(b) The number of units required is found from equation D-3:

0.012P log (D/d)

R'

,

L

or

0.012P lob (D/d)

L'

,

R

where P = 4000 ohm-centimeters, D = 5 feet, d = 2 inches or 0.166 foot, and R = 12 ohms. Thus,

(0.012)(4000 ohm&cm) log (5 ft/0.166 ft)

L'

12 ohms

48 log 30.1

L'

12

48 1.479

L'

12

L = 5.92 ft.

The nubmer of units is thus --

5.92/0.75 = 7.9 or 8 units.

For proper current distribution in the riser pipe, the anode units should not be placed too far apart. It is

generally considered that each anode unit protects a length along the riser pipe equal to 1 times the spacing

of the anode from the riser pipe wall. Therefore, for a riser height of 115 feet, spacing (center of anode to

tank wall) should be 2.5 feet. The riser length protected by one anode is 1.5 x 2.5 = 3.75 feet, so the number

of units required is 115/3.75=30.7 or 31 units. To satisfy the maximum anode discharge current for a G-2

anode

3.62 A

' 36.

0.1 A

Therefore, 36 anodes are needed instead of 31 or 8.

(c) To find the anode resistance using 36 anode units, use equation D-3:

0.012P log (D/d)

R'

,

L

where P = 4000 ohm-centimeters, D =5 feet, d = 2 inches or 0.166 foot, and L =36 x 9 inches =324 inches

or 27 feet. Thus,

(0.012)(4000 ohm&cm) log (5 ft/0.166 ft)

R'

27 ft

48 log 30.1

R'

27

R = 2.63 ohms

The L/d ratio for the riser anode string is 324/2 or 162; thus, no fringe factor correction is used.

(20) Find the voltage drop in the riser anode circuit.

(a) Electrical conductor to riser anodes. For a wire size No.2 AWG, 0.159 ohm per 1000 feet, and

estimated length 200 feet, the resistance (R) is

200 ft

0.159 ohm

R'

1000 ft

1000 ft