Difficulty: Medium
Correct Answer: 25 mm
Explanation:
Introduction / Context:
Pressure pipelines are often designed using the thin-cylinder hoop-stress approach with allowances for joint efficiency and corrosion. Riveted joints reduce effective strength; water hammer adds to static pressure. This problem consolidates those considerations.
Given Data / Assumptions:
Concept / Approach:
Use thin-cylinder hoop-stress relation including joint efficiency: t_req = (p_total * D) / (2 * f_allow * η) where p_total is the maximum operating pressure including surge. Add the corrosion allowance afterwards: t = t_req + c.
Step-by-Step Solution:
Compute total pressure: p_total ≈ 18 + 7.5 = 25.5 kg/cm^2.Compute required thickness without corrosion: t_req = (25.5 * 100) / (2 * 1000 * 0.6) = 2550 / 1200 ≈ 2.125 cm.Add corrosion allowance: t ≈ 2.125 + 0.3 = 2.425 cm ≈ 24.3 mm.Adopt nearest standard thickness → 25 mm.
Verification / Alternative check:
Using a slightly more precise static head conversion (≈17.7 kg/cm^2) yields t in the same vicinity; standardization still leads to 25 mm plates.
Why Other Options Are Wrong:
Common Pitfalls:
Omitting water-hammer allowance or forgetting to divide by joint efficiency significantly underestimates thickness.
Final Answer:
25 mm
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