OUTLINE FOR Chapter 2 REVIEW OF VECTOR RELATIONS (I)

AERODYNAMICS (W1-2-2) REVIEW OF VECTOR RELATIONS (II) • Scalar Fields (pressure, density, temperature..) • Gradients of Scalar Fields

Cartesian:

• Vector Fields (velocity..) Cylinerical: Spherical:

1. Its magnitude is the maximum rate of change of p per unit • Scalar and Vector Products length of the coordinate space at the given point. 2. Its direction is that of the maximum rate of change of p at the given point. • Directional derivatives:

AERODYNAMICS (W1-2-3) REVIEW OF VECTOR RELATIONS (III) • Divergence of a Vector Field => Scalar • Curl of a Vector Field => Vector

The physical meaning of divergence is the rate of change of the volume of a moving fluid element, per unit volume. <== see Chapter 2.3 for detail! The physical meaning of curl is twice of the angular velocity vector of a fluid element. REVIEW OF VECTOR RELATIONS (IV) Consider a vector field Relations between Line, Surface and Volume integrals: • Line integral • Line integral to Surface integral

Stokes theorem

• Surface integral • Surface integral to Volume integral Divergence theorem

Gradient theorem • Volume integral

AERODYNAMICS (W1_2_5) METHODS OF ANALYSIS

• System method  In mechanics courses.  Dealing with an easily identifiable rigid body. • Control volume method  In fluid mechanics course.  Difficult to focus attention on a fixed identifiable quantity of mass.  Dealing with the flow of fluids. System Method • A system is defined as a fixed, identifiable quantity of mass. • The boundaries separate the system from the surrounding. • The boundaries of the system may be fixed or movable. No mass crosses the system boundaries.

Piston-cylinder assembly: The gas in the cylinder is the system. If the gas is heated, the piston will lift the weight; The boundary of the system thus move. Heat and work may cross the boundaries, but the quantity of matter remain fixed. Control Volume Method

• Control Volume (CV) is an arbitrary volume in space through which the fluid flows. • The geometric boundary of the control volume is called the Control Surface (CS).

• The CS may be real or imaginary. • The CV may be at rest or in motion. METHODS OF DESCRIPTION

• Lagrangian description => System • Eulerian description => Control volume Lagrangian Description • Attention is focused on a material volume (MV) and follow individual fluid particle as it move. • The fluid particle is colored, tagged or identified. • Determining how the fluid properties associated with the particle change as a function of time.  Example: one attaches the temperature-measuring device to a particular fluid particle A and record that particle’s temperature as it moves about.

TA = TA(t)=T (xo,yo,zo, t)

where particle A passed through coordinate (xo,yo,zo) at to  The use of may such measuring devices moving with various fluid particles would provide the temperature of these fluid particles as a function of time. Eulerian Description

• Attention is focused on the fluid passing through a control volume (CV) fixed in the space. • Obtaining information about the flow in terms of what happens at the fixed points in space as the fluid flows past those points. • The fluid motion is given by completely prescribing the necessary properties as a functions of space and time.  Example: one attaches the temperature-measuring device to a particular point (x,y,z) and record the temperature at that point as a function of time. T = T ( x , y , z , t ) => field concept. The independent variables are the spatial coordinates ( x , y , z) and time t Field Representation of flow

• At a given instant in time, any fluid property ( such as density, pressure, velocity, and acceleration) can be described as a functions of the fluid’s location.This representation of fluid parameters as functions of the spatial coordinates is termed a field representation of flow. • The specific field representation may be different at different times, so that to describe a fluid flow we must determine the various parameter not only as functions of the spatial coordinates but also as a function of time. • EXAMPLE: Temperature field T = T ( x , y , z , t ) • EXAMPLE: Velocity field V = u(x, y, z,t) i + v(x, y, z,t) j + w(x, y, z,t) k Ball, with mass m, released from rest at a height y= 0. Air resistance is neglected FD=0. (a) Find the Lagrangian description of this velocity field. (b) Find the Eulerian description of this velocity field.

y Basic Laws

Analysis of any problem in fluid mechanics necessarily includes statement of the basic laws governing the fluid motion. The basic laws, which applicable to any fluid, are:

• Conservation of mass – Continuity Equation • Conservation of (angular) Momentum - Newton’s second law of motion. • Conservation of Energy The first law of thermodynamics BASIC LAWS FOR A SYSTEM - Conservation of Mass

• Conservation of Mass  Requiring that the mass, M, of the system be constant.

dM  DM D  = = ρdV = 0 ∫V (system) dt system Dt Dt

Where the mass of the system BASIC LAWS FOR A SYSTEM - Conservation of Momentum

• Newton’s Second Law  Stating that the sum of all external force acting on the system is equal to the time rate of change of linear momentum of the system.    dP  DP D  F =  = = VρdV  ∫V (system) dt system Dt Dt

Where P is the linear momentum of the system BASIC LAWS FOR A SYSTEM - Conservation of Energy • The First Law of Thermodynamics  Requiring that the energy of system be constant. δQ −δW = dE

  dE  DE D Q −W =  = = ( ρet dV ) ∫V (system) dt system Dt Dt

Where E is the total energy of the system and et is the total energy of the system per unit mass V 2 e = e + + gz e is specific internal energy, V the speed, and z t 2 the height of a particle having mass dm. How To Derive Control Volume Formulation

BASIC LAWS

System Method Control Volume Method

Governing Equation Governing Equation System Formulation Control Volume Formulation

Integral (large) control volume Reynolds Transport Theorem Differential (differential) control volume) Total (material) derivative Transformation between Lagrangian and Eulerian Description

• It is more nature to apply conservation laws by using Lagrangian description (ie. Material Volume). • However, the Eulerian description (ie. Control Volume) is preferred for solving most of problem in fluid mechanics. • The two descriptions are related and there are a transformation formula called Reynolds transport theorem and material derivative between Lagrangian and Eulerian descriptions. Reynolds Transport Theorem D ∂   ∫∫∫αdV = ∫∫∫αdV + ∫∫α V • dS Dt system ∂t CV CS

By converting the surface integral to volume integral by use of Gauss theorem    ∫∫α V • dS = ∫∫∫∇ • (αV )dV CS CV

Langragian derivative of a volume integral of a given property D ∂  ∂α  ∫∫∫αdV = ∫∫∫αdV + ∫∫∫∇ • (αV )dV = ∫∫∫( + ∇ • (αV ))dV Dt system ∂t CV CV CV ∂t

This is the fundamental relation between the rate of change of any arbitrary extensive property, α, of a system and the variations of this property associated with a control volume. Conservation of Mass D ∫∫∫ρdV = 0 • Basic Law for Conservation of Mass Dt system • The system and integral (large) control volume formulation ---- Reynolds Transport Theorem D ∂α  ∫∫∫αdV = ∫∫∫( + ∇ • (αV ))dV Dt system CV ∂t D ∂ρ  α=ρ ∫∫∫ρdV = ∫∫∫( + ∇ • (ρV ))dV = 0 Dt system CV ∂t

Continuity ∂ρ  Material derivative + ∇ • (ρV ) = 0 equation ∂ Dα ∂α  t = + u • ∇α A partial differential equation => velocity is Dt ∂t continuous ∂ρ  ∂ρ   Dρ  + ∇ • (ρV ) = +V •∇ρ + ρ(∇ •V ) = + ρ(∇ •V ) = 0 ∂t ∂t Dt How To Derive Control Volume Formulation

BASIC LAWS

System Method Control Volume Method

Governing Equation Governing Equation System Formulation Control Volume Formulation

Integral (large) control volume Reynolds Transport Theorem Differential (differential) control volume) Total (material) derivative Material Derivative (I) • Let α(x,y,z,t) be any field variable, e.g., ρ, T, V=(u,v,w), etc. (Eulerian description) • Observe a fluid particle for a time period δt as it flows (Langrangian description) • During the time period, the position of the fluid particle will change by amounts δx , δy , δz, while its vale of α will change by an amount ∂α ∂α ∂α ∂α δα = δt + δx + δy + δz ∂t ∂x ∂y ∂z δx δy δz • As one follow the fluid particle, ( , , ) = (u,v, w) So δt δt δt Dα δα ∂α ∂α δx ∂α δy ∂α δz ∂α ∂α ∂α ∂α = lim = + + + = + u + v + w Dt δt→0 δt ∂t ∂x δt ∂y δt ∂z δt ∂t ∂x ∂y ∂z which is called the material, total, or substantial derivative. Material Derivative (II) • Use the notation D/Dt to emphasize that the material derivative is the rate of change seen by an observer “following the fluid.” • The material derivative express a Langrangian derivative in terms of Eulerian derivatives. • In vector form,

Dα ∂α ∂α ∂α ∂α ∂α = + u + v + w = + (V •∇)α Dt ∂t ∂x ∂y ∂z ∂t EXAMPLE OF SUBSTANTIAL DERIVATIVE

The velocity flow field of a steady state flow is given by the equations: u=-x ; v=y The temperature of the field is described by the following expression: T(x,y,t)=xt+3xy Determine the time rate of change of temperature of a fluid element as it passes through the point (1, -2) at time t=6. y

convective derivative the time rate of change of DT ∂T ∂T ∂T x temperature of a fluid = + u + v element Dt ∂t ∂x ∂y

local derivative DT ∂T ∂T ∂T = + u + v = x + (−x)(t + 3y) + y(3x) =1+ (−1)(6 − 6) + (−2)(3) = −5 Dt ∂t ∂x ∂y Ball, with mass m, released from rest at a height y= 0. Air resistance is neglected FD=0. (a) From the Lagrangian description of this velocity field, find the acceleration of the ball. (b) From the Eulerian description of this velocity field, , find the acceleration of the ball.

y Conservation of Mass DM = 0 • Basic Law for Conservation of Mass Dt • The system and differential control volume formulation ---- Material derivative Dα ∂α  DM ∂M  = + u • ∇α α=M = + u • ∇M = 0 Dt ∂t Dt ∂t

∂(ρV )  ∂ρ ∂V   M = ρV + u • ∇(ρV ) = V + ρ + ρu • ∇V +Vu • ∇ρ = 0 ∂t ∂t ∂t ∂ρ ∂V   ∂ρ  ∂V  V + ρ + ρu • ∇V +Vu • ∇ρ = V ( + u • ∇ρ) + ρ( + u • ∇V ) = 0 ∂t ∂t ∂t ∂t Dρ ρ DV Dρ 1 DV Dρ  + = + ρ( ) = + ρ(∇ •V ) = 0 Dt V Dt Dt V Dt Dt

Volume dilation = divergence of velocity field  Physical Meaning of ∇ •V CONSERVATION OF MASS Rectangular Coordinate System • The differential equation for conservation of mass: The continuity equation

By “Del” operator

The continuity equation becomes

Dρ  ∂ρ   ∂ρ  + ρ(∇ •V ) = +V • ∇ρ + ρ(∇ •V ) = + ∇ • ρV = 0 Dt ∂t ∂t Description and Classification of Fluid Motions Continuity Equation for Incompressible Fluid Definition of Incompressible fluid: ρ = cosntant ? As a given fluid is followed, not only will its mass be observed to remain constant, but its volume, and hence its density, will be observed to remain constant.

Material derivative D (ρ) = 0 Dt

Follow a fluid particle

Continuity equation Dρ  ∂ρ  + ρ(∇ •V ) = + ∇ • ρV = 0 Dt ∂t  1 DV ∇ •V = = 0 Continuity eq. V Dt Stratified-Fluid Flow

• A fluid particle along the line ρ1 or ρ2 will have its density remain fixed at ρ=ρ1 or ρ=ρ2 D (ρ) = 0 Dt

Follow a fluid particle Stratified-fluid flow is considered to be incompressible, but ρ is not constant (ρ≠constant ) everywhere ie. ə ρ/ ə x ≠0, ə ρ/ əy≠0, Stratified-fluid flow may occurs in the ocean (owing to salinity variation) or in the atmosphere (owing to temperature variations). For 2D steady state stratified flow, the continuity equation should be ∂(ρu) ∂(ρv) + = 0 ∂x ∂y Summary of Continuity Equation

Integral form continuity equation: Steady and Unsteady flows

Unsteady: Steady:

Since the control volume is fixed in space, the time derivative can be placed inside the volume integral

Apply divergence theorem:

Integral form of continuity equation becomes: Compressible and Incompressible flows Compressible: D Incompressible: constant (ρ) = 0 Dt Differential form of continuity equation: OUTLINE FOR Chapter 2-2

AERODYNAMICS (W1-2-1) Basic Laws

Analysis of any problem in fluid mechanics necessarily includes statement of the basic laws governing the fluid motion. The basic laws, which applicable to any fluid, are:

• Conservation of mass – Continuity Equation • Conservation of (angular) Momentum - Newton’s second law of motion. • Conservation of Energy The first law of thermodynamics BASIC LAWS FOR A SYSTEM - Conservation of Momentum

• Newton’s Second Law  Stating that the sum of all external force acting on the system is equal to the time rate of change of linear momentum of the system.      dP  DP D F =  = = ∫∫∫VρdV dt system Dt Dt system

Where P is the linear momentum of the system • The forces act on fluid particles: – Body forces ( gravity, electromagnetic ).

– Surface forces ( pressure, viscous ).

Total viscous forces = Review of Momentum Equation (I) Momentum Equation

Physical principle: Time rate of change of momentum Force = time rate of change of momentum

Force: Body farces: gravity, electromagnetic forces, or any other forces which “act at a distance on the fluid inside V.

Surface forces: pressure and shear stress acting on the control surface S. The integral form of momentum equations

Total viscous forces =

AERODYNAMICS (W1-2-8) Review of Momentum Equation (II)

For

Divergence theorem Euler equations for steady inviscid flow

The differential form of momentum equations (Navier-Stokes equations)

Inviscid, Incompressible: constant 1 ∂P -1 u(∇ •V ) +V •∇u = − ρ ρ ∂x

-1 1 ∂P v(∇ •V ) +V •∇v = − ρ ρ ∂y -1 1 ∂P w(∇ •V ) +V •∇w = − For unsteady 3D flow, compressible or incompressible, viscous or inviscid ρ ρ ∂z AERODYNAMICS (W1-2-9) ∂u ∂u ∂u ∂P ∂2u ∂2u ρ( + u + v ) = − + ρf + µ( + ) ∂t ∂x ∂y ∂x x ∂x2 ∂y2 Du µ ∂u ∂u ma = ρdxdydz( ) = −dPdydz + f ρdxdydz + ( dydz + dxdz) = F Dt x ρ ∂x ∂y x

dP = P2 − P1

shear force = τ xxdydz +τ yxdxdz P1 P2 dz ∂u ∂u dy F = µ( dydz + dxdz) visoucs ∂x ∂y dx

Du dPdydz ∂u dydz ∂u dxdz dP ∂2u ∂2u ρ( ) = − + f ρ + µ( + ) = − + ρf + µ( + ) Dt dxdydz x ∂x dxdydz ∂y dxdydz dx x ∂x2 ∂y2 Du ∂u ∂u ∂u dP ∂2u ∂2u ρ( ) = ρ( + u + v ) = − + ρf + µ( + ) Dt ∂t ∂x ∂y dx x ∂x2 ∂y2 Dα ∂α  = + u • ∇α Dt ∂t Material derivative SUMMARY

For steady, incompressible, inviscid flow, no body force

Continuity equation:    V (x, y, z) = ui + vj + wk ∂u ∂v ∂w ∇ •V = + + = 0 ∂x ∂y ∂z

Momentum equation: ∂u ∂u ∂u 1 ∂P V •∇u = u + v + w = − ∂x ∂y ∂z ρ ∂x ∂v ∂v ∂v 1 ∂P V •∇v = u + v + w = − ∂x ∂y ∂z ρ ∂y

∂w ∂w ∂w 1 ∂P V •∇w = u + v + w = − ∂x ∂y ∂z ρ ∂z

AERODYNAMICS (W1-2-10) Vorticity

• Defining Vorticity ζ which is a measurement of the rotation of a fluid element as it moves in the flow field:     i j k       ∂ ∂ ∂  ζ = 2ω = curlV = ∇×V =   ∂x ∂y ∂z   u v w     ∂w ∂v   ∂u ∂w   ∂v ∂u   ζ = 2ω = i  −  + j −  + k −  = ∇×V   ∂y ∂z   ∂z ∂x   ∂x ∂y  • In cylindrical coordinates system:

   1 ∂Vz ∂Vθ    ∂Vr ∂Vz    1 ∂rVθ 1 ∂Vr  ∇×V = e  −  + eθ  −  + e  −  r  r ∂θ ∂z   ∂z ∂r  z  r ∂r r ∂θ  Fluid Rotation

       ∂w ∂v   ∂u ∂w   ∂v ∂u      ζ = ∇×V = 2ω = 2([(ωx )i + (ω y )j + (ωz )k ]= (i  −  + j −  + k −    ∂y ∂z   ∂z ∂x   ∂x ∂y 

1  ∂w ∂v  ≠   0 rotational ωx =  −  2  ∂y ∂z  = 0 Irrotational 1  ∂u ∂w  Rotational flow Irrotational flow ω =  −  y 2  ∂z ∂x  1  ∂v ∂u    ωZ =  −  2  ∂x ∂y  Irrotational and Rotational flows

    curlV = ∇×V = 0 curlV = ∇×V ≠ 0 Irrotational flow rotational flow W3 8 EXAMPLE OF VORTICITY

2 ( )2 ( ) CIRCULATION

Definition: C Circulation Lift

V∞ V ds From Stokes’ theorem Example: Circulation in a uniform flow

u = V∞ v = 0 Irrotational flow

Irrotational flow

For arbitrary close curve C

(irrotational) Stokes’ theorem STREAMFUNCTION AND VELOCITY POTENTIAL Streamfunction Ψ: Velocity Potential φ: definition definition automatically satisfy continuity equation automatically satisfy irrotational conditon streamfunction Ψ properties: Velocity potential φ properties:

1.

2. Relationship between streamfunction and velocity potential

= 0

AERODYNAMICS (W1-3-6) SUMMARY 1. Substantial derivative 5. Streamfunction Ψ:

substantial convective derivative derivative local derivative 2. Streamline

3. Vorticity 6. Velocity Potential φ:

≠ 0 rotational = 0 Irrotational

4. Circulation

Irrotational flow