The fluid is inviscid, soand the last term of the Navier – Stokes equation disappears.
The fluid is incompressible. This means we can divide each term of the Navier – Stokes equation byThe first term on the right becomes
The flow is steady, so
With these assumptions, the Navier – Stokes equation becomes
We now use the identityto obtainIf the force fieldis conservative we can writehence(1).
We can integrate this equation along a streamline. Suppose that s is a parameter for the streamline, thenis a tangent vector to the streamline. Take the dot product ofwith both sides of (1) to give
(2).
is perpendicular to bothandbutis parallel toso When we integrate (2) along a streamline the term on the right hand side returns 0. Since the streamline is arbitrary we havealong a streamline.
]]>We apply Bernoulli's equation along a streamline in the free surface of the liquid in a flat bottomed channel, givingwhereis the atmospheric pressure, assumed constant. Since the density is also constant we can subtractfrom each side, givingDividing bygiveswhereis called the specific energy.
Ifis constant thensoor
The last equation is a cubic in h, so may have 3 distinct real roots or one. If there are three roots for this particular equation, one of the roots is negative, so physically meaningless, so for givenandthere are either one or two possible values ofThere are three possible cases.
Iftwo flows are possible at different depthsandcorresponding to speedsandare possible.and sinceThe first is described as shallow and fast or supercritical, the second as deep and slow or subcritical.
Ifthe flow is unique. The liquid flows with depthand speedandare the critical speed and critical depth respectively.
Ifthen no flow is possible.
All the cases are illustrated below.
Since the critical depthoccurs at a minimum of the specific energy function, we can differentiateand equate to zero to find
The minimum value of the specific energy is then
]]>for water of depth (1)
at (2)
at (3)
whereis the velocity potential.
(1) can be solved by separation of variables technique. Assume(there will also be an arbitrary factor ofwhich we deal with later). (1) becomessince the left hand side is a function ofonly and the right hand side is a function ofonly, so both sides are equal to the same constant
Ifthe solution will be exponential, tending toasand the same problem occurs ifsoto givewhere
This equation has solutions
The corresponding equation foriswhich has solutionshence
atso
hence
In the same way we can solve (2) by the separation of variables method to findAssume (ignoring the factor)to giveas before, and as beforeelseasso putto give
Now writeandand sincewe haveso by picking a suitable point on the wave surface and a suitable time we have
]]>for water of depth (1)
at (2)
at (3)
(1) can be solved by separation of variables technique. Assume(there will also be an arbitrary factor ofwhich we deal with later). (1) becomessince the left hand side is a function ofonly and the right hand side is a function ofonly, so both sides are equal to the same constant
Ifthe solution will be exponential, tending toasand the same problem occurs ifsoto givewhere
This equation has solutions
The corresponding equation foriswhich has solutionshence
atsoand
In the same way we can solve (2) by the separation of variables method to findAssume (ignoring the factor)to giveas before, and as beforeelseasso putto give
Now writeandand sincewe haveso by picking a suitable point on the wave surface and a suitable time we have
]]>If the flow is irrotational we can define a velocity potentialsatisfyingIf the fluid is incompressible thenfor waves moving in thedirection withlabelling the depth of water, so Laplace's equation is satisfied.
If we assume that particles initially in the surface stay in the surface as the wave progresses, then the motion of a particle in the surface indicates the motion of the water surface. The vertical component of the velocity isAt the ocean bottom –for finite depth or for infinite depth - this must be zero, because water cannot move perpendicular to the surface.
A third equation can be derived by further considering the velocity potential. We know already that the velocity potential is unique only to within an additive constant, but is here time dependent , the constant here is actually a function of time and Bernoulli's equation becomes
Atwe can set(atmospheric pressure) and atThus the constant isand we may write Bernoulli's equation as
Subsequently the pressure at any time t the pressure at the surface isso
We can ignore the second order termand setto get
Differentiating this equation with respect to t gives
Now hence
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If the body force per unit volume isapplying Newton's Second Law to a cube of unit volume to obtain
We can also writewhereis the pressure in the fluid.
Equating these gives(1)
Consider the motion of a particle in the fluid from positionto position
Let the velocity atbe
The velocity atwill be
Then
Substitute this into (1) to obtain
We can define a vector fieldcalled the vorticity as
Use the identity
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