Marine Hydrodynamics - MIT Lecture Notes

13.021 Marine Hydrodynamics Lecture 1

1


Introduction Marine hydrodynamics is a large and diverse subject and only a few topics can be covered in an introductory course. Some course objectives to keep in mind throughout the semester are the following: Model testing similitude Effect of waves on resistance and ship motion Interaction between bodies and ideal fluids Viscosity and surface tension

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. Fluid mechanics is a way of looking at a group of particles without having to study each particle separately. A fluid at rest hydrostatics is a trivial case of fluid mechanics where no stresses due to fluid motion exist. Fluids have to be moving to be non-trivial. Fluid mechanics is fundamentally non-linear.

The mechanics of Fluids vs. Solids Most of us have taken some courses on solids or that relate to solids. Even those who havent can get an intuitive feel on some physical properties of a solid. Thus a comparison of solids and fluids will give some guidelines on which properties can be translated to fluids and on what terms.

Differences Fluids Fluids have no shape Fluids cannot sustain a shear force, i.e. a fluid is always in motion Stress is a function of the rate of strain, thus a fluid had a dynamic state The static properties of a fluid cannot be extended to dynamic properties. Solids Solids have a definite shape Solids can sustain a shear force; i.e. they remain static Stress is a function of strain, thus a solid maintains a static or quasi-static state. The static properties of a solid can be extended to dynamic properties.

Copyright 2001 MIT - Department of Ocean Engineering, All rights reserved.


2

Similarities The continuum hypothesis is used for both fluids and solids. The fundamental laws of mechanics apply to both fluids and solids. - Newtons law of motion (conservation of momentum) - Conservation of Mass - First law of thermodynamics (conservation of energy) The constitutive law relating stress and rate of strain also apply to both.

Sometimes it is hard to differentiate a liquid from a solid. This can be seen in the examples such as honey, jelly, paint,

Liquid vs. Gas Note that there are two separate terms that we are talking about here. Liquid and fluid. According to Websters Dictionary, a fluid is a body whose particles move easily among themselves. Fluid is a generic term, including liquids and gases as species. Water, air, and steam are fluids. 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. A liquid is generally incompressible and does not fill a volume by expanding into it. A gas on the other hand, is compressible and expands to fill any volume containing it.

Why is a liquid incompressible? The main difference between the study of hydrodynamics and the study of aerodynamics is the property if incompressibility. Hydrodynamic properties are generally incompressible while aerodynamic properties are compressible. Consider the following proof. Measuring the speed of sound in a medium will give a measure of compressibility of that medium. U: Characteristic fluid velocity C: Speed of sound in the medium M: Mach # U M C Cin air = 300 m/s air = 10 3 water

Cin water = 1200 m/s Cair 1 = C water 4



3

Typically, Uair >> Uwater Therefore Min air = O(1) compressible

Min water 0) nontrivial solutions, 1, 2, , JIn general, J < N, in fact, J = N K where K is the rank or dimension of (*)

Model Law: Instead of relating the N xis by (x1, x2, xN) = 0, relate the J s by F(1, 2, J) = 0, where J = N K < NFor similitude, we require (model)j = (prototype)j where j = 1, 2, , J If 2 problems have all the same js, they have similitude (in the j senses), so s serve as similarity parameters. Note: if is dimensionless, so is constant * , const, 1/ , etc If 1, 2 are dimensionless, so is 1 * 2 , 1 / 2, 1const1 * 2const2, etc In general, we want the set (not unique) of independent js, e.g. 1, 2, 3 1, 1 * 2, 3NOT: 1,

2, 1 * 2

Example: Application of Buckingham Theory Force on a smooth circular cylinder in steady incompressible fluid (no gravity)F U , D

xi = F, U, D, , x i = ci M m i Ll i T t i

N=5 P=3 N=5 U D 0 0 1 1 -1 0

P=3

mi i ti

F 1 1 -2

1 -3 0

0 2 -1

= F 1 U 2 D 3 4 5 i m i = 0 Dimensionless: i l i = 0 () t = 0 i i 1 2 0 = 0 3 4 0 5

0 0 1 0 1 1 1 3 2 1 2 1 0 0 1

1 0 0 1 0 0 1 1 4 2 0 0 1 2 1

1 2 0 = 0 3 4 0 5

K=3 J=PK=5-3=2 Two families of solutions for i for each fixed pair of (4, 5), exists a unique solution for (1, 2, 3)

consider pairs of (4 = 1, 5 = 0) and (4 = 0, 5 = 1), all other cases are linear combinations of these two. (1) 4 = 1, 5 = 0

1 0 0 1 1 0 1 1 2 = 4 0 0 1 2 3 1 1 2 = 2 2 3 1 = F U D 1 2 3 4 5

U 2 D 2 = F1 2

Conventionally, 1 21-1 1 = (2) 4 = 0, 5 = 1

F C d Drag coefficient U 2 D 2

1 0 0 1 0 0 1 1 2 = 2 0 0 1 1 3 1 0 2 = 1 1 3 2 = F 1 U 2 D 3 4 5 = UD UD R e Reynolds number or 1 = (2) or Cd = ( Re) UD F or ) =( 1 U 2 D 2 2

Conventionally, 2 2-1 2 =

Therefore

F(1, 2) = 0 F(Cd, Re) = 0 UD F )=0 F( , 1 U 2 D 2 2


13.021 - Marine Hydrodynamics Lecture 6

2.2 Similarity Parameters (from governing equations) Non-dimensionalize and normalize basic equations by scaling: Identify characteristic scales for the problem v v velocity U v = Uv * v v length L x = Lx * time T t = Tt * pressure po- pv p = (p o p v )p * All ()* quantities are dimensionless and normalized (i.e. O(1)),v v e.g. x = O(1)

Apply to governing equations: (also internal constitution, boundary conditions) continuity (incompressible flow):

v U v v v = v = 0, v = 0 LNavier-Stokes:

v v v 1 v v + (v )v = p + 2 v g j t v U v U 2 v * v v + (v )v = poL pv p + LU 2v gj 2 T t LL

2 divide through by U (order of magnitude of the convective inertia term)

v p p L v v v v gL 2v 2 j + ((v )v ) = o 2 v (p ) + U UL U UT t ~~~ ~~~ ~~~ ~~~~~

( )

Since all ()* terms are O(1), the coefficients ~~~ measure the relative importance of each term (as compared to the convective inertia term):v v Eulerian inertia t L = S = Strouhal number v v UT convective inertia ( v ) v

is a measure of transient behavior e.g. if T >> L , S > 1, no cavitation Alternatively, when cavitation is not a concern p = p o p

1 2

po U 2

= Eu = Euler number

pressure force inertia force

UL

= Re = Reynolds number

inertia force viscous force

Re >> 1, ignore viscosity

U2 U = gL gL

= Fr = Froude number

inertia force gravity force

v v v v Kinematic boundary conditions: v = U b v = U b

Dynamic boundary conditions: p = pa + p where p = 1 + 1 R1 R2

1 1 2 where p = p + + = ( p o p v )L R1 R 2 ( p o p v )L U 2 L a

u2L

= We = Weber number

inertial forces surface tension forces

note: L >> Ro usually



Alternatively, using physical arguments: forces acting on a fluid particle U 2 2 2 a) inertial forces mass acceleration L3 L = U L U 2 u b) viscous forces area L = UL y L { shear stress

( )

( )

3 c) gravitational forces mass gravity (L )g

2 d) pressure forces ( p o p v ) L

For similar streamlines, particles must be acted on by forces whose resultant is in the same direction at geosimilar points. Therefore, forces must be in the same ratios:

inertia U 2 L2 UL = = Re UL viscous2 U 2 L2 inertia U = Fr gravity gL3 = gL 2 1 1

1 inertia ( p p ) L2 p p 2 1o 2v 2 = 1o 2v = pressure 2 U L 2 U

1



Importance of Various Parameters e.g.g U

Govern flow similitude of different systems. Provide guidance and approximate the complex physical problem.

L

Parameters:

S=

P P L U 2L , = 1o 2v , We = UT 2 U gL , Re = UL

Fr = U

Force on the foil: F = F

(

1 2

U L

2 2

)

= C F S , 1 , We , Fr , Re1

(

1

)

(1) S = L / UT, change S with , We, Fr, Re fixed. F* transient Steady-State

=0 t S >> 1, unsteady effect is dominant.For S > T1, then steady-state since S> U1 cavitation (In practice, U > 30 m/sec to cause cavitation)

=O(1)

For steady non-cavitation flow (U U1)

F * = C F 0,0, We , Fr , Re1 = C F We , Fr , Re12 (3) W = U L o

(

1

)

(

1

)3 3

( ), e.g.U = 1 m s , = 0.07 N m (water-air 20 C), =10 kg/m

W = 1 for L1 = 10-4m For L >> L1, W and W-1 0 (Neglect surface tension effect) For steady, non-cavitation, non-surface tension effect,F * = C F 0,0,0, Fr , Re1 = C F Fr , Re1 (4) Fr = U

(

)

(

)

gL Froude # measures the effect of gravity.

For problems without dynamic boundary conditions(i.e. if free surface is absent) or if free surface effects are small (no waves), Fr is not important F* = C F R e 1

( )



e.g. (i)

(ii) Low speed (Fr = 0) (no wave)

U U closed channel no free surface g Free surface Wall

(iii) Large U No Wave

(iv) deeply submerged body h Fr = 0 No waves

U Fr g0

h U

In general F * = C F Fr , Re1 = C1 (Fr )+ = C2 Re1 Froudes Hypothesis Dynamic similarity requires (Re)1 = (Re)2, (Fr)1 = (Fr)2 for two geometrically similar systems U1 = U2 , L1 = L2 for the same and g. (5) Re = UL/ For steady, no , no W, no gravity effects, F * = C F Re1 F*

(

)

( )

( )

Sphere

Re (Re)cr, Turbulent flow R , ideal flow e.g. U = 10m/sec , L = 10m = 10-6m2/sec R = 108 R-1 = 10-8 Re

Plate

(Re)cr

For steady, no , no W, no gravity effect, ideal fluid: F = C F (0,0,0,0,0 ) = constant = 0 DAlemberts Paradox: No drag force on moving body.*


13.021 Marine Hydrodynamics, Fall 2004 Lecture 7


Chapter 3 Ideal Fluid FlowThe 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 ow. 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 vorticity.

Ideal Fluid Flow

Inviscid Fluid

=0 D = 0 or v = 0 Dt

Incompressible Flow ( 1.1)

+

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, Re = UL inertia forces = 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 = 106 m2 /s leading thus to typical Reynolds numbers with respect to U and L in the order of UL 106 >>> 1 0

Re =

1 viscous forces Re inertia forces

This means that viscous eects are > 1 U1 R 2

2

v 1

v 1

since

u v 0, 0 vortex ring r

/ L = constant

1 1 R = 2 2 = 2 ~ 0 x2D2 >0 Dt 4 123 4

u2 = 0

y x2

x x13 >0 u2 D2 > 0 >0 Dt 4 x3 123 4

vortex stretching rate

vortex turning rate

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 As the vortex tube length increases the diameter D becomes really small that big after all. Re Therefore IFCF is no longer valid.

Re is not

5

3.6 Potential FlowPotential Flow (P-Flow) is an ideal and irrotational uid ow Inviscid Fluid =0 Ideal Flow + Incompressible Flow v =0 + Irrotational Flow = 0 or = 0

P-Flow

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

x

x u >0 >0u=0

x u

1 (2D) Source (3D) 1 r = v 1 r 1 r2

ln r

(2D) Dipole (3D)

1 r 1 r2

1 r2 1 r3

Vortex

(2D)

1

1 r

28

13.021 Marine Hydrodynamics Lecture 11 3.9 Forces on a body undergoing steady translation DAlemberts paradox 3.9.1 Fixed bodies & translating bodies Galilean transformation.yy

o zFixed in space

x

o z

x

U

v O : v, , pU S

x = x + Ut

Fixed in translating space

v O' : v' , ' , pU

S

2 = 0

'2 ' = 0

v r v n = = U n = ( U ,0, 0) (n x , n y , n z ) n= Un x on body (t) v v 0 v as x 0

v ' v'n ' = = 0 on Body (not function of t ) n'v v ' ( U,0 ,0 ) v as x ' ' Ux

Galilean tranform: v v v( x, y, z, t ) = v' ( x ' = x Ut , y, z, t ) + ( U,0,0) ( x , y, z , t ) = ' ( x ' = x Ut , y, z, t ) + Ux ' Ux '+( x = x '+Ut , y, z, t ) = ' ( x' , y, z, t ) Pressure (no gravity)

1 p = v 2 + C o = C o 2 C o = C 'o

1 1 p = v'2 +C 'o = U 2 + C 'o 2 21 U 2 2

In O: unsteady flow 1 2 p s = v + C o t 2 { U2 x ' 2 = + ('+ Ux ') = U t { { x ' t t 0 U

In O: steady flow ' 1 p s = v'2 + C'o { t { 2 00

' = Co

1 1 p s = U 2 U 2 + C o = U 2 + C o 2 2

ps p =

1 U 2 2

ps p =Stagnation pressure

1 U 2 2

3.9.2 Forces v Total fluid force for ideal flow: F =No shear stress

B

pndSB

n

For potential flow: v 2 1 F = + + gy + c( t ) ndS t 2 B v For the hydrostatic (v 0 ) case:

B

v Fs =

( pgyn )dS = ( ) ( pgy )d = pg jGauss TheoremB

where =

n

dB

Archimedes Principle

Hydrodynamic Force: Fd =

v

t + 2 1B

2

n dS

v 1 For steady 0 motion: Fd = v 2 ndS 2 B t Example: Hydrodynamic force on 2D cylinder in steady translation2 v 2 2 Fd = ndl = r =a nad 2 2 B 0

nU S a B

x

v a 2 2 i Fx = F i = d r =a { n 2 0 cos

a = 2

2

0

cos d

2

a2 = U cos r + r r = a = v r 2 r =a

(

r =a

, v

r=a

)

, 1 = { r r r = a 4 4 0 r =a 1 2 3 2 U sin

= 4 U 2 sin 2 2

a Fx = 2

(0

2 1 2 2 2 d 4 U sin cos = U ( 2a ) 2 dsin 23 cos 1 { 2 { 2 123 diameter 0 even odd p p or 3 s projection w .r .t , 144 2442 3 4 42

)

Fx = 0 no forces ( symmetry fore-aft of the streamlines) 1 2 2 Similarly, Fy = U (2a )2 d sin sin = 0 2 02

0

v In fact, in general we find that F 0 on any 2D or 3D body DAlemberts paradox No hydrodynamic force* acts on a body moving with constant(1) translational velocity in an infinite(2), inviscid(3), irrotational(4) fluid.*

Note that the moment as measured in a local frame is not necessarily zero.

3.10 Lift due to circulation Example: force on a vortex in a uniform stream. U = Ux + 2y

U

Fy Fx

x

Consider a control surface in the form of a circle radius R centered at the point vortex. Then according to Newtons law: d 0 = M cv = (Fcs + Fv ) + M in dtFor steady flow

Control volume

MCV = Total (linear) momentum of control volume FCS = Hydrodynamic force on CS by surrounding fluids FV = Hydrodynamic force on CS by vortex = -(force on vortex by fluid) MIN = net flux of momentum in CV through CS force on vortex F = -FV = FCS + MINv =

2r

cos Momentum flux: u = U sin , v = 2r 2rVr = U x = U cos r2 2

U

(M x )IN = druVr = dr U sin U cos = 0 2r 0 0 (M y ) IN = drvVr = dr cos U cos 2r 0 02

2

U U = cos2 d = 2 0 2

2

p=

1 v2 v +C 2

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

= U2

U sin + r 2 r

2

(FCS )x (FCS )y

2

= drp( cos) = 00

1 U 2 = drp( sin ) = (r ) d sin = U 2 2 r 0 0 4 1 24 3

2

2

Finally, Fx = (FCS ) x + (M x )IN = 0

Fy = (FCS ) y + (M y )IN = U i.e. the fluid exerts a downward force U on the vortex Kutta-Jaukowski law: F = U v v v F = U



3.14 Lifting Surfaces3.14.1 2D Symmetric Streamlined Body 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 chord line mean camber line

U

angle of attack

Lift to U and Drag

to U

2

3.15 Potential Flow and Kutta ConditionFrom 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

=0

St=0 + U

S

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

Sfor later t U

S starting vortex left in wake

no

Kelvins theorem: d = 0 = 0 for t 0 if (t = 0) = 0 dt After a while the S 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 = U 2 l sin CL = |L|1 U 2 l 2

= 2 2 for small sin 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 = 1 U 2 Lspan ) are shown in the 2 following gure.

5

C

L This region independent of R, used only to get Kutta condition stall location f(R)

2

stall

O(5 o )

In experiments, CL < 2 for 3D foil - nite aspect ratio (nite span). With sharp leading edge, separation/stall to early.

sharp trailing edge round leading edge to forstall stalling to develop circulation

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 , (o )turbulent for Rex < 3.4 103 (o )laminar (o )turbulent for Rex 3.4 103 (o )laminar < (o )turbulent for Rex > 3.4 103Cf

C f L ~ RL

1

2

C fT ~ RL

1

5

Therefore, for most prototype scales: (Cf )turbulent > (Cf )laminar (o )turbulent > (o )laminar

~ 0.01 ln (RL) RL ~ 1.6 x 104

5

13.021 - Marine Hydrodynamics Lecture 19 Turbulent Boundary Layers: Roughness Effects So far, a smooth surface has been assumed. In practice, it is rarely so due to fouling, rust, rivets, etc.

Viscous sublayer Uo v k = characteristic roughness height

Equivalent sand roughness k Estimate k for actual surfaces e.g., ship hull For k < viscous sublayer thickness v, k does not affect the turbulent boundary layer significantly: C f C fsmooth ( Rl ) C f = C f ( Rl ), not k

( l)

Surface is hydraulically smooth for k < v For k >> v, :separation

k v

Drag due to separation = form drag >> viscous drag C f = C frough > C fsmooth C frough = C f k

( l ) , not ( R )l

Cf

k/l k/l = constant

C f rough

Rl

C f smooth

Recall for a bluff bodyC D f (R )

In summary: Important parameter is k / :k > 1 : rough ( x )

Therefore, for the same k, the smaller the , the more important the roughness k. Corollaries 1) Exactly scaled models (e.g. hydraulic models of rivers, harbors, etc) Samek ~ const. for model and prototype l k k l k 15 ~ Rl = l l k for Rl

For Rmodel Friction drag, therefore CD constant CP (within a regime) For a streamlined body: CD (R) = Cf (R) + CP Form drag (CP) not a function of Reynolds number within a regime. 1- Perform an experiment with a smooth model at Rm (Rm

Documents

Marine Hydrodynamics - MIT Lecture Notes