Page 123 - Water Engineering Hydraulics, Distribution and Treatment

Capacities of the four systems of Example 4.2
Table 4.14
Capacity of system (MGD)
B
Structure
C
A
D
1. River or well field
Maximum day
36.0
36.0
36.0
Maximum day
36.0
2. Conduit I
36.0
81.0
3. Conduit II
Maximum day
…
…
36.0
4. Conduit III
81.0
Maximum hour
81.0
81.0
81.0
48.0
…
Maximum day plus reserve
5. Low-lift pumps
48.0
…
6. High-lift pumps
108.0
…
Maximum hour plus reserve
108.0
…
48.0
…
Maximum day plus reserve
48.0
48.0
7. Treatment plant Required capacity 4.4 Variations or Patterns of Water Demand 36.0 101
8. Distribution system high-value district Maximum hour 81.0 81.0 81.0 81.0
Conversion factor: 1 MGD = 3.785 MLD.
Solution 2 (SI System):
Required capacities for waterworks systems of Fig. 4.4:
6
Average daily draft Q avg day = (568 Lpcd)(120,000)∕10 = 68.2MLD.
Maximum daily draft Q = coincident draft = 2 × Q = 2 × 68.2 = 136.4MLD.
max day avg day
MGD maximum hourly draft Q = 4.5 × Q = 4.5 × 68.2 = 306.9MLD.
max h avg day
Fire flow from Table 4.13:
2 2 2
Area of three floors = 3 × 12,000 ft ∕floor = 36,000 ft = 3,344.4m .
Flow for one fire = 5,000 gpm = 18,925 Lpm.
Flow for two simultaneous fires = 2 × 18,925 = 37,850 Lpm = 54.5MLD.
Coincident draft plus fire flow = 136.4 + 54.5 = 190.9MLD.
Provision for breakdowns and repair of pumps and water purification units by installing one reserve unit, as shown in Table 4.14
assuming total units = 3 + 1 = 4:
Low-lift pumps: 4∕3 × maximum daily draft = (4∕3) × 136.4 = 181.9 MLD.
High-lift pumps: 4∕3 × maximum hourly draft = (4∕3) × 306.9 = 409.2 MLD.
Treatment works: 4∕3 × maximum daily draft = 181.9 MLD.
The resultant capacities of systems components are summarized in Table 4.14.
EXAMPLE 4.3 ESTIMATION OF WATER DEMAND TO A RESIDENTIAL AREA
For management and design purposes, find the expected maximum day and peak hour demand and daily rates of water to be supplied
to a residential area of 200 houses with a gross housing density of three dwellings per acre (7.415 unit/ha), an average market value
of USD 125,000 per dwelling unit, and a potential evaporation of 0.28 in./d (7.11 mm/d) on the maximum day. Assume the site/time
factor F is equal to 1.0.
Solution 1 (US Customary System):
Average maximum and peak daily domestic demands per dwelling unit:
By Eq. (4.20a) Q = 157 + 3.46 MF = 157 + 3.46 × 125 × 1.0 = 590 gpud.
domestic
By Eq. (4.19a): A = 0.803 D −1.26 = 0.803∕3 1.26 = 0.20 acre per dwelling unit.
4
4
By Eq. (4.18a): Q = 1.63 × 10 A (E − P) = 1.63 × 10 × 0.20 × (0.28 − 0)
sprinkling
= 913 gpud excluding precipitation.