EI 11C103
CEMP-E
1 March 1997
This item will be reduced by the amount of water available during the period of the fire demand under
emergency conditions. The amount available under emergency conditions is the production of one well, so
this item becomes 260,000-(8.5 X 2 X 60 X 60) = 198,800 L.
Item 3:
50 percent of total daily domestic requirements plus the fire demand minus the production
of one well in 24 hours: 356,250 + 260,000 - (8.5 X 24 X 60 X 60) = -118,150 gallons.
The largest of the above items, 356,250 L, governs the total storage requirements. Storage of not less
than 400,000 L should be provided. In this case, it is suggested that a 400,000 L elevated tank be provided.
g. Water main sizes. The water distribution system will have mains of adequate size to meet peak
domestic demand (see TM 5-813-1/AFM 88-10, Vol. 1) and pressure requirements at all locations.
General design criteria for water mains is given in TM 5-813-5/AFM 88-10, Vol. 5.
B-3. EXAMPLE NO. 2: PERMANENT CAMP.
a. Effective population.
Nonresident: Negligible
Resident: 20,000
b. Water source. Surface supply from river.
c. Treatment. Coagulation, flocculation, sedimentation, filtration, and chlorination.
d. Required daily demand and fire flows.
Capacity factor:
1.15.
Design population:
20,000 X 1.15 = 23,000
Per capita allowance:
150 gal/day.
Special demand:
Irrigation of lawns and shrubbery estimated to
require as much as 400,000 gal/ day
Required daily demand:
(23,000 X 150) + 400,000 = 3,850,000 gal/day
equivalent to a rate of 2,674 gal/min
Maximum day demand:
2,674 X 2.5 = 6.684 gal/min
Fire fighting flows:
Fire Flow: 1500 gpm Duration: 2 hours = 1500 x 60 x 2 = 180,000 gals
(Exact value should be calculated in accordance with TM
5-813-6/AFM 88-10, Vol. 6).
Warehouse (40,000 ft , Type I construction, extra hazard occupancy):
2
2,800. gal/min for a duration of 3 hours = 2,800 x 60 x 3 = 504,000
gallons
Hospital (1,500 gal/min for a duration of 2 hours = 1,500 x 60 x 2 = 180,000 gallons
(Exact value of Fire flow required should be calculated in
accordance with TM 5-813-6/AFM 88-10, Vol. 6).
To meet the required daily demand, a treatment plant with pumping stations and appurtenances having a
rated capacity of approximately 4,000,000 gal/day would be provided.
e. Storage requirements.
Item 1:
50 percent of total daily domestic requirements =
(23,000 X 150) + 400,000
)))))))))))))))))))))))) 1,925,000 gallons
=
2
B-2
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