Frequent Questions

The equation for the surface area of the stainless steel coupon in Method 1110 is not clear to us. We have received widely differing answers from different laboratories for the analysis of the same waste and we believe it is because they calculated the ar

The apparent cause of the confusion is due, in part, to the age of Method 1110 (PDF, 6 pp, 36K), for corrosivity towards steel. It was written at a time when word processing programs did not have the ability to display equations as clearly as they do at present. However, going back to first principles, the equation is used to determine the surface area of the coupon. The area of a circle is determined from the familiar product of the radius of the circle, squared, and the constant Pi. Since the radius (r) is half of the diameter (d) of the circle, the area can expressed in the three ways shown below. 

Area = II (r) 2 = II (d/2) 2 = II (d 2/4) 

The total surface area of the coupon is the sum of the areas of the two faces of the coupon, plus the area around the outside edge, plus the area around the edge of the hole in the center of the coupon. Sec. 4.5 of Method 1110 contains an equation that illustrates adding these areas together to determine the total surface area. 

Some laboratories are apparently misreading the equation in Sec. 4.5 of Method 1110 and placing the two diameters in the first term of the expression in the denominator of the equation, not the numerator. Using the mathematical rules for the order of precedence of the operations of addition, subtraction, multiplication, and division, the equation shown in the method is correct. However, it is shown below in a format that is less subject to misunderstanding: 

A = [3.14 (D 2 - d 2)]/2  + (t)(3.14)(D)  + (t)(3.14)(d) 

 t = thickness 
D = diameter of the coupon 
d = diameter of the mounting hole 
and 3.14 is a simple approximation of the constant Pi.
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