² In analyzing fluid motion,
uSeeking to estimate of gross effects (mass flow, induced
force, energy change) over a finite region or control volume
à Chapter 3
uSeeking the point-by-point details of a flow pattern by
analyzing an infinitesimal region of the flow à Differential
equations, appropriate boundary conditions (Chapter 4)
² Why are we learning derivation of differential eqs.?
uThe equations reveal the basic dimensionless parameters
that govern fluid motion à Chapter 5
uA great number of useful solutions can be found if one
makes two simplifying assumptions: (1) steady flow and (2)
incompressible flow à Chapter 6
uThe assumption of “Frictionless flow” makes the Bernoulli
equation valid and yields a wide variety of idealized, or
perfect-fluid, possible solutions à Chapter 8
Instantaneous streamline pattern for flow past a rotating cylinder at
a Reynolds number of 1000. This pattern, which oscillates due to
vortex shedding, was obtained by CFD and agrees with flow
visualization experiments.
² Eulerian description
uConcerned with the field of flow
uEx) we compute the pressure field p(x,y,z,t) of the
flow pattern → appropriate for fluid mechanics
² Lagrangian description
uConcerned with a specific particle in the fluid
uEx) we compute the pressure changes p(t) that a
particle experiences as it moves through the field
uAppropriate for solid mechanics
uHowever, if you want to know the trajectory of a
particle in the fluid, you need this description
° The Velocity field
𝑉 𝑥, 𝑦, 𝑧, 𝑡 = 𝚤⃗𝑢 𝑥, 𝑦, 𝑧, 𝑡 + 𝚥⃗𝑣(𝑥, 𝑦, 𝑧, 𝑡)+𝑘𝑤(𝑥, 𝑦, 𝑧, 𝑡)
Substantial derivative or
material derivative
Given the Eulerian velocity vector field
𝑉 = 3𝑡⃗𝚤 + 𝑥𝑧⃗𝚥 + 𝑡𝑦 ! 𝑘
find the total acceleration of a particle.
}
Criteria of incompressibility:
𝑀𝑎 ≤ 0.3
𝑉
𝑉
𝑀𝑎 = =
𝑎
𝑘𝑅𝑇
a : speed of sound
V : flow velocity
}
Under what conditions does the velocity field
𝑉 = 𝑎" 𝑥 + 𝑏" 𝑦 + 𝑐" 𝑧 𝚤⃗ + (𝑎! 𝑥 + 𝑏! 𝑦 + 𝑐! 𝑧)⃗𝚥+(𝑎# 𝑥 + 𝑏# 𝑦 + 𝑐# 𝑧)𝑘
Where a1, b1, etc.=const, represent an incompressible flow that
conserves mass?
An incompressible velocity field is given by
𝑢 = 𝑎 𝑥! − 𝑦!
𝑣 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑤 = 𝑏
where a and b are constants. What must the form of the velocity
component v be?
}
}
A centrifugal impeller of 40-cm diameter is used to pump
hydrogen at 15oC and 1-atm pressure. Estimate the
maximum allowable impeller rotational speed to avoid
compressibility effects at the blade tips.
Face x
Face y
Face z
Navier-Stokes Equation
For incompressible flow
𝑥: 𝜌
𝑦: 𝜌
z: 𝜌
,-
,-
,-
,-
,2
,! -
,! -
,! -
+
+
+𝑢 +𝑣 +𝑤
= − + 𝜇( ! + ! + ! )+𝜌𝑔/
,.
,/
,0
,1
,/
,/
,0
,1
,3
,3
,3
,3
,2
,! 3
,! 3
,! 3
+𝑢 +𝑣 +𝑤
= − + 𝜇( ! + ! + ! )+𝜌𝑔0
,.
,/
,0
,1
,0
,/
,0
,1
,4
,4
,4
,4
,2
,! 4
,! 4
,! 4
,.
+𝑢
,/
+𝑣
,0
+𝑤
,1
=−
,1
+ 𝜇(
,/ !
,0 !
,1 !
)+𝜌𝑔1
Take the velocity field below
𝑢 = 𝑎 𝑥! − 𝑦!
𝑣 = −2𝑎𝑥𝑦
𝑤=0
And determine under what conditions it is a solution to the
Navier-Stokes momentum equations. Assuming that these
conditions are met, determine the resulting pressure
distribution when z is “up” (𝑔$ = 0, 𝑔% = 0, 𝑔& = −𝑔)
}
𝑧 = 𝜂(𝑥, 𝑦, 𝑡)
(kinematic boundary condition)
dh
¶h
¶h
¶h
=
+u
+v
dt
¶t
¶x
¶y
𝑤!"# = 𝑤$%&
Simplified free surface conditions
(open-channel flows in Chap. 10)
- Upper fluid merely exerts pressure on the
lower fluid.
- Shear and heat conduction are negligible.
wliq =
æ ¶T ö
æ ¶T ö
= çk
çk
÷
÷
è ¶z ø liq
è ¶z ø gas
𝛾
Effect of surface tension
¶h
¶t
or 𝑝!"# ≈ 𝑝$%&
æ ¶V ö
»0
ç
÷
è ¶z ø liq
Shear negligible
æ ¶T ö
»0
ç
÷
è ¶z ø liq
Heat conduction
negligible
æ ¶ 2h
¶ 2h ö
pliq » p gas - g ç
ç ¶x 2 + ¶y 2 ÷
÷
è
ø
Inviscid flow approximation
à Euler’s equation: it can be
integrated along a streamline to
obtain Bernoulli’s equation.
DV
r
= rg - Ñp
Dt
(Vn ) fluid = (Vn )wall
(The system is closed.)
V = Vwall
Solid surface:
Inlet or outlet:
Known V, p
Free surface: p » pa , w = ¶h ¶t
Vn = 0
Note: no tangential velocity
condition
Continuity and momentum are independent of T. Thus we solve
continuity and momentum equations entirely separately for the
pressure and velocity and later solve energy equation for temperature.
For steady incompressible laminar flow through a long tube,
the velocity distribution is given by
𝑟!
𝑣& = 𝑈 1 − !
𝑣' = 𝑣( = 0
𝑅
Where U is the maximum, or centerline, velocity and R is the
tube radius. If the wall temperature is constant at Tw and the
temperature T=T(r) only, find T(r) for this flow.
}
Couette flow between fixed and a moving plate
The Navier-Stokes equation for 2D (x,y)
incompressible flow
𝐷𝑉
𝜌
= −𝛻𝑝 + 𝜇𝛻 '𝑉 + 𝜌𝑔⃗
𝐷𝑡
Assumptions:
1. Fully developed
2. No y-direction velocity or pressure gradient
3. Gravity ignorable
4. Steady-state
5. No variation of velocity and pressure in xdirection
The x component of the NS equation
𝑥: 𝜌
𝑦: 𝜌
()
()
()
+
𝑢
+
𝑣
(*
(+
(,
(.
(.
(.
+
𝑢
+
𝑣
(*
(+
(,
(-
(! )
(! )
(-
(! .
(! .
= − (+ + 𝜇((+! + (,!)
= − (, + 𝜇((+! + (,!)
By the assumptions, the equation
reduces to
𝜕 '𝑢
= 0 or u = C/y + C'
𝜕𝑦 '
The boundary conditions at the
upper and lower plates (no-slip)
𝐴𝑡 𝑦 = +ℎ: 𝑢 = 𝑉 = 𝐶/ℎ + 𝐶'
𝐴𝑡 𝑦 = −ℎ: 𝑢 = 0 = 𝐶/ −ℎ + 𝐶'
𝑉
𝑉
𝐶/ =
,
𝐶' =
2ℎ
2
𝑢=
𝑉
𝑉
𝑦+
2ℎ
2
Poiseuille flow: Flow due to pressure gradient
between two fixed plates
The Navier-Stokes equation for 2D (x,y)
incompressible flow
𝐷𝑉
𝜌
= −𝛻𝑝 + 𝜇𝛻 '𝑉 + 𝜌𝑔⃗
𝐷𝑡
Assumptions:
1. Fully developed
2. No y-direction velocity or pressure gradient
3. Gravity ignorable
4. Steady-state
5. No variation of velocity in x-direction
6. Pressure gradient in x-direction constant
The x,y component of the NS equation
𝑥: 𝜌
𝑦: 𝜌
()
()
()
+
𝑢
+
𝑣
(*
(+
(,
(.
(.
(.
+
𝑢
+
𝑣
(*
(+
(,
(-
(! )
(! )
(-
(! .
(! .
= − (+ + 𝜇((+! + (,!)
= − (, + 𝜇((+! + (,!)
By the assumptions, the equation reduces to
𝜕𝑝
𝜕 '𝑢
−
+𝜇 ' =0
𝜕𝑥
𝜕𝑦
𝜕 '𝑢 𝜕𝑝
𝜇 '=
= const
𝜕𝑦
𝜕𝑥
1 𝜕𝑝 𝑦 '
𝑢=
+ 𝐶/𝑦 + 𝐶'
𝜇 𝜕𝑥 2
The boundary conditions at the
upper and lower plates (no-slip)
𝐴𝑡 𝑦 = +ℎ: 𝑢 = 0 𝑎𝑛𝑑 𝐶/ = 0
𝜕𝑝 ℎ'
𝐴𝑡 𝑦 = −ℎ: 𝑢 = 0 𝑎𝑛𝑑 𝐶' = −
𝜕𝑥 2𝜇
'
'
𝜕𝑝 ℎ
𝑦
𝑢=−
(1 − ' )
𝜕𝑥 2𝜇
ℎ
At y=0
(- 1!
𝑢0%+ = − (+ '2
Poiseuille flow: Fully developed laminar pipe
flow (in cylindrical coordinate)
The Navier-Stokes equation for 2D (x,y)
incompressible flow
𝐷𝑉
𝜌
= −𝛻𝑝 + 𝜇𝛻 '𝑉 + 𝜌𝑔⃗
𝑑𝑡
- Assumptions:
1.
2.
3.
4.
5.
6.
- The r,z component of the NS equation (in
cylindrical coordinate)
r:
z:
- The simplified z-momentum equation
- The boundary conditions
No slip at r=R
Finite velocity at r=0
- Final velocity solution of Poiseuille flow
- Volume flow rate (𝑄 = ∫ 𝑣3 𝑑𝐴)
The integral relation of energy conservation equation for fixed CV
𝜕
𝑝
𝑄̇ − 𝑊̇& − 𝑊.̇ =
r 𝑒𝜌𝑑𝑣 + r 𝑒 + 𝜌 𝑉 t 𝑛 𝑑𝐴
𝜕𝑡 45
𝜌
46
𝑊̇& =0 for infinitesimal CV
(no protruding shaft for very small volume)
Hence, for this tiny element
𝜕
𝜕
𝜕
𝜕
𝑄̇ − 𝑊.̇ =
𝜌𝑒 +
𝜌𝑢𝜁 +
𝜌𝑣𝜁 +
𝜌𝑤𝜁 𝑑𝑥𝑑𝑦𝑑𝑧
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
where 𝜁 = 𝑒 + 7 .
𝜕𝑒
𝜕𝜌
𝑝 𝜕
𝜕
𝑝
𝜌 +𝑒
+ (𝑒 + )
𝜌𝑢 + 𝜌𝑢
𝑒+
𝜕𝑡
𝜕𝑡
𝜌 𝜕𝑥
𝜕𝑥
𝜌
𝜕𝑒
𝑝 𝜕
𝜕
𝜕 𝑝
𝐷𝑒 𝜕𝑝𝑢
=𝜌 +
𝜌𝑢 + 𝜌𝑢
𝑒 + 𝜌𝑢
=𝜌
+
𝜕𝑡
𝜌 𝜕𝑥
𝜕𝑥
𝜕𝑥 𝜌
𝐷𝑡
𝜕𝑥
By eliminating the continuity term buried in the RHS
𝐷𝑒
𝑄̇ − 𝑊.̇ = (𝜌
+ 𝑉 t 𝛻𝑝 + 𝑝𝛻 t 𝑉)𝑑𝑥𝑑𝑦𝑑𝑧
𝐷𝑡
Fourier’s law of conduction
𝑞⃗ = −𝑘𝛻𝑇
(8
(8
(8
Or: 𝑞+ = −𝑘 (+ , 𝑞, = −𝑘 (, , 𝑞3 = −𝑘 (3
The rate of work done by viscous stresses =
the stress component X
its corresponding velocity component X the area
of the element face
𝑊.̇ = 𝑤+ 𝑑𝑦𝑑𝑧 (Left side of the volume)
𝑤ℎ𝑒𝑟𝑒 𝑤+ = − 𝑢𝜏++ + 𝑣𝜏,+ + 𝑤𝜏3+ (work in)
𝜕
𝑊.̇ = −[
𝑢𝜏++ + 𝑣𝜏,+ + 𝑤𝜏3+
𝜕𝑥
𝜕
+
𝑢𝜏+, + 𝑣𝜏,, + 𝑤𝜏3,
𝜕𝑦
(
𝑢𝜏+3 + 𝑣𝜏,3 + 𝑤𝜏33 ]𝑑𝑥𝑑𝑦𝑑𝑧
(3
= −𝛻 t 𝑉 t 𝜏"9 𝑑𝑥𝑑𝑦𝑑𝑧
Therefore, The rate of heat addition through conduction (for heat in(+), out (-))
𝜕
𝜕
𝜕
𝑄̇ = −
𝑞+ +
𝑞, +
𝑞 𝑑𝑥𝑑𝑦𝑑𝑧 = −𝛻 t 𝑞𝑑𝑥𝑑𝑦𝑑𝑧
⃗
𝜕𝑥
𝜕𝑦
𝜕𝑧 3
𝑄̇ = 𝛻 t (𝑘𝛻𝑇)𝑑𝑥𝑑𝑦𝑑𝑧
All together, the equation of energy conservation
𝑑𝑒
𝜌 + 𝑉 t 𝛻𝑝 + p𝛻 t 𝑉 = 𝛻 t 𝑘𝛻𝑇 + 𝛻 t 𝑉 t 𝜏"9
𝑑𝑡
After some modification
𝜌
𝐷 𝑢ˆ
+ 𝑝 𝛻 t 𝑉 = 𝛻 t 𝑘𝛻𝑇 + Φ
𝐷𝑡
() '
(. '
(: '
(.
() '
Where Φ = 𝜇[2 (+ + 2 (, + 2 (3 + (+ + (,
(:
(. '
()
(: '
+
+
+
+
]
(,
(3
(3
(+
(no heat source or radiation)
Internal energy/mass
k : Coefficient of thermal conductivity of the fluid
𝜌
𝐷 𝑢ˆ
k
+ 𝑝 𝛻 t 𝑉 = 𝛻 t 𝑘𝛻𝑇
+Φ
𝐷𝑡
(Internal E)
Φ
Temperature
k
𝐷𝑇
𝜌𝑐"
= 𝑘𝛻 # 𝑇 + Φ
𝐷𝑡
Φ
No convection (or no flow velocity) case:
Heat conduction equation
𝜕𝑇
𝜌𝑐"
= 𝑘𝛻 # 𝑇
𝜕𝑡
p
r =
RT
ˆ = ò cv dT » cvT + const
u
𝑧 = 𝜂(𝑥, 𝑦, 𝑡)
(kinematic boundary condition)
dh
¶h
¶h
¶h
=
+u
+v
dt
¶t
¶x
¶y
𝑤!"# = 𝑤$%&
Simplified free surface conditions
(open-channel flows in Chap. 10)
- Upper fluid merely exerts pressure on the
lower fluid.
- Shear and heat conduction are negligible.
wliq =
æ ¶T ö
æ ¶T ö
= çk
çk
÷
÷
è ¶z ø liq
è ¶z ø gas
𝛾
Effect of surface tension
¶h
¶t
or 𝑝!"# ≈ 𝑝$%&
æ ¶V ö
»0
ç
÷
è ¶z ø liq
Shear negligible
æ ¶T ö
»0
ç
÷
è ¶z ø liq
Heat conduction
negligible
æ ¶ 2h
¶ 2h ö
pliq » p gas - g ç
ç ¶x 2 + ¶y 2 ÷
÷
è
ø
Inviscid flow approximation
à Euler’s equation: it can be
integrated along a streamline to
obtain Bernoulli’s equation.
DV
r
= rg - Ñp
Dt
(Vn ) fluid = (Vn )wall
(The system is closed.)
V = Vwall
Solid surface:
Inlet or outlet:
Known V, p
Free surface: p » pa , w = ¶h ¶t
Vn = 0
Note: no tangential velocity
condition
Continuity and momentum are independent of T. Thus we solve
continuity and momentum equations entirely separately for the
pressure and velocity and later solve energy equation for temperature.
For steady incompressible laminar flow through a long tube,
the velocity distribution is given by
𝑟!
𝑣& = 𝑈 1 − !
𝑣' = 𝑣( = 0
𝑅
Where U is the maximum, or centerline, velocity and R is the
tube radius. If the wall temperature is constant at Tw and the
temperature T=T(r) only, find T(r) for this flow.
}
Take curl of the momentum equation
For 2-dimensional,
Stream function must be
defined such that:
Equation (4.87) is scalar and has only one variable.
Satisfies continuity eq.
(angular velocity)
(vorticy)
𝜕
𝜕𝜑
𝜕 𝜕𝜑
𝑘{
−
−
}
𝜕𝑥
𝜕𝑥
𝜕𝑦 𝜕𝑦
* It can be solved numerically.
Equations (4.86) and (4.87) reduce to
Relationship with streamline
For 2-dimensional, the streamline is , (From Sec. 1.11)
à We can plot lines of constant φ to give the
streamlines of the flow.
Relationship with the volume flow dQ
The volume flow dQ through an element ds of
control surface of unit depth:
The change in φ across the element is numerically equal to
the volume flow through the element. The volume flow
between any two streamlines in the flow field is equal to
the change in stream function between those streamlines.
Sign convention for flow in terms of change in stream function
If a stream function exists for the velocity field of
𝑢 = 𝑎 𝑥 ! − 𝑦 ! 𝑣 = −2𝑎𝑥𝑦 𝑤 = 0
find it, plot it, and interpret it
}
} Steady plane compressible flow
¶
(ru ) + ¶ (rv ) = 0
¶x
¶y
𝜕𝜑
𝜕𝜑
𝜌𝑢 =
, 𝜌𝑣 = −
𝜕𝑦
𝜕𝑥
} Incompressible plane flow in polar coordinates
1 ¶
(rvr ) + 1 ¶ (vq ) = 0 𝑟𝑣' = 𝜕𝜑 , 𝑣( = − 𝜕𝜑 → 𝑣' = 1 𝜕𝜑 , 𝑣( = − 𝜕𝜑
𝜕𝜃
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝑟
r ¶r
r ¶q
} Incompressible axisymmetric flow
𝜕
𝜕
𝜕
𝜕
1 ¶
¶
(rvr ) + (vz ) = 0 → 𝜕𝑟 𝑟𝑣' + 𝑟 𝜕𝑧 𝑣& = 0 → 𝜕𝑟 𝑟𝑣' + 𝜕𝑧 𝑟𝑣& = 0
r ¶r
¶z
𝑟𝑣' = −
𝜕𝜑
𝜕𝜑
1 𝜕𝜑
1 𝜕𝜑
, 𝑟𝑣& =
→ 𝑣' = −
, 𝑣& =
𝜕𝑧
𝜕𝑟
𝑟 𝜕𝑧
𝑟 𝜕𝑟
Investigate the stream function in polar coordinates
𝑅!
𝜑 = 𝑈𝑠𝑖𝑛𝜃(𝑟 − )
𝑟
Where U and R are constants, a velocity and a length,
respectively. Plot the streamlines. What does the flow
represents?
}
1 𝜕𝑤 𝜕𝑣
1 𝜕𝑢 𝜕𝑤
1 𝜕𝑣 𝜕𝑢
𝜔$ =
−
, 𝜔% =
−
, 𝜔& =
−
2 𝜕𝑦 𝜕𝑧
2 𝜕𝑧 𝜕𝑥
2 𝜕𝑥 𝜕𝑦
à irrotational
1 𝑑𝛼 𝑑𝛽
−
2 𝑑𝑡 𝑑𝑡
𝜕𝑣
𝑑𝛼 =
𝑑𝑡
𝜕𝑥
𝜕𝑢
𝑑𝛽 =
𝑑𝑡
𝜕𝑦
1 𝜕𝑣 𝜕𝑢
𝜔" =
−
2 𝜕𝑥 𝜕𝑦
𝜔" =
for irrotational flow,
, inviscid
𝑝
𝜌
1 #
𝑣
2
potential line
for incompressible flow in the xy plane,
stream line
(=
à
à
𝜕𝜑
𝜕𝜑
𝑑𝑥 −
𝑑𝑦)
𝜕𝑦
𝜕𝑥
potential line
and
are mutually orthogonal.
If a velocity potential exists for the velocity field
𝑢 = 𝑎 𝑥! − 𝑦!
𝑣 = −2𝑎𝑥𝑦
𝑤=0
Find it, plot it.
}
}
For the flow between parallel plates due to the pressure
gradient, compute (a) the wall shear stress, (b) the stream
function, (c) the vorticity, (d) the velocity potential, and the
average velocity.
𝛻𝑓 W 𝑑 𝑟⃗ = 𝑑𝑓
} 4.2
} 4.7
} 4.16
} 4.18
} 4.29
} 4.36
} 4.41
} 4.57
} 4.79