Lecture (part 1)
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.
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
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
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
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).
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
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.
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)
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,
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
d phi = 0 = d,x phi dx + d,y phi dy = u dx + v dy,
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.
Example 1: Consider problem 4.57. Determine the
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
Example 2: Find the stream function for the preceding
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.
Lecture (part 2):
1. Derivation of the continuity equation for cylindrical
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
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
(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.
Example 3: Consider problem 4.13. A two dimensional
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.