Marine Hydrodynamics, Fall 2004 Lecture 1

13.021 Marine Hydro dynamics Lecture 1

In tro duction Marine Hydrodynamics is the branch of Fluid Mechanics that studies the motion of incom6pressible uids (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 uids

Viscosity and boundary layers

Eect of waves on resistance and ship motion

Wh y study Marine Hydro dynamics? Studying Marine Hydrodynamics provides a greater understanding of a wide range of phe6nomena of considerable complexity involving uids. Another benet is that it allows pre6dictions to be made in many areas of practical importance.

1

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 havent can get an intuitive feeling about some physical properties of a solid. Thus a comparison of solids and uids will give some guidelines as to which properties can be translated to uids and on what terms.

Similarities

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

- Conservation of mass - Conservation of momentum (Newtons law of motion) - Conservation of energy (First law of thermodynamics)

2. Continuum hypothesis is used for both uids 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 105m VM 1015m3 . This corresponds to 3 1010 molecules of air in STP or 1013 molecules of water.

2

Dierences

1. Shape

- Solids have denite 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 ...)

Dierent constitutive laws are used for solids and dierent for uids.

- For solids Hookes 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 uids 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 uids at rest is referred to as Hydrostatics .

(Archimedes, c 200 BC)

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

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

3

Liquids vs. Gasses

Liquids and gasses are two categories of uids.

A uid 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 uids. [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 diuse readily, and the spontaneous tendency to become distributed uniformly throughout any container.[2]

In brief, a liquid is generally incompressible and does not ll a volume by expanding into it while on the other hand, a gas is compressible and expands to ll 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 dierence between the study of Hydrodynamics and the study of Aerodynamics is the property of incompressibility. In general hydrodynamic ows are treated as incompressible while aerodynamic ows are treated as compressible.

[1] Webster Dictionary

[2] American Heritage Dictionary

4

Why is a liquid ow incompressible?

It can be shown that the ratio of the characteristic uid velocities U in a ow to the speed of sound C in the medium gives a measure of compressibility of the medium for that particular ow. 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 uid velocities in water are signicantly smaller. Thus in the case of water, the Mach number is very small, indicating that water is virtually incompressible.

U : Characteristic uid ow 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 water 1kg/m3 = 103, it is harder to air 103kg/m3

move in water and therefore, typically, it is:

Uwater

Marine Hydrodynamics, Fall 2004Lecture 2

13.021 Marine Hydro dynamics Lecture 2

Chapte r 1 - Basic Equations 1.1 Description of a Flo w To dene a ow we use either the Lagrangian description or the Eulerian description.

Lagrangian description: Picture a uid o w where each uid 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 o w by recording the detailed histories of each uid 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 repre 0) nontrivial solutions, 1, 2, . . . , J . In general, J < N , in fact,

J = N K where K is the rank or dimension of the system of equations (1).

2

Mo del Law Instead of relating the N x i s by I (x1, x2, . . . xN ) = 0, relate the J s by

F ( 1, 2, . . . J ) = 0 , where J = N K < N For similitude, we require

( model) j = ( prototype) j where j = 1 , 2, . . . , J. If 2 problems have all the same j s, they have similitude (in the j senses), so s serve as similarit y parameters.

Note:

If is dimensionless, so is const, const , 1/ , etc. . .

If 1, 2 are dimensionless, so is 1 2 ,

12 , 1

const1 2

const2 , etc. . .

In general, we want the set (not unique) of independent j s, for e.g., 1, 2, 3 or 1, 1 2, 3, but not 1, 2, 1 2.

Example: Force on a smo oth circular cylinder in steady , incompressible o w Application of Buckinghams Theory.

F

U , D

Figure 1: Force on a smooth circular cylinder in steady incompressible uid (no gravit y)

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

3

xi : F,U,D, , N = 5 xi = ciM

mi Lli T ti P = 3

N = 5

F U D P = 3 mi 1 0 0 1 0

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

= F1 U2 D3 4 5

For to be non-dimensional, the set of equations

imi =0

ili =0

iti =0

has to be satised. The system of equations above after we substitute the values for the mis, lis and tis assume the form:

1

1 0 0 1 0 2 3

0

1 1 1 3 2 0 = 2 1 0 0 1 4 0

5

The rank of this system is K = 3, so we have j = 2 nontrivial solutions. Two families of solutions for i for each xed pair of (4, 5), exists a unique solution for (1, 2, 3). We consider the pairs (4 = 1, 5 = 0) and (4 = 0, 5 = 1), all other cases are linear combinations of these two.

4

Pair 4 = 1 and 5 = 0.

1 0 0 1 1 0 1 0 2 = 4

0 0 1 3 2which has solution

1 1

2 = 2 3 2

U 2D 2 1 = F 1 U 2 D 3 4 5 = F

Conventionally, 1 2 1 1 and 1 = F Cd, which is the Drag coecien t. 1 U 2D 2 2

2. Pair 4 = 0 and 5 = 1.

1 0 0 1 0 0 1 0 2 = 2

0 0 1 3 1which has solution

1 0

2 = 1 3 1

2 = F 1 U 2 D 3 4 5 = UD Conventionally, 2 2 1 , 2 = UD Re, which is the Reynolds number.

Therefore, we can write the following equivalent expressions for the non-dimensional inde7pendent parameters that describe this problem:

F ( 1, 2) = 0 or 1 = f ( 2) F (Cd, R e) = 0 or Cd = f (R e)

F UD F UD F ( , ) = 0 or = f ( )1/2 U 2D 2 1/2 U 2D 2

5

Appendix A

Dimensions of some uid properties

Quantities Dimensions

(MLT )

Angle none (M0L0T 0)

Length L L

Area A L2

Volume L3 Time t T

Velocity V LT1

Acceleration V LT2

Angular velocity T1

Density ML3

Momentum L MLT1 Volume ow rate Q L3T1 Mass ow rate Q MT1 Pressure p ML1T2

Stress ML1T2

Surface tension MT2

Force F MLT2

Moment M ML2T2

Energy E ML2T2

Power P ML2T3

Dynamic viscosity ML1T1

Kinematic viscosity L2T1

6

13.021 Marine Hydrodynamics, Fall 2004

Lecture 6

13.021 - Marine Hydro dynamics Lecture 6

2.2 Similarit y Parameters from Governing Equations and Boundary Conditions

In this paragraph we will see how we can specify the SPs for a problem that is governed by the Navier-Stokes equations. The SPs 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 ow 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) = U , where is such that 0 O(1). Then U can be chosen to be the characteristic velocit y of the o w and any velocity v can be written as:

v = Uv

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 quantit y normalized quantit y in terms of characteristic scale

Velocity U v v = Uv

Length L x x = Lx

tTime T t = T t

Pressure po pv p p = ( po pv)p

1

Non-dimensionalizing and normalizing the governing equations and boundary conditions

Substitute the dimensional quantities with their non-dimensional expressions (eg. substitute v with Uv , x with Lx, 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 SPs.

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

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 + 2v gj

t

U v U2 po pv U + (v ) v = p + ()2v gj

T t L U2 L2

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

L p gL v o pv + (v )v = (p) + 2v jUT t U2 UL U2

The coecients ( ) are SPs.

2

Since all the dimensionless and normalized terms () are of O(1), the SPs

( ) 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 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 U2 inertia force 2

UL inertia force Re = Reynolds number 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 + p = pa + + =

R1 R2 p=(popv )p

p

1 1 2 /

p = pa + + = pa + (po pv) L R1 R2 U2L

U2L We = Weber number inertial forces / surface tension forces

Some SPs used in hydrodynamics (the table is not exhaustive):

SP Denition

Reynolds number Re UL

inertia viscous Froude number Fr

U2

gL inertia gravity

Euler number Eu po

1 2 U2

pressure inertia Cavitation number popv 1

2 U2

pressure inertia Strouhal number S L

UT Eulerian inertia convective inertia

Weber number We U2L /

inertia surface tension

4

2.3 Similarity Parameters from Physical Arguments

Alternatively, we can obtain the same SPs by taking the dimensionless ratios of signicant ow quantities. Physical arguments are used to identify the signicant ow quantities. Here we obtain SPs from force ratios. We rst identify the types of dominant forces acting on the uid particles. The SPs are merely the ratios of those forces.

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

1.1 Inertial forces mass acceleration (L3) UL

2

= U2L2

u 1.2 Viscous forces area U

L (L2) = UL

y

shear stress

1.3 Gravitational forces mass gravity (L3)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 U2L2 UL = Reviscous UL

1/2 1/2inertia U2L2 U = Frgravity gL3 gL

1 inertia 1 (po pv)L2 po pv 2 = pressure 1

2 U2L2 1

2 U2

5

2.4 Importance of SPs

The SPs indicate whether dierent systems have similar ow properties. The SPs provide guidance in approximating complex physical problems.

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

,F g U

L

SPs for this problem:

L po pv U2L U UL S = , = 1 , We = , Fr = , Re = UT

2 U2 / gL

We dene the dimensionless force coecient:

F CF 1 U2L2

2

The force coecient must depend on the other SPs:

CF = CF (S, ,We, Fr, Re) or CF = CF S,

1,We 1, Fr, Re

1

Procedure We will rst study under what conditions each SP 0. We will estimate CF for the case that all of the SPs 0.

6

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

Change S keeping all other SPs (1 , We 1 , Fr, Re

1) xed.

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, unsteady eect is dominant.

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

L L S 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 ow:

CF = CF S 0, 1,We 1, Fr, Re 1

1CF = CF 1,We 1, Fr, Re

7

po pv2. Signicance of the cavitation number = 1 . U2

2

Change 1 keeping all other SPs (S 0, We 1 , Fr, Re 1) xed.

Some comments on cavitation

pv : Vapor pressure, the pressure at which water boils po pv : State of uid 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

Strongcavitation No cavitation

Strongcavitation No cavitation

inception

For > 1 1

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

The characteristic pressure is po = 105N/m2 and the vapor pressure is pv = 10

3N/m2 . Cavitation will not occur when:

1 U2 po pv1

U2L 3. Signicance of the Weber number We = .

/

Change We 1 keeping the other SPs (S 0, 1 0, Fr, Re 1) xed.

For We > 1 We 1 > 7 105 e 2U2L U Therefore for L >> 7 105 m it is We >> 1We

1 1 and surface tension eects can be neglected.

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

CF = CF 0, 0,We 1 0, Fr, Re 1

CF = CF Fr, Re 1

10

U 4. Signicance of the Froude number Fr = , which measures the gravity eects.

gh

Change Fr keeping the other SPs (S 0, 1 0, We 1 0, Re 1) xed.

Gravity eects, hydrostatic pressure do not create any ow (isotropic) nor do they change the ow dynamics unless Dynamic Boundary Conditions apply.

Gravity eects are not signicant when 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 gures (i - iv) illustrate cases where gravity eects are not signicant.

11

In any of those cases the gravity eects are insignicant 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 eects ow:

CF = CF 0, 0, 0, Fr 0 or Fr 1 0, Re 1

1CF = CF Re

A look ahead: Froudes Hypothesis

Froudes 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 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. Signicance of the Reynolds number Re = .

Change Re keeping the other SPs (S 0, 1 0, We 1 0, Fr 0 or Fr 1 0) xed.

Recall that for a steady, non-cavitating, non-surface tension, with no gravity eects ow:

CF = CF Re 1

CFCF

ReRe

Sphere

Plate

(Re)cr

Sphere

Plate

(Re)crLaminar Turbulent Transition

Re (Re)cr, Turbulent ow Re , Ideal uid

For example, a hydrofoil of cord length L = 1m travelling in water (kinematic viscosity = 101m2/s) with velocity U = 10m/s has a Reynolds number with respect to L:

Re = = 10

7 ideal uid, and Re 1 0 UL

13

Therefore for a steady, non-cavitating, non-surface tension, with no-gravity eects ow in an ideal uid:

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

DAlemberts Paradox No drag force on moving body in ideal uid.

14

Marine Hydrodynamics, Fall 2004Lecture 7

13.021 - Marine Hydro dynamics Lecture 7

Chapter 3 Ideal Fluid Flo w

The structure of Lecture 7 has as follows: In paragraph 3.0 we introduce the concept of inviscid uid and formulate the governing equations and boundary conditions for an ideal uid o w. In paragraph 3.1 we introduce the concept of circulation and state Kelvins theorem (a conservation law for angular momentum). In paragraph 3.2 we introduce the concept of vorticit y.

In viscid Fluid = 0

+Ideal Fluid Flo w D Incompressible Flo w ( 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 = =

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 water = 10

6m2/s leading thus to typical Reynolds numbers with respect to U and L in the order of

UL Re = 106 >>> 1

1 viscous forces 0 Re inertia forces

This means that viscous eects are

Governing Equations for Ideal Fluid Flow

Continuity Equation:

v = 0

Momentum (Navier-Stokes Euler) equations:

v 1 + v v = p gj

t

By neglecting the viscous stress term (2v) the Navier-Stokes equations reduce

to the Euler equations. (Careful not to confuse this with the Euler equation in

1.6).

The N-S equations are second order PDEs with respect to space

(2nd order in 2), thus: (a) require 2 kinematic boundary conditions, and (b)

produce smooth solutions in the velocity eld.

The Euler equations are rst order PDEs, thus: (a) require 1 kinematic bound

ary condition, and (b) may allow discontinuities in the velocity eld.

Boundary Conditions for Euler equations (Ideal Flow):

KBC:

v n = u n no ux + free (to) slip = Un given

Note: No slip condition v t = U t does not apply.

The no slip condition is required to ensure that the velocity gradients are nite and therefore the viscous stresses ij are nite.

But since = 0 the viscous stresses are identically zero (ij = = 0) and the velocity gradients can be innite. Or else the velocity eld need not be continuous.

3

u Viscousflow

w y < <

uU Inviscidflow 0 w y

DBC:

)dy(u

)0(u

y

p = . . . Pressure given on the boundary

Similarly to the argument for the KBC, viscous stresses ij cannot be specied on any boundary since = 0.

Summary of consequences neglecting viscous eects this far: Neglecting viscous eects is equivalent to setting the kinematic viscosity equal

to zero:

= 0

Setting = 0 inviscid uid

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

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

Setting = 0 all the viscous stresses ij = are identically 0. This may allow for innite velocity gradients.

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

4

3.1 Circulation Kelvins Theorem

3.1.1 Instantaneous circulation around any arbitrary closed contour C.

C xdv

v v

= v dx C tangential

velocity

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

5

3.1.2 Kelvins Theorem (KT) :

For ideal uid under conservative body forces,

d = 0 following any material contour C,

dt

i.e., 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 uid)

Kelvins Theorem is a statement of conservation of angular momentum under IFCF.

6

Example 1: Angular momentum of point mass.

v1 v2

m1 m2r2

r1

Angular momentum of point mass:

L = |r (mv)| = mvr = mr2

Conservation of angular momentum:

L = L

1 2

m1=m2 m1v1r1 = m2v2r2 = v1r1 = v2r2 or

r12 1 = r2

2 2

Conservation of angular momentum does not imply constant angular velocity:

Angular Momentum angular velocity

7

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

2r

2 1 1r

m V m

V

2 2

1 = dr1v1 = dr2v2 = 2

0 0

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

1 = v1 dx = v2 dx = 2 C1 C2

8

( ) ( ) ( )

3.2 Vorticity

3.2.1 Denition of Vorticity

v = w v i w u j + v u k = y z x z x y

Relationship of vorticity to circulation - Apply Stokes Theorem:

= d x = ( ndS = ndS Flux of vorticity out of S v v)

C any S covering C any S covering C

3.2.2 What is Vorticity?

For example, special case: 2D ow - w = 0; = 0; y = x = 0 and z

v u z =

x y

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

time t + t

jiu +

jiu +

jiu +

jiu + time t

v u = 0, = 0 z = 0 no vorticity

x y

9

(b) Pure Strain (no volume change):

Areat + t

Areat

No volume change Areat = Areat + t

u v u v = ; u = -v; = 0; = 0 z = 0

x y y x

10

(c) Angular deformation

r

=

dytu x y

=u +

r

r

dy

dx

t + ty i y jdy dy time

y = ( dx)tx time t

u i + dx dx j= 0 = x x

u v = x = y( for dx = = 0 only if dy)

y x

11

(d) Pure rotation with angular velocity

0=r

jdx=r

idy=r

dy

dx

ttime

tt +time

t

t

v u = ; = ; z = 2

x y

i.e. vorticity 2(angular velocity).

3.2.3 Irrotational Flow

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

0 everywhere 0 for any C Further on, if at t = to, the ow is irrotational, i.e., 0 for all C, then Kelvins theorem states that under IFCF, 0 for all C for all time t:

once irrotational, always irrotational

(Special case of Kelvins theorem)

12

Marine Hydrodynamics, Fall 2004Lecture 8

13.021 - Marine Hydro dynamics Lecture 8

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 o w.

r u

3.3 Prop erties of Vortex Structures 3.3.1 Vortex Structures

r

r

A vortex line is a line everywhere tangent to .

vortex line 2

2

1

1 1

r u2

A vortex tub e (lamen t) is a bundle of vortex lines. vortex tube

r 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 orices are

shown 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

v u

v

cross section v u

v

vv

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

= 0

= n dS = 0

V Divergence S vorticit y ux

Theorem

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

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

(Vorticity Flux)in = (Vorticity Flux)out ( n)in Ain = ( n)out Aout

0 = n v

(v n )out

(v n )in

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

r r

3

3.3.3 Conservation of Vorticity Flux

0 = 3 = v dx = ndS = 0 C3 S3

C1 C2 C1 C2

C3C1n 32n

1 = v dx = n1dS = n2dS = 2

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:

= 1A1 = 2A2

1 2

A1 A2

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

1

A1

2

A2 = 2A1 2 = 1/2

4

3.3.4 Vortex Structures are Material Structures

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

Am

Am

By denition,

n = 0 on An Then,

Am = x = v d nds = 0 Am Am

At time t + t, Am moves, and for an ideal uid under the inuence of conservative body forces, Kelvins theorem states that

Am = 0

So, n = 0 on Am still, i.e., Am still on the vortex tube. Therefore, the vortex tube is a material tube for an ideal uid under the inuence 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 uid.

5

3.3.5 Vortex stretching

Consider a small vortex lament of length L and radius R, where by denition is tangent to the tube.

AR

= A = constant (in time)

Stokes Kelvins Theorem Theorem

But tube is material with volume = AL = R2L = constant in time (continuity)

A = = = 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, increases. In other words,

Vortex stretching L (conservation of angular momentum) A and R (continuity)

6

Summary on Vortex Structures

r

R

r A

r

Vortex ring length L = 2 R [L]

Cross sectional area A = r 2 [L2]

Vortex ring volume = AL =

continuit y const [L3]

Vorticit y = v [T 1]

Circulation =

Kelvins theorem const

=

vorticit y ux through A A = const

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

Kelvins theorem + Continuity relate length ratios to , , U

r /L L

Ur = const U U as L U r

A/L L A = const as L

A

Example 1: Example 2:

r

A 22

1r

1A

r

1

1

1A 2

2A

1L

2L

r

2

A1 < A2 1 > 2

L1 < L2 A1 > A2 1 = 2 1 < 2 U1 < U2

Given From continuity only From Kelvins theorem From Kelvins theorem + continuity From Kelvins theorem + continuity

8

( )

( )

( )

( ) ( )

3.4 Bernoulli Equation for Steady ( = 0), Ideal( = 0),t Rotational ow

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

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

Steady, inviscid Euler equation (momentum equation):

v v = p + gy (1)

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 + gy2

v 0

v momentum (1) energy

Therefore, ( ) ( ) v2 p D v2 pv 2

+ + gy = 0 =

Dt 2 +

+ gy

streamline pathline i.e., v

2

2

+ p + gy = constant on a streamline

In general, v2

2

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

Assumptions: Ideal uid, Steady ow, Rotational in general.

9

3.4.1 Example: Contraction in Water or Wind Tunnel

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

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

( )2 1. From continuity: U1

( R1

2 ) = U2

( R2

2 ) U

2 =

R1 = 2 >> 1

U1 R2

2.

u Since = 0 vortex ring. = 0 ,

r

10

( ) ( )

[ [

1 2 2 R2 1 = =

)

( )

( )

13.021 Marine Hydrodynamics, Fall 2004 Lecture 9

13.021 - Marine Hydro dynamics Lecture 9

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 vorticit y equation and study some additional properties of vorticit y. In paragraph 3.6 we introduce the concept of potential o w and velocity potential. We formulate the governing equations and boundary conditions for potential o w and nally introduce the stream function.

3.5 Vorticit y Equation Return to viscous incompressible o w. The Navier-Stokes equations in vector form

v + v v =

p + gy + 2v

t By taking the curl of the Navier-Stokes equations we obtain the vorticit y equation. In detail and taking into account u we have

(Navier-Stokes) v + ( v v) = p + gy + ( 2v ) t

The rst term on the left side, for xed reference frames, becomes v

= ( v) = t t t

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

2v = 2

Applying the identit y scalar = 0 the pressure term vanishes, provided that the density is uniform ( )

( p + gy) = 0

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 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 ) = ( v)2

= (v ) ( ) v + ( v) + v ( ) =0 =0 since

incompressible (v)=0 uid

Putting everything together, we obtain the vorticity equation

D = ( ) v + 2

Dt

Comments-results obtained from the vorticity equation

Kelvins Theorem revisited - from vorticity equation:

If 0, then D = ( ) v, so if 0 everywhere at one time, 0 always.Dt

can be thought of as diusivity of vorticity (and momentum), i.e., once generated (on boundaries only) will spread/diuse in space if is present.

vv

vv

DDv v vv== 2v2v vv ++ ... D ... D

vv

== 22 vv ++ ......Dt DtDt Dt

2

T Diusion of vorticity is analogous to the heat equation: = K2T , where K is the t

heat diusivity.

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

In 2D space (x, y),

v = (u, v, 0) and 0

z

So, = v is to v ( is parallel to the z-axis). Then,

( ) v = x + y + z v 0, x y z 0 0

0

so in 2D we have

D = 2

Dt

If = 0, D = 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,

Di vi 2i

= j + Dt xj xj xj

vortex turning and stretching diusion

for example,

D2 u2 u2 u2 = 1 + 2 + 3 + diusion

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

43421 rateturningvortex

2

3

2 00 0

3 > > >

Dt

D

x

u

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

UD Re

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 uid ow

+

Inviscid Fluid = 0

Ideal Flow + Incompressible Flow = 0P-Flow v

Irrotational Flow = 0 = 0or

3.6.1 Velocity potential

For ideal ow under conservative body forces by Kelvins theorem if 0 at some time t, then 0 irrotational ow always. In this case the ow is P-Flow.

Given a vector eld v for which = v 0, there exists a potential function (scalar) - the velocity potential - denoted as , for which

v =

Note that v = 0 =

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

v (x, t) = (x, t) = , ,

x y z

6

(x)

u u

u = 0 > 0 < 0

x x u > 0 u < 0

from low to high

x

The velocity vector v is the gradient of the potential function , 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 = 2 = 0 Number of unknowns

Number of equations 2 = 0

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

p = f (v) = f () Solve for , then nd 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 ow.

Euler equation:

v v2 p+ v = + gy

t 2

Substituting v = and = 0 into Eulers equation above, we obtain

1 2 p + || = + gyt 2

or

1 2 p + || + + gy = 0,t 2

which implies that

1 2 p+ || + + gy = f(t)t 2

everywhere in the uid for unsteady, potential ow. The equation above can be written as

p = + 1 ||2 + gy + F (t)t 2

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

DO NOT CONFUSE WITH

BERNOULLI EQUATION FROM 3.4,

USED FOR STEADY, ROTATIONAL FLOW

8

)

( )

Summary: Bernoulli equationS for ideal ow.

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

1 2 p = v + gy + C()2

(b) For unsteady or steady irrotational ow everywhere in the uid:

1 2 p = + || + gy + F (t)t 2

(c) For hydrostatics, v 0, t

= 0:

p = gy + c hydrostatic pressure (Archimedes principle) (d) Steady and no gravity eect ( = 0, g 0):

t

v2 p = + c = ||2 + c Venturi pressure (created by velocity)

2 2

(e) Inertial, acceleration eect:

Eulerian inertia

p t

+ p v t

+

p

u

p + p x x

x

9

( )

(c) Boundary Conditions

KBC on an impervious boundary

v n = no ux across boundary = Unu n given n n Un given

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

1 2 + || + gy = given t 2

Note: On a free-surface p = patm.

10

( )

( ) ( )

( ) ( ) ( )

3.6.3 Stream function

Continuity: v = 0; Irrotationality: v = = 0 Velocity potential: v = , then v = () 0 for any , i.e.,

irrotationality is satised automatically. Required for continuity:

v = 2 = 0 Stream function dened by

v =

Then v = 0 for any , i.e., satises continuity automatically. Required for irrotationality:

v = 0 = 2 = 0 (1) still 3 unknown

=(x,y ,z )

For 2D and axisymmetric ows, is a scalar (stream functions are more useful for 2D and axisymmetric ows).

For 2D ow: v = (u, v, 0) and z 0.

i j k x y z

x y z

=

ki + j +v = = y z z xy x x y

Set x = y 0 and z = , then u = y ; v = x

So, for 2D:

= x + y + z 0 x y z

Then, from the irrotationality (see (1)) 2 = 0 and satises Laplaces equation.

11

) (

2D polar coordinates: v = (vr, v ) and z 0.

r r

x

y

v vzvr

er re ez r z

r z

=

1 1 z z 1

v = = r er e + r ezr r r r r

Again let r = 0 and z = , then

1 vr = and v =

r r

For 3D but axisymmetric ows, also reduces to (read JNN 4.6 for details).

12

Physical Meaning of . In 2D

u =

y and

v =

x

We dene

x x (x, t) = (x0, t) + v x0, t) + (udy vdx) nd = (

x0 x0

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

v x

tv

o xv

C

C n

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

= or = 0 v v ds = 0 n d =

C C C C CC S =0, continuity

Therefore, is unique because of continuity.

13

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

streamline

x 2 ( = ( + v n dx2) x1)

=0 x2 1 1 along a streamline

Therefore, 1 = 2, i.e., is a constant along any streamline. For example, on an impervious stationary body v n = 0, so = constant on the body is the appropriate boundary condition. If the body is moving v n = Un

= 0 + Un d on the boddy

given

= constant = 0 n

= given

u = 0

o

14

Flux = vx = uy.

Therefore, u = and v =

y x

streamline

streamline

-v

u

(x,y) (x +x, y)

+

(x, y + y)

15

Summary of velocity potential formulation vs. stream-function formulation for ideal ows

For irrotational ow use For incompressible ow use For P-Flow use or

velocity potential stream-function

denition continuity v = 0 irrotationality v = 0

v = 2 = 0 automatically satised

v = automatically satised

(

) =

(

) 2 = 0

2D: w = 0, z

= 0

continuity irrotationality

2 = 0 automatically satised

automatically satised 2 = 0 z :

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

Cartesian (x, y) u =

x u =

y

v = y v = x

Polar (r,) vr =

r

v = 1 r

vr = 1 r

v = r

Given or for 2D ow, use Cauchy-Riemann equations to nd the other:

e.g. If = xy, then = ?

u = = y = 1 2 = y + f1(x)

x y 2

1 v = = x = = x 2 + f2(y)

y x 2

16

1 = (y 2 x 2) + const 2

13.021 Marine Hydrodynamics, Fall 2004

Lecture 10

13.021 - Marine Hydrodynamics Lecture 10

3.7 Governing Equations and Boundary Conditions for P-Flow

3.7.1 Governing Equations for P-Flow

(a) Continuity 2 = 0 1

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

2||

3.7.2 Boundary Conditions for P-Flow

Types of Boundary Conditions:

(c) Kinematic Boundary Conditions - specify the ow velocity v at boundaries. = Un

n

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

p = t + () + gy + C (t) (prescribed)

2

1

The boundary conditions in more detail:

Kinematic Boundary Condition on an impermeable boundary (no ux condition)

v n = U n = Un = Given uid velocity boundary velocity nornal boundary velocity v=

n = Un

(n1

x1 + n2

x2 + n3

x3 ) = Un

n = Un

( )321 n,n,nn =v( )v

Uv

Dynamic Boundary Condition: In general, pressure is prescribed

1 2

p = t + () + gy + C (t) = Given 2

2

( ) )()2

1(

0

2

2

tCgypt

+++=

=

=+++

GIVEN)())(2

1(:DBC

19)(Lecture:KBC

surfaceFree

linearnon

2tCgy

t 321

GIVENUn

n==

:KBCboundarySolid

3.7.3 Summary: Boundary Value Problem for P-Flow

The aforementioned governing equations with the boundary conditions formulate the

Boundary Value Problem (BVP) for P-Flow.

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

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

3

3.8 Linear Superposition for Potential Flow

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

Potential function .

BonfUn

n==

Vin02 =

Stream function .

Vin02 =

=gonB

Linear Superposition: if 1, 2, . . . are harmonic functions, i.e., 2i = 0, then = ii, where i are constants, are also harmonic, and is the solution for the boundary

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

= (11 + 22 + . . .) = Un on B.

n n

The key is to combine known solution of the Laplace equation in such a way as to satisfy

the kinematic boundary conditions (KBC).

The same is true for the stream function . The K.B.C specify the value of on the

boundaries.

4

=

3.8.1 Examplex denote a unit-source ow with source at xi, i.e.,

1

ln

Let i

i x source x,xi (in 2D)x xi

2 1 (in 3D),= 4 x xithen nd mi such that

= i

mii( x) satises KBC on B

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

1xv

2xv

4xv

3x

v

Vin02 =

fn

=

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

5

3.9 - Laplace equation in dierent coordinate systems (cf Hildebrand 6.18) 3.9.1 Cartesian (x,y,z)

i j k

v = u, v, w , ,= =

x y z

2 2 2

ze

y

x

ye

xe

z

O

),,( zyxP

2 = x2

+ y2

+ z2

6

3.9.2 Cylindrical (r,,z)

2 2 2 r = x + y ,

= tan1(y/x)

er e ez

=

v = vr, v, vz r

2 1 1 2 = r2

+ r r

+ r2 2 z2

, 1

r

,

z

2 2 +

ze

P

y

x

ye

xe

z

O

),,( zr

r

1

(r )

1

1 2 2 2

r r r

2 = r r

rr

+ r 2

+ z2

7

3.9.3 Spherical (r,,)

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

= cos1(z/r) z = tan1(y/x)

er e e

v = = vr, v, v = ,r

2 2 1

2 = r2

+ r r

+ r2 sin

= r (cos )

1

r

,

1

r(sin )

sin +

1

r2 sin2

2

2

ze

P

y

x

ye

xe

z

O

),,( r

r

1 2

(r )2 r r r

1

1

1 2 2 = r2 r

r 2 r

+ r2 sin

sin

+ r2 sin2 2

8

3.10 Simple Potential ows

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

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

v = (U, V, 0)

v = (U, V,W )

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

3D: = Ux + V y + Wz + constant

2. Source (sink) ow

2D, Polar coordinates

1 1 2 2 =

r

+ , with r =

x2 + y2

2

r r r r 2

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

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

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

= m

2 ln r

= r

er = m

2r er vr = m

2r , v = 0

source(strengthm)

x

y

9

Net outward volume ux is C

v nds =

S

vds = S

vds C

v nds =

2 0

vrm

2r

rd = m source strength

C

S

S

n

x

y

If m < 0 sink. Source m at (x0, y0):

m

= ln (x x0)2 m

= (Stream function) 2

2 + (y y0)2

ln r (Potential function) = 2

m

y

x

=01

pi=

2mV

r

pi

=2m

10

2

3D: Spherical coordinates

1

2 2 = r2 r

rr

+

,

, , where r = x2 + y2 + z

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

r

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

m m =

4r vr =

r =

4r2 , v = 0, v = 0

Net outward volume ux is m vrdS = 4r 2 4r 2

= m (m < 0 for a sink )

11

3. 2D point vortex

2 = 1 r

r

r

r

+

1

r2 2

2

Another particular solution: = a + b. Verify that 2 = 0 except at r = 0.

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

=

2 vr =

r = 0, v =

1

r

=

2r and,

z = 1

r

r (rv) = 0 except at r = 0

Stream function:

= 2

ln r

Circulation:

C1

v d x =

C2

v d x +

C1C2

v d x

R R

S

z dS=0

=

2 0

2r rd =

vortex strength

12

4. Dipole (doublet ow)

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

2D dipole:

m

2 2

2 2

= ln (x a) + y ln (x + a) + y

2

2

lim = ln (x ) + y2a 0 2 =0 = 2ma constant x x

= 2 + y2

= 22 x 2 r

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

NOTE: dipole =

(unit source)

13

unitsource

x

unitsource

unitsource

x

= x cos + y sin

= cos cos + sin sin

2 x2 + y2 2 r

3D dipole:

where = 2ma xed1 1m = lim (x a)

(x + a)2

40a 2 + y2 + z2 + y2 + z2

1 x x

4

( )x 2 + y2 + z2

=0

= = 4 (x2 + y2 + z2)3/2

= 4 r3

3D dipole (doublet) of moment 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: = Ux + m

2 ln x2 + y2

DU

U

xm

stagnationpoint 0v =v

DividingStreamline

m x u = = U +

x 2 x2 + y2 m

u y=0 = U + , v y=0 = 0 | 2x | V = (u, v) = 0 at x = xs = m , y = 0

2U

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

U

15

3D: = Ux 4x

m 2 + y2 + z2

div.streamlines

stagnationpoint

m x u = = U +

x 4 (x2 + y2 + z2)3/2

m x u y=z=0 = U + 3 , v y=z=0 = 0, w y=z=0 = 0 | 4 |x| | |

m V = (u, v, w) = 0 at x = xs = , y = z = 0

4U

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

U

16

U SS x

y

+m-m

a

dividingstreamline(seethiswithPFLOW)

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

= Ux+

2 ln (x + a) + y2 ln (x a) + y2 = Ux+

4 ln

(x a)2 + y2

3D Rankine ovoid: m 1 1

= Ux 4

(x + a)2 + y2 + z2

(x a)2 + y2 + z2

17

For Rankine Ovoid,

m

x + a

u = = U + x 4 (x + a)2 + y2 + z23/2 (x a)2 x + ya 2 + z23/2

m

1 1

u =U + y=z=0 2| 4 (x + a)2 (x a)

m (4ax)

=U + 4 (x2 2)2

m

u|y=z=0 =0 at

x 2 a

2a2

= 4U

4ax

At x = 0,

m 2a 2 2 u = U +4 (a2 + R2)3/2

where R = y + z

Determine radius of body R0:

R0 2 uRdR = m

0

18

7. Stream + Dipole: circles and spheres

U r

x 2D: = Ux + = cos

Ur +

2r2 2r

x=rcos

The radial velocity is then

ur =

= cos U

.

r 2r2

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

2U

. This is the K.B.C. for a stationary circle of radius a. Therefore, for

= 2Ua2

the potential

= cos Ur +

2r

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

Flow past a circle (U, a).

19

2

= U cos r + a

r

2

V = 1 = U sin

1 + a

= 0,

= 2 , 3 2

2U

2U

r r2 = 0 at stagnation points

r=a V| = 2U sin = 2U at maximum tangential velocity

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

= Ur cos 1 +

4r3

y

x

r

z

U

cos 3D: = Ux +

4 r2

The radial velocity is then

vr = = cos

U

r 2r3

20

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

2U . This is the K.B.C.

for a stationary sphere of radius a. Therefore, choosing

= 2Ua3

the potential

= cos Ur +

2r

is the solution to ideal ow past a sphere of radius a.

Flow past a sphere (U, a).

3

a

= Ur cos 1 + 2r3

v = 1

r

= U sin

1 +

a3

2r3

v |r=a = 3U

2 sin

= 0 at = 0, = 3U

2 at =

2

x

3/2 U

3/2 U

21

8. 2D corner ow Velocity potential = r cos ; Stream function = r sin

(a) 2 =

2 + 1 + 1 2 = 0

r2 r r r2 2

(b)

ur = = r

1 cos r

1

u = = r1 sin r

u = 0 { or = 0} on = n, n = 0, 1, 2, . . .

i.e., on = 0 = 0, , 2 , . . . (0 2)

i. Interior corner ow stagnation point origin: > 1. For example,

= 1, 0 = 0, , 2, u = 1, v = 0

x

y

=0

22

(90o corner)

=0

=0

y2v,x2u2,23,,

2,0,2 0 ==pi

pipi

pi==

(120o corner)

=0, =0

=2pi/3, =0

=2pi, =0

=4pi/3, =0

120o120o120o pi

pipi== 2,

34,

32,0,23 0

23

ii. Exterior corner ow, |v| at origin:

< 1

0 = 0, only

For example,

= 1/2, 0 = 0, 2 (1/2 innite plate, ow around a tip)

Since we need 0 2, we therefore require 2, i.e., 1/2 only.

1/2 < 1 0 = 0,

=0, =0

=2pi, =0

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

=3pi/2, =0

=0, =0

24

Appendix A1: Summary of Simple Potential Flows

Cartesian Coordinate System

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

Uniform ow Ux + Vy + Wz Uy Vx

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

yyo xxo )

3D Source/Sink (m) at (xo, yo, zo) m 4 1q(xxo )2 +(yyo)2+(zzo)2

NA

Vortex () at (xo, yo) 2 arctan(

yyo xxo )

2 ln((x xo)2 + (y yo)2)

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

2 (xxo) cos +(yyo) sin

(xxo)2+(yyo)2 2

(yyo ) cos +(xxo) sin (xxo )2+(yyo)2

3D Dipole (+x) () at (xo, y0 , zo) 4 (xxo)

((xxo)2 +(yyo)2+(zzo)2)3/2 NA

25

Appendix A2: Summary of Simple Potential Flows

Cylindrical Coordinate System

Flow Streamlines Potential Stream function (r, , z) (r, )

Uniform ow Ur cos + V r sin + Wz Ur sin Vr cos

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

m 2

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

Vortex () at (xo, yo) 2 2 ln r

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

2 cos cos +sin sin

r 2

sin cos +cos sin r

3D Dipole (+x) () at (xo, yo, zo) 4 cos r2 NA

26

Appendix A3: Combination of Simple Potential Flows

Stream + Source

=

Rankine Half Body

(2D)

(3D)

= Ux + m 2 ln r xs = m 2U D = m U

= Ux m 4 1x2+y2+z2 xs =

m 4U

A = m U

Stream + Source + Sink

=

Rankine Closed Body

(2D)

(3D)

= Ux + m 2 ln((x + a)2 + y2) ln((x a)2 + y2)

= Ux + m 4 ( 1

(x+a)2+y2+z2 1

(xa)2+y2+z2 )

Stream + Dipole

=

Circle (Sphere) R = a

(2D)

(3D)

= Ux + x 2r2 if = 2a2U = U cos (r + a

2

r )

= Ux + cos 4r2 if = 2a3U = U cos (r + a

3

2r2 )

2D Corner Flow (2D) = Cr cos() = Cr sin() 0 = 0, n

27

Appendix B: Far Field Behavior of Simple Potential Flows

Far eld behavior

r >> 1 v =

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( )

13.021 Marine Hydrodynamics, Fall 2004

Lecture 11

13.021 - Marine Hydro dynamics Lecture 11

3.11 - Metho d of Images m

Poten tial for single source: = ln x2 + y2 2

m

Poten tial for source near a wall: = m ln x2 + ( y b)2 + l n x2 + ( y + b)2 2

b

b

Added source for

0= dy d

x

y

m

m

symmetry

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

1

( )

( )

Vortex near a wall (ground eect): = Ux+ tan1(y b ) tan1(y + b )2 x x

b

b

U

x

y

-

Added vortex for symmetry

d Verify that = 0 on the wall y = 0.

dy

2 2

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 satises the boundary condition on the wall ( = 0), and the degree it n

satises the boundary condition of no ux 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,

d1 ,

d1 2 .

2

More than one wall:

b'

b U

b'

b U

b'

Example 1:

Example 2: Example 3:

bb bb

- b'b' b'b'

b'b'b'b'

- b b b b

3.12 Forces on a body undergoing steady translation DAlemberts paradox

3.12.1 Fixed bodies & translating bodies - Galilean transformation.

y y

x Uox o

z z

Fixed in space Fixed in translating body

x = x` + Ut

3

Reference system O: v, , p Reference system O: v, , p

U

SO

X USO

X

2 = 0 v n =

n = U n = (U, 0, 0) (nx, ny, nz

= Unx on Body v 0 as |x| 0 as |x|

)

2 = 0 v n =

n = 0

v (U, 0, 0) as |x|

Ux as |x| Galilean transform:

v(x, y, z, t)

(x, y, z, t)

Ux + (x = x + Ut, y, z, t)

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

= (x = x Ut, y, z, t) + Ux = (x, y, z, t)

Pressure (no gravity)

p = 1 2 v2 + Co = Co

Co

= 1 2 v2 + C o = C

o 1 2 U2

= C o 1 2 U2 In O: unsteady ow

ps = t 1 2 v2 U2

+Co

t

= (

t 0

+ x

t U

x ) (

+ Ux) = U2

ps = U2 1 2 U2 + Co = 1 2 U2 + Co

ps p = 1 2 U2 stagnation pressure

In O: steady ow

ps =

t 0

1 2 v2

0

+C o = C o

ps p = 1 2 U2 stagnation pressure

4

( )

( )

( )

3.12.2 Forces

B

n

Total uid force for ideal ow (i.e., no shear stresses): F ndS= p B

For potential ow, substitute for p from Bernoulli:

1 2 F = + || + gy +c(t)ndS t 2 B

hydrodynamic hydrostatic force force

For the hydrostatic case v 0 :

F s = (gyn) dS = () (gy) d = gj where = d B Gauss outward B Archimedes Btheorem normal principle

We evaluate only the hydrodynamic force:

1 2F d = + || ndS t 2

B

For steady motion 0 :t

1 F d = 2 v ndS

2 B

5

(

( )

( )

(

)

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

n

B

SU a

x

2

0

F d =

B

) | |

2 i = a

2 2 2

2 ||nd = nad r=a

2 id ||Fx F n = 2 r=a 0 cos

= a 2

2

0

2|| cos dr=a Velocity potential for ow past a 2D cylinder:

2a = Ur cos 1 +

2r

Velocity vector on the 2D cylinder surface:

1

r

| = (vr| , v| ) =r=a r=a r=a ,r r=a

0 r=a 2U sin

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

||2 = 4U2 sin2 r=a

6

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

2 2

a ( ) (1 ) Fx = d 4U

2 sin2 cos = U2 (2a) 2 d sin2 cos = 0 2 2 odd 0 diameter 0 even 3 psp or w.r.t 2 , 2

projection 0

Therefore, Fx = 0 no horizontal force ( symmetry fore-aft of the streamlines). Similarly, 2

(1 ) Fy = U

2 (2a)2 d sin2 sin = 0 2

0

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

DAlemberts paradox:

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

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.

= Ux + = Ur cos +

2 2

U

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

d steady ow F = 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 uid.

F V = F = Hydrodynamic force exerted on the uid in the control volume from the vortex. F CS = Surface force (i.e., pressure) on the uid control surface.

M NET = Net linear momentum ux 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

Fy Fx

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

8

( )

( )

a. Net linear momentum ux 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

u = U sin

2r

v = cos 2r

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

x ur = U = U cos = V n

r

v = 2r

U

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

2 2

(MNET )x = druvr = dr U sin U cos = 0

2r 0 0

2 2

(MNET )y = drvvr = dr cos U cos

2r 0 0

2

U U = cos 2 d =

2 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 |v|2 + C 2

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

|v|2 =u 2 + v 2 = U sin + cos 2r 2r

2

=U2 U sin + r 2r

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

2

(FCS) = drp( cos ) = 0x 0

2 2 ( ) ( )

(FCS)y = drp( sin ) = 2 rU (r) d sin2 = 21 U 0

0

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

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

Fy = (FCS) + (My)IN = U y i.e., the uid exerts a downward force F = U on the vortex.

Kutta-Joukowski Law

2D : F = U 3D : F = U

Generalized Kutta-Joukowski Law:

n

F = U i i=1

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

10

Marine HydrodynamicsLecture 12

13.021 - Marine Hydro dynamics Lecture 12

3.14 Lifting Surfaces 3.14.1 2D Symmetric Streamlined Bo dy

No separation, even for large Reynolds numbers.

U

stream line

Viscous eects only in a thin boundary layer. Small Drag (only skin friction). No Lift.

1

3.14.2 Asymmetric Body

(a) Angle of attack ,

chord line

U

(b) or camber (x),

chord line mean camber line

U

(c) or both

amount of camber

U

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 ow past a body we obtain

P-Flow solution, innite velocity at trailing edge.

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

Circulatory ow only.

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

Figure 1: P-Flow solution plus circulatory ow.

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 ow separation at the trailing edge. As the foil moves, viscous eects streamline the ow 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

Kelvins theorem:

After a while the S

S

S

S

S

U

U

no

starting vortex left in wake

= 0

starting vortex due to separation (a real fluid effect, no infinite vel of potetial flow)

d = 0 = 0 for t 0 if (t = 0) = 0

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

4

3.15.2 How much S ?

Just enough so that the Kutta condition is satised, so that no separation occurs. For

example, consider a at plate of chord and angle of attack , as shown in the gure

below.

chord length

Simple P-Flow solution

= lU sin

L = U = U2l sin

|L |CL = = 2 sin 2 for small 1 U2l

2 only for small

However, notice that as 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 eects of stalling on the lift coecient (CL = U2

L span) are shown in the 1

2

following gure.

5

C L

This region independent of R,

used only to get Kutta

condition

stall location f(R)

stall 2

O(5o )

In experiments, CL < 2 for 3D foil - nite aspect ratio (nite 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 gure below.

y=yU(x)

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

yU ,

yL

Denitions

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

/2

= (x)dx /2

(x) U

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

= Ux + Free stream Disturbunce potential potential

The ow velocity can by expressed as

v = = (U + u, v)

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

u v u, v

Lift due to circulation

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

1 p p = |v|2 U2

2

p p = 1 (u U)2 + v 2 U2 = 1 (u 2 + v 2 2uU) 2 2 1 u v v

p p = uU( + 2)2 U U u

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 gure

below. (x)

)0,( +xu

0t

)0,(

xu

x

x

In this case we have

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

(x) u(x, 0) = (2)

2

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

l/2

L = U (x)dx = U l/2

=

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

/2

L = U = U(x)x L = U(x)x = U /2

L = U = U (x) x

x

= U

t 0 (x) x

x

10

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

x

L

y

2 l

2 l

cp x

M

L(x) = U(x)x

M = xL(x) = Ux(x)x /2

M = Ux(x)dx /2 M

=CM 1 U22 2

The center of pressure xcp, can be obtained by

M = Lxcp M

/

/

2

2 x(x)dx

xcp = = /2L /2 (x)dx

11

3.17 Simple Closed-Form Solutions for /

/

22 (x)dx from Linear

Theory

1. Flat plate at angle of attack , i.e., = x .

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

CL ,

/2

L/span = U (x)dx = = U 2 / 2

CL = L/span 1 U 2 2

CL = 2 ( exact nonlinear hydrofoil CL = 2 sin )

the moment coecien t CM , /2

M/ span = U x (x)dx = = 14 U 2 2 / 2 M/ spanCM = 1

U 2 2 2 CM = 12

and the center of pressure xcp

xcp = 41 i.e., at quarter chord

12

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

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

CL,

/2 L/span = U (x)dx = = 2U20

/2

CL = 4 0

, where 0 camber ratio

the moment coecient 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 = x + 0 1 .

0CL = CL + CL = 2 + 4

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

CL() = 2 + CL( = 0)

4 0 l

Lift coecient CL as a function of the angle of attack and l 0 .

In practice even if the camber is not parabolic, we still make use of the previous relations, i.e., CL( = 0) = 40/. Also note that the angle of attack for any camber is dened as

(/2) (/2) yU yL =

and 0 is determined from , where

= x.

14

13.021 Marine Hydrodynamics, Fall 2004

Lecture 13

13.021 - Marine Hydro dynamics Lecture 13

3.18 Unsteady Motion - Added Mass DAlem bert: ideal, irrotational , unbounded, steady.

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

n

U(t)

3D Dipole x

r

a

U(t)

K.B.C on sphere: = U(t) cos

r r = a

Solution: Simply a 3D dipole (no stream) 3a

= U(t) cos 2r 2 Check: r

= U(t) cos

r = a

1

( )

( ) ( )

Hydrodynamic force:

1 2Fx = + || nxdS t 2

B

On r = a,

= a3 1 U cos | = Ua cos

t 2r2 r=a 2 r=a 1 1 1 | = , , = U cos , U sin , 0 r=a r r r sin 2

||2 = U2 cos 2 + 1 U2 sin2 ; n = er, nx = cos r=a 4

dS = (ad) (2a sin )

B 0

x

a

adsina

2

( )

[ ]

( )

Finally, 1 2 Fx = () 2a2

d (sin ) cos

1 Ua cos + U cos 2 + 1 U2 sin2

2 2 4 0 nx

2||t

Fx = U(a3) d sin cos 2 + (U2)a2 d sin cos cos 2 + 1 sin2 4

0 0 2/3 = 0, Dalembert revisited 2

Fx = U(t) a3 3 Hydrodynamic Force Acceleration Fluid Density

Volume =1/2sphere

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

Fx = U(t) 2 a3 3

Comments:

If U = 0 Fx = 0, i.e., steady translation no force (DAlemberts Condition ok). Fx U with a () sign, i.e., the uid tends to resist the acceleration. [ ] has the units of (uid) mass ma Equation of Motion for a body of mass M that moves with velocity U :

M U = F = 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 uid 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 1, U5 2, U6 3: Rotational velocities

1

2

6

5

4

3

(2D) U1, U2: Translational velocities

U6 3: 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 = 1/2 = (mA) all other mij = 0

4

3.19.3 Added Masses of Simple 2D Geometries

Circle

a

1

2

m11 = m22 = = a2

Ellipse

m11 = a2 ,m22 = b

2

Plate 2

a

b 1

2

2a 1

m11 = a2 ,m22 = 0

5

Square

2

2a 1

2a

m11 = m22 4.754a2

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

(a) inscribed circle: mA = a2 = 3.14a2 .

2a

( )2 (b) circumscribed circle: mA = 2a = 6.28a

2 .

(2)a

Arithmetic mean of (a) + (b) 4.71a2 .

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 uid that is at rest at innity.

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

x1

x2

o

x3

)t(U v

)t(v

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

(t) = (1,2,3) (U4, U5, U6) , rotational velocity with respect to O

Consider a body with a 6 DoF motion ( ), and a xed 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 Uik 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 Uik 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

Einsteins 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 k 0 , Fj = U imji (as expected by denition 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 B where B is linear momentum (momentum from i coordinate into new xj direction).

(c) If k 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 coecients only on the diagonal, i.e. mij = ij . For a body motion given by U1 = t, U2 = t, and all other Ui, i = 0, the forces and moments on the body in terms of mi are:

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

Solution:

mij = ij

U1 = t U2 = t Ui = 0 i = 3, 4, 5, 6 k = 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): k=0 Fj = U imij EjklUikmil

Fj = U imij

k=0 Mj = U imi(j+3) EjklUikmi(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

2

U

1

2

33 (out of page)(out of page)

U1 =U cos

U2 = U sin

Consider steady translation motion: U = 0;k = 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 cos m22 m11

>0

Therefore, M3 > 0 for 0 < < /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 gure.

11

B

G H

Restoring moment (gH)sin. critical speed Ucr given by:

(g) H sin U2 sin cos (m22 m11)cr

H Ucr

Usually m22 >> m11,m22 . For small , cos 1. So, U2 gH or Fcr cr Ucr 1. Otherwise, control ns are required.

gH

12

13.021 Marine Hydrodynamics, Fall 2004

Lecture 14

13.021 - Marine Hydro dynamics Lecture 14

3.20 Some Prop erties of Added-Mass Coecien ts 1. mij = [function of geometry only]

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

2. Relationship to uid momentum.

F(t)

where we dene to denote the velocity potential that corresponds to unit velocity U = 1. In this case the velocity potential for an arbitrary velocity U is = U . The linear momentum L in the uid is given by

L = vdV = dV = + ndS

V V Greens B theorem 0 at

L x (t = T ) = U nxdS = U nxdS B B

The force exerted on the uid from the body is F (t) = ( mA U ) = mA U .

1

T T Newtons Law T dt [F (t)] = mAUdt = mA U ]0 = Lx (t = T )Lx (t = 0) = U nxdS

0 0 mAU B

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

K.B.C. = n = (U, 0, 0) n = Unx, = Unx U = Unx = nxn n n n

mA = dS n

B

For general 6 DOF:

jmji = i njdS = i dS = j uid momentum due to n

i body motion jforce/moment B potential due to body B idirection of motion moving with Ui=1

3. Symmetry of added mass matrix mij = mji.

( ) j

mji = i dS = i (j n)dS = (ij) dV n

B B Divergence VTheorem

= i j + i2jdV V =0

Therefore,

mji = i jdV = mij V

2

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

= Uii ; i = potential for Ui = 1

notation

1 K.E. = dV =

21 Uii Uj j dV

2 V V

1 =

2 UiUj i j dV = 12 mijUiUj

V

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

5. Symmetry simplies 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 coecients

m22 m26 0m33 m34 m35 mij = 0m44 m45 0 My m55 m66 Mz

U1 U2 U3 1 2 3

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 dierent coecients

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

Denitions

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

(b) mij = the 3D added mass coecient 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 coecient 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 coecients mij for a 3D slender body.

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

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

Discussion If the 1-axis is the longitudinal axis of the slender body, then the 3D added mass coecients mij are calculated by summing the added mass coecients 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 coecients 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 coecient 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

Putting everything together, we obtain

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