CEMP-E
EI 11C103
1 March 1997
APPENDIX B
TYPICAL DESIGN EXAMPLES
B-1. GENERAL. The following typical design examples illustrate procedures to be followed in the
determination of total capacity requirements for water storage facilities at Army or Air Force installations.
B-2. EXAMPLE NO. 1: Communications base; permanent construction
a. Effective population.
Nonresident: 400
Resident: 700
Total: 700 + (400 divided by 3) = 833
b. Water source. Wells on post; average yield 9.5 L/s each.
c. Treatment. Chlorination.
d. Required daily demand and fire flows.
Capacity factor:
1.5 (TM 5-813-I/AFM 88-10, Vol. 1)
Design population:
833 X 1.5 = 1,250
Per capital allowance:
570 L/d (TM 5-813-1/AFM 88-10, Vol. 1)
Special demands:
None
Required daily demand:
1,250 X 570 = 712,500 L/day,
equivalent to a rate of 8.25 L/s.
Firefighting flow:
32 L/s (500 gal/min) for 2 hours (Exact value should be calculated in
accordance with TM 5-813-6/AFM 88-10, Vol. 6)
Maximum day demand:
8.25 X 2.5 = 21 L/s.
e. Well requirements.
Total well yield: Assuming 24-hour/day well operation, one well has sufficient yield to meet the required
daily demand rate of 8.25 L/s. However, for firm production capability, it is necessary to have two wells,
each capable of 8.25 L/s.
Minimum pump requirement: The dependable output of the source of supply, i.e. the two wells, must be
equal to, or greater than, the required daily demand. Thus, each well should be equipped with a 8.5 L/s
pump. Two reliable sources of electric service should be provided, or one pump should be equipped with
both an electric motor and standby internal combustion engine. The size and number of distribution pumps
required are related to the type, size, and location of storage facilities. Provisions of elevated storage will
reduce the required pump capacity.
f. Storage requirement.
Item 1:
50 percent of total daily domestic requirements:
712,500 divided by 2 = 356,250 L.
Item 2:
Fire demand:
(32 + 8.25/2) X 60 X 60 X 2 = 260,000 L.
B-1
Previous Page
