Question

the inside diameter (I.D.) and outside diameter (O.D.) of a pope are shown in the figure. The wall thickness of the pope is the dimension labeled t. Calculate the wall thickness of the pipe if its I.D. is 0.599 in. and its O.D. is 1.315 in.

Answer 1

Given:

The inside diameter of the pope, I.D.=0.599 in.

The outside diameter of the pope, O.D.=1.315 in.

The inside radius of the pope is,

`IR=(ID)/(2)=(0.599)/(2)=0.2995\text{ in}`

The outside radius of the pope is,

`OR=(OD)/(2)=(1.315)/(2)=0.6575\text{ in}`

The wall thickness of the pope can be calculated as,

`t=OR-IR=0.6575-0.2995=0.358\text{ in}`

Therefore, the wall thickness of the pope is t=0.358 in.