The ThinCell incorporates 2X the effective anode area as the Tubular cell.
Typical wear of a tubular anode. This is due to only about ½ of the calculated anode area being effective in operation. This area is called “Effective Anode Area”. The back side of this anode is as new with no considerable wear. In a tubular cell;
≤60% of the Calculated anode area = Effective anode area (proven by FORD Motor Co.)
Shown here is a ThinCell anode, with 2.5 years of service in a high production E-Coat line. An even distribution of Current Density over the calculated Anode area lowers erosion.
In A ThinCell: Calculated Anode Area = Effective Anode Area.
FORD Motor Company proved in 1995 that calculating the entire circumference of the Tubular anode is not accurate for determining the active anode square footage (EFFECTIVE) used in anode area calculations.
Since that time most of the other automotive companies have come to the same conclusion. Their findings from extensive laboratory testing have proven that actually only up to 60% of the anode is truly used. The findings also have shown that as the anodes are spaced closer together this number lowers to 50% or less.
With less active (Effective) anode area of the anode used, the current density increases and wears the anode material causing pits and erosion of the metal. This wear issue is proven in the field. Tubular cells require that the anodes must be rotated with time. This raises the question as to why do the anodes need to be rotated, if the entire anode is used as calculated. The answer is obvious; the anode wears out in the area of electrical activity.
Since as FORD has proven, 40-50% of the anode is not actively participating in electro-deposition.
Another question that you may have is how can I paint with only 50-60% of the anode area that is in my tank? The answer is Painting the product is related to the amps produced and not the sqft of anode area. The amps are a function of the part(s) area to be painted and not the anode area.
This means that if the part requires 100 amps to be painted, the part will demand the 100 amps from the anode, whether the anode area is 20 sqft or 1 sqft. At 20 sqft the current density will be 5 amps/sqft. This (5amps/sqft) is the rating that the Tubular cells manufacturer’s claims as a maximum load. With only 1 sqft of anode area the current density is 20 amps/sqft. Both of the anodes will paint, but the smaller anode will erode very quickly and require replacement. This is evident on tubular anodes by the front face of the anode dissolving while the back is intact. This is the reason that tubular anodes require rotating. If the current density was distributed evenly around the anode, the anode would wear out uniformly.
Example: 2” tubular anode with 48” effective length anode area=
(3.14 x 2.375 (OD of anode) x 48” effective length)/ 144 = 2.48 calc sqft
A)2.48 x .60 (effective anode area proven by FORD) = 1.49 Actual effective
B)2.48 x .50 (effective anode area typical) = 1.24 Actual effective
LD8 ThinCell with 48” effective length
C)(8.25 width x 48” effective length)/144 = 2.75 actual effective
If the cells require pulling 12 amps then the current density would be
A)12 amps/1.49 = 8 amps/sqft B) 12 amps/1.24 = 9.6 amps/sqft C) 12/2.75 = 4.36 amps/sqft
Remember the Tubular cells are rated for current density at 5 amps/sqft Maximum and the ThinCell is rated at 10 amps/sqft Maximum.
Both A & B will be over the maximum current density rating and will erode the anode at a high rate. (C) The ThinCell will be below the maximum rate by 50% which mean that the production could be increased by double without the erosion issues of A & B, while also saving energy.