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FE
Rapid Preparation for the Civil
Fundamentals of Engineering Exam
CIVIL
REVIEW MANUAL
Michael R. Lindeburg, PE
Professional Publications, Inc. • Belmont, California
10
Hydraulics
. . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . .
1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Conservation Laws . . . . . . . . . . . . . . . . . . . . . .
3. Steady Incompressible Flow in Pipes and
Conduits . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4. Flow in Noncircular Conduits . . . . . . . . . . . .
5. Open Channel and Partial-Area Pipe
Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6. The Impulse-Momentum Principle . . . . . . . .
10-3
10-3
10-4
10-7
1. INTRODUCTION
. . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .
In a general sense, hydraulics is the study of the practical laws of fluid flow and resistance in pipes and open
channels. Hydraulic formulas are often developed from
experimentation, empirical factors, and curve fitting,
without an attempt to justify why the fluid behaves
the way it does.
2. CONSERVATION LAWS
m2
m
m1/2/s
m
m
m
–
N
–
m/s2
m
m
Ns
–
–
m
kg
kg/s
–
Pa
kgm/s
m3/sm
m3/s
m
–
s
m
m/s
m
m
. . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .
Equation 10.1 Through Eq. 10.3: Continuity
Equation
A1 v1 ¼ A2 v2
Q ¼ Av
_ ¼ Q ¼ Av
m
10:1
10:2
10:3
Description
Fluid mass is always conserved in fluid systems, regardless of the pipeline complexity, orientation of the flow,
and fluid. This single concept is often sufficient to solve
simple fluid problems.
_2
_1¼m
m
When applied to fluid flow, the conservation of mass law
is known as the continuity equation.
1 A1 v1 ¼ 2 A2 v2
If the fluid is incompressible, then 1 = 2. Equation 10.1,
then, is the continuity equation for incompressible flow.
Volumetric flow rate, Q, is defined as the product of
cross-sectional area and velocity, as shown in Eq. 10.2.
From Eq. 10.1 and Eq. 10.2, it follows that
Q1 ¼ Q2
Symbols
kinetic energy correction factor
specific weight
density
Subscripts
c
critical
h
hydraulic
H
hydraulic
–
N/m3
kg/m3
Various units are used for volumetric flow rate. MGD
(millions of gallons per day) and MGPCD (millions of
gallons per capita day) are units commonly used in
municipal water works problems. MMSCFD (millions
of standard cubic feet per day) may be used to express
gas flows.
Calculation of flow rates is often complicated by the
interdependence between flow rate and friction loss.
P P I
*
w w w . p p i 2 p a s s . c o m
Hydraulics/
Hydrologic Sys.
Nomenclature
A
area
B
channel width
C
Hazen-Williams coefficient
d
depth
D
diameter
E
specific energy
f
friction factor
F
force
Fr
Froude number
g
acceleration of gravity, 9.81
h
head
hf
head loss due to friction
I
impulse
k1
constant
K
constant
L
length
m
mass
_
m
mass flow rate
n
Manning’s roughness coefficient
p
pressure
P
momentum
q
flow per unit width
Q
flow rate
RH
hydraulic radius
S
slope
t
time
T
width at surface
v
velocity
y
depth
z
height above datum
10-1
10-1
10-2
F E
C I V I L
R E V I E W
M A N U A L
Each affects the other, so many pipe flow problems must
be solved iteratively. Usually, a reasonable friction factor is assumed and used to calculate an initial flow rate.
The flow rate establishes the flow velocity, from which a
revised friction factor can be determined.
Example
Example
The diameter of a water pipe gradually changes from
5 cm at the entrance, point A, to 15 cm at the exit,
point B. The exit is 5 m higher than the entrance. The
pressure is 700 kPa at the entrance and 664 kPa at the
exit. Friction between the water and the pipe walls is
negligible. The water density is 1000 kg/m3.
An incompressible fluid flows through a pipe with an
inner diameter of 10 cm at a velocity of 4 m/s. The pipe
contracts to an inner diameter of 8 cm. What is most
nearly the velocity of the fluid in the narrower pipe?
B
15 cm
(A) 4.7 m/s
(B) 5.0 m/s
5m
(C) 5.8 m/s
(D) 6.3 m/s
A
Solution
5 cm
The cross-sectional areas of the two pipes are
Hydraulics/
Hydrologic Sys.
A1 ¼
pD 21 pð10 cmÞ2
¼
¼ 78:54 cm2
4
4
A2 ¼
pD 22 pð8 cmÞ2
¼
¼ 50:27 cm2
4
4
What is most nearly the rate of discharge at the exit?
(A) 0.0035 m3/s
(B) 0.0064 m3/s
(C) 0.010 m3/s
(D) 0.018 m3/s
Use Eq. 10.1.
A1 v1 ¼ A2 v2
Solution
m
ð78:54 cm2 Þ 4
A v
s
v2 ¼ 1 1 ¼
50:27 cm2
A2
¼ 6:25 m=s ð6:3 m=sÞ
The answer is (D).
..............................................................................................................................
Equation 10.4 and Eq. 10.5: Bernoulli
Equation
Use the Bernoulli equation, Eq. 10.5, to find the velocity
at the exit.
v2
p2 v22
p
þ þ z2 ¼ 1 þ 1 þ z1
2g
2g
10:4
v2
p2 v22
p
þ þ z2g ¼ 1 þ 1 þ z1g
2
2
10:5
Description
The Bernoulli equation, also known as the field equation
or the energy equation, is an energy conservation equation that is valid for incompressible, frictionless flow.
The Bernoulli equation states that the total energy of
a fluid flowing without friction losses in a pipe is constant. The total energy possessed by the fluid is the sum
of its pressure, kinetic, and potential energies. In other
words, the Bernoulli equation states that the total head
at any two points is the same.
P P I
*
w w w . p p i 2 p a s s . c o m
First, find the relationship between the entrance and
exit velocities, v1 and v2, respectively. From Eq. 10.1
and substituting the pipe area equation,
A1 v1 ¼ A2 v2
2
2
pD1
pD2
v1 ¼
v2
4
4
!
!
D 22
ð15 cmÞ2 v2
v2 ¼
¼ 9v2
v1 ¼
D 21
ð5 cmÞ2
2
v2
ð9v2 Þ
p2 v22
p
p
þ þ z2g ¼ 1 þ 1 þ z1g ¼ 1 þ
þ z 1g
2
2
2
sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
p2 p1 gðz 2 z 1 Þ
þ
v2 ¼
49
49
vﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ﬃ
u
u ð664 kPa 700 kPaÞ 1000 Pa
u
u
kPa
u
kg
u
ð49Þ 1000 3
¼u
m
u
u
u
m
9:81 2 ð5 mÞ
t
s
þ
49
¼ 0:516 m=s
H Y D R A U L I C S
Multiply the velocity at the exit with the cross-sectional
area to get the rate of flow.
pD 22
v2
4
0
1
2
B pð15 cmÞ C
m
¼@ 2 A 0:516
s
cm
ð4Þ 100
m
¼ 0:0091 m3 =s ð0:010 m3 =sÞ
10-3
Solution
Use the energy equation, Eq. 10.7. Take point 1 at the
reservoir surface and point 2 at the pipe outlet.
Q ¼ A 2 v2 ¼
v2 p
v2
p1
þ z 1 þ 1 ¼ 2 þ z 2 þ 2 þ hf
g
2g g
2g
The pressure is atmospheric at the reservoir and the
outlet, so p1 = p2. The velocity at the reservoir surface
is v1 ≈ 0 m/s, so the equation reduces to
z1 ¼ z2 þ
The answer is (C).
v22
þ hf
2g
Solve for the velocity at the pipe outlet, v2.
3. STEADY INCOMPRESSIBLE FLOW IN
PIPES AND CONDUITS
. . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . .
Equation 10.6 and Eq. 10.7: Extended Field
Equation
pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2gðz 1 z 2 hf Þ
rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
m
¼ ð2Þ 9:81 2 ð200 m 180 m 18 mÞ
s
v2 ¼
v22
p1
p
þ z 1 þ ¼ 2 þ z 2 þ þ hf
2g
2g
10:6
v2 p
v2
p1
þ z 1 þ 1 ¼ 2 þ z 2 þ 2 þ hf
g
2g g
2g
10:7
Description
The extended field (or energy) equation, also known as
the steady-flow energy equation, for steady incompressible flow is shown in Eq. 10.6 and Eq. 10.7.
Hydraulics/
Hydrologic Sys.
¼ 6:26 m=s
v21
The flow rate out of the pipe outlet is
2
pD
Q ¼ v2 A ¼ v2
4
!
2
m pð1 mÞ
¼ 6:26
s
4
¼ 4:92 m3 =s
ð4:9 m3 =sÞ
The answer is (A).
The head loss due to friction is denoted by the symbol
hf.
..............................................................................................................................
If the cross-sectional area of the pipe is the same at
points 1 and 2, then v1 = v2 and v21 =2g ¼ v22 =2g. If the
elevation of the pipe is the same at points 1 and 2, then
z1 = z2. When analyzing discharge from reservoirs and
large tanks, it is common to use gauge pressures, so that
p1 = 0 at the surface. In addition, since the surface elevation changes slowly (or not at all) when drawing from
a large tank or reservoir, v1 = 0.
p1 p2 ¼ hf ¼ ghf
Equation 10.8: Pressure Drop
10:8
Description
For a pipe of constant cross-sectional area and constant
elevation, the pressure change (pressure drop) from one
point to another is given by Eq. 10.8.
4.
FLOW IN NONCIRCULAR CONDUITS
. . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .
Example
An open reservoir with a water surface level at an
elevation of 200 m drains through a 1 m diameter pipe
with the outlet at an elevation of 180 m. The pipe outlet
discharges to atmospheric pressure. The total head
losses in the pipe and fittings are 18 m. Assume steady
incompressible flow. The flow rate from the outlet is
most nearly
3
(A) 4.9 m /s
3
(B) 6.3 m /s
(C) 31 m3/s
(D) 39 m3/s
Equation 10.9: Hydraulic Radius
RH ¼
cross-sectional area D H
¼
wetted perimeter
4
10:9
Description
The hydraulic radius is defined as the area in flow
divided by the wetted perimeter.
The area in flow is the cross-sectional area of the fluid
flowing. When a fluid is flowing under pressure in a pipe
(i.e., pressure flow), the area in flow will be the internal
P P I
*
w w w . p p i 2 p a s s . c o m
10-4
F E
C I V I L
R E V I E W
M A N U A L
area of the pipe. However, the fluid may not completely
fill the pipe and may flow simply because of a sloped
surface (i.e., gravity flow or open channel flow).
The wetted perimeter is the length of the line representing the interface between the fluid and the pipe or
channel. It does not include the free surface length
(i.e., the interface between fluid and atmosphere).
For a circular pipe flowing completely full, the area in
flow is pR2. The wetted perimeter is the entire circumference, 2pR. The hydraulic radius in this case is half the
radius of the pipe.
RH ¼
What is most nearly the hydraulic radius of the flow?
(A) 1.5 cm
(B) 2.5 cm
(C) 3.0 cm
(D) 5.0 cm
Solution
The hydraulic radius is
cross-sectional area
wetted perimeter
3
ð8 cmÞ
ð12 cmÞ
4
¼
3
ð8 cmÞ þ 12 cm
ð2Þ
4
¼ 3:0 cm
RH ¼
pR
R D
¼ ¼
4
2pR 2
2
The hydraulic radius of a pipe flowing half full is also
R/2, since the flow area and wetted perimeter are both
halved.
Hydraulics/
Hydrologic Sys.
Many fluid, thermodynamic, and heat transfer processes
are dependent on the physical length of an object. The
general name for this controlling variable is characteristic dimension. The characteristic dimension in evaluating fluid flow is the hydraulic diameter (also known as
the equivalent diameter), DH. The hydraulic diameter
for a full-flowing circular pipe is simply its inside diameter. If the hydraulic radius of a noncircular duct is
known, it can be used to calculate the hydraulic
diameter.
DH ¼ 4RH ¼ 4 area in flow
wetted perimeter
The frictional energy loss by a fluid flowing in a rectangular, annular, or other noncircular duct can be calculated from the Darcy equation by using the hydraulic
diameter, DH, in place of the diameter, D. The friction
factor, f, is determined in any of the conventional
manners.
Example
The 8 cm 12 cm rectangular flume shown is filled to
three-quarters of its height.
The answer is (C).
5. OPEN CHANNEL AND PARTIAL-AREA
PIPE FLOW
. . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . .
An open channel is a fluid passageway that allows part
of the fluid to be exposed to the atmosphere. This type
of channel includes natural waterways, canals, culverts,
flumes, and pipes flowing under the influence of gravity
(as opposed to pressure conduits, which always flow
full). A reach is a straight section of open channel with
uniform shape, depth, slope, and flow quantity.
There are difficulties in evaluating open channel flow.
The unlimited geometric cross sections and variations in
roughness have contributed to a relatively small number
of scientific observations upon which to estimate the
required coefficients and exponents. Therefore, the
analysis of open channel flow is more empirical and less
exact than that of pressure conduit flow. This lack of
precision, however, is more than offset by the percentage
error in runoff calculations that generally precede the
channel calculations.
..............................................................................................................................
Equation 10.10: Manning’s Equation
2=3
v ¼ ðK =nÞRH S 1=2
10:10
Values
8 cm
SI units
K=1
customary U.S. units K = 1.486
concrete
n ≈ 0.013
Description
12 cm
P P I
*
w w w . p p i 2 p a s s . c o m
Manning’s equation has typically been used to estimate
the velocity of flow in any open channel. It depends on
the hydraulic radius, RH, the slope of the energy grade
line, S, and a dimensionless Manning’s roughness coefficient, n. A conversion constant, K, modifies the equation

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