TM 5-811-7
(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
D-22
Previous Page
