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Introduction to Fluid Mechanics Pipe Flow and Open Channel Flow

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ENGINEERS WITHOUT BORDERS. Open Channel Flow. Uniform Open Channel Flow is the hydraulic condition in ... Hydraulic Radius = A/P = R = BYn/(B 2Yn) B. Yn ... – PowerPoint PPT presentation

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Title: Introduction to Fluid Mechanics Pipe Flow and Open Channel Flow


1
ENGINEERS WITHOUT BORDERS
Introduction to Fluid Mechanics Pipe Flow and
Open Channel Flow
Ross Gordon rgordon_at_rice.edu
2
Open Channel Flow
Uniform Open Channel Flow is the hydraulic
condition in which the water depth and the
channel cross section do not change over some
reach of the channel Through experimental
observations and calculations, Mannings Equation
was developed to relate flow and channel geometry
to water depth. Knowing the flow in a channel,
you can solve for the water depth. Knowing the
maximum allowable depth, you can solve for the
maximum flow.
3
Open Channel Flow
Mannings equation is only accurate for cases
where the cross sections of a stream or channel
are uniform. Mannings equation works accurately
for man made channels, but for natural streams
and rivers, it can only be used as an
approximation.
4
Mannings Equation
Terms to know in the Mannings equation V
Channel Velocity
A Cross sectional area of the
channel P Wetted perimeter of
the channel R Hydraulic
Radius A/P
S Slope of the channel bottom (ft/ft or m/m)
n Mannings roughness coefficient
n 0.015 for
concrete n 0.03 for clean natural channel n
.01 for glass
Yn Normal depth (depth of uniform flow)
Area
Yn
Y
X
Wetted Perimeter
Slope S Y/X
5
Mannings Equation
V (1/n)R2/3v(S) for the metric system V
(1.49/n)R2/3v(S) for the English system Q
A(k/n)R2/3v(S) k is either 1 or 1.49
As you can see, Yn is not directly a part of
Mannings equation. However, A and R depend on
Yn. Therefore, the first step to solving any
Mannings equation problem, is to solve for the
geometrys cross sectional area and wetted
perimeter
For a rectangular Channel Area A B x
Yn Wetted Perimeter P B 2Yn Hydraulic
Radius A/P R BYn/(B2Yn)
Yn
B
6
Simple Mannings Example
A rectangular open concrete (n0.015) channel is
to be designed to carry a flow of 2.28 m3/s. The
slope is 0.006 m/m and the bottom width of the
channel is 2 meters.
Determine the
normal depth that will occur in this channel.
First, find A, P and R A 2Yn P 2 2Yn
R 2Yn/(2 2Yn) Next, apply Mannings
equation Q A(1/n)R2/3v(S) ? 2.28
(2Yn)x(1/0.015)x(2Yn/(2 2Yn))2/3xv(0.006)
Solving for Yn Yn 0.47 meters
Yn
2 m
7
The Trapezoidal Channel
House flooding occurs along Brays Bayou when
water overtops the banks. What flow is allowable
in Brays Bayou if it has the geometry shown below?
Slope S 0.001 ft/ft
25
T 20
Concrete Lined n 0.015
35
A, P and R for Trapezoidal Channels
A Yn(B Yn cot ?) P B (2Yn/sin ? ) R
(Yn(B Yn cot ?)) / (B (2Yn/sin ? ))
Yn
?
B
8
The Trapezoidal Channel
Slope S 0.0003 ft/ft
25
T 20
Concrete Lined n 0.015
35
A Yn(B Yn cot ?) A 25( 35 25 x cot(20))
2592 ft2 P B (2Yn/sin ? ) P 35 (2 x
25/sin(20)) 181.2 ft R (Yn(B Yn cot ?)) /
(B (2Yn/sin ? )) R 2592 / 181.2 14.3 ft
9
The Trapezoidal Channel
Slope S 0.0003 ft/ft
25
T 20
Concrete Lined n 0.015
35
A 2592 ft2 R 14.3 ft Q A(1.49/n)R2/3v(S) Q
2592 x (1.49 / .015) x 14.32/3 x v(.0003) Q
Max allowable Flow 26,273 cfs
10
Mannings Over Different Terrains
S .005 ft/ft
5
5
5
3
3
Grass n.03
Grass n.03
Concrete n.015
Estimate the flow rate for the above
channel? Hint Treat each different portion of
the channel separately. You must find an A, R, P
and Q for each section of the channel that has a
different roughness coefficient.
11
Mannings Over Different Terrains
S .005 ft/ft
5
5
5
3
3
Grass n.03
Grass n.03
Concrete n.015
The Grassy portions For each section

A 5 x 3 15 ft2 P 5 3 8
ft R 15 ft2/8 ft 1.88 ft Q
15(1.49/.03)1.882/3v(.005)
Q 80.24 cfs per
section ? For both sections… Q 2 x 80.24
160.48 cfs
12
Mannings Over Different Terrains
S .005 ft/ft
5
5
5
3
3
Grass n.03
Grass n.03
Concrete n.015
The Concrete portions
A 5 x
6 30 ft2 P 5 3 3 11 ft R 30
ft2/11 ft 2.72 ft Q 30(1.49/.015)2.722/3v(.005
)
Q 410.6 cfs For the entire channel… Q
410.6 129.3 540 cfs
13
Pipe Flow and the Energy Equation
For pipe flow, the Bernoulli equation alone is
not sufficient. Friction loss along the pipe,
and momentum loss through diameter changes and
corners take head (energy) out of a system that
theoretically conserves energy. Therefore, to
correctly calculate the flow and pressures in
pipe systems, the Bernoulli Equation must be
modified. P1/? V12/2g z1 P2/? V22/2g z2
Hmaj Hmin Major losses Hmaj Major losses
occur over the entire pipe, as the friction of
the fluid over the pipe walls removes energy from
the system. Each type of pipe as a friction
factor, f, associated with it.
Energy line with no losses
Hmaj
Energy line with major losses
1
2
14
Pipe Flow and the Energy Equation
Minor Losses Hmin Momentum losses in Pipe
diameter changes and in pipe bends are called
minor losses. Unlike major losses, minor losses
do not occur over the length of the pipe, but
only at points of momentum loss. Since Minor
losses occur at unique points along a pipe, to
find the total minor loss throughout a pipe, sum
all of the minor losses along the pipe. Each
type of bend, or narrowing has a loss
coefficient, KL to go with it.
Minor Losses
15
Major and Minor Losses
Major Losses Hmaj f x (L/D)(V2/2g) f
friction factor L pipe length D pipe
diameter V Velocity g gravity Minor
Losses Hmin KL(V2/2g) Kl sum of loss
coefficients V Velocity g gravity When
solving problems, the loss terms are added to the
system at the second point P1/? V12/2g z1
P2/? V22/2g z2 Hmaj Hmin
16
Loss Coefficients
Use this table
to find loss coefficients
17
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
If oil flows from the upper to lower reservoir at
a velocity of 1.58 m/s in the 15 cm diameter
smooth pipe, what is the elevation of the oil
surface in the upper reservoir? Include major
losses along the pipe, and the minor losses
associated with the entrance, the two bends, and
the outlet.
18
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
Apply Bernoullis equation between points 1 and
2 Assumptions P1 P2 Atmospheric 0 V1
V2 0 (large tank) 0 0 Z1 0 0 130m
Hmaj Hmin Hmaj (fxLxV2)/(Dx2g)(.035 x 197m x
(1.58m/s)2)/(.15 x 2 x 9.8m/s2) Hmaj 5.85m
19
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m 5.85m Hmin Hmin
2KbendV2/2g KentV2/2g KoutV2/2g From Loss
Coefficient table Kbend 0.19 Kent 0.5
Kout 1 Hmin (0.19x2 0.5 1) x
(1.582/2x9.8) Hmin 0.24 m
20
Pipe Flow Example
?oil 8.82 kN/m3 f .035
1
Z1 ?
2
Z2 130 m
60 m
Kout1
7 m
r/D 0
130 m
r/D 2
0 0 Z1 0 0 130m Hmaj Hmin 0 0
Z1 0 0 130m 5.85m 0.24m Z1 136.09
meters
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