<#11#>13.021 -- Marine Hydrodynamics<#11#>
<#12#>Lecture 10<#12#>
<#1542#>Copyright <#13#>©<#13#> 2001 MIT - Department of Ocean
Engineering,
All rights reserved.<#1542#>
13.021 - Marine Hydrodynamics
Lecture 10
Types of boundary conditions:
Consider a body moving in an infinite fluid domain, as illustrated by figure #fig:lfig101#19>.
<#20#>figure<#20#>
Figure:
Rigid body translating at constant speed #tex2html_wrap_inline1899# in an unbounded fluid domain.
| #figure21# |
We would like to know the flow field #tex2html_wrap_inline1901# generated by the body motion. How we can proceed?
We assume the fluid ideal and the flow irrotational. Under such assumptions, the velocity field #tex2html_wrap_inline1903# satisfies the equation
which implies that the velocity field #tex2html_wrap_inline1905# can be written as the gradient of a function #math2##tex2html_wrap_inline1907#, denoted as the potential function.
For an incompressible fluid, the velocity field satisfy the equation
which implies that the potential function #tex2html_wrap_inline1909# is the solution of the Laplace's equation
So far we know that the potential function #tex2html_wrap_inline1917# satisfies the Laplace's equation. Is this enough information to obtain #tex2html_wrap_inline1919#?
Laplace's equation has an infinite set of solutions, but which of these solutions satisfy our problem?
To answer this question, we need additional information. Where we can get this additional information about #tex2html_wrap_inline1921# or #tex2html_wrap_inline1923# ?
The surface of the body does not allow flow through it. This implies that the difference of the fluid velocity and body surface velocity in the direction normal to the body surface should be zero. This statement is expressed by the equation
and in terms of the potential function #tex2html_wrap_inline1927# we have
The other boundary lies at the infinite (unbounded fluid domain). Far from the body, the fluid is not affected by the body movement, so the velocity field at the infinite should be zero. This is expressed by the equation
and in terms of the potential function #tex2html_wrap_inline1929# we have
Consider a wave propagating over a body of water, and assume ideal fluid and irrotational flow. Water for flow speeds much smaller than the sound speed in water can be considered as an incompressible fluid. Under the asumptions above, the velocity field #tex2html_wrap_inline1931# for the flow can be written as the gradient of a potential function #math10##tex2html_wrap_inline1933#.
At solid boundaries, like the bottom, the kinematic boundary condition prescribes the flow velocity normal to the body surface, which is
since the bottom is at rest.
We can impose a kinematic boundary condition on the water surface. We assume that the water surface is at #math12##tex2html_wrap_inline1935# (#tex2html_wrap_inline1937# is the time parameterization), and the kinematic boundary condition in this case will be
where #math14##tex2html_wrap_inline1939#. With this expression for #tex2html_wrap_inline1941#, the condition above reduces to
Okay, now we have a kinematic boundary condition for the water surface, but there is a problem. We do not know before hand the position of the water surface, and we do not know the potential function #tex2html_wrap_inline1945#. We have only one equation, but two unknowns, #math16##tex2html_wrap_inline1947# and #math17##tex2html_wrap_inline1949#. We need another boundary condition for the potential function #math18##tex2html_wrap_inline1951# at the water surface.
At the boundaries of the fluid domain we can also specify stress (force per unit area) or force. For the case of ideal fluid, the stress is due only to the pressure. At interface of water and air, we impose that there is no pressure jump. This is a Dynamic Boundary Condition, since we prescribed force or stress (pressure) at the boundary.
The pressure in the air side of the interface is the atmospheric pressure #tex2html_wrap_inline1955#, and it can be assumed constant and independent of the water surface position. At the fluid side of the interface, the pressure is given by the Bernoulli equation, and chosing the integration constant #tex2html_wrap_inline1957# suitably, the dynamic boundary condition to be satisfied at the water surface is
Example: Body submerged in a body of water.
<#126#>figure<#126#>
#figure127#
In the absence of dynamic boundary
conditions, the boundary value problem for the potential function is linear.
If #math21##tex2html_wrap_inline1959# are harmonic functions (they satisfy Laplace's equation, i.e. #math22##tex2html_wrap_inline1961#), then #math23##tex2html_wrap_inline1963#, where #math24##tex2html_wrap_inline1965# are constants, are also harmonic, and is the solution for the boundary value problem provided the boundary conditions (kinematic boundary condition) are satisfied, i.e.
The key is to combine known solution of the Laplace equation in
such a way as to satisfy the K.B.C. (kinematic boundary condition).
The same is true for the stream function #tex2html_wrap_inline1967#. K.B.C.s
specify the value of #tex2html_wrap_inline1969# on the boundaries.
- Potential function #tex2html_wrap_inline1971#.
<#156#>figure<#156#>
#figure157#
- Stream function #tex2html_wrap_inline1973#.
<#160#>figure<#160#>
#figure161#
#math27##tex2html_wrap_inline1975# is a unit-source flow located at #math28##tex2html_wrap_inline1977#.
#alignat169#
A more general potential function can be written as the linear superposition of the unit-sources, as follows:
where the value of the constants #tex2html_wrap_inline1979# are found such that:
Caution: #math31##tex2html_wrap_inline1981# must be regular for #tex2html_wrap_inline1983# (fluid domain), so it is required that #math32##tex2html_wrap_inline1985#.
<#202#>figure<#202#>
Figure:
Note: #math33##tex2html_wrap_inline1987# are not in the fluid domain #tex2html_wrap_inline1989#.
| #figure203# |
#alignat211#
Potential function for 2D source of strength m at r = 0:
It satisfies #math36##tex2html_wrap_inline1991# (check Laplace's equation in polar coordinate in the keyword search utility), except at #math37##tex2html_wrap_inline1993# (so must exclude r = 0 from flow)
- <#3254#>1.<#3254#>
- Question: Derive the equations for the velocity field for the 2D source.
- <#3249#>(a)<#3249#>
- Hint: expression for the gradient in polar coordinates (use the keyword utility: coordinate system - velocity vector)
- <#3255#>2.<#3255#>
- Question: Evaluate the outward volume flux.
- <#3251#>(a)<#3251#>
- Hint: Consider a contour which contains the 2D source.
- <#3252#>(b)<#3252#>
- Hint: Use Gauss theorem to deform the contour into a small circle of radius #math38##tex2html_wrap_inline1995# around the source.
- <#3253#>(c)<#3253#>
- Hint: Evaluate the flux in the direction normal to the circle (radial velocity) and integrate along the circle.
If #math39##tex2html_wrap_inline1997# sink. Source with strength #tex2html_wrap_inline1999# located at #math40##tex2html_wrap_inline2001#:
#math42#
#displaymath1839#
<#257#>figure<#257#>
#figure258#
3D source - Spherical coordinates
A spherically symmetric solution: #math43##tex2html_wrap_inline2003#
(verify #math44##tex2html_wrap_inline2005# except at #tex2html_wrap_inline2007#)
Define 3D source of strength #tex2html_wrap_inline2009# located at #tex2html_wrap_inline2011#:
- <#3268#>1.<#3268#>
- Question: Derive the equations for the velocity field for the 3D source.
- <#3267#>(a)<#3267#>
- Hint: Expression for the gradient in Spherical coordinates (use the keyword utility: coordinate system - velocity vector)
- <#3269#>2.<#3269#>
- Question: Evaluate the net outward volume flux.
<#279#>figure<#279#>
#figure280#
#math46#
#displaymath1840#
#math47#
#displaymath1841#
<#318#>figure<#318#>
#figure319#
The Net outward flux is
Another particular solution: #math49##tex2html_wrap_inline2021#
(verify #math50##tex2html_wrap_inline2023# except at #tex2html_wrap_inline2025#)
Potential function for a point vortex of circulation
#tex2html_wrap_inline2027# at #tex2html_wrap_inline2029#:
Stream function:
- <#3290#>1.<#3290#>
- Question: Evaluate the radial velocity #tex2html_wrap_inline2031#, the tangential velocity #math53##tex2html_wrap_inline2033# and the vorticity #math54##tex2html_wrap_inline2035#.
- <#3284#>(a)<#3284#>
- Hint: velocity vector in polar coordinates (use the keyword utility: coordinate system - velocity vector).
- <#3285#>(b)<#3285#>
- Hint: write the vorticity #math55##tex2html_wrap_inline2037# in polar coordinates.
- <#3291#>2.<#3291#>
- Question: Evaluate the circulation #tex2html_wrap_inline2039# along an arbitrary closed contour containing the 2D vortex.
- <#3287#>(a)<#3287#>
- Hint: deform the closed contour to a circle containing the 2D vortex at the origin.
- <#3288#>(b)<#3288#>
- Hint: evaluate the velocity tangent to the circle.
- <#3289#>(c)<#3289#>
- Hint: integrate the tangent velocity along the circle.
Question: Evaluate the radial velocity #tex2html_wrap_inline2031#, the tangential velocity #math56##tex2html_wrap_inline2033# and the vorticity #math57##tex2html_wrap_inline2035#.
- <#3296#>1.<#3296#>
- Hint: velocity vector in polar coordinates (use the keyword utility: coordinate system - velocity vector).
- <#3297#>2.<#3297#>
- Hint: write the vorticity #math58##tex2html_wrap_inline2037# in polar coordinates.
Question: Evaluate the circulation #tex2html_wrap_inline2039# along an arbitrary closed contour containing the 2D vortex.
- <#3299#>1.<#3299#>
- Hint: deform the closed contour to a circle containing the 2D vortex at the origin.
- <#3300#>2.<#3300#>
- Hint: evaluate the velocity tangent to the circle.
- <#3301#>3.<#3301#>
- Hint: integrate the tangent velocity along the circle.
<#402#>figure<#402#>
#figure403#
A Dipole is a superposition of a
sink and a source with the same strength.
<#411#>figure<#411#>
#figure412#
2D dipole with orientation angle #tex2html_wrap_inline2061#.
#math63#
#displaymath1842#
<#424#>figure<#424#>
#figure425#
3D dipole at the origin oriented in the #tex2html_wrap_inline2063# direction.
- <#3324#>1.<#3324#>
- Question: derive the expression for a 2D dipole with orientation angle #tex2html_wrap_inline2065# from the superposition of a source and a sink.
- <#3316#>(a)<#3316#>
- Hint: write the superposition of a 2D source and a sink with strength #tex2html_wrap_inline2067# a distance #tex2html_wrap_inline2069# from each other.
- <#3317#>(b)<#3317#>
- Hint: take the limit #math65##tex2html_wrap_inline2071#.
- <#3325#>2.<#3325#>
- Question: derive the expression for a 3D dipole at the origin with orientation in the #tex2html_wrap_inline2073# direction.
- <#3322#>(a)<#3322#>
- Hint: write the superposition of a 3D source and a sink with strength #tex2html_wrap_inline2075# a distance #tex2html_wrap_inline2077# from each other.
- <#3323#>(b)<#3323#>
- Hint: take the limit #math66##tex2html_wrap_inline2079#.
We derive the expression for a 2D dipole from the superposition of a source and a sink with the same strength #tex2html_wrap_inline2083# and located a distance #tex2html_wrap_inline2085# appart from each other. The superposition of the source and sink follows:
We take the limit #math68##tex2html_wrap_inline2087#. To preserve a finite effect from the two singularities as they are brought togheter, their strength #tex2html_wrap_inline2089# must increase at the same time, for otherwise they will cancel out in the limit when they coincide. Thus it is necessary to make the product #tex2html_wrap_inline2091# a constant, with the result
#align453#
NOTE: dipole #math69##tex2html_wrap_inline2093#
(unit source)
We proceed the same way we did for the 2D dipole. We first write the superposition of a sorce and sink with the same strength #tex2html_wrap_inline2097# and a distance #tex2html_wrap_inline2099# appart from each other. Then, we take the limit #math70##tex2html_wrap_inline2101#. In this limit process we keep the #tex2html_wrap_inline2103# constant, otherwise the the source and sink cancel out in the limit.
#align473#
3D dipole (doublet) of moment #tex2html_wrap_inline2105# at the origin oriented in the
#tex2html_wrap_inline2107# direction.
It is the superposition of a uniform
stream of constant speed #tex2html_wrap_inline2109# and a source of strength
#tex2html_wrap_inline2111#.
<#505#>figure<#505#>
#figure506#
2D rankine half-body:
<#515#>figure<#515#>
#figure516#
3D Rankine half-body:
<#524#>figure<#524#>
#figure525#
- <#3365#>1.<#3365#>
- Question: find the position #math73##tex2html_wrap_inline2113# of the flow field stagnation point for the 2D Rankine half-body.
- <#3351#>(a)<#3351#>
- Hint: at the stagnation point #math74##tex2html_wrap_inline2115#.
- <#3352#>(b)<#3352#>
- Hint: obtain the expression for #tex2html_wrap_inline2117# and set it equal to #tex2html_wrap_inline2119#, and solve for #tex2html_wrap_inline2121# and #tex2html_wrap_inline2123#.
- <#3366#>2.<#3366#>
- Question: evaluate the width #tex2html_wrap_inline2125# of the 2D rankine half-body.
- <#3354#>(a)<#3354#>
- Hint: mass conservation
- <#3367#>3.<#3367#>
- Question: find the position #math75##tex2html_wrap_inline2127# of the flow field stagnation point for the 3D Rankine half-body.
- <#3361#>(a)<#3361#>
- Hint: at the stagnation point #math76##tex2html_wrap_inline2129#.
- <#3362#>(b)<#3362#>
- Hint: obtain the expression for #tex2html_wrap_inline2131# and set it equal to #tex2html_wrap_inline2133#, and solve for #tex2html_wrap_inline2135# and #tex2html_wrap_inline2137#.
- <#3368#>4.<#3368#>
- Question: evaluate the cross-sectional area #tex2html_wrap_inline2139# of the 3D rankine half-body.
- <#3364#>(a)<#3364#>
- Hint: mass conservation
From the expression for the potential, it is obvous that #tex2html_wrap_inline2143#. So, we need to find #tex2html_wrap_inline2145#. First, we obtain the #tex2html_wrap_inline2147# component of the velocity vector.
#math78#
#displaymath1843#
Now we set #tex2html_wrap_inline2149# and solve for #tex2html_wrap_inline2151#. We already know that #tex2html_wrap_inline2153#, so we set #tex2html_wrap_inline2155# in the expression for #tex2html_wrap_inline2157# to obtain #tex2html_wrap_inline2159#.
Therefore,
For large x, #tex2html_wrap_inline2163#, and #tex2html_wrap_inline2165# by continuity,
#math81##tex2html_wrap_inline2167#.
From the expression for the potential, it is obvous that #tex2html_wrap_inline2171# and #tex2html_wrap_inline2173#. So, we need to find #tex2html_wrap_inline2175#. First, we obtain the #tex2html_wrap_inline2177# component of the velocity vector.
#math83#
#displaymath1844#
Now we set #tex2html_wrap_inline2179# and solve for #tex2html_wrap_inline2181#. We already know that #math84##tex2html_wrap_inline2183#, so we set #tex2html_wrap_inline2185# in the expression for #tex2html_wrap_inline2187# to obtain #tex2html_wrap_inline2189#.
Therefore,
For large x, #tex2html_wrap_inline2193#, and #tex2html_wrap_inline2195# by continuity,
#math87##tex2html_wrap_inline2197#.
<#608#>figure<#608#>
#figure609#
To have a closed body, a necessary condition is to have
#math88##tex2html_wrap_inline2199#
2D Rankine ovoid:
#math89#
#displaymath1845#
3D Rankine ovoid:
#math90#
#displaymath1846#
- <#3415#>1.<#3415#>
- Question: Find the stagnation points for the 3D Rankine ovoid.
- <#3410#>(a)<#3410#>
- Hint: derive the velocity vector #math91##tex2html_wrap_inline2201#.
- <#3411#>(b)<#3411#>
- Hint: solve #math92##tex2html_wrap_inline2203# for #tex2html_wrap_inline2205#.
- <#3416#>2.<#3416#>
- Question: Find the radius #tex2html_wrap_inline2207# of the body.
- <#3414#>(a)<#3414#>
- mass flux at the half body equals #tex2html_wrap_inline2209#.
For the 3D Rankine Ovoid,
#align635#
and to obtain #tex2html_wrap_inline2211# we nee to solve the last equation above for #tex2html_wrap_inline2213#.
At #tex2html_wrap_inline2217#,
The radius of body R#tex2html_wrap_inline2219# is solution of
<#679#>figure<#679#>
#figure680#
2D:
#align684#
So #tex2html_wrap_inline2221# on #math95##tex2html_wrap_inline2223# (which is the K.B.C. for a stationary
circle radius a) or choose #math96##tex2html_wrap_inline2225#.
Steady flow past a circle (U,a):
<#700#>figure<#700#>
#figure701#
<#734#>figure<#734#>
#figure735#
Illustration of the points where the flow reaches
maximum speed around the circle.
3D:
<#739#>figure<#739#>
#figure740#
#math98#
#displaymath1847#
Steady flow past a sphere (U, a):
#math99#
#displaymath1848#
<#789#>figure<#789#>
#figure790#
Potential function:
#math100#
#displaymath1849#
Stream function:
#math101#
#displaymath1850#
Velocity vector components (polar coordinates):
#math102#
#displaymath1851#
- <#3446#>1.<#3446#>
- interior corner flow -- stagnation point origin: #tex2html_wrap_inline2227# 1.
e.g. #tex2html_wrap_inline2229# = 1, #math103##tex2html_wrap_inline2231#
<#810#>figure<#810#>
#figure811#
<#814#>figure<#814#>
#figure815#
<#818#>figure<#818#>
#figure819#
<#822#>figure<#822#>
#figure823#
- <#3447#>2.<#3447#>
- Exterior corner flow, #math104##tex2html_wrap_inline2233# at origin: #math105##tex2html_wrap_inline2239#). #math106##tex2html_wrap_inline2241# only. Since we
need #math107##tex2html_wrap_inline2243#, we therefore require #math108##tex2html_wrap_inline2245# i.e. #math109##tex2html_wrap_inline2247# only.
e.g. #math110##tex2html_wrap_inline2249#
(#math111##tex2html_wrap_inline2255# infinite plate,
flow around a tip)
<#837#>figure<#837#>
#figure838#
#math112##tex2html_wrap_inline2257#
(#tex2html_wrap_inline2259# exterior corner)
<#845#>figure<#845#>
#figure846#
- Source near wall:
#math113#
#displaymath1852#
<#858#>figure<#858#>
#figure859#
- <#3454#>1.<#3454#>
- Question: verify that the boundary condition of no flux on the wall is satisfied.
- <#3452#>(a)<#3452#>
- Hint: no flux trough the wall #math114##tex2html_wrap_inline2261# on #tex2html_wrap_inline2263#.
- <#3453#>(b)<#3453#>
- Hint: evaluate #math115##tex2html_wrap_inline2265#.
- Vortex near a wall: (ground effect)
#math116#
#displaymath1853#
<#880#>figure<#880#>
#figure881#
- <#3461#>1.<#3461#>
- Question: verify that the boundary condition of no flux on the wall is satisfied.
- <#3459#>(a)<#3459#>
- Hint: no flux trough the wall #math117##tex2html_wrap_inline2267# on #tex2html_wrap_inline2269#.
- <#3460#>(b)<#3460#>
- Hint: evaluate #math118##tex2html_wrap_inline2271#.
- Circle near a wall: (radius a)
#math119#
#displaymath1854#
<#898#>figure<#898#>
#figure899#
This solution satisfies the boundary condition on the
wall (#math120##tex2html_wrap_inline2273#), and the degree it
satisfies the boundary condition of no flux through the circle
boundary increases as the ratio #tex2html_wrap_inline2275#, i.e., the velocity due
to the image dipole small on the real circle for #tex2html_wrap_inline2277#. For a
2D dipole, #math121##tex2html_wrap_inline2279#
- <#3484#>1.<#3484#>
- Question: verify the boundary condition of no flux at the wall.
- <#3470#>(a)<#3470#>
- Hint: evaluate the component of the velocity vector in the #tex2html_wrap_inline2281# direction.
- <#3471#>(b)<#3471#>
- Hint: set #tex2html_wrap_inline2283# in the expression for the velocity vector in the #tex2html_wrap_inline2285# direction.
- <#3485#>2.<#3485#>
- Question: At the circle boundary the no flux boundary condition should be satisfied. Check to which degree is the boundary condition of no flux at the circle boundary satisfied.
- <#3479#>(a)<#3479#>
- Hint: Obtain the radial velocity vector at the circle boundary.
- <#3480#>(b)<#3480#>
- Hint: Obtain the velocity vectors in cartesian coordinates.
- <#3481#>(c)<#3481#>
- Hint: Change to polar coordinates: #math122##tex2html_wrap_inline2287# and #math123##tex2html_wrap_inline2289#.
- <#3482#>(d)<#3482#>
- Hint: Obtain the radial velocity vector from the the cartesian components of the velocity vector: #math124##tex2html_wrap_inline2291#.
- <#3483#>(e)<#3483#>
- Hint: Plot the radial velocity #tex2html_wrap_inline2293# for #tex2html_wrap_inline2295# (circle radius) as a function of #tex2html_wrap_inline2297# with #tex2html_wrap_inline2299# as a parameter.
- More than one wall:
<#926#>figure<#926#>
#figure927#
- <#3491#>1.<#3491#>
- Question: Use the method of images to derive the potential function for a circle in a free-stream of velocity #tex2html_wrap_inline2301# in the #tex2html_wrap_inline2303# direction, but between two walls parrallel to each other and to the free stream (first example in the figure above).
- <#3489#>(a)<#3489#>
- Hint: For a single wall, we already know the potential function given by the method of images. Write the image of this potential function with respect to the second wall.
- <#3490#>(b)<#3490#>
- Hint: Now the no flux condition is not satisfied in the first wall. Use the method of image again to satisfy the no-flux boundary condition at the first wall. Them, repeat the process since the b.c. of no-flux is not satisfied at the second wall. Keep repeating the process, which means keep adding singularities which are further and further from the body. At the end we have an infinite series.
We verify if a 2D source and its mirror image with respect to a wall located at #tex2html_wrap_inline2305# satisfies the no flux boundary condition at the wall. First we need to evaluate the component of the velocity vector in the #tex2html_wrap_inline2307# direction.
Next, we set #tex2html_wrap_inline2309#, and
as expected.
We verify if a 2D vortex and its mirror image with respect to a wall located at #tex2html_wrap_inline2311# satisfies the no flux boundary condition at the wall. First we need to evaluate the component of the velocity vector in the #tex2html_wrap_inline2313# direction.
Next, we set #tex2html_wrap_inline2315#, and
as expected.
We check if a 2D dipole oriented in the #tex2html_wrap_inline2317# direction in a free stream of speed #tex2html_wrap_inline2319# plus its mirror image with respect to a wall at #tex2html_wrap_inline2321# satisfy the no flux boundary condition at the wall. First, we obtain the #math130##tex2html_wrap_inline2323#, given by the equation
Next, we set #tex2html_wrap_inline2325#, and
First, we obtain the #tex2html_wrap_inline2327# and #tex2html_wrap_inline2329# components of the velocity vector.
and
Second, we make the change of variables
#align1036#
We substitute the expressions for #tex2html_wrap_inline2331# and #tex2html_wrap_inline2333# in terms of #tex2html_wrap_inline2335# and #tex2html_wrap_inline2337# in the equations for #tex2html_wrap_inline2339# and #tex2html_wrap_inline2341#. Then, we set #tex2html_wrap_inline2343#, and we define #math135##tex2html_wrap_inline2345# and #tex2html_wrap_inline2347#. Now we write
and
Third, evaluate the radial velocity #tex2html_wrap_inline2349# at the circles surface. The radial velocity #tex2html_wrap_inline2351# can be written in terms of the velocities #tex2html_wrap_inline2353# and #tex2html_wrap_inline2355# according to the equation
#align1074#
With this equation and the equations for #tex2html_wrap_inline2357# and #tex2html_wrap_inline2359# at the surface of the circle, we have
.
<#1118#>figure<#1118#>
#figure1119#
Reference system O: #math139##tex2html_wrap_inline2361#
<#1126#>figure<#1126#>
#figure1127#
#align1132#
Reference system O': #math140##tex2html_wrap_inline2367#
<#1151#>figure<#1151#>
#figure1152#
#align1157#
Galilean transform:
#align1171#
Pressure (no gravity)
#alignat1178#
In O: unsteady flow
#align1196#
In O`: steady flow
#align1221#
<#1234#>figure<#1234#>
#figure1235#
Total fluid force for
#math142#