In a similar manner to case A1, sol

In a similar manner to case A1, solving for D2 in the above
relation of ðCLÞ
opt=ðA22
Þ gives D2 ¼ 0.3696 m. Assuming again that a
schedule 80 steel pipe is required due to structural considerations,
a 18 in outside diameter pipe would be selected. For this pipe, the
wall thickness is 0.938 in, and hence the internal diameter is
16.124 in (409.5 mm). For this pipe diameter, the value of CL is
17.60. Again, this value can be used to determine the