Resevoir and Distribution System - Includes Hardy Cross Method and Some Ideas for brilliant Presentation Slides

7.RESEVOIR AND
DISTRIBUTION SYSTEM
Presented By
Saman 133, Sandhya 136,
Sameer 134, Sandip 137,
Sandeep 135, Sanish 138
Tutor
Asst. Prof. Shukra Raj Paudel
Department of Civil Engineering
IOE, Tribhuvan University
Date: 1st January,2018

Students will be able to learn the followings things:
Objectives of the Presentation
Reservoir and its types
Water consumption pattern
Calculate capacity of reservoir
Distribution system and its types
Advantages and disadvantages of each type
Layout of distribution system
Design steps of distribution system
2

Presentation Layouts
7.1 Systems of Supply
7.2 Reservoirs
7.3 Clean Water Reservoirs
7.4 Service reservoir
7.5 Water Consumption Pattern
7.6 Capacity Determination of Service Reservoirs
7.7 Layout of Distribution System
7.8 Design of Distribution System
7.9 Design of Pipe Network
3

Treatment
System
Reservoir and
Pipe Network
Consumers
• Distribution system is the system of reservoir and pipe
network through which water is distributed.
• It should be well planned to carry specific rate of flow
to meet water demand at requisite pressure.
7.0 Introduction
4

Service
Reser
voir
Network
of Pipes
Valves
Pumps
Fire
Hydrant
Service
Connec
tion
Public
Stand
post
Meters
To lift water from lower to higher
level
Distribution
System
7.0 Introduction(Contd)
To store water for equalization
of water and stabilizing of water
To convey water to the
community
To control and regulate supply
To deliver water during fireTo deliver water to households
To deliver water to those who
don’t have private connection
To measure quantity of water
supplied
Picture Source:
braunengineers.com
Picture Source:
acs.org
Picture Source:
Iedepot.ie
Picture Source:
Presentermedia.com
Picture Source:
Wpyz.info
Picture Source:
i.pinimg.com
Picture Source:
Presentermedia.com
Picture Source:
Presentermedia.com
5

Systems Of
Supply
Continuous
System
Intermittent
System
7.1.1 Continuous System
• Water is supplied to consumers 24*7.
• Best system and adequate water for fire fighting.
Wastage Of
Water
If People Have
no Civic Sense
If Leakage Exists
7.1 Systems of Supply
6

7.1.2 Intermittent System
• Water supplied only during fixed hours mainly
morning and evening.
• Supply time according to climate or seasons.
• Easy maintenance and repairs.
• Less water leakage.
• Adopted when
 Sufficient pressure is not available.
 Sufficient quantity of water is not available at source
• Should not be installed in permanent basis and
steps to be taken to change to continuous system
7

Large
numbers of
Valve
Fire
demand
Domestic
Storage
Pollution in
Supply
Greater size
of pipe
Wastage
from taps
Staff
Requirement
8

7.1.3 Comparison of Types of Distribution System
Parameter
Continuous
Supply
Intermittent Supply
Fire Demand
Can be met
within time
Could not be met within time
which may cause huge
damage
Domestic
Storage
No need
Need to store water for non
supply hours
Size of Pipes Smaller pipe
Larger pipe is required to
supply water for full day in
same time
Continuous Supply Intermittent Supply
Source: Gettyimages.co.uk
9

7.2. Reservoir
• A reservoir usually means an enlarged natural or
artificial lake, storage pond or impoundment created
using a dam or lock to store water.
10

7.3. Clear Water Reservoir
• To store treated and clear water.
• Provided at end of water treatment process.
• Should have capacity of 14 to 16 hours of daily
water demand.
• Located below ground so gravity convey water
from treatment unit to reservoir.
Treatment
Unit
Clear Water
Reservoir
Service or
Distribution
Reservoir
Water
Clear
11

7.4. Service Reservoir
• Used in distribution system to provide storage
to meet fluctuation in demand in water.
• To provide storage for fire fighting and
emergencies
• To stabilize pressure in distribution system
• Should be located near the community for
minimum head loss and supply time.
• Save cost of pipeline as smaller pipeline are
required in transmission line from source to
reservoir.
• Always covered to avoid contamination, algae
formation.
• Provisions for mosquito proof ventilation, water
level indicators.
Source: Gettyimages.co.uk 12

7.4.1 Types of Service Reservoir
• Surface Reservoir
• Elevated Reservoir • RCC
• Steel
• Stone masonry
• Brick Masonry
• Ferro Cement
• HDPE
• PVC
According to
Material Of
Construction
• Rectangular
• Circular
• Square
• Spherical
• Cylindrical
According to
Ground Situation
According to
Reservoir Shape
13

7.4.2 Construction of Service Reservoir
Surface Reservoir
• Circular or rectangular in shape.
• Constructed at ground level or below it so called ground
or non elevated reservoir.
• Treated water stored in it is pumped to elevated
reservoir for water supply.
• But if located at height, water supplied directly to
consumers, so placed as high as possible.
• Although treated water is placed in this reservoir, some
solids may be present and deposit at bottom which is
cleaned through washout pipes.
14

Inlet Pipe: Discharges water facing downward
so that pressure head is dissipated in the water
and walls are prevented from erosion.
Overflow Pipes: To maintain constant level of
water in reservoir.
Compartments: Usually two, for easy maintenance
and repair.
Control valves: Controls the water supply in
compartments.
Ventilators: Provided in roof slab for free circulation
of air over water surface.
Washout pipes: Provided at bottom to
occasionally clean out some solids called
depositional sludge.
Outlet Pipes: Placed slightly higher than
washout pipes.
Source: Dr. BN Kansakar,Water Supply
Engineering,2015 (pg.268)
15

Elevated Reservoir
• Constructed at en elevation so called overhead
tanks.
• Rectangular, circular or elliptical in shape.
Surface
Reservoir
Elevated
Reservoir
Consumers
PUMP
SUPPLY
Source: braunengineers.com
16

Elevated Reservoir
Inlet Pipe: For entry of
water
Outlet Pipes: For exit of
water
Overflow Pipe: For exit of
water above full supply
level
Ladders: To reach the top
and bottom of reservoir
for inspection.
Ventilators: For free air
circulation.
Manholes: For providing
entry inside of reservoir
for inspection
Washout pipe: For
removing water after
cleaning the reservoir
Level indicator: To
indicate the water level
from outside of reservoir
Lightning conductor: For
protection against
lightning. Source: Dr. BN Kansakar, Water Supply
17

Intz Tank
• An RCC tank
called intz tank is
commonly
adopted these
days.
• RCC Ring beam
and RCC bracing
are for structural
support and
strength.
Source: Dr. BN Kansakar,Water Supply
18

7.5. Water Consumption Pattern
• Refers to variation in water consumed with
respect to time.
• Depends on geographic location, people’s
habit, climatic conditions.
5:00 - 7:00
7:00 - 12:00
12:00 - 17:00
17:00 - 19:00
19:00 - 5:00
25
35
20
20
0
HOURS
According to DWSS Guideline
Daily Demand in Rural Area
19

7.6 Reservoir Capacity Determination
RESERVOIR
CAPACITY
BALANCING
RESERVE
BREAKDOWN
RESERVE
FIRE
RESERVE
20

7.6.1 Balancing or Equalizing Reserve
• Conventionally calculated
by mass curve and
hydrograph indicating
hourly consumption rate.
• Computers and
calculators are widely
used to calculate
balancing reserve.
• Demand of water varies with time but treated water
comes out at same rate.
• Balancing reserve is that quantity of water required
to store in the reservoir for balancing the variable
demand in distribution system.
Source: Astonet.com
21

7.6.2 Breakdown Reserve
• Breakdown means stoppage of water supply
due to some reasons like damages in pump.
• Breakdown may be minor or may takes weeks
to repair.
• Some amount of water is required to be stored
for such period. And such storage is called
breakdown reserve.
Generally,
25%
Total
Storage
Breakdown
Reserve
22

7.6.3 Fire Reserve
• Water reserved for fire-fighting purpose.
• R--Reserve Storage in liters
• F--Fire Demand in liters/minute
• P--Reserve Fire Pumping
Capacity in liters/minute
• T-- Duration of Fire in minutes
Source: cbsnews.com
23

Numerical Problems on Reservoir
• Requires reservoir if there is, in any day
Water Supply > Water demand
• Water Tapped from source should be greater
than water demand.
24

Q.A newly established town with population 1.2 million is to
be supplied with water daily at 45 lpcd. Water has to be
stored for fire demand at least 1% of total demand. The
variation in demand is as follows.
Determine total reservoir capacity assuming pumping to
be done at uniform rate and period of pumping is 5 AM to
10 AM and 5 PM to 8 PM in two shift.
Time Consumption %
5:00 to 7:00 25
7:00 to 12:00 35
12:00 to 17:00 20
17:00 to 19:00 20
19:00 to 5:00 0
Solution:
Population = 1.2*106
Per capita demand = 45 lpcd
Water Demand = 1.2*45
= 54 MLD
25

2 * 6.75 = 13.50
Inflow of reservoir
= Total Inflow/Pumping hour
= 54/8
= 6.75 ML/hr
Balance reserve = 10.80 million liters
Fire reserve = 1% of total demand = 0.54 million liters
Capacity of reservoir = 11.34 million liters
25 % * 54.00 =
13.50
Time Hours Consumption
%
Supply
(ML)
Demand
(ML)
Surplus
(ML)
Deficit
(ML)From To In Out
5:00 7:00 2 2 25 13.50 13.50 - -
7:00 12:00 3 5 35 20.25 18.90 1.35 -
12:00 17:00 0 5 20 0.00 10.80 - 10.80
17:00 19:00 2 2 20 13.50 10.80 2.70 -
19:00 5:00 1 10 0 6.75 0.00 6.75 -
Total 8 24 100 54.00 54.00 10.80 10.80
Pumping period Hours
5:00 to 10:00 5
17:00 to 20:00 3
Total 8
26

Q.A water is to be supplied to a village having daily water
demand of 66370 liters from a stream source with a safe
yield of 0.78 liters per second through 22 public stand
post. Assume the flow from each public stand post as
0.10 liters per sec. Take hourly consumption pattern as
follows. Determine the size of service reservoir.
Time Consumption %
5:00 to 7:00 25
7:00 to 12:00 35
12:00 to 17:00 20
17:00 to 19:00 20
19:00 to 5:00 0
Solution:
Water Demand
= 66370 liters/day
= 2765.42 liters/hr
Water tapped from source
= 0.78 lps
= 2808 liters/hr
Water tapped from source ≥
Water demand, Hence OK
27

Solution:
Water Demand = 66370 liters/day
Water tapped from source = 2808 liters/hr
2 * 2808 = 5616
25 % * 66370.0 =
16592.5
Time
Hours
Consum
ption %
Supply
(liters)
Demand
(liters)
Surplus
(liters)
Deficit
(liters)From To
5:00 7:00 2 25 5616.0 16592.5 - 10976.5
7:00 12:00 5 35 14040.0 23229.5 - 9189.5
12:00 17:00 5 20 14040.0 13274.0 766.0 -
17:00 19:00 2 20 5616.0 13274.0 - 7658.0
19:00 5:00 10 0 28080.0 0.0 28080 -
Total 24 100 67392.0 66370.0 28846.0 27824.0
Capacity of reservoir
= (10976.5 + 9189.5 – 766.0 + 7658.0)lts
= 27058.0 lts
28

Some Possible Cases
Surplus
(ML)
Deficit
(ML)
- 20.25
13.5 -
20.25 -
- 27.00
13.50 -
47.25 47.25
Surplus
(ML)
Deficit
(ML)
- 12420
- 6300
7200 -
- 6920
50400 -
57600 25640
Surplus
(ML)
Deficit
(ML)
- 12420
- 6300
7200 -
- 7920
50400 -
57600 26640
27.00
12420+6300-
7200+7920=
19440
1844013.5 + 20.25
= 33.75
12420+6300=
18720
29

7.7 Layout of Distribution System
Distribution System is a network of pipelines that
conveys water to the consumers in the community.
Layout depends on the layout of the roads in the
community.
30
Source: wrsc.comSource: images.yourstory.com

Requirements of Good Distribution System
31
No deterioration
of water in pipes
Water tight
Adequate
water for fire
fighting
No connection loss
during repairs
Supply with
sufficient
pressure head
Meter System

7.7.1. Dead End System(Tree or Branch System)
 Consist of one supply or trunk main from which
submains are taken
 Branches are taken from submains and service
connections are given to consumers
 Used in the
place
developed in
hazardous
manner.
Source:
Slidesharecdn.com
32

7.7.1. Dead End System (continued)
33
Advantages Disadvantages
1.Simple and easy design
calculation to determine
discharge and pressure
1.Stagnation of water and
accumulation of sediments
at dead ends
2.Cheap and economical
design of pipes considering
population
2.Scour valves and staffs are
required to remove stale
water and sediments
3.Simple laying of pipe 3.Repair of pipe requires to cut
it off completely
4.Less number of cutoff
valves are required
4.Discharge available for
fire fighting is less

7.7.2 Grid Iron System(Interlaced or Reticulation System)
Consist of interconnected mains, sub-mains and
branches forming number of closed loop
Continuous circulation of
water is possible through
whole distribution system
Source: Slidesharecdn.com
34

7.7.2 Grid Iron System (continued)
35
Advantages Disadvantages
1.Free water circulation without
stagnation or sediment deposit
and pollution
1.Requires large number
of cutoff valves
2.Water is available with min. head
loss
2.Requires longer pipe
length
3.Repairs only affect small
distribution area
3.Analysis of discharge,
velocity and pressure is
cumbersome
4. -Enough water available for fire
fighting in fire hydrant
4.Cost of pipe laying is
more

7.7.3 Ring System or Circular System
Layout of main pipe is laid to form closed ring (circular,
rectangular) around area to be served in the periphery
of blocks
 Sub mains take off from main pipelines and run on the
interior of area
 Suitable for towns
and cities having well
planned streets and
roads
Advantages and
disadvantages are
exactly like Grid Iron
System
Source: Slidesharecdn.com 36

7.7.4.Radial System
Reverse of Ring or Circular System with water
flowing towards outer periphery instead from it
Entire distribution area is divided to number of
small distribution zones
Distribution reservoir is provided at the center of
each zone.
Water from main pipe is conveyed to the
distribution reservoir of each zone
Water is supplied to radially laid distribution
pipes towards periphery
37

7.7.4.Radial System(contd….)
Advantage: Suitable for cities and towns having
roads laid out radially
Combination of any 2 or more of this is suitable
for cities and towns
Source: Slidesharecdn.com 38
Source: youtube.com

7.8 Design of distribution system
Determination of size of pipe used in distribution
system to carry required discharge under a
known pressure difference
39

7.8.1 Pipe Hydraulics
Continuity Equation
Q = A*V = (πd2/4)*V
In which Q = Discharge through pipe
A = Cross-Section area of pipe
d = Diameter of pipe
V = Velocity of flow in pipe
40

Bernoulli’s Equation
Energy loss in pipes
• Major Loss (caused by friction)
• Minor Loss (caused by change in velocity of flowing
fluid either in magnitude or direction)
Head
Loss
Energy Head
at Downstream
Energy
Head at
Upstream
(P
ρg
+
V2
2g
+ z)in
= (
P
ρg
+
V2
2g
+ z)out
+ hloss
41

1.Major Loss
Loss of energy due to friction can be determined by
using either of the formula:
42

(i) DarcyWeisbach formula
In which,hf=head loss due to friction in m
L=length of pipe in m;
d=diameter of pipe in m;
V=mean velocity of flow through pipe in m/s;
Q=discharge through pipe in m/s;
g=Acceleration due to gravity=9.81 m/s2
f=friction factor which is dimensionless.
Value of friction factor in Darcy-Weisbach Formula is
obtained using following equation given by Colebrook
and White: 43

Roughness increases with time according to
relation: k= k0+at
where, k0=Roughness of new pipe material,
k=Roughness at any time ‘t’,
α=Rate of increase of roughness with time
In which, k=roughness of the pipe material;
Re=Reynold’s number=
ν=kinematic viscosity of water
Darcy Weisbach formula(contd….)
Coolebrook and White equation:
44

(ii)Manning’s Formula
Mean velocity of flow is given as:
In which,
V=mean flow velocity in pipe in m/s,
R=hydraulic mean pipe depth
S=slope of Energy Grade line ,or head lost
per unit length of pipe
n=Manning’s roughness (or rugosity)
coefficient
45

3/4
22
4







d
LVn
hf
3/4
22
35.6
d
LVn
hf 
3/4
2
2
22
4
35.6
dd
LQn
hf







3/16
22
294.10
d
LQn
hf 
LShf *
L
h
S
f

If hf is the loss of head (in m) due to friction in a pipe of
length L(in m),then slope of Energy grade line is expressed
as:
46

(iii) Hazen Williams Formula
Mean velocity of flow V in m/s is given by:
If Hf = Loss of head (in m) due to friction in pipe of length
L, then slope of Energy Grade Line S is:
LShf *
852.1
87.4
852.1
367.1
68.10843.6













C
Q
dC
V
d
L
hfL
h
S
f

In which R = hydraulic mean depth of pipe in
m;
S = Slope of energy grade line ,or head
lost per unit length of pipe
C = Roughness Coefficient
47

2.Minor Loss:
Loss of energy due to other causes(except friction) is
classified as minor loss
I.Minor loss due to enlargement
v1 and v2 are mean
velocities of flow in
smaller and larger
sections of pipe
g
vv
hL
2
)21( 2


Source: tutorhelpdesk.com
48

II.Minor loss due to sudden contraction
g
v
hL
2
5.0
2

V2 is mean
velocity of flow in
smaller section of
pipe
Coefficient of
contraction=0.5
(generally)
Source:slideplayer.com 49

III.Minor loss at entrance of pipe
g
v
hL
2
5.0
2

V is the mean
velocity of flow in the
pipe
IV.Minor loss at exit of pipe
g
v
hL
2
2

V is the mean velocity of
flow at the exit section in
the pipe
Inflow (v)
Outflow (v)
Source:http://engineering-
references.sbainvent.com
Source:Vanoengineering-
wordpress.com
50

V.Minor loss due to bend:
VI.Minor loss due to pipe fittings:
V=Mean flow velocity in bend
K=Coefficient value depending on
• the angle of bend
• relative radius of curvature
(Radius of curvature of pipe axis/diameter of pipe)
V=Mean velocity of flow in bend
K=Coefficient value depending on
type of pipe fitting
g
v
khL
2
2

g
v
khL
2
2

Source:slideplayer.com
Source:tradekorea.com 51

VII.Minor loss due to gradual contraction or enlargement in
pipe:
V1 and V2 are the mean velocities of flow in smaller and
larger sections of pipe,
K is the coefficient the value of which depends in the
angle of convergence or divergence and on the ration of
smaller and larger cross sections of pipe
g
vv
khL
2
)21( 2


Source:slidesharecdn.com
Gradual Enlargement
Gradual Contraction
52

Parameters
Velocity
Pressure
Pipe
Size
7.8.2 Design Criteria:
 Required for system to work properly and smoothly
53

(a) Velocity:
Min. velocity=0.3m/s
Max. velocity=3.0m/s
54

(b)Pressure:
(c)Pipe Size:
 Nearest commercially available pipe size in higher
side is recommended
 Pipe diameters in mm are 15, 20, 25, 32, 40, 50,
65, 125, 150, 500,1000….upto 3000
Min. Pressure:5m
Desirable pressure: 15m
Max. Pressure:55m
55
Connection Type Recommended Pressure
Private Connection Min=15m
Without Private Connection Min=5m
At Stand Post

a) Maps and Surveys:
Topographical map of project area covering entire
distribution areas and all other water supply components is
required.
 Study of possibility of conveying the water either through
gravity or pumping is made
 2 types of surveying is involved:
• Technical or Detailed survey using survey equipments
like Level ,theodolite in possible route
• Social survey collecting data concerning water demand
and discharge available of project area
7.8.3 Design Steps(Contd)
57

c)Discharge in pipelines:
• Transmission line designed for maximum daily
demand
• Distribution system for maximum hourly demand
or peak factor which varies from 2 to 4.
• Discharge in each pipeline is computed according
to
• population density and per capita demand
• type and number of commercial capacity
• fire requirements ,etc
b)Tentative Layout:
• Includes showing layout of alignment of main sub
main and branches
• Includes position of intakes, water treatment plants,
pumping stations, valves, reservoirs etc..
58

d)Calculation of pipe diameters:
• Pipe diameter is based on head available at upstream
and downstream of pipe.
• Darcy Weisbach and Hazen Williams formulae are
used to calculate diameters
e)Computation of residual pressure and velocity:
• Residual pressure in distribution system is computed
using:
 pressure available at upstream points
and ground levels
 design discharge
 head loss in pipe.
After choosing pressure and diameter, velocity is
computed.
59

Design of Pipe Networks
Branched System Looped System
Source: slideshare.net Source: slideshare.net
60

A. Branched System
Population served by each section
Discharge to be carried by each
section
Allowable head loss in the pipe
Pipe diameter of pipe in each
section
Head loss in each pipe section
Residual pressure and velocity
Design
Criteria
C
o
m
p
u
t
e
61

63
d=0.919m d=0.100m
185 185
125
82.69
15
57.31
Residual Head
Head loss
Head At Q
d=0.710m d=0.800m
45
182.3
057
15
Residual Head
Head loss
Head At R
Include Residual
Head of PQ

45 45 45
182.3
057 102.1
447 34.4
56
15
95.1
61 162.
8497
Residual Head
Head loss
Head At R
64

Example: 7.15
A layout of rural water distribution system is shown below.
72

Looped System
• Not straight forward and simple as design of
branched system.
• Hardy cross method is widely used for it’s design
and analysis.
Source:Ftrameestruturas.blogspot.com
76

Hardy Cross Method
• An iterative method for determining the flow in
pipe network systems.
• Works under 3 Laws
1.) hf
= k* Qn
2.) ∑Q= 0 at any junction.
3.) ∑ hf = 0
Where,hf = Head loss in the pipe
Q=Quantity of water flowing
k and n are constants.
77

Solution:
We know, head loss HL=kQn
From Hazen Wiliam’s equation HL= 85.187.4
85.1
**68.10
Cd
QL
Hence,k= 85.187.4
68.10
Cd
L
n=1.85 And ∆Q=


Q
h
n
h
L
L
The values of assumed Q for all the pipes in the network with
their calculated k values are given in the table below.
78

Values of k and n for different
head loss formula
79

Example:7.17
A pipe network is shown below. Calculate the flows and head
loss in all the pipes .Use Hardy Cross method. Take Hazen
William’s coefficient as 100.
200lps
B D
A C
500lps
1000lps
300lps
L=100m
D=250m
D=400m
D=300m
D=300m
L=1000m
D=250m
L=1118m
L=500mL=500m
80

Methods Of Analysis
a) Balancing head by correcting assumed flows
b) Balancing flows by correcting assumed heads
A
B D
E
FQ Q
Q1
Q2
c
Here,
Correction
81

Pipe Diameter
(m)
Length(m) C k Assumed
Q(m3/s)
AB 0.40 500 100 92.37 0.7
BC 0.30 1118 100 838.36 0.1
AC 0.30 1000 100 749.88 0.3
BD 0.25 100 100 182.22 0.4
CD 0.25 500 100 911.12 0.1
There are 2 loops in network.
 Let Loop I is the loop formed by pipes AB,BC and AC
 Loop II is the loop formed by the pipes BD,BC and CD.
Four trials have been carried out.
85.187.4
)100(*)4.0(
)500(68.10
)(



Cd
L
ABk
AB
AB
300lps1000lps
200lps B D
A C
500lps
A=300lps
A=700lps
Loop I
Loop II
A=400lps
A=100lps
In each trial,
Modify
pipe
flow
until the
sum of
head loss
in a loop is
almost
zero.
Check
whether the
sum of head
loss of all the
pipes in a
loop is zero
Calculate
head loss
in each
pipe
Calculate
flow
correction
∆Q
82

Lo
op
Pipe K Q(m3/s) HL(m) HL/Q ∆Q
II
BD 182.22 0.4 33.451 83.628
-0.0040
BC 838.36 -0.1252 -17.947 143.347
CD 911.12 -0.1 -12.870 128.700
Total 2.634 355.675
Lo
op
I
AB 92.37 0.7 47.749 68.213
0.0252BC 838.36 0.1 11.842 118.420
AC 749.88 -0.3 -80.847 269.490
Total -21.256 456.123
First Trial
The results of each trial are given below:
(kAB=92.37)*(QAB=0.7)1.85
123.456*85.1
256.21



Q
h
n
h
Q
L
L
AB
Lo
op
I
AB 92.37 0.7252 50.997 70.294
0.0015BC 838.36 0.1292 19.022 147.229
AC 749.88 -0.2748 -68.734 250.124
Total 1.265 467.647
Second Trial =
83
0.1+0.0252+0.0040

Lo
op
I
AB 92.37 0.7252 50.997 70.294
0.0015BC 838.36 0.1292 19.022 147.229
AC 749.88 -0.2748 -68.734 250.124
Total 1.265 467.647
Second Trial
Lo
op
II
BD 182.22 0.396 32.835 82.917
0.0006
BC 838.36 -0.1277 -18.616 145.779
CD 911.12 -0.104 -13.838 133.058
Total 0.381 361.754 84

Loo
p
Pip
e
K Q(m3/s) HL(m) HL/Q ∆Q
II
BD 182.22 0.3954 32.743 82.810
0.0000
BC 838.36 -0.1281 -18.724 146.167
CD 911.12 -0.1046 -13.987 133.719
Total 0.032 362.696
Third Trial
Fourth Trial
Loo
p
Pip
e
I
AB 92.37 0.7235 50.757 70.155
0.00000BC 838.36 0.1281 18.724 146.167
AC 749.88 -0.2765 -69.523 251.439
Total -0.042 467.761
Loo
p
Pip
e
I
AB 92.37 0.7237 50.783 70.171
0.0002BC 838.36 0.1283 18.778 146.36
AC 749.88 -0.2763 -69.43 251.285
Total 0.131 467.816
Loo
p
Pip
e
II
BD 182.22 0.3954 32.743 82.810
0.0000
BC 838.36 -0.1281 -18.724 146.167
CD 911.12 -0.1046 -13.987 133.719
Total 0.032 362.696
85

Answer
0.2
B D
A C
0.5
1 0.3
0.7235
-0.2765
0.3954
-0.1046
I
II
= 0.2765
= 0.1046
0.1281 + 0.2765
= 0.3 + 0.1046
86

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