Lecture (part 1), Example problems

Example problem 1.
You may have wondered where the velocity profile
for laminar flow between two parallel plates came from in
homework problem 4.40. It was given as
u = [4 u_max y (h - y)]/h^2
where u is the velocity in the x direction, y = 0 starts at the
bottom plate, and y = h at the top plate. The transverse
velocity components v and w are zero, and the flow is
steady. This velocity profile comes from the Navier-Stokes
equation for x momentum
d,t u + u d,x u + v d,y u + w d,z u = - (d,x p)/rho + g_x +
nu del^2 u
where d,t etc. indicate partial differentiation.
The first term on the left is zero by the steady state
assumption. The second is zero because u does not change
with x. The third is zero because v is zero. The fourth is
zero because w = 0.
The streamwise pressure gradient is constant and
negative, so d,x p = - K. Gravity has no horizontal term, so
g_x = 0. The last term is nu [ d,x^2 u + d,y^2 u + d,z^2 u ]
= nu [d^2 u / dy^2 ], since u depends only on y. Thus
d^2 u / dy^2 = K/mu
is the ordinary differential equation that must be integrated.
The first integral is du/dy = - Ky/mu + C, where C is a
constant. The second integral gives u = - Ky^2/mu + Cy +
D, where D is a constant. By the no slip condition u = 0 at y
= 0 and y = h. From the first of these boundary conditions,
D = 0. By the second C = Kh/2mu . Thus;
u = [K/2mu] y [h - y] .
Now u = u_max at y = h/2, so [K/2mu] = - 4 u_max / h^2.
Substituting for [K/2mu] gives
u = [4 u_max y (h - y)]/h^2
which was the given parabolic velocity profile of the
homework problem.

Example problem.2.
Where did the expression eps = nu (du/dy)^2 come
from that was substituted into the energy equation in the
solution of 4.40 given in the last lecture?
Solution: The viscous dissipation rate per unit mass
eps is given by [tau dot dot e ]/ rho, where the viscous stress
tensor tau = 2 mu e . The rate of strain tensor e has
components eij = [d,xj ui + d,xi ui ]/2, but only e12 = e21 =
du/dy / 2 are not zero. Thus eps = 2 mu e^2 = 2 mu
[(du/dy)^2 /2 + (du/dy)^2 /2 ] = nu (du/dy)^2 , as expected.

More example problems.
Problems 4.56, 65, 69, 86.

4.56. The velocity potential phi = Kxy, K = constant gives
a set of hyperbolas symmetric about the x and y axes. The
orthogonal streamlines represent a stagnation point flow
against a plate tilted at an angle of 45 degrees up from the x
axis about the origin.

4.65. We calculated the z vorticity for this flow last time and
found it was zero. We also calculated the velocity potential
phi to be K theta.

4.69. Find the stream function and plot some streamlines
for the combination of a counterclockwise line vortex K at
(x,y) = (a,0) and an equal line vortex placed at (-a,0).
Solution: Because the equations of motion are linear,
we can add the individual stream functions for the two
vortices. The first is psi_1 = - K ln[(x-a)^2 + y^2]^1/2 ,
and the second is psi_2 = - K ln[(x+a)^2 + y^2]^1/2 .
Plotting this gives a "cat's-eyes" pattern.

4.86. SAE 10 oil has a parabolic velocity profile in laminar
flow between the plates, with average velocity = 2 u_max / 3
as shown in example 4.11, p 238 of the text. Thus v_ave =
K h^2 / 3 nu , where K is the pressure gradient and h = 4
mm is half the plate separation. For the oil rho = 870
kg/m^3, and mu = 0.104 kg/m s. From the manometer
reading the pressure difference for a meter distance
downstream is 7463 Pa. This is K. Solve for v_ave =
0.383 m/s. The flow rate Q = v_ave A = 0.383 x 0.008 =
0.00306 m^3/s m.

Lecture (part 2).

1. Irrotational (ideal) flows.
* Superposition of ideal flows.
**uniform flow
**counterclockwise vortex
2. Viscous flows.
3. Dimensional analysis.

Superposition of ideal flows.
Recall that the assumptions for the existence of ideal
flows are that 1. the flow is steady, 2. the fluid is constant
density, 3. the flow is two-dimensional, and 4. the flow is
irrotational. Assumptions 1-3 establish that a stream function
psi will exist that satisfies the continuity equation del dot v = 0.
Assumption 4 guarantees that the velocity can be expressed
as the gradient of a velocity potential phi. Because the
density is constant and the flow is steady, phi must satisfy
LaPlace's equation (substitute d,x phi and d,y phi in del dot v = 0).
del^2 phi = 0.
Because the flow is irrotational, the streamfunction psi also
satisfies LaPlace's equation (substitute d,y psi and - d,x psi
in w_z = d,x v - d,y u).
del^2 psi = 0.
Because LaPlace's equation is linear, any number of
solutions can be added to produce another solution.
If psi_1 and psi_2 are solutions of del^2 psi = 0, then
psi_3 = psi_1 + psi_2 is a solution, by subsitution. This is
the principle of superposition. Quite sophisticated ideal
flows can be constructed by combinations of the stream
functions and potential functions of simple ideal flows.
Many fluid mechanics books have been written devoted to
this process. Chapter 8 of White shows some of the most
useful applications and reviews some of the elementary
flows. Some are:
** Uniform stream iU; psi = Uy, phi = Ux
Check the given values of psi and phi by differentiation
to give U = d,x phi = d,y psi = u; 0 = d,y phi = - d,x psi = v.

** Line source or sink; psi = m @ , phi = m ln r
Consider a vertical source of length b emitting
Q vol/time. The velocity in the r direction v_r is Q/2 pi r b
= m/r, where m = Q/2 pi b, and v_@ = 0. From the
definitions of velocity components v_r and v_@ in terms
of derivatives of psi and phi, we can derive the given
expressions for phi and psi. Set v_r = m/r = d,@ psi / r.
Integrating gives psi = m@ + f(r). v_@ = -d,r psi = 0. Thus
psi = g(@) only, so f(r) is a constant.
In terms of the velocity potential phi, v_r = d,r phi =
m/r, so phi = m ln r + f(@). Substitute in v_@ = d,@ phi / r ,
giving 0 = d,@ f(@) = df/d@, an ordinary derivative, so f = const.

** Line vortex; psi = -K ln r , phi = K @ .
We proved this using u = -Kx/(x^2 + y^2)
and v = Ky/(x^2 + y^2). We can also prove it using
the Stokes theorem that the area integral of curl v equals
the line integral of v dot dl around the area in the counter-
clockwise direction. The line integral is defined as the
"circulation" gamma, so gamma / area equals the area
averaged vorticity. Note that all the vorticity is concentrated
at the origin, since the vortex flow is irrotational everywhere
else. The velocity v_@ = d,@ phi / r = K/r. Therefore
gamma = 2 pi r v_@ = 2 pi K
giving the physical interpretation of K as the circulation
of the vortex over 2 pi, just as the physical interpretation
of m for a source is Q/ 2 pi b.
In terms of psi, v_@ = - d,r psi . Thus,
d,r psi = - K/r , so psi = -K ln r + f(@). Substitute this
in v_r = d,@ psi / r to give df/d@ = 0, so f is a constant.

You might want to see some of these flows by observing
particle motions, streamlines and potential lines for various
combinations in the "Fluid Mechanics" computer program
provided with your text. You can change the parameters as
you like, and observe the effects.
The combination of a sink plus a vortex at the origin
thus has psi = -m @ - K ln r , and phi = -m ln r + K @ . The
flow is a good representation of the external flow induced by
a tornado.
Flow past a cylinder is represented by psi = U sin @
[r - R^2 / r ] as shown in Fig. E 4.8 p 222 of the text. This
is a uniform flow plus a doublet. A doublet is formed by
allowing two vortex lines of opposite sign to merge. If we
add a vortex we get lift, and this potential flow begins to
approximate the influence of a wing.
A uniform flow plus a source is described by psi =
Ur sin @ + m @ . This gives the Rankine half body flow of
Fig. 4.15, p 233. Adding a sink further down stream would
close the body. Merging the source and sink produces a
doublet. Lots of such combinations have been studied, as
you can see by looking ahead at Chapter 8.

Viscous flows.
Previously we discussed the laminar flow between
two fixed parallel plates separated by 2h with a pressure
gradient, and found a parabolic velocity profile. What
happens if the plates move? What happens if the sign of the
pressure gradient changes, or is zero?
The flow between long concentric cylinders with
various rotation rates is called a Couette flow, and can be
quite complex as the relative velocity of rotating cylinders
increases to for so-called Taylor vortices. The appropriate
coordinate system is cylindrical, and the conservation
equations are given in Appendix E, p 725. The steady
Couette flow with rotating cylinders is described by the theta
momentum equation
del^2 v_@ = v_@ / r^2 = r^-1 d,r [ r d,r v_@ ] .
This is solved by assuming the velocity has the form v_@ =
r^m and substituting, giving m = +,- 1. The solution is of
the form v_@ = C_1 r + C_2 r^-1. The constants are
determined by the boundary conditions at r = r_inner and

Dimensional analysis and similarity (Chapter 5)
The basis of dimensional analysis is the principle of
dimensional homogeneity for all physical equations. Many
problems of engineering interest are too complex to be
solved analytically, so experiments must be done to test
regularities in the relationships between possible
combinations of dimensional quantities. The Buckingham Pi
theorem is used to organize dimensional variables into the
minimum number of dimensionless ("Pi") groups. The theorem
states that if we have n relevant dimensional parameters that form
a dimensional matrix of rank r, then the number of dimensionless
"Pi" groups is only n-r. Because r is usually equal to the number
of fundamental dimensions (eg, 3 for MLt) this considerably reduces
the experimental effort needed to characterize the problem.