If a boiler has a bursting pressure of 1200 PSI and a MAWP of 200 PSI, what is the Factor of Safety?

Prepare for the New Jersey Third Grade Steam Engineer License Exam. Utilize flashcards and multiple choice questions, with hints and explanations for each question. Maximize your readiness for the test!

To determine the Factor of Safety, you can use the formula:

Factor of Safety = Bursting Pressure / Maximum Allowable Working Pressure (MAWP).

In this case, the bursting pressure of the boiler is 1200 PSI, and the MAWP is 200 PSI. Plugging these values into the formula gives:

Factor of Safety = 1200 PSI / 200 PSI = 6.

Thus, the correct answer is 6. The Factor of Safety is a critical calculation in engineering as it ensures that a system can handle unexpected loads or conditions beyond the normal working parameters, aiming to provide a margin of safety to prevent catastrophic failures. It reflects how much stronger the system is than it needs to be for its intended use, and a Factor of Safety of 6 indicates a robust design that can endure significant stress while operating within safe limits.

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