Safety control method for pressure pipeline having incomplete welding flaw and bearing internal pressure, bending moment and torsion complex loads
A technology for complex loads and pressure pipelines, applied in the field of safety control of pressure pipelines with incomplete penetration defects, to achieve high calculation accuracy, ensure safe operation, and ensure safety
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Embodiment 1
[0082] Example 1. There is a pressure pipeline containing a non-penetration defect. Its design parameters are: the material is 20# steel (σ s =365MPa, σ b =507MPa), the specification is Φ89×4(R 0 =44.5mm, R i = 40.5mm, wall thickness T = 4mm), after non-destructive testing, a non-penetration defect 1 was found, the size of the in-penetration defect 1 is: the actual maximum circumferential length l = 128mm, the actual deepest depth h = 2.1 mm, width B=2mm. In actual work, the pressure P=3.2MPa.
[0083] The evaluation process is as follows:
[0084] 1) To determine whether this method is applicable:
[0085] ①Whether the pipe material is 20# steel: Yes;
[0086] ②Pipe diameter ratio k=R 0 / R i =44.5 / 40.5=1.1, satisfying 1.1≤k≤1.2;
[0087] Conclusion: the above two conditions are met at the same time, so go to step 2);
[0088] 2) Obtain the regularized parameters of incomplete penetration defects:
[0089] Regularized circumferential length L of incomplete penetrat...
Embodiment 2
[0122] Embodiment 2, a pressure pipeline containing incomplete penetration defects, the pipeline material is 20# steel (σ s =365MPa, σ b =507MPa), the pipe specification is Φ57×3.5(R 0 =28.5mm, R i =25mm, wall thickness T=3.5mm), after non-destructive testing, a non-penetration defect 1 was found, the size of the non-penetration defect 1 is: the actual maximum circumferential length l = 141mm, the actual deepest depth h = 1.6 mm, width B=2mm. In actual work, the pressure P=6.3MPa.
[0123] The evaluation process is as follows:
[0124] 1) To determine whether this method is applicable:
[0125] ①Whether the pipe material is 20# steel: Yes;
[0126] ②Pipe diameter ratio k=R 0 / R i =28.5 / 25=1.14, satisfying 1.1≤k≤1.2;
[0127] Conclusion: the above two conditions are met at the same time, so go to step 2);
[0128] 2) Obtain the regularized parameters of incomplete penetration defects:
[0129] Regularized circumferential length L of incomplete penetration defects: ...
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