1. Integral Relations for Control Volumes (Chapter 3 of
White).
Conservation of momentum (non-inertial frames).
Example (earth).
Mechanical energy (the Bernoulli equation).
2. Examples.
3. Homework hints

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Conservation of momentum.
============
Last time we derived the Reynolds transport
theorem relating the rate of change of a conserved fluid
property B for a system control volume to that for the same
initial control volume, but moving in an arbitrary fashion
(or remaining fixed). This is useful because most
conservation laws refer to systems, but many control
volumes needed in practical problems are not for systems.
We saw that the conservation of momentum for a control
volume is dP/dt|_CVsystem = dP/dt|_CVgeneral + int [ rho
v v_r dot dA] gives the sum of the forces on the left hand
side from Newton's law. For fixed control volumes in
steady state the sum of the forces is just int [ rho v v dot
dA] . Moving the area integral to the left hand side of the
equation changes its interpretation from that of an outward
momentum flux integral ( rho v is momentum per unit
volume and v dot dA is the local volume flux out of the
CV) to that of an integrated inertial force due to the inertial
stress tensor - rho v v. For example, the forces required to
keep a rocket stationary on a test stand are equal and
opposite to the inertial forces acting on the nozzle surface
area - rho v v dot A. A rocket in outer space with no
forces moves by causing exhaust gases to move rapidly in
the opposite direction, so that the net momentum of the
rocket system does not change.

============
Conservation of momentum in accelerating coordinate
frames.
============
Strictly speaking, Newton's law does not work in
coordinate systems fixed to the earth because earth-bound
coordinate systems are accelerating due to the earth's
rotation with respect to inertial coordinates. The rate of
change of momentum is equal to the sum of the forces on a
system (Newton's law) only in an "inertial" coordinate
system moving at constant velocity without rotation with
respect to the universe. The difference between
accelerations in the two systems can be considered
(fictitious) forces in the accelerating coordinate system,
just as gravitational force can be simulated by centripetal
acceleration in a space station. The time derivative of
vector B_I in an inertial coordinate system,
dB_I/dt|_I = dB_I/dt|_R + Omega cross B_I
in a rotating coordinate system, where Omega is the
angular velocity of the rotating coordinate system and the
origins of the inertial and rotating coordinate systems are
the same. Thus,
dr/dt|_I = dr/dt|_R + Omega cross r ,
where r is the same position vector in both the rotating and
inertial coordinate systems at the initial time, or
v_I = v_R + Omega cross r,
where
v_I = dr/dt|_I
and
v_R = dr/dt|_R.
The acceleration of the position r in the inertial frame
a_I = dv_I/dt|_I = dv_I/dt|_R + Omega cross v_I.
Substituting the expression
v_I = v_R + Omega cross r
gives
a_I = a_R + dOmega/dt cross r + 2 Omega cross
v_R + Omega cross ( Omega cross r ). Newton's law
states that a_I = sum of forces. Thus,
a_R = sum of forces - [dOmega/dt cross r + 2
Omega cross v_R + Omega cross ( Omega cross r )],
where all the accelerations in the brackets are "fictitious
forces" when placed on the left hand side of the
conservation of momentum equation. Such forces are
required to compensate for the coordinate system
accelerations.
The most famous "fictitious force" is the Coriolis
force - 2 Omega cross v_R, which accounts for the
direction of wind motions about high and low pressure
points in the northern and southern hemispheres of the
earth's atmosphere. Horizontal motions in the north are
deflected to the right by Coriolis force, and to the left in the
south, because the earth spins from west to east.
We saw examples in hydrostatics of the role of
accelerations. Without accelerations the forces per unit
mass in a fluid with v = 0 were - (grad p)/ rho + g = 0
where g is gravitational force per unit mass and - g is the
gravitational acceleration. With acceleration a, then - (grad
p)/ rho + g = a. Clearly we could consider the acceleration
a as a force per unit mass -a if we moved it to the left hand
side of this equation. Note that numerous accelerations are
neglected in applying Newton's law to laboratory (earth-
bound) coordinate systems...the rotation of the earth on its
axis, about the centers of gravity with the moon and the
sun and the center of the galaxy, etc. ... because they are
small compared to the forces of interest.
*******
Example: How large are "fictitious forces" due to the
earth's rotation?
Coriolis force = - 2 Omega cross v. For horizontal
velocities, this force is largest near the poles. Omega is 2
7x10^4 m/s is required to produce 1 g of Coriolis
acceleration at the north pole (artillery shell trajectories at
high latitudes are indeed affected significantly). A racing
car at 100 m/s is deflected by only 0.0014 g by Coriolis
forces at the north or south pole, and not at all at the
equator. The centrifugal force Omega cross ( Omega cross
r ) due to the earth's rotation is twice this at the equator,
with a value of 0.003 g (r is about 6 megameters).
*******

=============
Conservation of mechanical energy for a stream tube
(Bernoulli's equation).
=============
Consider the momentum equation for a Newtonian
fluid with constant physical properties (including density
and viscosity) in the form:
part v/ part t = - grad B + v cross omega + div ( tau
/ rho )
where tau is the viscous stress tensor 2 mu e , B is the
Bernoulli group, and e is the rate of strain tensor. We can
derive the mechanical energy equation; that is, an equation
for the rate of change of the kinetic energy per unit mass
v^2/2, by taking the dot product of the equation with the
velocity v. The left hand side becomes
part (v^2 / 2) / part t
which is what we want.
The first term on the right becomes
- div (v B) + B div v = - div (v B)
because div v = 0 from the continuity equation with rho =
constant.
The second term on the right becomes
v dot ( v cross omega ) = 0
because v cross omega is perpendicular to v.
The third term on the right becomes
v dot div ( tau / rho ) = div [ v dot tau / rho ] - [ tau
/ rho ] dot dot [ grad v ]
where grad v is the velocity gradient tensor. The term [ tau
/ rho ] dot dot [ grad v ] is the viscous dissipation rate
epsilon per unit mass, and we can show that eps = 2 nu
e^2, so eps is always positive. Combining all terms gives
part (v^2 / 2) / part t = - div (v B) + div [ v dot tau
/ rho ] - eps,
which is the mechanical energy equation we need.
The famous "Bernoulli equation" can be found by
applying this equation to the steady flow of a constant
property fluid in a stream tube. A stream tube is formed
from a cylinder of adjacent stream lines. Take the entrance
to the tube to be point 1 and the exit as point 2. The kinetic
energy KE in the tube is the integral of rho v^2 / 2 over the
control volume. The derivative of KE with time is found
from Leibnitz' Rule and is equal to zero by our steady state
assumption, so
d KE /dt = int [ rho part (v^2 / 2) / part t dV] + int
[ rho v^2 / 2 v_S dot dA] = 0
where the surface integral is zero for the fixed control
volume because v_S = 0.
Now we can substitute the kinetic energy equation
for part (v^2 / 2) / part t in the volume integral, and convert
two of the resulting integrals to surface integrals by the
divergence theorem
- int [ rho vB dot dA ] + int [ v dot tau dot dA ] -
int [ eps dV] = 0
where the first integral is the flow of B into the control
volume, the second is rate of viscous work being done on
the control volume per unit time, and the third is minus the
rate that mechanical energy is being lost within the control
volume per unit time due to internal viscous friction.
Recall B = p/ rho + v^2 / 2 + gz has the units of energy per
unit mass (it is specific enthalpy minus specific internal
energy).
The first term is just (B_1 - B_2)(dm/dt), where
dm/dt is the mass flow rate through the stream tube. The
second term is -dW_s/dt, the rate of "shaft work" being
done on the surroundings per unit time. The third term is
-dW_l/dt, the "lost work" per unit time. Dividing by dm/dt
gives the Bernoulli equation
B_1 = B_2 + gh_s + gh_f
where the shaft work and lost work terms are expressed as
stream tube, h_s is positive if a turbine is encountered and
positive shaft work is extracted, and h_s is negative if a
pump is encountered. The frictional head loss h_f is
always positive. If no work is done or extracted, and there
are no losses, then the Bernoulli group B is constant and
"where the speed is greatest the pressure is least".

********
Example: Consider the case of a tank of water with an exit
port a distance h below the surface. What is the velocity of
the fluid exiting the port?
Apply Bernoulli's equation to the streamline from
the surface to the port. There are no losses or shaft work
terms, so B is constant.
B_1 = B_2
where B_1 = p_a / rho + 0 + 0 and B_2 = p_a / rho + v^2
/ 2 + gz_2. Now z_2 is -h if z_1 = 0. Therefore v =
(2gh)^1/2 (ans).
*********

********
Example: How does a pitot tube work? What is the
pressure at the entrance of an impact tube used to measure
velocity by comparison to the static pressure in the
surroundings?
Consider the streamline that stagnates at the
entrance to the impact tube. Again, B_1 = B_2. The z
values are the same at 2 and 1. Thus the pressure
difference p_2 - p_1 = rho v^2 / 2, where v=0 at 2 and v at
1, and can be detected with a manometer between the
points.
********

********
Homework hints
********
3.5 Plot the given data and show that the turbulent velocity
profile is different from a parabolic profile expected for
laminar flow. Integrate graphically to get Q = 0.006 m^3/s.

3.10 Integrate the enthalpy flux over the velocity profile.

3.38 Use the RTT for mass conservation to show V=V_o/2h.

3.43 Use the RTT for momentum and mass for a CV
enclosing the pipe. The forces in the x direction include
pressure forces and flange forces on the inlet and exit.
These equal int[rho v v dot dA] for the inlet and exit,
which are both in the minus x direction! Summing gives
F_flange = -750N in the x direction and 14 N in the y
direction.

3.102 Apply RTT for mass and momentum for a CV
enclosing the hydraulic jump from h_1 to h_2. Neglect
wall shear, so the only forces are due to hydrostatic pressure.
From the mass balance rho V_1 h_1 b = rho V_2 h_2 b.
From the mass-mom. balances, rho g h_1 h_1 b / 2 -
rho g h_2 h_2 b / 2 = rho V_1 h_1 b [ V_2 - V_1]. Some
algebra gives h_2 / h_1 = -1/2 + (1/2)sqrt[1+8V_1^2 / (gh_1)].

1. Integral Relations for Control Volumes (Chapter 3 of
White).
Conservation of energy...the first law of
thermodynamics for fluids.
The total specific energy of a fluid particle.
The general mechanical energy equation of a fluid
particle.
The internal energy of a fluid particle.
2. Examples.
3. Homework hints.
Design project D3.1.

============
Conservation of energy.
============
Above we derived a form of the conservation of
mechanical energy equation for a constant density fluid
partial ke /partial t = - div vB + div (v dot tau / rho )
- eps
where ke = v^2 / 2 is the kinetic energy per unit mass, B is
p/rho + ke + pe, tau is the viscous stress tensor 2 mu e (for
a Newtonian fluid), and eps is the viscous dissipation rate
per unit mass (tau / rho ) dot dot grad v =
eps = 2 nu e dot dot e. This last
expression results from the identity
grad v = e + Omega
where e is the rate of strain tensor with components (
partial v_i / partial x_j + partial v_j / partial x_i )/2 , and
Omega is the rotation tensor ( partial v_i / partial x_j -
partial v_j / partial x_i )/2. Clearly e is symmetric because
e_ij = e_ji and Omega is antisymmetric because Omega_ij
= - Omega_ji. Substituting 2 mu e for tau in eps = (tau /
rho ) dot dot grad v gives
eps = 2 nu e dot dot e
because e dot dot Omega = 0 (this is proved by reversing
the dummy indices and substituting e_ij = e_ji and
Omega_ij = - Omega_ji, so that e dot dot Omega = - e dot
dot Omega). We see that eps must always be positive if nu
is positive (as it must be), since e dot dot e = e^2 is
positive.
We applied this equation to the steady flow of a
constant property fluid in a streamtube and found the
Bernoulli equation
B_1 = B_2 + gh_s + gh_f
where h_s is the head loss due to viscous working on the
surroundings, and h_f is the "lost work" head loss due to
irreversible internal friction.
Now let us apply the first law of thermodynamics
to a system of fluid contained by a system control volume
to derive an equation for the specific energy, derive
another form of the mechanical energy equation without
assuming the density is constant, and subtract the two to
interpret the viscous dissipation rate eps. According to the
first law, the heat flow dQ/dt into the system minus dW/dt
done by the system on the surroundings equals the rate of
change of the total energy dE/dt within the system.
The heat flow is int [ k grad T dot dA] = int [ div (k
grad T) dV], where k is the thermal conductivity of the
fluid and T is the temperature.
The work done on the surroundings per unit time is
the pressure work rate dW_p/dt = int [ p v dot dA] = int [
div pv dV] plus the viscous work rate dW_v/dt = - int [ v
dot tau dot dA] = - int [ div (v dot tau ) dV] .
The rate of change of energy in the system dE/dt =
(d/dt) int [ rho e dV] = int [ (partial rho e / partial t) dV] +
int [ rho e v dot dA] = int [ {(partial rho e / partial t) + div (
rho e v )} dV] , where e is the specific total energy u + ke
+ pe and u is the internal energy per unit mass.
Collecting these terms in the first law of
thermodynamics gives
int [ {(partial rho e / partial t) + div ( rho e v ) - div
(v dot tau) + div (pv) - div (kgradT) } = 0.
Now the system control volume was chosen
internal point, so that the {.} term becomes constant by the
continuum hypothesis. Factoring {.} out of the integral
shows that it must be zero at every point in the fluid (since
the remaining integral is finite), or identically zero,
according to the first law of thermodynamics. Thus
(partial rho e / partial t) + div ( rho e v ) - div (v dot
tau) + div (pv) - div (kgradT) = 0
is the total energy equation for a fluid.
Expanding the first two terms gives (from the
continuity equation)
rho De/Dt - div (v dot tau) + div (pv) - div
which shows how the total energy per unit volume of a
fluid particle changes with time.
Previously we found the following form for the
momentum conservation of a fluid particle
rho Dv/Dt - div tau + grad p - rho g = 0
where g is the gravitational force - grad (gz), and z is up.
Taking the dot product of the velocity v with this equation
gives
rho D(ke + pe)/Dt - v dot div tau + v dot grad p = 0
which is the mechanical energy equation for a fluid particle
with no assumptions of constant properties. Subtracting
this from the total energy equation for a fluid particle,
using the definition of e, gives
rho Du/Dt - tau dot dot grad v + p div v - div
where u is the internal energy. Thus the internal energy of
a fluid particle per unit volume increases with positive rho
eps = tau dot dot grad v, increases due to heating at the rate
div kgradT (note that this is positive for a cold spot,
negative for a hot spot), and increases if -p div v is
positive (corresponding to compression of the fluid particle
from the continuity equation).

==============
2. Examples.
==============

*********
Example 6.1. Consider the solid fuel rocket of problem
3.35. The nozzle diameter is 18 cm and the exit velocity is
1150 m/s. Take a fixed control volume around the rocket,
perpendicular to the flow out of the nozzle. The Reynolds
transport theorem for the mass of the control volume gives
zero on the left hand side (for the rate of change of the
mass of the system) and int [ partial rho / partial t dV] + int
[ rho v dot dA] on the right hand side, where int [ rho v
dot dA] is the mass flow rate out of the control volume at
the nozzle. The molecular weight is given as 28, the gas
constant R is 8313/28 = 297 m^2/s^2 K for the gas.
Hence the exit gas density is
rho_exit = p/RT = 90,000 Pa / (297)x(750 K) =
0.404 kg/m^3.
The rocket mass increases at rate int [ partial rho / partial t
dV] equal to minus the flow rate of mass out int [ rho v dot
dA]. Both are negative. int [ rho v dot dA] = rho_exit
A_exit V_exit = (0.404)(pi/4)(0.18^2)(1150) = 11.8 kg/s.
Therefore the mass increase of the rocket is - 11.8 kg/s
(ans).
********

********
Example 6.2. Consider a uniform flow past a cylinder
with a V shaped wake, as in problem 3.44. This problem
is similar to the von Karman integral momentum theory for
a boundary layer. Take a control volume of width b about
the cylinder, with streamline surfaces separated by +,- L at
the downstream position, and +,- H upstream. The
velocity is uniform with value U upstream, and u =
(U/2)(1 + y/L) for positive y and (U/2)(1 - y/L) for
negative y in the wake, and U outside, downstream.
Apply the conservation of mass to this fixed
control volume, using the Reynolds transport theorem.
The rate of change of mass for the system is zero, and for
the control volume also zero for this steady flow problem.
This leaves the mass flux integral int [ rho v dot dA] = 0.
Only the upstream and downstream surfaces have nonzero
integrals. Upstream we have - 2 int [ rho U b dy] from y
= 0 to H, giving - 2 rho U b L. Downstream we have + 2
int [ rho (U/2)(1 + y/L) b dy] from y = 0 to L, giving H =
3L/4.
Now apply the conservation of x momentum. The
rate of change of momentum for the system is the sum of
the forces, equal to the negative of the drag force on the
cylinder - F_drag. This equals the rate of change of
momentum in the control volume, plus the surface integral
of the momentum flux int [ rho v v dot dA] which is
nonzero only for the upstream and downstream surfaces.
Upstream we have - 2 int [ rho U^2 b dy] from y = 0 to H.
Downstream we have + 2 int [ rho {(U/2)(1 + y/L)}^2 b
dy. Combining these and H = 3L/4 gives - (3/2) rho U^2
L b upstream and (7/6) rho U^2 L b downstream. Hence
F_drag = (1/3) rho U^2 L b, giving a drag coefficient C_D
= F_drag/rho U^2 L b of 1/3 (ans).

==============
3. Homework Hints.
==============
3.110. Apply the RTT for conservation of angular momentum.
The sum of the torques equals the integral of r X rho v v dot dA
for a control volume surrounding the sprinkler and perpendicular
to the two nozzles.

3.127. Apply the RTT for conservation of energy to the river.
The heat capacity of water is about 4280 J/kg K. The waste heat
of 55MW equals the mass flow rate of the river times the heat
capacity times the temperature difference, giving 23.2 deg C for
the downstream temperature.

3.142. Apply the Bernoulli equation to a steamline passing from
the lower reservoir surface, through the pump, to the upper reservoir
surface at 50 ft higher elevation. The frictional head
h_f = 27 V^2 / 2g from 3.141, and the shaft work head is minus
the pump head 300 - 50Q. Q = VA, so substitution gives a
quadratic equation in Q. The solution is 2.57 ft^3 / s. The pump
power is rho g Q h_p / eta to give 67 hp.

3.160. The air cushion vehicle is lifted by the pressure difference
between that inside and out, times the skirt area, so from the known
weight the inside pressure must be 1768 above atmospheric. From
Bernoulli's equation across the skirt we get the skirt velocity 30.6
m/s. The power is Q delta p, or 54000 W.

========
Design project D3.1.
========
The problem is to select a pump with minimum cost
with given performance characteristics. It must deliver no
less than 1.0 ft^3 / s of water and rotate no slower than 10
rev/s. The cost is assumed to be proportional to the power
input to the pump. The relevant dimensional parameters of
the pump performance are the pump work rate per mass flow
rate gh_p, the volume flow rate Q, the water density rho, the
shaft rotation rate n, the impeller diameter D, the power to
the water rho Q g h_p , and the power input to the pump.
The relevant dimensionless parameters for the pump are
functions of the dimensionless volume flow rate zeta =
Q/nD^3 . The dimensionless pump head is phi = gh_p / n^2
D^2 = 6.04 - 161 zeta . The efficiency eta = 70 zeta - 91500
zeta^3. The dimensionless power to the water pi_W = P_W
/ (rho n^3 D^5) = zeta phi . The dimensionless power input
P_I is pi_I / eta which should be minimized as a function of
zeta . Since Q and n are fixed, the pump impeller diameter to
give minimum power can be determined.
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