Tube-sheet ligament: in shell-and-tube exchangers, the minimum clear metal between adjacent tube holes (“ligament”) should be what fraction of tube outside diameter, but in no case less than 4.5 mm?
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AEqual to the outside diameter
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BOne-half of the outside diameter
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COne-fourth of the outside diameter
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DThree-fourths of the outside diameter
Answer
Correct Answer: One-fourth of the outside diameter
Explanation
Introduction / Context:The tube sheet of a shell-and-tube exchanger is drilled with a precise pattern of holes. The remaining web of material between neighboring holes is called the ligament. Adequate ligament prevents weakening of the tube sheet and provides sealing strength for tube-to-tubesheet joints.
Given Data / Assumptions:
- Conventional exchangers with drilled tube sheets and standard tube pitches.
- Tube outside diameter (OD) is known; ligament = pitch − OD.
- Design follows common industry practice; detailed code checks still apply.
Concept / Approach:Too small a ligament raises risks of leakage, plastic deformation, and cracking around tube holes. A widely used preliminary rule is to keep the minimum ligament at roughly one-fourth of tube OD, subject to a hard minimum (about 4.5 mm) to preserve structural integrity and manufacturability for small tubes.
Step-by-Step Solution:
Define ligament = tube pitch − tube OD.Apply the rule of thumb: ligament ≥ 0.25 * OD.Impose absolute minimum: ligament ≥ 4.5 mm irrespective of OD-based fraction.Verification / Alternative check:Fabrication standards and layout practices ensure adequate bridge strength around holes; FEA or code formulas can refine this further for high load or thermal cycling services.
Why Other Options Are Wrong:
- Equal to 1.0× OD or 0.75× OD: overly conservative and impractical for common pitches.
- 0.5× OD: more conservative than necessary in many services, increasing shell diameter and cost.
Common Pitfalls:Ignoring corrosion allowance (which effectively reduces ligament); using very small pitch that complicates rolling/welding; not checking for tube-sheet bending stresses under differential pressure.
Final Answer:One-fourth of the outside diameter