐶𝑉 ∇ ∙ 𝐛𝑑𝑉 = ∫𝐶𝑆 𝐛 ∙ 𝐧𝑑𝐴
𝑑𝐵sys
𝑑𝑡
𝜕
= ∫𝐶𝑉 [ (𝛽𝜌) + ∇ ∙ (𝛽𝜌𝐮)] 𝑑𝑉
𝜕𝑡
Since CV fixed and arbitrary lim
𝑑𝑉→0
3) Steady Flow:
𝜕
𝜕𝑡
gives differential equation.
=0
𝑑𝐵sys
𝑑𝑡
= ∫𝐶𝑆 𝛽𝜌𝐮 ∙ 𝐧𝑑𝐴
4) Uniform flow across discrete CS (steady or unsteady)
∫𝐶𝑆 𝛽𝜌𝐮 ∙ 𝐧𝑑𝐴 = ∑𝐶𝑆 𝛽𝜌𝐮 ∙ 𝐧𝑑𝐴 (− inlet, +outlet)
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Jianming Yang
Chapters 3&4
5
Fall 2012
3 Continuity Equation
𝐵 = 𝑀 = mass of system; 𝛽 = 1
𝑑𝑀
𝑑𝑡
= 0 by definition, system = fixed amount of mass
3.1Integral Form
𝑑𝑀
𝑑𝑡
−
=0=
𝑑
∫
𝑑𝑡 𝐶𝑉
𝑑
∫
𝑑𝑡 𝐶𝑉
𝜌𝑑𝑉 + ∫𝐶𝑆 𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
𝜌𝑑𝑉 = ∫𝐶𝑆 𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
Rate of decrease of mass in CV = net rate of mass outflow across CS.
Note simplifications for non-deforming CV, fixed CV, steady flow, and uniform flow
across discrete CS
Incompressible Fluid: 𝜌 = const.
−
𝑑
∫
𝑑𝑡 𝐶𝑉
𝑑𝑉 = ∫𝐶𝑆 𝐮𝑟 ∙ 𝐧𝑑𝐴
“conservation of volume”
058:0160
Jianming Yang
Fall 2012
Special case of CV form continuity equation:
Fixed CV
𝜕𝜌
∫𝐶𝑉 𝜕𝑡 𝑑𝑉 + ∫𝐶𝑆 𝜌𝐮 ∙ 𝐧𝑑𝐴 = 0
and uniform flow over discrete inlet/outlet
𝜕𝜌
∫𝐶𝑉 𝜕𝑡 𝑑𝑉 + ∑ 𝜌𝐮 ∙ 𝐧𝐴 = 0
and steady flow
∑ 𝜌𝐮 ∙ 𝐧𝐴 = 0
Or
− ∑(𝜌𝑢𝐴)in + ∑(𝜌𝑢𝐴)out = 0
𝜌𝑢𝐴 = 𝜌𝑄 = 𝑚̇
⇒ ∑(𝑚̇)in = ∑(𝑚̇)out
and incompressible flow
− ∑ 𝑄in + ∑ 𝑄out = 0
if non-uniform flow over discrete inlet/outlet
𝑄𝐶𝑆𝑖 = ∫𝐶𝑆 𝐮 ∙ 𝐧𝑑𝐴 = (𝑢𝑎𝑣 𝐴)𝐶𝑆𝑖
𝑢𝑎𝑣 =
1
∫ 𝐮 ∙ 𝐧𝑑𝐴
𝐴 𝐶𝑆
Chapters 3&4
6
058:0160
Jianming Yang
Chapters 3&4
7
Fall 2012
3.2Differential Form
𝜕𝜌
𝜕𝑡
𝜕𝜌
𝜕𝑡
𝐷𝜌
+ ∇ ∙ (𝜌𝐮) = 0
+ 𝜌∇ ∙ 𝐮 + 𝐮 ∙ ∇ρ = 0
+ 𝜌∇ ∙ 𝐮 = 0
𝐷𝑡
1 𝐷𝜌
𝜌 𝐷𝑡
1 𝐷𝜌
𝜌 𝐷𝑡
+∇∙𝐮=0
is the rate of change of 𝜌 per unit 𝜌
∇∙𝐮=
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
is the rate of change of volume 𝑉 per unit volume
It is called the continuity equation since the implication is that 𝜌 and 𝐮 are continuous
functions of 𝐱.
Incompressible Fluid: 𝜌 = const.
∇∙𝐮=
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
=0
The rate of change of volume 𝑉 per unit volume
𝑀 = 𝜌𝑉
⇒
𝑑𝑀
𝑑𝑡
=𝜌
𝑑𝑉
𝑑𝑡
+𝑉
𝑑𝜌
𝑑𝑡
=0
⇒
1 𝐷𝜌
𝜌 𝐷𝑡
=−
1 𝐷𝑉
𝑉 𝐷𝑡
058:0160
Jianming Yang
Chapters 3&4
8
Fall 2012
4 Momentum Equation
RTT:
𝑑𝐵sys
𝑑𝑡
=
𝑑
∫
𝑑𝑡 𝐶𝑉
𝛽𝜌𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
Momentum: 𝐵 = 𝑀𝐮, 𝛽 = 𝐮
4.1Integral Form
𝑑 (𝑀𝐮)
𝑑𝑡
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 =
= ∑𝐅 =
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 + ∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
rate of change of momentum in CV
∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴 = rate of outflux of momentum across CS
∑ 𝐅 = 𝐅𝐵 + 𝐅𝑆 = vector sum of all body forces acting on entire CV and surface
forces acting on entire CS.
𝐅𝐵 = Body forces, which act on entire CV of fluid due to external force field such as
gravity or electrostatic or magnetic forces. Force per unit volume.
𝐅𝑆 = Surface forces, which act on entire CS due to normal (pressure and viscous
stress) and tangential (viscous stresses) stresses. Force per unit area.
When CS cuts through solids, 𝐅𝑆 may also include 𝐅𝑅 = reaction forces, e.g., reaction
force required to hold nozzle or bend when CS cuts through bolts holding nozzle/bend in
place.
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Jianming Yang
Chapters 3&4
9
Fall 2012
4.2Differential Form
𝜕
∫𝐶𝑉 [𝜕𝑡 (𝐮𝜌) + ∇ ∙ (𝐮𝜌𝐮)] 𝑑𝑉 = ∑ 𝐅
where
𝜕
𝜕𝑡
(𝐮𝜌) = 𝐮
𝜕𝜌
𝜕𝑡
+𝜌
𝜕𝐮
𝜕𝑡
, and 𝐮𝜌𝐮 = 𝜌𝐮𝐮 is a tensor
∇ ∙ (𝐮𝜌𝐮) = ∇ ∙ (𝜌𝐮𝐮) =
𝜕
𝜕𝑥
𝜕𝜌
(𝜌𝑢𝐮) +
𝜕
𝜕𝑦
(𝜌𝑣𝐮)
𝜕
𝜕𝑧
(𝜌𝑤𝐮) = 𝐮∇ ∙ (𝜌𝐮) + 𝜌𝐮 ∙ ∇𝐮
𝜕𝐮
∫𝐶𝑉 [𝐮 ( 𝜕𝑡 + ∇ ∙ (𝜌𝐮)) + 𝜌 ( 𝜕𝑡 + 𝐮 ∙ ∇𝐮)] 𝑑𝑉 = ∑ 𝐅
Since the continuity equation:
𝐷𝐮
∫𝐶𝑉 𝜌 𝐷𝑡 𝑑𝑉 = ∑ 𝐅
𝜕𝜌
+ ∇ ∙ (𝜌𝐮) = 0 and
𝜕𝑡
𝐷𝐮
⇒ 𝜌
𝐷𝑡
= ∑𝐟
𝜕𝐮
𝜕𝑡
+ 𝐮 ∙ ∇𝐮 =
𝐷𝐮
𝐷𝑡
per elemental fluid volume
𝜌𝐚 = 𝐟𝑏 + 𝐟𝑠
𝐟𝑏 = body force per unit volume
𝐟𝑠 = surface force per unit volume
4.2.1 Body forces
Body forces are due to external fields such as gravity or magnetic fields. Here we only
consider a gravitational field; that is,
∑ 𝐅body = 𝑑𝐅grav = 𝜌𝐠𝑑𝑥𝑑𝑦𝑑𝑧
or,
and 𝐠 = −𝑔𝐤, i.e., 𝐟body = −𝜌𝑔𝐤.
058:0160
Jianming Yang
4.2.2
Fall 2012
Chapters 3&4
10
Surface Forces
Surface forces are due to the stresses that act on the sides of the control surfaces
Considering the torque produced on the fluid element by the various stresses’
components and its rotational dynamics in the limit 𝑑𝑉 ⟶ 0, it can be shown stresses
themselves cause no rotation of a fluid point and the stress tensor is symmetric, i.e.,
𝜎𝑖𝑗 = 𝜎𝑗𝑖
Decomposition of the stress:
𝜎𝑖𝑗 = −𝑝𝛿𝑖𝑗 + 𝜏𝑖𝑗
−𝑝 + 𝜏𝑥𝑥
𝜏𝑥𝑦
𝜏𝑥𝑧
−𝑝 + 𝜏𝑦𝑦
𝜏𝑦𝑧 ]
= [ 𝜏𝑦𝑥
𝜏𝑧𝑥
𝜏𝑧𝑦
−𝑝 + 𝜏𝑧𝑧
𝐟𝑠 = 𝐟𝑝 + 𝐟𝜏
As shown before, for 𝑝 alone it is not
the stresses themselves that cause a
net force but their gradients. Recall
𝐟𝑝 = −∇𝑝 based on 1st-order Taylor
series. 𝐟𝜏 is more complex since 𝜏𝑖𝑗 is
a 2nd-order tensor, but similarly as for
𝑝, the force is due to stress gradients and are derived based on 1st order Taylor series.
058:0160
Jianming Yang
Chapters 3&4
11
Fall 2012
Resultant stress on each face:
𝛔𝑥 = 𝜎𝑥𝑥 𝐞𝑥 + 𝜎𝑥𝑦 𝐞𝑦 + 𝜎𝑥𝑧 𝐞𝑧
𝛔𝑦 = 𝜎𝑦𝑥 𝐞𝑥 + 𝜎𝑦𝑦 𝐞𝑦 + 𝜎𝑦𝑧 𝐞𝑧
𝛔𝑧 = 𝜎𝑧𝑥 𝐞𝑥 + 𝜎𝑦𝑦 𝐞𝑦 + 𝜎𝑧𝑧 𝐞𝑧
𝐅𝑠 = [
𝜕
𝜕𝑥
𝐟𝑠 =
𝐟𝑠 = [
+[
+[
𝜕
𝜕𝑥
𝜕
𝜕𝑥
𝜕
𝜕𝑥
𝜕
(𝛔𝑥 ) +
𝜕
𝜕𝑥
𝜕𝑦
(𝛔𝑦 ) +
(𝛔𝑥 ) +
(𝜎𝑥𝑥 ) +
(𝜎𝑥𝑦 ) +
(𝜎𝑥𝑧 ) +
(𝑓𝑠 )𝑥 =
(𝑓𝑠 )𝑦 =
𝜕
𝜕𝑥
𝜕
𝜕𝑥
𝜕
𝜕
𝜕𝑦
𝜕
𝜕𝑦
𝜕
𝜕𝑦
𝜕
𝜕𝑦
𝜕
𝜕𝑧
(𝛔𝑧 )] 𝑑𝑥𝑑𝑦𝑑𝑧
(𝛔𝑦 ) +
(𝜎𝑦𝑦 ) +
(𝜎𝑥𝑥 ) +
(𝜎𝑥𝑦 ) +
𝜕
𝜕𝑦
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝜕
(𝜎𝑦𝑥 ) +
(𝜎𝑦𝑧 ) +
𝜕
(𝜎𝑧𝑥 )] 𝐞𝑥
𝜕𝑧
𝜕
𝜕𝑧
𝜕
𝜕𝑧
(𝛔𝑧 )
(𝜎𝑧𝑦 )] 𝐞𝑦
(𝜎𝑧𝑧 )] 𝐞𝑧 𝐟𝑠 = ∇ ∙ 𝛔 =
(𝜎𝑦𝑥 ) +
(𝜎𝑦𝑦 ) +
𝜕
𝜕𝑧
𝜕
𝜕𝑧
𝜕
𝜕
𝜕𝑥𝑗
(𝜎𝑗𝑖 )
⇒
(𝜎𝑧𝑥 )
(𝜎𝑧𝑦 )
{ (𝑓𝑠 )𝑧 = 𝜕𝑥 (𝜎𝑥𝑧 ) + 𝜕𝑦 (𝜎𝑦𝑧 ) + 𝜕𝑧 (𝜎𝑧𝑧 )
Putting together the above results,
𝜌𝐚 = 𝜌
𝐷𝐮
𝐷𝑡
= 𝜌𝐠 + ∇ ∙ 𝛔
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Jianming Yang
Chapters 3&4
12
Fall 2012
4.2.3 Strain and Rotation Rates
Next, we need to relate the stresses 𝜎𝑖𝑗 to the fluid motion, i.e. the velocity field. To this
end, we examine the relative motion between two neighboring fluid particles at 𝐱 and 𝐱 +
𝑑𝐱, respectively.
At 𝐱 + 𝑑𝐱, the first-order Taylor series of the
fluid velocity is
𝐮(𝐱 + 𝑑𝐱) = 𝐮(𝐱) + 𝑑𝐮 = 𝐮(𝐱) + ∇𝐮 ∙ 𝑑𝐱
𝑑𝐮 = ∇𝐮 ∙ 𝑑𝐱 =
𝜕𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑥
𝜕𝑣
𝜕𝑦
𝜕𝑣
𝜕𝑧
𝜕𝑣
𝜕𝑥
𝜕𝑤
𝜕𝑥
𝜕𝑤
𝜕𝑧
𝜕𝑤
[ 𝜕𝑥
𝜕𝑦
𝜕𝑧 ]
𝑑𝑥
𝜕𝑢
[𝑑𝑦] = 𝑖 𝑑𝑥𝑗
𝜕𝑥𝑗
𝑑𝑧
The velocity gradient (deformation rate) tensor
𝜕𝑢𝑖
𝜕𝑥𝑗
1 𝜕𝑢𝑖
= (
2 𝜕𝑥𝑗
+
𝜕𝑢𝑗
𝜕𝑥𝑖
1 𝜕𝑢𝑖
)+ (
2 𝜕𝑥𝑗
−
𝜕𝑢𝑗
𝜕𝑥𝑖
1
) = 𝑆𝑖𝑗 + 𝑅𝑖𝑗
2
𝑆𝑖𝑗 is the strain rate tensor, the symmetric part of the velocity gradient tensor, 𝑆𝑖𝑗 = 𝑆𝑗𝑖
𝑅𝑖𝑗 is the rotation tensor, which is antisymmetric, 𝑅𝑖𝑗 = −𝑅𝑗𝑖
058:0160
Jianming Yang
Chapters 3&4
13
Fall 2012
𝜕𝑢
Strain rate tensor
4.2.4
𝑆𝑖𝑗 =
1 𝜕𝑢
𝜕𝑥
1 𝜕𝑣
𝜕𝑢
(
2 𝜕𝑥
1 𝜕𝑤
+
(
)
𝜕𝑦
𝜕𝑢
[2 ( 𝜕𝑥 + 𝜕𝑧 )
Linear Strain Rates
+
𝜕𝑣
2 𝜕𝑦
𝜕𝑥
𝜕𝑣
)
𝜕𝑦
1 𝜕𝑤
𝜕𝑣
(
2 𝜕𝑦
+
𝜕𝑧
)
1 𝜕𝑢
(
+
(
+
2 𝜕𝑧
1 𝜕𝑣
𝜕𝑤
𝜕𝑥
𝜕𝑤
2 𝜕𝑧
𝜕𝑦
𝜕𝑤
𝜕𝑧
)
)
]
The diagonal terms of the strain rate tensor represent elongation and contraction per
unit length in the various coordinate directions.
Take 𝑆11 as an example, the rate of change of fluid element length in the 𝑥-direction per
unit length in this direction is
1 𝐷(𝛿𝑥)
𝛿𝑥
𝐷𝑡
= lim
1
𝑑𝑡→0 𝑑𝑡
(
𝐴′ 𝐵′ −𝐴𝐵
𝐴𝐵
) = lim
1
𝑑𝑡→0 𝛿𝑥𝑑𝑡
(𝛿𝑥 +
𝜕𝑢
𝜕𝑥
𝛿𝑥𝑑𝑡 − 𝛿𝑥) =
𝜕𝑢
𝜕𝑥
The sum of the diagonal terms of 𝑆𝑖𝑗 is the volumetric strain rate or bulk strain rate.
𝑆𝑖𝑖 =
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
=∇∙𝐮=
1 𝐷𝑉
𝑉 𝐷𝑡
It specifies the rate of volume change
per unit volume and it does not depend
on the orientation of the coordinate
system.
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Angular Strain Rates
For 𝑆𝑖𝑗 ’s first off-diagonal component, 𝑆12 = 𝑆21 , the average rate at which the initially
perpendicular segments 𝛿𝑥 and 𝛿𝑦 rotate toward each other is
1 𝐷(𝛼+𝛽 )
2
= lim
𝐷𝑡
1
[
1
(
𝜕𝑢
𝑑𝑡→0 2𝑑𝑡 𝛿𝑦 𝜕𝑦
𝛿𝑦𝑑𝑡) +
1
(
𝜕𝑣
𝛿𝑥 𝜕𝑥
1 𝜕𝑢
𝛿𝑥𝑑𝑡)] = (
2 𝜕𝑦
+
𝜕𝑣
𝜕𝑥
) = 𝑆12 = 𝑆21
The off-diagonal terms of 𝑆𝑖𝑗 represent the average rate at which line segments initially
parallel to the 𝑖- and 𝑗-directions rotate toward each other (shear deformations).
𝑆𝑖𝑗 is zero for any rigid body motion
composed of translation at a spatially
uniform velocity 𝐔 and rotation at a
constant rate 𝛀. Thus, 𝑆𝑖𝑗 is independent
of the frame of reference
Rates of shear deformations
1 𝜕𝑢
(
+
(
+
(
+
2 𝜕𝑦
1 𝜕𝑢
2 𝜕𝑧
1 𝜕𝑣
2 𝜕𝑧
𝜕𝑣
) = distortion w.r.t. (𝑥, 𝑦) plane
𝜕𝑥
𝜕𝑤
𝜕𝑥
𝜕𝑤
𝜕𝑦
) = distortion w.r.t. (𝑥, 𝑧) plane
) = distortion w.r.t. (𝑦, 𝑧) plane
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4.2.6 Rotation Rates
Rotation tensor
𝜕𝑢
𝜕𝑣
0
−
𝜕𝑦
𝑅𝑖𝑗 =
𝜕𝑣
𝜕𝑥
𝜕𝑤
−
−
𝜕𝑢
𝜕𝑥
𝜕𝑢
𝜕𝑧
𝜕𝑣
0
𝜕𝑦
𝜕𝑢
𝜕𝑤
−
𝜕𝑧
−
−
𝜕𝑤
𝜕𝑥
𝜕𝑤
𝜕𝑦
𝜕𝑣
0 ]
[ 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧
𝑅𝑖𝑗 is antisymmetric: 𝑅𝑖𝑗 = −𝑅𝑗𝑖
For the motion of an initially square fluid
element in the (𝑥, 𝑦) -plane when 𝜕𝑢/𝜕𝑦
and 𝜕𝑣/𝜕𝑥 are nonzero and unequal so
that 𝑅12 = −𝑅21 ≠ 0. In this situation, the fluid element translates and deforms in the
(𝑥, 𝑦)-plane, and rotates about the third coordinate axis. The average rotation rate is
1 𝐷(−𝛼+𝛽 )
2
𝐷𝑡
= lim
1
𝑑𝑡→0 2𝑑𝑡
1
𝜕𝑢
2
𝜕𝑦
= (−
[−
+
1
(
𝜕𝑢
𝛿𝑦 𝜕𝑦
𝜕𝑣
𝜕𝑥
𝛿𝑦𝑑𝑡) +
)=−
𝑅12
2
=
1
(
𝜕𝑣
𝛿𝑥 𝜕𝑥
𝛿𝑥𝑑𝑡)]
𝑅21
2
𝑅𝑖𝑗 represents twice the fluid element rotation rate. This means that 𝑅𝑖𝑗 depends on the
(rotating or non-rotating) frame of reference in which they are determined.
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The three independent elements of 𝑅𝑖𝑗 are related with the vorticity vector, 𝛚 = ∇ × 𝐮
0
−𝜔3 𝜔2
0
−𝜔1 ]
𝑅𝑖𝑗 = −𝜀𝑖𝑗𝑘 (∇ × 𝐮)𝑘 = −𝜀𝑖𝑗𝑘 𝜔𝑘 = [ 𝜔3
−𝜔2 𝜔1
0
𝜕𝑤
𝜕𝑣
𝜕𝑢
𝜕𝑤
𝜕𝑣
𝜕𝑢
where 𝜔1 =
− , 𝜔2 = − , 𝜔3 = −
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑥
𝜕𝑥
𝜕𝑦
Fluid motion is called irrotational if 𝛚 = ∇ × 𝐮 = 0.
Circulation 𝛤 is the amount of fluid rotation within a closed contour (or circuit) 𝐶.
𝛤 = ∮𝐶 𝐮 ∙ 𝑑𝐬 = ∫𝐴 𝛚 ∙ 𝐧𝑑𝐴
4.2.7
Decomposition of Motion
𝐮(𝐱 + 𝑑𝐱) = 𝐮(𝐱) + 𝑑𝐮 = 𝐮(𝐱) + ∇𝐮 ∙ 𝑑𝐱
1
= 𝐮(𝐱) + 𝐒 ∙ 𝑑𝐱 + 𝛚 × 𝑑𝐱
2
1
1
𝑑𝑢𝑖 = (𝑆𝑖𝑗 − 𝜀𝑖𝑗𝑘 𝜔𝑘 ) 𝑑𝑥𝑗 = 𝑆𝑖𝑗 𝑑𝑥𝑗 + (𝛚 × 𝑑𝐱)𝑖
2
2
where 𝜀𝑖𝑗𝑘 𝜔𝑘 𝑑𝑥𝑗 is the 𝑖 -component of the cross
product −𝛚 × 𝑑𝐱. Because the velocity at a distance 𝐱
from the axis of rotation of a rigid body rotating at
1
angular velocity 𝛀 is 𝛀 × 𝐱. Thus, (𝛚 × 𝑑𝐱)𝑖 represents the velocity of point 𝑃 relative
2
to point 𝑂 because of an angular velocity of 𝛚⁄2.
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Thus, general motion consists of:
1) pure translation described by 𝐮
2) volumetric dilatation described by 𝑆𝑖𝑖
3) distortion in shape described by 𝑆𝑖𝑗 , 𝑖 ≠ 𝑗
4) rigid-body rotation described by 𝛚⁄2
4.2.8 Constitutive Equation for a Newtonian Fluid
It is now necessary to make certain postulates concerning the relationship between the
fluid stress tensor (𝜎𝑖𝑗 ) and deformation rate tensor (𝜕𝑢𝑖 ⁄𝜕𝑥𝑗 ). These postulates are
based on physical reasoning and experimental observations and have been verified
experimentally even for extreme conditions. For a Newtonian fluid:
1) When the fluid is at rest the stress is hydrostatic and the pressure is the
thermodynamic pressure
2) 𝜎𝑖𝑗 is linearly related to 𝜕𝑢𝑖 ⁄𝜕𝑥𝑗 and only depends on 𝜕𝑢𝑖 ⁄𝜕𝑥𝑗 .
3) Since there is no shearing action in rigid body rotation, it causes no shear stress.
4) There is no preferred direction in the fluid, so that the fluid properties are point
functions (condition of isotropy).
Using statements 1-3:
𝜎𝑖𝑗 = −𝑝𝛿𝑖𝑗 + 𝐾𝑖𝑗𝑚𝑛 𝑆𝑚𝑛
𝐾𝑖𝑗𝑚𝑛 = 4th-order tensor with 81 components such that each of the 9 components of 𝜎𝑖𝑗 is
linearly related to all 9 components of 𝑆𝑖𝑗 .
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However, statement (4) requires that the fluid has no directional preference, i.e. 𝜎𝑖𝑗 is
independent of rotation of coordinate system, which means 𝐾𝑖𝑗𝑚𝑛 is an isotropic tensor =
even order tensor made up of products of 𝛿𝑖𝑗 .
𝐾𝑖𝑗𝑚𝑛 = 𝜆𝛿𝑖𝑗 𝛿𝑚𝑛 + 𝜇𝛿𝑖𝑚 𝛿𝑗𝑛 + 𝛾𝛿𝑖𝑛 𝛿𝑗𝑚
𝜆, 𝜇, 𝛾 are scalars.
Lastly, the symmetry condition 𝜎𝑖𝑗 = 𝜎𝑗𝑖 requires,
𝐾𝑖𝑗𝑚𝑛 = 𝐾𝑗𝑖𝑚𝑛
𝜎𝑖𝑗 = −𝑝𝛿𝑖𝑗 + 𝐾𝑖𝑗𝑚𝑛 𝑆𝑚𝑛
⇒
⇒
𝛾=𝜇
𝜎𝑖𝑗 = −𝑝𝛿𝑖𝑗 + 2𝜇𝑆𝑖𝑗 + 𝜆𝑆𝑚𝑚 𝛿𝑖𝑗
𝜆 and 𝜇 can be further related if one considers mean normal stress vs. thermodynamic 𝑝.
𝜎𝑖𝑖 = −3𝑝 + (2𝜇 + 3𝜆)𝑆𝑚𝑚 = −3𝑝 + (2𝜇 + 3𝜆)∇ ∙ 𝐮
1
2
3
3
𝑝 = − 𝜎𝑖𝑖 + ( 𝜇 + 𝜆) ∇ ∙ 𝐮
1
Mean normal stress: 𝑝̅ = − 𝜎𝑖𝑖
3
2
𝑝 − 𝑝̅ = ( 𝜇 + 𝜆) ∇ ∙ 𝐮
3
Incompressible flow: 𝑝 = 𝑝̅ and absolute pressure is indeterminate since there is no
equation of state for 𝑝. Equations of motion determine ∇𝑝.
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Compressible flow: 𝑝 ≠ 𝑝̅ and 𝜇𝑣 = 2𝜇⁄3 + 𝜆, coefficient of bulk viscosity must be
1 𝐷𝜌
determined; however, it is a very difficult measurement requiring large ∇ ∙ 𝐮 = −
=
𝜌 𝐷𝑡
1 𝐷𝑉
𝑉 𝐷𝑡
, e.g., within shock waves.
2
𝜆+ 𝜇 =0
Stokes assumption:
3
is found to be accurate in many situations.
𝑝 = 𝑝̅
2
𝜎𝑖𝑗 = − (𝑝 + 𝜇∇ ∙ 𝐮) 𝛿𝑖𝑗 + 2𝜇𝑆𝑖𝑗
3
Generalization of 𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
for 3D flow,
𝜎𝑖𝑗 = 𝜇 (
𝜕𝑢𝑖
𝜕𝑥𝑗
+
𝜕𝑢𝑗
𝜕𝑥𝑖
)
𝑖≠𝑗
relates shear stress to strain rate.
2
𝜕𝑢
3
𝜕𝑥
𝜎𝑥𝑥 = − (𝑝 + 𝜇∇ ∙ 𝐮) + 2𝜇
1
𝜕𝑢
3
𝜕𝑥
= −𝑝 + 2𝜇 (− ∇ ∙ 𝐮 +
)
where the normal viscous stress 2𝜇[− (∇ ∙ 𝐮)/3 + 𝜕𝑢/𝜕𝑥 ] is the difference between the
extension rate in the 𝑥 direction and average expansion at a point. Only differences from
1
the average ∇ ∙ 𝐮 generate normal viscous stresses. For incompressible fluids, average
3
= 0, i.e. ∇ ∙ 𝐮 = 0.
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4.2.9 Non-Newtonian fluids
𝜏𝑖𝑗 ∝ 𝑆𝑖𝑗 for small strain rates, which works well for air, water, etc. Newtonian fluids
𝑛
𝑆𝑖𝑗
𝜏𝑖𝑗 ∝
effect.
+
𝜕𝑆𝑖𝑗
𝜕𝑡
non-Newtonian viscoelastic materials: nonlinear relation and history
Non-Newtonian fluids include:
(1) Polymers molecules with large molecular weights and form long chains coiled
together in spongy ball shapes that deform under shear.
(2) Emulsions and slurries containing suspended particles such as blood and
water/clay
4.2.10 Navier Stokes Momentum Equation
𝜌𝐚 = 𝜌
𝜌
𝐷𝑢𝑗
𝐷𝑡
= 𝜌𝑔𝑗 −
𝜕𝑝
𝜕𝑥𝑗
+
𝜕
𝜕𝑥𝑖
𝐷𝐮
𝐷𝑡
[𝜇 (
= 𝜌𝐠 + ∇ ∙ 𝛔
𝜕𝑢𝑗
𝜕𝑥𝑖
+
𝜕𝑢𝑖
𝜕𝑥𝑗
2
𝜕𝑢𝑘
3
𝜕𝑥𝑘
) + (𝜇𝑣 − 𝜇)
𝛿𝑖𝑗 ]
Recall 𝜇 = 𝜇(𝑇) and 𝜇 increases with temperature for gases, decreases with temperature
for liquids, but if its assumed that 𝜇 = const.:
𝜌
𝐷𝑢𝑗
𝐷𝑡
= 𝜌𝑔𝑗 −
𝜕𝑝
𝜕𝑥𝑗
+𝜇
𝜕
𝜕𝑥𝑖
(
𝜕𝑢𝑗
𝜕𝑥𝑖
)+𝜇
𝜕
(
𝜕𝑢𝑖
𝜕𝑥𝑗 𝜕𝑥𝑖
2
) + (𝜇𝑣 − 𝜇)
3
𝜕
(
𝜕𝑢𝑘
𝜕𝑥𝑗 𝜕𝑥𝑘
)
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For compressible flow
𝜌
𝐷𝑢𝑗
𝐷𝑡
𝜕𝑝
= 𝜌𝑔𝑗 −
𝜕𝑥𝑗
For incompressible flow ∇ ∙ 𝐮 =
𝜌
or,
+𝜇
𝜕𝑢𝑘
𝜕𝑥𝑘
𝐷𝑢𝑗
𝐷𝑡
𝜌
𝐷𝐮
𝐷𝑡
𝜕
𝜕𝑥𝑖
(
𝜕𝑢𝑗
𝜕𝑥𝑖
1
) + (𝜇𝑣 + 𝜇)
3
𝜕
(
𝜕𝑢𝑘
𝜕𝑥𝑗 𝜕𝑥𝑘
)
=0
= 𝜌𝑔𝑗 −
𝜕𝑝
𝜕𝑥𝑗
+𝜇
𝜕
𝜕𝑥𝑖
(
𝜕𝑢𝑗
𝜕𝑥𝑖
)
= 𝜌𝐠 − ∇𝑝 + 𝜇∇2 𝐮
For 𝜇 = 0, Euler Equation
𝜌
𝐷𝐮
𝐷𝑡
= 𝜌𝐠 − ∇𝑝
NS equations for constant 𝜌, 𝜇, the piezometric pressure: 𝑝̂ = 𝑝 + 𝜌𝑔𝑧
−∇𝑝̂ = 𝜌𝐠 − ∇𝑝
𝜌
𝜌[
𝜕𝐮
𝜕𝑡
𝐷𝐮
𝐷𝑡
= −∇𝑝̂ + 𝜇∇2 𝐮
+ 𝐮 ∙ ∇𝐮] = −∇𝑝̂ + 𝜇∇2 𝐮
Non-linear 2nd-order PDE, as is the case for 𝜌, 𝜇 not constant.
Combine with ∇ ∙ 𝐮 for 4 equations for 4 unknowns 𝐮, 𝑝 and can be, albeit difficult,
solved subject to initial and boundary conditions for 𝐮, 𝑝 at 𝑡 = 𝑡0 and on all boundaries
i.e. “well posed” Initial Boundary Value Problem (IBVP).
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4.2.11 Application of Differential Momentum Equation
1. Navier-Stokes equation valid both laminar and turbulent flow
𝑈𝛿
2. Laminar flow: 𝑅𝑒crit =
≤ 1000, 𝑅𝑒 > 𝑅𝑒crit instability
𝜐
3. Turbulent flow 𝑅𝑒transition ≥ (10~20)𝑅𝑒crit
Random motion superimposed on mean coherent structures.
Cascade: energy from large scale dissipates at smallest scales due to viscosity.
Kolmogorov hypothesis for smallest scales
4. No exact solutions for turbulent flow: RANS, DES, LES, DNS (all CFD)
5. 80 exact solutions for simple laminar flows are mostly linear 𝐮 ∙ ∇𝐮 = 0
a. Couette flow = shear driven
b. Steady duct flow = Poiseuille flow
c. Unsteady duct flow
d. Unsteady moving walls
e. Asymptotic suction
f. Wind-driven flows
g. Similarity solutions. etc.
6. Also many exact solutions for low Re Stokes and high Re BL approximations
7. Can also use CFD for non-simple laminar flows
8. AFD or CFD requires well posed IBVP; therefore, exact solutions are useful for
setup of IBVP, physics, and verification CFD
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4.3Application of CV Momentum Equation
∑𝐅 =
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 + ∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
𝐅 = 𝐅𝐵 + 𝐅𝑆
Note:
1. Vector equation
2. 𝐧 = outward unit normal: 𝐮𝑟 ∙ 𝐧 < 0: inlet, > 0 outlet
3. 1D Momentum flux
∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴 = ∑(𝑚̇𝑖 𝐮𝑖 )out − ∑(𝑚̇𝑖 𝐮𝑖 )in
where 𝑚̇𝑖 = 𝜌𝑖 𝑢𝑛𝑖 𝐴𝑖 , 𝐮𝑖 , 𝜌𝑖 are assumed uniform over discrete inlets and outlets
∑𝐅 =
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 + ∑(𝑚̇𝑖 𝐮𝑖 )out − ∑(𝑚̇𝑖 𝐮𝑖 )in
∑ 𝐅: net force on CV
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉: time rate of change of momentum in CV
∑(𝑚̇𝑖 𝐮𝑖 )out : outlet momentum flux
∑(𝑚̇𝑖 𝐮𝑖 )in : inlet momentum flux
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4. Momentum flux correlation factors
2
∫𝐶𝑆 𝐮𝜌𝐮 ∙ 𝐧𝑑𝐴 = 𝜌 ∫𝐶𝑆 𝑢2 𝑑𝐴 = 𝛽𝜌𝐴𝑈𝑎𝑣
1
2
𝑢
where 𝛽 = ∫𝐶𝑆 ( ) 𝑑𝐴,
𝐴
𝑈
𝑎𝑣
1
𝑈𝑎𝑣 = ∫𝐶𝑆 𝑢𝑑𝐴 = 𝑄 ⁄𝐴,
𝐴
𝜌 ∫𝐶𝑆 𝑢2 𝑑𝐴 is the axial flow with non-uniform velocity profile
Laminar pipe flow:
𝑢 = 𝑈0 (1 −
𝑟2
𝑅2
𝑟 1/2
) ≈ 𝑈0 (1 − )
𝑅
𝑈𝑎𝑣 = 0.53𝑈0 , 𝛽 =
,
4
3
Turbulent pipe flow:
𝑟 𝑚 1
𝑢 ≈ 𝑈0 (1 − ) , ≤ 𝑚 ≤
𝑅
9
2
𝑈,
1+𝑚)(2+𝑚) 0
𝑈𝑎𝑣 = (
𝛽=
(1+𝑚)2 (2+𝑚)2
,
2(1+2𝑚)(2+2𝑚)
1
5
1
𝑚 = , 𝑈𝑎𝑣 = 0.82𝑈0
7
1
𝑚 = , 𝛽 = 1.02
7
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Jianming Yang
Chapters 3&4
25
Fall 2012
5. Constant 𝑝 causes no force; Therefore, use
𝑝gage = 𝑝atm − 𝑝abs
𝐅𝑝 = − ∫𝐶𝑆 𝑝𝐧𝑑𝐴 = − ∫𝐶𝑉 ∇𝑝𝑑𝑉 = 0, for 𝑝 = const.
6. At jet exit to atmosphere 𝑝gage = 0.
7. Choose CV carefully with CS normal to flow and indicating coordinate system and
∑ 𝐅 on CV similar as free body diagram used in dynamics.
8. Many applications, usually with continuity and energy equations. Careful practice is
needed for mastery.
a. Steady and unsteady developing and fully developed pipe flow
b. Emptying or filling tanks
c. Forces on transitions
d. Forces on fixed and moving vanes
e. Hydraulic jump
f. Boundary Layer and bluff body drag
g. Rocket or jet propulsion
h. Nozzle
i. Propeller
j. Water-hammer
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Jianming Yang
Chapters 3&4
26
Fall 2012
9. Relative inertial coorindates
The CV moves at a constant velocity 𝐮𝑐𝑠 with respect to the absolute inertial
coordinates. If 𝐮𝑟 represents the velocity in the relative intertial coordinates that
move together with the CV, then 𝐮𝑟 = 𝐮 − 𝐮𝑐𝑠
Reynolds transport theorem for an arbitrary moving deforming CV:
𝑑𝐵sys
𝑑𝑡
=
𝑑
∫
𝑑𝑡 𝐶𝑉
𝛽𝜌𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
For a non-deforming CV moving at constant velocity, RTT for incompressible flow:
𝑑𝐵sys
𝑑𝑡
Conservation of mass:
= 𝜌 ∫𝐶𝑉
𝜕𝛽
𝜕𝑡
𝑑𝑉 + 𝜌 ∫𝐶𝑆 𝛽𝐮𝑟 ∙ 𝐧𝑑𝐴
𝐵sys = 𝑀, 𝛽 = 1
For steady flow: ∫𝐶𝑆 𝐮𝑟 ∙ 𝐧𝑑𝐴 = 0
Conservation of momentum: 𝐵sys
𝑑 [𝑀(𝐮𝑟 +𝐮𝑐𝑠 )]
𝑑𝑡
= ∑ 𝐅 = 𝜌 ∫𝐶𝑉
= 𝑀(𝐮𝑟 + 𝐮𝑐𝑠 ), 𝛽 =
𝜕(𝐮𝑟 +𝐮𝑐𝑠 )
𝜕𝑡
𝑑𝐵sys
𝑑𝑀
= 𝐮𝑟 + 𝐮𝑐𝑠
𝑑𝑉 + 𝜌 ∫𝐶𝑆 (𝐮𝑟 + 𝐮𝑐𝑠 )𝐮𝑟 ∙ 𝐧𝑑𝐴
For steady flow with the use of continuity:
∑ 𝐅 = 𝜌 ∫𝐶𝑆 (𝐮𝑟 + 𝐮𝑐𝑠 )𝐮𝑟 ∙ 𝐧𝑑𝐴 = 𝜌 ∫𝐶𝑆 𝐮𝑟 𝐮𝑟 ∙ 𝐧𝑑𝐴 + 𝜌𝐮𝑐𝑠 ∫𝐶𝑆 𝐮𝑟 ∙ 𝐧𝑑𝐴
= 𝜌 ∫𝐶𝑆 𝐮𝑟 𝐮𝑟 ∙ 𝐧𝑑𝐴
(∫𝐶𝑆 𝐮𝑟 ∙ 𝐧𝑑𝐴 = 0)
058:0160
Jianming Yang
Fall 2012
4.4CV continuity equation for steady incompressible flow
One inlet and outlet 𝐴 = const.
𝜌 ∫in 𝐮 ∙ 𝐧𝑑𝐴 = 𝜌 ∫out 𝐮 ∙ 𝐧𝑑𝐴 = 𝑚̇ = 𝜌𝑄
𝑄in = 𝑄out
(𝑈𝑎𝑣 𝐴)in = (𝑈𝑎𝑣 𝐴)out
For 𝐴 = const., (𝑈𝑎𝑣 )in = (𝑈𝑎𝑣 )out
∑ 𝐅 = 𝜌 ∫in 𝐮(𝐮 ∙ 𝐧)𝑑𝐴 + 𝜌 ∫out 𝐮(𝐮 ∙ 𝐧)𝑑𝐴
Pipe:
∑ 𝐹𝑥 = 𝜌 ∫in 𝑢(𝐮 ∙ 𝐧)𝑑𝐴 + 𝜌 ∫out 𝑢(𝐮 ∙ 𝐧)𝑑𝐴
2 )
2
= −𝜌(𝛽𝐴𝑈𝑎𝑣
in + 𝜌(𝛽𝐴𝑈𝑎𝑣 )out
= 𝜌𝑄𝑈𝑎𝑣 (𝛽out − 𝛽in )
Change in shape of velocity profile
Vane:
∑ 𝐅 = 𝑚̇(𝑈out − 𝑈in )
|𝑈out | = |𝑈in |
∑ 𝐹𝑥 = 𝑚̇(𝑈out − 𝑈in ) = 𝑚̇(−2𝑈in )
Change in direction of velocity
Chapters 3&4
27
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Jianming Yang
Fall 2012
5 Review of Control Volume Method
5.1Volume and Mass Rate of Flow
The volume flow 𝑄 passing through a surface (imaginary) defined in the flow
𝑑𝑉 = 𝑢𝑑𝑡𝑑𝐴 cos 𝜃 = (𝐮 ∙ 𝐧)𝑑𝑡𝑑𝐴
𝐧 is the outward normal unit vector.
𝐮 ∙ 𝐧 > 0, outflow
𝐮 ∙ 𝐧 < 0, inflow
Total volume rate of Flow 𝑄
𝑄 = ∫𝑆 (𝐮 ∙ 𝐧)𝑑𝐴 = ∫𝑆 𝑢𝑛 𝑑𝐴
Mass flow:
𝑚̇ = ∫𝑆 𝜌(𝐮 ∙ 𝐧)𝑑𝐴 = ∫𝑆 𝜌𝑢𝑛 𝑑𝐴
One-dimensional approximation if constant density and velocity over 𝑆
𝑚̇ = 𝜌𝑄 = 𝜌𝐴𝑈
Chapters 3&4
28
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Jianming Yang
Fall 2012
5.2Conservation of Mass
Fixed control volume, uniform flow over discrete inlet/outlet, and steady flow
− ∑(𝜌𝑢𝐴)in + ∑(𝜌𝑢𝐴)out = 0
𝜌𝑢𝐴 = 𝜌𝑄 = 𝑚̇
⇒ ∑(𝑚̇)in = ∑(𝑚̇)out
Constant density:
− ∑ 𝑄in + ∑ 𝑄out = 0
If non-uniform flow over discrete inlet/outlet
𝑄𝐶𝑆𝑖 = ∫𝐶𝑆 𝐮 ∙ 𝐧𝑑𝐴 = (𝑢𝑎𝑣 𝐴)𝐶𝑆𝑖
1
𝑢𝑎𝑣 = ∫𝐶𝑆 𝐮 ∙ 𝐧𝑑𝐴
𝐴
One inlet and one outlet:
𝜌 ∫in 𝐮 ∙ 𝐧𝑑𝐴 = 𝜌 ∫out 𝐮 ∙ 𝐧𝑑𝐴 = 𝑚̇ = 𝜌𝑄
𝑄in = 𝑄out
(𝑈𝑎𝑣 𝐴)in = (𝑈𝑎𝑣 𝐴)out
𝐴 = const.,
(𝑈𝑎𝑣 )in = (𝑈𝑎𝑣 )out
Chapters 3&4
29
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Jianming Yang
Chapters 3&4
30
Fall 2012
5.3Conservation of Linear Momentum
1. Vector equation, forces and momentum terms and directional.
2. u has a sign depending on its direction. ρu ∙ n < 0: inlet, > 0 outlet
3. One-dimensional approximation and momentum flux correlation factor β
2
𝜌 ∫𝐶𝑆 𝐮𝐮 ∙ 𝐧𝑑𝐴 = 𝜌 ∫𝐶𝑆 𝑢2 𝑑𝐴 = 𝛽𝜌𝐴𝑈𝑎𝑣
1
𝑢
2
𝛽 = ∫𝐶𝑆 ( ) 𝑑𝐴,
𝐴
𝑈
𝑎𝑣
1
𝑈𝑎𝑣 = ∫𝐶𝑆 𝑢𝑑𝐴 = 𝑄 ⁄𝐴,
𝐴
4. Constant p causes no force; Therefore, use pgage = patm − pabs
5. At jet exit to atmosphere pgage = 0.
6. Choose CV with CS normal to flow and indicating coordinate system
7. Relative inertial coorindates, for steady flow:
∫𝐶𝑆 𝐮𝑟 ∙ 𝐧𝑑𝐴 = 0
∑ 𝐅 = 𝜌 ∫𝐶𝑆 𝐮𝑟 𝐮𝑟 ∙ 𝐧𝑑𝐴
058:0160
Jianming Yang
Chapters 3&4
31
Fall 2012
5.4The Bernoulli Equation
Steady frictionless incompressible flow:
𝑝2 −𝑝1
𝜌
or
1
+ (𝑢22 − 𝑢12 ) + 𝑔(𝑧2 − 𝑧1 ) = 0
2
𝑝1
𝜌
1
𝑝2
2
𝜌
+ 𝑢12 + 𝑔𝑧1 =
1
+ 𝑢22 + 𝑔𝑧2 = const.
2
along a streamline.
Restrictions:
1. Steady flow;
2. Incompressible flow;
3. Frictionless flow: major assumption as solid walls and mixing have friction effects;
4. Flow along a single streamline: in most cases, flow is irrotational and one constant.
Some tips:
1. No control volume analysis required.
2. Select two points along a given streamline.
3. Reference points for applying the Bernoulli equation:
a. Free surface;
b. Pipe centerline;
c. Jet exit: atmospheric pressure.
4. Surface condition for a large tank: zero velocity; atmospheric pressure.
5. Continuity relation is almost always involved.
058:0160
Jianming Yang
Fall 2012
Chapters 3&4
32
058:0160
Jianming Yang
Chapters 3&4
33
Fall 2012
Steady flow, fixed CV, one inlet uniform flow and one outlet non-uniform flow
Continuity:
0  U 0 R2   umax 1  r R  2 rdr
R
17
0
𝑅
𝑟 1⁄7
2𝜋𝑢max ∫ (1 − ) 𝑟𝑑𝑟
𝑅
0
=
𝑅
2𝜋𝑢max ∫0 𝑅2 [(1
𝑟 1⁄7
− )
𝑅
= 2𝜋𝑢max 𝑅2 [
7
15
𝑟
𝑟 1⁄7
𝑅
𝑅
(1 − ) − (1 − )
𝑟 15⁄7
(1 − )
𝑅
= 2𝜋𝑢max 𝑅2 [0 − (
7
15
7
8
𝑅
− (1 − )
7
49
8
60
0  U 0 R 2  umax
U0
49

umax 60
𝑅
𝑟 8⁄7
− )] = 2𝜋𝑅2
49 2
R
60
𝑟
] 𝑑 (1 − )
𝑅
]|
0
𝑢max
058:0160
Jianming Yang
Fall 2012
Chapters 3&4
34
058:0160
Jianming Yang
Chapters 3&4
35
Fall 2012
Unsteady flow, deforming CV, one inlet one outlet uniform flow
First draw the CV sketch as the right figure
Continuity equation:
0
d
  d  Q1  Q2
dt CV
d
d2
d2
0
 d   V1
 V2

dt CV
4
4
∀(𝑡) = 𝐴𝐻(𝑡) + ∀𝑔𝑟𝑎𝑦
𝜋𝐷 2
=
𝐻(𝑡) + ∀𝑔𝑟𝑎𝑦
4
𝑑
𝑑
∫ 𝜌𝑑∀ = 𝑑𝑡 ∫𝐴𝐻(𝑡)+∀
𝑑𝑡 ∀
=
𝐻(𝑡)
𝜌𝐴𝑑ℎ
∫
𝑑𝑡 0
𝑑
0=
𝜌𝜋𝐷2 𝑑𝐻 (𝑡)
4
𝑑𝐻(𝑡)
𝑑𝑡
𝑡=
𝑑𝑡
𝑑 2
+
+
𝑑
𝑔𝑟𝑎𝑦
∫
𝑑𝑡 ∀𝑔𝑟𝑎𝑦
𝜌𝜋𝑑 2
4
𝜌𝑑∀
𝜌𝑑∀ = 𝜌𝐴
(𝑉2 − 𝑉1 )
= ( ) (𝑉1 − 𝑉2 ) = 0.0153 𝑚/𝑠
𝐷
𝐻
0.0153
=
.7
0.0153
= 46 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝑑𝐻(𝑡)
𝑑𝑡
+0
058:0160
Jianming Yang
Chapters 3&4
36
Fall 2012
Steady flow, one inlet and two exits with uniform flow
0  Q1  Q2  Q3
Q3  Q1  Q2 
 d 2
4
V1  V2 
 D2
 dh

4
dt  
2
Q  d
V1  V2 
4
2
D
dh  
d
  
V1  V2 
058:0160
Jianming Yang
Fall 2012
Chapters 3&4
37
058:0160
Jianming Yang
Chapters 3&4
38
Fall 2012
First relate 𝑢max to 𝑈0 using continuity equation
Laminar flow: 𝑢2 = 𝑢max (1 −
𝑟2
) ≈ 𝑢max (1 − )
𝑅2
𝑟 𝑚
𝑅
𝑟 1⁄2
𝑅
0 = −𝑈0 𝜋𝑅2 + ∫0 𝑢max (1 − ) 2𝜋𝑟𝑑𝑟
𝑈0 =
1
𝑅
𝑟 𝑚
𝑅
(1 − ) 2𝜋𝑟𝑑𝑟 = 𝑈𝑎𝑣
∫ 𝑢
𝜋𝑅 2 0 max
𝑅
𝑅
𝑟 𝑚𝑟 𝑟
2𝑢max ∫0 (1 − )
𝑑 = 𝑈𝑎𝑣
𝑅
𝑅 𝑅
𝑅
𝑟 𝑚
𝑟
𝑟
2𝑢max ∫0 (1 − ) (− ) 𝑑 (1 − ) = 𝑈𝑎𝑣
𝑅
𝑅
𝑅
𝑚
𝑅
𝑟
𝑟
𝑟 𝑚
𝑟
2𝑢max ∫0 [(1 − ) (1 − ) − (1 − ) ] 𝑑 (1 − ) =
𝑅
𝑅
𝑅
𝑅
𝑚+1
𝑚
𝑅
𝑟
𝑟
𝑟
2𝑢max ∫0 [(1 − )
− (1 − ) ] 𝑑 (1 − ) = 𝑈𝑎𝑣
𝑅
𝑅
𝑅
𝑅
1
𝑟 𝑚+2
1
𝑟 𝑚+1
2𝑢max [
(1 − )
−
(1 − )
]| = 𝑈𝑎𝑣
𝑚+2
𝑅
𝑚+1
𝑅
0
1
1
2
2𝑢max (−
+
) = 𝑢max (
= 𝑈𝑎𝑣
𝑚+2
𝑚+1
𝑚+2)(𝑚+1)
2
8
8
Laminar: 𝑚 = 1⁄2 , (
=
,
𝑈
=
𝑢
→
0
𝑚+2)(𝑚+1)
15
15 max
2
49
49
Turbulent: 𝑚 = 1⁄7 , (
=
,
𝑈
=
𝑢
→
0
𝑚+2)(𝑚+1)
60
60 max
𝑈𝑎𝑣
𝑢max =
𝑢max =
15
8
60
49
𝑈0
𝑈0
058:0160
Jianming Yang
Chapters 3&4
39
Fall 2012
Use 𝑢2 = 𝑢max (1 −
𝑟2
𝑅2
) for laminar flow: 𝑢max = 2𝑈0
1
2
𝑢
Momentum flux correction factor: 𝛽 = ∫ ( ) 𝑑𝐴
𝐴
𝑈
𝑎𝑣
Laminar flow:
𝛽=
=
=
𝑅
[𝑢max (1
∫
𝜋𝑅 2 0
1
𝑅
2
(
)
2𝑈
(1
∫
0
2
𝜋𝑅 0
1
8
𝑟2
𝑅
2
2[
−
−
−2
𝑟4
𝑟6
𝑅
2𝑅
6𝑅
0
2 +
2
𝑟2
𝑅
2 )]
𝑟2
𝑅2
+
1
2
𝑈𝑎𝑣
𝑟4
𝑅 2
𝑢
(1
∫
𝜋𝑅 2 0 max
1
2𝜋𝑟𝑑𝑟 =
)
1
2
𝑅 2 𝑈𝑎𝑣
2𝜋𝑟𝑑𝑟 =
𝑅
(𝑟
∫
𝑅2 0
8
−
−2
2
𝑟2
𝑅
𝑟3
𝑅2
2)
+
1
2
𝑈𝑎𝑣
𝑟5
𝑅4
2𝜋𝑟𝑑𝑟,
) 𝑑𝑟 (𝑈0 = 𝑈𝑎𝑣 )
4
4 ]| =
3
Turbulent flow:
𝛽=
=
𝑅
[𝑢max (1
∫
𝜋𝑅 2 0
1
2
𝑟 1⁄7
1
− )
𝑅
]
2
𝑈𝑎𝑣
2
𝑅 60
𝑟 2⁄7 𝑟 𝑟
𝑑
2 ∫0 (49 𝑈0 ) (1 − 𝑅 )
𝑈𝑎𝑣
𝑅 𝑅
60 2 𝑅
𝑟 2⁄7
𝑟
2
= 2 ( ) ∫0 [(1 − )
49
𝑅
60 2
7
𝑟 16⁄7
49
16
𝑅
= 2 ( ) [ (1 − )
2𝜋𝑟𝑑𝑟 =
60 2
= 2( )
49
𝑅
∫0 (1
𝑟 2⁄7
(1 − ) − (1 − )
𝑅
𝑅 2
𝑢
(1
∫
𝜋𝑅 2 0 max
1
7
𝑅
𝑅
𝑟 9⁄7
9
𝑅
− (1 − )
𝑟 2⁄7
− )
𝑅
− )
𝑅
2
𝑈𝑎𝑣
2𝜋𝑟𝑑𝑟,
𝑟
𝑟
𝑅
𝑅
(− ) 𝑑 (1 − )
𝑟
] 𝑑 (1 − )
𝑅
60 2
7
49
16
]| = 2 ( ) [− (
0
𝑟 2⁄7 1
7
− )] = 1.02
9
058:0160
Jianming Yang
Chapters 3&4
40
Fall 2012
Second, calculate 𝐹 using momentum equation:
𝐹 = wall drag force = 𝜏𝑤 2𝜋𝑅∆𝑥 (force fluid on wall)
−𝐹 = force wall on fluid
𝑅
∑ 𝐹𝑥 = (𝑝1 − 𝑝2 )𝜋𝑅2 − 𝐹 = ∫0 𝑢2 (𝜌𝑢2 2𝜋𝑟𝑑𝑟) − 𝑈0 (𝜌𝜋𝑅2 𝑈0 )
𝑅
𝐹 = (𝑝1 − 𝑝2 )𝜋𝑅2 + 𝜌𝜋𝑅2 𝑈02 − ∫0 𝜌𝑢22 2𝜋𝑟𝑑𝑟
2
𝐹 = (𝑝1 − 𝑝2 )𝜋𝑅2 + 𝜌𝜋𝑅2 𝑈02 − 𝛽𝜌𝜋𝑅2 𝑈𝑎𝑣
𝐹 = (𝑝1 − 𝑝2 )𝜋𝑅2 + (1 − 𝛽)𝜌𝜋𝑅2 𝑈02
Laminar flow:
𝑅
2
(𝛽𝜌𝐴𝑈𝑎𝑣
= ∫0 𝜌𝑢22 2𝜋𝑟𝑑𝑟)
(𝑈0 = 𝑈𝑎𝑣 ) from continuity
4
1
3
3
𝛽 = , 𝐹lam = (𝑝1 − 𝑝2 )𝜋𝑅2 − 𝜌𝜋𝑅2 𝑈02
Turbulent flow:
𝛽 = 1.02, 𝐹tur = (𝑝1 − 𝑝2 )𝜋𝑅2 − 0.02𝜌𝜋𝑅2 𝑈02
058:0160
Jianming Yang
Chapters 3&4
41
Fall 2012
Reconsider the problem for fully developed flow:
Continuity:
−𝑚̇in + 𝑚̇out = 0
𝑚̇ = 𝑚̇in = 𝑚̇out
or
𝑄 = const., 𝑈𝑎𝑣 = const.
Momentum:
∑ 𝐹𝑥 = (𝑝1 − 𝑝2 )𝜋𝑅2 − 𝐹 = ∑ 𝜌𝑢𝐮 ∙ 𝐀
= 𝜌𝑢1 (−𝑢1 𝐴) + 𝜌𝑢2 (𝑢2 𝐴) = 𝜌𝑄 (𝑢2 − 𝑢1 ) = 0
(𝑝1 − 𝑝2 )𝜋𝑅2 − 𝜏𝑤 2𝜋𝑅∆𝑥 = 0
𝑅
𝑑𝑝
2
𝑑𝑥
𝜏𝑤 = (−
)
Or, for smaller CV,
𝑟
𝑑𝑝
2
𝑑𝑥
𝑟 < 𝑅, 𝜏 = (−
)
It is valid for laminar or turbulent flow, but here we assume the flow is laminar
058:0160
Jianming Yang
Fall 2012
𝜏=𝜇
𝑑𝑢
𝑑𝑟
Chapters 3&4
42
𝑑𝑢
= −𝜇
𝑑𝑦
𝑟
=−
𝑢=−
2𝜇
𝑟2
4𝜇
(−
(−
𝑑𝑢
𝑑𝑟
𝑑𝑝
𝑑𝑥
𝑑𝑝
𝑑𝑥
𝑢=−
4𝜇
𝑢max = −
𝑄=
𝑅2
4𝜇
𝑈𝑎𝑣 =
𝐴
𝑅
=
𝜏𝑤 = (−
𝑓=
𝑑𝑝
𝑑𝑥
(−
2
8𝜏𝑤
2
𝜌𝑈𝑎𝑣
8𝜇
𝑑𝑝
𝑑𝑥
=
𝑑𝑥
→
(−
𝑅2
2
(𝑦 = 𝑅 − 𝑟) wall coordinate
)
)+𝑐
𝑑𝑝
𝑑𝑥
(−
)
𝑑𝑝
𝜋𝑅 4
8𝜇
)=
(−
𝑑𝑝
𝑑𝑥
)
=
𝑢(𝑟) = 𝑢max (1 −
𝑑𝑝
𝑑𝑥
)
𝑢max
𝑑𝑥
2
𝑅 8𝜇𝑈𝑎𝑣
2
𝜌𝑅𝑈𝑎𝑣
4𝜇
(−
→
=
)= (
32𝜇
𝑐=
𝑅2
)
𝑟
∫0 𝑢(𝑟)2𝜋𝑟𝑑𝑟
𝑄
𝑑𝑝
)
𝑢(𝑟 = 𝑅) = 0
𝑅 2 −𝑟 2
𝑟
= (−
𝑅2
64𝜇
)=
𝜌𝑈𝑎𝑣 𝐷
=
4𝜇𝑈𝑎𝑣
𝑅
64
𝑅𝑒
(𝑅𝑒 =
𝜌𝑈𝑎𝑣 𝐷
𝜇
)
Exact solution of NS for laminar fully developed pipe flow
𝑟2
𝑅2
)
058:0160
Jianming Yang
Fall 2012
Chapters 3&4
43
058:0160
Jianming Yang
Chapters 3&4
44
Fall 2012
Assume relative inertial coordinates with non-deforming CV, i.e., CV moves at constant
translational non-accelerating: 𝐕𝑐𝑠 = 𝑉𝑐 𝐢
Then 𝐕𝑟 = 𝐕 − 𝐕𝑐𝑠 . Also assume steady flow 𝜕⁄𝜕𝑡 = 0 with 𝜌 = const. and neglect
gravity effect.
Continuity:
∫𝐶𝑆 𝐕𝑟 ∙ 𝐧𝑑𝐴 = 0 ⇒
−|𝑽𝑟1 |𝐴1 + |𝑽𝑟2 |𝐴2 = 0
Here for the normal component of 𝑽r , we have |𝑉r 𝑛 | = |𝑽r |
|𝑽𝑟1 |𝐴1 = |𝑽𝑟2 |𝐴2
|𝑽𝑟1 | = |𝑽𝑟2 |
(𝐴1 = 𝐴2 = 𝐴𝑗 )
|𝑽𝑟1 | = |𝑽𝑟2 | = (𝑉𝑗 −𝑉𝑐 ) (𝐕𝑟 = 𝐕 − 𝐕𝑐𝑠 )
𝑽𝑟1 = (𝑉𝑗 −𝑉𝑐 )𝐢
Momentum:
∑ 𝐅 = 𝜌 ∫𝐶𝑆 𝐕𝑟 𝐕𝑟 ∙ 𝐧𝑑𝐴
−𝐹𝑥 = 𝜌 ∫𝐶𝑆 𝑉𝑟 𝑥 𝐕𝑟 ∙ 𝐧𝑑𝐴
−𝐹𝑥 = 𝜌𝑉𝑟1 𝑥 (−|𝑽𝑟1 |𝐴1 ) + 𝜌𝑉𝑟2 𝑥 (|𝑽𝑟2 |𝐴2 )
−𝐹𝑥 = 𝜌(𝑉𝑗 −𝑉𝑐 )[−(𝑉𝑗 −𝑉𝑐 )𝐴𝑗 ] + 𝜌(𝑉𝑗 −𝑉𝑐 ) cos 𝜃 (𝑉𝑗 −𝑉𝑐 )𝐴𝑗
058:0160
Jianming Yang
Chapters 3&4
45
Fall 2012
Force:
2
𝐹𝑥 = 𝜌(𝑉𝑗 −𝑉𝑐 ) 𝐴𝑗 (1 − cos 𝜃)
Power
2
𝑃 = 𝑉𝑐 𝐹𝑥 = 𝑉𝑐 𝜌(𝑉𝑗 −𝑉𝑐 ) 𝐴𝑗 (1 − cos 𝜃)
Maximum force:
𝐹𝑥 max = 𝜌𝑉𝑗2 𝐴𝑗 (1 − cos 𝜃) (𝑉𝑐 = 0)
Maximum power:
𝑑𝑃
𝑑𝑉𝑐
=0
𝑑𝑃
𝑑𝑉𝑐
𝑑
𝑑𝑉𝑐
=
2
𝑑
𝑑𝑉𝑐
[𝑉𝑐 𝜌(𝑉𝑗 −𝑉𝑐 ) 𝐴𝑗 (1 − cos 𝜃)] = 0
[(𝑉𝑐3 − 2𝑉𝑗 𝑉𝑐2 + 𝑉𝑗2 𝑉𝑐 )𝜌𝐴𝑗 (1 − cos 𝜃)] = 0
(3𝑉𝑐2 − 4𝑉𝑗 𝑉𝑐 + 𝑉𝑗2 ) = 0
𝑉𝑐 =
4𝑉𝑗 ±√16𝑉𝑗2 −12𝑉𝑗2
6
=
4𝑉𝑗 ±2𝑉𝑗
6
1
𝑉𝑐 = 𝑉𝑗
3
𝑃max =
𝑉𝑗
3
2
2
𝜌 ( 𝑉𝑗 ) 𝐴𝑗 (1 − cos 𝜃 ) =
3
4
27
𝑉𝑗3 𝜌𝐴𝑗 (1 − cos 𝜃)
058:0160
Jianming Yang
Fall 2012
Chapters 3&4
46
058:0160
Jianming Yang
Chapters 3&4
47
Fall 2012
Assume gravity force is negligible and the jet cross section area does not change after
striking the bucket. Taking moving CV at speed 𝐕𝑐𝑠 = 𝛺𝑅𝐢 enclosing jet and bucket:
Use relative inertial coordinates
Continuity: 𝜌 ∫𝐶𝑆 𝐕𝑟 ∙ 𝐧𝑑𝐴 = 0
−𝑚̇r1 + 𝑚r2 = 0 → 𝑚̇𝑟 = 𝑚̇r1 = 𝑚̇r2
−|𝑽r1 |𝐴1 + |𝑽r2 |𝐴2 = 0 (|𝑉r 𝑛 | = |𝑽r |)
|𝑽r1 | = |𝑽r2 |
𝑽r1
Inlet
(𝐴1 = 𝐴2 = 𝐴𝑗 )
= −𝑽r2 from figure
(𝐕𝑟 = 𝐕 − 𝐕𝑐𝑠 )
𝑽r1 = (𝑉𝑗 − 𝛺𝑅)𝐢
Outlet
Momentum:
𝑽r2 = −(𝑉𝑗 − 𝛺𝑅)𝐢
𝐧 = −𝐢
(𝐕𝑟 = 𝐕 − 𝐕𝑐𝑠 )
𝐧 = −𝐢
∑ 𝐅 = 𝜌 ∫𝐶𝑆 𝐕𝑟 𝐕𝑟 ∙ 𝐧𝑑𝐴
−𝐹x = 𝜌𝑉r1 𝑥 (−𝐴1 |𝑽r1 |) + 𝜌𝑉r2 𝑥 (𝐴2 |𝑽r2 |)
𝐹x = 𝜌𝑉r1 𝑥 𝐴1 |𝑽r1 | − 𝜌𝑉r2 𝑥 𝐴2 |𝑽r2 |
𝐹x = 𝜌(𝑉𝑗 − 𝛺𝑅)𝐴𝑗 (𝑉𝑗 − 𝛺𝑅) − 𝜌[−(𝑉𝑗 − 𝛺𝑅)]𝐴2 (𝑉𝑗 − 𝛺𝑅)
Here we assume |𝑽r2 | = (𝑉𝑗 − 𝛺𝑅) ≥ 0, or, 𝑉𝑗 ≥ 𝛺𝑅.
𝐹x = 2𝜌𝐴𝑗 (𝑉𝑗 − 𝛺𝑅)
2
058:0160
Jianming Yang
Chapters 3&4
48
Fall 2012
Power:
𝑃 = 𝛺𝑅𝐹x = 2𝜌𝐴𝑗 𝛺𝑅(𝑉𝑗 − 𝛺𝑅)
Maximum power:
2
𝑑𝑃⁄𝑑𝛺 = 0
1
𝛺𝑅 = 𝑉𝑗
3
→
𝑃max =
8
27
𝜌𝐴𝑗 𝑉𝑗3
If infinite number of buckets:
𝑚̇ = 𝜌𝐴𝑗 𝑉𝑗
that is, all jet mass flow results in work
Force
𝐹x = 2𝜌𝐴𝑗 𝑉𝑗 (𝑉𝑗 − 𝛺𝑅)
Power
𝑃 = 𝛺𝑅𝐹x = 2𝜌𝐴𝑗 𝑉𝑗 (𝑉𝑗 − 𝛺𝑅)
Maximum power
𝑑𝑃
𝑑𝛺
1
= 0 for 𝛺𝑅 = 𝑉𝑗
2
→
1
𝑃max = 𝜌𝐴𝑗 𝑉𝑗3
2
058:0160
Jianming Yang
Chapters 3&4
49
Fall 2012
Solution 2: absolute inertial coordinates
Continuity: from solution 1: 𝑽r1 = −𝑽r2 = (𝑉𝑗 − 𝛺𝑅)
Express in the absolute inertial coordinates
(𝑽a1 − 𝛺𝑅𝐢) = −(𝑽a2 − 𝛺𝑅𝐢)
Inlet
𝑽a1 = 𝑉𝑗 𝐢
𝐧 = −𝐢
Outlet
𝑽a2 = −(𝑉𝑗 − 2𝛺𝑅)𝐢
Note: 𝑚̇a1 ≠ 𝑚̇𝑎2 !
Momentum:
∑𝐅 =
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐧 = −𝐢
𝐮𝜌𝑑𝑉 + ∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
Moving control volume but momentum in the control volume does not change with time
∑ 𝐅 = ∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
−𝐹x = 𝜌𝑉a1 𝑥 (−𝐴𝑗 |𝑽r1 |) + 𝜌𝑉a2 𝑥 (𝐴𝑗 |𝑽r2 |)
𝐹x = −𝜌𝑉𝑗 [−𝐴𝑗 (𝑉𝑗 − 𝛺𝑅)] − 𝜌[−(𝑉𝑗 − 2𝛺𝑅)][𝐴𝑗 (𝑉𝑗 − 𝛺𝑅)]
𝐹x = 2𝜌𝐴𝑗 (𝑉𝑗 − 𝛺𝑅)
2
058:0160
Jianming Yang
Chapters 3&4
50
Fall 2012
6 Non-inertial Reference Frame
Thus far we have assumed use of an inertial reference frame (i.e. fixed with respect to the
distant stars in deriving the CV and differential form of the momentum equation).
However, in many cases non-inertial reference frames are useful (e.g. rotational
machinery, vehicle dynamics, geophysical applications, etc).
In a noninertial frame of reference:
Continuity equation is unchanged.
Momentum equation must be modified.
Reference frame O′1′2′3′ translates at velocity
d𝐗(t)⁄dt = 𝐔(t) and rotates at angular
velocity 𝛀(t) with respect to a stationary
reference frame O123. The vectors 𝐔 and 𝛀
may be resolved in either frame. t = t′.
A fluid particle P can be located in the rotating frame 𝐱′ or in the stationary frame 𝐱, and
𝐱 = 𝐗 + 𝐱′. The velocity 𝐮 of the fluid particle is:
𝐮=
=𝐔+
𝑑𝑥1′
𝐞1′
𝑑𝑡
+
𝑑𝐱
𝑑𝑡
𝑑𝑥2′
=
𝐞′2
𝑑𝑡
𝑑𝐗
𝑑𝑡
+
+
𝑑𝐱 ′
𝑑𝑡
𝑑𝑥3′
=𝐔+
𝐞′3
𝑑𝑡
+
𝑑
𝑑𝑡
′
′ 𝑑𝐞1
𝑥1
𝑑𝑡
(𝑥1′ 𝐞1′ + 𝑥2′ 𝐞′2 + 𝑥3′ 𝐞′3 )
+
′
𝑑𝐞
𝑥2′ 2
𝑑𝑡
′
𝑑𝐞
+ 𝑥3′ 3
𝑑𝑡
= 𝐔 + 𝐮′ + 𝛀 × 𝐱′
058:0160
Jianming Yang
Chapters 3&4
51
Fall 2012
′
𝑑𝐞
𝑥1′ 1
𝑑𝑡
=𝛀×
(𝑥1′ 𝐞1′ )
⇒
𝑑𝐞′1
= 𝛀 × 𝐞1′
𝑑𝑡
is from the geometric construction of the cross
product for 𝐞1′ .
In 𝑑𝑡, the rotation of O′1′2′3′ causes 𝐞1′ to trace a
small portion of a cone with radius sin 𝛼:
The magnitude of the change in 𝐞1′ is |𝑑𝐞1′ | =
sin 𝛼 |𝛀|𝑑𝑡, so 𝑑 |𝐞1′ |⁄𝑑𝑡 = sin 𝛼 |𝛀|, which is equal
to the magnitude of 𝛀 × 𝐞1′ .
The direction of 𝑑𝐞1′ ⁄𝑑𝑡 is perpendicular to 𝛀
and 𝐞1′ , which is the direction of 𝛀 × 𝐞1′ .
Thus, by geometric construction, 𝑑𝐞1′ ⁄𝑑𝑡 = 𝛀 ×
𝐞1′ , and 𝑑𝐞′𝑖 ⁄𝑑𝑡 = 𝛀 × 𝐞′𝑖 .
The acceleration 𝐚 of a fluid particle at P, take the time derivative of the velocity 𝐮:
𝐚=
𝑑𝐮
𝑑𝑡
=
=
𝑑𝐔
𝑑𝑡
=
𝑑𝐔
𝑑𝑡
𝑑
𝑑𝑡
(𝐔 + 𝐮′ + 𝛀 × 𝐱′) =
𝑑𝐔
𝑑𝑡
𝑑𝑡
+
𝑑
𝑑𝑡
(𝐮′ ) +
𝑑
𝑑𝛀
𝑑𝑡
+ 𝐚′ + 2𝛀 × 𝐮′ +
(𝛀 × 𝐱′)
𝑑
𝑑𝑡
𝑑
𝑑𝑡
𝑑𝑡
+ (𝐚′ + 𝛀 × 𝐮′ ) + [ (𝛀) × 𝐱 ′ + 𝛀 ×
+ (𝐚′ + 𝛀 × 𝐮′ ) + {(
=
𝑑𝐔
(𝐱 ′ )]
× 𝐱 ′ + 𝛀 × 𝛀 × 𝐱 ′ ) + 𝛀 × [𝐮′ + (𝛀 × 𝐱 ′ )]}
𝑑𝛀
𝑑𝑡
× 𝐱 ′ + 𝛀 × (𝛀 × 𝐱 ′ )
(𝛀 × 𝛀 = 0)
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Jianming Yang
Chapters 3&4
52
Fall 2012
In fluid mechanics, the acceleration 𝐚 of fluid particles is denoted 𝐷𝐮⁄𝐷𝑡:
𝐷𝐮
( )
𝐷𝑡 𝑂123
=(
𝐷 ′ 𝐮′
𝐷𝑡
)
+
′
𝑂′ 1′ 2′ 3
𝑑𝐔
𝑑𝑡
+ 2𝛀 × 𝐮′ +
𝑑𝛀
𝑑𝑡
× 𝐱 ′ + 𝛀 × (𝛀 × 𝐱 ′ )
The momentum equation of an incompressible flow in a noninertial reference frame:
𝜌(
𝐷′ 𝐮′
𝐷𝑡
)
𝑂′ 1′ 2′ 3′
= −∇′ 𝑝 + 𝜌 [𝐠 −
𝑑𝐔
𝑑𝑡
− 2𝛀 × 𝐮′ −
𝑑𝛀
𝑑𝑡
2
× 𝐱 ′ − 𝛀 × (𝛀 × 𝐱 ′ )] + 𝜇∇′ 𝐮′
2
= −∇′ 𝑝 + 𝜌[𝐠 − 𝐚rel ] + 𝜇∇′ 𝐮′
where the primes denote differentiation, velocity, and position in the noninertial frame.
Thermodynamic variables and the net viscous stress are independent of the reference
frame. The primary effect of a noninertial frame is the addition of extra body force terms
due to the relative acceleration:
𝐚rel =
𝑑𝐔
𝑑𝑡
+ 2𝛀 × 𝐮′ +
𝑑𝛀
𝑑𝑡
× 𝐱 ′ + 𝛀 × (𝛀 × 𝐱 ′ )
𝑑𝐔⁄𝑑𝑡 = 𝑑 2 𝐗⁄𝑑𝑡 2 is the acceleration of O′ with respect to O. It provides the
apparent force that pushes occupants back into their seats when a vehicle accelerates.
2𝛀 × 𝐮′ is the Coriolis acceleration, which depends on the velocity, not on position.
𝑑𝛀
× 𝐱′ is the acceleration caused by angular acceleration of the noninertial frame
𝑑𝑡
𝛀 × (𝛀 × 𝐱 ′ ) is the centripetal acceleration, which depends strongly on the rotation
rate and the distance of the fluid particle from the axis of rotation.
058:0160
Jianming Yang
Chapters 3&4
53
Fall 2012
In geophysical flows:
𝑑𝐔⁄𝑑𝑡 ≈ 0 (earth not accelerating relative to distant stars)
𝑑𝛀⁄𝑑𝑡 ≈ 0 (for earth)
𝛀 × (𝛀 × 𝐱 ′ ) ≈ 0 (can be added to 𝐠)
2𝛀 × 𝐮′ is the most important!
𝑉0
Rossby number 𝑅𝑜 = (𝑉02 /𝐿)⁄(Ω𝑉0 ) = Ω𝐿
, if 𝑅𝑜 < 1, Coriolis term is larger than the
inertia terms and is important. The Coriolis acceleration is responsible for wind
circulation patterns around centers of high and low pressure in the earth’s atmosphere.
Control volume form of Momentum equation for non-inertial reference frame:
∑ 𝐅 − ∫𝐶𝑉 𝐚rel 𝜌𝑑𝑉 =
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 + ∫𝐶𝑆 𝐮𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
where 𝐮 is the fluid velocity in the non-inertial reference frame (without prime), 𝐮𝑟 is the
fluid velocity relative to the control volume in the non-inertial reference frame.
Usually, the control volume is fixed in a non-inertial reference frame:
∑ 𝐅 − ∫𝐶𝑉 𝐚rel 𝜌𝑑𝑉 =
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 + ∫𝐶𝑆 𝐮𝜌𝐮 ∙ 𝐧𝑑𝐴
Both equations in the non-inertial reference frame have the “same” right-hand sides as
those in the inertial reference frame. Only difference is the “apparent” force ∫𝐶𝑉 𝐚rel 𝜌𝑑𝑉
on the left-hand side.
058:0160
Jianming Yang
Fall 2012
7 Curvilinear coordinate systems
Position vector: 𝐱
Velocity vector: 𝐮
Gradient of a scalar 𝜓: ∇𝜓
Laplacian of a scalar 𝜓: ∇2 𝜓 = ∇ ∙ ∇𝜓
Divergence of a vector: ∇ ∙ 𝐮
Gradient of a vector: ∇𝐮
Curl of a vector, vorticity: 𝛚 = ∇ × 𝐮
Laplacian of a vector: ∇2 𝐮 = ∇ ∙ ∇𝐮 = ∇(∇ ∙ 𝐮) − ∇ × (∇ × 𝐮)
Divergence of a tensor: ∇ ∙ 𝛕
Chapters 3&4
54
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Fall 2012
7.1Cartesian Coordinates (Kundu et al., Fluid Mechanics)
Chapters 3&4
55
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Fall 2012
7.2Cylindrical Coordinates (Kundu et al., Fluid Mechanics)
Chapters 3&4
56
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Fall 2012
7.3Spherical Coordinates (Kundu et al., Fluid Mechanics)
Chapters 3&4
57
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Chapters 3&4
58
Fall 2012
7.4General Orthogonal Coordinates (Bird et al., Transport Phenomena)
The relation between Cartesian coordinates 𝑥𝑖 and the curvilinear coordinates 𝑞𝛼 is
𝑥1 = 𝑥1 (𝑞1 , 𝑞2 , 𝑞3 )
{𝑥2 = 𝑥2 (𝑞1 , 𝑞2 , 𝑞3 ) or 𝑥𝑖 = 𝑥𝑖 (𝑞𝛼 )
𝑥3 = 𝑥3 (𝑞1 , 𝑞2 , 𝑞3 )
These can be solved for the 𝑞𝛼 to get the inverse relations 𝑞𝛼 = 𝑞𝛼 (𝑥𝑖 ). Then the unit
vectors 𝐞𝑖 in rectangular coordinates and the 𝐞𝛼 in curvilinear coordinates are related:
𝐞𝛼 = ∑𝑖 ℎ𝛼 (
{
𝐞𝑖 = ∑𝛼 ℎ𝛼 (
𝜕𝑞𝛼
𝜕𝑥𝑖
𝜕𝑞𝛼
𝜕𝑥𝑖
) 𝐞𝑖 = ∑𝑖
) 𝐞𝛼 =
1
(
𝜕𝑥𝑖
) 𝐞𝑖
ℎ𝛼 𝜕𝑞𝛼
1 𝜕𝑥
∑𝛼 ( 𝑖 ) 𝐞𝛼
ℎ𝛼 𝜕𝑞𝛼
in which the “scale factors” ℎ𝛼 are given by
ℎ𝛼2 = ∑𝑖 (
𝜕𝑥𝑖 2
−1
𝜕𝑞𝛼 2
𝜕𝑞𝛼
𝜕𝑥𝑖
) = [∑𝑖 (
) ]
The spatial derivatives of the unit vectors 𝐞𝛼 can then be found to be
𝜕𝐞𝛼
𝜕𝑞𝛽
=
𝐞𝛽 𝜕ℎ𝛽
ℎ𝛼 𝜕𝑞𝛼
− 𝛿𝛼𝛽 ∑3𝛾=1
𝐞𝛾 𝜕ℎ𝛼
ℎ𝛾 𝜕𝑞𝛾
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Chapters 3&4
59
Fall 2012
The ∇-operator is
∇= ∑𝛼
𝐞𝛼 𝜕
ℎ𝛼 𝜕𝑞𝛼
The divergence of a vector:
∇∙𝐮=
1
ℎ1 ℎ2 ℎ3
𝜕
∑𝛼
𝜕𝑞𝛼
(
ℎ1 ℎ2 ℎ3
𝑢𝛼 )
ℎ𝛼
The Laplacian of a scalar:
∇2 𝜓 =
1
ℎ1 ℎ2 ℎ3
∑𝛼
𝜕
𝜕𝑞𝛼
(
ℎ1 ℎ2 ℎ3 𝜕𝜓
2
ℎ𝛼
𝜕𝑞𝛼
)
The curl of a vector:
∇×𝐮=
1
ℎ1 ℎ2 ℎ3
|
ℎ1 𝐞1
ℎ2 𝐞2
ℎ3 𝐞3
𝜕
𝜕
𝜕
𝜕𝑞1
𝜕𝑞2
𝜕𝑞3
|
ℎ1 𝑢1 ℎ2 𝑢2 ℎ3 𝑢3
in which the unit vectors are those belonging to the curvilinear coordinate system.
Volume element:
𝑑𝑉 = ℎ1 ℎ2 ℎ3 𝑑𝑞1 𝑑𝑞2 𝑑𝑞3
Surface element:
here 𝑑𝑆𝛼𝛽
𝑑𝑆𝛼𝛽 = ℎ𝛼 ℎ𝛽 𝑑𝑞𝛼 𝑑𝑞𝛽 (𝛼 ≠ 𝛽)
is a surface element on a surface of constant 𝛾, where 𝛾 ≠ 𝛼 and 𝛾 ≠ 𝛽
058:0160
Jianming Yang
Chapters 3&4
60
Fall 2012
Derive the expression for ∇ ∙ 𝐮 in cylindrical coordinates
Write ∇ and 𝐮 in cylindrical coordinates:
∇ ∙ 𝐮 = (𝐞𝑟
= (𝐞𝑟
+ (𝐞𝜃
+ (𝐞𝑧
Use:
𝜕𝐞𝑟
𝜕𝑟
=
𝜕
𝜕𝑟
1 𝜕
+ 𝐞𝜃
∙ 𝐞𝑟 𝑢𝑟 +
𝜕𝑟
1 𝜕
∙ 𝐞𝑟 𝑢𝑟 + 𝐞𝑧
𝜕𝐞𝜃
=
𝜕𝑟
𝜕𝐞𝑧
∇ ∙ 𝐮 = (𝐞𝑟 ∙ 𝐞𝑟
+ (𝐞𝜃 ∙ 𝐞𝑟
+ (𝐞𝑧 ∙ 𝐞𝑟
1 𝜕𝑢𝑟
𝑟 𝜕𝜃
𝜕𝑢𝑟
𝜕𝑧
=
𝜕𝑟
+ 𝐞𝑧
𝑟 𝜕𝜃
𝜕
𝐞𝑟 ∙
𝜕𝑟
∙ 𝐞𝑟 𝑢𝑟 + 𝐞𝜃
𝑟 𝜕𝜃
𝜕
𝜕𝑧
𝜕
𝑟 𝜕𝜃
𝜕𝑧
𝜕𝐞𝑟
𝜕𝑢𝑟
𝜕𝑟
𝜕𝑧
) ∙ (𝐞𝑟 𝑢𝑟 + 𝐞𝜃 𝑢𝜃 + 𝐞𝑧 𝑢𝑧 )
𝜕
∙ 𝐞𝜃 𝑢𝜃 + 𝐞𝑧
=
+ 𝐞𝜃 ∙ 𝐞𝜃
𝜕𝐞𝜃
𝜕𝑧
𝑢𝑟
𝑟
𝜕𝑢𝜃
𝜕𝑧
𝜕𝑟
=
𝜕𝐞𝑧
𝜕𝑧
𝜕𝑢𝜃
𝜕𝑟
𝜕
𝜕𝑧
=
𝜕𝑢𝑟
𝜕𝑟
+(
𝑢𝑟
𝑟
+
1 𝜕𝑢𝜃
+ 𝐞𝜃 ∙ 𝐞𝜃
+ 𝐞𝑧 ∙ 𝐞𝑧
𝑟 𝜕𝜃
)+
𝜕𝑢𝑧
𝜕𝑧
1 𝜕
𝑟 𝜕𝜃
∙ 𝐞𝑧 𝑢𝑧 )
∙ 𝐞𝑧 𝑢𝑧 )
𝜕𝐞𝑧
𝜕𝜃
= 0,
+ 𝐞𝑟 ∙ 𝐞𝑧
Since 𝐞𝑟 ∙ 𝐞𝑟 = 1, 𝐞𝑟 ∙ 𝐞𝜃 = 0, and so on
∇∙𝐮=
∙ 𝐞𝑧 𝑢𝑧 )
∙ 𝐞𝜃 𝑢𝜃 + 𝐞𝜃
+ 𝐞𝑟 ∙ 𝐞𝜃
+ 𝐞𝑧 ∙ 𝐞𝜃
𝜕𝑧
𝐞𝜃 𝑢𝜃 + 𝐞𝑟
1 𝜕
𝜕
𝜕
𝜕𝑢𝑧
1 𝜕𝑢𝜃
𝑟 𝜕𝜃
𝜕𝑢𝑧
𝜕𝑧
)
𝜕𝑟
𝜕𝐞𝑟
𝜕𝜃
= 𝐞𝜃 ,
𝜕𝐞𝜃
𝜕𝜃
= −𝐞𝑟
)
+ 𝐞𝜃 ∙ (−𝐞𝑟 )
𝑢𝜃
𝑟
+ 𝐞𝜃 ∙ 𝐞𝑧
1 𝜕𝑢𝑧
𝑟 𝜕𝜃
)
058:0160
Jianming Yang
Chapters 3&4
61
Fall 2012
Find the 𝑟-component of ∇ ∙ 𝛕 in spherical coordinates
Use the ∇-operator in spherical coordinates:
∇ ∙ 𝛕 = (𝐞𝑟
𝜕
𝜕𝑟
+ 𝐞𝜃
1 𝜕
𝑟 𝜕𝜃
+ 𝐞𝜙
1
𝜕
𝑟 sin 𝜃 𝜕𝜙
)
∙ (𝐞𝑟 𝐞𝑟 𝜏𝑟𝑟 + 𝐞𝑟 𝐞𝜃 𝜏𝑟𝜃 + 𝐞𝑟 𝐞𝜙 𝜏𝑟𝜙
+𝐞𝜃 𝐞𝑟 𝜏𝜃𝑟 + 𝐞𝜃 𝐞𝜃 𝜏𝜃𝜃 + 𝐞𝜃 𝐞𝜙 𝜏𝜃𝜙
+𝐞𝜙 𝐞𝑟 𝜏𝜙𝑟 + 𝐞𝜙 𝐞𝜃 𝜏𝜙𝜃 + 𝐞𝜙 𝐞𝜙 𝜏𝜙𝜙 )
Use:
𝐞𝑖 ∙ 𝐞𝑗 𝐞𝑘 = 𝛿𝑖𝑗 𝐞𝑘 ,
𝜕𝐞𝑟
𝜕𝑟
𝜕𝐞𝑟
𝜕𝜃
𝜕𝐞𝑟
𝜕𝜙
=
𝜕𝐞𝜃
𝜕𝑟
= 𝐞𝜃 ,
=
𝜕𝐞𝜙
𝜕𝑟
𝜕𝐞𝜃
𝜕𝜃
= 0,
= −𝐞𝑟 ,
= 𝐞𝜙 sin 𝜃 ,
𝜕𝐞𝜃
𝜕𝜙
𝜕𝐞𝜙
𝜕𝜃
= 0,
= 𝐞𝜙 cos 𝜃 ,
𝜕𝐞𝜙
𝜕𝜙
= −𝐞𝑟 sin 𝜃 − 𝐞𝜃 cos 𝜃
058:0160
Jianming Yang
Chapters 3&4
62
Fall 2012
Only consider terms that contribute to the coefficient of 𝐞𝑟
𝐞𝑟
𝐞𝜃
𝐞𝜙
𝜕
∙ 𝐞𝑟 𝐞𝑟 𝜏𝑟𝑟 = 𝐞𝑟 ∙ 𝐞𝑟 𝐞𝑟
𝜕𝑟
1 𝜕
𝑟 𝜕𝜃
1
∙ 𝐞𝜃 𝐞𝑟 𝜏𝜃𝑟 = 𝐞𝜃 ∙
𝜕𝜏𝑟𝑟
= 𝐞𝑟
𝜕𝑟
1 𝜕𝜏𝜃𝑟
𝐞𝜃 𝐞𝑟
𝑟 𝜕𝜃
𝜕𝜏𝑟𝑟
𝜕𝑟
+ other terms
𝜕
𝜕𝜏𝜙𝑟
1
∙ 𝐞𝜙 𝐞𝑟 𝜏𝜙𝑟 = 𝐞𝜙 ∙ 𝐞𝜙 𝐞𝑟
𝑟 sin 𝜃 𝜕𝜙
𝑟 sin 𝜃 𝜕𝜙
1 𝜕
𝜏
𝜕𝐞
𝜏
𝐞𝜃
∙ 𝐞𝑟 𝐞𝑟 𝜏𝑟𝑟 = 𝑟𝑟 [𝐞𝜃 ∙ 𝑟 𝐞𝑟 ] + 𝑟𝑟 [𝐞𝜃 ∙
𝑟 𝜕𝜃
𝑟
𝜕𝜃
𝑟
𝐞𝜙
1
𝜕
∙ 𝐞𝑟 𝐞𝑟 𝜏𝑟𝑟 =
𝑟 sin 𝜃 𝜕𝜙
1 𝜕
𝐞𝜃
∙ 𝐞𝜃 𝐞𝜃 𝜏𝜃𝜃
𝑟 𝜕𝜃
𝐞𝜙
𝐞𝜙
1
𝜕
𝑟 sin 𝜃 𝜕𝜙
1
𝜕
𝑟 sin 𝜃 𝜕𝜙
𝜏𝑟𝑟
𝑟 sin 𝜃
𝜏𝜃𝜃
= 𝐞𝑟 (−
∙ 𝐞𝜃 𝐞𝑟 𝜏𝜃𝑟 = 𝐞𝑟
[𝐞𝜙 ∙
𝜕𝐞𝑟
𝜕𝜙
𝐞𝑟 ] =
+ other terms
𝐞𝑟
𝜕𝐞𝑟
𝜕𝜃
𝜏𝑟𝑟
𝑟 sin 𝜃
]=
𝜏𝑟𝑟
𝑟
[𝐞𝜃 ∙ 𝐞𝜃 𝐞𝑟 ] = 𝐞𝑟
[𝐞𝜙 ∙ 𝐞𝜙 sin 𝜃 𝐞𝑟 ] = 𝐞𝑟
) + other terms
𝑟
𝜏𝜃𝑟 cos 𝜃
𝑟 sin 𝜃
𝜏𝜙𝜙
∙ 𝐞𝜙 𝐞𝜙 𝜏𝜙𝜙 = 𝐞𝑟 (−
𝑟
) + other terms
Combining the above results:
(∇ ∙ 𝛕)𝑟 =
1 𝜕
𝑟 2 𝜕𝑟
(𝑟 2 𝜏𝑟𝑟 ) +
𝜏𝜃𝑟
𝑟
cot 𝜃 +
1 𝜕𝜏𝜃𝑟
𝑟 𝜕𝜃
+
1
𝜕𝜏𝜙𝑟
𝑟 sin 𝜃 𝜕𝜙
−
𝜏𝜃𝜃 +𝜏𝜙𝜙
𝑟
𝜏𝑟𝑟
𝑟
𝜏𝑟𝑟
𝑟
058:0160
Jianming Yang
Chapters 3&4
63
Fall 2012
8 Energy Equation
𝑑𝐵sys
RTT:
𝑑𝑡
=
𝑑
∫
𝑑𝑡 𝐶𝑉
𝛽𝜌𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
Energy:
𝐵=𝐸
Energy per unit mass: 𝛽 = 𝑒 = 𝑑𝐸 ⁄𝑑𝑚
8.1Integral Form (fixed CV)
Conservation of energy:
𝑑𝐸
𝑑𝑡
𝑑𝐸
𝑑𝑡
dE
dt
= ∫𝐶𝑉
𝜕
𝜕𝑡
= 𝑄̇ − 𝑊̇
(𝑒𝜌)𝑑𝑉 + ∫𝐶𝑆 𝑒𝜌𝐮 ∙ 𝐧𝑑𝐴 = 𝑄̇ − 𝑊̇
Rate of change of 𝐸
:
𝜕
∫𝐶𝑉 𝜕𝑡 (𝑒𝜌)𝑑𝑉:
Rate of change of 𝐸 in CV
∫𝐶𝑆 𝑒𝜌𝐮 ∙ 𝐧𝑑𝐴:
Rate of outflux of 𝐸 across CS
𝑄̇ :
Rate of heat added to CV
𝑊̇ :
Rate of work done by CV
058:0160
Jianming Yang
Chapters 3&4
64
Fall 2012
Total energy per unit mass:
1
𝑒 = 𝑢̂ + 𝑢2 + 𝑔𝑧 = internal energy + kinetic energy + potential energy
2
Rate of heat added to CV:
𝑄̇ = conduction + convection + radiation
Rate of work done by CV:
𝑊̇ = 𝑊̇𝑠 + 𝑊̇𝑝 + 𝑊̇𝑣
= pump/turbine shaft work + pressure force work + viscous force work
Pressure force work:
𝑑𝑊̇𝑝 = −(𝑝𝑑𝐴)𝑢𝑛,inward = −𝑝(−𝐧 ∙ 𝐮)𝑑𝐴 = pressure force times velocity
𝑊̇𝑝 = ∫𝐶𝑆 𝑝(𝐮 ∙ 𝐧)𝑑𝐴
Viscous force work:
𝑑𝑊̇𝑣 = −𝛕𝑑𝐴 ∙ 𝐮 = viscous force times velocity
𝑊̇𝑣 = − ∫𝐶𝑆 𝛕 ∙ 𝐮𝑑𝐴
Therefore,
𝜕
𝑝
𝑄̇ − 𝑊̇𝑠 − 𝑊̇𝑣 = ∫𝐶𝑉 (𝑒𝜌)𝑑𝑉 + ∫𝐶𝑆 (𝑒 + ) 𝜌𝐮 ∙ 𝐧𝑑𝐴
𝜕𝑡
𝜌
058:0160
Jianming Yang
Chapters 3&4
65
Fall 2012
For our purposes, we are interested in steady flow with one inlet and one outlet. Also
𝑊̇𝑣 ~0 in most cases; since, 𝐮 = 0 at solid surface; on inlet and outlet only 𝜏𝑛𝑛 𝑢𝑛 𝑑𝐴~0;
or for 𝐮 ≠ 0 and along a streamline outside the boundary layer, viscous work is
negligible.
Assume parallel flow with constant p⁄ρ + gz (i.e., hydrostatic pressure variation) and û
over inlet and outlet.
𝑝
𝜌
𝑄̇ − 𝑊̇𝑠 = (𝑢̂ + + 𝑔𝑧) ∫inlet&outlet 𝜌(𝐮 ∙ 𝐧)𝑑𝐴 + ∫inlet&outlet 𝑢2 (𝐮 ∙ 𝐧)𝑑𝐴
𝜌
2
𝜌
𝑝
𝑄̇ − 𝑊̇𝑠 = (𝑢̂ + + 𝑔𝑧) (−𝑚̇in ) − ∫inlet 𝑢3 𝑑𝐴
𝜌
2
𝑝
𝜌
+ (𝑢̂ + + 𝑔𝑧) ( 𝑚̇out ) + ∫outlet 𝑢3 𝑑𝐴
𝜌
2
Define kinetic energy correction factor
1
𝑢
3
𝛼 = ∫A ( ) 𝑑𝐴
𝐴
𝑈
𝑎𝑣
→
𝜌
∫
2 𝐴
Laminar pipe flow: 𝑢 = 𝑈0 (1 −
2(
𝑢 𝐮 ∙ 𝐧)𝑑𝐴 = 𝛼
2
𝑈,
1+𝑚)(2+𝑚) 0
2
𝑟2
4
𝑅
3
1
𝑟 𝑚 1
1
𝑅
5
9
𝑚 = , 𝑈𝑎𝑣 = 0.82𝑈0
7
𝑚̇
(𝑚̇ = 𝜌𝐴𝑈𝑎𝑣 )
2 ), 𝑈𝑎𝑣 = 0.5𝑈0 , 𝛽 = , 𝛼 = 2
Turbulent pipe flow: 𝑢 ≈ 𝑈0 (1 − ) , ≤ 𝑚 ≤
𝑈𝑎𝑣 = (
2
𝑈𝑎𝑣
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𝛽=
(1+𝑚)2 (2+𝑚)2
,
2(1+2𝑚)(2+2𝑚)
𝑚 = , 𝛽 = 1.02
𝛼=
(1+𝑚)3 (2+𝑚)3
,
4(1+3𝑚)(2+3𝑚)
𝑚 = , 𝛼 = 1.058
𝑄̇
𝑚̇
−
𝑊̇𝑠
𝑚̇
1
7
1
7
𝑝
2
𝑈𝑎𝑣
𝜌
2
= (𝑢̂ + + 𝑔𝑧 + 𝛼
)
𝑝
2
𝑈𝑎𝑣
𝜌
2
− (𝑢̂ + + 𝑔𝑧 + 𝛼
outlet
)
inlet
Let in = 1, out = 2, 𝑈 = 𝑈𝑎𝑣 , and divide by 𝑔
𝑝1
𝛼 2
𝑝2
𝛼 2
+
𝑈 + 𝑧1 + ℎ𝑝 =
+
𝑈 + 𝑧2 + ℎ𝑡 + ℎ𝐿
𝜌𝑔 2𝑔 1
𝜌𝑔 2𝑔 2
𝑊̇𝑝
𝑊̇𝑠
𝑊̇𝑡
=
−
= ℎ𝑡 − ℎ𝑝
𝑔𝑚̇
𝑔𝑚̇ 𝑔𝑚̇
1
𝑄̇
ℎ𝐿 = (𝑢̂2 − 𝑢̂1 ) −
𝑔
𝑚̇𝑔
Assuming no heat transfer, mechanical energy is converted to thermal energy through
viscosity and can not be recovered; therefore, it is referred to as head loss ℎ𝐿 ≥ 0, which
can be shown from 2nd law of thermodynamics.
Steady, 1D energy equation can be considered as modified Bernoulli equation for
head ℎ𝑝 , ℎ𝑡 , and ℎ𝐿 .
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8.2Differential Form of Energy Equation
𝑑𝐸
𝑑𝑡
= ∫𝐶𝑉
𝜕
𝜕𝑡
(𝑒𝜌)𝑑𝑉 + ∫𝐶𝑆 𝑒𝜌𝐮 ∙ 𝐧𝑑𝐴 = 𝑄̇ − 𝑊̇
Guass’ Theorem:
𝑑𝐸
𝑑𝑡
𝜕
𝜕𝑡
= ∫𝐶𝑉
𝜕
𝜕𝑡
[(𝑒𝜌) + ∇ ∙ (𝑒𝜌𝐮)]𝑑𝑉 = 𝑄̇ − 𝑊̇
[(𝑒𝜌) + ∇ ∙ (𝑒𝜌𝐮)] = 𝜌
𝜕𝑒
𝜕𝑡
+𝑒
𝜕𝜌
𝜕𝑡
+ 𝑒∇ ∙ (𝜌𝐮) + 𝜌𝐮 ∙ ∇𝑒
Continuity equation:
𝜕𝜌
𝜕𝑡
𝜕
𝜕𝑡
+ ∇ ∙ (𝜌𝐮) = 0
[(𝑒𝜌) + ∇ ∙ (𝑒𝜌𝐮)] = 𝜌
𝐷𝑒
𝐷𝑡
= 𝜌(
𝜕𝑒
𝜕𝑡
+ 𝐮 ∙ ∇𝑒)
Energy per unit mass
1
1
2
2
𝑒 = 𝑢̂ + 𝑢2 + 𝑔𝑧 = 𝑢̂ + 𝑢2 − 𝑔 ∙ 𝑟
𝜌
𝐷𝑒
𝐷𝑡
= 𝜌(
̂
𝐷𝑢
𝐷𝑡
+𝐮∙
𝐷𝐮
𝐷𝑡
− 𝐮 ∙ 𝐠) = (𝑄̇ − 𝑊̇ )⁄𝑉 = 𝑞̇ − 𝑤̇
Heat added to fluid per unit mass (Fourier’s Law):
𝑞̇ = −∇ ∙ 𝐪 = −∇ ∙ (−𝑘∇𝑇) = ∇ ∙ (𝑘∇𝑇)
Work done fluid per unit mass
𝑤̇ = −∇ ∙ (𝐮 ∙ 𝛔) = −𝐮 ∙ (∇ ∙ 𝛔) − 𝛔 ∶ ∇𝐮
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Momentum equation:
𝜌
𝐷𝐮
𝐷𝑡
= 𝜌𝐠 + ∇ ∙ 𝛔
−𝐮 ∙ (∇ ∙ 𝛔) = −𝐮 ∙ (𝜌
𝐷𝐮
𝐷𝑡
− 𝜌𝐠)
Internal energy:
𝜌
𝜌
̂
𝐷𝑢
𝐷𝑡
̂
𝐷𝑢
𝐷𝑡
= ∇ ∙ (𝑘∇𝑇) + 𝛔 ∶ ∇𝐮
= ∇ ∙ (𝑘∇𝑇) + 𝛕 ∶ ∇𝐮 − 𝑝∇ ∙ 𝐮
(𝛔 = 𝛕 − 𝑝𝐈)
Continuity equation:
1 𝐷𝜌
𝜌 𝐷𝑡
+∇∙𝐮=0
→
∇∙𝐮=−
−𝑝∇ ∙ 𝐮 =
𝜌
𝐷
𝐷𝑡
𝑝 𝐷𝜌
𝜌 𝐷𝑡
1 𝐷𝜌
𝜌 𝐷𝑡
𝐷 𝑝
= −𝜌
( )+
𝑝
𝐷𝑡 𝜌
𝐷𝑝
𝜌
𝐷𝑡
(𝑢̂ + ) = ∇ ∙ (𝑘∇𝑇) +
𝑝
𝜌
𝐷ℎ
𝐷𝑡
= ∇ ∙ (𝑘∇𝑇) +
Viscous dissipation function:
𝛷 = 𝛕 ∶ ∇𝐮
𝐷𝑝
𝐷𝑡
𝐷𝑡
+ 𝛕 ∶ ∇𝐮
Enthalpy ℎ = 𝑢̂ + :
𝜌
𝐷𝑝
+𝛷
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9 Summary GDE
9.1Compressible Non-constant Property Fluid Flow
𝜕𝜌
Continuity:
𝜕𝑡
𝜌
Momentum:
𝐷𝐮
𝐷𝑡
= 𝜌𝐠 − ∇𝑝 + ∇ ∙ 𝛕
𝜌
Energy:
or
+ ∇ ∙ (𝜌𝐮) = 0
𝜌
𝐷ℎ
𝐷𝑡
̂
𝐷𝑢
𝐷𝑡
2
(𝜏𝑖𝑗 = 𝜇𝑆𝑖𝑗 + 𝛿𝑖𝑗 (𝜇𝑣 − 𝜇) ∇ ∙ 𝐮)
= ∇ ∙ (𝑘∇𝑇) +
3
𝐷𝑝
𝐷𝑡
+𝛷
= ∇ ∙ (𝑘∇𝑇) + 𝛷 − 𝑝∇ ∙ 𝐮
Primary variables:
𝑝, 𝐮, 𝑇
Auxiliary relations: 𝜌 = 𝜌 (𝑝, 𝑇), 𝜇 = 𝜇 (𝑝, 𝑇), ℎ = ℎ (𝑝, 𝑇), 𝑘 = 𝑘 (𝑝, 𝑇)
Restrictive Assumptions:
1) Continuum
2) Newtonian fluids
3) Thermodynamic equilibrium
4) Body force 𝐟 = −𝜌𝑔𝐤
5) heat conduction follows Fourier’s law
6) no internal heat sources
For compressible flow, 𝜌 is solved from continuity equation, 𝑇 from energy equation, and
𝑝 = (𝜌, 𝑇) from equation of state (eg, ideal gas law).
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9.2Incompressible Constant Property Fluid Flow
∇∙𝐮=0
Continuity:
𝐷𝐮
Momentum:
𝜌
Energy:
𝜌𝑐𝑣
𝐷𝑡
= 𝜌𝐠 − ∇𝑝 + 𝜇∇2 𝐮
𝐷𝑇
𝐷𝑡
= 𝑘∇2 𝑇 + 𝛷
(𝛷 = 𝜏𝑖𝑗
𝜕𝑢𝑖
𝜕𝑥𝑗
)
The following approximations are applied in the above energy equation:
𝑑𝑢̂ = 𝑐𝑣 𝑑𝑇
𝑐𝑣 , 𝜇, 𝑘, 𝜌 ≈ const.
̂
𝐷𝑢
𝐷𝑇
𝜌 = ∇ ∙ (𝑘∇𝑇) + 𝛷 − 𝑝∇ ∙ 𝐮
→ 𝜌𝑐𝑣 = 𝑘∇2 𝑇 + 𝛷
𝐷𝑡
𝐷𝑡
For static fluid or 𝐮 small, dissipation and convective terms become negligible:
𝜌𝑐𝑝
𝜕𝑇
𝜕𝑡
= 𝑘∇2 𝑇
which is the heat conduction equation (also valid for solids)
For incompressible flow with constant 𝜌, the continuity and momentum euqations are
uncoupled from energy equation; therefore, the energy equation can be solved separately
using solution from them to get 𝑇 . The momentum equation contains ∇𝑝 such that
reference 𝑝 is arbitrary (i.e. for incompressible flow, there is no connection between 𝑝
and 𝜌). The connection is between ∇𝑝 and ∇ ∙ 𝐮 = 0, i.e. a solution for 𝑝 requires ∇ ∙ 𝐮 =
0.
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9.3Approximate Models
9.3.1 Stokes Flow
For low 𝑅𝑒 = 𝑈𝐿⁄𝜐 ≪ 1, 𝐮 ∙ ∇𝐮 ≈ 𝟎
∇∙𝐮=0
Continuity:
𝜕𝐮
Momentum:
𝜕𝑡
𝟏
= − ∇𝑝 + 𝜐∇2 𝐮
𝜌
Linear, “elliptic” equation. Most exact solutions for NS equations
𝜕𝐮
𝟏
𝜕𝑡
𝜌
∇ ∙ ( ) = ∇ ∙ (− ∇𝑝 + 𝜐∇2 𝐮)
→ ∇2 𝑝 = 0
9.3.2 Boundary Layer Equations
For high 𝑅𝑒 ≫ 1 and attached boundary layers or fully developed free shear flows
(wakes, jets, mixing layers), 𝑥 is the streamwise direction:
𝜕
𝑉 ≪ 𝑈,
𝜕𝑈
𝜕𝑥
𝜕𝑈
𝜕𝑡
+
𝜕𝑥
𝜕𝑉
𝜕𝑦
+𝑈
≪
𝜕
𝜕𝑦
,
𝜕𝑝
𝜕𝑦
= 0 , and for free shear flows
𝜕𝑝
𝜕𝑥
= 0.
=0
𝜕𝑈
𝜕𝑥
+𝑉
𝜕𝑈
𝜕𝑦
=−
𝜕𝑝
𝜕𝑥
+𝜐
𝜕2 𝑈
𝜕𝑦 2
Non-linear, “parabolic” equation. Many exact solutions; similarity methods
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Inviscid Flow
Continuity:
𝜕𝜌
+ ∇ ∙ (𝜌𝐮) = 0
𝜕𝑡
𝐷𝐮
Euler equation:
𝜌
Energy:
𝜌
𝐷𝑡
𝐷ℎ
𝐷𝑡
= 𝜌𝐠 − ∇𝑝
nonlinear, “hyperbolic” equation
= ∇ ∙ (𝑘∇𝑇) +
𝐷𝑝
𝐷𝑡
𝑝, 𝐮, 𝑇 are unknowns, 𝜌, ℎ, 𝑘 = 𝑓 (𝑝, 𝑇)
9.3.4 Inviscid, Incompressible, Irrotational
∇×𝐮=0
⇒ 𝐮 = ∇𝛹
∇×𝐮=0
⇒ ∇2 𝛹 = 0 linear, “elliptic” equation
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The horizontal distance of the frictionless jet is
𝐿 = 𝑡𝑡𝑟𝑎𝑣𝑒𝑙 𝑉2 cos 𝜃
𝑡𝑡𝑟𝑎𝑣𝑒𝑙 = 2𝑉2 sin 𝜃 /𝑔
(𝑉 = 𝑔𝑡 in free fall)
𝐿 = 2𝑉2 sin 𝜃 × 𝑉2 cos 𝜃/𝑔 = 𝑉22 sin 2𝜃 /𝑔
For maximum horizontal distance
𝑑𝐿
𝑑𝜃
=
2𝑉22 cos 2𝜃
𝑔
=0
→
𝜃 = 45°
The jet reaches a maximum height of ℎ𝑚𝑎𝑥 = 25 m
𝑔 𝑡𝑡𝑟𝑎𝑣𝑒𝑙 2
ℎ𝑚𝑎𝑥 = (
2
2
) =
1
2𝑔
1
(𝑉2 sin 𝜃)2
(ℎ = 𝑔𝑡 2 in free fall)
2
Therefore, 𝑉2 sin 𝜃 = √2𝑔ℎ𝑚𝑎𝑥 , or the jet exit velocity is
𝑉2 =
√2𝑔ℎ𝑚𝑎𝑥
sin 𝜃
=
√2×9.81×25
sin 45°
≈ 31.32 𝑚/𝑠
The steady flow, 1D energy equation for the piping system between 1 and 2 is
𝑝1
𝜌𝑔
+
or: 0 +
1
2𝑔
𝑉12 + 𝑧1 =
0
𝑝2
𝜌𝑔
+
+ 15 = 0 +
1
𝑉22
+ 𝑧2 + ℎ𝐿 − ℎ𝑝
(2×9.81)
+ 2 + 6.5 −ℎ𝑝
2𝑔
(31.32)2
𝜋
→
ℎ𝑝 ≈ 43.5 𝑚
𝑃𝑝𝑢𝑚𝑝 = 𝛾𝑄ℎ𝑝 = 𝛾𝐴2 𝑉22 ℎ𝑝 = (9790) [ × (0.05)2 × 31.32] × 43.5 ≈ 26200 𝑊
4
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(a) First suppose 2D problem: D1 and D2 denotes width in y instead of diameter and we
take unit in z direction
Momentum equation for the CV shown:
∑ 𝐅 = 𝜌 ∫𝐶𝑆 𝐕𝐕 ∙ 𝐧𝑑𝐴
𝐹𝑥 = −𝐹 = 𝜌𝑉2 (−𝑉2 𝐴2 ) = SG𝜌water 𝑉2 (−𝑉2 𝐴2 )
(0.79 × 989) × 𝑉2 𝑉2 × (0.02 × 1) = 425𝑁
𝑉2 = 5.22 𝑚/𝑠,
𝑚̇ = 𝜌𝑉2 𝐴2 = 81.6 𝑘𝑔/𝑠
Continuity equation between points 1 and 2
V1 A1  V2 A2  V1  V2
D2
 2.09m / s
D1
Bernoulli neglect 𝑔, 𝑝2 = 𝑝𝑎𝑡𝑚
1
1
V12  p2  V22
2
2
1
p1  p2   V22  V12  
2
p1 
p1  101, 000 
hL=0, z=constant
.79  998
(5.222  2.092 )
2
p1  110,020 Pa
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(b) For the round jet given in the problem statement
𝐹𝑥 = −𝐹 = 𝜌𝑉2 (−𝑉2 𝐴2 ) = SG𝜌water 𝑉2 (−𝑉2 𝐴2 )
𝜋
(0.79 × 989) × 𝑉2 𝑉2 × × (0.02)2 = 425𝑁
4
𝑉2 = 41.4 𝑚/𝑠,
𝑚̇ = 𝜌𝑉2 𝐴2 = 10.3 𝑘𝑔/𝑠
Continuity equation between points 1 and 2
V1 A1  V2 A2  V1  V2
D 
V1 A1  V2 A2  V1  V2  2 
 D1 
2
V1  41.4  
5
D2
 2.09m / s
D1
2
2
V1  6.63 m / s
Bernoulli neglect g, p2=pa
p1 
1
1
V12  p2  V22
2
2
1
p1  p2   V22  V12 
2
p  760,000 Pa
1
hL=0, z=constant
 p1  101, 000 
.79  998
(41.42  6.632 )
2
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10 Conservation of Angular Momentum
𝑑𝐵sys
RTT:
𝑑𝑡
=
𝑑
∫
𝑑𝑡 𝐶𝑉
𝛽𝜌𝑑𝑉 + ∫𝐶𝑆 𝛽𝜌𝐮𝑟 ∙ 𝐧𝑑𝐴
Angular momentum of system about the origin 𝑂 of the inertial coordinate system:
𝐵 = 𝐇𝑂 = ∫𝑠𝑦𝑠 𝐫 × 𝐮𝑑𝑚
Angular momentum per unit mass: 𝛽 =
10.1
𝑑𝐇𝑂
𝑑𝑚
=𝐫×𝐮
Integral Form
𝑑𝐇𝑂
𝑑𝑡
=
𝑑
∫
𝑑𝑡 𝐶𝑉
(𝐫 × 𝐮)𝜌𝑑𝑉 + ∫𝐶𝑆 (𝐫 × 𝐮)𝜌𝐮 ∙ 𝐧𝑑𝐴 = ∑ 𝐌𝑂 = ∑(𝐫 × 𝐅)𝑂
MO : vector sum all external moments applied CV due to both 𝐅𝐵 and 𝐅𝑆
Momentum equation:
𝑑 (𝑀𝐮)
𝑑𝑡
=
𝑑
∫
𝑑𝑡 𝐶𝑉
𝐮𝜌𝑑𝑉 + ∫𝐶𝑆 𝐮𝜌𝐮 ∙ 𝐧𝑑𝐴 = ∑ 𝐅
Moment form of momentum equation (not a new conservation law!)
For uniform flow across discrete inlet/outlet:
∫𝐶𝑆 (𝐫 × 𝐮)𝜌𝐮 ∙ 𝐧𝑑𝐴 = ∑(𝐫 × 𝐕)out 𝑚̇out − ∑(𝐫 × 𝐕)in 𝑚̇in
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Differential Form
𝑑
∫
𝑑𝑡 𝐶𝑉
(𝐫 × 𝐮)𝜌𝑑𝑉 + ∫𝐶𝑆 (𝐫 × 𝐮)𝜌𝐮 ∙ 𝐧𝑑𝐴 = ∑ 𝐌𝑂
Apply CV form for fixed CV:
𝐼
𝑑2 𝜃
1
1
𝜕𝜏𝑥𝑦
2
1
2
1
𝜕𝑥
𝜕𝜏𝑦𝑥
2
2
𝜕𝑦
= 𝑑𝑥 (𝜏𝑥𝑦 𝑑𝑦𝑑𝑧) + 𝑑𝑥 [(𝜏𝑥𝑦 +
𝑑𝑡 2
− 𝑑𝑦 (𝜏𝑦𝑥 𝑑𝑥𝑑𝑧) − 𝑑𝑦 [(𝜏𝑦𝑥 +
= (𝜏𝑥𝑦 +
1 𝜕𝜏𝑥𝑦
2 𝜕𝑥
𝑑𝑥 − 𝜏𝑦𝑥 −
1 𝜕𝜏𝑦𝑥
2 𝜕𝑦
𝑑𝑥) 𝑑𝑦𝑑𝑧]
𝑑𝑦) 𝑑𝑥𝑑𝑧]
𝑑𝑦) 𝑑𝑥𝑑𝑦𝑑𝑧
Neglect higher-order terms, and give the moment of inertia 𝐼
𝐼
𝑑2 𝜃
𝑑𝑡 2
1
12
=
1
12
2
(𝜌𝑑𝑥𝑑𝑦𝑑𝑧)(𝑑𝑥
𝜌(𝑑𝑥 +
2
𝑑 𝜃
𝑑𝑦 2 ) 2
𝑑𝑡
2
[
1
𝑑𝑥→0,𝑑𝑦→0 12
= (𝜏𝑥𝑦 − 𝜏𝑦𝑥 )𝑑𝑥𝑑𝑦𝑑𝑧
= 𝜏𝑥𝑦 − 𝜏𝑦𝑥
Assuming the angular acceleration
lim
2
𝑑 𝜃
+ 𝑑𝑦 2 ) 2
𝑑𝑡
𝜌(𝑑𝑥 2 +
𝑑2 𝜃
is finite
𝑑𝑡 2
2
2) 𝑑 𝜃
𝑑𝑦
]=
𝑑𝑡 2
0 = 𝜏𝑥𝑦 − 𝜏𝑦𝑥
Therefore
Similarly
𝜏𝑥𝑧
𝜏𝑥𝑦 = 𝜏𝑦𝑥
= 𝜏𝑧𝑥 , 𝜏𝑧𝑦 = 𝜏𝑦𝑧
The stress tensor is symmetric, i.e., stresses themselves cause no rotation
There is no separate differatial equation for angular momentum to remember.
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11 Boundary Conditions for Governing Differential Equations
Summary of GDEs:
𝜕𝜌
Continuity:
𝜕𝑡
𝜌
Momentum:
Energy:
𝜌
̂
𝐷𝑢
𝐷𝑡
𝐷𝐮
𝐷𝑡
+ ∇ ∙ (𝜌𝐮) = 0
= 𝜌𝐠 − ∇𝑝 + ∇ ∙ 𝛕
= ∇ ∙ (𝑘∇𝑇) + 𝛷 − 𝑝∇ ∙ 𝐮
A system of 2nd-order nonlinear PDE, which require IC/BC, together with equations of
state, for their solution, depending on physical problem and appropriate approximations.
First, if the flow is unsteady, there must be an initial condition or initial spatial
distribution known for each variable.
Types of Boundaries:
1. Solid Surface
2. Inlet/outer/exit
3. Interface
Internal flow is a flow for which the fluid is confined by a surface, e.g., pipe flow.
External flow is such a flow that boundary layers develop freely, without constraints
imposed by adjacent surfaces, e.g., air flow around an airplane.
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11.1
Solid Surface
No relative motion or temperature difference between liquid and solid.
𝐮fluid = 𝐮solid , 𝑇fluid = 𝑇solid
Exception is for contact line.
11.2
Inlet /outer /oulet
a) inlet: 𝐮, 𝑝, 𝑇 specified, e.g., 𝑇 = const., 𝐮 = 𝑈𝐢, 𝑝 = 0
b) outer: 𝐮, 𝑝, 𝑇 specified, e.g., 𝑇 = const., 𝐮 = 𝑈𝐢, 𝑝 = 0
c) exit: depends on the problem, but often use, 𝜕 2 𝑢⁄𝜕𝑥 2 = 0, i.e. zero streamwise
diffusion for external flow; and periodic for fully developed internal flow.
11.3
Idealized gas/liquid interface
11.3.1 Kinematic FSBC
Just as with solid surface, there can be no relative velocity across interface (i.e. exact
condition for liquid/liquid and gas/gas or gas/liquid non-mixing fluids).
𝐮1 = 𝐮2
Normal velocity:
𝐮1 ∙ 𝐧 = 𝐮2 ∙ 𝐧 = 𝑢𝑛1 = 𝑢𝑛2
required by KFSBC.
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Tangential should also match, but
usually due to different approximations
used in fluid 1 or 2. Often, in fact, motions
in gas are neglected and therefore 𝐮 is not
continuous in the tangential direction.
If the free surface is a single value
function given by function:
𝑧 = 𝜂(𝑥, 𝑦, 𝑡)
Velocity:
𝑤liquid = 𝑤gas =
𝑑𝜂
𝑑𝑡
=
𝜕𝜂
𝜕𝑡
+𝑢
𝜕𝜂
𝜕𝑥
+𝑣
𝜕𝜂
𝜕𝑦
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11.3.2 Dynamic FSBC
Viscous shear stresses must balance:
(𝜏𝑧𝑦 )
= (𝜏𝑧𝑦 )
liquid
gas
, (𝜏𝑧𝑥 )liquid = (𝜏𝑧𝑥 )gas
Neglect ing the viscous normal stresses, the pressures must balance at the interface except
for surface tension effect:
𝑝liquid = 𝑝gas − 𝜎𝜅
Many approximations, e.g., inviscid approximation:
patm = 0
Small slope:
∂η
∂x
~
∂η
∂y
~0
Small normal velocity gradient:
∂w
∂x
~
∂w
∂y
~
∂w
∂z
~0
p = 0, or, piezometric preesure distribution: p̂ = ρgz
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Incompressible flow
𝜕𝑢
𝜕𝑥
𝜕𝑣
𝜕𝑦
=−
𝜕𝑢
𝜕𝑥
= −𝑈
𝜕𝛿
𝜕𝑥
𝜕𝑣
𝜕𝑦
=
𝜕
𝜕𝑥
𝜕
𝑦3
𝛿
𝛿3
(2 − 2
𝜕𝑣
𝜕𝑦
𝑦4
+
𝛿4
=0
) = −𝑈 (−2
1
(𝐶𝑥 1/2 ) = 𝐶𝑥 −1/2 =
𝜕𝑥
𝑦
= 𝑈 (2
𝑦
+
2
𝛿2
−6
𝑦3
𝛿4
+4
𝑦4
)
𝛿
𝛿 5 2𝑥
1
2𝑥
𝑈 𝑦
𝑦
𝛿2
𝐶𝑥 1/2 =
= ( −3
𝑥
+6
𝛿
𝑦3
𝛿3
𝑦3
𝛿4
𝛿
−4
𝑦 4 𝜕𝛿
2𝑥
+2
𝑦4
𝛿4
)
(a) The expression of 𝑣 velocity component:
𝑣=
𝑦𝑈 𝑦
∫0 𝑥 (𝛿
−3
𝑦3
𝛿3
+2
(b) For maximum value of 𝑣, we have
𝜕𝑣
𝜕𝑦
𝑈 𝑦
𝑦3
𝑥
𝛿3
= ( −3
𝛿
+2
𝑦4
𝛿4
𝑦4
𝛿4
𝜕𝑣
𝑈 𝑦2
) 𝑑𝑦 = (
𝜕𝑦
𝑥
2𝛿
−
3 𝑦4
4 𝛿3
+
2 𝑦5
5 𝛿4
)
=0
)=
𝑈𝑦
𝑥𝛿
(1 − 3
𝑦2
𝛿2
+2
𝑦3
𝛿3
)=0
𝑦
Apparently = 1, or 𝑦 = 𝛿 is a (actually the only) solution
𝛿
𝑣max
𝑈 𝛿2 3 𝛿4 2 𝛿5
𝑈𝛿 1 3 2
(
)
=𝑣 𝛿 = ( −
+
)=
( − + )
𝑥 2𝛿 4 𝛿 3 5 𝛿 4
𝑥 2 4 5
=
3 𝑈𝛿
20 𝑥
=
3 3×0.011
20
1
≈ 0.0050 𝑚/𝑠
)
𝛿 5 𝜕𝑥
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Steady flow and only the z-component of NS equation is relevant:
𝜕
𝜕𝑡
= 0, 𝑢 = 0, 𝑣 = 0,
𝜕
𝜕𝑦
= 0,
𝜕
𝜕𝑧
𝜕2 𝑤
= 0, 𝑔𝑥 = 0, 𝑔𝑦 = 0,
NS equation reduces to
𝜌𝑔𝑧 + 𝜇
No-slip wall BC:
Integrate
𝑤(−ℎ) = 𝑤(ℎ) = 0
𝜕2 𝑤
𝜕𝑥 2
=−
𝜌𝑔𝑧
𝜇
→
Apply BC: 𝑤(−ℎ) = −
𝜌𝑔𝑧
2𝜇
𝜕𝑤
𝜕𝑥
2
=−
𝜌𝑔𝑧
𝜇
𝜕𝑥 2
𝑥 + 𝑐1
=0
→
𝑤=−
ℎ − 𝑐1 ℎ + 𝑐2 = 0, 𝑤(ℎ) = −
𝑐1 = 0, 𝑐2 =
𝑤(𝑥 ) =
𝜌𝑔𝑧
𝜌𝑔𝑧
2𝜇
𝜌𝑔𝑧
2𝜇
2
ℎ + 𝑐1 ℎ + 𝑐2 = 0
ℎ2
2𝜇
𝜌𝑔
𝜌𝑔
𝜌𝑔
− 𝑧 𝑥 2 + 𝑧 ℎ2 = 𝑧 (ℎ2
2𝜇
2𝜇
2𝜇
𝑥 2 + 𝑐1 𝑥 + 𝑐2
− 𝑥 2)
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12 Exact Solutions of NS Equations
12.1
Couette Flow
Couette flow is the flow between two parallel plates. Here, one plate is at rest and the
other is moving with a velocity U. Let us assume the plates are infinitely large in z
direction, so the z dependence is not there.
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12.2
Flow between Two Concentric Rotating Cylinders
Consider flow in the annulus of two cylinders, where 𝑟1 and 𝑟2 are the radii of inner and
outer cylinders, respectively, and the cylinders move with different rotational speeds 𝜔1
and 𝜔2 respectively
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IC (Initial condition): steady flow:
𝜕
𝜕𝑡
=0
BC: Fully developed laminar flow, i.e.,
𝑢𝑧 = 0,
𝜕
𝜕𝜃
= 0,
𝜕
𝜕𝑧
=0
No-slip at the wall, 𝑟 = 𝑟1 , 𝑢𝑟 = 0, 𝑢𝜃 = 𝜔1 𝑟1 ; 𝑟 = 𝑟2 , 𝑢𝑟 = 0, 𝑢𝜃 = 𝜔2 𝑟2
Continuity equation:
1 𝜕
𝑟 𝜕𝑟
(𝑟𝑢𝑟 ) +
1 𝜕𝑢𝜃
𝑟 𝜕𝜃
+
𝜕𝑢𝑧
=0
𝜕𝑧
1 𝜕
𝑟 𝜕𝑟
(𝑟𝑢𝑟 ) = 0
→
𝑟𝑢𝑟 = 𝑐
𝑟 = 𝑟1 , 𝑢𝑟 = 0; 𝑟 = 𝑟2 , 𝑢𝑟 = 0
→
𝑢𝑟 = 0
r-momentum:
𝜕𝑝
𝜕𝑟
=𝜌
2
𝑢𝜃
𝑑𝑝
𝑟
𝑑𝑟
=𝜌
2
𝑢𝜃
𝑟
This equation shows that the centrifugal force is supplied by the radial pressure, exerted
by the wall of the enclosure on the fluid, or, it describes the radial pressure distribution.
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𝜃-momentum:
𝜐[
1 𝜕
𝑟 𝜕𝑟
𝜕 2 𝑢𝜃
𝜕𝑟 2
𝑑
(𝑟
+
[
𝜕𝑢𝜃
𝜕𝑟
1 𝜕𝑢𝜃
𝑟 𝜕𝑟
1 𝑑
𝑑𝑟 𝑟 𝑑𝑟
1 𝑑
𝑟 𝑑𝑟
)−
−
𝑢𝜃
𝑟2
𝑢𝜃
𝑟2
=0
(𝑟𝑢𝜃 )] = 0
(𝑟𝑢𝜃 ) = 𝐴
𝑟𝑢𝜃 =
𝑑
(𝑟𝑢𝜃 )
𝑑𝑟
𝐴 2
𝑟 +𝐵
2
𝐴
𝐵
𝑢𝜃 = 𝑟 +
2
BC:
]=0
𝑟
𝑟 = 𝑟1 , 𝑢𝜃 = 𝜔1 𝑟1 ; 𝑟 = 𝑟2 , 𝑢𝜃 = 𝜔2 𝑟2
𝐴
𝐵
2
𝐴
𝑟1
𝐵
2
𝑟2
𝑢𝜃 (𝑟 = 𝑟1 ) = 𝑟1 +
𝑢𝜃 (𝑟 = 𝑟2 ) = 𝑟2 +
= 𝜔1 𝑟1
= 𝜔2 𝑟2
= 𝐴𝑟
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𝐴=
𝐵=
2
𝑟22 −𝑟12
𝑟12 𝑟22
𝑟22 −𝑟12
(𝜔2 𝑟22 − 𝜔1 𝑟12 )
(𝜔1 − 𝜔2 )
1
1
2
2)
2 2(
(
)
𝑢𝜃 = 2
[ 𝜔2 𝑟2 − 𝜔1 𝑟1 𝑟 + 𝑟1 𝑟2 𝜔1 − 𝜔2 ]
𝑟
𝑟2 − 𝑟12
Special cases:
𝜔1 → 0, 𝑟1 → 0, 𝑢𝜃 = 𝜔2 𝑟, steady rotation of cylinder filled with fluid under
rigid body rotation.
𝑟12 𝜔1
𝜔2 = 0, 𝑟2 → ∞, 𝑢𝜃 =
, potential vortex driven by rotating cylinder with no𝑟
slip boundary condition.
Small clearance, 𝑟2 − 𝑟1 ≪ 𝑟1 , 𝜔2 = 0, 𝑢𝜃 = 𝜔1 𝑟1
Stress and torque:
𝜏𝑟𝜃 = 𝜇 (
𝜕𝑢𝜃
𝜕𝑟
+
1 𝜕𝑢𝑟
𝑟 𝜕𝜃
Here:
𝜏𝑟𝜃 = 𝜇 (
Or
𝜏𝑟𝜃 =
𝜕𝑢𝜃
−
𝑢𝜃
𝜕𝑟
𝑟
𝑑 𝑢
𝜇𝑟 ( 𝜃 )
𝑑𝑟 𝑟
)
−
𝑢𝜃
𝑟
)
𝑟−𝑟1
𝑟2 −𝑟1
, linear couette flow.
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𝜏𝑟𝜃 = 𝜇𝑟
𝜏𝑟𝜃 = 2𝜇
Now
Chapters 3&4
96
𝑑
1
{
1
2
2
2 2
2 [(𝜔2 𝑟2 − 𝜔1 𝑟1 ) + 𝑟1 𝑟2 (𝜔1 − 𝜔2 ) 2 ]}
𝑑𝑟 𝑟22 −𝑟1
𝑟
𝑟12 𝑟22
1
𝑟22 −𝑟1
𝑟2
2 (𝜔2 − 𝜔1 )
𝑟 = 𝑟1 , 𝜏𝑟𝜃 |𝑟=𝑟1 = 2𝜇
𝑟 = 𝑟2 , 𝜏𝑟𝜃 |𝑟=𝑟2 = 2𝜇
𝑟12 𝑟22
𝑟22 −𝑟12
𝑟12 𝑟22
𝑟22 −𝑟12
(𝜔2 − 𝜔1 )
(𝜔2 − 𝜔1 )
1
𝑟12
1
𝑟22
= 2𝜇
= 2𝜇
𝑟22
𝑟22 −𝑟12
𝑟12
𝑟22 −𝑟12
(𝜔2 − 𝜔1 )
(𝜔2 − 𝜔1 )
When the inner cylinder is at rest and the outer cylinder rotates, the torque transmitted by
the outer cylinder to the fluid is
𝑇2 = 2𝜇
𝑟12
𝑟22 −𝑟12
𝜔2 × 2𝜋𝑟2 𝑙 × 𝑟2
𝑇2 = 4𝜋𝜇𝑙
𝑟12 𝑟22
𝑟22 −𝑟12
𝜔2
Where 𝑙 is the length of the cylinder.
The moment 𝑇1 , with which the fluid acts on the inner cylinder has the same magnitude.
If the angular velocity of the external cylinder and the moment acting on the inner
cylinder are measured, the coefficient of viscosity can be evaluated by the above
equation.
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12.3
Two Immicsible Liquid Films Flowing down an Inclined Wall
Consider two immiscible liquid films flowing down an inclined wal under the influence
of gravity. Assume that the flow is fully established and the thicknesses of the films are
ℎ𝑎 and ℎ𝑏 , respectively, and that the atmospheric pressure is 𝑝0 and the viscosity of air is
much smaller than those of the liquids. Solve for the velocity distributions in each of the
liquids.
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For both liquids a and b
𝑔𝑥 = 𝑔 cos 𝜃; 𝑔𝑦 = −𝑔 sin 𝜃; 𝑔𝑧 = 0
IC (Initial condition): steady flow:
𝜕
𝜕𝑡
=0
BC: Fully developed 1D flow, i.e.,
𝐮 = 𝑢(𝑦)𝐢, 𝑣 = 0, 𝑤 = 0,
𝜕
𝜕𝑥
= 0,
𝜕
𝜕𝑧
=0
No-slip at the wall, 𝑦 = 0, 𝑢𝑎 (0) = 0
Kinematic Condition at the interface:
Continuous velocity at the interface, 𝑦 = ℎ𝑎 : 𝑢𝑎 (ℎ𝑎 ) = 𝑢𝑏 (ℎ𝑎 )
Dynamic Condition at the interface: (continuous stresses)
Continuous shear stress at the interface, 𝑦 = ℎ𝑎 :
𝑎
𝑏
𝜏𝑦𝑥
= 𝜏𝑦𝑥
𝜇𝑎
𝜕𝑢𝑎
𝜕𝑦
= 𝜇𝑏
𝜕𝑢𝑏
𝜕𝑦
Neglect shear stress at the air/liquid interface, 𝑦 = ℎ𝑏 :
𝑏
𝑎𝑖𝑟
𝜏𝑦𝑥
= 𝜏𝑦𝑥
=0
𝜇𝑏
𝜕𝑢𝑏
𝜕𝑦
=0
Continuous normal stress and neglect viscous normal stress at air/liquid interface
𝑦 = ℎ𝑏 : 𝑝𝑏 = 𝑝0
𝑦 = ℎ𝑎 : 𝑝𝑎 = 𝑝𝑏
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Continuity equation:
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
𝜕𝑢𝑎
𝜕𝑥
= 0 for both a and b
=
𝜕𝑢𝑏
𝜕𝑥
=0
x-momentum:
𝜌𝑔𝑥 = 𝜌𝑔 cos 𝜃 = −
𝜕𝜏𝑦𝑥
𝜕𝑦
𝜇
=−
𝜕2 𝑢
𝜕𝑦 2
𝜕
𝜕𝑦
(𝜇
𝜕𝑢
𝜕𝑦
) = −𝜇
𝜕2 𝑢
𝜕𝑦 2
for both a and b
= −𝜌𝑔 cos 𝜃
Integrate for both a and b with 𝜐 = 𝜇/𝜌
𝑢𝑎 = −
𝑢𝑏 = −
𝑔 cos 𝜃
2𝜐𝑎
𝑔 cos 𝜃
2𝜐𝑏
𝑦 2 + 𝑐1 𝑦 + 𝑐2
𝑦 2 + 𝑐3 𝑦 + 𝑐4
BC:
𝑢𝑎 (𝑦 = 0) = 0 → 𝑐2 = 0
𝜇𝑏
𝜕𝑢𝑏
|
𝜕𝑦 𝑦=ℎ
𝑏
=0
→
𝜕𝑢𝑏
|
𝜕𝑦 𝑦=ℎ
𝑏
=0
→
𝑐3 =
𝑔 cos 𝜃
𝜐𝑏
ℎ𝑏
058:0160
Jianming Yang
𝜇𝑎
Chapters 3&4
100
Fall 2012
𝜕𝑢𝑎
|
𝜕𝑦 𝑦=ℎ
𝑎
= 𝜇𝑏
𝜕𝑢𝑏
|
𝜕𝑦 𝑦=ℎ
𝑎
→
𝜇𝑎 (−
𝑔 cos 𝜃
𝜐𝑎
ℎ𝑎 + 𝑐1 ) = 𝜇𝑏 (−
𝑔 cos 𝜃
𝜐𝑏
ℎ𝑎 + 𝑐3 )
(−𝜌𝑎 𝑔 cos 𝜃 ℎ𝑎 + 𝑐1 𝜌𝑎 𝜐𝑎 ) = (−𝜌𝑏 𝑔 cos 𝜃 ℎ𝑎 + 𝜌𝑏 𝑔 cos 𝜃 ℎ𝑏 )
1
(𝜌𝑎 𝑔 cos 𝜃 ℎ𝑎 − 𝜌𝑏 𝑔 cos 𝜃 ℎ𝑎 + 𝜌𝑏 𝑔 cos 𝜃 ℎ𝑏 )
𝑐1 =
𝜌𝑎 𝜐𝑎
𝑐1 =
𝑢𝑎 (ℎ𝑎 ) = 𝑢𝑏 (ℎ𝑎 )
𝑐4 = −
𝑔 cos 𝜃
2𝜐𝑎
𝑐4 = 𝑔 cos 𝜃
𝑐4 = 𝑔 cos 𝜃
ℎ𝑎2 + {
2
ℎ𝑎
ℎ𝑎 𝑔 cos 𝜃
𝜐𝑎
𝜐𝑎
−
[1 +
𝜌𝑏 ℎ𝑏
(
𝜐𝑎 2𝜐𝑏
2
2 1
ℎ𝑎
1 𝜐𝑎
𝜌𝑎 ℎ𝑎
[ +
𝜐𝑎 2
2 𝜐𝑏
−
𝜌𝑏
𝜌𝑎
+(
𝜌𝑏
𝜌𝑎
𝑢𝑎 =
𝑢𝑎 = −
𝑔 cos 𝜃
2𝜐𝑎
= 𝑔 cos 𝜃
2
ℎ𝑎
𝜐𝑎
𝜌𝑏 ℎ𝑏
(
𝜌𝑎 ℎ𝑎
− 1)]
−
−
𝑔 cos 𝜃
− 1)]} ℎ𝑎 +
𝜌𝑎 ℎ𝑎
𝜌𝑏 ℎ𝑏
− +1+
[1 +
𝑔 cos 𝜃 2
𝑔 cos 𝜃 2
ℎ𝑎 + 𝑐1 ℎ𝑎 = −
ℎ𝑎 + 𝑐3 ℎ𝑎 + 𝑐4
2𝜐𝑎
2𝜐𝑏
1
(
𝜐𝑎
→
ℎ𝑎 𝑔 cos 𝜃
𝜌𝑏
𝜌𝑎
−
𝜐𝑎 ℎ𝑏
𝜐𝑏 ℎ𝑎
𝜐𝑎 ℎ𝑏
)
ℎ𝑎2 − [
2𝜐𝑏
)
]
𝜐𝑏 ℎ𝑎
𝑔 cos 𝜃 2
−
𝑦 + 𝑐1 𝑦 + 𝑐2
2𝜐𝑎
ℎ𝑎 𝑔 cos 𝜃
𝜌𝑏 ℎ𝑏
2
𝑦 +
{[1 +
𝜐𝑎
𝜌𝑏 ℎ𝑏
(
𝑔 cos 𝜃
𝜌𝑎 ℎ𝑎
[1 +
(
𝜌𝑎 ℎ𝑎
− 1)] 𝑦
𝑦 2
𝑦
1
ℎ𝑎
2 ℎ𝑎
− 1)] ( ) − ( ) }
𝜐𝑏
ℎ𝑏 ] ℎ𝑎
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Jianming Yang
Chapters 3&4
101
Fall 2012
𝑢𝑏 = −
𝑢𝑏 = −
𝑔 cos 𝜃
2𝜐𝑏
𝑢𝑏 = 𝑔 cos 𝜃
2
𝑦 +
2
ℎ𝑎
𝜐𝑎
[−
𝑔 cos 𝜃
𝜐𝑏
1 𝜐𝑎
𝑔 cos 𝜃
2𝜐𝑏
𝑦 2 + 𝑐3 𝑦 + 𝑐4
2 1
ℎ𝑎
1 𝜐𝑎
𝜐𝑎 2
2 𝜐𝑏
𝑦
1
1 𝜐𝑎
𝜐𝑏 ℎ𝑎 ℎ𝑎
2
2 𝜐𝑏
ℎ𝑏 𝑦 + 𝑔 cos 𝜃
𝑦 2
( ) +
2 𝜐𝑏 ℎ𝑎
𝜐𝑎 ℎ𝑏
[ +
( )+ +
−
−
𝜌𝑏
𝜌𝑎
𝜌𝑏
𝜌𝑎
+(
𝜌𝑏
𝜌𝑎
+(
𝜌𝑏
𝜌𝑎
−
−
𝜐𝑎 ℎ𝑏
)
𝜐𝑏 ℎ𝑎
]
𝜐𝑎 ℎ𝑏
)
𝜐𝑏 ℎ𝑎
]
z-momentum:
0=0
y-momentum:
𝜌𝑔𝑦 = −𝜌𝑔 sin 𝜃 =
𝜕𝑝
𝜕𝑧
for both a and b
𝑝 = −𝜌𝑔 sin 𝜃 𝑦 + 𝑓 (𝑥 )
BC:
𝑦 = ℎ𝑏 : 𝑝𝑏 = −𝜌𝑏 𝑔 sin 𝜃 ℎ𝑏 + 𝑓(𝑥 ) = 𝑝0 → 𝑓(𝑥 ) = 𝑝0 + 𝜌𝑏 𝑔 sin 𝜃 ℎ𝑏
𝑝 = 𝑝0 + 𝜌𝑏 𝑔 sin 𝜃 (ℎ𝑏 − 𝑦) ℎ𝑎 ≤ 𝑦 ≤ ℎ𝑏
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Jianming Yang
Chapters 3&4
102
Fall 2012
𝑦 = ℎ𝑎 : 𝑝𝑎 = −𝜌𝑎 𝑔 sin 𝜃 ℎ𝑎 + 𝑓 (𝑥 ) = 𝑝0 + 𝜌𝑏 𝑔 sin 𝜃 (ℎ𝑏 − ℎ𝑎 )
→ 𝑓(𝑥 ) = 𝑝0 + 𝜌𝑏 𝑔 sin 𝜃 (ℎ𝑏 − ℎ𝑎 ) + 𝜌𝑎 𝑔 sin 𝜃 ℎ𝑎
𝑝 = 𝜌𝑎 𝑔 sin 𝜃 (ℎ𝑎 − 𝑦) + 𝜌𝑏 𝑔 sin 𝜃 (ℎ𝑏 − ℎ𝑎 ) + 𝑝0 0 ≤ 𝑦 ≤ ℎ𝑎
𝑝(0) = 𝜌𝑎 𝑔 sin 𝜃 ℎ𝑎 + 𝜌𝑏 𝑔 sin 𝜃 (ℎ𝑏 − ℎ𝑎 ) + 𝑝0 > 𝑝0
𝜋
𝜃 = , 𝑝(0) = 𝜌𝑎 𝑔ℎ𝑎 + 𝜌𝑏 𝑔(ℎ𝑏 − ℎ𝑎 ) + 𝑝0
2
Also, cos 𝜃 = 0, 𝑢𝑎 = 0, 𝑢𝑏 = 0 (no flow)
𝜃 = 0, 𝑝(0) = 𝑝0
Single layer of liquid film:
ℎ𝑏 = ℎ𝑎 = ℎ
𝑢𝑎 = 𝑔 cos 𝜃
2
ℎ𝑎
𝜐𝑎
{[1 +
𝜌𝑏 ℎ𝑏
(
𝜌𝑎 ℎ𝑎
𝑢 = 𝑔 cos 𝜃
ℎ2
𝜐
𝑦
1
𝑦 2
− 1)] ( ) − ( ) }
𝑦
ℎ𝑎
1 𝑦 2
ℎ
2 ℎ
{( ) − ( ) }
2 ℎ𝑎
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Jianming Yang
Chapters 3&4
103
Fall 2012
13 Vorticity Theorems
The incompressible flow momentum equations focus attention on 𝐮 and 𝑝 and explain the
flow pattern in terms of inertia, pressure, gravity, and viscous forces. Alternatively, one
can focus attention on 𝛚 and explain the flow pattern in terms of the rate of change,
deforming, and diffusion of 𝛚 by way of the vorticity equation. As will be shown, the
existence of 𝛚 generally indicates the viscous effects are important since fluid particles
can only be set into rotation by viscous forces. Thus, the importance of this topic is to
demonstrate that under most circumstances, an inviscid flow can also be considered
irrotational.
13.1
Vorticity Kinematics
Vorticity vector, 𝛚 = ∇ × 𝐮, or, 𝜔𝑖 = 𝜀𝑖𝑗𝑘
and
𝜔1 =
𝜔2 =
𝜔3 =
𝜕𝑤
𝜕𝑦
𝜕𝑢
𝜕𝑧
𝜕𝑣
𝜕𝑥
−
−
−
𝜕𝑣
,
𝜕𝑧
𝜕𝑤
𝜕𝑥
𝜕𝑢
𝜕𝑦
,
𝜕𝑢𝑗
𝜕𝑥𝑘
,
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Jianming Yang
Fall 2012
Chapters 3&4
104
A quantity intimately tied with vorticity is the circulation: Circulation 𝛤 is the amount of
fluid rotation within a closed contour (or circuit) 𝐶.
𝛤 = ∮𝐶 𝐮 ∙ 𝑑𝐬 = ∮𝐶 ∇ × 𝐮 ∙ 𝐧𝑑𝐴 = ∫𝐴 𝛚 ∙ 𝐧𝑑𝐴
(Stokes Theorem: ∮𝐶 𝐚 ∙ 𝑑𝐬 = ∮𝐶 ∇ × 𝐚 ∙ 𝐧𝑑𝐴)
Fluid motion is called irrotational if 𝛚 = ∇ × 𝐮 = 0 (𝛤 = 0).
Vortex line = lines which are everywhere
tangent to the vorticity vector.
Next, we shall see that vorticity and vortex
lines must obey certain properties known as
the Helmholtz vorticity theorems, which
have great physical significance.
The first is the result of its very definition:
∇ ∙ 𝛚 = ∇ ∙ (∇ × 𝐮) = 0
(vector identity: The divergence of the curl of any vector field is always zero)
i.e. the vorticity is divergence-free, which means that there can be no sources or sinks of
vorticity within the fluid itself.
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Jianming Yang
Chapters 3&4
105
Fall 2012
Helmholtz Theorem #1: a vortex line cannot end in the fluid. It must form a closed path
(smoke ring), end at a boundary, solid or free surface, or go to infinity.
The second follows from the first and using the
divergence theorem:
∫𝑉 ∇ ∙ 𝛚𝑑𝑉 = ∫𝐴 𝛚 ∙ 𝐧𝑑𝐴 = 0
Application to a vortex tube results in the following
∫𝐴 𝛚 ∙ 𝐧𝑑𝐴 + ∫𝐴 𝛚 ∙ 𝐧𝑑𝐴 = 0
1
or – 𝛤1 + 𝛤2 = 0
2
𝛤1 = 𝛤2
𝛤1 = − ∫𝐴 𝛚 ∙ 𝐧𝑑𝐴 (minus sign due to outward normal),
1
𝛤2 = ∫𝐴 𝛚 ∙ 𝐧𝑑𝐴
2
Helmholtz Theorem #2: The circulation around a given vortex line (i.e., the strength of
the vortex tube) is constant along its length.
This result can be put in the form of a simple one-dimensional incompressible continuity
equation. Define 𝜔1 and 𝜔2 as the average vorticity across 𝐴1 and 𝐴1 , respectively
𝜔1 𝐴1 = 𝜔2 𝐴2
which relates the vorticity strength to the cross sectional area changes of the tube.
058:0160
Jianming Yang
Chapters 3&4
106
Fall 2012
13.2
Vortex dynamics
Consider the substantial derivative of the circulation assuming incompressible flow and
conservative body forces
𝐷𝛤
=
𝐷𝑡
𝐷
∮ 𝐮 ∙ 𝑑𝐱 = ∮𝐶
𝐷𝑡 𝐶
𝐷𝐮
𝐷𝑡
∙ 𝑑𝐱 + ∮𝐶 𝐮 ∙
𝐷
𝐷𝑡
𝑑𝐱
From the N-S equations we have
𝐷𝐮
𝐷𝑡
Also,
𝐷
𝐷𝑡
𝐷𝛤
𝐷𝑡
𝐷𝛤
𝐷𝑡
𝟏
𝑝
𝝆
𝜌
= 𝐠 − ∇𝑝 + 𝜐∇2 𝐮 = −∇ ( + 𝑔𝑧) + 𝜐∇2 𝐮
𝑑𝐱 = 𝑑
𝐷𝐱
𝐷𝑡
= 𝑑𝐮
𝑝
= ∮𝐶 −∇ ( + 𝑔𝑧) ∙ 𝑑𝐱 + ∮𝐶 𝜐∇2 𝐮 ∙ 𝑑𝐱 + ∮𝐶 𝐮 ∙ 𝑑𝐮
𝜌
𝑝
= ∮𝐶 [−𝑑 ( + 𝑔𝑧) + 12𝑑𝑢2 ] + ∮𝐶 𝜐∇2 𝐮 ∙ 𝑑𝐱
𝜌
𝑝
∮𝐶 [−𝑑 (𝜌 + 𝑔𝑧) + 12𝑑𝑢2 ] = ∮𝐶 [−
𝑑𝑝
𝜌
− 𝑔𝑑𝑧 + 12𝑑𝑢2 ] = 0
Since the integration is around a closed contour and 𝑝, 𝑧, and 𝑢 are single-valued.
𝐷𝛤
𝐷𝑡
= ∮𝐶 𝜐∇2 𝐮 ∙ 𝑑𝐱 = −𝜐 ∮𝐶 ∇ × 𝛚 ∙ 𝑑𝐱
(∇ × (∇ × 𝐮) = ∇(∇ ∙ 𝐮) − ∇2 𝐮)
Implication: The circulation around a material loop of particles changes only if the net
viscous force on those particles gives a nonzero integral.
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Jianming Yang
Fall 2012
Chapters 3&4
107
If 𝜐 = 0 or 𝛚 = 0 (i.e., inviscid or irrotational flow, respectively) then
𝐷𝛤
𝐷𝑡
=0
The circulation of a material loop never changes.
Kelvins Circulation Theorem: for an ideal fluid (i.e. inviscid and incompressible) acted
upon by conservative forces (eg, gravity) the circulation is constant about any closed
material contour moving with the fluid.
Which leads to:
Helmholtz Theorem #3: No fluid particle can have rotation if it did not originally rotate.
Or, equivalently, in the absence of rotational forces, a fluid that is initially irrotational
remains irrotational. In general, we can conclude that vortices are preserved as time
passes. Only through the action of viscosity can they decay or disappear.
Kelvins Circulation Theorem and Helmholtz Theorem #3 are very important in the
study of inviscid flow. The important conclusion is reached that a fluid that is initially
irrotational remains irrotational, which is the justification for ideal-flow theory.
In a real viscous fluid, vorticity is generated by viscous forces. Viscous forces are
large near solid surfaces as a result of the no-slip condition. On the surface there is a
direct relationship between the viscous shear stress and the vorticity.
058:0160
Jianming Yang
Chapters 3&4
108
Fall 2012
13.2.1 Vorticity Generation at a Wall
Consider a 1-D flow near a wall:
The viscous shear stress is given by:
𝜏𝑥 = 𝜇
𝑑𝑢
𝑑𝑦
= 𝜇(
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
) = −𝜇𝜔𝑧
i.e., the wall vorticity is directly proportional
to the wall shear stress.
NOTE: In this case, the only component of 𝛚
is 𝜔𝑧 . Actually, this is a general result in that
it can be shown that 𝛚 surface is
perpendicular to the limiting streamline.
This analysis can be easily extended for general 3D flow:
𝜏𝑖 = 𝜏𝑖𝑗 𝑛𝑗 = −𝜇𝑅𝑖𝑗 𝑛𝑗
at a fixed solid wall, where 𝑅𝑖𝑗 is the rotation tensor.
It is true since at a wall with coordinate 𝑥2 ,
𝜕
𝜕𝑥1
=
𝜕
𝜕𝑥3
= 0 and from continuity
𝜕𝑣
𝜕𝑥2
=0
Once vorticity is generated, its subsequent behavior is governed by the vorticity equation.
058:0160
Jianming Yang
Chapters 3&4
109
Fall 2012
13.2.2 Vorticity Transport Equation
Navier-Stokes equation for incompressible flow with constant density and viscosity:
𝜕𝐮
𝜕𝑡
𝑝
+ 𝐮 ∙ ∇𝐮 = −∇ ( ) + 𝜐∇2 𝐮
𝜌
Body force (gravity) can be combined into 𝑝.
The vorticity equation is obtained by taking the curl of this equation.
∇×(
𝜕(∇×𝐮)
𝜕𝑡
𝜕𝐮
𝜕𝑡
𝑝
+ 𝐮 ∙ ∇𝐮) = ∇ × [−∇ ( ) + 𝜐∇2 𝐮]
𝜌
𝑝
+ ∇ × (𝐮 ∙ ∇𝐮) = −∇ × ∇ ( ) + 𝜐∇ × [∇(∇ ∙ 𝐮) − ∇ × (∇ × 𝐮)]
𝜌
2
Vector calculus identity: ∇ 𝐮 = ∇(∇ ∙ 𝐮) − ∇ × (∇ × 𝐮)
𝜕𝛚
𝜕𝑡
+ ∇ × (𝐮 ∙ ∇𝐮) = −𝜐∇ × [∇ × 𝛚]
𝜕𝛚
𝜕𝑡
𝜕𝛚
𝜕𝑡
(∇ × ∇𝐚 = 0, ∇ ∙ 𝐮 = 𝟎)
1
+ ∇ × [ ∇(𝐮 ∙ 𝐮) − 𝐮 × 𝛚] = −𝜐∇ × [∇ × 𝛚]
2
− ∇ × (𝐮 × 𝛚) = −𝜐∇ × [∇ × 𝛚] (∇ × ∇𝐚 = 0)
𝜕𝛚
𝜕𝑡
− ∇ × (𝐮 × 𝛚) = 𝜐[∇2 𝛚 − ∇(∇ ∙ 𝛚)]
Vector calculus identity: ∇2 𝛚 − ∇(∇ ∙ 𝛚) = −∇ × (∇ × 𝛚)
𝜕𝛚
𝜕𝑡
− ∇ × (𝐮 × 𝛚) = 𝜐∇2 𝛚
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Chapters 3&4
110
Fall 2012
∇ × (𝐮 × 𝛚) = 𝐮(∇ ∙ 𝛚) − 𝛚(∇ ∙ 𝐮) + (𝛚 ∙ ∇)𝐮 − (𝐮 ∙ ∇)𝛚, ∇ ∙ 𝛚 = 0, ∇ ∙ 𝐮 = 𝟎
𝜕𝛚
𝜕𝑡
𝜕𝛚
𝜕𝑡
− (𝛚 ∙ ∇)𝐮 + (𝐮 ∙ ∇)𝛚 = 𝜐∇2 𝛚
+ (𝐮 ∙ ∇)𝛚 = (𝛚 ∙ ∇)𝐮 + 𝜐∇2 𝛚
𝐷𝛚
𝐷𝑡
= (𝛚 ∙ ∇)𝐮 + 𝜐∇2 𝛚
𝐷𝛚
𝐷𝑡
: rate of change of 𝛚
(𝜔 ∙ 𝛻)𝑢: rate of deforming vortex lines
𝜐∇2 𝛚: rate of viscous diffuision of 𝛚
𝜕𝜔𝑥
𝜕𝑡
𝜕𝜔𝑦
𝜕𝑡
𝜕𝜔𝑧
𝜕𝑡
+𝑢
+𝑢
+𝑢
𝜕𝜔𝑥
𝜕𝑥
𝜕𝜔𝑦
𝜕𝑥
𝜕𝜔𝑧
𝜕𝑥
+𝑣
+𝑣
𝜔𝑥
Note:
𝜕𝜔𝑥
+𝑣
𝜕𝑦
𝜕𝜔𝑦
𝜕𝑦
𝜕𝜔𝑧
𝜕𝑦
𝜕𝑢
𝜕𝑥
+𝑤
+𝑤
+𝑤
𝜕𝜔𝑥
𝜕𝑧
𝜕𝜔𝑦
𝜕𝑧
𝜕𝜔𝑧
𝜕𝑧
= 𝜔𝑥
= 𝜔𝑥
= 𝜔𝑥
: stretching; 𝜔𝑦
𝜕𝑢
𝜕𝑥
𝜕𝑣
𝜕𝑥
𝜕𝑤
𝜕𝑥
𝜕𝑢
𝜕𝑦
𝜕𝑢
+ 𝜔𝑦
+ 𝜔𝑦
+ 𝜔𝑦
+ 𝜔𝑧
𝜕𝑦
𝜕𝑣
𝜕𝑦
𝜕𝑤
𝜕𝑢
𝜕𝑧
𝜕𝑦
+ 𝜔𝑧
+ 𝜔𝑧
+ 𝜔𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑤
𝜕𝑧
+ 𝜐∇2 𝜔𝑥
+ 𝜐∇2 𝜔𝑦
+ 𝜐∇2 𝜔𝑧
: turning
(1) Equation does not involve 𝑝 explicitly
(2) for 2-D flow (𝛚 ∙ ∇)𝐮 since 𝜔 is perpendicular to 𝐮 and there can be no
𝐷𝛚
deformation of 𝜔, i.e.,
= 𝜐∇2 𝛚
𝐷𝑡
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13.2.3 Pressure Poisson Equation
To determine the pressure field in terms of the vorticity:
∇∙(
𝜕𝐮
𝜕𝑡
𝑝
+ 𝐮 ∙ ∇𝐮) = ∇ ∙ [−∇ ( ) + 𝜐∇2 𝐮]
𝜌
𝑝
1
𝜌
2
∇2 ( ) = −∇ ∙ (𝐮 ∙ ∇𝐮) = − ∇2 (𝐮 ∙ 𝐮) + 𝐮 ∙ ∇2 𝐮 + 𝛚 ∙ 𝛚
does not explicitly depend on viscosity
Pressure Poisson equation in vector form:
 p
 2       V V 

1

      V  V   V     V  
2

1
V V    V  V   V     V 
2
1
  2  V  V      V  ω 
2
1
  2  V  V   ω    V   V    ω 
2
1
  2  V  V   ω  ω  V      V 
2
1
   2  V  V   ω  ω  V    2 V     V 


2
1
   2  V  V   V  2 V  ω  ω
2
  a  b   b    a   a   b 


  a  b   b    a   a   b 
   a   2a    a 
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Pressure Poisson equation in tensor form:
 p
1
 2      2  V  V   V  2 V  ω  ω
2

 2  uk ek 
1 2
 u j e j    uk ek     ui ei  

   V     V 

2 xi xi 
x j x j

 2 uk
u  
u 
1 2

u j uk  jk   ui ik 
   ijk k ei     lmn n el 

2 xi xi
x j x j 
x j  
xm 
2

 2ui
u u 
1  u ju j 

 ui
   ijk  lmn k n   ei  el 
2 xi xi
x j x j 
x j xm 


 2ui
uk un 
1   

u
u

u








j j 
i
 ijk lmn x x  il
2 xi  xi

x

x
j
j
j
m 


u j 
 2ui
u u
1  

  jm kn   jn km  k n
 2u j
  ui
2 xi 
xi 
x j x j
x j xm
 2ui
uk un
uk un
  u j 

u

u






 j
 i
jm kn
jn km
xi  xi 
x j x j
x j xm
x j xm
 u j u j
 2u j 
 2ui
uk uk uk u j
 
uj

u


 i
 x x

x

x

x

x

x

x
x j xk
i
i
i 
j
j
j
j
 i

uk u j
x j xk
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13.3
Kinematic Decomposition of flow fields
Previously, we discussed the decomposition of fluid motion into translation, rotation, and
deformation. This was done locally for a fluid element. Now we shall see that a global
decomposition is possible.
Helmholtz’s Decomposition: any continuous and finite vector field can be expressed as
the sum of the gradient of a scalar function 𝜃 plus the curl of a zero-divergence vector 𝐚.
The vector 𝐚 vanishes identically if the original vector field is irrotational.
𝐮 = 𝐮𝜔 + 𝐮𝜃
𝛚 = ∇ × 𝐮𝜔
0 = ∇ × 𝐮𝜃
→
𝐮𝜃 = ∇𝜃
If ∇ ∙ 𝐮 = ∇ ∙ 𝐮𝜔 + ∇ ∙ 𝐮𝜃 = 0
Then
∇2 θ = 0
The GDE for θ is the Laplace Equation
And
uω = ∇ × a
Since
∇ ∙ (∇ × a) = 0
∇ × uω = 𝛚 = ∇ × ∇ × a = −∇2 a + ∇(∇ ∙ a)
i.e.
∇2 a = −𝛚
a=
The solution of this equation is
Thus,
uω = −
Which is known as the Biot-Savart law.
1
𝛚
1
𝐑×𝛚
∫ 𝑑𝑉
4𝜋 |𝐑|
𝑑𝑉
∫
4𝜋 |𝐑|𝟑
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The Biot-Savart law can be used to compute the velocity field induced by a known
vorticity field. It has many useful applications, including in ideal flow theory (e.g., when
applied to line vortices and vortex sheets it forms the basis of computing the velocity
field in vortex-lattice and vortex-sheet lifting-surface methods).
The important conclusion from the Helmholtz decomposition is that any incompressible
flow can be thought of as the vector sum of rotational and irrotational components. Thus,
a solution for irrotational part 𝐮𝜃 represents at least part of an exact solution. Under
certain conditions, high Re flow about slender bodies with attached thin boundary layer
and wake, uω is small over much of the flow field such that 𝐮𝜃 is a good approximation
to u. This is probably the strongest justification for ideal-flow theory. (incompressible,
inviscid, and irrotational flow).
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The vorticity components in cylindrical coordinates are
𝜔𝑟 = 0, 𝜔𝜃 = 0, 𝜔𝑧 = 0
So the flow is irrotational.
The flow is incompressible, steady, and irrotational, (also, the free surface is a
streamline), so the Bernoulli equation holds at the surface, where 𝑝 = 𝑝atm
1 2
1 2
𝑝atm + 𝜌𝑢𝑟=∞ + 𝜌𝑔𝐻 = 𝑝atm + 𝜌𝑢𝑟=𝑅 + 𝜌𝑔𝑧𝑐
2
2
Introduce 𝑢𝑟=∞ = 0 and 𝑢𝑟=𝑅 = 𝐾𝑅
1 2 2
𝜌𝑔𝐻 = 𝜌𝐾 𝑅 + 𝜌𝑔𝑧𝑐
2
𝐾 2 𝑅2
𝑧𝑐 = 𝐻 −
2𝑔
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