Dimenson of the water system
We have already calculated the designing water demand. We have to
define the great consumer. We consider a great consumer if it has 100 m3/day or
more consumption. First we have to draw our one-loop main conduits line.
There is a ring-mains on our area. The conduits must be laid down on the area of
common use. We don’t lay it on the private area, and below the buildings. We
have to divide this main ring into some parts, about 7-8 parts. These parts join
with node. We set the nodes at the breakpoint (but it isn’ t necessary), and at the
branch joint, and at the charging node. (1.1. Figure).
1.1. Figure: The one-loop main
We have to take on the nodes by every great consumer (for example the
industrie, the public institution), because they have concentrated consumption in
our modell. We consider the communal consumption an equipartition water
consumption on our ring-mains. We have to burden this equipartition
consumption upon the nodes. We carry out this distribution on the basis of
conduits length or the area which belongs to the current node. We distribute the
consumption according to the length in our example. You can see this
calculating in the table 1.1..
Branches
1-2
2-3
3-4
4-5
5-6
6-7
7-8
8-9
9-2
sum
Lenght
(m)
250
500
1000
500
500
500
500
1000
1500
6250
Q
Q node 1 Q node 2
branches
(m3/d)
16,6
8,3
8,3
33,3
16,6
16,6
66,5
33,3
33,3
33,3
16,6
16,6
33,3
16,6
16,6
33,3
16,6
16,6
33,3
16,6
16,6
66,5
33,3
33,3
99,8
49,9
49,9
415,8
Qdweller =
415,8 m3/d
q = 0,066528 q m3/d.m
Node
1
2
3
4
5
6
7
8
9
Node
1
2
3
4
5
6
7
8
9
szum
Branch
1-2 2-3 3-4 4-5 5-6 6-7 7-8 8-9 9-2
8,3
8,3 16,6
49,9
16,6 33,3
33,3 16,6
16,6 16,6
16,6 16,6
16,6 16,6
16,6 33,3
33,3 49,9
Node
(m3/d)
8,3
74,8
49,9
49,9
33,3
33,3
33,3
49,9
83,2
Qdweller Qpublic ins. Qind. Soc. Qpeople Qind. Tech.
m3/d
m3/d
m3/d
m3/d
m3/d
8,3
8,3
74,8
74,8
49,9
49,9
49,9
49,9
33,3
300
333,3
33,3
33,3
33,3
33,3
49,9
49,9
83,2
200
283,2
1100
416
300
200
916
1100
2016
Qfire
l/min
600
1.1. table: Consumption of node
The water system has typical state, for example maximum and minimum
consumption, fire event, burst in a water pipe. This event influences the
behaviour of the water system. Every state has other pressure value and the flow
quantity and the transporting direction. We design our system for the hourly
peak demand. We multiply the daily people water demand by 8 per cent. That
means the following The hourly consumption is 8 per cent of daily peak. We
check our system in other states. The checking states are next:
 Minimum demand (1%)
 Average demand (4,17%)+ fire water
We have to compute the nodes consumption in these states (Table 1.2.).
Q people Qind tech
m3/d
m3/d
8,3
0
74,8
0
49,9
0
49,9
0
333,3
0
33,3
0
33,3
0
49,9
0
283,2
1100
Q fire
l/perc
Q people Qind tech
Node
m3/d
m3/d
1
8,3
0
2
74,8
0
3
49,9
0
4
49,9
0
5
333,3
0
6
33,3
0
7
33,3
0
8
49,9
0
9
283,2
1100
Qcons
Q fire
l/perc
Q people Qind tech
Node
m3/d
m3/d
1
8,3
0
2
74,8
0
3
49,9
0
4
49,9
0
5
333,3
0
6
33,3
0
7
33,3
0
8
49,9
0
9
283,2
1100
Qcons
Q fire
l/perc
Node
1
2
3
4
5
6
7
8
9
Qcons
0
0
0
0
0
0
600
0
0
0
0
0
0
0
0
600
0
0
0
0
0
0
0
0
600
0
0
Q people Qind. tech
Qfire
l/s
0,02
0,21
0,14
0,14
0,92
0,09
0,09
0,14
0,79
12,73
Qcsp
0,02
0,21
0,14
0,14
0,92
0,09
0,09
0,14
13,52
15,27
peak consumption (8 %)
Q people Qind. tech
Qfire
Qcsp
l/s
0,18
0,18
1,66
1,66
1,11
1,11
1,11
1,11
7,40
7,40
0,74
0,74
0,74
0,74
1,11
1,11
6,29
12,73
19,02
33,07
check (4.17 %)+fire
Q people Qind. tech
Qfire
l/s
0,10
0,87
0,58
0,58
3,86
0,39
0,39
10,00
0,58
3,28
12,73
Qcsp
1.2. Table: Consumption of node in the states
0,10
0,87
0,58
0,58
3,86
0,39
10,39
0,58
16,01
33,33
When we distributed the consumption upon the nodes we can modelling
the ring-mains. We use the Kirchoff's lows (First and second).
First Kirchoff's law: The sum of the waters quantities, what flows in the
node and out the node that is null.
Second Kirchoff's law: We take on a circuity direction in our ring-mean.
The sums of the signed flow pressure loss is null along the ring-mean. The
formula of flow pressure loss is in the next:
hv =
L
λ * --- *
D
v2
-------2*g
v=Q/A
hv =
L
*
16 * λ
2
---------------- * Q
5
2
D *π *2*g
hv= L * c * Q2
hv= C * Q2
Where:
D: Diameter of pipe (m)
L: lenght of conduit (m)
v : velocity (m/s)
g : 9,81 m/s2
hv: local loss (in the pipe)
Q: water flow (m3/s)
The „c” values, what belongs to different pipe diameter you can see on
table 1.3..
λ= 0,033
Diameter
c
(mm)
100
150
200
250
300
400
500
600
800
1000
1200
1400
1600
1800
2000
272,667353531
35,906811988
8,520854798
2,792113700
1,122087875
0,266276712
0,087253553
0,035065246
0,008321147
0,002726674
0,001095789
0,000506982
0,000260036
0,000144301
0,000085209
1.3. Table: „c” value
First we calculate the nodes consumption. Than we take on a flow
direction as we wish, but we have one regulation the first Kirchoff's law the low
of node. So we get the flow quantity figure 1.2.. (We can make a branch from
the ring, we cut a conduit between two nodes. So we can calculate the directions
of the flows. The flow of the cut branch is null)
1.2. Figure: The beginning flow directions
We take on the diameter of conduits.
The diameter of each pipes can be calculated based on the maximum flow.
We assume an optimal velocity of 1 m/s. We can calculate the diameters from
the following expression, where only the D is unknown.
Q= v * D2 * pi / 4
The minimum diameter is 100mm, and the fabricated diameter sizes are
increasing with 25mm. We choose the closest value to the calculated one.
So we can calculate the flow pressure loss. We analyse our ring that is it
adequate for the second Kirchoff's law. We define the failure of the sum of the
pressure flow loss. Then we correct with dq all the pipes flows.
Dq= - ∑CQ2 / ∑CQ
Branches Lenght Diameter main-ring
(m)
(mm)
1-2
250
200 branch
2-3
500
150
3-4
1000
100
4-5
500
100
5-6
500
100 one-loop
6-7
500
100
7-8
500
100
8-9
1000
200
9-2
1500
200
Sign
Then we calculate the flows pressure loss again, until the failure will null,
this is the Cross method. Finally we get the right flow direction, and quantity.
Than we check the ring, is it adequate for the pressure demand. We can get the
pressure at the nodes, if we start from the connection node where the pressure is
known, what is about 45-60 meter (4,5-6,0 bar) and we add the signed flow
pressure loss on the path. The pressures must be higher as 20 meter or buildings
height + 10 meter at the node. The water velocity must be about 0,5-1,50 m/s.
The smallest allowed conduit is 100 mm in the water system. If the water system
is not adequate for the pressure demand or water velocity, we have to change the
conduit (diameter). We can see the cross method on table 1.4..
1
1
1
1
1
1
1
1
1
C=c*l
[s2/m5]
2130
17953
272667
136334
136334
136334
136334
8521
12781
Q[m3/s]
0,0329
0,0156
0,0145
0,0134
0,0060
0,0053
0,0045
0,0034
-0,0156
abs(CQ)
C*Q*abs(Q)
SCQ
140,0900
2,3032
560,5018
4,3747
560,5018
7908,4370
57,3440 8468,9388
3652,1266
24,4584 12121,0654
1634,4027
4,8984 13755,4681
1433,0081
3,7656 15188,4762
1231,6135
2,7815 16420,0898
58,0951
0,0990 16478,1849
399,0291
-3,1144 16877,2140
SCQ2
Dq
4,3747
61,7187
86,1771
91,0756
94,8411
97,6227
97,7217
94,6073
-0,0056
failure
correction
water
quantity
2. iteration
Q (m3/s)
0,0329
0,0100
0,0089
0,0078
0,0004
-0,0004
-0,0011
-0,0022
-0,0212
Q[m3/s]
0,0329
0,0100
0,0089
0,0078
0,0004
-0,0004
-0,0011
-0,0022
-0,0212
abs(CQ)
359,2217
4851,4957
2123,6560
105,9321
95,4625
296,8571
37,4343
542,3232
C*Q*abs(Q)
SCQ
1,7969
21,5803
8,2700
0,0206
-0,0167
-0,1616
-0,0411
-5,7528
359,2217
5210,7174
7334,3734
7440,3055
7535,7680
7832,6251
7870,0594
8412,3826
SCQ2
1,7969
23,3772
31,6472
31,6678
31,6511
31,4895
31,4484
25,6955
Dq
-0,0031
3. iteration
Q (m3/s)
0,0329
0,0069
0,0058
0,0047
-0,0027
-0,0034
-0,0041
-0,0053
-0,0243
Q[m3/s]
0,0329
0,0069
0,0058
0,0047
-0,0027
-0,0034
-0,0041
-0,0053
-0,0243
abs(CQ)
249,5448
3185,7772
1290,7967
726,9272
928,3218
1129,7163
89,4880
620,4038
C*Q*abs(Q)
0,8671
9,3055
3,0553
-0,9690
-1,5803
-2,3403
-0,2350
-7,5286
SCQ
249,5448
3435,3220
4726,1187
5453,0459
6381,3676
7511,0840
7600,5720
8220,9758
SCQ2
0,8671
10,1726
13,2279
12,2589
10,6786
8,3383
8,1033
0,5747
Dq
-0,0001
4. iteration
Q (m3/s)
0,0329
0,0069
0,0058
0,0047
-0,0027
-0,0035
-0,0042
-0,0053
-0,0243
Q[m3/s]
0,0329
0,0069
0,0058
0,0047
-0,0027
-0,0035
-0,0042
-0,0053
-0,0243
abs(CQ) C*Q*abs(Q)
247,0345
3147,6522
1271,7342
745,9897
947,3843
1148,7789
90,6794
622,1909
0,8498
9,0841
2,9657
-1,0205
-1,6458
-2,4200
-0,2413
-7,5720
SCQ
247,0345
3394,6867
4666,4209
5412,4106
6359,7948
7508,5737
7599,2531
8221,4440
SCQ2
Dq
0,8498
9,9339
12,8996
11,8791
10,2332
7,8133
7,5720
0,0000
0,0000
1.4. Table: The Cross-method
The solution of the calculation is showed by table 1.5.. We can see the
right flow direction, the local flows pressure loss and the water velocity in the
branches.
Branches Lenght Diameter Q (m3/s)
(m)
(mm)
1-2
2-3
3-4
4-5
5-6
6-7
7-8
8-9
9-2
250
500
1000
500
500
500
500
1000
1500
200
150
100
100
100
100
100
200
200
0,0329
0,0069
0,0058
0,0047
-0,0027
-0,0034
-0,0041
-0,0053
-0,0243
Q (l/s)
32,88
6,95
5,84
4,73
-2,67
-3,40
-4,14
-5,25
-24,27
water
velocity
(m/s)
1,05
0,39
0,74
0,60
-0,34
-0,43
-0,53
-0,17
-0,77
local loss
C*Q*abs(Q)
(m)
2,3032
0,8671
9,3055
3,0553
-0,9690
-1,5803
-2,3403
-0,2350
-7,5286
1.5. Table: The solution of calculation
specific
pressure
flow loss
%o
9,2
1,7
9,3
6,1
-1,9
-3,2
-4,7
-0,2
-5,0
We have to represent our solution on our system. We have to make a lettering,
we have to write down on your branches:




The right flow direction
The water velocity or quantity of water flow
The relative pressure at nodes
The diameter of conduits