Transient hydraulics in pipelines: The “critical time” for development of water hammer corresponds to which travel time of the pressure wave?

Difficulty: Easy

Correct Answer: the wave to travel from the valve to the reservoir and back

Explanation:


Introduction / Context:
Water hammer (surge) arises when flow is suddenly altered, sending a pressure wave that travels at acoustic velocity in the liquid–pipe system. The concept of “critical time” is crucial for evaluating whether a given valve-closure schedule will cause maximum surge.


Given Data / Assumptions:

  • Single pipeline between valve and reservoir/tank.
  • Wave speed a; pipeline length L.
  • Elastic water and pipe wall properties define a.


Concept / Approach:

The reflection of the wave at a reservoir boundary and its return to the valve controls transient magnitude. The critical time equals the round-trip travel time 2L/a. Closure faster than this is “sudden,” producing near-Joukowsky surge; slower closure spreads acceleration and reduces peak pressure.


Step-by-Step Solution:

Compute travel time one-way: t1 = L / a.Round-trip (critical) time: tc = 2L / a.Compare valve-closure time tv: if tv ≤ tc, expect near-maximum surge.


Verification / Alternative check:

Transient analysis software and classical water-hammer charts confirm that the most severe pressure rise occurs when closure completes within the round-trip time.


Why Other Options Are Wrong:

  • “Closing the valve” is an action, not the definition of critical time.
  • One-way travel underestimates the condition that sets peak surge.
  • “Entire network once” is irrelevant in a single-pipe context.


Common Pitfalls:

  • Ignoring elastic wave speed reduction due to pipe wall flexibility.
  • Overlooking boundary conditions (air vessel, surge tank) that change reflections.


Final Answer:

the wave to travel from the valve to the reservoir and back.

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