TM 5-811-7
(a) For the top section (T)--
AT = 2 B rx (approximately),
where r == 28 feet (tank radius), x = 10 feet. Thus,
AT =2 x 3.1416 x 28 ft x 10 ft
A= 1759 sq ft.
(b) For the center section (C)--
Ac =2 B rh,
(eq D-5)
where r = 28 feet (tank radius) and h = 11 feet. Thus,
` Ac = 2 x 3.1415 x 28 ft x 11 ft
Ac = 1935 sq ft.
(c) For the bottom section (b)--
r a 2 % r 2,
AB ' 2
(eq D-6)
where r = 28 feet (tank radius) and a 14 feet. Thus,
AB ' 2 3.1416 28 ft 14 ft 2 % 28 ft 2,
AB = 3894 sq ft.
(d) Therefore, the total wetted area of the tank bowl is--
AT + AC +AB or 7588 sq ft.
(2) Find the riser pipe*s area using equation D-7:
Ar = 2 B rRhR,
(eq D-7)
where rR = 2.5 feet (riser radius) and hR = 115 feet (riser height). Thus,
AR = 2 x 3.1416 x 2.5 ft x 115 ft
AR = 1806 sq ft
(3) Find the maximum design current for the tank:
IT = 2.0 mA/sq ft x 7588 sq ft
IT = 15,176 mA or 15.2 A.
(4) Find the maximum design current for the riser:
IR = 2.0 mA/sq ft x 1806 sq ft
Ir 3612 mA or 3.62 A.
(5) Find the minimum weight of tank anode material needed (eq C-1 from appendix C):
YSI
W'
,
E
where Y = 10 years, S = 1.0 pound per ampere-year, E = 0.50, and I = 15.2 amperes. Thus,
(10 yr)(1.0 lb/A&yr)(15.2A)
W'
,
0.50
W = 304 lb.
D-14
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