Pipe NetworksPipe Networks

Dr. Kristoph-Dietrich KinzliDepartment of Environmental and Civil Engineering

Florida Gulf Coast University

CWR 4540 C Fall 2011

Sept 8, 2011

Magdeburg Water Bridge Wasserstraßenkreuz opened in October 2003 connects the Elbe-Havel Kanal to the

Mittellandkanal, crossing over the Elbe River. longest navigable aqueduct in the world, with a

total length of 918 meters (3,012 ft) previously 12-kilometre (7.5 mi) detour,

offloading due to water levels in Elbe 1350 tons vs 800 tons 500 million Euros – 6 years 24,000 tons of steel and 68,000 cubic meters of

concrete

Magdeburg Water Bridge

Magdeburg Water Bridge

Magdeburg Water Bridge

Magdeburg Water Bridge

Magdeburg Water Bridge

Magdeburg Water Bridge

Magdeburg Water Bridge

Introduction – Pipe Networks The purpose of a water distribution network is to

supply the system’s users with the amount of water demanded and to supply this water with adequate pressure under various loading conditions.

Water distribution system have three major components; pumping station, distribution storage and distribution piping.

Performance criteria are generally minimum flow rates and pressure (why?)

1. Apply continuity and energy equations to pipe system networks

2. Analyze pipe flow and pressure in a network

3. Illustrate the iterative solution of the loop equations using the Hardy Cross Method

4. Develop a spreadsheet to solve a simple water distribution system using the Hardy-Cross method

Today’s Learning Objectives

The energy equation is written for each pipeline in the network as Nodal Method:

pm hQ

Q

gA

QQK

D

fLhh

212

2

h1 : head at the upstream end of a pipe (m) h2 : head at the downstream end of a pipe (m) hp : head added by pumps in the pipeline (m) A : cross-sectional area of a pipe (m2)

Friction loss

local losses

Positive flow direction is from node 1 to node 2 Application is limited to relatively simple networks

Example 2.9 (Chin, pg 41-43)

The high-pressure ductile-iron pipeline shown in Figure 2.10 becomes divided at point B and rejoins at point C. The pipeline characteristics are given in the following tables. If the flowrate in Pipe 1 is 2 m3/s and the pressure at point A is 900 kPa, calculate the pressure at point D. Assume that the flows are fully turbulent in all pipes.

D

k

f

S

7.3log2

1ks

Pressure(P)

Flowrate(Q)

Step 1: Calculate ks/D and f for each pipe

Step 2: Calculate the total energy head at location A, hA

Step 3: Write down the energy equations for each pipe

Step 4: Use continuity equations at two pipe junctions (Qin = Qout1 + Qout2)

Step 5: Determine Q of each pipe and hD

Step 6: Determine PD using the calculated hD

Example 2.9 (Chin, pg 41-43) – cont.

In-class activity (Example problem)

The pipe system shown in the figure below connects reservoirs that have an elevation difference of 20 m. This pipe system consists of 200 m of 50-cm concrete pipe (pipe A), that branches into 400 m of 20-cm pipe (pipe B) and 400 m of 40-cm pipe (pipe C) in parallel. Pipes B and C join into a single 50-cm pipe that is 500 m long (pipe D). For f = 0.030 in all the pipes, what is the flow rate in each pipe of the system?

Hardy Cross Method (Cross, 1936) The Hardy Cross method is a simple technique for hand

solution of the loop system of equations in pipe networks  The flowrate in each pipe is adjusted iteratively until all

equations are satisfied.  The method is based on two primary physical laws:

The sum of pipe flows into and out of a node equals the flow entering or leaving the system through the node.

the algebraic sum of the head losses within each loop is equal to zero.

0)( ,

1

,

jp

n

j

jL hhhL,j : head loss in pipe j of a loop hp,j : head added by any pumps in pipe jn : the number of pipes in a loop

The total flow entering each point (node) must equal to the total flow leaving that joint (node).

ΣQin = ΣQout

The flow in a pipe must follow the pipe’s friction law for the pipe.

hf = kQn

n is coefficient depend on equation used The algebra sum of the head loss in the pipe

network must be zero.

Σ hf = 0

For analysis, basic criteria should be followed:

General equation for losses, hf = rQn (SI units) Darcy-Weisbach

Hazen Williams

For analysis, basic criteria should be followed:

25

2

122Q

D

Lf

g

V

D

Lfh f

85.185.187.4

85.1

17.1

67.1082.6 Q

CD

L

C

V

D

Lh

HHf

Q : flow rate (m3/s),

D : pipe diameter (m)

L : length of pipe (m)

f : friction factorCH : H-W

coefficient

General equation for losses, hf = rQn (US units) Darcy-Weisbach

Hazen William

For analysis, basic criteria should be followed:

25

2

7.392Q

D

Lf

g

V

D

Lfh f

85.185.187.4

85.1

17.1

73.482.6 Q

CD

L

C

V

D

Lh

HHf

Q : flow rate (ft3/s), D : pipe diameter

(ft)L : length of pipe

(ft)f : friction factorCH : H-W

coefficient

Hardy-Cross Method (Derivation)

na

nf

QQr

rQh

n

j

n

jaj

n

j

n

jajaj

Qnr

QQr

Q

1

1

,

1

1

,,

For Closed Loop:

])()(

!2

1[ 221 nn

ana

na QQQ

nnQnQQr

n=2.0, Darcy-Weisbach n=1.85, Hazen-Williams

QrnQQr na

na 1

If ΔQ is small, higher order terms in ΔQ can be neglected

QQQ a Qa : assumed flowrate ΔQ : error in the

assumption Q : actual flowrate

QQrnQQrn

a

n

a 11

0, jLh

Step 1: Build up system configuration and assume a reasonable distribution of flows the pipe network. but Q=0 at each node.

Step 2: Calculate head loss of each pipe section with D-W or H-W equ.

Step 3: Calculate (rQ|Q|n-1) and (n r |Q|n-1) in each loop of the pipe network.

Step 4: Get ∆Q value for correction

Step 5: Calculate the new flow rate Qnew = Q + ∆Q

Step 6: Repeat this step (3 ~ 5) until ∆Q = 0

Step 7: Check the mass and energy balance.

Step 8: Compute the pressure distribution in the network and check on the pressure requirement.

Steps for Q distribution using Hardy Cross method:

n

j

n

jaj

n

j

n

jajaj

Qnr

QQr

Q

1

1

,

1

1

,,

Example 2.10 (Chin, pg 46-48)

Compute the distribution of flows in the pipe network shown in Figure 2.11 (a), where the head loss in each pipe is given by , and the relative values of r are shown in Figure 2.11 (a). The flows are taken as dimensionless for the sake of illustration.

2QrhL

Given condition

Assumed Q and direction

In-class activity (Pipe networks)- cont.

100 cfs

25 cfs 50cfs

25 cfs

r = 3 r =

4

r = 1 r = 4

r = 5

r =

2

Loop ILoop

II

In-class activity (Pipe networks)- cont.

100 cfs

25 cfs 50cfs

25 cfs

r = 3 r =

4

r = 1 r = 4

r = 5

r =

2

61.7 cfs

12

.1

cfs

38.3 cfs

35.2 cfs

61.7 cfs

14

.8

cfs

Loop ILoop

II

In-class activity (Pipe networks) 2QrhL

1.5 cfs

1 cfs

r = 15r

= 1

1r = 8

r = 6 r = 25

r =

8Determine the flow rate in each pipe of the system with direction, with the data shown!

0.5 cfs