Static Equipment Interview Questions Updated

Q: Given a cylindrical vessel: internal pressure 3.5 barg, design temp 60°C, internal diameter 1.2 m, using SA-516 Gr70 with allowable stress S=138 MPa, corrosion allowance 2 mm, joint efficiency E=0.85. Calculate required minimum shell thickness (ignore nozzle openings). A: Use thin‑wall formula for cylindrical shell per ASME: t = (P R) / (S E - 0.6 P) ; convert units: P = 3.5 bar = 0.35 MPa; R = 0.6 m. Compute: numerator = 0.35 0.6 = 0.21 MPa·m; denominator = 138 0.85 - 0.6 0.35 = 117.3 - 0.21 = 117.09 MPa. t = 0.21 / 117.09 = 0.001794 m = 1.79 mm. Add corrosion allowance 2 mm → 3.79 mm. Add minimum fabrication allowance/weld/rounding (use 6 mm minimum for practical manufacture per code) → use 6 mm shell thickness.

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The tubes are rigidly attached to the tube sheets at both ends. Q: Given a cylindrical vessel: internal pressure 3