PIPE NETWORKS 1.0
Introduction
Analysis of pipe networks networks is important important in the design of of water distribution systems. systems. Such analysis can be performed by applying Bernoulli’s equation and the continuity equation but, for large networks, the solution of these equations involves lengthy calculations. Ex: Analysis of the pipe network shown below requires the solution of of 13 nonlinear simultaneous equations.
Hence the analysis of pipe networks is usually based on other methods such as Head Balancing Method, Quantity Balancing Method, Electrical Analogue Method etc.
1.1
Flow in Pipe Networks
The flow in a pipe network has to satisfy two conditions: i
Continuity of flow at a node At a node, node, Total inflow inflow
=
Total outflow outflow
If inflow is taken as positive (+) and outflow is taken as negative (-), this condition can be expressed as
Q 0 where
………………………………………...……………………………. (1) Q = Flow rate
Ex: For the node A shown below 1 m /s
5 m /s A
4 m /s
3 m /s 3 m /s
Q +5 m3/s + 3 m3/s –1 m3/s – 4 m3/s – 3 m3/s = 0 1
ii
The head difference between any two points in the network is the same for all paths between them. If the flow rate Q and the corresponding head loss H L in clockwise sense are taken as positive (+), for a closed loop this condition can be expressed as
H 0 ………………………………………………………………..……. (2) L
Ex: For the closed loop ABCD in the pipe network shown below, A
B
D
C
H (H ) L
L AB
(H L )BC (H L )CD (H L )DA
H A H B H B H C H C H D H D H A
0 In addition, the head loss in a pipe can be expressed in the form H L KQ n ………………………………………..…………………………………… (3)
where K and n are constants for a given pipe. Ex: Local loss H L k 1 K
8k 1 2
π gD
4
v 2 2g
k 1
Q
2g πD 2
2
8k 1 Q 2 KQ n 2 4 π gD 4
and n = 2
Friction loss: Darcy-Weisbach equation: 2
Q 8 λL 2 2 5 Q 2 KQ n H L λ D 2g 2gD D 4 gD L v 2
K
8 λL
2
gD 5
λL
and n = 2
2
Friction loss: Hazen-Williams equation: v 0.355CD0.63 (H L / L )0.54 H L
K
6.79 L V 1.85 1.85
C
10.61L C 1.85 D 3.17
where
1.2
D
1.17
10.61 L Q 1.85 1.85
C
D
3.17
KQ n
and n = 1.85
k 1 = g = D = λ = L = C =
loss coefficient acceleration due to gravity pipe diameter friction factor pipe length Hazen-Williams coefficient
Head Balancing Method
This method, also known as loop method and presented by Hardy-Cross in 1936, can be used when the inflows and outflows of a network are known and the heads at nodes and flow rates in pipes of the network are required to be determined. Step 1
Flow rates Q satisfying the continuity of flow at nodes (equation (1)) are assumed for all the pipes in the network.
Step 2
Head losses H L are calculated (equation (3)) and the condition H L 0 (equation (2)) is checked for all the loops in the network.
Step 3
If
H 0 , assumed Q values are taken to be correct values.
If
H 0 , correction ΔQ is applied for assumed Q values.
L
L
ΔQ is calculated as follows:
H L KQ n
With assumed flow rates Q,
[KQ
with corrected flow rates Q+ ΔQ ,
3
n
] 0 and
[K (Q ΔQ)
n
]0
[KQ
n
[KQ
n
KnQ n 1 ΔQ K
n(n 1)
Q n 2 ΔQ 2 ... K ΔQ n ] 0
2 As ΔQ << Q , neglecting higher powers of ΔQ ,
KnQ n1ΔQ] 0
KQ KQ ΔQ nKQ n KQ
n
n 1
n 1
KQ n 1
As H L KQ n ,
ΔQ
n
H L Q
H H n Q L
L
For a pipe shared by two loops, corrections from both loops are applied. Step 4
Steps 2 and 3 are repeated until ΔQ is negligible.
It has been found that this method converges rapidly although a high storage capacity in computing is required. It is convenient to perform the calculation in tabular form as illustrated by example (1). Example (1)
Determine the flow rates in the pipes of the network shown in Figure Ex(1). The dimensions of the pipes are as given in Table Ex(1a). The friction factor λ for all pipes can be taken as 0.0 2 and the local losses can be neglected. Pipe AB AD BC CD CF DE EF
Length (m) 700 160 160 700 160 160 700 Table Ex(1a)
4
Diameter (m) 0.20 0.25 0.15 0.25 0.25 0.15 0.15
0.05 m /s
0.008 m /s A
B
D
C
E 1
2
Figure Ex(1)
F
0.042 m /s
Check total inflow and outflow. Inflow
=
0.05 m 3/s
Outflow
=
0.008 m 3/s + 0.042 m 3/s = 0.05 m 3/s = Inflow
Express head loss H L KQ n determine K and n for all pipes. 2
Q 8 λL H L H f λ Q 2 KQ n 2 2 5 D 2g 2gD πD 4 π gD L v 2
K
8 λL 2
π gD
5
λL
and n = 2
Prepare Table Ex(1b) and complete columns (1) and (2).
3
Assume flow rates Q for all pipes. Note: Sign convention: In a loop, clockwise Q and H L are positive (+) Let Q AB = 0.02 m 3/s QBC = 0.012 m 3/s and Q AD = 0.03 m 3/s Let QDC = 0.016 m3/s QCF = 0.028 m 3/s and QDE = 0.014 m 3/s= QEF i.e. complete column (3) in Table Ex(1b).
4
Calculate H L KQ n for all pipes. i.e. complete column (4)-from columns (2) and (3)-in Table Ex(1b).
5
5
Check the condition
H 0 for both loops. L
H 1.4 0 and for loop DCFE H 3.155 0 H for both loops. Apply correction ΔQ H n Q For loop ABCD
L
L
L
L
To determine ΔQ , calculate H L /Q for all pipes in a loop and obtain
H L Q
I.e. complete column (5) in Table Ex(1b).
For loop ABCD, ΔQ = -(1.4)/(2×141.12) m 3/s = -0.00496 m 3/s For loop DCFE, ΔQ = -(-3.155)/(2×288.66) m 3/s = 0.00546 m 3/s For pipe CD in loop ABCD,
= -0.00496 m 3/s + (-0.00546 m 3/s)
ΔQ
= -0.01042 m 3/s For pipe DC in loop DCFE ,
= 0.00546 m 3/s + (+0.00496 m 3/s)
ΔQ
= 0.01042 m 3/s Apply correction ΔQ for assumed Q values. i.e. complete column (6) in Table Ex(1b).
7
Repeat steps 4 to 6 until ΔQ can be neglected. i.e. complete columns (7), (8), (9),… in Table Ex(1b) until ΔQ is negligible.
6
(1)
(2)
Pipe
D C B A p o o L
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
K (s /m )
Q (m /s) (assumed)
H L (m)
H L /Q 2 (s/m )
Q (m /s) (corrected)
H L (m)
H L /Q 2 (s/m )
Q (m /s) (corrected)
H L (m)
AB
3615
+0.020
+1.446
72.30
+0.01504
+0.818
54.39
BC
3482
+0.012
+0.501
41.75
+0.00704
+0.173
24.57
CD
1185
-0.016
-0.303
18.94
-0.02642
-0.827
31.30
DA
271
-0.030
-0.244
8.13
-0.03496
-0.331
9.47
1.4
141.12
-0.167
119.73
2
5
3
E F C D p o o L
3
ΔQ =-1.4/(2×141.2)=-0.00496 m /s
ΔQ =-(-0.167)/(2×119.73)=0.00069 m /s
For CD, ΔQ =-0.00496+(-0.00546) 3 = -0.01042 m /s
For CD, ΔQ =0.00069+(-0.00058) 3 = 0.00011 m /s
DC
1185
+0.016
+0.303
18.94
+0.02642
+0.827
31.3
CF
271
+0.028
+0.212
7.57
+0.03346
+0.303
9.06
FE
15233
-0.014
-2.986
213.29
-0.00854
-1.110
129.98
ED
3482
-0.014
-0.682
48.71
-0.00854
-0.254
29.7
-3.153
288.66
-0.234
200.04
3
ΔQ =-(-3.153)/(2×288.66)=0.00546 m /s
ΔQ =-(-0.234)/(2×200.04)=0.00058 m3/s
For DC, ΔQ =0.00546+(0.00496) 3 = 0.01042 m /s
For DC, ΔQ =0.00058 +(-0.00069) 3 = -0.00011 m /s Table Ex(1b) 7
1.3
Quantity Balancing Method
This method, also known as the nodal method and presented by Cornish in 1939, can be used when the heads at various points in a pipe network are known and the flow rates in pipes are required to be determined. Step 1
The unknown heads H at nodes are assumed.
Step 2
The flow rates Q ( = (H L / K )1/ n ) are calculated and the condition
Q 0 (equation (1)) is checked for all junctions in the network. Step 3
Q 0 , assumed H values are taken as correct. If Q 0 , correction ΔH is applied for assumed H values.
If
ΔH is calculated as follows:
1.3
Quantity Balancing Method
This method, also known as the nodal method and presented by Cornish in 1939, can be used when the heads at various points in a pipe network are known and the flow rates in pipes are required to be determined. Step 1
The unknown heads H at nodes are assumed.
Step 2
The flow rates Q ( = (H L / K )1/ n ) are calculated and the condition
Q 0 (equation (1)) is checked for all junctions in the network. Step 3
Q 0 , assumed H values are taken as correct. If Q 0 , correction ΔH is applied for assumed H values.
If
ΔH is calculated as follows:
Let ΔQ be the change in Q due to the change ΔH in H .
Q 0 and With corrected heads H + ΔH , (Q ΔQ) 0 With assumed heads H ,
Considering a pipe in the network as shown below, H 0
Q H
H 0 H H L KQ n Differentiating with respect to Q, 0 dH dH L KnQ n 1dQ
dH dH L n(KQ n 1 )dQ As H L KQ n
KQ n 1 H L / Q
dH dH L n(H L / Q)dQ dQ
Q nH L
dH or ΔQ
and
Q nH L
8
ΔH
As
(Q ΔQ) 0 ,
Q
Q nH
ΔH
L
ΔH
n
Q
Q / H
0
and ΔH L ΔH
L
Step 4
n
Q
Q / H L
Steps 2 and 3 are repeated until ΔH is negligible.
It has been found that this method does not converge rapidly but a high storage capacity in computing is not required. It is convenient to perform the calculation in tabular form as illustrated by example (2). Example (2)
Determine the flow rates in the pipes of the network shown in Figure Ex(2). The dimensions of the pipes are as given in Table Ex(2a). The friction factor λ for all pipes can be taken as 0.04 and the local losses can be neglected Pipe AJ JB JC
Length (m) 1500 800 400
Diameter (m) 0.30 0.25 0.20
Table Ex2(a) A
J
60 m
B 30 m
C 15 m
Figure Ex2
9
1
Express head loss H L KQ n determine K and n for all pipes. 2
Q 8 λL 2 n H L H f λ 2 2 5 Q KQ D 2g 2gD πD 4 π gD L v 2
K
8 λL 2
π gD
5
λL
and n = 2
i.e. prepare Table Ex (2b) and complete columns (1) and (2).
2
Assume head at node J = H = 35 m and calculate H L for all pipes. Note: sign convention: inflows and corresponding head losses (+) i.e. complete column (3) in the Table Ex (2b) .
3
Calculate Q for all pipes. 1/ n
H L kQ n and Q H L / K
i.e. complete column (4) in the Table Ex (2b) .
4
Check the condition At J,
Q 0 at node J.
Q 0.0018 m
3
/s 0
Apply correction ΔH for H at J. To determine ΔH calculate Q/H L for all pipes meeting at J and obtain
Q L
H
i.e. complete column (5) in the Table Ex (2b). ΔH = - 2×(-0.0018)/0.01648 m = 0.218 m
Apply correction ΔH for the assumed H value. i.e. complete column (6) in the Table Ex (2b).
7
Repeat steps 4 to 6 until ΔH can be neglected. i.e. complete columns (7), (8), (9),… in the Table Ex (2b) negligible.
10
until
ΔH is
(1) Pipe
(2) 2
(3) 5
(4)
Q (H L / K ) 3 (m /s)
(5)
(6)
(7)
Q/H L (s/m )
H L (m)
Q (H L / K )1/ n 3 (m /s)
1/ n
2
K ( s /m )
H L (m)
AJ
2040
+25
+0.1107
0.004428
+25.218
+0.1112
JB
2708
-5
-0.0429
0.00858
-4.782
-0.04202
JC
4131
-20
-0.0696
0.00348
-19.782
-0.0018
0.01648
-0.0692 -0.00002 0
H L
2(0.0018) /(0.01648) 0.218m Table Ex (2b)
11
References:
1
Chadwick A and Morfett J, Hydraulics in Civil and Environmental Engineering, Third Edition, E & FN SPON,1999
2
Fox J A, An Introduction to Engineering Fluid Mechanics, Second Edition, Macmillan, 1977
3
Douglas J F, Gasiorek J M and Swaffield J A, Fluid Mechanics, Third Edition, Addison-Wesley, 1999
References:
1
Chadwick A and Morfett J, Hydraulics in Civil and Environmental Engineering, Third Edition, E & FN SPON,1999
2
Fox J A, An Introduction to Engineering Fluid Mechanics, Second Edition, Macmillan, 1977
3
Douglas J F, Gasiorek J M and Swaffield J A, Fluid Mechanics, Third Edition, Addison-Wesley, 1999
12
Q1
Determine the flow rates in the network shown in Figure Q1. The head loss H L in pipes can be expressed as H L = KQ2 . The K values and the elevation of nodes above datum are shown in Tables Q1(a) and Q1(b). If the pressure head at node F must be not less than 20 m, calculate the lowest pressure head permissible at node A. 0.2 m /s
0.1 m3/s
0.1 m /s
A
B
C
D
E
F
0.1 m /s
0.2 m /s
0.1 m /s
Figure Q1 Pipe
AB
BC
AD
BE
CF
DE
EF
K (s /m5)
250
200
150
150
150
200
200
2
Table Q1(a) Node
A
B
C
D
E
F
Elevation (m)
30
27
20
25
23
18
Table Q1(b)
Q2
Four reservoirs A, B, C and D are connected by pipes to a common junction J as shown in Figure Q2. The head loss H L in pipes can be expressed as H L = KQ2 . The water levels in reservoirs and the K values are given in Table Q2. Determine the flow rates in pipes. State the assumptions made in the analysis.
A
B
J
C
Figure Q2
13
D
Reservoir A B C D
Water level above datum (m) 50 45 40 30 Table Q2
14
K for pipe connecting reservoir to J (s 2/m5) 4.0 3.0 2.0 2.0
