We have 16 gauge steel sheet, with 0.066" holes on 0.125' staggered centers. Then:
b = 0.125ff
d = 0.066ff
t = 0.0625'f
A = (0.125)3 x 0.87
= 13.6 x 10-3 sq in
n = 1/A = 73.6 holes/sq in
a = b - d
= 0.125 -0.066
= 0.059"
P = [7r(0.066)2/4 x 13.6 x 10-3] x 100
= 26.16%
TI = [73.6 x (0.066)2/0.0625 x (0.059)2]
= 1474;
or:
TI = [0.04 x 25.16/Tr x (0.0625) x (0.059)2]
= 1472.
Again, the agreement between the two values of TI is very good. But notice that the 10-kHz-attenuation has increased to 2.9 dB, much more than the attenuation of Example 1, despite the fact that the open area for this example is 53% greater than in the earlier case!
Example 2.