Water hammer basics: The minimum valve-closing time to qualify as “gradual closure” (so that the full Joukowsky rise does not occur) is expressed in terms of pipe length L and wave speed C as what?

Difficulty: Medium

Correct Answer: t_g = 2L / C

Explanation:


Introduction / Context:

Water hammer is a transient pressure surge caused by rapid changes in flow. The characteristic time for a pressure wave to travel the pipe length and return (a round trip) governs whether a closure is considered “sudden” or “gradual” for surge calculations.


Given Data / Assumptions:

  • Rigid-pipe approximation with wave speed C.
  • One-dimensional transient behavior; reflection at reservoir/valve ends.
  • Valve located at one end of a line of length L.


Concept / Approach:

If the valve-closing time t_c is less than 2L/C, the closure is “sudden”, and the Joukowsky equation Δp = ρ C ΔV applies for the maximum surge. If t_c ≥ 2L/C, the closure is “gradual”; the maximum head rise reduces and scales with t_c. Thus, 2L/C is the dividing line between sudden and gradual closures.


Step-by-Step Solution:

Compute the wave round-trip time: t_round = 2L/C.Define criterion: t_c ≥ t_round ⇒ gradual closure.Therefore, the minimum time for “gradual” classification is t_g = 2L/C.


Verification / Alternative check:

Method of characteristics solutions and classic water-hammer charts corroborate the 2L/C threshold for end-valve systems.


Why Other Options Are Wrong:

  • L/C (one-way travel) underestimates the interaction time.
  • Other multiples (4L/C, L/2C) do not reflect the physics of wave reflection timing.


Common Pitfalls:

  • Using local pipe segment length instead of the full length to the reflecting boundary.


Final Answer:

t_g = 2L / C

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