The first exam is after the break today.
It will be open book, closed notes, except for one page
(both sides) of handwritten material that you may want to
prepare (no Xeroxes of anything please). The exam will
cover Chapters 1, 2 and 3 of White, plus the lectures.


1. Integral Relations for Control Volumes (Chapter 3 of
The rate-of-strain and rotation tensors.
The deformation rate and angular velocity of a fluid
Relationship of vorticity to angular velocity.
2. Differential relations for a fluid particle (Chapter 4 of
The conservation of vorticity equation.
3. Midterm review.

The velocity near a fluid particle.
In the derivation of the momentum and energy
equations of fluid systems and control volumes within
fluids we have frequently encountered the velocity gradient
tensor and the related rate-of-strain tensor. In order to
understand the physical significance of these quantities it is
useful to consider the velocity field in the neighborhood of
a fluid particle. By a fluid particle we mean a small system
control volume that is spherical at time t. What happens to
the system control volume (fluid particle) as a function of
time? What about the fluid in its vicinity? How big can
the system be and still be considered a fluid particle?
Clearly there is a point in the system control
volume where the velocity is zero, corresponding to the
center of mass. We choose this point as the origin of the
coordinate system. The velocity dv at a distance dx away
from the origin is given by the chain rule
dv_i = ( partial v_i / partial x_j ) dx_j ,
dvi = vi,j dxj
introducing the "comma" notation for partial
differentiation. Decompose the velocity dv into
components parallel and perpendicular to dx; that is, dv||
and dv#, respectively. Decompose the velocity gradient
tensor into symmetric and antisymmetric components e and
vi,j = (vi,j + vj,i)/2 + (vi,j - vj,i)/2 = eij + Oij
eij = (vi,j + vj,i)/2 = eji
Oij = (vi,j - vj,i)/2 = - Oji
dv = (e + O) dot dx = dv|| + dv# .
Now take the dot product of dx with dv , making these
substitutions for dv
dx dot dv = dx dot dv|| + dx dot dv# = dx dot (e +
O) dot dx
dx dot dv# = 0
because the vectors are perpendicular, and
dx dot O dot dx = 0
because O is antisymmetric (expand out, reverse dummy
indices, use definition of antisymmetric tensor to show that
dx dot O dot dx = - dx dot O dot dx ; therefore zero).
dx dot dv|| = dx dot e dot dx
dv|| = e dot dx.
But since
dv = dv|| + dv# = e dot dx + O dot dx
we see that
dv# = O dot dx
by subtracting the expression for dv|| from dv.
However, the perpendicular velocity dv# is related
to the angular velocity W by the expression
dv# = W cross dx = O dot dx.
where the i componet of dv# is
dv#i = eijk Wj dxk
using the alternating tensor eijk representation of the cross
product C = AxB
Ci = eijk Aj Bk : eijk = 1 for ijk = 123, 231, 312 ;
eijk = -1 for ijk = 132, 321, 213 ; eijk = 0 for ijk = 111,
112, 113, 221, 222, 223, 331, 332 and 333.
This notation is identical to using the determinant definition
of the cross product components. An important identity
we need is
eijk eklm = dil djm - dim djl
where d is the identity tensor with components
dij = dji = 1 for ij = 11, 22, 33 and dij = 0 for ij =
12, 13, 23.
Now, compute the quantity w cross dx for comparison
with W cross dx = O dot dx, where w is the vorticity
vector defined as del cross v. Thus,
(w cross dx)i = eijk wj dxk = eijk ejlm vm,l dxk =
ekij ejlm vm,l dxk = (dkl dim - dkm dil) vm,l dxk = (vi,k -
vi,k) dxk = 2 Oij dxj = 2 (W cross dx)i .
But if A cross B = C and D cross B = C it follows that A =
D. Therefore,
w = 2 W ;
that is, the vorticity vector is twice the angular velocity
vector of any fluid particle. This is an important physical
interpretation of the vorticity field.

Example: How large can the fluid particle be before the
flow is unstable?
Solution: The stability of a flow is determined by
the ratio of the inertial vortex force term v cross w to the
viscous force nu del^2 v, known as the Reynolds number
Re, where Re = u L / nu, u is a characteristic velocity
scale, L is the length scale and nu is the kinematic
viscosity. If Re > Re_crit the flow is unstable and may
become turbulent. In the expression Re = u L / nu
u = e x
L = x
where e = (e:e)^1/2 is the magnitude, or rms value, of the
rate of strain tensor; so,
Re = e x^2 / nu = Re_crit
where x = x_crit, the largest size fluid particle with stable
flow. Thus,
x_crit = (nu/e)^1/2 (Re_crit)^1/2.
The rms rate of strain e is related to the viscous dissipation
rate eps by the equation eps = 2 nu e:e, so e = (eps/2
nu)^1/2. Therefore the critical length x_crit is proportional
to (nu^3/eps)^1/4 = L_K, the Kolmogorov length scale,
first introduced by A. N. Kolmogorov in a classic paper of
turbulence theory in 1941. Velocities on scales smaller
than L_K are stabilized by viscous forces. Those on larger
scales are unstable, and can break up into a growing
cascade of turbulent eddies. A universal critical Reynolds
number Re_crit of about 25-100 has been determined

Conservation of vorticity equation.
Recall the conservation of momentum equation for
a constant density and viscosity fluid in the form
partial v / partial t = - grad B + v cross w + nu
del^2 v
where B is the Bernoulli group. Taking curl = del cross of
this equation gives on the left
(curl v,t )i= eijk d,j vk,t = d,t eijk d,j vk = d,t eijk
vk,j = wi,t = partial w_i / partial t
which is the rate of change of the i component of vorticity.
The first term on the right, - curl grad B, is zero because
the curl of the gradient of any function is equal to zero.
This is proved by reversing the order of partial
differentiation, reversing two dummy indices, changing
the sign of eijk by reversing two indices, and noting that
the only quantity equal to its own negative is zero
(curl grad B)i = eijk d,j d,k B = eijk d,k d,j B =
eikj d,j d,k B = - eijk d,j d,k B = 0
where d,j is the partial derivative with respect to x_j.
The second term on the right curl (v cross w) has i
[curl (v cross w)]i = eijk d,j eklm vl wm = [dil djm
- dim djl] [ vl d,j wm + wm d,j vl ] = vi d,j wj + wj d,j vi -
vj d,j wi - wi d,j vj
where the divergence of the vorticity d,j wj is zero in all
cases (proved as an exercise below), and the divergence of
the velocity d,j vj is zero for constant density fluids (from
the continuity equation). Thus,
[curl (v cross w)]i = wj d,j vi - vj d,j wi
for incompressible fluids. The first term on the right can
be further simplified because the velocity gradient tensor
d,j vi = eij + Oij and wj Oij = 0 because the local rotation
cannot rotate itself (xj Oij was shown above to be [w cross
x /2]i, or v#i, which is zero for x in the direction of w).
[curl (v cross w)]i = wj eij - vj d,j wi
where eij is the rate of strain tensor and wj eij is the rate of
stretching of vortex lines. The final viscous term is
[curl nu del^2 v ]i = nu del^2 curl vi = nu del^2 wi
Combining all terms gives the vorticity equation for
constant density, constant viscosity Newtonian fluids
d,t wi + vj d,j wi = wj eij + nu del^2 wi
or in vector form
********** D w / Dt = w dot e + nu del^2 w ********.
We see that the vorticity of a fluid particle increases with
time due to vortex stretching and decreases due to viscous

Midterm review
The first midterm examination covers the material
of the first three chapters of White, plus the lectures. You
should be able to derive the continuity equation by either
the shrinking system control volume method or the
shrinking fixed finite volume element method, and the
momentum and energy equations by the shrinking system
control volume method. Exam questions are often taken
from those given at the end of the chapters, so read over all
the problems and think about how you would solve them
(or just solve them). Show all of your work. You must
include explanations in words of what you are doing as
well as writing down equations. Make it very clear that
you understand all aspects of your problem solutions. Be
sure to write down all units and check all equations for
dimensional consistency and reasonableness.

Example: Problem 2.42 describes a two fluid manometer
system. Small pressure differences between reservoirs of
fluid 1 are measured by the height difference h between
levels of fluid 2 in the manometer, where rho_2 is greater
than rho_1 and the levels of fluid 1 in the reservoirs are
Solution: The vertical momentum equation is dp/dz
= - rho g, or dp = - rho g dz. Integrating this equation
from reservoir B to reservoir A gives
p_A - p_B = rho_1 g (h_1 - h) + rho_2 g h+ rho_2
g h_3 - rho_2 g h_3 - rho_1 g h_1
p_A - p_B = (rho_2 - rho_1) g h.
Thus, if (rho_2 - rho_1) is very small, h will be very large
for a given p_A - p_B, so the manometer is very sensitive.

Example: Problem 3.149 has a cylindrical jet of alcohol
striking a vertical plate. A force of 425 N is required to
hold the plate stationary. Assuming there are no losses in
the nozzle, estimate (a) the mass flow rate of alcohol and
(b) the absolute pressure in the 5 cm diameter tube before
the 2 cm nozzle exit. The SG of alcohol is 0.79 and the
density of water is 998 kg/m^3.
Solution: Take a control volume enclosing the
impinging jet and the plate, and consider the conservation
of x momentum for the fixed control volume, where x is in
the direction of the jet. The Reynolds transport theorem
for the control volume is dPx /dt |system = dPx /dt |CV +
int [ rho v v dot dA ]. These terms are
dPx /dt |system = - F , the force on the plate.
dPx /dt |CV = 0 , for this steady state flow.
int [ rho v v dot dA ] = - rho v^2 A_jet
F = rho v^2 A_jet .
We know all the terms in this equation except the jet
velocity v. F = 425 N. rho = 0.79x998 kg/m^3. A_jet =
pi x (0.02)^2 / 4, to give v = 41.4 m/s. The mass flow
rate dm/dt = rho A v = 10.3 kg/s (ans. a).
We can find the mass flow in the tube from the
integral conservation of mass, and the pressure from
Bernoulli's equation. Consider a control volume enclosing
the tube and the jet. Reynolds transport theorem for mass
is dM/dt|system = dM/dt|CV + int [ rho v dot dA], where
the terms are
dM/dt|system = 0 by the conservation of mass.
dM/dt|CV = 0 for this steady state flow.
int [ rho v dot dA] = - rho v_1 A_1 + rho v_2 A_2,
where A_1 is the 5 cm diameter cross section and A_2 is 2
cm. Hence v_1 = 41.4 m/s (2/5)^2 = 6.63 m/s. There are
no viscous losses or viscous working terms in the nozzle,
so B_1 = B_2 along all streamlines
B_1 = p_1/rho + v_1^2 / 2 + g z = B_2 = p_2/rho
+ v_2^2 / 2 + g z
where z_1 = z_2 = z, p_2 = p_atm = 101,000 Pa, so p_1
= 760,000 Pa (ans. b).
Exercise: Show that div curl v = 0.
d,j wj = d,j ejkl d,k vl = ejkl d,j d,k vl
but we can reverse the order of the partial derivatives
d,j wj = ejkl d,k d,j vl
and because j and k are dummy indices
d,j wj = ekjl d,j d,k vl = - ejkl d,j d,k vl ,
so that
d,j wj = - d,j wj = 0. QED.