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COLLEGE OF ENGINEERING KWAME NKRUMAH UNIVERSITY OF SCIENCE AND TECHNOLOGY
PRELIMINARY READER
in
CE 356 HYDRAULIC ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
S. N. ODAI FEBRUARY 2007
DR S. N. ODAI - KNUST HYDRAULICS (CE 356)
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TABLE OF CONTENTS
CHAPTER ONE INTRODUCTION TO HYDRAULICS ........................................................................ 2
1.1 DEFINITION...................................................................................................................................................... 2 1.2 HISTORY OF HYDRAULICS .............................................................................................................................. 2
CHAPTER TWO OPEN-CHANNEL FLOW (FREE SURFACE FLOW) ............................................... 3
2.1 DEFINITION...................................................................................................................................................... 3 2.2 CLASSIFICATION OF FLOWS ......................................................................................................................... 3 2.3 VELOCITY DISTRIBUTION ................................................................................................................................ 3 2.4 OPEN CHANNEL UNIFORM FLOW FORMULAS ............................................................................................... 4 2.5 HYDRAULIC DESIGN OF OPEN CHANNELS .................................................................................................... 5 2.6 MOST EFFICIENT CROSS SECTIONS (BEST HYDRAULIC CROSS SECTIONS) ............................................. 16 2.7 UNLINED AND LINED CHANNELS .................................................................................................................. 18 2.8 NOMOGRAPHIC DESIGN OF OPEN CHANNELS ............................................................................................ 21
CHAPTER THREE NON-UNIFORM FLOW ..........................................................................................23
3.1 RAPIDLY VARIED FLOWS ............................................................................................................................. 23 3.1.1 Energy Principle ..............................................................................................................23 3.1.2 Specific Energy ................................................................................................................25 3.1.3 Flow Measurements ........................................................................................................27 3.1.4 Critical Flow .....................................................................................................................31
3.2 FLOW DEPTH FOR MAXIMUM DISCHARGE AT A GIVEN SPECIFIC ENERGY .................................................. 35 3.3 THE MOMENTUM EQUATION......................................................................................................................... 37
3.3.1 The Hydraulic Jump ........................................................................................................37 3.4 GRADUALLY VARIED FLOW .......................................................................................................................... 42
3.4.1 Governing Equations ......................................................................................................42 3.4.2 Classification of Channel Slopes ...................................................................................44 3.4.3 Principles for determining the surface profiles ...........................................................44
CHAPTER FOUR PIPES AND PIPE NETWORKS ..............................................................................51
4.1 DEFINITION.................................................................................................................................................... 51 4.2 LAMINAR FLOW AND TURBULENT FLOW ..................................................................................................... 51 4.3 ENERGY EQUATION OF PIPE FLOW ............................................................................................................. 52 4.4 CONTINUITY EQUATION ............................................................................................................................... 52 4.5 EVALUATION OF HEAD LOSS DUE TO FRICTION ......................................................................................... 53 4.6 MINOR HEAD LOSSES ................................................................................................................................... 56 4.7 PIPELINES WITH PUMPS AND TURBINES ....................................................................................................... 58 4.8 PIPES IN SERIES ............................................................................................................................................... 58 4.9 PIPES IN PARALLEL ...................................................................................................................................... 59 4.10 WATER HAMMER ....................................................................................................................................... 65 4.11 PIPE NETWORKS ........................................................................................................................................ 67
CHAPTER FIVE HYDRODYNAMIC MACHINES ...............................................................................72
5.1 HYDRODYNAMIC MACHINES ......................................................................................................................... 72 5.2 PUMPS CLASSIFICATION ............................................................................................................................... 72
5.2.1 Introduction to Centrifugal Pumps ................................................................................73 5.2.2 Introduction to Vertical Pumps ......................................................................................74 5.2.3 Introduction to Positive Displacement ..........................................................................74
5.3 THE CENTRIFUGAL PUMP .............................................................................................................................. 75 5.4 WORKING OF A CENTRIFUGAL PUMP............................................................................................................ 80 5.5 SPECIFIC SPEED OF PUMPS ......................................................................................................................... 80 5.6 RELATIONS FOR GEOMETRICALLY SIMILAR PUMPS ................................................................................... 81 5.7 RELATIONS FOR ALTERATION IN THE SAME PUMP ...................................................................................... 81 5.8 HEAD DEVELOPED AND POWER REQUIRED .................................................................................................. 82 5.9 CAVITATION AND NET POSITIVE SUCTION HEAD ........................................................................................ 85 5.10 PERFORMANCE OF CENTRIFUGAL PUMPS................................................................................................... 88 5.11 SINGLE PUMP AND PIPELINE SYSTEM ......................................................................................................... 90 5.12 MULTIPLE PUMP SYSTEM ............................................................................................................................ 91
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CHAPTER ONE INTRODUCTION TO HYDRAULICS 1.1 Definition The term hydraulics refers generally to the study of the behaviour of Liquids. The word hydraulics comes from the Greek word hydraulikos meaning water. It is the study of the mechanical behaviour of water in physical systems and processes. It involves flows in open channels, conduits, porous media, sediments and other contaminants transported with water.
Table 1.1 Differences between Fluid Mechanics and Hydraulics
Fluid Mechanics Hydraulics
1. Theoretical 1. Empirical
2. Refers to both liquids and gasses 2. Refers to liquids (often refers to Water)
3. Compressible and incompressible Flows
3. Incompressible flows
Objective of this study: The application of engineering principles and methods to the planning, control, transportation, conservation and utilization of water. Scope of study: The present study will cover the following broad topics or chapters.
1. Free surface flow (Open Channel Flow) 2. Pipe Flow and Pipe Networks 3. Hydrodynamic Machines (Emphasis on Pumps) 4. Dimensional Analysis and Hydraulic Similitude
Liquids are transported from one location to another using natural or constructed conveyance structures and flow passages. The flow passages may have cross sections that are open or close at the top. The structures with the closed tops are referred to as closed conduits and those with open tops are called open channels. 1.2 History of Hydraulics Early civilizations developed in regions where an abundance of water could be distributed over fairly flat land for irrigation In Egypt, to augment the flow of irrigation water during the low flow season, there are signs that one of the early rulers, King Menes (about 3000 B. C.) had a masonry dam built across the Nile. This dam was used to divert the river into a canal, thus, to irrigate part of the adjoining and lands. Civilization in Mesopotamia (Iraq) started about the same time as in Egypt (3000 B. C.). The Euphrates and Tigris rivers formed a network of channels before finally emptying into the Persian Gulf. Furthermore, the people of the area built many canals for irrigating crops, draining swamps, and water transportation. Early hydraulic engineering in this area included developing flood protection works and dam construction. Ancient ruins in the valleys of the Indus River in Asia and the Yellow River in China reveal evidence of water systems developed at least 3000 years ago. From about 200 B. C. to 50 A. D., the Romans developed elaborate water-supply systems throughout their empire. It is reported that aqueducts supplied Rome with about 200 million gallons of water daily.
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CHAPTER TWO OPEN-CHANNEL FLOW (FREE SURFACE FLOW) 2.1 Definition The flow in an open channel or closed conduit having a free surface is referred to as free-surface flow or open channel flow (open channel Hydraulics). If there is no free surface and the conduit is flowing full, then the flow is called pipe flow, or pressurised flow. Open channel flow occurs when a liquid flowing due to gravity has a free surface, and the liquid is not under pressure other than that caused by its own weight and by atmospheric pressure. For example, tunnels, pipes and aqueducts are closed conduits whereas rivers, streams, estuaries, etc, are open channels.
Common examples of open channel flows occur in rivers, canals, storm water drains, and irrigation canals. A channel with constant shape and slope is known as a prismatic channel.
Fig 2.1 Various Sections of Open Channels
2.2 Classification of Flows
Steady and unsteady flows (it is w.r.t. time)
Uniform and non-uniform flows (w.r.t. distance) also called varied flows
Laminar and turbulent flows (w.r.t movement of liquid particles)
Sub critical, Supercritical, and critical flows. (w.r.t. Froude Number)----- gDv
2.3 Velocity Distribution
The flow velocity in a channel section usually varies from one point to another. This is due to shear stress (resistance, friction) at the bottom and the sides of the channel and due to the presence of free surface.
Fig 2.2 Velocity Distribution
Curves created by joining points of equal velocities are called isovels. Velocity distribution in an open channel is not axisymetric due to friction at the walls
°
0.5
°
°
2.0
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2.4 Open Channel Uniform Flow Formulas 2.5
Fig 2.3 Open Channel Flow bottom Slopes
Flow in most channels is turbulent, and laminar flow in open channels is very rare. Laminar open-channel flow is known to exist, however usually where the sheers of water flow over the ground or where it is created deliberately. The fact that a stream surface appears smooth and glassy to the observer is by no means an indication that the flow is Laminar: most probably, it indicates that the surface velocity is lower than that required for waves to form.
Fig 2.4 Open Channel
Where T=Top width of the channel; t=width of water surface for depth h; h= flow depth in channel; D=Depth of channel after free board is added; b=bottom width; c=length of wetted sides of channel; θ= angle b/n sloping side and the horizontal
b
c
t
T
D h
θ
Free board
b) Positive slope, S>0 c) zero slope, S=0 d) negative slope, S<0
y
V Velocity Distribution Curve
V
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Area of flow: 2
)( htbA
Wetted Perimeter (P): The sum of the length of that part of the channel sides and bottom, which are in contact with water
bcP 2 and c= )21( mh
Hydraulic radiusP
AR
Hydraulic Slope (S) is the ratio of vertical drop to length of channel travelled
L
HS D
And the Velocity S
The flow velocity by the Chezy formula is given by RSCV
Where C = Chezy coefficient is difficult to determine. An alternative form is the Manning‟s formula given by
V= SRn
2/13/21
Where n= Manning roughness coefficient. Freeboard The vertical distance from the top of the channel (retaining banks) to the water surface (highest anticipated) at the design condition. This distance should be sufficient to prevent waves or fluctuation in the water surface from overflowing the sides. There is no universal rule for the determination of the freeboard, since wave action or water surface fluctuation in a channel may be created by many uncontrollable causes. However, freeboards varying from 5% to 30% of the depth of flow are commonly used in design. Table 2.1 Some recommended values for freeboard
Channel flow rate (m3/s) 2~10 1~2 0.5~1.0 <0.5
Recommended freeboard (m) 0.4~0.6 0.35 0.25 0.15
2.5 Hydraulic Design of Open Channels Designs of channel with flow through various cross-sections will be discussed in this section. There are two main types of design problems encountered in open channel flow studies. They are
Given the depth, determine flow rate
Given the flow rate, determine depth
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Flow through rectangular and trapezoidal sections EXAMPLE 1 Water flows in a rectangular concrete open channel of slope 0.0028, width 12.0m, and flow depth 2.5m. If n=0.013 determine flow velocity and flow rate. Solution
mx
P
AR 765.1
5.20.125.2
0.125.2
Given S=0.0028, n=0.013
2/13/21SR
nv = sm
x/945.5
013.0
)0028.0()765.1(1 2/13/2
VAQ = 2.5 x 12 x 5.945 = 178m3/s
EXAMPLE 2 Water flows in the symmetrical trapezoidal channel lined with asphalt. S=0.001, n=0.015, determine flow rate.
Fig 2.5 Symmetrical trapezoidal channel
Solution
Given S = 0.001, n = 0.015, m = 3
hmhbmhbhA )(2 ;
22 )1(2h
mbP
A= (16 + 3 x 4.5) x 4.5 = 132.75m2, P = 44.46m
PAR = 2.987m
AVQ = smx
SRn
A/297.580
015.0
)001.0()987.2(8.132 32/13/2
2/13/2
θ
4.5m
b=16.0
3
1
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Graphical Method Assume values of d and calculate A, P, R and then Q. The two values of d that bracket the given Q should also bracket the value of drequired.
h (m) A (m2) P (m) R (m) Q (m3/s)
h1 A1 P1 R1 Q1 h2 A2 P2 R2 Q2 „ „ „ „ „ „ „ „ „ „ hn An Pn Rn Qn
Usually 4~5 trials should bracket the required value.
Fig 2.6 Graphical method of determining flow rate
Example 1: How deep will water flow at the rate of 240m3/s in a rectangular channel 20m wide laid on a slope of 0.0001? Take n = 0.015. Solution Employing the Manning's Formula
21
321
SARn
Q
hP
hA
220
20
240 = 21
32
0001.0220
20240
h
h
n
h
12 = 001.0220
20 32
h
h
n
h
1200 = 3
2
220
20
015.0
h
hh
18 = 3
2
220
20
h
hh
0
hreqd
hd (m)
Q (m3/s)
Qgiven
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h
h
AAh
h
h
hh
220818.3
220
2037.76
220
2018
25
25
23
23
h ≈ 7m
Example 2: How wide must a rectangular channel be constructed in order to carry 500m3/s of water at a depth of 6m on a slope of 0.0004 and n = 0.010? Solution Using Manning's Formula
bhbP
bbhA
122
6
b
bR
12
6
Q = 21
32
SRn
A
b
bb
b
bb
b
bb
12
6957.268
12
6667.41
0004.012
6
01.0
6500
23
32
21
32
= b
b
12
6 25
ABb
bb
1282.44
25
b ≈ 17.8 m
h AA
10 7.906 1 0.045
5 1.863 8 5.028 7 3.813
7.01 3.824
b AB
10 14.37
20 55.9
15 32.27
18 45.82
17 41.08
17.8 44.857
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Flow through Compound Channels
A compound channel may be defined as a channel in which various sub areas have different flow properties; e.g. surface roughness, n, flow area, A. A natural river stream having over bank (flood plain) flow during a flood is a typical example of a compound section. In calculating the flow through a compound channel the conveyance is found to be of great help. Channel Conveyance
A = bh+mh2=(b+mh) h A (h)
P = b+2[1+m2] 1/2h P (h)
R =
P
A R (h)
C = 6/11R
n C (h)
K = CAR K (h)
For Q = KS, it is clear that only S is independent of h. K is the conveyance of the channel section given by
K=32 /R
n
A
S
Q
EXAMPLE 1: Determine the normal discharge for the channel shown in Fig 2.7a. The water surface in the left flood plain is 42m wide and that in the right flood plain is 26m. The manning roughness value for the flood plain is 0.10, while the Manning roughness value for the channel is 0.05 for the channel. The longitudinal slope of the channel is 0.0005. Solution:
Fig 2.7 a
12m 30m 15m 16m 10m
12m 12m
2m 2m
5m
n2=0.05 n1=0.10 n3=0.10
5m
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Left over bank Channel Right over bank
Area of flow (A)
2
1
72
22
1230
m
A
Wetted Perimeter
m
yxbP
2.42
21230 22
22
1
Hydraulic Radius
P
AR
mR 71.12.42
721
2
2
213
239527
m
A
m
P
41
512215 22
2
mR 20.541
2132
2
3
42
221
m
A
m
P
2.26
21016 22
3
mR 6.12.26
423
Conveyance
n
ARK
32
sm
K
/030,1
10.0
71.172
3
1
32
sm
K
/790,12
05.0
2.5213
3
1
32
sm
K
/575
10.0
6.142
3
1
32
Discharge 21
21
0005.0575790,12030,1 oi SKQ = 322m3/s
EXAMPLE 2: A compound channel is shown in Fig 2.7b. It has the following dimensions: m1 = 1, n1 = 0.03, b1 = 10m, d1 = 2m. The upper part of the channel has the following dimensions m2 = 1.5, n2 = 0.0225, b2 = 5m, d2 = 1.5m. The channel bottom slope is So = 0.0003. Determine the (i) channel discharge (ii) Average velocity at the cross section. Solution:
Fig 2.7 b
b2=5m
d2=1.5m b3=10m b1=10m
n1=0.03
d1=2m d1=2m
n3=0.03 n2=0.0225
1
1
1
1
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Dividing the channel section a -a and b-b, gives three channels. So=0.0003 Determine the Areas
2
21
222
22
2232
22
2111
2
11111
38.6319.92452
19.95.12
5.15.15
2
455.121210212102
mAAA
mhm
hbAA
mhhmbhmhbA
Determine the Perimeters
mmhbPP
mmhbP
7.75.115.151
64.1511221012
22
22232
22
1111
Determine the hydraulic Radius
mP
ARR
mP
AR
19.17.7
19.9
88.264.15
45
2
232
1
11
Chezy Coefficient for each section
smRn
CC
smRn
C
/8.4519.10225.0
11
/76.3988.203.0
11
21
61
61
21
61
61
232
1
1
1
Conveyance for each section
sm
SKKKSKQ
smRACKK
smRACK
ooi
/49.68
0003.04594593036
/45919.119.98.45
/303688.24576.39
3
321
3
22232
3
1111
The velocity at the section is
smA
Qv /08.1
38.63
49.68
Flow through channel of circular section There are two main flow conditions in this kind of cross –section.
(i) flow depth less than radius (ii) flow depth greater than radius
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(i) Flow depth less than radius
Area of sector ADCB = π R2
o360
2
Area of triangle AOC 22
1
2
12 22 SinRCosSinRRCosRSinx
Fig 2.8 Channel of circular section
Area of flow = Area of sector – Area of triangle
22
1
360
2 2
0
2 SinRRA
The perimeter,
0360
22
RP
The hydraulic radius
0
2
0
2
360
22
22
1
360
2
R
SinRR
P
ARh
045
290
RSinR
SRn
ASRACQ hh
32
(ii) Flow depth greater than radius
Area of flow = Area of (circle + - sector)
=
oRSinRR
360
22
2
1 222
h
R
D
C
O
B
A
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Perimeter = Perimeter of (circle – sector) = 2R - 2R
o360
2 =
0902
R
0
0
222
902
360
22
2
1
R
RSinRR
P
ARh
00
0
454
9022
9022
360
2222
RRSinRRRSinR
Rh
SRn
ASRACQ hh
32
EXAMPLE 1: A circular sewer 1m in diameter conveys a discharge of water at a depth of 0.2m. If the sector is laid at a slope of 1 in 500 find the rate of flow. Take C= 60 Solution
011 13.535.0
3.0
Cos
R
hRCos
Area of flow =
22
1
180
2
0
2 SinRR
= 0.2319 – 0.12 = 0.1119m2
Perimeter of flow =
0360
22
R = 0.928m
Hydraulic radius Rh = mP
A1206.0
928.0
1119.0
5001206.01119.060 xxSRCAQ h = 0.1124m3/S
Example 2: The depth of water in a circular brick lined conduit 1.8m in diameter is to be 1.5m and the flow rate 2.16 x 105 m3/day. Find the gradient of the conduit. Take C = 67. Solution
C
O
B
A
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011 19.489.0
6.0
Cos
R
RhCos
Area of flow =
0
222
1802
2
1 RSinRR
= 2.5457 + 0.4025 – 0.6735 = 2.2747m2
Perimeter P
018012
R
= 4.133m
Hydraulic Radius Rh = 133.4
2747.2 = 0.55m
From Q = RSAC
S2042
11089644
5504489175
256 4
22
2
xxxRCA
Q.
..
.
Example 3: A sewer is laid on a slope of 0.0020 and is to carry 83.5m3/s, when the pipe flow 0.9 full. What size of pipe should be used? Take n = 0.015 Solution
Area of flow =
0
222
360
22
2
1 RSinRR and
2
DR
22222
0
2
0
222
744801206248022180
14720
284
mDDDSinDD
DSinDD
A ...
R
011 87.365.0
4.0
5.0
5.09.0
Cos
D
DDCos
Perimeter, DmDRP 502180
1180
200
.
A
B
h=0.9D D
C
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DD
D
P
ARh 2980
502
74480 2
..
.
mD
D
D
DxDx
SRn
AQ
1.8
53.266
3323.0565.88
0002.0298.07448.0015.0
15.83
38
38
21
322
32
D D8/3
8 256
9 350.466
8.5 300.92
8.2 273.42 8.1 264.62
EXAMPLE 4: A 50cm diameter concrete pipe on a slope of 0.002 is to be used for conveying water of a flow rate of 0.04m3/s. If n=0.013, determine the flow depth. Solution
CA
B
D
D = 50 cm
E
2222 25.025.0
25.0
hhrrCEAE
hhrBE
= ABE = EBC = ArcCos (25.0
25.0 h)
25.0
25.0cos001091.0
360
1
25.0
25.0cos2
4
2 har
har
x
dAreaABCDA
Free surface water subtends an angle of 2 at the centre B.
(Area) ABEA = (Area) BCEB = 2
])25.0(25.0[()25.0( 22 hh
(Area) AECDA = (Area) ABCDA-2 (Area) ABEA
(Area) AECDA =
2
))25.0(25.0(()25.0(2
25.0
25.0)001091.0(
22 hhharcCos
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P = Length of arc ADC = 0
2
360
25.0
25.0.
harcCosd
=
250
2500087280
.
..
harcCos
2/13/21SR
nV
and therefore Q=VA and Q= 2/13/2 SRn
A
Now we have A & P and Q, n, S given. Given that n=0.013, S=0.002
0.04m3/s= 2/13/2 )002.0(013.0
RA
AR2/3=0.01163m3/s
Since
3/2
3/2
P
AR
P
AR
smP
A/.
/
/3
32
35
011630
Substituting values into expressions,
sm
harcCos
hhh
arcCos
/.
.
.).(
).(.).(.
.).(
/
/
3
35
32
22
011630
250
2500087270
250250250250
2500010910
This equation is difficult to solve but a trial-and-error solution yields h=0.166m.
If h=r, A=
42
1 2d and P= d
2
1 then R=
4
d
P
A
Hence Q= 2/13/2 SRn
A= 21
322
48
/
/
Sd
n
d
2.6 Most Efficient Cross Sections (Best Hydraulic Cross Sections) Some channel cross sections are more efficient then others in that they provide more flow for a given wetted perimeter. When a channel is constructed, the excavation, and possibly the lining must be paid for. The best hydraulic section or most efficient cross section for an open channel is the one that will have the greatest capacity for a given slope, area, and roughness coefficient.
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The most efficient cross section should be the cheapest because it has the smallest wetted perimeter and would require the least amount of lining material or surface finishing but may not necessarily have the minimum cost of construction.
Q=2132 // SR
n
A= Q=
21
32
35/
/S
nP
A
For a trapezoidal section the area
2mhbhA
mh
Ab
and the perimeter
)21(2 mhbP
)(12 2 hfmhmhh
AP
From the above, it is known that when A and m are constants, then P varies with h. If these parameters remain constant, it is clear from the equation, that Q will be largest when the perimeter is smallest. Taking
212 mhmh
h
A
dh
d
dh
dP
2
212 mm
h
A
Substitute hmhbA into dh
dp expression to yield.
dh
dp
2
2
2
12 mmh
mhbh
2122 mmh
b
For minimum value, 002
2
dh
Pdand
dh
dP
mmh
b 212
h
b is the best hydraulic section ratio of width to depth say
.)(12 2 monlyoffunctionamfmmh
bbestbest
m 0 0.25 0.5 0.75 1.00 1.25 1.50 1.75 2.00 2.50 3.00
best 2 1.56 1.24 1.00 0.83 0.70 0.61 0.53 0.47 0.39 0.32
e.g. for m=0, rectangular cross section, best = 2 = h
b b = 2h
Generally, the most efficient of all cross sections is a semicircle because it has the smallest wetted perimeter for a given area.
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2.7 Unlined and Lined Channels Unlined earth channels are often found in irrigation projects as conveyance systems on the farms. The advantages of earth unlined channels include the facts that: - they are understood and accepted by farmers - they can be built and maintained by unskilled labour - they do not require special equipment or materials - construction materials are locally available Construction of Earth Channels – Some facts
1. They should be built with stable side slopes and with banks strong enough to carry the required flow safely.
2. They should have ample capacity to carry the design flow at non-erosive velocities. 3. Side slopes should be flat enough so that the banks will neither cake in nor slide when
they are saturated with water. 4. For stiff clay, steep slopes up to ½ to 1 are possible. 5. Loose sand should have flat slopes of about 2 to 1. 6. Channels that are constructed higher than the surrounding field level should have banks
large enough to withstand damage by seepage or trampling. Earth channels are lined with impervious materials to prevent excessive seepage and growth of weeds. This is important because if not, a large portion of the water harnessed at high cost, through the canal network or through wells and pumps, is lost by seepage from unlined conveyance systems. In permeable soils like sand and sandy loam, the losses in earth channels may be as high as 20~40% of the water delivered to the channel. Water losses in unlined channels may occur by: - seepage - breaches along the channel through rat holes - ponding of water in depressions and irregularities in the channel section - evaporation The length of the channel affects the quantity of water lost by seepage and evaporation. One of the main problems in the use of unlined channels is the control of weeds. Weeds in a channel obstruct the flow of water. If the weeds are allowed to grow up to maturity, their seeds may spread over the farm through the irrigation water. Thus unlined channels require continuous maintenance to: - control weed growth - repair damage by livestock rodents - control erosion Generally, there are several other problems associated with sharing water for irrigation from unlined channels. Usually the farmer or user at the upstream end gets nearly the full supply of water due to less seepage. However, the user at the tail end of the channel gets comparatively much less water due to seepage. Thus providing lining in the channels gives nearly equal distribution of water amongst all farmers. Well mixed and well made cement concrete lining and single layer bricks or stones laid in cement, provide virtually water-proof channel lining. Other materials which are rather susceptible to damage and thus uneconomical include bituminous mixtures, soil cement, chemical sealants, polythene film and impervious earth materials.
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Table: Typical Values of Manning’s n
Materials Manning’s n
Earth channels 0.023 ~ 0.04
Lined channels
concrete 0.015
masonry 0.017 ~ 0.03
metal, smooth 0.011 ~ 0.015
wooden 0.011 ~ 0.014
vegetated waterway 0.02 ~ 0.04
Pipes
cast iron 0.012 ~ 0.013
clay or concrete drain tile 0.011
steel 0.015 ~ 0.017
vitrified sewer pipe 0.013 ~ 0.015
Importance of Lining of Earth Channels
To avoid excessive loss of water by seepage
To avoid piping through or under banks
To provide needed stability
To avoid erosion
To avoid water logging of adjacent lands
To promote the continued movements of sediments
To facilitate cleaning
To promote economy by a reduction in excavation
To reduce flow resistance
To aid in the control of weeds and aquatic growths Some materials for lining:
Concrete
Brick or stone masonry
Asphalt lining
Compacted earth lining
pre-cast concrete Permissible Slope of Earth Channels The natural slope of the land is usually the deciding factor in determining the channel bed slope. The steeper the channel, the more will be the velocity and the more the discharge for the same cross-section. But high slopes result in high velocities which cause erosion. An earth channel should have a gradient of about 0.1 %. However, silting may occur if the channel has gradient less than 0.05%. When bed slopes of channels should be determined, the velocity should be checked so that it does not exceed a certain maximum – thus avoiding erosion.
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Permissible Velocities for Various Soil Textures
Soil type Maximum permissible velocity (m/s)
Bare channels
sand and silt 0.45
loam, sandy loam, silt loam 0.6
clay loam 0.65
Vegetated channels
poor vegetation 0.9
fair vegetation 1.2
good vegetation 1.5
Where earth channels are to be used on steep slopes, it is necessary to control the gradients and thus the velocity by constructing drop structures or by lining the channel bed. Structures for Controlling Erosion in Channels (Irrigation) Often it is necessary to build (irrigation) channels on land slopes so steep that the water will attain erosive velocities. Severe erosion will occur in earth channels if structures to control the slope are not provided. Drop structures and chute drops are used to prevent erosion in channels. Drop Structures Drop structures are used to discharge water in a channel from higher level to a lower one. They may be open type drops or pipe drops. Open drop structures can be made of timber, concrete, or brick or stone masonry. Timber is usually not preferred due to its short life.
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Drop structures often set up eddy currents in the (irrigation) stream and these currents cause erosion of the channel section immediately downstream from the structure. Stones or brick placed over a length of 1 to 2 metres from the structure help to prevent channel erosion near drop structures. When earth channels are to be constructed on steep slope, it is necessary to construct a series of drop structures to flatten the channel slope. Chute Spillways Chute spillways carry the flow down steep slopes through a lined channel rather than by dropping the water in free overfall. A chute spillway consists of an inlet, channel section and an outlet. The structure may be made of concrete, or stone, or bricks laid in cement mortar. A low cost chute spillway can be made of pre-cast concrete channel sections with a stilling basin at the outlet. A check gate is often installed at the inlet to regulate the water level in the upstream channel. The depth of the stilling basin is about 10 – 20 cm below the bed level of the downstream channel. When the high velocity water is slowed down to a low velocity in a stilling basin, there is a sudden rise in the depth of flow which is known as a hydraulic jump. Thus the height of the walls in the downstream channel should be increased as compared to the channel section of the chute spillway in order to prevent overflow. Water Control and Diversion Structures Water control and diversion structures are necessary to give easy and effective control of irrigation water on the farm. Good control will reduce the labour required to irrigate and check erosion and water loss. The structures include check gates, portable check dams, diversion boxes, turnout boxes, siphons and pipe turnouts. 2.8 Nomographic Design of Open Channels The nomograph allows fast determination of the mean velocity of flow when the values of R, S, and n are given. It can also be used to determine the value of anyone of the factors R, S, n and v when any three of the factors are known. To use the nomograph, a line is drawn to join the S and n values on the respective scales, and passing through the pivot line. The point of the intersection of this line with the pivot line is the pivot point. A line originating from the known value on the R scale and passing through the pivot point when extended to meet the velocity scale provides the required value of the velocity.
Drop points
Fig 2.
Check gate
10~20 cm
shaft
1~2 m
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Assignment
1. Derive from first principle the chezy equation for uniform flow.
2. A channel 5m wide at the top and 2m deep has sides sloping 2 vertically in 1 horizontally. The slope of the channel is 1 in 1000. Find the volume rate of flow when the depth of water is constant at 1m. Take c = 53 in si units. What will be the depth of water if the flow were to be doubled? [ q = 4.79m3/s, h = 1.6m]
3. A trapezoidal channel is to be designed to carry 280m3/minute of water. Determine the cross – sectional dimensions of the channel if the slope is 1 in 1600, side slopes 45o and the cross – section is to be a minimum, take c = 50 in si units. [h = 1.53m, b = 1.27m]
4. A 2.0m diameter concrete pipe on a slope of 0.005 is to carry water at a normal depth of 1.5m. Determine the flow velocity and flow rate. Take n=0.014. Also determine q when ho=0.75 and h=1.0m
5. Prove that the best hydraulic section for a rectangular channel
6. Determine the best section for a semi-circular cross section.
7. Find the best dimension for a rectangular channel to carry a flow of 0.5m3/s at a velocity of 1.2m2/s.
Fig 1.
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CHAPTER THREE NON-UNIFORM FLOW In considering uniform flow in chapter 2 it was assumed that successive cross-sections and corresponding mean velocities were everywhere the same and that the loss of head in friction was equal to the fall of the channel bed, so that bed, water surface and energy gradient were parallel. In non-uniform flow, none of these conditions apply. Depth may vary from section to section and the energy gradient, water surface and bed would no longer be parallel. Non-uniform flows are produced by changes in the channel geometry, while changing from one uniform flow to another. There are two types of non-uniform flows Rapidly varied flow, in which the change in depth takes place over a short distance, hence
friction, can be neglected. Gradually varied flow, in which the change in depth extends over a long distance. 3.1 Rapidly Varied Flows Rapidly varied flow occurs whenever there is a sudden change in the geometry of the channel or in the regime of the flow. Typical examples of the first type of flow include flow through regions of rapidly-varied cross section, e.g. venture flume and broad crested weirs. The second type is usually associated with the hydraulic jump phenomenon in which flow with high velocity and small depth is rapidly changed to flow with low velocity and large depth. The regime of flow is defined by the Froude Number. In regions of rapidly varied flow, the water surface profile changes suddenly and therefore has pronounced curvature. Therefore, the assumptions of parallel streamlines and hydrostatic pressure distribution which are used for uniform flow and gradually varied flow do not apply. Solutions to rapidly varied flow problems rely on the energy and momentum equations. 3.1.1 Energy Principle
Fig 3.1 Energy Diagram
Datum
Water surface (slope=Sw)
Channel bottom (slope=So)
Energy gradeline (slope=Sf)
Z1
h1
Z2
h2
v2
1/ 2g
v2
2/ 2g
1
2
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The total energy possessed by a body (volume) of water flowing in an open channel is given by
2
2mvmgHEtotal (in kWh)
Dividing through by mg, (per unit weight) yields
g
vHEtotal
2
2
(m)
The kinetic energy component (in m) is given by
g
vEkinetic
2
2
A volume of water V (m3) positioned at elevation Z (m) with flow depth h (m), possesses the following amount of potential energy:
HhZEpotential
The total energy in an open channel flowing with water at depth h is given by:
g
vhZEtotal
2
2
Writing the energy equation between two sections (sections 1 and 2) in the channel gives the Bernoulli‟s equation
LHg
vhZ
g
vhZ
22
2
222
2
111
Where Z=elevation of the channel bottom above an arbitrary datum; h = the depth of flow; v=average velocity; HL= head loss between sections 1 and 2. But note that because rapidly varied flow occurs within a short distance, HL = 0 In the special case of steady uniform flow V1=V2, and h1=h2; and Z1 = Z2 + HL
Fig 3.2 Example of Rapidly Varied Flow – Gated Flow
EGL
g
v
2
2
h1
g
v
2
2
h2
supercritical
flow subcritical
flow
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3.1.2 Specific Energy The energy with respect to the channel bottom is the sum of the flow depth and the velocity head at the section. This is called the specific energy of the section and it is given by
g
vhE
2
2
Let us consider a steady non-uniform flow. Let the width of the channel be b, and the steady rate of flow Q. Then the discharge per unit width q will be:
b
Qq = constant (Since Q = const, and b = const.)
v is the velocity at the section given by
A
Q
bh
Qv , therefore
2
2
2gA
QhE
At various sections of the channel the depth of flow will change with corresponding change in velocity so that the product 'vh' is constant at all sections. At any section,
2
2
2
22 1
222 hg
qh
gh
qh
g
vhE
2
2
2gA
Qh
For a given value of q, the specific energy head is a function of the depth of flow
E = E1 + E2 where
E1 = h – static (potential) energy head, and
E2 =
2
2 1
2 hg
q– kinetic energy head
When the depth of flow is plotted against the specific energy for a given channel section and discharge, a specific energy curve is obtained. Studying the Specific Energy Curve BCD, we find that
(i) The Specific Energy Head, first decreases with increase in depth and reaches a minimum value of C. (Supercritical flow zone)
(ii) Further increase in depth causes a corresponding increase in the Specific Energy. (Sub-critical flow zone)
The depth h corresponding to point C at which the specific energy is a minimum is called the critical depth. For each value of the specific energy head there are two possible flow depths. Consider the line GHI. The specific energy for this condition is EI = OG, but the flow depth may be either GH or GI. The depth GH = h1 is less than the critical depth hc, while the depth GI = h2 is greater than the critical depth. The flow is supercritical when h1 < hc, and it is subcritical when h2 > hc. When the depth of flow is hc, the flow is called critical flow, and the velocity, vc, is called
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critical velocity. The depths h1 and h2 are known as alternate depths. The critical depth corresponding to a given flow rate can be determined as presented in the next section. O
Example of rapidly varied flow The discharge in a rectangular channel of width 5 m and maximum depth 2 m is 10m3/s. The normal depth of flow is 1.25m. Determine the depth of flow downstream of a section in which the bed rises by 0.2 m over a distance of 1 m. Solution
Assuming frictional losses are negligible, then the following equation applies,
Depth of flow, h
E2 Curve E = y
B
E = E1 + E2 Curve
Sub-critical Zone
Super critical zone
Specific Energy, E
I
C
H
G
Emin
E
D h1
hc
h2
45o
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zg
vh
g
vh
22
2
22
2
11
zEE SS 21
In this case,
m
gg
VhES 38.1
2
25.15
10
25.12
2
2
111
m
ghh
g
hhES 38.1
2
2
2
5
10
2
2
2
2
2
2
22
2.0z
Hence
2.02
238.1
2
2
2
2 gh
h
Or
2
2
2
22
218.1
ghh
This is a cubic equation for h2, but the correct solution in this case is about 0.9 m. This is used as the initial estimate in a trial-and-error solution, as follows:
H2
(M)
2
222 /2 ghhES
(M)
0.9 1.15
1.0 1.2
0.96 1.18
Hence the solution is h2 = 0.96 m 3.1.3 Flow Measurements A gauging flume is a device for measuring flow in a channel. We have the Parshall flume, the standing wave flume and the Venturi flume. We will treat the last one. The Venturi flume A channel section at which there is a unique relationship between the depth and discharge is referred to as a control.
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The figure below shows the elevation and plan of a venturi flume. Let B, H1, v1 be the normal breadth, flow depth, and flow velocity at the entrance to the flume. Let b, H2, v2 be the corresponding parameters for the throat.
At the throat, the velocity v2 is greater than v1. Hence there will be a drop in water level at the throat, as the total energy head practically remains the same. Due to continuity of flow,
2211 VbHVBHQ
21, bHaBHA
21 aVAVQ
21 VA
aV
By Bernoulli‟s equation
g
VH
g
VH
22
2
22
2
11
21
2
1
2
2
22HH
g
V
g
V
)(2 21
2
1
2
2 HHgVV
)(2 21
2
22
22
2 HHgVA
aV
)(2)1( 212
22
2 HHgA
aV
g
v
2
2
1
H1 H2
g
v
2
2
2
Constriction
b B v v
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)(2)( 212
222
2 HHgA
aAV
)(2 2122
2
2 HHgaA
AV
The discharge, Q = aV2
This is the theoretical discharge through the venturi flume. In actual cases the discharge obtained is slightly less than the above value due to losses in the flume. Hence
)(2 2122
HHgaA
CdaAQ
Cd = coefficient of discharge, usually less than 1.0. Cd is usually = 0.95 – 0.99
The Broad Crested Weir A broad-crested weir consists of an obstruction in the form of a raised portion of the bed extending across the full width of the channel with a flat upper surface or crest sufficiently broad in the direction of flow for the surface of the liquid to become parallel to the crest. The upstream edge is rounded to avoid separation losses, which will occur at sharp edges.
The flow upstream is sub-critical and there is a free fall downstream. Since there is no restraining force on the liquid, the discharge over the weir will be the maximum possible and flow over the weir will take place at the critical depth. For rectangular channels, it is known that the critical depth
32
2
gb
Qhc
3
cghbQ
Since Ehc3
2
21
3
27
8
EgbQ = 1.705b 2
3
E
The specific energy measured on the crest assuming no losses is
g
VHE
2
2
1
g
v
2
2
1
H 1v
L
h
2v
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If the depth upstream is large compared with the depth over the weir then g
V
2
2
1 is negligible and
Q becomes
Q = 1.705 2
3
bH
Thus a single measurement of the upstream depth H is sufficient to determine Q. In actual cases the theoretical value of Q is higher than the practical value, therefore a discharge coefficient is introduced to obtain
Q = 2
3
705.1 CdbH (Cd = 0.9 ~ 0.97)
Alternative derivation method Applying Bernoulli‟s equation to the water surface on the crest and upstream of the crest
g
vh
g
vH
22
2
2
2
1
It is assumed that Hg
v
2
2
1 .
)(22
2
2
2 hHgvg
vhH
But Q = AV therefore discharge over the crest is given by
)(2 hHghbQ theoretical discharge
Theoretical discharge has discharge coefficient dc applied to it. Thus
)(2 hHghbcQ d
The thickness of the water at the downstream end would adjust itself in such a way so as to make the discharge a maximum. The relation between H and h for maximum discharge can be obtained by alternating Q with respect to h
)(2 hHghbcdh
d
dh
dQd
For maximum discharge, the term 32 hHh must be a maximum
HhhHhhHhdh
d
3
2320 232
23
max3
2
3
2)
3
2(2
3
2bHc
gHHgHbcQ dd
23
705.1 bHcQ d 97.0~9.0dc
Same results as using the critical depth approach
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3.1.4 Critical Flow Given the specific energy head
2
22
22 gh
qh
g
vhE
For minimum E,
3
2
1
0
gh
q
dh
dE
dh
dE
3
2
g
qh = critical depth
But b
32
2
gb
Qhc = critical depth
Again from above
3232ghvhghq and ghv 2
g
vh
2
critical depth
At minimum energy we have
Emin = c
c
c
c
c hh
hg
vh
2
3
22
2
OR g
v
g
v
g
v
g
vh cccc
c
2222
2
3
22
min3
2Ehc
Also corresponding to critical flow
12
c
c
gh
v and the Froude number of critical depth for a rectangular channel is given by
1
c
c
gh
v
Hence for critical flow, the Froude number, Fr = 1. For non-rectangular channel (Trapezoidal)
2
2
2gA
QhE
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Now dh
dA
gA
Q
dh
dE3
2
1
But depthmeaneiB
AhandB
dh
dA.,
A
B
gA
Q
dh
dE2
2
1 gh
v
A
B
g
v
dh
dE22
11
2
g
AQ
B
A
At the critical state of flow the specific energy is a minimum, thus
hencedh
dE,0
.22
22
channelprismaticnonforhg
vh
g
v
This is the criterion for critical flow, which states that at the critical state of flow, the velocity head is equal to half the mean depth. Also we can write it in terms of the Froude number.
1
C
rgh
vF Critical flow
1C
rgh
vF Sub-critical flow
1C
rgh
vF Supercritical flow
The critical velocity is given
C
ghv
The critical depth can be determined by solving the equation:
g
Q
B
Bh
g
Q
B
A c
2323
;
g
BQ
hg
QBh cc
2
32
23 )(
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If the shape of the channel cross section is prismatic, the critical depth, hc, will be constant throughout for a given discharge (since Q, B and g are constants in the equation)
3
2
g
BQ
hc
is independent of channel slope for a given flow rate.
When h0=hc chezy equation becomes
cccc SRCAQ
and Manning‟s equation becomes
n
SRAQ ccc
2132
Hence
2
32
2
32
c
c
cc
nc
R
nV
RA
QS
Example The triangular channel shown below is to carry water at a flow rate of 10m3/s. if n=0.012, determine the
i. Critical depth ii. Critical velocity iii. Critical slope
Solution i. Put hc = critical depth
determine A in hc
A=0.5Bhc
But B=6h=6hc A=0.5*6hc*hc=3hc
2 g=9.81m/s2
1
3 3
1
B
h
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Q=10m3/s
Using the equation g
Q
B
A23
6hc/(3hc
2) 3=9.81/102
4.5hc
5=100/9.81
hc5=2.265
hc=5 265.2 = 1.178m
ii. A=3hc2=4.163m2
Vc=Q/A=10/4.163=2.402m/s
iii. Sc=
2
3/2
c
c
R
nv
cc hmhP 10212 2 = 7.45m
R = A/P = 4.163/7.45=0.5588m
Sc = [0.012*2.402/(0.5588) 2/3] 2 = 0.00181
Study Example A trapezoidal channel with bottom width b=10m, side slope m=1.5, the flow rate is 50m3/s. determine the critical flow depth hc, Vc, and Sc. [Hint: B=b+2mh, A=(b+mh)h, A3/B=Q2/g = constant, work to obtain A3/B=255]
h
h B A A3 A3/B
1 13 11.5 117 2 16 1098.5
1.5 14.5 18.375 427.9 „ „ „ „ „ „ „ „
1.2 13.6 14.16 208.76
A3/B 255
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3.2 Flow depth for Maximum discharge at a given Specific Energy The Specific Energy Head at any section is given by
g
vhE
2
2
hEgv 2
But Q = bhv 3222 hEhgbhEgbhQ
For Q to be maximum Eh2 – h3 must be maximum
Eh
hEh
hEhdh
d
3
2
032
0
2
32
But g
vhE
2
2
When h = ,2
3,
3
2horEE we obtain
g
vh
g
vh
g
Vhh
222
2222
3
or g
vh
2
and Frgh
v 1
Hence for maximum discharge condition, Fr = 1 is also the condition of critical flow. Thus the critical depth may be released as the depth that gives the maximum discharge for a specific energy.
Example 1 The specific energy for a 3 metre wide channel is to be 3m, what would be the maximum possible discharge? Solution E= 3m, b=3
hc = 2/3E = 2/3 x 3 = 2m
3
cghbQ = 3x(9.81x23)0.5
= 26.57m3/s
smxghV cc /429.4281.9
Qmax = AVc = (3x2) x 4.429 = 26.574m3/s Example 2 A rectangular channel 4m wide discharges water at a rate of 16m3/s. if the Specific energy is 2.25m, find the possible depths.
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Solution
Q = qb = vbh hhh
q
bh
Qv
4
4
16
Specific energy head, E = 25.22
2
g
vh
25.28155.0
25.281.92
16
2
2
hhE
hhE
x
trial and error gives y1 = 0.73m and y2 = 2.06m
The critical depth is obtained as 3
2
g
qh 3
2
81.9
4 =1.18m
Example 3 Water flows at the rate of 16m3/s in a channel of 10m wide at a velocity of 1.6m/s. Calculate the specific energy head. Find also the critical depth, the critical velocity and the minimum value of the Specific Energy Head corresponding to this discharge. Solution Q = 16m3/s, b=10m, v=1.6m
Q = hbv h = mxbv
Q1
6.110
16
the specific energy head
m
xg
vhE 1305.1
81.92
6.11
2
22
Critical depth mg
q
g
bQ
hc 639.081.9
6.13
2
3
23
2
Critical velocity mxghV cc 504.2639.081.9
The minimum Specific Energy E min
mxg
VhE c
c 9585.081.92
504.2639.0
2min
22
OR
mhE c 9585.02
3min
Example A channel 5m wide conveys a discharge of 10m3/s of water. Plot graphs of the following:
(i) Static Energy Head (ii) Kinetic energy Head (iii) Specific Energy Head
For values of the depth of flow, find also the critical depth and the minimum specific energy head, corresponding to this discharge.
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3.3 The Momentum Equation 3.3.1 The Hydraulic Jump Under special conditions a rapidly flowing stream of liquid in an open channel suddenly changes to a slowly flowing stream with a large cross-sectional area and a sudden rise in elevation of the liquid surface. When the flow is changing from supercritical to subcritical flow, then the phenomenon of hydraulic jump is said to occur. In effect the rapidly flowing liquid expands and converts kinetic energy into potential energy and losses. At the jump location, there is a sharp discontinuity in the water surface and considerable amount of energy is dissipated due to turbulence. That is why only the energy equation cannot be used for its analysis. Hydraulic jump can be used 1. To dissipate excessive energy – spillways 2. To provide control section 3. For aeration of drinking water 4. For thorough mixing of chemicals in water Example of hydraulic jump forming in a chute canal or spillway
A hydraulic jump is formed whenever supercritical flow changes to sub-critical flow. Thus, in the
upstream section (S1 > Sc), h1 < hc V1 > Vc i.e. supercritical flow occurs; and in the
downstream section (S2 < Sc), h2>hc V2 < Vc i.e. sub-critical flow occurs. But at critical flow conditions, S = Sc, h = hc and V = Vc When a jump occurs there is a change of momentum since the flow is slowed down. The force producing the change is the difference in hydrostatic pressures resulting from the change of depth.
Fig. Free body diagram for analysis of hydraulic jump
v1 y1 F1
v2
y2
F2
hc
h1
h2 hc S1>S2
S2<Sc
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For rectangular channel, the forces are given by
22
2
111
bghgAhF
2
2
2
2
bghF
Momentum is given by
tQVM '
Rate of change of momentum
bh
Q
bh
QQQVQV
t
MM
21
1212 ''
Net force in the x-direction
22
2
2
2
121
bghbghFF
From the Newton‟s second law
1221 MMFF
12
2
2
2
1
11
2 hhb
qhh
gb Substituting q = Q/b
12
22
2
2
1 11
2 hhg
qhh
21
21
2
21212
1
hh
hh
g
qhhhh
21
2
21
1
2
1
hhg
qhh (A)
Note that 22112211 hvhvqhVbhVQ
The discharge per unit width q of channel through a jump can be determined, if the sequent depths are known, and it‟s given by
2/1
2121 )(
2
hgh
hhq
Substituting 11hvq into (A)
g
hVhh
h 1
2
112
2
2
Dividing through by h1
2 gives
1
2
1122
1
2 2
gh
Vhh
h
h
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2
1
1
2
1
1
2
1
2 22
1 Frgh
V
h
h
h
h
We obtain a quadratic in (h2/h1) as 02 2
1
1
2
2
1
2
Fr
h
h
h
h
And using a
acbb
2
42 to solve yields the required depths
Because of the energy dissipated in the jump, h2 is not the alternate depth of h1.Thus h2 is actually less than the alternate depth, therefore y1 and y2 are called sequent depths or conjugate depths. Solution of the above equation yields the sequent depths as
1812
2
11
2 Frh
h
This equation specifies a relationship between the upstream and downstream depths of the jump in terms of Fr1. Proceeding similarly, we can derive the following equation in terms of Fr2:
1812
2
22
1 Frh
h
Noting that q = vh, the energy dissipation in a hydraulic jump is obtained by
2
2
2
22
1
2
12122 gh
qh
gh
qhEEE
2
2
2
1
2
21
11
2 hhg
qhhE
122
2
2
1
2 11
2hh
hhg
qE
122
2
2
1
2
1
2
2
2
2hh
hh
hh
g
qE
(B)
But we know that
21
2
21
1
2
1
hhg
qhh (A)
Substitute (A) into (B)
12
2
1
2
2
21
21
4hhhh
hh
hhE
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21
2
1121212
4
4
hh
hhhhhhhE
21
2
2
2
1
2
2
2
1
2
1
2
212
4
42
hh
hhhhhhhhE
=
21
2
2
2
1
2
1
2
212
4
2
hh
hhhhhh
21
2
1
2
212
44 hh
hhhhE
The energy lost in a hydraulic jump is given by
21
3
12
4
)(
hh
hhE
Example A hydraulic jump is formed in a 5-m wide outlet at a short distance downstream of a control gate. If the flwo depths just upstream and downstream of the gate are 10m and 2m, respectively, and the outlet discharge is 150m3/s, determine:
a) flow depth downstream of the jump b) Head loss in the jump c) Thrust on the gate
Solution Assume there are no losses in the flow through the gate. Given Q = 150 m3/s, B = 5m, h1 = 10 m, h2 = 2 m.
h2
h1
∆E
h
E
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q = 150 /5 = 30 m3/s
v2 =q/h2 = 30/2 =15 m/s
47.1181.92
152
2
2
22
2
gh
vFr
a) Depth downstream of jump: h3
m
Frh
63.8
147.11812
2
1812
22
b) Head loss in the jump: E
m
gh
qh
gh
qh
EE
22.4
22 2
3
2
32
2
2
2
32
OR using the equation 21
3
12
4
)(
hh
hhE
gives
mE 22.404.69
43.291
63.8*2*4
)263.8( 3
1
1
2
2
3
3
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3.4 Gradually varied flow Gradually varied flow is a steady non-uniform flow in which the depth, area, roughness, bottom slope, and hydraulic radius change very slowly (if at all) along the channel. The basic assumption required is that the head-loss rate at a given section is given by the Manning formula for the same depth and discharge, regardless of trends in depth 3.4.1 Governing Equations It is known from section 3.1 that the total energy at a channel cross section is given by
E = Z+h+v2/2g = Z + h + 2
2
2gA
Q
)1
(2 2
2
Adx
d
g
Q
dx
dh
dx
dz
dx
dE
now
dx
dA
AdA
d
Adx
d)
1()
1(
22
dx
dh
dh
dA
AdA
d
Adx
d)
1()
1(
22 [ B
dh
dA ]
dx
dh
A
B3
2
dx
dh
A
B
g
Q
dx
dh
dx
dz
dx
dE3
2
)1(3
2
gA
BQ
dx
dh
dx
dz
dx
dE
)/(1 32gABQ
dx
dz
dx
dE
dx
dh
Datum
Water surface (slope=Sw)
Channel bottom (slope=So)
Energy gradeline (slope=Sf)
Z1
h1
Z2
h2
V21/ 2g
V22/ 2g = 2
2
2gA
Q
1
2
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21 Fr
SS
dx
dh fo
)/(1 32
gABQ
SS
dx
dh fo
21 r
fo
F
SS
dx
dh
By definition of Sdx
dzandS
dx
dE
34
22
n
o
h
vnS
34
22
h
nvS f
gh
vFr
22
The negative sign with Sf and So indicate that both H and z decreases as x increases. Also
222
3
2rF
BgA
AQ
gA
BQ
Consider an irregular channel
The incremental area
2/
)2(2
1
2
1
dhdBBdhdA
dhdBBdA
dhdBBBdA
dh
dAB
BdhdA
dhdBBut
0
B+dB
B
dA
dh
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3.4.2 Classification of Channel Slopes
Channel slope: So < 0 negative slope, hn is non existent
So = 0 horizontal slope, hn = So > 0 positive slope, hn is existent in different forms Positive slope: So > Sc steep slope, hn < hc supercritical flow So = Sc critical slope, hn = hc critical flow So < Sc mild slope hn > hc sub-critical flow 3.4.3 Principles for determining the surface profiles
1. When 0dx
dhor positive. In this case the depth of flow increases with distance. This occurs
when
(a) fo ss and 21 rF the water surface has a
concave profile up and it is called backwater curve
(b) fo ss and 21 rF / / / / / / / / / / / / /
2. When 0dx
dh or negative. In this case the depth of flow decreases with distance. This
occurs when
(a) fo ss and 21 rF the water surface has a
convex profile up and it
(b) fo ss and 21 rF / / / / / / / / / / / / / is called drawdown curve
Approach of the surface profile to the normal depth, critical depth lines and the channel bottom
1. As hhn (uniform flow), Sf-So~0 i.e. SfSo. From equation (A) 0dx
dh and provided
Fr1
2. As hhc, Fr1(critical flow) and the denominator tends to zero. Therefore dx
dh. [For
dh/dx to approach infinity means x0, and h is big]. Thus the water-surface profile approaches the CDL vertically. Physically it is impossible. It is assumed that the approach is very steep.
3. As h, V0. Consequently both Fr and Sf tend to zero. From Equation (A) it implies
0Sdx
dh for very large values of h. since So is assumed to be very small, we may say that
the water surface profile almost becomes horizontal as h becomes large. [But we know that
V=R2/3S1/2/n for So 0, V0, h , Fr0, Sf0]
4. As h0
a. Chezy formula Sf=Q2/C2A2h assuming R~h yields
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)(
)(232
2322
BQgBhC
QhBCSgB
dx
dh o
As h0, the 2
0lim
C
g
dx
dh
h
as h0 the profile has a positive finite value, and it is a
function of the Chezy constant, C. b. Manning formula
Sf =
2
32
Ah
Qn assume R~h
)(
)(233/4
223/42
BQgAh
nQAhSgA
dx
dh o
0
lim0
valueA
dx
dy
h
No!
Surface Profiles There are 12 different types of surface profiles: 3 for M; 3 for S; 2 for C (zone 2 does not exist,
since hn = hc); 2 for H (zone 1 does not exist since hn = ); and 2 for A (zone 1 does not exist since hn does not exist).
Zone 1: region above both lines Zone 2: region between two lines Zone 3: region between lower line and channel bottom
By considering the signs of the numerator and denominator of 21 r
fo
F
SS
dx
dh
, we can make
quantitative observations about various water-surface profiles. (A). 1. Sf > So if h < hn 2. Sf < So if h > hn
hese two inequalities will help determine the sign of the numerator. (B) 1. Fr >1 if h < hc; 2. Fr<1 if h > hc These two inequalities will also help determine the sign of the denominator. Example for Mild Slopes Zone 1: h > hn > hc Zone 2: hn > h > hc Zone 3: hn > hc > h
Zone 3
Zone 2
Zone 1
NDL or CDL
CDL or NDL
B
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Mild slope Zone 1
h > hn Sf < So numerator is positive
h > hc Fr < 1 denominator is positive
dx
dh
h increases as x increases. h hn asymptotically; water surface becomes horizontal as h increases. Zone 2
h < hn Sf > So numerator is negative
h > hc Fr < 1 denominator is positive
dx
dh
h decreases as x increases. hhn asymptotically; and hhc steeply. Zone 3
h < hn Sf > So numerator is negative
h < hc Fr > 1 denominator is negative
dx
dh
h increases as x increases. h hc steeply and approaches the channel bottom at a finite positive slope. Steep slopes Zone 1
h > hn Sf < So numerator is positive
h > hc Fr < 1 denominator is positive
dx
dh
h increases as x increases. h hc steeply; water surface becomes horizontal as h increases. Zone 2
h > hn Sf < So numerator is positive
h < hc Fr > 1 denominator is negative
dx
dh
h decreases as x increases. h hn asymptotically; and hhc steeply. Zone 3
h < hn Sf > So numerator is negative
h < hc Fr > 1 denominator is negative
dx
dh
NDL
CDL 3M
2M
1M
Zone 2
Zone 3
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h increases as x increases. h hn asymptotically and approaches the channel bottom at a finite positive slope.
Adverse Slope Zone 1 is non existent Zone 2
So is negative So – Sf < 0 and Numerator is negative
h > hc Fr < 1 denominator is positive
dx
dy drawdown
Zone 3
h < hc Fr > 1 denominator is negative
So – Sf < 0 Numerator is negative
dx
dy backwater
Critical Slope
Zone 1
h > hn (= hc) Sf < So numerator is positive
h > hc (= hn) 1rF denominator is positive
dx
dy backwater
Zone 2 is non-existent
Zone 1
Zone 2
Zone 3
CDL
NDL
S1
S3
S2
A3
A2
Zone 2
Zone 3
CDL
Zone 1
Zone 3
C1
C3 NDL/CDL
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Zone 3
h < hn Sf > So numerator is negative
h < hc 1rF denominator is negative
dx
dy backwater
Horizontal Slope
Zone 1 non existent Zone 2
So = 0, So –Sf < 0 numerator is negative
h > hc Fr < 1 denominator is positive
dx
dy draw down
Zone 3 So –Sf < 0 numerator is negative
h < hc 1rF denominator is negative
dx
dy backwater
Example A rectangular channel 6m wide has a bed slope of 1 in 2000 and under original conditions the depth is 1m. A dam was placed across the channel, increasing the depth at the dam site to 1.4m. Calculate the depth of flow at 150m upstream, assuming that the flow remains unchanged and C in Chezy formula remains constant at 60. Solution When the depth of flow is 1m
Area of flow bhA = 6 x 1 6m2
Wetted Perimeter bhP 2 = 2 + 6 = 8m
The hydraulic radius is given by
8
6
P
AR
Channel Bed Slope 2000
1S
2H
3H
ny
cy
CDL
Zone 2
Zone 3
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Therefore velocity RSv
2000
1
8
660v
smv /162.1 When the depth of flow is 1.4m
mR
mP
mA
smvbh
bhV
vAvA
9545.08.8
4.8
8.84.126
4.864.1
/83.0162.14.1
0.1
2
2
2
2
1
2
12
2211
21
But 2
2
RC
VS f and
gh
VFr
22
The rate of change of depth with distance is given by
000316.09498.0
0003.0
0502.01
0002.00005.0
40.181.9
83.01
609545.0
83.0
2000
1
11
2
2
2
2
2
2
2
gh
V
RC
VS
Fr
SS
dx
dh ofo
Assuming the above rate of change of depth to be uniform, change in depth in a distance of 150m
mh 047.0000316.0150
Depth of flow 150m upstream of dam = 1.4 – 0.047 = 1.353
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sitedamthefromupstreammxxh
h
xh
h
m
m
8.1265
8.126500016.0
4.0
4.00.14.1
Concept of lake created as a result of building a dam. Example Sketch the water-surface profile in the channels connecting the two reservoirs, as shown in the figure below. The bottom slope of channel 1 is steep and that of channel 2 is mild. Solution Compute the critical and normal depths for each channel. Then plot the critical-depth line (marked as CDL in the Fig (b)) and the normal-depth line (marked as NDL in Fig (b)) The water depth at the channel entrance is equal to the critical depth, since the water level in the upstream reservoir is above the CDL of channel 1. Let us mark this water level at the channel entrance by a dot. The water level at the downstream end is lower than the CDL at the downstream end of channel 2. Therefore, the water surface passes through the CDL approximately three to four times the critical depth upstream of the entrance to the downstream reservoir. Let us again mark this water level at the downstream end by a dot, as shown in Fig (b). In channel 1 the water surface at the entrance passes through the critical depth and then it tends to the normal depth. Thus, we have an S2 profile in cannel 1. The flow decelerates downstream of the junction of channels 1 and 2 because of mild slope. Hence, the flow depth keeps on increasing until it intersects the CDL. Approximately at this location, a hydraulic jump is formed. The water surface follows the M2 profile downstream of the jump. Detailed calculations are required to determine the exact location of the jump.
Fig. Example Water-surface profiles
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CHAPTER FOUR PIPES AND PIPE NETWORKS Sometimes referred to as pressure conduits, pressure flow systems, or flow in closed conduits. 4.1 Definition In closed conduit flow, the conduits flow full, and the fluid is under pressure. Majority of closed conduits have circular cross sections (such as pipes), hence the name pipe flow. It is also referred to as pressurised flow. Examples of pressurised flow in practice are shown below.
pump
pressure conduit
Irrigation and Domestic pumping
Mountain
Intake
A Spillw ay
Tunnel
Intake
pow er station headrace canal
Hydropow er development
4.2 Laminar Flow and Turbulent Flow Laminar Flow: adjacent fluid layers move at the same velocity. Turbulent Flow: adjacent fluid layers move at different velocities and paths of individual fluid particles do cross and intersect each other. The friction factor relation depends on the state of flow, which is classified according to the Reynolds number. For pipes the diameter is used as a characteristics dimension and the Reynolds number is given by
DvRe -----Dimensionless [
]
Where D =internal diameter of pipe (m); v = average velocity of fluid flow (m/s); v = kinematic
viscosity of fluid (m2/s); = viscosity (kg/ms); = density of fluid (kg/m3)
Type of Flow Value of Re
Laminar <2000
Transition to turbulent (critical region) 2000~4000
Turbulent >4000
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4.3 Energy Equation of pipe Flow The energy equation between sections 1 and 2 is
losshg
VPZ
gVP
Z 22
2
222
2
111
Usually α = 1.0
hf is the head loss along the pipe due to friction. The energy gradient Sf=hf/L. Usually additional losses resulting from valves, fittings, bends and so on are known as the minor losses, hm, and have to be included when present. In that case, hf in the energy equation is replaced by the total head loss hloss. Since minor head losses are localised, the energy grade line, represented by hloss/L will have breaks wherever the minor losses occur.
P/ = pressure head (m); V2/2g = velocity head (m); Z=static head (m); hf = Frictional head loss between two sections (m); HL = hf + hm = hloss. 4.4 Continuity Equation The volume of water leaving section 1 will be the same arriving at section 2 if water is not taken out or added in between these two sections. Thus 1. Q1 = Q2 = Q A1V1 = A2V2
2
2
21
2
12
2
2
1
2
1
44VDVDV
DV
D
or (D1/D2)
2=V2/V1
2. V1=Q/A1 or V2=Q/A2 Example
From the large reservoir shown above, water flows at a rate of 10m3/s through a pipe 1m in diameter. Determine the loss of head in the system. Solution
Q=10m3/s, D=1m A= 4/2D =0.78m2
V = Q/A = 10/0.78 = 12.74m/s Writing the energy equation
fhg
VPZ
gVP
Z 22
2
222
2
111
RD
D
D
44
2
1
2
40m
1m
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Datum passes through point 2 Z2=0 For reservoir the surface area is considered to be big hence V1=0 Since pressure is atmospheric at points 1 and 2, P1=P2=0
40 + 0 + 0 = 0 + 0 + (12.74) 2/(2*9.81) + hf hence hf = 40 - 8.280 = 31.72m 4.5 Evaluation of Head Loss Due to Friction Darcy-Weisbach Equation The Darcy-Weisbach equation is the most general formula in the pipe flow application. It is an
empirical formula. According to Chezy‟s formula V=C(RS). Since S=hf/L, R=D/4 for pipe and
treating C=(8g/f) the Chezy‟s equation reduces to
fg
V
D
Lh f
2
2
and this is Darcy – Weisbach eqn.
where f = friction factor (dimensionless); L = length of conduit (m); D = internal diameter of pipe (m); V = mean velocity of flow in the pipe (m/s) f is used in American practice and 4f found in old British practice but this has changed in all modern books. NB: Wherever you see f, note that the factor 4 is already incorporated Nikuradse’s Experiments 1. For laminar flow, the friction factor is a function of the Reynolds number only. It is given by
f=64/Re (dimensionless) 2. In the critical region of Re between 2000~4000 the flow alternates between laminar and
turbulent regimes. Any friction factor relation cannot be applied with certainty in this region. 3. In the turbulence regime, the friction factor is a function of the Reynolds number as well as
the relative roughness of the pipe surface. I.e. f=f (Re,/D).
130
21045
f
Re
4 104 105 106
1504
11014
1120
1252
161
x10-2 Crit ical region
RoughKD
Smooth Pipe
Nikuradse’s Experimental Results
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Nikuradse used smoother pipe coated with sand grains of uniform size K (average diameter of sand grains). The uniform character of the sand grains used in Nikuradse‟s tests produces a dip in the f-versus-Re curve before reaching a constant value of f. However, tests on commercial pipes where the roughness is somewhat random reveal that no such dip occurs. Based on Nikuradse‟s experimental results, the following relations are obtained for determining friction factor in the turbulent region.
fDf Re
51.2
7.3log2
1
For very smooth pipes, /D is very small and the equation reduces to
ff Re
51.2log2
1
For fully rough pipe in a turbulent regime, Re is very big and the equation reduces to
Df 7.3log2
1
A modification of the equation was proposed by Jain in 1976 as
9.0Re
72.5
7.3log2
1
Df
f is eliminated, therefore iteration is also not necessary. A. Another common formula for head loss in pipes that has found almost exclusive usage in water supply engineering is the Hazen William equation given by
54.063.0849.0 SCRV SI units
There is a comparison between Chezy formula and Hazen Williams‟s formula. For R=1 and S=1/1000, the Hazen Williams coefficient C becomes equal to Chezy C. Recent studies have shown that errors up to 39% could occur if the formula is used indiscriminately. because the multiplying factor is supposes to change for different R and S the Hazen William C is considered to be related to pipe material only, where as it must also
depend on pipe diameter, velocity, and viscosity, similar to the friction factor of Darcy- Weisbach.
Now in textbooks a diagram prepared by Moody is used. It is called Moody diagram. He prepared a diagram between the friction factor versus the Reynolds number and the relative roughness; this can be used to determine f. The Moody’s Diagram By plotting data for commercial pipe from a number of sources, Moody developed a design chart similar to that shown in the figure below. In the figure below, the variable ks is the symbol used to denote the equivalent sand roughness. That is, a pipe that has the same resistance characteristics at high Re values as a sand-roughened pipe of the same size is said to have a
DR S. N. ODAI - KNUST HYDRAULICS (CE 356)
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size of roughness equivalent to that of the sand-roughened pipe. The figure gives approximate values of ks and ks/D for various kinds of pipe.
In the figure above, the abscissa (labelled at the bottom) is the Reynolds number, Re, and the ordinate (labelled at the left) is the resistance coefficient f. Each solid curve is for a constant relative roughness, ks/D, and the values of ks/D are given on the right at the end of each curve. To find f, given Re and ks/D, go to the right to find the correct relative-roughness curve; then look at the bottom of the chart to find the given value of Re and, with this value of Re, move vertically upward until you reach the given ks/D curve. Finally, from this point, move horizontally to the left scale to read the value of f. If the curve for the given value of ks/D is not plotted in the figure above, simply find the proper position on the graph by interpolation between curves of ks/D, which bracket the given ks/D. For some problems, it is convenient to enter the figure above using a value of the parameter Ref1/2. This parameter is useful when hf and ks/D are known but the velocity, V, is not. Basically three types of problems are involved with uniform flow in a single pipe.
1. Determine the head loss, given the kind and size of pipe along with the flow rate. 2. Determine the flow rate, given the head, kind, and size of pipe. 3. Determine the size of pipe needed to carry the flow, given the kind of pipe, head, and
flow rate. In the first type of problem, the Reynolds number and ks/D are first computed and then f is read from the figure, after which the head loss is obtained by the use of the following equation:
g
V
D
Lfh f
2
2
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4.6 Minor Head Losses In addition to the continuous head loss along the pipe length due to friction, local head losses occur at changes in pipe section, at bends, valves, fittings, entrance to or exit from a conduit. These losses may be neglected for long pipes but are significant for less than about 30m long pipes. Since pipe lengths in water supply and wastewater plants are generally short, minor losses are important. The minor loss is expressed in terms of the applicable velocity head or considered proportional to the kinematic energy. Thus,
g
VKhm
2
2
where k=loss coefficient, V mean velocity (m/s); g=9.81m/s2 Entrance Exit Others are: Sudden expansion/contraction, gradual conical expansion/contraction, bends etc All these minor or local losses should be considered in design. Example Two reservoirs are connected by a 200m long cast iron pipeline, as shown below. If the pipeline is to convey a discharge of 2m3/s at 15.6oC, what is the size of the pipeline required?
1
2
Entrance loss
Hydraulic grade line Energy line
Bend loss
Exit loss
Valve loss
Datum
20m
Re-entrant K=0.8
Sharp edged K=0.4~0.5
Slightly rounded K=0.2~0.25
Well rounded K=0.05
all K=1.0
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Item K
Entrance loss 0.5
Exit loss 1.0
valve 10.0
Two 90o bends (0.9x2) 1.8
Total 13.3
Solution 1. Applying the energy equation to points 1 and 2 with respect to point 2 as the datum
losshg
VPZ
gVP
Z 22
2
222
2
111
20 + 0 + 0 = 0 + 0 + 0 + hloss
hloss=20m
1. Friction loss
gD
Q
D
fl
g
V
D
Lfh f
2]4/[2 22
22
5
2
08.12 D
LQf
2. Minor losses
4
2
22
22
08.12)4/(22 D
KQ
D
Q
g
K
g
KVhm
3. hloss = hf + hm = 5
2
08.12 D
LQf +
4
2
08.12 D
KQ = 20
2008.12
)2(3.13
08.12
)2)(200(4
2
5
2
DD
f
508.12
3.13
08.12
20045
DD
f
04.603.1320045
DD
f
200f + 13.3D = 60.04D5 60.04D5 - 200f - 13.3D = 0 5. First trial, assume f=0.03~0.05. Substitute f into equation
60.04D5 - 13.3D - 6 = 0 D=-****
thus V = 4Q/D2 = 4*2/D2 = ****m/s
Re=VD/ =*****/(1.21*10-5) =
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Now for cast iron = 8.0 * 10-4
/D = 8.0 * 10-4/****
using Re and /D, f = *** (from Moody diagram) First revision: substitute f in 60.04D5 - 200f - 13.3D = 0 Solve (by trial and error), to obtain D=****
Thus V = 4Q/D2 = 8/D2 = ****m/s
Re=VD/ =*****/(1.21*10-5) =
/D = 8.0 * 10-4/****
f = **** (from Moody diagram) Continue till f stabilizes. Hence D = 4.7 Pipelines with Pumps and Turbines When there is a pump then energy is being added, ha When there is a turbine then energy is being removed, hr The term representing energy added should be added on the left and the term representing energy removed should be subtracted on the left.
the Bernoulli‟s equation becomes
Lhg
VPZhrha
gVP
Z 22
2
222
2
111
Note A pump with power P, can raise a liquid of specific weight γ, flow rate Q through a height of ha.
= 9.81 kN/m3; Q m3/s; H m. (H could be ha or hr)
ha = Pa/Q; hr = Pr/Q. Usually one is added at a time, so it is either ha or hr. 4.8 Pipes in Series Q1= Q2 = Q3= Q4 = Q hf = hf1 + hf2 + hf3 + hf4 + ---
This is similar to electricity, where Q (constant) replaces I (constant), hf = hfi replaces R=Ri
in kW aQHP
hf1 hf2 hf3
Q2
D2
Q3
D3
L1
Q1
D1 Compound Pipelines
L2 L3
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Pipes in Parallel
Pipe Discharge from a reservoir For large area reservoir vo = 0
Vo 0 (1)
hm v2/2g EGL hf Z1 HGL v2/2g A (2) The figure shows a pipe of uniform cross – section leading from a reservoir and discharging free into atmosphere. Applying Bernoulli‟s equation to sections (1) and (2) yields,
losso h
g
vpz
g
vpz
22
2
22
2
11
fm
fm
hhg
vz
hhg
vz
2
20000
2
1
2
1
But for entrance in which pipe is flush with reservoir,
)50.1(2
2
25.0
2
1
2
2
fD
L
g
vz
fgD
Lvh
g
vh
f
m
Q Q
Q3,D3,hf3
Q2,D2,hf2
Q1,D1,hf1
Q = Q1 + Q2 + Q3 hf = hf1 = hf2 = hf3
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z1 represents the difference of water levels in the two reservoirs. It is the total head loss in the system.
Note: The hydraulic grade line is at a distance of g
v
2
2
below the energy gradient line. As the
liquid flows from A to B, there is a loss of head due to friction, which is fh . At the entrance there
is head loss mh . The exit loss is neglected if it is assumed that the velocity head is due tog
v
2
2
.
For very long pipes the termD
fL is very large compared to 1.50, therefore the entrance and exit
losses (1.5g
v
2
2
) may be neglected. Usually when the length of the pipe is greater than 1000D,
only the frictional loss need be considered. Example 1 A pipe 20cm in diameter and 4m long conveys water at a velocity of 2.5m/s. Find the head lost in friction.
(a) Using the formula dg
flvh f
2
2
taking f = 0.0242
(b) Using the formula RSCV , taking C = 57
Solution
(a) mgd
flvh f 73.1
20.0*81.9*2
5.2*45*0242.0
2
22
(b)4
dR ,
L
hS
f , 57C ,
4
222 d
L
hCSRCVSRCV
f
mdC
LVh f 73.1
57*2.0
45*5.2*442
2
2
2
Example 2 Water is discharged from a large reservoir to atmosphere through a 10cm diameter and 500m long pipe. Find the discharge if the outlet is 15m below the free surface of water in the reservoir.
Assume the entry to the pipe as sharp. Take 04.0f
Solution.
From
D
fl
g
vHz 5.1
2
2
1
15 =
10.0
500*04.05.1
2
2
g
v
15 = 201.5g
v
2
2
sm
AvQ
smv
/10*5.9
209.1*1.0*4
/209.1
33
2
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Pipe connecting two Reservoirs
The figure shows a pipe of uniform cross-section connecting two reservoirs with liquid surface at different elevations. The liquid flows from the higher reservoir to the lower reservoir. At point A there is loss of head at entrance (hm). The frictional head loss takes place throughout the pipe length. At the exit, there is a head loss of v2/2g. Writing the Bernoulli‟s equation gives
g
v
D
LfH
g
v
g
vf
gD
Lv
g
vhmhfH
hH
hg
vpZ
g
vpZ
loss
loss
25.1
25.0
222
00000
22
2
2222
2
222
2
111
Example of Pipes in series: Q1. Two reservoirs are connected by a pipeline consisting of two pipes, one of 15cm diameter and length 6m, while the other of diameter 22.5cm and 16m length. If the difference of water levels in the two reservoirs is 6m, calculate the discharge and sketch the hydraulic as well as the energy grade line. Take f = 0.04
From continuity equation Q1 = Q2
2
2
2
1
2
2
2
1
2
1 25.215
5.22
44vvvv
Dv
D
A.
V2
2g
hL
hf H
B
E.G.L
H.G.L
V2
2g
Hm
hL1
hf1 hL2
hf2
H = 6m
V2
2g
(1) (2)
H.G.L
entry exit hm1 + hm2
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i. Loss of head at entrance g
v
g
v
g
vhm
253.2
2
25.25.0
25.0
2
2
2
2
2
11
ii. Loss of head due to friction g
v
gD
vLfhf
2*
15.0
6*04.0
2
2
1
1
2
111
= g
v
g
v
21.8
225.2
5.0
24.0 1
2
22
iii. Loss due to sudden enlargement
g
vvhm
2
2
212
g
v
g
vhm
256.1
2
125.22
2
2
2
2
2
iv. Loss due to friction g
v
gD
vLfhf
2*
225.0
16*04.0
2
2
2
2
2
222
g
v
284.2
2
2
v. Loss of head at exit = g
v
2
2
2
But know that 62
2
22211
g
vhfhmhfhmhH loss
62
184.256.11.853.2
2
2 g
v
smv /71.22
smvAQ /108.071.2*225.0*4
32
22
EX2 Parallel Example: Two reservoirs are connected by 2 pipes of the same length laid in parallel. The diameters of the pipes are 10cm and 30cm. If the discharge through the 10cm pipe is 0.01m3/s, what will be the discharge through the 30cm pipe? Assume that „f‟‟ is the same for both pipes Solution: For such problems, it is more convenient to express the Darcy-Weisbach equation in terms of discharge as
g
Q
D
Lf
gD
Q
D
Lf
g
v
D
Lfhf
2**
16
2
1*
42
2
5222
22
As the pipes are in parallel
2
25
2
15
2
25
2
2
2
15
1
2
21
30.0
1
1.0
1
*1
*2
16*
1*
2
16
QDg
fLQ
Dg
fL
hfhf
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2
1
2
2 243QQ
But Q1 = 0.01
smQ
Q
/156.0
0243.001.0243
3
2
22
2
Siphons Sometimes it may become necessary to provide a pipeline over an obstacle like a ridge or small hill and then to a lower level. Quite a part of the pipe line may be situated not only above the hydraulic gradient but also above the water level of the supply reservoir. Such a pipe is called a siphon. OR When a pipe is laid in such a manner that part of it is above the hydraulic gradient line, it is called a siphon.
The pressure head at any point along the pipe axis is equal to the distance between the HGL and the axis. It follows that the pressures at points C and E are zero, i.e., the pressure is atmospheric. The pressure in the pipe CDE, where the pipeline is above the hydraulic grade line, is negative. The minimum pressure will be at summit point D where the vertical distance between the point and the HGL is maximum. Application of siphons
i. Transmission of water from one reservoir to another separated by a ridge ii. To empty a tank not provided with any outlet iii. To take out water from a channel
Applying Bernoulli‟s equation yields (between the surfaces of two reservoirs)
g
vhfhmH
2
2
1
Also applying Bernoulli‟s equation to points G and D with datum at G the reservoir level gives,
11
22
22hmhf
g
VPZ
g
VPZ DD
DGG
G
11
2
2000 hmhf
g
VPh DD
D
H hf
hm1
E.G.L
H.G.L
hf1 hm1
A
B
V2
2g C
E
V2
2g
D
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11
2
2hmhf
g
Vh
P DD
D
HL
LH
L
Lhf 11
1 [equivalent pipe length]
g
VH
L
Lhd
g
V
g
VH
L
Lh
P DDD
D
2
5.1
25.0
2
2
1
22
1
Example A pipe of 1m diameter connects two reservoirs having a difference of level of 6m. The total length of the pipe is 800m and rises to a maximum height of 3m above the level of water in the higher reservoir at a distance of 200m from the entrance. Find the discharge in the pipe and the pressure at the highest point. Take f= 0.04, and neglect minor losses.
Solution Neglecting minor losses implies
smVAQ
smfL
DgHV
g
V
D
LfhfH
/51.192.114
/92.104.0800
681.90.122
2
32
2
Loss of head at point C
mg
V
D
Lfhf 5.1
81.92
92.1
1
20004.0
2
22
1
OR
mHHL
Lhf 5.16
800
200
800
200
2
11
A
B
H = 6m
3m
C
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Applying Bernoulli‟s equation between A and C and taking datum at level A,
1
2
222
2
111
22hf
g
VPZ
g
VPZ
5.1
81.92
92.13000
2
CP
188.05.13 CP
= - 4.69m of water This is the negative gauge (vacuum) pressure at point C. The absolute pressure at C
mabsolutePc 61.569.43.10)(
of water (absolute)
4.10 Water Hammer The interim stage when a flow changes from one steady state condition to another steady state condition is known as the transient (unsteady) state of flow. In conduits and open channels, such conditions occur when the flow is accelerated due to sudden closing or opening of the controlling valves, starting or stopping of pumps, rejecting or accepting of the load by a hydraulic turbines or similar situations of sudden increase or decrease in flows. The variations in velocity result in change of momentum. The fluid is subjected to an impulse force equivalent to the rate of change of momentum according to Newton‟s second law. An appreciable increase of pressure occurs with respect to time due to this impulse force. This pressure fluctuation is called water hammer (or oil hammer) because a hammering noise is usually associated with this phenomenon. More commonly, this is now referred to as hydraulic transients. The system design should be adequate to withstand both the normal static pressure and the maximum rise in pressure due to hydraulic transient. Definition When a liquid flowing in a pipeline is abruptly stopped by the closing of a valve, dynamic energy is converted to elastic energy and a series of positive and negative pressure waves travel back and forth in the pipe until they are damped out by friction. This phenomenon is known as water hammer.
Wave of increase pressure
H c
0t<L/c
t=0+ Ho
Vo V=0
2
C=speed of propagation of pressure or wave Liquid is compressed. A wave of increased pressure travels in the upstream direction
Local pressure
L
t=0 Steady state Ho=P/
Vo
1
Valve
At t=0, V is decreased to zero, causing an
increase in pressure to (P+P)= (H+H)
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H
Wave front
4
c
L/Ct<2L/C
t=L/C+
Vo V=0
Wave front
H
t=L/C
V=0
3
At the end of step 1 the pressure in the pipe is much higher than the pressure in the reservoir. To relieve pressure in the pipe, water begins to flow from the pipe into the reservoir.
H 6
Wave front
C 2L/Ct<3L/C
t=2L/C+
Vo V=0
Sudden stoppage causes pressure to drop below normal level. This sends a wave of negative pressure again upstream towards the reservoir.
Wave of decreased pressure
Wave front
H
t=3L/C
V=0
7
At this point the pressure in the pipeline drops below that in the reservoir, therefore fluid flows from the reservoir into the pipe.
H 8
Wave front
C
3L/Ct<4L/C
t=3L/C+
Vo V=0
t=2L/C
Wave front
Vo
5
9
Wave front
L
t=4L/C
Ho
Vo
Since the pressure in the reservoir remains unchanged, and the pressure in the pipe is much higher than that in the reservoir, the fluid in the pipe begins to discharge in the reverse into the reservoir.
When the decompressed wave arrives at the valve, the reversed flow cannot proceed further so the fluid cannot flow back to the reservoir. A negative pressure is generated at the valve which produces a negative shockwave. This in turn travels towards the reservoir
If the system is assumed to be frictionless, the pressure wave will travel back and forth in the pipeline indefinitely with the same flow conditions being repeated every 4L/c seconds. The time interval, 4L/c seconds, after which conditions are repeated is referred to as the theoretical period of the pipeline.
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4.11 Pipe Networks A water supply distribution system consists of a complex network of interconnected pipes, service reservoirs and pumps which deliver water from the treatment plant to the consumer. Water demand is highly variable, both by day and season. Supply, by contrast, is normally constant; consequently, the distribution system must include storage elements, and must be capable of flexible operation. Water pressures within the system are normally kept above a minimum of about 15m head. In addition to new distribution systems, there is a common need for improvement of existing (often ageing) systems. The two main network configurations used are
Branching system Ring main system (Grid)
The structure of the branching type of water distribution network is similar to a tree.
Fig. Branching pattern pipe network Grid Systems The ring system is preferred over the branching system because it prevents the occurrence of “dead ends” with the consequent risk of stagnant water and permits more flexible operation, particularly when repairs must be carried out.
Fig. Grid pattern pipe network The distinguished feature of the grid system is that all of the pipes are interconnected and there are no dead ends. In the grid system, water can reach a given point of withdrawal from several directions. The grid system overcomes all of the difficulties of the branching system discussed previously. One disadvantage of the grid system is that the determination of the pipe sizes is somewhat more complicated.
Main
Service mains
Sub-mains
Building connections
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Pipe Network The city water supply system consists of several loops and branches of pipes. The system is known as pipe network. The solution of pipe network is very time consuming. Prof. Hardy Cross developed an ingenious method of successive approximation. By this method, the distribution of discharge amongst various pipes can be easily obtained. Fig shows a network of pipes
A pipe network must satisfy the following 3 basic conditions:
(1) At any junction the total inflow must be equal to the total outflow. (2) The algebraic sum of the head losses around any closed circuit is zero. (3) The head loss equation must be satisfied for each pipe.
The head loss in any element (pipe) of the system may be expressed as
n
f KQh - - - - - - - - - - - - - - - - (1)
where hf = head loss or energy loss in the pipe element (m); Q = flow in that element (m3/s); K = constant depending on pipe diameter, length, type and condition; n = 1.85 to 2 normally, depending on equation used. (Darcy-Weisbach equation, n = 2) Procedure 1) Assume a reasonable distribution of flow in various pipes satisfying condition (1).
2) Compute the head loss ( fh ) in each pipe, using the equation given in condition (3).
3) Divide the network into a number of closed circuits so that each pipe is included in at least one circuit.
4) Compute the algebraic sum of the head losses in each circuit ( fh ). Take suitable sign
convention. Unless the assumed distribution of flow happens to be correct, fh is not
zero and the assumed discharge needs correction. 5) Revise the assumed flow by applying the correction ∆Q obtained as follows: For any pipe in a loop of the system, the actual flows will differ from an assumed flow by an amount ∆Q: Q = Qo + ∆Q - - - - - - - - - - - - (2) Where Q = actual flow in pipe; Qo = assumed flow; ∆Q = required discharge Substituting (2) into (1) gives
D E F
A B C
E F J
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n
o
n
f QQkkQh
Using binomial expansion and neglecting higher terms with the assumption that ∆Q is small compared to Q; and also for a closed loop, the sum of the head losses about the loop must be equal to zero,
0 fh
Therefore ∑ K (Qn
o + nQ n-1∆Q) = 0
11 no
n
no
n
KQn
KQ
nKQ
KQQ
1no
f
KQn
h
It must be noted that in the numerator the algebraic sum is taken (signs are considered). In the denominator, the arithmetic summation is done, without considering the sign. While applying the correction, the sign of the correction obtained from the above equation must be considered. Since some pipes are common to more than one circuit, more than one correction will be applied to such pipes. After the corrections have been applied, new values of assumed discharge are obtained. 6) Assume a discharge as found in step 5. Repeat the procedure till the correction become
negligibly small. Examples: Q1. Find the discharge in each pipe of the network shown below.
Solution: The assumed distribution is shown in fig (b). The corrected flow after the first iteration for the top horizontal (AB) is determined as 15 + 11.06 = 26.06 and for the diagonal (AC) as 35 + (-21.17) + (-11.06) = 2.77. Fig (c) shows the flow after one correction and fig (d) the values after four corrections. The corresponding changes in discharges after the third iteration are:
169.00079.0 21 QandQ ; while the corresponding changes in discharges after the
fourth iteration are: 0003.00013.0 21 QandQ . It is evident that after the fourth iteration
step, the values of the discharge corrections approach zero, implying the calculated values are almost stable and the results acceptable.
50
100
K=
6
K=5
20
30
K
=2
A B
C D
K=1
K =3
2
1
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(b) (c)
(d)
First iteration
CIRCUIT PIPE 2kQh f fh 1n
onkQ 1n
onkQ
1n
o
f
nkQ
hQ
(1) DA
AC
CD
6 X 702 = 29400
3 X 352 = 3675
-5 X 302 = -4500
28575
3003052
2103532
8407062
1350 1350
28575 = -21.17
(2) AB
BC
CA 57483.133
2450352
255151
2
2
2
-2799
8383.1332
1403522
301512
253 06.11
253
2799
30
50
B 50
20 A
D C
70
30
35
15
35
100
20
100 30 D C
B A
48.8
3
51.17
2.77
26.06
23
.94
100
A B 50
30
20
D C
47
.73
52.27
29.24 -1.51
20
.76
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Second iteration
CIRCUIT PIPE 2kQh f fh 1n
onkQ 1nonkQ
1n
o
f
nkQ
hQ
(1) DA
AC
CD
6X48.832 = 14308
3 X2.772 = 23
-5X51.172=-13090
1241
51117.5152
1777.232
58683.4862
1114
1114
1241 =- 1.114
(2) AB
BC
CA 8656.13
114694.232
67906.261
2
2
2
-475
10656.132
9694.2322
5206.2612
158 006.3
158
475
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CHAPTER FIVE HYDRODYNAMIC MACHINES 5.1 Hydrodynamic machines A hydrodynamic machine is a device in which mechanical energy is transferred from the liquid flowing through the machine to its operating member (turbines) or from the operating member of the machine to the liquid flowing through it (pump). 5.2 Pumps classification Pumps are classified according to the way in which energy is imparted to the fluid. The basic methods are (1) volumetric displacement, (2) addition of kinetic energy, and (3) use of electromagnetic force. As shown above, there are several different types of pumps, but the types the civil/hydraulic engineer will encounter most frequently are classified into two main categories: Turbo-hydraulic (kinetic) pumps and the Positive-displacement (static) pumps. Pumps in which displacement is accomplished mechanically are called positive displacement pumps. The analysis of these pumps involves purely mechanical concepts and does not require detail knowledge of hydraulics; therefore it is not considered in detail in this study. Our discussion will focus on the kinetic pumps, in which kinetic energy is imparted to the fluid by means of a rapidly rotating impeller. Kinetic pumps include mainly the centrifugal pumps and vertical pumps. Its analysis involves hydraulic principles. The centrifugal pumps are the most common types of kinetic pumps used and they are the most common used in water and wastewater works because they have lower capital and maintenance costs while giving high power.
Pump Classification
Pumps
Positive displacement pumps
Kinetic pumps
Vertical pumps
Centrifugal pumps
Reciprocating pumps
Rotary pumps
Overhung Impeller
Impeller Between bearings
Closed coupled
Submersible
Axial split
Radial split
Lineshaft pumps
Submersible Pumps
Plunger/piston
Lobe
Progressive cavity
Screw
Horizontally mounted axial flow pumps
Separately coupled
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Turbo-hydraulic (kinetic) pumps – Centrifugal Pump
Positive-displacement (static) pumps
5.2.1 Introduction to Centrifugal Pumps A centrifugal pump consists of two main parts: 1. The rotating element called the impeller and mounted on a rotating shaft 2. The housing/casing encloses the rotating impeller and seals the pressurised liquid inside;
and also has suction and discharge openings for the main flow path.
impeller
volute
casing
entrance rotat ing shaft
volute
outlet
Parts of Centrifugal Pump
entrance
outlet
casing/housing
impeller
vanes
volute
Diffuse
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Depending on the direction of water flow around the impeller, the centrifugal pump may be classified as Radial flow pump – centrifugal Axial flow pump – propeller Mixed flow pump – centrifugal
5.2.2 Introduction to Vertical Pumps Vertical pumps were originally developed for borehole pumping and consequently the outside diameter is limited to the size of the borehole. Vertical pumps have proved to be very versatile and are today used in many applications not related to well pumping.
Lineshaft Vertical Pump Submersible Borehole Pump
5.2.3 Introduction to Positive Displacement Positive displacement (PD) pumps generally have lower discharge capacity and higher pressure when compared with kinetic pumps. They are often used when pumping viscous fluids which cannot be handled using centrifugal pumps. However, PD pumps are sometimes used for pumping water, in particular low discharge borehole applications. Rotary pumps Rotary pumps operate in a circular motion and displace a constant amount of liquid with each revolution of the pump shaft. In general, this is accomplished by pumping elements (e.g., gears, lobes, vanes, screws) moving in such a way as to expand volumes to allow liquid to enter the pump.
Electric Motor
6 Pump Impellers
Inflow
Outflow
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Examples of Rotary Positive Displacement Pumps Table. Summary of Pump type and Application Parameter Centrifugal Pumps Reciprocating
Pumps Rotary Pumps
Optimum Discharge and Pressure Applications
Medium-High Discharge/Low-Medium Pressure
Low Discharge, High Pressure
Low-Medium Capacity, Low-Medium Pressure
Maximum Discharge 50,000+ m3/hr 2,000+ m3/hr 2,000+ m3/hr
Low Discharge Capability
No Yes Yes
Maximum Pressure 4,000+ m head 50,000+ m head 3000+ m head
Requires Relief Valve No Yes Yes
Smooth or Pulsating Flow
Smooth Pulsating Smooth
Variable or Constant Flow
Variable Constant Constant
Self Priming No Yes Yes
Space Considerations Requires Less Requires More Requires Less
Costs Lower Capital Lower Maintenance High Power
Higher Capital Higher Maintenance Lower Power
Lower Capital Lower Maintenance Lower Power
Fluid Handling Suitable for wide range fluids from clean, clear, non abrasive fluids to fluids with abrasive high-solid content. Not suitable for high viscosity fluids.
Suitable for clean, clear, non-abrasive fluids. Can be adapted for abrasive slurry service. Suitable for high viscosity fluids.
Requires clean, clear, non-abrasive fluid due to close tolerances. Optimum performance with high viscosity fluids
5.3 The centrifugal pump The pump is driven by power from an external source, usually an electric motor. The wheel of the centrifugal pump on which the vanes are fitted is known as the impeller. The liquid enters the pump at the centre (eye). The impeller gives a centrifugal head to the liquid in the pump and the liquid leaves the pump at the outer periphery with a high pressure and velocity. Part of the velocity is also converted to pressure, as the liquid leaves the pump. The high pressure developed in the centrifugal pump may be used to raise the liquid from a lower level to a higher level or to increase the pressure in the system. The rotary motion of the impeller creates a centrifugal force that enables the liquid to enter the pump at the low-pressure region near the centre of the impeller and to move along the direction
A: 3-Lobe Pump B: Screw Pump
(Progressive Cavity) C: Double Screw Pump
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of the impeller vane towards the higher-pressure region near the outside of the housing surrounding the impeller as shown in the figure below. The housing is designed with a gradually expanding spiral shape so that the entering liquid is led towards the discharge pipe with minimum loss, in which the kinematic energy in the liquid is converted into pressure energy.
Radial Flow Pump Axial Flow Pump
A centrifugal pump consists of the following components Impeller: The wheel fitted with a series of backward curved vanes (or blades) is known as
impeller. The impeller is mounted on a shaft which is coupled to an electric motor. There are three main types of impellers.
a) Closed impeller: The vanes are completely closed by plates on both sides. Thus this
type of impeller is used when the liquid to be pumped is relatively free from debris so that the passage is not choked. This type has very high efficiency as it provides a smooth passage for the liquid.
b) Open impeller: the vanes are open on both sides. They have neither the crown plate nor the base plate. This type of impeller is used when the liquid contains a large amount of debris – wastewater.
c) Semi-open impeller: In this type there is a plate on the base, and there is no crown plate. This type of impeller is used when the liquid contains small amounts of debris.
The impeller is the main rotating part that provides the centrifugal acceleration to the fluid. They are often classified in many ways: based on major direction of flow with reference to the axis of rotation; on suction type; and on mechanical construction. Based on major direction of flow with reference to the axis of rotation radial flow axial flow mixed flow Radial flow and mixed flow pumps are commonly referred to as centrifugal pumps while axial flow pumps are called propeller pumps. Radial and mixed flow impellers may be either open or closed, while the axial is open. Generally axial flow pumps have about two to four blades, and hence, large unobstructed passages that permit handling of liquid containing debris without clogging.
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Radial Flow Pump In a radial flow pump, fluid enters at the centre of the impeller and is out along the impeller blades in a direction at right angles to the pump shaft (axis)
Radial Flow Pump
Axial Flow Pump The impellers are shaped so as to move water in the axial direction only. In an axial flow pump, the impeller pushes the liquid in a direction parallel to the pump axis or shaft (axial direction). Axial flow pumps are sometimes referred to as propeller pump as they operate in a similar manner to that of a boat
Flow Centrifugal Pump Mixed Flow Pump A mixed flow pump has characteristics of both radial and axial flow pumps. As liquid flows through the impeller of a mixed flow pump, the impeller blades force the liquid to move out in both radial and axial directions (away from the pump shaft)
Mixed Flow Centrifugal Pump
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Based on suction type
Single-suction: Liquid inlet on one side.
Double-suction: Liquid inlet to the impeller symmetrically from both sides. Based on mechanical construction
Closed: Shrouds or sidewall enclosing the vanes.
Open: No shrouds or wall to enclose the vanes.
Semi-open or vortex type.
Impeller type based on mechanical construction
Closed impellers require wear rings and these wear rings present another maintenance problem. Open and semi-open impellers are less likely to clog, but need manual adjustment to the volute or back-plate to get the proper impeller setting and prevent internal re-circulation. Vortex pump impellers are great for solids and "stringy" materials but they are up to 50% less efficient than conventional designs. The number of impellers determines the number of stages of the pump. A single stage pump has one impeller only and is best for low head service. A two-stage pump has two impellers in series for medium head service. A multi-stage pump has three or more impellers in series for high head service.
Centrifugal Pumps (i) give satisfactory and economic service and (ii) are better suited than other pumps (rotary and reciprocating) for the pumping of dirty liquids (better for water and wastewater works – sewage treatment plants). Other parts of a centrifugal pump Other parts of a centrifugal pump include Casing: It is the airtight chamber covering the impeller. There are the volute, volute with a
vortex chamber and a diffuser pump. In a volute pump, the casing is of a spiral shape. This type of casing is also known as volute casing. The area of flow gradually increases from the impeller outlet to the delivery pipe. Thus the kinetic energy is converted into pressured energy.
Suction pipe: The suction pipe connects the supply reservoir with the pump inlet. The lower end of the pipe is fitted with a non-return foot valve. The foot valve does not permit the liquid to drain out of the suction pipe when the pump is not working. This also helps in priming for water supply pumps; a strainer is usually fitted at the lower end so that only relatively clear liquid enters the suction pipe. The upper end of the suction pipe is connected to the centre of the impeller known as the eye of the pump.
Delivery pipe: The delivery pipe connects the outlet of the casing to the delivery reservoir. A valve is provided on the delivery pipe to regulate the supply of water. The valve is usually close to the pump.
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Setup of a centrifugal pump
Detailed parts of a centrifugal pump
To delivery reservoir
Delivery pipe
Casing
Supply reservoir
Foot value
Strainer
Vanes
Impeller
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5.4 Working of a centrifugal pump Fluid enters the pump near the axis of an impeller rotating at high speed. The fluid is thrown radially outward into the pump casing. A partial vacuum is created that continuously draws more fluid into the pump. The key idea is that the energy created by the centrifugal force is kinetic energy. The amount of energy given to the liquid is proportional to the velocity at the edge or vane tip of the impeller. The faster the impeller revolves or the bigger the impeller, the higher will be the velocity of the liquid at the vane tip and the greater the energy imparted to the liquid. Volute centrifugal pumps can pump liquids containing solid particles, but, when pumping liquids containing more than a small amount of vapour, their suction is broken by cavitation. Volute centrifugal pumps operate best when pumping relatively non-viscous liquids and their capacity is greatly reduced when used to pump viscous liquids. In practice most pumps used for drinking water supply are of the radial type. A centrifugal pump should be primed before it is started. Priming consists of filling the casing with water so that air trapped in the pump does not hinder its operation and reduce its efficiency. Priming is done by filling the pump with water from an outside source while permitting the displaced air to escape through an exhaust valve. It may be noted that had there been no foot valve in the suction pipe, the entire liquid poured into the priming funnel would have gone to the supply reservoir, and priming would not have been complete.
Large pumps are primed by vacuum pumps. Sometimes, a special priming reservoir containing the liquid is provided on the suction pipe. By directing the flow from this reservoir, it is possible to prime the pump. A pump located below the source of supply (submersible pumps) will not ordinarily require priming although some air may be trapped in the casing of pumps mounted on horizontal shafts. The following procedure is adopted when operating the pump. The delivery valve is closed and the priming of the pump is done While the delivery valve is closed, the external energy is supplied to the pump shaft. It is done
by starting the coupled electric motor. This causes an increase in the impeller pressure. The delivery valve is then opened. The liquid starts flowing into the delivery pipe. A partial vacuum is created at the eye of the centrifugal pump due to the centrifugal action.
The liquid rushes from the supply reservoir to the pump due to the pressure difference at the two ends of the suction pipe.
As the impeller continues to run, more and more water is made available to the centrifugal pump at the eye. The impeller increases the energy of the liquid and delivers it to the reservoir.
While stopping the pump, the delivery valve should be first closed otherwise there may be some backflow from the reservoir.
5.5 Specific Speed of Pumps It represents the speed of a pump under a head of 1m while delivering a discharge of 1m3/s. All geometrically similar pumps have the same specific speed. Specific speeds are determined by the operating characteristics at the point of maximum efficiency. Three parameters usually important in selecting centrifugal pumps are discharge (Q) to be delivered (pumped), head (H) to be delivered (overcome), and impeller speed (N). To aid in analysing pump problems and selecting pumps, these three parameters are often combined into another dimensionless parameter known as the specific speed, given by
4/3H
QNNs
Where Ns=specific speed; N=rotational speed (rotative impeller speed) in rpm; Q discharge in lit/sec (m3/s); H = head in m.
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Actually, specific speed is not really a speed. Rather it is a useful pump selection parameter that includes the effect of discharge, head, and rotative impeller speed. The specific speed is a fixed parameter for all pumps operating under dynamic conditions that are geometrically similar (homologous) to one another. The parameter is suitable for grouping pump with respect to the similarity of their design and to compare the performance of the pump of different design. Two pump impellers having the same shape have the same specific speed although their sizes may differ. 5.6 Relations for Geometrically Similar Pumps The relations that relate the parameter of geometrically similar pumps are known as the affinity Laws. These are useful in predicting the performance of a pump from tests on a model pump or homologous pump. From dimensional analysis considerations, the affinity laws are
3
1
2
1
2
1
2
D
D
N
N
Q
Q
2
1
2
2
1
2
1
2
D
D
N
N
H
H Dimensionless
5
1
2
3
1
2
1
2
D
D
N
N
P
P
Two units that are geometrically similar and have similar vector diagrams are said to be Homologous
5.7 Relations for Alteration in the same pump For a given pump operating at a given speed, there are definite relationships among parameters, known as the performance characteristics. If the pump size is altered or speed is changed, the same relations do not hold. For the velocity triangle at the exit from the impeller to remain the same before and after the alteration in the pump diameter, the following relations should apply
1
2
1
2
1
2
D
D
N
N
Q
Q
2
1
2
2
1
2
1
2
D
D
N
N
H
H Dimensionless
3
1
2
3
1
2
1
2
D
D
N
N
P
P
The efficiency is considered constant with change in speed and diameter in the relations above. However, when only speed is changed for the same diameter, the relations are as follows
3/1
1
2
2/1
1
2
1
2
1
2
P
P
H
H
Q
Q
N
N Dimensionless
For changed diameter at the same speed, the relations are as follows
3/1
1
2
2/1
1
2
1
2
2
1
P
P
H
H
Q
Q
D
D Dimensionless
The above three equations are used to determine the revised characteristics of a pump for a desired change in speed or diameter.
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5.8 Head developed and power required Pump Heads 1. Static Suction Lift (hs): The vertical distance from the water level in the source tank to the
centreline of the pump. If the pump is located at a lower level than the source tank, the static suction lift is negative.
2. Static discharge head (hd): The vertical distance from the centreline of the pump to the water level in the discharge tank.
3. Total static head (Hs): the sum of the static suction lift and the static discharge head, which is equal to the difference between the water levels of discharge and source tanks.
4. Total dynamic head (TDH): the sum of the total static head and the friction and minor losses. This is commonly known as the Total Head.
The static power to be overcome by a pump is given by:
QhhQHP dsss
where hs = (static) suction head; hd= (static) discharge head; Hs= total static head, which is the difference between the supply reservoir and delivery reservoir.
g
kv
g
kvhhHH sd
fdfssm22
22
lossss hH
D
DkQfLQH
5
22
12
where Hm = the manometric head; hfs & hfd = frictional losses in suction and delivery pipes. The plot of the above equation Hm versus Q is known as the system head curve. This curve representing the behaviour of the pipe system is important in the selection of a pump. If the velocity head in the delivery and suction pipes are neglected then,
dfsfsm hhHH
Hm represents the head against which the pump has to work. It is ultimately the head developed by the pump.
Head Hsys = Hs + hloss
Losses
Total dynamic Head, Hm (m)
Total Static Head (Hs)
Q (m3/s)
TYPICAL SYSTEM HEAD CURVE
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Total stat ic head
Static discharge head
Static Suction Lift
Static Suction Lif t
Stat ic discharge head
Total stat ic head
Total stat ic head
Static discharge head
Pump Power and Efficiencies The pump is driven by power from an external source, usually a motor. The shaft which is connected to the impeller receives a power Psh (shaft power) from the motor. Psh impeller (Pi) Output (P)
Motor Shaft Pump 1. Mechanical Efficiency. Because of mechanical losses like friction in the bearing, the power imparted to the impeller is less than the shaft power, that is Psh = Pi + Mechanical loss (∆Pi) Psh = Pi + ∆Pi Pi = Psh - ∆Pi.
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The ratio of the impeller power to the shaft power is called the mechanical efficiency, ηm. Thus,
ii
i
sh
i
mPP
P
P
P
<1
2. Manometric Efficiency.
The power imparted by the impeller is ii QHP
Where Hi is the head imparted by the impeller, and Q is the flow through the impeller. The power delivered by the pump is less than the impeller power because of hydraulic losses in the impeller and the casing. The power delivered by the pump is known as the water power, given by P = Pi –∆Pm. (manometric losses)
QhhQHQHci ffim
where hfi = losses in impeller, hfc= losses in casing. The actual head delivered by the pump is known as the manometric head and it can be measured by a differential manometer installed across the inlet and exit of the impeller. The ratio of the waterpower to the impeller power is known as the manometric efficiency and it is given by
i
m
i
fifi
i
manoH
H
H
hhH
P
Pc
< 1
3. Volumetric Efficiency. It is the ratio of the discharge from the pump (Q) to the discharge flowing through the impeller (Q+∆Q). The difference is due to leakage through the shaft and the casing, meant to lubricate and cool the packing and prevent it from burning out.
Qv
≈ 1
ηv is slightly less than one. 4. Overall efficiency is given by
Q
P
P
P
P
ish
i
vmanomo
Usually ηv is close to one, therefore unless otherwise stated ηv is taken equal to one and the overall efficiency is
mmano
sh
oP
P
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5.9 Cavitation and Net Positive Suction Head The term ‘cavitation’ comes from the Latin word cavus, which means a hollow space or a cavity. Webster‟s Dictionary defines the word „cavitation‟ as the rapid formation and collapse of cavities in a flowing liquid in regions of very low pressure. Cavitation is one of the most serious problems encountered in the operation of pumps as it can cause permanent damage and reduce the performance of the pump and it should not occur throughout its operating capacity range because cavitation causes: Erosion of metal from the impeller Limitation in the head against which a pump can work Reduction in the capacity of pump Noise and vibrations during operation with possible eventual drop in operational efficiency.
.
E
C B
A
hs
hd D
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Fig. Effects of cavitation
Vaporous cavitation is the most common form of cavitation found in process plants. Generally it occurs due to insufficiency of the available NPSH or internal recirculation phenomenon. The extent of the cavitation damage can range from a relatively minor amount of pitting after years of service to catastrophic failure in a relatively short period of time.
Net Positive Suction Head (NPSH) It is defined as the net head in metres of liquid that is required to make the liquid flow through the suction pipe from the supply reservoir to the impeller. The term NPSH is frequently used in the pump industry. The minimum NPSH depends on the pump design, its speed and discharge. Its value is usually given by the manufacturer. Apply Bernoulli‟s equation to section A and B gives
02
2
fss
ss hhg
kvP
Thus the negative head at the inlet of the impeller is given by
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fss
ss hhg
kvP
2
2
But the absolute pressure at the point is given by
fss
saas hhg
kvPPP
2
2
where Pa is the atmospheric pressure. Vapour pressure is the pressure required to keep a liquid in a liquid state. If the pressure applied to the surface of the liquid is not enough to keep the molecules close together, the molecules will be free to separate and roam around as vapour. The vapour pressure is dependent upon the temperature of the liquid. Higher the temperature, higher will be the vapour pressure. For no cavitation to occur, the absolute pressure at the point B (eye) should be equal to or greater than the vapour pressure (Pv).
vsa PPP
or
fss
sav hhg
kvPP
2
2
or
v
sf
sa
s
Ph
g
kvPh
2
2
hs gives the maximum suction lift. If hs is installed such that
v
fs
sa
s
Ph
g
kvPh
2
2
then cavitation will occur.
Thus the expression
fss
sva hhg
kvPP
2
2
should be a positive finite value.
This value is the NPSH given by
g
kvhh
PPNPSH s
fss
va
2
2
In terms of manometric head Hm, NPSH is expressed as NPSH = σHm Where σ = Thoma‟s cavitation number
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m
fss
sva
H
hhg
kvPP
2
2
Cavitation will occur if Thoma‟s number is less than the critical value σc given by
34
1000103.0
Nsc
where Ns is the specific speed of the pump. 5.10 Performance of Centrifugal Pumps For a given pump at a speed, there are definite relationships among the pump discharge capacity, head, power, and efficiency. These relations are derived from the actual tests on a given pump or a similar unit and are usually depicted graphically by the pump characteristic curves, comprising Pumping head against discharge Efficiency against discharge Power input against discharge The important feature of the curves is that, as head increases the discharge (capacity) decreases. These curves are supplied by the manufacturer. And since a pump casing can accommodate impellers of several sizes, the manufacturer supplies a series of sets of curves drawn on the graph corresponding to various sizes of impellers, which can be derived by use of the affinity laws explained previously. At a given speed, a pump is rated at the head and discharge, which gives the maximum efficiency, referred to as the best efficiency. The characteristic curves, particularly the head-discharge curve is important in pump selection.
The pump curves relate flow rate and head developed by the pump at different impeller sizes and rotational speeds. The centrifugal pump operation should conform to the pump curves supplied by the manufacturer. A pump is designed to work under design speed, discharge and head. When the pump runs at conditions different from the design conditions, its performance is quite different. In order to predict varying conditions of speed, discharge and head, tests are usually performed. The results of these tests are plotted in the form of characteristic curves. These curves are very useful for predicting the performance of pumps under different condition of speed, discharge and head. There are about four different curves namely; Operating characteristic curves Main characteristic curves Constant efficiency curves and Constant head and constant discharge curves.
Pump
Suction Tank
Q
Common Manifold
Pipe
Suction Pipework Discharge Pipework
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The operating characteristic curves are obtained by running the pump at the design speed of the driving motor. The pump is at design speed and the discharge is varied. The design discharge and head are obtained from the corresponding curves where efficiency is highest. Question: A centrifugal pump, impeller 0.5m, when running at 750rev/min gave on test the following performance characteristics:
Q (m3/min) 0 7 14 21 28 35 42 49 56
H (m) 40.0 40.6 40.4 39.3 38.0 33.6 25.6 14.5 0
η (%) 0 41 60 74 83 83 74 51 0
1. Predict the performance of a geometrically similar pump of 0.35m diameter and running at
1450rev/min. plot both sets of characteristics. 2. Predict the performance if only the diameter of the same pump is changed from 0.5m to
0.35m 3. Predict the performance if only the speed of the same pump is changed from 750rev/min to
1450rev/min. SOLUTION Let suffix 1 refers to the 0.5m diameter pump and suffix 2 refers to the 0.35m diameter pump.
From
3
2
1
2
1
2
1
D
D
N
N
Q
Q
3
1
2
1
212
D
D
N
NQQ =
3
15.0
35.0
750
1450
Q = 0.663Q1
P
Hm
Q
Design discharge
De
sig
n D
ep
th
Hm
ηo
P
ηo
ηo = efficiency; Hm = head; P = Power.
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From
2
2
1
2
2
1
2
1
D
D
N
N
H
H
2
1
2
2
1
212
D
D
N
NHH =
22
15.0
35.0
750
1450
H = 1.83H1
The values of Q1 and H1 are given by the table above. Therefore by multiplying them by the multipliers calculated above, Q2 and H2 may be tabulated. These together with some values of efficiency (is constant) constituted the predicted characteristics of pump 2 as follows;
Q(m3/min) 0 4.64 9.28 13.92 18.56 23.2 27.8 32.5 37.0
H(m) 73.2 74.3 74.0 71.9 69.5 61.5 46.8 26.5 0
η(%) 0 41 60 74 83 83 74 51 0
Solution
3
1
212
D
DQQ =
3
15.0
35.0
Q = 0.343Q1
2
1
212
D
DHH =
2
15.0
35.0
H =0.49H1
1
212
N
NQQ =
750
14501Q =1.933Q1
2
1
212
N
NHH =
2
1750
1450
H = 3.738H1
5.11 Single Pump and Pipeline System Determination of pump operating condition for single pumps
Pump H-Q Curve
Pump Operating Point
Efficiency
80
70
60
50
40
30
Effi
cien
cy, %System H-Q Curve
10
20
30
40
50
60
70
80
00 50 100 150 200 250 300 350
Discharge, m /hr3
Hea
d, m
hst
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The suitability of a given pump for a certain known piping system is determined by superimposing the system head curve of the piping system on the head-capacity characteristic curve of the pump. The intersection point of the two curves indicates the operating point (i.e. the head and discharge of a given pump). The point on any specific system H-Q curve at which a single speed pump must operate is determined by superimposing the pump H-Q. However if the efficiency of the pump is too low at this point another pump must be considered. 5.12 Multiple Pump System A single pump is suitable within a narrow range of head and discharge in proximity of the optimum pump efficiency. However in a piping system the discharge and head requirements may vary considerably at different times. Usually a variable-speed motor can accommodate this variation. Pumps in Series Pumps are used in series in a system where substantial head changes take place without appreciable difference in the discharge (i.e. the system head curve is steep). In series, each pump has the same discharge. H = HA + HB Q = QA = QB
BBAA
BA
HH
HH
//
)( BA HHQp
The composite head characteristic curve is prepared by adding the ordinates (heads) of all the pumps for the same values of discharge. The intersection point of the composite head characteristic curve and he system curve provides the operating condition. Pumps in Parallel The parallel pumps are useful for systems with considerable discharge variations with no appreciable head change. In parallel, each pump has the same head The following relations apply
0.0
5.0
10.0
15.0
20.0
25.0
30.0
0.0 100.0 200.0 300.0 400.0 500.0
Discharge, m3/hr
He
ad
, m P1
P2
P1
P2
P1+P2 (Parallel)
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H = HA = HB Q = QA + QB The composite head characteristic curve is obtained by summing up abscissas (discharges) of all the pumps for the same values of head as shown below
0.0
5.0
10.0
15.0
20.0
25.0
30.0
0.0 100.0 200.0 300.0 400.0 500.0 600.0
Discharge, m3/hr
Head
, m One Pump (P1, P2)
Two Pumps in Parallel (P1+P2)
P1
P2