Question 17.6: The header mentioned in Example 17.4 is subjected to bending...

p = 3in.

T_{0} =0.25 in.             d_{0}=0.875 in.

b_{0} = 3 − 0.875 = 2.125in.

T_{1} = 0.125in.             d_{1} = 0.9375in.

b_{1} = 3 − 0.9375 = 2.0625in.

T_{2} = 0.25in.            d_{2} = 0.875in.

b_{2} = 3 − 0.875 = 2.125in.

T_{3} = 0.125in.             d_{3} = 0.9375in.

b_{3} = 3 − 0.9375 = 2.0625in.

T_{4} = 0.5in.                d_{4} = 0.875in.

b_{4} = 3 − 0.875 = 2.125in.

\Sigma A X = (2.125)(0.25) × (0.125 + 0.125 + 0.25 + 0.125 + 0.5) + (2.0625)(0.125) × (0.0625 + 0.25 + 0.125 + 0.5) + (2.125)(0.25)(0.125 + 0.125 + 0.5) + (2.0625)(0.125)(0.0625 + 0.5) + (2.125)(0.5)(0.25)

\Sigma A X = 1.6484

\Sigma A = (2.125)(0.25)+(2.0625)(0.125) + (2.125)(0.25)+(2.0625)(0.125) + (2.125)(0.5) = 2.6406

\bar{X}=\frac{1.6484}{2.6406} = 0.6243

I = 0.3451in. ^{4}

c = (the larger of 0.6243 or 1.25 − 0.6243)

= 0.6257 in.

The equivalent diameter is equal to

D_{E}=3-\frac{(6)(0.3451)}{(1.25)^{2}(0.6257)} =3-2.118=0.882 in.

Thus,

e_{ b }=\frac{3-0.882}{3} =0.706

Comparing this with the efficiency from Example 17.4, we see that this is a case where the grooves for expanding the tube have little to do with the efficiency, because e_{ m } =0.704 and e_{ b }  =0.706.