Fluid Mechanics 101

A Skeleton Guide

J. E. Shepherd Aeronautics and Mechanical Engineering California Institute of Technology Pasadena, CA USA 91125

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 1.1 Control Volume Statements ...... 1 1.2 Reynolds Transport Theorem ...... 2 1.3 Integral Equations ...... 2 1.3.1 Simple Control Volumes ...... 3 1.3.2 Steady Momentum Balance ...... 3 1.4 Vector Calculus ...... 3 1.4.1 Vector Identities ...... 3 1.4.2 Curvilinear Coordinates ...... 3 1.4.3 Gauss’ Divergence Theorem ...... 4 1.4.4 Stokes’ Theorem ...... 4 1.4.5 Div, Grad and Curl ...... 4 1.4.6 Specific Coordinates ...... 5 1.5 Differential Relations ...... 5 1.5.1 Conservation form ...... 5 1.6 Convective Form ...... 6 1.7 Divergence of Viscous Stress ...... 7 1.8 Euler Equations ...... 7 1.9 Bernoulli Equation ...... 7 1.10 Vorticity ...... 8 1.11 Dimensional Analysis ...... 9

2 Thermodynamics 11 2.1 Thermodynamic potentials and fundamental relations ...... 11 2.2 Maxwell relations ...... 11 2.3 Various defined quantities ...... 12 2.4 v(P, s) relation ...... 13 2.5 Equation of State Construction ...... 13

3 Compressible Flow 15 3.1 Steady Flow ...... 15 3.1.1 Streamlines and Total Properties ...... 15 3.2 Quasi-One Dimensional Flow ...... 15 3.2.1 Isentropic Flow ...... 16 3.3 Heat and Friction ...... 18 3.3.1 Fanno Flow ...... 18 3.3.2 Rayleigh Flow ...... 18 3.4 Shock Jump Conditions ...... 19 3.4.1 Lab frame (moving shock) versions ...... 19 3.5 Perfect Gas Results ...... 20 3.6 Reflected Shock Waves ...... 20 3.7 Detonation Waves ...... 22 3.8 Perfect-Gas, 2-γ Model ...... 22 3.8.1 2-γ Solution ...... 23 3.8.2 High-Explosives ...... 23 3.9 Weak shock waves ...... 25 3.10 Acoustics ...... 26 3.11 Multipole Expansion ...... 27 3.12 Baffled (surface) source ...... 28 3.13 1-D Unsteady Flow ...... 29 3.14 2-D Steady Flow ...... 31 3.14.1 Oblique Shock Waves ...... 31

i 3.14.2 Weak Oblique Waves ...... 31 3.14.3 Prandtl-Meyer Expansion ...... 32 3.14.4 Inviscid Flow ...... 32 3.14.5 Potential Flow ...... 32 3.14.6 Natural Coordinates ...... 33 3.14.7 Method of Characteristics ...... 33

4 Incompressible, Inviscid Flow 34 4.1 Velocity Field Decomposition ...... 34 4.2 Solutions of Laplace’s Equation ...... 34 4.3 Boundary Conditions ...... 35 4.4 Streamfunction ...... 35 4.4.1 2-D Cartesian Flows ...... 36 4.4.2 Cylindrical Polar Coordinates ...... 37 4.4.3 Spherical Polar Coordinates ...... 38 4.5 Simple Flows ...... 38 4.6 Vorticity ...... 40 4.7 Key Ideas about Vorticity ...... 41 4.8 Unsteady Potential Flow ...... 42 4.9 Complex Variable Methods ...... 42 4.9.1 Mapping Methods ...... 44 4.10 Airfoil Theory ...... 44 4.11 Thin-Wing Theory ...... 45 4.11.1 Thickness Solution ...... 46 4.11.2 Camber Case ...... 48 4.12 Axisymmetric Slender Bodies ...... 49 4.13 Wing Theory ...... 50

5 Viscous Flow 52 5.1 Scaling ...... 52 5.2 Two-Dimensional Flow ...... 53 5.3 Parallel Flow ...... 53 5.3.1 Steady Flows ...... 54 5.3.2 Poiseuille Flow ...... 56 5.3.3 Rayleigh Problem ...... 57 5.4 Boundary Layers ...... 57 5.4.1 Blasius Flow ...... 59 5.4.2 Falkner-Skan Flow ...... 59 5.5 K´arm´anIntegral Relations ...... 60 5.6 Thwaites’ Method ...... 60 5.7 Laminar Separation ...... 61 5.8 Compressible Boundary Layers ...... 61 5.8.1 Transformations and Approximations ...... 61 5.8.2 Energy Equation ...... 62 5.8.3 Moving Shock Waves ...... 64 5.8.4 Weak Shock Wave Structure ...... 64 5.9 Creeping Flow ...... 66

A Famous Numbers 69

B Books on Fluid Mechanics 71

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: d Z ρ dV = 0 (1) dt Ω Momentum conservation: d Z ρu dV = F (2) dt Ω Forces: Z Z F = ρG dV + T dA (3) Ω ∂Ω Surface traction forces T = −P nˆ + τ · nˆ = T · nˆ (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  τ = 2µ D − δ D + µ δ D implicit sum on j (7) ik ik 3 ik jj v ik jj

Deformation tensor   1 ∂ui ∂uk 1 T
Dik = + or ∇u + ∇u (8) 2 ∂xk ∂xi 2 Energy conservation: d Z  |u|2  ρ e + dV = Q˙ + W˙ (9) dt Ω 2 Work: Z Z W˙ = ρG · u dV + T · u dA (10) Ω ∂Ω Heat: Z Q˙ = − q · nˆ dA (11) ∂Ω

heat flux q, thermal conductivity k and thermal radiation qr

q = −k∇T + qr (12)

Entropy inequality (2nd Law of Thermodynamics):

d Z Z q · nˆ ρs dV ≥ − dA (13) dt Ω ∂Ω T 1 FUNDAMENTALS 2

1.2 Reynolds Transport Theorem The multi-dimensional analog of Leibniz’s theorem: d Z Z ∂φ Z φ(x, t) dV = dV + φuV · nˆ dA (14) dt V (t) V (t) ∂t ∂V The transport theorem proper. Material volume Ω, arbitrary volume V . d Z d Z Z φ dV = φ dV + φ(u − uV ) · nˆ dA (15) dt Ω dt V ∂V

1.3 Integral Equations The equations of motions can be rewritten with Reynolds Transport Theorem to apply to an (almost) arbi- trary moving control volume. Beware of noninertial reference frames and the apparent forces or accelerations that such systems will introduce. Moving control volume: d Z Z ρdV + ρ (u − uV ) · nˆ dA = 0 (16) dt V ∂V d Z Z Z Z ρudV + ρu (u − uV ) · nˆ dA = ρG dV + T dA (17) dt V ∂V V ∂V

d Z  |u|2  Z  |u|2  ρ e + dV + ρ e + (u − uV ) · nˆ dA = dt V 2 ∂V 2 Z Z Z ρG · u dV + T · u dA − q · nˆ dA (18) V ∂V ∂V

d Z Z Z q ρsdV + ρs (u − uV ) · nˆ dA + · nˆ dA ≥ 0 (19) dt V ∂V ∂V T Stationary control volume: d Z Z ρdV + ρu · nˆ dA = 0 (20) dt V ∂V d Z Z Z Z ρudV + ρuu · nˆ dA = ρG dV + T dA (21) dt V ∂V V ∂V

d Z  |u|2  Z  |u|2  ρ e + dV + ρ e + u · nˆ dA = dt V 2 ∂V 2 Z Z Z ρG · u dV + T · u dA − q · nˆ dA (22) V ∂V ∂V

d Z Z Z q ρsdV + ρsu · nˆ dA + · nˆ dA ≥ 0 (23) dt V ∂V ∂V T 1 FUNDAMENTALS 3

1.3.1 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

I J dM d Z X X = ρdV = A m˙ − A m˙ (24) dt dt i i j j V i=1 j=1

where Ai is the cross-sectional area of the ith connection andm ˙ i = ρiui is the mass flow rate per unit area through this connection. The energy equation for this same situation is

Z  2  I  2  dE d |u| X |ui| = ρ e + + gz dV = A m˙ h + + gz dt dt 2 i i i 2 i V i=1 J  2  X |uj| − A m˙ h + + gz + Q˙ + W˙ (25) j j j 2 j j=1

where Q˙ is the thermal energy (heat) transferred into the control volume and W˙ is the mechanical work done on the fluid inside the control volume.

1.3.2 Steady Momentum Balance For a stationary control volume, the steady momentum equation can be written as Z Z Z Z ρuu · nˆ dA + P nˆ dA = ρG dV + τ · nˆ dA + Fext (26) ∂V ∂V V ∂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 Vector Calculus 1.4.1 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 (27) ∇(A · B) = (B · ∇)A + (A · ∇)B + B × (∇ × A) + A × (∇ × B) (28) ∇ × (∇φ) = 0 (29) ∇ · (∇ × A) = 0 (30) ∇ × (∇ × A) = ∇(∇ · A) − ∇2A (31) ∇ × (φA) = ∇φ × A + φ∇ × A (32)

1.4.2 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)

The scale factors hi are defined by

dr = h1dx1e1 + h2dx2e2 + h3dx3e3 (35) or

∂r hi ≡ (36) ∂xi The unit of arc length in this coordinate system is ds2 = dr · dr:

2 2 2 2 2 2 2 ds = h1 dx1 + h2 dx2 + h3 dx3 (37) The unit of differential volume is

dV = h1h2h3 dx1 dx2 dx3 (38)

1.4.3 Gauss’ Divergence Theorem For a vector or tensor field F, the following relationship holds: Z Z ∇ · F dV ≡ F · nˆ dA (39) V ∂V This leads to the simple interpretation of the divergence as the following limit 1 Z ∇ · F ≡ lim F · nˆ dA (40) V →0 V ∂V A useful variation on the divergence theorem is Z Z (∇ × F) dV ≡ nˆ × F dA (41) V ∂V This leads to the simple interpretation of the curl as 1 Z ∇ × F ≡ lim nˆ × F dA (42) V →0 V ∂V

1.4.4 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) · nˆ dA ≡ F · dr (43) S ∂S

1.4.5 Div, Grad and Curl The gradient operator ∇ or grad for a scalar field ψ is 1 ∂ψ 1 ∂ψ 1 ∂ψ ∇ψ = e1 + e2 + e3 (44) h1 ∂x1 h2 ∂x2 h3 ∂x3 A simple interpretation of the gradient operator is in terms of the differential of a function in a direction aˆ

daˆψ = lim ψ(x + da) − ψ(x) = ∇ψ · da (45) da→0 1 FUNDAMENTALS 5

The divergence operator ∇· or div is

1  ∂ ∂ ∂  ∇ · F = (h2h3F1) + (h3h1F2) + (h1h2F3) (46) h1h2h3 ∂x1 ∂x2 ∂x3 The curl operator ∇× or curl is

h e h e h e 1 1 1 2 2 3 3 ∇ × F = ∂ ∂ ∂ (47) ∂x1 ∂x2 ∂x3 h1h2h3 h1F1 h2F2 h3F3 The components of the curl are:

  e1 ∂ ∂ ∇ × F = (h3F3) − (h2F2) h2h3 ∂x2 ∂x3   e2 ∂ ∂ + (h1F1) − (h3F3) h3h1 ∂x3 ∂x1   e3 ∂ ∂ + (h2F2) − (h1F1) (48) h1h2 ∂x1 ∂x2

The Laplacian operator ∇2 for a scalar field ψ is

1  ∂ h h ∂ψ ∂ h h ∂ψ ∂ h h ∂ψ  ∇2ψ = ( 2 3 ) + ( 3 1 ) + ( 1 2 ) (49) h1h2h3 ∂x1 h1 ∂x1 ∂x2 h2 ∂x2 ∂x3 h3 ∂x3

1.4.6 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

Spherical (r, φ, θ) r sin φ cos θ r sin φ sin θ r cos φ 1 r r sin φ

Parabolic Cylindrical √ 1 2 2 2 2 (u, v, z) 2 (u − v ) uv z u + v h1 1

Paraboloidal √ 1 2 2 2 2 (u, v, φ) uv cos φ uv sin φ 2 (u − v ) u + v h1 uv

Elliptic Cylindrical p 2 2 (u, v, z) a cosh u cos v a sinh u sin v z a sinh u + sin v h1 1

Prolate Spheroidal p 2 2 (ξ, η, φ) a sinh ξ sin η cos φ a sinh ξ sin η sin φ a cosh ξ cos η a sinh ξ + sin η h1 a sinh ξ sin η

1.5 Differential Relations 1.5.1 Conservation form The equations are first written in conservation form ∂ density + ∇ · flux = source (50) ∂t 1 FUNDAMENTALS 6 for a fixed (Eulerian) control volume in an inertial reference frame by using the divergence theorem.

∂ρ + ∇ · (ρu) = 0 (51) ∂t ∂ (ρu) + ∇ · (ρuu − T) = ρG (52) ∂t ∂  |u|2    |u|2   ρ e + + ∇ · ρu e + − T · u + q = ρG · u (53) ∂t 2 2 ∂  q  (ρs) + ∇ · ρus + ≥ 0 (54) ∂t T 1.6 Convective Form This form uses the convective or material derivative D ∂ = + u · ∇ (55) Dt ∂t

Dρ = −ρ∇ · u (56) Dt Du ρ = −∇P + ∇ · τ + ρG (57) Dt D  |u|2  ρ e + = ∇ · (T · u) − ∇ · q + ρG · u (58) Dt 2 Ds  q  ρ ≥ −∇ · (59) Dt T Alternate forms of the energy equation:

D  |u|2  ρ e + = −∇ · (P u) + ∇ · (τ · u) − ∇ · q + ρG · u (60) Dt 2 Formulation using enthalpy h = e + P/ρ

D  |u|2  ∂P ρ h + = + ∇ · (τ · u) − ∇ · q + ρG · u (61) Dt 2 ∂t

Mechanical energy equation

D |u|2 ρ = − (u · ∇) P + u · ∇ · τ + ρG · u (62) Dt 2 Thermal energy equation De Dv = −P + vτ :∇u − v∇ · q (63) Dt Dt Dissipation

∂ui Υ = τ :∇u = τik sum on i and k (64) ∂xk Entropy

Ds  q  Υ ∇T 2 ρ = −∇ · + + k (65) Dt T T T 1 FUNDAMENTALS 7

1.7 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  ∇ · τ = µ∇2u + µ + µ ∇(∇ · u) (66) v 3

1.8 Euler Equations Inviscid, no heat transfer, no body forces.

Dρ = −ρ∇ · u (67) Dt Du ρ = −∇P (68) Dt D  |u|2  ∂P ρ h + = (69) Dt 2 ∂t Ds ≥ 0 (70) Dt 1.9 Bernoulli Equation Consider the unsteady energy equation in the form

D  |u|2  ∂P ρ h + = + ∇ · (τ · u) − ∇ · q + ρG · u (71) Dt 2 ∂t 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

D  |u|2  ∂P ρ h + + Φ = + ∇ · (τ · u) − ∇ · q (73) Dt 2 ∂t

The Bernoulli constant is |u|2 H = h + + Φ (74) 2 In the absence of unsteadiness, viscous forces and heat transfer we have

 |u|2  u · ∇ h + + Φ = 0 (75) 2

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

|u|2 P + ρ + ρ gz = constant (76) ◦ 2 ◦ 1 FUNDAMENTALS 8

1.10 Vorticity Vorticity is defined as ω ≡ ∇ × u (77) and the vector identities can be used to obtain |u|2 (u · ∇)u = ∇( ) − u × (∇ × u) (78) 2 The momentum equation can be reformulated to read:

 |u|2  ∂u ∇ · τ ∇H = ∇ h + + Φ = − + u × ω + T ∇s + (79) 2 ∂t ρ 1 FUNDAMENTALS 9

1.11 Dimensional Analysis Fundamental Dimensions L length meter (m) M mass kilogram (kg) T time second (s) θ temperature Kelvin (K) I current Ampere (A) Some derived dimensional units force Newton (N) MLT −2 pressure Pascal (Pa) ML−1T −2 bar = 105 Pa energy Joule (J) ML2T −2 frequency Hertz (Hz) T −1 power Watt (W) ML2T −3 viscosity (µ) Poise (P) ML−1T −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

Xi Πi = α1 α2 αr X1 X2 ··· Xr that can be formed is p ≥ n − f

Conventional Dimensionless Numbers Reynolds Re ρUL/µ Mach Ma U/c Prandtl P r µcP /k = ν/κ Strouhal St L/UT Knudsen Kn Λ/L Peclet P e UL/κ Schmidt Sc ν/D Lewis Le 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

Air Water shear viscosity µ (kg/ms) 1.8×10−5 1.00×10−3 kinematic viscosity ν (m2/s) 1.5×10−5 1.0×10−6 thermal conductivity k (W/mK) 2.54×10−2 0.589 thermal diffusivity κ (m2/s) 2.1×10−5 1.4×10−7 specific heat cp (J/kgK) 1004. 4182. sound speed c (m/s) 343.3 1484 density ρ (kg/m3) 1.2 998. gas constant R (m2/s2K) 287 462. thermal expansion α (K−1) 3.3×10−4 2.1×10−4 −1 −6 −10 isentropic compressibility κs (Pa ) 7.01×10 4.5×10

Prandtl number P r .72 7.1 Fundamental derivative Γ 1.205 4.4 ratio of specific heats γ 1.4 1.007 Gr¨uneisen coefficient G 0.40 0.11 2 THERMODYNAMICS 11

2 Thermodynamics 2.1 Thermodynamic potentials and fundamental relations

energy e(s, v) de = T ds − P dv (80) enthalpy h(s, P ) = e + P v dh = T ds + v dP (81) Helmholtz f(T, v) = e − T s df = −s dT − P dv (82) Gibbs g(T,P ) = e − T s + P v dg = −s dT + v dP (83)

2.2 Maxwell relations

∂T  ∂P  = − (84) ∂v s ∂s v ∂T  ∂v  = (85) ∂P s ∂s P ∂s  ∂P  = (86) ∂v T ∂T v ∂s  ∂v  = − (87) ∂P T ∂T P Calculus identities:

∂F  ∂F  F (x, y, . . . ) dF = dx + dy + ... (88) ∂x y,z,... ∂y x,z,...

∂f  ∂x ∂y = − x (89) ∂f  ∂y f ∂x y ∂x  1 = (90) ∂f  ∂f y ∂x y 2 THERMODYNAMICS 12

2.3 Various defined quantities

∂e  specific heat at constant volume cv ≡ (91) ∂T v ∂h  specific heat at constant pressure cp ≡ (92) ∂T P c ratio of specific heats γ ≡ p (93) cv s ∂P  sound speed c ≡ (94) ∂ρ s 1 ∂v  coefficient of thermal expansion α ≡ (95) v ∂T P 1 ∂v  isothermal compressibility KT ≡ − (96) v ∂P T 1 ∂v  1 isentropic compressibility Ks ≡ − = 2 (97) v ∂P s ρc Specific heat relationships

∂P  ∂P  KT = γKs or = γ (98) ∂v s ∂v T ∂P   ∂v 2 cp − cv = −T (99) ∂v T ∂T P Fundamental derivative

c4 ∂2v  Γ ≡ 3 2 (100) 2v ∂P s v3 ∂2P  = 2 2 (101) 2c ∂v s  ∂c  = 1 + ρc (102) ∂P s 1 v2 ∂2h  = 2 2 + 1 (103) 2 c ∂v s Sound speed (squared)

∂P  c2 ≡ (104) ∂ρ s ∂P  = −v2 (105) ∂v s v = (106) Ks v = γ (107) Kt Gr¨uneisen Coefficient 2 THERMODYNAMICS 13

vα G ≡ (108) cvKT ∂P  = v (109) ∂e v vα = (110) cpKs v ∂T  = − (111) T ∂v s

2.4 v(P, s) relation

dv 2 T ds = −Ks dP + Γ(Ks dP ) + α + ... (112) v cp dP  dP 2 T ds = − + Γ + G + ... (113) ρc2 ρc2 c2

2.5 Equation of State Construction

Given cv(v, T ) and P (v, T ), integrate  ∂P   de = cv dT + T − P dv (114) ∂T v c ∂P  ds = v dT + dv (115) T ∂T v along two paths: I: variable T , fixed ρ and II: variable ρ, fixed T . Energy: ! Z T Z ρ ∂P  dρ e = e◦ + cv(T, ρ◦) dT + P − T 2 (116) T◦ ρ◦ ∂T ρ ρ | {z } | {z } I II

Ideal gas limit ρ◦ → 0,

ig lim cv(T, ρ◦) = cv (T ) (117) ρ◦→0 The ideal gas limit of I is the ideal gas internal energy

Z T ig ig e (T ) = cv (T ) dT (118) T◦ Ideal gas limit of II is the residual function

Z ρ  ! r ∂P dρ e (ρ, T ) = P − T 2 (119) 0 ∂T ρ ρ and the complete expression for internal energy is

ig r e(ρ, T ) = e◦ + e (T ) + e (ρ, T ) (120) Entropy: Z T Z ρ  ! cv(T, ρ◦) ∂P dρ s = s◦ + dT + − 2 (121) T◦ T ρ◦ ∂T ρ ρ | {z } | {z } I II 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

Z T cig(T ) Z ρ dρ sig = v dT − R (122) T◦ T ρ◦ ρ 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 sr(ρ, T ) = R − (123) 0 ρ ∂T ρ ρ and the complete expression for entropy is

ig r s(ρ, T ) = s◦ + s (ρ, T ) + s (ρ, T ) (124) 3 COMPRESSIBLE FLOW 15

3 Compressible Flow 3.1 Steady 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 ∆ρ = c−2∆P ∼ − ρM 2 (125) s 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 Dρ u · ∇u2 1 1 ∂u2 1 ∂P  = −∇ · 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 X = x when t = 0 (127) dt ◦ which in Cartesian coordinates yields dx dx dx 1 = 2 = 3 (128) u1 u2 u3 Total enthalpy is constant along streamlines in adiabatic, steady, inviscid flow

|u|2 h = h + = constant (129) t 2 Velocity along a streamline is given by the energy equation: p u = |u| = 2(ht − h) (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 Adiabatic, frictionless flow:

d(ρuA) = 0 (132) ρudu = −dP (133) u2 h + = constant or dh = −udu (134) 2 ds ≥ 0 (135) 3 COMPRESSIBLE FLOW 16

3.2.1 Isentropic Flow If ds = 0, then

(dρ)2 dP = c2dρ + c2 (Γ − 1) + ... (136) ρ For isentropic flow, the quasi-one-dimensional equations can be written in terms of the Mach number as:

1 dρ M 2 1 dA = (137) ρ dx 1 − M 2 A dx 1 dP M 2 1 dA = (138) ρc2 dx 1 − M 2 A dx 1 du 1 1 dA = − (139) u dx 1 − M 2 A dx 1 dM 1 + (Γ − 1)M 2 1 dA = − (140) M dx 1 − M 2 A dx 1 dh M 2 1 dA = (141) c2 dx 1 − M 2 A dx At a throat, the gradient in Mach number is:

dM 2 Γ d2A = (142) dx 2A dx2

Constant-Γ Gas If the value of Γ is assumed to be constant, the quasi-one dimensional equations can be integrated to yield:

ρ 1 t = 1 + (Γ − 1)M 2
2(Γ−1) (143) ρ Γ−1 c ρ  1/2 t = t = 1 + (Γ − 1)M 2
(144) c ρ h t = 1 + (Γ − 1)M 2
(145) h  M 2 1/2 u = c (146) t 1 + (Γ − 1)M 2 Γ A 1 1 + (Γ − 1)M 2  2(Γ−1) = (147) A∗ M Γ

 2Γ−1  P − Pt 1 2
− 2(Γ−1) 2 = 1 + (Γ − 1)M − 1 (148) ρtct 2Γ − 1 (149)

Ideal Gas For an ideal gas P = ρRT and e = e(T ) only. In that case, we have

Z T Z T cP (T ) h(T ) = e + RT = h◦ + cv(T ) dT, s = s◦ + dT − R ln(P/P◦) (150) T◦ T◦ T and you can show that Γ is given by: γ + 1 γ − 1 T dγ Γig = + (151) 2 2 γ dT 3 COMPRESSIBLE FLOW 17

Perfect or Constant-γ Gas Perfect gas results for isentropic flow can be derived from the equation of state γR P = ρRT h = c T c = (152) p p γ − 1 the value of Γ for a perfect gas, γ + 1 Γpg = (153) 2 the energy integral,  γ − 1  T = T 1 + M 2 (154) t 2 and the expression for entropy T s − so = cp ln − R ln P/Po (155) To or T s − so = cv ln − R ln ρ/ρo To

T γ − 1 t = 1 + M 2 (156) T 2 γ P T  γ−1 t = t (157) P T 1 ρ T  γ−1 t = t (158) ρ T

Mach Number–Area Relationship

γ+1 A 1  2  γ − 1  2(γ−1) = 1 + M 2 (159) A∗ M γ + 1 2 Choked flow mass flux

γ+1  2  2(γ−1) M˙ = c ρ A∗ (160) γ + 1 t t or

γ+1 √  2  2(γ−1) P M˙ = γ √ t A∗ γ + 1 RTt Velocity-Mach number relationship M u = ct q (161) γ−1 2 1 + 2 M Alternative reference speeds

rγ + 1 rγ + 1 c = c∗ u = c∗ (162) t 2 max γ − 1 3 COMPRESSIBLE FLOW 18

3.3 Heat and Friction Constant-area, steady flow with friction F and heat addition Q

ρu =m ˙ = constant (163) ρudu + dP = −F dx (164) dh + udu = Qdx (165) 1  F  ds = Q + dx (166) T ρ

F is the frictional stress per unit length of the duct. In terms of the Fanning friction factor f 2 F = fρu2 (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 isq ˙, then we have q˙ 4 Q = (168) ρu D

3.3.1 Fanno Flow Constant-area, adiabatic, steady flow with friction only:

ρu =m ˙ = constant (169) ρudu + dP = −F dx (170) u2 h + = h = constant (171) 2 t (172)

2 2 Change in entropy with volume along Fanno line, h + 1/2m ˙ v =ht

ds  c2 − u2 T = (173) dv F anno v(1 + G)

3.3.2 Rayleigh Flow Constant-area, steady flow with heat transfer only:

ρu =m ˙ = constant (174) P + ρu2 = I (175) dh + udu = Qdx (176) (177)

Change in entropy with volume along Rayleigh line, P +m ˙ 2v = I

ds  c2 − u2 T = (178) dv Rayleigh vG 3 COMPRESSIBLE FLOW 19

3.4 Shock Jump Conditions The basic jump conditions,

ρ1w1 = ρ2w2 (179) 2 2 P1 + ρ1w1 = P2 + ρ2w2 (180) w2 w2 h + 1 = h + 2 (181) 1 2 2 2 s2 ≥ s1 (182) or defining [f] ≡ f2 - f1

[ρw] = 0 (183) P + ρw2 = 0 (184)  w2  h + = 0 (185) 2 [s] ≥ 0 (186) The Rayleigh line:

P2 − P1 2 2 = −(ρ1w1) = −(ρ2w2) (187) v2 − v1 or [P ] = −(ρw)2 (188) [v] Rankine-Hugoniot relation:

h2 − h1 = (P2 − P1)(v2 + v1)/2 or e2 − e1 = (P2 + P1)(v1 − v2)/2 (189) Velocity-P v relation

2 p [w] = −[P ][v] or w2 − w1 = − −(P2 − P1)(v2 − v1) (190) Alternate relations useful for numerical solution

  2 ρ1 P2 = P1 + ρ1w1 1 − (191) ρ2 "  2# 1 2 ρ1 h2 = h1 + w1 1 − (192) 2 ρ2

3.4.1 Lab frame (moving shock) versions Shock velocity

w1 = Us (193) Particle (fluid) velocity in laboratory frame

w2 = Us − up (194) Jump conditions

ρ2 (Us − up) = ρ1Us (195)

P2 = P1 + ρ1Usup (196)

h2 = h1 + up (Us − up/2) (197) 3 COMPRESSIBLE FLOW 20

Kinetic energy:

u2 1 p = (P − P )(v − v ) 2 2 2 1 1 2 3.5 Perfect Gas Results

[P ] 2γ 2
= M1 − 1 (198) P1 γ + 1 [w] 2  1  = − M1 − (199) c1 γ + 1 M1 [v] 2  1  = − 1 − 2 (200) v1 γ + 1 M1 [s] P = − ln t2 (201) R Pt1 γ   γ−1 γ + 1 2 P 1 M1 t2 =  2  (202) 1  γ − 1  Pt1  2γ γ − 1 γ−1 2 M 2 − 1 + M1 γ + 1 1 γ + 1 2

Shock adiabat or Hugoniot: γ + 1 v − 2 P2 γ − 1 v = 1 (203) P γ + 1 v2 1 − 1 γ − 1 v1 Some alternatives

P2 2γ 2
= 1 + M1 − 1 (204) P1 γ + 1 2γ γ − 1 = M 2 − (205) γ + 1 1 γ + 1 ρ2 γ + 1 = 2 (206) ρ1 γ − 1 + 2/M1 2 M 2 + 1 γ − 1 M 2 = (207) 2 2γ M 2 − 1 γ − 1 1 Prandtl’s relation

∗2 w1w2 = c (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 h , h = h + (209) γ + 1 t t 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 = ρ (210) 3 − 1 ρ2

Pressure P3 behind reflected shock:

2 ρ3u2 P3 = P2 + ρ (211) 3 − 1 ρ2

Enthalpy h3 behind reflected shock:

ρ3 2 + 1 u2 ρ2 h3 = h2 + ρ (212) 2 3 − 1 ρ2 Perfect gas result for incident shock waves: P (3γ − 1) 2 − (γ − 1) P3 P = 1 (213) P P2 2 (γ − 1) + (γ + 1) P1 3 COMPRESSIBLE FLOW 22

3.7 Detonation Waves Jump conditions:

ρ1w1 = ρ2w2 (214) 2 2 P1 + ρ1w1 = P2 + ρ2w2 (215) w2 w2 h + 1 = h + 2 (216) 1 2 2 2 s2 ≥ s1 (217)

3.8 Perfect-Gas, 2-γ Model Perfect gas with energy release q, different values of γ and R in reactants and products.

h1 = cp1T (218)

h2 = cp2T − q (219)

P1 = ρ1R1T1 (220)

P2 = ρ2R2T2 (221) γ1R1 cp1 = (222) γ1 − 1 γ2R2 cp2 = (223) γ2 − 1 (224)

Substitute into the jump conditions to yield:

2 P2 1 + γ1M1 = 2 (225) P1 1 + γ2M2 2 2 v2 γ2M2 1 + γ1M1 = 2 2 (226) v1 γ1M1 1 + γ2M2 1 1 q + M 2 + T γ R γ − 1 2 1 c2 2 = 1 1 1 1 (227) T1 γ2R2 1 1 2 + M2 γ2 − 1 2

Chapman-Jouguet Conditions Isentrope, Hugoniot and Rayleigh lines are all tangent at the CJ point

P − P ∂P  ∂P  CJ 1 = = (228) vCJ − V1 ∂v Hugoniot ∂v s which implies that the product velocity is sonic relative to the wave

w2,CJ = c2 (229) Entropy variation along adiabat 1 ds = (v − v)2 dm ˙ 2 (230) 2T 1 3 COMPRESSIBLE FLOW 23

Jouguet’s Rule " # w2 − c2  G  ∂P  ∆P 2 = 1 − (v1 − v) − (231) v 2v ∂v Hug ∆v where G is the Gr´uniesen 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 + (232) 2γ1(γ1 − 1) 2γ1(γ1 − 1)

where the parameter H is the nondimensional energy release

(γ − 1)(γ + 1)q H = 2 2 (233) 2γ1R1T1 CJ pressure P γ M 2 + 1 CJ = 1 CJ (234) P1 γ2 + 1 CJ density 2 ρCJ γ1(γ2 + 1)MCJ = 2 (235) ρ1 γ2(1 + γ1MCJ ) CJ temperature T P R ρ CJ = CJ 1 1 (236) T1 P1 R2ρCJ

Strong detonation approximation MCJ  1

q 2 UCJ ≈ 2(γ2 − 1)q (237)

γ2 + 1 ρCJ ≈ ρ1 (238) γ2 1 2 PCJ ≈ ρ1UCJ (239) γ2 + 1 (240)

3.8.2 High-Explosives For high-explosives, the same jump conditions apply but the ideal gas equation of state is no longer appro- priate 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 vk = 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, γs,CJ vCJ = v1 (244) γs,CJ + 1 and from momentum conservation, with PCJ  P1, we have

2 ρ1UCJ PCJ = (245) γs,CJ + 1 3 COMPRESSIBLE FLOW 25

3.9 Weak shock waves Nondimensional pressure jump [P ] Π = (246) ρc2 A useful version of the jump conditions (exact):

[w] 2 [v] [w] [v] Π = −M1 = −M1 = M1 (247) c1 v1 c1 v1 Thermodynamic expansions:

[v] = −Π + ΓΠ2 + O(Π)3 (248) v1 [v] [v]2 Π = − + Γ + O ([v])3 (249) v1 v1 Linearized jump conditions:

[w] Γ − = Π − Π2 + O(Π)3 (250) c1 2 Γ [w] [w]2 M1 = 1 − + O (251) 2 c1 c1 Γ M = 1 + Π + O(Π)2 (252) 1 2 Γ M = 1 − Π + O(Π)2 (253) 2 2 [c] = (Γ − 1)Π + O(Π)2 (254) c1 M1 − 1 ≈ 1 − M2 (255)

Prandtl’s relation 1 1 c∗ ≈ w + [w] or ≈ w − [w] (256) 1 2 2 2 Change in entropy for weak waves:

 3 T [s] 1 3 1 [v] 2 = ΓΠ + ... or = − Γ + ... (257) c1 6 6 v 3 COMPRESSIBLE FLOW 26

3.10 Acoustics Simple waves ∆P = c2∆ρ (258) ∆P = ±ρc∆u (259) + for right-moving waves, - for left-moving waves Acoustic Potential φ

u = ∇φ (260) ∂φ P 0 = −ρ (261) o ∂t 0 ρo ∂φ ρ = − 2 (262) co ∂t Potential Equation

2 2 1 ∂ φ ∇ φ − 2 2 = 0 (263) co ∂t d’Alembert’s solution for planar (1D) waves

φ = f(x − cot) + g(x + cot) (264)

Acoustic Impedance The specific acoustic impedance of a medium is defined as

P 0 z = (265) |u| 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)

2z1 ut/ui = (266) z2 + z1 0 0 2z2 Pt /Pi = (267) z2 + z1 Harmonic waves (planar) ω 2π 2π φ = A exp i(wt − kx) + B exp i(wt + kx) c = k = ω = = 2πf (268) k λ T Spherical waves

f(t − r/c) g(t + r/c) φ = + (269) r r Spherical source strength Q,[Q] = L3T −1

2 Q(t) = lim 4πr ur (270) r→0 potential function

Q(t − r/c) φ(r, t) = − (271) 4πr 3 COMPRESSIBLE FLOW 27

Energy flux

Φ = P 0u (272) Acoustic intensity for harmonic waves

0 1 Z T P 2 I =< Φ >= Φ dt = rms (273) T 0 ρc Decibel scale of acoustic intensity

−12 2 dB = 10 log10(I/Iref ) Iref = 10 W/m (274) or

0 0 0 −10 dB = 20 log10(Prms/Pref ) Pref = 2 × 10 atm (275) Cylindrical waves, q source strength per unit length [q] = L2T −1

1 Z t−r/c q(η) dη φ(r, t) = − (276) p 2 2 2 2π −∞ (t − η) − r /c or

1 Z ∞ φ(r, t) = − q(t − r/c cosh ξ) dξ (277) 2π 0

3.11 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| (278) 4π Vs R Harmonic source q = f(x) exp(−iωt) Potential function 1 Z exp i(kR − ωt) φ(x, t) = − f(xs) dVs (279) 4π Vs R 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: ∞ n   exp ikR X (−xs · ∇x) exp ikr = (280) R n! r n=0 Series expansion of potential φ = φ0 + φ1 + φ2 + ... (281) Monopole term Z 0 exp i(kr − ωt) φ (x, t) = − f(xs)dVs (282) 4πr Vs 3 COMPRESSIBLE FLOW 28

Dipole term ikx · D  i  φ1(x, t) = 1 + exp i(kr − ωt) (283) 4πr2 kr Dipole moment vector D Z D = xsf(xs)dVs (284) Vs Quadrupole term k2  3i 3  X φ2(x, t) = 1 + − exp i(kr − ωt) x x Q (285) 4πr3 kr k2r2 i j ij i,j

Quadrupole moments Qik 1 Z Qij = xs,ixs,jf(xs)dVs (286) 2 Vs

3.12 Baffled (surface) source Rayleigh’s formula for the potential

1 Z u (x , t − R/c) φ = − n s dA (287) 2π R Normal component of the source surface velocity

un = u · nˆ (288)

Harmonic source

un = f(x) exp(−iwt)

Fraunhofer conditions |xs| ≤ a a a  1 λ r Approximate solution: exp i(kr − wt) Z φ = − f(xs) exp iκ · xsdA 2πr As 3 COMPRESSIBLE FLOW 29

3.13 1-D Unsteady Flow The primitive variable version of the equations is:

∂ρ + ∇ · (ρu) = 0 (289) ∂t ∂ρu + ∇ · (ρuu) = −∇P (290) ∂t ∂  u2   u2  ρ e + + ∇ · ρu(h + ) = 0 (291) ∂t 2 2 ∂s + ∇ · (us) ≥ 0 (292) ∂t (293) Alternative version

1 Dρ = −∇ · u (294) ρ Dt Du ρ = −∇P (295) Dt D  u2  ∂P ρ h + = (296) Dt 2 ∂t Ds ≥ 0 (297) Dt The characteristic version of the equations for isentropic flow (s = constant) is: d dx (u ± F ) = 0 on C± : = u ± c (298) dt dt This is equivalent to: ∂ ∂ (u ± F ) + (u ± c) (u ± F ) = 0 (299) ∂t ∂x Riemann invariants: Z c Z d P Z d c F = d ρ = = (300) ρ ρc Γ − 1 Bending of characteristics: d Γ (u + c) = (301) dP ρc For an ideal gas: 2c F = (302) γ − 1 Pressure-velocity relationship for expansion waves moving to the right into state (1), final state (2) with velocity u2 < 0.

2γ   γ−1 P2 γ − 1 u2 −2c1 = 1 + < u2 ≤ 0 (303) 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 c1 = 1 + 1 +  u2 > 0 (304) P1 4 c1 γ + 1 u2 3 COMPRESSIBLE FLOW 30

Shock Tube Performance

−2γ4    γ −1   P4 c1 γ4 − 1 1 4 2γ1 2
= 1 − Ms − 1 + Ms − 1 (305) P1 c4 γ + 1 Ms γ1 + 1

Limiting shock Mach number for P4/P1 → ∞

c4 γ1 + 1 Ms → (306) c1 γ4 − 1 3 COMPRESSIBLE FLOW 31

3.14 2-D Steady Flow 3.14.1 Oblique Shock Waves Geometry:

w1 = u1 sin β (307)

w2 = u2 sin(β − θ) (308)

v = u1 cos β = u2 cos(β − θ) (309)

ρ w tan β 2 = 1 = (310) ρ1 w2 tan(β − θ) Shock Polar

[w] M tan θ − = 1 (311) c1 cos β(1 + tan β tan θ) 2 [P ] M1 tan θ 2 = (312) ρ1c1 cot β + tan θ Real fluid results

w2 = f(w1) normal shock jump conditions (313) −1 β = sin (w1/u1) (314) ! w θ = β − tan−1 2 (315) p 2 2 u1 − w1 Perfect gas result

2 cot β M 2 sin2 β − 1
tan θ = 1 (316) 2 2 2
(γ + 1)M1 − 2 M1 sin β − 1 Mach angle 1 µ = sin−1 (317) M

3.14.2 Weak Oblique Waves Results are all for C+ family of waves, take θ → -θ for C− family.

Γ 1 [w] [w]2 β = µ − 1 + O (318) 2 p 2 c c M1 − 1 1 1 p 2  2 M1 − 1 [w] [w] θ = − 2 + O (319) M1 c1 c1 [P ] M 2 = 1 θ + O(θ)2 (320) ρ c2 p 2 1 1 M1 − 1 T [s] Γ M 6 1 = 1 1 θ3 + O(θ)4 (321) 2 2 3/2 c1 6 (M1 − 1) Perfect Gas Results

[P ] γM 2 = 1 θ + O(θ)2 (322) P p 2 1 M1 − 1 3 COMPRESSIBLE FLOW 32

3.14.3 Prandtl-Meyer Expansion q 2 d u1 d θ = − M1 − 1 (323) u1 Function ω, d θ = -d ω √ M 2 − 1 d M d ω ≡ (324) 1 + (Γ − 1)M 2 M Perfect gas result

rγ + 1 rγ − 1  p ω(M) = tan−1 (M 2 − 1) − tan−1 M 2 − 1 (325) γ − 1 γ + 1 Maximum turning angle

π rγ + 1  ω = − 1 (326) max 2 γ − 1

3.14.4 Inviscid Flow Crocco-Vaszonyi Relation

∂u u2 + (∇ × u) × u = T ∇S − ∇(h + ) (327) ∂t 2

3.14.5 Potential Flow Steady, homoentropic, homoenthalpic, inviscid:

∇ · (ρu) = 0 (328) ∇ × u = 0 (329) u2 h + = constant (330) 2

or with u = ∇φ = (φx, φy)

2 2 2 2 (φx − c )φxx + (φy − c )φyy + 2φxφyφxy = 0 (331) Linearized potential flow:

0 u = U∞ + φx (332) 0 v = φy (333) 2
0 0 0 = M∞ − 1 φxx − φyy (334) Wave equation solution

p 2 0 λ = M∞ − 1 φ = f(x − λy) + g(x + λy) (335) Boundary conditions for slender 2-D (Cartesian) bodies y(x)

U dy  f 0(ξ) = − ∞ y ≥ 0 (336) λ dx ξ Prandtl-Glauert Scaling for subsonic flows

inc p 2 2 inc φ(x, y) = φ (x, 1 − M∞y) ∇ φ = 0 (337) 3 COMPRESSIBLE FLOW 33

Prandtl-Glauert Rule

inc Cp Cp = (338) p 2 1 − M∞

3.14.6 Natural Coordinates

∂ ∂ ∂ = cos θ − sin θ (339) ∂x ∂s ∂n ∂ ∂ ∂ = sin θ + cos θ (340) ∂y ∂s ∂n u = U cos θ (341) v = U sin θ (342) The transformed equations of motion are:

∂ρU ∂θ + ρU = 0 (343) ∂s ∂n ∂U ∂P ρU + = 0 (344) ∂s ∂s ∂θ ∂P ρU 2 + = 0 (345) ∂s ∂n ∂θ ∂U ω = U − = 0 (346) z ∂s ∂n Curvature of stream lines, R = radius of curvature ∂θ 1 = (347) ∂s R Vorticity production

1 ∂Po (T − To) ∂S ωz = − + (348) Uρo ∂n U ∂n Elimination of pressure dP = c2dρ

∂U ∂θ (M 2 − 1) − U = 0 (349) ∂s ∂n ∂U ∂θ − U = 0 (350) ∂n ∂s

3.14.7 Method of Characteristics

∂ 1 ∂ (ω − θ) + √ (ω − θ) = 0 (351) ∂s M 2 − 1 ∂n ∂ 1 ∂ (ω + θ) − √ (ω + θ) = 0 (352) ∂s M 2 − 1 ∂n (353) Characteristic directions dn 1 C± = ±√ = ± tan µ (354) ds M 2 − 1 Invariants

J ± = θ ∓ ω = constant on C± (355) 4 INCOMPRESSIBLE, INVISCID FLOW 34

4 Incompressible, Inviscid Flow 4.1 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 = 0 (357)

∇ · ue = e(x, t) volume source distribution (358)

This is satisfied by deriving ue from a velocity potential φ

ue = ∇φ (359) ∇2φ = e(x, t) (360)

Rotational Flow Define the rotational part of the flow by:

∇ · uv = 0 (361)

∇ × uv = ω(x, t) vorticity source distribution (362)

This is satisfied by deriving uv from a vector potential B

uv = ∇ × B (363) ∇ · B = 0 choice of gauge (364) ∇2B = −ω(x, t) (365)

4.2 Solutions of Laplace’s Equation The equation ∇2φ = 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) Ωξ For a infinite domain, Green’s function is the solution to

∇2G = δ(x − ξ) (367) 1 1 1 G = − = − (368) 4π |x − ξ| 4πr r = |r| r = x − ξ (369)

This leads to the following solutions for the potentials

1 Z e(ξ, t) φ(x, t) = − dV (370) 4π r ξ Ωξ 1 Z ω(ξ, t) B(x, t) = dV (371) 4π r ξ Ωξ 4 INCOMPRESSIBLE, INVISCID FLOW 35

The velocity fields are

1 Z re(ξ, t) u (x, t) = dV (372) e 4π r3 ξ Ωξ 1 Z r × ω(ξ, t) u (x, t) = − dV (373) v 4π r3 ξ Ωξ

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

0 u = ue + uv + u (374)

4.3 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 nˆ normal to the boundary surface.

u · nˆ = u∂Ω · nˆ on the surface ∂Ω (375) In particular, if the surface is stationary, the normal component of velocity must vanish on the surface

u · nˆ = 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 · ˆt 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 · nˆ = u2 · nˆ = u∂Ω · nˆ (379) and the interface has to be in mechanical equilibrium (in the absence of surface forces such as interfacial tension)

P1 = P2 (380)

4.4 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 36

4.4.1 2-D Cartesian Flows Compressible In a steady two-dimensional compressible flow: ∂ρu ∂ρv ∇ · ρu = 0 u = (u, v) x = (x, y) + = 0 (381) ∂x ∂y The streamfunction is: 1 ∂ψ 1 ∂ψ u = v = − (382) ρ ∂y ρ ∂x

Incompressible The density ρ is a constant ∂u ∂v ∇ · u = 0 u = (u, v) x = (x, y) + = 0 (383) ∂x ∂y The streamfunction defined by ∂ψ ∂ψ u = v = − (384) ∂y ∂x will identically satisfy the continuity equation as long as

∂2ψ ∂2ψ − = 0 (385) ∂x∂y ∂y∂x

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 ∇ψ nˆ = (386) ψ |∇ψ|

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 = u · nˆLdl = −v dx + u dy (387) Z Z dψ = ψ2 − ψ1 = u · nˆLdl = volume flux across L (388) L 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 ∂ψ ∂ψ B = (B ,B ,B ) = (0, 0, ψ) u = ∇ × B = xˆ − yˆ (389) x y z ∂y ∂x and the equation that the streamfunction has to satisfy will be

∂2ψ ∂2ψ ∇2ψ = + = −ω (390) ∂x2 ∂y2 z where the z-component of vorticity is ∂v ∂u ω = − (391) z ∂x ∂y

A special case of this is irrotational flow with ωz = 0. 4 INCOMPRESSIBLE, INVISCID FLOW 37

4.4.2 Cylindrical Polar Coordinates

In cylindrical polar coordinates (r, θ, z) with u = (ur, uθ, uz)

x = r cos θ (392) y = r sin θ (393) z = z (394)

u = ur cos θ − uθ sin θ (395)

v = ur sin θ + uθ cos θ (396)

w = uz (397) The continuity equation is 1 ∂ru 1 ∂u ∂u ∇ · u = 0 = r + θ + z (398) r ∂r r ∂θ ∂z

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 u = (400) r r ∂θ ∂ψ u = − (401) θ ∂r The only nonzero component of vorticity is 1 ∂ru 1 ∂u ω = θ − r (402) z r ∂r r ∂θ and the stream function satisfies

1 ∂  ∂ψ  1 ∂ 1 ∂ψ  r + = −ω (403) r ∂r ∂r r ∂θ r ∂θ z

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 u = − (405) r r ∂z 1 ∂ψ u = (406) z r ∂r The only nonzero vorticity component is ∂u ∂u ω = r − z (407) θ ∂z ∂r The stream function satisfies

∂ 1 ∂ψ  ∂ 1 ∂ψ  + = −ω (408) ∂z r ∂z ∂r r ∂r θ 4 INCOMPRESSIBLE, INVISCID FLOW 38

4.4.3 Spherical Polar Coordinates This coordinate system (r, φ, θ) results in the continuity equation 1 ∂ 1 ∂u 1 ∂ r2u
+ θ + (u sin φ) = 0 (409) r2 ∂r r r sin φ ∂θ r sin φ ∂φ φ 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  B = 0, 0, (410) 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:

1 ∂ψ u = (411) r r2 sin φ ∂φ 1 ∂ψ u = − (412) φ r sin φ ∂r The only non-zero vorticity component is: 1 ∂ru 1 ∂u ω = φ − r (413) θ r ∂r r ∂φ The stream function satisfies

1 ∂  1 ∂ψ  1 ∂  1 ∂ψ  + = −ω (414) r ∂r sin φ ∂r r ∂φ r2 sin φ ∂φ θ

4.5 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

∇2φ = 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 = Uxˆ, the streamfunction is

ψ = Uy (417) In spherical polar coordinates, Stokes streamfunction is

Ur2 ψ = sin2 φ U = Uzˆ (418) 2 4 INCOMPRESSIBLE, INVISCID FLOW 39

Source Distributions Single source of strength Q(t) located at point ξ1. The meaning of Q is the volume of fluid per unit time introduced or removed at point ξ1.

2 lim 4πr1u · ˆr1 = Q(t) r1 = x − ξ1 e = Q(t)δ(x − ξ1) (419) r1→0 which leads to the solution:

Q(t) r1Q(t) ˆr1Q(t) φ = − u = 3 = 2 (420) 4πr1 4πr1 4πr1 For multiple sources, add the individual solutions

k 1 X riQi u = − (421) 4π r3 i=1 i In spherical polar cordinates, Stokes’ stream function for a single source of strength Q at the origin is Q ψ = − 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. q q q u = u ˆr u = φ = ln r ψ = θ (423) r r 2πr 2π 2π

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 dˆ, then the dipole moment vector can be defined as

d = µdˆ (425) The dipole potential for spherical (3-D) sources is d · r φ = − (426) 4πr3 and the resulting velocity field is

1 3d · r d  u = r − (427) 4π r5 r3 If the dipole is aligned with the z-axis, Stokes’ stream function is

µ sin2 φ ψ = (428) 4πr and the velocity components are

µ cos φ u = (429) r 2πr3 µ sin φ u = (430) φ 4πr3 The dipole potential for 2-D source-sink pairs is µ cos θ φ = − (431) 2π r 4 INCOMPRESSIBLE, INVISCID FLOW 40

and the stream function is µ sin θ ψ = (432) 2π r The velocity components are

µ cos θ u = (433) r 2π r2 µ sin θ u = (434) θ 2π r2

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: source + uniform flow sphere: dipole (3-D) + uniform flow cylinder: dipole (2-D) + uniform flow closed-body: sources & sinks + uniform flow

4.6 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:

Dω ∇ · τ  = (ω · ∇)u − ω(∇ · u) + ∇T × ∇s + ∇ × (436) Dt ρ The cross products of the thermodynamic derivatives can be written as ∇P × ∇ρ ∇T × ∇s = ∇P × ∇v = − (437) ρ2 which is known as the baroclinic torque. For incompressible, homogeneous flow, the viscous term can be written ν∇2ω and the incompressible vorticity transport equation for a homogeneous fluid is Dω = (ω · ∇)u + ν∇2ω (438) Dt

Circulation The circulation Γ is defined as I Z Γ = u · dl = ω · nˆ dA (439) ∂Ω 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. dx dy dz = = (440) ωx ωy ωz The collection of vortex lines passing through a simple curve C form a vortex tube. On the surface of the vortex tube, we have nˆ · 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 Γ r × dl du = − (441) 4π r3

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 ω = zˆΓδ(r) u = (442) θ 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)

Γi y − yi u = − 2 2 (443) 2π (x − xi) + (y − yi) Γi x − xi v = 2 2 (444) 2π (x − xi) + (y − yi) (445)

Or setting Γ = zˆΓ

Γi × ri ui = 2 (446) 2πri

where ri = i - xi. The streamfunction for the line vortex is found by integration to be Γ ψ = − i ln r (447) i 2π i 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:

n X Γi × ri u = (448) 2πr2 i=1 i

4.7 Key Ideas about Vorticity 1. Vorticity can be visualized as local rotation within the fluid. The local angular frequency of rotation about the direction nˆ is

uθ 1 |ω · nˆ| fnˆ = lim = r→0 2πr 2π 2 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 · nˆ dA = constant tube However, the circulation can change with time due to viscous forces, baroclinic torque or nonconser- vative 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 I dP Z ∇P × ∇ρ ˆ = − = − 2 · ndA (450) Dt ∂Ω ρ Ω ρ Viscous fluids have an additional contribution due to the diffusion of vorticity into or out of the tube.

4.8 Unsteady Potential Flow Bernoulli’s equation for unsteady potential flow

∂ U 2 |∇φ|2 P − P = −ρ (φ − φ ) + ρ ∞ − ρ (451) ∞ ∂t ∞ 2 2

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 F = (m + M·) (452) ext dt where M is the induced mass tensor. For a sphere (3-D) or a cylinder (2-D), the induced mass is simply M = miI.

2 m = πa3ρ (453) i,sphere 3 2 mi,cylinder = πa ρ (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 φ = − (456) r dt Integration of the momentum equation in spherical coordinates yields the Rayleigh equation

d2R 3 dR2 P (R) − P R + = ∞ (457) dt2 2 dt ρ

4.9 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 (458) and the complex velocity w is defined as 4 INCOMPRESSIBLE, INVISCID FLOW 43

dF w = u − iv = (459) dz NB sign of v-term! The complex potential is an analytic function and the derivatives satisfy the Cauchy- Riemann conditions

∂φ ∂ψ = (460) ∂x ∂y ∂φ ∂ψ = − (461) ∂y ∂x which implies that both ∇2φ = 0 and ∇2ψ = 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 q F = ln(z − z ) 2π ◦

3. Line vortex of strength Γ located at z◦

Γ F = −i ln(z − z ) 2π ◦

4. Source doublet (dipole) at z◦ oriented along +x axis µ F = − 2π(z − z◦)

5. Vortex doublet at z◦ oriented along +x axis

iλ F = 2π(z − z◦) 6. Stagnation point

F = Cz2

7. Exterior corner flow

F = Czn 1/2 ≤ n ≤ 1

8. Interior corner flow, angle α

π F = Czn 1 ≤ n = α 9. Circular cylinder at origin, radius a, uniform flow U at x = ±∞

a2 F = U(z + ) z 4 INCOMPRESSIBLE, INVISCID FLOW 44

4.9.1 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 isw ˜ and is related to the z-plane velocity by dF w dF w˜ w˜ = = or w = = (462) dζ dζ dz dz dz dζ In order to have well behaved values of w, requirew ˜ =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 (463) 2 body 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

1 I  M = − ρ< zw2 dz (465) 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

 a2 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 Γ sin θs = (467) 4πU∞a The lift L is given by

L = −ρU∞Γ (468) The pressure coefficient on the surface of the cylinder is

P − P 4Γ  Γ 2 C = ∞ = 1 − 4 sin2 θ + sin θ − (469) P 1 2 2 ρU∞ 2πaU∞ 2πaU∞

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

ζ2 z = ζ + T (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 z 2 ζ = ± − ζ2 (472) 2 2 T

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 L a C = = 8π sin(α + β) (474) L 1 2 2 ρU∞c c 4.11 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 = U∞ cos α + ut + uc (475)

v = U∞ sin α + vt + vc (476)

ut = ∇φt (477)

uc = ∇φc (478) (479)

2 where α is the angle of attack and ∇ 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+(x) for upper surface 0 ≤ x ≤ c (480)

y = Y−(x) for lower surface 0 ≤ x ≤ c (481) and can be represented by a thickness function f(x) and a camber function g(x).

f(x) = Y+(x) − Y−(x) (482) 1 g(x) = [Y (x) + Y (x)] (483) 2 + − 4 INCOMPRESSIBLE, INVISCID FLOW 46

The profiles of the upper and lower surface can be expressed in terms of f and g as

1 Y (x) = g(x) + f(x) upper surface (484) + 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 t h α ∼ ∼  1 and u , v << U (486) c c i i ∞

Boundary Conditions The exact slip boundary condition for inviscid flow on the body is: dY v = (487) dx u (x,Y (x)) The linearized version of this is:

dY± vt + vc = α + lim (488) y→±0 dx U∞ (x,y) with cos α ≈ 1, and sin α ≈ α. This can be written as

 1  v (x, 0+) + v (x, 0+) = U g0 + f 0 − αU (489) t c ∞ 2 ∞  1  v (x, 0−) + v (x, 0−) = U g0 − f 0 − αU (490) t c ∞ 2 ∞ where f 0 = df/dx and g0 = dg/dx. The boundary conditions are then divided between the thickness and camber disturbance flows as follows:

1 v = ± U f 0 for y → ±0 (491) t 2 ∞ 0 vc = U∞(g − α) for y → ±0 (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 2 2 φt(x, y) = ln(y + (x − ξ) )q(ξ) dξ (493) 2π 0 The velocity components are:

1 Z c (x − ξ)q(ξ) dξ ut = 2 2 (494) 2π 0 y + (x − ξ) 1 Z c yq(ξ) dξ vt = 2 2 (495) 2π 0 y + (x − ξ) Apply the linearized boundary condition to obtain

1 df 1 Z c yq(ξ) dξ ± U∞ = lim 2 2 (496) 2 dx y→±0 2π 0 y + (x − ξ) 4 INCOMPRESSIBLE, INVISCID FLOW 47

Delta Function Representation The limit of the integrand is one of the representations of the Dirac delta function 1 y lim = ±δ(x − ξ) (497) y→±0 π y2 + (x − ξ)2 where  0 x 6= ξ Z +∞ δ(x − ξ) = f(ξ)δ(x − ξ) dξ = f(x) (498) ∞ x = ξ −∞

Source Distribution This leads to the source distribution df q(x) = U (499) ∞ dx and the solution for the velocity field is

Z c 0 ∂φt U∞ (x − ξ)f (ξ) dξ ut = = 2 2 (500) ∂x 2π 0 y + (x − ξ) Z c 0 ∂φt U∞ yf (ξ) dξ vt = = 2 2 (501) ∂y 2π 0 y + (x − ξ) The velocity components satisfy the following relationships across the surface of the wing

[u] = u(x, 0+) − u(x, 0−) = 0 (502) [v] = v(x, 0+) − v(x, 0−) = q(x) (503)

Pressure Coefficient The pressure coefficient is defined to be

P − P u2 + v2 C = ∞ = 1 − (504) P 1 2 2 2 ρU∞ U∞ The linearized version of this is:

ut + uc CP ≈ −2 (505) U∞ For the pure thickness case, then we have the following result:

1 Z c f 0(ξ) dξ CP ≈ − (506) π 0 (x − ξ) 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 a Z x− Z a  P g(ξ) dξ = lim g(ξ) dξ + g(ξ) dξ (507) 0 →0 0 x+ Important principal value integrals are

Z c dξ  x  P = ln (508) 0 (x − ξ) x − c and √ √ Z c ξ−1/2dξ 1  c + x P = √ ln √ √ (509) 0 (x − ξ) x c − x 4 INCOMPRESSIBLE, INVISCID FLOW 48

A generalization to other powers can be obtained by the recursion relation

Z c ξndξ Z c ξn−1dξ cn P = xP − (510) 0 (x − ξ) 0 (x − ξ) n A special case can be found for the transformed variables cos θ = 1 - 2ξ/c

Z π cos nθdθ sin nθ P = π o (511) 0 cos θ − cos θo sin θo

4.11.2 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   1 −1 y φc = γ(ξ) tan dξ (512) 2π 0 x − ξ The velocity components are:

Z c ∂φc 1 yγ(ξ) dξ uc = = − 2 2 (513) ∂x 2π 0 y + (x − ξ) Z c ∂φc 1 (x − ξ)γ(ξ) dξ vc = = 2 2 (514) ∂y 2π 0 y + (x − ξ) The u component of velocity on the surface of the wing is

γ(x) lim uc(x, y) = u(x, ±0) = ∓ (515) y→±0 2 Apply the linearized boundary condition to obtain the following integral equation for the vorticity distribution γ

 dg  1 Z c γ(ξ) dξ U∞ − α = P dξ (516) dx 2π 0 (x − ξ) The total circulation Γ is given by Z c Γ = 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) (518) [v] = v(x, 0+) − v(x, 0−) = 0 (519)

Kutta Condition 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: " # c − x1/2 1 Z c g0(ξ)  ξ 1/2 γ(x) = −2U∞ α + P dξ (521) x π 0 x − ξ c − ξ The pressure coefficient for the pure camber case is

γ(x) CP = ± for y → ±0 (522) U∞ The integrals can be computed exactly for several special cases, which can be expressed most conveniently using the transformation 2x 2ξ z = − 1 ρ = − 1 (523) c c

Z 1 1 r1 + ρ P dρ = −π (524) −1 z − ρ 1 − ρ

Z 1 p1 − ρ2 P dρ = πz (525) −1 z − ρ Z 1 1 P dρ = 0 (526) p 2 −1 1 − ρ (z − ρ) Z 1 ρ P dρ = −π (527) p 2 −1 1 − ρ (z − ρ) Z 1 ρ2 P dρ = −πz (528) p 2 −1 1 − ρ (z − ρ) Higher powers of the numerator can be evaluated from the recursion relation:

Z 1 ρn Z 1 ρn−1 π 1(3) ··· (n − 2) P dρ = zP dρ − [1 − (−1)n] (529) p 2 p 2 −1 1 − ρ (z − ρ) −1 1 − ρ (z − ρ) 2 2(4) ··· (n − 1)

4.12 Axisymmetric Slender Bodies Disturbance potential solution using source distribution on x-axis:

1 Z c f(ξ) dξ φ(x, r) = − (530) p 2 2 4π 0 (x − ξ) + r Velocity components:

∂φ 1 Z c (x − ξ)f(ξ) dξ u = U + = (531) ∞ 3/2 ∂x 4π 0 [(x − ξ)2 + r2] ∂φ 1 Z c rf(ξ) dξ v = = (532) 3/2 ∂r 4π 0 [(x − ξ)2 + r2] (533)

Exact boundary condition on body R(x) v dR = (534) u (x,R(x)) dx 4 INCOMPRESSIBLE, INVISCID FLOW 50

Linearized boundary condition, first approximation: dR v(x, r = R) = U (535) ∞ dx Extrapolation to x axis: dR lim(2πrv) = 2πR U∞ (536) r→0 dx Source strength dR f(x) = U 2πR = U A0(x) A(x) = πR2(x) (537) ∞ dx ∞ Pressure coefficient

2u dR2 CP ≈ − − (538) U∞ dx

4.13 Wing Theory Wing span is −b/2 < y < +b/2. The section lift coefficient, L0 = lift per unit span

L0 C0 (y) = = m (y)(α − α − α (y)) (539) L 1 2 ◦ i ◦ 2 ρU∞c(y) Induced angle of attack, w = downwash velocity   −1 w w αi = tan ≈ (540) U∞ U∞ Induced drag

Di = ρU∞Γαi (541) Load distribution Γ(y), bound circulation at span location y 1 Γ(y) = m U c(y)(α − α − α (y)) (542) 2 ◦ ∞ i ◦ Trailing vortex sheet strength dΓ γ = − (543) dy Downwash velocity

1 Z +b/2 γ(ξ) dξ w = P (544) 4pi −b/2 ξ − y Integral equation for load distribution " # 1 1 Z +b/2 Γ0(ξ) dξ Γ(y) = m◦(y)U∞c(y) α − α◦(y) − P (545) 2 4piU∞ −b/2 ξ − y Boundary conditions b b Γ( ) = Γ(− ) = 0 (546) 2 2 Elliptic load distribution, constant downwash, induced angle of attack 4 INCOMPRESSIBLE, INVISCID FLOW 51

 21/2  y  Γs Γs Γ(y) = Γs 1 − w = αi = (547) 2b 2b 2U∞ Lift

πb2 L = ρU Γ (548) ∞ s 4 Induced drag (minimum for elliptic loading)

1 L2 D = (549) i 1 2 2 π 2 ρU∞b Induced drag coefficient

C2 b C = L AR = b2/S ≈ (550) D,i πAR c 5 VISCOUS FLOW 52

5 Viscous Flow

Equations of motion in cartesian tensor form (without body forces) are: Conservation of mass: ∂ρ ∂ρu + k = 0 (551) ∂t ∂xk Momentum equation:

∂ui ∂ui ∂P ∂τik ρ + ρuk = − + (i = 1, 2, 3) (552) ∂t ∂xk ∂xi ∂xk Viscous stress tensor   ∂ui ∂uk ∂uj τik = µ + + λδik sum on j (553) ∂xk ∂xi ∂xj Lam´e’s constant 2 λ = µ − µ (554) v 3 Energy equation, total enthalpy form:

∂ht ∂ht ∂P ∂τkiui ∂qi ρ + ρuk = + − sum on i and k (555) ∂t ∂xk ∂t ∂xk ∂xi Thermal energy form

∂h ∂h ∂P ∂P ∂ui ∂qi ρ + ρuk = + uk + τik − sum on i and k (556) ∂t ∂xk ∂t ∂xk ∂xk ∂xi or alternatively

∂e ∂e ∂uk ∂ui ∂qi ρ + ρuk = −P + τik − sum on i and k (557) ∂t ∂xk ∂xk ∂xk ∂xi Dissipation function

∂ui Υ = τik (558) ∂xk Fourier’s law ∂T qi = −k (559) ∂xi

5.1 Scaling Reference conditions are

velocity U◦ length L time T density ρ◦ viscosity µ◦ thermal conductivity k◦ 5 VISCOUS FLOW 53

Inertial flow Limit of vanishing viscosity, µ → 0. Nondimensional statement:

ρ◦U◦L 2 Reynolds number Re =  1 P ∼ ρ◦U◦ (560) µ◦ Nondimensional momentum equation Du 1 ρ = −∇P + ∇ · τ (561) Dt Re Limiting case, Re → ∞, inviscid flow Du ρ = −∇P (562) Dt

Viscous flow Limit of vanishing density, ρ → 0. Nondimensional statement: ρ U L µ U Reynolds number Re = ◦ ◦  1 P ∼ ◦ ◦ (563) µ◦ L Nondimensional momentum equation Du Reρ = −∇P + ∇ · τ (564) Dt Limiting case, Re → 0, creeping flow.

∇P = ∇ · τ (565)

5.2 Two-Dimensional Flow For a viscous flow in two-space dimensions (x, y) the components of the viscous stress tensor in cartersian coordinates are         2µ ∂u + λ ∂u + ∂v µ ∂u + ∂v τxx τxy ∂x ∂x ∂y ∂y ∂x τ = =       (566) τyx τyy ∂u ∂v ∂v ∂u ∂v µ ∂y + ∂x 2µ ∂y + λ ∂x + ∂y Dissipation function " # ∂u2 ∂v 2 ∂u ∂v 2 ∂u ∂v 2 Υ = µ 2 + 2 + + + λ + (567) ∂x ∂y ∂y ∂x ∂x ∂y

5.3 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

∂u ∂P ∂  ∂u ρ = − + µ (569) ∂t ∂x ∂y ∂y ∂P 0 = − (570) ∂y We conclude from the y-momentum equation that P = P (x) only. Energy equation

∂e ∂e ∂u2 ∂  ∂T  ∂  ∂T  ρ + ρu = µ + k + k (571) ∂t ∂x ∂y ∂x ∂x ∂y ∂y 5 VISCOUS FLOW 54

5.3.1 Steady Flows ∂ In these flows ∂t = 0 and inertia plays no role. Shear stress is either constant or varies only due to imposed axial pressure gradients.

Couette Flow A special case are flows in which pressure gradients are absent ∂P = 0 (572) ∂x ∂ and the properties strictly depend only on the y coordinate, these flows have ∂x = 0. The shear stress is constant in these flows ∂u τ = µ = τ (573) xy ∂y w 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 !−1 Z y dy0 Z H dy0 u(y) = τw 0 τw = U 0 (575) o µ(y ) 0 µ(y ) The dissipation is balanced by thermal conduction in the y direction.

∂u2 ∂  ∂T  µ = − k (576) ∂y ∂y ∂y Using the constant shear stress condition, we have the following energy integral ∂T uτ − q = −q = constant q = −k q = q(y = 0) (577) w ∂y w This relationship can be further investigated by defining the Prandtl number c µ ν k P r = P = κ = (578) k κ ρcP For gases, P r ∼ 0.7, approximately independent of temperature. The Eucken relation is a useful approxi- mation that only depends on the ratio of specific heats γ 4γ P r ≈ (579) 7.08γ − 1.80 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 = cpd T , the energy equation can be integrated to obtain the Crocco-Busemann relation

2 u qw h − hw + P r = − P r u (580) 2 τw

For constant cP , this is

2 u qw P r T = Tw − P r − u (581) 2cP τw cP 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 h = h(T ) ≡ h(T ) + P r U 2 (582) r r H 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

1 U 2 Tr = Te + P r (583) 2 cP Contrast with the adiabatic stagnation temperature

1 U 2 Tt = Te + (584) 2 cP The recovery factor is defined as T − T r = r e (585) Tt − Te

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

qw Tw = Tr + P rU (586) cP τw

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 q St = w (587) ρUcP (Tw − Tr)

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 τ C = w (588) f 1 2 2 ρU For Couette flow, C St = f (589) 2P r This relationship between skin friction and heat transfer is the Reynolds analogy.

Constant properties If µ and k are constant, then the velocity profile is linear: U τ τ = µ u = w y (590) w H µ The skin friction coefficient is 2 ρUH C = Re = (591) f Re µ 5 VISCOUS FLOW 56

5.3.2 Poiseuille Flow ∂P If an axial pressure gradient is present, ∂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

∂P ∂2u 0 = − + µ (592) ∂x ∂y2 With the boundary conditions u(0) = u(H) = 0, this can be integrated to yield the velocity distribution

∂P H2 y  y  u = − 1 − (593) ∂x 2µ H H and the wall shear stress ∂P H τ = − (594) w ∂x 2

Pipe Flow The same situation for a round channel, a pipe of radius R, reduces to 1 ∂ ∂u 1 ∂P r = (595) r ∂r ∂r µ ∂x which integrates to the velocity distribution 1 ∂P u = − R2 − r2
(596) 4µ ∂x and a wall shear stress of ∂P R τ = − (597) w ∂x 2 The total volume flow rate is

∂P πR4 Q = − (598) ∂x 8µ The skin friction coefficient is traditionally based on the mean speedu ¯ and using the pipe diameter d = 2R as the scale length.

Q ∂P R2 u¯ = = − (599) πR2 ∂x 8µ and is equal to τw 16 ρud¯ Cf = 2 = Red = (600) 1/2ρu¯ Red µ In terms of the Darcy friction factor,

8τw 64 Λ = 2 = (601) ρu¯ Red Turbulent flow in smooth pipes is correlated by Prandtl’s formula 1  √  √ = 2.0 log Red Λ − 0.8 (602) Λ or the simpler curvefit

−2.5 Λ = 1.02 (log Red) (603) 5 VISCOUS FLOW 57

5.3.3 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

∂u ∂2u = ν (604) ∂t ∂y2 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 y η u = Uf(η) η = √ f 00 + f 0 = 0 (606) νt 2 The solution is the complementary error function

η 2 Z s f = erfc( ) erfc(s) = 1 − erf(s) erf(s) = √ exp(−x2) dx (607) 2 π 0 Shear stress at the wall µU τw = −√ (608) πνt Vorticity

∂u U η2 ω = − = √ exp(− ) (609) ∂y πνt 4 Vorticity thickness

1 Z ∞ √ δω = ω(y, t) dy = πνt (610) ω◦ 0

5.4 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 (611) y ∼ δ (612) u ∼ U (613) δ U v ∼ U ∼ (614) L Re1/2 L δ ∼ (615) Re1/2

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→0ue. Equations are 5 VISCOUS FLOW 58

Boundary Layer Equations The unsteady, compressible boundary-layer equations are:

∂ρ ∂ρu ∂ρv + + = 0 (616) ∂t ∂x ∂y ∂u ∂u ∂u ∂P ∂τ ρ + ρu + ρv = − + xy (617) ∂t ∂x ∂y ∂x ∂y ∂P 0 = − (618) ∂y ∂h ∂h ∂h ∂P ∂ ρ t + ρu t + ρv t = + (uτ − q ) (619) ∂t ∂x ∂y ∂t ∂y xy y

Thickness Measures 99% velocity thickness

δ.99 = y(u = .99ue) (620)

Displacement thickness

Z ∞  ρu  δ∗ = 1 − dy (621) 0 ρeue Momentum thickness

Z ∞ ρu  u  θ = 1 − dy (622) 0 ρeue ue

Displacement Velocity Near the boundary layer, the external flow produces a vertical velocity ve which can be estimated by continuity to be ∂ρ u ρ v ≈ −y e e (623) e e ∂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 d ρ v∗ = (ρ u δ∗) (624) e dx e e The boundary layer influence on the outer flow can therefore by visualized as a source distribution producing an equivalent displacement.

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 + = 0 (626) ∂x ∂y ∂u ∂u ∂u ∂2u u + v = u e + ν (627) ∂x ∂y e ∂x ∂y2 ∂e ∂e ∂u2 ∂2T ρu + ρv = µ + k (628) ∂x ∂y ∂y ∂y2 5 VISCOUS FLOW 59

5.4.1 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.

y r2νx η = δ(x) = (629) δ(x) U Define a stream function ∂ψ ∂ψ u = v = − ψ = δ(x)Uf(η) (630) ∂y ∂x to obtain the Blasius equation

f 000 + ff 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 0.664 ρUx C = Re = (632) f 1/2 x Rex µ The various thickness measures are: 5.0x 1.7208x 0.664x δ = δ∗ = θ = (633) .99 1/2 1/2 1/2 Rex Rex 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 dδ∗ 0.861U v∗ ∼ U = (634) 1/2 dx Rex which agrees with direct computation from the stream function ∂ψ U v∗ = lim − = lim √ (ηf 0(η) − f(η)) (635) η→∞ η→∞ 1/2 ∂x 2Rex where by numerical computation

lim f = η − η∗ η∗ = 1.21678 f 0(∞) = 1 (636) η→∞

5.4.2 Falkner-Skan Flow

m For flows of the type ue = Cx , 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 η = y/δ(x) δ = ψ = ue(x)δ(x)f(η) (637) (m + 1)ue(x) The resulting equation for the function f is 2m f 000 + ff 00 + β 1 − f 02
= 0 β = (638) m + 1 Some cases 5 VISCOUS FLOW 60

m flow -.0904 separating < 0 retarded flows, expansion corner 0 flat plate, zero pressure gradient 1 stagnation point 0 < accelerated flows, wedges -2 doublet near a wall -1 point sink

5.5 K´arm´anIntegral Relations Integration of the momentum equation for incompressible flow results in C dθ θ du δ∗ f = + (2 + H) e H = (639) 2 dx ue dx θ The K´arm´an-Pohlhausen technique consists of assuming a Blausius-type similarity profile for the velocity y u = u (x)f(η) η = (640) e 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

f(0) = 0 (641) δτ f 0(0) = w (642) µue δ2 du f 00(0) = − e (643) ν dx f 000(0) = 0 (644) f(1) = 1 (645) f n>1(1) = 0 (646)

This results in an ordinary differential equation for δ as a function of x.

5.6 Thwaites’ Method Rewrite the K´arm´anintegral equation as

u dθ2 θ2 du θτ e = 2(S − (2 + H)λ) λ = e S = w (647) ν dx ν dx µue Thwaites’ 1949 correlation

2(S − (H + 2)λ) ≈ 0.45 − 6λ (648) K´arm´anintegral equation   d λ 0 due ue 0 = 0.45 − 6λ ue = (649) dx ue dx Approximate solution Z x 2 0.45ν 5 θ = 6 ue dx (650) ue 0 Correlation functions S(λ) and H(λ) 5 VISCOUS FLOW 61

µu τ = e S(λ) δ∗ = θH(λ) (651) w θ 5.7 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 τw,sep = µ = 0 (652) ∂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 + = 0 (654) ∂x ∂y ∂u ∂u ∂P ∂ ∂u ρu + ρv = − e + µ (655) ∂x ∂y ∂x ∂y ∂y ∂h ∂h ∂u2 ∂ ∂T ρu + ρv = µ + k (656) ∂x ∂y ∂y ∂y ∂y

5.8.1 Transformations and Approximations Modified stream function ∂Ψ ∂Ψ ρu = ρ ρv = −ρ (657) ◦ ∂y ◦ ∂x Density-weighted y-coordinate (Howarth-Doronitsyn-Stewartson) Z ρ Y = dy X = x (658) ρ◦ Derivative transformation ∂ ρ ∂ ∂ ∂ ∂Y ∂ = = + (659) ∂y ρ◦ ∂Y ∂x ∂X ∂x ∂Y Chapman-Rubesin parameter, enthalpy-temperature relation ρµ C = dh = cp dT (660) ρ◦µ◦ Boundary layer equations

∂Ψ ∂2Ψ ∂Ψ ∂2Ψ ∂ ∂2Ψ − = ν C (661) ∂Y ∂XY ∂X ∂Y 2 o ∂Y ∂Y 2 ∂Ψ ∂h ∂Ψ ∂h ∂  C ∂h   ∂2Ψ 2 − = ν + ν C (662) ∂Y ∂X ∂X ∂Y ◦ ∂Y P r ∂Y ◦ ∂Y 2

Similarity variable 5 VISCOUS FLOW 62

y r2ν x η = δ = ◦ (663) δ(x) U Streamfunction ansatz for zero pressure gradient

Ψ = Uδf(η) h = h◦g(η) (664) Similarity function equations

(Cf 00)0 + ff 00 = 0 (665) C ( g0)0 + fg0 = −CEc (f 00)2 (666) P r where the Eckert number is

U 2 Ec = = (γ − 1)M 2 for perfect gases (667) h◦ Transport property approximation c µ C = 1 ρµ = ρ µ P r = P = constant (668) ◦ ◦ k Approximate equation set:

f 000 + ff 00 = 0 (669) g00 + P rfg0 = −P rEc (f 00)2 (670)

5.8.2 Energy Equation Integration of the energy equation results in the integral relationship for heat flux at the wall Z ∞   ∂ ρu ht qw = (ρeueht,eΘh)Θh = − 1 dy (671) ∂x 0 ρeue ht,e

where Θh is the energy thickness. The recovery factor r determines the wall enthalpy in adiabatic flow, 1 h = h (q = 0) = h + r u2 (672) r w w ∞ 2 ∞ The recovery factor is found to be an increasing function of the Prandtl number. In gases,

r ' P r1/2 laminar boundary layers (673) r ' P r1/3 turbulent boundary layers

Unity Prandtl Number For P r = 1, the energy equation is

∂h ∂h ∂  ∂h  u2 ρu t + ρv t = µ t h = h + e (674) ∂x ∂y ∂y ∂y t e 2 This has as a solution in adiabatic flow

u2 u2 h = h + e = h + ∞ = constant for q = 0 (675) t e 2 ∞ 2 w Therefore, the recovery enthalpy is 5 VISCOUS FLOW 63

u2 h = h + ∞ (676) r ∞ 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

  2  2  u u∞ u h = h∞ + [hw − hr] 1 − + 1 − 2 (677) u∞ 2 u∞ The Stanton number can be derived from this result in the form of Reynolds analogy C St = f (678) 2 The generalization of this to other Prandtl numbers that is valid for laminar and turbulent boundary layers in gases is C St ' f (679) 2P r2/3

General Prandtl Number For similarity solutions, the nondimensional enthalpy can be found by inte- gration of the energy equation, simplest when C = 1, and P r = constant.

g00 + P r fg0 = −P r Ec (f 00)2 (680) This equation can be integrated exactly to yield

" 0 # Z η Z η Z η (f 00(ξ))2 dξ g(η) = g(0) + g0(0) F (η0; P r) dη0 − P r Ec F (η0; P r) dη0 (681) 0 0 0 F (ξ; P r) where

Z η Z η0 ! F (η; P r) = exp −P r f(ξ) dξ dη0 (682) 0 0 and the boundary conditions yield

hw ∂T k he ρ 0 g(0) = qw = −k = − g (0) (683) he ∂y w cP δ ρe This results in a recovery factor of " # Z ∞ Z η (f 00(ξ))2 dξ r = 2P r F (η; P r) dη (684) 0 0 F (ξ; P r) 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

q G(P r) Z ∞ −1 St ≡ w = G(P r) = F (η; P r) dη (686) √ 1/2 ρeue(hw − hr) 2P rRex 0 and for a flat plate boundary layer G can be approximated as

G ≈ 0.4969P r1/3 0.1 ≤ P r ≤ 3.0 (687) so that the Stanton number for flat plate gas flow is approximately 0.33206 St = (688) 2/3 1/2 P r Rex 5 VISCOUS FLOW 64

Coordinate stretching The physical coordinate can be computed from the transformed similarity variable and the velocity profile

r u Z η ρ y ∞ = ∞ dη (689) 2ν∞x 0 ρ 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 ∞ = (690) ρ T∞ For the case of P r = 1 and a perfect gas, the temperature profile is Z η T γ − 1 2 02
0 = 1 + M∞ 1 − f dη (691) T∞ 2 0 0 where u = u∞f (η). The coordinate transformation is then r Z η u∞ γ − 1 2 02
0 y = η + M∞ 1 − f dη (692) 2ν∞x 2 0 If we suppose that the viscosity varies as µ ∼ T ω, then the skin friction coefficient is √ 2f 00(0) 1 Cf = 1−ω (693) 1/2 γ−1 2
Rex 1 + 2 M∞

5.8.3 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 u w f(0) = 0, f 0(0) = w = 1 f 0(∞) = 1 (694) ue w2 This results in a negative displacement thickness.

5.8.4 Weak Shock Wave Structure ∂ ∂ In contrast to the usual Boundary layer equations, here ∂y = ∂z = 0, and only derivatives in the x direction are considered.

ρu = ρ1u1 (695) 4 ∂u P + ρ u u − µ0 = P + ρ u2 (696) 1 1 3 ∂x 1 1 1 2 0 2 u 4 µ ∂u k ∂T u1 h + − u − = h1 + (697) 2 3 ρ1u1 ∂x ρ1u1 ∂x 2 where 3 µ0 = µ + µ (698) 4 v Entropy creation by gradients: " # 1 Z +∞ 4 µ0 ∂u2  1 ∂T 2 s2 − s1 = + k dx (699) ρ1u1 −∞ 3 T ∂x T ∂x Weak shock thickness estimate based on maximum slope: 5 VISCOUS FLOW 65

0 8µ 1 0 3 ∆m = µ = µ + µv (700) 3ρc M1n − 1 4 For a perfect gas (γ = constant), the mean free path can be estimated as

πγ 1/2 µ Λ = (701) 2 ρc and the shock thickness for γ = 1.4, µv = 0, is 1.8Λ ∆m = (702) (M1n − 1) 5 VISCOUS FLOW 66

5.9 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∇2u or ∇P = −µ∇ × ω (704) Applying the divergence and curl operations to these equations yields

∇2P = 0 or ∇2ω = 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 = Ug(x/L, geometry) (706) then the pressure distribution can be obtained by integrating the momentum equation to get

ρU 2 ρUL P = f(x/L, geometry) ReL = (707) ReL µ If the pressure is specified at the boundaries,

P = ρU 2f(x/L, geometry) (708) then the velocity will be given by

u = UReLg(x/L, geometry) (709) 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

∇4ψ = 0 (710)

Stokes Sphere Flow The force on a moving body in viscous flow is Z Z F = τ · nˆ dA − P nˆ dA (711) ∂Ω ∂Ω Estimating the magnitude of the integrals, the force in a particular direction will have the magnitude

F = CµUL (712) The constant C will in general depend on the shape of the body, the direction xˆ 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:

1 ∂ψ u = (713) r r2 sin φ ∂φ 1 ∂ψ u = − (714) φ r sin φ ∂r The analog of the biharmonic equation is 5 VISCOUS FLOW 67

 ∂2 sin φ ∂  1 ∂ 2 + ψ = 0 (715) ∂r2 r2 ∂φ sin φ ∂φ The boundary conditions at the surface of the sphere are: ∂ψ ∂ψ ψ = 0 = 0 = 0 r = a (716) ∂r ∂φ and the flow approaches a uniform flow far from the sphere

Ur2 lim ψ = sin2 φ (717) r→∞ 2 The solution is

U a 3r 2r2  ψ = a2 sin2 φ − + (718) 4 r a a2 The pressure on the body is found by integrating the momentum equation 3µaU P = P − cos φ (719) ∞ 2r2 and the force (drag) is directed opposite to the direction of motion of the sphere with magnitude D 24 ρU2a D = 6πµUa C ≡ = Re = (720) D 1/2ρU 2πa2 Re µ For a thin disk of radius a moving normal to the freestream the drag is

D = 16πµUa (721) and moving parallel to the freestream 32 D = µUa (722) 3

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 (723) ∂u ∂u ρ + ρU = −∇P + µ∇2u (724) ∂t ∂x This results in a corrected drag law (the flow now has a wake) for the sphere

24  3Re 9  C = 1 + + Re2 ln Re + ... (725) D Re 16 160

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 + = 0 (727) ∂t ∂x ∂P ∂ ∂u 0 = − + µ (728) ∂x ∂y ∂y ∂P 0 = − (729) ∂y 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

h2 ∂P y  y   y  u = − 1 − + U 1 − (730) 2µ ∂x h h h Combining this result with the continuity equation yields the Reynolds lubrication equation

1 ∂  ∂P  ∂h ∂h h3 = 6U + 12 (731) µ ∂x ∂x ∂x ∂t

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 α  1 (732) ◦ L The pressure is given by

   ∗  2  µUL h◦ h h◦ P − P◦ = 2 6 − 1 − 3 2 − 1 (733) αh◦ h h◦ h where h∗ is the gap height at the location of the pressure maximum

h∗ 1 − α α α2 = 2 ≈ 1 − − + O(α3) (734) h◦ 2 − α 2 4 and the maximum pressure is approximately

3 µUL 2 Pmax − P◦ ≈ α 2 + O(α ) (735) 4 h◦ A FAMOUS NUMBERS 69

A Famous Numbers

Fundamental Physical Constants

8 co speed of light in a vacuum 2.998×10 m/s −12 2 o permittivity of the vacuum 8.854×10 C /kg-m −7 µo permeability of the vacuum 4π×10 H/m h Planck constant 6.626×10−34 J-s k Boltzmann constant 1.381×10−23 J/K 23 No Avogadro number 6.022×10 molecules/mol e charge on electron 1.602×10−19 C amu atomic mass unit 1.661×10−27 kg −31 me electron mass 9.109×10 kg −27 mp proton mass 1.673×10 kg G universal gravitational constant 6.673×10−11 m3/kg-s2 σ Stefan-Boltzmann constant 5.670×10−8 W/m2K4

Astronautics

2 go gravitational acceleration at earth’s surface 9.807 m/s RE radius of earth 6378 km 24 ME mass of earth 5.976×10 kg 30 MS mass of sun 1.99×10 kg 8 au mean earth-sun distance 1.496×10 km mass of moon 7.349×1022 kg mean earth-moon distance 3.844×105 km

Ideal Gas Stuff

R˜ Universal gas constant 8314.5 J/kmol-K 8.3145 J/mol-K 82.06 cm3-atm/mol-K 1.9872 cal/mol-K mechanical equivalent of heat 4.186 J/cal volume of 1 kmol at 273.15 K and 1 atm 22.41 m3 number of molecules at 298.15 K and 1 atm 2.46×1025 m3 collision frequency at 273.15 K and 1 atm 4.3×109 s−1 mean free path in N2 at 273.15 K and 1 atm 74 nm

Consistent with the 1998 CODATA adjustment of the funadamental physical constants A FAMOUS NUMBERS 70

Our Atmosphere

composition (mol fractions) 0.7809 N2 0.2095 O2 0.0093 Ar 0.0003 CO2

Sea level

P pressure 1.01325×105 Pa ρ density 1.225 kg/m3 T temperature 288.15 K c sound speed 340.29 m/s R gas constant 287.05 m2/s2-K W molar mass 28.96 kg/kmol µ viscosity (absolute) 1.79×10−5 kg/m-s k thermal conductivity 2.54×10−3 W/m-K cp heat capacity 1.0 kJ/kg-K

30 kft

P pressure 3.014×104 Pa ρ density 0.458 kg/m3 T temperature 228.7 K c sound speed 303.2 m/s

Based on the U.S. Standard Atmosphere, 1976 NOAA-S/T 76-1562, 1976.

Unit Conversions

1 m ≡ 3.28 ft 0.3048 ft ≡ 1 m 1 lb (force) ≡ 4.452 N 1 lb (mass) ≡ 0.454 kg 1 btu ≡ 1055 J 1 hp ≡ 745.7 W 1 hp ≡ 550 ft-lbf /s 1 mile (land) ≡ 1.609 km 1 mph ≡ 0.447 m/s 1 mile (nautical) ≡ 1.852 km B BOOKS ON FLUID MECHANICS 71

B 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