The Transparency Approach
Example 3:
Again, we have 16 gauge steel, but with 7/64" holes on 3/16" staggered centers.
b = 0.1875"
d = 0.109"
t = 0.0625"
A = (0.1875)2 x 0.87
= 30.6 x 10-3 sq in
n = 1/A = 32.7 holes/sq in
a = b - d
= 0.0785"
p = [1f(0.109)2/4 x 30.6 x 10-3] x 100
= 30.49%
Then:
n = [32.7 x (0.109)2/(0.0625) x (0.0785)2]
= 1009;
or:
TI = [0.04 x 30.49/'71' x (0.0625) x (0.0785)2
= 1008.
Here, even with a percent open area greater than 30%, the 10-kHz-attenuation has increased to 3.3 dB.
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