Let the tensile strength of the material be σ, the pressure be P, and the outer diameter of the pipeline be D; Wall thickness δ=(P*D)/(2*σ/S) where s is the safety factor; Because p is less than 7MPa, S chooses S = 8；; P is less than 17.5MPa, and s and s are taken as S = 6；; P is greater than 17.5MPa, and s and s are selected as S = 4；; We choose the safety factor as S = 6；; The tensile strength of 20 steel is 4 10MPa, so the pressure that the pipeline can bear is p = (2 * σ/s * δ)/d = (2 * 410/6 * 2.8)/(10+6) = 26. MPa。

& gt let 17.5, then the safety factor is S=4, then the pressure that the pipeline can bear is p = (2 * σ/s * δ)/d = (2 * 414 * 2.8)/(10+6) = MPa。