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F"igure 1 hsows a street map 0faa commumty WIt present popu anoncommuni . hI' 0f 1,000 thatEy.
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is to be served entirely by public standpipes. The purpose of this example is toP
illustrate the four steps of design using a conventional approach that employs a desk calculator.Figure 1: Street Map of Worked Example
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Background Data
Present Population 1,000
Average per capita flow 100 l/c/d
Peaking factor 3
Town growth rate 2 %/year
Design period 20 years
Unaccounted for losses 20 0/0
Number of persons per standpipe 100
Maximum level of elevated tank 16 m
Minimum level of elevated tank 12 m
Ground elevation, all nodes 0 m
M;n;mnm llllowahle nressure 5 m\~
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Layout
With 1,000 persons in town and 100 persons per standpipe, there will be 10 standpipes. If the population is more or less evenly distributed throughout the town, the standpipes should be evenly spaced. They should be located along streets where users have easy access, near street comers if possible. Figure 1 shows one arrangement of the standpipes, which are denoted by node numbers.
This network has a single source at the elevated tank (node 11). The task now is to connect the standpipe nodes to the source node keeping total pipe length as short as possible, laying pipes in streets, and selecting routes where the greatest number of houses are located so that in the future when the system is upgraded to individual connections, these houses will be able to connect. Figure 2 shows the resulting layout. This network has 1 source node (No. 11), 10 demand nodes (Nos.
1-10),3 junction nodes (Nos. 12-14), and 13 pipes. In a branched network, the number of pipes is always 1 less than the number of nodes.
Figure 2: Network Layout of Worked Example
Flows
The peak hourly design flow for this network is calculated in the slide show and above. It is 557 mt/d which is equivalent to 6.5 litres/second.
Since this is a small town with only domestic demand to be served and since each standpipe serves the same number of persons, the flow at each demand node (Nos.
1-10) is assumed to be identical, namely 0.65 litres/second, and the inflow at node No. 11 is 6.5 litres/second. Note that the network is being designed to meet peak hourly demands.
With a takeoff flow of 0.65 Ips at each demand node, it is an easy matter to calculate the flows in the 13 pipes of the network. It is preferable to start the
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calculations at the terminal ends and work toward the source. For example, the flow in pipe No. 13 is 0.65, in No. 12 it is 1.30, etc. The list of pipe flows and their lengths is as follows:
Pipe No. Length (m) Flow (Ips)
1 1~ ~~
From the Background Data, the water elevation in the tank varies between 12 and 16 m. At the time of peak hourly demand, the level should be approximately midway in the tank. Hence, the inlet pressure is 14 m.
The minimum target pressure is 5 m. Because the network is flat, this pressure should occur at the terminal nodes of the network. That is, if the network is well designed, the pressure at each of the 6 terminal nodes (Nos I, 2, 6, 7, 8 and 10) should be about 5 m.
Diameters
Using the method in the slide show, the first task is to calculate the minimum hydraulic gradient. This means finding the longest branch and dividing its length into the available head, which in this case is 14 - 5
=
9 m.This network has 6 branches, one for each terminal node. The branches can be designated by the terminal node numbers. For example, branch No. 6 includes pipes No. I, 2, 7 and 8, and branch No.8 includes pipes No. I, 2, 10 and 11. The branches with their pipe numbers and total lengths are:
Branch Pipe Numbers Total Length (m)
6
The branch with terminal node No. 10 is longest, and its average hydraulic gradient is 9 m: 925 m
=
0.00973..
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Using the Hazen Williams equation and this gradient, the diameter of each pipe in the network can be calculated based on its design flow. The Hazen Williams equation is:
Q
=
3.7 X 10-6C 1)2.63(H/L)OS4 Q ::::flows, IpsC = roughness co-efficient = 130 D
=
diameter, mmH/L
=
hydraulic gradient=
0.00973Substituting these values into this equation yields:
Q
=
3.94 x 10-5D2.63 Rearranging by solving for diameter:Q
=
47.28 QO.38Using this equation, the diameter of each pipe can be calculated. For many of the pipes, the diameter will not be a commercial size. Hence, the exact diameter will have to be rounded up rather than down, but this is a matter of judgement. The resulting diameters in mm are:
Exact
Pipe No. Diameter Rounded mm
1 96 100
This is an initial estimate of diameters. Note that all of them were rounded down to commercial sizes except those for pipes No. 1 and 2. If all diameters had been rounded up including pipes No.3 to 13, there would be no question about the feasibility of this design; all node pressures would be above the minimum target value because the gradient would be less than 0.00973 which was used to calculate the exact diameters. Because eleven of the diameters were rounded down,
however, a check must be made to verify that the proposed design is feasible.
The actual headloss should be calculated in each pipe, and if the minimum target pressure of at least 5 m is not obtained at each terminal node, then diameters should be adjusted until this is achieved. There are 6 branches in this network;
only two of them will be checked in this worked example. The others can be checked as an exercise.
Consider the branch with terminal node No.1; its pipes are Nos. 1,2,3,4 and 6.
The Hazen Williams equation can be rearranged to solve for pipe headloss (H) as a function of flow (Q), diameter (D), and length (L).The equation (for C = 130) is:
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.
H = 1.39 X 106 QI.86L/D4.87
Now, using the initial rounded diameters, pipe flows and pipe lengths, the headloss in each pipe can be calculated. For the branch with terminal node No.1, the actual headloss would be:
Pipe No. 1 2 3 4 6
Q(Ips) 6.50 5.85 1.95 1.30 0.65
L (m) 165 210 225 95 165
D (mm) 100 100 50 50 38
H (m) 1.33 1.39 5.73 1.14 2.09
The total headloss in this branch is the sum of the individual pipe losses which is 11.68 m. Hence, with a pressure at the inlet of 14 m, the pressure at terminal node No.1 is 14.0 - 11.68
=
2.34 m, which is below the minimum pressure target of 5 m. Hence the diameters for this branch are infeasible.Now let's check the branch with terminal node No. 10; the pipes in this branch are Nos. 1,2, 10, 12 and 13. Using the same approach, the total headloss is calculated to be 11.51 m, which means that the pressure at node No. 10 would be
14.0 - 11.51 = 2.49 m. Again, the design is infeasible because the pressure is below the minimum target of 5m. Both of these infeasibilities occurred because of rounding eleven of the pipes down instead of up. It is important to note that this happened despite what appears to be substantial increases that were made by rounding up pipe Nos. 1 and 2.
The question now is how can some diameters be changed to make the design feasible? Which pipes should be adjusted? The general rule to follow in cases where diameters need to be enlarged is: leave pipes near the ends small and enlarge them near the source. In cases where diameters are too large and must be reduced, the rule is reversed: leave pipes near the source large and reduce those nearer the ends. This rule will generally result in lowering the cost.
In the case of these branches, it was pipes fairly close to the source that were rounded down that caused the infeasibility, namely pipes No.3 and 10. Since diameters need to be enlarged, the above rule suggests that these are the pipes whose diameters should be changed. Assume that the next larger commercial size above 50 mm is 75 mm. Hence, it is proposed that pipes No.3 and 10 both be increased to 75 mm. The resulting design is:
Pipe No. Diameter
Using the Hazen Williams equation, the headloss in pipe No.3 with 75 mm diameter would be 0.79 m, and in pipe No. 10 and 75 mrn, the headloss would be
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0.66 m. Hence, these changes would increase the pressures at terminal nodes No. 1 and 10 to 7.26 m and 5.88 m, respectively. As expected, the design is feasible.
To conclude this example, let us calculate the cost of pipe in the network using the diameters listed above. Recent prices in pesos per metre for constructing pipe in South Atnerica are shown below; the table also shows the length of each different diameter pipe in the proposed network. Total pipe cost is 1.269 million pesos.
Diameter(mm) Price (P1m) Length (m) Cost (103P)
38 300 870 261
50 440 485 213
75 800 380 304
100 1310 375 491