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M O M E N T U M A N A LY S I S
O F F L O W S Y S T E M S
W
hen dealing with engineering problems, it is desirable to obtain
fast and accurate solutions at minimal cost. Most engineering
problems, including those associated with fluid flow, can be ana-
lyzed using one of three basic approaches: differential, experimental, and
control volume. In differential approaches , the problem is formulated accu-
rately using differential quantities, but the solution of the resulting differen-
tial equations is difficult, usually requiring the use of numerical methods
with extensive computer codes. Experimental approaches complemented
with dimensional analysis are highly accurate, but they are typically time-
consuming and expensive. The finite control volume approach described in
this chapter is remarkably fast and simple and usually gives answers that are
sufficiently accurate for most engineering purposes. Therefore, despite the
approximations involved, the basic finite control volume analysis performed
with a paper and pencil has always been an indispensable tool for engineers.
In Chap. 5, the control volume mass and energy analysis of fluid flow
systems was presented. In this chapter, we present the finite control volume
momentum analysis of fluid flow problems. First we give an overview of
Newton’s laws and the conservation relations for linear and angular momen-
tum. Then using the Reynolds transport theorem, we develop the linear
momentum and angular momentum equations for control volumes and use
them to determine the forces and torques associated with fluid flow.
227
CHAPTER
OBJECTIVES
When you finish reading this chapter, you
should be able to
Identify the various kinds of
forces and moments acting on
a control volume
Use control volume analysis to
determine the forces associated
with fluid flow
Use control volume analysis to
determine the moments caused
by fluid flow and the torque
transmitted
228 FLUID MECHANICS
6–1 ^ NEWTON’S LAWS AND CONSERVATION OF MOMENTUM
Newton’s laws are relations between motions of bodies and the forces act-
ing on them. Newton’s first law states that a body at rest remains at rest,
and a body in motion remains in motion at the same velocity in a straight
path when the net force acting on it is zero. Therefore, a body tends to pre-
serve its state of inertia. Newton’s second law states that the acceleration of
a body is proportional to the net force acting on it and is inversely propor-
tional to its mass. Newton’s third law states that when a body exerts a force
on a second body, the second body exerts an equal and opposite force on
the first. Therefore, the direction of an exposed reaction force depends on
the body taken as the system.
For a rigid body of mass m , Newton’s second law is expressed as
Newton’s second law : (6–1)
where F
→
is the net force acting on the body and a → is the acceleration of the
body under the influence of F
→
The product of the mass and the velocity of a body is called the linear
momentum or just the momentum of the body. The momentum of a rigid
body of mass m moving with a velocity V
→
is mV
→
(Fig. 6–1). Then Newton’s
second law expressed in Eq. 6–1 can also be stated as the rate of change of
the momentum of a body is equal to the net force acting on the body (Fig.
6–2). This statement is more in line with Newton’s original statement of the
second law, and it is more appropriate for use in fluid mechanics when
studying the forces generated as a result of velocity changes of fluid
streams. Therefore, in fluid mechanics, Newton’s second law is usually
referred to as the linear momentum equation.
The momentum of a system remains constant when the net force acting
on it is zero, and thus the momentum of such systems is conserved. This is
known as the conservation of momentum principle. This principle has
proven to be a very useful tool when analyzing collisions such as those
between balls; between balls and rackets, bats, or clubs; and between atoms
or subatomic particles; and explosions such as those that occur in rockets,
missiles, and guns. The momentum of a loaded rifle, for example, must be
zero after shooting since it is zero before shooting, and thus the rifle must
have a momentum equal to that of the bullet in the opposite direction so that
the vector sum of the two is zero.
Note that force, acceleration, velocity, and momentum are vector quanti-
ties, and as such they have direction as well as magnitude. Also, momentum
is a constant multiple of velocity, and thus the direction of momentum is the
direction of velocity. Any vector equation can be written in scalar form for a
specified direction using magnitudes, e.g., Fx max d ( mVx )/ dt in the x -
direction.
The counterpart of Newton’s second law for rotating rigid bodies is
expressed as M
→
I a
→
, where M
→
is the net moment or torque applied on the
body, I is the moment of inertia of the body about the axis of rotation, and a→
is the angular acceleration. It can also be expressed in terms of the rate of
change of angular momentum dH
→
/ dt as
F
→ ma → m
dV
→
dt
d ( mV
→ ) dt
V
mV
m
m
→
→
FIGURE 6–
Linear momentum is the product of
mass and velocity, and its direction
is the direction of velocity.
Net force
Rate of change of momentum
= ma = m dt dt
m
FIGURE 6–
Newton’s second law is also expressed
as the rate of change of the momentum
of a body is equal to the net force
acting on it.
230 FLUID MECHANICS
velocity to use is the velocity of the exhaust gases relative to the nozzle exit
plane, that is, the relative velocity V
→
r. Since the entire control volume moves
at velocity V
→
CV, the relative velocity becomes^ V
→
r ^ V
→
V
→
CV, where^ V
→
is
the absolute velocity of the exhaust gases, i.e., the velocity relative to a
fixed point on earth. Note that V
→
r is the fluid velocity expressed relative to a
coordinate system moving with the control volume. Also, this is a vector
equation, and velocities in opposite directions have opposite signs. For
example, if the airplane is cruising at 500 km/h to the left, and the velocity
of the exhaust gases is 800 km/h to the right relative to the ground, the
velocity of the exhaust gases relative to the nozzle exit is
That is, the exhaust gases leave the nozzle at 1300 km/h to the right relative
to the nozzle exit (in the direction opposite to that of the airplane); this is
the velocity that should be used when evaluating the outflow of exhaust
gases through the control surface (Fig. 6–4 b ). Note that the exhaust gases
would appear motionless to an observer on the ground if the relative veloc-
ity were equal in magnitude to the airplane velocity.
When analyzing the purging of exhaust gases from a reciprocating inter-
nal combustion engine, a wise choice for the control volume is one that
comprises the space between the top of the piston and the cylinder head
(Fig. 6–4 c ). This is a deforming control volume, since part of the control
surface moves relative to other parts. The relative velocity for an inlet or
outlet on the deforming part of a control surface (there are no such inlets
or outlets in Fig. 6–4 c ) is then given by V
→
r ^ V
→
V
→
CS where^ V
→
is the
absolute fluid velocity and V
→
CS is the control surface velocity, both relative
to a fixed point outside the control volume. Note that V
→
CS ^ V
→
CV for mov-
ing but nondeforming control volumes, and V
→
CS ^ V
→
CV ^ 0 for fixed ones.
6–3 ^ FORCES ACTING ON A CONTROL VOLUME
The forces acting on a control volume consist of body forces that act
throughout the entire body of the control volume (such as gravity, electric,
and magnetic forces) and surface forces that act on the control surface (such
as pressure and viscous forces and reaction forces at points of contact).
In control volume analysis, the sum of all forces acting on the control vol-
ume at a particular instant in time is represented by F
→
and is expressed as
Total force acting on control volume : (6–4)
Body forces act on each volumetric portion of the control volume. The body
force acting on a differential element of fluid of volume dV within the con-
trol volume is shown in Fig. 6–5, and we must perform a volume integral to
account for the net body force on the entire control volume. Surface forces
act on each portion of the control surface. A differential surface element of
area dA and unit outward normal n
→
on the control surface is shown in Fig.
6–5, along with the surface force acting on it. We must perform an area
integral to obtain the net surface force acting on the entire control surface.
As sketched, the surface force may act in a direction independent of that of
the outward normal vector.
a F
→ (^) a F
→ body ^ a F
→ surface
V
→ r ^ V
→ V
→ CV ^800 i
→ ( 500 i
→ ) 1300 i
→ km/h
V
V
( a )
( b )
( c )
CV
V V
V CV
r
r
Moving control volume
Deforming control volume
Fixed control volume
x
x
y
V CS
→ → → → → →
FIGURE 6–
Examples of ( a ) fixed, ( b ) moving,
and ( c ) deforming control volumes.
231 CHAPTER 6
The most common body force is that of gravity, which exerts a down-
ward force on every differential element of the control volume. While other
body forces, such as electric and magnetic forces, may be important in some
analyses, we consider only gravitational forces here.
The differential body force dF
→
body ^ dF
→
gravity acting on the small fluid ele-
ment shown in Fig. 6–6 is simply its weight,
Gravitational force acting on a fluid element : (6–5)
where r is the average density of the element and g
→
is the gravitational vec-
tor. In Cartesian coordinates we adopt the convention that g →^ acts in the neg-
ative z -direction, as in Fig. 6–6, so that
Gravitational vector in Cartesian coordinates : (6–6)
Note that the coordinate axes in Fig. 6–6 have been rotated from their usual
orientation so that the gravity vector acts downward in the z -direction. On
earth at sea level, the gravitational constant g is equal to 9.807 m/s^2. Since
gravity is the only body force being considered, integration of Eq. 6–
yields
Total body force acting on control volume : (6–7)
Surface forces are not as simple to analyze since they consist of both nor-
mal and tangential components. Furthermore, while the physical force act-
ing on a surface is independent of orientation of the coordinate axes, the
description of the force in terms of its coordinate components changes with
orientation (Fig. 6–7). In addition, we are rarely fortunate enough to have
each of the control surfaces aligned with one of the coordinate axes. While
not desiring to delve too deeply into tensor algebra, we are forced to define
a second-order tensor called the stress tensor s ij in order to adequately
describe the surface stresses at a point in the flow,
Stress tensor in Cartesian coordinates : (6–8)
The diagonal components of the stress tensor, s xx , s yy , and s zz , are called
normal stresses; they are composed of pressure (which always acts
inwardly normal) and viscous stresses. Viscous stresses are discussed in
more detail in Chap. 9. The off-diagonal components, s xy , s zx , etc., are
called shear stresses; since pressure can act only normal to a surface, shear
stresses are composed entirely of viscous stresses.
When the face is not parallel to one of the coordinate axes, mathematical
laws for axes rotation and tensors can be used to calculate the normal and
tangential components acting at the face. In addition, an alternate notation
called tensor notation is convenient when working with tensors but is usu-
ally reserved for graduate studies. (For a more in-depth analysis of tensors
and tensor notation see, for example, Kundu, 1990.)
In Eq. 6–8, s ij is defined as the stress (force per unit area) in the j -direction
acting on a face whose normal is in the i -direction. Note that i and j are
merely indices of the tensor and are not the same as unit vectors i
→
and j
→
. For
example, s xy is defined as positive for the stress pointing in the y -direction
s ij £
s xx s yx s zx
s xy s yy s zy
s xz s yz s zz
a F
→ body ^ CV
r g → dV m CV g →
g → gk
→
dF
→ gravity ^ r g
→ dV
body
Control volume (CV)
Control surface (CS)
n
dF dF surface
dA
d V
→ →
→
FIGURE 6–
The total force acting on a control
volume is composed of body forces
and surface forces; body force is
shown on a differential volume
element, and surface force is shown
on a differential surface element.
g
z , k dF body^ = dF gravity^ =^ r g d^ V
y , j x , i
dy
dz
dx
→
→ → →
→
→
→
d V , r
FIGURE 6–
The gravitational force acting on a
differential volume element of fluid is
equal to its weight; the axes have been
rotated so that the gravity vector acts
downward in the negative z -direction.
233 CHAPTER 6
only the forces that are to be determined (such as reaction forces) and a
minimum number of other forces.
Only external forces are considered in the analysis. The internal forces
(such as the pressure force between a fluid and the inner surfaces of the
flow section) are not considered in a control volume analysis unless they are
exposed by passing the control surface through that area.
A common simplication in the application of Newton’s laws of motion is
to subtract the atmospheric pressure and work with gage pressures. This is
because atmospheric pressure acts in all directions, and its effect cancels out
in every direction (Fig. 6–9). This means we can also ignore the pressure
forces at outlet sections where the fluid is discharged to the atmosphere
since the discharge pressures in such cases will be very near atmospheric
pressure at subsonic velocities.
As an example of how to wisely choose a control volume, consider con-
trol volume analysis of water flowing steadily through a faucet with a par-
tially closed gate valve spigot (Fig. 6–10). It is desired to calculate the net
force on the flange to ensure that the flange bolts are strong enough. There
are many possible choices for the control volume. Some engineers restrict
their control volumes to the fluid itself, as indicated by CV A (the colored
control volume). With this control volume, there are pressure forces that
vary along the control surface, there are viscous forces along the pipe wall
and at locations inside the valve, and there is a body force, namely, the
weight of the water in the control volume. Fortunately, to calculate the net
force on the flange, we do not need to integrate the pressure and viscous
stresses all along the control surface. Instead, we can lump the unknown
pressure and viscous forces together into one reaction force, representing
the net force of the walls on the water. This force, plus the weight of the
faucet and the water, is equal to the net force on the flange. (We must be
very careful with our signs, of course.)
When choosing a control volume, you are not limited to the fluid alone.
Often it is more convenient to slice the control surface through solid objects
such as walls, struts, or bolts as illustrated by CV B (the gray control vol-
ume) in Fig. 6–10. A control volume may even surround an entire object,
like the one shown here. Control volume B is a wise choice because we are
not concerned with any details of the flow or even the geometry inside the
control volume. For the case of CV B, we assign a net reaction force acting
at the portions of the control surface that slice through the flange. Then, the
only other things we need to know are the gage pressure of the water at
the flange (the inlet to the control volume) and the weights of the water and
the faucet assembly. The pressure everywhere else along the control surface
is atmospheric (zero gage pressure) and cancels out. This problem is revis-
ited in Section 6–4, Example 6–7.
6–4 ^ THE LINEAR MOMENTUM EQUATION
Newton’s second law for a system of mass m subjected to a net force F
→
is
expressed as
a F^ (6–13)
→ ma → m
dV
→
dt
d dt
( mV
→ )
dy
y
x z
dz
dx
s xz
s xx
s xy
s yz
s yy
s yx
s zy s zx s zz
FIGURE 6–
Components of the stress tensor in
Cartesian coordinates on the right,
top, and front faces.
FR
P 1
W
P atm
P atm
P 1 (gage) With atmospheric pressure considered
With atmospheric pressure cancelled out
FR
W
FIGURE 6–
Atmospheric pressure acts in all
directions, and thus it can be ignored
when performing force balances since
its effect cancels out in every direction.
W faucet
W water
CV B
Out
Spigot
In
Bolts
x
z
CV A
FIGURE 6–
Cross section through a faucet
assembly, illustrating the importance
of choosing a control volume wisely;
CV B is much easier to work with
than CV A.
234 FLUID MECHANICS
where mV
→
is the linear momentum of the system. Noting that both the den-
sity and velocity may change from point to point within the system, New-
ton’s second law can be expressed more generally as
(6–14)
where d m r dV is the mass of a differential volume element dV , and
r V
→
dV is its momentum. Therefore, Newton’s second law can be stated as
the sum of all external forces acting on a system is equal to the time rate of
change of linear momentum of the system. This statement is valid for a
coordinate system that is at rest or moves with a constant velocity, called an
inertial coordinate system or inertial reference frame. Accelerating systems
such as aircraft during takeoff are best analyzed using noninertial (or accel-
erating) coordinate systems fixed to the aircraft. Note that Eq. 6–14 is a
vector relation, and thus the quantities F
→
and V
→
have direction as well as
magnitude.
Equation 6–14 is for a given mass of a solid or fluid and is of limited use
in fluid mechanics since most flow systems are analyzed using control vol-
umes. The Reynolds transport theorem developed in Section 4–5 provides
the necessary tools to shift from the system formulation to the control vol-
ume formulation. Setting b V
→
and thus B mV
→
, the Reynolds transport
theorem can be expressed for linear momentum as (Fig. 6–11)
(6–15)
But the left-hand side of this equation is, from Eq. 6–13, equal to F
→
. Sub-
stituting, the general form of the linear momentum equation that applies to
fixed, moving, or deforming control volumes is obtained to be
General : (6–16)
which can be stated as
Here V
→
r ^ V
→
V
→
CS is the fluid velocity relative to the control surface (for
use in mass flow rate calculations at all locations where the fluid crosses the
control surface), and V
→
is the fluid velocity as viewed from an inertial refer-
ence frame. The product r( V
→
r ·^ n
→
) dA represents the mass flow rate through
area element dA into or out of the control volume.
For a fixed control volume (no motion or deformation of control volume),
V
→
r ^ V
→
and the linear momentum equation becomes
Fixed CV : (6–17)
Note that the momentum equation is a vector equation , and thus each term
should be treated as a vector. Also, the components of this equation can be
resolved along orthogonal coordinates (such as x, y , and z in the Cartesian
a F
→
d dt CV r V
→ dV (^) CS
r V
→ ( V
→ n → ) dA
The sum of all external forces acting on a CV
The time rate of change of the linear momentum of the contents of the CV
The net flow rate of linear momentum out of the control surface by mass flow
a F
→
d dt CV
r V
→ dV (^) CS
r V
→ ( V
→ r ^ n
→ ) dA
d ( mV
→ )sys dt
d dt CV
r V
→ dV (^) CS
r V
→ ( V
→ r ^ n
→ ) dA
a F
→
d dt sys
r V
→ dV = r b d V+
B = mV
dB sys dt V
d dt (^) CV r b ( (^) r · n ) dA CS
= r V d V+
d ( mV )sys dt V
d dt (^) CV r V ( (^) r · n ) dA CS
b = V b = V
→ →
→ →
→ → → → →
→
FIGURE 6–
The linear momentum equation
is obtained by replacing B in the
Reynolds transport theorem by the
momentum mV
→
, and b by the
momentum per unit mass V
→
236 FLUID MECHANICS
dimensionless correction factor b, called the momentum-flux correction
factor, is required, as first shown by the French scientist Joseph Boussinesq
(1842–1929). The algebraic form of Eq. 6–17 for a fixed control volume is
then written as
(6–21)
where a unique value of momentum-flux correction factor is applied to each
inlet and outlet in the control surface. Note that b 1 for the case of uni-
form flow over an inlet or outlet, as in Fig. 6–14. For the general case, we
define b such that the integral form of the momentum flux into or out of the
control surface at an inlet or outlet of cross-sectional area Ac can be
expressed in terms of mass flow rate m
through the inlet or outlet and aver-
age velocity V
→
avg through the inlet or outlet,
Momentum flux across an inlet or outlet : (6–22)
For the case in which density is uniform over the inlet or outlet and V
→
is in
the same direction as V
→
avg over the inlet or outlet, we solve Eq. 6–22 for^ b,
(6–23)
where we have substituted r V avg Ac for m·^ in the denominator. The densities
cancel and since V avg is constant, it can be brought inside the integral. Fur-
thermore, if the control surface slices normal to the inlet or outlet area, we
have ( V
→
· n → ) dAc V dAc. Then, Eq. 6–23 simplifies to
Momentum-flux correction factor : (6–24)
It turns out that for any velocity profile you can imagine, b is always greater
than or equal to unity.
b
Ac Ac
a
V
V avg
b
2 dAc
b
Ac
r V ( V
→ n^ → ) dAc
m
V avg
Ac
r V ( V
→ n^ → ) dAc
r V avg AcV avg
Ac
r V
→ ( V
→ n → ) dAc b m
V
→ avg
a F
→
d dt CV
r V
→ dV (^) a out
b m
V
→ avg ^ a in
b m
V
→ avg
V avg V avg
CV CV CV
Nozzle
( a ) ( b ) ( c )
V avg
FIGURE 6–
Examples of inlets or outlets
in which the uniform flow
approximation is reasonable:
( a ) the well-rounded entrance to
a pipe, ( b ) the entrance to a wind
tunnel test section, and ( c ) a slice
through a free water jet in air.
237 CHAPTER 6
EXAMPLE 6–1 Momentum-Flux Correction Factor
for Laminar Pipe Flow
Consider laminar flow through a very long straight section of round pipe. It is shown in Chap. 8 that the velocity profile through a cross-sectional area of the pipe is parabolic (Fig. 6–15), with the axial velocity component given by
(1)
where R is the radius of the inner wall of the pipe and V avg is the average velocity. Calculate the momentum-flux correction factor through a cross sec- tion of the pipe for the case in which the pipe flow represents an outlet of the control volume, as sketched in Fig. 6–15.
SOLUTION For a given velocity distribution we are to calculate the momen-
tum-flux correction factor. Assumptions 1 The flow is incompressible and steady. 2 The control volume slices through the pipe normal to the pipe axis, as sketched in Fig. 6–15. Analysis We substitute the given velocity profile for V in Eq. 6–24 and inte- grate, noting that dAc 2 p r dr ,
(2)
Defining a new integration variable y 1 r^2 / R^2 and thus dy 2 r dr / R^2 (also, y 1 at r 0, and y 0 at r R ) and performing the integration, the momentum-flux correction factor for fully developed laminar flow becomes
Laminar flow : (3)
Discussion We have calculated b for an outlet, but the same result would have been obtained if we had considered the cross section of the pipe as an inlet to the control volume.
From Example 6–1 we see that b is not very close to unity for fully devel-
oped laminar pipe flow, and ignoring b could potentially lead to significant
error. If we were to perform the same kind of integration as in Example 6–
but for fully developed turbulent rather than laminar pipe flow, we would
find that b ranges from about 1.01 to 1.04. Since these values are so close
to unity, many practicing engineers completely disregard the momentum-
flux correction factor. While the neglect of b in turbulent flow calculations
may have an insignificant effect on the final results, it is wise to keep it in
our equations. Doing so not only improves the accuracy of our calculations,
but reminds us to include the momentum-flux correction factor when solv-
ing laminar flow control volume problems.
For turbulent flow b may have an insignificant effect at inlets and outlets, but for laminar flow b may be important and should not be neglected. It is wise to include b in all momentum control volume problems.
b (^4)
0
1
y^2 dy 4 c
y^3 3
d
0
1
b
Ac Ac
a
V
V avg
b
2 dAc
p R^2
R
0
a 1
r^2 R^2
b
2 2 p r dr
V 2 V avg a 1
r^2 R^2
b
V avg
R^ V
r
CV
FIGURE 6–
Velocity profile over a cross section
of a pipe in which the flow is fully
developed and laminar.
239 CHAPTER 6
When the mass m of the control volume remains nearly constant, the first
term of the Eq. 6–28 simply becomes mass times acceleration since
Therefore, the control volume in this case can be treated as a solid body, with
a net force or thrust of
Thrust: (6–29)
acting on the body. This approach can be used to determine the linear accel-
eration of space vehicles when a rocket is fired (Fig. 6–19).
EXAMPLE 6–2 The Force to Hold a Deflector Elbow in Place
A reducing elbow is used to deflect water flow at a rate of 14 kg/s in a hori- zontal pipe upward 30° while accelerating it (Fig. 6–20). The elbow dis- charges water into the atmosphere. The cross-sectional area of the elbow is 113 cm^2 at the inlet and 7 cm^2 at the outlet. The elevation difference between the centers of the outlet and the inlet is 30 cm. The weight of the elbow and the water in it is considered to be negligible. Determine ( a ) the gage pressure at the center of the inlet of the elbow and ( b ) the anchoring force needed to hold the elbow in place.
SOLUTION A reducing elbow deflects water upward and discharges it to the
atmosphere. The pressure at the inlet of the elbow and the force needed to hold the elbow in place are to be determined. Assumptions 1 The flow is steady, and the frictional effects are negligible. 2 The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The flow is turbulent and fully developed at both the inlet and outlet of the control volume, and we take the momentum-flux correction factor to be b 1.03. Properties We take the density of water to be 1000 kg/m^3. Analysis ( a ) We take the elbow as the control volume and designate the inlet by 1 and the outlet by 2. We also take the x- and z- coordinates as shown. The continuity equation for this one-inlet, one-outlet, steady-flow sys- tem is m
1 ^ m
2 ^ m
14 kg/s. Noting that m
r AV , the inlet and outlet velocities of water are
V 2
m
r A 2
14 kg/s (1000 kg/m^3 )(7 10 ^4 m^2 )
20.0 m/s
V 1
m
r A 1
14 kg/s (1000 kg/m^3 )(0.0113 m^2 )
1.24 m/s
F
→
body ^ m body a
→
a
in
b m
V
→
a
out
b m
V
→
d ( mV
→ )CV dt
m CV
dV
→ CV dt
( ma → )CV
L = 2 m V 0 = 2000 m/s
FIGURE 6–
The thrust needed to lift the space
shuttle is generated by the rocket
engines as a result of momentum
change of the fuel as it is accelerated
from about zero to an exit speed of
about 2000 m/s after combustion.
NASA
FRz FRx
P atm
30 °
30 cm
P 1,gage
z
x
1
2
·
mV^ · 1
mV 2 →
→
FIGURE 6–
Schematic for Example 6–2.
240 FLUID MECHANICS
We use the Bernoulli equation (Chap. 5) as a first approximation to calculate the pressure. In Chap. 8 we will learn how to account for frictional losses along the walls. Taking the center of the inlet cross section as the reference level ( z 1 0) and noting that P 2 P atm, the Bernoulli equation for a stream- line going through the center of the elbow is expressed as
( b ) The momentum equation for steady one-dimensional flow is
We let the x - and z -components of the anchoring force of the elbow be FRx and FRz , and assume them to be in the positive direction. We also use gage pressure since the atmospheric pressure acts on the entire control surface. Then the momentum equations along the x- and z- axes become
Solving for FRx and FRz , and substituting the given values,
The negative result for FRx indicates that the assumed direction is wrong, and it should be reversed. Therefore, FRx acts in the negative x -direction. Discussion There is a nonzero pressure distribution along the inside walls of the elbow, but since the control volume is outside the elbow, these pressures do not appear in our analysis. The actual value of P 1, gage will be higher than that calculated here because of frictional and other irreversible losses in the elbow.
EXAMPLE 6–3 The Force to Hold a Reversing Elbow in Place
The deflector elbow in Example 6–2 is replaced by a reversing elbow such that the fluid makes a 180° U-turn before it is discharged, as shown in Fig. 6–21.
FRz b m
V 2 sin u (1.03)(14 kg/s)(20 sin 30 m/s)a
1 N
1 kg m/s^2
b 144 N
232 2285 2053 N
(202,200 N/m^2 )(0.0113 m^2 )
1.03(14 kg/s) 3 (20 cos 30 1.24) m/s4a
1 N
1 kg m/s^2
b
FRx b m
( V 2 cos u V 1 ) P 1, gage A 1
FRz b m
V 2 sin u
FRx P 1, gage A 1 b m
V 2 cos u b m
V 1
a F
→ (^) a out
b m
V
→ (^) a in
b m
V
→
P 1, gage 202.2 kN/m^2 202.2 kPa (gage)
a
(20 m/s)^2 (1.24 m/s)^2 2(9.81 m/s^2 )
0.3 0 b a 1 kN 1000 kg m/s^2
b
P 1 P atm (1000 kg/m^3 )(9.81 m/s^2 )
P 1 P 2 r g a
V^22 V^21
2 g
z 2 z 1 b
P 1
r g
V^21
2 g
z 1
P 2
r g
V^22
2 g
z 2
FRz FRx
P atm
P 1,gage
1
2
mV^ · 2
mV^ · 1
→
→
FIGURE 6–
Schematic for Example 6–3.
effect on the horizontal reaction force. 5 The effect of the momentum-flux correction factor is negligible, and thus b 1. Analysis We draw the control volume for this problem such that it contains the entire plate and cuts through the water jet and the support bar normally. The momentum equation for steady one-dimensional flow is given as
Writing it for this problem along the x -direction (without forgetting the nega- tive sign for forces and velocities in the negative x -direction) and noting that V 1, x V 1 and V 2, x 0 gives
Substituting the given values,
Therefore, the support must apply a 200-N horizontal force (equivalent to the weight of about a 20-kg mass) in the negative x -direction (the opposite direction of the water jet) to hold the plate in place. Discussion The plate absorbs the full brunt of the momentum of the water jet since the x -direction momentum at the outlet of the control volume is zero. If the control volume were drawn instead along the interface between the water and the plate, there would be additional (unknown) pressure forces in the analysis. By cutting the control volume through the support, we avoid having to deal with this additional complexity. This is an example of a “wise” choice of control volume.
EXAMPLE 6–5 Power Generation and Wind Loading
of a Wind Turbine
A wind generator with a 30-ft-diameter blade span has a cut-in wind speed (minimum speed for power generation) of 7 mph, at which velocity the tur- bine generates 0.4 kW of electric power (Fig. 6–23). Determine ( a ) the effi- ciency of the wind turbine–generator unit and ( b ) the horizontal force exerted by the wind on the supporting mast of the wind turbine. What is the effect of doubling the wind velocity to 14 mph on power generation and the force exerted? Assume the efficiency remains the same, and take the density of air to be 0.076 lbm/ft^3.
SOLUTION The power generation and loading of a wind turbine are to be
analyzed. The efficiency and the force exerted on the mast are to be deter- mined, and the effects of doubling the wind velocity are to be investigated. Assumptions 1 The wind flow is steady and incompressible. 2 The efficiency of the turbine–generator is independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is con- verted to thermal energy. 4 The average velocity of air through the wind tur- bine is the same as the wind velocity (actually, it is considerably less—see the discussion that follows the example). 5 The wind flow is uniform and thus the momentum-flux correction factor is b 1.
FR b m
V
→ 1 ^ (1)(10 kg/s)(20 m/s)a^
1 N
1 kg m/s^2
b 200 N
FR 0 b m
V
→ 1
a F
→ (^) a out
b m
V
→ (^) a in
b m
V
→
242 FLUID MECHANICS
1 2
P atm
P atm
mV^ · 2 mV^ · 1
FR
Streamline
x
→ →
FIGURE 6–
Schematic for Example 6–5.
Properties The density of air is given to be 0.076 lbm/ft^3. Analysis Kinetic energy is a mechanical form of energy, and thus it can be converted to work entirely. Therefore, the power potential of the wind is proportional to its kinetic energy, which is V^2 /2 per unit mass, and thus the maximum power is m
V^2 /2 for a given mass flow rate:
Therefore, the available power to the wind turbine is 1.225 kW at the wind velocity of 7 mph. Then the turbine–generator efficiency becomes
(or 32.7% )
( b ) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted to electric power leaves the wind turbine as outgoing kinetic energy. Noting that the mass flow rate remains constant, the exit velocity is determined to be
or
We draw a control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet and the entire con- trol surface is at atmospheric pressure. The momentum equation for steady one-dimensional flow is given as
Writing it along the x -direction and noting that b 1, V (^) 1, x V (^) 1 , and V (^) 2, x V (^) 2 give
Substituting the known values gives
31.5 lbf
FR m
( V 2 V 1 ) (551.7 lbm/s)(8.43 10.27 ft/s) a
1 lbf 32.2 lbm ft/s^2
b
FR m
V 2 m
V 1 m
( V 2 V 1 )
a F
→ (^) a out
b m
V
→ (^) a in
b m
V
→
V 2 V 1 21 hwind turbine (10.27 ft/s) 21 0.327 8.43 ft/s
m
ke 2 m
ke 1 (1 hwind turbine) → m
# V^^22
m
# V^^21
(1 hwind turbine)
hwind turbine
W
act W
max
0.4 kW 1.225 kW
1.225 kW
(551.7 lbm/s)
(10.27 ft/s)^2 2
a
1 lbf 32.2 lbm ft/s^2
b a
1 kW 737.56 lbf ft/s
b
W
max ^ m
ke 1 m
# V^
2 1 2
m
r 1 V 1 A 1 r 1 V 1
p D^2 4
(0.076 lbm/ft^3 )(10.27 ft/s)
p(30 ft)^2 4
551.7 lbm/s
V 1 (7 mph)a
1.4667 ft/s 1 mph b 10.27 ft/s
243 CHAPTER 6
effects and losses are neglected, the power generated by a wind turbine is simply the difference between the incoming and the outgoing kinetic energies:
Dividing this by the available power of the wind W
max ^ m
V^21 /2 gives the effi- ciency of the wind turbine in terms of a ,
The value of a that maximizes the efficiency is determined by setting the derivative of hwind turbine with respect to a equal to zero and solving for a. It gives a 1/3. Substituting this value into the efficiency relation just pre- sented gives hwind turbine 16/27 0.593, which is the upper limit for the efficiency of wind turbines and propellers. This is known as the Betz limit. The efficiency of actual wind turbines is about half of this ideal value.
EXAMPLE 6–6 Repositioning of a Satellite
An orbiting satellite has a mass of m sat 5000 kg and is traveling at a con- stant velocity of V (^) 0. To alter its orbit, an attached rocket discharges mf 100 kg of gases from the reaction of solid fuel at a velocity Vf 3000 m/s relative to the satellite in a direction opposite to V 0 (Fig. 6–25). The fuel discharge rate is constant for 2 s. Determine ( a ) the acceleration of the satellite during this 2-s period, ( b ) the change of velocity of the satellite dur- ing this time period, and ( c ) the thrust exerted on the satellite.
SOLUTION The rocket of a satellite is fired in the opposite direction to
motion. The acceleration, the velocity change, and the thrust are to be determined. Assumptions 1 The flow of combustion gases is steady and one-dimensional during the firing period. 2 There are no external forces acting on the satel- lite, and the effect of the pressure force at the nozzle exit is negligible. 3 The mass of discharged fuel is negligible relative to the mass of the satel- lite, and thus the satellite may be treated as a solid body with a constant mass. 4 The nozzle is well-designed such that the effect of the momentum- flux correction factor is negligible, and thus b 1. Analysis ( a ) We choose a reference frame in which the control volume moves with the satellite. Then the velocities of fluid streams become simply their velocities relative to the moving body. We take the direction of motion of the satellite as the positive direction along the x -axis. There are no exter- nal forces acting on the satellite and its mass is nearly constant. Therefore, the satellite can be treated as a solid body with constant mass, and the momentum equation in this case is simply Eq. 6–28,
d ( mV
→ )CV dt (^) a out
b m
V
→ (^) a in
b m
V
→ → m sat
dV
→ sat dt m
fV
→ f
hwind turbine
W
W
max
2 r AV^31 a (1 a )^2 (r AV 1 ) V^21 /
2 r AV^31 a (1 a )^2
W
m
(ke 1 ke 2 )
m
( V^21 V^22 )
r AV 1 (1 a ) 3 V^21 V^21 (1 2 a )^24 2
245 CHAPTER 6
V 0
Vf (^) Satellite m sat
x CS
→ →
FIGURE 6–
Schematic for Example 6–6.
Noting that the motion is on a straight line and the discharged gases move in the negative x -direction, we can write the momentum equation using mag- nitudes as
Substituting, the acceleration of the satellite during the first 2 s is deter- mined to be
( b ) Knowing acceleration, which is constant, the velocity change of the satel- lite during the first 2 s is determined from the definition of acceleration a sat dV sat / dt to be
( c ) The thrust exerted on the satellite is, from Eq. 6–29,
Discussion Note that if this satellite were attached somewhere, it would exert a force of 150 kN (equivalent to the weight of 15 tons of mass) to its support. This can be verified by taking the satellite as the system and apply- ing the momentum equation.
EXAMPLE 6–7 Net Force on a Flange
Water flows at a rate of 18.5 gal/min through a flanged faucet with a par- tially closed gate valve spigot (Fig. 6–26). The inner diameter of the pipe at the location of the flange is 0.780 in ( 0.0650 ft), and the pressure at that location is measured to be 13.0 psig. The total weight of the faucet assembly plus the water within it is 12.8 lbf. Calculate the net force on the flange.
SOLUTION Water flow through a flanged faucet is considered. The net force
acting on the flange is to be calculated. Assumptions 1 The flow is steady and incompressible. 2 The flow at the inlet and at the outlet is turbulent and fully developed so that the momentum- flux correction factor is about 1.03. 3 The pipe diameter at the outlet of the faucet is the same as that at the flange. Properties The density of water at room temperature is 62.3 lbm/ft^3. Analysis We choose the faucet and its immediate surroundings as the con- trol volume, as shown in Fig. 6–26 along with all the forces acting on it. These forces include the weight of the water and the weight of the faucet assembly, the gage pressure force at the inlet to the control volume, and the
F sat 0 m
f ( Vf )^ (100/2 kg/s)(3000 m/s)a^
1 kN 1000 kg m/s^2
b 150 kN
dV sat a sat dt → V sat a sat t (30 m/s^2 )(2 s) 60 m/s
a sat
dV sat dt
mf / t m sat
Vf
(100 kg)/(2 s) 5000 kg
(3000 m/s) 30 m/s^2
m sat
dV sat dt
m
f Vf →^
dV sat dt
m^ #^ f m sat Vf
mf / t m sat Vf
246 FLUID MECHANICS
W faucet
W water
P 1,gage
CV
Out
Spigot
Flange
x
z
In
FR
FIGURE 6–
Control volume for Example 6–7 with
all forces shown; gage pressure is used
for convenience.
