1. The conservation of vorticity equation (variable density,

incompressible, constant nu).

2. Irrotational flows, w = 0. The velocity potential phi (v =

grad phi ).

3. Conditions for the existence of a stream function psi

(steady, 2-D, incompressible).

*Relationship to streamlines.

*Relationship to volume flow rate.

*Relationship to velocity potential phi.

4. Examples 1 and 2 of ideal flows with both phi and psi.

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Conservation of vorticity.

============

Previously we found the conservation of vorticity

equation by assuming constant density flow of a constant

viscosity Newtonian fluid:

D,t w = w dot e + nu del^2 w.

Thus the vorticity (twice the angular velocity) of a fluid

particle increases with time due to stretching of the vortex

lines in the direction of the vorticity vector w and decreases

due to viscous diffusion of vorticity away from the strongest

vortex lines, with diffusivity nu = mu/rho. Diffusivities

have units length^2 / time: for example; for temperature the

diffusivity is alpha in the equation

D,t T = alpha del^2 T

where alpha is the thermal diffusivity k/ rho c_p, and k is the

thermal conductivity. Again, alpha has units length^2 / time.

Just as points of high temperature always cool off due to

thermal diffusion, points of strong vorticity that are not

stretching will always be slowing down due to viscous

torques, and points that have no vorticity can never get any

except by viscous diffusion from those that have. Particles

that acquire vorticity from walls develop inertial vortex

forces v cross w away from the walls. Particles with the

same strong vorticity try to spin around each other, and

those with opposite vorticity move perpendicular to the

direction between them. Vorticity tends to strongly

concentrate into a small volume fraction of a rotational flow

because of the vortex stretching term...the more vorticity

there is the faster the vorticity grows. Irrotational, constant

density flows tend to stay irrotational unless they get near

walls.

However, suppose the flow is incompressible, but

the density is not constant. This is like flows in the interior

of the atmosphere or ocean. In this case vorticity can be

produced in the interior of the fluid because density gradients

may be tilted. Assume (for simplicity) that the viscosity nu

is constant

D,t v = d,t v + ( v dot del ) v = - (del p) / rho + g +

nu del^2 v

where del is the gradient operator d,x, we get a new vorticity

equation by taking the curl of this momentum equation. The

first term (for the i component)

curl d,t vi = eijk d,j d,t vi = d,t eijk d,j vi = d,t wi

is just the partial of the i component of vorticity wi with

respect to time.

The curl of the second term on the left requires the

identity (proved below)

( v dot del ) v = grad (v^2 / 2) - v cross w

so

curl [ grad (v^2 / 2) - v cross w ]i = 0 - curl [v cross

w]i = - eijk d,j eklm vl wm = - [dil djm - dim djl] [ d,j (vl

wm)] = - [dil djm - dim djl] [ wm d,j (vl) + vl d,j (wm)] = -

[ wj d,j (vi) + vi d,j (wj)] + [ wi d,j (vj) + vj d,j (wi)] = - [

wj d,j (vi)] + [ vj d,j (wi)] = - wj d,j eij + vj d,j wi

We have dropped d,j vj by the incompressibility

assumption, and we previously proved d,j wj always equals

zero. The pressure term gives

- curl [ (grad p) rho ]i = - eijk d,j [ d,k p / rho ] = -

(1/ rho ) eijk d,j d,k p - eijk d,k p d,j (1/ rho ) = 0 - eijk d,j

(1/ rho ) d,k p = - [grad (1/ rho ) cross grad p]i = [grad rho

cross grad p]i / rho^2.

The viscous term is

[curl nu del^2 v ]i = nu eijk d,j del^2 vk = nu del^2

eijk d,j vk = nu del^2 wi.

Putting all the terms together gives

D,t w = w dot e + [grad rho cross grad p]/ rho^2 +

nu del^2 w

where we see that a new term [grad rho cross grad p]/ rho^2

appears. In a gravitational field the pressure gradient will be

nearly vertical and down, but the density gradient may not

since any disturbance will tilt the density surfaces. This

"baroclinic" term leads to vorticity, shear, and turbulence

formation on strong density gradient surfaces that are tilted

by internal wave fields in the ocean and atmosphere

interiors, so flows in these regions are not irrotational, and

the density gradients are smoothed by the turbulent mixing.

**********

Exercise: Prove (v dot del) v = grad (v^2 / 2) - v cross w.

Expand [ v cross w ]i = eijk vj eklm d,l vm = [dil

djm - dim djl] [vj d,l vm] = vj d,i vj - vj d,j vi = d,i (v^2 / 2)

- vj d,j vi . Rearranging gives (v dot del) v = grad (v^2 / 2)

- v cross w (QED).

*********

==============

Irrotational flows.

==============

If the vorticity of a flow is everywhere zero, the flow

is by definition irrotational. Because w = curl v = 0

everywhere, and because

w = curl grad phi = 0,

there must exist a function phi such that

v = grad phi.

Irrotational flows are therefore sometimes called "potential

flows". If the flow is also incompressible, so that

d,j vj = 0

then

d,j d,j phi = del^2 phi = 0

which is the LaPlace's equation, with many solutions in

physics. Thus, for every solved steady heat transfer

problem (for example) with del^2 T = 0, there exists a

corresponding potential flow solution.

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Conditions for the existence of a stream function.

=============

Stream functions permit the elimination of the

continuity equation. Generally, the existence of a stream

function psi requires that the flow be steady,

incompressible, and two dimensional (not necessarily

irrotational) so that only two terms remain in the continuity

equation. Substitution of the stream function velocities

satisfy the equation. Choosing x and y as the two directions

u = d,y psi ; v = - d,x psi

where psi (x,y) is the stream function.

Substituting u = d,y psi ; v = - d,x psi in the

incompressible continuity equation d,x u + d,y v = 0 gives

d,x d,y psi - d,y d,x psi = d,x d,y psi - d,x d,y

psi = 0

which satisfies the continuity equation as required.

The only possible component of vorticity is in the z

direction (be sure you see why), where

wz = d,x v - d,y u (this is proved below)

so

wz = - d,x d,x psi - d,y d,y psi = - del^2 psi .

Clearly, if the flow were irrotational then psi satisfies

LaPlace' equation and we have lots of solutions.

If the flow is not irrotational, we have a general

equation for psi from the vorticity equation in two

dimensions for steady flow. Recall that

d,t w + ( v dot del ) w = w dot e + nu del^2 w.

The first term on the left is zero because the flow is steady.

The first term on the right is zero because there is no

stretching of vortex lines in the z direction. For our steady,

incompressible, 2-D flow

u d,x wz + v d,y wz = nu del^2 wz

so (changing the sign of both sides, substituting for u, v,

wz)

d,y psi d,x (del^2 psi) - d,x psi d,y (del^2 psi)

= nu del^4 psi

which is the general criterion for the stream function psi

(x,y). This is a fourth order, linear equation that can be

rather easily solved by numerical methods given four

boundary conditions for psi. Known volocities on two

bounding surfaces give four boundary conditions on psi.

************

Proof: wz = d,x v - d,y u .

wz is equivalent to w3. w3 = e3ij d,i vj = e312 d,1

v2 + e321 d,2 v1 = d,1 v2 - d,2 v1 QED.

************

===========

Relationship to streamlines.

===========

The equation for a streamline in 2-D flow is dx/u =

dy/v. Substituting the expression for the velocity

components in terms of the stream function psi gives

udy - vdx = 0 = d,y psi dy + d,x psi dx = d psi

so psi is constant along a streamline.

============

Relationship to volume flow rate.

============

The volume flow rate dQ across an area element of

length ds in the x,y plane and width b in the z direction is

v dot dA = (i u + j v ) dot ds b = (i d,y psi - j d,x

psi ) dot (i d,s y - j d,s x) ds b = [ d,x psi dx + d,y psi

dy ] b = dpsi b

Therefore the flow rate per unit width dQ/b = dpsi.

When streamlines get close together, the flow rate across any

perpendicular line ds is large.

============

Relationship to velocity potential.

============

If a flow is not only incompressible, steady, and two

dimensional (so a stream function exists), but is also

irrotational (so a velocity potential exists), then

u = d,x phi = d,y psi ; v = d,y phi = - d,x psi

for the velocity components. The functions phi and psi are

orthogonal. This follows because a line of constant phi

satisfies

d phi = 0 = d,x phi dx + d,y phi dy = u dx + v dy,

so

dy/dx|phi = - u/v,

but the slope of a line of constant psi (a streamline) is

dy/dx|psi = v/u,

which is the orthogonal direction.

Therefore such flows have orthogonal nets of iso-phi

and iso-psi lines in the x,y plane.

***********

velocity potential phi (x,y) for the flow field (Example 1.10)

where u = Kx and v = -Ky, where K is a constant.

Solution: Set u = d,x phi = Kx, and v = d,y phi = -

Ky. Integrating the first equation gives phi = K x^2 / 2 +

f(y). Integrating the second equation gives phi = - K y^2 / 2

+ g(x). Thus, phi = K(x^2 - y^2)/2 . This is a set of

hyperbolas. The streamfunction forms another set of

orthogonal hyperbolas.

***********

************

example.

Solution: Set u = d,y psi = Kx, and v = - d,x psi

= -Ky. Integrating the first equation gives psi = Kxy +

f(x). Integrating the second equation gives psi = Kyx +

g(y). Therefore g(y) = f(x) = constant, so psi = Kxy +

const. This is a set of hyperbolas perpendicular to those for

constant phi. To find the direction of flow, calculate sample

velocities. For example, for point (x,y) = (1,1) the velocity

(u,v) = (K,-K), which has magnitude 1.414 K and direction

down and to the right.

************

1. Derivation of the continuity equation for cylindrical

coordinates.

2. The stream function psi for cylindrical coordinates.

3. Example 3.

============

Continuity equation for cylindrical coordinates.

============

Many interesting ideal flows occur in circumstances

where it is natural to use cylindrical polar coordinates

(r,@,z); for example, steady, incompressible, irrotational,

inviscid flow about a vortex line. Can a stream function be

found that will satisfy the continuity equation in cylindrical

coordinates? What is the continuity equation in cylindrical

coordinates? What are cylindrical coordinates?

Cylindrical coordinates (r,@,z) are related to

Cartesian coordinates by the following

x = r cos @ ; y = r sin @ ; z = z

where @ is the angle measured counterclockwise from the x

axis in the x,y plane to the r distance from the z axis.

We derive the continuity equation by considering a

finite element with sides dr, r d@ , and dz. The velocity

components normal to the r, @, and z faces of the element

are vr, v@, and vz. The continuity equation is derived by

application of the following mass budget for the volume dV

= dr r d@ dz over the time interval dt, starting from time t

accumulation + (outward flow rate - inward flow

rate)dt = 0 .

The accumulation of mass is

rho dV|t+dt - rho dV|t .

The outward mass flow rate is

rho vr r d@ dz |r+dr + rho v@ dr dz |@+d@ + rho

vz dr r d@ |z+dz .

The inward mass flow rate is

rho vr r d@ dz |r + rho v@ dr dz |@ + rho vz dr r

d@ |z .

Combining these gives

rho dV|t+dt - rho dV|t + [(rho vr r d@ dz |r+dr + rho

v@ dr dz |@+d@ + rho vz dr r d@ |z+dz) - (rho vr r d@ dz

|r + rho v@ dr dz |@ + rho vz dr r d@ |z )]dt .

Dividing by dV dt and taking the limit as dV and dt approach

zero gives (by the definitions of the various partial

derivatives)

d,t rho + (1/r) d,r ( rho vr r) + (1/r) d,@ ( rho v@) +

d,z ( rho vz) = 0

where rho is the density. This is the continuity equation in

cylindrical coordinates. Partial differentiation is indicated by

d,t , d,r , d,@ and d,z .

============

Stream function in cylindrical coordinates.

============

Suppose the flow is constant density, steady, and

two dimensional (the usual conditions for the existence of a

stream function). The continuity equation derived above

becomes

(1/r) d,r (vr r) + (1/r) d,@ (v@) = 0

dropping the unsteady term, canceling out rho, and dropping

the z term. Consider the possible form for the stream

function psi (r,@)

vr = (1/r) d,@ psi ; v@ = - d,r psi .

Substituting these definitions into the continuity equation

shows that it is satisfied.

**********

velocity field is given by u = - Ky/(x^2 + y^2) and v =

Kx/(x^2 + y^2). Does this field satisfy incompressible

continuity? Transform these velocities to polar components

vr and v@. What might the flow represent?

Solution: First check continuity; that is, whether d,x

u + d,y v = 0. It is satisfied because d,x u = 2Kxy/(x^2 +

y^2)^2 and d,y v = - 2Kyx/(x^2 + y^2)^2 . Now transform

to polar coordinates

vr = u d,r (r cos@) + v d,r (r sin@) = [- Ky/(x^2 +

y^2)] cos@ + [Kx/(x^2 + y^2)] sin@ = [- Ky/(x^2 + y^2)]

(x/r) + [Kx/(x^2 + y^2)] (y/r) = 0.

v@ = u (1/r)d,@ (r cos@) + v (1/r)d,@ (r sin@) =

[- Ky/(x^2 + y^2)] [(1/r)(- r sin@] + [Kx/(x^2 + y^2)]

[(1/r)(r cos@)] = [Ky/(x^2 + y^2)] [(y/r)] + [Kx/(x^2 +

y^2)] [(x/r)] = K/r.

This is the flow around a vortex line along the z axis.

Now substitute vr and v@ into the cylindrical polar

continuity equation (1/r) d,r (vr r) + (1/r) d,@ (v@) = 0.

The equation is satisfied.

**********

**********

Example: Does a potential function exist for the flow of problem

4.13? Solution: First compute the vorticity wz=d,x v - d,y u to

show w=0. Then integrate the equations from grad phi = (d,r phi,

d,@ phi / r)=(vr,v@) to give phi = K@ + const.

**********