Pipe networks analysis_modified [compatibility mode]

1CEE F312: Hydraulic Engineering By Prof. Ajit Pratap Singh BITS Pilani

Pipe Networks AnalysisPipe Networks Analysis

By

Prof. Ajit Pratap Singh

Civil Engineering Department

BITS Pilani

�We have three types of problems in our day

to day life

(1) Individual Pipes: Connects two points in the

distribution system say from a reservoir to thepoint of use (studied in last chapter).

(2) Branching System: Branch system is used insmall community projects (We’ll study in thischapter).

(3) Complete Pipe Network: It may consist ofSeveral nodes, loops, sources etc. (We’ll study inthis chapter).

Branching SystemBranching System Pipe networksPipe networks

�When a group of interconnected pipes

forming several loops or circuits, then it is

called as networks of pipes.

�Aim: To determine the distribution of flow

through the various pipes of the network

such that all the conditions of flow are

satisfied and all the circuits are then

balanced.

Branched and Looped NetworkBranched and Looped Network Conditions ???Conditions ???

�As per the principle of continuity, flow into each

junction = flow out of each junction

� In each loop, the loss of head due to flow in

clockwise direction must be equal to the loss of

head due to flow in anti-clockwise direction

�Darcy-Weisbach equation must be satisfied for

flow in each pipe and minor losses may be

neglected if the pipe lengths are large.


Networks of PipesNetworks of Pipes

�____ __________ at all nodesA

0.32 m3/s 0.28 m3/s

?

b

Mass conservation

Water Distribution System

Assumption


Assumption

� Each point in the system can only have one _______

� The pressure change from 1 to 2 by path amust equal the pressure change from 1 to 2 by path b

a

p1

γ+

V12

2g+ z1 =

p2

γ+

V22

2g+ z2 + hL

p2

γ−

p1

γ=

V1a

2

2g+ z1 −

V2 a

2

2g− z2 − hLa

b

1 2pressure

Same for path b!

hLa

= hLb

a

b

1 2Pressure change by path a


Assumption


Assumption

�Pipe diameters are constant or K.E. is small

�Model withdrawals as occurring at nodes so V is constant between nodes

Or sum of head loss around loop is _____.zero

(Need a sign convention)

V1a

2

2g+ z1 −

V2a

2

2g− z2 − hLa

=V1b

2

2g+ z1 −

V2 b

2

2g− z2 − hLb

Networks of PipesNetworks of Pipes

�The relationship between head

loss and discharge must be

maintained for each pipe

�Darcy-Weisbach equation

�Exponential friction formula

�_____________

b

a

1 2

Hazen-Williams

�As per Darcy-Weisbach equation, loss of

head through any pipe discharging at the

rate Q can be

�Where r is proportionality factor depending

upon length of pipe, diameter of pipe,

friction factor etc.

..(I) Qr Q 12.10D

fL

D4

π2g

fLQh nn

5

5

2

2

f =

=

=

� As per Hazen William equation, loss of head through any

pipe discharging at the rate Q can be

� Where n is an exponent having a numerical value ranging

from 1.72 to 2.00 depending upon the relationship used for

it. L and D are in meters. If D is in mm

� Hardy cross method, Newton Raphason, Method, Linear

Graph Theory.

n1.852

4.871.852f Qr Q DC

10.62Lh =

=

n1.852

4.871.852

15

f Qr Q DC

L 4.351x10h =

=


Hardy Cross MethodHardy Cross Method

� The method was first published in

November 1936 by Hardy Cross, a

structural engineering professor at the

University of Illinois at Urbana–

Champaign. The Hardy Cross

method is an adaptation of the

Moment distribution method, which

was also developed by Hardy Cross

as a way to determine the moments in

indeterminate structures.

Hardy Cross 1936

Hardy Cross Method (Example) Hardy Cross Method (Example) Hardy Cross MethodHardy Cross Method

� Assume a most suitable distribution of flow that satisfiescontinuity at each junction.

� With the assumed Q, compute the head-loss for each pipeusing equation (I).

� Consider different loops or circuits and compute net headloss around each circuit by taking clockwise flows aspositive and anti-clockwise flows as negative

� Note that for a correct distribution of flow the net head-lossaround each circuit should be equal to zero i.e. circuitshould be balanced.

� Calculate correction factor using

∑

∑∑∑

=−=∆

Q

hn

h-

Qn r

Qr Q

f

f

1-n

0

n

0

Hardy Cross Method…Hardy Cross Method…

� It should be noted that any pipe that appears in two looploops has a negative sign in one loop equation and apositive in the other loop equation . This convention mustalso be considered when updating flows for each pipewhile applying correction in the next iteration.

� In a Pipe Network Systems, the Number of equationsrequired (Neq) can be obtained by

� Neq = Number of equations required

� Nj= number of junction nodes

� Nl= number of loops

� Nf= number of Fixed Grade Nodes (e.g. elevated

reservoirs)

Network AnalysisNetwork Analysis

Find the flows in the loop given the inflows and outflows.The pipes are all 25 cm cast iron (k=0.26 mm).

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

200 m

100 m



�Assign a flow to each pipe link

�Flow into each junction must equal flow out

of the junction

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

0.320.00

0.10

0.04

arbitrary


�Calculate the head loss in each pipe

f=0.02 for Re>200000 hf =

8 fL

gD5π 2

Q 2

fh kQ Q=

339)25.0)(8.9(

)200)(02.0(8

251 =

=

πk

k1,k3=339k2,k4=169

A B

C D0.10 m3/s

0.32 m3/s 0.28 m3/s

0.14 m3/s

1

4 2

3

hf1= 34.7m

hf2= 0.222m

hf3

= −3.39m

hf4

= −0.00m

hfi

i=1

4

∑ = 31.53m

Sign convention +CW

2

5

s

m


�The head loss around the loop isn’t zero

�Need to change the flow around the loop

� the ___________ flow is too great (head loss is

positive)

� reduce the clockwise flow to reduce the head loss

�Solution techniques

�Hardy Cross loop-balancing (___________ _________)

�Use a numeric solver (Solver in Excel) to find a change

in flow that will give zero head loss around the loop

�Use Network Analysis software (EPANET)

clockwise

optimizes correction

Network ElementsNetwork Elements

�Reservoirs: infinite source, elevation is not affected by demand

�Tanks: specific geometry, mass conservation applies

�Pumps: need a relationship between flow and head

�Controls

�Check valve (CV)

�Pressure relief valve

�Pressure reducing valve (PRV)

�Pressure sustaining valve (PSV)

�Flow control valve (FCV)

Reservoirs, and PumpsReservoirs, and Pumps

�Impounding Reservoirs:

These type of reservoirs are constructed

directly on streams to store water during

periods of high runoff and release it for use

during periods of low runoff

�Service and Balancing Reservoirs:

Service Reservoirs Service Reservoirs

� Service reservoirs, also termed storage reservoirs, or distribution reservoirs are provided near consumption centers to supply water when required at sufficient pressure.

� These reservoirs fill when the rate of filtration or pumping exceeds the rate of demand and empty under the reverse conditions. This allows the pumps and the treatment plant to operate at a constant rate hereby reducing their capacity.

� These reservoirs also provide fire storage for immediate use in large quantities for fire-fighting purposes and also provide emergency storage during failure of intake, supply conduit or power.


Balancing Reservoirs Balancing Reservoirs

� Balancing reservoirs, also termed euqalising reservoirs are provided to equalize pressures in the distribution system and thereby to reduce fluctuations in pressures caused by fluctuating demand.

� They also raise pressures at remote points far from storage reservoirs and pumping stations and thus provide improved service during periods of peak demand.

� When located near pumping stations, they equalize the head on pumps. This results in better pump selection and allows them to operate at higher efficiencies.

� Incidentally, service reservoirs also serve as balancing reservoirs and balancing reservoirs as service reservoirs.

Service and balancing reservoirsService and balancing reservoirs

�Service and balancing reservoirs may be surfacereservoirs or elevated reservoirs.

�They may be connected to distribution networkssuch that water enters to and leaves from areservoir through the same pipe so that thereservoir is floating on the system;

�Or by separate inlet and outlet pipes. In the lattercase the reservoir is not floating on the system. Itreceives water from one system through inlet pipeand in turn supplies it to another system throughoutlet pipe.

Problem 1Problem 1

� Pipes 1 and 2 connect reservoir 1 [water level (W.L.) = 120.00 m], and reservoir 2 (W.L. = 100.00 m), respectively, to a junction point J as shown in Fig. A third pipe, pipe 3, starts from the junction point J and discharges into the atmosphere at point 3. The length, diameter and C coefficient values of the pipes are: Pipe 1 – 300 m, 300 mm, 100; pipe 2 – 250 m, 200 mm, 130; and pipe 3 – 120 m, 300 mm, 100. Determine the discharges in pipes 1 and 2 and the hydraulic gradient level (HGL) values at points J and 3, when the outflow at point 3 is: (1) 0.2 m3/s; and (2) 0.4 m3/s.

�As per Hazen William equation, loss of

head through any pipe discharging at the

rate Q can be

� If D is in mm, as per Hazen William equation, loss of head

through any pipe discharging at the rate Q can be

n1.852

4.871.852

15

f Qr Q DC

L 4.351x10h =

=


� If D is in mm, as per Hazen William equation, loss of head

through any pipe discharging at the rate Q can be given.

61.148300x 100

x2004.351x10R

90.493200x 130

x1504.351x10R

91.222300x 100

x3004.351x10R

4.871.852

15

3

4.871.852

15

2

4.871.852

15

1

=

=

=

=

=

=

� If q1 = q3 = 0.2720 m3/sec, reservoir 2 neither supplies nor

receives water

� If q3 > 0.2720 m3/sec, reservoir 2 supplies water and the

flow in pipe 2 is from 2 to J

� Similarly, If q3 < 0.2720 m3/sec, reservoir 2 receives water

and the flow in pipe 2 is from J to 2

� Reservoir 2 either supplies or receives water through the

same pipe 2 and thus reservoir 2 is floating on the system.

Problem 2Problem 2 PumpsPumps

�A pump is provided in a pipeline to supply extra

head to lift water from a lower level to a higher

level.

�May be provided externally to serve as supply

pumps to provide water from external sources to

the pipe networks

�Or may be provided internally within a network as

booster pumps to boost up pressure at some points

within the system.

Pumps…Pumps…

�Typical pump installation lifting water from

a lower-level reservoir 1 to a high-level

reservoir 2 is shown in Fig. Writing

Bernoulli’s equation, Eq. for points 1 and 2,

∑+

∑+++=+++

losses

minor

losses

friction

2g

V

γ

pZh

2g

V

γ

pZ

2

222p

2

111

In which hp = net head delivered by the pump. The velocity heads

are small and therefore they can be neglected. Further, p1 = p2 (=

the atmospheric pressure).


Pumps…Pumps…

( ) ( )

Lsp

12p

hhh

OR

lossesminor andfriction ZZh

+=

∑+−=

In which hs = static head (= Z2 – Z1) and hL = total

head loss.

• The sum of the static head and total head loss is

termed the system head.

System Head-Discharge Curve System Head-Discharge Curve

� If water levels in the two reservoirs remain

unchanged, the static head remains constant.

�However, the total head loss, which comprises of

frictional head losses and minor head losses,

varies with the discharge Qp through the pump.

Thus, the system head hT can be expressed as

)f(Qhh psT +=

�A curve showing the relationship between

the system head hT and the discharge Qp is

known as the system head-discharge curve.

A typical system head-discharge curve is

shown in Fig.

Pump Head-Discharge Curve Pump Head-Discharge Curve

� At a constant rotational speed, a pump has a unique relationship between the net delivered head hp and its capacity, i.e., the discharge Qp, supplied by it.

� The curve showing this relationship is known as pump head-discharge curve. A typical pump head-discharge curve is shown in Fig. in which the system head-discharge curve of earlier fig. is also shown.

� The point of intersection of the system head-discharge curve and the pump head-discharge curve is the actual operating point having discharge Qp that gives hp = hT

�Head-discharge curves for different types

of pumps are shown in Fig. below:

� The pump head-discharge curve in fig. (a) is strictly monotonically decreasing, i.e., the head decreases as the discharge increases. Thus, for a given head there is only one value for the discharge (point a). Such a curve is a stable head-discharge curve.

� The head-discharge curves in Figs. (b) and (c), which are obtained from centrifugal pumps and half-axial pumps, respectively, are not monotonically decreasing. Thus, for the same head, two discharges points [b and c in Fig. (b)] or three discharges [points d, e and f in Fig. (c)] may exist. Such curves having two or more discharges for the same head are unstable head-discharge curves.

kiran

Highlight

kiran

Highlight


� In practice, improvement in the system design with provision of pumps having stable head-discharge curves (except perhaps at very small percentage of discharge) can ensure a single operating point and thus a unique steady-state flow condition.

� Therefore, it is quite common in pipe network analysis to assume that the pumps have stable head-discharge curves.

� A distribution network having several pumps having unstable head-discharge curves would have multiple operating points, and the analysis and operation of such networks would be uncertain.

Head-Discharge Relationships Head-Discharge Relationships

� hp as a Function of Qp

It is common to use a relationship

(I) HBQAQh op

2

pp ++=

(II) 2A

BQG pp +=

so that Eq. (I) becomes (III) 4A

BHAGh

2

0

2

pp

−+=

In which A, B and H0 = constants, determined by fitting above Eq. to three points taken from the expected working range of the pump head-discharge curve. To improve the convergence characteristics in network analysis, Jeppson and Travallaee (1975) suggested a transformation

The Convergence is improved because the exponent 2 of Gp in Eq. (II) is close to

the value of n in general head loss relationship of a pipe, Eq. n

f RQh =

�If Hp and Rp are constants and n is an

exponent where Hp can be considered as

shut off head i.e. the maximum head that

can be provided by the pump as Qp

approaches zero, and Rp as the resistance

constant of the pump, a relationship of the

form can be given as

(IV) QRHh n

pppp −=

ExampleExample

Consider typical discharge-head data-0.0 m3/s: 18.0 m, 0.1 m3/s:

16.5 m, 0.2 m3/s: 14.7 m, 0.3 m3/s: 11.9 m, and 0.4 m3/s: 8.0 m.

The corresponding head-discharge curve is shown in Fig. (a);

and the corresponding discharge-head curve is shown in fig. (b).

It is seen that both curves are concave. Assuming the working

range of the pump to be 0.1-0.3 m3/s, the analytical relationships

described earlier can be obtained:

SolutionSolution

(IV) eq from formula)HW for 1.852 (n 49.20Q17.19h

(IV) eq from formulae) Mannig andDW for 2 (n ,57.35Q17.04h

49.95Q17.8h

(III) eq and (II) eq from 0.03QGin which ,50G17.345h

(I) eq from 50Q3Q17.3h

1.852

pp

2

pp

1.866

pp

pp

2

pp

2

ppp

=−=

=−=

−=

+=−=

−−=

ProblemProblem

A pump lifts water from a sump (W.L. = 100.00 m) to an elevated reservoir (W.L. = 110.00 m) through a 200-m long, 300-mm diameter pipe (k = 0.26 mm). Using the pump data of Fig. below and taking kinematic viscosity ν = 1 mm2/s for water, determine the discharge to the elevated reservoir. Use the Swamee-Jain friction-factor relationship for Darcy-Weisbach formula.

kiran

Highlight


Swamee and Jain in 1986 proposed the following

explicit relations that are accurate to within 2% of

the Moody chart

Swamee and Jain in 1986 proposed the following

explicit relations that are accurate to within 2% of

the Moody chart

8

262

9.0

5

2

L103Re3000

10k/D10

Q

D4.62

3.7D

kln

gD

LQ1.07h

×<<

<<

+=

−−−

υ

2000Re hgD

L3.17

3.7D

kln

L

hgD0.965Q

0.5

L

3

20.5

L

5

>

+

−=

υ

8

260.04

5.2

9.4

4.75

L

21.25

103Re5000

10k/D10

gh

LυQ

gh

LQk0.66D

×<<

<<

+

=

−−

2

0.910

10

10

Re

5.74

3.7

k/Dlog

0.25f :FormulaJain Swamee

3.7

k/Dlog 2.0

f

1 :Formula Prandtl

fRe

2.51

3.7

k/Dlog 2.0

f

1:formulaColebrook

+

=

−=

+−=

ProblemProblem

A pump lifts water from a low level reservoir (W.L. = 80.00 m) to an elevated reservoir (W.L. = 90.00 m) as shown in following. The suction pipe is 30 m long, 300 mm in diameter, and the delivery pipe is 150 m long, 250 mm in diameter. Both pipes are new cast iron with k = 0.26 mm. Several appurtenances with their K values are located on the pipeline as shown in the figure. The pump head-discharge data are also given in the following Table. Develop a system head –discharge curve and determine the steady state head and flow. Use Swami-Jain relationship for determining f for Darcy-Weisbach equation. Take kinematic viscosity ν = 1.3 mm2/s for water.

Pump Head Discharge DataPump Head Discharge Data

Discharge, m3/sec Head, m

0.00 30.8

0.02 30.9

0.04 30.8

0.06 29.0

0.08 27.0

0.10 23.0

0.12 17.3

0.14 9.0

Using Swamee-Jain EquationUsing Swamee-Jain Equation

Dischargem3/ sec f1 f2 ∑hf ∑hminor

Total hL hstatic

Total Head

0.00 - - - 0.000 0.000 10.00 10.000

0.02 0.02294 0.02349 0.129 0.039 0.168 10.00 10.168

0.04 0.02132 0.02197 0.481 0.157 0.638 10.00 10.638

0.06 0.02067 0.02137 1.052 0.353 1.405 10.00 11.405

0.08 0.02031 0.02104 1.842 0.627 2.468 10.00 12.468

0.10 0.02009 0.02083 2.849 0.979 3.828 10.00 13.828

0.12 0.01993 0.02069 4.073 1.410 5.484 10.00 15.484

0.14 0.01981 0.02058 5.515 1.919 7.435 10.00 17.435


Head-Discharge

0.00

5.00

10.00

15.00

20.00

25.00

30.00

35.00

0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16

Discharge

He

ad

System Head-Discharge

Pump Head-Discharge

�Analyze the single source, three demand

node, two loop network shown in Figure

below. The head loss in a pipe is given by h

= RQ1.852:

Multiple Source Networks with

known Pipe Resistances

Multiple Source Networks with

known Pipe Resistances

� The total head lost equation in such pseudo loop could be written as by

considering heads of both the reservoirs:

� The water level between the two tanks/reservoirs should balance the

summation of head lost through the pseudo loop.

ProblemProblem

� Analyze the network (i.e. determine unknown heads and

discharges in the network) shown in Figure below using

the Darcy-Weisbach formula for the head loss in pipes and

the Swamee-Jain explicit relationship for the friction

coefficient i.e. . The pipe details are as given in the Table

below. Three points on the head-discharge curve for the

pump have the following discharge to head characteristics:

0.1 m3/s – 11.89 m; 0.2 m3/s – 11.01 m; and 0.3 m3/s –

9.37 m, respectively. Take kinematic viscosity (v) of water

at 200C, which is equal to 1.00 10-6 m2/s.

ProblemProblem DataData

Pipe Length (m) Diameter

(mm)

Roughness

height (mm)

(1) (2) (3) (4)

1 300 300 0.20

2 400 200 0.25

3 400 150 0.25

4 150 150 0.25

5 150 200 0.25

6 150 150 0.30

7 500 350 0.20


Solution: Iteration 1Solution: Iteration 1

c Pipe x 1Qs 1fx 1Rx Rx 2Rx

(1) (2) (3) (4) (5) (6) (7) (8)

I 2 0.08 0.0214 2207 14.13 353.2

4 0.02 0.0237 3865 1.55 154.6 0.0118

3 0.06 0.0229 9947 -35.81 1193.7

-20.13 1701.5

II 4 - - - -1.55 154.6

5 0.06 0.0215 835 -3.00 100.2 -0.0031

6 0.04 0.0241 3933 6.29 314.7

1.74 569.5

III 7 0.22 0.0179 141 6.81 61.9

5 - - - 3.00 100.2

2 - - - -14.13 353.2

1 0.22 0.0184 187 -9.07 82.64 0.0370

Pump 0.1847a - 38.2b 10.75c 14.11d

- - - - -20.00e -

-22.64 611.9

∑∑

SolutionSolution

�Based on the Pump head-discharge cureve, we can

transform the pump head-discharge equation by

�The first-trial discharges in pipe satisfying the

node-flow continuity equations at nodes 3,…, 6

are assumed as follows: Q1 = 0.22 m3/s, 1Q2 = 0.08

m3/s, 1Q3 = 0.06 m3/s, 1Q4 = 0.02 m3/s, 1Q5 = 0.06

m3/s, 1Q6 = 0.04 m3/s and 1Q7 = 0.22 m3/s.

2

2

2.387.212

2.3805.12

ppp

pp

QQh

OR

Qh

−+=

−=

�The final solution is as follows: f1 =

0.01852, R1 = 189.05, Q1 = 0.1619 m3/s;

f12= 0.02167, R2 = 2238, Q2 = 0.0508 m3/s;

f3 = 0.02325, R3 = 10123, Q3 = 0.0311 m3/s;

f4 = 0.02322, R4 = 3791, Q4 = 0.0324 m3/s;

f5 = 0.02125, R5 = 823.3, Q5 = 0.1016 m3/s;

f6 = 0.0232, R16= 3905, Q16= 0.0565 m3/s; f7

= 0.01776, R7 = 139.7, Q7 = 0.2781 m3/s;

and hp = 11.436 m.

� For iteration I, as 1Q1 = 1Qp = 0.22 m3/s, we use Eq. (5).

The pump can be made an integral part of pipe 1 with

modified resistance constant = 187 + 32.8 = 219.8. The

head at the sump becomes 80 + 12.32 = 92.32 m.

Subsequently, when Q1 becomes less than 0.2 m3/s, we use

Eq. (4). To include the pump effect, the pipe resistnace

constant is increased by 29.3, and the sump head becomes

92.18 m. This gives practically the same final solution.

�If the head-discharge relationship given by

Eq. (5.7) is selected to represent the pump

data, and taking n = 2 since the DW formula

is used, we get

3.0Q0.2for 8.3232.12

2.0Q0.1for 3.2918.12

p

2

p

2

<<−=

<<−=

pp

pp

Qh

and

Qh

(ii) Branching System(ii) Branching System

� The advantages are relatively few joints and thesystem is easy to build and design.

� The disadvantages are that sediments mayaccumulate at dead ends of the pipe. Secondly, ifthere is pipe bursts, a total cut off for zone beyondfailure results.

� This means that in case of bursts, the system will becut off.

� Also there is limitations in adding to the systembeyond a certain point.

� Because of these disadvantages, branch system isused in small community projects.

kiran

Sticky Note

f=?

kiran

Highlight

kiran

Highlight

kiran

Sticky Note

how?

kiran

Sticky Note

?


Solution: Computation TableSolution: Computation Table

Pipe

Sect.

Flow

(lires/s

econd)

Length

(m)

Pipe

Dia

mm

Head

Loss

(m/1000

m

Flow Vel

m/s

Head

Loss

(m)

Elev. of

hydr.

Grade

(m)

Groun

d level

elev

(m)

Pres

s

Hea

d

(m)

Rem.

AB 5.0 700 80 26 0.95 18.2 A

264.475

B

246.275

A-219

B-189

45.4

75

57.2

7 m

O.K

BC 3.0 825 65 27 0.88 22.27

5

B

246.275

C 224

B-189

C-219

57.2

7 m

5.0

m

Just

O.K

ChartChart An Alternative Approach….An Alternative Approach….

An alternative approach to avoid using the flow

correction equation given by Hardy Cross is to

directly use the Excel Solver to obtain the

suitable loop flow correction ΔQ that is required

to make the summation of the head losses

across any loop equal zero.

ExampleExample

Find the flows in the loop given the inflows and

outflows. The pipes are all 250 mm cast iron

(ε=0.26 mm)

� Assign a flow to each pipe link

� Flow into each junction must equal flow out of the

junction

kiran

Sticky Note


�Calculate the head loss in each pipe

� f = 0.02 for Re > 200000

�g = 9.81 m/s2

�D = 0.25 m

� The head loss around the loop isn’t zero

� Need to change the flow around the loop

� The clockwise flow is too great (head loss is positive)

� Reduce the clockwise flow to reduce the head loss

Solution TechniqueSolution Technique

� Use a numeric solver (Solver in Excel) to find a

change in flow that will give zero head loss

around the loop

�We must first input this problem into an Excel

Worksheet in a format that Solver can

understand.

Set up a spread sheetSet up a spread sheet

The numbers in bold were entered, the other cells are

calculations

Pipe f L (m) D (m) KQ0

(m3/s)hf (m)

1 0.02 200 0.25 338.4396 0.32 34.656

2 0.02 100 0.25 169.2198 0.04 0.271

3 0.02 200 0.25 338.4396 -0.1 -3.384

4 0.02 100 0.25 169.2198 0 0.000

Ʃ hf = 31.543

Now calculate Q0 and hf

Now calculate Qnew and hf new

for initially ∆Q is 0


Solver in ExcelSolver in Excel

�Now we can actually load the model into

Solver. Once we invoke Solver, we are

presented with the Solver Parameters

Dialog Box. Simply

Parameters Dialog Box

Solver Parameters Dialog Box

Enter the cell location of the Set Cell in the Set Cell box

Enter the cell locations of the Changing Cells in the By Changing Variable Cells box.

the OK button

Click on the Solve button to initiate the solution algorithm.Solver will then present you with a status box called the SolverResults Box. Click on Keep Solver Solution and then onthe OK button


After solving problem in Solver we get above results

∆Q -0.1231

Pipe f L (m) D (m) KQ0

(m3/s)hf (m)

Qnew

(m3/s)hf (m)

1 0.02 200 0.25 338.4396 0.32 34.656 0.1969 13.116

2 0.02 100 0.25 169.2198 0.04 0.271 -0.0831 1.170

3 0.02 200 0.25 338.4396 -0.10 -3.384 -0.2231 -16.851

4 0.02 100 0.25 169.2198 0.00 0.000 -0.1231 2.566

Ʃ hf = 31.543 0.000

Final Result Final Result

�You can refer following website for a Video

of explaining how to use solver for HCM

�http://www.youtube.com/watch?v=dQyYw

xrBTLE

Limitations of Hardy Cross Method (HCM)Limitations of Hardy Cross Method (HCM)

� You need first to guess the initial flow in all pipes and the

initial pipe flow should satisfy the continuity equations at

each node;

� It could take long period to converge especially for big

systems;

� Some times it fails to converge;

� Original method was restricted to closed looped systems;

� Original method did not simulate pumps and valves;

� Its coding manipulation is not in the matrices form

� Pipes 1 and 2 connect reservoir 1 [water level (W.L.) = 90.00 m which is

working like a sump], and reservoir 2 (W.L. = 100.00 m), respectively, to a

junction point J as shown in Fig. below. A third pipe, pipe 3, starts from the

junction point J and discharges into the atmosphere at point 3. A pump is also

fitted in pipe 1 as shown in the figure, characteristics curve of which may be

obtained by considering typical discharge-head data-0.0 m3/s: 18.0m, 0.1 m3/s:

16.5 m, 0.2 m3/s: 14.7 m, 0.3 m3/s: 11.9 m, and 0.4 m3/s: 8.0 m (You may

develop the corresponding head-discharge curve or the corresponding

discharge-head curve using MS Excel). It is seen that both curves are concave.

The length, diameter and CH-W coefficient values of the pipes are: Pipe 1 – 300

m, 300 mm, 100; pipe 2 – 250 m, 200 mm, 130; and pipe 3 – 120 m, 300 mm,

100. Determine the discharges in pipes 1 and 2 and the hydraulic gradient level

(HGL) values at points J and 3, when the outflow at point 3 is 0.1 m3/s. For

the pipe diameter in millimeters (D), pipe length in meters (L) and discharge in

m3/s, the resistance constant for a pipe may calculated using

and . 4.871.852WH

15

pipeDC

L104.351 R

×

××=

−1.852f RQh =

Characteristic Curves Characteristic Curves

� To completely describe the performance of pumps, a group of curves – head-capacity curve, power-capacity curve and efficiency-capacity curve – are used

� These curves known as pump characteristic curves are obtained from pumping tests and can be obtained from the manufactures.

� If P = power in kilowatts; γ= specific weight of the liquid in newtons per cubic metre; Q = discharge in cubic metres per second; hp = head in metres; and η = efficiency, then the power delivered by a pump is given by

1000η

γQhP

p=


ValvesValves

� Valves and the head losses through them can be

considered.

� Though all valves cause loss of head when flow takes

place through them, some valves are simple as far as their

inclusion in pipe network analysis is concerned.

� For example, gate valves, and such other valves control the

magnitude of flow through them. They can be considered

in the analysis by using appropriate head loss coefficients

as discussed in the previous problem depending upon how

far open the valves are.

ValvesValves

� It is assumed that the coefficients remain same regardless

of the Magnitude of the discharge and the direction of flow

through them.

� Some valves, on the other hand, have multiple actions.

They control the direction of flow through them, regulate

the flow to a constant value, and reduce the water pressure

to a present value. Check valves, flow-control valves and

pressure reducing valves are examples of this type and

require special treatment in the analysis.

� A check valve allows flow in one direction only. It opens

when the flow is in the desired direction and closes when

the flow is in the opposite direction, thus preventing flow.

Check Valves Check Valves

� Check valves are mainly used in suction and delivery sides of pumps.

It holds water in the suction pipe and thus obviates or reduces the

priming effort.

� Generally pumps are primed by filling them with water before they can

operate. The objective of priming is to remove a sufficient amount of

air from the pump and suction line to permit atmospheric pressure and

submergence pressure to cause water to flow into the pump when

pressure at the eye of the impeller is reduced below atmospheric as the

impeller rotates.

� Different situations arising due to the presence of a check valve:

�(a) When Hi > HJ, flow takes from i to j and flow in any pipe x

would be Qx > 0

�(b) When Hi <= HJ, the check valves closes and prevents flow from j

to i, so that flow in any pipe (x) be Qx = 0.

Check ValveCheck Valve

�Valve only allows flow in one direction

�The valve automatically closes when flow

begins to reverse

closedopen

Pressure Relief ValvePressure Relief Valve

Valve will begin to open when pressure in the pipeline ________ a set pressure (determined by force on the spring).

pipelineclosed

relief flow

open

exceeds

Low pipeline pressure High pipeline pressure

Where high pressure could cause an explosion (boilers, water heaters, …)


Pressure Regulating ValvePressure Regulating Valve

Valve will begin to open when the pressure ___________ is _________ than the setpoint pressure (determined by the force of the spring).

sets maximum pressure downstream

closed open

lessdownstream

High downstream pressure Low downstream pressure

Similar function to pressure break tank

Pressure Sustaining ValvePressure Sustaining Valve

Valve will begin to open when the pressure ________ is _________ than the setpoint pressure (determined by the force of the spring).

sets minimum pressure upstream

closed open

upstream greater

Low upstream pressure High upstream pressure

Similar to pressure relief valve

Flow control valve (FCV)Flow control valve (FCV)

�Limits the ____ ___ through the valve to a specified value, in a specified direction

�Commonly used to limit the maximum flow to a value that will not adversely affect the provider’s system

flow rate

ValvesValves

�Solve the earlier problem by assuming that

a check value (K=2.5) is fitted in pipe 2 as

shown in Fig. Take resistance factor R for

check valve

( )129.10

0.209.81π

2.58

gDπ

8KR

4242=

××

×==

WATER HAMMER

� When water flowing in a long pipe is suddenly brought to rest

by closing the valve or by any similar cause then there will

be sudden rise in the pressure of the moving water being

destroyed.

� This causes a wave of high pressure to be transmitted along

the pipe which creates noise called as KNOCKING.

� This phenomenon of sudden rise in pressure in the pipe is

known as water hammer or hammer blow.

When water is moving quickly through a pipe and the faucet is

turned off, the water is forced to stop. But instead of coming to

an abrupt stop, it bounces against the closed valve, creating a

wave of pressure that moves back and forth within the pipe.

The rate at which the wave bounces is determined by the pipe's

resonant frequency, a natural frequency of vibration that is, in

turn, determined by several factors, including its length and

shape.


�The rise in pressure in some cases may be so high that the pipe may even

burst and therefore essential to take into account this pressure rise in the

design of pipes.

The magnitude of the pressure rise depends on the following factors

- The speed with which the valve is closed.

- The velocity of flow

- The length of the pipe

- The elastic properties of pipe material as well as the flowing fluid.

The valve may be closed either gradually or instantaneously.

�The rise in pressure in some cases may be so high that the pipe may even

burst and therefore essential to take into account this pressure rise in the

design of pipes.

The magnitude of the pressure rise depends on the following factors

- The speed with which the valve is closed.

- The velocity of flow

- The length of the pipe

- The elastic properties of pipe material as well as the flowing fluid.

The valve may be closed either gradually or instantaneously.

According to the nature of closure the expressions for

calculating the pressure head due to water hammer may be

developed .

1.GRADUAL CLOSURE OF VALVE:

Let the pipe of length ‘L’ and cross sectional area ‘a’ carrying

a liquid flowing with a velocity. Due to gradual closure of the

valve let the liquid is brought to rest in time ‘t’ seconds

Therefore the total mass of liquid contained in the pipe is

w (aL) /g

It is assumed that rate of closure of valve is so adjusted that liquid column in

the pipe is brought to rest with a uniform retardation from an initial velocity to

zero in ‘t’ seconds .

The rate of retardation = (V / t)

The axial force available for producing retardation is

inertia pressure * cross sectional area

= pi*a

But the force = mass*retardation

pi*a = (waL)V / (gt)

=> (pi / w) = (LV / gt)

where (pi / w) = Inertia head

2. PRESSURE GROWTH DUE TO QUICK CLOSURE OF

THE VALVE IN AN ELASTIC PIPE

2. PRESSURE GROWTH DUE TO QUICK CLOSURE OF

THE VALVE IN AN ELASTIC PIPE

DistanceDistance

First truck stoppedFirst truck stopped

All trucks stoppedAll trucks stopped

As shown in the figure if the head of a loosely coupled trucks is

suddenly checked the following trucks still keep on moving until

all the gap between them has been covered and then whole train

will come to rest i.e. last truck will travel certain distance even

after the first truck has stopped.

(A)

Normal flow condition of the liquid in the pipe with normal

velocity ‘V’ under normal pressure ‘p’ at the moment just before

the valve at the end ‘B’ is instantaneously shut.

(B)

pi/w

M B

Condition immediately after the valve at B is closed. Thus a wave

of inertia pressure has begun to travel with velocity Vo in the

upstream direction which results in compressing the water and

expanding the pipe. Mean while the liquid on the left of

advancing wave continues to move as nothing has happened to it.


©

HGL

Vo

M B

Wave advanced further on the upstream side with a velocity ‘Vo’

(D)

M M’ B

dL D+dD

Now liquid in the entire pipe is at rest under the pressure p + pi.

The whole of pipe having been distended under this pressure.

If dt is the time required to stop the liquid column. So in this time the liquid

column moves through a distance dL.

Let dqc= the volume by which the liquid is compressed due to pressure pi

dqe = the additional volume provided by stretching of the pipe walls under

pressure pi.

dD = the resulting increase in the pipe.

dt = time for pressure wave to traverse the pipe.

T = the thickness of wall.

E = Young’s modulus for the material of the pipe.

ft = tensile hoop stress in the pipe walls due to pressure pi.

1/m = poisson’s ratio

The pressure rise due to quick closure of the valve causes radial

expansion of the walls of an elastic pipe. The longitudinal and

circumferintial stresses are produced. The strain in the direction of hoop stress or circumferential strain is

given by

dD/D = ft(1- (1/2m)) / E

as circumferential or hoop stress ft = piD/2T

dQ = dqe+ dqc

on substituting the values in the above expression the inertia

pressure is given by

pi = wLV/gdt

−+

=

=

2m

11

TE

D

k

1

w

g

Vp

gp

wLV dt

i

i

If elasticity of the pipe material is not considered then the term

D/TE(1- 1/2m) may be neglected. Then the equation becomes

Pi = V (wK/g)

The above equation represents the maximum pressure rise for

instantaneous closure of the valve . However maximum pressure

rise is independent of dimensions of the pipe but depends on the

initial velocity of flow of liquid and its physical characteristics.

The velocity of the compressive wave Vo = L/dt. Again if the

term D/TE(1 – 1/2m) is neglected then Vo = gK/w

ρ

k

w

gkV

2m

11

TE

D

k

1

w

g V

o

o

==

−+=

Which is identical with the velocity of the propagation of sound wave in the

liquid.

Rapid closure is meant any time t <= 2L/Vo

Valve closure is said to be gradual if t > 2L/Vo

where Vo is the velocity of pressure wave.

By substituting Vo in the expression for Pi the expression for the inertia head

may be obtained.

hi = Pi/w = (V Vo)/g

This is known as Allievi formula.

Further as an approximation if poisson’s ratio is not considered then

circumferential strain is given by


Pi = V/ g/w(1/K + D/TE)

If the elasticity of the pipe material is not considered the term D/TE

may be neglected.

SURGE TANK:

The surge tank is a device to protect the pipe line from bursting due

to pressure rise when the water in the pipe is suddenly brought to

rest. This device is generally fitted in the pipe line which feeds

water to the turbine. The surge tank consists of a storage reservoir

fitted to the pipe line of a hydraulic power plant. When for any

reason the lower end

Pi = V/ g/w(1/K + D/TE)

If the elasticity of the pipe material is not considered the term D/TE

may be neglected.

SURGE TANK:

The surge tank is a device to protect the pipe line from bursting due

to pressure rise when the water in the pipe is suddenly brought to

rest. This device is generally fitted in the pipe line which feeds

water to the turbine. The surge tank consists of a storage reservoir

fitted to the pipe line of a hydraulic power plant. When for any

reason the lower end

If the valve of pipe is closed, the quantity of water which is rejected

by the pipe line is allowed to flow into the surge tank. By this

arrangement the pipe line is relieved to high inertia pressure.

If the valve of pipe is closed, the quantity of water which is rejected

by the pipe line is allowed to flow into the surge tank. By this

arrangement the pipe line is relieved to high inertia pressure.

Pipeline systems:

Pipe networks

Pipeline systems:

Pipe networks

�Water distribution systems for municipalities

�Multiple sources and multiple sinks connected

with an interconnected network of pipes.

�Computer solutions!

�KYpipes

�WaterCAD

�EPANET http://www.epa.gov/ORD/NRMRL/wswrd/epanet.html

Pipes in ParallelPipes in Parallel

A B

Q1

Qtotal

energy

proportion

�Find discharge given pressure at A and B

�______& ____ equation

�add flows

�Find head loss given the total flow

�assume a discharge Q1’ through pipe 1

�solve for head loss using the assumed discharge

�using the calculated head loss to find Q2’

�assume that the actual flow is divided in the same

_________ as the assumed flow

Q2

S-J

Pressure Break TanksPressure Break Tanks

� In the developing world small water supplies in

mountainous regions can develop too much

pressure for the PVC pipe.

�They don’t want to use PRVs because they are too

expensive and are prone to failure.

�Pressure break tanks have an inlet, an outlet, and

an overflow.

� Is there a better solution?

Pipeline DesignPipeline Design

�The selection of pipes is an economictradeoff between large diameter which will

give high capital cost and low friction lossesand low pumping costs (if there is pumping)OR small diameter, which will involve low

capital cost, more head losses and morepumping cost.

�Energy cost is a function of head losses whilepipe cost is a function of diameter.


Allowable Head LossesAllowable Head Losses

� (i) Allow 1 m (for big pipes) to 10 m (small pipes)head loss per 1000 m of mainline

� (ii) Using velocity as criteria as head loss effects isrelated to velocity.

� Normal practice in water supply for irrigation is tokeep velocity within 0.6 to 1.5 m/s. Above that, therecan be ‘water hammer’ or high rates of corrosion.Water hammer is transient high pressure waves dueto rapid valve closure. Below 0.6 m/s, there may besilting or sediment deposition.

� Pipe diameter can be chosen using head losses andvelocity using charts or equations.

Pipe Layout: Types of Distribution SystemsPipe Layout: Types of Distribution Systems

�(i) Individual Pipes: Connectstwo points in the distribution

system say from a reservoir tothe point of use.

Example 1: A reservoir is situated 65 m vertically above some farm buildings.

The length of pipe required to lead water from the reservoir is 750 m and thepressure required at the buildings is 30 m head. Rate of flow required is 2 m3/h

(2000 litre/hr).Solution:Solution:

� If the head available due to the height of thereservoir is 65 m, and the pressure head needed atthe buildings is 30 m, the head available forovercoming friction is 65 – 30 = 35 m being thedifference in head between the ends of the pipe.

� The equivalent length of the pipe is:

� Actual length (750 m) + 10% (75 m)

= 825 m

� Plus (say) 1 tap + 2 stop taps = 135 m

� Total = 1060 m

Solution ConcludedSolution Concluded

�The hydraulic gradient is Pressure difference/equivalent length = 35/1060 = 1/30

�Since the maximum head is 65 m, a Class C(9 bar or 90 m) pipe is required, and referringto Chart provided, it can be seen that a 32mm nominal (internal) diameter Class C lowdensity polythene pipe would satisfy theserequirements.

�Velocity is about 0.8 m/s which is acceptable(within 0.6 and 1.5 m/s).

Thank You