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Water Flow in Pipes
The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Hydraulics - ECIV 3322
Chapter 3Chapter 3
2
3.1 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.
3
• 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.
4
Difference between open-channel flow and the pipe flow
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.
5
Types of Flow Steady and Unsteady flow The flow parameters such as velocity (v), pressure (P)
and density () of a fluid flow are independent of time in a steady flow. In unsteady flow they are independent.
0ooo ,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
6
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
7
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.
8
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
9
3.2 Reynolds Experiment
Reynolds performed a very carefully prepared pipe flow experiment.
10
Increasing flow
velocity
11
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
12
Reynolds number
VDVD
NR
whereV: 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]
13
14
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
15
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
16
Example 3.1 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
17
3.3 Forces in Pipe Flow
• Cross section and elevation of the pipe are varied along the axial direction of the flow.
18
)(.. '22'11 massfluidfluxmassdVoldVol
Conservation law of mass
Mass enters the control volume
Mass leaves the control volume
QVAVAdt
dSA
dt
dSA
dt
dVol
dt
dVol
.......
..
22112
21
1
'22'11
QVAVA 2211 ..
For Incompressible and Steady flows:
Continuity equation for Incompressible Steady flow
19
Apply Newton’s Second Law:
t
VMVM
dt
VdMaMF
12
xxx WFAPAPF 2211
)(.
)(.
)(.
.
12
12
12
zzz
yyy
xxx
VVQF
VVQF
VVQF
rateflowmassQtMbut
Fx is the axial direction force exerted on the control volume by the wall of the pipe.
)(. 12
VVQF
Conservation of moment equation
20
dA= 40 mm, dB= 20 mm, PA= 500,000 N/m2, Q=0.01m3/sec. Determine the reaction force at the hinge.
Example 3.2
21
3.4 Energy Head in Pipe Flow
Water flow in pipes may contain energy in three basic forms:
1- Kinetic energy,
2- potential energy,
3- pressure energy.
22
Consider the control volume: • In time interval dt:
- Water particles at sec.1-1 move to sec. 1`-1` with velocity V1.
- Water particles at sec.2-2 move to sec. 2`-2` with velocity V2.
• To satisfy continuity equation:
dtVAPdsAP ..... 111111
• The work done by the pressure force
dtVAdtVA .... 2211
dtVAPdsAP ..... 222222
-ve sign because P2 is in the opposite direction to distance traveled ds2
……. on section 1-1
……. on section 2-2
23
• The work done by the gravity force :
).(.. 2111 hhdtVAg
)(...2
1.
2
1.
2
1 21
2211
21
22 VVdtVAVMVM
)(..2
1).(...... 2
12
22121 VVdtQhhdtQgdtQPdtQP
• The kinetic energy:
The total work done by all forces is equal to the change in kinetic energy:
Dividing both sides by gQdt
22
22
11
21
22h
P
g
Vh
P
g
V
Bernoulli Equation
Energy per unit weight of waterOR: Energy Head
24
Energy head and Head loss in pipe flow
25
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:
26
Example 3.3 &3.4
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.
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
24
24
Without calculation sketch the (E.G.L) and (H.G.L)
30
Basic components of a typical pipe system
31
Calculation of Head (Energy) Losses:
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.
3.5 Losses of Head due to Friction
• 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
33
Major losses formulas• Several formulas have been developed in the past.
Some of these formulas have faithfully been used in various hydraulic engineering practices.
1. Darcy-Weisbach formula
2. The Hazen -Williams Formula
3. The Manning Formula
4. The Chezy Formula
5. The Strickler Formula
34
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).
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
The friction factor is function of different terms:
D
eVDF
D
eVDF
D
eNFf R ,,,
Renold number Relative roughness
36
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
37
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:
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
Friction Factor f
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
Moody diagram
• A convenient chart was prepared by Lewis F. Moody A convenient chart was prepared by Lewis F. Moody and commonly called the Moody diagram of friction and commonly called the Moody diagram of friction factors for pipe flow,factors for pipe flow, There are 4 zones of pipe flow in the chart:
• A laminar flow zone where f is simple linear function of NR
• 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 NR 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
40
Laminar
Marks Reynolds Number independence
42
Typical values of the absolute roughness (e) are given in table 3.1
43
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
Problems (head loss)
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 ?
45
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
Example 1
Type 1:Given the kind and size of pipe and the flow rate head loss ?
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
Example 2
Type 1:Given the kind and size of pipe and the flow rate head loss ?
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
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
Example 3
Type 2:Given the kind and size of pipe and the head loss flow rate ?
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
Example 3.5Compute 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 ?
• 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
flow rate ?
Determines relative roughness e/D
2/12/3 2
L
ghDfN f
R
Type 2. Given the kind and size of pipe and the head loss
Given and e/D we can determine f (Moody diagram)fNR
Use Darcy-Weisbach to determine velocity and flow rate
Alternative Method for solution of Type 2 problems
Because V is unknown we cannot calculate the Reynolds number
However, if we know the friction loss hf, we can use the Darcy-Weisbach equation to write:
hf fL
D
V 2
2g
V 2ghf
L
1/ 2D
f
1/ 2
We also know that:
Re VD
Re 1
f 1/ 2
D 3 / 2
2ghf
L
1/ 2
unknowns
2/12/32/1 2
L
ghDfN f
R
Can be calculated based on available data
Quantity plotted along the top of the Moody diagram
Moody Diagram
Lam
inar flo
w
Smooth pipes
Transitionally rough pipes
Fully rough pipes
Res
ista
nce
Co
effi
cien
t f
Reynolds number
Rel
ativ
e ro
ug
hn
ess
e/D
2/12/32/1 2
L
ghDfN f
R
Example 3.5Compute 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
f = 0.0121
Example (type 2)
H
L
H = 4 m, L = 200 m, and D = 0.05 m
What is the discharge through the galvanized iron pipe?
Table : Galvanized iron pipe: e = 0.15 mm e/D = 0.00015/0.05 = 0.003
= 10-6 m2/s
We can write the energy equation between the water surface in the reservoir and the free jet at the end of the pipe:
Lhg
Vh
p
g
Vh
p
22
2
22
2
2
11
1
g
V
D
Lf
g
V
2200040
22
2
fDL
f
gV
40001
5.78
1
422
1
2
Example (continued)Assume Initial value for f : fo = 0.026
Initial estimate for V: m/sec 865.0026.040001
5.78
V
Calculate the Reynolds number 44 103.4105 VDV
NR
Updated the value of f from the Moody diagram f1 = 0.029
m/sec 819.0029.040001
5.78
V
442 101.4105 VDV
NR
0 0.026 0.865 4.3104
1 0.029 0.819 4.1104
2 0.0294 0.814 4.07104
3 0.0294
Iteration f V NR
Convergence
Solution:
V2 0.814 m/s
Q VA 0.814 0.052
41.60 10 3 m3 /s
e/D = 0.003
Initial estimate for fA good initial estimate is to pick the f value that is valid for a fully rough pipe with the specified relative roughness
fo = 0.026
Solution of Type 3 problems-uniform flow in a single pipe
Given the kind of pipe, the head loss and flow rate size of pipe ?
Determines equivalent roughness e
Problem?Without D we cannot calculate the relative roughness e/D, NR, or fNR
Solution procedure: Iterate on f and D
1. Use the Darcy Weisbach equation and guess an initial value for f2. Solve for D3. Calculate e/D4. Calculate NR
5. Update f6. Solve for D7. If new D different from old D go to step 3, otherwise done
Example (Type 3)A pipeline is designed to carry crude oil (S = 0.93, = 10-5 m2/s) with a discharge of 0.10 m3/s and a head loss per kilometer of 50 m. What diameter of steel pipe is needed? Available pipe diameters are 20, 22, and 24 cm.
From Table 3.1 : Steel pipe: e = 0.045 mm
Darcy-Weisbach:
g
V
D
Lfh f 2
2
2
2
542
22
2
2
1614
22 g
fLQ
DDg
Q
D
Lf
gAQ
D
Lfh f
5/1
2
2
2
16
fhg
fLQD
5/15/1
5/1
2
2
440.05081.92
10.0100016ffD
Make an initial guess for f : fo = 0.015
D 0.440 0.0151/ 5 0.190 m
Now we can calculate the relative roughness and the Reynolds number:
33
2108.66
1107.12
144
DD
QD
D
QD
A
QVDNR
00024.010045.0 3
DD
e
update ff = 0.021
e/D = 0.00024
Updated estimate for f
f1 = 0.021
0 0.015 0.190 66.8103 0.00024
Iteration f D NR e/D
Example Cont’d5/1440.0 fD
update f
DNR
1107.12 3
From moody diagram, updated estimated for f :
f1 = 0.021 D = 0.203 m3105.62 RN
00023.0D
e
1 0.021 0.203 62.5103 0.00023
2 0.021 Convergence
Solution:
D = 0.203
Use next larger commercial size:
D = 22 cm
Example 3.6Estimate 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:
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.
Empirical Formulas 1• Hazen-Williams
UnitsSISRCV hHW54.063.085.0
tCoefficien iamsHazen Will
44
P wetted
A wetted Radius hydraulic
2
CL
hS
D
D
D
R
HW
f
h
UnitsSI
0.71 852.1
87.4852.1 QDC
Lh
HW
f
Sim
pli
fied
UnitsBritishSRCV
mVcmD
hHW54.063.0318.1
sec/0.35
tCoefficien iamsHazen WillHWC
68
tCoefficien iamsHazen WillHWC
69
081.0
sec/0.3
V
VCC
mVWhen
oHoH
Where:CH = corrected valueCHo = value from tableVo = velocity at CHo
V = actual velocity
Empirical Formulas 2
70
Manning Formula
• This formula has extensively been used for open channel designs
• It is also quite commonly used for pipe flows
71
• Manning
tCoefficien M
4 P wetted
A wetted Radius hydraulic
anningnL
hS
D
R
f
h
Sim
pli
fied
UnitsSI
0.3133.5
2
D
nQLh f
2/13/21SR
nV h
72
• 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
73
74
The Chezy Formula
V C R Sh 1 2 1 2/ /
2
4
C
V
D
Lh f
where C = Chezy coefficient
75
• 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
.
(.
)
76
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
77
Relations between the coefficients in Chezy, Manning , Darcy , and Strickler formulas.
nkstr
1
6/1hstr RkC
g
Rfn h
8
3/1
Example
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
79
Minor losses
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
80
• 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,
Losses due to contractionA 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.
g
Vkh cc 2
22Along centerline
Along wall
Value of the coefficient Kc for sudden contraction
VV22
83
Head Loss Due to a Sudden Contraction
h KV
gL L 22
2
g
VhL 2
5.02
2
Head losses due to pipe contraction may be greatly reduced by introducing a gradual pipe transition gradual pipe transition known as a confusor confusor
g
V'k'h cc 2
22
'kc
85
Head Loss Due to Gradual Contraction (reducer or nozzle)
g
VVKh LL 2
21
22
100 200 300 400
KL 0.2 0.28 0.32 0.35
A different set of data is :
Losses due to Enlargement
g
VVhE 2
)( 221
A sudden EnlargementA sudden Enlargement in a pipe
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
88
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 :
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
90
Head Loss Due to Gradual Enlargement (conical diffuser)
g
VVKh LL 2
22
21
100 200 300 400
KL 0.39 0.80 1.00 1.06
91
Gibson tests
92
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
Different pipe inlets
increasing loss coefficient
94
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
95
Another Typical values for various amount of rounding of the lip
96
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
97
Head Loss Due to Bends in Pipes
R/D 1 2 4 6 10 16 20
Kb 0.35 0.19 0.17 0.22 0.32 0.38 0.42
g
Vkh bb 2
2
98
Miter bends
For situations in which space is limited,
99
Head Loss Due to Pipe Fittings (valves, elbows, bends, and tees)
h KV
gv v2
2
100
101
The loss coefficient for elbows, bends, and tees
Loss coefficients for pipe components (Table)
Minor loss coefficients (Table)
Minor loss calculation using equivalent pipe length
f
DkL l
e
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)
Example In 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
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
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
Example
A pipe enlarge suddenly from D1=240mm to D2=480mm. the H.G.L rises by 10 cm calculate the flow in the pipe
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
L
L
/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
111
• Note that the above values are average typical values, actual values will depend on the make (manufacturer) of the components.
• See:– Catalogs – Hydraulic handbooks !!