Question 17.4: Tubes are expanded into a rectangular header that is 1.25 in...
Tubes are expanded into a rectangular header that is 1.25 in. thick. The holes are 0.875 in. diameter with two 0.03125 in. deep grooves in the hole for holding power. The grooves are 0.125 in. high with 0.25 in. spacing between them. The top groove is 0.25 in. from the top edge. What is the membrane ligament efficiency if the openings are on 3 in. centers?
Calculate the equivalent diameter D_{E} using Eq. (17.4):
D_{ E }=\frac{1}{t}\left(d_{0} T_{0}+d_{1} T_{1}+d_{2} T_{2}+\cdots+d_{n} T_{n}\right) (17.4)
p = 3 in.
d_{0} = 0.875in. T_{0} = 0.25in.
d_{1} = 0.9375in. T_{1} = 0.125in.
d_{2} = 0.875in. T_{2} = 0.25in.
d_{3} = 0.9375in. T_{3} = 0.125in.
d_{4} = 0.875in. T_{4} = 0.50in.
D_{ E }=\frac{1}{1.25} (0.875 × 0.25 + 0.9375 × 0.125 + 0.875 × 0.25+0.9375×0.125+0.875×0.5) = 0.888.