Air Receiver Wall Thinning and a Weld Defect Found at Inspection
An air receiver in a foundry is 1.2 m inside diameter and 3.6 m long, vertical, designed for 10 bar gauge at 60 °C in SA-516 Gr 70 plate with an allowable stress of 138 MPa. The longitudinal seam is a single-V butt weld with spot radiography (E = 0.85) and the vessel carries a 1.5 mm corrosion allowance. Original shell thickness was 10 mm. At the five-yearly inspection a UT survey gives a minimum reading of 7.2 mm over most of the bottom third of the shell, with a 200 mm × 200 mm patch reading 6.9 mm. The external coating is sound, but the internal surface shows general pitting, and the thickness has fallen from 8.1 mm at the previous survey. The compressor runs continuously and the receiver supplies instrument air; the plant can give up no more than 24 hours of outage.
1. Calculate the minimum required thickness and the remaining life. Can the receiver stay in service at 10 bar?
Read the worked answer
Formula and inputs. For a cylindrical shell under internal pressure, t = PR/(SE − 0.6P) + CA. P = 1.0 MPa, R = 600 mm, S = 138 MPa at 60 °C, CA = 1.5 mm.
Away from the long seam (E = 1.0). t_req = (1.0 × 600)/(138 − 0.6) + 1.5 = 600/137.4 + 1.5 = 4.37 + 1.5 = 5.87 mm.
At the long seam (E = 0.85). t_req = 600/(117.3 − 0.6) + 1.5 = 5.14 + 1.5 = 6.64 mm.
The correct value to apply depends on where the thinned patch sits relative to the seam. If the 6.9 mm patch straddles or lies within one weld-thickness of the longitudinal seam, use 6.64 mm; if it is plain plate, use 5.87 mm. This distinction must be resolved by mapping the readings, not assumed.
Remaining life. Using the general 7.2 mm reading and the lower requirement: (7.2 − 5.87)/0.3 = 4.4 years. Using the 6.9 mm patch: (6.9 − 5.87)/0.3 = 3.4 years. If the patch is at the seam, (6.9 − 6.64)/0.3 = 0.9 years - under one year, which changes the answer entirely.
Current MAWP of the thinned patch. P = SE(t − CA)/(R + 0.6(t − CA)) = 138 × 5.4/(600 + 3.24) = 745/603 = 1.235 MPa, i.e. 12.4 bar, above the 10 bar operating pressure.
Judgement. With the plain-plate assumption the receiver may continue at 10 bar, and there is no need to derate immediately. But the inspection interval must be cut to half the remaining life - about 18 months, not five years - and the 200 × 200 patch must be mapped and monitored at each visit.
2. What is the root cause of the thinning, and would you repair now or defer? Set out the conditions for continued service.
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Root cause. Thinning concentrated in the bottom third of a vertical receiver is condensate attack, not general atmospheric corrosion. The external coating is sound, so the mechanism is internal: water carryover from the compressor and inadequate condensate drainage. Look for a failed or undersized trap, a drain point above the lowest point, or a dryer that is bypassed or saturated. Pitting rather than uniform loss suggests the water is sitting in a layer rather than draining freely - check the direction of fall and whether the drain is at the true low point.
Repair now or defer? Defer, but on conditions, not on optimism. The measured 6.9 mm patch retains a margin above the required thickness, so a 24-hour outage would be spent for no immediate safety gain and would be better used for a permanent fix. The decision would flip if the patch were at the seam, if the corrosion rate were accelerating beyond 0.3 mm/year, or if any cracking were found at the nozzle welds.
Conditions for continued service.
- Reduce MAWP to 8 bar if the inspector is not satisfied with the seam proximity question - a 20% derating costs compressor efficiency but buys time honestly.
- Confirm the corrosion rate from the two surveys (8.1 → 6.9 mm in five years is about 0.24 mm/yr; the quoted 0.3 mm/yr is conservative and should be used for planning).
- Fix the drainage: inspect and replace the trap, add a second drain at the true low point, verify the dryer dew point, and add an automatic drain with a strainer.
- Apply an internal coating or a corrosion inhibitor if the condensate cannot be eliminated, but only after the surface is cleaned and the deposit removed.
- Shorten the next UT survey to 18 months, covering a grid over the bottom third and all nozzle welds.
- Plan a bottom-plate renewal or weld build-up at the next major turnaround, with a qualified WPS and post-weld examination.
3. At the next inspection a 25 mm long crack is found in the longitudinal seam. What is the disposition, and what does the repair involve before the vessel returns to service?
Read the worked answer
Disposition: reject and repair. No length is acceptable. Radiographic acceptance standards for this class of vessel permit certain elongated slag inclusions up to a length limit, but cracks and lack of fusion are rejectable irrespective of size, because they are sharp-tipped planar defects that grow by fatigue under pressure cycling. A 25 mm crack in a longitudinal seam is also oriented in the worst possible direction - the hoop stress is twice the longitudinal stress, so the seam is the highest-stress weld in the vessel.
Before repair. Establish the crack depth and extent: grind or use ultrasonic shear-wave examination to determine whether it is through-wall or partially embedded, and whether it has propagated into the adjacent plate. Check for hydrogen cracking - if the vessel was welded without adequate preheat or with damp electrodes, additional cracks should be assumed until proven otherwise. Investigate the cause: a crack in a seam that has survived for years usually reflects either a fabrication defect that has grown by fatigue, or a new source of cyclic loading such as a compressor short-cycling on a tight pressure band.
Repair sequence.
- Remove the defect by grinding to sound metal, with the excavated length at least twice the depth and with a smooth profile; verify by magnetic particle or dye penetrant that no trace remains.
- If the excavation exceeds the permitted amount, weld-repair using a qualified WPS and qualified welders, with preheat appropriate to the plate grade and thickness. Post-weld heat treatment is generally not mandatory at 10 mm in this material, but it should be evaluated if the joint is highly restrained or the steel shows high hardness.
- Full radiography of the repaired seam (not spot), plus MPI or DPI of the ground areas.
- Hydrostatic test at 1.3 × MAWP, i.e. 13 bar, holding for the specified period and inspecting all seams and connections.
- Record the repair, retain the radiographs, and obtain re-certification from the inspecting authority before return to service.
Judgement point. The 24-hour outage constraint does not survive a through-wall crack in a longitudinal seam. The plant should be told that the alternative to a planned repair is an unplanned rupture of a compressed-air vessel, and instrument-air backup should be arranged while the work is done.