For Hardy-Cross Analysis - University of Alabamarpitt.eng.ua.edu/Class/Water Resources Engineering/M4b Water... · Steps for Setting Up and Solving a Water Distribution System using

1

Module 4b: Water Distribution System

Design

Hardy Cross M

ethod

Robert Pitt

University of Alabam

a

and

Shirley Clark

Penn State -Harrisburg

Hardy Cross M

ethod

•Used in design and analysis of water

distribution systems for many years..

•Based on the hydraulic form

ulas we reviewed

earlier in the term

.

For Hardy-Cross Analysis:

•Water is actually rem

oved from the distribution system

of a

city at a very large number of points.

•Its is not reasonable to attem

pt to analyze a system with this

degree of detail

•Rather, individual flows are concentrated at a sm

aller number

of points, commonly at the intersection of streets.

•The distribution system can then be considered to consist of a

network of nodes (corresponding to points of concentrated

flow withdrawal) and links (pipes connecting the nodes).

•The estimated water consumption of the areas contained

within the links is distributed to the appropriate nodes

Netw

ork

mo

del o

verl

aid

on

aeri

al p

ho

tog

rap

h

(Walski, et al.2004 figure 3.4)

2

Skele

ton

izati

on

-A

n a

ll-l

ink n

etw

ork


Min

imal skele

ton

izati

on


Mo

dera

te s

kele

ton

izati

on


Maxim

um

skele

ton

izati

on


3


Cu

sto

mers

mu

st

be s

erv

ed

fro

m s

ep

ara

te p

ressu

re z

on

es

Pro

file

of

pre

ssu

re z

on

es


Pre

ssu

re z

on

e t

op

og

rap

hic

map


Imp

ort

an

t ta

nk

ele

vati

on

s


4

Sch

em

ati

c n

etw

ork

illu

str

ati

ng

th

e u

se o

f a p

ressu

re

red

ucin

g v

alv

e


Hardy-C

ross M

ethod of Water Distribution

Design

•Definitions

–Pipe sections or links

are the most abundant elem

ents in the

network.

•These sections are constant in diameter and m

ay contain fittingsand

other appurtenances.

•Pipes are the largest capital investm

ent in the distribution system

.

–Noderefers to either end of a pipe.

•Two categories of nodes are junction nodes

and fixed-grade nodes.

•Nodes where the inflow or outflow is known are referred to as

junction nodes. These nodes have lumped dem

and, which m

ay vary

with tim

e.

•Nodes to which a reservoir is attached are referred to as fixed-grade

nodes. These nodes can take the form

of tanks or large constant-

pressure m

ains.

Steps for Setting Up and Solving a W

ater Distribution

System

using the Hardy-Cross M

ethod

1.

Set up grid network to resem

ble planned flow

distribution pattern.

Steps for the Hardy-Cross M

ethod

2.

Calculate water use on each street (including fire dem

and

on the street where it should be located).

0.0

0.0

0.0

0.0

No buildings

11

0.080

0.052

0.080

0.052

8 A

10

0.070

0.045

0.070

0.045

7 A

9

0.020

0.013

0.020

0.013

2 A

8

0.100

0.065

0.100

0.065

10 A

7

0.070

0.045

0.070

0.045

7 A

6

0.020

0.013

0.020

0.013

2 A

5

0.086

0.056

0.086

0.056

3 A, 1 O, 1 C

4

0.18

0.12

0.18

0.12

18 A

3

0.030

0.019

0.030

0.019

3 A

2

3.10

2.00

0.092

0.059

5 A, 1 S

1**

ft3/sec

MGD

ft3/sec

MGD

With Fire Dem

and (worst

building)

Without Fire Dem

and

Building

Description

Street

Number

5


ethod

3.

Add up the flow used in the neighborhood without fire

dem

and and distribute it out the nodes where known

outflow is required. Repeat for fire dem

and.

Total without Fire Dem

and = 0.75 cfs

Influent = 2.5 M

GD = 3.87 cfs

Left Over to Other Neighborhoods = 3.12 cfs

Distribute 50/50 to two outflow nodes = 1.56 cfs

(arbitrary for this problem –

would be based on known

“downstream

”requirem

ents).


ethod

4.

Assume internally consistent distribution of flow,

i.e., at any given node and for the overall water

distribution system

:

Σflow entering node = Σ

flow leaving node


ethod

5.

For the inflow node,

split the flow among the

pipes leaving that node

(there will be no

additional outflow since

no water has been used

by the neighborhood as

yet).

Inflow = Σ

Outflows


ethod

6.

For each of the pipes

leaving the inflow

node, put the water

dem

and for that

street at the node at

the end of the pipe.

6


ethod

7.

For the above node and the next pipes in the distribution

system

, subtract the water used on the street (and

aggregated at the node) from the water flowing down the

pipe. Pass the remaining water along to one or more of

the pipes connected to that node.

8.

Repeat Steps 6 and 7 for each pipe and node in the

distribution system. The calculation can be checked by

seeing if the total water outflow from the system equals

the total inflow to the system, as well as checking each

node to see if inflow equals outflow.


ethod

9a. Check each node to see if inflow equals outflow.


ethod

9b. Check each node to see if inflow equals outflow.


ethod

10.

Select initial pipe sizes (assume a velocity of 3 ft/sec for norm

al flow with no fire

dem

and). W

ith a known/assumed flow and an assumed velocity, use the

continuity equation (Q = VA) to calculate the cross-sectional area of flow. (w

hen

conducting computer design, set diameters to m

inim

um allowable diameters for

each type of neighborhood according to local regulations)

5.9

5.9

7.2

8.2

3.7

7.5

7.8

10.9

7.5

4.2

10.9

Diameter

(in)

Actual D

(in)

Area (ft2)

60.49

0.19

3.0

0.562

11

60.49

0.19

3.0

0.562

10

80.60

0.28

3.0

0.852

9

10

0.68

0.37

3.0

1.100

8

40.31

0.07

3.0

0.220

7

80.63

0.31

3.0

0.922

6

80.65

0.33

3.0

1.000

5

12

0.91

0.64

3.0

1.930

4

80.63

0.31

3.0

0.922

3

60.35

0.10

3.0

0.288

2

12

0.91

0.65

3.0

1.94

1

Diameter

(ft)

Velocity

(ft/sec)

Flow (ft3/sec)

Pipe

Number

7


ethod

11.

Determine the convention for flow. Generally, clockwise

flows are positive and counter-clockwise flows are negative.


ethod

12.

Paying attention to sign (+/-), compute the head loss in each

elem

ent/pipe of the system (such as by using Darcy-

Weisbach or Hazen-W

illiam

s).

= ===

− −−−

= ===

− −−−

g

V

DLf

h

Weisbach

Darcy

CD

QL

h

Williams

Hazen

LL

2

432

.0

2

85

.1

63

.2


ethod

13.

Compute the sum of the head losses around each

loop (carrying the appropriate sign throughout the

calculation).

14.

Compute the quantity, head loss/flow (hL/Q), for

each element/pipe (note that the signs cancel out,

leaving a positive number).

15.

Compute the sum of the (h

L/Q)sfor each loop.


ethod

16.

Compute the correction for each loop.

where

n = 1.85

∑ ∑∑∑∑ ∑∑∑− −−−

= ===

loop

L

loopL Qh

n

h

Q∆

8


ethod

17.

Apply the correction for each pipe in the loop that is

not shared with another loop.

Q1= Q

0+ ∆Q

18.

For those pipes that are shared, apply the following

correction equation (continuing to carry all the

appropriate signs on the flow):

Q1= Q

0+ ∆Q

loopin–∆Q

shared

loop


ethod

19.

Reiterate until corrections are sufficiently small (10 –15% or

less of sm

allest flow in system), or until oscillation occurs.

20.

Calculate velocities in each pipe and compare to standards to

ensure that sufficient velocity (and pressure) are available in

each pipe. Adjust pipe sizes to reduce or increase velocities

as needed.

21.

Repeat all the above steps until a satisfactory solution is

obtained.

22.

Apply fire flow and other conditions that m

ay be critical and

reevaluate.

Exam

ple for the Hardy-Cross M

ethod

(From M

cGhee, Water Supply and Sewerage, Sixth Edition)

Exam


ethod

Convert units to U.S. Customary units:

9

Exam


ethod

•Insert data into spreadsheet for Hardy-Cross (solve using

Hazen-W

illiam

s).

•ASSUME: Pipes are 20-year old cast iron, so C = 100.

3.29

10

2625

AB

-2.6

12

3609

DA

-5.54

16

3281

CD

6.23

18

4921

BC

Flow

0(ft3/sec)

Pipe Diameter

(in)

Pipe Length (ft)

Pipe Section


ethod

•Paying attention to sign (+/-), compute the head loss in each

elem

ent/pipe of the system by using Hazen-W

illiam

s (check

that the sign for the head loss is the same as the sign for the

flow).

85

.1

63

.2

432

.0

=

−

CD

QL

h

Williams

Hazen

L

Exam


ethod

•Calculate head loss using Hazen-W

illiam

s.

3.29

-2.6

-5.54

6.23

Flow

0(ft3/sec)

54.39

10

2625

AB

-19.93

12

3609

DA

-18.11

16

3281

CD

19.03

18

4921

BC

hL(ft)

Pipe Diameter

(in)

Pipe Length

(ft)

Pipe

Section

Exam


ethod

•Calculate h

L/Q for each pipe (all of these ratios have

positive signs, as the negative values for hLand Q

cancel out).

16.53

54.39

3.29

AB

7.66

-19.93

-2.6

DA

3.27

-18.11

-5.54

CD

3.05

19.03

6.23

BC

hL/Q (sec/ft2)

hL(ft)

Flow

0(ft3/sec)

Pipe Section

10

Exam


ethod

•Calculate head loss using Hazen-W

illiam

s and column totals:

16.53

54.39

AB

Σ(h

L/Q) = 30.51

ΣhL= 35.38

7.66

-19.93

DA

3.27

-18.11

CD

3.05

19.03

BC

hL/Q (sec/ft2)

hL(ft)

Pipe Section

Exam


ethod

•Calculate the correction factor for each pipe in the loop.

where n = 1.85

∑ ∑∑∑∑ ∑∑∑− −−−

= ===

loop

L

loopL Qh

n

h

Q∆

= -(35.38)/1.85(30.51) = -0.627

Exam


ethod

•Calculate the new

flows for each pipe using the following

equation:

Q1= Q

0+ ∆Q

2.66

-0.627

3.29

AB

-3.23

-0.627

-2.6

DA

-6.17

-0.627

-5.54

CD

5.60

-0.627

6.23

BC

Flow

1(ft3/sec)

∆Q (ft3/sec)

Flow

0(ft3/sec)

Pipe Section

Exam


ethod

HA

RD

Y C

RO

SS

ME

TH

OD

FO

R W

AT

ER

SU

PP

LY

DIS

TR

IBU

TIO

N

Tri

al

I

Pipe Section

Pipe Length

(ft)

Pipe Diameter

(in)

Flow

0

(ft3/sec)

HL (ft)

HL/Q

(sec/ft2)

nΣ(H

L/Q)

(sec/ft2)

Σ(H

L) (ft)

∆Q (ft3/sec)

Flow

1

(ft3/sec)

BC

4921

18

6.2

19.03

3.05

-0.627

5.60

CD

3281

16

-5.5

-18.11

3.27

-0.627

-6.17

DA

3609

12

-2.6

-19.93

7.66

-0.627

-3.23

BA

2625

10

3.3

54.39

16.53

-0.627

2.66

Tri

al

2

Pipe Section

Pipe Length

(ft)

Pipe Diameter

(in)

Flow

1

(ft3/sec)

HL (ft)

HL/Q

(sec/ft2)

nΣ(H

L/Q)

(sec/ft2)

Σ(H

L) (ft)

∆Q (ft3/sec)

Flow

2

(ft3/sec)

BC

4921

18

5.60

15.64

2.79

-0.012

5.59

CD

3281

16

-6.17

-22.08

3.58

-0.012

-6.18

DA

3609

12

-3.23

-29.71

9.21

-0.012

-3.24

BA

2625

10

2.66

36.79

13.81

-0.012

2.65

Tri

al

3

Pipe Section

Pipe Length

(ft)

Pipe Diameter

(in)

Flow

2

(ft3/sec)

HL (ft)

HL/Q

(sec/ft2)

nΣ(H

L/Q)

(sec/ft2)

Σ(H

L) (ft)

∆Q (ft3/sec)

Flow

3

(ft3/sec)

BC

4921

18

5.59

15.58

2.79

0.000

5.59

CD

3281

16

-6.18

-22.16

3.59

0.000

-6.18

DA

3609

12

-3.24

-29.91

9.24

0.000

-3.24

BA

2625

10

2.65

36.49

13.76

0.000

2.65

54.34

0.00

56.46

35.38

54.38

0.64

11

Final Flows for the Hardy-Cross Example

Pressure Water Distribution System

The pressure at any node can be calculated by starting with

a known pressure at one node and subtracting the absolute

values of the head losses along the links in the direction of

flow.

In this example, assume that the pressure head at node C

is 100 ft. and the pressure head at node A is desired.

There are two paths between the known and unknown

nodes for this example and both should be examined: CB

and BA or CD and DA.

In the first case: 100 ft. –15.58 ft. –36.49 ft. = 47.93 ft.

And in the second: 100 ft. –22.16 ft. –29.91 ft. = 47.93 ft.

Exte

nd

ed

peri

od

sim

ula

tio

n (

EP

S)

run

s s

ho

win

g lo

w

pre

ssu

re d

ue t

o e

lev

ati

on

or

syste

m c

ap

acit

y p

rob

lem


Pre

ssu

re c

om

pari

so

n f

or

6-,

8-,

12-,

an

d 1

6-

inch

pip

es


12

Head

lo

ss c

om

pari

so

n f

or

6-,

8-,

12-,

an

d 1

6-

inch

pip

es


Chemical Reactions in Water

Distribution Systems

•Many of the water distribution m

odels include water

quality subcomponents.

•These are used to determine the age of the water in

the distribution system

and the mixing of water from

different sources at the different locations.

•This inform

ation is used to calculate the resulting

concentrations of conservative and nonconservative

compounds in the water.

•Knowledge of the fate of disinfectants in the

distribution system

s is very important to ensure safe

drinking water: we need to ensure that sufficient

concentrations of the disinfectants exist at all

locations in the system

.

Co

ncep

tual illu

str

ati

on

of

co

ncen

trati

on

vs. ti

me f

or

zero

,

firs

t, a

nd

seco

nd

-ord

er

decay r

ea

cti

on

s

(Walski, et al. 2001 figure 2.23)

Dis

infe

cta

nt

reacti

on

s o

ccu

rrin

g w

ith

in a

typ

ical

dis

trib

uti

on

syste

m p

ipe

N.O.M. = natural organic matter


13

Another Example for the Hardy-Cross Method

EPANetWater Distribution Model

Documents

For Hardy-Cross Analysis - University of Alabamarpitt.eng.ua.edu/Class/Water Resources Engineering/M4b Water... · Steps for Setting Up and Solving a Water Distribution System using