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Transcript of Cengel Numerical
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Special Case: Incompressible FlowThe conservation of mass relations can be simplified even further when the
fluid is incompressible, which is usually the case for liquids. Canceling the
density from both sides of the general steady-flow relation gives
Steady, incompressible flow: (522)
For single-stream steady-flow systems it becomes
Steady, incompressible flow (single stream): (523)
It should always be kept in mind that there is no such thing as a conserva-
tion of volume principle. Therefore, the volume flow rates into and out of a
steady-flow device may be different. The volume flow rate at the outlet of
an air compressor is much less than that at the inlet even though the mass
flow rate of air through the compressor is constant (Fig. 511). This is due
to the higher density of air at the compressor exit. For steady flow of liq-
uids, however, the volume flow rates, as well as the mass flow rates, remain
constant since liquids are essentially incompressible (constant-density) sub-
stances. Water flow through the nozzle of a garden hose is an example ofthe latter case.
The conservation of mass principle is based on experimental observations
and requires every bit of mass to be accounted for during a process. If you
can balance your checkbook (by keeping track of deposits and withdrawals,
or by simply observing the conservation of money principle), you should
have no difficulty applying the conservation of mass principle to engineer-
ing systems.
EXAMPLE 51 Water Flow through a Garden Hose Nozzle
A garden hose attached with a nozzle is used to fill a 10-gal bucket. Theinner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle
exit (Fig. 512). If it takes 50 s to fill the bucket with water, determine
(a) the volume and mass flow rates of water through the hose, and (b) the
average velocity of water at the nozzle exit.
SOLUTION A garden hose is used to fill a water bucket. The volume andmass flow rates of water and the exit velocity are to be determined.
Assumptions 1 Water is an incompressible substance. 2 Flow through the
hose is steady. 3 There is no waste of water by splashing.
Properties We take the density of water to be 1000 kg/m3 1 kg/L.
Analysis (a) Noting that 10 gal of water are discharged in 50 s, the volume
and mass flow rates of water are
(b) The cross-sectional area of the nozzle exit is
Ae pr2e p(0.4 cm)
2 0.5027 cm2 0.5027 104m2
m#
rV#
(1 kg/L)(0.757 L/s) 0.757 kg/s
V#
V
t
10 gal
50 sa3.7854 L
1 galb 0.757 L/s
V#
1 V#
2 V1A1 V2A2
ain
V#
aout
V# (m3/s)
178FLUID MECHANICS
m1 = 2 kg/s
Air
compressor
m2 = 2 kg/s
V2 = 0.8 m3/s
V1 = 1.4 m3/s
FIGURE 511
During a steady-flow process,
volume flow rates are not necessarily
conserved although mass flow
rates are.
Nozzle
BucketGardenhose
FIGURE 512
Schematic for Example 51.
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The volume flow rate through the hose and the nozzle is constant. Then the
average velocity of water at the nozzle exit becomes
Discussion It can be shown that the average velocity in the hose is 2.4 m/s.
Therefore, the nozzle increases the water velocity by over six times.
EXAMPLE 52 Discharge of Water from a Tank
A 4-ft-high, 3-ft-diameter cylindrical water tank whose top is open to the
atmosphere is initially filled with water. Now the discharge plug near the bot-
tom of the tank is pulled out, and a water jet whose diameter is 0.5 in
streams out (Fig. 513). The average velocity of the jet is given by
V , where h is the height of water in the tank measured from the
center of the hole (a variable) and g is the gravitational acceleration. Deter-
mine how long it will take for the water level in the tank to drop to 2 ft from
the bottom.
SOLUTION The plug near the bottom of a water tank is pulled out. Thetime it will take for half of the water in the tank to empty is to be deter-
mined.
Assumptions 1 Water is an incompressible substance. 2 The distance
between the bottom of the tank and the center of the hole is negligible com-
pared to the total water height. 3 The gravitational acceleration is 32.2 ft/s2.
Analysis We take the volume occupied by water as the control volume. The
size of the control volume decreases in this case as the water level drops,
and thus this is a variable control volume. (We could also treat this as a
fixed control volume that consists of the interior volume of the tank by disre-
garding the air that replaces the space vacated by the water.) This is obvi-
ously an unsteady-flow problem since the properties (such as the amount ofmass) within the control volume change with time.
The conservation of mass relation for a control volume undergoing any
process is given in the rate form as
(1)
During this process no mass enters the control volume (m.
in 0), and the
mass flow rate of discharged water can be expressed as
(2)
where Ajet pD2jet/4 is the cross-sectional area of the jet, which is constant.
Noting that the density of water is constant, the mass of water in the tank at
any time is(3)
where Atank pD2tank/4 is the base area of the cylindrical tank. Substituting
Eqs. 2 and 3 into the mass balance relation (Eq. 1) gives
r22ghAjet d(rAtankh)
dt r22gh(pD2jet/4)
r(pD2tank/4) dh
dt
mCV rV rAtankh
m#
out (rVA)out r22ghAjet
m#
in m#
out dmCV
dt
12gh
Ve V
#
Ae
0.757 L/s
0.5027 104 m2a 1 m3
1000 Lb 15.1 m/s
179CHAPTER 5
Water
Air
0 Dtank
Djeth2
h0
h
FIGURE 51
Schematic for Example 52
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Canceling the densities and other common terms and separating the vari-
ables give
Integrating from t 0 at which h h0 to t tat which h h2 gives
Substituting, the time of discharge is determined to be
Therefore, half of the tank will be emptied in 12.6 min after the discharge
hole is unplugged.
Discussion Using the same relation with h2 0 gives t 43.1 min for the
discharge of the entire amount of water in the tank. Therefore, emptying the
bottom half of the tank takes much longer than emptying the top half. This
is due to the decrease in the average discharge velocity of water with
decreasing h.
53 MECHANICAL ENERGY AND EFFICIENCY
Many fluid systems are designed to transport a fluid from one location to
another at a specified flow rate, velocity, and elevation difference, and the
system may generate mechanical work in a turbine or it may consume
mechanical work in a pump or fan during this process. These systems do
not involve the conversion of nuclear, chemical, or thermal energy to
mechanical energy. Also, they do not involve any heat transfer in any signif-
icant amount, and they operate essentially at constant temperature. Suchsystems can be analyzed conveniently by considering the mechanical forms
of energy only and the frictional effects that cause the mechanical energy to
be lost (i.e., to be converted to thermal energy that usually cannot be used
for any useful purpose).
The mechanical energy can be defined as the form of energy that can be
converted to mechanical work completely and directly by an ideal mechani-
cal device such as an ideal turbine. Kinetic and potential energies are the
familiar forms of mechanical energy. Thermal energy is not mechanical
energy, however, since it cannot be converted to work directly and com-
pletely (the second law of thermodynamics).
A pump transfers mechanical energy to a fluid by raising its pressure, and
a turbine extracts mechanical energy from a fluid by dropping its pressure.Therefore, the pressure of a flowing fluid is also associated with its mechan-
ical energy. In fact, the pressure unit Pa is equivalent to Pa N/m2
N m/m3 J/m3, which is energy per unit volume, and the product Pv or
its equivalent P/rhas the unit J/kg, which is energy per unit mass. Note that
pressure itself is not a form of energy. But a pressure force acting on a fluid
through a distance produces work, called flow work, in the amount of P/r
per unit mass. Flow work is expressed in terms of fluid properties, and it is
convenient to view it as part of the energy of a flowing fluid and call it flow
t24 ft 22 ft
232.2/2 ft/s2a3 12 in
0.5 inb2 757 s 12.6 min
t
0
dt D2tank
D2jet22g
h2
h0
dh
2h t
2h0 2h2
2g/2aDtankDjet
b2
dt D2tank
D2jet
dh
22gh
180FLUID MECHANICS
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occurs in EGL and HGL whenever mechanical energy is removed from
the fluid (by a turbine, for example), as shown in Fig. 536.
The pressure (gage) of a fluid is zero at locations where the HGL
intersects the fluid. The pressure in a flow section that lies above the HGL
is negative, and the pressure in a section that lies below the HGL is
positive (Fig. 537). Therefore, an accurate drawing of a piping system
and the HGL can be used to determine the regions where the pressure inthe pipe is negative (below the atmospheric pressure).
The last remark enables us to avoid situations in which the pressure drops
below the vapor pressure of the liquid (which causes cavitation, as dis-
cussed in Chap. 2). Proper consideration is necessary in the placement of a
liquid pump to ensure that the suction side pressure does not fall too low,
especially at elevated temperatures where vapor pressure is higher than it is
at low temperatures.
Now we examine Fig. 534 more closely. At point 0 (at the liquid surface),
EGL and HGL are even with the liquid surface since there is no flow there.
HGL decreases rapidly as the liquid accelerates into the pipe; however, EGL
decreases very slowly through the well-rounded pipe inlet. EGL declines con-tinually along the flow direction due to friction and other irreversible losses in
the flow. EGL cannot increase in the flow direction unless energy is supplied
to the fluid. HGL can rise or fall in the flow direction, but can never exceed
EGL. HGL rises in the diffuser section as the velocity decreases, and the sta-
tic pressure recovers somewhat; the total pressure does notrecover, however,
and EGL decreases through the diffuser. The difference between EGL and
HGL is V21/2g at point 1, and V22/2g at point 2. Since V1 V2, the difference
between the two grade lines is larger at point 1 than at point 2. The downward
slope of both grade lines is larger for the smaller diameter section of pipe
since the frictional head loss is greater. Finally, HGL decays to the liquid sur-
face at the outlet since the pressure there is atmospheric. However, EGL is
still higher than HGL by the amount V2
2/2g since V3
V2 at the outlet.
55 APPLICATIONS OF THEBERNOULLI EQUATION
In Section 54, we discussed the fundamental aspects of the Bernoulli equa-
tion. In this section, we demonstrate its use in a wide range of applications
through examples.
EXAMPLE 55 Spraying Water into the Air
Water is flowing from a hose attached to a water main at 400 kPa gage (Fig.538). A child places his thumb to cover most of the hose outlet, causing a
thin jet of high-speed water to emerge. If the hose is held upward, what is
the maximum height that the jet could achieve?
SOLUTION Water from a hose attached to the water main is sprayed intothe air. The maximum height the water jet can rise is to be determined.
Assumptions 1 The flow exiting into the air is steady, incompressible, and
irrotational (so that the Bernoulli equation is applicable). 2 The water pressure
194FLUID MECHANICS
PumpTurbine
EGL
HGL
Wpump Wturbine
FIGURE 536
A steep jump occurs in EGL and HGL
whenever mechanical energy is added
to the fluid by a pump, and a steep drop
occurs whenever mechanical energy is
removed from the fluid by a turbine.
Negative P
P= 0
P= 0HGL
Positive P
Positive P
FIGURE 537
The pressure (gage) of a fluid is zero
at locations where the HGL intersects
the fluid, and the pressure is negative
(vacuum) in a flow section that lies
above the HGL.
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in the hose near the outlet is equal to the water main pressure. 3 The surface
tension effects are negligible. 4 The friction between the water and air is neg-
ligible. 5 The irreversibilities that may occur at the outlet of the hose due to
abrupt expansion are negligible.
Properties We take the density of water to be 1000 kg/m3.
Analysis This problem involves the conversion of flow, kinetic, and potential
energies to each other without involving any pumps, turbines, and wastefulcomponents with large frictional losses, and thus it is suitable for the use of
the Bernoulli equation. The water height will be maximum under the stated
assumptions. The velocity inside the hose is relatively low (V1 0) and wetake the hose outlet as the reference level (z1 0). At the top of the water
trajectory V2 0, and atmospheric pressure pertains. Then the Bernoulli
equation simplifies to
Solving for z2 and substituting,
Therefore, the water jet can rise as high as 40.8 m into the sky in this case.
Discussion The result obtained by the Bernoulli equation represents the
upper limit and should be interpreted accordingly. It tells us that the water
cannot possibly rise more than 40.8 m, and, in all likelihood, the rise will be
much less than 40.8 m due to irreversible losses that we neglected.
EXAMPLE 56 Water Discharge from a Large Tank
A large tank open to the atmosphere is filled with water to a height of 5 m
from the outlet tap (Fig. 539). A tap near the bottom of the tank is now
opened, and water flows out from the smooth and rounded outlet. Determine
the water velocity at the outlet.
SOLUTION A tap near the bottom of a tank is opened. The exit velocity ofwater from the tank is to be determined.
Assumptions 1 The flow is incompressible and irrotational (except very close
to the walls). 2 The water drains slowly enough that the flow can be approxi-
mated as steady (actually quasi-steady when the tank begins to drain).
Analysis This problem involves the conversion of flow, kinetic, and potential
energies to each other without involving any pumps, turbines, and wasteful
components with large frictional losses, and thus it is suitable for the use of
the Bernoulli equation. We take point 1 to be at the free surface of water so
that P1 Patm (open to the atmosphere), V1 0 (the tank is large relative tothe outlet), and z1 5 m and z2 0 (we take the reference level at the
center of the outlet). Also, P2 Patm (water discharges into the atmosphere).
Then the Bernoulli equation simplifies to
P1
rg
V21
2g z1
P2
rg
V22
2gz2 z1
V22
2g
40.8 m
z2 P1 Patm
rg
P1, gage
rg
400 kPa
(1000 kg/m3
)(9.81 m/s2
) a1000 N/m2
1 kPa b a1 kg m/s2
1 N b
P1
rg
V21
2g z1
P2
rg
V 22
2g z2
P1
rg
Patm
rg z2
195CHAPTER 5
Water jet
2
Hose
0
z
1
FIGURE 53
Schematic for Example 55
Water5m
1
2
V2
FIGURE 53
Schematic for Example 56
0 0 0Q Q
0 0Q
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Solving for V2 and substituting,
The relation V is called the Toricelli equation.
Therefore, the water leaves the tank with an initial velocity of 9.9 m/s.
This is the same velocity that would manifest if a solid were dropped a dis-
tance of 5 m in the absence of air friction drag. (What would the velocity beif the tap were at the bottom of the tank instead of on the side?)
Discussion If the orifice were sharp-edged instead of rounded, then the flow
would be disturbed, and the velocity would be less than 9.9 m/s, especially
near the edges. Care must be exercised when attempting to apply the
Bernoulli equation to situations where abrupt expansions or contractions
occur since the friction and flow disturbance in such cases may not be neg-
ligible.
EXAMPLE 57 Siphoning Out Gasoline from a Fuel Tank
During a trip to the beach (Patm
1 atm
101.3 kPa), a car runs out ofgasoline, and it becomes necessary to siphon gas out of the car of a Good
Samaritan (Fig. 540). The siphon is a small-diameter hose, and to start the
siphon it is necessary to insert one siphon end in the full gas tank, fill the
hose with gasoline via suction, and then place the other end in a gas can
below the level of the gas tank. The difference in pressure between point 1
(at the free surface of the gasoline in the tank) and point 2 (at the outlet of
the tube) causes the liquid to flow from the higher to the lower elevation.
Point 2 is located 0.75 m below point 1 in this case, and point 3 is located
2 m above point 1. The siphon diameter is 4 mm, and frictional losses in
the siphon are to be disregarded. Determine (a) the minimum time to with-
draw 4 L of gasoline from the tank to the can and (b) the pressure at point
3. The density of gasoline is 750 kg/m3.
SOLUTION Gasoline is to be siphoned from a tank. The minimum time ittakes to withdraw 4 L of gasoline and the pressure at the highest point in
the system are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Even though the
Bernoulli equation is not valid through the pipe because of frictional losses,
we employ the Bernoulli equation anyway in order to obtain a best-case esti-
mate. 3 The change in the gasoline surface level inside the tank is negligible
compared to elevations z1 and z2 during the siphoning period.
Properties The density of gasoline is given to be 750 kg/m3.
Analysis (a) We take point 1 to be at the free surface of gasoline in the
tank so that P1 Patm (open to the atmosphere), V1 0 (the tank is largerelative to the tube diameter), and z2 0 (point 2 is taken as the reference
level). Also, P2 Patm (gasoline discharges into the atmosphere). Then the
Bernoulli equation simplifies to
Solving for V2 and substituting,
V2 22gz1 22(9.81 m/s2)(0.75 m) 3.84 m/s
P1
rg
V21
2g z1
P2
rg
V22
2gz2 z1
V22
2g
12gz
V2 22gz1 22(9.81 m/s2)(5 m) 9.9 m/s
196FLUID MECHANICS
0 0Q
0
z
z3
z1
z2
Gastank
0.75 m
2 m
Gasoline
siphoning
tube
Gas can
1
2
3
FIGURE 540
Schematic for Example 57.
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The cross-sectional area of the tube and the flow rate of gasoline are
Then the time needed to siphon 4 L of gasoline becomes
(b) The pressure at point 3 can be determined by writing the Bernoulli equa-
tion between points 2 and 3. Noting that V2 V3 (conservation of mass), z2 0, and P2 Patm,
Solving for P3 and substituting,
Discussion The siphoning time is determined by neglecting frictional
effects, and thus this is the minimum timerequired. In reality, the time will
be longer than 53.1 s because of friction between the gasoline and the tube
surface. Also, the pressure at point 3 is below the atmospheric pressure. If
the elevation difference between points 1 and 3 is too high, the pressure at
point 3 may drop below the vapor pressure of gasoline at the gasoline tem-
perature, and some gasoline may evaporate (cavitate). The vapor then may
form a pocket at the top and halt the flow of gasoline.
EXAMPLE 58 Velocity Measurement by a Pitot Tube
A piezometer and a Pitot tube are tapped into a horizontal water pipe, as
shown in Fig. 541, to measure static and stagnation (static dynamic)
pressures. For the indicated water column heights, determine the velocity at
the center of the pipe.
SOLUTION The static and stagnation pressures in a horizontal pipe aremeasured. The velocity at the center of the pipe is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Points 1 and 2 are
close enough together that the irreversible energy loss between these two
points is negligible, and thus we can use the Bernoulli equation.
Analysis We take points 1 and 2 along the centerline of the pipe, with point1 directly under the piezometer and point 2 at the tip of the Pitot tube. This
is a steady flow with straight and parallel streamlines, and the gage pres-
sures at points 1 and 2 can be expressed as
P2 rg(h1 h2 h3)
P1 rg(h1 h2)
81.1 kPa
101.3 kPa
(750 kg/m
3
)(9.81 m/s
2
)(2.75 m)a1 N
1 kg m/s2b a1 kPa
1000 N/m2b
P3 Patm rgz3
P2
rg
V22
2gz2
P3
rg
V23
2gz3
Patm
rg
P3
rgz3
t VV
# 4 L0.0753 L/s
53.1 s
V#
V2A (3.84 m/s)(1.96 105m2) 7.53 105m3/s 0.0753 L/s
A pD2/4 p(5 103m)2/4 1.96 105m2
197CHAPTER 5
0
h3= 12 cm
h2= 7 cm
h1= 3 cm
Stagnation
point
Water1 2
V1
FIGURE 54Schematic for Example 5
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Noting that point 2 is a stagnation point and thus V2 0 and z1 z2, the
application of the Bernoulli equation between points 1 and 2 gives
Substituting the P1 and P2 expressions gives
Solving for V1 and substituting,
Discussion Note that to determine the flow velocity, all we need is to mea-
sure the height of the excess fluid column in the Pitot tube.
EXAMPLE 59 The Rise of the Ocean Due to a Hurricane
A hurricane is a tropical storm formed over the ocean by low atmospheric
pressures. As a hurricane approaches land, inordinate ocean swells (very
high tides) accompany the hurricane. A Class-5 hurricane features winds in
excess of 155 mph, although the wind velocity at the center eye is very
low.
Figure 542 depicts a hurricane hovering over the ocean swell below. The
atmospheric pressure 200 mi from the eye is 30.0 in Hg (at point 1, gener-
ally normal for the ocean) and the winds are calm. The hurricane atmo-
spheric pressure at the eye of the storm is 22.0 in Hg. Estimate the ocean
swell at (a) the eye of the hurricane at point 3 and (b) point 2, where the
wind velocity is 155 mph. Take the density of seawater and mercury to be
64 lbm/ft3 and 848 lbm/ft3, respectively, and the density of air at normal
sea-level temperature and pressure to be 0.076 lbm/ft3.
SOLUTION A hurricane is moving over the ocean. The amount of oceanswell at the eye and at active regions of the hurricane are to be determined.
Assumptions 1 The airflow within the hurricane is steady, incompressible,
and irrotational (so that the Bernoulli equation is applicable). (This is cer-
tainly a very questionable assumption for a highly turbulent flow, but it is jus-
tified in the solution.) 2 The effect of water drifted into the air is negligible.
Properties The densities of air at normal conditions, seawater, and mercury
are given to be 0.076 lbm/ft3, 64 lbm/ft3, and 848 lbm/ft3, respectively.
Analysis (a) Reduced atmospheric pressure over the water causes the water
to rise. Thus, decreased pressure at point 2 relative to point 1 causes the
ocean water to rise at point 2. The same is true at point 3, where the storm air
velocity is negligible. The pressure difference given in terms of the mercury
column height can be expressed in terms of the seawater column height by
P (rgh)Hg (rgh)sw hsw rHg
rswhHg
V1 22gh3 22(9.81 m/s2)(0.12 m) 1.53 m/s
V21
2g
P2 P1rg
rg(h1 h2 h3) rg(h1 h2)
rg h3
P1
rg
V21
2gz1
P2
rg
V22
2g z2
V21
2g
P2 P1
rg
198FLUID MECHANICS
0Q
Calm
ocean
level
Ocean
HurricaneEye
1
2
3
h1h2
A B
FIGURE 542
Schematic for Example 59. The
vertical scale is greatly exaggerated.
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Then the pressure difference between points 1 and 3 in terms of the seawa-
ter column height becomes
which is equivalent to the storm surge at the eye of the hurricanesince thewind velocity there is negligible and there are no dynamic effects.
(b) To determine the additional rise of ocean water at point 2 due to the high
winds at that point, we write the Bernoulli equation between points A and B,
which are on top of the points 2 and 3, respectively. Noting that VB 0 (theeye region of the hurricane is relatively calm) and zA zB (both points are
on the same horizontal line), the Bernoulli equation simplifies to
Substituting,
where r is the density of air in the hurricane. Noting that the density of an
ideal gas at constant temperature is proportional to absolute pressure and
the density of air at the normal atmospheric pressure of 14.7 psia 30 inHg is 0.076 lbm/ft3, the density of air in the hurricane is
Using the relation developed above in part (a), the seawater column height
equivalent to 803 ft of air column height is determined to be
Therefore, the pressure at point 2 is 0.70 ft seawater column lower than the
pressure at point 3 due to the high wind velocities, causing the ocean to rise
an additional 0.70 ft. Then the total storm surge at point 2 becomes
h2 h1 hdynamic 8.83 0.70 9.53 ft
Discussion This problem involves highly turbulent flow and the intense
breakdown of the streamlines, and thus the applicability of the Bernoulli
equation in part (b) is questionable. Furthermore, the flow in the eye of the
storm is notirrotational, and the Bernoulli equation constant changes across
streamlines (see Chap. 10). The Bernoulli analysis can be thought of as the
limiting, ideal case, and shows that the rise of seawater due to high-velocity
winds cannot be more than 0.70 ft.
The wind power of hurricanes is not the only cause of damage to coastal
areas. Ocean flooding and erosion from excessive tides is just as serious, as
are high waves generated by the storm turbulence and energy.
hdynamic rair
rswhair a0.056 lbm/f t
3
64 lbm/ft3b (803 ft) 0.70 ft
rair Pair
Patm airratm air a22 in Hg
30 in Hgb (0.076 lbm/ft3) 0.056 lbm/f t3
PB PArg
V2A2g
(155 mph)2
2(32.2 ft/s2)a1.4667 ft/s
1 mph b 2 803 ft
PA
rg
V2A
2g zA
PB
rg
V2B
2g zB
PB PA
rg
V2A
2g
h1 rHg
rswhHg a848 lbm/ft
3
64 lbm/ft3b3(30 22) in Hg4a 1 f t
12 inb 8.83 ft
199CHAPTER 5
0Q
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EXAMPLE 510 Bernoulli Equation for Compressible Flow
Derive the Bernoulli equation when the compressibility effects are not negli-
gible for an ideal gas undergoing (a) an isothermal process and (b) an isen-
tropic process.
SOLUTION The Bernoulli equation for compressible flow is to be obtainedfor an ideal gas for isothermal and isentropic processes.
Assumptions 1 The flow is steady and frictional effects are negligible. 2 The
fluid is an ideal gas, so the relation P rRT is applicable. 3 The specific
heats are constant so that P/rk constant during an isentropic process.
Analysis (a) When the compressibility effects are significant and the flow
cannot be assumed to be incompressible, the Bernoulli equation is given by
Eq. 540 as
(1)
The compressibility effects can be properly accounted for by expressing r in
terms of pressure, and then performing the integration dP/r in Eq. 1. But
this requires a relation between Pand r for the process. For the isothermalexpansion or compression of an ideal gas, the integral in Eq. 1 can be per-
formed easily by noting that T constant and substituting r P/RT. It
gives
Substituting into Eq. 1 gives the desired relation,
Isothermal process: (2)
(b) A more practical case of compressible flow is the isentropic flow of ideal
gasesthrough equipment that involves high-speed fluid flow such as nozzles,
diffusers, and the passages between turbine blades. Isentropic (i.e.,reversible and adiabatic) flow is closely approximated by these devices, and
it is characterized by the relation P/rk C constant, where k is the spe-
cific heat ratio of the gas. Solving for r from P/rk Cgives r C1/kP1/k.
Performing the integration,
(3)
Substituting, the Bernoulli equation for steady, isentropic, compressible flow
of an ideal gas becomes
Isentropic flow: (4a)
or
(4b)
A common practical situation involves the acceleration of a gas from rest
(stagnation conditions at state 1) with negligible change in elevation. In
that case we have z1 z2 and V1 0. Noting that r P/RT for ideal
a kk 1
bP1r1
V212
gz1 a kk 1bP2r2
V222
gz2
a kk 1
bPR
V2
2 gz constant
dPr C1/kP1/k dP C1/k P1/k1
1/k 1
P1/k
rP1/k1
1/k 1a k
k 1bPr
RTlnP V2
2 gz constant
dPr dPP/RT RTln P
dPr V2
2 gz constant (along a streaml ine)
200FLUID MECHANICS
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gases, P/rk constant for isentropic flow, and the Mach number is defined
as Ma V/cwhere is the local speed of sound for ideal gases,
Eq. 4b simplifies to
(4c)
where state 1 is the stagnation state and state 2 is any state along the flow.Discussion It can be shown that the results obtained using the compress-
ible and incompressible equations deviate no more than 2 percent when the
Mach number is less than 0.3. Therefore, the flow of an ideal gas can be
considered to be incompressible when Ma 0.3. For atmospheric air at
normal conditions, this corresponds to a flow speed of about 100 m/s or
360 km/h, which covers our range of interest.
56 GENERAL ENERGY EQUATION
One of the most fundamental laws in nature is the first law of thermody-
namics, also known as the conservation of energy principle, which pro-vides a sound basis for studying the relationships among the various formsof energy and energy interactions. It states that energy can be neither cre-ated nor destroyed during a process; it can only change forms. Therefore,every bit of energy must be accounted for during a process.
A rock falli ng off a cli ff , for example, picks up speed as a result of itspotential energy being converted to kinetic energy (Fig. 543). Experimen-tal data show that the decrease in potential energy equals the increase inki neti c energy when the air resistance is negli gibl e, thus confi rmi ng the con-servation of energy principle. The conservation of energy principle alsoforms the backbone of the diet industry: a person who has a greater energyinput (food) than energy output (exercise) will gain weight (store energy in
the form of fat), and a person who has a smaller energy i nput than outputwill lose weight. The change in the energy content of a system is equal tothe dif ference between the energy input and the energy output, and theconservation of energy principle for any system can be expressed simply asEin Eout E.
The transfer of any quantity (such as mass, momentum, and energy) isrecognizedat the boundaryas the quantity crosses the boundary. A quantityis said to enter a system if it crosses the boundary from the outside to theinside, and to exit the system if it moves in the reverse direction. A quantitythat moves from one location to another within a system is not consideredas a transferred quantity in an analysis since it does not enter or exit the sys-tem. Therefore, it is important to specif y the system and thus clearly iden-
tify its boundaries before an engineering analysis is performed.The energy content of a fixed quantity of mass (a closed system) can be
changed by two mechanisms:heat transfer Qandwork transfer W. Then theconservation of energy for a fixed quantity of mass can be expressed in rateform as (Fig. 544)
(549)Q#
net in W#
net in dEsys
dt or Q
#
net in W#
net in d
dtsys
re dV
P1P2
c1 ak 12 b M a22d
k/(k1)
c 1kRT
201CHAPTER 5
PE1= 10 m
K E1= 0
PE2= 7 km K E2= 3 k
z
FIGURE 54
Energy cannot be created o
destroyed duri ng a procesit can only change form
Wshaft, i n = 6 k
= 18 kJ
Qout
= 3 kJ
Qin
= 15 kJ
E= (15 3) + 6
FIGURE 54
The energy change of a systeduring a process is equal to the nwork and heat transfer between th
system and its surrounding
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239CHAPTER 6
When the mass mof the control volume remains nearly constant, the fi rstterm of the Eq. 628 simply becomes mass times acceleration since
Therefore, the control volume in this case can be treated as a solid body, wi tha net force or thrust of
Thrust (629)
acting on the body. This approach can be used to determine the linear accel-eration of space vehicles when a rocket is fired (Fig. 619).
EXAMPLE 62 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. 620). The elbow dis-
charges water into the atmosphere. The cross-sectional area of the elbow is
113 cm2 at the inlet and 7 cm2 at the outlet. The elevation difference
between the centers of the outlet and the inlet is 30 cm. The weight of theelbow 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 theatmosphere. 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/m3.
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
. rAV, the inlet and outlet
velocities of water are
V2 m
#
rA2
14 kg/s
(1000 kg/m3)(7 10 4m2) 20.0 m/s
V1 m
#
rA1
14 kg/s
(1000 kg/m3)(0.0113 m2) 1.24 m/s
F
body mbodya
ain
bm#V
aout
bm#V
d(mV
)CV
dt mCV
dV
CV
dt (ma
)CV
= 2 m
V0= 2000 m/s
FIGURE 61
The thrust needed to lift the spacshuttle is generated by the rockeengines as a result of momentum
change of the fuel as it is acceleratefrom about zero to an exit speed o
about 2000 m/s after combustionNAS
FRz
FRx
Patm
30
30 cm
P1,gage
z
x
1
2
mV1
mV2
FIGURE 62
Schematic for Example 62
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240FLUID 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 (z1 0) and noting that P2 Patm, 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 FRxand 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 P1, gage will be higher than
that calculated here because of frictional and other irreversible losses in theelbow.
EXAMPLE 63 The Force to Hold a Reversing Elbow in Place
The deflector elbow in Example 62 is replaced by a reversing elbow such that
the fluid makes a 180 U-turn before it is discharged, as shown in Fig. 621.
FRz bm#V2sin u (1.03)(14 kg/s)(20 sin 30m/s)a 1 N
1 kg m/s2b 144 N
232 2285 2053 N
(202,200 N/m2)(0.0113 m2)
1.03(14 kg/s)3(20 cos 30 1.24) m/s4a1 N
1 kg m/s2b
FRx bm#(V2cos u V1) P1, gageA 1
FRz bm#V2sin u
FRx P1, gageA1 bm#V2cos u bm
#V1
a F
aout bm#
V
ain bm#
V
P1, gage 202.2 kN/m2
202.2 kPa (gage)
a(20 m/s)2 (1.24 m/s)22(9.81 m/s2)
0.3 0b a 1 kN1000 kg m/s2
bP1 Patm (1000 kg/m
3)(9.81 m/s2)
P1 P2 rgaV22 V
21
2g z2 z1b
P1
rgV2
12g
z1 P
2rg
V22
2g z2
FRzFRx
Patm
P1,gage
1
2
mV2
mV1
FIGURE 621
Schematic f or Example 63.
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The elevation difference between the centers of the inlet and the exit sec-
tions is still 0.3 m. Determine the anchoring force needed to hold the elbow
in place.
SOLUTION The inlet and the outlet velocities and the pressure at the inletof the elbow remain the same, but the vertical component of the anchoring
force at the connection of the elbow to the pipe is zero in this case ( FRz 0)since there is no other force or momentum flux in the vertical direction (we
are neglecting the weight of the elbow and the water). The horizontal compo-
nent of the anchoring force is determined from the momentum equation
written in the x-direction. Noting that the outlet velocity is negative since it
is in the negative x-direction, we have
Solving for FRx and substituting the known values,
Therefore, the horizontal force on the flange is 2591 N acting in the nega-
tive x-direction (the elbow is trying to separate from the pipe). This force is
equivalent to the weight of about 260 kg mass, and thus the connectors
(such as bolts) used must be strong enough to withstand this force.
Discussion The reaction force in the x-direction is larger than that of Exam-
ple 62 since the walls turn the water over a much greater angle. If the
reversing elbow is replaced by a straight nozzle (like one used by firefighters)
such that water is discharged in the positive x-direction, the momentum
equation in the x-direction becomes
since both V1 and V2 are in the positive x-direction. This shows the impor-tance of using the correct sign (positive if in the positive direction and nega-
tive if in the opposite direction) for velocities and forces.
EXAMPLE 64 Water Jet Striking a Stationary Plate
Water is accelerated by a nozzle to an average speed of 20 m/s, and strikes
a stationary vertical plate at a rate of 10 kg/s with a normal velocity of
20 m/s (Fig. 622). After the strike, the water stream splatters off in all
directions in the plane of the plate. Determine the force needed to prevent
the plate from moving horizontally due to the water stream.
SOLUTION A water jet strikes a vertical stationary plate normally. The forceneeded to hold the plate in place is to be determined.
Assumptions 1 The flow of water at nozzle outlet is steady. 2 The water
splatters in directions normal to the approach direction of the water jet. 3
The water jet is exposed to the atmosphere, and thus the pressure of the
water jet and the splattered water leaving the control volume is atmospheric
pressure, which is disregarded since it acts on the entire system. 4 The ver-
tical forces and momentum fluxes are not considered since they have no
FRx P1, gageA1 bm#V2 bm
#V1 FRx bm
#(V2 V1) P1, gageA1
306 2285 2591 N
(1.03)(14 kg/s)3(20 1.24) m/s4a 1 N1 kg m/s2
b (202,200 N/m2)(0.0113 m2)FRx bm
#(V2 V1) P1, gageA1
FRx P1, gageA1 b2m#(V2) b1m
#V1 bm
#(V2 V1)
241CHAPTER 6
z
x
Patm
In
Out
V1
V2
1
FIGURE 62
Schematic for Example 64
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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
V1, x V1 and V2, 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 65 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. 623). Determine (a) the effi-
ciency of the wind turbinegenerator 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/ft3.
SOLUTION The power generation and loading of a wind turbine are to beanalyzed. 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 turbinegenerator 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 lesssee
the discussion that follows the example). 5 The wind flow is uniform and
thus the momentum-flux correction factor is b 1.
FR bm#V
1 (1)(10 kg/s)(20 m/s)a 1 N1 kg m/s2
b 200 N
FR 0 bm#V
1
a F
aout bm#V
ain bm#V
242FLUID MECHANICS
1 2
Patm
Patm
mV2
mV1
FR
Streamline
x
FIGURE 623
Schematic f or Example 65.
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Properties The density of air is given to be 0.076 lbm/ft3.
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 V2/2 per unit mass, and thus
the maximum power is m.V2/2 for a given mass flow rate:
Therefore, the available power to the wind turbine is 1.225 kW at the windvelocity of 7 mph. Then the turbinegenerator 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, V1, x V1, and V2, xV2 give
Substituting the known values gives
31.5 lbf
FR m#(V2 V1) (551.7 l bm/s)(8.43 10.27 ft/s) a 1 lbf
32.2 lbm ft/s2b
FR m# V2 m# V1 m# (V2 V1)
a F
aout
bm#V
ain
bm#V
V2 V121 hwind turbine (10.27 ft /s)21 0.327 8.43 ft/s
m#ke2 m
#ke1(1 hwind turbine) m
# V22
2 m
# V21
2(1 hwind turbine)
hwind turbine W
#
act
W#
max
0.4 kW
1.225 kW 0.327
1.225 kW
(551.7 lbm/s)(10.27 ft/s)2
2 a 1 lbf
32.2 lbm ft/s2b a 1 kW
737.56 lbf ft/sb
W#
max m#ke1 m
# V21
2
m#
r1V1A1 r1V1pD2
4 (0.076 lbm/ft3)(10.27 ft/s)
p(30 ft)2
4 551.7 lbm/s
V1 (7 mph)a1.4667 ft/s1 mph b 10.27 ft/s
243CHAPTER 6
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The negative sign indicates that the reaction force acts in the negative x-
direction, as expected. Then the force exerted by the wind on the mast
becomes Fmast FR 31.5 lbf.
The power generated is proportional to V3 since the mass flow rate is pro-
portional to V and the kinetic energy to V2. Therefore, doubling the wind
velocity to 14 mph will increase the power generation by a factor of 23 8
to 0.4 8 3.2 kW. The force exerted by the wind on the support mast isproportional to V2. Therefore, doubling the wind velocity to 14 mph will
increase the wind force by a factor of 22 4 to 31.5 4 126 lbf.
Discussion To gain more insight into the operation of devices with propellers
or turbines such as helicopters, wind turbines, hydraulic turbines, and turbo-
fan engines, we reconsider the wind turbine and draw two streamlines, as
shown in Fig. 624. (In the case of power-consuming devices such as a fan
and a helicopter, the streamlines converge rather than diverge since the exit
velocity will be higher and thus the exit area will be lower.) The upper and
lower streamlines can be considered to form an imaginary duct for the
flow of air through the turbine. Sections 1 and 2 are sufficiently far from the
turbine so that P1 P2 Patm. The momentum equation for this large con-
trol volume between sections 1 and 2 was obtained to be
(1)
The smaller control volume between sections 3 and 4 encloses the turbine,
and A3 A4 A and V3 V4 since it is so slim. The turbine is a device
that causes a pressure change, and thus the pressures P3 and P4 are differ-
ent. The momentum equation applied to the smaller control volume gives
(2)
The Bernoulli equation is not applicable between sections 1 and 2 since the
path crosses a turbine, but it is applicable separately between sections 1
and 3 and sections 4 and 2:
Adding these two equations and noting that z1 z2 z3 z4, V3 V4,
and P1 P2 Patm gives
(3)
Substituting m.
rAV3 into Eq. 1 and then combining it with Eqs. 2 and 3
gives
(4)
Thus we conclude that the average velocity of a fluid through a turbine is the
arithmetic average of the upstream and downstream velocities. Of course, the
validity of this result is limited by the applicability of the Bernoulli equation.
Now back to the wind turbine. The velocity through the turbine can be
expressed as V3 V1(1 a), where a 1 since V3 V1. Combining this
expression with Eq. 4 gives V2 V1(1 2a). Also, the mass flow rate
through the turbine becomes m.
rAV3 rAV1(1 a). When the frictional
V3 V1 V2
2
V22 V21
2
P4 P3
r
P1
rg
V21
2g z1
P3
rg
V23
2g z3
and
P4
rg
V24
2g z4
P2
rg
V22
2g z2
FR P3 A P4 A 0 FR (P4 P3)A
FR m# (V2 V1)
244FLUID MECHANICS
1 3 4 2
Patm
Patm
V2V1
A
V3 =V4
Streamline
Streamline
Windturbine
FIGURE 624
The large and small control volumesfor the analysis of a wind turbinebounded by upper and lowerstreamlines.
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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.V21/2 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 aequal 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 66 Repositioning of a Satellite
An orbiting satellite has a mass of msat 5000 kg and is traveling at a con-
stant velocity of V0. 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 V0 (Fig. 625). 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 tomotion. 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. 628,
0 d(mV
)CV
dt a
out
bm#V
ain
bm#V
msat
dV
sat
dt m
#fV
f
hwind turbine W
#
W#
max
2rAV31a(1 a)
2
(rAV1)V21/2
2rAV31a(1 a)2
W#
m#(ke1 ke2)
m#(V21 V
22)
2
rAV1(1 a)3V21 V21(1 2a)242
245CHAPTER 6
V0
Vf Satellitemsat
xCS
FIGURE 62
Schematic for Example 66
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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 asat dVsat/dtto be
(c) The thrust exerted on the satellite is, from Eq. 629,
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 67 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. 626). 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 forceacting 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/ft3.
Analysis We choose the faucet and its immediate surroundings as the con-
trol volume, as shown in Fig. 626 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
Fsat 0 m#
f(Vf) (100/2 kg/s)(3000 m/s)a 1 kN1000 kg m/s2
b 150 kN
dVsat asat dt Vsat asatt (30 m/s2)(2 s) 60 m/s
asat dVsat
dt
mf/t
msatVf
(100 kg)/(2 s)
5000 kg(3000 m/s) 30 m/s2
msatdVsat
dt m
#fVf
dVsat
dt
m#
f
msatVf
mf/t
msatVf
246FLUID MECHANICS
Wfaucet
Wwater
P1,gage
CV
Out
Spigot
Flange
x
z
In
FR
FIGURE 626
Control volume for Example 67 wi thall forces shown; gage pressure is usedfor convenience.
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net force of the flange on the control volume, which we call F
R. We use gage
pressure for convenience since the gage pressure on the rest of the control
surface is zero (atmospheric pressure). Note that the pressure through the
outlet of the control volume is also atmospheric since we are assuming
incompressible flow; hence, the gage pressure is also zero through the outlet.
We now apply the control volume conservation laws. Conservation of mass
is trivial here since there is only one inlet and one outlet; namely, the massflow rate into the control volume is equal to the mass flow rate out of the
control volume. Also, the outflow and inflow average velocities are identical
since the inner diameter is constant and the water is incompressible, and
are determined to be
Also,
Next we apply the momentum equation for steady flow,
We let the x- and z-components of the force acting on the flange be FRx and
FRz, and assume them to be in the positive directions. The magnitude of the
velocity in the x-direction is V1 at the inlet, but zero at the outlet. The
magnitude of the velocity in the z-direction is zero at the inlet, but V2 at
the outlet. Also, the weight of the faucet assembly and the water within it
acts in the z-direction as a body force. No pressure or viscous forces act on
the chosen control volume in the z-direction.
The momentum equations along the x- and z-directions become
Solving for FRx and FRz, and substituting the given values,
Then the net force of the flange on the control volume can be expressed in
vector form as
F
R FRxi
FRzk
7.20 i
11.8 k
lbf
(2.568 lbm/s)(12.42 f t/s)a 1 lbf32.2 lbm ft/s2
b 12.8 lbf 11.8 lbfFRz m
#V2 Wfaucetwater
7.20 lbf
(2.568 lbm/s)(12.42 f t/s)a 1 lbf32.2 lbm ft/s2
b (13 lbf/in2)p(0.780 in)24
FRx m#V1 P1, gageA1
FRz Wfaucet Wwater m#(V2) 0
FRx P1, gageA1 0 m#(V1)
a F
aout
bm#V
ain
bm#V
m#
rV#
(62.3 lbm/f t3)(18.5 gal/mi n)a0.1337 ft31 gal
b a1 min60 s
b 2.568 lbm/s
V2 V1 V V
#
A c
V#
pD2/4
18.5 gal/mi n
p(0.065 ft)2/4a0.1337 ft3
1 gal b a1 min
60 sb 12.42 ft/s
247CHAPTER 6
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From Newtons third law, the force the faucet assembly exerts on the flange
is the negative of F
R,
Discussion The faucet assembly pulls to the right and down; this agrees
with our intuition. Namely, the water exerts a high pressure at the inlet, butthe outlet pressure is atmospheric. In addition, the momentum of the water
at the inlet in the x-direction is lost in the turn, causing an additional force
to the right on the pipe walls. The faucet assembly weighs much more than
the momentum effect of the water, so we expect the force to be downward.
Note that labeling forces such as faucet on flange clarifies the direction of
the force.
65
REVIEW OF ROTATIONAL MOTIONAND ANGULAR MOMENTUM
The motion of a rigid body can be considered to be the combination of thetranslational motion of its center of mass and rotational motion about itscenter of mass. The translational motion can be analyzed using the linearmomentum equation, Eq. 616. Now w e discuss the rotational motionamotion during which all points in the body move in circles about the axis ofrotation. Rotational motion is described with angular quantities such as theangular distance u, angular velocity v, and angular acceleration a.
The amount of rotation of a point in a body is expressed in terms of theangleu swept by a line of length r that connects the point to the axis of rota-tion and is perpendicular to the axis. The angle u is expressed in radians (rad),which is the arc length corresponding to u on a circle of unit radius. Notingthat the circumference of a circle of radius r is 2pr, the angular distance trav-eled by any point in a rigid body during a complete rotation is 2p rad. Thephysical distance traveled by a point along its circular path is l ur, where ris the normal distance of the point from the axis of rotation and u is the angu-lar distance in rad. Note that 1 rad corresponds to 360/(2p) 57.3.
Angular velocity v is the angular distance traveled per unit time, andangular acceleration a is the rate of change of angular velocity. They areexpressed as (Fig. 627),
(630)
or
(631)
whereV is the linear velocity and at is the linear acceleration in the tangen-tial direction for a point located at a distance r from the axis of rotation.Note that v anda are the same for all points of a rotating rigi d body, but Vandat are not (they are proportional to r).
V rv and at ra
v du
dt
d(l/r)
dt
1
r
dl
dt
V
r and a dv
dt
d2u
dt2
1
r
dV
dt
at
r
F
faucet on fl ange F
R 7.20
11.8k
lbf
248FLUID MECHANICS
v
v= =
du
dt
u
r
r
V= rv
V
r
FIGURE 627
The relations between angulardistance u, angular velocity v,and linear velocity V.
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EXAMPLE 81 Flow Rates in Horizontal and Inclined Pipes
Oil at 20C (r 888 kg/m3 and m 0.800 kg/m s) is flowing steadily
through a 5-cm-diameter 40-m-long pipe (Fig. 817). The pressure at the
pipe inlet and outlet are measured to be 745 and 97 kPa, respectively.
Determine the flow rate of oil through the pipe assuming the pipe is (a) hor-
izontal, (b) inclined 15 upward, (c) inclined 15 downward. Also verify thatthe flow through the pipe is laminar.
SOLUTION The pressure readings at the inlet and outlet of a pipe are given.The flow rates are to be determined for three different orientations, and the
flow is to be shown to be laminar.
Assumptions 1 The flow is steady and incompressible. 2 The entrance
effects are negligible, and thus the flow is fully developed. 3 The pipe
involves no components such as bends, valves, and connectors. 4 The piping
section involves no work devices such as a pump or a turbine.
Properties The density and dynamic viscosity of oil are given to be r
888 kg/m3 and m 0.800 kg/m s, respectively.
Analysis The pressure drop across the pipe and the pipe cross-sectional
area are
(a) The flow rate for all three cases can be determined from Eq. 834,
where u is the angle the pipe makes with the horizontal. For the horizontal
case, u 0 and thus sin u 0. Therefore,
0.00311 m3/s
V#
horiz
PpD4
128mL
(648 k Pa)p(0.05 m)4
128(0.800 kg/m s)(40 m)a1000 N/m2
1 kPa b a1 kg m/s2
1 N b
V#
(P rgL sin u)pD4
128mL
A c pD2/4 p(0.05 m)2/4 0.001963 m2
P P1 P2 745 97 648 kPa
333CHAPTER 8
(b) For uphill flow with an inclination of 15, we have u 15, and
(c) For downhill flow with an inclination of 15, we have u 15, and
0.00354 m3/s
[648,000 Pa (888 kg/m3)(9.81 m/s2)(40 m)sin (15)]p(0.05 m)4
128(0.800 kg/m s)(40 m) a1 kg m/s2
1 Pa m2b
V#
downhill (P rgL sin u)pD4
128mL
0.00267 m3/s
[648,000 Pa (888 kg/m3)(9.81 m/s2)(40 m)sin 15]p(0.05 m)4
128(0.800 kg/m s)(40 m) a1 kg m/s2
1 Pa m2b
V#
uphill (P rgL sin u)pD4
128mL
+15
15
Horizontal
FIGURE 81
Schematic for Example 8
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The flow rate is the highest for the downhill flow case, as expected. The
average fluid velocity and the Reynolds number in this case are
which is much less than 2300. Therefore, the flow is laminar for all three
cases and the analysis is valid.
Discussion Note that the flow is driven by the combined effect of pressure
difference and gravity. As can be seen from the flow rates we calculated,
gravity opposes uphill flow, but enhances downhill flow. Gravity has no effect
on the flow rate in the horizontal case. Downhill flow can occur even in the
absence of an applied pressure difference. For the case of P1 P2 97 kPa
(i.e., no applied pressure difference), the pressure throughout the entire pipe
would remain constant at 97 Pa, and the fluid would flow through the pipe at
a rate of 0.00043 m3/s under the influence of gravity. The flow rate increases
as the tilt angle of the pipe from the horizontal is increased in the negative
direction and would reach its maximum value when the pipe is vertical.
EXAMPLE 82 Pressure Drop and Head Loss in a Pipe
Water at 40F (r 62.42 lbm/ft3 and m 1.038 103 lbm/ft s) is
flowing through a 0.12-in- ( 0.010 ft) diameter 30-ft-long horizontal pipe
steadily at an average velocity of 3.0 ft/s (Fig. 818). Determine (a) the head
loss, (b) the pressure drop, and (c) the pumping power requirement to over-
come this pressure drop.
SOLUTION The average flow velocity in a pipe is given. The head loss, thepressure drop, and the pumping power are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance
effects are negligible, and thus the flow is fully developed. 3 The pipe
involves no components such as bends, valves, and connectors.
Properties The density and dynamic viscosity of water are given to be r
62.42 lbm/ft3 and m 1.038 103 lbm/ft s, respectively.
Analysis (a) First we need to determine the flow regime. The Reynolds num-
ber is
which is less than 2300. Therefore, the flow is laminar. Then the friction
factor and the head loss become
(b) Noting that the pipe is horizontal and its diameter is constant, the pres-
sure drop in the pipe is due entirely to the frictional losses and is equivalent
to the pressure loss,
hL fL
D
V2avg
2g 0.0355
30 ft
0.01 ft
(3 ft/ s)2
2(32.2 ft/s2) 14.9 ft
f64Re
64
1803 0.0355
Re rVavgD
m
(62.42 lbm/ft3)(3 ft/s)(0.01 ft)
1.038 10 3lbm/ft s 1803
Re
rVavgD
m
(888 kg/m3)(1.80 m/s)(0.05 m)
0.800 kg/m s 100
Vavg V
#
A c
0.00354 m3/s
0.001963 m2 1.80 m/s
334FLUID MECHANICS
3.0 ft/s
30 ft
0.12 in
FIGURE 818Schematic f or Example 82.
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(c) The volume flow rate and the pumping power requirements are
Therefore, power input in the amount of 0.30 W is needed to overcome the
frictional losses in the flow due to viscosity.
Discussion The pressure rise provided by a pump is often listed by a pump
manufacturer in units of head (Chap. 14). Thus, the pump in this flow needs to
provide 14.9 ft of water head in order to overcome the irreversible head loss.
85
TURBULENT FLOW IN PIPESM ost f lows encountered in engineering practice are turbulent, and thus i t isimportant to understand how turbulence affects wall shear stress. However,turbulent f low is a complex mechanism dominated by f luctuations, anddespite tremendous amounts of work done in thi s area by researchers, thetheory of turbulent f low remains largely undeveloped. Therefore, we mustrely on experiments and the empirical or semi-empirical correlations devel-oped for various situations.
Turbulent flow is characterized by random and rapid fluctuations ofswi rli ng regions of f luid, call ed eddies, throughout the flow. These fluctua-tions provide an additional mechanism for momentum and energy transfer.In laminar f low, fl uid parti cles fl ow in an orderly manner along pathli nes,
and momentum and energy are transferred across streamlines by moleculardif fusion. In turbulent fl ow, the swi rl ing eddies transport mass, momentum,and energy to other regions of flow much more rapidly than molecular dif-fusion, greatly enhancing mass, momentum, and heat transfer. A s a result,turbulent f low is associated with much higher values of fr icti on, heat trans-fer, and mass transfer coeffi cients (Fig. 819).
Even when the average f low is steady, the eddy moti on in turbulent fl owcauses signif icant fluctuations in the values of veloci ty, temperature, pres-sure, and even density (in compressible f low). Figure 820 shows the varia-tion of the instantaneous velocity component uwith time at a specified loca-tion, as can be measured wi th a hot-wi re anemometer probe or othersensitive device. We observe that the instantaneous values of the velocity
f luctuate about an average value, which suggests that the veloci ty can beexpressed as the sum of an average value uand a fluctuating component u,
(835)
This is also the case for other properties such as the velocity component vin the y-direction, and thus v v v,P P
P, and T T
T. The
average value of a property at some location is determined by averaging itover a time interval that is sufficiently large so that the time average levelsoff to a constant. Therefore, the time average of f luctuating components is
u u u
W#
pump V#
P (0.000236 ft3/s)(929 lbf /f t2)a 1 W0.737 lbf ft/s
b 0.30 WV
#
Vavg A c Vavg(pD2
/4) (3 ft/s)[p(0.01 ft)2
/4] 0.000236 f t3
/s
929 lbf/ft2 6.45 psi
P PL fL
D
rV2avg
2 0.0355
30 ft
0.01 ft
(62.42 lbm/ft3)(3 ft/s)2
2 a 1 lbf
32.2 lbm ft/s2b
335CHAPTER 8
(a) Before turbulence
2 2 2 2 2
55
7
12
7
12
7
12
7
12
7
12
5 5 5
(b) Af ter turbulence
12 2 5 7 5
122
7
2
7
5
1
2
12
7
5
12
5 7 2
FIGURE 81
The intense mixi ng in turbulent fl o
brings fluid particles at differenmomentums into close contact anthus enhances momentum t ransfe
u
uu
Time,
FIGURE 82
Fluctuations of the velocitcomponentuwith time at a specifie
location in turbulent f low