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```Fluid Mechanics 101
A Skeleton Guide
J. E. Shepherd
Aeronautics and Mechanical Engineering
California Institute of Technology
1995-present - Revised December 16, 2007
Foreword
This is guide is intended for students in Ae101 “Fluid Mechanics”, the class on the fundamentals of fluid
mechanics that all first year graduate students take in Aeronautics and Mechanical Engineering at Caltech.
It contains all the essential formulas grouped into sections roughly corresponding to the order in which the
material is taught when I give the course (I have done this now five times beginning in 1995). This is not a
text book on the subject or even a set of lecture notes. The document is incomplete as description of fluid
mechanics and entire subject areas such as free surface flows, buoyancy, turbulent flows, etc., are missing
(some of these elements are in Brad Sturtevant’s class notes which cover much of the same ground but are
more expository). It is simply a collection of what I view as essential formulas for most of the class. The
need for this typeset formulary grew out of my poor chalk board work and the many mistakes that happen
when I lecture. Several generations of students have chased the errors out but please bring any that remain
to my attention.
JES December 16, 2007
Contents
1 Fundamentals
1.1 Control Volume Statements . . . .
1.2 Reynolds Transport Theorem . . .
1.3 Integral Equations . . . . . . . . .
1.3.1 Simple Control Volumes . .
1.4 Vector Calculus . . . . . . . . . . .
1.4.1 Vector Identities . . . . . .
1.4.2 Curvilinear Coordinates . .
1.4.3 Gauss’ Divergence Theorem
1.4.4 Stokes’ Theorem . . . . . .
1.4.5 Div, Grad and Curl . . . .
1.4.6 Specific Coordinates . . . .
1.5 Differential Relations . . . . . . . .
1.5.1 Conservation form . . . . .
1.6 Convective Form . . . . . . . . . .
1.7 Divergence of Viscous Stress . . . .
1.8 Euler Equations . . . . . . . . . . .
1.9 Bernoulli Equation . . . . . . . . .
1.10 Vorticity . . . . . . . . . . . . . . .
1.11 Dimensional Analysis . . . . . . . .
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1
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9
2 Thermodynamics
2.1 Thermodynamic potentials and fundamental relations
2.2 Maxwell relations . . . . . . . . . . . . . . . . . . . . .
2.3 Various defined quantities . . . . . . . . . . . . . . . .
2.4 v(P, s) relation . . . . . . . . . . . . . . . . . . . . . .
2.5 Equation of State Construction . . . . . . . . . . . . .
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11
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3 Compressible Flow
3.1 Steady Flow . . . . . . . . . . . . . . . . .
3.1.1 Streamlines and Total Properties .
3.2 Quasi-One Dimensional Flow . . . . . . .
3.2.1 Isentropic Flow . . . . . . . . . . .
3.3 Heat and Friction . . . . . . . . . . . . . .
3.3.1 Fanno Flow . . . . . . . . . . . . .
3.3.2 Rayleigh Flow . . . . . . . . . . .
3.4 Shock Jump Conditions . . . . . . . . . .
3.4.1 Lab frame (moving shock) versions
3.5 Perfect Gas Results . . . . . . . . . . . . .
3.6 Reflected Shock Waves . . . . . . . . . . .
3.7 Detonation Waves . . . . . . . . . . . . .
3.8 Perfect-Gas, 2-γ Model . . . . . . . . . . .
3.8.1 2-γ Solution . . . . . . . . . . . . .
3.8.2 High-Explosives . . . . . . . . . . .
3.9 Weak shock waves . . . . . . . . . . . . .
3.10 Acoustics . . . . . . . . . . . . . . . . . .
3.11 Multipole Expansion . . . . . . . . . . . .
3.12 Baffled (surface) source . . . . . . . . . .
3.13 1-D Unsteady Flow . . . . . . . . . . . . .
3.14 2-D Steady Flow . . . . . . . . . . . . . .
3.14.1 Oblique Shock Waves . . . . . . .
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3.14.2
3.14.3
3.14.4
3.14.5
3.14.6
3.14.7
Weak Oblique Waves . . .
Prandtl-Meyer Expansion
Inviscid Flow . . . . . . .
Potential Flow . . . . . .
Natural Coordinates . . .
Method of Characteristics
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31
32
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33
4 Incompressible, Inviscid Flow
4.1 Velocity Field Decomposition . . . .
4.2 Solutions of Laplace’s Equation . . .
4.3 Boundary Conditions . . . . . . . . .
4.4 Streamfunction . . . . . . . . . . . .
4.4.1 2-D Cartesian Flows . . . . .
4.4.2 Cylindrical Polar Coordinates
4.4.3 Spherical Polar Coordinates .
4.5 Simple Flows . . . . . . . . . . . . .
4.6 Vorticity . . . . . . . . . . . . . . . .
4.7 Key Ideas about Vorticity . . . . . .
4.8 Unsteady Potential Flow . . . . . . .
4.9 Complex Variable Methods . . . . .
4.9.1 Mapping Methods . . . . . .
4.10 Airfoil Theory . . . . . . . . . . . . .
4.11 Thin-Wing Theory . . . . . . . . . .
4.11.1 Thickness Solution . . . . . .
4.11.2 Camber Case . . . . . . . . .
4.12 Axisymmetric Slender Bodies . . . .
4.13 Wing Theory . . . . . . . . . . . . .
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34
34
34
35
35
36
37
38
38
40
41
42
42
44
44
45
46
48
49
50
5 Viscous Flow
5.1 Scaling . . . . . . . . . . . . . . . . . . . . . .
5.2 Two-Dimensional Flow . . . . . . . . . . . . .
5.3 Parallel Flow . . . . . . . . . . . . . . . . . .
5.3.1 Steady Flows . . . . . . . . . . . . . .
5.3.2 Poiseuille Flow . . . . . . . . . . . . .
5.3.3 Rayleigh Problem . . . . . . . . . . .
5.4 Boundary Layers . . . . . . . . . . . . . . . .
5.4.1 Blasius Flow . . . . . . . . . . . . . .
5.4.2 Falkner-Skan Flow . . . . . . . . . . .
5.5 Kármán Integral Relations . . . . . . . . . . .
5.6 Thwaites’ Method . . . . . . . . . . . . . . .
5.7 Laminar Separation . . . . . . . . . . . . . .
5.8 Compressible Boundary Layers . . . . . . . .
5.8.1 Transformations and Approximations
5.8.2 Energy Equation . . . . . . . . . . . .
5.8.3 Moving Shock Waves . . . . . . . . . .
5.8.4 Weak Shock Wave Structure . . . . .
5.9 Creeping Flow . . . . . . . . . . . . . . . . .
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52
52
53
53
54
56
57
57
59
59
60
60
61
61
61
62
64
64
66
A Famous Numbers
69
B
71
Books on Fluid Mechanics
ii
1
FUNDAMENTALS
1
1
Fundamentals
1.1
Control Volume Statements
Ω is a material volume, V is an arbitrary control volume, ∂Ω indicates the surface of the volume.
mass conservation:
Z
d
ρ dV = 0
dt Ω
(1)
Momentum conservation:
d
dt
Forces:
Z
ρu dV = F
Z
F=
(2)
Ω
Z
ρG dV +
T dA
Ω
(3)
∂Ω
Surface traction forces
T = −P n̂ + τ · n̂ = T · n̂
(4)
Stress tensor T
T = −P I + τ
or Tik = −P δik + τik
(5)
where I is the unit tensor, which in cartesian coordinates is
I = δik
(6)
Viscous stress tensor, shear viscosity µ, bulk viscosity µv
1
τik = 2µ Dik − δik Djj + µv δik Djj
3
Deformation tensor
Dik
1
=
2
∂ui
∂uk
+
∂xk
∂xi
implicit sum on j
1
∇u + ∇uT
2
or
(7)
(8)
Energy conservation:
d
dt
|u|2
ρ e+
2
Ω
Z
Work:
dV = Q̇ + Ẇ
Z
(9)
Z
ρG · u dV +
Ẇ =
Ω
T · u dA
(10)
∂Ω
Heat:
Z
Q̇ = −
q · n̂ dA
(11)
∂Ω
heat flux q, thermal conductivity k and thermal radiation qr
q = −k∇T + qr
Entropy inequality (2nd Law of Thermodynamics):
Z
Z
q · n̂
d
ρs dV ≥ −
dA
dt Ω
∂Ω T
(12)
(13)
1
FUNDAMENTALS
1.2
2
Reynolds Transport Theorem
The multi-dimensional analog of Leibniz’s theorem:
Z
Z
Z
d
∂φ
φ(x, t) dV =
dV +
φuV · n̂ dA
dt V (t)
V (t) ∂t
∂V
The transport theorem proper. Material volume Ω, arbitrary volume V .
Z
Z
Z
d
d
φ dV =
φ dV +
φ(u − uV ) · n̂ dA
dt Ω
dt V
∂V
1.3
(14)
(15)
Integral Equations
The equations of motions can be rewritten with Reynolds Transport Theorem to apply to an (almost) arbitrary moving control volume. Beware of noninertial reference frames and the apparent forces or accelerations
that such systems will introduce.
Moving control volume:
Z
Z
d
ρdV +
ρ (u − uV ) · n̂ dA = 0
(16)
dt V
∂V
Z
Z
Z
Z
d
ρudV +
ρu (u − uV ) · n̂ dA =
ρG dV +
T dA
(17)
dt V
∂V
V
∂V
d
dt
Z
V
Z
|u|2
|u|2
ρ e+
dV +
ρ e+
(u − uV ) · n̂ dA =
2
2
∂V
Z
Z
Z
ρG · u dV +
T · u dA −
q · n̂ dA
V
d
dt
Z
∂V
Z
Z
ρs (u − uV ) · n̂ dA +
ρsdV +
V
(18)
∂V
∂V
∂V
q
· n̂ dA ≥ 0
T
(19)
Stationary control volume:
d
dt
d
dt
d
dt
Z
Z
V
Z
Z
ρu · n̂ dA = 0
ρdV +
V
Z
Z
Z
ρuu · n̂ dA =
ρudV +
V
(20)
∂V
∂V
ρG dV +
V
T dA
(21)
∂V
Z
|u|2
|u|2
ρ e+
dV +
ρ e+
u · n̂ dA =
2
2
∂V
Z
Z
Z
ρG · u dV +
T · u dA −
q · n̂ dA
V
d
dt
Z
∂V
Z
V
Z
ρsu · n̂ dA +
ρsdV +
∂V
(22)
∂V
∂V
q
· n̂ dA ≥ 0
T
(23)
1
FUNDAMENTALS
1.3.1
3
Simple Control Volumes
Consider a stationary control volume V with i = 1, 2, . . ., I connections or openings through which there is
fluid flowing in and j = 1, 2, . . ., J connections through which the fluid is following out. At the inflow and
outflow stations, further suppose that we can define average or effective uniform properties hi , ρi , ui of the
fluid. Then the mass conservation equation is
d
dM
=
dt
dt
Z
ρdV =
V
I
X
Ai ṁi −
i=1
J
X
Aj ṁj
(24)
j=1
where Ai is the cross-sectional area of the ith connection and ṁi = ρi ui is the mass flow rate per unit area
through this connection. The energy equation for this same situation is
d
dE
=
dt
dt
Z
V
|u|2
ρ e+
+ gz dV
2
=
I
X
|ui |2
+ gzi
Ai ṁi hi +
2
i=1
J
X
|uj |2
+ gzj + Q̇ + Ẇ
−
Aj ṁj hj +
2
j=1
(25)
where Q̇ is the thermal energy (heat) transferred into the control volume and Ẇ is the mechanical work
done on the fluid inside the control volume.
1.3.2
For a stationary control volume, the steady momentum equation can be written as
Z
Z
Z
Z
ρuu · n̂ dA +
P n̂ dA =
ρG dV +
τ · n̂ dA + Fext
∂V
∂V
V
(26)
∂V
where Fext are the external forces required to keep objects in contact with the flow in force equilibrium.
These reaction forces are only needed if the control volume includes stationary objects or surfaces. For a
control volume completely within the fluid, Fext = 0.
1.4
1.4.1
Vector Calculus
Vector Identities
If A and B are two differentiable vector fields A(x), B(x) and φ is a differentiable scalar field φ(x), then
the following identities hold:
∇ × (A × B) = (B · ∇)A − (A · ∇)B − (∇ · A)B + (∇ · B)A
∇(A · B) = (B · ∇)A + (A · ∇)B + B × (∇ × A) + A × (∇ × B)
∇ × (∇φ) = 0
∇ · (∇ × A) = 0
∇ × (∇ × A) = ∇(∇ · A) − ∇2 A
∇ × (φA) = ∇φ × A + φ∇ × A
1.4.2
(27)
(28)
(29)
(30)
(31)
(32)
Curvilinear Coordinates
Scale factors Consider an orthogonal curvilinear coordinate system (x1 , x2 , x3 ) defined by a triad of unit
vectors (e1 , e2 , e3 ), which satisfy the orthogonality condition:
ei · ek = δik
(33)
1
FUNDAMENTALS
4
and form a right-handed coordinate system
e3 = e1 × e2
(34)
dr = h1 dx1 e1 + h2 dx2 e2 + h3 dx3 e3
(35)
∂r hi ≡ ∂xi (36)
The scale factors hi are defined by
or
The unit of arc length in this coordinate system is ds2 = dr · dr:
ds2 = h21 dx21 + h22 dx22 + h23 dx23
(37)
dV = h1 h2 h3 dx1 dx2 dx3
(38)
The unit of differential volume is
1.4.3
Gauss’ Divergence Theorem
For a vector or tensor field F, the following relationship holds:
Z
Z
∇ · F dV ≡
F · n̂ dA
V
(39)
∂V
This leads to the simple interpretation of the divergence as the following limit
Z
1
F · n̂ dA
∇ · F ≡ lim
V →0 V
∂V
A useful variation on the divergence theorem is
Z
Z
(∇ × F) dV ≡
V
(40)
n̂ × F dA
(41)
n̂ × F dA
(42)
∂V
This leads to the simple interpretation of the curl as
1
V →0 V
Z
∇ × F ≡ lim
1.4.4
∂V
Stokes’ Theorem
For a vector or tensor field F, the following relationship holds on an open, two-sided surface S bounded by
a closed, non-intersecting curve ∂S:
Z
Z
(∇ × F) · n̂ dA ≡
F · dr
(43)
S
1.4.5
∂S
The gradient operator ∇ or grad for a scalar field ψ is
∇ψ =
1 ∂ψ
1 ∂ψ
1 ∂ψ
e1 +
e2 +
e3
h1 ∂x1
h2 ∂x2
h3 ∂x3
(44)
A simple interpretation of the gradient operator is in terms of the differential of a function in a direction â
dâ ψ = lim ψ(x + da) − ψ(x) = ∇ψ · da
da→0
(45)
1
FUNDAMENTALS
5
The divergence operator ∇· or div is
1
∂
∂
∂
∇·F=
(h2 h3 F1 ) +
(h3 h1 F2 ) +
(h1 h2 F3 )
h1 h2 h3 ∂x1
∂x2
∂x3
(46)
The curl operator ∇× or curl is
1
∇×F=
h1 h2 h3
h 1 e1
∂
∂x1
h 1 F1
h3 e3 ∂
∂x3 h3 F3 h 2 e2
∂
∂x2
h 2 F2
(47)
The components of the curl are:
e1
∂
∂
∇×F =
(h3 F3 ) −
(h2 F2 )
h2 h3 ∂x2
∂x3
e2
∂
∂
+
(h1 F1 ) −
(h3 F3 )
h3 h1 ∂x3
∂x1
∂
e3
∂
(h2 F2 ) −
(h1 F1 )
+
h1 h2 ∂x1
∂x2
(48)
The Laplacian operator ∇2 for a scalar field ψ is
1
∂ h2 h3 ∂ψ
∂ h3 h1 ∂ψ
∂ h1 h2 ∂ψ
2
∇ ψ=
(
)+
(
)+
(
)
h1 h2 h3 ∂x1 h1 ∂x1
∂x2 h2 ∂x2
∂x3 h3 ∂x3
1.4.6
(49)
Specific Coordinates
(x1 , x2 , x3 )
x
y
z
h1
h2
h3
Cartesian
(x, y, z)
x
y
z
1
1
1
Cylindrical
(r, θ, z)
r sin θ
r cos θ
z
1
r
1
r sin φ cos θ
r sin φ sin θ
r cos φ
1
r
r sin φ
uv
z
u2 + v 2
h1
1
u2 + v 2
h1
uv
p
sinh2 u + sin2 v
h1
1
p
sinh2 ξ + sin2 η
h1
a sinh ξ sin η
Spherical
(r, φ, θ)
Parabolic Cylindrical
1
(u, v, z)
(u2 − v 2 )
2
Paraboloidal
(u, v, φ)
uv cos φ
Elliptic Cylindrical
(u, v, z)
a cosh u cos v
Prolate Spheroidal
(ξ, η, φ)
a sinh ξ sin η cos φ
1.5
1.5.1
uv sin φ
1
(u2
2
√
√
− v2 )
a sinh u sin v
z
a
a sinh ξ sin η sin φ
a cosh ξ cos η
a
Differential Relations
Conservation form
The equations are first written in conservation form
∂
density + ∇ · flux = source
∂t
(50)
1
FUNDAMENTALS
6
for a fixed (Eulerian) control volume in an inertial reference frame by using the divergence theorem.
∂ρ
+ ∇ · (ρu)
∂t
=
0
∂
(ρu) + ∇ · (ρuu − T) = ρG
∂t
∂
|u|2
|u|2
ρ e+
+ ∇ · ρu e +
−T·u+q
= ρG · u
∂t
2
2
∂
q
≥ 0
(ρs) + ∇ · ρus +
∂t
T
1.6
(51)
(52)
(53)
(54)
Convective Form
This form uses the convective or material derivative
Dρ
Dt
Du
ρ
Dt
D
|u|2
ρ
e+
Dt
2
Ds
ρ
Dt
D
∂
=
+u·∇
Dt
∂t
(55)
= −ρ∇ · u
(56)
= −∇P + ∇ · τ + ρG
(57)
= ∇ · (T · u) − ∇ · q + ρG · u
(58)
≥
−∇ ·
q
(59)
T
Alternate forms of the energy equation:
D
|u|2
ρ
e+
= −∇ · (P u) + ∇ · (τ · u) − ∇ · q + ρG · u
Dt
2
Formulation using enthalpy h = e + P/ρ
|u|2
∂P
D
h+
=
+ ∇ · (τ · u) − ∇ · q + ρG · u
ρ
Dt
2
∂t
(60)
(61)
Mechanical energy equation
ρ
D |u|2
= − (u · ∇) P + u · ∇ · τ + ρG · u
Dt 2
(62)
De
Dv
= −P
+ vτ :∇u − v∇ · q
Dt
Dt
(63)
Thermal energy equation
Dissipation
Υ = τ :∇u = τik
∂ui
∂xk
sum on i and k
(64)
Entropy
ρ
q Υ
Ds
= −∇ ·
+ +k
Dt
T
T
∇T
T
2
(65)
1
FUNDAMENTALS
1.7
7
Divergence of Viscous Stress
For a fluid with constant µ and µv , the divergence of the viscous stress in Cartesian coordinates can be
reduced to:
1
(66)
∇ · τ = µ∇2 u + µv + µ ∇(∇ · u)
3
1.8
Euler Equations
Inviscid, no heat transfer, no body forces.
Dρ
Dt
Du
ρ
Dt
D
|u|2
ρ
h+
Dt
2
Ds
Dt
1.9
= −ρ∇ · u
(67)
= −∇P
(68)
=
∂P
∂t
≥ 0
(69)
(70)
Bernoulli Equation
Consider the unsteady energy equation in the form
D
|u|2
∂P
ρ
h+
=
+ ∇ · (τ · u) − ∇ · q + ρG · u
Dt
2
∂t
(71)
and further suppose that the external force field G is conservative and can be derived from a potential Φ as
G = −∇Φ
(72)
then if Φ(x) only, we have
ρ
D
Dt
h+
|u|2
+Φ
2
=
∂P
+ ∇ · (τ · u) − ∇ · q
∂t
(73)
The Bernoulli constant is
|u|2
+Φ
2
In the absence of unsteadiness, viscous forces and heat transfer we have
|u|2
u·∇ h+
+Φ =0
2
H =h+
(74)
(75)
Or
H◦ = constant on streamlines
For the ordinary case of isentropic flow of an incompressible fluid dh = dP/ρ◦ in a uniform gravitational
field Φ = g(z − z◦ ), we have the standard result
P + ρ◦
|u|2
+ ρ◦ gz = constant
2
(76)
1
FUNDAMENTALS
1.10
8
Vorticity
Vorticity is defined as
ω ≡∇×u
(77)
and the vector identities can be used to obtain
(u · ∇)u = ∇(
|u|2
) − u × (∇ × u)
2
The momentum equation can be reformulated to read:
∂u
∇·τ
|u|2
+Φ =−
+ u × ω + T ∇s +
∇H = ∇ h +
2
∂t
ρ
(78)
(79)
1
FUNDAMENTALS
1.11
9
Dimensional Analysis
Fundamental Dimensions
L
M
T
θ
I
length
mass
time
temperature
current
meter (m)
kilogram (kg)
second (s)
Kelvin (K)
Ampere (A)
Some derived dimensional units
force
pressure
Newton (N)
Pascal (Pa)
bar = 105 Pa
Joule (J)
Hertz (Hz)
Watt (W)
Poise (P)
energy
frequency
power
viscosity (µ)
M LT −2
M L−1 T −2
M L2 T −2
T −1
M L2 T −3
M L−1 T −1
Pi Theorem Given n dimensional variables X1 , X2 , . . ., Xn , and f independent fundamental dimensions
(at most 5) involved in the problem:
1. The number of dimensionally independent variables r is
r≤f
2. The number p = n - r of dimensionless variables Πi
Πi =
Xi
· · · Xrαr
X1α1 X2α2
that can be formed is
p≥n−f
Conventional Dimensionless Numbers
Reynolds
Mach
Prandtl
Strouhal
Knudsen
Peclet
Schmidt
Lewis
Re
Ma
Pr
St
Kn
Pe
Sc
Le
ρU L/µ
U/c
µcP /k = ν/κ
L/U T
Λ/L
U L/κ
ν/D
D/κ
Reference conditions: U , velocity; µ, vicosity; D, mass diffusivity; k, thermal conductivity; L, length scale;
T , time scale; c, sound speed; Λ, mean free path; cP , specific heat at constant pressure.
Parameters for Air and Water Values given for nominal standard conditions 20 C and 1 bar.
1
FUNDAMENTALS
10
shear viscosity
kinematic viscosity
thermal conductivity
thermal diffusivity
specific heat
sound speed
density
gas constant
thermal expansion
isentropic compressibility
µ
ν
k
κ
cp
c
ρ
R
α
κs
Prandtl number
Fundamental derivative
ratio of specific heats
Grüneisen coefficient
Pr
Γ
γ
G
(kg/ms)
(m2 /s)
(W/mK)
(m2 /s)
(J/kgK)
(m/s)
(kg/m3 )
(m2 /s2 K)
(K−1 )
(Pa−1 )
Air
1.8×10−5
1.5×10−5
2.54×10−2
2.1×10−5
1004.
343.3
1.2
287
3.3×10−4
7.01×10−6
Water
1.00×10−3
1.0×10−6
0.589
1.4×10−7
4182.
1484
998.
462.
2.1×10−4
4.5×10−10
.72
1.205
1.4
0.40
7.1
4.4
1.007
0.11
2
THERMODYNAMICS
2
2.1
11
Thermodynamics
Thermodynamic potentials and fundamental relations
energy
e(s, v)
de
enthalpy h(s, P )
dh
Helmholtz f (T, v)
df
Gibbs g(T, P )
dg
2.2
T ds − P dv
e + Pv
T ds + v dP
e − Ts
−s dT − P dv
e − Ts + Pv
−s dT + v dP
=
=
=
=
=
=
=
(80)
(81)
(82)
(83)
Maxwell relations
∂T
∂v s
∂T
∂P s
∂s
∂v T
∂s
∂P T
∂P
∂s v
∂v
=
∂s P
∂P
=
∂T v
∂v
= −
∂T P
= −
(84)
(85)
(86)
(87)
Calculus identities:
F (x, y, . . . )
dF =
∂F
∂x
∂x
∂y
dx +
y,z,...
∂x
∂f
=−
f
=
y
∂f
∂y
∂f
∂x
dy + . . .
(88)
x,z,...
x
(89)
y
1
∂f
∂x
∂F
∂y
(90)
y
2
THERMODYNAMICS
2.3
12
Various defined quantities
specific heat at constant volume
cv
≡
specific heat at constant pressure
cp
≡
γ
≡
ratio of specific heats
c ≡
sound speed
α
≡
isothermal compressibility
KT
≡
isentropic compressibility
Ks
≡
coefficient of thermal expansion
∂e
∂T v
∂h
∂T P
cp
c
sv
∂P
∂ρ s
1 ∂v
v ∂T P
1 ∂v
−
v ∂P T
1 ∂v
1
−
= 2
v ∂P s
ρc
(91)
(92)
(93)
(94)
(95)
(96)
(97)
Specific heat relationships
∂P
∂v T
s
2
∂P
∂v
cp − cv = −T
∂v T ∂T P
KT = γKs
or
∂P
∂v
=γ
(98)
(99)
Fundamental derivative
Γ ≡
=
=
=
c4 ∂ 2 v
2v 3 ∂P 2 s
v3 ∂ 2 P
2c2 ∂v 2 s
∂c
1 + ρc
∂P
2 2 s
1 v
∂ h
+1
2 c2 ∂v 2 s
(100)
(101)
(102)
(103)
Sound speed (squared)
c2
≡
∂P
∂ρ
= −v 2
v
Ks
v
= γ
Kt
=
Grüneisen Coefficient
(104)
s
∂P
∂v
(105)
s
(106)
(107)
2
THERMODYNAMICS
13
vα
cv KT
∂P
= v
∂e v
vα
=
cp Ks
v ∂T
= −
T ∂v s
≡
G
2.4
(109)
(110)
(111)
v(P, s) relation
dv
v
= −Ks dP + Γ(Ks dP )2 + α
= −
2.5
(108)
dP
+Γ
ρc2
dP
ρc2
2
+G
T ds
+ ...
cp
T ds
+ ...
c2
(112)
(113)
Equation of State Construction
Given cv (v, T ) and P (v, T ), integrate
∂P
de = cv dT + T
− P dv
∂T v
∂P
cv
dT +
dv
ds =
T
∂T v
(114)
(115)
along two paths: I: variable T , fixed ρ and II: variable ρ, fixed T .
Energy:
!
Z ρ
Z T
∂P
dρ
e = e◦ +
P −T
cv (T, ρ◦ ) dT +
∂T
ρ2
ρ
T
ρ
{z
} | ◦
| ◦
{z
}
I
(116)
II
Ideal gas limit ρ◦ → 0,
lim cv (T, ρ◦ ) = cig
v (T )
(117)
ρ◦ →0
The ideal gas limit of I is the ideal gas internal energy
Z
ig
T
e (T ) =
cig
v (T ) dT
(118)
T◦
Ideal gas limit of II is the residual function
Z
r
ρ
e (ρ, T ) =
0
∂P
P −T
∂T
!
ρ
dρ
ρ2
(119)
and the complete expression for internal energy is
e(ρ, T ) = e◦ + eig (T ) + er (ρ, T )
(120)
Entropy:
Z
T
s = s◦ +
T◦
|
Z ρ
cv (T, ρ◦ )
dT +
T
ρ
{z
} | ◦
I
∂P
−
∂T
{z
II
!
ρ
dρ
ρ2
}
(121)
2
THERMODYNAMICS
14
The ideal gas limit ρ◦ → 0 has to be carried out slightly differently since the ideal gas entropy, unlike the
internal energy, is a function of density and is singular at ρ = 0. Define
sig =
Z
T
T◦
cig
v (T )
dT − R
T
Z
ρ
ρ◦
dρ
ρ
(122)
where the second integral on the RHS is R ln ρ◦ /ρ. Then compute the residual function by substracting the
singular part before carrying out the integration
!
Z ρ
1 ∂P
dρ
r
s (ρ, T ) =
R−
(123)
ρ ∂T ρ
ρ
0
and the complete expression for entropy is
s(ρ, T ) = s◦ + sig (ρ, T ) + sr (ρ, T )
(124)
3
COMPRESSIBLE FLOW
3
3.1
15
Compressible Flow
A steady flow must be considered as compressible when the Mach number M = u/c is sufficiently large. In
an isentropic flow, the change in density produced by a speed u can be estimated as
1
∆ρs = c−2 ∆P ∼ − ρM 2
(125)
2
from the energy equation discussed below and the fundamental relation of thermodynamics.
If the flow is unsteady, then the change in the density along the pathlines for inviscid flows without body
forces is
1 1 ∂u2
u · ∇u2
1 ∂P
1 Dρ
−
= −∇ · u = −
−
(126)
ρ Dt
2c2
c2 2 ∂t
ρ ∂t
This first term is the steady flow condition ∼ M 2 . The second set of terms in the square braces are the
unsteady contributions. These will be significant when the time scale T is comparable to the acoustic transit
time L/c◦ , i.e., T ∼ Lco .
3.1.1
Streamlines and Total Properties
Stream lines X(t; x◦ ) are defined by
dX
=u
dt
which in Cartesian coordinates yields
X = x◦
when t = 0
dx1
dx2
dx3
=
=
u1
u2
u3
(127)
(128)
|u|2
= constant
2
Velocity along a streamline is given by the energy equation:
p
u = |u| = 2(ht − h)
ht = h +
(129)
(130)
Total properties are defined in terms of total enthalpy and an idealized isentropic deceleration process along
a streamline. Total pressure is defined by
Pt ≡ P (s◦ , ht )
(131)
Other total properties Tt , ρt , etc. can be computed from the equation of state.
3.2
Quasi-One Dimensional Flow
d(ρuA) = 0
ρudu = −dP
u2
h+
= constant or
2
ds ≥ 0
(132)
(133)
dh = −udu
(134)
(135)
3
COMPRESSIBLE FLOW
3.2.1
16
Isentropic Flow
If ds = 0, then
dP = c2 dρ + c2 (Γ − 1)
(dρ)2
+ ...
ρ
(136)
For isentropic flow, the quasi-one-dimensional equations can be written in terms of the Mach number as:
1 dρ
ρ dx
1 dP
ρc2 dx
1 du
u dx
1 dM
M dx
1 dh
c2 dx
=
=
=
=
=
M 2 1 dA
1 − M 2 A dx
M 2 1 dA
1 − M 2 A dx
1 dA
1
−
2
1 − M A dx
1 + (Γ − 1)M 2 1 dA
−
1 − M2
A dx
2
1 dA
M
1 − M 2 A dx
(137)
(138)
(139)
(140)
(141)
At a throat, the gradient in Mach number is:
dM
dx
2
=
Γ d2 A
2A dx2
(142)
Constant-Γ Gas If the value of Γ is assumed to be constant, the quasi-one dimensional equations can be
integrated to yield:
ρt
ρ
ct
c
ht
h
u
A
A∗
P − Pt
ρt c2t
1 + (Γ − 1)M 2
=
=
ρt
ρ
Γ−1
1
2(Γ−1)
(143)
= 1 + (Γ − 1)M 2
1/2
1 + (Γ − 1)M 2
1/2
M2
= ct
1 + (Γ − 1)M 2
Γ
1 1 + (Γ − 1)M 2 2(Γ−1)
=
M
Γ
− 2Γ−1
1
=
1 + (Γ − 1)M 2 2(Γ−1) − 1
2Γ − 1
=
(144)
(145)
(146)
(147)
(148)
(149)
Ideal Gas For an ideal gas P = ρRT and e = e(T ) only. In that case, we have
Z
T
h(T ) = e + RT = h◦ +
Z
cv (T ) dT,
T
s = s◦ +
T◦
T◦
cP (T )
dT − R ln(P/P◦ )
T
(150)
and you can show that Γ is given by:
Γig =
γ + 1 γ − 1 T dγ
+
2
2 γ dT
(151)
3
COMPRESSIBLE FLOW
17
Perfect or Constant-γ Gas Perfect gas results for isentropic flow can be derived from the equation of
state
P = ρRT
h = cp T
cp =
γR
γ−1
(152)
the value of Γ for a perfect gas,
γ+1
2
Γpg =
(153)
the energy integral,
Tt = T
1+
γ−1 2
M
2
(154)
and the expression for entropy
s − so = cp ln
T
− R ln P/Po
To
s − so = cv ln
T
− R ln ρ/ρo
To
(155)
or
γ−1 2
1+
M
2
γ
γ−1
Tt
=
T
1
γ−1
Tt
=
T
Tt
T
Pt
P
=
ρt
ρ
(156)
(157)
(158)
Mach Number–Area Relationship
A
1
=
A∗
M
2
γ+1
γ−1 2
M
1+
2
γ+1
2(γ−1)
(159)
Choked flow mass flux
Ṁ =
2
γ+1
γ+1
2(γ−1)
ct ρt A∗
(160)
or
Ṁ =
√
γ
2
γ+1
γ+1
2(γ−1)
P
√ t A∗
RTt
Velocity-Mach number relationship
M
u = ct q
1+
(161)
γ−1
2
2 M
Alternative reference speeds
r
∗
ct = c
γ+1
2
∗
umax = c
r
γ+1
γ−1
(162)
3
COMPRESSIBLE FLOW
3.3
18
Heat and Friction
ρu
ρudu + dP
dh + udu
= ṁ = constant
= −F dx
= Qdx
1
F
ds =
Q+
dx
T
ρ
(163)
(164)
(165)
(166)
F is the frictional stress per unit length of the duct. In terms of the Fanning friction factor f
2
f ρu2
(167)
D
where D is the hydraulic diameter of the duct D = 4×area/perimeter. Note that the conventional D’Arcy
or Moody friction factor λ = 4 f .
Q is the energy addition as heat per unit mass and unit length of the duct. If the heat flux into the fluid
is q̇, then we have
q̇ 4
Q=
(168)
ρu D
F =
3.3.1
Fanno Flow
ρu
ρudu + dP
u2
h+
2
= ṁ = constant
= −F dx
(169)
(170)
= ht = constant
(171)
(172)
Change in entropy with volume along Fanno line, h + 1/2ṁ2 v 2 =ht
c2 − u2
ds
=
T
dv F anno
v(1 + G)
3.3.2
(173)
Rayleigh Flow
Constant-area, steady flow with heat transfer only:
ρu
P + ρu2
dh + udu
= ṁ = constant
= I
= Qdx
Change in entropy with volume along Rayleigh line, P + ṁ2 v = I
ds
c2 − u2
T
=
dv Rayleigh
vG
(174)
(175)
(176)
(177)
(178)
3
COMPRESSIBLE FLOW
3.4
19
Shock Jump Conditions
The basic jump conditions,
ρ1 w1
P1 + ρ1 w12
w2
h1 + 1
2
s2
= ρ2 w2
= P2 + ρ2 w22
w2
= h2 + 2
2
≥ s1
(179)
(180)
(181)
(182)
or defining [f ] ≡ f2 - f1
[ρw]
P + ρw2
w2
h+
2
[s]
= 0
= 0
(183)
(184)
=
0
(185)
≥ 0
(186)
The Rayleigh line:
P2 − P1
= −(ρ1 w1 )2 = −(ρ2 w2 )2
v2 − v1
(187)
[P ]
= −(ρw)2
[v]
(188)
or
Rankine-Hugoniot relation:
h2 − h1 = (P2 − P1 )(v2 + v1 )/2
or
e2 − e1 = (P2 + P1 )(v1 − v2 )/2
(189)
Velocity-P v relation
[w]2 = −[P ][v]
p
w2 − w1 = − −(P2 − P1 )(v2 − v1 )
or
(190)
Alternate relations useful for numerical solution
P2
h2
3.4.1
ρ1
= P1 + ρ1 w12 1 −
ρ2
"
2 #
1
ρ1
= h1 + w12 1 −
2
ρ2
(191)
(192)
Lab frame (moving shock) versions
Shock velocity
w1 = Us
(193)
w2 = Us − up
(194)
ρ2 (Us − up ) = ρ1 Us
P2 = P1 + ρ1 Us up
h2 = h1 + up (Us − up /2)
(195)
(196)
(197)
Particle (fluid) velocity in laboratory frame
Jump conditions
3
COMPRESSIBLE FLOW
20
Kinetic energy:
u2p
1
= (P2 − P1 )(v1 − v2 )
2
2
3.5
Perfect Gas Results
[P ]
P1
[w]
c1
[v]
v1
[s]
R
2γ
M12 − 1
γ+1
2
1
= −
M1 −
γ+1
M1
2
1
= −
1− 2
γ+1
M1
Pt2
= − ln
Pt1
(198)
=
(199)
(200)
(201)
γ
 γ−1
γ+1 2
M1


2

γ − 1 2
1+
M1
2

Pt2
Pt1
1
=
2γ
γ−1
M2 −
γ+1 1
γ+1
1
γ−1
(202)
γ + 1 v2
−
P2
γ − 1 v1
=
γ + 1 v2
P1
−1
γ − 1 v1
(203)
Some alternatives
P2
P1
=
=
ρ2
ρ1
=
M22
=
2γ
M12 − 1
γ+1
2γ
γ−1
M12 −
γ+1
γ+1
γ+1
γ − 1 + 2/M12
2
M12 +
γ−1
2γ
M2 − 1
γ−1 1
1+
(204)
(205)
(206)
(207)
Prandtl’s relation
w1 w2 = c∗2
(208)
where c∗ is the sound speed at a sonic point obtained in a fictitious isentropic process in the upstream flow.
r
γ−1
w2
∗
c = 2
ht ,
ht = h +
(209)
γ+1
2
3.6
Reflected Shock Waves
Reflected shock velocity UR in terms of the velocity u2 and density ρ2 behind the incident shock or detonation
wave, and the density ρ3 behind the reflected shock.
3
COMPRESSIBLE FLOW
21
u2
UR = ρ3
−1
ρ2
(210)
Pressure P3 behind reflected shock:
ρ3 u 2
P 3 = P 2 + ρ3 2
−1
ρ2
(211)
ρ3
+1
u22 ρ2
h3 = h2 +
2 ρ3 − 1
ρ2
(212)
Enthalpy h3 behind reflected shock:
Perfect gas result for incident shock waves:
P3
=
P2
P2
− (γ − 1)
P1
P2
(γ − 1)
+ (γ + 1)
P1
(3γ − 1)
(213)
3
COMPRESSIBLE FLOW
3.7
22
Detonation Waves
Jump conditions:
ρ1 w1
P1 + ρ1 w12
w2
h1 + 1
2
s2
3.8
= ρ2 w2
= P2 + ρ2 w22
w2
= h2 + 2
2
≥ s1
(214)
(215)
(216)
(217)
Perfect-Gas, 2-γ Model
Perfect gas with energy release q, different values of γ and R in reactants and products.
h1
h2
P1
P2
cp1
cp2
=
=
=
=
cp1 T
cp2 T − q
ρ1 R 1 T 1
ρ2 R 2 T 2
γ1 R1
=
γ1 − 1
γ2 R2
=
γ2 − 1
(218)
(219)
(220)
(221)
(222)
(223)
(224)
Substitute into the jump conditions to yield:
P2
1 + γ1 M12
=
P1
1 + γ2 M22
(225)
v2
γ2 M22 1 + γ1 M12
=
v1
γ1 M12 1 + γ2 M22
(226)
1
q
1
+ M2 + 2
T2
γ1 R1 γ1 − 1 2 1
c1
=
1
1
T1
γ2 R2
+ M22
γ2 − 1 2
(227)
Chapman-Jouguet Conditions Isentrope, Hugoniot and Rayleigh lines are all tangent at the CJ point
∂P
∂P
PCJ − P1
=
=
(228)
vCJ − V1
∂v Hugoniot
∂v s
which implies that the product velocity is sonic relative to the wave
w2,CJ = c2
(229)
1
(v1 − v)2 dṁ2
2T
(230)
ds =
3
COMPRESSIBLE FLOW
23
Jouguet’s Rule
#
"
G
w2 − c2
∂P
∆P
= 1 − (v1 − v)
−
v2
2v
∂v Hug
∆v
(231)
where G is the Grúniesen coefficient.
The flow downstream of a detonation is subsonic relative to the wave for points above the CJ state and
supersonic for states below.
3.8.1
2-γ Solution
Mach Number for upper CJ (detonation) point
s
s
(γ1 + γ2 )(γ2 − 1)
(γ2 − γ1 )(γ2 + 1)
MCJ = H +
+ H+
2γ1 (γ1 − 1)
2γ1 (γ1 − 1)
(232)
where the parameter H is the nondimensional energy release
H=
(γ2 − 1)(γ2 + 1)q
2γ1 R1 T1
(233)
CJ pressure
2
+1
PCJ
γ1 MCJ
=
P1
γ2 + 1
(234)
2
ρCJ
γ1 (γ2 + 1)MCJ
=
2 )
ρ1
γ2 (1 + γ1 MCJ
(235)
TCJ
PCJ R1 ρ1
=
T1
P1 R2 ρCJ
(236)
CJ density
CJ temperature
Strong detonation approximation MCJ 1
UCJ
≈
ρCJ
≈
PCJ
≈
q
2(γ22 − 1)q
γ2 + 1
ρ1
γ2
1
2
ρ1 UCJ
γ2 + 1
(237)
(238)
(239)
(240)
3.8.2
High-Explosives
For high-explosives, the same jump conditions apply but the ideal gas equation of state is no longer appropriate for the products. A simple way to deal with this problem is through the nondimensional slope γs of
the principal isentrope, i.e., the isentrope passing through the CJ point:
v ∂P
γs ≡ −
(241)
P ∂v s
Note that for a perfect gas, γs is identical to γ = cp /cv , the ratio of specific heats. In general, if the principal
isentrope can be expressed as a power law:
P v k = constant
(242)
3
COMPRESSIBLE FLOW
24
then γs = k. For high explosive products, γs ≈ 3. From the definition of the CJ point, we have that the
slope of the Rayleigh line and isentrope are equal at the CJ point:
∂P
PCJ − P1
PCJ
=
=−
γs,CJ
(243)
∂v s
vCJ − V1
vCJ
From the mass conservation equation,
vCJ = v1
γs,CJ
γs,CJ + 1
(244)
and from momentum conservation, with PCJ P1 , we have
PCJ =
2
ρ1 UCJ
γs,CJ + 1
(245)
3
COMPRESSIBLE FLOW
3.9
25
Weak shock waves
Nondimensional pressure jump
Π=
[P ]
ρc2
(246)
A useful version of the jump conditions (exact):
Π = −M1
[w]
[v]
= −M12
c1
v1
[w]
[v]
= M1
c1
v1
(247)
Thermodynamic expansions:
[v]
v1
= −Π + ΓΠ2 + O(Π)3
2
[v]
[v]
3
+ O ([v])
+Γ
Π = −
v1
v1
(248)
(249)
Linearized jump conditions:
−
[w]
c1
=
M1
=
M1
=
M2
=
Γ 2
Π + O(Π)3
2
2
Γ [w]
[w]
1−
+O
2 c1
c1
Γ
1 + Π + O(Π)2
2
Γ
1 − Π + O(Π)2
2
Π−
[c]
= (Γ − 1)Π + O(Π)2
c1
M1 − 1 ≈ 1 − M2
Prandtl’s relation
1
c∗ ≈ w1 + [w]
2
or
(250)
(251)
(252)
(253)
(254)
(255)
1
≈ w2 − [w]
2
(256)
Change in entropy for weak waves:
T [s]
1
= ΓΠ3 + . . .
c21
6
1
or = − Γ
6
[v]
v
3
+ ...
(257)
3
COMPRESSIBLE FLOW
3.10
26
Acoustics
Simple waves
∆P = c2 ∆ρ
(258)
∆P = ±ρc∆u
(259)
+ for right-moving waves, - for left-moving waves
Acoustic Potential φ
u = ∇φ
P0
ρ0
(260)
∂φ
= −ρo
∂t
ρo ∂φ
= − 2
co ∂t
(261)
(262)
Potential Equation
∇2 φ −
1 ∂2φ
=0
c2o ∂t2
(263)
d’Alembert’s solution for planar (1D) waves
φ = f (x − co t) + g(x + co t)
Acoustic Impedance
(264)
The specific acoustic impedance of a medium is defined as
z=
P0
|u|
(265)
For a planar wavefront in a homogeneous medium z = ±ρc, depending on the direction of propagation.
Transmission coefficients A plane wave in medium 1 is normally incident on an interface with medium
2. Incident (i) and transmitted wave (t)
ut /ui
=
Pt0 /Pi0
=
2z1
z2 + z1
2z2
z2 + z1
(266)
(267)
Harmonic waves (planar)
φ = A exp i(wt − kx) + B exp i(wt + kx)
c=
ω
k
k=
2π
λ
ω=
2π
= 2πf
T
(268)
Spherical waves
φ=
f (t − r/c) g(t + r/c)
+
r
r
(269)
Q(t) = lim 4πr2 ur
(270)
Spherical source strength Q, [Q] = L3 T −1
r→0
potential function
φ(r, t) = −
Q(t − r/c)
4πr
(271)
3
COMPRESSIBLE FLOW
27
Energy flux
Φ = P 0u
(272)
Acoustic intensity for harmonic waves
1
I =< Φ >=
T
Z
T
0
0
P 2
Φ dt = rms
ρc
(273)
Decibel scale of acoustic intensity
Iref = 10−12
dB = 10 log10 (I/Iref )
W/m2
(274)
or
0
0
dB = 20 log10 (Prms
/Pref
)
0
Pref
= 2 × 10−10
2
Cylindrical waves, q source strength per unit length [q] = L T
1
φ(r, t) = −
2π
Z
t−r/c
−∞
atm
(275)
−1
q(η) dη
p
(t − η)2 − r2 /c2
(276)
or
φ(r, t) = −
3.11
1
2π
∞
Z
q(t − r/c cosh ξ) dξ
(277)
0
Multipole Expansion
Potential from a distribution of volume sources, strength q per unit source volume
Z
1
q(xs , t − R/c)
φ(x, t) = −
dVs
R = |x − xs |
4π Vs
R
(278)
Harmonic source
q = f (x) exp(−iωt)
Potential function
1
φ(x, t) = −
4π
Z
f (xs )
Vs
exp i(kR − ωt)
dVs
R
(279)
Compact source approximation:
1. source distribution is in bounded region around the origin xs < a,
and small a r = |x|
2. source distribution is compact ka << 1 or λ a, so that the phase factor exp ikR does not vary too
much across the source
Multipole expansion:
∞
exp ikR X (−xs · ∇x )n exp ikr
=
R
n!
r
n=0
(280)
φ = φ0 + φ1 + φ2 + . . .
(281)
Series expansion of potential
Monopole term
exp i(kr − ωt)
φ (x, t) = −
4πr
0
Z
f (xs )dVs
Vs
(282)
3
COMPRESSIBLE FLOW
28
Dipole term
φ1 (x, t) =
ikx · D
4πr2
i
kr
1+
exp i(kr − ωt)
(283)
Dipole moment vector D
Z
D=
xs f (xs )dVs
(284)
Vs
k2
φ (x, t) =
4πr3
2
3.12
3i
3
1+
−
kr k 2 r2
1
Qij =
2
exp i(kr − ωt)
X
xi xj Qij
(285)
i,j
Z
xs,i xs,j f (xs )dVs
(286)
un (xs , t − R/c)
dA
R
(287)
Vs
Baffled (surface) source
Rayleigh’s formula for the potential
1
φ=−
2π
Z
Normal component of the source surface velocity
un = u · n̂
Harmonic source
un = f (x) exp(−iwt)
Fraunhofer conditions |xs | ≤ a
aa
1
λr
Approximate solution:
exp i(kr − wt)
φ=−
2πr
Z
f (xs ) exp iκ · xs dA
As
(288)
3
COMPRESSIBLE FLOW
3.13
29
The primitive variable version of the equations is:
∂ρ
+ ∇ · (ρu)
∂t
∂ρu
+ ∇ · (ρuu)
∂t
∂
u2
u2
ρ e+
+ ∇ · ρu(h + )
∂t
2
2
∂s
+ ∇ · (us)
∂t
=
0
(289)
= −∇P
(290)
=
0
(291)
≥ 0
(292)
(293)
Alternative version
1 Dρ
= −∇ · u
ρ Dt
Du
ρ
= −∇P
Dt
D
u2
∂P
ρ
h+
=
Dt
2
∂t
Ds
≥ 0
Dt
The characteristic version of the equations for isentropic flow (s = constant) is:
d
(u ± F ) = 0
dt
on C ± :
dx
=u±c
dt
(294)
(295)
(296)
(297)
(298)
This is equivalent to:
∂
∂
(u ± F ) + (u ± c)
(u ± F ) = 0
∂t
∂x
(299)
Riemann invariants:
Z
F =
c
dρ=
ρ
Z
dP
=
ρc
Z
dc
Γ−1
(300)
Bending of characteristics:
d
Γ
(u + c) =
dP
ρc
(301)
For an ideal gas:
2c
(302)
γ−1
Pressure-velocity relationship for expansion waves moving to the right into state (1), final state (2) with
velocity u2 < 0.
F =
2γ
P2
γ − 1 u2 γ−1
−2c1
= 1+
< u2 ≤ 0
P1
2 c1
γ−1
Shock waves moving to the right into state (1), final state (2) with velocity u2 > 0.


s
2
2
[P ]
γ(γ + 1) u2
4
c
1
1 + 1 +

=
u2 > 0
P1
4
c1
γ + 1 u2
(303)
(304)
3
COMPRESSIBLE FLOW
30
Shock Tube Performance
4 −2γ
γ4 −1
2γ1
P4
c1 γ4 − 1
1
1+
= 1−
Ms −
Ms2 − 1
P1
c4 γ + 1
Ms
γ1 + 1
(305)
Limiting shock Mach number for P4 /P1 → ∞
Ms →
c4 γ1 + 1
c1 γ4 − 1
(306)
3
COMPRESSIBLE FLOW
3.14
3.14.1
31
Oblique Shock Waves
Geometry:
w1
w2
v
= u1 sin β
= u2 sin(β − θ)
= u1 cos β = u2 cos(β − θ)
(307)
(308)
(309)
ρ2
w1
tan β
=
=
ρ1
w2
tan(β − θ)
(310)
Shock Polar
−
[w]
M1 tan θ
=
c1
cos β(1 + tan β tan θ)
M12 tan θ
[P ]
=
2
ρ1 c1
cot β + tan θ
(311)
(312)
Real fluid results
w2
β
= f (w1 )
normal shock jump conditions
−1
= sin (w1 /u1 )
!
w
2
= β − tan−1 p 2
u1 − w12
θ
(313)
(314)
(315)
Perfect gas result
tan θ =
2 cot β M12 sin2 β − 1
(γ + 1)M12 − 2 M12 sin2 β − 1
(316)
Mach angle
µ = sin−1
3.14.2
1
M
(317)
Weak Oblique Waves
Results are all for C + family of waves, take θ → -θ for C − family.
β
θ
[P ]
ρ1 c21
T1 [s]
c21
2
Γ1
1
[w]
[w]
p
+
O
2
c1
M12 − 1 c1
p
2
M12 − 1 [w]
[w]
+O
= −
2
M1
c1
c1
2
M
= p 21
θ + O(θ)2
M1 − 1
= µ−
=
Γ1
M16
θ3 + O(θ)4
6 (M12 − 1)3/2
(318)
(319)
(320)
(321)
Perfect Gas Results
[P ]
γM 2
= p 2 1 θ + O(θ)2
P1
M1 − 1
(322)
3
COMPRESSIBLE FLOW
3.14.3
32
Prandtl-Meyer Expansion
q
d u1
d θ = − M12 − 1
u1
(323)
Function ω, d θ = -d ω
√
M2 − 1 d M
1 + (Γ − 1)M 2 M
dω≡
(324)
Perfect gas result
r
ω(M ) =
γ+1
tan−1
γ−1
r
p
γ−1
(M 2 − 1) − tan−1 M 2 − 1
γ+1
(325)
Maximum turning angle
ωmax
3.14.4
π
=
2
r
γ+1
−1
γ−1
(326)
Inviscid Flow
Crocco-Vaszonyi Relation
u2
∂u
+ (∇ × u) × u = T ∇S − ∇(h + )
∂t
2
3.14.5
(327)
Potential Flow
∇ · (ρu) = 0
∇×u = 0
u2
= constant
h+
2
(328)
(329)
(φ2x − c2 )φxx + (φ2y − c2 )φyy + 2φx φy φxy = 0
(331)
(330)
or with u = ∇φ = (φx , φy )
Linearized potential flow:
u
v
= U∞ + φ0x
= φ0y
0
=
(332)
(333)
2
M∞
− 1 φ0xx − φ0yy
(334)
Wave equation solution
λ=
p
2 −1
M∞
φ0 = f (x − λy) + g(x + λy)
Boundary conditions for slender 2-D (Cartesian) bodies y(x)
U∞ dy
f 0 (ξ) = −
λ dx ξ
y≥0
(335)
(336)
Prandtl-Glauert Scaling for subsonic flows
φ(x, y) = φinc (x,
p
2 y)
1 − M∞
∇2 φinc = 0
(337)
3
COMPRESSIBLE FLOW
33
Prandtl-Glauert Rule
Cp = p
3.14.6
Cpinc
(338)
2
1 − M∞
Natural Coordinates
∂
∂x
∂
∂y
u
v
∂
∂
− sin θ
∂s
∂n
∂
∂
= sin θ
+ cos θ
∂s
∂n
= U cos θ
= U sin θ
=
cos θ
(339)
(340)
(341)
(342)
The transformed equations of motion are:
∂ρU
∂θ
+ ρU
∂s
∂n
∂U
∂P
ρU
+
∂s
∂s
∂P
∂θ
+
ρU 2
∂s
∂n
∂θ ∂U
−
ωz = U
∂s
∂n
Curvature of stream lines, R = radius of curvature
=
0
(343)
=
0
(344)
=
0
(345)
=
0
(346)
∂θ
1
=
∂s
R
(347)
Vorticity production
ωz = −
1 ∂Po
(T − To ) ∂S
+
U ρo ∂n
U
∂n
(348)
Elimination of pressure dP = c2 dρ
(M 2 − 1)
3.14.7
∂θ
∂U
−U
∂s
∂n
∂U
∂θ
−U
∂n
∂s
=
0
(349)
=
0
(350)
Method of Characteristics
∂
1
∂
(ω − θ) + √
(ω − θ)
2
∂s
M − 1 ∂n
∂
1
∂
(ω + θ) − √
(ω + θ)
2
∂s
∂n
M −1
=
0
(351)
=
0
(352)
(353)
Characteristic directions
C±
dn
1
= ±√
= ± tan µ
ds
M2 − 1
(354)
Invariants
J ± = θ ∓ ω = constant on C ±
(355)
4
INCOMPRESSIBLE, INVISCID FLOW
4
4.1
34
Incompressible, Inviscid Flow
Velocity Field Decomposition
Split the velocity field into two parts: irrotational ue , and rotational (vortical) uv .
u = ue + uv
(356)
Irrotational Flow Define the irrotational portion of the flow by the following two conditions:
∇ × ue
∇ · ue
= 0
= e(x, t)
volume source distribution
(357)
(358)
This is satisfied by deriving ue from a velocity potential φ
ue = ∇φ
∇2 φ = e(x, t)
(359)
(360)
Rotational Flow Define the rotational part of the flow by:
∇ · uv
∇ × uv
= 0
= ω(x, t)
vorticity source distribution
(361)
(362)
This is satisfied by deriving uv from a vector potential B
uv = ∇ × B
∇·B = 0
choice of gauge
2
∇ B = −ω(x, t)
4.2
(363)
(364)
(365)
Solutions of Laplace’s Equation
The equation ∇2 φ = e is known as Laplaces equation and can be solved by the technique of Green’s functions:
Z
φ(x, t) =
G(x|ξ)e(ξ, t)dVξ
(366)
Ω
ξ
For a infinite domain, Green’s function is the solution to
∇2 G
G
r
= δ(x − ξ)
1
1
1
= −
=−
4π |x − ξ|
4πr
= |r|
r=x−ξ
(367)
(368)
(369)
This leads to the following solutions for the potentials
Z
1
e(ξ, t)
dVξ
4π Ω
r
ξ
Z
1
ω(ξ, t)
dVξ
4π Ω
r
ξ
φ(x, t)
= −
(370)
B(x, t)
=
(371)
4
INCOMPRESSIBLE, INVISCID FLOW
35
The velocity fields are
ue (x, t)
=
1
4π
Z
1
= −
4π
uv (x, t)
Ω
ξ
Z
Ω
re(ξ, t)
dVξ
r3
ξ
r × ω(ξ, t)
dVξ
r3
(372)
(373)
If the domain is finite or there are surfaces (stationary or moving bodies, free surfaces, boundaries), then
an additional component of velocity, u0 , must be added to insure that the boundary conditions (described
subsequently) are satisfied. This additional component will be a source-free, ∇ · u0 = 0, irrotational ∇ × u0
= 0 field. The general solution for the velocity field will then be
u = ue + uv + u0
4.3
(374)
Boundary Conditions
Solid Boundaries In general, at an impermeable boundary ∂Ω, there is no relative motion between the
fluid and boundary in the local direction n̂ normal to the boundary surface.
u · n̂ = u∂Ω · n̂ on the surface ∂Ω
(375)
In particular, if the surface is stationary, the normal component of velocity must vanish on the surface
u · n̂ = 0
on a stationary surface ∂Ω
(376)
For an ideal or inviscid fluid, there is no restriction on the velocity tangential to the boundary, slip boundary
conditions.
u · t̂
arbitrary on the surface ∂Ω
(377)
For a real or viscous fluid, the tangential component is zero, since the relative velocity between fluid and
surface must vanish, the no-slip condition.
u=0
on the surface ∂Ω
(378)
Fluid Boundaries At an internal or free surface of an ideal fluid, the normal components of the velocity
have to be equal on each side of the surface
u1 · n̂ = u2 · n̂ = u∂Ω · n̂
(379)
and the interface has to be in mechanical equilibrium (in the absence of surface forces such as interfacial
tension)
P1 = P2
4.4
(380)
Streamfunction
The vector potential in flows that are two dimensional or have certain symmetries can be simplified to one
component that can be represented as a scalar function known as the streamfunction ψ. The exact form of
the streamfunction depends on the nature of the symmetry and related system of coordinates.
4
INCOMPRESSIBLE, INVISCID FLOW
4.4.1
36
2-D Cartesian Flows
Compressible
In a steady two-dimensional compressible flow:
∇ · ρu = 0
u = (u, v)
x = (x, y)
∂ρu ∂ρv
+
=0
∂x
∂y
(381)
The streamfunction is:
u=
1 ∂ψ
ρ ∂y
v=−
1 ∂ψ
ρ ∂x
(382)
Incompressible The density ρ is a constant
∇·u=0
u = (u, v)
x = (x, y)
∂u ∂v
+
=0
∂x ∂y
(383)
The streamfunction defined by
u=
∂ψ
∂y
v=−
∂ψ
∂x
(384)
will identically satisfy the continuity equation as long as
∂2ψ
∂2ψ
−
=0
∂x∂y ∂y∂x
(385)
which is always true as long as the function ψ(x, y) has continuous 2nd derivatives.
Stream lines (or surfaces in 3-D flows) are defined by ψ = constant. The normal to the stream surface is
n̂ψ =
∇ψ
|∇ψ|
(386)
Integration of the differential of the stream function along a path L connecting points x1 and x2 in the plane
can be interpreted as volume flux across the path
dψ
Z
dψ
= u · n̂L dl = −v dx + u dy
Z
= ψ2 − ψ1 =
u · n̂L dl = volume flux across
L
(387)
L
(388)
L
where ψ1 = ψ(x1 ) and ψ2 = ψ(x2 ). For compressible flows, the difference in the streamfunction can be
interpreted as the mass flux rather than the volume flux.
For this flow, the streamfunction is exactly the nonzero component of the vector potential
u = ∇ × B = x̂
B = (Bx , By , Bz ) = (0, 0, ψ)
∂ψ
∂ψ
− ŷ
∂y
∂x
(389)
and the equation that the streamfunction has to satisfy will be
∇2 ψ =
∂2ψ ∂2ψ
+
= −ωz
∂x2
∂y 2
(390)
where the z-component of vorticity is
ωz =
∂u
∂v
−
∂x ∂y
A special case of this is irrotational flow with ωz = 0.
(391)
4
INCOMPRESSIBLE, INVISCID FLOW
4.4.2
37
Cylindrical Polar Coordinates
In cylindrical polar coordinates (r, θ, z) with u = (ur , uθ , uz )
x
y
z
u
v
w
r cos θ
r sin θ
z
ur cos θ − uθ sin θ
ur sin θ + uθ cos θ
uz
(392)
(393)
(394)
(395)
(396)
(397)
1 ∂rur
1 ∂uθ
∂uz
+
+
r ∂r
r ∂θ
∂z
(398)
=
=
=
=
=
=
The continuity equation is
∇·u=0=
Translational Symmetry in z The results given above for 2-D incompressible flow have translational
symmetry in z such that ∂/∂z = 0. These can be rewritten in terms of the streamfunction ψ(r, θ) where
B = (0, 0, ψ)
(399)
The velocity components are
1 ∂ψ
r ∂θ
∂ψ
= −
∂r
ur
(400)
=
uθ
(401)
The only nonzero component of vorticity is
ωz =
1 ∂ur
1 ∂ruθ
−
r ∂r
r ∂θ
(402)
and the stream function satisfies
1 ∂
r ∂r
∂ψ
r
∂r
1 ∂
+
r ∂θ
1 ∂ψ
r ∂θ
= −ωz
(403)
Rotational Symmetry in θ If the flow has rotational symmetry in θ, such that ∂/∂θ = 0, then the
stream function can be defined as
ψ
B = 0, , 0
(404)
r
and the velocity components are:
1 ∂ψ
r ∂z
1 ∂ψ
r ∂r
ur
= −
(405)
uz
=
(406)
ωθ =
∂ur
∂uz
−
∂z
∂r
The only nonzero vorticity component is
(407)
The stream function satisfies
∂
∂z
1 ∂ψ
r ∂z
+
∂
∂r
1 ∂ψ
r ∂r
= −ωθ
(408)
4
INCOMPRESSIBLE, INVISCID FLOW
4.4.3
38
Spherical Polar Coordinates
This coordinate system (r, φ, θ) results in the continuity equation
1 ∂
1 ∂uθ
1
∂
r 2 ur +
+
(uφ sin φ) = 0
2
r ∂r
r sin φ ∂θ
r sin φ ∂φ
(409)
Note that the r coordinate in this system is defined differently than in the cylindrical polar system discussed
previously. If we denote by r0 the radial distance from the z-axis in the cylindrical polar coordinates, then
r0 = r sin φ. With symmetry in the θ direction ∂/∂θ, the following Stokes’ stream function can be defined
ψ
(410)
B = 0, 0,
r sin φ
Note that this stream function is identical to that used in the previous discussion of the case of rotational
symmetry in θ for the cylindrical polar coordinate system if we account for the reordering of the vector
components and the differences in the definitions of the radial coordinates.
The velocity components are:
∂ψ
1
r2 sin φ ∂φ
1 ∂ψ
= −
r sin φ ∂r
ur
(411)
=
uφ
(412)
The only non-zero vorticity component is:
ωθ =
1 ∂ruφ
1 ∂ur
−
r ∂r
r ∂φ
(413)
The stream function satisfies
1 ∂
r ∂r
4.5
1 ∂ψ
sin φ ∂r
+
1 ∂
r ∂φ
1
∂ψ
2
r sin φ ∂φ
= −ωθ
(414)
Simple Flows
The simplest flows are source-free and irrotational, which can be derived by a potential that satisfies the
Laplace equation, a special case of ue
∇2 φ = 0
∇·u=0
(415)
In the case of flows, that contain sources and sinks or other singularities, this equation holds everywhere
except at those singular points.
Uniform Flow The simplest solution is a uniform flow U:
φ=U·x
u = U = constant
(416)
In 2-D cartesian coordinates with U = U x̂, the streamfunction is
ψ = Uy
(417)
In spherical polar coordinates, Stokes streamfunction is
ψ=
U r2
sin2 φ
2
U = U ẑ
(418)
4
INCOMPRESSIBLE, INVISCID FLOW
39
Source Distributions Single source of strength Q(t) located at point ξ 1 . The meaning of Q is the volume
of fluid per unit time introduced or removed at point ξ 1 .
lim 4πr12 u · r̂1 = Q(t)
r1 = x − ξ 1
r1 →0
e = Q(t)δ(x − ξ 1 )
(419)
φ=−
Q(t)
4πr1
u=
r̂1 Q(t)
r1 Q(t)
=
4πr13
4πr12
(420)
For multiple sources, add the individual solutions
u=−
k
1 X ri Qi
4π i=1 ri3
(421)
In spherical polar cordinates, Stokes’ stream function for a single source of strength Q at the origin is
Q
cos φ
(422)
4π
For a 2-D flow, the source strength q is the volume flux per unit length or area per unit time since the
source can be thought of as a line source.
ψ=−
u = ur r̂
q
2πr
ur =
φ=
q
ln r
2π
ψ=
q
θ
2π
(423)
Dipole Consider a source-sink pair of equal strength Q located a distance δ apart. The limiting process
δ→0
Q→∞
δQ → µ
(424)
defines a dipole of strength µ. If the direction from the sink to the source is d̂, then the dipole moment
vector can be defined as
d = µd̂
(425)
The dipole potential for spherical (3-D) sources is
d·r
4πr3
(426)
1 3d · r
d
r
−
4π
r5
r3
(427)
φ=−
and the resulting velocity field is
u=
If the dipole is aligned with the z-axis, Stokes’ stream function is
ψ=
µ sin2 φ
4πr
(428)
and the velocity components are
ur
=
uφ
=
µ cos φ
2πr3
µ sin φ
4πr3
(429)
(430)
The dipole potential for 2-D source-sink pairs is
φ=−
µ cos θ
2π r
(431)
4
INCOMPRESSIBLE, INVISCID FLOW
40
and the stream function is
ψ=
µ sin θ
2π r
(432)
The velocity components are
ur
=
uθ
=
µ cos θ
2π r2
µ sin θ
2π r2
(433)
(434)
Combinations More complex flows can be built up by superposition of the flows discussed above. In
particular, flows over bodies can be found as follows:
half-body:
sphere:
cylinder:
closed-body:
4.6
source + uniform flow
dipole (3-D) + uniform flow
dipole (2-D) + uniform flow
sources & sinks + uniform flow
Vorticity
Vorticity fields are divergence free In general, we have ∇ · (∇ × A) ≡ 0 so that the vorticity ω = ∇ × u,
satisfies
∇·ω ≡0
(435)
Transport The vorticity transport equation can be obtained from the curl of the momentum equation:
∇·τ
Dω
= (ω · ∇)u − ω(∇ · u) + ∇T × ∇s + ∇ ×
(436)
Dt
ρ
The cross products of the thermodynamic derivatives can be written as
∇T × ∇s = ∇P × ∇v = −
∇P × ∇ρ
ρ2
(437)
which is known as the baroclinic torque.
For incompressible, homogeneous flow, the viscous term can be written ν∇2 ω and the incompressible
vorticity transport equation for a homogeneous fluid is
Dω
= (ω · ∇)u + ν∇2 ω
Dt
Circulation The circulation Γ is defined as
I
Γ=
(438)
Z
u · dl =
∂Ω
ω · n̂ dA
(439)
Ω
where is Ω is a simple surface bounded by a closed curve ∂Ω.
Vortex Lines and Tubes A vortex line is a curve drawn tangent to the vorticity vectors at each point in
the flow.
dy
dz
dx
=
=
ωx
ωy
ωz
(440)
The collection of vortex lines passing through a simple curve C form a vortex tube. On the surface of the
vortex tube, we have n̂ · ω =0.
4
INCOMPRESSIBLE, INVISCID FLOW
41
A vortex tube of vanishing area is a vortex filament, which is characterized by a circulation Γ. The
contribution du to the velocity field due to an element dl of a vortex filament is given by the Biot Savart
Law
du = −
Γ r × dl
4π r3
(441)
Line vortex A potential vortex has a singular vorticity field and purely azimuthal velocity field. For a
single vortex located at the origin of a two-dimensional flow
Γ
(442)
2πr
For a line vortex of strength Γi located at (xi , yi ), the velocity field at point (x, y) can be obtained by
transforming the above result to get velocity components (u, v)
ω = ẑΓδ(r)
uθ =
y − yi
Γi
2π (x − xi )2 + (y − yi )2
x − xi
Γi
2π (x − xi )2 + (y − yi )2
u
= −
(443)
v
=
(444)
(445)
Or setting Γ = ẑΓ
ui =
Γi × ri
2πri2
(446)
where ri = i - xi .
The streamfunction for the line vortex is found by integration to be
Γi
ln ri
(447)
2π
For a system of n vortices, the velocity field can be obtained by superposition of the individual contributions
to the velocity from each vortex. In the absence of boundaries or other surfaces:
ψi = −
u=
n
X
Γi × ri
i=1
4.7
2πri2
(448)
1. Vorticity can be visualized as local rotation within the fluid. The local angular frequency of rotation
1 |ω · n̂|
uθ
=
r→0 2πr
2π 2
fn̂ = lim
2. Vorticity cannot begin or end within the fluid.
∇·ω =0
3. The circulation is constant along a vortex tube or filament at a given instant in time
Z
ω · n̂ dA = constant
tube
However, the circulation can change with time due to viscous forces, baroclinic torque or nonconservative external forces. A vortex tube does not have a fixed identity in a time-dependent flow.
4
INCOMPRESSIBLE, INVISCID FLOW
42
4. Thompson’s or Kelvin’s theorem Vortex filaments move with the fluid and the circulation is constant
for an inviscid, homogeneous fluid subject only to conservative body forces.
DΓ
=0
(449)
Dt
Bjerknes theorem If the fluid is inviscid but inhomgeneous, ρ(x, t), then the circulation will change due
to the baroclinic torque ∇P × ∇ρ:
DΓ
=−
Dt
I
∂Ω
dP
=−
ρ
Z
Ω
∇P × ∇ρ
· n̂dA
ρ2
(450)
Viscous fluids have an additional contribution due to the diffusion of vorticity into or out of the tube.
4.8
Bernoulli’s equation for unsteady potential flow
P − P∞ = −ρ
∂
U2
|∇φ|2
(φ − φ∞ ) + ρ ∞ − ρ
∂t
2
2
(451)
Induced Mass If the external force Fext is applied to a body of mass M , then the acceleration of the
body dU/dt is determined by
dU
(452)
dt
where M is the induced mass tensor. For a sphere (3-D) or a cylinder (2-D), the induced mass is simply M
= mi I.
Fext = (m + M·)
mi,sphere
mi,cylinder
2 3
πa ρ
3
= πa2 ρ
=
(453)
(454)
(455)
Bubble Oscillations The motion of a bubble of gas within an incompressible fluid can be described by
unsteady potential flow in the limit of small-amplitude, low-frequency oscillations. The potential is given by
the 3-D source solution. For a bubble of radius R(t), the potential is
R2 (t) dR
r dt
Integration of the momentum equation in spherical coordinates yields the Rayleigh equation
φ=−
d2 R 3
R 2 +
dt
2
4.9
dR
dt
2
=
P (R) − P∞
ρ
(456)
(457)
Complex Variable Methods
Two dimensional potential flow problems can be solved in the complex plane
z = x + iy = r exp(iθ) = r cos θ + ir sin θ
The complex potential is defined as
F (z) = φ + iψ
and the complex velocity w is defined as
(458)
4
INCOMPRESSIBLE, INVISCID FLOW
43
dF
(459)
dz
NB sign of v-term! The complex potential is an analytic function and the derivatives satisfy the CauchyRiemann conditions
w = u − iv =
∂φ
∂x
∂φ
∂y
∂ψ
∂y
∂ψ
= −
∂x
=
(460)
(461)
which implies that both ∇2 φ = 0 and ∇2 ψ = 0, i.e., the real and imaginary parts of an analytic function
represent irrotational, potential flows.
Examples
1. Uniform flow u = (U∞ , V∞ )
F = (U∞ − iV∞ )z
2. Line source of strength q located at zo
F =
q
ln(z − z◦ )
2π
3. Line vortex of strength Γ located at z◦
F = −i
Γ
ln(z − z◦ )
2π
4. Source doublet (dipole) at z◦ oriented along +x axis
F =−
µ
2π(z − z◦ )
5. Vortex doublet at z◦ oriented along +x axis
F =
iλ
2π(z − z◦ )
6. Stagnation point
F = Cz 2
7. Exterior corner flow
F = Cz n
1/2 ≤ n ≤ 1
8. Interior corner flow, angle α
F = Cz n
1≤n=
π
α
9. Circular cylinder at origin, radius a, uniform flow U at x = ±∞
F = U (z +
a2
)
z
4
INCOMPRESSIBLE, INVISCID FLOW
4.9.1
44
Mapping Methods
A flow in the ζ plane can be mapped into the z plane using an analytic function z = f (ζ). An analytic function
is a conformal map, preserving angles between geometric features such as streamlines and isopotentials as
long as df /dz does not vanish. The velocity in the ζ-plane is w̃ and is related to the z-plane velocity by
w̃ =
w
dF
=
dζ
dζ
dz
or
w=
w̃
dF
=
dz
dz
dζ
(462)
In order to have well behaved values of w, require w̃ =0 at point where dz/dζ vanishes.
Blasius’ Theorem
The force on a cylindrical (2-D) body in a potential flow is given by
I
i
D − iL = ρ
w2 dz
2 body
(463)
For rigid bodies
D=0
L = −ρU∞ Γ
(464)
where the lift is perpendicular to the direction of fluid motion at ∞. The moment of force about the origin
is
I
1
2
zw dz
(465)
M = − ρ<
2
body
4.10
Airfoil Theory
Rotating Cylinder The streamfunction for a uniform flow U∞ over a cylinder of radius a with a bound
vortex of strength Γ is
a 2 Γ
r
ψ = U∞ r sin θ 1 −
−
ln( )
(466)
r
2π
a
The stagnation points on the surface of the cylinder can be found at
sin θs =
Γ
4πU∞ a
(467)
The lift L is given by
L = −ρU∞ Γ
(468)
The pressure coefficient on the surface of the cylinder is
CP =
4Γ
P − P∞
= 1 − 4 sin2 θ +
sin θ −
1
2
2πaU
ρU
∞
∞
2
Γ
2πaU∞
2
(469)
Generalized Cylinder Flow If the flow at infinity is at angle α w.r.t. the x-axis, the complex potential
for flow over a cylinder of radius a, center µ and bound circulation Γ is:
a2 exp(iα)
Γ
z−µ
F (z) = U exp(−iα)(z − µ) +
−i
ln(
)
(470)
z−µ
2π
a
4
INCOMPRESSIBLE, INVISCID FLOW
45
Joukowski Transformation The transformation
ζT2
(471)
ζ
is the Joukowski transformation, which will map a cylinder of radius ζT in the ζ-plane to a line segment y
=0, −2ζT ≤ x ≤ 2ζT . Use this together with the generalized cylinder flow in the ζ plane to produce the flow
for a Joukowski arifoil at an angle of attack. The inverse transformation is
r z 2
z
− ζT2
(472)
ζ= ±
2
2
z=ζ+
Kutta Condition The flow at the trailing edge of an airfoil must leave smoothly without any singularities.
There are two special cases:
• For a finite-angle trailing edge in potential flow, the trailing edge must be a stagnation point.
• For a cusp (zero angle) trailing edge in potential flow, the velocity can be finite but must be equal on
the two sides of the separating streamline.
Application to Joukowski airfoil: Locating the stagnation point at ζT = µ + a exp −iβ, the circulation
is determined to be:
Γκ = −4πaU∞ sin(α + β)
(473)
and the lift coefficient is
CL =
4.11
L
1
2
2 ρU∞ c
= 8π
a
c
sin(α + β)
(474)
Thin-Wing Theory
The flow consists of the superposition of the free stream flow and an irrotational velocity field derived from
disturbance potentials φt and φc associated with the thickness and camber functions.
u
v
ut
uc
=
=
=
=
U∞ cos α + ut + uc
U∞ sin α + vt + vc
∇φt
∇φc
(475)
(476)
(477)
(478)
(479)
where α is the angle of attack and ∇2 φi = 0.
Geometry A thin, two-dimensional, wing-like body can be represented by two surfaces displaced slightly
about a wing chord aligned with the x-axis, 0 ≤ x ≤ c. The upper (+) and lower (−) surfaces of the wing
are given by
y
y
= Y+ (x) for upper surface 0 ≤ x ≤ c
= Y− (x) for lower surface 0 ≤ x ≤ c
(480)
(481)
and can be represented by a thickness function f (x) and a camber function g(x).
f (x)
g(x)
= Y+ (x) − Y− (x)
1
=
[Y+ (x) + Y− (x)]
2
(482)
(483)
4
INCOMPRESSIBLE, INVISCID FLOW
46
The profiles of the upper and lower surface can be expressed in terms of f and g as
1
(484)
= g(x) + f (x) upper surface
2
1
Y− (x) = g(x) − f (x) lower surface
(485)
2
The maximum thickness t = O(f ), the maximum camber h = O(g), and the angle of attack are all
considered to be small in this analysis
Y+ (x)
α∼
t
h
∼ 1
c
c
and ui , vi << U∞
(486)
Boundary Conditions The exact slip boundary condition for inviscid flow on the body is:
dY
v = dx
u (x,Y (x))
(487)
dY±
vt + vc = α + lim
y→±0
dx
U∞ (x,y)
(488)
The linearized version of this is:
with cos α ≈ 1, and sin α ≈ α. This can be written as
vt (x, 0+) + vc (x, 0+)
vt (x, 0−) + vc (x, 0−)
= U∞ g 0 +
= U∞ g 0 −
1 0
f − αU∞
2
1 0
f − αU∞
2
(489)
(490)
where f 0 = df /dx and g 0 = dg/dx.
The boundary conditions are then divided between the thickness and camber disturbance flows as follows:
vt
vc
1
= ± U∞ f 0 for y → ±0
2
= U∞ (g 0 − α) for y → ±0
(491)
(492)
In addition, the disturbance velocities have to vanish far from the body.
4.11.1
Thickness Solution
The potential φt for the pure thickness case, which can be interpreted as a symmetric body at zero angle
of attack, can be calculated by the superposition of sources of strength q dx using the general solution for
potential flow
Z c
1
φt (x, y) =
ln(y 2 + (x − ξ)2 )q(ξ) dξ
(493)
2π 0
The velocity components are:
ut
=
vt
=
Z c
1
(x − ξ)q(ξ) dξ
2π 0 y 2 + (x − ξ)2
Z c
1
yq(ξ) dξ
2π 0 y 2 + (x − ξ)2
(494)
(495)
Apply the linearized boundary condition to obtain
1
df
1
± U∞
= lim
2
dx y→±0 2π
Z
0
c
y2
yq(ξ) dξ
+ (x − ξ)2
(496)
4
INCOMPRESSIBLE, INVISCID FLOW
47
Delta Function Representation The limit of the integrand is one of the representations of the Dirac
delta function
lim
y→±0
1
y
= ±δ(x − ξ)
π y 2 + (x − ξ)2
where
δ(x − ξ) =
Source Distribution
Z
0 x 6= ξ
∞ x=ξ
(497)
+∞
f (ξ)δ(x − ξ) dξ = f (x)
(498)
−∞
This leads to the source distribution
q(x) = U∞
df
dx
(499)
and the solution for the velocity field is
ut
=
vt
=
Z
∂φt
U∞ c (x − ξ)f 0 (ξ) dξ
=
∂x
2π 0 y 2 + (x − ξ)2
Z
∂φt
U∞ c yf 0 (ξ) dξ
=
∂y
2π 0 y 2 + (x − ξ)2
(500)
(501)
The velocity components satisfy the following relationships across the surface of the wing
[u] = u(x, 0+) − u(x, 0−) = 0
[v] = v(x, 0+) − v(x, 0−) = q(x)
Pressure Coefficient
(502)
(503)
The pressure coefficient is defined to be
P − P∞
u2 + v 2
=1−
1
2
2
U∞
2 ρU∞
CP =
(504)
The linearized version of this is:
ut + uc
U∞
For the pure thickness case, then we have the following result:
Z
1 c f 0 (ξ) dξ
CP ≈ −
π 0 (x − ξ)
CP ≈ −2
(505)
(506)
The integral is to be evaluated in the sense of the Principal value interpretation.
Principal Value Integrals If an integral has an integrand g that is singular at ξ = x, the principal value
or finite part is defined as
Z x−
Z a
Z a
P
g(ξ) dξ = lim
g(ξ) dξ +
g(ξ) dξ
(507)
→0
0
0
x+
Important principal value integrals are
Z
P
0
c
dξ
= ln
(x − ξ)
x
x−c
(508)
and
Z
P
0
c
ξ −1/2 dξ
1
= √ ln
(x − ξ)
x
√
√ c+ x
√
√
c− x
(509)
4
INCOMPRESSIBLE, INVISCID FLOW
48
A generalization to other powers can be obtained by the recursion relation
Z c n
Z c n−1
ξ dξ
ξ
dξ
cn
P
= xP
−
n
0 (x − ξ)
0 (x − ξ)
A special case can be found for the transformed variables cos θ = 1 - 2ξ/c
Z π
cos nθdθ
sin nθo
=π
P
cos
θ
−
cos
θ
sin θo
o
0
4.11.2
(510)
(511)
Camber Case
The camber case alone accounts for the lift (non-zero α) and the camber. The potential φc for the pure
camber case can be represented as a superposition of potential vortices of strength γ(x) dx along the chord
of the wing:
Z c
y
1
−1
γ(ξ) tan
dξ
(512)
φc =
2π 0
x−ξ
The velocity components are:
uc
=
vc
=
Z c
1
yγ(ξ) dξ
∂φc
=−
2
∂x
2π 0 y + (x − ξ)2
Z c
∂φc
1
(x − ξ)γ(ξ) dξ
=
∂y
2π 0 y 2 + (x − ξ)2
(513)
(514)
The u component of velocity on the surface of the wing is
lim uc (x, y) = u(x, ±0) = ∓
y→±0
γ(x)
2
(515)
Apply the linearized boundary condition to obtain the following integral equation for the vorticity distribution
γ
Z c
1
γ(ξ) dξ
dg
−α =
P
dξ
(516)
U∞
dx
2π
0 (x − ξ)
The total circulation Γ is given by
Z
Γ=
c
γ(ξ) dξ
(517)
0
The velocity components satisfy the following relationships across the surface of the wing
[u] = u(x, 0+) − u(x, 0−) = −γ(x)
[v] = v(x, 0+) − v(x, 0−) = 0
Kutta Condition
(518)
(519)
The Kutta condition at the trailing edge of a sharp-edged airfoil reduces to
γ(x = c) = 0
(520)
4
INCOMPRESSIBLE, INVISCID FLOW
49
Vorticity Distribution The integral equation for the vorticity distribution can be solved explicity. A
solution that satisfies the Kutta boundary condition is:
1/2 "
1/2 #
Z c 0
1
c−x
g (ξ)
ξ
α+ P
dξ
(521)
γ(x) = −2U∞
x
π
c−ξ
0 x−ξ
The pressure coefficient for the pure camber case is
CP = ±
γ(x)
U∞
for y → ±0
(522)
The integrals can be computed exactly for several special cases, which can be expressed most conveniently
using the transformation
2x
2ξ
−1
ρ=
−1
c
c
r
Z 1
1
1+ρ
P
dρ = −π
z
−
ρ
1−ρ
−1
Z 1 p
1 − ρ2
dρ = πz
P
z−ρ
−1
Z 1
1
p
dρ = 0
P
1 − ρ2 (z − ρ)
−1
Z 1
ρ
p
dρ = −π
P
1 − ρ2 (z − ρ)
−1
Z 1
ρ2
p
P
dρ = −πz
1 − ρ2 (z − ρ)
−1
z=
(523)
(524)
(525)
(526)
(527)
(528)
Higher powers of the numerator can be evaluated from the recursion relation:
Z
1
P
−1
4.12
ρn
Z
p
1 − ρ2 (z − ρ)
1
dρ = zP
−1
π
1(3) · · · (n − 2)
ρn−1
p
dρ − [1 − (−1)n ]
2
2
2(4) · · · (n − 1)
1 − ρ (z − ρ)
(529)
Axisymmetric Slender Bodies
Disturbance potential solution using source distribution on x-axis:
Z c
1
f (ξ) dξ
p
φ(x, r) = −
4π 0
(x − ξ)2 + r2
(530)
Velocity components:
Z c
1
∂φ
(x − ξ)f (ξ) dξ
=
∂x
4π 0 [(x − ξ)2 + r2 ]3/2
Z c
∂φ
1
rf (ξ) dξ
=
∂r
4π 0 [(x − ξ)2 + r2 ]3/2
u
= U∞ +
(531)
v
=
(532)
(533)
Exact boundary condition on body R(x)
v dR
=
u (x,R(x))
dx
(534)
4
INCOMPRESSIBLE, INVISCID FLOW
50
Linearized boundary condition, first approximation:
dR
dx
(535)
dR
U∞
dx
(536)
v(x, r = R) = U∞
Extrapolation to x axis:
lim (2πrv) = 2πR
r→0
Source strength
f (x) = U∞ 2πR
dR
= U∞ A0 (x)
dx
A(x) = πR2 (x)
(537)
Pressure coefficient
2u
CP ≈ −
−
U∞
4.13
dR
dx
2
(538)
Wing Theory
Wing span is −b/2 < y < +b/2. The section lift coefficient, L0 = lift per unit span
CL0 (y) =
L0
1
2
2 ρU∞ c(y)
= m◦ (y) (α − αi − α◦ (y))
(539)
Induced angle of attack, w = downwash velocity
αi = tan−1
w
U∞
≈
w
U∞
(540)
Induced drag
Di = ρU∞ Γαi
(541)
Load distribution Γ(y), bound circulation at span location y
Γ(y) =
1
m◦ U∞ c(y) (α − αi − α◦ (y))
2
(542)
Trailing vortex sheet strength
γ=−
dΓ
dy
(543)
Downwash velocity
w=
1
P
4pi
Z
+b/2
−b/2
γ(ξ) dξ
ξ−y
(544)
"
#
Z +b/2 0
1
1
Γ (ξ) dξ
Γ(y) = m◦ (y)U∞ c(y) α − α◦ (y) −
P
2
4piU∞
ξ−y
−b/2
(545)
Boundary conditions
b
b
Γ( ) = Γ(− ) = 0
2
2
Elliptic load distribution, constant downwash, induced angle of attack
(546)
4
INCOMPRESSIBLE, INVISCID FLOW
51
y 2 1/2
Γ(y) = Γs 1 −
2b
w=
Γs
2b
αi =
Γs
2U∞
(547)
Lift
πb2
4
(548)
L2
1
1
2 b2
π 2 ρU∞
(549)
L = ρU∞ Γs
Di =
Induced drag coefficient
CD,i =
CL2
πAR
AR = b2 /S ≈
b
c
(550)
5
VISCOUS FLOW
5
52
Viscous Flow
Equations of motion in cartesian tensor form (without body forces) are:
Conservation of mass:
∂ρ ∂ρuk
+
=0
∂t
∂xk
(551)
Momentum equation:
ρ
∂ui
∂P
∂τik
∂ui
+ ρuk
=−
+
∂t
∂xk
∂xi
∂xk
(i = 1, 2, 3)
(552)
sum on j
(553)
Viscous stress tensor
τik = µ
Lamé’s constant
∂ui
∂uk
+
∂xk
∂xi
+ λδik
∂uj
∂xj
2
λ = µv − µ
3
(554)
Energy equation, total enthalpy form:
ρ
∂τki ui
∂ht
∂P
∂qi
∂ht
+ ρuk
+
=
−
∂t
∂xk
∂t
∂xk
∂xi
sum on i and k
(555)
Thermal energy form
ρ
∂h
∂h
∂P
∂P
∂ui
∂qi
+ ρuk
+ uk
=
+ τik
−
∂t
∂xk
∂t
∂xk
∂xk
∂xi
sum on i and k
(556)
or alternatively
ρ
∂e
∂e
∂uk
∂qi
∂ui
+ ρuk
= −P
+ τik
−
∂t
∂xk
∂xk
∂xk
∂xi
sum on i and k
(557)
Dissipation function
Υ = τik
∂ui
∂xk
(558)
qi = −k
∂T
∂xi
(559)
Fourier’s law
5.1
Scaling
Reference conditions are
velocity
length
time
density
viscosity
thermal conductivity
U◦
L
T
ρ◦
µ◦
k◦
5
VISCOUS FLOW
53
Inertial flow Limit of vanishing viscosity, µ → 0. Nondimensional statement:
Reynolds number
Re =
ρ◦ U◦ L
1 P ∼ ρ◦ U◦2
µ◦
(560)
Nondimensional momentum equation
ρ
Du
1
= −∇P +
∇·τ
Dt
Re
(561)
Du
= −∇P
Dt
(562)
Limiting case, Re → ∞, inviscid flow
ρ
Viscous flow Limit of vanishing density, ρ → 0. Nondimensional statement:
Reynolds number
Re =
ρ◦ U◦ L
1
µ◦
P ∼
µ◦ U◦
L
(563)
Nondimensional momentum equation
Reρ
Du
= −∇P + ∇ · τ
Dt
(564)
Limiting case, Re → 0, creeping flow.
∇P = ∇ · τ
5.2
(565)
Two-Dimensional Flow
For a viscous flow in two-space dimensions (x, y) the components of the viscous stress tensor in cartersian
coordinates are


∂u
∂v
∂u
∂v
∂u
+
λ
+
µ
+
2µ
τxx τxy
∂x ∂x ∂y
∂y ∂x

(566)
τ =
=
τyx τyy
µ ∂u + ∂v
2µ ∂v + λ ∂u + ∂v
∂y
∂x
∂y
∂x
∂y
Dissipation function
" 2 2 #
2
2
∂u
∂v
∂u ∂v
∂u ∂v
Υ=µ 2
+
+
+2
+
+λ
∂x
∂y
∂y
∂x
∂x ∂y
5.3
(567)
Parallel Flow
The simplest case of viscous flow is parallel flow,
(u, v) = (u(y, t), 0) ⇒ ∇ · u = 0 ⇒ ρ = ρ(y)
only
(568)
Momentum equation
ρ
∂u
∂t
0
∂P
∂
+
∂x
∂y
∂P
= −
∂y
= −
µ
∂u
∂y
(569)
(570)
We conclude from the y-momentum equation that P = P (x) only.
Energy equation
ρ
∂e
∂e
+ ρu
=µ
∂t
∂x
∂u
∂y
2
+
∂
∂x
k
∂T
∂x
+
∂
∂y
k
∂T
∂y
(571)
5
VISCOUS FLOW
5.3.1
54
∂
In these flows ∂t
= 0 and inertia plays no role. Shear stress is either constant or varies only due to imposed
Couette Flow A special case are flows in which pressure gradients are absent
∂P
=0
∂x
(572)
and the properties strictly depend only on the y coordinate, these flows have
constant in these flows
τxy = µ
∂
∂x
= 0. The shear stress is
∂u
= τw
∂y
(573)
The motion is produced by friction at the moving boundaries
u(y = H) = U
u(y = 0) = 0
(574)
and given the viscosity µ(y) the velocity profile and shear stress τw can be determined by integration
Z
u(y) = τw
o
y
dy 0
µ(y 0 )
H
Z
τw = U
0
dy 0
µ(y 0 )
!−1
(575)
The dissipation is balanced by thermal conduction in the y direction.
µ
∂u
∂y
2
=−
∂
∂y
k
∂T
∂y
(576)
Using the constant shear stress condition, we have the following energy integral
uτ − q = −qw = constant
q = −k
∂T
∂y
qw = q(y = 0)
(577)
This relationship can be further investigated by defining the Prandtl number
Pr =
cP µ
ν
=
k
κ
κ=
k
ρcP
(578)
For gases, P r ∼ 0.7, approximately independent of temperature. The Eucken relation is a useful approximation that only depends on the ratio of specific heats γ
Pr ≈
4γ
7.08γ − 1.80
(579)
For many gases, both viscosity and conductivity can be approximated by power laws µ ∼ T n , k ∼ T m where
the exponents n and m range between 0.65 to 1.4 depending on the substance.
Constant Prandtl Number Assuming P r = constant and using d h = cp d T , the energy equation can
be integrated to obtain the Crocco-Busemann relation
h − hw + P r
u2
qw
= − Pr u
2
τw
(580)
u2
qw P r
−
u
2cP
τw cP
(581)
For constant cP , this is
T = Tw − P r
5
VISCOUS FLOW
55
Recovery Temperature If the lower wall (y = 0) is insulated qw = 0, then the temperature at y = 0 is
defined to be the recovery temperature. In terms of the conditions at the upper plate (y = H), this defines
a recovery enthalpy
1
hr = h(Tr ) ≡ h(TH ) + P r U 2
(582)
2
If the heat capacity cP = constant and we use the conventional boundary layer notation, for which TH =
Te , the temperature at the outer edge of the boundary layer
Tr = Te + P r
1 U2
2 cP
(583)
Contrast with the adiabatic stagnation temperature
Tt = Te +
1 U2
2 cP
(584)
The recovery factor is defined as
r=
Tr − Te
Tt − Te
(585)
In Couette flow, r = P r. The wall temperature is lower than the adiabatic stagnation temperature Tt when
P r < 1, due to thermal conduction removing energy faster than it is being generated by viscous dissipation.
If P r > 1, then viscous dissipation generates heat faster than it can be conducted away from the wall and
Tr > Tt .
Reynolds Analogy If the wall is not adiabatic, then the heat flux at the lower wall may significantly
change the temperature profile. In particular the lower wall temperature (for cp = constant) is
Tw = Tr +
qw
P rU
cP τw
(586)
In order to heat the fluid qw > 0, the lower wall must be hotter than the recovery temperature.
The heat transfer from the wall can be expressed as a heat transfer coefficient or Stanton number
St =
qw
ρU cP (Tw − Tr )
(587)
where qw is the heat flux from the wall into the fluid, which is positive when heat is being added to the fluid.
The Stanton number is proportional to the skin friction coefficient
Cf =
τw
1
2
ρU
2
(588)
For Couette flow,
Cf
2P r
This relationship between skin friction and heat transfer is the Reynolds analogy.
St =
(589)
Constant properties If µ and k are constant, then the velocity profile is linear:
τw = µ
U
H
u=
τw
y
µ
(590)
ρU H
µ
(591)
The skin friction coefficient is
Cf =
2
Re
Re =
5
VISCOUS FLOW
5.3.2
56
Poiseuille Flow
If an axial pressure gradient is present, ∂P
∂x < 0, then the shear stress will vary across the channel and fluid
motion will result even when the walls are stationary. In that case, the shear stress balances the pressure
drop. This is the usual situation in industrial pipe and channel flows. For the simple case of constant µ
0=−
∂2u
∂P
+µ 2
∂x
∂y
(592)
With the boundary conditions u(0) = u(H) = 0, this can be integrated to yield the velocity distribution
u=−
∂P H 2 y y
1−
∂x 2µ H
H
(593)
∂P H
∂x 2
(594)
and the wall shear stress
τw = −
Pipe Flow The same situation for a round channel, a pipe of radius R, reduces to
1 ∂P
1 ∂ ∂u
r
=
r ∂r ∂r
µ ∂x
(595)
which integrates to the velocity distribution
u=−
1 ∂P
R2 − r 2
4µ ∂x
(596)
∂P R
∂x 2
(597)
∂P πR4
∂x 8µ
(598)
and a wall shear stress of
τw = −
The total volume flow rate is
Q=−
The skin friction coefficient is traditionally based on the mean speed ū and using the pipe diameter d = 2R
as the scale length.
ū =
∂P R2
Q
=
−
πR2
∂x 8µ
(599)
and is equal to
Cf =
τw
16
=
1/2ρū2
Red
Red =
ρūd
µ
(600)
In terms of the Darcy friction factor,
Λ=
8τw
64
=
2
ρū
Red
(601)
Turbulent flow in smooth pipes is correlated by Prandtl’s formula
√ 1
√ = 2.0 log Red Λ − 0.8
Λ
(602)
or the simpler curvefit
−2.5
Λ = 1.02 (log Red )
(603)
5
VISCOUS FLOW
5.3.3
57
Rayleigh Problem
Also known as Stokes’ first problem. Another variant of parallel flow is unsteady flow with no gradients in
the x direction. The Rayleigh problem is to determine the motion above an infinite (−∞ < x < ∞) plate
impulsively accelerated parallel to itself.
The x-momentum equation (for constant µ) is
∂2u
∂u
=ν 2
∂t
∂y
(604)
The boundary conditions are
u(y, t = 0) = 0
u(y = 0, t > 0) = U
(605)
The problem is self similar and in terms of the similarity variable η, the solution is
η
f 00 + f 0 = 0
2
y
η=√
νt
u = U f (η)
(606)
The solution is the complementary error function
η
f = erfc( )
2
erfc(s) = 1 − erf(s)
2
erf(s) = √
π
Z
s
exp(−x2 ) dx
(607)
0
Shear stress at the wall
µU
τw = − √
πνt
(608)
Vorticity
ω=−
U
η2
∂u
=√
exp(− )
∂y
4
πνt
(609)
Vorticity thickness
1
δω =
ω◦
5.4
Z
∞
ω(y, t) dy =
√
πνt
(610)
0
Boundary Layers
For streamline bodies without separation, viscous effects are confined to a thin layer y ≤ δ, when the
Reynolds number is sufficiently high, Re 1.
Scaling
x ∼ L
y ∼ δ
u ∼ U
U
δ
U∼
v ∼
L
Re1/2
L
δ ∼
Re1/2
(611)
(612)
(613)
(614)
(615)
Exterior or outer flow, ue . Re → ∞, slip boundary conditions. Equations are inviscid flow equations of
motion.
Interior or inner flow , ui . Finite Re but δ L, noslip boundary conditions ui (y = 0) = 0, matching to
outer flow, limy→∞ ui = limy→0 ue . Equations are
5
VISCOUS FLOW
58
Boundary Layer Equations The unsteady, compressible boundary-layer equations are:
∂ρ ∂ρu ∂ρv
+
+
∂t
∂x
∂y
∂u
∂u
∂u
ρ
+ ρu
+ ρv
∂t
∂x
∂y
=
(616)
0
∂P
∂τxy
+
∂x
∂y
∂P
0 = −
∂y
∂ht
∂ht
∂ht
∂P
∂
ρ
+ ρu
+ ρv
=
+
(uτxy − qy )
∂t
∂x
∂y
∂t
∂y
= −
(617)
(618)
(619)
Thickness Measures 99% velocity thickness
δ.99 = y(u = .99ue )
(620)
Displacement thickness
Z
∗
∞
δ =
0
ρu
1−
ρe u e
dy
(621)
Momentum thickness
∞
Z
θ=
0
ρu
ρe ue
1−
u
ue
dy
(622)
Displacement Velocity Near the boundary layer, the external flow produces a vertical velocity ve which
can be estimated by continuity to be
∂ρe ue
(623)
∂x
The boundary layer displaces the outer flow, producing a vertical velocity v far from the surface which differs
from ve by the amount v ∗
ρe ve ≈ −y
d
(ρe ue δ ∗ )
(624)
dx
The boundary layer influence on the outer flow can therefore by visualized as a source distribution producing
an equivalent displacement.
ρe v∗ =
Steady Incompressible Boundary layers The pressure gradient can be replaced by using Bernoulli’s
equation in the outer flow
∂P
∂ue = −ρue
(625)
∂x
∂x y=0
For constant µ and k, the equations are
∂u ∂v
+
∂x ∂y
∂u
∂u
u
+v
∂x
∂y
ρu
∂e
∂e
+ ρv
∂x
∂y
=
0
∂ue
∂2u
+ν 2
∂x
∂y
2
∂u
∂2T
= µ
+k 2
∂y
∂y
= ue
(626)
(627)
(628)
5
VISCOUS FLOW
5.4.1
59
Blasius Flow
The steady flow ue = U over a semi-infinite flat plate (0 ≤ x < ∞) with no pressure gradient can be solved
by a similarity transformation for the case of isothermal, incompressible flow.
r
2νx
y
δ(x) =
(629)
η=
δ(x)
U
Define a stream function
u=
∂ψ
∂y
v=−
∂ψ
∂x
ψ = δ(x)U f (η)
(630)
to obtain the Blasius equation
f 000 + f f 00 = 0
f (0) = f 0 (0) = 0
f 0 (∞) = 1
(631)
Numerical solution yields f 00 (0) = 0.469600 for a skin friction coefficient of
Cf =
0.664
1/2
Rex
Rex =
ρU x
µ
(632)
The various thickness measures are:
δ.99 =
5.0x
δ∗ =
1/2
Rex
1.7208x
1/2
Rex
θ=
0.664x
1/2
(633)
Rex
The displacement is equivalent to that produced by a slender body of thickness δ ∗ (x). The vertical velocity
outside the boundary layer (y → ∞) is
v∗ ∼ U
dδ ∗
0.861U
=
1/2
dx
Rex
(634)
which agrees with direct computation from the stream function
v ∗ = lim −
η→∞
∂ψ
U
= lim √
(ηf 0 (η) − f (η))
1/2
η→∞
∂x
2Rex
(635)
where by numerical computation
lim f = η − η ∗
η→∞
5.4.2
η ∗ = 1.21678
f 0 (∞) = 1
(636)
Falkner-Skan Flow
For flows of the type ue = Cxm , i.e., external flows representing flow over an exterior or interior corner of
angle α = πm/(m + 1) , similarity solutions to the boundary layer equations can be obtained. Define the
similarity variable and streamfunction similar to Blasius case
s
2νx
ψ = ue (x)δ(x)f (η)
(637)
η = y/δ(x) δ =
(m + 1)ue (x)
The resulting equation for the function f is
f 000 + f f 00 + β 1 − f 02 = 0
Some cases
β=
2m
m+1
(638)
5
VISCOUS FLOW
m
-.0904
<0
0
1
0<
-2
-1
5.5
60
flow
separating
retarded flows, expansion corner
stagnation point
accelerated flows, wedges
doublet near a wall
point sink
Kármán Integral Relations
Integration of the momentum equation for incompressible flow results in
dθ
θ due
Cf
=
+ (2 + H)
2
dx
ue dx
H=
δ∗
θ
(639)
The Kármán-Pohlhausen technique consists of assuming a Blausius-type similarity profile for the velocity
y
(640)
δ
where δ locates a definite outer edge of the boundary layer. Matching the boundary layer solution
smoothly to the outer flow at η = 1 and satisfying the noslip condition at η = 0, results in the following
conditions on f
u = ue (x)f (η) η =
(641)
f (0)
= 0
δτw
f 0 (0) =
µue
δ 2 due
f 00 (0) = −
ν dx
f 000 (0) = 0
f (1) = 1
n>1
f
(1) = 0
(642)
(643)
(644)
(645)
(646)
This results in an ordinary differential equation for δ as a function of x.
5.6
Thwaites’ Method
Rewrite the Kármán integral equation as
ue dθ2
= 2(S − (2 + H)λ)
ν dx
λ=
θ2 due
ν dx
S=
θτw
µue
(647)
Thwaites’ 1949 correlation
2(S − (H + 2)λ) ≈ 0.45 − 6λ
(648)
Kármán integral equation
d
ue
dx
λ
u0e
= 0.45 − 6λ
u0e =
due
dx
(649)
Approximate solution
θ2 =
Correlation functions S(λ) and H(λ)
0.45ν
u6e
Z
0
x
u5e dx
(650)
5
VISCOUS FLOW
61
τw =
5.7
µue
S(λ)
θ
δ ∗ = θH(λ)
(651)
Laminar Separation
Seperation of the boundary layer from the wall and the creation of a recirculating flow region occurs when
the shear stress vanishes.
∂u =0
(652)
τw,sep = µ
∂y y=0,x=xsep
For laminar boundary layers, this occurs when a sufficiently long region of adverse pressure gradient dP/ dx >
0 exists.
2
δ.99
dP '5
λsep ' −0.0931
(653)
µue dx sep
5.8
Compressible Boundary Layers
Steady, compressible, two-dimensional boundary layer equations:
∂ρu ∂ρv
+
∂x
∂y
∂u
∂u
+ ρv
ρu
∂x
∂y
ρu
5.8.1
∂h
∂h
+ ρv
∂x
∂y
=
0
(654)
∂Pe
∂ ∂u
+
µ
∂x
∂y ∂y
2
∂u
∂ ∂T
= µ
+
k
∂y
∂y ∂y
= −
(655)
(656)
Transformations and Approximations
Modified stream function
ρu = ρ◦
∂Ψ
∂y
ρv = −ρ◦
∂Ψ
∂x
Density-weighted y-coordinate (Howarth-Doronitsyn-Stewartson)
Z
ρ
Y =
dy
X=x
ρ◦
(657)
(658)
Derivative transformation
ρ ∂
∂
=
∂y
ρ◦ ∂Y
∂
∂
∂Y ∂
=
+
∂x
∂X
∂x ∂Y
(659)
Chapman-Rubesin parameter, enthalpy-temperature relation
C=
ρµ
ρ◦ µ◦
dh = cp dT
(660)
Boundary layer equations
∂Ψ ∂ 2 Ψ
∂Ψ ∂ 2 Ψ
−
∂Y ∂XY
∂X ∂Y 2
∂Ψ ∂h
∂Ψ ∂h
−
∂Y ∂X
∂X ∂Y
Similarity variable
∂ ∂2Ψ
C
∂Y ∂Y 2
2 2
∂
C ∂h
∂ Ψ
= ν◦
+ ν◦ C
∂Y P r ∂Y
∂Y 2
= νo
(661)
(662)
5
VISCOUS FLOW
62
r
y
η=
δ(x)
δ=
2ν◦ x
U
(663)
Streamfunction ansatz for zero pressure gradient
Ψ = U δf (η)
h = h◦ g(η)
(664)
Similarity function equations
(Cf 00 )0 + f f 00
C
( g 0 )0 + f g 0
Pr
=
(665)
0
2
= −CEc (f 00 )
(666)
where the Eckert number is
Ec =
U2
= (γ − 1)M 2
h◦
for perfect gases
(667)
Transport property approximation
C = 1 ρµ = ρ◦ µ◦
Pr =
cP µ
= constant
k
(668)
Approximate equation set:
f 000 + f f 00
00
g + P rf g
5.8.2
0
=
0
(669)
00 2
= −P rEc (f )
(670)
Energy Equation
Integration of the energy equation results in the integral relationship for heat flux at the wall
Z ∞
∂
ρu
ht
qw =
(ρe ue ht,e Θh )
Θh =
− 1 dy
∂x
ρe ue ht,e
0
(671)
where Θh is the energy thickness.
The recovery factor r determines the wall enthalpy in adiabatic flow,
1
hr = hw (qw = 0) = h∞ + r u2∞
2
The recovery factor is found to be an increasing function of the Prandtl number. In gases,
r ' P r1/2
r ' P r1/3
laminar boundary layers
turbulent boundary layers
Unity Prandtl Number For P r = 1, the energy equation is
∂ht
∂ht
∂
∂ht
u2
ρu
+ ρv
=
µ
ht = he + e
∂x
∂y
∂y
∂y
2
(672)
(673)
(674)
This has as a solution in adiabatic flow
ht = he +
Therefore, the recovery enthalpy is
u2e
u2
= h∞ + ∞ =
2
2
constant for qw = 0
(675)
5
VISCOUS FLOW
63
u2∞
(676)
2
From the exact correspondence to the x-momentum equation, the general, qw 6= 0, solution is ht = a + bu.
This leads directly to the Crocco integral
u2∞
u2
u
(677)
+
1− 2
h = h∞ + [hw − hr ] 1 −
u∞
2
u∞
The Stanton number can be derived from this result in the form of Reynolds analogy
hr = h∞ +
Cf
(678)
2
The generalization of this to other Prandtl numbers that is valid for laminar and turbulent boundary layers
in gases is
St =
St '
Cf
2P r2/3
(679)
General Prandtl Number For similarity solutions, the nondimensional enthalpy can be found by integration of the energy equation, simplest when C = 1, and P r = constant.
2
g 00 + P r f g 0 = −P r Ec (f 00 )
(680)
This equation can be integrated exactly to yield
0
Z
η
0
0
F (η ; P r) dη − P r Ec
g(η) = g(0) + g (0)
"Z
η
Z
0
η0
F (η ; P r)
0
0
0
2
(f 00 (ξ)) dξ
F (ξ; P r)
#
dη 0
(681)
where
η
Z
η0
Z
exp −P r
F (η; P r) =
0
!
dη 0
f (ξ) dξ
(682)
0
and the boundary conditions yield
∂T k he ρ 0
qw = −k
=−
g (0)
∂y w
cP δ ρe
hw
g(0) =
he
(683)
This results in a recovery factor of
Z
r = 2P r
"Z
∞
η
F (η; P r)
0
0
2
(f 00 (ξ)) dξ
F (ξ; P r)
#
dη
(684)
which for a laminar flat plate boundary layer has the approximate value
r ≈ P r1/2
0.1 ≤ P r ≤ 3.0
(685)
The Stanton number is
qw
G(P r)
St ≡
=√
G(P r) =
1/2
ρe ue (hw − hr )
2P rRex
and for a flat plate boundary layer G can be approximated as
G ≈ 0.4969P r1/3
Z
−1
∞
F (η; P r) dη
(686)
0
0.1 ≤ P r ≤ 3.0
(687)
so that the Stanton number for flat plate gas flow is approximately
St =
0.33206
1/2
P r2/3 Rex
(688)
5
VISCOUS FLOW
Coordinate stretching
and the velocity profile
64
The physical coordinate can be computed from the transformed similarity variable
r
y
u∞
=
2ν∞ x
Z
η
0
ρ∞
dη
ρ
(689)
The density profile can be computed from the temperature profile since the pressure is constant across the
boundary layer. For an ideal gas
ρ∞
T
=
ρ
T∞
(690)
For the case of P r = 1 and a perfect gas, the temperature profile is
Z η
T
γ−1 2
1 − f 02 dη 0
=1+
M∞
T∞
2
0
(691)
where u = u∞ f 0 (η). The coordinate transformation is then
r
Z η
u∞
γ−1 2
1 − f 02 dη 0
=η+
M∞
y
2ν∞ x
2
0
(692)
If we suppose that the viscosity varies as µ ∼ T ω , then the skin friction coefficient is
√ 00
1
2f (0)
Cf =
1−ω
1/2
γ−1
2
Rex
1 + 2 M∞
5.8.3
(693)
Moving Shock Waves
For a moving shock wave, the boundary conditions in the shock fixed frame are that the wall is moving with
the upstream velocity w1 and the freestream condition is w2 . If the reference velocity is w2 , then boundary
conditions on f are
f 0 (0) =
f (0) = 0,
uw
w1
=
ue
w2
f 0 (∞) = 1
(694)
This results in a negative displacement thickness.
5.8.4
Weak Shock Wave Structure
In contrast to the usual Boundary layer equations, here
are considered.
∂
∂y
=
∂
∂z
ρu
4 0 ∂u
P + ρ1 u 1 u − µ
3 ∂x
u2
4 µ0 ∂u
k ∂T
h+
−
u
−
2
3 ρ1 u1 ∂x ρ1 u1 ∂x
= 0, and only derivatives in the x direction
= ρ1 u 1
(695)
= P1 + ρ1 u21
(696)
= h1 +
u21
2
(697)
where
3
µ0 = µ + µv
4
(698)
1
s2 − s1 =
ρ1 u 1
Z
+∞
−∞
"
4 µ0
3T
∂u
∂x
Weak shock thickness estimate based on maximum slope:
2
+k
1 ∂T
T ∂x
2 #
dx
(699)
5
VISCOUS FLOW
65
∆m =
8µ0
1
3ρc M1n − 1
3
µ0 = µ + µv
4
(700)
For a perfect gas (γ = constant), the mean free path can be estimated as
Λ=
πγ 1/2 µ
2
ρc
(701)
1.8Λ
(M1n − 1)
(702)
and the shock thickness for γ = 1.4, µv = 0, is
∆m =
5
VISCOUS FLOW
5.9
66
Creeping Flow
In the limit of zero inertia, the flow is described by Stokes approximation to the momentum equation
∇P = ∇ · τ
(703)
If the viscosity and density are constant this is equivalent to
∇P = µ∇2 u
∇P = −µ∇ × ω
or
(704)
Applying the divergence and curl operations to these equations yields
∇2 P = 0
or
∇2 ω = 0
(705)
The Reynolds number enters solely through the boundary conditions. Consider a flow with characteristic
velocity U , lateral dimension L and viscosity µ. If the velocity is specified at the boundaries,
u = U g(x/L, geometry)
(706)
then the pressure distribution can be obtained by integrating the momentum equation to get
P =
ρU 2
f (x/L, geometry)
ReL
ReL =
ρU L
µ
(707)
If the pressure is specified at the boundaries,
P = ρU 2 f (x/L, geometry)
(708)
u = U ReL g(x/L, geometry)
(709)
then the velocity will be given by
For flows in two space dimensions, a streamfunction ψ can be used to satisfy the continuity equation.
In cartesian coordinates, the streamfunction for Stokes flow of a constant viscosity fluid will satisfy the
Biharmonic equation
∇4 ψ = 0
(710)
Stokes Sphere Flow The force on a moving body in viscous flow is
Z
Z
F=
τ · n̂ dA −
P n̂ dA
∂Ω
(711)
∂Ω
Estimating the magnitude of the integrals, the force in a particular direction will have the magnitude
F = CµU L
(712)
The constant C will in general depend on the shape of the body, the direction x̂ of the force and the motion
of the body.
For a sphere, the flow can be solved by using Stokes axisymmetric streamfunction ψ. The velocity
components are:
ur
uφ
The analog of the biharmonic equation is
∂ψ
1
sin φ ∂φ
1 ∂ψ
= −
r sin φ ∂r
=
r2
(713)
(714)
5
VISCOUS FLOW
67
sin φ ∂
∂2
+ 2
∂r2
r ∂φ
1 ∂
sin φ ∂φ
2
ψ=0
(715)
The boundary conditions at the surface of the sphere are:
ψ=0
∂ψ
=0
∂r
∂ψ
=0
∂φ
r=a
(716)
and the flow approaches a uniform flow far from the sphere
U r2
sin2 φ
2
lim ψ =
r→∞
(717)
The solution is
U
ψ = a2 sin2 φ
4
a 3r 2r2
−
+ 2
r
a
a
(718)
The pressure on the body is found by integrating the momentum equation
3µaU
cos φ
2r2
and the force (drag) is directed opposite to the direction of motion of the sphere with magnitude
P = P∞ −
D = 6πµU a
CD ≡
D
24
=
2
2
1/2ρU πa
Re
Re =
ρU 2a
µ
(719)
(720)
For a thin disk of radius a moving normal to the freestream the drag is
D = 16πµU a
(721)
32
µU a
3
(722)
and moving parallel to the freestream
D=
Oseen’s Approximation The inertial terms neglected in Stokes’ approximation become significant at a
distance r ∼ a/Re. The Oseen equations are a uniform approximation for incompressible viscous flow over
a body. If the mean flow at large distances from body is U in direction x, then the Oseen equations are:
∇·u = 0
∂u
∂u
ρ
+ ρU
= −∇P + µ∇2 u
∂t
∂x
This results in a corrected drag law (the flow now has a wake) for the sphere
24
3Re
9
2
CD =
1+
+
Re ln Re + . . .
Re
16
160
(723)
(724)
(725)
Reynolds Lubrication Theory Incompressible flow in a two-dimensional channel with a slowly-varying
height h(x) and length L can be treated as a “boundary layer”-like flow if
L ∂h
∂h
1
which implies that
v≈u
(726)
h ∂x
∂x
The thin-layer or lubrication equations result when the channel is very thin h/L → 0, and viscous forces
dominate inertia Re 1.
5
VISCOUS FLOW
68
∂ρh ∂ρhu
+
∂t
∂x
0
0
=
0
∂ ∂u
∂P
+
µ
∂x
∂y ∂y
∂P
= −
∂y
= −
(727)
(728)
(729)
For a constant property flow, the velocity is given at any point in the channel by the Couette-Poiseuille
expression of parallel flow if the lower boundary is moving with velocity U and the upper boundary is at
most moving in the y direction
u=−
y
y
h2 ∂P y 1−
+U 1−
2µ ∂x h
h
h
Combining this result with the continuity equation yields the Reynolds lubrication equation
1 ∂
∂h
∂h
3 ∂P
+ 12
h
= 6U
µ ∂x
∂x
∂x
∂t
(730)
(731)
For a slipper pad bearing, the pressure is equal to the ambient value P◦ at x = 0 and x = L and the gap
height h is
x
h = h◦ 1 − α
α1
(732)
L
The pressure is given by
h∗ h2◦
h◦
µU L
−1 −3
−1
(733)
6
P − P◦ =
αh2◦
h
h◦ h2
where h∗ is the gap height at the location of the pressure maximum
h∗
α α2
1−α
≈1− −
+ O(α3 )
=2
h◦
2−α
2
4
(734)
and the maximum pressure is approximately
Pmax − P◦ ≈
3 µU L
α
+ O(α2 )
4 h2◦
(735)
A
A
FAMOUS NUMBERS
69
Famous Numbers
Fundamental Physical Constants
co
o
µo
h
k
No
e
amu
me
mp
G
σ
speed of light in a vacuum
permittivity of the vacuum
permeability of the vacuum
Planck constant
Boltzmann constant
charge on electron
atomic mass unit
electron mass
proton mass
universal gravitational constant
Stefan-Boltzmann constant
2.998×108
8.854×10−12
4π×10−7
6.626×10−34
1.381×10−23
6.022×1023
1.602×10−19
1.661×10−27
9.109×10−31
1.673×10−27
6.673×10−11
5.670×10−8
m/s
C2 /kg-m
H/m
J-s
J/K
molecules/mol
C
kg
kg
kg
m3 /kg-s2
W/m2 K4
Astronautics
go
RE
ME
MS
au
gravitational acceleration at earth’s surface
mass of earth
mass of sun
mean earth-sun distance
mass of moon
mean earth-moon distance
9.807
6378
5.976×1024
1.99×1030
1.496×108
7.349×1022
3.844×105
m/s2
km
kg
kg
km
kg
km
8314.5
8.3145
82.06
1.9872
4.186
22.41
2.46×1025
4.3×109
74
J/kmol-K
J/mol-K
cm3 -atm/mol-K
cal/mol-K
J/cal
m3
m3
s−1
nm
Ideal Gas Stuff
R̃
Universal gas constant
mechanical equivalent of heat
volume of 1 kmol at 273.15 K and 1 atm
number of molecules at 298.15 K and 1 atm
collision frequency at 273.15 K and 1 atm
mean free path in N2 at 273.15 K and 1 atm
A
FAMOUS NUMBERS
70
Our Atmosphere
composition (mol fractions)
0.7809
0.2095
0.0093
0.0003
N2
O2
Ar
CO2
Sea level
P
ρ
T
c
R
W
µ
k
cp
1.01325×105
1.225
288.15
340.29
287.05
28.96
1.79×10−5
2.54×10−3
1.0
pressure
density
temperature
sound speed
gas constant
molar mass
viscosity (absolute)
thermal conductivity
heat capacity
Pa
kg/m3
K
m/s
m2 /s2 -K
kg/kmol
kg/m-s
W/m-K
kJ/kg-K
30 kft
P
ρ
T
c
3.014×104
0.458
228.7
303.2
pressure
density
temperature
sound speed
Based on the U.S. Standard Atmosphere, 1976 NOAA-S/T 76-1562, 1976.
Unit Conversions
1m
0.3048 ft
1 lb (force)
1 lb (mass)
1 btu
1 hp
1 hp
1 mile (land)
1 mph
1 mile (nautical)
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
3.28 ft
1m
4.452 N
0.454 kg
1055 J
745.7 W
550 ft-lbf /s
1.609 km
0.447 m/s
1.852 km
Pa
kg/m3
K
m/s
B
B
BOOKS ON FLUID MECHANICS
71
Books on Fluid Mechanics
There are a very large number of books on fluid mechanics. The following are a selected texts that cover
various topics. The call numbers (LOC system) are those used in the Caltech library system.
General Fluid Dynamics
1. Tritton, D. J. Physical Fluid Dynamics QC151.T74 1988 (available in paperback)
2. Panton, R. L. Incompressible Flow TA357.P29 1995.
3. Batchelor, G. K. An Introduction to Fluid Dynamics QA911.B33 1973 (available in paperback)
4. Landau, L. D. and Lifshitz, E. M. Fluid Mechanics QA901.L283 1989
Compressible Flow
1. Anderson, J. D. Modern Compressible Flow: with historical perspective, QA911.A6 1990.
2. Liepmann, H. A. and Roshko, A. Elements of Gasdynamics, QC168.L5 (available now as a Dover
paperback)
3. Thompson, P. A. Compressible Fluid Dynamics QC168.T5 1984. (Available from RPI bookstore
- see JES web page for information on ordering and errata for older editions)
Viscous Flow and Boundary Layers
1. Sherman, F. Viscous Flow TA357.S453 1990
2. Schlichting, H. Boundary Layer Theory TL574.B6 S283 1968
3. White, F.M. Viscous Fluid Flow 2nd Ed. 1991 QA929.W48
Acoustics
1. Pierce, A.D. Acoustics: an introduction to its physical principles and applications QC225.15
P52 1989
Aerodynamics
1. Kuethe, A.M. and Chow, C.-Y. Foundations of Aerodynamics TL570.K76 1998
2. Karamcheti, K. Principles of Ideal-Fluid Aerodynamics QA930.K74
Rotating Flow
1. Vanyo, J. P. Rotating Fluids in Engineering and Science TA357.V36 2000 (Dover paperback)
Thermodynamics
1. D. Kondepudi and I. PrigogineModern Thermodynamics, Wiley, 1998.
2. M. Abbott and H. van Ness Thermodynamics Schaum’s Outline Series. TJ265.A19 1989
3. Adkins, C. J. Equilibrium Thermodynamics QC318.T47 A34 1983
Some Data Tables
1. NACA 1135, Equations, Tables and Charts for Compressible Flow
2. Reynolds, W.C. Thermodynamic Properties in SI QC311.3 .R38
B
BOOKS ON FLUID MECHANICS
72
3. Poling, B. E., Prausnitz, J. M., and O’Connell, J. P. The Properties of Gases and Liquids.
TP243.P62 2000.
Vector Analysis
1. Aris, R. Vectors, Tensors, and the Basic Equations of Fluid Mechanics QA911.A69 1989
2. Spiegel, M.R. Vector Analysis Schaum’s Outline Series Not in CIT libraries
Dimensional Analysis
1. Hornung, H. G. Dimensional Analysis: Examples of the use of Symmetry, Dover, 2006.
2. Bridgman, P. W. Dimensional Analysis QC39.B7 1922.
3. Sedov, L. I. Similarity and Dimensional Methods in Mechanics QC20.7 .D55 S42
Visualization
1. van Dyke, M. An Album of Fluid Motion TA357.V35
2. Lugt, H. Vortex Flow in Nature and Technology QC159.L8413 1983.
3. Pozrikidis, C. Little Book of Streamlines QA911.P66 1999
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