Page 211 - Water Engineering Hydraulics, Distribution and Treatment

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6.6 Office Studies of Pipe Networks
domestic draft, and (b) the fire demand (Table 4.13).
Domestic use decreases progressively from section
select the sizes and routes of pipes that will offset the
to section, as population or industry is left behind;
deficiency. General familiarity with the community
and studies of the network plan will aid judgment.
fire demand remains the same until the high-value
district has been passed, after which it drops to a
Some existing small pipes may have to be removed
figure applicable to the type of outskirt area.
to make way for larger mains.
3. Estimate the distribution system capacity at each sec-
7. Determine the size of the equivalent pipe for the
reinforced system and calculate the velocity of flow.
tion across the piping. To do this (a) tabulate the num-
ber of pipes of each size cut; count only pipes that
Excessive velocities may make for dangerous water
hammer. They should be avoided, if necessary, by
deliver water in the general direction of flow; and (b)
determine the average available hydraulic gradient or
lowering the hydraulic gradients actually called into
play.
frictional resistance, which depends on the pressure 6. For the available, or desirable, hydraulic gradient,
to be maintained in the system and the allowable pipe 8. Check important pressure requirements against the
velocity. Ordinarily, hydraulic gradients lie between plan of the reinforced network.
1‰ (per thousand) and 3‰, and velocities range from
2 to 5 ft/s (0.60–1.5 m/s).
4. For the available, or desirable, hydraulic gradient,
The method of sections is particularly useful (a) in pre-
determine the capacities of existing pipes and sum
liminary studies of large and complicated distribution sys-
them for total capacity.
tems, (b) as a check on other methods of analysis, and
5. Calculate the deficiency or difference between
(c) as a basis for further investigations and more exact
required and existing capacity.
calculations.
EXAMPLE 6.2 ANALYSIS OF A WATER NETWORK USING THE SECTIONS METHOD
Analyze the network of Fig. 6.10 by sectioning. The hydraulic gradient available within the network proper is estimated to lie
close to 2‰. The value of C in the Hazen–Williams formula is assumed to be 100, and the domestic (coincident) draft, in this
case, only 150 gpcd (568 Lpcd). The fire demand is taken from Table 4.13. Assume the population for each section as follows:
section a-a, 16,000; section b-b, 16,000; section c-c, 14,700; section d-d, 8,000, and section e-e, 3,000. Also assume that the type of
building construction in the high-value district is combustible and unprotected and that the maximum total surface area per building
2
2
is 20,000 ft (1,858 m ). The area downstream of the high-value district is residential with one- and two-family dwellings having a
2
2
maximum area of 6,000 ft (557.4 m ).
Solution:
Calculations are shown only for the first three sections.
1. Section a-a population = 16,000:
6
(a) Demand: domestic = 16,000 × 150/10 = 2.4 MGD (9.1 MLD); fire (from Table 4.13) = 3,750 gpm = 5.4 MGD
(20.4 MLD); total = 7.8 MGD (29.5 MLD).
(b) Existing pipes: one 24 in. (600 mm); capacity = 6.0MGD (22.7 MLD).
(c) Deficiency: 7.8 − 6.0 = 1.8MGD or (29.5 − 22.7 = 6.8 MLD).
(d) If no pipes are added, the 24 in. (600 mm) pipe must carry 7.8 MGD (29.5 MLD).
This it will do with a loss of head of 3.2‰ at a velocity of 3.8 ft/s (1.16 m/s).
2. Section b-b population and flow as in section a-a:
(a) Total demand = 7.8MGD (29.5 MLD).
(b) Existing pipes: two 20 in. (500 mm) at 3.7 MGD = 7.4MGD (28.0 MLD).
(c) Deficiency = 7.8 − 7.4 = 0.4MGD or(29.5 − 28.0 = 1.5 MLD).
(d) If no pipes are added, existing pipes will carry 7.8 MGD (29.5 MLD) with a loss of head of 2.2‰, at a velocity of 2.8 ft/s
(0.85 m/s).
3. Section c-c population = 14,700:
6
(a) Demands: domestic = 14,700 × 150/10 = 2.2 MGD (8.3 MLD); fire = 5.4 MGD (20.4 MLD); total = 7.6 MGD
(28.7 MLD).
(b) Existing pipes: one 20 in. (500 mm) at 3.7 MGD; two 12 in. (300 mm) at 1.0MGD = 2.0 MGD; five 6 in. (150 mm) at
0.16 MGD = 0.8 MGD; total = 6.5MGD (24.6 MLD).
(c) Deficiency = 7.6 − 6.5 = 1.1MGD or (28.7 − 24.6 = 4.1 MLD).