1. Conservation of momentum equations, various forms.
a* From last time, momentum per unit volume and
forces per unit volume.
b* Same, in terms of total stress tensor.
c* Rearranged, to get substantive derivative of
d* Assuming constant rho and mu, to give Navier
Stokes equation.
e* Rearranged Navier Stokes, to reveal the Bernoulli
group B and the inertial-vortex force vxw.
2. Hydrostatics exercises.

Conservation of momentum equations.
Different problems in fluid mechanics require
different forms of the conservation of momentum equations,
and the different forms give useful physical interpretations
and insights. From the conservation of momentum principle
applied to a system control volume, we found that
(1) partial rho v / partial t + div ( rho v v ) + grad p - div
[ tau ] - rho g = 0
everywhere in a continuous fluid. Moving the last four
terms to the right hand side
a* partial rho v / partial t = - grad p - div ( rho v v )+
div [ tau ] + rho g .
The terms on the right are pressure, inertial, viscous, and
gravity forces per unit volume, and the term on the left is the
rate of change of momentum per unit volume. In terms of
the total stress tensor T,
b* partial rho v / partial t = div [ T ] + rho g.
where T = - p d - rho v v + tau is the sum of the pressure,
inertial and viscous stress tensors, respectively.
This shows why local velocities change. What about
an individual fluid particle? For this we need the substantive
derivative Dv/Dt. Go back to (1) and rearrange the two left

partial rho v / partial t = rho partial v / partial t + v
partial rho / partial t

div ( rho v v ) = rho (v dot grad) v + v div ( rho v ).

Adding these together gives:

rho partial v / partial t + v partial rho / partial t +
rho (v dot grad) v + v div ( rho v ) = rho Dv/Dt

from the continuity equation and the definition of Dv/Dt.
c* rho Dv/Dt = - grad p + div [ tau ] + rho g
which shows that the momentum per unit volume of a fluid
particle changes due to pressure, viscous and gravity forces
per unit volume.
The most popular form of the momentum equations
is the "Navier-Stokes" version, for a Newtonian fluid with
constant density rho and viscosity coefficient mu.
From c* and these assumptions, dividing by rho and
bringing it inside derivatives:

partial v /partial t + (v dot grad) v = - grad (p/ rho) +
g + div [ tau / rho ] .

However, div [ tau / rho ] = div [ 2 mu e / rho ] = nu del^2
v from the continuity equation for constant density div v = 0.
Please prove as an exercise. Thus,

d* partial v /partial t + (v dot grad) v = - grad (p/ rho) +
g + nu del^2 v

which are the Navier-Stokes equations...the most commonly
quoted form of the momentum equations in fluid mechanics.
Let us rearrange the NS equations by using the
vector identity (v dot grad) v = grad (v^2)/2 - vxw, where w
= curl v is the vorticity vector. We will prove this later, and
show that the vorticity is twice the angular velocity of a fluid
particle. Also, write the gravity vector as the gradient of the
gravitational potential -gz. Then

e* partial v /partial t = - grad B + vxw + nu del^2 v

where B is the Bernoulli group B = p/ rho + (v^2)/2 + gz,
and vxw and nu del^2 v are the inertial vortex force and
viscous force per unit mass, respectively.
This is a particularly useful form of the momentum
equations. For steady, irrotational, non-viscous flows
(inviscid) we see from e* that B is a constant. This is the
basis for statements like "where the speed is greatest the
pressure is least" that you may have heard in high school to
explain why airplanes fly. Later, we will use this equation
to derive the conservation of mechanical energy equation.
The non-linear term vxw is the cause of turbulence.
Perturbed vortex sheets induce such forces in the direction of
the perturbations. These amplify the perturbations to form
the eddies of turbulence. Note that viscosity does not cause
turbulence as stated on page 1 of the textbook, it
damps it out. For turbulence to exist, forces due to vxw
must be larger than forces due to nu del^2 v. The ratio
(vxw/ nu del^2 v) is the Reynolds number Re, which must
be larger than some critical value for turbulence to exist.

Hydrostatics exercises.
In problem 2.18, 1 meter of SAE 30 oil is over 2
meters of water, which is over 3 meters of fluid x, which is
over 1/2 meter of mercury in a tank. The pressure at the top
of the tank is 101.33 kPa, and the bottom pressure is 242
kPa. What is the specific gravity of fluid X?
Integrate the hydrostatic equation dp/dz = - rho g
from the bottom to the top of the tank. Thus, int [dp] =
p_bottom - p_top = sum gamma_i delta z_i , where gamma
= rho g for the various fluids. Table A.3 has the densities at
20 C that are needed. Substituting:
242000 Pa = 101330 + (8720)(1.0) + (9790)(2.0) +
gamma_X (3.0) + (133100)(0.5),
to give gamma_X = 15273 N/m^3. The specific gravity
rho_X / rho_water = gamma_X / gamma_water =
15273/9790 = 1.56 (ans.).
In problem 2.146 a helium balloon floats in a tank of
water. If the tank accelerates to the right at 5 m/s^2, which
way will the balloon string lean?
The equations of motion for the accelerating system

rho a = - grad p + rho g

so grad p = rho (g - a) in the water as well as in the helium
of the balloon. Because g is down and a is to the right, then
g - a is down to the left, which is the direction of the
pressure gradients in both the fluids. Because the density in
the balloon is so much less than that of water, hydrostatic
forces will develop forcing the balloon up and right, that
must be balanced by string forces down and to the left. You
can see this happening in a water vortex. Any bubbles move
toward the vortex center, in the direction of the acceleration.
A bubble in a bottle of liquid in space will also move in the
direction of any acceleration of the bottle. It is as though
gravity were turned on in the opposite direction.
The angle of the tether is tan^-1 (a/g) = tan^-1
(5/9.81) = 27 degrees (ans), independent of the size of the
balloon or the density difference between fluids.

1. Integral Relations for Control Volumes (Chapter 3 of
Derivation of the Reynolds' Transport Theorem.
Application to the conservation of mass.
Conservation of momentum.
Angular momentum.
2. Control-volume exercises.
3. Homework hints.

Integral Relations for Control Volumes.
Suppose we have some conserved quantity B, with
concentration beta per unit mass. Presumably we have a
physical law describing how B varies with time for a
system (a specified quantity of matter. For example, if B
is the mass of a system, then beta is 1 and the time
variation of B is zero. Often we want to know how B
varies with time for leaky volumes that contain different
systems at various times. The connection between the time
variation of B for a system and B for a different (leaky)
control volume is called the Reynolds Transport Theorem.

System control volume
B is the integral of rho beta over the control volume
of interest. The time rate of change of B is given by
Leibnitz' rule for differentiating integrals:

(1) dB/dt|syst = int [ rho beta dV] + int [ rho beta v dot

where the surface velocity v_S is replaced by the fluid
velocity v because the surface of a system control volume
is not "leaky".
Arbitrarily moving control volume
Now consider the time rate of change of B for an
arbitrary moving control volume that happens to coincide
exactly with the system control volume above at time t.
The surface of this control volume is generally not moving
with the fluid velocity v, but with velocity v_S. The time
rate of change of B for this arbitrarily moving control
volume is:

(2) dB/dt|CV = int [ rho beta dV] + int [ rho beta v_S
dot dA] .

Reynolds transport theorem
Subtracting (2) from (1) gives the Reynolds Transport

(3) dB/dt|syst = dB/dt|CV + int [ rho beta v_r dot dA]

where the two volume integrals cancel because they are
integrals with identical integrands over the same volume,
and v_r = (v - v_S) is the relative velocity of the fluid of
the system with respect to the arbitrarily moving control
Notice that if the control volume is fixed, the
relative velocity equals the fluid velocity. Also, if the
control volume surface moves with the fluid, the area
integral in (3) vanishes showing that dB/dt|syst =
dB/dt|CV, as expected.

Conservation of mass.
As an illustration, consider the integral
conservation of mass for a fixed control volume, where
fluid enters and leaves through a number of ports of
different sizes. Assume the fluid has constant density
everywhere, and all the velocities are constant in time, and
perpendicular to and constant over the port areas. The total
mass in both the system and fixed control volumes is a
constant M, where M is the integral of rho over the volume
(beta = 1). Thus, the Reynolds Transport Theorem of (3)
becomes 0 = 0 + int [ rho v dot dA], since v_r = v - 0.
Since rho was assumed constant everywhere, the area
integral reduces to the sum of the mass flows out minus
the sum of the mass flows into the control volume equals
If the densities of the port fluids had been different,
the negative of the area integral would be dB/dt|CV -
dB/dt|syst = dM/dt|CV - dM/dt|syst = dM/dt|CV - 0, which
is the rate of accumulation of mass in the fixed control

Conservation of Momentum.
For B = P, the total momentum, then beta is the
velocity v (momentum per unit mass) in (3). The left hand
side is the sum of all forces on the system by Newton's
second law of motion. The right hand side depends on the
assumptions given for the arbitrarily moving control
volume. If it is fixed, and the flow rates and density
distributions are constant then the momentum is constant in
the control volume (first term zero). The area integral is int
[ rho v v dot dA], which is minus the sum of all the inertial
forces on the fixed control volume due to the inertial stress
tensor - rho v v dotted into the various areas through
which there is flow from or into the fixed control volume.
Moving this term to the left hand side shows that the rate
of change of the momentum in a fixed control volume
equals the sum of the forces on the system contained, plus
the inertial forces due to flows through the surface.

Example 3.11 (p134 text): The von Karman
momentum integral theory for boundary layers illustrates
the application of integral conservation methods for both
momentum and mass. The control volume chosen about a
segment of the leading edge boundary layer of a plate
parallel to a uniform flow of velocity U. The boundary
layer is delta thick at a distance x = L. The streamlines
intersecting y = delta were at y = h at x = 0. The control
volume is bounded at the top by these streamlines, at the
bottom by the wall fluid, and by surfaces normal to the
flow u = U at x = 0 and u = u(y) at x = L.
Integral conservation of x momentum using (3)
gives the sum of forces on the left side, zero for the time
derivative of momentum on the right, and the surface
integral of rho v v dot dA only over the upstream and
downstream surfaces because v dot dA = 0 for all other
CV surfaces.
The forces in the x direction on the system CV
consist only of -D, where D is the frictional drag on the
plate. Note that the pressure is constant throughout the
boundary layer and in the external flow. A constant
pressure surrounding a control volume exerts no force
(this follows from the divergence theorem...the volume
integral of grad p = 0 if p = constant). Viscous forces are
zero except at the bottom of CV. Gravity acts in the y
direction and is irrelevant. Thus:
-D = 0 + int [ rho v v dot b dy]_x=0 + int [ rho v v
dot b dy]_x=L = - rho b UU h + rho b int [ u u dy]_x=L.
From the integral conservation of mass for the
control volume:
0 = 0 + int [ rho v dot dA]_x=0 + int [ rho v dot
dA}_x=L = - rho U h b + int [ rho u(y) b dy]_x=L
rho U h b = int [ rho u(y) b dy ]_x=L,
which is substituted into the momentum equation, to give:
D = rho U^2 b delta int [ u/U{1- u/U}],
which is the von Karman momentum integral equation.
The integral term is a drag coefficient because it is a drag
force D divided by rho U^2 times the projected area of the
boundary layer. Von Karman showed the integral was
2/15 assuming a parabolic velocity profile, which as within
10% of the exact (Blasius) velocity profile value discussed
in Chapter 7 on boundary layer theory.
The parabolic velocity profile is obtained by setting
u(y) = a + by + cy^2, with the boundary conditions that u
= 0 at y = 0 (no slip assumption), u = U at y = delta, and
du/dy = 0 at y = delta, giving u(y) = U(2 eta - eta^2),
where eta = y/delta. Substituting u/U = (2 eta - eta^2) into
the integral gives 2/15.
Conservation of angular momentum.
If we take the cross product of the integral
momentum equation with the distance r from an origin,
then we get the total angular momentum of the material
about this origin. For a fixed control volume, the left hand
side becomes the sum of the torques of the forces about the
origin. The right hand side is the rate of change of the
integrated r cross v dm values (beta is r cross v), plus the
flux of angular momentum out of the control volume.

2. Homework hints.

2.13. Integrate dp = - rho g dz from bottom to top through
both the water and the gasoline to get two equations in two
unknowns (the pressure difference and the height h of the
third fluid under the gasoline). Eliminate the pressure
difference and solve for h = 1.52 m.
2.28. Use equation 2.27 and the temperature and pressure
ratios to estimate the lapse rate B, giving 0.00899 deg F /
ft. Estimate the altitude z = delta T / B = 5400 ft.
2.44. The manometer only indicates pressure losses in the
pipe due to friction (392 psf).
Example: 2.91. The bolts must resist the weight of the missing
water from a cylinder surrounding the dome and standpipe
(90700 N).
Example: 2.113. The weight of the spar buoy equals the weight of
the displaced water (W_steel = 5.4 lbf).