COLLEGE OF ENGINEERING - Webs · 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

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)

1

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


2

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.


3

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


4

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


5

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


6

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


7

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


8

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


9

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


10

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


11

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


12

(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


13

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


14

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


15

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


16

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.


17

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.


18

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.


19

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.


20

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.


21

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


22

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.


23

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


24

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


25

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


26

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


27

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.


28

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


29

)(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


30

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


31

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

Qq

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


32

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 )(


33

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


34

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


35

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.


36

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.


37

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


38

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


39

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


40

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


41

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


42

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


43

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


44

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


45

)(

)(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


46

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


dx

dh

h decreases as x increases. hhn asymptotically; and hhc steeply. Zone 3


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



dx

dh

h increases as x increases. h hc steeply; water surface becomes horizontal as h increases. Zone 2



dx

dh

h decreases as x increases. h hn asymptotically; and hhc steeply. Zone 3



dx

dh

NDL

CDL 3M

2M

1M

Zone 2

Zone 3


47

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


dx

dy drawdown

Zone 3


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


48

Zone 3


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


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


49

Therefore velocity RSv

2000

1

8

660v

smv /162.1 When the depth of flow is 1.4m

mR

mP

mA

smvbh

bhV

vAvA

QQ

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


50

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


51

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


52

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


53

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


54

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


55

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


56

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


57

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) =


58

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


59

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


60

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


61

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


62

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

QQ

QDg

fLQ

Dg

fL

hfhf


63

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


64

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


65

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)


66

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.


67

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


68

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


69

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


70

(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


71

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


72

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


73

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


74

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


75

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


76

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.


77

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


78

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.


79

Setup of a centrifugal pump

Detailed parts of a centrifugal pump

To delivery reservoir

Delivery pipe

Casing

Supply reservoir

Foot value

Strainer

Vanes

Impeller


80

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.


81

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.


82

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


83

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.


84

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.

QQ

Qv

≈ 1

ηv is slightly less than one. 4. Overall efficiency is given by

QQ

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


85

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


86

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


87

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


88

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


89

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.


90

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


91

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)


92

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

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