chapter 1: water flow in pipes university of palestine engineering hydraulics 2 nd semester...
TRANSCRIPT
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CHAPTER 1:Water Flow in Pipes
CHAPTER 1:Water Flow in Pipes
University of PalestineEngineering Hydraulics2nd semester 2010-2011
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Description of A Pipe Flow
• Water pipes in our homes and the distribution
system
• Pipes carry hydraulic fluid to various components
of vehicles and machines
• Natural systems of “pipes” that carry blood
throughout our body and air into and out of our
lungs.
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• Pipe Flow: refers to a full water flow in a closed conduits or circular cross section under a certain pressure gradient.
• The pipe flow at any cross section can be described by:
cross section (A), elevation (h), measured with respect to a
horizontal reference datum. pressure (P), varies from one point to another,
for a given cross section variation is neglected The flow velocity (v), v = Q/A.
Description of A Pipe Flow
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Pipe flow• The pipe is completely filled with the fluid being transported.
• The main driving force is likely to be a pressure gradient along the pipe.
Open-channel flow• Water flows without completely filling the pipe.
• Gravity alone is the driving force, the water flows down a hill.
Difference between open-channel flow and the pipe flow
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Steady and Unsteady flow The flow parameters such as velocity (v),
pressure (P) and density (r) of a fluid flow are independent of time in a steady flow. In unsteady flow they are independent. 0
ooo ,z,yxtvFor a steady flow
0ooo ,z,yxtvFor an unsteady flow
If the variations in any fluid’s parameters are small, the average is constant, then the fluid is considered to be steady
Types of Flow
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Uniform and non-uniform flow
A flow is uniform if the flow characteristics at any given instant remain the same at different points in the direction of flow, otherwise it is termed as non-uniform flow.
0ot
svFor a uniform flow
For a non-uniform flow 0ot
sv
Types of Flow
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Examples:
The flow through a long uniform pipe diameter at a constant rate is steady uniform flow.
The flow through a long uniform pipe diameter at a varying rate is unsteady uniform flow.
The flow through a diverging pipe diameter at a constant rate is a steady non-uniform flow.
The flow through a diverging pipe diameter at a varying rate is an unsteady non-uniform flow.
Types of Flow
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Laminar and turbulent flowLaminar flow:
Turbulent flow:
The fluid particles move along smooth well defined path or streamlines that are parallel, thus particles move in laminas or layers, smoothly gliding over each other.
The fluid particles do not move in orderly manner and they occupy different relative positions in successive cross-sections. There is a small fluctuation in magnitude and direction of the velocity of the fluid particles
Transitional flowThe flow occurs between laminar and turbulent flow
Types of Flow
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Reynolds Experiment Reynolds performed a very carefully prepared
pipe flow experiment.
Types of Flow
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Increasing flow velocity
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Reynolds Experiment• Reynold found that transition from laminar to
turbulent flow in a pipe depends not only on the velocity, but only on the pipe diameter and the viscosity of the fluid.
• This relationship between these variables is commonly known as Reynolds number (NR)
ForcesViscous
ForcesInertialVDVDNR
It can be shown that the Reynolds number is a measure of the ratio of the inertial forces to the viscous forces in the flow
FI ma AFV
Types of Flow
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Reynolds number
VDVD
NR
where V: mean velocity in the pipe [L/T]
D: pipe diameter [L]: density of flowing fluid
[M/L3]: dynamic viscosity [M/LT]: kinematic viscosity [L2/T]
Types of Flow
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Types of Flow
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Flow laminar when NR < Critical NR
Flow turbulent when NR > Critical
NR
It has been found by many experiments that for flows in circular pipes, the critical Reynolds number is about 2000
The transition from laminar to turbulent flow does
not always happened at NR = 2000 but varies due to
experiments conditions….….this known as
transitional range
Types of Flow
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Laminar flows characterized by:
• low velocities•small length scales
•high kinematic
viscosities
•NR < Critical NR
•Viscous forces are
dominant.
Turbulent flows characterized by
•high velocities
• large length scales
• low kinematic
viscosities
•NR > Critical NR
• Inertial forces are
dominant
Laminar Vs. Turbulent flows
Types of Flow
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Example 1 40 mm diameter circular pipe carries water at
20oC. Calculate the largest flow rate (Q) which laminar flow can be expected.
mD 04.0
CTat o20101 6
sec/1028.6)04.0(4
05.0. 352 mAVQ
sec/05.02000101
)04.0(2000
6mV
VVDNR
Types of Flow
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Water flow in pipes may contain energy in three basic forms:
1- Kinetic energy.
2- potential energy.
3- pressure energy.
22
22
11
21
22h
P
g
Vh
P
g
V
Bernoulli Equation
Energy per unit weight of waterOR: Energy Head
Energy Head in Pipe Flow
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Energy head and Head loss in pipe flow
Energy Head in Pipe Flow
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11
21
1 2h
P
g
VH
22
22
2 2h
P
g
VH
Kinetic head
Elevation head
Pressure head
Energy head
= + +
LhhP
g
Vh
P
g
V 2
22
21
12
1
22
Notice that:• In reality, certain amount of energy loss (hL) occurs when the water mass flow from one section to another.• The energy relationship between two sections can be written as:
Energy Head in Pipe Flow
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Example
Energy Head in Pipe Flow
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ExampleIn the figure shown:Where the discharge through the system is 0.05 m3/s, the total losses through the pipe is 10 v2/2g where v is the velocity of water in 0.15 m diameter pipe, the water in the final outlet exposed to atmosphere.
Energy Head in Pipe Flow
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Calculate the required height (h =?)below the tank
mh
h
hzg
V
g
pz
g
V
g
p
smV
smV
L
AQ
AQ
147.2181.9*2
83.21020
81.9*2
366.60)5(00
22
/366.610.0
05.0
/83.215.0
05.0
22
2
222
1
211
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Energy Head in Pipe Flow
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Without calculation sketch the (E.G.L) and (H.G.L)
Energy Head in Pipe Flow
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Basic components of a typical pipe system
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In General:When a fluid is flowing through a pipe, the fluid
experiences some resistance due to which some of energy (head) of fluid is lost.
Energy Losses(Head losses)
Major Losses Minor losses
loss of head due to pipe friction and to viscous dissipation in flowing water
Loss due to the change of the velocity of the flowing fluid in the magnitude or in direction as it moves through fitting like Valves, Tees, Bends and Reducers.
Calculation of Head (Energy) Losses
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Part A:Major Losses
Head Losses in Pipelines
Calculation of Head (Energy) Losses
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• Energy loss through friction in the length of pipeline is commonly termed the major loss hf
• This is the loss of head due to pipe friction and to the viscous dissipation in flowing water.
• Several studies have been found the resistance to flow in a pipe is:
- Independent of pressure under which the water flows- Linearly proportional to the pipe length, L- Inversely proportional to some water power of the pipe
diameter D- Proportional to some power of the mean velocity, V- Related to the roughness of the pipe, if the flow is
turbulent
Losses of Head due to Friction
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Energy Head & Head loss in pipe flow
Losses of Head due to Friction
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Several formulas have been developed in
the past. Some of these formulas have
faithfully been used in various hydraulic
engineering practices.
1. Darcy-Weisbach ( f )
2. Hazen-William (CHW)
3. Manning (n)
4. The Chezy Formula
5. The Strickler Formula
Major losses formulas
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The resistance to flow in a pipe is a function of:
• The pipe length, L
• The pipe diameter, D
• The mean velocity, V
• The properties of the fluid .
• The roughness of the pipe, (the flow is turbulent).
Major losses formulas
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Darcy-Weisbach Equation
25
22
8
2
Dg
QLf
g
V
D
LfhL
Where: f is the friction factorL is pipe lengthD is pipe diameterQ is the flow ratehL is the loss due to friction
It is conveniently expressed in terms of velocity (kinetic) head in the pipe
D
VDF
D
VDF
DRFf e
,,,
The friction factor is function of different terms:
Renold number Relative roughness
Re
s
NR
ke
Major losses formulas
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Friction Factor: (f )
• For Laminar flow: (NR < 2000) [depends only on Reynolds’ number and not on the surface roughness]
RN
64f
• For turbulent flow in smooth pipes (e/D = 0) with 4000 < NR < 105 is
4/1
316.0
RNf
Major losses formulas
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2.51log2
1 fN
fR
510e
7.3log21
RNfor
D
f• Colebrook-White Equation for f
fND
e
f R
51.2
7.3ln86.0
1
For turbulent flow ( NR > 4000 ) with e/D > 0.0, the friction factor can be founded from:• Th.von Karman formulas:
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Major losses formulas
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There are other Equation such as Karman Equation see Text book 2
Pandtle - Colebrook Equation
510
713log21
eRfor
use
D.
f
Major losses formulas
There is some difficulty in solving this equationSo, Miller improve an initial value for f , (fo)
2
9.0
74.5
7.3log25.0
R
oND
ef
The value of ffoo can be use directly as ff if: 26
83
101101
101104-
R
D e
N
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e7.1'
ee 7.108.0 '
RNf
64
pipe wall
e
51.2log2
110
fN
fR
e
pipe wall
transitionallyrough
e
pipe wall
rough
f independent of relative roughness e/D
f independent of NR
f varies with NR and e/D
turbulent flow
NR > 4000
laminar flow
NR < 2000
e08.0'
e
D
f7.3log2
110
fN
De
f R
51.2
7.3log2
110
Colebrook formula
The thickness of the laminar sublayer decrease with an increase in NR
Smooth
Friction Factor f
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A convenient chart was prepared by Lewis F. Moody and commonly called the Moody diagram of friction Moody diagram of friction factors for pipefactors for pipe flow, There are 4 zones of pipe flow in the chart:
• A laminar flow zone where f is simple linear function of Re
• A critical zone (shaded) where values are uncertain because the flow might be neither laminar nor truly turbulent
• A transition zone where f is a function of both Re and relative roughness
• A zone of fully developed turbulence where the value of f depends solely on the relative roughness and independent of the Reynolds Number
Moody diagram
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Moody diagram
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Laminar
Marks Reynolds Number independence
critical
Transition
Moody diagram
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Notes:
• Colebrook formula
is valid for the entire nonlaminar range (4000 < Re < 108) of the Moody chart
12
3 7
2 51
f
e D
f
log
/
.
.
Re
In fact , the Moody chart is a graphical representation of this equation
Moody diagram
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Bonus:
Find the theoretical formulation for friction factor for laminar flow.
Re
64f
Moody diagram
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Typical values of the absolute roughness (e) are given
Absolute roughness
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Materials Roughness
Absolute roughness
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Three types of problems for uniform flow in a single pipe:
Type 1:Given the kind and size of pipe and the flow rate head loss ?
Type 2:Given the kind and size of pipe and the head loss
flow rate ?
Type 3:Given the kind of pipe, the head loss and flow rate
size of pipe ?
Problems (head loss)
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Problems type I (head loss)
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The water flow in Asphalted cast Iron pipe (e = 0.12mm) has a diameter 20cm at 20oC. Is 0.05 m3/s. determine the losses due to friction per 1 km
1.59m/sm0.2π/4
/s0.05mV
22
3
56
26
1015.33148521001.1
2.059.1
0006.0200
12.0
12.0
/sm101.01υ20
VD
N
mm
mm
D
e
mme
CT
R
o
f = 0.018 Moody
m
m/s.
.
m.
m,.
g
V
D
Lfh f
55.11
8192
591
200
00010180
2 2
22
Type 1:Given the kind and size of pipe and the flow rate head loss ?
Example 2
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The water flow in commercial steel pipe (e = 0.045mm) has a diameter 0.5m at 20oC. Q=0.4 m3/s. determine the losses due to friction per 1 km
sm
A
QV / 037.2
45.0
4.02
013.0
109105.0
045.0
10012.110006.1
037.25.0
10006.15.4220
10497
5.42
10497
53
66
65.1
6
5.1
6
f
D
e
N
T
Moody
R
kmmh f / 5.581.92
037.2
5.0
1000013.0
2
Type 1:Given the kind and size of pipe and the flow rate head loss ?
Example 3
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fRD
k
f e
s 51.2
7.3ln86.0
1
Use other methods to solve f
01334.0
10012.1
74.5
7.3
109log25.0
74.5
7.3log25.0
2
9.06
52
9.0
e
so
R
Dkf
678.866.8
01334.0
51.2
7.3
109ln86.0
01334.0
1 5
eR
1 -Cole brook
kmmh f / 5.581.92
037.2
5.0
100001334.0
2
Example 3-cont.
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2- Pandtle - Colebrook Equation
solutionlastf
.f
D.
f
01334.0012.0
045.0
500713log2
1
713log21
Example 3-cont.
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Problems type II (head loss)
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Method for solution of Type 2 problems
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Re
lati
ve
ro
ug
hn
es
s e
/D
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Example 4:
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Example 4:
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Cast iron pipe (e = 0.26), length = 2 km, diameter = 0.3m. Determine the max. flow rate Q , If the allowable maximum head loss = 4.6m. T=10oC
10135.0
81.923.0
20006.4
2
2
2
2
fV
Vf
g
V
D
LfhF
00009.01067.8103.0
26.0
210296.21031.1
3.0
1031.15.4210
10497
5.42
10497
53
66
65.1
6
5.1
6
D
e
VV
N
T
R
Type 2:Given the kind and size of pipe and the head loss flow rate ?
Example 5:
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02.0
1067.8
10668.2
m/s 16.101.0
4
52
1
f
D
e
N
Vf
Moody
Req
eq
021.0
1067.8
10886.1
m/s 82.002.0
4
52
1
f
D
e
N
Vf
Moody
Req
eq
10135.02 f
V
210296.2 6 VNR
Trial 1
Trial 2
V= 0.82 m/s , Q = V*A = 0.058 m3/s
Another solution?
Example 5:cont.
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Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best condition carrying water at 10oC. It is allowed to have a head loss of 2m/km of pipe length.
hf fL
D
V 2
2g
V 2ghf
L
1/ 2D
f
1/ 2
fV
Vf
12.0
)81.9(23
10002 2
2
Table 3.1 : wood stave pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm
Solution 1:
0001.03
3.0
D
e
At T= 10oC, = 1.31x10-6 m2/sec VVVD
NR .1029.21031.1
3 6
6
Type 2:Given the kind and size of pipe and the head loss flow rate ?
Example 6:
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• Solve by trial and error:• Iteration 1:• Assume f = 0.02 sec/45.2
02.0
12.02 mVV
66 106.545.2.1029.2 RN
From moody Diagram: 0122.0f
Iteration 2:update f = 0.0122 sec/14.3
0122.0
12.02 mVV
66 102.714.3.1029.2 RN
From moody Diagram: 0122.00121.0 f
0 0.02 2.45 5.6106
1 0.0122 3.14 7.2106
2 0.0121
Iteration f V NR
Convergence
Solution:
/sm 2.2724
3.15.3
m/s 15.3
3
2
2
VAQ
V
Another solution?64
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Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best condition carrying water at 10oC. It is allowed to have a head loss of 2m/km of pipe length.
5
6
232/12/3
1062.91000
)3)(81.9(2
1031.1
)3(2
L
ghDfN f
R
Table 3.1 : wood pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm
Solution 2:
0001.03
3.0
D
e
Type 2: Given the kind and size of pipe and the head loss flow rate?
At T= 10oC, = 1.31x10-6 m2/sec
From moody Diagram: 0121.0f
sec/15.32
2
2/12/12
mf
D
L
ghV
g
V
D
Lfh f
f
/sm 2.2724
3.15.3,
3
2
VAQ
Example 6:Another solution?
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f = 0.0121
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Problems type III (head loss)
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Example 7:
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Example 7:cont.
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Estimate the size of a uniform, horizontal welded-steel pipe installed to carry 14 ft3/sec of water of 70oF (20oC). The allowable pressure loss is 17 ft/mi of pipe length.
From Table : Steel pipe: ks = 0.046 mm
Darcy-Weisbach:
hL fLD
V 2
2g
Q VA
hL fLD
QA
2
2gf
LD
Q 2
2g42
2D 4 1
D 5
16fLQ 2
2g 2
5/1
2
28
Lhg
fLQD
afff
D 5/15/1
5/1
2
2
33.41781.9
1452808
Let D = 2.5 ft, then V = Q/A = 2.85 ft/sec
Now by knowing the relative roughness and the Reynolds number:
55
10*6.610*08.1
5.2*85.2
VDNR
0012.05.2
003.0
D
e
We get f =0.021
Solution 2:
Example 8:
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A better estimate of D can be obtained by substituting the latter values into equation a, which gives
ftfD 0.2021.0*33.433.4 5/15/1
A new iteration provide V = 4.46 ft/secNR = 8.3 x 105 e/D = 0.0015f = 0.022, andD = 2.0 ft.More iterations will produce the same results.
Example 8:cont.
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Example 9:
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• Hazen-Williams
54.063.085.0 SRCV hHW
tCoefficien iamsHazen Will
4 P wetted
A wetted Radius hydraulic
CL
hS
D
R
HW
f
h
UnitSI
0.71 852.1
87.4852.1 QDC
Lh
HW
f
Sim
pli
fied
UnitsBritishSRCV
mVcmD
hHW54.063.0318.1
sec/0.35
Empirical Formulas 1
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Empirical Formulas 1
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tCoefficien iamsHazen WillHWC
Empirical Formulas 1
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Empirical Formulas 1
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79
081.0
sec/0.3
V
VCC
mVWhen
oHoH
Where:CH = corrected valueCHo = value from tableVo = velocity at CHo
V = actual velocity
Empirical Formulas 1
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• Manning
tCoefficien M
4 P wetted
A wetted Radius hydraulic
anningnL
hS
D
R
f
h
Sim
pli
fied
UnitSI
0.3133.5
2
D
nQLh f
2/13/21SR
nV h
• This formula has extensively been used for open channel designs. It is also quite commonly used for pipe flows
Empirical Formulas 2
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81
• n = Manning coefficient of roughness (See Table)• Rh and S are as defined for Hazen-William
formula.
Vn
R Sh1 2 3 1 2/ /
3/16
223.10
D
QLnh f
2233.1
35.6 VnD
Lh f
Empirical Formulas 2
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82
Empirical Formulas 2
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tCoefficien Manningn
Empirical Formulas 2
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84
The Chezy Formula
V C R Sh 1 2 1 2/ /
2
4
C
V
D
Lh f
where C = Chezy coefficient
Empirical Formulas 3
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85
• It can be shown that this formula, for circular pipes, is equivalent to Darcy’s formula with the value for
[f is Darcy Weisbeich coefficient]
• The following formula has been proposed for the value of C:
[n is the Manning coefficient]
Cg
f
8
C S n
S
n
Rh
230 00155 1
1 230 00155
.
(.
)
Empirical Formulas 3
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86
The Strickler Formula:
V k R Sstr h 2 3 1 2/ /
2
33.135.6
strf k
V
D
Lh
where kstr is known as the Strickler coefficient.
Comparing Manning formula and Strickler formula, we can see that
1
nkstr
Empirical Formulas 4
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87
Relations between the coefficients in Chezy, Manning , Darcy , and Strickler formulas.
nkstr
1
6/1hstr RkC
g
Rfn h
8
3/1
Empirical Formulas
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New Cast Iron (CHW = 130, n = 0.011) has length = 6 km
and diameter = 30cm. Q= 0.32 m3/s, T=30o. Calculate the head loss due to friction using:
a) Hazen-William Method
b) Manning Method
33332030130
6000710
710
85218748521
8521
8748521
m . .
. h
Q DC
L.h
...f
.
..HW
f
m
.
.. .h
D
nQ L h
.f
.f
47030
32001106000310
3.10
335
2
335
2
Example 10
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Part B:Minor Losses
Head Losses in Pipelines
Calculation of Head (Energy) Losses
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• Additional losses due to entries and exits, fittings and valves are traditionally referred to as minor losses
2
22
22 gA
Qk
g
Vkh LLm
Minor Losses
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91
It is due to the change of the velocity of the flowing fluid in the magnitude or in direction [turbulence within bulk flow as it moves through and fitting]
Flow pattern through a valve
Minor Losses
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92
• The minor losses occurs du to :• Valves • Tees• Bends• Reducers• Valves• And other appurtenances
• It has the common form
2
22
22 gA
Qk
g
Vkh LLm
can be the dominant cause of head loss in shorter pipelines
“minor” compared to friction losses in long pipelines but,
Minor Losses
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93
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Losses due to contraction
g
Vkh cc 2
22
A sudden contractionA sudden contraction in a pipe usually causes a marked drop in pressure in the pipe due to both the increase in velocity and the loss of energy to turbulence.
Along centerline
Along wall
Minor Losses
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Value of the coefficient Kc for sudden contraction
VV22
Minor Losses
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96
Head Loss Due to a Sudden Contraction
h KV
gL L 22
2
g
VhL 2
5.02
2
Minor Losses
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Head losses due to pipe contraction may be greatly reduced by introducing a gradual pipe gradual pipe transition transition known as a confusor confusor
g
V'k'h cc 2
22
'kc
Minor Losses
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98
Head Loss Due to Gradual Contraction (reducer or nozzle)
g
VVKh LL 2
21
22
100200300400
KL0.20.280.320.35
A different set of data is:
Minor Losses
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Losses due to Enlargement
g
VVhE 2
)( 221
A sudden EnlargementA sudden Enlargement in a pipe
Minor Losses
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Head losses due to pipe enlargement may be greatly reduced by introducing a gradual pipe gradual pipe transition transition known as a diffusor diffusor
g
VV'k'h EE 2
22
21
Minor Losses
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Note that the drop in the energy line is much larger than in the case of a contraction
abrupt expansion
gradual expansion
smaller head loss than in the case of an abrupt expansion
Minor Losses
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102
Head Loss Due to a Sudden Enlargement
h KV
gL L 12
2
KA
AL
1 1
2
2
h
V V
gL 1 2
2
2
or:
Minor Losses
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103
Head Loss Due to Gradual Enlargement (conical diffuser)
g
VVKh LL 2
22
21
100200300400
KL0.390.801.001.06
Minor Losses
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104
Gibson tests Minor Losses
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Loss due to pipe entranceGeneral formula for head loss at the entrance of a pipe is also expressed in term of velocity head of the pipe
g
VKh entent 2
2
Minor Losses
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Different pipe inlets
increasing loss coefficient
Minor Losses
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107
Head Loss at the Entrance of a Pipe (flow leaving a tank)
Reentrant)embeded(
KL = 0.8
Sharpedge
KL = 0.5
Wellrounded
KL = 0.04
SlightlyroundedKL = 0.2
h KV
gL L2
2
Minor Losses
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108
Another Typical values for various amount of rounding of the lip
Minor Losses
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Loss at pipe exit (discharge head loss)
In this case the entire velocity head of the pipe flow is dissipated and that the discharge loss is
g
Vhexit 2
2
Minor Losses
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110
Head Loss at the Exit of a Pipe (flow entering a tank)
hV
gL 2
2
the entire kinetic energy of the exiting fluid (velocity V1) is dissipated through viscous effects as the stream of fluid mixes with the fluid in the tank and eventually comes to rest (V2 = 0).
KL = 1.0 KL = 1.0
KL = 1.0 KL = 1.0
Minor Losses
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Loss of head in pipe bends
g
Vkh bb 2
2
Minor Losses
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112
Miter bends
For situations in which space is limited,
Minor Losses
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Loss of head through valves
g
VKh vv 2
22
Minor Losses
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Minor Losses
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115
The loss coefficient for elbows, bends, and teesMinor Losses
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Loss coefficients for pipe components (Table)
Minor Losses
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Minor loss coefficients (Table)
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Minor loss calculation using equivalent pipe length
f
DkL l
e
Minor Losses
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119
• Note that the above values are average typical values, actual values will depend on the make (manufacturer) of the components.
• See:– Catalogs – Hydraulic handbooks !!
Minor Losses
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Energy and hydraulic grade lines
Unless local effects are of particular interests the changes in the EGL and HGL are often shown as abrupt changes (even though the loss
occurs over some distance)
Minor Losses
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Example 11In the figure shown two new cast iron pipes in series, D1 =0.6m , D2 =0.4m length of the two pipes is 300m, level at A =80m , Q = 0.5m3/s (T=10oC).there are a sudden contraction between Pipe 1 and 2, and Sharp entrance at pipe 1.Fine the water level at B
e = 0.26mmv = 1.31×10-
6Q = 0.5 m3/s
Minor Losses
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exitcentffL
fBA
hhhhhh
hZZ
21
g
Vk
g
Vk
g
Vk
g
V
D
Lf
g
V
D
Lfh exitcentL 22222
22
22
21
22
2
22
21
1
11
01800170
000650000430600
26.0
102211018
sec98340
4
50sec771
604
50
21
11
6222
5111
222
211
.f .f
,.D
, .D
,.υ
DV R , .
υ
DVR
, m/..
π.
A
Q, V m/.
.π
.
A
QV
moodymoody
ee
1 ,27.0 ,5.0 exitcent hhh
Solution
Minor Losses
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m.g
.
g
..
g
..
g
. .
. .
g
. .
. .h f
36132
983
2
983270
2
77150
2
983
40
3000180
2
771
60
3000170
222
22
ZB = 80 – 13.36 = 66.64 m
g
Vk
g
Vk
g
Vk
g
V
D
Lf
g
V
D
Lfh exitcentL 22222
22
22
21
22
2
22
21
1
11
Minor Losses
123
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Example 12A pipe enlarge suddenly from D1=240mm to D2=480mm. the H.G.L rises by 10 cm calculate the flow in the pipe
Minor Losses
124
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Solution
125
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smAVQsmV
g
V
g
VV
g
V
g
V
VV
VV
AVAV
g
VV
g
V
g
V
zg
pz
g
ph
g
V
g
V
hzg
V
g
pz
g
V
g
p
e
e
/103.048.057.0/57.0
1.02
6
1.02
4
22
16
4
48.024.0
1.0222
22
22
324222
22
2
222
22
2
21
242
241
2211
2
212
22
1
11
22
22
21
2
222
1
211
Solution
Minor Losses
126
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http://www.haestad.com/library/books/awdm/online/wwhelp/wwhimpl/java/html/wwhelp.htm
127