(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.