chapter 2: flow through single &combined pipelines university of palestine engineering...
TRANSCRIPT
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CHAPTER 2: Flow through single &combined Pipelines
CHAPTER 2: Flow through single &combined Pipelines
University of PalestineEngineering Hydraulics2nd semester 2010-2011
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ContentPipelines in series &
parallel
Pipelines with negative
pressure
Branching pipe systems
Power in pipelines
Flow through single &combined Pipelines
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Part APipelines in series &
parallel Pipelines with
negative pressure
Flow through single &combined Pipelines
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Any water conveying system may include the following elements:
•pipes (in series, pipes in parallel)•elbows •valves •other devices.
• If all elements are connected in series, The arrangement is known as a pipeline.
• Otherwise, it is known as a pipe network.
Introduction
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How to solve flow problems
• Calculate the total head loss (major and minor) using the methods of chapter 2
• Apply the energy equation (Bernoulli’s equation)
This technique can be applied for different systems.
Introduction
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Flow Through A Single Pipe (simple pipe flow)• A simple pipe flow: It is a
•flow takes place in one pipe •having a constant diameter •with no branches.
• This system may include bends, valves, pumps and so on.
Introduction
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Simple pipe flow
(1)
(2)
Introduction
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To solve such system:
• Apply Bernoulli’s equation
• where
pL hhzg
VPz
g
VP 2
222
1
211
22
(1)
(2)
g
VK
g
V
D
fLhhh LmfL 22
22
For the same material and constant diameter (same f , same V) we can write:
L
TotalmfL K
D
fL
g
Vhhh
2
2
Introduction
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Example 1• Determine the difference in the elevations between the
water surfaces in the two tanks which are connected by a horizontal pipe of diameter 30 cm and length 400 m. The rate of flow of water through the pipe is 300 liters/sec. Assume sharp-edged entrance and exit for the pipe. Take the value of f = 0.032. Also, draw the HGL and EGL.
Z1 Z
Introduction
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Compound Pipe flow • When two or more pipes with
different diameters are connected together head to tail (in series) or connected to two common nodes (in parallel)
The system is called compound pipe flow
Introduction
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• pipes of different lengths and different diameters connected end to end (in series) to form a pipeline
Flow Through Pipes in Series
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• Discharge:The discharge through each pipe is the same
• Head loss: The difference in liquid surface levels is equal to the sum of the total head loss in the pipes:
332211 VAVAVAQ
LBBB
AAA hz
g
VPz
g
VP
22
22
332211 VAVAVAQ
Flow Through Pipes in Series
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LBBB
AAA hz
g
VPz
g
VP
22
22
Hhzz LBA
Where
4
1
3
1 jmj
ifiL hhh
g
VK
g
VK
g
VK
g
VK
g
V
D
Lfh exitenlcent
i
i
i
iiL 22222
23
22
22
21
3
1
2
Flow Through Pipes in Series
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Pipelines in Series
nQQQQ 21
LnLLL ....hhhh 21
Flow Through Pipes in Series
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Example 2• Two new cast-iron pipes in series connect two reservoirs. Both pipes
are 300 m long and have diameters of 0.6 m and 0.4 m, respectively. • The elevation of water surface in reservoir A is 80 m. The discharge
of 10o C water from reservoir A to reservoir B is 0.5 m3/sec. • Find the elevation of the surface of reservoir B. • Assume a sudden contraction at the junction and a square-edge
entrance.
Flow Through Pipes in Series
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• If a main pipe divides into two or more branches and again join together downstream to form a single pipe, then the branched pipes are said to be connected in parallel (compound pipes).
• Points A and B are called nodes.
Flow Through Parallel Pipes
n
iiQQ
1
LnLLLL ....hhhhh 321
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• Discharge:
• Head loss: the head loss for each branch is the same
3
1321
iiQQQQQ
Q1, L1, D1, f1
Q2, L2, D2, f2
Q3, L3, D3, f3
321 fffL hhhh
g
V
D
Lf
g
V
D
Lf
g
V
D
Lf
222
23
3
33
22
2
22
21
1
11
Flow Through Parallel Pipes
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Example 3:Determine the flow in each pipe and the flow in the main pipe if Head loss between A & B is 2m & f=0.01
Solution
/sm...π
AVQ
m/s.V.
V
..
g
V.
D
Lf
332111
1
21
21
1
1
1015350620404
5062
28192040
25010
22
221 ff hh
/sm.QQQ
/sm...π
Q
m/s.V.
V
..
g
V.
D
Lf
3321
3322
2
22
22
2
2
10178
1002555720504
55728192050
30010
22
Flow through single &combined Pipelines
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Example 4
The following figure shows pipe system from cast iron steel. The main pipe diameter is 0.2 m with length 4m at the end of this pipe a Gate Valve is fixed as shown. The second pipe has diameter 0.12m with length 6.4m, this pipe connected to two bends R/D = 2.0 and a globe valve. Total Q in the system = 0.26 m3/s at T=10oC. Determine Q in each pipe at fully open valves.
Flow through single &combined Pipelines
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22
031402
20 m.
.πAa
22
011302
120 m.
.πAb
ba
babbaa
hh
V.V.VAV A m.
QQQ
0113003140260 3
21
g
V.
g
V
D
Lfh aa
a
aaa 2
1502
22
g
V
g
V.
g
V
D
Lfh bbb
b
bbb 2
102
19022
222
Solution
Flow through single &combined Pipelines
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g
V.
.
.f
g
V.
.f b
ba
a 210380
120
46
2150
20
4 22
22 38103353 15020 bbaa V.f.V.f 0255.0
0185.0
b
a
f
f
22 3810025503353 1500185020 ba V...V..
ba V.V 7194
m/s.V
m/s.V
b
a
6301
6937
V.V.VAV A m. bbbbaa 01130)719.4(03140260 3
/s m...VAQ
/s m...VAQ
bbb
aaa
3
3
0180630101130
2420693703140
Flow through single &combined Pipelines
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Example 5Determine the flow rate in each pipe (f=0.015)Also, if the two pipes are replaced with one pipe of the same length determine the diameter which give the same flow.
Flow through single &combined Pipelines
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Flow through single &combined Pipelines
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Flow through single &combined Pipelines
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Group work Example
• Four pipes connected in parallel as shown. The following details are given:
PipeL (m)D (mm)f
12002000.020
23002500.018
31503000.015
41002000.020
• If ZA = 150 m , ZB = 144m, determine the discharge in each pipe ( assume PA=PB = Patm)
Flow through single &combined Pipelines
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Group work Example Two reservoirs with a difference in water levels of 180 m and are connected
by a 64 km long pipe of 600 mm diameter and f of 0.015. Determine the discharge through the pipe. In order to increase this discharge by 50%, another pipe of the same diameter is to be laid from the lower reservoir for part of the length and connected to the first pipe (see figure below). Determine the length of additional pipe required.
=180mQN QN1
QN2
Flow through single &combined Pipelines
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• Long pipelines laid to transport water from one reservoir to another over a large distance usually follow the natural contour of the land.
• A section of the pipeline may be raised to an elevation that is above the local hydraulic gradient line (siphon phenomena) as shown:
(siphon phenomena)
Pipe line with negative Pressure
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Definition: It is a long bent pipe which is used to transfer
liquid from a reservoir at a higher elevation to another reservoir at a lower level when the two reservoirs are separated by a hill or high ground
Occasionally, a section of the pipeline may be raised to an elevation that is
above the local HGL.
(siphon phenomena)
Pipe line with negative Pressure
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Siphon happened in the following cases:• To carry water from one reservoir to
another reservoir separated by a hill or high ground level.
• To take out the liquid from a tank which is not having outlet
• To empty a channel not provided with any outlet sluice.
Pipe line with negative Pressure
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Characteristics of this system
• Point “S” is known as the summit.• All Points above the HGL have pressure
less than atmospheric (negative value)• If the absolute pressure is used then
the atmospheric absolute pressure = 10.33 m
• It is important to maintain pressure at all points ( above H.G.L.) in a pipeline above the vapor pressure of water (not be less than zero Absolute )
Pipe line with negative Pressure
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A S
LSS
SA
LSSS
AAA
hP
g
VZZ
hZP
g
VZ
P
g
V
2
222
22
-ve valueMust be -ve value ) below the atmospheric pressure(
Negative pressure exists in the pipelines wherever the pipe line is raised above the hydraulic gradient line (between P & Q)
Pipe line with negative Pressure
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The negative pressure at the summit point can reach theoretically -10.3 m water head (gauge pressure) and zero (absolute pressure) But in the practice water contains dissolved gasses that will vaporize before -10.3 m water head which reduces the pipe flow cross section. Generally, this pressure reach to -7.6m water head (gauge pressure) and 2.7m (absolute pressure)
Pipe line with negative Pressure
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Example 1Siphon pipe between two pipe has diameter of 20cm and length 500m as shown. The difference between reservoir levels is 20m. The distance between reservoir A and summit point S is 100m. Calculate the flow in the system and the pressure head at summit. f=0.02
Pipe line with negative Pressure
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Solution
Pipe line with negative Pressure
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• Pumps may be needed in a pipeline to lift water from a lower elevation or simply to boost the rate of flow. Pump operation adds energy to water in the pipeline by boosting the pressure head
• The computation of pump installation in a pipeline is usually carried out by separating the pipeline system into two sequential parts, the suction side and discharge side.
Pumps
Pipe line with negative Pressure
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LsRP hHHH
Pumps design will be discussed in details in next chapters
Pipe line with negative Pressure
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Part B
Branching pipe systems
Flow through single &combined Pipelines
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Branching in pipes occur when water is brought by pipes to a junction when more than two pipes meet. This system must simultaneously satisfy two basic conditions:1 – the total amount of water brought by pipes to a junction must equal to that carried away from the junction by other pipes.
2 – all pipes that meet at the junction must share the same pressure at the junction. Pressure at point J = P
0Q
Branching pipe systems
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Three-reservoirs problem
)Branching System(
How we can demonstrate the hydraulics of branching pipe System??
by the classical three-reservoirs problem
Branching pipe systems
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This system must satisfy:
Q3 = Q1 + Q2
2( All pipes that meet at junction “J” must share the same pressure at the junction.
1( The quantity of water brought to junction “J” is equal to the quantity of water taken away from the junction:
Flow Direction????
Branching pipe systems
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Type 1:• given the lengths , diameters, and materials of all
pipes involved; D1 , D2 , D3 , L1 , L2 , L3 , and e or f
• given the water elevation in each of the three reservoirs, Z1 , Z2 , Z3
• determine the discharges to or from each reservoir, Q1 , Q2 ,and Q3 .
This types of problems are most conveniently solved by trail and
error
Consider a case where three reservoirs are connected by a branched-pipe system .
There are two types of Branching pipe problem:
Branching pipe systems
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• First assume a piezometric surface elevation, P , at the junction. • This assumed elevation gives the head losses hf1, hf2, and hf3
• From this set of head losses and the given pipe diameters, lengths, and material, the trail computation gives a set of values for discharges Q1 , Q2 ,and Q3 .
• If the assumed elevation P is correct, the computed Q’s should satisfy:
• Otherwise, a new elevation P is assumed for the second trail. • The computation of another set of Q’s is performed until the above
condition is satisfied.
Q Q Q Q 1 2 3 0
Branching pipe systems
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Note:• It is helpful to plot the computed trail values of P
against .• The resulting difference may be either plus or
minus for each trail. • However, with values obtained from three trails, a
curve may be plotted as shown in the next example.
The correct discharge is indicated by the intersection of the curve with the
vertical axis.
Branching pipe systems
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The problem is to determine the discharge in each pipe and the pressure head at the junction (point J). Four unknowns: Q in each pipe and P at JGiven:All pipes parametersAll Tanks elevation
The solution based on:1- Assuming the pressure at J and then estimate the value of hf in each pipe2- Calculate the flow in each pipe and check3- try the last procedure until
Q
0Q
Branching pipe systems
Type 1:
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Example 1
PipeCJBJAJ
Length m200040001000
Diameter cm405030
f0.0220.0210.024
In the following figure determine the flow in each pipe
Branching pipe systems
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Trial 1 ZP= 110mApplying Bernoulli Equation between A , J :
g
V
g
V
D
LfZZ PA 23.0
1000024.0110120
2.
21
21
1
11
V1 = 1.57 m/s , Q1 = 0.111 m3/s
g
V
g
V
D
LfZZ BP 25.0
4000021.0100110
2.
22
22
2
22
V2 = 1.08 m/s , Q2 = - 0.212 m3/s
Applying Bernoulli Equation between B , J :
Branching pipe systemsExample 1.cont.
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g
V
g
V
D
LfZZ CP 24.0
2000022.080110
2.
23
23
3
33
Applying Bernoulli Equation between C , J :
V3 = 2.313 m/s , Q2 = - 0.291 m3/s
0392.0291.0212.0111.0321 QQQQ
Branching pipe systemsExample 1.cont.
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Trial 2 ZP= 100m
0/ 08.0237.00157.0 3321 smQQQQ
Trial 3ZP= 90m
0/ 324.0168.03.0192.0 3321 smQQQQ
Example 1.cont.
Branching pipe systems
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Draw the relationship between and P Q
99m Pat 0 Q
Branching pipe systemsExample 1.cont.
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Type 2:
• Given the lengths , diameters, and materials of all pipes involved; D1 , D2 , D3 , L1 , L2 , L3 , and e or f
• Given the water elevation in any two reservoirs, Z1 and Z2 (for example)
• Given the flow rate from any one of the reservoirs,Q1 or Q2 or Q3
• Determine the elevation of the third reservoir Z3 (for example) and the rest of Q’s
This types of problems can be solved by simply using:
• Bernoulli’s equation for each pipe • Continuity equation at the junction.
Branching pipe systems
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The problem is to determine the flow in two pipes and the pressure head at the Junction J. Three unknown:Q in two pipe and P at JGiven:All pipes parametersQ in a pipetwo Tanks elevationThe solution based on:1- Determine the pressure at J by calculate hf in pipe with known Q2- Estimate the value of hf in the other pipe2- Calculate the flow in the other pipes
Type 2:
Branching pipe systems
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Example 2
In the following figure determine the flow in pipe BJ & pipe CJ. Also, determine the water elevation in tank C
Branching pipe systems
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m.Z
.
.
..Z
g
V.
D
LfZZ
m/s..
π.
A
QV
P
PPA
47536
8192
8490
30
1200024040
2
849030
4
060
221
1
11
21
11
Solution
/sm 0.0203Q0.645m/sV
9.812
V
0.2
6000.02436.47538
2g
V.
D
LZZ
322
22
22
2
22PB
f
Applying Bernoulli Equation between B , J :
Applying Bernoulli Equation between A , J :
Branching pipe systemsExample 2 cont.
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mZ
gg
V
D
LfZZ
c
CP
265.32
2
136.1
3.0
800024.0 Z- 6.4753
2.
2
c
23
3
33
Applying Bernoulli Equation between C , J :
smQQQ
QQQQ
/ 0803.00203.006.0
03
213
321
sm
A
QV / 136.1
3.04
0803.0
23
33
Branching pipe systemsExample 2 cont.
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Group Work 1
ACAB
ACAB
BDBC
BDBCAB
VV
VV
QQQ
125.1
3.024.0
0
24
24
smQsmV
smQQsmV
VV
VV
g
V
g
V
hh
ABAB
BDBCBC
BCBC
BCAB
ABAB
BCAB
/31.0/5.2
/155.0/2.2
10 816.0)125.1(55.2
10 7.155.2
1023.0
100001.0
24.0
200001.0
10
3
3
22
22
01.0f
Find the flow in each pipe
Flow through single &combined Pipelines
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Group Work 2Determine the flow ifa) The valve is closedb) The valve is open and the flow through the small pipe = 100L/S
Flow through single &combined Pipelines
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Part C
Power Transmission Through Pipes
Flow through single &combined Pipelines
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• Power is transmitted through pipes by the water (or other liquids) flowing through them.
• The power transmitted depends upon:
(a) the weight of the liquid flowing through the pipe
(b) the total head available at the end of the pipe.
Power Transmission Through Pipes
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• What is the power available at the end B of the pipe?
• What is the condition for maximum transmission of power?
Power Transmission Through Pipes
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Total head (energy per unit weight) H of fluid is given by:
time
Weightx
weight
Energy
time
EnergyPower
ZP
g
VH
2
2
QQgtime
Weight
Therefore: Power Q HUnits of power:
N . m/s = Watt745.7 Watt = 1 HP (horse
power)
Power Transmission Through Pipes
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For the system shown in the figure, the following can be stated:
mf
m
f
hhHγ Q
γ Q h
γ Q h
γ Q H
PowerExit At
lossminor todue dissipatedPower
friction todue dissipatedPower
Power EntranceAt
Power Transmission Through Pipes
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Calculate the max transported power through pipe line
f
π
π
f
mf
hg
V
D
f LH
g
V
D
f L HDγ
dV
dP
g
V
D
Lf HVDγ P
VAQhHγ Q
hhHγ Q
32
..3
2..30at Max.
2
P
lossminor neglect
PExit At
2
22
4
32
4
The max transported power through pipe line at 3
H h f
Power Transmission Through Pipes
Power transmitted through a pipe is maximum when the loss of head due
of the total head at the inlet 3
1to friction equal 62
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Maximum Efficiency of Transmission of Power:
Efficiency of power transmission is defined as
inlet at the suppliedPower
outlet at the availablePower
H
hhH
QH
hhHQ mfmf ][][
or
H
hH f ][
Maximum efficiency of power transmission occurs when3
Hh f
%67.663
2]
3[
max
H
HH
)If we neglect minor losses(
Power Transmission Through Pipes
%67.661003max
H
HH
H
hHη
H
hhH
γQH
hhHγQη
f
mfmf
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ExamplePipe line has length 3500m and Diameter 0.5m is used to transport Power Energy using water. Total head at entrance = 500m. Determine the maximum power at the Exit. F = 0.024
fout h Hγ QP
mH
h f 3
500
3at Power Max.
g
V
..
g
V
D
Lfh f 230
35000240
2
22
m/s 3.417V
/s m...AVQ π 32
4 24150417330
Power Transmission Through Pipes
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HP.
tt) N.m/s (Wa
..
HgQ
HHgQ
hHγQP f
10597745
789785789785
500241508191000
3
32
32
Power Transmission Through Pipes
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http://www.haestad.com/library/books/awdm/online/wwhelp/wwhimpl/java/html/wwhelp.htm
66