marine hydrodynamics mit notes

13.021 – Marine Hydrodynamics, Fall 2004 Lecture 1

13.021 – Marine Hydrodynamics Lecture 1

Introduction

Marine Hydrodynamics is the branch of Fluid Mechanics that studies the motion of incompressible fluids (liquids) and the forces acting on solid bodies immersed in them.

Marine hydrodynamics is a large and diverse subject and only a limited number of topics can be covered in an introductory course. The topics that will be covered throughout the semester include:

• Model testing and similitude

• Interaction between bodies and ideal fluids

• Viscosity and boundary layers

• Effect of waves on resistance and ship motion

Why study Marine Hydrodynamics?

Studying Marine Hydrodynamics provides a greater understanding of a wide range of phenomena of considerable complexity involving fluids. Another benefit is that it allows predictions to be made in many areas of practical importance.

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Fluids vs. Solids

In brief, Fluid Mechanics studies the kinematics and dynamics of a group of particles without having to study each particle separately.

Most of us have taken some courses on solids or related to solids. Even those who haven’t can get an intuitive feeling about some physical properties of a solid. Thus a comparison of solids and fluids will give some guidelines as to which properties can be translated to fluids and on what terms.

Similarities

1. Fundamental laws of mechanics apply to both fluids and solids

- Conservation of mass - Conservation of momentum (Newton’s law of motion) - Conservation of energy (First law of thermodynamics)

2. Continuum hypothesis is used for both fluids and solids

loca

l v

alu

e

mea

sure

d p

rop

erty

length scale

“particle” size

variations due to variations due to

molecular fluctuations varying flow

O(10-10 – 10-8)m

e.g. The smallest measurement scales are in the order of �M ∼ 10−5m → VM ∼ 10−15m3 . This corresponds to ∼ 3 × 1010 molecules of air in STP or ∼ 1013 molecules of water.

2

••

� ��

� �

Differences

1. Shape

- Solids have definite shape - Fluids have no preferred shape

2. Constitutive laws

Constitutive laws are empirical formulas that relate certain unknown variables.The constitutive laws used in 13.021 relate:dynamics (force, stress ...) to kinematics (position, displacement, velocity ...)Different constitutive laws are used for solids and different for fluids.

- For solids Hooke’s law is used to relate stress and strain

stress � �� = f (strain) � �� force/area relative displacement/length �

N/m2 �

[m/m]

• For solid mechanics ‘statics’ is a dominant aspect

- For fluids stress is related to rate of strain

stress = f (rate of strain)

force/area velocity gradient

N/m2 [1/s]

• •• Fluids at rest cannot sustain shear force. Fluids have to be moving to be non-trivial.

The branch of Fluid Mechanics that studies fluids at rest is referred to as ‘Hydrostatics ’.(Archimedes, c 200 BC)

Hydrostatics study the trivial case where no stresses due to fluid motion exist.

Sometimes distinction between liquids and solids is not a sharp one(honey, jelly, paint, . . . ). Fortunately most common fluids, such as air and water are very close to ”ideal” fluids.

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Liquids vs. Gasses

Liquids and gasses are two categories of fluids.

A fluid is ‘a body whose particles move easily among themselves. Fluid is a generic term, including liquids and gasses as species. Water, air, and steam are fluids.’ [1].

A liquid is ‘Being in such a state that the component parts move freely among themselves, but do not tend to separate from each other as the particles of gases and vapors do; neither solid nor aeriform.’[1]

A gas is ‘The state of matter distinguished from the solid and liquid states by relatively low density and viscosity, relatively great expansion and contraction with changes in pressure and temperature, the ability to diffuse readily, and the spontaneous tendency to become distributed uniformly throughout any container.’[2]

In brief, a liquid is generally incompressible and does not fill a volume by expanding into it while on the other hand, a gas is compressible and expands to fill any volume containing it.

The science that studies the dynamics of liquids is referred to as ‘Hydrodynamics’, while the science that studies the dynamics of gasses is referred to as ‘Aerodynamics’.

The main difference between the study of ‘Hydrodynamics’ and the study of ‘Aerodynamics’ is the property of incompressibility. In general hydrodynamic flows are treated as incompressible while aerodynamic flows are treated as compressible.

[1] Webster Dictionary

[2] American Heritage Dictionary

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Why is a liquid flow incompressible?

It can be shown that the ratio of the characteristic fluid velocities U in a flow to the speed of sound C in the medium gives a measure of compressibility of the medium for that particular flow. This ratio is called the Mach number M . Although the speed of sound in water is of comparable magnitude to the speed of sound in air, the characteristic fluid velocities in water are significantly smaller. Thus in the case of water, the Mach number is very small, indicating that water is virtually incompressible.

U : Characteristic fluid flow velocity

C : Speed of sound in the medium U

M ≡ : Mach number C

The average speed of sound in air and water is:

Cair ∼ 300m/s = 984ft/sec = 583knots

Cwater ∼ 1200m/s = 3, 937ft/sec = 2, 333knots

Therefore the average ratio of the speed of sound in water to air is Cwater ∼ 4. Further on, Cair

because the average water to air density ratio is ρwater ∼ 1kg/m3

= 103, it is ‘harder’ to ρair 103kg/m3

move in water and therefore, typically, it is:

Uwater<<Uair

giving thus typical values of Mach numbers in the order of:

Mair ∼ O(1) ⇒ COMPRESSIBLE flow

Mwater<<1 ⇒ INCOMPRESSIBLE flow

Note: An incompressible flow does not mean constant density.

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13.021 – Marine Hydrodynamics, Fall 2004Lecture 2


Chapter 1 - Basic Equations

1.1 Description of a Flow

To define a flow we use either the ‘Lagrangian’ description or the ‘Eulerian’ description.

• Lagrangian description: Picture a fluid flow where each fluid particle caries its own properties such as density, momentum, etc. As the particle advances its properties may change in time. The procedure of describing the entire flow by recording the detailed histories of each fluid particle is the Lagrangian description. A neutrally buoyant probe is an example of a Lagrangian measuring device.

The particle properties density, velocity, pressure, . . . can be mathematically represented as follows: ρp(t), �vp(t), pp(t), . . .

The Lagrangian description is simple to understand: conservation of mass and Newton’s laws apply directly to each fluid particle . However, it is computationally expensive to keep track of the trajectories of all the fluid particles in a flow and therefore the Lagrangian description is used only in some numerical simulations.

)(tp υr

p

Lagrangian description; snapshot

1

• Eulerian description: Rather than following each fluid particle we can record the evolution of the flow properties at every point in space as time varies. This is the Eulerian description. It is a field description. A probe fixed in space is an example of an Eulerian measuring device.

This means that the flow properties at a specified location depend on the location and on time. For example, the density, velocity, pressure, . . . can be mathematically represented as follows: �v(�x, t), p(�x, t), ρ(�x, t), . . .

The aforementioned locations are described in coordinate systems. In 13.021 we use the cartesian, cylindrical and spherical coordinate systems.

The Eulerian description is harder to understand: how do we apply the conservation laws? However, it turns out that it is mathematically simpler to apply. For this reason, in Fluid Mechanics we use mainly the Eulerian description.

),( tx r

y

x

),( tx rrυ

Eulerian description; Cartesian grid

2

1.2 Flow visualization - Flow lines

• Streamline: A line everywhere tangent to the fluid velocity �v at a given instant (flow snapshot). It is a strictly Eulerian concept.

• Streakline: Instantaneous locus of all fluid particles that have passed a given point (snapshot of certain fluid particles).

• Pathline: The trajectory of a given particle P in time. The photograph analogy would be a long time exposure of a marked particle. It is a strictly Lagrangian concept.

Can you tell whether any of the following figures ( [1] Van Dyke, An Album of Fluid Motion 1982 (p.52, 100)) show streamlines/streaklines/pathlines?

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1.3 Some Quantities of Interest

• Einstein Notation

– Range convention: Whenever a subscript appears only once in a term, the subscript takes all possible values. E.g. in 3D space:

xi (i = 1, 2, 3) → x1, x2, x3

– Summation convention: Whenever a subscript appears twice in the same term the repeated index is summed over the index parameter space. E.g. in 3D space:

aibi = a1b1 + a2b2 + a3b3 (i = 1, 2, 3)

Non repeated subscripts remain fixed during the summation. E.g. in 3D space ai = xij nj denotes three equations, one for each i = 1, 2, 3 and j is the dummy index.

Note 1: To avoid confusion between fixed and repeated indices or different repeated indices, etc, no index can be repeated more than twice.

Note 2: Number of free indices shows how many quantities are represented by a single term.

Note 3: If the equation looks like this: (ui) (xi) , the indices are not summed.

– Comma convention: A subscript comma followed by an index indicates partial differentiation with respect to each coordinate. Summation and range conventions apply to indices following a comma as well. E.g. in 3D space:

∂ui ∂u1 ∂u2 ∂u3 ui,i = = + +

∂xi ∂x1 ∂x2 ∂x3

• Scalars, Vectors and Tensors

Scalars magnitude

Vectors (aixi) magnitude direction

density ρ (�x, t) pressure p (�x, t)

velocity �v (�x, t) /momentum mass flux

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Tensors (aij ) magnitude direction orientation momentum flux stress τij (�x, t)

1.4 Concept and Consequences of Continuous Flow

For a fluid flow to be continuous, we require that the velocity �v(�x, t) be a finite and continuous function of �x and t.

vi.e. ∇ · �v and ∂� are finite but not necessarily continuous. ∂t

Since ∇ · �v and ∂�v < ∞, there is no infinite acceleration i.e. no infinite forces , which is ∂t

physically consistent.

1.4.1 Consequences of Continuous Flow

• Material volume remains material. No segment of fluid can be joined or broken apart.

• Material surface remains material. The interface between two material volumes always exists.

• Material line remains material. The interface of two material surfaces always exists.

Material surface

fluid a

fluid b

• Material neighbors remain neighbors. To prove this mathematically, we must prove that, given two particles, the distance between them at time t is small, and the distance between them at time t + δt is still small.

5

�

∂�vAssumptions At time t, assume a continuous flow (∇ · �v, ∂t << ∞) with fluid velocity

�v(�x, t). Two arbitrary particles are located at �x and �x + δ�x(t), respectively.

Result If δ�x(t) ≡ δ�x → 0 then δ�x(t + T ) → 0, for all subsequent times t + T .

Proof v x v + {υr (x

v )δt}

x

δx v δx

v (t + δt)

x v +δ x

v x v + r

δ x v +

r{(υ (x v ) + δ x

v ⋅∇υ (x v ))δt}

After a small time δt:

• The particle initially located at �x will have travelled a distance �v(�x)δt and at time t + δt will be located at �x + {�v(�x)δt}.

• The particle initially located at �x+δ�x will have travelled a distance (�v(�x) + δ�x · ∇�v(�x)) δt. (Show this using Taylor Series Expansion about(�x, t)). Therefore after a small time δt this particle will be located at �x + δ�x + {(�v(�x) + δ�x · ∇�v(�x))δt}.

• The difference in position δ�x(t + δt) between the two particles after a small time δt will be:

δ�x(t + δt) = �x + δ�x + {(�v(�x) + δ�x · ∇�v(�x))δt} − (�x + {�v(�x)δt}) ⇒ δ�x(t + δt) = δ�x + (δ�x · ∇�v(�x))δt ∝ δ�x

Therefore δ�x(t + δt) ∝ δ�x because ∇�v is finite (from continuous flow assumption). Thus, if δ�x → 0 , then δ�x(t + δt) → 0. In fact, for any subsequent time t + T :

t+T

δ�x(t + T ) ∝ δ�x + δ�x · ∇�vdt ∝ δ�x, t

and δ�x(t+T ) → 0 as δ�x → 0. In other words the particles will never be an infinite distance apart. Thus, if the flow is continuous two particles that are neighbors will always remain neighbors.

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(xf r

), ttt δυ δ ++ r

p p

Df(−→ xp(t), t) = lim

f(−→ xp + −→ vp δt, t + δt) − f(−→ xp, t) (1) (f

ttx pp υ δrr

( ) +

Dt δt→0 δt particle p

), tx p

r

)(tx r

1.5 Material/Substantial/Total Time Derivative: D/Dt

A material derivative is the time derivative – rate of change – of a property following a fluid particle ‘p’. The material derivative is a Lagrangian concept.

By expressing the material derivative in terms of Eulerian quantities we will be able to apply the conservation laws in the Eulerian reference frame.

Consider an Eulerian quantity f(�x, t). The time rate of change of f as experienced by a →particle ‘p’ travelling with velocity −vp is the substantial derivative of f and is given by:

→ →Performing a Taylor Series Expansion about (− , t) and taking into account that − δt = δ�x,xp vp

we obtain:

−→ − ∂f(�x, t)→f(xp + vp δt, t + δt) = f(�x, t) + δt + δ�x · ∇f(�x, t) + O(δ2)(higher order terms)(2)∂t

From Eq.(1, 2) we see that the substantial derivative of f as experienced by a particle →travelling with −vp is given by:

Df ∂f −→ = + vp · ∇f Dt ∂t

The generalized notation:

D ∂ −→≡ + vp · ∇ Dt ∂t ��

Lagrangian Eulerian

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Example 1: Material derivative of a fluid property G� (�x, t) as experienced by a fluid particle.

Let ‘p’ denote a fluid particle. A fluid particle is always travelling with the local fluid velocity �vp(t) = �v(�xp, t). The material derivative of a fluid property G� (�x, t) as experienced by this fluid particle is given by:

D �G =

∂ �G + �v · ∇ �G

Dt ∂t � �� Convective

Lagragian rate of change

Eulerian rate of change

rate of change

Example 2: Material derivative of the fluid velocity �v(�x, t) as experienced by a fluid particle. This is the Lagrangian acceleration of a particle and is the acceleration that appears in Newton’s laws. It is therefore evident that its Eulerian representation will be used in the Eulerian reference frame.

Let ‘p’ denote a fluid particle. A fluid particle is always travelling with the local fluid D�v(�x,t)velocity �vp(t) = �v(�xp, t). The Lagrangian acceleration

Dt as experienced by this fluid particle is given by:

D�v ∂�v = + �v · ∇�v

Dt ∂t � �� Convective

Lagragian Eulerian acceleration

acceleration acceleration

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1.6 Difference Between Lagrangian Time Derivative and Eulerian Time Derivative

Example 1: Consider an Eulerian quantity, temperature, in a room at points A and B where the temperature is different at each point.

Point A: 10o Point C: ∂∂ T

t Point B: 1o

At a fixed in space point C, the temperature rate of change is ∂T which is an Eulerian time ∂t

derivative.

Example 2: Consider the same example as above: an Eulerian quantity, temperature, in a room at points A and B where the temperature varies with time.

DT ∂ T = + υ

v ⋅ ∇ Tfly

Point A: 10o Dt ∂ t Point B: 1o

Following a fly from point A to B, the Lagrangian time derivative would need to include DT ∂Tthe temperature gradient as both time and position changes: Dt =

∂t + �vfly · ∇T

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1.6.1 Concept of a Steady Flow ( ∂ ≡ 0)∂t

A steady flow is a strictly Eulerian concept.

Assume a steady flow where the flow is observed from a fixed position. This is like watching∂ Dfrom a river bank, i.e. ∂t = 0 . Be careful not to confuse this with

Dt which is more like following a twig in the water. Note that D = 0 does not mean steady since the flow could

Dt speed up at some points and slow down at others.

∂ = 0 ∂t

Dρ1.6.2 Concept of an Incompressible Flow ( Dt ≡ 0)

An incompressible flow is a strictly Lagrangian concept.

Assume a flow where the density of each fluid particle is constant in time. Be careful not to confuse this with ∂ρ = 0, which means that the density at a particular point in the flow

∂t is constant and would allow particles to change density as they flow from point to point. Also, do not confuse this with ρ = const , which for example does not allow a flow of two incompressible fluids.

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1.7 Stress Tensor

1.7.1 Stress Tensor τij

The stress (force per unit area) at a point in a fluid needs nine components to be completely specified, since each component of the stress must be defined not only by the direction in which it acts but also the orientation of the surface upon which it is acting.

The first index i specifies the direction in which the stress component acts, and the second index j identifies the orientation of the surface upon which it is acting. Therefore, the ith

component of the force acting on a surface whose outward normal points in the jth direction is τij .

X1

X2

X3

31

11

21

22

12 32

13

23

33

Figure 1: Shear stresses on an infinitesimal cube whose surfaces are parallel to the coordinate system.

1

∑

∑

∑

X2 2

A1

X3

3

A2

1

A3

area A0

X1

nP

Q

R

Figure 2: Infinitesimal body with surface PQR that is not perpendicular to any of the Cartesian axis.

Consider an infinitesimal body at rest with a surface PQR that is not perpendicular to any of the Cartesian axis. The unit normal vector to the surface PQR is n = n1x1 +n2x2 +n3x3. The area of the surface = A0, and the area of each surface perpendicular to Xi is Ai = A0ni, for i = 1, 2, 3.

Newton’s law: Fi = (volume force)i for i = 1, 2, 3 on all 4 faces

Note: If δ is the typical dimension of the body : surface forces ∼ δ2

: volume forces ∼ δ3

An example of surface forces is the shear force and an example of volumetric forces is the gravity force. At equilibrium, the surface forces and volumetric forces are in balance. As the body gets smaller, the mass of the body goes to zero, which makes the volumetric forces equal to zero and leaving the sum of the surface forces equal zero. So, as δ → 0, all4faces Fi = 0 for i = 1, 2, 3 and ∴ τiA0 = τi1A1 + τi2A2 + τi3A3 = τij Aj . But the area of each surface ⊥ to Xi is Ai = A0ni. Therefore τiA0 = τij Aj = τij (A0nj ), where τij Aj is the notation (represents the sum of all components). Thus τi = τij nj for i = 1, 2, 3, where τi is the component of stress in the ith direction on a surface with a normal �n . We call τ i the stress vector and we call τij the stress matrix or tensor.

2

︷︸︸︷

1.7.2 Example: Pascal’s Law for Hydrostatics In a static fluid, the stress vector cannot be different for different directions of the surface normal since there is no preferred direction in the fluid. Therefore, at any point in the fluid, the stress vector must have the same direction as the normal vector �n and the same magnitude for all directions of �n .

no summation

Pascal’s Law for hydrostatics: τij = − (pi) (δij )

⎡ ⎤ −p1 0 0 ⎣ ⎦τ = 0 −p2 0

0 0 −p3

where pi is the pressure acting perpendicular to the ith surface. If p0 is the pressure acting perpendicular to the surface PQR, then τi = −nip0 , but:

τi = τij nj = −(pi)δij nj = −(pi)(ni)

Therefore po = pi , i = 1, 2, 3 and �n is arbitrary.

3

∑

∑

1.7.3 Symmetry of the Stress TensorTo prove the symmetry of the stress tensor we follow the steps:

j

o i

ji

ij

ji

ij

Figure 3: Material element under tangential stress.

1. The of surface forces = body forces + mass× acceleration. Assume no symmetry. Balance of the forces in the ith direction gives:

(δ)(τij )TOP − (δ)(τij )BOTTOM = O(δ2),

since surface forces are ∼ δ2, where the O(δ2) terms include the body forces per unit depth. Then, as δ → 0, (τij )TOP = (τij )BOTTOM .

2. The of surface torque = body moment + angular acceleration. Assume no symmetry. Balance of moments about o gives:

(τjiδ)δ − (τij δ)δ = O(δ3),

since the body moment is proportional to δ3. As δ → 0 , τij = τji.

4

∫∫∫

∫∫∫

1.8 Mass and Momentum Conservation

Consider a material volume ϑm and recall that a material volume is a fixed mass of material. A material volume always encloses the same fluid particles despite a change in size, position, volume or surface area over time.

1.8.1 Mass ConservationThe mass inside the material volume is:

M(ϑm) = ρdϑ

ϑm(t)

Sm(t)

)t(m ϑ

Figure 4: Material volume ϑm(t) with surface Sm(t).

Therefore the time rate of increase of mass inside the material volume is:

d d M(ϑm) = ρdϑ = 0,

dt dt ϑm (t)

which is the integral form of mass conservation for the material volume ϑm.

5

∫∫∫

∫∫∫ ∫∫∫ ∫∫

∫∫∫ ∫∫

1.8.2 Momentum Conservation The fluid velocity inside the material volume in the ith direction is denoted as ui. Linear momentum of the material volume in the ith direction is

ρuidϑ

ϑm(t)

Newton’s law of motion: The time rate of change of momentum of the fluid in the material control volume must equal the sum of all the forces acting on the fluid in that volume. Thus:

d (momentum)i =(body force)i + (surface force)i

dt d

ρuidϑ = Fidϑ + τij nj dS dt ︸︷︷︸

ϑm(t) ϑm(t) Sm(t) τi

∫∫∫ ∫∫ Divergence Theorems For vectors: ∇ · �vdϑ = ⊂⊃ �v.n dS ︸︷︷︸︸︷︷︸

ϑ ∂vj S vjnj ∂xj

∂τijFor tensors: dϑ = ⊂⊃ τij nj dS

∂xj ϑ S

Using the divergence theorems we obtain

∫∫∫ ∫∫∫ ( ) d ∂τij

ρuidϑ = Fi + dϑ dt ∂xj

ϑm(t) ϑm(t)

which is the integral form of momentum conservation for the material volume ϑm.

6

∫∫∫

1.8.3 Kinematic Transport TheoremsConsider a flow through some moving control volume ϑ(t) during a small time interval Δt.Let f (�x, t) be any (Eulerian) fluid property per unit volume of fluid (e.g. mass, momentum,etc.). Consider the integral I(t):

I(t) = f (�x, t) dϑ

ϑ(t)

According to the definition of the derivative, we can write

d I(t + Δt) − I(t)I(t) = lim

dt Δt→0 Δt ⎧ ⎫ ⎪∫∫∫ ∫∫∫ ⎪

1 ⎨ ⎬ = lim f(�x, t + Δt)dϑ − f(�x, t)dϑ

Δt→0 Δt ⎪ ⎪ ⎩ ⎭ ϑ(t+Δt) ϑ(t)

S(t+Δt)

)tt( Δ+ϑ

)t(ϑ S(t)

Figure 5: Control volume ϑ and its bounding surface S at instants t and t + Δt.

7

∫∫∫ ∫∫∫ ∫∫∫

∫∫∫ ∫∫

Next, we consider the steps

1. Taylor series expansion of f(�x, t + Δt) about (�x, t).

∂f f(�x, t + Δt) = f(�x, t) + Δt (�x, t) + O((Δt)2)

∂t

2. dϑ = dϑ + dϑ

ϑ(t+Δt) ϑ(t) Δϑ

where, dϑ = [Un(�x, t)Δt] dS and Un(�x, t) is the normal velocity of S(t).

Δϑ S(t)

S(t)

dS 2

n )t(Ot)t,x(U Δ+Δv

S(t+Δt)

Figure 6: Element of the surface S at instants t and t + Δt.

Putting everything together:

⎧ ⎫ ⎪∫∫∫ ∫∫∫ ∫∫ ∫∫∫ ⎪

d 1 ⎨ ∂f ⎬ I(t) = lim dϑf + Δt dϑ + Δt dSUnf − dϑf + O(Δt)2 (1)

dt Δt→0 Δt ⎪ ∂t ⎪ ⎩ ⎭ ϑ(t) ϑ(t) S(t) ϑ(t)

8

∫∫∫ ∫∫∫ ∫∫

∫∫∫ ∫∫∫ ∫∫

∫∫∫ ∫∫

From Equation (1) we obtain the Kinematic Transport Theorem (KTT), which is equivalent to Leibnitz rule in 3D.

d ∂f(�x, t)f(�x, t)dϑ = dϑ + f(�x, t)Un(�x, t)dS

dt ∂t ϑ(t) ϑ(t) S(t)

For the special case that the control volume is a material volume it is ϑ(t) = ϑm(t) and Un

= �v n v is the fluid particle velocity. The Kinematic Transport Theorem (KTT), · ˆ, where �then takes the form

d ∂f(�x, t)f(�x, t)dϑ = dϑ + f(�x, t)(�v n· ˆ)dS

dt ∂t ︸︷︷︸ ϑm(t) ϑm(t) Sm(t) f(vini)

(Einstein Notation)

Using the divergence theorem,

∇ · α�dϑ = ⊂⊃ α� n· ˆdS ︸︷︷︸︸︷︷︸ ∂ αiniϑ ∂xi

αi S

we obtain the 1st Kinematic Transport Theorem (KTT)

∫∫∫ ∫∫∫ [ ] d ∂f(�x, t)

f (�x, t) dϑ = + ∇ · (f�v) dϑ,dt ∂t ︸︷︷︸

∂ϑm(t) ϑm(t) ∂xi

(fvi)

where f is some fluid property per unit volume.

9

︸︷︷︸

1.8.4 Continuity Equation for Incompressible Flow

• Differential form of conservation of mass for all fluids Let the fluid property per unit volume that appears in the 1st KTT be mass per unit volume ( f = ρ):

∫∫∫ ∫∫∫ [ ] d ∂ρ

0 = ρdϑ = + ∇ · (ρ�v) dϑ ↑ dt ↑ ∂t

conservation ϑm(t) 1stKTT ϑm(t)of mass

But since ϑm is arbitrary the integrand must be ≡ 0 everywhere.

Therefore:

∂ρ + ∇ · (ρ�v) = 0

∂t ∂ρ

+ [�v · ∇ρ + ρ∇ · �v] = 0∂t

Dρ Dt

Leading to the differential form of

Dρ Conservation of Mass: + ρ∇ · �v = 0

Dt

10

• Continuity equation ≡ Conservation of mass for incompressible flow In general it is ρ = ρ(p, T, . . .), but we consider the special case of an incompressible flow, i.e. Dρ = 0 (Lecture 2).

Dt

Note: For a flow to be incompressible, the density of the entire flow need not be constant (ρ(� x, t) = const). As an example consider a flow of more than one incompressible fluids, like water and oil, as illustrated in the picture below.

Constant ρ

fluid particle

oil water fluid particle ρ1

ρ2

Figure 7: Interface of two fluids (oil-water)

Since for incompressible flows Dρ = 0, substituting into the differential form of the Dt

conservation of mass we obtain the

∂viContinuity Equation: ∇ · �v ≡ = 0

∂xi ︸︷︷︸ rate of volume dilatation

11

∫∫∫ ∫∫∫

1.8.5 Euler’s Equation (Differential Form of Conservation of Momentum)

• 2nd Kinematic Transport Theorem ≡ 1st KTT + differential form of conservation of mass for all fluids. If G = fluid property per unit mass, then ρG = fluid property per unit volume

∫∫∫ ∫∫∫ [ ] d ∂

ρGdϑ = (ρG) + ∇ · (ρG�v) dϑ dt ↑ ∂t

ϑm(t) 1stKTT ϑm(t) ⎡ ⎤

∫∫∫ ⎢ ( ) ( )⎥ ⎢ ∂ρ ∂G ⎥ ⎢ ⎥= G + ∇ · ρ�v + ρ + �v · ∇G dϑ ⎢ ∂t ∂t ⎥ ⎣ ⎦ ϑm(t) ︸︷︷︸︸︷︷︸

DG=0 from mass conservation = Dt

The 2nd Kinematic Transport Theorem (KTT) follows:

d DG ρGdϑ = ρ dϑ

dt Dt ϑm ϑm

Note: The 2nd KTT is obtained from the 1st KTT (mathematical identity) and the only assumption used is that mass is conserved.

12

( )

• Euler’s Equation

We consider G as the ith momentum per unit mass (vi). Then,

∫∫∫ ( ) ∫∫∫ ∫∫∫ Fi +

∂τij dϑ =

d ρvidϑ =

Dviρ dϑ

∂xj ↑ dt ↑ Dt ϑm(t) conservation ϑm(t) 2ndKTT ϑm(t)

of momentum

But ϑm(t) is an arbitrary material volume, therefore the integral identity gives

Euler’s equation:

⎛ ⎞

Dvi ⎜∂vi ⎟ ∂τij⎜ ⎟ρ ≡ ρ + �v · ∇vi = Fi + Dt ⎝ ∂t ︸︷︷︸ ⎠ ∂xj

∂vivj ∂xj

And in vector tensor form:

ρD�v ≡ ρ

∂�v + �v · ∇�v = F� + ∇ · τ

Dt ∂t

13


13.021 - Marine Hydrodynamics Lecture 4

Introduction

Governing Equations so far:

Knowns Equations # Unknowns #

density ρ(�x, t)

body force Fi

Continuity

(conservation of mass)

Euler

(conservation of momentum)

1

3

velocities vi(�x, t)

stresses τij (�x, t)

3

©6

4 9

© 3 of the 9 unknowns of the stress tensor are eliminated by symmetry

The number of unknowns (9) is > than the number of equations (4), i.e. we don’t have closure. We need constitutive laws to relate the kinematics vi to the dynamics τij .

1

�

1.9 Newtonian Fluids

1. Consider a fluid at rest (vi ≡ 0). Then according to Pascal’s Law:

τij = −psδij (Pascal’s Law)

⎡ ⎤ −ps 0 0 ⎦τ = ⎣ 0 −ps 0

0 0 −ps

where ps is the hydrostatic pressure and δij is the Kroenecker delta function, equal to 1 if i = j and 0 if i = j.

2. Consider a fluid in motion. The fluid stress is defined as:

τij ≡ −pδij + τij ︸︷︷︸︸︷︷︸ isotropic components all non-isotropic components

on diagonal both on/off diagonal

where p is the thermodynamic pressure and τij are the dynamic stresses. It should be emphasized that −pδij includes all the isotropic components of the stress tensor on the diagonal, while τij represents all the non-isotropic components, which may or may not be on the diagonal (shear and normal stresses). The dynamic stresses τij is related to the velocity gradients by empirical relations.

Experiments with a wide class of ‘Newtonian’ fluids showed that the dynamic stresses are proportional to the rate of strain.

τij

Newtonian Fluid

Fluid

∂uk

∂x m

2

︸︷︷︸

( ) ( )

( ) ( )

︸︷︷︸

( )

τij ≈ linear function of the ( rate of strain ≡ velocity gradient) ︷︸︸︷︷︸︸︷ ∂ ∂X ∂ ∂X ∂uk

= ∂t ∂x ∂x ∂t ∂xm

u

∂uki.e. τij ≈ αijkm i, j, k, m = 1, 2, 3 ︸︷︷︸ ∂xm

34=81 empirical coefficients

(constants for Newtonian fluids)

For isotropic fluids, this reduces to:

∂ui ∂uj ∂ulτij = μ + + λ ,

∂xj ∂xi ∂xl

∇·�v

The fluid properties in the previous relation are:

• μ - (coefficient of) dynamic viscosity.

• λ - bulk elasticity, ‘second’ coefficient of viscosity

∂uiFor incompressible flow, we have shown that ∇ · �v = = 0.

∂xi

Therefore, for an incompressible, isotropic, Newtonian fluid the dynamic or viscous stresses τij are expressed as:

∂ui ∂ujτij = μ +

∂xj ∂xi

3

)

)

)

(

(

(

1.9.1 Discussion on viscous stresses τij

1. Verify that for �v = 0 we recover Pascal’s law.∂ui

Proof: �v = 0 ⇒ = 0 ⇒ τij = 0 ⇒ τij = −pδij + 0 ⇒ hydrostatic conditions ∂xj

2. Verify that τij = τji.

∂ui ∂ujProof: τij = μ + = τji ⇒ symmetry of stress tensor

∂xj ∂xi

3. When τij , i = j the viscous stress is a normal stress and is given by:

∂uiτii = 2μ

∂xi

The normal viscous stresses τii are the diagonal terms of the viscous stress tensor.

The τii in general are not isotropic.

4. When ˆ � j the viscous stress is a shear stress and is given by: τij , i =

∂ui ∂ujτij = μ +

∂xj ∂xi

The shear viscous stresses ˆ � j are the off diagonal terms of the viscous stress τij , i = tensor.

⎤⎡

5. A 2D viscous tensor has the form: μ⎢⎢⎣

∂u ∂u ∂v 2 + ∂x ∂y ∂x

∂u ∂v ∂v + 2

∂y ∂x ∂y

⎥⎥⎦

6. Notation 1: The viscous stresses τij are often referred to (somewhat confusingly) as shear stresses, despite the fact that when i = j the viscous stress is a normal stress.

∂ui ∂uj7. Notation 2: Often τij ↔ τij and τij is used to denote μ ∂xj

+ ∂xi

.

4

︸︷︷︸

︸︷︷︸

( )

( )

( ( ) )

1.10 Navier-Stokes equations(for Incompressible, Newtonian Fluid)

Equations # Unknowns # Continuity 1 velocities vi(�x, t) 3 Euler 3 stresses τij (�x, t) 6 Newtonian 6 pressure p(�x, t) 1 fluid symmetry

10 10

closure

To form the Navier-Stokes equations for incompressible, Newtonian fluids, we first substitute the equation for the stress tensor for a Newtonian fluid, i.e.

∂ui ∂ujτij = −pδij + τij = −pδij + μ +

↑ ∂xj ∂xiNewtonian Fluid

into Euler’s equation:

Dui ∂τijρ = Fi +

Dt ∂xj

∂p ∂ ∂ui ∂uj= Fi − + μ +

∂xi ∂xj ∂xj ∂xi

∂p ∂2ui ∂ ∂uj= Fi − + μ +

∂xi ∂x2 j ∂xi ∂xj

Continuity = 0

Therefore the Navier-Stokes equations for an incompressible, Newtonian fluid in cartesian coordinates are given as:

Dui ∂ui ∂ui 1 ∂p ∂2ui 1 = + uj = − + ν

2 + Fi Tensor form Dt ∂t ∂xj ρ ∂xi ∂xj ρ

D�v ∂�v 1 1= + �v · ∇�v = − ∇p + ν∇2�v + F� Vector form

Dt ∂t ρ ρ

where ν ≡ μρ denoted as the kinematic viscosity [ L2/T ].

5

• Unknowns and governing equations for incompressible, Newtonian fluids

Equations # Unknowns #

Continuity 1 pressure p(�x, t) 1

Navier-Stokes 3 velocities vi(�x, t) 3

4 4

• Values of constants (density, dynamic and kinematic viscosity) used in 13.021

water air units

density ρ 103 1 [kg/m3]

dynamic viscosity μ 10−3 10−2 [kg/ms]

kinematic viscosity ν 10−6 10−5 [m2/s]

• Notation 1: The Continuity and the Navier-Stokes equations form the Governing Equations for incompressible, Newtonian fluids.

Continuity + Navier-Stokes = Governing Equations

Dvi 1 ∂p 2Notation 2: Alternatively, we refer to each equation Dt = −

ρ ∂xi + ν∇ vi + 1

ρ Fi

as the ith Momentum Equation. In this case, the Continuity and the Momentum equations form the Navier-Stokes System of Equations.

Continuity + Momentum Equations = Navier-Stokes System of Equations

Both notations are equivalent, and in this text it will be made clear from the context when the term Navier-Stokes refers to the Momentum Equations or to the System of Governing Equations.

6

1.11 Boundary Conditions

In the previous paragraphs we formulated the governing equations that describe the flow of an incompressible, Newtonian fluid. The governing equations (N-S) are a system of partial differential equations (PDE’s). This 4 × 4 system of equations describes all the incompressible flows, from rain droplets to surface waves.

One of the reasons this system of equations provides such different solutions lies on the variety of the imposed boundary conditions. To complete the description of this problem it is imperative that we specify appropriate boundary conditions. For the N-S equations we need to specify ‘Kinematic Boundary Conditions’ and ‘Dynamic Boundary Conditions’.

1. Kinematic Boundary Conditions specify the boundary kinematics (position, velocity, . . . ). On an impermeable solid boundary, velocity of the fluid = velocity of the body. i.e. velocity continuity.

�v = �u ‘no-slip’ + ‘no-flux’ boundary condition

where �v is the fluid velocity at the body and �u is the body surface velocity

· ˆ = � ˆ• �v n u n· ‘no flux’ ← continuous flow

�v t· ˆ u t ‘no slip’ ← finite shear stress • = � ·

v v

u v

7

2. Dynamic Boundary Conditions specify the boundary dynamics ( pressure, sheer stress, . . . ).

Stress continuity: p = p ′ + p interfaceτij = τij

′ + τij, interface

The most common example of interfacial stress is surface tension.

τp interface, ij interface

τij ' p'

τ ijp

1.12 Surface Tension

• Notation: Σ [Tension force / Length] ≡ [Surface energy / Area].

• Surface tension is due to the intermolecular attraction forces in the fluid.

• At the interface of two fluids, surface tension implies in a pressure jump across the interface. Σ gives rise to Δp across an interface.

• For a water/air interface: Σ = 0.07 N/m. This is a function of temperature, impurities etc. . .

• 2D Example: dθ dθ

cos · Δp Rdθ = 2Σsin ≈ 2Σdθ· 2 2 2 ︸︷︷︸︸︷︷︸

≈1 ≈ dθ 2

∴ Δp = Σ R

Higher curvature implies in higher pressure jump at the interface.

8

( )

∮ ∫ ∫

p

Σ Σ

p’=p+Δp

dθ

R

dθ/2

• 3D Example: Compound curvature

1 1 Δp = + Σ

R1 R2

where R1 and R2 are the principle radii of curvature.

1.13 Body Forces – Gravity

• Conservative forces

F� = −∇ϕ for some ϕ,

where ϕ is the force potential.

2 2

F� · d�x = 0 or F� · d�x = − ∇ϕ · d�x = ϕ(�x1) − ϕ(�x2) 1 1

9

︸︷︷︸

• A special case of a conservative force is gravity F� = −ρgk.

– In this case the gravitational potential is given by ϕg = ρgz. Therefore:

F� = ∇(−ϕg) = ∇(−ρgz) = ∇ps

≡ hydrostatic pressure ps

– Substitute in Navier-Stokes equation

ρD�v

= −∇p + F� +ρν∇2�v Dt ︸︷︷︸

body force

= −∇p + ∇(−ρgz) + ρν∇2�v

– Define: total pressure ≡ hydrostatic pressure + hydrodynamic pressure p = ps + pd

p = −ρgz + pd =⇒

pd = p + ρgz

– Re-write Navier-Stokes:

D�v ( ) ρ = −∇ p + ρgz + ρν∇2�v

Dt ︸︷︷︸ pd = p + ρgz

D�v ρ = −∇pd + ρν∇2�v

Dt

Therefore:

– Presence of gravity body force is equivalent to replacing the total pressure by a dynamic pressure (pd = p − ps = p + ρgz) in the Navier-Stokes(N-S) equation.

– Solve the N-S equation with pd. To calculate the total pressure p simply add back the hydrostatic component p = pd + ps = pd − ρgz.

10



Chapter 2 - Similitude (Keyword: EQUAL RATIOS)

Similitude: Similarity of behavior for different systems with equal similarity parameters.

Prototype ↔ Model(real world) (physical/ analytical/ numerical . . . experiments)

Similitude Similarity Parameters (SP’s) Geometric Similitude Length ratios, angles Kinematic Similitude Displacement ratios, velocity ratios Dynamic Similitude Force ratios, stress ratios, pressure ratios

. . .Internal Constitution Similitude ρ, νBoundary Condition Similitude

. . .

For similitude we require that the similarity parameters SP’s (eg. angles, length ratios, velocity ratios, etc) are equal for the model and the real world.

Example 1 Two similar triangles have equal angles or equal length ratios. In this case the two triangles have geometric similitude.

Example 2 For the flow around a model ship to be similar to the flow around the prototype ship, both model and prototype need to have equal angles and equal length and force ratios. In this case the model and the prototype have geometric and dynamic similitude.

1

︷︸︸︷

︸︷︷︸

2.1 Dimensional Analysis (DA) to Obtain SP’s

2.1.1 Buckingham’s π theoryReduce number of variables → derive dimensionally homogeneous relationships.

1. Specify (all) the (say N) relevant variables (dependent or independent): x1, x2, . . . xN

e.g. time, force, fluid density, distance. . .We want to relate the xi’s to each other I( x1, x2, . . . xN ) = 0

2. Identify (all) the (say P) relevant basic physical units (“dimensions”) e.g. M,L,T (P = 3) [temperature, charge, . . . ].

3. Let π = x1 α1 x2

α2 . . . xNαN be a dimensionless quantity formed from the xi’s. Suppose

xi = CiMmi Lli T ti , i = 1, 2, . . . , N

where the Ci are dimensionless constants. For example, if x1 = KE = 12 MV 2 =

1 M1L2T−2 (kinetic energy), we have that C1 = 1 ,m1 = 1, l1 = 2, t1 = −2. Then 2 2

π = (Cα1 Cα2 . . . CαN )Mα1m1+α2m2+...+αN mN Lα1l1+α2l2+...+αN lN Tα1t1+α2t2+...+αN tN 1 2 N

For π to be dimensionless, we require

N

αimi = 0 P αili = 0 aP × N system of Linear Equations (1)

αiti = 0

Σ notation Since (1) is homogeneous, it always has a trivial solution,

αi ≡ 0, i = 1, 2, . . . , N (i.e. π is constant)

There are 2 possibilities:

(a) (1) has no nontrivial solution (only solution is π = constant, i.e. independent of xi’s), which implies that the N variable xi, i = 1, 2, . . . , N are Dimensionally Independent (DI), i.e. they are ‘unrelated’ and ‘irrelevant’ to the problem.

(b) (1) has J(J > 0) nontrivial solutions, π1, π2, . . . , πJ . In general, J < N , in fact,J = N − K where K is the rank or ‘dimension’ of the system of equations (1).

2

⎫ ⎪⎪⎪⎪⎪⎬

⎧ ⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎭

⎪⎪⎪⎪⎪⎩

2.1.2 Model Law Instead of relating the N xi’s by I(x1, x2, . . . xN ) = 0, relate the J π’s by

F (π1, π2, . . . πJ ) = 0, where J = N − K < N

For similitude, we require

(πmodel)j = (πprototype)j where j = 1, 2, . . . , J.

If 2 problems have all the same πj’s, they have similitude (in the πj senses), so π’s serve as similarity parameters.

Note:

• If π is dimensionless, so is π × const, πconst , 1/π , etc. . .

• If π1, π2 are dimensionless, so is π1 × π2 , ππ1

2 , π1

const1 × π2 const2 , etc. . .

In general, we want the set (not unique) of independent πj ’s, for e.g., π1, π2, π3 or π1, π1

× π2, π3, but not π1, π2, π1 × π2.

Example: Force on a smooth circular cylinder in steady, incompressible flow Application of Buckingham’s π Theory.

F

Uρ,ν D

Figure 1: Force on a smooth circular cylinder in steady incompressible fluid (no gravity)

A Fluid Mechanician found that the relevant dimensional quantities required to evaluate the force F on the cylinder from the fluid are: the diameter of the cylinder D, the fluid velocity U , the fluid density ρ and the kinematic viscosity of the fluid ν. Evaluate the non-dimensional independent parameters that describe this problem.

3

︷︸︸︷

xi : F, U,D, ρ, ν → N = 5

xi = ciMmi Lli T ti → P = 3

N = 5

F U D ρ ν P = 3 mi 1 0 0 1 0

li 1 1 1 -3 2 ti -2 -1 0 0 -1

π = Fα1 Uα2 Dα3 ρα4 να5

For π to be non-dimensional, the set of equations

αimi =0

αili =0

αiti =0

has to be satisfied. The system of equations above after we substitute the values for the mi’s, li’s and ti’s assume the form:

⎞⎛α1 ⎞⎛⎞⎛

1 0 0 1 0 α2

α3

0⎜⎜⎜⎜

⎟⎟⎟⎟1 1 1 −3 2 0⎝ ⎠ =⎝ ⎠

−2 −1 0 0 −1 α4 0⎝ ⎠

α5

The rank of this system is K = 3, so we have j = 2 nontrivial solutions. Two families of solutions for αi for each fixed pair of (α4, α5), exists a unique solution for (α1, α2, α3). We consider the pairs (α4 = 1, α5 = 0) and (α4 = 0, α5 = 1), all other cases are linear combinations of these two.

4

1. Pair α4 = 1 and α5 = 0.

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 0 0 α1 −1

⎝ 0 1 0 ⎠ ⎝ α2 ⎠ = ⎝ 4 ⎠

0 0 1 α3 2

which has solution

⎛ ⎞ ⎛ ⎞α1 −1

⎝ α2 ⎠ = ⎝ 2 ⎠

α3 2

ρU2D2

∴ π1 = F α1 Uα2 Dα3 ρα4 να5 = F

Conventionally, π1 → 2π1 −1 and ∴ π1 = F ≡ Cd, which is the Drag coefficient. 1 ρU2D2

2

2. Pair α4 = 0 and α5 = 1.

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 0 0 α1 0

⎝ 0 1 0 ⎠ ⎝ α2 ⎠ = ⎝ −2 ⎠

0 0 1 α3 −1

which has solution

⎛ ⎞ ⎛ ⎞α1 0

⎝ α2 ⎠ = ⎝ −1 ⎠

α3 −1

ν∴ π2 = F α1 Uα2 Dα3 ρα4 υα5 = UD

Conventionally, π2 → π2 −1 , ∴ π2 = UD ≡ Re, which is the Reynolds number.

ν

Therefore, we can write the following equivalent expressions for the non-dimensional independent parameters that describe this problem:

F (π1, π2) = 0 or π1 = f(π2)

F (Cd, Re) = 0 or Cd = f(Re) F UD F UD

F ( , ) = 0 or = f( )1/2 ρU2D2 ν 1/2 ρU2D2 ν

5

Appendix A

Dimensions of some fluid properties

Quantities Dimensions

(MLT )

Angle θ none (M0L0T 0)

Length L L

Area A L2

Volume ∀ L3

Time t T

Velocity V LT−1

Acceleration V LT−2

Angular velocity ω T−1

Density ρ ML−3

Momentum L MLT−1

Volume flow rate Q L3T−1

Mass flow rate Q MT−1

Pressure p ML−1T−2

Stress τ ML−1T−2

Surface tension Σ MT−2

Force F MLT−2

Moment M ML2T−2

Energy E ML2T−2

Power P ML2T−3

Dynamic viscosity μ ML−1T−1

Kinematic viscosity ν L2T−1

6

�



2.2 Similarity Parameters from Governing Equations and Boundary Conditions

In this paragraph we will see how we can specify the SP’s for a problem that is governed by the Navier-Stokes equations. The SP’s are obtained by scaling, non-dimensionalizing and normalizing the governing equations and boundary conditions.

1. Scaling First step is to identify the characteristic scales of the problem.

For example: Assume a flow where the velocity magnitude at any point in space or time |� x, t) is about equal to a velocity U , i.e. |v(� |v(� | � x, t) = α U , where α is such that 0 ≤ α ∼ O(1). Then U can be chosen to be the characteristic velocity of the flow and any velocity �v can be written as:

�v = U�v �

where it is evident that �v� is:

(a) dimensionless (no units), and

(b) normalized (|�v�| ∼ O(1)).

Similarly we can specify characteristic length, time, pressure etc scales:

Characteristic scale Dimensionless and Dimensional quantity

normalized quantity in terms of characteristic scale

Velocity U �v� �v = U�v�

Length L �x� �x = L�x�

t�Time T t = Tt�

Pressure po − pv p p = (po − pv)p�

1

2. Non-dimensionalizing and normalizing the governing equations and boundary conditions

Substitute the dimensional quantities with their non-dimensional expressions (eg. substitute �v with U�v� , �x with L�x�, etc) into the governing equations, and boundary conditions. The linearly independent, non-dimensional ratios between the characteristic quantities (eg. U , L, T , po − pv) are the SP’s.

(a) Substitute into the Continuity equation (incompressible flow)

∇ · �v = 0 ⇒ U � �∇ · �v = 0 ⇒ L ∇� · �v � = 0

Where all the ()� quantities are dimensionless and normalized (i.e., O(1)), vfor example, ∂��

= O(1).∂x�

(b) Substitute into the Navier-Stokes (momentum) equations

∂�v 1 + (�v · ∇) �v = − ∇p + ν∇2�v − gj ⇒

∂t ρ

U ∂�v� U2 po − pv νU + (�v � · ∇�) �v � = − ∇� p � + (∇�)2�v � − gj

T ∂t� L ρU2 L2

divide through by UL 2 , i.e., order of magnitude of the convective inertia term ⇒

�L �

∂��

p� ν �� gL v o − pv � � � � + (�v · ∇)�v = − (∇p) + ∇2�v − j

UT ∂t ρU2 UL U2

The coefficients ( � ) are SP’s.

2

�

�

Since all the dimensionless and normalized terms ()� are of O(1), the SP’s

( � ) measure the relative importance of each term compared to the convective inertia. Namely,

L Eulerian inertia ∂�v

• ≡ S = Strouhal number ∼ ∼ ∂t

UT convective inertia (�v · ∇)�v

The Strouhal number S is a measure of transient behavior.

For example assume a ship of length L that has been travelling with velocity U for time T . If the T is much larger than the time required to travel a ship length, then we can assume that the ship has reached a steady-state.

L << T ⇒

U L

= S << 1 ⇒ UT

∂�v ignore → assume steady-state

∂t

• po − pv ≡ σ = cavitation number. 1 ρU2

2

The cavitation number σ is a measure of the likelihood of cavitation.

If σ >> 1, no cavitation. If cavitation is not a concern we can choose po

as a characteristic pressure scale, and non-dimensionalize the pressure p as p = pop

• po ≡ Eu = Euler number ∼ pressure force

1 ρU2 inertia force 2

UL inertia force • ≡ Re = Reynold’s number ∼ ν viscous force

If Re >> 1, ignore viscosity.

U2 U � � 1

• = √ ≡ Fr = Froude number ∼ inertia force 2

gL gL gravity force

3

� � �

� � �

� �

� ��

��

(c) Substitute into the kinematic boundary conditions

�u = U� boundary ⇒

�u = Uboundary

(d) Substitute into the dynamic boundary conditions

1 1 � R=LR�

p = pa + Δp = pa + + =⇒ R1 R2 p=(po−pv )p�

Δp

� � � � � 1 1 2 /ρ

p = pa + + = pa + (po − pv) L R1

� R2 � σ U2L

U2L • � ≡ We = Weber number ∼ inertial forces /ρ surface tension forces

• Some SP’s used in hydrodynamics (the table is not exhaustive):

SP Definition

Reynold’s number Re UL ν ∼ inertia

viscous

Froude number Fr

� U2

gL ∼ inertia gravity

Euler number Eu po

1 2 ρU2 ∼ pressure

inertia

Cavitation number σ po−pv 1 2 ρU2 ∼ pressure

inertia

Strouhal number S L UT ∼ Eulerian inertia

convective inertia

Weber number We U2L Σ/ρ ∼ inertia

surface tension

4

��

� �

2.3 Similarity Parameters from Physical Arguments

Alternatively, we can obtain the same SP’s by taking the dimensionless ratios of significant flow quantities. Physical arguments are used to identify the significant flow quantities. Here we obtain SP’s from force ratios. We first identify the types of dominant forces acting on the fluid particles. The SP’s are merely the ratios of those forces.

1. Identify the type of forces that act on a fluid particle:

1.1 Inertial forces ∼ mass × acceleration ∼ (ρL3) UL

2 = ρU2L2

∂u � � 1.2 Viscous forces ∼ μ × area ∼ μU

L (L2) = μUL ∂y

shear stress

1.3 Gravitational forces ∼ mass × gravity ∼ (ρL3)g

1.4 Pressure forces ∼ (po − pv)L2

2. For similar streamlines, particles must be acted on forces whose resultants are in the same direction at geosimilar points. Therefore, the following force ratios must be equal:

inertia ρU2L2 UL • ∼ = ≡ Reviscous μUL ν

� �1/2 � �1/2inertia ρU2L2 U • ∼ = √ ≡ Frgravity ρgL3 gL

� 1 inertia �−1

(po − pv)L2 po − pv • 2 ∼ = ≡ σ

pressure 12 ρU2L2 1

2 ρU2

5

� �

2.4 Importance of SP’s

• The SP’s indicate whether different systems have similar flow properties.

• The SP’s provide guidance in approximating complex physical problems.

Example A hydrofoil of length L is submerged in a known fluid (density ρ, kinematic viscosity ν). Given that the hydrofoil is travelling with velocity U and the gravitational acceleration is g, determine the hydrodynamic force F on the hydrofoil.

ρ ,νF g U

L

SP’s for this problem:

L po − pv U2L U UL S = , σ = 1 , We = � , Fr = √ , Re =

UT2 ρU2 /ρ gL ν

We define the dimensionless force coefficient:

F CF ≡ 1 ρU2L2

2

The force coefficient must depend on the other SP’s:

CF = CF �

(S, σ,We, Fr, Re) or CF = CF S, σ−1,We −1, Fr, Re

−1

Procedure We will first study under what conditions each SP → 0. We will estimate CF for the case that all of the SP’s → 0.

6

� �

� �

1. Significance of the Strouhal number S = L/UT .

Change S keeping all other SP’s (σ−1 , We −1 , Fr, Re

−1) fixed.

Steady-State

transientCF

Steady-State

transientCF

S-1 = UT / LS-1 = UT / L S~O(S~O 1)(1)

Exact position of the cutExact position of the cut depends on the problem anddepends on the problem and the quantities of interest.the quantities of interest.

∂For S << 1, assume steady-state: ∂t = 0

For S >> 1, unsteady effect is dominant.

For example, for the case L = 10m and U = 10m/s we can neglect the unsteady effects when:

L L S << 1 ⇒ << 1 ⇒ T >> ⇒ T >> 1s

UT U

Therefore for T >> 1s we can approximate S 1 and we can assume steady state. In the case of a steady flow:

CF = CF S 0, σ−1,We −1, Fr, Re

−1 ⇒

∼ −1CF = CF σ−1, We −1, Fr, Re

7

po − pv2. Significance of the cavitation number σ = 1 .

ρU2 2

Change σ−1 keeping all other SP’s (S 0, We −1 , Fr, Re

−1) fixed.

Some comments on cavitation pv : Vapor pressure, the pressure at which water boils po ≤ pv : State of fluid changes from liquid to gas ⇒ CAVITATION Consequences : Unsteady → Vibration of structures, which may lead to fatigue, etc

Unstable → Sudden cavity collapses → Large force acting on the structure surface → Surface erosion

CFCCFF

Strong cavitation No cavitation

σ

Strongcavitation No cavitation

σ

Strongcavitation No cavitation

σσinception

For σ << 1, cavitation occurs.

For σ >> 1 ⇒ σ−1 << 1, cavitation will not occur.

In general cavitation occurs when we have large velocities, or when po ∼ pv

8

�

� �

� �

For example, assume a hydrofoil travelling in water of density ρ = 103kg/m3 .

The characteristic pressure is po = 105N/m2 and the vapor pressure is pv = 103N/m2 . Cavitation will not occur when:

1 ρU2 po − pvσ−1 << 1 ⇒ 2 << 1 ⇒ U << 1 ⇒ U << 14m/s

po − pv 2 ρ

Therefore for U << 14m/s it is σ >> 1 ⇒ σ−1 0 and cavitation will not occur.

In the case of a steady, non-cavitating flow:

CF = CF 0, σ−1 0,We −1, Fr, Re

−1 ⇒

∼ −1CF = CF We −1, Fr, Re

9

�

� �

� �

U2L 3. Significance of the Weber number We = .

Σ/ρ

Change We −1 keeping the other SP’s (S 0, σ−1 0, Fr, Re

−1) fixed.

For We << 1, surface tension is significant.

For We >> 1 ⇒ We −1 << 1, surface tension is not significant.

For example, assume a hydrofoil travelling with velocity U = 1m/s near an air/water interface (water density ρ = 103kg/m3, surface tension coefficient = 0.07N/m).

Surface tension can be neglected when: Σ Σ

W−1 << 1 ⇒ ρ << 1 ⇒ L >> ρ ⇒ · mL >> 7 10−5

e 2U2L U

Therefore for L >> 7 10−5 m it is We >> 1We −1 1 and surface tension effects can ·

be neglected.

So in the case of a steady, non-cavitating, non-surface tension flow:

CF = CF 0, 0,We −1 0, Fr, Re

−1 ⇒

∼CF = CF Fr, Re −1

10

U 4. Significance of the Froude number Fr = √ , which measures the ‘gravity effects’.

gh

Change Fr keeping the other SP’s (S 0, σ−1 0, We −1 0, Re

−1) fixed.

‘Gravity effects’, hydrostatic pressure do not create any flow (isotropic) nor do they change the flow dynamics unless Dynamic Boundary Conditions apply.

√ √ ‘Gravity effects’ are not significant when U << gh ⇒ Fr 0, or U >> gh ⇒ Fr

−1 0. Physically, this is the case when the free surface is

• absent or

• far away or

• not disturbed, i.e., no wave generation.

The following figures (i - iv) illustrate cases where gravity effects are not significant.

11

� �

� �

In any of those cases the gravity effects are insignificant and equivalently Fr is not important (i.e. Fr 0 or Fr

−1 0).

So in the case of a steady, non-cavitating, non-surface tension, with no gravity effects flow:

CF = CF 0, 0, 0, Fr 0 or Fr −1 0, R−

e 1 ⇒

∼ −1CF = CF Re

A look ahead: Froude’s Hypothesis

Froude’s Hypothesis states that

CF = CF (Fr, Re) = C1 (Fr) + C2 (Re)

Therefore dynamic similarity requires

(Re)1 = (Re)2, and

(Fr)1 = (Fr)2

Example: Show that if ν and g are kept constant, two systems (1, 2) can be both geometrically and dynamically similar only if:

L1 = L2, and

U1 = U2

12

� �

UL 5. Significance of the Reynolds number Re = .

ν

Change Re keeping the other SP’s (S 0, σ−1 0, We −1 0, Fr 0 or Fr

−1 0) fixed.

Recall that for a steady, non-cavitating, non-surface tension, with no gravity effects flow:

CF = CF Re −1

CFCF

ReRe

Sphere

Plate

(Re)cr

Sphere

Plate

(Re)crLaminar Turbulent

Transition

Re << 1, Stokes flow (creeping flow) Re < (Re)cr, Laminar flow Re > (Re)cr, Turbulent flow Re → ∞, Ideal fluid

For example, a hydrofoil of cord length L = 1m travelling in water (kinematic viscosity ν = 10−1m2/s) with velocity U = 10m/s has a Reynolds number with respect to L:

ν Re = = 107 → ideal fluid, and Re

−1 0 UL

13

Therefore for a steady, non-cavitating, non-surface tension, with no-gravity effects flow in an ideal fluid:

CF = CF (0, 0, 0, 0, 0) = constant = 0

→D’Alembert’s Paradox No drag force on moving body in ideal fluid.

14



Chapter 3 – Ideal Fluid Flow

The structure of Lecture 7 has as follows: In paragraph 3.0 we introduce the concept of inviscid fluid and formulate the governing equations and boundary conditions for an ideal fluid flow. In paragraph 3.1 we introduce the concept of circulation and state Kelvin’s theorem (a conservation law for angular momentum). In paragraph 3.2 we introduce the concept of vorticity.

⎧ Inviscid Fluid ν = 0⎪ ⎨

+Ideal Fluid Flow ≡ ⎪ Dρ ⎩ Incompressible Flow (§ 1.1) = 0 or ∇ · �v = 0 Dt

1

3.0 Governing Equations and Boundary Conditions for Ideal Flow

• Inviscid Fluid, Ideal Flow

Recall Reynolds number is a qualitative measure of the importance of viscous forces compared to inertia forces,

UL inertia forces Re = =

ν viscous forces

For many marine hydrodynamics problems studied in 13.021 the characteristic lengths and velocities are L ≥ 1m and U ≥ 1m/s respectively. The kinematic viscosity in water is νwater = 10−6m2/s leading thus to typical Reynolds numbers with respect to U and L in the order of

UL Re = ≥ 106 >>> 1 ⇒

ν

1 viscous forces ∼ � 0 Re inertia forces

This means that viscous effects are << compared to inertial effects - or confined within very small regions. In other words, for many marine hydrodynamics problems, viscous effects can be neglected for the bulk of the flow.

Neglecting viscous effects is equivalent to setting the kinematic viscosity ν = 0, but

ν = 0 ⇔ inviscid fluid

Therefore, for the typical marine hydrodynamics problems we assume

incompressible flow + inviscid fluid ≡ ideal fluid flow

which turns out to be a good approximation for many problems.

2

� �

︸︷︷︸

• Governing Equations for Ideal Fluid Flow

– Continuity Equation:

∇ · �v = 0

– Momentum (Navier-Stokes ⇒ Euler) equations:

∂�v 1 + �v · ∇�v = − ∇p − gj

∂t ρ

By neglecting the viscous stress term (ν∇2�v) the Navier-Stokes equations reduceto the Euler equations. (Careful not to confuse this with the Euler equation in§1.6).

The N-S equations are second order PDE’s with respect to space

(2nd order in ∇2), thus: (a) require 2 kinematic boundary conditions, and (b)produce smooth solutions in the velocity field.

The Euler equations are first order PDE’s, thus: (a) require 1 kinematic boundary condition, and (b) may allow discontinuities in the velocity field.

• Boundary Conditions for Euler equations (Ideal Flow):

– KBC:

v n = u n ← ‘no flux’ + free (to) slip · · ˆ = Un

given

Note: ‘No slip’ condition �v t· ˆ= U� · t does not apply.

The ‘no slip’ condition is required to ensure that the velocity gradients are finite and therefore the viscous stresses τij are finite.

But since ν = 0 the viscous stresses are identically zero (τij ∝ μ = ρν = 0) and the velocity gradients can be infinite. Or else the velocity field need not be continuous.

3

∂u Viscousflow τ ∝ μ

w ∂y < ∞ < ∞

∂uU Inviscidflow τ ∝0

w ∂y

– DBC:

)dy(u

)0(u

y

p = . . . Pressure given on the boundary

Similarly to the argument for the KBC, viscous stresses τij cannot be specified on any boundary since ν = 0.

• Summary of consequences neglecting viscous effects this far:

– Neglecting viscous effects is equivalent to setting the kinematic viscosity equal to zero:

ν = 0

Setting ν = 0 ⇐⇒ inviscid fluid

– Setting ν = 0 the viscous term in the Navier-Stokes equations drops out and we obtain the Euler equations.

The Euler equations are 1st order PDE’s in space, thus (a) require only one boundary condition for the velocity and (b) may allow for velocity jumps.

– Setting ν = 0 all the viscous stresses τij ∝ μ = ρν are identically 0. This may allow for infinite velocity gradients.

This affects (a) the KBC, allowing free slip, and (b) the DBC, where no viscous stresses can be specified on any boundary.

4

∫

︸︷︷︸

3.1 Circulation – Kelvin’s Theorem

3.1.1 Γ ≡ Instantaneous circulation around any arbitrary closed contour C.

C xdv v

v

Γ = � · �v dx C tangential

velocity

The circulation Γ is an Eulerian idea and is instantaneous, a ‘snapshot’.

5

3.1.2 Kelvin’s Theorem (KT) :

For ideal fluid under conservative body forces,

dΓ = 0 following any material contour C,

dt

i.e., Γ remains constant under for Ideal Fluid under Conservative Forces (IFCF).

This is a statement of conservation of angular momentum.

(Mathematical Proof: cf JNN pp 103)

Kinematics of a small deformable body:

(a) Uniform translation → Linear momentum

(b) Rigid body rotation → Angular momentum

(c) Pure strain→ No linear or angular momentum involved (no change in volume

(d) Volume dilatation

For Ideal Fluid under Conservative body Forces:

(a) Linear momentum → Can change

(b) Angular momentum → By K.T., cannot change

(c) Pure strain→ Can change

(d) Volume dilatation → Not allowed (incompressible fluid)

Kelvin’s Theorem is a statement of conservation of angular momentum under IFCF.

6

∣ ∣ ∣∣ ∣ ∣

∣ ∣ ∣∣ ∣ ∣

∣ ∣ ∣∣ ∣ ∣

Example 1: Angular momentum of point mass.

v1 v2 m1

m2r2

r1

θ

Angular momentum of point mass:

L� = |�r × (m�v)| = mvr = mr2θ

Conservation of angular momentum:

L� = L�1 2

m1=m2 m1v1r1 = m2v2r2 =⇒

v1r1 = v2r2 or

r12θ

1 = r22θ

2

Conservation of angular momentum does not imply constant angular velocity:

Angular Momentum � angular velocity ω�

7

∫ ∫

∫ ∫

Example 2: Conservation of circulation around a shrinking circular material volume Vm.

2r 2υ

1υ 1r

m V m V

2π 2π

Γ1 = dθr1v1 = dθr2v2 = Γ2

0 0

Example 3: Conservation of circulation around a shrinking arbitrary material volume Vm, Cm.

Γ1 = �v1 · d�x = �v2 · d�x = Γ2 C1 C2

8

( ) ( ) ( )

3.2 Vorticity

3.2.1 Definition of Vorticity

� v = ∂w −

∂v i −

∂w − ∂u

j + ∂v −

∂u kω = ∇× �

∂y ∂z ∂x ∂z ∂x ∂y

Relationship of vorticity to circulation - Apply Stokes’ Theorem:

∮ ∫∫ ∫∫ Γ = � d� x = (∇× � · ndS = � ndS ≡ Flux of vorticity out of S v · v) ˆ ω · ˆ

C any S ‘covering’ C any S ‘covering’ C

3.2.2 What is Vorticity?

For example, special case: 2D flow - w = 0; ∂ = 0; ωy = ωx = 0 and ∂z

∂v ∂u ωz = −

∂x ∂y

(a) Translation: u = constant, v = constant

time t + Δt

jiu ˆˆ υ+

jiu ˆˆ υ+

jiu ˆˆ υ+

jiu ˆˆ υ+ time t

∂v ∂u = 0, = 0 ⇒ ωz = 0 → no vorticity

∂x ∂y

9

(b) Pure Strain (no volume change):

Areat + Δt

Areat

No volume change Areat = Areat + Δt

∂u ∂v ∂u ∂v = − ; u = -v; = 0; = 0 ⇒ ωz = 0

∂x ∂y ∂y ∂x

10

(c) Angular deformation

rυ

= ⎛⎜⎝

dy⎞⎟⎠Δt∂uδ x ∂y

= ∂u + ∂υ

υr υr

dy

dx

t + Δt∂yi ∂y

jdy ˆ dy ˆ time

δ y = (∂υ dx)Δt∂x time t

∂u i + ∂υ ∂ ∂dx dx j= 0 = x x

∂u ∂v � = → δx = δy( for dx =ω = 0 only if dy)

∂y ∂x

11

�

(d) Pure rotation with angular velocity Ω

0=υr jdxΩ=υr

idyΩ−=υr

dy

dx

ttime

tt Δ+time

Ω

tΔΩ

tΔΩ

∂v ∂u = Ω; = −Ω; ωz = 2Ω

∂x ∂y

i.e. vorticity ∝ 2(angular velocity).

3.2.3 Irrotational Flow

A flow is irrotational if the vorticity is zero everywhere or if the circulation is zero along any arbitrary closed contour:

ω ≡ 0 everywhere ⇔ Γ ≡ 0 for any C

Further on, if at t = to, the flow is irrotational, i.e., Γ ≡ 0 for all C, then Kelvin’s theorem states that under IFCF, Γ ≡ 0 for all C for all time t:

once irrotational, always irrotational

(Special case of Kelvin’s theorem)

12



In Lecture 8, paragraph 3.3 we discuss some properties of vortex structures. In paragraph 3.4 we deduce the Bernoulli equation for ideal, steady flow.

r u

3.3 Properties of Vortex Structures

3.3.1 Vortex Structures

r

r

• A vortex line is a line everywhere tangent to �ω.

vortex line

ω2

Ω2

ω1

Ω11

r u2

• A vortex tube (filament) is a bundle of vortex lines.

vortex tube

ωr u r

vortex lines

1

• A vortex ring is a closed vortex tube.

A sketch and two pictures of the production of vortex rings from orifices areshown in Figures 1, 2, and 3 below.

(Figures 2,3: Van Dyke, An Album of Fluid Motion 1982 p.66, 71)

side view

v u

U

v u

Γ

v u

ωv

cross section v u ωv

ωvv u

U

Figure 1: Sketch of vortex ring production

2

∫∫∫ ∫∫ ︸︷︷︸

3.3.2 No Net Flux of Vorticity Through a Closed Surface

Calculus identity, for any vector �v:

∇ · (∇× �v) = 0 ⇒ ︸︷︷︸ �ω

∇ · �ω = 0 ⇒

∇ · �ω = © ω� · n dS = 0 ⇑

V Divergence S vorticity fluxTheorem

i.e. The net vorticity flux through a closed surface is zero.

(a) No net vorticity flux through a vortex tube:

(Vorticity Flux)in = (Vorticity Flux)out ⇒ (�ω · n)in δAin = (ω� · n)out δAout

0ˆ =⋅ω n v

(ωv ⋅ n)out

(ωv ⋅ n)in

(b) Vorticity cannot stop anywhere in the fluid. It either traverses the fluid beginning or ending on a boundary or closes on itself (vortex ring).

ωr rω

3

∮ ∫∫

3.3.3 Conservation of Vorticity Flux

0 = Γ3 = �v · d�x = � ndS = 0ω · ˆC3 S3

C1 C2 C1 C2

C3C1n 3

2n

∮ ∫∫ ∫∫ Γ1 = �v · d�x = ω� · n1dS = ω� · n2dS = Γ2

C1 S1 S2

Therefore, circulation is the same in all circuits embracing the same vortex tube. For the special case of a vortex tube with ‘small’ area:

Γ = ω1A1 = ω2A2

ω1 ω2

A1 A2

An application of the equation above is displayed in the figure below:

ω1

A1

ω2

A2 = 2A1

ω2 = ω1/2

4

∮ ∫∫

3.3.4 Vortex Structures are Material Structures

Consider a material patch Am on a vortex tube at time t.

Am

∂Am

By definition,

�ω · n = 0 on An

Then,

Γ∂Am = � x = ω · ˆv · d� � nds = 0

∂Am Am

At time t + Δt, Am moves, and for an ideal fluid under the influence of conservative body forces, Kelvin’s theorem states that

Γ∂Am = 0

So, �ω · n = 0 on Am still, i.e., Am still on the vortex tube. Therefore, the vortex tube is a material tube for an ideal fluid under the influence of conservative forces. In the same manner it can be shown that a vortex line is a material line, i.e., it moves with the fluid.

5

3.3.5 Vortex stretching

Consider a small vortex filament of length L and radius R, where by definition �ω is tangent to the tube.

AR

Γ = ωA = constant (in time) ↑ ↑

Stokes Kelvins’ Theorem Theorem

But tube is material with volume = AL = πR2L = constant in time (continuity)

Γ ωA ω∴ = = = constant Volume LA L

As a vortex stretches, L increases, and since the volume is constant (from continuity), A and R decrease, and due to the conservation of the angular momentum, ω increases. In other words,

Vortex stretching ⇔ L ↑ ⇒ ω ↑ (conservation of angular momentum)

⇒ A and R ↑ (continuity)

6

3.3.6 Summary on Vortex Structures

ωr

R

r Aωr

Γ Γ

Vortex ring length L = 2πR [L]

Cross sectional area A = πr2 [L2]

Vortex ring volume ∀ = AL = ⇑

continuity

const [L3]

Vorticity �ω = ∇× �v [T−1]

Circulation Γ = ⇑

Kelvin’s theorem

const

Γ = ⇑

vorticity flux through A

ωA = const

Γ ∝ Ur = const

[L2T−1]

[L2T−1]

[L2T−1]

7

√

Continuity relates length ratios

∀ = LA = const

⎧ ⎪⎪⎪⎪⎪⎪⎪⎨

A ∝ ∀ L

∴ as L ↑ A ↓

⎪⎪⎪⎪⎪⎪⎪⎩ r ∝

√ ∀ ∴ as L ↑ r ↓ L

Kelvin’s theorem + Continuity relate length ratios to Γ, ω, U

√Γ r∝ ∀/L L

Ur ∝ Γ = const → U ∝ → U ∝ Γ ∴ as L ↑ U ↑ r ∀

Γ A∝∀/L L ωA ∝ Γ = const → ω ∝ → ω ∝ Γ ∴ as L ↑ ω ↑

A ∀

Example 1: Example 2:

ωr A 22

1ωr

Γ

1A

ωr1

1Γ

1A 2Γ

2A

1L 2L

ωr 2

A1 < A2

ω1 > ω2

L1 < L2

A1 > A2

Γ1 = Γ2

ω1 < ω2

U1 < U2

Given From continuity only From Kelvin’s theorem From Kelvin’s theorem + continuity From Kelvin’s theorem + continuity

8

( ) ⇀

( )

( )

( ) ( )

︸︷︷︸

3.4 Bernoulli Equation for Steady ( ∂ = 0), Ideal(ν = 0),∂t Rotational flow

p = f v Viscous flow: Navier-Stokes’ Equations (Vector Equations)

p = f(|�v|) Ideal flow: Bernoulli Equation (Scalar equation)

Steady, inviscid Euler equation (momentum equation):

�v · ∇�v = −∇ p

+ gy (1)ρ

From Vector Calculus we have

∇ (�u · �v) = (�u · ∇) �v + (�v · ∇) �u + �u × (∇× �v) + �v × (∇× �u) ⇒

∇(1 |�v|2) = �v · ∇�v + �v × (∇× �v) ⇒2

2

�v · ∇�v = ∇ v − �v × (∇× �v) where v 2 ≡ �v · �v = |�v|2

2

From the previous identity and Equation (1) we obtain

2

�v · (1) → �v · ∇ v − �v · �v × (∇× �v) = −�v · ∇

p + gy

2 ︸︷︷︸ ρ ⊥�v

0

�v · momentum (1) → energy

Therefore, ( ) ( ) v2 p D v2 p�v · ∇ 2 +

ρ + gy = 0 = Dt 2 +

ρ + gy

streamline pathline i.e., v

2

2 +

ρp + gy = constant on a streamline

In general, v2

2 + p

ρ + gy = F (Ψ) where Ψ is a tag for a particular streamline.

Assumptions: Ideal fluid, Steady flow, Rotational in general.

9

3.4.1 Example: Contraction in Water or Wind Tunnel

Contraction Ratio: γ = R1/R2 >> 1 ( γ = O(10) for wind tunnel ; γ = O(5) for water tunnel)

¯ ¯Let U1 and U2 denote the average velocities at sections 1 and 2 respectively.

¯ ( )2

1. From continuity: U1

( πR1

2 )

= U2

( πR2

2 ) →

U¯

2 =

R1 = γ2 >> 1

U1 R2

2.

∂u Since ω = 0 → vortex ring. = 0 , �

∂r

10

( ) ( )

[ [

⎧ ⎪⎪⎪⎪⎨

ω1 ω2 ω2 R2 1 = → = ∼ << 1

2πR1 2πR2 ω1 R1 γ ω/L = constant ⇒⎪⎪⎪⎪⎩

∂u ∂u ∂u since ω ∼ → <<

∂r ∂r 2 ∂r 1

i.e.,

Section 2 Section 1

¯ ¯3. Near the center, let U1 = U1 (1 + ε1) and U2 = U2 (1 + ε2) where ε1 and ε2 measure the relative velocity fluctuations. Apply the Bernoulli equation along a reference average streamline

P1 + 1 ρU12 = P2 + 1 ρU2

2 (2)2 2

Apply Bernoulli Equation to a particular streamline

¯P1 + 21 ρ U1 (1 + ε1) = P2 +

]2 ]21 U2 (1 + ε2)ρ2

From (2) and (3) we obtain

ε1U12 = ε2U2

2 + O(ε2) → ε2 ∼

U12

∼ 1

<< 1 ε1 U2

2 γ4

11

( )

( )

( )



Lecture 9 is structured as follows: In paragraph 3.5 we return to the full Navier-Stokes equations (unsteady, viscous momentum equations) to deduce the vorticity equation and study some additional properties of vorticity. In paragraph 3.6 we introduce the concept of potential flow and velocity potential. We formulate the governing equations and boundary conditions for potential flow and finally introduce the stream function.

3.5 Vorticity Equation

Return to viscous incompressible flow. The Navier-Stokes equations in vector form

∂�v + �v · ∇�v = −∇

p + gy + ν∇2�v

∂t ρ

By taking the curl of the Navier-Stokes equations we obtain the vorticity equation. In detail and taking into account ∇× �u ≡ ω� we have

∇× (Navier-Stokes) → ∇× ∂�v

+ ∇× (�v · ∇�v) = −∇×∇ p

+ gy + ∇× ( ν∇2�v

)

∂t ρ

The first term on the left side, for fixed reference frames, becomes

∂�v ∂ ∂�ω ∇× = (∇× �v) =∂t ∂t ∂t

In the same manner the last term on the right side becomes

∇× ν∇2�v = ν∇2ω�

Applying the identity ∇ × ∇ · scalar = 0 the pressure term vanishes, provided that the density is uniform ( )

∇× ∇(p

+ gy) = 0 ρ

1

( )

( )

The inertia term �v · ∇�v, as shown in Lecture 8, §3.4, can be rewritten as

1 v2

�v · ∇�v = ∇ (�v · �v) − �v × (∇× �v) = ∇ − �v × �ω where v 2 ≡ |�v|2 = �v · �v 2 2

and then the second term on the left side can be rewritten as

2v ∇× (�v · ∇) �v = ∇×∇ −∇× (�v × �ω) = ∇× (�ω × �v)2

= (�v · ∇) �ω − (�ω · ∇) �v + �ω (∇ · �v) + �v (∇ · �ω) ︸︷︷︸︸︷︷︸ =0 =0 since

incompressible ∇·(∇×�v)=0 fluid

Putting everything together, we obtain the vorticity equation

D�ω = (ω� · ∇) �v + ν∇2�ω

Dt

Comments-results obtained from the vorticity equation

• Kelvin’s Theorem revisited - from vorticity equation:

If ν ≡ 0, then D�ω = (�ω · ∇) �v, so if ω� ≡ 0 everywhere at one time, ω� ≡ 0 always.Dt

• ν can be thought of as diffusivity of vorticity (and momentum), i.e., �ω once generated (on boundaries only) will spread/diffuse in space if ν is present.

ωvωvωvωv

DDv v vv

== υυ∇∇2v2v vv ++ ... D ... Dωvωv == υυ∇∇2ω2 vωv ++ ......

Dt DtDt Dt

2

∂T • Diffusion of vorticity is analogous to the heat equation: = K∇2T , where K is the ∂t

heat diffusivity.

Numerical example for ν ∼ 1 mm2/s. For diffusion time t = 1 second, diffusion (√ ) distance L ∼ O νt ∼ O (mm). For diffusion distance L = 1cm, the necessary diffusion time is t ∼ O (L2/ν) ∼ O(10)sec.

• In 2D space (x, y),

∂ �v = (u, v, 0) and ≡ 0

∂z

So, ω� = ∇× �v is ⊥ to �v (ω� is parallel to the z-axis). Then,⎛ ⎞

⎜ ∂ ∂ ∂ ⎟(�ω · ∇) �v = ⎝ ωx + ωy + ωz ⎠ �v ≡ 0, ︸︷︷︸ ∂x ︸︷︷︸ ∂y ∂z

0 0 ︸︷︷︸

0

so in 2D we have

D�ω = ν∇2ω�

Dt

If ν = 0, D�ω = 0, i.e., in 2D following a particle the angular velocity is conserved. Dt

Reason: In 2D space the length of a vortex tube cannot change due to continuity.

3

• In 3D space,

Dωi ∂vi ∂2ωi = ωj + ν

Dt ∂xj ∂xj ∂xj ︸︷︷︸︸︷︷︸ vortex turning and stretching diffusion

for example,

Dω2 ∂u2 ∂u2 ∂u2 = ω1 + ω2 + ω3 + diffusion

Dt ∂x1 ∂x2 ∂x3 ︸︷︷︸︸︷︷︸︸︷︷︸ vortex turning vortex stretching vortex turning

1xx ≡

2xy ≡

3xz ≡

1xx ≡

2xy ≡

3xz ≡

02 =u

0dy 2

2 > ∂ ∂

x

u

dy

43421 ratestretching vortex

2

2

2 00 >⇒>∂ ∂

Dt

D

x

u ω 43421

rateturningvortex

2

3

2 00 0

3 >⎯⎯ →⎯>∂ ∂ >

Dt

D

x

u ωω

02 =u

0dz 3

2 >∂ ∂

x

u

dz

4

3.5.1 Example: Pile on a River

Scouring

What really happens as length of the vortex tube L increases?

IFCF is no longer a valid assumption.

Why?

Ideal flow assumption implies that the inertia forces are much larger than the viscous effects. The Reynolds number, with respect to the vortex tube diameter D is given by

UD Re ∼

ν As the vortex tube length increases ⇒ the diameter D becomes really small ⇒ Re is not that big after all.

Therefore IFCF is no longer valid.

5

) (

3.6 Potential Flow

Potential Flow (P-Flow) is an ideal and irrotational fluid flow

+

⎫ Inviscid Fluid ν = 0 ⎬

Ideal Flow + ⎭ ∇ ·≡ Incompressible Flow � = 0P-Flow v

Irrotational Flow �ω = 0 Γ = 0or

⎧ ⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

3.6.1 Velocity potential

For ideal flow under conservative body forces by Kelvin’s theorem if �ω ≡ 0 at some time t, then �ω ≡ 0 ≡ irrotational flow always. In this case the flow is P-Flow.

Given a vector field �v for which �ω = ∇ × �v ≡ 0, there exists a potential function (scalar) - the velocity potential - denoted as φ, for which

�v = ∇φ

Note that � v = ∇×∇φ ≡ 0ω = ∇× �

for any φ, so irrotational flow guaranteed automatically. At a point �x and time t, the velocity vector �v(�x, t) in cartesian coordinates in terms of the potential function φ(�x, t) is given by

∂φ ∂φ ∂φ �v (�x, t) = ∇φ (�x, t) = , ,

∂x ∂y ∂z

6

φ (x)

u u

∂φ u = 0 ∂φ> 0 < 0 ∂x ∂x u > 0 u < 0

from low φ ⎯⎯→ to high φ

x

The velocity vector �v is the gradient of the potential function φ, so it always points towards higher values of the potential function.

3.6.2 Governing Equations and Boundary Conditions for Potential Flow

(a) Continuity

∇ · �v = 0 = ∇ · ∇φ ⇒ ∇2φ = 0

Number of unknowns → φ

Number of equations → ∇2φ = 0

Therefore we have closure. In addition, the velocity potential φ and the pressure p are decoupled. The velocity potential φ can be solved independently first, and after φ is obtained we can evaluate the pressure p.

p = f (�v) = f (∇φ) → Solve for φ, then find pressure.

7

( ) ( )

( ) ( ) ( )

{ }

[ ]

(b) Bernoulli equation for P-Flow

This is a scalar equation for the pressure under the assumption of P-Flow for steady or unsteady flow.

Euler equation:

∂�v v2 p+ ∇ − �v × �ω = −∇ + gy

∂t 2 ρ

Substituting �v = ∇φ and �ω = 0 into Euler’s equation above, we obtain

∂φ 1 2 p∇ + ∇ |∇φ| = −∇ + gy∂t 2 ρ

or

∂φ 1 2 p∇ + |∇φ| + + gy = 0,∂t 2 ρ

which implies that

∂φ 1 2 p+ |∇φ| + + gy = f(t)

∂t 2 ρ

everywhere in the fluid for unsteady, potential flow. The equation above can be written as

p = −ρ ∂φ

+1 |∇φ|2 + gy + F (t)

∂t 2

which is the Bernoulli equation for unsteady or steady potential flow.

DO NOT CONFUSE WITHBERNOULLI EQUATION FROM § 3.4,

USED FOR STEADY, ROTATIONAL FLOW

8

( )

( )

Summary: Bernoulli equationS for ideal flow.

(a) For steady rotational or irrotational flow along streamline:

1 2 p = −ρ v + gy + C(ψ)2

(b) For unsteady or steady irrotational flow everywhere in the fluid:

∂φ 1 2 p = −ρ + |∇φ| + gy + F (t)∂t 2

∂(c) For hydrostatics, �v ≡ 0, ∂t = 0:

p = −ρgy + c ← hydrostatic pressure (Archimedes’ principle)

(d) Steady and no gravity effect ( ∂ = 0, g ≡ 0):∂t

ρv2 ρ p = − + c = − |∇φ|2 + c ← Venturi pressure (created by velocity)

2 2

(e) Inertial, acceleration effect:

Eulerian inertia ︷︸︸︷ ∂φ

p ∼ − ρ ∂t

+ · · ·

∂ ∇p ∼ − ρ �v ∂t

· · · +

p

u

p + ∂p δx ∂x

δx

9

( )

(c) Boundary Conditions

• KBC on an impervious boundary

∂φ �v n· ˆ = � · ˆ no flux across boundary ⇒ = Unu n given ︸︷︷︸︸︷︷︸ ∂n n·∇φ Un given

• DBC: specify pressure at the boundary, i.e.,

∂φ 1 2−ρ + |∇φ| + gy = given ∂t 2

Note: On a free-surface p = patm.

10

︸︷︷

( )

( ) ( )

︸

�

∣ ∣ ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ ∣

( ) ( ) ( )

3.6.3 Stream function

• Continuity: ∇ · �v = 0; Irrotationality: ∇× �v = �ω = 0

• Velocity potential: �v = ∇φ, then ∇ × �v = ∇ × (∇φ) ≡ 0 for any φ, i.e., irrotationality is satisfied automatically. Required for continuity:

∇ · �v = ∇2φ = 0

• Stream function ψ� defined by

�v = ∇× ψ�

Then ∇·�v = ∇· ∇× ψ� ≡ 0 for any ψ�, i.e., satisfies continuity automatically.

Required for irrotationality:

∇× �v = 0 ⇒ ∇× ∇× ψ� = ∇ ∇ · ψ� −∇2ψ� = 0 (1)

still 3 unknown ψ�=(ψx,ψy ,ψz )

• For 2D and axisymmetric flows, ψ� is a scalar ψ (stream functions are more ‘useful’ for 2D and axisymmetric flows).

∂For 2D flow: �v = (u, v, 0) and ∂z ≡ 0.

i j k ∂ ∂ ∂ ∂x ∂y ∂z

ψx ψy ψz

=∂ ∂ ∂ ∂

ki + j +�v = ∇× ψ = − ψy −ψz ψz ψx∂y ∂x ∂x ∂y

∂ψ ∂ψSet ψx = ψy ≡ 0 and ψz = ψ, then u = ∂y ; v = −

∂x

So, for 2D:

∂ ∂ ∂ ∇ · ψ� = ψx + ψy + ψz ≡ 0 ∂x ∂y ∂z

Then, from the irrotationality (see (1)) ⇒ ∇2ψ = 0 and ψ satisfies Laplace’s equation.

11

∣∣∣∣∣∣

∣∣∣∣∣∣�

︷︸︸︷ ) (

∂• 2D polar coordinates: �v = (vr, vθ ) and ∂z ≡ 0.

ê êr

r

x

y

v vzvr θ

er reθ ez ∂ ∂ ∂ ∂r ∂θ ∂z ψr ψθ ψz

=

︷︸︸︷︷︸︸︷ 1 1 ∂ψz ∂ψz 1 ∂ ∂

�v = ∇× ψ = − rψθ −er eθ + ψr ezr ∂θ ∂r r ∂r ∂θr

Again let ψr = ψθ ≡ 0 and ψz = ψ , then

1 ∂ψ ∂ψ vr = and vθ = −

r ∂θ ∂r

• For 3D but axisymmetric flows, ψ� also reduces to ψ (read JNN 4.6 for details).

12

︸︷︷︸

︸︷︷︸

∫

• Physical Meaning of ψ.

In 2D

u = ∂ψ ∂y

and ∂ψ

v = − ∂x

We define

∫ �x ∫ �x

ψ(�x, t) = ψ(�x0, t) + v · ˆ x0, t) + (udy − vdx)� nd = ψ(� �x0 �x0

total volume flux from left to right accross a curve C between �x and �x0

v x

tv

o xv

C’

C n

For ψ to be single-valued, must be path independent.

∫ ∫ ∫ ∫ ∮ ∫∫ = or − = 0 −→ v ˆ v ds = 0� · n d = ∇ · �

C C� C C� C−C� S =0, continuity

Therefore, ψ is unique because of continuity.

13

∫

︸︷︷︸

Let �x1, �x2 be two points on a given streamline (�v · n = 0 on streamline)

streamline

�x∫ 2

ψ (� = ψ (� + �v · n dx2) x1) ︸︷︷︸︸︷︷︸︸︷︷︸ =0ψ ψ �x2 1 1 along a

streamline

Therefore, ψ1 = ψ2, i.e., ψ is a constant along any streamline. For example, on an impervious stationary body �v · n = 0, so ψ = constant on the body is the appropriate boundary condition. If the body is moving �v · n = Un

ψ = ψ0 + Un d on the boddy given

∂φψ = constant ≡ = 0 ∂n

ψ = given

u = 0

o ψ

14

Flux Δψ = −vΔx = uΔy.

∂ψ ∂ψTherefore, u = and v = −

∂y ∂x

streamline

streamline

-v

u

(x,y) (x +Δx, y)

ψ ψ + Δψ

(x, y + Δy)

15

Summary of velocity potential formulation vs. stream-function formulation for ideal flows⎧⎨

⎩

For irrotational flow use φ For incompressible flow use ψ For P-Flow use φ or ψ

⎫⎬

⎭

velocity potential stream-function

definition continuity ∇ · �v = 0

irrotationality ∇× �v = 0

�v = ∇φ ∇2φ = 0

automatically satisfied

�v = ∇× �ψ automatically satisfied

∇× ( ∇× �ψ

) = ∇

( ∇ · �ψ

) −∇2 �ψ = 0

2D: w = 0, ∂ ∂z = 0

continuity irrotationality

∇2φ = 0 automatically satisfied

automatically satisfied 2ψ = 0ψ ≡ ψz : ∇

Cauchy-Riemann equations for (φ, ψ) = (real, imaginary) part of an analytic complex function of z = x + iy

Cartesian (x, y) u = ∂φ

∂x u = ∂ψ ∂y

v = ∂φ ∂y v = −∂ψ

∂x

Polar (r,θ) vr = ∂φ

∂r

vθ = 1 r

∂φ ∂θ

vr = 1 r

∂ψ ∂θ

vθ = −∂ψ ∂r

Given φ or ψ for 2D flow, use Cauchy-Riemann equations to find the other:

e.g. If φ = xy, then ψ = ?

∂φ ∂ψ u = = y = →

1 2ψ = y + f1(x)∂x ∂y 2

∂φ ∂ψ 1 v = = x = − → ψ = − x 2 + f2(y)

∂y ∂x 2

16

⎫ ⎪⎪⎪⎬

⎪⎪⎪⎭

1 ⇒ ψ = (y 2 − x 2) + const 2



3.7 Governing Equations and Boundary Conditions for P-Flow

3.7.1 Governing Equations for P-Flow

(a) Continuity � 2φ = 0

�1

�(b) Bernoulli for P-Flow (steady or unsteady) p = −ρ φt + 2 + gy + C(t)

2|�φ|

3.7.2 Boundary Conditions for P-Flow

Types of Boundary Conditions:

∂φ (c) Kinematic Boundary Conditions - specify the flow velocity �v at boundaries. = Un

∂n

(d) Dynamic Boundary Conditions - specify force F� or pressure p at flow boundary. �1 2

�

p = −ρ φt + (�φ) + gy + C (t) (prescribed) 2

1

The boundary conditions in more detail:

Kinematic Boundary Condition on an impermeable boundary (no flux condition) •

�v n = U� n = Un = Given �� · �� · ��fluid velocity boundary velocity nornal boundary velocity

�v=�φ

�φ n = Un· ⇒

∂ ∂ ∂ (n1

∂x1 + n2

∂x2 + n3

∂x3 )φ = Un ⇒

∂φ ∂n

= Un

( )321 n,n,nn =v

( )v

Uv

Dynamic Boundary Condition: In general, pressure is prescribed •

�1 2

�

p = −ρ φt + (�φ) + gy + C (t) = Given 2

2

( ) )()2

1(

0

2

2

tCgypt

++∇+−=

=∇

φφρ

φ

=++∇+−−

GIVEN)())(2

1(:DBC

19)(Lecture:KBC

surfaceFree

linearnon

2tCgy

t 321φφρ

GIVENUn

n==

∂∂φ

:KBCboundarySolid

3.7.3 Summary: Boundary Value Problem for P-Flow

The aforementioned governing equations with the boundary conditions formulate theBoundary Value Problem (BVP) for P-Flow.

The general BVP for P-Flow is sketched in the following figure.

It must be pointed out that this BVP is satisfied instantaneously.

3

3.8 Linear Superposition for Potential Flow

In the absence of dynamic boundary conditions, the potential flow boundary value problem is linear.

Potential function φ. •

BonfUn n ==

∂φ∂

Vin02 =φ∇

Stream function ψ. •

Vin02 =ψ∇

ψ=g on B

Linear Superposition: if φ1, φ2, . . . are harmonic functions, i.e., �2φi = 0, then φ = � αiφi, where αi are constants, are also harmonic, and is the solution for the boundary

value problem provided the kinematic boundary conditions are satisfied, i.e.,

∂φ ∂ = (α1φ1 + α2φ2 + . . .) = Un on B.

∂n ∂n The key is to combine known solution of the Laplace equation in such a way as to satisfythe kinematic boundary conditions (KBC).The same is true for the stream function ψ. The K.B.C specify the value of ψ on theboundaries.

4

� �

� ��

= � � ��

� �

3.8.1 Example

�x�

denote a unit-source flow with source at xi, i.e.,

1ln

��

Let φi

φi

�x� ≡ φsource x,xi (in 2D)x −xi

2π ��−1 (in 3D),= −

�4π

��x − xi

then find mi such that

φ = �

i

miφi(� x) satisfies KBC on B

Caution: φ must be regular for x ∈ V , so it is required that �x /∈ V .

1xv

•

2xv•

4xv•

3xv•

Vin02 =φ∇

fn

=∂Φ∂

Figure 1: Note: �xj , j = 1, . . . , 4 are not in the fluid domain V .

5

3.9 - Laplace equation in different coordinate systems (cf Hildebrand §6.18)

3.9.1 Cartesian (x,y,z)

ˆ ˆ�

i j k � �

∂φ �v = u, v, w , ,= �φ =

∂x ∂y ∂z

∂2φ ∂2φ ∂2φ

∂φ ∂φ �

ze

y

x

ye

xe

z

O

),,( zyxP

� 2φ = ∂x2

+ ∂y2

+ ∂z2

6

�

3.9.2 Cylindrical (r,θ,z)

2 2 2 r = x + y ,

θ = tan−1(y/x)

�er eθ ez

� =

�∂φ

v = vr, vθ, vz ∂r

∂2φ 1 ∂φ 1 � 2φ = ∂r2

+ r ∂r

+ r2 ∂θ2 ∂z2

⇔

, 1 r

∂φ ∂θ

, ∂φ ∂z

�

∂2φ ∂2φ +

ze

P

y

x

θy

e

xe

z

O

),,( zr θ

r

� 1 ∂φ

��∂φ

�(r )

1 ∂φ �

∂φ �

1 ∂2φ ∂2φ 2

r ∂r ∂r

� 2φ = r ∂r

r∂r

+ r ∂θ2

+ ∂z2

7

3.9.3 Spherical (r,θ,ϕ)

2 2 2 2 r = x + y + z ,

θ = cos−1(z/r) z ⇔

ϕ = tan−1(y/x)

�

�er eθ eϕ

� �∂φ

v = �φ = vr, vθ, vϕ = ,∂r

∂2φ 2 ∂φ 1 ∂ �

� 2φ = ∂r2

+ r ∂r

+ r2 sin θ ∂θ

= r (cos θ)

1 r

∂φ ∂θ

, 1

r(sin θ) ∂φ ∂ϕ

�

sin θ ∂φ

�

+ ∂θ

1

r2 sin2 θ

∂2φ ∂ϕ2

⇔

ze

P

y

x

ye

xe

z

O

),,( φθr

r

φ

θ

1 2 ∂φ

� ∂

�� (r )2 ∂r ∂r r

1 ∂ �

∂φ �

1 ∂ �

∂φ �

1 ∂2φ � 2φ = r2 ∂r

r 2

∂r +

r2 sin θ ∂θ sin θ

∂θ +

r2 sin2 θ ∂ϕ2

8

�

�

�

3.10 Simple Potential flows

1. Uniform Stream �2(ax + by + cz + d) = 0

1D: φ = Ux + constant ψ = Uy + constant; v = (U, 0, 0)

v = (U, V, 0)

v = (U, V, W )

2D: φ = Ux + V y + constant ψ = Uy − V x + constant;

3D: φ = Ux + V y + Wz + constant

2. Source (sink) flow

2D, Polar coordinates

1 ∂ ∂ 1 ∂2 2 =

�

r

�

+ , with r = �

x2 + y2 2

� r ∂r ∂r r ∂θ2

An axisymmetric solution: φ = a ln r + b. Verify that it satisfies �2φ = 0, except at r =

�x2 + y2 = 0. Therefor, r = 0 must be excluded from the flow.

Define 2D source of strength m at r= 0:

φ = m 2π

ln r

�φ = ∂φ ∂r

er = m

2πr er ⇐⇒ vr =

m 2πr

, vθ = 0

source (strength m)

x

y

9

Net outward volume flux is

�

C

� v · nds =

��

S � · �vds =

��

Sε

� · �vds

�

Cε

� v · nds =

2π�

0 vr��m

2πrε

rεdθ = m�� source

strength

C

S

ε

Sε

n

x

y

If m < 0 sink. Source m at (x0, y0):⇒

m �

φ = ln (x − x0)2π m

φ = (Stream function) 2π

2 + (y − y0)2

ln r (Potential function) ψ = θ ←→ 2π m

y

x

θ

ψ = 0 1

π=

2

mVr

θπ

=Ψ2

m

10

2

3D: Spherical coordinates

1 ∂ �

∂ � �

∂ ∂ �

2 2 � = r2 ∂r

r∂r

+ ∂θ

, ∂ϕ

, · · · , where r = �

x2 + y2 + z

a A spherically symmetric solution: φ = + b. Verify �2φ = 0 except at r = 0.

r

Define a 3D source of strength m at r = 0. Then

m ∂φ m φ = −

4πr ⇐⇒ vr =

∂r =

4πr2 , vθ = 0, vϕ = 0

Net outward volume flux is

�� m � vrdS = 4πrε

2 · 4πrε

2 = m (m < 0 for a sink )

11

3. 2D point vortex

� 2 = 1 r

∂ ∂r

�

r ∂ ∂r

�

+ 1 r2

∂2

∂θ2

Another particular solution: φ = aθ + b. Verify that �2φ = 0 except at r = 0.

Define the potential for a point vortex of circulation Γ at r = 0. Then

φ = Γ 2π

θ ⇐⇒ vr = ∂φ ∂r

= 0, vθ = 1 r

∂φ ∂θ

= Γ

2πr and,

ωz = 1 r

∂ ∂r

(rvθ) = 0 except at r = 0

Stream function:

ψ = − Γ 2π

ln r

Circulation:

�

C1

� v · d�

x = �

C2

� v · d�

x + �

C1−C2

� v · d�

x

� �� R R S

ωz dS=0

=

2π�

0

Γ 2πr

rdθ = Γ��vortex

strength

12

��

4. Dipole (doublet flow)

A dipole is a superposition of a sink and a source with the same strength.

2D dipole:

m � �

2 2

�2 2

�φ = ln (x − a) + y − ln (x + a) + y

2π µ ∂

�2

��lim φ = ln (x − ξ) + y2

a 0 2π ∂ξ→ξ=0

µ = 2ma constant

µ x µ x = −

2 + y2 = −

22π x 2π r

2D dipole (doublet) of moment µ at the origin oriented in the +x direction.

∂NOTE: dipole = µ∂ξ (unit source)

13

��

ξ

α unit source

x

ξ

α unit source

ξ

α unit source

x

φ = −µ x cos α + y sin α

= −µ cos θ cos α + sin θ sin α

2π x2 + y2 2π r

3D dipole:

⎛⎝

⎞⎠ where µ = 2ma fixed

1 1mφ = lim �

(x − a)�

(x + a)2− −

4π0a 2 + y2 + z2 + y2 + z2→

µ ∂ 1 µ x µ x4π ∂ξ

�( ξ)−x 2 + y2 + z2

ξ=0

= − = −4π (x2 + y2 + z2)3/2

= −4π r3

3D dipole (doublet) of moment µ at the origin oriented in the +x direction.

14

U

m

5. Stream and source: Rankine half-body

It is the superposition of a uniform stream of constant speed U and a source of strength m.

2D: φ = Ux + m 2π

ln �

x2 + y2

DU

U

x m

stagnation point 0v =v

Dividing Streamline

∂φ m x u = = U +

∂x 2π x2 + y2

m u y=0 = U + , v y=0 = 0 |

2πx| ⇒

V� = (u, v) = 0 at x = xs = − m

, y = 0 2πU

m For large x, u U , and UD = m by continuity D = . → ⇒

U

15

3D: φ = Ux − 4π

�x

m 2 + y2 + z2

div. streamlines

stagnation point

∂φ m x u = = U +

∂x 4π (x2 + y2 + z2)3/2

m x u y=z=0 = U + 3 , v y=z=0 = 0, w y=z=0 = 0 |

4π |x| | | ⇒

� m

V� = (u, v, w) = 0 at x = xs = − , y = z = 0 4πU

m For large x, u U and UA = m by continuity A = . → ⇒

U

16

U SS x

y

+m -m

a

dividing streamline (see this with PFLOW)

6. Stream + source/sink pair: Rankine closed bodies

To have a closed body, a necessary condition is to have �

min body = 0

2D Rankine ovoid:

2m � �

2 �

2 �

m �

(x + a)2 + y�

φ = Ux+2π

ln (x + a) + y2 − ln (x − a) + y2 = Ux+4π

ln (x − a)2 + y2

3D Rankine ovoid: ⎡ ⎤

m 1 1φ = Ux −

4π ⎣�

(x + a)2 + y2 + z2 − �

(x − a)2 + y2 + z2

⎦

17

For Rankine Ovoid,

∂φ m �

x + a �

u = = U + ∂x 4π �

(x + a)2 + y2 + z2�3/2

− �(x − a)2

x

+

−

y

a

2 + z2�3/2

m �

1 1 �

u =U + y=z=0 2|4π (x + a)2 −

(x − a)m (−4ax)

=U + 4π (x2 2)2

mu|y=z=0 =0 at

�x 2 − a

− 2�a2

= �

4πU

� 4ax

At x = 0,

m 2a 2 2 u = U +4π (a2 + R2)3/2

where R = y + z

Determine radius of body R0:

�R0

2π uRdR = m

0

18

7. Stream + Dipole: circles and spheres

U µ

r

θ

µx µ2D: φ = Ux + = cos θ

�Ur +

�

2πr2 2πr x=r

↑cos θ

The radial velocity is then

ur = ∂φ

= cos θ �U −

µ � .

∂r 2πr2

Setting the radial velocity vr = 0 on r = a we obtain a = �

µ 2πU . This is the K.B.C.

for a stationary circle of radius a. Therefore, for

µ = 2πUa2

the potential

φ = cos θ �Ur +

µ 2πr

�

is the solution to ideal flow past a circle of radius a.

• Flow past a circle (U, a).

19

2 φ = U cos θ

�r + a

�

r 2

Vθ = 1 ∂φ = −U sin θ �1 + a

�

θ = 0, π θ = π

2 , 3π 2

2U

2U

θ

r ∂θ r2 � = 0 at − stagnation points

r=a Vθ| = −2U sin θ = �2U at − maximum tangential velocity

Illustration of the points where the flow reaches maximum speed around the circle.

= Ur cos θ �1 +

µ 4πr3

�

y

x

r

z

U

µ

θ

µ cos θ 3D: φ = Ux +

4π r2

The radial velocity is then

∂φ µ vr = = cos θ

�U −

�

∂r 2πr3

20

3 µSetting the radial velocity vr = 0 on r = a we obtain a = �

2πU . This is the K.B.C. for a stationary sphere of radius a. Therefore, choosing

µ = 2πUa3

the potential µ

φ = cos θ �Ur +

�

2πr is the solution to ideal flow past a sphere of radius a.

• Flow past a sphere (U, a).

3�

a�

φ = Ur cos θ 1 + 2r3

vθ = 1 r

∂φ ∂θ

= −U sin θ

�1 +

a3

2r3

�

vθ |r=a = − 3U 2

sin θ

� = 0 at θ = 0, π = −3U

2 at θ = π 2

xθ

3/2 U

3/2 U

21

8. 2D corner flow Velocity potential φ = r α cos αθ; Stream function ψ = r α sin αθ

(a) �2φ = �

∂2 + 1 ∂ + 1 ∂2

� φ = 0

∂r2 r ∂r r2 ∂θ2

(b)

∂φ ur = = αrα−1 cos αθ

∂r1 ∂φ

uθ = = −αrα−1 sin αθ r ∂θ

∴ uθ = 0 { or ψ = 0} on αθ = nπ, n = 0, ±1, ±2, . . .

i.e., on θ = θ0 = 0, π α ,

2π α , . . . (θ0 ≤ 2π)

i. Interior corner flow – stagnation point origin: α > 1. For example,

α = 1, θ0 = 0, π, 2π, u = 1, v = 0

x

y

ψ = 0

22

(90o corner)

ψ = 0

ψ = 0

y2v,x2u2,2

3,,

2,0,2 0 −==ππππ=θ=α

(120o corner)

θ=0, ψ = 0

θ=2π/3, ψ = 0

θ=2π, ψ = 0

θ=4π/3, ψ = 0

120o

120o

120o πππ=θ=α 2,3

4,

3

2,0,23 0

23

ii. Exterior corner flow, |v| → ∞ at origin:

α < 1 π

θ0 = 0, only

α

For example,

α = 1/2, θ0 = 0, 2π (1/2 infinite plate, flow around a tip)

α

Since we need θ0 ≤ 2π, we therefore require π α ≤ 2π, i.e., α ≥ 1/2 only.

1/2 ≤ α < 1

θ0 = 0, π

θ=0, ψ = 0

θ=2π, ψ = 0

α = 2/3, θ0 = 0, 3π 2 (90o exterior corner)

θ=3π/2, ψ = 0

θ=0, ψ = 0

24

Appendix A1: Summary of Simple Potential Flows

Cartesian Coordinate System

Flow Streamlines Potential Stream function φ(x, y, z) ψ(x, y)

Uniform flow U∞x + V∞y + W∞z U∞y − V∞x

2D Source/Sink (m) at (xo, yo) m 2π ln((x − xo)2 + (y − yo)2) m

2π arctan( y−yo x−xo

)

3D Source/Sink (m) at (xo, yo, zo) − m 4π

1q(x−xo )2 +(y−yo)2+(z−zo)2

NA

Vortex (Γ) at (xo, yo) Γ 2π arctan( y−yo

x−xo ) − Γ

2π ln((x − xo)2 + (y − yo)2)

2D Dipole (µ) at (xo, yo) at an angle α

α− µ

2π (x−xo) cos α+(y−yo) sin α

(x−xo)2+(y−yo)2 µ 2π

(y−yo ) cos α+(x−xo) sin α (x−xo )2+(y−yo)2

3D Dipole (+x) (µ) at (xo, y0 , zo) − µ 4π

(x−xo)

((x−xo)2 +(y−yo)2+(z−zo)2)3/2 NA

25

Appendix A2: Summary of Simple Potential Flows

Cylindrical Coordinate System

Flow Streamlines Potential Stream function φ(r, θ, z) ψ(r, θ)

Uniform flow U∞r cos θ + V∞ r sin θ + W∞z U∞r sin θ − V∞r cos θ

2D Source/Sink (m) at (xo, yo) m 2π ln r m

2π θ

3D Source/Sink (m) at (xo, yo, zo) − m 4πr NA

Vortex (Γ) at (xo, yo) Γ 2π θ − Γ

2π ln r

2D Dipole (µ) at (xo, yo) at an angle α

α− µ

2π cos θ cos α+sin θ sin α

r µ 2π

sin θ cos α+cos θ sin α r

3D Dipole (+x) (µ) at (xo, yo, zo) − µ 4π

cos θ r2 NA

26

Appendix A3: Combination of Simple Potential Flows

Stream + Source

=

Rankine Half Body

(2D)

(3D)

φ = U∞x + m 2π ln r xs = − m

2πU∞ D = m

U∞

φ = U∞x − m 4π

1√x2+y2+z2

xs = −�

m 4πU∞

A = m U∞

Stream + Source + Sink

=

Rankine Closed Body

(2D)

(3D)

φ = U∞x + m 2π

�ln((x + a)2 + y2) − ln((x − a)2 + y2)

�

φ = U∞x + m 4π (

1√(x+a)2+y2+z2

− 1√(x−a)2+y2+z2

)

Stream + Dipole

=

Circle (Sphere) R = a

(2D)

(3D)

φ = U∞x + µx 2πr2 if µ = 2πa2U∞ φ = U∞ cos θ(r + a 2

r )

φ = U∞x + µ cos θ 4πr2 if µ = 2πa3U∞ φ = U∞ cos θ(r + a 3

2r2 )

2D Corner Flow (2D) φ = Crα cos(αθ) ψ = Crα sin(αθ) θ0 = 0, nπ α

27

Appendix B: Far Field Behavior of Simple Potential Flows

Far field behavior

r >> 1 φ �v = �φ

Source

(2D)

(3D)

∼ ln r

∼ 1 r

∼ 1 r

∼ 1 r2

Dipole

(2D)

(3D)

∼ 1 r

∼ 1 r2

∼ 1 r2

∼ 1 r3

Vortex (2D) ∼ 1 ∼ 1 r

28

m

m

( √ √ )



3.11 - Method of Images m √

• Potential for single source: φ = ln x2 + y2

2π

m

• Potential for source near a wall: φ = m

ln x2 + (y − b)2 + ln x2 + (y + b)2

2π

b

b

Added source for

0= φ dy

d x

y

m

m

symmetry

Note: Be sure to verify that the boundary conditions are satisfied by symmetry or by calculus for φ (y) = φ (−y).

1

( )

( )

• Vortex near a wall (ground effect): φ = Ux+ Γ

tan−1(y − b

) − tan−1(y + b

)2π x x

b

b

Γ

U

x

y

-Γ

Added vortex for symmetry

dφ Verify that = 0 on the wall y = 0.

dy

2 2

φ ∼ a a • Circle of radius a near a wall: = Ux 1 + + x2 + (y − b)2 x2 + (y + b)2

b

bU y

x

y

This solution satisfies the boundary condition on the wall (∂φ = 0), and the degree it ∂n

satisfies the boundary condition of no flux through the circle boundary increases as the ratio b/a >> 1, i.e., the velocity due to the image dipole small on the real circle for b >> a. For a 2D dipole, φ ∼

d1 , ∇φ ∼

d1 2 .

2

• More than one wall:

b'

b U

b'

b U

b'

Example 1:

Example 2: Example 3:

bb bb

-Γ Γ b'

b' b'b'

b'b'b'b'

Γ -Γ b b b b

3.12 Forces on a body undergoing steady translation “D’Alembert’s paradox”

3.12.1 Fixed bodies & translating bodies - Galilean transformation.

y y’

x’ Uo’x o z’ z

Fixed in space Fixed in translating body

x = x` + Ut

3

Reference system O: �v, φ, p Reference system O’: �v′, φ′, p′

U

SO

X USO’

X’

∇2φ = 0 ⇀ v · n = ∂φ

∂n = �U · n = (U, 0, 0) · (nx, ny, nz

= Unx on Body ⇀ v → 0 as |�x| → ∞

φ → 0 as |�x| → ∞

)

∇2φ′ = 0

⇀ v ′ · n′ = ∂φ′

∂n = 0

⇀ v ′ → (−U, 0, 0) as |�x′| → ∞

φ′ → −Ux′ as |�x′| → ∞

Galilean transform: ⇀ v(x, y, z, t)

φ(x, y, z, t)

−Ux′ + φ(x = x′ + Ut, y, z, t)

= ⇀ v ′ (x′ = x − Ut, y, z, t) + (U, 0, 0)

= φ′(x′ = x − Ut, y, z, t) + Ux′ ⇒

= φ′(x′, y, z, t)

Pressure (no gravity)

p∞ = −1 2 ρv2 + Co = Co

∴ Co

= −1 2 ρv′2 + C ′

o = C ′ o − 1

2 ρU2

= C ′ o − 1

2 ρU2

In O: unsteady flow

ps = −ρ∂φ ∂t − 1

2 ρ v2 ︸︷︷︸

U2

+Co

∂φ ∂t = (

∂ ∂t ︸︷︷︸ 0

+ ∂x′

∂t ︸︷︷︸ −U

∂ ∂x′ ) (φ′ + Ux′) = −U2

∴ ps = ρU2 − 1 2 ρU2 + Co = 1

2 ρU2 + Co

ps − p∞ = 1 2 ρU2 stagnation pressure

In O’: steady flow

ps = −ρ ∂φ′

∂t ︸︷︷︸ 0

−1 2 ρ v′2

︸︷︷︸ 0

+C ′ o = C ′

o

ps − p∞ = 1 2 ρU2 stagnation pressure

4

∫∫

⇀

⇀

⇀

⇀

∫∫

( )

∫∫ ∫∫∫ ∫∫∫

∫∫ ( )

( )

∫∫

3.12.2 Forces

B

n

Total fluid force for ideal flow (i.e., no shear stresses): F� ndS= © pˆ

B

For potential flow, substitute for p from Bernoulli: ⎛ ⎞

⎜ ⎟ ⇀ ⎜∂φ 1 2 ⎟F = ©−ρ ⎜ + |∇φ| + gy +c(t)⎟ ndSˆ ⎝ ∂t 2 ⎠ ︸︷︷︸︸︷︷︸

B hydrodynamic hydrostatic

force force

For the hydrostatic case v ≡ φ ≡ 0 :

F s = © (−ρgyn) dS = (−) ∇ (−ρgy) dυ = ρg∀j where ∀ = dυ ↑ ︸︷︷︸ ↑

B Gauss outward υB Archimedes υBtheorem normal principle

We evaluate only the hydrodynamic force:

∂φ 1 2F d = −ρ + |∇φ| ndS ∂t 2

B

For steady motion ∂φ ≡ 0 :∂t

1 F d = − ρ 2 ˆv ndS

2 B

5

⇀ ∫ (

∣ ∣ ∣ ∣ ∣ ∣ ∣

∫ ( )

⇀

∫

( )

( ∣ ∣ ∣ ∣)

︸︷︷︸︸︷︷︸

∣ ∣

3.12.3 Example Hydrodynamic force on 2D cylinder in a steady uniform stream.

n

B

SU

a

x

2π

0 F d =

B

)ρ |∇ |φ

∫2π

i = −ρa

2 2− 2

−ρ2 |∇φ|nd� =ˆ nadθ r=a

2 idθ |∇φ|Fx F n· ·=2 r=a ︸︷︷︸

0 − cos θ

= ρa 2

2π

0

2|∇φ| cos θdθr=a

Velocity potential for flow past a 2D cylinder:

2a φ = Ur cos θ 1 +

2r

Velocity vector on the 2D cylinder surface:

∂φ 1 ∂φr ∂θ

∇φ| = (vr| , vθ| ) =r=a r=a r=a ,∂r r=a 0 r=a −2U sin θ

Square of the velocity vector on the 2D cylinder surface:

|∇φ|2 = 4U2 sin2 θ r=a

6

∫ ∫

∫

Finally, the hydrodynamic force on the 2D cylinder is given by

2π 2π ρa ( ) (1 )

Fx = dθ 4U2 sin2 θ cos θ = ρU2 (2a) 2 dθ sin2 θ cos θ = 0 2 2 ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ odd 0 diameter 0 even

π 3π ps−p∞ or

w.r.t 2 , 2

projection ︸︷︷︸ ≡0

Therefore, Fx = 0 ⇒ no horizontal force ( symmetry fore-aft of the streamlines). Similarly,

2π (1 )

Fy = ρU2 (2a)2 dθ sin2 θ sin θ = 0 2

0

In fact, in general we find that F� ≡ 0, on any 2D or 3D body.

D’Alembert’s “paradox”:

No hydrodynamic force∗ acts on a body moving with steady translational (no circulation) velocity in an infinite, inviscid, irrotational fluid.

∗ The moment as measured in a local frame is not necessarily zero.

7

3.13 Lift due to Circulation

3.13.1 Example Hydrodynamic force on a vortex in a uniform stream.

Γ Γ φ = Ux + θ = Ur cos θ + θ

2π 2π

U Γ

Consider a control surface in the form of a circle of radius r centered at the point vortex. Then according to Newton’s law:

� d � steady flow ΣF = LCV −→

dt (F�

V + F� CS) + M�

NET = 0 ⇔ F� ≡ − F� V = F�

CS + M� NET

Where,

F� = Hydrodynamic force exerted on the vortex from the fluid.

F� V = − F� = Hydrodynamic force exerted on the fluid in the control volume from the vortex.

F� CS = Surface force (i.e., pressure) on the fluid control surface.

M� NET = Net linear momentum flux in the control volume through the control surface.

d L� CV = Rate of change of the total linear momentum in the control volume.

dt

Control volume

x

U

y

θFy

Fx

Γ

The hydrodynamic force on the vortex is F� = F� CS + M�

IN

8

∫ ∫ ( )

∫ ∫ ( )

∫

a. Net linear momentum flux in the control volume through the control surfaces, M� NET .

Recall that the control surface has the form of a circle of radius r centered at the point vortex.

a.1 The velocity components on the control surface are

Γu = U − sin θ

2πr Γ

v = cos θ 2πr

The radial velocity on the control surface is therefore, given by

∂x ⇀

ur = U = U cos θ = V · n∂r

Γ vθ =

2πr

U

θ

a.2 The net horizontal and vertical momentum fluxes through the control surface are given by

2π 2π

Γ (MNET )x = − ρ dθruvr = − ρ dθr U − sin θ U cos θ = 0

2πr 0 0

2π 2π

Γ (MNET )y = − ρ dθrvvr = − ρ dθr cos θ U cos θ

2πr 0 0

2π ρUΓ ρUΓ

= − cos 2 θdθ = − 2π 2

0

9

( ) ( )

( )

∫

∫

︸︷︷︸

( ) ∑

b. Pressure force on the control surface, F� CS.

b.1 From Bernoulli, the pressure on the control surface is

p = − 1 ρ |�v|2 + C

2

b.2 The velocity | |�v 2 on the control surface is given by 2 2

Γ Γ |�v|2 =u 2 + v 2 = U − sin θ + cos θ 2πr 2πr

2Γ Γ

=U2 − U sin θ + πr 2πr

b.3 Integrate the pressure along the control surface to obtain F� CS

2π

(FCS) = dθrp(− cos θ) = 0x 0

2π 2π∫ ( ) ( )

(FCS)y = dθrp(− sin θ) = −2 ρ −Γ

πrU (−r) dθ sin2 θ = −

21 ρUΓ

0 0

π

c. Finally, the force on the vortex F� is given by

Fx = (FCS)x + (Mx)IN = 0

Fy = (FCS) + (My)IN = −ρUΓ y

i.e., the fluid exerts a downward force F = −ρUΓ on the vortex.

Kutta-Joukowski Law

2D : F = −ρUΓ

3D : F� = ρU� × �Γ

Generalized Kutta-Joukowski Law:

n

F� = ρU� × �Γi

i=1

where F� is the total force on a system of n vortices in a free stream with speed U� .

10

13.021 – Marine HydrodynamicsLecture 12


3.14 Lifting Surfaces

3.14.1 2D Symmetric Streamlined Body

No separation, even for large Reynolds numbers.

U

stream line

• Viscous effects only in a thin boundary layer.

• Small Drag (only skin friction).

• No Lift.

1

3.14.2 Asymmetric Body

(a) Angle of attack α,

chord line

U α

(b) or camber η(x),

chord line mean camber line

U

(c) or both

amount of camber

U α

angle of attack

mean camber line chord line

Lift ⊥ to U� and Drag ‖ to U�

2

3.15 Potential Flow and Kutta Condition

From the P-Flow solution for flow past a body we obtain

P-Flow solution, infinite velocity at trailing edge.

Note that (a) the solution is not unique - we can always superimpose a circulatory flow without violating the boundary conditions, and (b) the velocity at the trailing edge → ∞. We must therefore, impose the Kutta condition, which states that the ‘flow leaves tangentially the trailing edge, i.e., the velocity at the trailing edge is finite’. To satisfy the Kutta condition we need to add circulation.

Circulatory flow only.

Superimposing the P-Flow solution plus circulatory flow, we obtain

Figure 1: P-Flow solution plus circulatory flow.

3

3.15.1 Why Kutta condition?

Consider a control volume as illustrated below. At t = 0, the foil is at rest (top control volume). It starts moving impulsively with speed U (middle control volume). At t = 0+ , a starting vortex is created due to flow separation at the trailing edge. As the foil moves, viscous effects streamline the flow at the trailing edge (no separation for later t), and the starting vortex is left in the wake (bottom control volume).

t = 0

+t = 0

for later t

Kelvin’s theorem:

After a while the ΓS

Γ Γ

Γ Γ

S

S

S

S

U

U

no Γ

starting vortex left in wake

Γ = 0

starting vortex due to separation (a real fluid effect,

no infinite vel of potetial flow)

dΓ = 0 → Γ = 0 for t ≥ 0 if Γ(t = 0) = 0

dt in the wake is far behind and we recover Figure 1.

4

3.15.2 How much ΓS ?Just enough so that the Kutta condition is satisfied, so that no separation occurs. Forexample, consider a flat plate of chord � and angle of attack α, as shown in the figurebelow.

chord length

Simple P-Flow solution

Γ = πlU sin α

L = ρUΓ = ρU2πl sin α

|L� |CL = = 2π sin α ≈ 2πα for small α1 ρU2l � ��

2 only for small α

However, notice that as α increases, separation occurs close to the leading edge.

Excessive angle of attack leads to separation at the leading edge.

When the angle of attack exceeds a certain value (depends on the wing geometry) stall occurs. The effects of stalling on the lift coefficient (CL =

ρU2 L span) are shown in the 1

2

following figure.

5

C L

This region independent of R,

ν used only to get Kutta

condition

stall location f(R)

stall 2π

α

O(5o )

• In experiments, CL < 2πα for 3D foil - finite aspect ratio (finite span).

• With sharp leading edge, separation/stall to early.

round leading edge to forstall

sharp trailing edge

to develop circulation

stalling

6

3.16 Thin Wing, Small Angle of Attack

• Assumptions

– Flow: Steady, P-Flow.

– Wing: Let yU (x), yL(x) denote the upper and lower vertical camber coordinates, respectively. Also, let x = �/2, x = −�/2 denote the horizontal coordinates of the leading and trailing edge, respectively, as shown in the figure below.

y=yU(x)

For thin wing, at a small angle of attack it isyU

, yL

<< 1 � �

dyU dyL , << 1

dx dx

The problem is then linear and superposition applies. Let η(x) denote the camber line

1 t(x) η(x) = (yU (x) + yL(x)),2

and t(x) denote the half-thickness η(x) Camber line

t(x)

1 t(x) = (yU (x) − yL(x)).

2

For linearized theory, i.e. thin wing at small AoA, the lift on the wing depends only on the camber line but not on the wing thickness. Therefore, for the following analysis we approximate the wing by the camber line only and ignore the wing thickness.

7

�

• Definitions

In general, the lift on the wing is due to the total circulation Γ around the wing. This total circulation can be given in terms due to a distribution of circulation γ(x) (Units: [LT −1]) inside the wing, i.e.,

�/2

Γ = γ(x)dx −�/2

γ (x) Γ

U

Noting that superposition applies, let the total potential Φ for this flow be expressed as the sum of two potentials

Φ = −Ux + φ � �� Free stream Disturbunce potential potential

The flow velocity can by expressed as

�v = ∇Φ = (−U + u, v)

where (u, v) are given by ∇φ = (u, v) and denote the velocity disturbance, due to the presence of the wing. For linearized wing we can assume

u v u, v << U ⇒ , << 1

U U

Consider a flow property q, such as velocity, pressure etc. Then let qU = q(x, 0+) and qL = q(x, 0−) denote the values of q at the upper and lower wing surfaces, respectively.

8

��

� �

� � � � � �

• Lift due to circulation

Applying Bernoulli equation for steady, inviscid, rotational flow, along a streamline from ∞ to a point on the wing, we obtain

1 � � p − p∞ = − ρ |�v|2 − U2 ⇒

2

p − p∞ = − 1 ρ (u − U)2 + v 2 − U2 = −

1 ρ(u 2 + v 2 − 2uU) ⇒

2 2 1 u v v

p − p∞ = − ρuU( + −2)2 U U u ��

<<1 <<1 ∼1

Dropping terms of order Uu ,

Uv << 1 we obtained the linearized Bernoulli equation

for thin wing at small AoA p − p∞ = ρuU

Integrating the pressure along the wing surface, we obtain an expression for the total lift L on the wing

l/2 ��

L = (p − p∞)nydS = p(x, 0−) − p∞ − p(x, 0+) − p∞ dx

−l/2

l/2 l/2

L = p(x, 0−) − p(x, 0+) dx = ρU u(x, 0−) − u(x, 0+) dx (1)

−l/2 −l/2

9

�

� ��

�

To obtain the total lift on the wing we will seek an expression for u(x, 0±).

Consider a closed contour on the wing, of negligible thickness, as shown in the figurebelow. γ (x)

)0,( +xu

0→t

)0,( −xu

xδ

x

In this case we have

γ(x)δx = |u(x, 0+)|δx + u(x, 0−)δx ⇒ γ(x) = |u(x, 0+)| + u(x, 0−)

For small u/U we can argue that u(x, 0+) ∼= −u(x, 0−), and obtain

γ(x) u(x, 0±) = ∓ (2)

2

From Equations (1), and (2) the total lift can be expressed as l/2

L = ρU γ(x)dx = ρUΓ −l/2

=Γ

The same result can be obtained from the Kutta-Joukowski law (for nonlinear foil) �/2

δL = ρUδΓ = ρUγ(x)δx ⇒ L = ρUγ(x)δx = ρUΓ −�/2

δ L = ρU δΓ = ρUγ (x)δ x

x δ =Γ

U t → 0

γ (x)δ x

xδ

10

�

• Moment, with respect to mid-chord, due to circulation

x

L y

2 l

2 l

cp x M

δL(x) = ρUγ(x)δx

δM = xδL(x) = ρUxγ(x)δx ⇒ �/2

M = ρUxγ(x)dx ⇒ −�/2

M =CM 1 ρU2�2

2

The center of pressure xcp, can be obtained by

M = Lxcp ⇒

M � −�/

�/

2

2 xγ(x)dx xcp = = � �/2L

−�/2 γ(x)dx

11

�

3.17 Simple Closed-Form Solutions for � −�/

�/

22 γ(x)dx from Linear

Theory

1. Flat plate at angle of attack α, i.e., η = αx.

Linear lifting theory gives γ(x), which can be integrated to give the lift coefficientCL,

� �/2

L/span = ρU γ(x)dx = · · · = ρU2�πα ⇒ −�/2

CL = L/span 1 ρU2�

⇒ 2

CL = 2πα ( exact nonlinear hydrofoil CL = 2π sin α)

the moment coefficient CM ,

�/2

M/span = ρU xγ(x)dx = · · · = 14 ρU2�2πα ⇒

−�/2

M/spanCM = 1 ⇒

ρU2�2 2

CM = 12 πα

and the center of pressure xcp

xcp = 41 � i.e., at quarter chord

12

2. Parabolic camber η = η0{1 − (2x )2}, at zero AoA α = 0. l

Linear lifting theory gives γ(x), which can be integrated to give the lift coefficientCL,

� �/2

L/span = ρU γ(x)dx = · · · = 2ρU2πη0 ⇒ −�/2

CL = 4π η0

� , where

η0

� ≡ ‘camber ratio’

the moment coefficient CM ,

M/span = 0 (from symmetry) ⇒

CM = 0

and the center of pressure xcp

xcp = 0

13

�

�

� ��

� �

� � �2 � 2x

3. Linear superposition: Both AoA and camber η = αx + η0 1 − .

η0CL = CLα + CLη = 2πα + 4π

We can also write the previous relation in a more general form

CL(α) = 2πα + CL(α = 0)

≡ 4π η0 l

Lift coefficient CL as a function of the angle of attack α and ηl 0 .

In practice even if the camber is not parabolic, we still make use of the previous relations, i.e., CL(α = 0) ∼= 4πη0/�. Also note that the angle of attack for any camber is defined as

η(�/2) − η(−�/2) yU − yLα ≡ =

and η0 is determined from η∗, where

η ∗ = η − αx.

14

∣ ∣ ∣ ∣

∣ ∣ ∣ ∣



3.18 Unsteady Motion - Added Mass

D’Alembert: ideal, irrotational, unbounded, steady.

Example Force on a sphere accelerating (U = U(t), unsteady) in an unbounded fluid that is at at rest at infinity.

θϕ

n

U(t)

3D Dipole x

r

a

U(t)

∂φK.B.C on sphere: = U(t) cos θ

∂r r=a

Solution: Simply a 3D dipole (no stream)

3a φ = −U(t) cos θ

2r2

∂φ Check:

∂r= U(t) cos θ

r=a

1

∫∫ ( )

∣

( ) ( )

∣ ∫∫ ∫

Hydrodynamic force:

∂φ 1 2Fx = −ρ + |∇φ| nxdS ∂t 2

B

On r = a,

∂φ ∣∣ = − ˙ a3 1 ˙U cos θ| = − Ua cos θ

∂t ∣ 2r2 r=a 2 r=a

∂φ 1 ∂φ 1 ∂φ 1 ∇φ| = , , = U cos θ, U sin θ, 0 r=a ∂r r ∂θ r sin θ ∂ϕ 2

|∇φ|2∣ = U2 cos 2 θ +1 U2 sin2 θ; n = −er, nx = − cos θ

r=a 4 π

dS = (adθ) (2πa sin θ)

B 0

x

a θ

θad

θsina

2

∫ ∫ ( )

[ ]

( )

Finally, ⎡ ⎤ ⎛ ⎞ ⎛ ⎞

⎠ ˙ 1 ⎜ 2 ⎟Fx = (−ρ) 2πa2 ∫π

dθ (sin θ) ⎝− cos θ ⎢⎢⎢ −

1 Ua cos θ +

⎜ U cos 2 θ +

1 U2 sin2 θ

⎟⎥⎥⎥

︸︷︷︸ ⎣ 2 2 ⎝ 4 ⎠⎦0 ︸︷︷︸︸︷︷︸nx ∂φ 2|∇φ|∂t

π π

Fx = −U(ρa3)π dθ sin θ cos 2 θ + (ρU2)πa2 dθ sin θ cos θ cos 2 θ + 1

sin2 θ 4

0 0 ︸︷︷︸︸︷︷︸ 2/3 = 0, D’alembert revisited ⎡ ⎤

⎢ 2 ⎥Fx = − U(t) ⎢ ρ πa3 ⎥

︸︷︷︸︸︷︷︸ ⎣ ︸︷︷︸ 3 ⎦ ︸︷︷︸ Hydrodynamic Force Acceleration Fluid Density

Volume =1/2∀sphere

Thus the Hydrodynamic Force on a sphere of diameter a moving with velocity U(t) in an unbounded fluid of density ρ is given by

Fx = −U(t) ρ 2 πa3

3

Comments:

• If U = 0 → Fx = 0, i.e., steady translation → no force (D’Alembert’s Condition ok).

• Fx ∝ U with a (−) sign, i.e., the fluid tends to ‘resist’ the acceleration.

• [· · · ] has the units of (fluid) mass ≡ ma

• Equation of Motion for a body of mass M that moves with velocity U :

M U = ΣF = FH + FB = − U ma + FB ⇔ ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ Body mass Hydrodynamic force All other forces on body Fluid mass

pndSS

(M + ma) U = FB

i.e., the presence of fluid around the body acts as an added or virtual mass to the body.

3

3.19 General 6 Degrees of Freedom Motions

3.19.1 Notation Review

(3D) U1, U2, U3: Translational velocities

U4 ≡ Ω1, U5 ≡ Ω2, U6 ≡ Ω3: Rotational velocities

1

2

6

5

4

3

(2D) U1, U2: Translational velocities

U6 ≡ Ω3: Rotational velocity

U3 = U4 = U5 = 0

2

6 1

3.19.2 Added Mass Tensor (matrix)

mij ; i, j = 1, 2, 3, 4, 5, 6

mij : associated with force on body in i direction due to unit acceleration in j direction. For example, for a sphere:

m11 = m22 = m33 = 1/2ρ∀ = (mA) all other mij = 0

4

3.19.3 Added Masses of Simple 2D Geometries

• Circle

a

1

2

m11 = m22 = ρ∀ = ρπa2

• Ellipse

m11 = ρπa2 ,m22 = ρπb2

• Plate

2

a

b 1

2

2a 1

m11 = ρπa2 ,m22 = 0

5

• Square

2

2a 1

2a

m11 = m22 ≈ 4.754ρa2

A reasonable approximation to estimate the added mass of a 2D body is to use the displaced mass (ρ∀) of an ‘equivalent cylinder’ of the same lateral dimension or one that ‘rounds off’ the body. For example, consider a square and approximate with an

(a) inscribed circle: mA = ρπa2 = 3.14ρa2 .

2a

(√ )2 (b) circumscribed circle: mA = ρπ 2a = 6.28ρa2 .

(√2)a

Arithmetic mean of (a) + (b) ≈ 4.71ρa2 .

6

�

3.19.4 Generalized Forces and Moments

In this paragraph we are looking at the most general case where forces and moments are induced on rigid body moving with 6 DoF motions, in an unbounded fluid that is at rest at infinity.

Body fixed reference frame, i.e., OX1X2X3 is fixed on the body.

x1

x2

o

x3

)t(U v

)t(Ωv

U� (t) = (U1, U2, U3) , translational velocity

Ω(t) = (Ω1, Ω2, Ω3) ≡ (U4, U5, U6) , rotational velocity with respect to O

Consider a body with a 6 DoF motion (� Ω), and a fixed reference frame OX1X2X3.U, �

Then the hydrodynamic forces and moments with respect to O are given by the following relations (JNN §4.13)

• Forces

Fj = −U imji − Ejkl UiΩk mli with i = 1, 2, 3, 4, 5, 6 1. 2.

and j, k, l = 1, 2, 3

• Moments

Mj = −U imj+3,i − Ejkl UiΩk ml+3,i − Ejkl UkUi mli with i = 1, 2, 3, 4, 5, 6 3. 2. 3.

and j, k, l = 1, 2, 3

7

⇀ ⇀ ⇀

︸︷︷︸

Einstein’s Σ notation applies.

Ejkl = ‘alternating tensor’ =

⎧ ⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

0 if any j, k, l are equal 1 ifj, k, l are in cyclic order, i.e.,

(1, 2, 3), (2, 3, 1), or (3, 1, 2) −1 ifj, k, l are not in cyclic order i.e.,

(1, 3, 2), (2, 1, 3), (3, 2, 1)

Note:

(a) if Ωk ≡ 0 , Fj = −U imji (as expected by definition of mij ).

Also if U i ≡ 0, then Fj = 0 for any Ui, no force in steady translation.

(b) Bl ∼ Uimli ‘added momentum’ due to rotation of axes.

Then all the terms marked as 2. are proportional to ∼ Ω × B where B is linear momentum (momentum from i coordinate into new xj direction).

(c) If Ωk ≡ 0 : Mj = −U m m − E U U m .i j+3,i ij jkl k i li

even with U Mj �˙ =0, =0 due to this term

Moment on a body due to pure steady translation – ‘Munk’ moment.

8

︸︷︷︸

︸︷︷︸

3.19.5 Example Generalized motions, forces and moments.

A certain body has non-zero added mass coefficients only on the diagonal, i.e. mij = δij . For a body motion given by U1 = t, U2 = − t, and all other Ui, Ωi = 0, the forces and moments on the body in terms of mi are:

F1 = , F2 = , F3 = ,M1 = ,M2 = ,M3 =

Solution:

mij = δij

U1 = t U2 = − t Ui = 0 i = 3, 4, 5, 6 Ωk = 0 k = 1, 2, 3

U 1 = 1 U

2 = − 1 U i = 0 i = 3, 4, 5, 6

Use the relations from (JNN § 4.13):

Ωk=0 Fj = − U

imij − EjklUiΩkmil −→

Fj = − U imij

Ωk=0 Mj = − U

imi(j+3) − EjklUiΩkmi(l+3) − EjklUkUimli −→

Mj = − U imi(j+3) − EjklUkUimli

where i = 1, 2, 3, 4, 5, 6 and j, k, l = 1, 2, 3

For F1, F2, F3 use the previous relationship for Fj with j = 1, 2, 3 respectively:

F1 = − U 1 m11 − U

2 m21 − U 3 m31 − U

4 m41 − U 5 m51 − U

6 m61 → F1 = − m11 ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ =1 =0 =0 =0 =0 =0

Check F2 = − U

2 m22 → F2 = m22

=−1

Check ˙F3 = − U3 m33 → F3 = 0 =0

9

( )

( )

For M1,M2, M3 use the previous relationship for Mj with j = 1, 2, 3 respectively:

M1 = −U imi(1+3) − E1klUkUimli

= −U imi4 − E1klUkUimli

= −U 1 m14 −U

2 m24 −U 3 m34 − U

4 m44 − U 5 m54 −U

6 m64︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ =0 =0 =0 =0 =0 =0

−E123U2 U1 m13 +U2 m23 + U3 m33 + U4 m43 +U5 m53 +U6 m63 ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ =0 =0 =0 =0 =0 =0

−E132 U3 U1 m12 + U2 m22 + U3 m32 +U4 m42 +U5 m52 +U6 m62 → ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ =0 =0 =−1 =0 =0 =0 =0

M2 = −U imi5 − E2klUkUimli

= U 5m55 − E231U3Uim1i − E213U1Uim3i

= −E213U1U3m33 → M2 = 0

M3 = −U imi6 − E3klUkUimli

= U 6m66 − E312 U1Uim2i − E321 U2Uim1i ︸︷︷︸︸︷︷︸

+1 −1

= − U1 U2 m22 + U2 U1 m11 → M3 = t2(m22 − m11) ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ t −t −t t

M1 = 0

10

︸︷︷︸

3.19.6 Example Munk Moment on a 2D submarine in steady translation

U

1

θ

2

U

1

θ

2

33 (out of page)(out of page)

U1 =U cos θ

U2 = − U sin θ

Consider steady translation motion: U = 0; Ωk = 0. Then

M3 = −E3klUkUimli

For a 2D body, m3i = mi3 = 0, also U3 = 0, i, k, l = 1, 2. This implies that:

M3 = − E312 U1 (U1m21 + U2m22) − E321 U2 (U1m11 + U2m12) ︸︷︷︸︸︷︷︸ =1 =−1

= −U1U2 (m22 − m11) ⎛ ⎞

⎠= U2 sin θ cos θ ⎝m22 − m11

>0

Therefore, M3 > 0 for 0 < θ < π/2 (‘Bow up’). Therefore, a submarine under forward motion is unstable in pitch (yaw). For example, a small bow-up tends to grow with time, and control surfaces are needed as shown in the following figure.

11

B

G H

• Restoring moment ≈ (ρg∀H)sinθ.

• critical speed Ucr given by:

(ρg∀) H sin θ ≥ U2 sin θ cos θ (m22 − m11)cr

H Ucr

θ

θ

Usually m22 >> m11,m22 ≈ ρ∀. For small θ, cos θ ≈ 1. So, U2 ≤ gH or Fcr ≡cr √Ucr ≤ 1. Otherwise, control fins are required.

gH

12

∫∫∫ ∫∫∫ ∫ ∫

∫∫ ∫∫∫



3.20 Some Properties of Added-Mass Coefficients

1. mij = ρ·[function of geometry only]

F, M = U, Ω][linear function of mij ] × [function of instantaneous U, ˙ not of motion history

2. Relationship to fluid momentum.

F(t)

where we define Φ to denote the velocity potential that corresponds to unit velocity U = 1. In this case the velocity potential φ for an arbitrary velocity U is φ = UΦ.

The linear momentum L� in the fluid is given by

L� = vdV = ρ∇φdV = + ρφˆρ� ndS ↑

V V Green’s B ∞ theorem ︸︷︷︸

φ→0 at ∞

Lx (t = T ) = ρUΦnxdS = U ρΦnxdS

B B

The force exerted on the fluid from the body is −F (t) = −(−mAU) = mAU .

1

∫∫∫

︸︷︷︸

∫∫

∫∫ ∫∫

∫∫∫

︸︷︷︸

T T∫ ∫ Newton’s Law ∫∫ ˙ T ↓

dt [−F (t)] = mAUdt = mA U ]0 = Lx (t = T )−Lx (t = 0) = U ρΦnxdS

0 0 mAU B

Therefore, mA = total fluid momentum for a body moving at U = 1 (regardless of how we get there from rest) = fluid momentum per unit velocity of body.

∂Φ K.B.C. ∂φ = ∇φ · n = (U, 0, 0) · n = Unx,

∂φ = Unx ⇒ ∂UΦ = Unx ⇒ = nx∂n ∂n ∂n ∂n

∂Φ∴ mA = ρ Φ dS ∂n

B

For general 6 DOF:

∂Φjmji = ρ Φi njdS = ρ Φi dS = j fluid momentum due to ︸︷︷︸︸︷︷︸ ∂n

i body motion j−force/moment B potential due to body B i−direction of motion moving with Ui=1

3. Symmetry of added mass matrix mij = mji.

∫∫ ( ) ∫∫ ∫∫∫ ∂Φj

mji = ρ Φi dS = ρ Φi (∇Φj · n)dS = ρ ∇ · (Φi∇Φj) dV ∂n ↑

B B Divergence VTheorem ⎛ ⎞

= ρ ⎝∇Φi · ∇Φj + Φi∇2Φj⎠dV

V =0

Therefore,

mji = ρ ∇Φi · ∇ΦjdV = mij

V

2

︸︷︷︸

∫∫∫

∫∫∫

4. Relationship to the kinetic energy of the fluid. For a general 6 DoF body motion Ui = (U1, U2, . . . , U6),

φ = UiΦi ; Φi = potential for Ui = 1

notation

1 ∫∫∫ K.E. = ρ ∇φ · ∇φdV =

21 ρ Ui∇Φi · Uj ∇Φj dV

2 V V

1 =

2 ρUiUj ∇Φi · ∇Φj dV = 1

2 mijUiUj

V

K.E. depends only on mij and instantaneous Ui.

5. Symmetry simplifies mij . From 36 → 21 → ‘?’. Choose such coordinate system symmetry

that some mij = 0 by symmetry.

Example 1 Port-starboard symmetry.

⎤⎡ m11 m12 0 0 0 m16 Fx

0 0 0 Fy

Fz

Mx 12 independent coefficients

⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥

m22 m26

0m33 m34 m35 mij = 0m44 m45

0 My ⎣ ⎦m55

m66 Mz

U1 U2 U3 Ω1 Ω2 Ω3

3

Example 2 Rotational or axi-symmetry with respect to x1 axis.

⎤⎡ m11 0 0 0 0 0

m22 0 0 0 m35

m22 0 m35 0 0 0 0

⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥

where m22 = m33, m55 = m66

and m26 = m35, so 4 different coefficients mij =

⎣ m55 0 m55

⎦

Exercise How about 3 planes of symmetry (e.g. a cuboid); a cube; a sphere?? Work out the details.

4

∑

3.21 Slender Body Approximation

Definitions

(a) Slender Body = a body whose characteristic length in the longitudinal direction is considerably larger than the body’s characteristic length in the other two directions.

(b) mij = the 3D added mass coefficient in the ith direction due to a unit acceleration in the jth direction. The subscripts i, j run from 1 to 6.

(c) Mkl = the 2D added mass coefficient in the kth direction due to a unit acceleration in the lth direction. The subscripts k, l take the values 2,3 and 4.

x2x5

x2

x1 xx 34

x6

Goal To estimate the added mass coefficients mij for a 3D slender body.

Idea Estimate mij of a slender 3D body using the 2D sectional added mass coefficients (strip-wise Mkl). In particular, for simple shapes like long cylinders, we will use known 2D coefficients to find unknown 3D coefficients.

mij = [Mkl(x) contributions] 3D 2D

Discussion If the 1-axis is the longitudinal axis of the slender body, then the 3D added mass coefficients mij are calculated by summing the added mass coefficients of all the thin slices which are perpendicular to the 1-axis, Mkl. This means that forces in 1-direction cannot be obtained by slender body theory.

5

3x

O

x

L

4x

∫

∫ ∫

∫

Procedure In order to calculate the 3D added mass coefficients mij we need to:

1. Determine the 2D acceleration of each crossection for a unit acceleration in the ith

direction,

2. Multiply the 2D acceleration by the appropriate 2D added mass coefficient to get the force on that section in the jth direction, and

3. Integrate these forces over the length of the body.

Examples

• Sway force due to sway acceleration Assume a unit sway acceleration u3 = 1 and all other uj , u j = 0, with j = 1, 2, 4, 5, 6. It then follows from the expressions for the generalized forces and moments (Lecture 12, JNN §4.13) that the sway force on the body is given by

f3 = −m33u 3 = −m33 ⇔ m33 = −f3 = − F3(x)dx L

A unit 3 acceleration in 3D results to a unit acceleration in the 3 direction of each 2D ‘slice’ (U

3 = u 3 = 1). The hydrodynamic force on each slice is then given by

F3(x) = −M33(x)U 3 = −M33(x)

Putting everything together, we obtain

m33 = − −M33(x)dx = M33(x)dx L L

• Sway force due to yaw acceleration Assume a unit yaw acceleration u5 = 1 and all other uj , uj = 0, with j = 1, 2, 3, 4, 6. It then follows from the expressions for the generalized forces and moments that the sway force on the body is given by

f3 = −m35u 5 = −m35 ⇔ m35 = −f3 = − F3(x)dx L

For each 2D ‘slice’, a distance x from the origin, a unit 5 acceleration in 3D, results to a unit acceleration in the -3 direction times the moment arm x (U

3 = −xu 5 = −x). The hydrodynamic force on each slice is then given by

F3(x) = −M33(x)U 3 = xM33(x)

6

∫

∫

∫


m35 = − xM33(x)dx L

• Yaw moment due to yaw acceleration Assume a unit yaw acceleration u5 = 1 and all other uj , uj = 0, with j = 1, 2, 3, 4, 6. It then follows from the expressions for the generalized forces and moments that the yaw force on the body is given by

f5 = −m55u 5 = −m55 ⇔ m55 = −f5 = − F5(x)dx L

For each 2D ‘slice’, a distance x from the origin, a unit 5 acceleration in 3D, results to a unit acceleration in the -3 direction times the moment arm x (U

3 = −xu 5 = −x). The hydrodynamic force on each slice is then given by

F3(x) = −M33(x)U 3 = xM33(x)

However, each force F3(x) produces a negative moment at the origin about the 5 axis

F5(x) = −xF3(x)


m55 = x 2M33(x)dx L

In the same manner we can estimate the remaining added mass coefficients mij - noting that added mass coefficients related to the 1-axis cannot be obtained by slender body theory.

7

∫ ∫ ∫ ∫ ∫

In summary, the 3D added mass coefficients are shown in the following table. The empty boxes may be filled in by symmetry.

⎤

⎥⎥⎥⎥⎥ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

m22 = M22dx L

m23 = M23dx L

m24 = M24dx L

m25 = −xM23dx L

m26 = xM22dx L

m33 = M33dx ∫

L m34 = M34dx

∫

L m35 = −xM33dx

∫

L m36 = xM32dx

∫

L

m44 = M44dx ∫

L m45 = −xM34dx

∫

L m46 = xM24dx

∫

L

m55 = x2M33dx ∫

L m56 = −x2M32dx

∫

L

m66 = x2M22dx ∫

L

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

8

︸︷︷︸

∣

( )

∣

∫ ∫

( )

︸︷︷︸

3.22 Buoyancy Effects Due to Accelerating Flow

Example Force on a stationary sphere in a fluid that is accelerated against it.

⎛ ⎞

3⎜ a ⎟ φ (r, θ, t) = U (t) ⎜r + ⎟ cos θ ⎝ 2r2 ⎠

dipole for sphere

∂φ∣ ˙ 3a ∣ = U cos θ

∂t ∣ 2 r=a

3 ∇φ| = 0, − U sin θ, 0 r=a 2 1 ∣ 9 |∇φ|2 = U2 sin2 θ 2 r=a 8

Then,

∫π [ ]

Fx = (−ρ) ( 2πr2

) dθ sin θ (− cos θ) U

3a cos θ +

9 U2 sin2 θ

2 8 0

π π

= U3πρa3 dθ sin θ cos 2 θ +ρU2 9πa 2 dθ cos θ sin3 θ 4

0 0 ︸︷︷︸︸︷︷︸ =2/3 =0

˙Fx = U ρ 2πa3

4 πa3ρ + 2 πa3ρ3 3

⇓=ρ∀

9

︸︷︷︸

⇀ ∫∫ ∫∫∫ ∫∫∫

Part of Fx is due to the pressure gradient which must be present to cause the fluid to accelerate:

∂U ∂U ∂U ∂U 1 ∂p x-momentum, noting U = U (t) : + U + v + w = − (ignore gravity)

∂t ∂x ∂y ∂z ρ ∂x ︸︷︷︸︸︷︷︸ 0 0 0

dp = −ρU for uniform (1D) accelerated flow

dx

Force on the body due to the pressure field

F = pndSˆ = − ∇pdV ; Fx = − ∂p

dV = ρ∀U ∂x

B VB VB

‘Buoyancy’ force due to pressure gradient = ρ∀U

10

⇀

( )

Analogue: Buoyancy force due to hydrostatic pressure gradient. Gravitational acceleration g ↔ U = fluid acceleration.

ps = −ρgy

∇ps = −ρgj → F s = −ρg∀j Archimedes principle

Summary: Total force on a fixed sphere in an accelerated flow

Fx = U ρ∀ + = U2 Um(1)m(1)

˙ 3 ρ∀ = 3 ˙ ︸︷︷︸︸︷︷︸

Buoyancy added mass 1 ρ∀2

In general, for any body in an accelerated flow:

˙Fx = Fbuoyany + Um(1),

where m(1) is the added mass in still water (from now on, m)

= − ˙Fx Um for body acceleration with U

Added mass coefficient

m cm =

ρ∀

in the presence of accelerated flow Cm = 1 + cm

11

Appendix A: More examples on symmetry of added mass tensor

•Symmetry with respect to Y (= “X-Z” plane symmetry) 12 non-zero, independent coefficients

•Symmetry with respect to X and Y (= “Y-Z” and “X-Z” plane symmetry) 7 non-zero, independent coefficients

12

•Axisymmetric with respect to X-axis 4 non-zero, independent coefficients

•Axisymmetric with respect to X axis and X (=“Y-Z” plane symmetry) 3 non-zero, independent coefficients

13



Chapter 4 - �Real Fluid Effects (ν = 0)

Potential Flow Theory → Drag = 0.

Observed experiment (real fluid ν << 1 but =� 0) → Drag �= 0.

In particular the total drag measured on a body is regarded as the sum of two components: the pressure or form drag, and the skin friction or viscous drag.

Total Drag = Pressure Drag + Skin Friction Drag Profile Drag or Form Drag or Viscous Drag

Drag Force due to Pressure � �� pnds

� Drag Force due to Viscous Stresses � ��

τ tds �

S S

where n and t are the normal and tangential unit vectors on the body surface respectively. The pressure and the viscous stresses on the body surface are p and τ respectively.

The form drag is evaluated by integrating the pressure along the surface of the body. For bluff bodies that create large wakes the form drag is ∼ total drag.

The skin friction drag is evaluated by integrating the viscous stresses on and along the body boundary. For streamlined bodies that do not create appreciable wakes, friction drag is dominant.

1

� ��

4.1 Form Drag

4.1.1 Form Drag on a Bluff Body Consider a sphere of diameter d:

D (Drag)

ρ

ν U

d

If no DBC apply then we have seen from Dimensional Analysis that the drag coefficient is a function of the Reynolds number only:

CD = CD(Re)

The drag coefficient CD is defined with respect to the body’s projected area S:

D D = =CD 1 ρU2S 1 ρU2 πd2/42 2

Projected area

The Reynolds number Re is defined with respect to the body’s diameter d:

Ud Re =

ν The following graph shows the dependence of CD on Re as measured from numerous experiments on spheres.

CD

0.5

0.25

Re 3x105

Figure 1: Drag coefficient (CD) for a sphere for Re > 102 .

2

i) Bluff Body → Form Drag

For a bluff body (examples: sphere, cylinder, flat plate, etc.) there is appreciable flow separation and a wake is formed downstream of the body. The pressure within the wake is significantly smaller than that upstream of the body. Therefore the integral of the pressure along the body boundary (= form drag) does not evaluate to zero as predicted by P-Flow.

In general, for bluff bodies form drag >> friction drag

Flow separation

Real Flow Potential Flow

Assume: pressure in the wake ∼ p∞ (pressure at infinity)

pressure on the upstream boundary of the body ∼ ps (stagnation pressure)

Then:

� � �1 � D = CD (Projected/frontal area)(ps − p∞) = CDS ρU2

�� ↑ 2 Friction Coefficient S Bernoulli to be determined

3

ii) CD = CD(Re) → Regime Dependence

In general, for typical bluff bodies such as spheres, it is found that CD = CD(Re) has a form similar to that shown in Figure (1). This means that CD has a ‘regime dependence’ on Re.

There are two main regimes of interest:

∼• Laminar regime: R e no separation < Re < R e critical = ( 3 × 105) for a smooth sphere with smooth inflow

Laminar boundary layer

Drag

Separation pt

Separation pt

Stagnation pt

No Stagnation pt

Laminar boundary layer

• Wide wake Wake • Early separation Width ~ Diameter • ‘Large’ CD =O(1)

• Turbulent regime: Re > R e critical

Turbulent boundary layer

Separation pt

Separation pt

Stagnation pt

No Stagnation pt Wake • Narrow wake Width ~ Diameter/2 • Delayed separation

• Smaller CD


4

iii) Cylinder The drag coefficient for a cylinder is defined as:D/L

CD = 1 ρU2d2

U

ρdL

D

CD

3x105

Figure 2: Drag coefficient (CD) for a cylinder for Re > 102

iv) Bodies with Fixed Separation Points

For bodies with fixed separation points, the drag coefficient is roughly constant, i.e., does not depend on Re. For example, for a flat plate or disc CD ≈ 1.2

1.2

0.6

Re

Separation pt

Separation pt

5

⇀

��

�

4.1.2 Boundary Layers

� �∗ L ∂v �

⇀ ⇀�∗ ν �

2⇀�∗

+ v · ∇v = ... + ∇ v UT ∂t UL

1 ReL

For most flows of interest to us ReL >> 1, i.e., viscosity can be ignored if U, L govern the problem, thus potential flow can be assumed. In the context of potential flow theory, drag = 0! Potential flow (no τij) allows slip at boundary, but in reality, the no-slip condition applies on the boundaries. Otherwise, if ν =� 0 and a free-slip KBC is imposed then τ ∼ ν ∂u → ∞ at the boundary.

∂y

Prandtl: There is a length scale δ (boundary layer thickness δ << L) over which velocity goes from zero on the wall to the potential flow velocity U outside the boundary layer.

L

U δ<<L y

x

u=0

U u=U

Estimate δ: Inside the boundary layer, viscous effects are of the same order as the inertial effects.

∂2U ∂U U U2 ν δ2 δ ν 1 ν ∼ U → ν ∼ → ∼ → ∼ = � << 1 As ReL ↑, δ ↓

∂y2 ∂x δ2 L UL L2 L UL ReL

Generally: ReL >> 1, Lδ << 1, thus potential flow is good outside a very thin bound

ary layer (i.e., provided no separation - a real fluid effect). For Reynolds number not >> 1(Re ∼ O(1)), then thick boundary layer ( δ ∼ O(L)) and Prandtl’s boundary layer idea not useful. If separation occurs, then boundary layer idea is not valid.

6

4.1.3 Boundary Layers and Flow Separation

Boundary layers help understand flow separation.

Example for flow past a circle.Outside the boundary layer P-Flow is valid. Let capital U denote the potential flowtangential velocity on the circle and let x denote the distance along the circle surface (i.e.,x = body coordinate).

From the steady inviscid x-momentum equation (steady Euler) along the body boundary (y = 0, V = 0), we obtain :

dU 1 dpU = − (1)

dx ρ dx

Note 1: Equation (1) is used frequently in boundary layer theory.Note 2: From Equation (1) → for flow past a flat plate dp = 0 along the plate.

dx

:U dU = − 1 dp

P − Flow solution on body y=0

dx dx

dU > 0dx

dp < 0dx

x

y

0=U0=U

0U

υu,:layer boundary

ν

δ

0

0

<

>

dx

dp

dx

dU

0

0

>

<

dx

dp

dx

dU

max UU =

Acceleration dU dx < 0 Deceleration

‘Favorable’ pressure gradient dp dx > 0 ‘Adverse’ pressure gradient

7

X3

X=X1y y

X2 > X1

X1 X4

X2

X5

P

p

u

p

U1

P

v u ωv U2 > U1ω

P > p Flow is being pushed to attach

X3 > X2 X4 > X3 y y

P

U4 =ωv

U3

P

p

uu U3 U2 ∂u

0, 0= τ3 > 0 ∂y τ4 = 0

X4 is defined as the point of separation

X5 > X4y

u U4

P

U5

Separated Flow τ5 < 0 Flow reversal

8

⇀ ⇀

⇀

⇀

A better way to think about separation is in terms of diffusion of vorticity.

ω=0 outside B.L. y y

ω2 ω(y)

P

ω added to fluid

ω1 ω(y)

P

ω removed from fluid

ω3 ω4 = 0

by diffusion

Think of vorticity as heat; ω(y) is equivalent to a temperature distribution. Note:

DV Dω = ... + ν∇2V and = ... + ν∇2ω

Dt Dt

9


D (drag)x

F

(lift)y

F

0U


4.1.4 Vortex Shedding and Vortex Induced Vibrations Consider a steady flow Uo over a bluff body with diameter D.

We would expect the average forces to be:

Fy

Fx

t

F

However, the measured oscillatory forces are:

Average

Average

t

F

Fx

Fy

• The measured drag Fx is found to oscillate about a non-zero mean value with frequency 2f .

• The measured lift Fy is found to oscillate about a zero mean value with frequency f .

• f = ω/2π is the frequency of vortex shedding or Strouhal frequency.

1

Uo D

Von

Karman

Street

Fy

Fx

Reason: Flow separation leads to vortex shedding. The vortices are shed in a staggered array, within an unsteady non-symmetric wake called von Karman Street. The frequency of vortex shedding is the Strouhal frequency and is a function of Uo, D, and ν.

i) Strouhal Number We define the (dimensionless) Strouhal number S ≡

Strouhal frequency ��f D

U0 .

The Strouhal number S has a regime dependence on the Re number S = S(Re).

105 106 107

0.22

0.3

S(Re)

Re

For a cylinder:

Laminar flow S ∼ 0.22•

Turbulent flow S ∼ 0.3•

ii) Drag and Lift The drag and lift coefficients CD and CL are functions of the correlation length.

For ‘∞’ correlation length:

If the cylinder is fixed, CL ∼ O(1) comparable to CD. •

• If the cylinder is free to move, as the Strouhal frequency fS approaches one of the cylinder’s natural frequencies fn, ‘lock-in’ occurs. Therefore, if one natural frequency is close to the Strouhal Frequency fn ∼ fS , we have large amplitude motions Vortex Induced Vibration (VIV).⇒

2

L

b D Uo

4.2 Drag on a Very Streamlined Body

UL ReL ≡

ν

Cf ≡ D

1 2 ρU2 (Lb)��

S=wetted area one side of plate

Cf = Cf (ReL , L/b)

Cf

0.01 Laminar

105 Re

0.001

Turbulent

106

= ν ∂u �

Unlike a bluff body, Cf is a strong function of ReL since D is proportional to ν �τ

∂y .

See an example of Cf versus ReL for a flat plate in the figure below.

Skin friction coefficient as a function of the Re for a flat plate

• ReL depends on plate smoothness, ambient turbulence, . . .

In general, Cf ’s are much smaller than CD’s (Cf /CD ∼ O(0.1) to O(0.01)). Therefore, • designing streamlined bodies allows minimal separation and smaller form drag at the expense of friction drag.

• In general, for streamlined bodies CTotal Drag is a combination of CD (Re) and Cf (ReL ),

1

� �and the total drag is D =

2 ρU2 CD S + Cf Aw , where CD has a regime frontal area wetted area

dependence on Re and Cf is a continuous function ReL .

3

4.3 Known Solutions of the Navier-Stokes Equations

4.3.1 Boundary Value Problem

Navier-Stokes’: •

∂�v 1 1 ∂t

+ (�v · �) �v = −ρ�p + ν� 2�v +

ρf�

Conservation of mass: •

� · �v = 0

• Boundary conditions on solid boundaries “no-slip”:

�v = U�

Equations very difficult to solve, analytic solution only for a few very special cases (usually when � v = 0. . . ) v · ��

4

4.3.2 Steady Laminar Flow Between 2 Long Parallel Plates: Plane Couette Flow

y

h

z

x

U

Steady, viscous, incompressible flow between two infinite plates. The flow is driven by a pressure gradient in x and/or motion of the upper plate with velocity U parallel to the x-axis. Neglect gravity.

Assumptions Governing Equations Boundary Conditions

i. Steady Flow: ∂ ∂t = 0

ii. (x, z) >> h: ∂�v ∂x = ∂�v

∂z = 0

iii. Pressure: independent of z

Continuity: ∂u ∂x + ∂v

∂y + ∂w ∂z = 0

NS: ∂�v ∂t + �v · ��v = −1

ρ �p + ν�2�v

�v = (0, 0, 0) on y = 0

�v = (0, 0, 0) on y = h

Continuity

∂u ∂v ∂w ∂v + + = 0 = 0 v = v(x, z) v = 0 (1)

∂x ∂y ∂z⇒

∂y ⇒ ⇒

�� BC: v(x,

↑0,z)=0

=0, from assumption ii

5

��

Momentum x

∂u ∂u ∂u ∂u 1 ∂p �

∂2u ∂2u ∂2u �

+u + v + w = − + ν + + ∂t ∂x �� ∂y ∂z ρ ∂x ∂x2 ∂y2 ∂z2

⇒=0, (1)��

=0, ��

ii

�� ii

��i =0, =0, ii =0, =0, ii

∂2u 1 ∂p ν = (2)

∂y2 ρ ∂x

Momentum y

∂v 1 ∂p +� v 2 v

∂tv · �

=0

��,

= −ρ ∂y

+ ν� ��(1)

⇒(1) =0,

=0, i

∂p ∂p dp= 0 p = p(x) and = (3)

∂y ⇒

∂x dx↑assumption iii

Momentum z

∂w ∂w ∂u ∂w 1 ∂p �

∂2w ∂2w ∂2w �

+u + v + w = − +ν + + ∂t ∂x �� ∂y ∂z ρ ∂z ∂x2 ∂y2 ∂z2

⇒=0, (1)��

ii

�� iii

�� =0

��, �ii=0, i =0, =0, ii =0, =0, ii

∂2w = 0 w = ay + b w = 0 (4)

∂y2 ⇒ ⇒

↑w(x,0,z)=0 w(x,h,z)=0

From Equations (1), (4)

∂u du �v = (u, 0, 0). Also u = u(y) and = (5)⇒

∂y dy↑assumption ii

From Equations (2), (3, and (5)

d2u 1 d2p 1 �

dp �

2 1 �

dp �

y dy2

= ρν dx2

⇒ u = −2µ

−dx

y +C1y+C2 ⇒ u = − (h − y)y + U 2µ dx h

µ=↑ρν u(x,0

↑,z)=0

w(x,h,z)=U

6

y

)(yu

0)( == Uhu

0)0( =u

p p

0)( >−dx

dp

profileParabolic

h

y

)(yu

Uhu =)(

0)0( =u

0)( =−dx

dp

profileLinear

h

U

Special cases for Couette flow • 1 dp dp Px−Px+Lu(y) = 2µ (h − y)y(−

dx ) + U hy , where (−

dx ) = L

dp dpI. U = 0, � −

� > 0 II. U = 0� ,

� − �

= 0dx dx

Velocity•u(y) = 1 dp ) u(y) = U y

2µ (h − y)y(−dx h

Max velocity•dpumax = u(h/2) =

8hµ

2 (−

dx ) umax = U

Volume flow rate •

h3 dpQ = �

0 h u(y)dy =

8µ (−dx ) Q = h U2

Average velocity•Q h2 dpu = h =

6µ (−dx ) u = U 2

Viscous stress on bottom plate (skin friction)•du

�� Uτw = µ du �� = h dp �

> 0 τw = µ dy = µ

hdy 2

� − dx

y=0 y=0

7

0,0)(,0 >>−> Gdx

dpU

wτ

Uhu =)(

h

U

0,0)(,0 <<−> Gdx

dpU

flowback

Uhu =)( U

wτ

III. U �= 0, � − dp

dx

� �= 0

� − dp dx

� > 0

� − dp dx

� < 0

•Viscous stress on bottom plate (skin friction)

τw = h 2

� − dp dx

� + µ U

h

τw < = >0 when (−

dp dx

) < = > −

2µU h2

, in which case the flow is

� attached insipient separated

8

0,0)(,0 >>−> Gdx

dpU

Uhu =)(

h

U

0,0)(,0 <<−> Gdx

dpU 01,0)(,0 =⇒−=<−>

wG

dx

dpU τ

Uhu =)( U UUhu =)(

flowback

dpFor the general case of U = 0 and � � − �

= 0, �dx

τw = h� −

dp � + µ

U 2 dx h

We define a Dimensionless Pressure Gradient G

G ≡ h2 � −

dp �

2µU dx

such that

G > 0 denotes a favorable pressure gradient •

G < 0 denotes an adverse pressure gradient •

G = −1 denotes an incipient flow •

G < −1 denotes a separated or back-flow •

Lessons learned in § 4.3.2:

1. Reviewed how to simplify the Navier-Stokes equations.

2. Obtained one solution to the Navier-Stokes equations.

3. Realized that once the Navier-Stokes are solved we know everything.

In the next paragraph we are going to study one more solution to the Navier-Stokes equation, in polar coordinates.

9

x∂

L

y

a x

z

θ

x∂

L

y

a x

z

θ

� ��

4.3.3 Steady Laminar Flow in a Pipe: Poiseuille Flow

r = a

Vx(r)

Steady, laminar pipe flow. KBC: vx(a) = 0 (no slip) and 2 2 dvx(r = y2 + z , �v = (vx, vr, vθ)) dr (0) = 0 (symmetry).

Assumptions Governing Equations Boundary Conditions

i. Steady Flow: ∂ ∂t = 0

ii. (x, z) >> h: ∂�v ∂x = ∂�v

∂θ = 0

⇒ �v = �v(r)

iii. Pressure: independent of θ

Continuity: 1 r

∂rvr ∂r + 1

r ∂vθ ∂θ + ∂vx

∂x = 0

NS: In polar coordinates (see SAH pp.74)

vx(r = a) = 0 no-slip

dvx dr |r=0 = 0 symmetry

Following a procedure similar to that for plane Couette flow (left as an exercise) we can show that

1 dp �

1 d �

dvx ��

vr = vθ = 0, vx = vx(r), p = p(x), and = ν r ρ dx r dr dr

2r component of �in cylindrical coordinates

After applying the boundary conditions we find:

1 �

dp 2 vx(r) = 4µ

−dx

� �a − r 2

�

Therefore the volume flow rate is given by2π a

Q = �

dθ �

rdrvx(r) = π

a 4� −

dp �

8µ dx0 0

and the skin friction evaluates to ∂vr ∂vx ∂vx a dp

τw = τx(−r) = −τxy = −µ ( ∂x

+ ∂r

)

��r=a

= −µ ∂r

��r=a

⇒ τw =2

� − dx

�

10

x

y

0U

xwithincreasesthickness

layerboundary hx >>forflowCouette

h

4.4 Boundary Layer Growth Over an Infinite Flat Plate for Unsteady Flow

Boundary layer thickness is related to the area where the viscosity and vorticity effects are diffused.

For a flow over an infinite flat plate, the boundary layer thickness increases unless it is constrained in the y direction and/or by time (unsteady flow).

1. Steady flow, constrained in y

For a steady flow past a flat plate, the boundary layer thickness increases with x. If the flow is constrained in y, eventually the viscous effects are diffused along the entire cross section and the flow becomes invariant in the streamwise direction.

In paragraphs 4.3.2 and 4.3.3, we studied two cases of steady laminar viscous flows, where the viscous effects had diffused along the entire cross section.

� Couette

� � h

�Steady flow, we assumed that viscous effects diffused through entire .

Poiseuille a

11

� ��

2. Unsteady flow, unconstrained in y

Consider the simplest example of an infinite plate in unsteady motion.

z

y

x

U(t)

Assumptions �p = 0, ∂�v = ∂�v = 0 �v = �v (y, t)∂x ∂z ⇒

Can show v = w = 0 and u = u(y, t).

Finally, from u momentum (Navier-Stokes in x) we obtain

⎛ ⎞

∂u ∂u ∂u ∂u 1 ∂p ∂2u ∂2u ∂2u ∂t

+ u ∂x

+ v∂y

+ w∂z

= −ρ ∂x

+ν ⎜∂x2

+ ∂y2

+ ∂z2

⎟⎝ ⎠ ⇒=0��

=0

�� =0

� ��=0

��=0

��=0

∂u ∂2u = ν ‘ momentum ’ diffusion equation (6)

∂t ∂y2 velocity (heat)

Equation (6) is:

� first order PDE in time requires 1 Initial Condition →

� second order PDE in y requires 2 Boundary Conditions →

- u(y, t) = U(t) at y = 0, for t > 0 - u(y, t) 0 as y →∞→

From Equation (6), we observe that the flow over a moving flat plate is due to viscous dissipation only.

12

4.5.1 Sinusoidally Oscillating Plate

i. Evaluation of the Velocity Profile for Stokes Boundary Layer

The flow over an oscillating flat plate is referred to as ‘Stokes Boundary Layer’.

Recall that eiα = cos α + i sin α where α is real.

Assume that the plate is oscillating with U(t) = Uo cos ωt = Real {Uoeiωt}. From linear

theory, it is known that the fluid velocity must have the form

u (y, t) = Real �f (y) e iωt

� , (7)

where f(y) is the unknown complex (magnitude & phase) amplitude of oscillation.

To obtain an expression for f(y), simply substitute (7) in (6). This leads to:

d2f iωf = ν (8)

dy2

Equation (8) is a 2nd order ODE for f(y). The general solution is

(1+i) �√

ω/2ν � y + C2e

−(1+i) �√

ω/2ν � y

f (y) = C1e (9)

The velocity profile is obtained from Equations (7), (9) after we apply the Boundary Conditions.

u(y, t) must be bounded as y →∞⇔ C1 = 0 �

u (y, t) = Uo(e−y√

2ων ) cos

� − y

� ω

+ ωt�

u(y = 0, t) = U(t) f(y = 0) = Uo C2 = Uo 2ν⇔ ⇔

Stokes Boundary Layer

13

oU

yu )(

y

νω2

y

e−

e/1

δδ ≡e/1

11−

λ =

2πδ

ii. Some Calculations for the Stokes Boundary Layer

Once the velocity profile is evaluated, we know everything about the flow.

Stokes Boundary Layer. Velocity ratio uU(y

o

) as a function of the distance from the plate y.

Observe:

2ν )Uo 2ν

u(y, t)= (e−y

√ ω

cos � − y

� ω

+ ωt�

(10) Exponentially decaying

� �� envelope Oscillating component

SBL thickness

The ratio Uu o

is composed of an exponentially decaying part thickness of SBL decays →exponentially with y. We define various parameters that can be used as measures of the SBL thickness:

We define δ1/e as the distance y from the plate where u(δ

U1

o

/e) = 1

e . Substituting• �2νinto (10), we find that δ ≡ δ1/e = ω

�2ν u(λ)The oscillating component has wave length λ = 2πω = 2πδ. At λ,

Uo = 0.002. • ∼

We define δ1% as the distance y from the plate, where u(δ1%) = 1%. Substituting • Uo

into (10), we find that δ1% = − ln(u(δ1%) )�

2ν = 4.6δ.Uo ω

∼

14

Numerical examples:

For oscillating plate in water (ν = 10−6m2/s= 1mm2/s) we have

δ1% 4.6 √

T ∼= = 2.6�

Tin sec

�� √π ��

in mm

T = 2π ω δ1%

1s 3mm

10s ≤1cm

Excursion length and SBL

The plate undergoes a motion of amplitude A.

˙ UoX = A sin(ωt) U = X = Aω cos(ωt) ω = ⇒ �� ⇒

A Uo

Comparing the SBL thickness ∼ δ with A, we find

δ �

ν/ω �

νA/Uo �

ν 1 = =

A ∼

A A UoA ∼ �

ReA↑Uoω= A

Skin friction

The skin friction on the plate is given by

τw = µ ∂u

�� = . . . = µUo

� ω �

sin ωt − cos ωt�

∂y y=0 2ν

The maximum skin friction on the wall is �

ω |τw|max = µUo ν

and occurs at ωt = 34 π , 7

4 π , · · ·

15

U(t)

t

Uo

4.5.2 Impulsively Started Plate

Recall Equation (6) that describes the the flow u(y, t) over an infinite flat plate undergoing unsteady motion.

∂u ∂2u = ν

∂t ∂y2

For an impulsively started plate, the Boundary Conditions are:

u(o, t) = Uo

�

for t > 0, i.e. u(y, 0) = 0 u(∞, t) = 0

Notice that the problem stated by Equation (6) with the above Boundary Conditions has no explicit time scale. In this case it is standard procedure to (a) use Dimensional Analysis to find the similarity parameters of the problem, and (b) look for solution in terms of the similarity parameters:

u u = f (Uo, y, t, ν) ⇒

U

u

o = f

�

2√y

νt

� ⇒ = f(η) Self similar solution Uo↑

DA � ≡��

η �

similarity parameter

The velocity profile is thus given by�:

u 2 �η

= erfc (η) = 1 − erf (η) = 1 − e−α2 dα

Complementaryerror function

Uo � �� √π

0

16

� �

�

� Hints on obtaining the solution:

η = 2√y

νt

∂u =ν ∂2 u

∂ ∂η ∂ y ∂ ∂t ∂y2 d(u/Uo) d2(u/Uo) ∂t =

∂t ∂η = −4t√

νt ∂η −→ −η dη

= dη2

−→ . . .

∂2 ∂2

��2nd

= �

∂η �2 ∂2 1 order ODE = ∂y2 ∂y ∂η2 4νt ∂η2

⎫⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎭

Boundary layer thickness In the same manner as for the SBL, we define various parameters that can be used to measure the boundary layer thickness:

δ ≡ 2√

νt. = 0.16. • At y = δ −→ uU(δ

o

) ∼

δ1% ∼= 1.82δ. •

Excursion length and boundary layer thickness At time t, the plate has travelled a distance L = Uot t =

UL o . →

Comparing the boundary layer thickness ∼ δ with L, we find

δ √

νt �

νL/Uo ν 1L ∼ = =

L ∼�

ReL L L Uo

Skin friction The skin friction on the plate is given by

∂u��

Uoτw = µ

∂y y=0

= . . . = −µ√πνt

17



4.6 Laminar Boundary Layers

Uo

L

u, v viscous flow δ

y

x

U potential flow

4.6.1 Assumptions

∂ • 2D flow: w, ≡ 0 and u (x, y) , v (x, y) , p (x, y) , U (x, y).∂z

∂• Steady flow: ≡ 0. ∂t

• For δ << L, use local (body) coordinates x, y, with x tangential to the body and y normal to the body.

• u ≡ tangential and v ≡ normal to the body, viscous flow velocities (used inside the boundary layer).

• U, V ≡ potential flow velocities (used outside the boundary layer).

1

( )

( )

4.6.2 Governing Equations

• Continuity

∂u ∂v + = 0 (1)

∂x ∂y

• Navier-Stokes:

∂u ∂u 1 ∂p ∂2u ∂2u u + v = − + ν + (2)∂x ∂y ρ ∂x ∂x2 ∂y2

∂v ∂v 1 ∂p ∂2v ∂2v u + v = − + ν + (3)∂x ∂y ρ ∂y ∂x2 ∂y2

4.6.3 Boundary Conditions

• KBC Inside the boundary layer:

No-slip u(x, y = 0) = 0

No-flux v(x, y = 0) = 0

Outside the boundary layer the velocity has to match the P-Flow solution.Let y� ≡ y/δ, y ∗ ≡ y/L, and x ∗ ≡ x/L. Outside the boundary layer y� → ∞

∗but y → 0. We can write for the tangential and normal velocities u(x ∗ , y� → ∞) = U(x ∗ , y ∗ → 0) ⇒ u(x ∗ , y� → ∞) = U(x ∗ , 0), and v(x ∗ , y� → ∞) = V (x ∗ , y ∗ → 0) ⇒ v(x ∗ , y� → ∞) = V (x ∗ , 0) = 0

No-flux P-Flow

In short:u(x, y � → ∞) = U(x, 0)

v(x, y � → ∞) = 0

• DBC As y� → ∞, the pressure has to match the P-Flow solution. The x-momentum

∗equation at y = 0 gives

∂U ∂U 1 dp ∂2U dp ∂U U + V = − + ν ⇒ = −ρU

∂x ↓ ∂y ρ dx ↓ ∂y2 dx ∂x 0

0

2

︸︷︷︸︸︷︷︸

√

( )

4.6.4 Boundary Layer Approximation

Assume that ReL >> 1, then (u, v) is confined to a thin layer of thickness δ (x) << L. For flows within this boundary layer, the appropriate order-of-magnitude scaling / normalization is:

Variable Scale Normalization

u U u = Uu ∗

x L x = Lx∗

y δ y = δy∗

v V =? v = Vv ∗

• Non-dimensionalize the continuity, Equation (1), to relate V to U

( )∗ ( )∗ ( ) U ∂u V ∂v δ + = 0 =⇒ V = O U

L ∂x δ ∂y L

• Non-dimensionalize the x-momentum, Equation (2), to compare δ with L ⎡ ⎤

∗

U2 ( ∂u

)∗ UV (

∂u )∗

1 ∂p νU ⎢⎢ δ2 ( ∂2u

) ( ∂2u

)∗⎥⎥ u + v = − + ⎢ + ⎥

L ∂x δ ∂y ρ ∂x δ2 ⎣L2 ∂x2 ∂y2 ⎦ O(U2/L) ignore

The inertial effects are of comparable magnitude to the viscous effects when:

U2 νU δ ν 1 ∼ =⇒ ∼ = << 1 L δ2 L UL ReL

The pressure gradient ∂p must be of comparable magnitude to the inertial effects ∂x

∂p U2

= O ρ ∂x L

3

( )

( ) ( )

• Non-dimensionalize the y-momentum, Equation (3), to compare ∂p to ∂p ∂y ∂x

( )∗ ( )∗ ( )∗ ( )∗ UV ∂v V2 ∂v 1 ∂p νV ∂2v νV ∂2v u + v = − + +

L ∂x δ ∂y ρ ∂y L2 ∂x2 δ2 ∂y2 ︸︷︷︸︸︷︷︸︸︷︷︸︸︷︷︸ O( U2 δ O( U2 δ O( U2 δ3 U 2 δ

L L ) L L )

L L3 ) O( L L )

The pressure gradient ∂p must be of comparable magnitude to the inertial effects ∂y

∂p U2 δ = O ρ

∂y L L

Comparing the magnitude of ∂p to ∂p we observe ∂x ∂y

∂p U2 δ ∂p U2

= O ρ while = O ρ =⇒ ∂y L L ∂x L ∂p ∂p ∂p

<< =⇒ ≈ 0 =⇒ ∂y ∂x ∂y

p = p(x)

• Note:

- From continuity it was shown that V/U ∼ O(δ/L) ⇒ v << u, inside the boundary layer.

- It was shown that ∂p = 0, p = p(x) inside the boundary layer. This means that ∂y

the pressure across the boundary layer is constant and equal to the pressure outside the boundary layer imposed by the external P-Flow.

4

︸︷︷︸

︸︷︷︸

∫ ( )

∫ ( )

4.6.5 Summary of Dimensional BVP

Governing equations for 2D, steady, laminar boundary layer

∂u ∂vContinuity : + = 0

∂x ∂y ∂u ∂u 1 dp ∂2u

x-momentum : u + v = − +ν ∂x ∂y ρ dx ∂y2

UdU/dx, y=0

∂p y - momentum : = 0

∂y

Boundary Conditions

KBC

At y=0 : u(x, 0) = 0

v(x, 0) = 0

At y/δ → ∞ : u(x, y/δ → ∞) = U(x, 0)

v(x, y/δ → ∞) = 0

DBC dp ∂U IN the b.l. 1

= −ρU or p(x) = C − ρU2(x, 0)dx ∂x 2

Bernoulli for the P-Flow at y =0

4.6.6 Definitions ∞

Displacement thickness δ ∗ ≡ 1 − u

dyU

0

∞ u u

Momentum thickness θ ≡ 1 − dyU U

0

5

o

Physical Meaning of δ∗ and θ

Assume a 2D steady flow over a flat plate.

Recall for steady flow over flat plate dp = 0 and pressure p = const.dx

Choose a control volume ([0, x] × [0, y/δ → ∞]) as shown in the figure below.

y /δ → ∞ CVQ

U o

o U U

u(y)

0 x

Layer Boundary

Flow -P

1

2

3

4

CV for steady flow over a flat plate.

Control Volume ‘book-keeping’

Surface n �v �v · n �v(�v · n) −pn

1© −i Uo i −Uo −U2 o i pi

2© −j 0 0 0 pj

3© i u(x, y)i + v(x, y)j u(x, y) u2(x, y)i + u(x, y)v(x, y)j −pi

4© j Uo i + v(x, y)j v(x, y) v(x, y)Uo i + v2(x, y)j −pj

6

︸︷︷︸

∮ ∫ ∫ ∫

︸︷︷︸

∮ ∑

︸︷︷︸

∫ )

∑ ∫ ( ) ∑

︸︷︷︸

Conservation of mass, for steady CV

∞ ∞ x

�v · ndS = 0 ⇒ − Uody′ + u(x, y ′)dy′ + v(x ′ˆ , y)dx′ = 0 ⇒ 1234 0 0 0

Q ∫ ∞ ∫ ∞ ∫ ∞ ∫ ∞ ( )

Q = U0δ ∗

0 0 0 0 Uo Q = Uody′ − udy′ = (Uo − u)dy′ = Uo 1 −

u dy′ ⇒

δ∗

where ()′ are the dummy variables.

Conservation of momentum in x, for steady CV

u(�v n· ˆ)dS = Fx ⇒ 1234

∫ ∞ ∫ ∞ ∫ x ∫ ∞ ∫ ∞ ∑ −Uo

2dy′ + u 2(x, y ′)dy′ + v(x ′ , y)Uodx′ = pdy′ − pdy′ + Fx,friction ⇒ 0 0 0 0 0

∫ ∞ ∫ ∞ ∫ x ∑−Uo

2dy′ + u 2(x, y ′)dy′ + Uo v(x ′ , y)dx′ = Fx,friction ⇒0 0 0

Q ∫ ∞ ∫ ∞ ∫ ∞ ∑

−Uo 2dy′ + u 2(x, y ′)dy′ + Uo (Uo − u)dy′ = Fx,friction ⇒

0 0 0

∫ ∞ ( ) ∑ − Uo

2 + u 2 + Uo 2 − Uou dy′ = Fx,friction ⇒

0

∞ ( u2 u ∑

Uo 2

0 Uo 2 −

Uo dy′ = Fx,friction ⇒

∞

Fx,friction = −Uo 2 u

1 − u

dy′ ⇒ Fx,friction = −Uo 2θ

0 Uo Uo

θ

7

︸︷︷︸

( )

√ √

4.7 Steady Flow over a Flat Plate: Blasius’ Laminar Boundary Layer

y

xL

Uo

Steady flow over a flat plate: BLBL

4.7.1 Derivation of BLBL

dp • Assumptions Steady, 2D flow. Flow over flat plate → U = U0, V = 0, = 0 dx

• LBL governing equations

∂u ∂v + = 0

∂x ∂y ∂u ∂u ∂2u

u + v = ν ∂x ∂y ∂y2

• Boundary conditions u = v = 0 on y = 0

y v → V = 0, u → Uo outside the BL, i.e., >> 1

δ • Solution Mathematical solution in terms of similarity parameters.

u Uo y η νx and η ≡ y ⇔ = √ ⇔ y = η

U νx x Rx Uo

Similarity solution must have the form

u (x, y) = F (η)

Uo

self similar solution

8

︸︷︷︸

√ √

︸︷︷︸ √

√

√/√

√ √

√

(

)(

︸︷︷︸

We can obtain a PDE for F by substituting into the governing equations. The PDE has no-known analytical solution. However, Blasius provided a numerical solution. Once again, once the velocity profile is evaluated we know everything about the flow.

4.7.2 Summary of BLBL Properties: δ, δ0.99, δ∗, θ, τo, D, Cf

u (x, y) Uo νx y η y ≡ ηF (η) ; η = y ; ;= =

Uo νx Uo x Rxevaluated

numerically local R#

νxδ ≡

Uo

, i.e., η.99 = 4.9

⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

∼ νx = 4.9 √

δ ∝ x, δ ∝ 1 Uo δ.99

Uo ⎬

νx δ ν 1∝ = √x Uox Rx

∼δ ∗ = 1.72⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪

η ∗ , i.e., = 1.72Uo

θ ∼ νx = 0.664

Uo ⎭

⎫Uox

)−1/2 ⎪⎪⎪⎪⎪⎬τo ≡ τw

∼= 0.332ρU2 o

1τo ∝ √ν

x

= 0.332 ρU2 Ro −1/2

x

⎪⎪⎪⎪⎪⎭ τo ∝ Uo 3/2

local R#

9

Total drag on plate L x B

∫L ( )−1/2

D = B τodx ∼= 0.664 ( ρUo

2 ) (BL)

UoL ⇒ D ∝√

L, D ∝ U3/2 ︸︷︷︸ ν width 0 ︸︷︷︸

−1/2ReL

Friction (drag) coefficient:

D 1.328 1 1∼Cf = = √ ⇒ Cf ∝ √ , Cf ∝ √1 (ρU2) (BL) ReL L U2 o

C f

Layer BoundaryLaminarBlasius

L

f C Re

328.1≅

Layer BoundaryTurbulent

L Re

008.0

310

5103×

⎧

2.3)(JNNplateflatfor

Layer BoundaryTurbulent

5

f C

⎪Rex ~ 3×10 Transition at ⎨

⎪Re ~ 600⎩ δ

Skin friction coefficient as a function of Re.

A look ahead: Turbulent Boundary Layers

Observe form the previous figure that the function Cf, laminar(Re) for a laminar boundary layer is different from the function Cf, turbulent(Re) for a turbulent boundary layer for flow over a flat plate.

Turbulent boundary layers will be discussed in proceeding Lecture.

10

4.8 Laminar Boundary Layers for Flow Over a Body of General Geometry

The velocity profile given in BLBL is the exact velocity profile for a steady, laminar flow over a flat plate. What is the velocity profile for a flow over any arbitrary body? In general it is dp/dx �= 0 and the boundary layer governing equations cannot be easily solved as was the case for the BLBL. In this paragraph we will describe a typical approximative procedure used to solve the problem of flow over a body of general geometry.

1. Solve P-Flow outside B ≡ B0

2. Solve boundary layer equations (with ∇P term) → get δ∗(x)

3. From B0 + δ∗ → B

4. Repeat steps (1) to (3) until nochange

U

y x

L

.constU

0P

≠

≠∇

B0

• von Karman’s zeroth moment integral equation

τ0 d ( ) dU = U2(x)θ(x) + δ ∗ (x)U(x) (4)

ρ dx dx

• Approximate solution method due to Polthausen for general geometry (dp/dx �= 0) using von Karman’s momentum integrals.

The basic idea is the following: we assume an approximate velocity profile (e.g. linear,4th order polynomial, . . .) in terms of an unknown parameter δ(x). From the velocityprofile we can immediately calculate δ∗, θ and τo as functions of δ(x) and the P-Flowvelocity U(x).

Independently from the boundary layer approximation, we obtain the P-Flow solution outside the boundary layer U(x), dU .

dx dUUpon substitution of δ∗, θ, τo, U(x), dx in von Karman’s moment integral equation(s)

we form an ODE for δ in terms of x.

11

∣ ( )

∣ ︸︷︷︸

{

[ ]

• Example for a 4th order polynomial Polthausen velocity profile

Polthausen profiles - a family of profiles as a function of a single parameter Λ(x) (shape function factor).

� Assume an approximate velocity profile, say a 4th order polynomial:

u (x, y) y ( y )2 ( y )3 ( y )4 = a (x) + b (x) + c (x) + d (x) (5)

U (x, 0) δ δ δ δ

There can be no constant term in (5) for the no-slip BC to be satisfied y = 0, i.e, u(x, 0) = 0. We use three BC’s at y = δ

u ∂u ∂2u = 1, = 0, = 0, at y = δ (6)

U ∂y ∂y2

From (6) in (5), we re-write the coefficients a(x), b(x), c(x) and d(x) in terms of Λ(x)

a = 2 + Λ/6, b = −Λ/2, c = −2 + Λ/2, d = 1 − Λ/6

� To specify the approximate velocity profile u(x,y) in terms of a single unknown U(x,0)

parameter δ we use the x-momentum equation at y = 0, where u = v = 0

∂u ∂u ∂U ∂2u ∣ 1 dU δ2 dU δ2(x) u + v = U + ν ∣ ⇒ b = − ⇒ Λ(x) = ↓ ∂x ↓ ∂y ∂x ∂y2 2 dx ν dx ν ︸︷︷︸ y=00 0

− 1 dp

ν 2bUρ dx δ2

dU Λ > 0 : favorable pressure gradient Observe: Λ ∝ ⇒

dx Λ < 0 : adverse pressure gradient

Putting everything together:

u (x, y) (y ) (y )3 (y )4= 2 − 2 + +

U (x, 0) δ δ δ dU δ2 1(y ) 1(y )2 1(y )3 1(y )4

+ − + − dx ν 6 δ 2 δ 2 δ 6 δ

12

∫ ( )

∣ ( )

u(x,y)� Once the approximate velocity profile U(x,0) is given in terms of a single unknown

parameter δ(x), then δ∗ , θ and τo are evaluated

∫∞ ( ) ( u ) 3 1 (dU δ2 ) δ ∗ = 1 − dy = δ −

U 10 120 dx ν 0

δ u ( u ) 37 1 (dU δ2 ) 1 (dU δ2 )2

θ = 1 − dy = δ − − U U 315 945 dx ν 9072 dx ν

0

∂u ∣ μU 1(dU δ2 ) τo = μ ∣ = 2 +

∂y ∣ y=0 δ 6 dx ν

Notes:

- Incipient flow (τo = 0) for Λ = −12. However, recall that once the flow is separated the boundary layer theory is no longer valid.

- For dU = 0 → Λ = 0 Pohlhausen profile differs from Blasius LBL only by a dx

few percent.

� After we solve the P-Flow and determine U(x), dU we substitute everything into dx

von Karman’s momentum integral equation (4) to obtain

dδ 1 dU d2U/dx2

= g(δ) + h(δ)dx U dx dU/dx

where g, h are known rational polynomial functions of δ. This is an ODE for δ = δ(x) where U, dU , d

2U are specified from the P-Flow dx dx2

solution.General procedure:

1. Make a reasonable approximation in the form of (5), 2. Apply sufficient BC’s at y = δ, and the x-momentum at y = 0 to reduce (5)

as a function a single unknown δ, 3. Determine U(x) from P-Flow, and 4. Finally substitute into Von Karman’s equation to form an ODE for δ(x).

Solve either analytically or numerically to determine the boundary layer growth as a function of x.

13

︸︷︷︸

︸︷︷︸

′

¯ ′

′

′ ′



4.9 Turbulent Flow – Reynolds Stress

Assume a flow �v with a time scale T . Let τ denote a time scale τ << T . We can then write for each component of the velocity

ui = u i + ui (1)

where by definition

1 ∫ τ

u i = uidt τ 0

It immediately follows that

∂ ∂ui u = ui − u i = u i − u i = 0, also u i etc.=i ∂x ∂x

Substitute Eq. (1) into continuity and average over τ , i.e., take ( )

+ ∂ui = 0, =⇒

∂ui ∂u i ∂u i = 0

∂xi =

∂xi ∂xi ∂xi

↓ 0

∂ui ∂u i ∂ui , =⇒∂ui = 0 ∂xi

but = 0 = +∂xi ∂xi ∂xi

↓ 0 , just shown

1

′

︸︷︷︸

′( ) ′ ′ ′

′ ′ ′ ′

︸︷︷︸

′ ′

︸︷︷︸

′ ′

′ ′

Substitute Eq. (1) into the momentum equations and take ( )

∂ui ∂ui 1 ∂τij 1 ∂p 2+ uj = = − + ν∇ ui∂t ∂xj ρ ∂xj ρ ∂xi

⎧ ν∇2 = ν∇2ui ui⎨∂ui ∂u i ∂ui ; similarly +=

∂t ∂t ∂t ⎩ ∂p = ∂ (p + p′) = ∂p ∂xi ∂xi0

etc.∂xi

∂ui +uj∂xj ∂xj

∂ui ∂ ∂u i ∂u i ∂ ui+ uj∂xj ∂xj

′(¯ +u ui i∂xj ∂xj

′¯ + uj j ) = u j + u juj = u︸︷︷︸︸︷︷︸

0 0

but from continuity we have

′uj

∂ujui − ui∂xj ∂xj

0→by continuity

ui

∂ ∂ j′ iu = u

∂xj

and thus we finally obtain

∂u i ∂u i 1 ∂p ∂ + uj = − + ν∇2 u i − uj

∂u i ∂u i 1 ∂ [ ] + uj = τij − ρuiuj∂t ∂xj ρ ∂xj

∂t ∂xj ρ ∂xi ∂xj

1 ∂ ρ ∂xj

τij

Reynolds averaged N-S equation:

Reynolds stress: τRij ≡ −ρuiuj

2

o

4.10 Turbulent Boundary Layer Over a Smooth Flat Plate

We have already seen that the function of the friction coefficient Cf (ReL ) differs for laminar and turbulent flows. In this paragraph we will discuss the case of a turbulent boundary layer. Following a procedure similar to that for flow past a body of general geometry, we will use an approximate velocity profile, obtain the P-Flow solution and eventually substitute everything into von Karman’s momentum integral equation. The velocity profiles used in practice are either empirical ((1/7)th power) or semi-empirical (logarithmic) laws.

δ

y

Uo u U

δ y

log

1/7

oU

ulog

4.10.1 (1/7)th Power Velocity Profile Law

Let the velocity profile be determined by the following empirical law

u ( y )1/7 = (2)

Uo δ

where δ = δ(x) is to be determined.

From equation (2) we can obtain directly δ∗and θ

δ δ ∗ =

8 7

δ ∼θ = = 0.0972 δ 72

However, we need to use an additional empirical law to determine the skin friction. From Blasius’ law of friction for pipes we obtain an expression for τo

τo (

Uo

)−1/4δ

= 0.0227 ρU2 νo

3

( )

From P-Flow for flow past a flat plate we have U(x) = U0 = const, and dp/dx = 0

Substituting δ∗, θ, τo, Uo into von Karman’s moment equation

τo d(

Uoδ )−1/4

7 dδ = (θ) =⇒ 0.0227 =

ρU2 dx ν 72 dxo

This is a 1st order ODE for δ. One BC is required. We assume that the the flow is tripped at x = 0, i.e., at x = 0 the flow is already turbulent. Further on, we assume that the turbulent boundary layer starts at x = 0, i.e., δ(0) = 0. It follows that

δ (x) ∼ (

Uox )−1/5

δ ∼= 0.373x =⇒ = 0.373R−1/5

ν x ex

Compare:

Laminar Boundary Layer Turbulent Boundary Layer (1/7th power law) √ δ (x) ∝ x δ (x) ∝ x4/5

δ∗ ∼ √ νx δ∗ ∼

( νx4

)1/5 = 1.72 = 0.047

Uo Uo

Once the profile has been determined we can evaluate the friction drag

D = 0.036 ρU2 BL R−1/5 o eL

Thus, the friction coefficient for turbulent (tripped and/or ReL > 5 × 105) flow over a flat plate is

Cf = D

= 0.073R−1/5

1ρU2BL eL

o2

4.10.2 Logarithmic Velocity Profile Law

If the velocity profile is determined by the semi-empirical logarithmic velocity profile law, following an approach similar to that for the 1/7th power law, we obtain Schoenherr’s formula for the friction coefficient

0.242 √ = log10 (ReL Cf )Cf

4

4.10.3 Summary of Boundary Layer Over a Flat Plate

Laminar BL (Blasius) Turbulent BL (1/7th power law)

δ x ∝ R−1/2

ex

δ ∗ = 1.72xR−1/2 ex

∝ √

x

δ x ∝ R−1/5

ex

δ ∗ = 0.047xR−1/5 ex

∝ x 4/5

τo = 0.332ρU2 o R

−1/2 ex

τo = 0.0227ρU2 o R

−1/4 eδ

τo = 0.02297ρU2 o R

−1/5 ex

D = 0.664ρU2 0 (BL)R−1/2

eL D = 0.03625ρU2

0 (BL)R−1/5 eL

Cf ≡ D

ρU2 o (BL)

= 1.328R−1/2 eL

Cf ≡ D

ρU2 o (BL)

= 0.0725R−1/5 eL

For τo, the cross-over is at Rex ∼ 3.4 x 103, i.e.,

(τo)laminar > (τo)turbulent for Rex < 3.4 × 103 Cf

(τo)laminar ∼ (τo)turbulent for Rex ∼ 3.4 × 103

(τo)laminar < (τo)turbulent for Rex > 3.4 × 103

C fL ~ RL

−

C f ~ RL − 5

1

T

~ 0.01 Therefore, for most prototype scales:

2 1

ln (RL) (Cf )turbulent > (Cf )laminar

RL ~ 1.6 x 104

(τo)turbulent > (τo)laminar

5



Turbulent Boundary Layers: Roughness Effects

So far, we have assumed a ‘hydraulically smooth’ surface. In practice, it is rarely so, due to fouling, rust, rivets, etc. . . .

Uo

Viscous sublayer

δv k = characteristic roughness height

To account for roughness we first define an ‘equivalent sand roughness’ coefficient k (units: [L]), a measure of the characteristic roughness height.

The parameter that determines the significance of the roughness k is the ratio

k δ

We thus distinguish the following two cases, depending of the value of the ratio kδ on the

actual surface - e.g., ship hull.

1. Hydraulically smooth surface For k < δv << δ, where δv is the viscous sub-layer thickness, k does not affect the turbulent boundary layer significantly.

k << 1 ⇒ Cf � Cf , smooth ⇒ Cf = Cf (ReL )δ

1

2. Hydraulically rough surface For k >> δ >> δv, the flow will resemble what is sketched in the following figure.

separation

δv k

In terms of sand grains: each sand grain can be thought of as a bluff body. The flow, thus separates downstream of each sand grain. Recalling that drag due to ‘separation’ = form drag >> viscous drag we can approximate the friction drag as the resultant drag due to the separation behind each sand grain.

k k δ

>> 1 ⇒ Cf ≡ Cf , rough ⇒ Cf �� = Cf ( , ReL )L

weak dependence

k/l ↑ Cf C

D ≠ F (ReL )

k/l = constant

C f rough

RL

C fsmooth

Cf , rough has only a weak dependence on ReL , since for bluff bodies CD � )= F (ReL

2

� �

� � � �

In summary The important parameter is k/δ:

k << 1 : hydraulically smooth

δ (x)

k >> 1 : rough

δ (x)

Therefore, for the same k, the smaller the δ, the more important the roughness k.

4.11.1 Corollaries

1. Exactly scaled models (e.g. hydraulic models of rivers, harbors, etc. . . )

k Same relative roughness: ∼ const for model and prototype

L

k k L k = ∼ 1/5ReLδ L δ L

k ↑ for ReL ↑ δ

For Re model << Re prototype :

k k <

δ δ m p

(Cf )m < (Cf )p

3

��

� ��

2. Roughness Allowance. Often, the model is hydraulically smooth while the prototype is rough. In practice, the roughness of the prototype surface is accounted for ‘indirectly’.

Cf

ΔCf remains

Uk = Rk = constantν

constant with Rl RL

C fsmooth

Uk • For the same ship (Re same), different k gives different Rek = . ν

• For a given Rek , the friction coefficient Cf is increased by almost a constant for Uk

= Rek = const over a wide range of ReL . ν • If the model is hydraulically smooth, can we account for the roughness of the

prototype? Notice that ΔCf = ΔCf (Rek ) has only a weak dependence on ReL . We can therefore, run an experiment using hydraulically smooth model, and add ΔCf

to the final friction coefficient for the prototype

Cf (ReL ) = Cf smooth + ΔCf (Rek )

not (R )eL

Gross estimate: For ships, we typically use ΔCf = 0.0004.

k Rek ∼ RekReality: = � � = 4/5

=⇒ δ δ/L ReL ReL

∼ReL −1/5

k ↓ as ReL ↑, i.e., ΔCf smaller for larger ReL . δ

• Hughes’ Method Adjust for ReL dependence of Cfrough .

Cfrough = Cfsmooth

(1 + γ) =⇒ ΔCf = γCfsmooth (ReL )

i.e., As ReL ↑, ΔCf ↓.

4

Chapter 5 - Model Testing.

5.1 Steady Flow Past General Bodies

- In general, CD = CD(Re).

- For bluff bodies

Form drag >> Friction drag ⇒ CD ≈ const ≡ CP (within a regime)

Recall that the form drag (CP ) has only regime dependence on Reynold’s number, i.e, its NOT a function of Reynold’s number within a regime.

- For streamlined bodies CD(Re) = Cf (Re) + CP

5.1.1 Steps followed in model testing:

(a) Perform an experiment with a smooth model at ReM (ReM << ReS ) and obtain the model drag CDM .

(b) Calculate CPM = CDM − CfM(ReM ) = CPS = CP ; CDM measured, CfM(ReM ) calculated.

(c) Calculate CDS = CP + CfS(ReS )

(d) Add ΔCf for roughness if needed.

R

Cf (Rship)

CP measured

CD predicted

Cf (Rship) calculated

Cf (Rm) calculated

Rm Rship

5

�

Caution: In an experiment, the boundary layer must be in the same regime (i.e., turbulent) as the prototype. Therefore turbulence stimulator(s) must be added.

R

CP turbulent regime

Turbulent CfLaminar Cf

MODEL U


TBL LBL TBL

� �

to be triggered here

5.1.2 Drag on a ship hull For bodies near the free surface, the Froude number Fr is important, due to wave effects. Therefore CD = CD(Re, Fr). In general the ra-

Re gL3

tio = . It is impossible to easily scale both Re and Fr. For exampleFr ν

Re Lm 1 νm gm= constant and = ⇒ = 0.032 or = 1000!

Fr Lp 10 νp gp

This makes ship model testing seem unfeasible. Froude’s Hypothesis proves to be invaluable for model testing

calculate measure indirectly � ��

CD(Re, Fr) ≈ Cf (Re) + CR(Fr) � ��

Cf for flat plate residual drag of equivalent wetted area

In words, Froude’s Hypothesis assumes that the drag coefficient consists of two parts, Cf that is a known function of Re, and CR, a residual drag that depends on Fr number only and not on Re. Since Cf (Re) ∼ Cf (Re)flat plate, we need to run experiments to (indirectly) get CR(Fr).

Thus, for ship model testing we require Froude similitude to measure CR(Fr), while Cf (Re) is estimated theoretically.

6

��

��

�

�

��

� �

5.1.3 OUTLINE OF PROCEDURE FOR FROUDE MODEL TESTING(S ≡ ‘SHIP’ M ≡‘MODEL’; in general νS = νM , and ρS �� = ρM )

1. Given US , calculate: FrS = US / gLS = FrM

2. For Froude similitude, tow model at: UM = FrS gLM

3. Measure total resistance (drag) of model: Measure DM

DM4. Calculate total drag coefficient for model: CDM =

0.5ρM U2 SMM

wetted area

0.075 5. Use ITTC line to calculate Cf (ReM ): Cf (ReM ) =

(log10 ReM − 2)2

6. Calculate residual drag of model: CRM = CDM − Cf (ReM )

7. Froude’s Hypothesis: CRM (ReM , Fr) = CRM (Fr) = CRS (Fr) = CR(Fr)

0.075 8. Use ITTC line to calculate Cf (ReS ): Cf (ReS ) =

(log10 ReS − 2)2

9. Calculate total drag coefficient for ship: CDS = CR(Fr) + Cf (ReS ) + ΔCf

∼= 0.0004 typical value

10. Calculate the total drag of ship: DS = CDS · 0.5ρSUS 2 SS

wetted area

11. Calculate the power for the ship: PS = DS US

12. Repeat for a series of US

7

⇀



Chapter 6 - Water Waves

6.1 Exact (Nonlinear) Governing Equations for Surface Gravity Waves, Assuming Potential Flow

Free surface definition

,(xB

0),,,(or),,( == tzyxFtzxy η

x

y

y

z

Unknown variables

Velocity field:

Position of free surface:

Pressure field:

Governing equations

Continuity:

Bernoulli for P-Flow:

Far way, no disturbance:

y, z, t) = 0

v (x, y, z, t) = ∇φ (x, y, z, t)

y = η (x, z, t) or F (x, y, z, t) = 0

p (x, y, z, t)

∇2φ = 0 y < η or F < 0 ∂φ 1 2 p−pa+ |∇φ| + + gy = 0; y < η or F < 0∂t 2 ρ

∂φ/∂t, ∇φ → 0 and p = pa − ρgy ︸︷︷︸︸︷︷︸ atmospheric hydrostatic

1

( )( )

︸︷︷︸

︸︷︷︸

Boundary Conditions

1. On an impervious boundary B (x, y, z, t) = 0, we have KBC:

�v n· · ˆ = ∂φ � (� · n x, t) = Unˆ = ∇φ n = U x, t) ˆ (� on B = 0 ∂n

Alternatively: a particle P on B remains on B, i.e., B is a material surface. For example if P is on B at t = t0, P stays on B for all t.

B(�xP , t0) = 0, then B(�xP (t), t) = 0 for all t,

so that, following P B is always 0.

DB ∂B ∴ = + (∇φ · ∇) B = 0 on B = 0 Dt ∂t

For example, for a flat bottom at y = −h ⇒ B = y + h = 0 ⇒

DB ∂φ ∂ ∂φ = (y + h) = 0 ⇒ = 0 on B = y + h = 0

Dt ∂y ∂y ∂y

=1

2. On the free surface, y = η or F = y − η(x, z, t) = 0 we have KBC and DBC.

KBC: free surface is a material surface, no normal velocity relative to the free surface. A particle on the free surface remains on the free surface for all times.

DF D ∂φ ∂η ∂φ ∂η ∂φ ∂η Dt

= 0 = Dt

(y − η) = ∂y ︸︷︷︸

− ∂t

− ∂x ∂x ︸︷︷︸

− ∂z ∂z ︸︷︷︸

on y = η ︸︷︷︸ still

vertical slope slope unknown

of f.s. of f.s. velocity

DBC: p = pa on y = η or F = 0. Apply Bernoulli equation at y = η:

∂φ 1 2+ |∇φ| + g η = pa on y = η ∂t 2 ︸︷︷︸

still unknown non-linear term

2

︸︷︷︸

6.2 Linearized (Airy) Wave Theory

Assume small wave amplitude compared to wavelength, i.e., small free surface slope

A << 1

λ

SWL

crest

wavelength Water depth h

trough

Wave height H

λ

Wave amplitude A

Wave period T

H =

A/2

Consequently

φ η , << 1

λ2/T λ

We keep only linear terms in φ, η.

∂ For example: () | = () + η ()| + . . . Taylor series y=η y=0 y=0 ︸︷︷︸ ∂y

keep discard

3

{ }

{ }

∣ ∣ ∣ ∣

{ }

{ }

6.2.1 BVP In this paragraph we state the Boundary Value Problem for linear (Airy) waves.

∂ 2 φ ∂φ+ g = 0 ∂t 2 ∂y

y = 0

∇ 2 φ = 0

y = -h

∂φ = 0 ∂y

Finite depth h = const Infinite depth

GE: ∇2φ = 0, −h < y < 0 ∇2φ = 0, y < 0

BKBC: ∂φ ∂y = 0, y = −h ∇φ → 0, y → −∞

FSKBC:

FSDBK:

∂φ ∂y = ∂η

∂t , y = 0 ∂φ ∂t + gη = 0, y = 0

⎫ ⎬

⎭ → ∂2φ

∂t2 + g ∂φ ∂y = 0

Introducing the notation {} for infinite depth we can rewrite the BVP:

Constant finite depth h {Infinite depth}∇2φ = 0, −h < y < 0 ∇2φ = 0, y < 0 (1)

∂φ ∂y

= 0, y = −h {∇φ → 0, y → −∞} (2)

∂2φ ∂φ ∂2φ ∂φ + g = 0, y = 0 + g = 0, y = 0 (3)

∂t2 ∂y ∂t2 ∂y

Given φ calculate:

1 ∂φ∣ 1 ∂φ ∣ η (x, t) = − ∣ η (x, t) = − ∣ (4)

g ∂t y=0 g ∂t y=0

∂φ ∂φ p − pa = −ρ − ρgy p − pa = −ρ − ρgy (5)

∂t ︸︷︷︸ ∂t ︸︷︷︸︸︷︷︸︸︷︷︸ hydrostatic hydrostatic

dynamic dynamic

4

{ }

[ ( )]

6.2.2 Solution Solution of 2D periodic plane progressive waves, applying separation of variables.

We seek solutions to Equation (1) of the form eiωt with respect to time. Using the KBC (2), after some algebra we find φ. Upon substitution in Equation (4) we can also obtain η.

gA cosh k (y + h) gA kyφ = sin (kx − ωt) φ = sin (kx − ωt) eω cosh kh ω ⎧ ⎫

⎨ ⎬ η = A cos (kx − ωt) η = A cos (kx − ωt)︸︷︷︸︸︷︷︸ ⎩ ⎭

using (4) using(4)

where A is the wave amplitude A = H/2.

Exercise Verify that the obtained values for φ and η satisfy Equations (1), (2), and (4).

6.2.3 Review on plane progressive waves

(a) At t = 0 (say), η = A cos kx → periodic in x with wavelength: λ = 2π/k Units of λ : [L]

λK = wavenumber = 2 /λ [L-1]kx

(b) At x = 0 (say), η = A cos ωt → periodic in t with period: T = 2π/ω Units of T : [T ]

T

ω = frequency = 2 /T [T-1], e.g. rad/sec t

(c) η = A cos k x − ωk t ω

[ L ]

Units of : k T

ω (ω) Following a point with velocity , i.e., xp = t + const, the phase of η does

k k ω λ

not change, i.e., = ≡ Vp ≡ phase velocity. k T

5

{ }

√ { √ }

6.2.4 Dispersion Relation

So far, any ω, k combination is allowed. However, recall that we still have not made use of the FSBC Equation (3). Upon substitution of φ in Equation (3) we find that the following relation between h, k, and ω must hold:

∂2φ ∂φ + g = 0 −→ − ω2 cosh kh + gk sinh kh = 0 ⇒ ω2 = gk tanh kh

∂t2 ∂y ↑ gAφ= ω sin(kx−ωt)f(z)

• This is the Dispersion Relation

ω2 = gk tanh kh ω2 = gk (6)

Given h, the Dispersion Relation (6) provides a unique relation between ω and k, i.e., ω = ω(k; h) or k = k(ω; h).

• Proof

1

kh

C ω2h C ≡ = (kh) tanh (kh)

g ︸︷︷︸ from (6)

tanh kh C

= tanh kh kh

kh → obtain unique solution for k kh =f(c)

• Comments

- General As ω ↑ then k ↑ , or equivalently as T ↑ then λ ↑ .

λ ω g g - Phase speed Vp ≡ = = tanh kh Vp =

T k k k

Therefore as T ↑ or as λ ↑ , then Vp ↑ , i.e., longer waves are ‘faster’ in terms of phase speed.

- Water depth effect For waves the same k (or λ), at different water depths, as h ↑ then Vp ↑ , i.e., for fixed k Vp is fastest in deep water.

- Frequency dispersion Observe that Vp = Vp(k) or Vp(ω). This means that waves of different frequencies, have different phase speeds, i.e., frequency dispersion.

6

︷︸︸︷

︸︷︷︸

{

6.2.5 Solutions to the Dispersion Relation : ω2 = gk tanh kh

Property of tanh kh:

long waves shallow water

sinh kh 1 − e−2kh ∼ kh for kh << 1. In practice h < λ/20 tanh kh = = =

cosh kh 1 + e−2kh 1 for kh >∼ 3. In practice h > λ 2

short waves deep water

Shallow water waves

or long waves

Intermediate depth

or wavelength

Deep water waves

or short waves

kh << 1

∼ h < λ/20

Need to solve ω2 = gk tanh kh

given ω, h for k

(given k, h for ω - easy!)

kh >> 1

∼ h > λ/2

ω2 ∼= gk · kh → ω = √

gh k

λ = √

gh T

(a) Use tables or graphs (e.g.JNN fig.6.3)

ω2 = gk tanh kh = gk∞

⇒ k∞

k =

λ λ

= Vp

Vp∞ = tanh kh

∞

ω2 = gk

λ = g 2π

T 2

( λ(in ft.) ≈ 5.12T 2 (in sec.)

)

(b) Use numerical approximation

(hand calculator, about 4 decimals )

i. Calculate C = ω2h/g

ii. If C > 2: ”deeper” ⇒

kh ≈ C(1 + 2e−2C − 12e−4C + . . .)

If C < 2: ”shallower” ⇒

kh ≈ √

C(1 + 0.169C + 0.031C2 + . . .)

No frequency dispersion

Vp = √

gh

Frequency dispersion

Vp =

√ g k

tanh kh

Frequency dispersion

Vp =

√ g 2π

λ

7

6.3 Characteristics of a Linear Plane Progressive Wave

2π k =

λ MWL 2π

ω = T

H = 2A

Define U ≡ ωA

Linear Solution:

η = A cos (kx − ωt) ; φ = Ag cosh k (y + h)

sin (kx − ωt) , where ω2 = gk tanh kh ω cosh kh

6.3.1 Velocity field

A Vp

h

x

y

λ η(x,t) = y

Velocity on free surface �v(x, y = 0, t)

u(x, 0, t) ≡ Uo = Aω 1

tanh kh cos (kx − ωt) v(x, 0, t) ≡ Vo = Aω sin (kx − ωt) =

∂η ∂t

Velocity field �v(x, y, t)

u = ∂φ ∂x

= Agk ω

cosh k (y + h) cosh kh

cos (kx − ωt)

= Aω ︸︷︷︸ U

cosh k (y + h) sinh kh

cos (kx − ωt) ⇒

u Uo

= cosh k (y + h)

cosh kh

⎧ ⎪⎨

⎪⎩

∼ eky deep water

∼ 1 shallow water

v = ∂φ ∂y

= Agk ω

sinh k (y + h) cosh kh

sin (kx − ωt)

= Aω ︸︷︷︸ U

sinh k (y + h) sinh kh

sin (kx − ωt) ⇒

v Vo

= sinh k (y + h)

sinh kh

⎧ ⎪⎨

⎪⎩

∼ eky deep water

∼ 1 + y h shallow water

• u is in phase with η • v is out of phase with η

8

Velocity field �v(x, y)

Shallow water Intermediate water Deep water

9

6.3.2 Pressure field

• Total pressure p = pd − ρgy.

• Dynamic pressure pd = −ρ∂φ .∂t

• Dynamic pressure on free surface pd(x, y = 0, t) ≡ pdo

Pressure field

Shallow water Intermediate water Deep water

pd = ρgη pd = ρgA cosh k (y + h)

cosh kh cos (kx − ωt)

= ρg cosh k (y + h)

cosh kh η

pd = ρgekyη

pd

pdo

same picture as u Uo

pd(−h) pdo

= 1 (no decay) pd(−h)

pdo

= 1

cosh kh pd (−h)

pdo

= e −ky

p = ρg(η − y) ︸︷︷︸ “hydrostatic” approximation

p = ρg ( ηeky − y

)

1<<kh

y

dp )( hp −

y

o dp

ghV p =

y x

)( hp −

∇

1>>kh

y

od

p )( hp −

y

o dp

π λ

2 g

V p =

y x

)( hp −

∇

Pressure field in deep water Pressure field in shallow water

10

︸︷︷︸

∫ ∫

∫ ∫

6.3.3 Particle Orbits (‘Lagrangian’ concept)

Let xp(t), yp(t) denote the position of particle P at time t.

Let (x; y) denote the mean position of particle P.

The position P can be rewritten as xp(t) = x + x′(t), yp(t) = y + y′(t), where (x′(t), y′(t)) denotes the departure of P from the mean position.

In the same manner let �v ≡ � x, ¯v(¯ y, t) denote the velocity at the mean position and �vp ≡ �v(xp, yp, t) denote the velocity at P.

),( yx

(x )',' y P (x , y ) �vp = �v(x + x ′ , y + y ′, t) =⇒

P P TSE

∂�v ′ ∂�v ′�vp = � x, x, ¯ + (x, y, t¯ ) yv (¯ y, t) + (¯ y, t) x + . . . ⇒ ∂x ∂y

ignore - linear theory

∼ v�vp = �

To estimate the position of P, we need to evaluate (x′(t), y′(t)):

x ′ = dt u (¯ y, t) = cosh k (y + h)

¯ − ωt) ⇒x, ¯ dt ωA cos (kxsinh kh

cosh k (y + h) = −A sin (kx − ωt)

sinh kh

y ′ = x, ¯ dt ωA sinh k (y + h)

dt v (¯ y, t) = sin (kx − ωt) ⇒ sinh kh

sinh k (y + h) = A cos (kx − ωt)

sinh kh

Check: On y = 0, y′ = A cos (kx − ωt) = η, i.e., the vertical motion of a free surface particle (in linear theory) coincides with the vertical free surface motion.

It can be shown that the particle motion satisfies

x′2 y′2 (xp − x)2 (yp − y)2

+ = 1 ⇔ + = 1 a2 b2 a2 b2

cosh k (y + h) sinh k (y + h)where a = A and b = A , i.e., the particle orbits form

sinh kh sinh kh closed ellipses with horizontal and vertical axes a and b.

11

crest

(a) deep water kh >> 1: a = b = Ae

circular orbits with radii Ae decreasing

exponentially with depth

ky

ky

V p

A

A

kyAe

trough

A

(b) shallow water kh << 1:

A y a = = const. ; b = A(1+ )

kh h

decreases linearly

Vp = gh

with depth A/kh

V p

V p

A

P

QQ S S

(c) Intermediate depth

R R R

λ

6.3.4 Summary of Plane Progressive Wave Characteristics

f(y) Deep water/ short waves

kh > π (say)

Shallow water/ long waves

kh << 1

cosh k(y+h) cosh kh = f1 (y) ∼

e.g.pd

eky 1

cosh k(y+h) sinh kh = f2 (y) ∼

e.g.u, a eky 1

kh

sinh k(y+h) sinh kh = f3 (y) ∼

e.g. v, b eky 1 + y

h

13

C (x) = cos (kx − ωt)

(in phase with η)

S (x) = sin (kx − ωt)

(out of phase with η)

η A = C (x)

u Aω = C (x) f2 (y) v

Aω = S (x) f3 (y)

pd ρgA = C (x) f1 (y)

y� A = C (x) f3 (y) x�

A = −S (x) f2 (y)

a A = f2 (y) b

A = f3 (y)

b

a

14

︷︸︸︷

√



6.4 Superposition of Linear Plane Progressive Waves

1. Oblique Plane Waves

v k

kz

z kx

Vp k v

= (k , k z )x

θ x

(Looking up the y-axis from below the surface)

Consider wave propagation at an angle θ to the x-axis

�k �·x

η =A cos(kx cos θ + kz sin θ −ωt) = A cos (kxx + kzz − ωt)

gA cosh k (y + h)φ = sin (kx cos θ + kz sin θ − ωt)

ω cosh kh ω2 =gk tanh kh; kx = k cos θ, kz = k sin θ, k = kx

2 + kz 2

1

y

∣ ∣ ∣ ∣

2. Standing Waves

+ Same A, k, ω, no phase shift

η =A cos (kx − ωt) + A cos (−kx − ωt) = 2A cos kx cos ωt

2gA cosh k (y + h)φ = − cos kx sin ωt

ω cosh kh

90o at all times

t = 0, T, 2T, …

L,3

, TT

t = 53 TTT

node

y

x

am

pli

tud

e 2

A

antinode t = , , L2 24 4 4

∂η ∂φ nπ nλ ∼ = · · · sin kx = 0 at x = 0, = ∂x ∂x k 2

∂φ∂x

Therefore, = 0. To obtain a standing wave, it is necessary to have perfectx=0

reflection at the wall at x = 0. AR

Define the reflection coefficient as R ≡ (≤ 1).AI

y

AI = AR

x ARR = = 1

AI

2

θ

3. Oblique Standing Waves

ηI =A cos (kx cos θ + kz sin θ − ωt)

ηR =A cos (kx cos (π − θ) + kz sin (π − θ) − ωt)

z

θ θ

Iη

Rη Rθθθθ

Iθθθθx θR = π − θI

Note: same A, R = 1.

kxx kz z−ωt ︷︸︸︷︷︸︸︷ ηT = ηI + ηR = 2A cos (kx cos θ) cos (kz sin θ − ωt) ︸︷︷︸︸︷︷︸

standing wave in x propagating wave in z

and

2π 2π ω λx = ;

k cos θ VPx = 0; λz = ;

k sin θ VPz =

k sin θ

Check:

∂φ ∂η ∂x

∼ ∂x

∼ · · · sin (kx cos θ) = 0 on x = 0

3

{ }

{ }

{ ( )}

4. Partial Reflection

+

Iη Rη

ηI =AI cos (kx − ωt) = AI Re e i(kx−ωt)

ηR =AR cos (kx + ωt + δ) = AIRe R e −i(kx+ωt)

R: Complex reflection coefficient

ARR = |R| e −iδ , |R| =

AI

ηT =ηI + ηR = AIRe e i(kx−ωt) 1 + Re −2ikx

|ηT |2 =A2 [ 1 + |R|2 + 2 |R| cos (2kx + δ)

] I

free surface

2 λ

∇

|ηT

| wave envelope

AI

1+ | R |2

x

node antinode

At node,

|ηT | = |ηT | = AI (1 − |R|) at cos (2kx + δ) = −1 or 2kx + δ = (2n + 1) πmin

At antinode,

|ηT | = |ηT | = AI (1 + |R|) at cos (2kx + δ) = 1 or 2kx + δ = 2nπ max

λ 2kL = 2π so L =

2

| | max minR = |ηT | − |ηT |

= |R (k)||ηT | + |ηT |max min

4

( )

( )

{ [ ]}

5. Wave Group

2 waves, same amplitude A and direction, but ω and k very close to each other.

VP1

η1 =� Aei(k1x−ω1t)

η2 =� Aei(k2x−ω2t)

VP2

ω1,2 =ω1,2 (k1,2) and VP1 ≈ VP2

ηT = η1 + η2 = � Aei(k1x−ω1t) 1 + e i(δkx−δωt) with δk = k2 − k1 and δω = ω2 − ω1

A2 kg δ π =λ 2

21 PP VV ≈

Vg

ω π = 2T δω

π = 2 gT

2π = λ1 ≈ λ2k1

⎫ | | = 2 | | A when δkx − δωt = 2nπ ⎬ηT max δω xg = Vgt, δkVgt−(δω) t = 0 then Vg = ⎭ δk|ηT | = 0 when δkx − δωt = (2n + 1) πmin

5

∣ ∣ ∣ ∣

( (

︸︷︷︸︸︷︷

)

︸

In the limit,

dω δk, δω → 0, Vg =

dk,

k1≈k2≈k

and since

ω2 = gk tanh kh ⇒

ω) 1 2khVg = 1 +

k 2 sinh 2kh Vp n

(a) deep water kh >> 1Vg 1 n = = Vp 2

(b) shallow water kh << 1

n = V

V g

p = 1 (no dispersion)

(c) intermediate depth

1 < n < 12

⎫ ⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎬

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎭

PV

Vg

Disappear Appear

Vg ≤ Vp

6

PE

KE

[ ]

6.5 Wave Energy - Energy Associated with Wave Motion.

For a single plane progressive wave:

Energy per unit surface area of wave

• Potential energy PE • Kinetic energy KE

PE without wave =

0∫

−h

ρgydy = −1 2 ρgh2

PE with wave

η∫

−h

ρgydy = 1 2 ρg (η2 − h2)

PEwave = 1 2 ρgη2 = 1

2 ρgA2 cos2 (kx − ωt)

KEwave =

η∫

−h

dy 1 2 ρ (u2 + v2)

Deep water = · · · = 1 4 ρgA2

︸︷︷︸ KE const in x,t

to leading order

Finite depth = · · ·

Average energy over one period or one wavelength

PEwave = 1 4 ρgA2 KEwave = 1

4 ρgA2 at any h

• Total wave energy in deep water:

E = PE + KE = 12 ρgA2 cos2 (kx − ωt) + 1

2

• Average wave energy E (over 1 period or 1 wavelength) for any water depth:

E = 12 ρgA2[ 1

2 + 1

2 ] = 1

2 ρgA2 = Es, ↑ ↑

PE KE

Es ≡ Specific Energy: total average wave energy per unit surface area.

• Linear waves: PE = KE = 1Es Vp2 (equipartition).

• Nonlinear waves: KE > PE. Vp

x

x

sE

E

½ =EKE 2 1=

EPE 2 1=

)(cos txEPE 2 s ω−= k1

Recall: cos2 x = 1 + 1 cos 2x2 2

7

∫ ( )

︸︷︷︸

︸︷︷︸

6.6 Energy Propagation - Group Velocity

SVp

x

Varea persEE =

Consider a fixed control volume V to the right of ‘screen’ S. Conservation of energy:

dW dE = = J

dt dt ︸︷︷︸︸︷︷︸︸︷︷︸ rate of work done on S rate of change of energy in V energy flux left to right

where

η dφ ∂φ

J- = pu dy with p = −ρ + gy and u = dt ∂x

−h ( ) (ω ) [ ( )]

1 2khJ- = 12 ρgA2

2 1 + sinh 2kh = E (nVp) = EVg ︸︷︷︸ k ︸︷︷︸

E nVp

Vg

e.g. A = 3m, T = 10 sec → J- = 400KW /m

8

1F FF2F

( )

∣ ∣ ∣ ∣

( )

( )

6.7 Equation of Energy Conservation

Δx

21 x

1 2

= ( )E = E( )x , x

h = h(x)

J-1 − J-2 Δt = ΔEΔx

J-2 = J-1 + ∂J∂x

Δx +· · · 1

∂E ∂J+ = 0, but J- = VgE

∂t ∂x ∂E ∂

+ VgE = 0 ∂t ∂x

∂E1. = 0, VgE = constant in x for any h(x).

∂t

2. Vg = constant (i.e., constant depth, δk << k)

∂ ∂ + Vg E = 0, so E = E (x − Vgt) or A = A (x − Vgt)

∂t ∂x

i.e., wave packet moves at Vg.

9

F

+L

︷︸︸︷ /

∣ ( )∣ ∣ ( ) ( )

√

6.8 Steady Ship Waves, Wave Resistance

2A

2 2 1 gAE ρ=

Vp = U

( )( )2 2 1

2 1 gAUEVg ρ==

L

0=E ahead of ship

U

D

x = 0

C.V.

• Ship wave resistance drag Dw

Rate of work done = rate of energy increase d ( )

DwU + J- = EL = EU dt

deep water

Dw = 1

(EU − EU 2 ) = 12 E = 1

4 ρgA2 ⇒ Dw ∝ A2

force / length U energy / area

• Amplitude of generated waves

The amplitude A depends on U and the ship geometry. Let � ≡ effective length.

-+-l

L

To approximate the wave amplitude A superimpose a bow wave (ηb) and a stern wave (ηs).

ηb = a cos (kx) and ηs = −a cos (k (x + �))

ηT = ηb + ηs

A = |ηT | = 2a ∣sin 1 k� ← envelope amplitude max 2

Dw = 14 ρgA2 = ρga2 sin2 1

2 k� ⇒ Dw = ρga2 sin2 12 U

g� 2

• Wavelength of generated waves To obtain the wave length, observe that the phase speed of the waves must equal U . For deep water, we therefore have

ω deep g U2

Vp = U ⇒ = U −→ = U , or λ = 2π water k k g

10

√

( ) ( )

• Summary Steady ship waves in deep water.

U = ship speed

g g U2

Vp = k

= U ; so k = U2

and λ = 2πg

L =ship length, � ∼ L ( ) 1 1∼ ∼Dw =ρga2 sin2

21

Ug�

2 = ρga2 sin2

2F 2 = ρga2 sin2

2F 2 rL rL

1 Fl = ≈ 0.56 ⇒

π max at:

0.56hull

2 gaρ gU ≈ 56.0 ≅l D

w

F = l

gL ⇒ U ∝ Lhull

1

U , where l ≤ L

0 gl 1

π Increasing U

Small speed U

• Short waves • Significant wave cancellation • D ~ small w

11

� � �

( )



6.9 Wave Forces on a Body

UP

U = ωA U� ωA�

Re = = ν ν

UT AωT A Kc = = = 2π

F A � h CF = = f , , Re, , roughness, . . .

ρgA�2 λ λ λ︸︷︷︸︸︷︷︸ Wave Diffraction

steepness parameter

1

6.9.1 Types of Forces

1. Viscous forces Form drag, viscous drag = f(Re, Kc, roughness, . . .).

• Form drag (CD)

Associated primarily with flow separation - normal stresses.

��

• Friction drag (CF )

Associated with skin friction τw, i.e., F ∼ ∫∫

τwdS. body

(wetted surface)

b.l.ττττωωωωU

2

∫∫

︸︷︷︸

)

}

︸︸

∫∫ ∫∫∫

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ︸︷︷︸

2. Inertial forces Froude-Krylov forces, diffraction forces, radiation forces.

Forces arising from potential flow wave theory,

F = ndS, where p = −ρ (∂φ 1 |∇φ|2 pˆ + gy +

∂t 2body

(wetted surface) =0, for linear theory, small amplitude waves

For linear theory, the velocity potential φ and the pressure p can be decomposed to

φ = φI + φD + φR ︸︷︷︸︸︷︷︸︸︷︷︸ Incident wave Diffracted wave Radiated wave potential (a) potential (b.1) potential (b.2)

p ∂φI ∂φD ∂φR− = + + + gyρ ∂t ∂t ∂t

(a) Incident wave potential

• Froude-Krylov Force approximation When � << λ, the incident wave field is not significantly modified by the presence of the body, therefore ignore φD and φR.

Froude-Krylov approximation:

φ ≈ φI ∫∫ (

∂φI )

(∂φI ) = −ρ + gy ndS ← can calculate knowing (incident) ⇒ FFK ˆ wave kinematics (and body geometry) p ≈ −ρ + gy ∂t

∂t body ︷︷ surface ≡ pI

• Mathematical approximation After applying the divergence theorem, the FFK

can be rewritten as FFK = − pI ˆ ∇pId∀.ndS = − body body

surface volume

If the body dimensions are very small comparable to the wave length, we can assume that ∇pI is approximately constant through the body volume ∀ and ‘pull’ the ∇pI out of the integral. Thus, the FFK can be approximated as

( ) ∫∫∫ ( ) ∼FFK = −∇pI ∣ d∀ = ∀ −∇pI ∣

at body at body body body

center volume center volume

The last relation is particularly useful for small bodies of non-trivial geometry for 13.021, that is all bodies that do not have a rectangular cross section.

3

�

� � �

�

� ∫∫ ( )

��

� �

�

∫∫ ( )

(b) Diffraction and Radiation Forces

(b.1) Diffraction or scattering force When � ≮< λ, the wave field near the body will be affected even if the body is stationary, so that no-flux B.C. is satisfied.

�� φ��

φ��

φ��

� φ�� ∂φ∂

φ��

φ��

= � = (φ � + φ� )∂� ∂� ∂φ ∂φ

� �� = − � ← �� ∂� ∂�

FD = ∂φD

ˆ−ρ ndS ∂t

body surface

(b.2) Radiation Force - added mass and damping coefficient Even in the absence of an incident wave, a body in motion creates waves and hence inertial wave forces.

� φ��

φ��

� � ∂φ � = � ⋅ ��

� ∂�

∂φR ndS Uj −FR = −ρ ˆ = − mij dij Uj

∂t ︸︷︷︸︸︷︷︸ body added wave

surface mass radiation damping

4

�

� ( ) ( )

6.9.2 Important parameters

(1)Kc = UT = 2π A�

⎫ ⎪⎪⎪⎬

⎪⎪⎪⎭

Interrelated through maximum wave steepness

A ≤ 0.07λ

A � ≤ 0.07(2)diffraction parameterλ � λ

• If Kc ≤ 1: no appreciable flow separation, viscous effect confined to boundary layer (hence small), solve problem via potential theory. In addition, depending on the value of the ratio

λ� ,

– If λ� << 1, ignore diffraction , wave effects in radiation problem (i.e., dij ≈

0,mij ≈ mij infinite fluid added mass). F-K approximation might be used, calculate FFK .

– If � >> 1/5, must consider wave diffraction, radiation (A ≤ 0.07 ≤ 0.035).λ � �/λ

• If Kc >> 1: separation important, viscous forces can not be neglected. Further on if

� 0.07 � ≤ so << 1 ignore diffraction, i.e., the Froude-Krylov approximation is valid. λ A/� λ

1 F = ρ�2 U(t) |U(t)|CD(Re)

2 ︸︷︷︸ relative velocity

• Intermediate Kc - both viscous and inertial effects important, use Morrison’s formula.

F =1 ρ�2U(t)|U(t)|CD(Re) + ρ�3 ˙ (Re, KcUCm )

2

5

• Summary

Limiting case: wave breaking occurs

I

II III

I. Use: CD and F − K approximation.

II. Use: CF and F − K approximation.

III. CD is not important and F − K approximation is not valid.

6

marine hydrodynamics mit notes

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